% The following TeXfile comprises part of

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

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

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

%        %   \amstexsupportedfalse

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

%        \input amstex
%        \documentstyle{amsppt}

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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


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

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

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

   \Centerline{#1}

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

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

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

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

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

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

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

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

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

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

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

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

%End of lines for missing AmSTeX

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

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

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

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

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


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

%hyphenations
\hyphenation{Archi-me-dean}

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

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

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

%Headline assignments

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

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

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

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

\newdimen\pagetotalplus
\newdimen\pagetotalminus

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  \Centerline{#2}

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

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

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

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

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

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

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

\frfilename{mt39.tex} 
\versiondate{17.11.10} 
\copyrightdate{2010} 
 
\def\chaptername{Measurable algebras} 
 
\newchapter{39} 
 
In the final chapter of this volume, I present results connected with 
the following question:  which algebras can appear as the underlying 
Boolean algebras of measure algebras?   Put in this form, there is a 
trivial answer (391A).   The proper question is rather:  which algebras 
can appear as the underlying Boolean algebras of semi-finite measure 
algebras?   This is easily reducible to the question:  which algebras 
can appear as the underlying Boolean algebras of probability algebras? 
Now in one sense Maharam's theorem (\S332) gives us the answer exactly: 
they are the countable simple products of the measure algebras of 
$\{0,1\}^{\kappa}$ for cardinals $\kappa$.   But if we approach from 
another direction, things are more interesting.   Probability algebras 
share a very large number of very special properties.   Can we find a 
selection of these properties which will be sufficient to force an 
abstract Boolean algebra to be a probability 
algebra when endowed with a suitable functional? 
 
No fully satisfying answer to this question is known.   But in exploring 
the possibilities we encounter some interesting and important ideas. 
In \S391 I discuss algebras which have strictly positive additive 
real-valued functionals;  for such algebras, weak 
$(\sigma,\infty)$-distributivity is necessary and sufficient for the 
existence of a measure;  so we are led to look for conditions sufficient 
to ensure that there is a strictly positive additive functional.   A 
slightly different approach lies through the concept of `submeasure'. 
Submeasures arise naturally in the theories of topological Boolean 
algebras (393J), topological Riesz spaces (393K) and vector measures 
(394P), and on any given algebra there is a strictly positive 
`uniformly exhaustive' 
submeasure iff there is a strictly positive additive functional;  this 
is the Kalton-Roberts theorem (392F).    
 
Submeasures in general are common, but correspondingly limited in what they 
can tell us about a structure in the absence of further properties.   
Uniformly exhaustive submeasures are not far from additive functionals. 
An intermediate class, the `exhaustive' submeasures, has been intensively 
studied, originally in the hope that they might lead to characterizations 
of measurable algebras, but more recently for their own sake.   Just as 
additive functionals lead to measurable algebras, totally 
finite exhaustive submeasures lead to `Maharam algebras' (\S393).   For 
many years it was not known whether every exhaustive submeasure was 
uniformly
exhaustive (equivalently, whether every Maharam algebra was a measurable 
algebra);  an example was eventually found by M.Talagrand, and is presented 
in \S394.
 
In \S395, I look at a characterization  
of measurable algebras in terms of the special properties 
which the automorphism group of a measure algebra must have (Kawada's 
theorem, 395Q).   \S396 complements the previous section by looking 
briefly at the subgroups of an automorphism group $\Aut\frak A$ which 
can appear as groups of measure-preserving automorphisms. 
 
\discrpage 
 
\frfilename{mt391.tex}
\versiondate{5.9.07}
\copyrightdate{2000}

\def\chaptername{Measurable algebras}
\def\sectionname{Kelley's theorem}

\newsection{391}

In this section I introduce the notion of `measurable algebra' (391B),
which will be the subject of the whole chapter once the trivial
construction of 391A has been dealt with.   I show that for \wsid\
algebras countable additivity can be left to look after itself, and all
we need to find is a strictly positive finitely additive functional
(391D).   I give Kelley's criterion for the existence of such a
functional (391H-391J).

\leader{391A}{Proposition} Let $\frak A$ be any Dedekind
$\sigma$-complete Boolean algebra.   Then there is a function
$\bar\mu:\frak A\to[0,\infty]$ such that $(\frak A,\bar\mu)$ is a
measure algebra.

\proof{ Set $\bar\mu 0=0$, $\bar\mu a=\infty$ for
$a\in\frak A\setminus\{0\}$.
}%end of proof of 391A

\leader{391B}{Definition (a)} I will call a Boolean algebra $\frak A$ {\bf
measurable} if there is a functional
$\bar\mu:\frak A\to\coint{0,\infty}$ such that $(\frak A,\bar\mu)$ is a
totally finite measure algebra.

\cmmnt{In this case, if $\bar\mu\ne 0$, then it has a scalar multiple
with total mass $1$.   So a Boolean algebra $\frak A$ is measurable iff
either it is $\{0\}$ or there is a functional $\bar\mu$ such that
$(\frak A,\bar\mu)$ is a probability algebra.
}%end of comment

\spheader 391Bb\dvAnew{2007}
I will call a Boolean algebra $\frak A$ {\bf chargeable} if
there is an additive functional $\nu:\frak A\to\coint{0,\infty}$ which is
{\bf strictly positive}, that is, $\nu a>0$ for every non-zero
$a\in\frak A$.

\cmmnt{Of course a measurable algebra is chargeable.}

\spheader 391Bc\dvAnew{2007}
I will call a Boolean algebra {\bf nowhere measurable} if
none of its non-zero principal ideals are measurable algebras.

\vleader{48pt}{391C}{Proposition} Let $\frak A$ be a Boolean algebra.

(a) The following are equiveridical:  (i) there is a functional
$\bar\mu:\frak A\to[0,\infty]$ such that $(\frak A,\bar\mu)$ is a
semi-finite measure algebra;  (ii) $\frak A$ is Dedekind
$\sigma$-complete and $\{a:a\in\frak A,\,\frak A_a$ is measurable$\}$ is
order-dense in $\frak A$\cmmnt{, writing $\frak A_a$ for the principal
ideal generated by $a$}.

(b) The following are equiveridical:  (i) there is a functional
$\bar\mu:\frak A\to[0,\infty]$ such that $(\frak A,\bar\mu)$ is a
localizable measure algebra;  (ii) $\frak A$ is Dedekind
complete and $\{a:a\in\frak A,\,\frak A_a$ is measurable$\}$ is
order-dense in $\frak A$.

\proof{{\bf (a)} (i)$\Rightarrow$(ii):  if $(\frak A,\bar\mu)$ is a
semi-finite measure algebra, then $\frak A^f=\{a:\bar\mu a<\infty\}$ is
order-dense in $\frak A$ and $\frak A_a$ is measurable for every
$a\in\frak A^f$.

(ii)$\Rightarrow$(i):  setting $D=\{a:a\in\frak A,\,\frak A_a$ is
measurable$\}$, $D$ is order-dense, so there is a partition of unity
$C\subseteq D$ (313K).   For each $c\in C$, choose $\bar\mu_c$ such
that $(\frak A_c,\bar\mu_c)$ is a totally finite measure algebra.   Set
$\bar\mu a=\sum_{c\in C}\bar\mu_c(a\Bcap c)$ for every $a\in\frak A$;
then it is easy to check that $(\frak A,\bar\mu)$ is a semi-finite
measure algebra.
\medskip

{\bf (b)} Follows immediately.
}%end of proof of 391C

\leader{391D}{Theorem}\cmmnt{ ({\smc Kantorovich Vulikh \& Pinsker 50})}
% statement 1.44 page 428;  possibly taken from a 1947 paper by Pinsker
Let $\frak A$ be a Boolean algebra.   Then the following are equiveridical:

(i) $\frak A$ is measurable;

(ii) $\frak A$ is Dedekind $\sigma$-complete, \wsid\ and chargeable.

\proof{{\bf (i)$\Rightarrow$(ii)} Put the definition together with
322C(b)-(c) (for Dedekind completeness) and 322F (for weak
$(\sigma,\infty)$-distributivity).

\medskip

{\bf (ii)$\Rightarrow$(i)}
Given that (ii) is satisfied, let $M$ be the $L$-space of bounded additive
functionals on $\frak A$, $M_{\tau}\subseteq M$ the band of completely
additive functionals, and $P_{\tau}:M\to M_{\tau}$ the band projection
(362Bd).   Let $\nu:\frak A\to\coint{0,\infty}$ be a strictly positive
additive functional, and set $\bar\mu=P_{\tau}(\nu)$.
Then $\bar\mu$ is strictly
positive.   \Prf\ If $c\in\frak A$ is non-zero, there is an
upwards-directed set $A$, with supremum $c$, such that
$\bar\mu c=\sup_{a\in A}\nu c$ (362D);  as $\nu$ is strictly positive and
$A$ contains a non-zero element, $\bar\mu c>0$.\ \QeD\  Of course $\bar\mu$
is countably additive, so witnesses that $\frak A$ is measurable.
}%end of proof of 391D

\leader{391E}{}\cmmnt{ Thus we are led naturally to the question:
which Boolean algebras carry strictly positive {\it finitely} additive
functionals?   The Hahn-Banach theorem, suitably applied, gives some
sort of answer to this question.   For the sake of applications later
on, I give two general results on the existence of additive functionals
related to given functionals.

\medskip

\noindent}{\bf Theorem} Let $\frak A$ be a Boolean algebra, not $\{0\}$,
and $\phi:\frak A\to[0,1]$ a functional.   Then the following are
equiveridical:

(i) there is a finitely additive functional $\nu:\frak A\to[0,1]$ such
that $\nu 1=1$ and $\nu a\le\phi a$ for every $a\in\frak A$;

(ii) whenever $\langle a_i\rangle_{i\in I}$ is a finite indexed family
in $\frak A$, $m\in\Bbb N$ and $\sum_{i\in I}\chi a_i\ge m\chi 1$ in
$S=S(\frak A)$\cmmnt{ (definition:  361A)}, then
$\sum_{i\in I}\phi a_i\ge m$.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} If $\nu:\frak A\to[0,1]$ is a
finitely additive functional such that $\nu 1=1$ and $\nu a\le\phi a$
for every $a\in\frak A$, let $h:S\to\Bbb R$ be the positive linear
functional corresponding to $\nu$ (361G).   Now if $\familyiI{a_i}$ is a
finite family in $\frak A$ and $\sum_{i\in I}\chi a_i\ge m\chi 1$, then

$$\eqalign{\sum_{i\in I}\phi a_i
&\ge\sum_{i\in I}\nu a_i
=\sum_{i\in I}h(\chi a_i)\cr
&=h(\sum_{i\in I}\chi a_i)
\ge h(m\chi 1)
=m.\cr}$$

\noindent As $\familyiI{a_i}$ is arbitrary, (ii) is true.

\medskip

{\bf (b)(ii)$\Rightarrow$(i)} Now suppose that $\phi$ satisfies (ii).
For $u\in S$, set

\Centerline{$p(u)
=\inf\{\sum_{i=0}^n\alpha_i\phi a_i:
a_0,\ldots,a_n\in\frak A,\,\alpha_0,\ldots,\alpha_n\ge 0,\,
\sum_{i=0}^n\alpha_i\chi a_i\ge u\}$.}

\noindent Then it is easy to check that $p(u+v)\le p(u)+p(v)$ for all
$u$, $v\in S$, and that $p(\alpha u)=\alpha p(u)$ for all $u\in S$,
$\alpha\ge 0$.   Also $p(\chi 1)\ge 1$.   \Prf\Quer\ If not, there are
$a_0,\ldots,a_n\in\frak A$ and $\alpha_0,\ldots,\alpha_n\ge 0$ such that
$\chi 1\le\sum_{i=0}^n\alpha_i\chi a_i$ but
$\sum_{i=0}^n\alpha_i\phi a_i<1$.   Increasing each $\alpha_i$ slightly
if necessary, we may suppose that every $\alpha_i$ is rational;  let
$m\ge 1$ and $k_0,\ldots,k_n\in\Bbb N$ be such that $\alpha_i=k_i/m$ for
each $i\le n$.

Set $K=\{(i,j):0\le i\le n,\,1\le j\le k_i\}$, and for $(i,j)\in K$ set
$a_{ij}=a_i$.   Then

\Centerline{$\sum_{(i,j)\in K}\chi a_{ij}
=\sum_{i=0}^nk_i\chi a_i
=m\sum_{i=0}^n\alpha_i\chi a_i
\ge m\chi 1$,}

\noindent but

\Centerline{$\sum_{(i,j)\in K}\phi a_{ij}
=\sum_{i=0}^nk_i\phi a_i
=m\sum_{i=0}^n\alpha_i\phi a_i
<m$,}

\noindent which is supposed to be impossible.\ \Bang\Qed

By the Hahn-Banach theorem, in the form 3A5Aa, there is a linear
functional $h:S\to\Bbb R$ such that $h(\chi 1)=p(\chi 1)\ge 1$ and
$h(u)\le p(u)$ for
every $u\in S$.   In particular, $h(\chi a)\le\phi b$ whenever
$a\Bsubseteq b\in\frak A$.   Set $\nu a=h(\chi a)$ for $a\in\frak A$;
then $\nu:\frak A\to\coint{0,\infty}$ is an additive functional,
$\nu 1\ge 1$ and
$\nu a\le\phi b$ whenever $a\Bsubseteq b$ in $\frak A$.   We do not know
whether $\nu$ is positive, but if we define $\nu^+$ as in 362Ab, we
shall have a non-negative additive functional such that

\Centerline{$\nu^+a=\sup_{b\Bsubseteq a}\nu b\le\phi a$}

\noindent for every $a\in\frak A$, and

\Centerline{$1\le\nu 1\le\nu^+1\le\phi 1\le 1$,}

\noindent so $\nu^+$ witnesses that (i) is true.
}%end of proof of 391E

\leader{391F}{Theorem} Let $\frak A$ be a Boolean algebra, not $\{0\}$,
and $\psi:A\to[0,1]$ a functional, where $A\subseteq\frak A$.   Then the
following are equiveridical:

(i) there is a finitely additive functional
$\nu:\frak A\to[0,1]$ such that $\nu 1=1$ and $\nu a\ge\psi a$ for every
$a\in A$;

(ii) whenever $\langle a_i\rangle_{i\in I}$ is a finite indexed family
in $A$, there is a set $J\subseteq I$ such that
$\#(J)\ge\sum_{i\in I}\psi a_i$ and $\inf_{i\in J}a_i\ne 0$.

\cmmnt{\medskip

\noindent{\bf Remark} In (ii) here, we may have to interpret the infimum
of the empty set in $\frak A$ as $1$.
}%end of comment

\proof{{\bf (a)} We apply 391E to $\phi$, where

$$\eqalign{\phi a&=1-\psi(1\Bsetminus a)\text{ if }a\in\frak A
  \text{ and }1\Bsetminus a\in A,\cr
&=1\text{ for other }a\in\frak A.\cr}$$

\medskip

{\bf (b)} Suppose that (i) here is true of $\psi$.   Then 391E(i) is true
of $\phi$.   \Prf\ Let $\nu:\frak A\to[0,1]$ be an additive
functional such that $\nu 1=1$ and $\nu a\ge\psi a$ for every
$a\in\frak A$.   If $a\in\frak A$ and $1\Bsetminus a\in A$, then

\Centerline{$\nu a=1-\nu(1\Bsetminus a)\le 1-\psi(1\Bsetminus a)=\phi a$;}

\noindent for other $a\in\frak A$, $\nu a\le 1=\phi a$.\ \Qed

\medskip

{\bf (c)} Suppose that 391E(i) is true of $\phi$.   Then (i) here is true
of $\psi$.   \Prf\ There is an additive functional $\nu:\frak A\to[0,1]$
such that $\nu 1=1$ and $\nu a\le\phi a$ for every $a\in\frak A$;  in this
case, for $a\in A$,

\Centerline{$\nu a=1-\nu(1\Bsetminus a)\ge 1-\phi(1\Bsetminus a)=\psi a$.
\Qed}

\medskip

{\bf (d)} Suppose that (ii) here is true of $\psi$, and that
$\langle a_i\rangle_{i\in I}$ is a finite family
in $\frak A$ such that $\sum_{i\in I}\chi a_i\ge m\chi 1$, while
$\sum_{i\in I}\phi a_i=\beta$.   Set
$K=\{i:i\in I$, $1\Bsetminus a_i\in A\}$.

\Centerline{$\sum_{i\in K}\psi(1\Bsetminus a_i)
=\sum_{i\in K}(1-\phi a_i)
=\#(K)-\sum_{i\in I}\phi a_i+\#(I\setminus K)
=\#(I)-\beta$,}

\noindent so there is a set $J\subseteq K$ such that
$\#(J)\ge\#(I)-\beta$ and $\inf_{i\in J}(1\Bsetminus a_i)=c\ne 0$.   Now
$c\Bcap a_i=0$ for $i\in J$, so

\Centerline{$m\chi c\le\sum_{i\in I}\chi(a_i\Bcap c)
=\sum_{i\in I\setminus J}\chi(a_i\Bcap c)\le\#(I\setminus J)\chi c$}

\noindent and $m\le\#(I)-\#(J)\le\beta$.   As $\familyiI{a_i}$ is
arbitrary, 391E(ii) is true of $\phi$.

\medskip

{\bf (e)} Suppose that 391E(ii) is true of $\phi$, and that
$\familyiI{a_i}$ is a family in $A$.   Set

\Centerline{$\beta=\sum_{i\in I}\phi(1\Bsetminus a_i)
=\#(I)-\sum_{i\in I}\psi a_i$}

\noindent and let $k$ be the least integer greater than $\beta$.   Since
$\sum_{i\in I}\phi(1\Bsetminus a_i)<k$,
$\sum_{i\in I}\chi(1\Bsetminus a_i)\not\ge k\chi 1$, that is,
$\sum_{i\in I}\chi a_i\not\le(\#(I)-k)\chi 1$.   But this means that
there must be some $J\subseteq I$ such
that $\#(J)>\#(I)-k$ and $\inf_{i\in J}a_i\ne 0$.   Now

\Centerline{$\sum_{i\in I}\psi a_i=\#(I)-\beta\le\#(I)-(k-1)\le\#(J)$.}

\noindent As $\familyiI{a_i}$ is arbitrary, (ii) here is true of $\psi$.

\medskip

{\bf (f)} Since we know that 391E(i)$\Leftrightarrow$391E(ii),
we can conclude that (i) and (ii) here are equiveridical.
}%end of proof of 391F

\leader{391G}{Corollary} Let $\frak A$ be a Boolean algebra, $\frak B$ a
subalgebra of $\frak A$, and $\nu_0:\frak B\to\Bbb R$ a non-negative
finitely additive functional.   Then there is a non-negative finitely
additive functional $\nu:\frak A\to\Bbb R$ extending $\nu_0$.

\proof{{\bf (a)} Suppose first that $\nu_01=1$.   Set $\psi b=\nu_0b$
for every $b\in\frak B$.   Then $\psi$ must satisfy the condition (ii)
of 391F when regarded as a functional defined on a subset of $\frak B$;
but this means that
it satisfies the same condition when regarded as a functional
defined on a subset of
$\frak A$.   So there is a non-negative finitely additive functional
$\nu:\frak A\to\Bbb R$ such that $\nu 1=1$ and $\nu b\ge\nu_0b$ for
every $b\in\frak B$.   In this case

\Centerline{$\nu b=1-\nu(1\Bsetminus b)\le 1-\nu_0(1\Bsetminus
b)=\nu_0b\le\nu b$}

\noindent for every $b\in\frak B$, so $\nu$ extends $\nu_0$.

\medskip

{\bf (b)} For the general case, if $\nu_01=0$ then $\nu_0$ must be the
zero functional on $\frak B$, so we can take $\nu$ to be the zero functional
on $\frak A$;  and if $\nu_01=\gamma>0$, we apply (a) to
$\gamma^{-1}\nu_0$.
}%end of proof of 391G

\leader{391H}{Definition} Let $\frak A$ be a Boolean algebra, and
$A\subseteq\frak A\setminus\{0\}$ any non-empty set.   The {\bf
intersection number} of $A$ is the largest $\delta\ge 0$ such that
whenever $\langle a_i\rangle_{i\in I}$ is a finite family in $A$, with
$I\ne\emptyset$, there is a $J\subseteq I$ such that
$\#(J)\ge\delta\#(I)$ and $\inf_{i\in J}a_i\ne 0$.

\cmmnt{\medskip

\noindent{\bf Remarks (a)} It is essential to note that in the
definition here the $\langle a_i\rangle_{i\in I}$ are indexed families,
with repetitions allowed;  see 391Xi.

\medskip

{\bf (b)} I spoke perhaps rather glibly of `the largest $\delta$ such
that $\ldots$';  you may prefer to write

\Centerline{$\delta=\inf\{\sup_{\emptyset\ne
J\subseteq\{0,\ldots,n\},\inf_{j\in J}a_j\ne
0}\Bover{\#(J)}{n+1}:a_0,\ldots,a_n\in A\}$.}
}%end of comment

\leader{391I}{Proposition} Let $\frak A$ be a Boolean algebra and
$A\subseteq\frak A\setminus\{0\}$ any non-empty set.   Write $C$ for the
set of non-negative finitely additive functionals $\nu:\frak A\to[0,1]$
such that $\nu 1=1$.    Then the intersection number of $A$ is precisely
$\max_{\nu\in C}\inf_{a\in A}\nu a$.

\proof{ Write $\delta$ for the intersection number of $A$, and $\delta'$
for $\sup_{\nu\in C}\inf_{a\in A}\nu a$.

\medskip

{\bf (a)} For any $\gamma<\delta'$, we can
find a $\nu\in C$ such that $\nu a\ge\gamma$ for every $a\in A$.   So if
we set $\psi a=\gamma$ for every $a\in A$, $\psi$ satisfies condition
(i) of 391F.   But this means that if $\langle a_i\rangle_{i\in I}$ is
any finite family in $A$, there must be a $J\subseteq I$ such that
$\inf_{i\in J}a_i\ne
0$ and $\#(J)\ge\gamma\#(I)$.   Accordingly $\gamma\le\delta$;  as
$\gamma$ is arbitrary, $\delta'\le\delta$.

\medskip

{\bf (b)} Define $\psi:A\to[0,1]$ by setting $\psi a=\delta$ for every
$a\in A$.   If $\langle a_i\rangle_{i\in I}$ is a finite indexed family
in $A$, there is a $J\subseteq I$ such that $\#(J)\ge\delta\#(I)$ and
$\inf_{i\in
J}a_i\ne 0$;  but $\delta\#(I)=\sum_{i\in I}\psi a_i$, so this means
that condition (ii) of 391F is satisfied.   So there is a $\nu\in C$
such that
$\nu a\ge\delta$ for every $a\in A$;  and $\nu$ witnesses not only that
$\delta'\ge\delta$, but that the supremum is a maximum.
}%end of proof of 391I

\leader{391J}{Theorem} Let $\frak A$ be a Boolean algebra.   Then the
following are equiveridical:

(i) $\frak A$ is chargeable;

(ii) either $\frak A=\{0\}$ or $\frak A\setminus\{0\}$ is expressible as
a countable union of sets with non-zero intersection numbers.

\proof{{\bf (i)$\Rightarrow$(ii)} If there is a strictly positive
finitely additive functional $\nu$ on $\frak A$, and $\frak A\ne\{0\}$,
set $A_n=\{a:\nu a\ge 2^{-n}\nu 1\}$ for every $n\in\Bbb N$;  then
(applying 391I to the functional $\bover1{\nu 1}\nu$) we see that every
$A_n$ has intersection number at least $2^{-n}$, while $\frak
A\setminus\{0\}=\bigcup_{n\in\Bbb N}A_n$ because $\nu$ is strictly
positive, so (ii) is satisfied.

\medskip

{\bf (ii)$\Rightarrow$(i)} If $\frak A\setminus\{0\}$ is expressible as
$\bigcup_{n\in\Bbb N}A_n$, where each $A_n$ has intersection number
$\delta_n>0$, then for each $n$ choose a finitely additive functional
$\nu_n$ on $\frak A$ such that $\nu_n1=1$ and $\nu_na\ge\delta_n$
for every $a\in A_n$.   Setting $\nu a=\sum_{n=0}^{\infty}2^{-n}\nu_na$
for
every $a\in\frak A$, $\nu$ is a strictly positive additive functional on
$\frak A$, and (i) is true.
}%end of proof of 391J

\leader{391K}{Corollary} Let $\frak A$ be a Boolean algebra.   Then
$\frak A$ is measurable iff it is Dedekind $\sigma$-complete and
\wsid\ and either $\frak A=\{0\}$ or
$\frak A\setminus\{0\}$ is expressible as a countable union of sets with
non-zero intersection numbers.

\proof{ Put 391D and 391J together.
}%end of proof of 391K

\leader{391L}{}\dvAnew{2015}\cmmnt{ When we come to study the
structure of measurable algebras in later volumes, it will be
convenient to have the following facts on the table.

\medskip

\noindent}{\bf Proposition} (a) If $\frak A$ is a measurable algebra, all
its principal ideals and $\sigma$-subalgebras are, in themselves,
measurable algebras.
%for 548J

(b) The simple product of countably many measurable algebras is a
measurable algebra.

(c) If $\frak A$ is a measurable algebra, $\frak B$ is a Boolean algebra
and $\pi:\frak A\to\frak B$ is a surjective order-continuous Boolean
homomorphism, then $\frak B$ is a measurable algebra, isomorphic to a
principal ideal of $\frak A$.
%for 548E 548J

\proof{{\bf (a)} Use 322H and 322Na.

\medskip

{\bf (b)} Let $\familyiI{\frak A_i}$ be a
countable family of measurable algebras with simple product $\frak A$.
For each $i\in I$ let
$\bar\mu_i$ be such that $(\frak A_i,\bar\mu_i)$ is a measure algebra
and $\bar\mu_i1_{\frak A_i}\le 1$.
Let $f:I\to\Bbb N$ be an injection.   For $a=\familyiI{a_i}\in\frak A$, set
$\bar\mu a=\sum_{i\in I}2^{-f(i)}\bar\mu_ia_i$.   Then $(\frak A,\bar\mu)$
is a measure algebra (see 322La);  as $\bar\mu 1_{\frak A}\le 2$,
$\frak A$ is a measurable algebra.

\medskip

{\bf (c)} Consider the kernel $I=\{a:\pi a=0\}$ of $\pi$.   By 313Pa,
$I$ is order-closed.   Because $\frak A$ is Dedekind complete,
$c=\sup I$ is defined in $\frak A$;  as $I$ is upwards-directed,
$c\in I$ and $I$ is the principal ideal generated by $c$.
Let $\frak A_{1\Bsetminus c}$ be the principal ideal generated
by $1\Bsetminus c$.   Then $I\cap\frak A_{1\Bsetminus c}=\{0\}$ so
$\pi\restr\frak A_{1\Bsetminus c}$ is injective.   We are supposing that

\Centerline{$\frak B
=\{\pi a:a\in\frak A\}
=\{\pi(a\Bcap c)\Bcup\pi(a\Bsetminus c):a\in\frak A\}
=\{\pi(a\Bsetminus c):a\in\frak A\}
=\pi[\frak A_{1\Bsetminus c}]$.}

\noindent So $\pi\restr\frak A_{1\Bsetminus c}$ is an isomorphism
between $\frak A_{1\Bsetminus c}$ and $\frak B$.   But
$\frak A_{1\Bsetminus c}$ is a measurable algebra, by (a), so
$\frak B$ is a measurable algebra.
}%end of proof of 391L

\exercises{
\leader{391X}{Basic exercises (a)}
%\spheader 391Xa
Show
that a chargeable Boolean algebra is ccc, so is Dedekind complete iff it
is Dedekind $\sigma$-complete.
%391D

\spheader 391Xb Show (i) that any subalgebra of a chargeable Boolean
algebra is chargeable (ii) that a countable simple product of chargeable
Boolean algebras is chargeable (iii) that any free product of chargeable
Boolean algebras is chargeable.
%391Xa

\spheader 391Xc(i) Let $\frak A$ be a Boolean algebra with a chargeable
order-dense subalgebra.   Show that $\frak A$ is chargeable.   (ii) Show
that the
Dedekind completion of a chargeable Boolean algebra is chargeable.
%391Xa

\spheader 391Xd(i) Show that the algebra of open-and-closed subsets of
$\{0,1\}^{I}$ is chargeable for any set $I$.   (ii) Show that the
regular open algebra of $\Bbb R$ is chargeable.
%391Xa

\spheader 391Xe(i) Show that any principal ideal of a chargeable
Boolean algebra is chargeable.   (ii) Let $\frak A$ be a chargeable
Boolean algebra and $\Cal I$ an order-closed ideal of $\frak A$.   Show
that $\frak A/\Cal I$ is chargeable.
%391Xa

\sqheader 391Xf Show that a Boolean algebra is chargeable iff it is
isomorphic to a subalgebra of a measurable algebra.   \Hint{324O, 392H.}
%391Xa

\spheader 391Xg Let $\frak A$ be a Boolean algebra.   Show that the
following are equiveridical:  (i) $\frak A$ is chargeable and \wsid;  (ii)
there is a strictly positive countably additive functional on $\frak A$;
(iii) there is a strictly positively completely additive functional on
$\frak A$.
%391D

\spheader 391Xh Explain how to use the Hahn-Banach theorem to prove
391G directly, without passing through 391F.   \Hint{$S(\frak B)$ can be
regarded as a subspace of $S(\frak A)$.}
%391G

\sqheader 391Xi Take $X=\{0,1,2,3\}$, $\frak A=\Cal PX$,
$A=\{\{0,1\},\{0,2\},\{0,3\},\{1,2,3\}\}$.
Show that the intersection number of $A$ is $\bover35$.   \Hint{use
391I.}   Show that if $a_0,\ldots,a_n$ are {\it distinct} members of $A$
then there is a set $J\subseteq\{0,\ldots,n\}$, with
$\#(J)\ge\bover23(n+1)$, such that $\inf_{j\in J}a_j\ne 0$.
%391H

\spheader 391Xj
Let $\frak A$ be a Boolean algebra.   For non-empty
$A\subseteq\frak A\setminus\{0\}$ write $\delta(A)$ for the intersection
number of $A$.   Show that for any non-empty
$A\subseteq\frak A\setminus\{0\}$,
$\delta(A)=\inf\{\delta(I):I$ is a non-empty finite subset of $A\}$.
%391H

\spheader 391Xk Let $\frak A$ be a Boolean algebra, not $\{0\}$.   For
$a_0,\ldots,a_n\in\frak A$ set
$t(a_0,\ldots,a_n)=\max\{m:m\in\Bbb N$, $m\chi 1\le\sum_{i=0}^n\chi a_i\}$.
Let $A\subseteq\frak A$ be non-empty.   Show that

$$\eqalign{\sup\{\Bover1{n+1}&t(a_0,\ldots,a_n):a_0,\ldots,a_n\in A\}\cr
&=\min\{\sup_{a\in A}\nu a:\nu\text{ is a non-negative additive
functional on }\frak A,\,\nu 1=1\}.\cr}$$

\noindent(This is the {\bf Kelley covering number} of $A$.)
%391I

\spheader 391Xl Let $\frak A$ be a Boolean algebra.   (i) Show that the
following are equiveridical:  ($\alpha$) there is a functional $\bar\mu$
such that $(\frak A,\bar\mu)$ is a localizable measure algebra;  ($\beta$)
$L^{\infty}(\frak A)$ is a perfect Riesz space (definition:  356J).
(ii) Show that in this case $\frak A$ is a measurable algebra iff it is
ccc.
%391C

\leader{391Y}{Further exercises (a)}
%\spheader 391Ya
Show that in 391D and 391K we can replace `weakly
$(\sigma,\infty)$-distributive' by `weakly $\sigma$-distributive'.
%391D

\spheader 391Yb Show that $\Cal P\Bbb N$ is
chargeable but that the quotient algebra
$\Cal P\Bbb N/[\Bbb N]^{<\omega}$ is not ccc, therefore not chargeable.
%391Xa

\spheader 391Yc(i)
Show that if $X$ is a separable topological space, then its regular
open algebra is chargeable.
(ii) Let $\langle X_i\rangle_{i\in I}$ be any family of
topological spaces with chargeable regular open algebras.   Show that
their product has a chargeable regular open algebra.
%391Xa

\spheader 391Yd Let $\mu$ be Lebesgue measure on $[0,1]$, and $\Sigma$ its
domain.   Let $\Cal A$ be a non-empty family of non-empty subsets of $X$,
with intersection number $\delta$,
and let $\Cal W$ be the family of those sets
$W\in\Cal PX\tensorhat\Sigma$ such that $W^{-1}[\{t\}]\in\Cal A$ for every
$t\in[0,1]$.   Set
$\alpha=\inf_{W\in\Cal W}\sup_{x\in X}\mu W[\{x\}]$.   (i) Show that
$\alpha\le\delta$.   (ii) Give an example in which $\alpha<\delta$.

\spheader 391Ye\dvAnew{2011}
Let $\frak A$ be a Boolean algebra, $\frak B$ a subalgebra
of $\frak A$, $U$ a linear space and $\nu_0:\frak B\to U$ an additive
functional.   Show that there is an additive functional
$\nu:\frak A\to U$ extending $\nu_0$.   \Hint{361F.}
%391G out of order query
}%end of exercises

\cmmnt{\Notesheader{391} By the standards of this volume, this is an
easy section;  I note that I have hardly called on anything after
Chapter 32, except for a reference to the construction $S(\frak A)$ in
\S361.   I do ask for a bit of functional analysis (the Hahn-Banach
theorem) in 391E.

391J-391K are due to {\smc Kelley 59};
condition (ii) of 391J is called {\bf Kelley's criterion}.   It provides
some sort of answer to the question `which Boolean algebras carry
strictly positive finitely additive functionals?', but leaves quite open
the possibility that there is some more abstract criterion which is also
necessary and sufficient.   It is indeed a non-trivial exercise to find
any ccc Boolean algebra which does not carry a strictly positive
finitely additive functional.   The first example published seems to
have been that of {\smc Gaifman 64}, which is described in
{\smc Comfort \& Negrepontis 82}.   But for the purposes of this book
Gaifman's example
has been superseded by Talagrand's example, presented in \S394.

Kelley's criterion is a little unsatisfying.
It is undoubtedly important (see 392F
below), but at the same time the structure of the criterion -- a special
sequence of subsets of $\frak A$ -- is rather close to the structure of
the conclusion;  after all, one is, or can be represented by, a function
from $\frak A\setminus\{0\}$ to $\Bbb N$, while the other is a function
from $\frak A$ to $\Bbb R$.   Also the actual intersection number of a
family $A\subseteq\frak A\setminus\{0\}$ can be hard to calculate;  as
often as not, the best method is to look at the additive functionals
on $\frak A$ (see 391Xi).
}%end of notes

\discrpage

\frfilename{mt392.tex}
\versiondate{11.2.08}
\copyrightdate{2008}

\def\chaptername{Measurable algebras}
\def\sectionname{Submeasures}

\newsection{392}

In \S391 I looked at what we can deduce if a Boolean algebra carries a
strictly positive finitely additive functional.   There are important
contexts in which we find ourselves with subadditive, rather than
additive, functionals, and these are what I wish to investigate here.
It turns out that, once we have found the right hypotheses, such
functionals can also provide a criterion for measurability of an algebra
(392G below).   The argument runs through a new idea, using a result in
finite combinatorics (392D).

At the end of the section I include notes on metrics associated with
submeasures (392H) and on products of submeasures (392K).

\leader{392A}{Definition}\dvArevised{2007}
Let $\frak A$ be a Boolean algebra.   A
{\bf submeasure} on $\frak A$ is a functional
$\nu:\frak A\to[0,\infty]$ such that

\inset{$\nu 0 = 0$,

$\nu a\le\nu b$ whenever $a\Bsubseteq b$,

$\nu(a\Bcup b)\le\nu a+\nu b$ for all $a$, $b\in\frak A$.}

\leader{392B}{}\cmmnt{ The following list mostly repeats
ideas we have already used in the context of measures;  but (b) and (c)
are new, and will be the basis of this section.

\medskip

\noindent}{\bf Definitions} Let $\frak A$ be a Boolean algebra and
$\nu:\frak A\to[0,\infty]$ a submeasure.

(a) $\nu$ is {\bf strictly positive} if $\nu a>0$ for every $a\ne 0$.

(b) $\nu$ is {\bf exhaustive} if $\lim_{n\to\infty}\nu a_n=0$ for every
disjoint sequence $\sequencen{a_n}$ in $\frak A$.

(c) $\nu$ is {\bf uniformly exhaustive} if for every $\epsilon>0$ there
is an $n\in\Bbb N$ such that there is no disjoint family
$a_0,\ldots,a_n$ with $\nu a_i\ge\epsilon$ for every $i\le n$.

(d) $\nu$ is {\bf totally finite} if $\nu 1<\infty$.

(e) $\nu$ is {\bf unital} if $\nu 1=1$.

(f) $\nu$ is {\bf atomless} if whenever $a\in\frak A$ and $\nu a>0$ there
is a $b\Bsubseteq a$ such that $\nu b>0$ and $\nu(a\Bsetminus b)>0$.
%G\lowcy\'nski 08 differs

(g)\dvAformerly{393D} If $\nuprime$ is another submeasure on $\frak A$, then $\nuprime$ is
{\bf absolutely continuous} with respect to $\nu$ if for every
$\epsilon>0$ there is a
$\delta>0$ such that $\nuprime a\le\epsilon$ whenever $\nu a\le\delta$.

\leader{392C}{Proposition} Let $\frak A$ be a Boolean algebra.

(a) If there is an exhaustive strictly positive submeasure on $\frak A$,
then $\frak A$ is ccc.

(b) A uniformly exhaustive submeasure on $\frak A$ is exhaustive.

(c) Any non-negative additive functional on $\frak A$ is a uniformly
exhaustive submeasure.

\proof{ These are all elementary.   If $\nu:\frak A\to[0,\infty]$
is an exhaustive strictly positive submeasure, and
$\langle a_i\rangle_{i\in I}$ is a disjoint family in
$\frak A\setminus\{0\}$,
then $\{i:\nu a_i\ge 2^{-n}\}$ must be finite for each $n$, so $I$ is
countable.   (Cf.\ 322G.)   If $\nu:\frak A\to[0,\infty]$ is a
uniformly exhaustive submeasure and $\sequencen{a_n}$ is disjoint in
$\frak A$, then $\{i:\nu a_i\ge 2^{-n}\}$ is finite for each $n$, so
$\lim_{i\to\infty}\nu a_i=0$.   If $\nu:\frak A\to\coint{0,\infty}$ is a
non-negative additive functional, it is a submeasure, by
326Ba and 326Bf.   If $\epsilon>0$, then take
$n\ge\bover1{\epsilon}\nu 1$;  if $a_0,\ldots,a_n$ are disjoint, then
$\sum_{i=0}^n\nu a_i\le\nu 1$, so $\min_{i\le n}\nu a_i<\epsilon$.
}%end of proof of 392C

\leader{392D}{Lemma} Suppose that $k$, $l$, $m\in\Bbb N$ are such that
$3\le k\le l\le m$ and $18mk\le l^2$.   Let $L$, $M$ be sets with
$l$, $m$ members
respectively.   Then there is a set $R\subseteq M\times L$ such that (i)
each vertical section of $R$ has just three members (ii)
$\#(R[E])\ge\#(E)$ whenever $E\in[M]^{\le k}$;  so that for every
$E\in[M]^{\le k}$ there is an injective function $f:E\to L$ such that
$(x,f(x))\in R$ for every $x\in E$.

\cmmnt{\medskip

\noindent{\bf Recall} that $[M]^{\le k}=\{I:I\subseteq M,\,\#(I)\le k\}$,
$[M]^k=\{I:I\subseteq M,\,\#(I)=k\}$
(3A1J).}%end of comment

\proof{{\bf (a)} We need to know that $n!\ge 3^{-n}n^n$ for every
$n\in\Bbb N$;  this is immediate from the inequality

\Centerline{$\sum_{i=2}^n\ln i\ge\int_1^n\ln x\,dx=n\ln n-n+1$ for every
$n\ge 2$.}

\wheader{392D}{4}{2}{2}{36pt}

{\bf (b)} Let $\Omega$ be the set of those $R\subseteq M\times L$ such
that each vertical section of $R$ has just three members, so that

\Centerline{$\#(\Omega)=\#([L]^3)^m
=\bigl(\Bover{l!}{3!(l-3)!}\bigr)^m$.}

\noindent Let us regard $\Omega$ as a probability space with the
uniform probability.

If $F\in[L]^n$, where $3\le n\le k$, and $x\in M$, then

$$\eqalignno{\Pr(R[\{x\}]\subseteq F)
&=\Bover{\#([F]^3)}{\#([L]^3)}\cr
\noalign{\noindent (because $R[\{x\}]$ is a random member of $[L]^3$)}
&=\Bover{n(n-1)(n-2)}{l(l-1)(l-2)}\le\Bover{n^3}{l^3}\cr}$$

\noindent as $n<l$.   So if $E\in[M]^{n}$ and $F\in[L]^n$, then

$$\eqalignno{\Pr(R[E]\subseteq F)
&=\prod_{x\in E}\Pr(R[\{x\}]\subseteq F)\cr
\noalign{\noindent (because the sets $R[\{x\}]$ are chosen
independently)}
&\le\Bover{n^{3n}}{l^{3n}}.\cr}$$

Accordingly

$$\eqalignno{\Pr(&\text{there is an }E\subseteq M
  \text{ such that }\#(R[E])<\#(E)\le k)\cr
&\le\Pr(\text{there is a non-empty }E\subseteq M
  \text{ such that }\#(R[E])\le\#(E)\le k)\cr
&=\Pr(\text{there is an }E\subseteq M
  \text{ such that }3\le\#(R[E])\le\#(E)\le k)\cr
\noalign{\noindent (because if $E\ne\emptyset$ then $\#(R[E])\ge 3$)}
&\le\sum_{n=3}^{k}\sum_{E\in[M]^{n}}\sum_{F\in[L]^n}
  \Pr(R[E]\subseteq F)
\le\sum_{n=3}^{k}\#([M]^{n})\#([L]^n)\Bover{n^{3n}}{l^{3n}}\cr
&=\sum_{n=3}^{k}\Bover{m!}{n!(m-n)!}\Bover{l!}{n!(l-n)!}
  \Bover{n^{3n}}{l^{3n}}
\le\sum_{n=3}^{k}\Bover{m^{n}l^nn^{3n}}{n!n!l^{3n}}
\le\sum_{n=3}^k\Bover{m^nn^n3^{2n}}{l^{2n}}\cr
\noalign{\noindent (using (a))}
&=\sum_{n=3}^k\bigl(\Bover{9mn}{l^2}\bigr)^n
\le\sum_{n=3}^k\Bover1{2^n}
<1.\cr}$$

\noindent There must therefore be some $R\in\Omega$ such that
$\#(R[E])\ge\#(E)$ whenever $E\subseteq M$ and $\#(E)\le k$.

\medskip

{\bf (c)} If now $E\in [M]^{\le k}$,
the restriction $R_E=R\cap(E\times L)$ has the property that
$\#(R_E[I])\ge\#(I)$ for every $I\subseteq E$.   By Hall's Marriage
Lemma (3A1K) there is an injective function $f:E\to L$ such that
$(x,f(x))\in R_E\subseteq R$ for every $x\in E$.
}%end of proof of 392D

\cmmnt{\medskip

\noindent{\bf Remark} Of course this argument can be widely generalized;
see references in {\smc Kalton \& Roberts 83}.
}%end of comment

\leader{392E}{Lemma} Let $\frak A$ be a Boolean algebra and
$\nu:\frak A\to[0,\infty]$ a uniformly exhaustive submeasure.
Then for any
$\epsilon\in\ocint{0,\nu 1}$ the set $A=\{a:\nu a\ge\epsilon\}$ has
intersection number greater than $0$.

\proof{{\bf (a)} To begin with (down to the end of (d) below), suppose that
$\nu 1=1$.
Because $\nu$ is uniformly exhaustive, there is an $r\ge 1$ such
that whenever $\langle c_i\rangle_{i\in I}$ is a disjoint family in
$\frak A$ then $\#(\{i:\nu c_i>\bover15\epsilon\})\le r$, so that
$\sum_{i\in I}\nu c_i\le r+\bover15\epsilon\#(I)$.   Set
$\delta=\epsilon/5r$, $\eta=\bover1{74}\delta^2$, so that

\Centerline{$\delta-\eta\ge\bover1{18}(\delta-\eta)^2
\ge\bover1{18}(\delta^2-2\eta)=4\eta$.}

\medskip

{\bf (b)} Let $\familyiI{a_i}$ be a non-empty finite family in $A$.
Let $m$ be any multiple of $\#(I)$ greater than or equal to $1/\eta$.
Then there are integers $k$, $l$ such that

\Centerline{$3\eta\le\Bover{k}{m}\le 4\eta
\le\Bover1{18}(\delta-\eta)^2$,
\quad$\delta-\eta\le\Bover{l}{m}\le\delta$,}

\noindent in which case

\Centerline{$3\le k\le l\le m$,
\quad$18mk\le m^2(\delta-\eta)^2\le l^2$.}

\medskip

{\bf (c)} Take a set $M$ of the form $I\times S$ where $\#(S)=m/\#(I)$,
so that $\#(M)=m$.   For $x=(i,s)\in M$ set $d_x=a_i$.   Let $L$ be a
set with $l$ members.   By 392D, there is a set $R\subseteq M\times L$
such that every vertical section of $R$ has just three members and
whenever $E\in[M]^{\le k}$ there is an injective function $f_E:E\to L$
such that $(x,f_E(x))\in R$ for every $x\in E$.

For $E\subseteq M$ set

\Centerline{$b_E
=\inf_{x\in E}d_x\Bsetminus\sup_{x\in M\setminus E}d_x$,}

\noindent so that $\langle b_E\rangle_{E\subseteq M}$ is a partition of
unity in $\frak A$.   For $x\in M$ and $j\in L$ set

\Centerline{$c_{xj}=\sup\{b_E:x\in E\in[M]^{\le k},\,f_E(x)=j\}$.}

\noindent If $x$, $y$ are distinct members of $M$ and $j\in L$ then

\Centerline{$c_{xj}\Bcap c_{yj}
=\sup\{b_E:x$, $y\in E\in[M]^{\le k},\,f_E(x)=f_E(y)=j\}=0$,}

\noindent because every $f_E$ is injective.   Set

\Centerline{$m_j=\#(\{x:x\in M,\,c_{xj}\ne 0\})$}

\noindent for each $j\in L$.   Note that $c_{xj}=0$ if $(x,j)\notin R$,
so $\sum_{j\in L}m_j\le\#(R)=3m$.

We have

\Centerline{$\sum_{x\in M}\nu c_{xj}\le r+\Bover15\epsilon m_j$}

\noindent for each $j$, by the choice of $r$;  so

$$\eqalign{\sum_{x\in M,j\in L}\nu c_{xj}
&\le rl+\Bover15\epsilon\sum_{j\in L}m_j
\le rl+\Bover35m\epsilon\cr
&\le (r\delta+\Bover35\epsilon)m
=\Bover45\epsilon m
<\epsilon m\cr}$$

\noindent by the choice of $l$ and $\delta$.   There must therefore be
some $x\in M$ such that

\Centerline{$\nu(\sup_{j\in L}c_{xj})
\le\sum_{j\in L}\nu c_{xj}
<\epsilon\le\nu d_x$,}

\noindent and $d_x$ cannot be included in

\Centerline{$\sup_{j\in L}c_{xj}=\sup\{b_E:x\in E\in[M]^{\le k}\}$.}

\noindent But as $\sup\{b_E:x\in E\subseteq M\}$ is just $d_x$, there
must be an $E\subseteq M$, with cardinal greater than $k$, such that
$b_E\ne 0$.

Recall now that $M=I\times S$, and that

\Centerline{$k\ge 3\eta m=3\eta\#(I)\#(S)$.}

\noindent The set
$J=\{i:\,\Exists s$, $(i,s)\in E\}$ must therefore have more than
$3\eta\#(I)$ members,
since $E\subseteq J\times S$.   But also $d_{(i,s)}=a_i$ for each
$(i,s)\in E$, so that $\inf_{i\in J}a_i\Bsupseteq b_E\ne 0$.

\medskip

{\bf (d)} As $\familyiI{a_i}$ is arbitrary, the intersection number of
$A$ is at least $3\eta>0$.

\medskip

{\bf (e)} This completes the proof in the case in which $\nu 1=1$.   If
$\nu 1=0$ the result is vacuous.   If $\nu 1>0$, set
$\nu'a=\Bover{\min(\nu a,1)}{\min(\nu 1,1)}$ for each $a$;  then
it is easy to check that $\nu'$ is a uniformly
exhaustive submeasure with $\nu'1=1$, and

\Centerline{$\{a:\nu a\ge\epsilon\}
\subseteq\{a:\nu'a\ge\Bover{\min(\epsilon,1)}{\min(\nu 1,1)}\}$}

\noindent has non-zero intersection number for any
$\epsilon\in\ocint{0,\nu 1}$.
So the result is true in the generality stated.
}%end of proof of 392E

\leader{392F}{Theorem} Let $\frak A$ be a Boolean algebra with a
strictly positive uniformly exhaustive submeasure.   Then $\frak A$
is chargeable\cmmnt{, that is, has
a strictly positive finitely additive functional}.

\proof{ If $\frak A=\{0\}$ this is trivial.   Otherwise, let
$\nu:\frak A\to[0,\infty]$ be a strictly positive uniformly
exhaustive
submeasure.   For each $n$, $A_n=\{a:\nu a\ge\min(2^{-n},\nu 1)\}$ has
intersection number greater than $0$, and
$\bigcup_{n\in\Bbb N}A_n=\frak A\setminus\{0\}$ because $\nu$ is
strictly positive;  so $\frak A$ has a
strictly positive finitely additive functional, by Kelley's theorem
(391J).
}%end of proof of 392F

\leader{392G}{Corollary} Let $\frak A$ be a Boolean algebra.   Then it is
measurable iff it is \wsid\ and Dedekind $\sigma$-complete and has a
strictly positive uniformly exhaustive submeasure.

\proof{ Put 391D and 392F together.
}%end of proof of 392G

\leader{392H}{}\cmmnt{ This completes the main work of
this section.
However it will be convenient later to have some more facts available
which belong to the same group of ideas.

\medskip

\noindent}{\bf Metrics from submeasures:  Proposition} Let
$\frak A$ be a Boolean algebra and $\nu$ a
strictly positive totally finite submeasure on $\frak A$.

(a) We have a metric $\rho$ on $\frak A$ defined by the formula

\Centerline{$\rho(a,b)=\nu(a\Bsymmdiff b)$}

\noindent for all $a$, $b\in\frak A$.

(b) The Boolean operations $\Bcup$, $\Bcap$, $\Bsymmdiff$, $\Bsetminus$
and the function $\nu:\frak A\to\Bbb R$ are all uniformly continuous
for $\rho$.

(c) The metric space completion $(\widehat{\frak A},\hat\rho)$ of
$(\frak A,\rho)$ is a Boolean algebra under the natural continuous
extensions of the Boolean operations, and $\nu$ has a unique continuous
extension $\hat\nu$ to $\widehat{\frak A}$ which is again a strictly
positive submeasure.

(d) If $\nu$ is additive, then $(\widehat{\frak A},\hat\nu)$ is a
totally finite measure algebra.

\proof{{\bf (a)-(b)} This is just a generalization of 323A-323B;
essentially the same formulae can be used.   For the triangle inequality
for $\rho$, we have
$a\Bsymmdiff c\Bsubseteq(a\Bsymmdiff b)\Bcup(b\Bsymmdiff c)$, so

\Centerline{$\rho(a,c)
=\nu(a\Bsymmdiff c)
\le\nu(a\Bsymmdiff b)+\nu(b\Bsymmdiff c)
=\rho(a,b)+\rho(b,c)$.}

\noindent For the uniform continuity of the Boolean operations, we have

\Centerline{$(b\star c)\Bsymmdiff(b'\star c')
\Bsubseteq (b\Bsymmdiff b')\cup(c\Bsymmdiff c')$}

\noindent so that

\Centerline{$\rho(b\star c,b'\star c')\le\rho(b,b')+\rho(c,c')$}

\noindent for each of the operations $\star=\Bcup$, $\Bcap$, $\Bsetminus$
and $\Bsymmdiff$ and all $b$, $c$, $b'$, $c'\in\frak A$.
For the uniform continuity of the function $\nu$ itself, we have

\Centerline{$\nu b\le\nu c+\nu(b\Bsetminus c)\le\nu c+\rho(b,c)$,}

\noindent so that $|\nu b-\nu c|\le\rho(b,c)$.

\medskip

{\bf (c)} $\frak A\times\frak A$ is a dense subset of
$\widehat{\frak A}\times\widehat{\frak A}$, so the Boolean operations on
$\frak A$,
regarded as uniformly continuous functions from $\frak A\times\frak A$
to $\frak A\subseteq\widehat{\frak A}$, have unique extensions to
continuous binary operations on $\widehat{\frak A}$ (3A4G).   If we look at

\Centerline{$A=\{(a,b,c):a\Bsymmdiff(b\Bsymmdiff c)
=(a\Bsymmdiff b)\Bsymmdiff c\}$,}

\noindent this is a closed subset of
$\widehat{\frak A}\times\widehat{\frak A}\times\widehat{\frak A}$,
because the maps
$(a,b,c)\mapsto a\Bsymmdiff(b\Bsymmdiff c)$,
$(a,b,c)\mapsto(a\Bsymmdiff b)\Bsymmdiff c$ are continuous and the topology of $\widehat{\frak A}$
is Hausdorff;  since $A$ includes the dense set
$\frak A\times\frak A\times\frak A$, it is the whole of
$\widehat{\frak A}\times\widehat{\frak A}\times\widehat{\frak A}$, that is,
$a\Bsymmdiff(b\Bsymmdiff c)=(a\Bsymmdiff b)\Bsymmdiff c$ for all $a$,
$b$, $c\in\widehat{\frak A}$.   All the other identities we need to show
that $\widehat{\frak A}$ is a Boolean algebra can be confirmed by the
same method.   Of course $\frak A$ is now a subalgebra of
$\widehat{\frak A}$.

Because $\nu:\frak A\to\coint{0,\infty}$ is uniformly continuous, it has
a unique continuous extension
$\hat\nu:\widehat{\frak A}\to\coint{0,\infty}$.   We have

\Centerline{$\hat\nu 0=0$,
\quad$\hat\nu a\le\hat\nu(a\Bcup b)\le\hat\nu a+\hat\nu b$,
\quad$\hat\nu a=\hat\rho(a,0)$}

\noindent for every $a$, $b\in\frak A$ and therefore for every
$a$, $b\in\widehat{\frak A}$, so $\hat\nu$ is a submeasure on
$\widehat{\frak A}$, and

\Centerline{$\hat\nu a=0\Longrightarrow\hat\rho(a,0)=0
\Longrightarrow a=0$,}

\noindent so $\hat\nu$ is strictly positive.

\medskip

{\bf (d)} We have $\nu(a\Bcup b)+\nu(a\Bcap b)=\nu a+\nu b$ for all $a$,
$b\in\frak A$;  because all the operations are continuous,
$\hat\nu(a\Bcup b)+\hat\nu(a\Bcap b)=\hat\nu a+\hat\nu b$ for all $a$,
$b\in\widehat{\frak A}$.   In particular, since $\hat\nu 0=0$,
$\hat\nu$ is additive.
Next, if $\sequencen{a_n}$ is a non-decreasing sequence in
$\widehat{\frak A}$,
$\hat\rho(a_m\Bsymmdiff a_n)=|\hat\nu a_m-\hat\nu a_n|$ for all $m$, $n\in\Bbb N$,
and $\sequencen{a_n}$ is $\hat\rho$-Cauchy, therefore convergent to some
$a\in\widehat{\frak A}$.
Since

\Centerline{$a\Bcap a_n=\lim_{m\to\infty}a_m\Bcap a_n=a_n$}

\noindent for each $n$,
$a\Bsupseteq a_n$ for every $n$.   If $b\in\widehat{\frak A}$ is any upper
bound for $\{a_n:n\in\Bbb N\}$, then

\Centerline{$b\Bcap a=\lim_{n\to\infty}b\Bcap a_n
=\lim_{n\to\infty}a_n=a$}

\noindent and $b\Bsupseteq a$;  thus $a$ is the least upper bound
of $\{a_n:n\in\Bbb N\}$.

So, first, if $\sequencen{b_n}$ is any sequence in $\widehat{\frak A}$, and
we set $a_n=\sup_{i\le n}b_i$ for each $n$, $\sup_{n\in\Bbb N}a_n$ is
defined and must be equal to $\sup_{n\in\Bbb N}b_n$;  accordingly $\widehat{\frak A}$
is Dedekind $\sigma$-complete.   Next, if $\sequencen{b_n}$ is a disjoint
sequence in $\widehat{\frak A}$, and again we set $a_n=\sup_{i\le n}b_i$ for each $n$,
$a=\sup_{n\in\Bbb N}a_n=\sup_{n\in\Bbb N}b_n$, we shall have

\Centerline{$\hat\nu a
=\lim_{n\to\infty}\hat\nu a_n
=\lim_{n\to\infty}\sum_{i=0}^n\hat\nu b_i
=\sum_{n=0}^{\infty}\hat\nu b_n$;}

\noindent which means that $\hat\nu$ is countably additive, and
$(\widehat{\frak A},\hat\nu)$ is a measure algebra.
}%end of proof of 392H

\leader{392I}{Corollary} Let $\frak A$ be a Boolean algebra and $\nu$ a
non-negative additive functional on $\frak A$.   Then there are a totally
finite measure algebra $(\frak C,\bar\mu)$ and a Boolean homomorphism
$\pi:\frak A\to\frak C$ such that $\nu a=\bar\mu(\pi a)$ for every
$a\in\frak A$.

\proof{ Set $I=\{a:\nu a=0\}$;  then $I\normalsubgroup\frak A$, so we can
form the quotient algebra $\frak B=\frak A/I$ (312L);  let
$\pi:\frak A\to\frak B$ be the canonical map.   As in part (b) of the proof
of 321H, we have an additive functional $\mu:\frak B\to\coint{0,\infty}$
such that $\mu(\pi a)=\nu a$ for every $a\in\frak A$, and (as in 321H)
$\mu$ is strictly positive.   Take
$(\frak C,\bar\mu)$ to be $(\widehat{\frak B},\hat\mu)$ as in 392Hd, so
that $(\frak C,\bar\mu)$ is a totally finite measure algebra.   If we now
think of $\pi$ as a map from $\frak A$ to $\frak C$, it will still be a
Boolean homomorphism, and

\Centerline{$\nu a=\mu(\pi a)=\bar\mu(\pi a)$}

\noindent for every $a\in\frak A$.
}%end of proof of 392I

\leader{392J}{Proposition}\dvAnew{2008}
Let $\frak A$ be a Boolean algebra, $\nu$
an exhaustive submeasure on $\frak A$, and $\sequencen{a_n}$ a sequence in
$\frak A$ such that $\inf_{n\in\Bbb N}\nu a_n>0$.   Then there is an
infinite $I\subseteq\Bbb N$ such that $\nu(\inf_{i\in I\cap n}a_i)>0$ for
every $n\in\Bbb N$.

\cmmnt{\medskip

\noindent{\bf Remark} In the formula $I\cap n$ I am identifying $n$ with
the set of its predecessors, as in 3A1H.
}

\proof{ For finite $J\subseteq\Bbb N$ set $b_J=\inf_{i\in J}a_i$.   Let
$\Cal J$ be the family of those $J\in[\Bbb N]^{<\omega}$ such that
$\limsup_{n\to\infty}\nu(a_n\Bcap b_J)>0$.

\Quer\ Suppose, if possible, that there is no strictly increasing sequence
in $\Cal J$.   Then $\Cal J$ must have a maximal element $J$ say.
Set $a'_n=a_n\Bcap b_J$ for $n\in\Bbb N$ and
$\delta=\limsup_{n\to\infty}\nu a'_n>0$.   For any $n\in\Bbb N\setminus J$,
$J\cup\{n\}\notin\Cal J$ so

\Centerline{$\lim_{m\to\infty}a'_m\Bcap a'_n
=\lim_{m\to\infty}a_m\Bcap b_{J\cup\{n\}}=0$.}

\noindent We can therefore choose inductively a sequence $\sequencen{k_n}$
such that

\Centerline{$k_n>\sup J$,
\quad$\nu a'_{k_n}\ge\Bover34\delta$,
\quad$\nu(a'_{k_n}\Bcap a'_{k_i})\le 2^{-i-2}\delta$
for every $i<n$}

\noindent for each $n\in\Bbb N$.   Now set
$b_n=a_{k_n}\Bsetminus\sup_{i<n}a_{k_i}$ for each $n$.   Then
$\sequencen{b_n}$ is disjoint.   Also

\Centerline{$\Bover34\delta
\le\nu a_{k_n}
\le\nu b_n+\sum_{i=0}^{n-1}\nu(a_{k_n}\Bcap a_{k_i})
\le\nu b_n+\sum_{i=0}^{n-1}2^{-i-2}\delta
\le\nu b_n+\Bover12\delta$}

\noindent and $\nu b_n\ge\bover14\delta$ for every $n$;  which is
impossible.\ \Bang

There must therefore be a strictly increasing sequence $\sequencen{J_n}$ in
$\Cal J$.   Set $I=\bigcup_{n\in\Bbb N}J_n$.   If $n\in\Bbb N$, there is an
$m\in\Bbb N$ such that $I\cap n\subseteq J_m$ and
$\nu(\inf_{i\in I\cap n}a_i)\ge\nu b_{J_m}>0$.   So we have an appropriate
$I$.
}%end of proof of 392J

\leader{*392K}{Products of submeasures}\dvAnew{2008} \cmmnt{There seems
to be no fully satisfying general construction for
products of submeasures.   However the
following method has some interesting features.

\medskip

}{\bf (a)} Let $\frak A$ and $\frak B$ be Boolean algebras with
submeasures $\mu$, $\nu$ respectively.   On the free product
$\frak A\otimes\frak B$\cmmnt{ (\S315)}, we have a
functional $\mu\ltimes\nu$ defined by saying that whenever
$c\in\frak A\otimes\frak B$ is of the form
$\sup_{i\in I}a_i\otimes b_i$ where $\familyiI{a_i}$ is a finite
partition of unity in $\frak A$, then

$$\eqalign{(\mu\ltimes\nu)(c)
&=\min_{J\subseteq I}
  \max(\{\mu(\sup_{i\in J}a_i)\}\cup\{\nu b_i:i\in I\setminus J\})\cr
&=\min\{\epsilon:\epsilon\in[0,\infty],
  \,\mu(\sup\{a_i:i\in I,\,\nu b_i>\epsilon\})
\le\epsilon\}.\cr}$$

\prooflet{\noindent\Prf\ Every $c\in\frak A\otimes\frak B$ 
can be expressed in
this form (315Oa).   Of course this can be done in
many different ways.   But if
$c=\sup_{j\in J}a'_j\otimes b'_j$ is another expression of the same
kind, then $b_i=b'_j$ whenever $a_i\Bcap a'_j\ne 0$.   So

$$\eqalign{\sup\{a_i:i\in I,\,\nu b_i>\epsilon\}
&=\sup\{a_i\Bcap a'_j:i\in I,\,j\in J,\,a_i\Bcap a'_j\ne 0,\,
  \nu b_i>\epsilon\}\cr
&=\sup\{a_i\Bcap a'_j:i\in I,\,j\in J,\,a_i\Bcap a'_j\ne 0,\,
  \nu b'_j>\epsilon\}\cr
&=\sup\{a'_j:j\in J,\,\nu b'_j>\epsilon\}\cr}$$

\noindent for any $\epsilon$, and the two calculations for $\mu\ltimes\nu$
give the same result.\ \Qed}

\cmmnt{Note that }$(\mu\ltimes\nu)(a\otimes b)=\min(\mu a,\nu b)$ for all
$a\in\frak A$ and $b\in\frak B$.

\spheader 392Kb In the context of (a), $\mu\ltimes\nu$ is a submeasure.

\prooflet{\Prf\ By definition, $(\mu\ltimes\nu)c\ge 0$ for every
$c\in\frak A\otimes\frak B$;  and if $c=0$ then $c=1\otimes 0$ and
$(\mu\ltimes\nu)c=0$.

If $c$, $c'$ are two members of $\frak A\otimes\frak B$, express them
in the forms
$c=\sup_{i\in I}a_i\otimes b_i$ and $c'=\sup_{j\in J}a'_j\otimes
b'_j$ where $\familyiI{a_i}$ and
$\family{j}{J}{a'_j}$ are partitions of unity in $\frak A$.   Set
$K=\{(i,j):a_i\Bcap a'_j\ne 0\}\subseteq I\times J$,
$a''_{ij}=a_i\Bcap a'_j$ for $(i,j)\in K$;  then $\langle
a''_{ij}\rangle_{(i,j)\in K}$ is a partition of unity in $\frak A$,
$c=\sup_{(i,j)\in K}a''_{ij}\otimes b_i$ and
$c'=\sup_{(i,j)\in K}a''_{ij}\otimes b'_j$.   Set
$\alpha=(\mu\ltimes\nu)c$, $\beta=(\mu\ltimes\nu)c'$,
$L=\{(i,j):(i,j)\in K$, $\nu b_i>\alpha\}$, $L'=\{(i,j):(i,j)\in K$,
$\nu b'_j>\beta\}$,
$e=\sup\{a_{ij}:(i,j)\in L\}$ and $e'=\sup\{a_{ij}:(i,j)\in L'\}$;
then $\mu e\le\alpha$ and
$\mu e'\le\beta$.   So $\mu(e\Bcup e')\le\alpha+\beta$;  but
$e\Bcup e'=\sup_{(i,j)\in L\cup L'}a''_{ij}$ and

\Centerline{$\nu(b_i\Bcup b'_j)\le\nu b_i+\nu b'_j\le\alpha+\beta$}

\noindent for all $(i,j)\in K\setminus(L\cup L')$.   So
$(\mu\ltimes\nu)(c\Bcup c')\le\alpha+\beta$.

If $c\Bsubseteq c'$, then $b_i\Bsubseteq b'_j$ for every $(i,j)\in
K$.   So $\nu b_i\le\beta$ for every $(i,j)\in K\setminus L'$ and
$(\mu\ltimes\nu)c\le\beta$.

Thus $\mu\ltimes\nu$ is subadditive and order-preserving and is a
submeasure.\ \Qed}

\spheader 392Kc I note that only in exceptional cases will $\mu\ltimes\nu$
be matched with $\nu\ltimes\mu$ by the canonical isomorphism between
$\frak A\otimes\frak B$ and $\frak B\otimes\frak A$;  this product is
not `commutative'.   \cmmnt{(See 392Yc.)}
It is however `associative', in the following sense.
Let $(\frak A_1,\mu_1)$, $(\frak A_2,\mu_2)$, $(\frak A_3,\mu_3)$ be
Boolean algebras endowed with submeasures.   Set

\Centerline{$\lambda_{12}=\mu_1\ltimes\mu_2$,
\quad$\lambda_{(12)3}=\lambda_{12}\ltimes\mu_3$,
\quad$\lambda_{23}=\mu_2\ltimes\mu_3$,
\quad$\lambda_{1(23)}=\mu_1\ltimes\lambda_{23}$.}

\noindent Then the canonical isomorphisms between
$(\frak A_1\otimes\frak A_2)\otimes\frak A_3$,
$\frak A_1\otimes\frak A_2\otimes\frak A_3$
and $\frak A_1\otimes(\frak A_2\otimes\frak A_3)$\cmmnt{ (315L)}
identify $\lambda_{(12)3}$ with $\lambda_{1(23)}$.

\prooflet{\Prf\ Take any $d\in\frak A_1\otimes\frak A_2\otimes\frak A_3$.
Express $d$ as
$\sup_{i\in I}a_i\otimes e_i$ where $\familyiI{a_i}$ is a partition
of unity in $\frak A_1$ and $e_i\in\frak A_2\otimes\frak A_3$ for
each $i$;  express each $e_i$ as
$\sup_{j\in J_i}b_{ij}\otimes c_{ij}$ where $\family{j}{J_i}{b_{ij}}$
is a partition of unity in $\frak A_2$ and $c_{ij}\in\frak A_3$ for
$i\in I$, $j\in J_i$.   In this case,
$\langle a_i\otimes b_{ij}\rangle_{i\in I,j\in J_i}$ is a partition
of unity in $\frak A_1\otimes\frak A_2$ and
$d=\sup_{i\in I,j\in J_i}a_i\otimes b_{ij}\otimes c_{ij}$.

Let $\epsilon>0$.   For $i\in I$, set $J'_i=\{j:j\in J_i$,
$\mu_3c_{ij}>\epsilon\}$,
$e'_i=\sup_{j\in J'_i}b_{ij}$.   Then
$\lambda_{23}(\sup_{j\in J_i}b_{ij}\otimes c_{ij})\le\epsilon$ iff
$\mu_2e'_i\le\epsilon$.
Set $I'=\{i:\mu_2e'_i>\epsilon\}$;  then
$\lambda_{1(23)}d\le\epsilon$ iff
$\mu_1(\sup_{i\in I'}a_i)\le\epsilon$.   From the other direction, set
$f=\sup\{a_i\otimes b_{ij}:i\in I$, $j\in J'_i\}$;  then
$\lambda_{(12)3}d\le\epsilon$ iff $\lambda_{12}f\le\epsilon$.   But
$f=\sup_{i\in I}a_i\otimes e'_i$, so
$\lambda_{12}f\le\epsilon$ iff $\mu_1(\sup_{i\in I'}a_i)\le\epsilon$.

As $\epsilon$ and $d$ are arbitrary,
$\lambda_{(12)3}=\lambda_{1(23)}$, as claimed.\ \Qed}

\spheader 392Kd If $\mu$, $\mu'$ are submeasures on $\frak A$,
$\nu$ and $\nuprime$ are submeasures on $\frak B$,
$\mu$ is absolutely continuous
with respect to $\mu'$ and $\nu$ is
absolutely continuous with respect to $\nuprime$, then
$\mu\ltimes\nu$ is absolutely continuous with respect to
$\mu'\otimes\nuprime$.   \prooflet{\Prf\ For any $\epsilon>0$ there is a
$\delta>0$ such that
$\mu a\le\epsilon$ whenever $\mu'a\le\delta$ and $\nu b\le\epsilon$
whenever $\nuprime b\le\delta$.   If now $c\in\frak A\otimes\frak B$ and
$(\mu'\ltimes\nuprime)(c)\le\delta$, we have an expression
$c=\sup_{i\in I}a_i\otimes b_i$ and a set $J\subseteq I$ such that
$\familyiI{a_i}$ is a partition of unity,
$\mu'(\sup_{i\in J}a_i)\le\delta$ and $\nuprime b_i\le\delta$ for every
$i\in I\setminus J$;  so $\mu(\sup_{i\in J}a_i)\le\epsilon$,
$\nu b_i\le\epsilon$ for every
$i\in I\setminus J$ and $(\mu\ltimes\nu)(c)\le\epsilon$.\ \Qed}

\spheader 392Ke If $\mu$ and $\nu$ are exhaustive, so is
$\mu\ltimes\nu$.   \prooflet{\Prf\ Let $\sequencen{c_n}$ be a sequence in
$\frak A\otimes\frak B$ such that $(\mu\ltimes\nu)c_n>\epsilon>0$ for
every $n$.   For each $n$, express $c_n$ as
$\sup_{i\in I_n}a_{ni}\otimes b_{ni}$ where $\family{i}{I_n}{a_{ni}}$
is a partition of unity;  set $I'_n=\{i:i\in I_n$,
$\nu b_{ni}>\epsilon\}$,
$a_n=\sup_{i\in I'_n}a_{ni}$;  then
$\mu a_n>\epsilon$.   By 392J,
there is an infinite $J\subseteq\Bbb N$ such that
$\inf_{i\in J\cap n}a_i\ne 0$ for every $n\in\Bbb N$.
Let $Z$ be the Stone space of $\frak A$, and write $\widehat{a}\subseteq Z$
for the open-and-closed set corresponding to $a\in\frak A$;  then
there is a $z\in\bigcap_{n\in J}\widehat{a}_n$.   For every $n\in J$
there is an $i_n\in I'_n$ such
that $z\in\widehat{a}_{n,i_n}$.   But now observe that
$\nu b_{n,i_n}>\epsilon$ for every $n\in J$, so there must be distinct
$m$, $n\in J$ such that
$b_{m,i_m}\Bcap b_{n,i_n}\ne 0$;  as $a_{m,i_m}\Bcap a_{n,i_n}$ is
also non-zero, $c_m\Bcap c_n\ne 0$.
As $\sequencen{c_n}$ is arbitrary, $\mu\ltimes\nu$ is exhaustive.\ \Qed}

\spheader 392Kf We can extend the construction to infinite products, as
follows.   Let $I$ be a totally ordered set and
$\familyiI{(\frak A_i,\mu_i)}$ a family of Boolean
algebras endowed with unital submeasures.   For a finite set
$J=\{i_0,\ldots,i_n\}$ where $i_0<\ldots<i_n$ in $I$, let $\lambda_J$
be the product submeasure
$(.(\mu_{i_0}\ltimes\mu_{i_1})\ltimes\ldots)\ltimes\mu_{i_n}$ on
$\frak C_J=\bigotimes_{j\in J}\frak A_j$;  for definiteness, on
$\frak C_{\emptyset}=\{0,1\}$
take $\lambda_{\emptyset}$ to be the unital submeasure, while
$\frak C_{\{i\}}=\frak A_i$ and $\lambda_{\{i\}}=\mu_i$ for each $i\in I$.
Using (c) repeatedly, we see that if $J$, $K\in[I]^{<\omega}$ and
$j<k$ for every $j\in J$,
$k\in K$, then the identification of $\frak C_{J\cup K}$ with
$\frak C_J\otimes\frak C_K$\cmmnt{ (315L)} matches $\lambda_{J\cup K}$
with $\lambda_J\ltimes\lambda_K$.
Moreover, if $K\in[I]^{<\omega}$ and $J$ is any subset of $K$\cmmnt{ (not
necessarily an initial segment)} and
$\varepsilon_{JK}:\frak C_J\to\frak C_K$ is the canonical embedding
corresponding to the identification of $\frak C_K$ with
$\frak C_J\otimes\frak C_{K\setminus J}$, then
$\lambda_J=\lambda_K\varepsilon_{JK}$;  this also is an easy
induction on $\#(K)$.   What this means is that for any subset $M$ of
$I$ we have a submeasure $\lambda_M$ on
$\frak C_M=\bigcup\{\varepsilon_{JM}\frak C_J:J\in[M]^{<\omega}\}$,
being the unique functional such that
$\lambda_M\varepsilon_{JM}=\lambda_J$ for every
$J\in[M]^{<\omega}$.   Finally, if
$L$, $M$ are subsets of $I$ with $l<m$ for every $l\in L$ and
$m\in M$, then $\lambda_{L\cup M}$ can be identified with
$\lambda_L\ltimes\lambda_M$.

\spheader 392Kg I should perhaps have remarked already that if $\mu$ and
$\nu$, in (a), are additive and unital,
then we have an additive function $\lambda'$ on $\frak A\otimes\frak B$
such that
$\lambda'(a\otimes b)=\mu a\cdot\nu b$ for every $a\in\frak A$ and
$b\in\frak B$\cmmnt{ (326E)}.   Now, setting $\lambda=\mu\ltimes\nu$,
each of $\lambda$, $\lambda'$ is absolutely continuous with respect to the
other.   \prooflet{\Prf\ If $c\in\frak A\otimes\frak B$, express $c$ as
$\sup_{i\in I}a_i\otimes b_i$ where
$\familyiI{a_i}$ is a finite partition of unity.
Then $\mu(\sup\{a_i:\nu b_i>\lambda c\})\le\lambda c$, so
$\lambda'c=\sum_{i\in I}\mu a_i\cdot\nu b_i$ is at most $2\lambda c$.
On the other hand,
$\mu(\sup\{a_i:\nu b_i>\sqrt{\lambda'c}\})\le\sqrt{\lambda'c}$, so
$\lambda c\le\sqrt{\lambda'c}$.\ \Qed}

\exercises{
\leader{392X}{Basic exercises (a)}
%\spheader 392Xa
Show that the first two clauses of the definition 392A can be replaced
by `$\nu a\le\nu(a\Bcup b)\le\nu a+\nu b$ whenever $a\Bcap b=0$'.
%392A

\spheader 392Xb Let $\frak A$ be any Boolean algebra and $\nu$ a
finite-valued submeasure on $\frak A$.   (i) Show that
$\nu$ is order-continuous iff whenever
$A\subseteq\frak A$ is non-empty, downwards-directed and has infimum
$0$, then $\inf_{a\in A}\nu a=0$.   (ii) Show that in this case
$\nu$ is exhaustive.   \Hint{if $\sequencen{a_n}$ is disjoint, then
$\bigcup_{n\in\Bbb N}\{b:b\Bsupseteq a_i$ for every $i\ge n\}$ has
infimum $0$.}
%392C

\spheader 392Xc Let $\frak A$ be a Boolean algebra and $\mu$, $\nu$ two
strictly positive submeasures on $\frak A$, each of which is absolutely
continuous with respect to the other.   Show that they induce uniformly
equivalent metrics on $\frak A$ (392H), so that both give the same metric
completion of $\frak A$.
%392H

\spheader 392Xd Let $\frak A$, $\frak B$ be Boolean algebras with
uniformly exhaustive submeasures $\mu$, $\nu$ respectively.
Show that $\mu\ltimes\nu$ is uniformly exhaustive.
%392K

\leader{392Y}{Further exercises (a)}
%\spheader 392Ya
Let $\frak A$ be a Boolean algebra and
$\lambda:\frak A\to[0,1]$ a functional such that $\lambda 0=0$
and $\lambda a\le\lambda(a\Bcup b)\le 2\max(\lambda a,\lambda b)$ for all
$a$, $b\in\frak A$.   Show that there is a submeasure $\nu$ on $\frak A$
such that $\bover12\lambda\le\nu\le\lambda$.
%\Hint{consider first the
%case in which every non-zero value of $\lambda$ is of the form $2^{-k}$.}
%392A

\spheader 392Yb (T.Jech)
Show that a Boolean algebra $\frak A$ is chargeable
iff there are sequences $\sequencen{A_n}$ and
$\sequencen{k_n}$ such that
($\alpha$) $\bigcup_{n\in\Bbb N}A_n=\frak A\setminus\{0\}$
($\beta$) whenever $a$, $b\in\frak A$, $n\in\Bbb N$
and $a\Bcup b\in A_n$ then at least
one of $a$, $b$ belongs to $A_{n+1}$ ($\gamma$) if $n\in\Bbb N$ then
$k_n\in\Bbb N$ and if $a_0,\ldots,a_{k_n}\in\frak A$ are disjoint
then some $a_j$ does not belong to $A_n$.
%\Hint{392Ya.}
%392F 392Ya

\spheader 392Yc I will say that a submeasure $\nu$ on a Boolean algebra
$\frak A$ is {\bf properly atomless} if for every $\epsilon>0$ there is a
finite partition $A$ of unity in $\frak A$ such that $\nu a\le\epsilon$ for
every $a\in A$.   (Compare 326F.)
(i) Show that if $\frak A$ and $\frak B$ are Boolean algebras
with submeasures $\mu$, $\nu$ respectively, we have a functional
$\mu\rtimes\nu:\frak A\otimes\frak B\to[0,\infty]$ defined by saying that

\Centerline{$(\mu\rtimes\nu)(\sup_{i\in I}a_i\otimes b_i)
=\min_{J\subseteq I}
  \max(\{\nu(\sup_{i\in J}b_i)\}\cup\{\mu a_i:i\in I\setminus J\})$}

\noindent whenever $\familyiI{b_i}$ is a finite partition of unity in
$\frak B$ and $a_i\in\frak A$ for each $i\in I$.   (ii) Show that if
$\mu$ is a non-zero properly atomless submeasure,
$\nu$ is a submeasure, and
$\mu\ltimes\nu$ is absolutely continuous with respect to $\mu\rtimes\nu$,
then $\nu$ is uniformly exhaustive.
%392K

\spheader 392Yd (See 328H.)
Let $(I,\le)$ be a non-empty upwards-directed partially
ordered set, and $\familyiI{(\frak A_i,\bar\mu_i)}$
a family of probability algebras;  suppose
that $\pi_{ji}:\frak A_i\to\frak A_j$ is a measure-preserving
Boolean homomorphism whenever $i\le j$, and that
$\pi_{ki}=\pi_{kj}\pi_{ji}$ whenever $i\le j\le k$.
(i) Let $\Cal F$ be the filter

\Centerline{$\{A:A\subseteq I$, there is some $i\in I$ such that $j\in A$
whenever $i\le j\}$,}

\noindent and set
$\nu\familyiI{a_i}=\limsup_{i\to\Cal F}\bar\mu_ia_i$ for
$\familyiI{a_i}\in\prod_{i\in I}\frak A_i$.   Show that $\nu$ is a
submeasure on $\frak A=\prod_{i\in I}\frak A_i$.   (ii) Let $\Cal J$ be the
ideal $\{d:\nu d=0\}$ of $\frak A$, and $\frak D$ the quotient algebra
$\frak A/\Cal J$.   Show that we have a strictly positive
unital submeasure
$\bar\nu$ on $\frak D$ such that $\bar\nu d^{\ssbullet}=\nu d$ for every
$d\in\frak A$, and that $\frak D$ is complete under the metric defined by
$\bar\nu$.   (iii) Show that for each $i\in I$ we have a Boolean
homomorphism $\pi_i:\frak A_i\to\frak D$ defined by setting
$\pi_ia=\family{j}{I}{a_j}^{\ssbullet}$, where $a_j=\pi_{ji}a$ if
$j\ge i$, $0_{\frak A_j}$ otherwise, and that $\bar\nu\pi_i=\bar\mu_i$.
Show that $\pi_i=\pi_j\phi_{ji}$ whenever $i\le j$.
(iv) Show that
$\frak D_0=\bigcup_{i\in I}\pi_i[\frak A_i]$ is a subalgebra of $\frak D$,
and that $\bar\nu\restrp\frak D_0$ is additive.   (v) Let $\frak C$ be the
closure of $\frak D_0$ in $\frak D$, and set
$\bar\lambda=\bar\nu\restrp\frak C$.   Show that $(\frak C,\bar\lambda)$ is
a probability algebra.
(vi) Now suppose that $(\frak B,\bar\nu)$ is a probability algebra, and
that for each $i\in I$ we are given a measure-preserving
Boolean homomorphism $\phi_i:\frak A_i\to\frak B$ such that
$\phi_i=\phi_j\pi_{ji}$ whenever $i\le j$.   Show that
there is a unique measure-preserving Boolean homomorphism
$\phi:\frak C\to\frak B$ such that $\phi\pi_i=\phi_i$ for every $i\in I$.
%328H 392H out of order query

\spheader 392Ye Let $\frak A$ be a Boolean algebra and
$\nu:\frak A\to[0,\infty]$ a submeasure.   Show that
the following are equiveridical:   (i) $\nu$ is uniformly exhaustive;
(ii) whenever
$\sequencen{a_n}$ is a sequence in $\frak A$ such that
$\inf_{n\in\Bbb N}\nu a_n>0$, there is a set
$I\subseteq\Bbb N$, not of zero asymptotic density, such that
$a_i\Bcap a_j\ne 0$ for all $i$, $j\in I$;  (iii)
whenever $\sequencen{a_n}$ is a sequence in $\frak A$ such that
$\inf_{n\in\Bbb N}\nu a_n>0$, there is a set
$I\subseteq\Bbb N$, not of zero asymptotic density, such that
$\nu(\inf_{i\in I,i\le n}a_i)>0$ for every $n\in\Bbb N$.
%392J out of order query
}%end of exercises

\cmmnt{\Notesheader{392} Much of the first part of this section is a
matter of generalizing earlier arguments.   Thus 392C ought by now to be
very easy, while 392Xb recalls the elementary theory of $\tau$-additive
functionals.

The new ideas are in the combinatorics of 392D-392E.
I have cast 392D in the form of an argument in probability theory.   Of
course there is nothing here but simple counting, since the probability
measure simply puts the same mass on each point of $\Omega$, and every
statement of the form `$\Pr(R\,\ldots)\le\ldots$' is just a matter of
counting the elements $R$ of $\Omega$ with the given property.   But I
think many of us find that the probabilistic language makes the
calculations more natural;  in particular, we can use intuitions
associated with the notion of independence of events.   Indeed I
strongly recommend the method.   It has been used to very great effect
in the last sixty years in a wide variety of combinatorial
problems.
392F and 392G together constitute the {\bf Kalton-Roberts theorem}
({\smc Kalton \& Roberts 83}).

}%end of notes

\discrpage

\frfilename{mt393.tex}
\versiondate{11.5.08}
\copyrightdate{2007}

\def\chaptername{Measurable algebras}
\def\sectionname{Maharam submeasures}

\newsection{393}

Continuing our exploration of variations on the idea of `measurable
algebra', we come to sequentially order-continuous submeasures.
These are associated with `Maharam algebras' (393E), which share a great
many properties with measurable algebras;  for instance, the existence of a
standard topology defined by the algebraic structure (393G).   This
topology is intimately connected with the order*-convergence of sequences
introduced in \S367 (393L).   We can indeed characterize Maharam algebras
in terms of properties of
the order-sequential topology defined by this convergence
(393Q).

\leader{393A}{Definition}\dvAformerly{392G}
Let $\frak A$ be a Boolean algebra.   A {\bf
Maharam submeasure} or {\bf continuous outer measure} on $\frak A$ is a
totally finite submeasure $\nu:\frak A\to\coint{0,\infty}$ such that
$\lim_{n\to\infty}\nu a_n=0$ whenever $\sequencen{a_n}$ is a
non-increasing sequence in $\frak A$ with infimum $0$.

\leader{393B}{Lemma}\dvAformerly{392H}
Let $\frak A$ be a Boolean algebra and $\nu$ a
Maharam submeasure on $\frak A$.

(a) $\nu$ is sequentially order-continuous.

(b) $\nu$ is `countably subadditive', that is, whenever
$\sequencen{a_n}$ is a sequence in $\frak A$ and $a\in\frak A$ is such
that $a=\sup_{n\in\Bbb N}a\Bcap a_n$, then $\nu
a\le\sum_{n=0}^{\infty}\nu a_n$.

(c) If $\frak A$ is Dedekind $\sigma$-complete, then $\nu$ is
exhaustive.

\proof{{\bf (a)} (Of course $\nu$ is an order-preserving function, by
the definition of `submeasure';  so we can apply the ordinary definition
of `sequentially order-continuous' in 313Hb.)  (i) If $\sequencen{a_n}$
is a non-decreasing sequence in $\frak A$ with supremum $a$, then
$\sequencen{a_n\Bsetminus a}$ is a non-increasing sequence with infimum
$0$, so $\lim_{n\to\infty}\nu(a_n\Bsetminus a)=0$;  but as

\Centerline{$\nu a_n\le\nu a\le\nu a_n+\nu(a\Bsetminus a_n)$}

\noindent for every $n$, it follows that $\nu a=\lim_{n\to\infty}\nu
a_n$.   (ii) If $\sequencen{a_n}$ is a non-increasing sequence in $\frak
A$ with infimum $a$, then

\Centerline{$\nu a\le\nu a_n\le\nu a+\nu(a_n\Bsetminus a)\to \nu a$}

\noindent as $n\to\infty$.

\medskip

{\bf (b)} Set $b_n=\sup_{i\le n}a\Bcap a_i$;  then $\nu
b_n\le\sum_{i=0}^n\nu a_i$ for each $n$ (inducing on $n$), so that

\Centerline{$\nu a=\lim_{n\to\infty}\nu b_n\le\sum_{i=0}^{\infty}\nu
a_i$.}

\medskip

{\bf (c)} If $\sequencen{a_n}$ is a disjoint sequence in $\frak A$, set
$b_n=\sup_{i\ge n}a_i$ for each $n$;  then $\inf_{n\in\Bbb N}b_n=0$, so

\Centerline{$\limsup_{n\to\infty}\nu a_n\le\lim_{n\to\infty}\nu b_n=0$.}

}%end of proof of 393B

\leader{393C}{Proposition} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra and $\nu$ a strictly positive Maharam submeasure on
$\frak A$.   Then $\frak A$ is ccc, Dedekind complete and \wsid, and
$\nu$ is order-continuous.

\proof{ By 393Bc, $\nu$ is exhaustive;  by 392Ca, $\frak A$ is ccc;  by
316Fa, $\frak A$ is Dedekind complete;
by 316Fc and 393Ba, $\nu$ is order-continuous

Now suppose that we have a sequence $\sequencen{A_n}$ of non-empty
downwards-directed subsets of $\frak A$, all with infimum $0$.   Let $B$
be the set

\Centerline{$\{b:b\in\frak A,\,\Forall \,n\in\Bbb
N\,\,\exists\,\,a\in A_n$ such that $a\Bsubseteq b\}$.}

\noindent As $\nu$ is order-continuous,
$\inf_{a\in A_n}\nu a=0$ for each $n$.   Given $\epsilon>0$, we can
choose $\sequencen{a_n}$ such that $a_n\in A_n$ and
$\nu a_n\le 2^{-n}\epsilon$
for each $n$;  now $b=\sup_{n\in\Bbb N}a_n$ belongs to $B$ and
$\nu b\le\sum_{n=0}^{\infty}\nu a_n\le 2\epsilon$.  Thus
$\inf_{b\in B}\nu b=0$.   Since $\nu$ is strictly
positive, $\inf B=0$.   As $\sequencen{A_n}$ is arbitrary, $\frak A$ is
\wsid.
}%end of proof of 393C

\leader{393D}{Theorem} Let $\frak A$ be a Boolean algebra.   Then it is
measurable iff it is Dedekind $\sigma$-complete and carries a uniformly
exhaustive strictly positive Maharam submeasure.

\proof{ If $\frak A$ is measurable, it surely satisfies the conditions,
since any totally finite measure on $\frak A$ is also a uniformly
exhaustive strictly positive Maharam submeasure.   If $\frak A$
satisfies the conditions, then it is \wsid, by 393C, so 392G gives the
result.
}%end of proof of 393D

\leader{393E}{Maharam algebras\dvAnew{2008} (a) Definition} A
{\bf Maharam algebra} is a
Dedekind $\sigma$-complete Boolean algebra $\frak A$ such that there is a
strictly positive Maharam submeasure on $\frak A$.

\spheader 393Eb Every measurable
algebra is a Maharam algebra, while every Maharam algebra is ccc and
\wsid\cmmnt{ (393C)}, therefore Dedekind complete.
A Maharam algebra $\frak A$ is measurable iff there is a strictly
positive uniformly exhaustive submeasure on $\frak A$.
\prooflet{(Put 393C and 392G together again.)}

\spheader 393Ec{\bf (i)}
A principal ideal in a Maharam algebra is a Maharam
algebra;  an order-closed subalgebra of a Maharam algebra is a Maharam
algebra.  \prooflet{\Prf\ Let $\frak A$ be a Maharam algebra and $\frak B$
{\it either} a principal ideal of $\frak A$ {\it or} an order-closed
subalgebra of $\frak A$.   Because $\frak A$ is Dedekind complete, so is
$\frak B$ (314Ea).   Let $\nu:\frak A\to\coint{0,\infty}$ be a
strictly positive Maharam submeasure.   Then $\nu\restr\frak B$ is a
strictly positive Maharam submeasure on $\frak A$, so $\frak B$ is a
Maharam algebra.\ \Qed}

\medskip

\quad{\bf (ii)} The simple product of a countable family of Maharam
algebras is a Maharam algebra.  \prooflet{\Prf\ Let $\familyiI{\frak A_i}$
be a countable family of Maharam algebras and $\frak A$ its simple product.
Then $\frak A$ is Dedekind complete (315De).   For each $i\in I$, let
$\nu_i$ be a strictly positive Maharam submeasure on $\frak A_i$.   Let
$\familyiI{\epsilon_i}$ be a family of strictly positive real numbers such
that $\sum_{i\in I}\epsilon_i$ is finite.   Set
$\nu a=\sum_{i\in I}\min(\epsilon_i,\nu_ia(i))$ for $a\in\frak A$;
then $\nu$ is a strictly positive Maharam submeasure on $\frak A$, so
$\frak A$ is a Maharam algebra.\ \Qed}

\leader{393F}{Lemma}\dvAformerly{393E}
Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra and $\nu$, $\nuprime$ two Maharam submeasures on $\frak A$
such that $\nu a=0$ whenever $\nuprime a=0$.   Then $\nu$ is absolutely
continuous with respect to $\nuprime$.

\proof{ (Compare 232Ba.)   \Quer\ Otherwise, we can find a sequence
$\sequencen{a_n}$ in
$\frak A$ such that $\nuprime a_n\le 2^{-n}$ for every $n$, but
$\epsilon=\inf_{n\in\Bbb N}\nu a_n>0$.   Set $b_n=\sup_{i\ge n}a_i$ for
each $n$, $b=\inf_{n\in\Bbb N}b_n$.   Then
$\nuprime b_n\le\sum_{i=n}^{\infty}2^{-i}\le 2^{-n+1}$ for each $n$
(393Bb),
so $\nuprime b=0$;  but $\nu b_n\ge\epsilon$ for each $n$, so
$\nu b\ge\epsilon$ (393Ba), contrary to the hypothesis.\ \Bang
}%end of proof of 393F

\leader{393G}{Proposition}\dvAnew{2008}
Let $\frak A$ be a Maharam algebra,
and $\nu$ and
$\nuprime$ two strictly positive Maharam submeasures on $\frak A$.
Then the
metrics they induce on $\frak A$ are uniformly equivalent, so we have a
topology and uniformity on $\frak A$ which we may call the
{\bf Maharam-algebra topology} and the {\bf Maharam-algebra uniformity}.

\proof{ By 393F, $\nu$ and $\nuprime$ are mutually absolutely continuous;
translating this with the formula of 392Ha, we see that the metrics are
uniformly equivalent, so induce the same topology and uniformity.   As
$\frak A$ does have a strictly positive Maharam submeasure, we may use it
to define the Maharam-algebra topology and uniformity of $\frak A$.
}%end of proof of 393G

\leader{393H}{Proposition}\dvAformerly{393B}
Let $\frak A$ be a Boolean algebra, and $\nu$
an exhaustive strictly positive totally finite submeasure on $\frak A$.
Let $\widehat{\frak A}$ be the metric completion of $\frak A$\cmmnt{, as
described in 392H}, and $\hat\nu$ the continuous extension of $\nu$ to
$\widehat{\frak A}$.   Then $\hat\nu$ is a Maharam submeasure, so
$\widehat{\frak A}$ is a Maharam algebra.

\proof{ (Compare 392Hd.)

\medskip

{\bf (a)} The point is that any non-increasing sequence
$\sequencen{a_n}$ in $\widehat{\frak A}$ is a Cauchy sequence for the
metric $\hat\rho$.   \Prf\ Let $\epsilon>0$.   For each $n\in\Bbb N$,
choose $b_n\in\frak A$ such that
$\hat\rho(a_n,b_n)\le 2^{-n}\epsilon$, and set $c_n=\inf_{i\le n}b_i$.
Then

\Centerline{$\hat\rho(a_n,c_n)
=\hat\rho(\inf_{i\le n}a_i,\inf_{i\le n}b_i)
\le\sum_{i=0}^n\hat\rho(a_i,b_i)
\le 2\epsilon$}

\noindent for every $n$.   Choose $\sequence{k}{n(k)}$ inductively so
that, for each $k\in\Bbb N$, $n(k+1)\ge n(k)$ and

\Centerline{$\nu(c_{n(k)}\Bsetminus c_{n(k+1)})
\ge\sup_{i\ge n(k)}\nu(c_{n(k)}\Bsetminus c_i)-\epsilon$.}

\noindent Then $\sequence{k}{c_{n(k)}\Bsetminus c_{n(k+1)}}$ is a
disjoint sequence in $\frak A$, so

$$\eqalign{\limsup_{k\to\infty}\sup_{i\ge n(k)}\hat\rho(a_{n(k)},a_i)
&\le 4\epsilon
  +\limsup_{k\to\infty}\sup_{i\ge n(k)}\hat\rho(c_{n(k)},c_i)\cr
&=4\epsilon
  +\limsup_{k\to\infty}\sup_{i\ge n(k)}\nu(c_{n(k)}\Bsetminus c_i)\cr
&\le 4\epsilon+\limsup_{k\to\infty}
    \nu(c_{n(k)}\Bsetminus c_{n(k+1)})+\epsilon
=5\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\sequencen{a_n}$ is Cauchy.\ \Qed

\medskip

{\bf (b)} It follows that $\widehat{\frak A}$ is Dedekind
$\sigma$-complete.   \Prf\ If $\sequencen{a_n}$ is any sequence in
$\widehat{\frak A}$, $\sequencen{b_n}=\sequencen{\inf_{i\le n}a_i}$ is a
Cauchy sequence with a limit $b\in\widehat{\frak A}$.   For any
$k\in\Bbb N$,

\Centerline{$\hat\nu(b\Bsetminus a_k)
=\lim_{n\to\infty}\hat\nu(b_n\Bsetminus a_k)=0$,}

\noindent so $b\Bsubseteq a_k$, because $\hat\nu$ is strictly positive.
While if $c\in\widehat{\frak A}$ is a lower bound for $\{a_n:n\in\Bbb
N\}$, we have $c\Bsubseteq b_n$ for every $n$, so

\Centerline{$\hat\nu(c\Bsetminus b)
=\lim_{n\to\infty}\hat\nu(c\Bsetminus b_n)=0$}

\noindent and $c\Bsubseteq b$.   Thus $b=\inf_{n\in\Bbb N}a_n$;  as
$\sequencen{a_n}$ is arbitrary, $\widehat{\frak A}$ is Dedekind
$\sigma$-complete (314Bc).\ \Qed

\medskip

{\bf (c)} We find also that $\hat\nu$ is a Maharam submeasure,
because if $\sequencen{a_n}$ is a non-increasing sequence in
$\widehat{\frak A}$ with infimum $0$, it must have a limit $a$ which (as
in (b) above) must be its infimum, that is, $a=0$;  consequently

\Centerline{$\lim_{n\to\infty}\hat\nu a_n=\hat\nu a=0$.}

\medskip

{\bf (d)} It follows at once that $\hat\nu$ is exhaustive
(393Bc), so that $\widehat{\frak A}$ is ccc (392Ca) and Dedekind
complete (316Fa).
}%end of proof of 393H

\leader{393I}{Proposition}\dvAnew{2008}
Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra and $\nu$ an atomless Maharam submeasure on $\frak A$.
Then for every $\epsilon>0$
there is a finite partition $C$ of unity in $\frak A$ such that
$\nu c\le\epsilon$ for every $c\in C$.

\proof{ Let $A\subseteq\frak A$ be a maximal disjoint set such that
$0<\nu a\le\epsilon$ for every $a\in A$.   As $\nu$ is exhaustive
(393Bc), $A$ is countable.   Set $c=1\Bsetminus\sup A$.
\Quer\ If $\nu c>0$, then
(because $\nu$ is atomless) we can choose inductively a sequence
$\sequencen{b_n}$ such that $b_0=c$,
$b_{n+1}\Bsubseteq b_n$, $\nu b_{n+1}>0$ and
$\nu(b_n\Bsetminus b_{n+1})>0$ for every $n\in\Bbb N$.   But now
$\sequencen{b_n\Bsetminus b_{n+1}}$ is a disjoint sequence of elements of
non-zero submeasure, so one of them has submeasure in $\ocint{0,\epsilon}$
and ought to have been added to $A$.\ \Bang

If $A$ is finite, we can set $C=A\cup\{c\}$ and stop.   Otherwise,
enumerate $A$ as $\sequencen{a_n}$ and set $c_n=\sup_{i\ge n}a_i$ for each
$n$;  then $\lim_{n\to\infty}\nu c_n=0$, so there is an $n$ such that
$\nu c_n\le\epsilon$, and we can set $C=\{a_i:i<n\}\cup\{c_n\Bcup c\}$.
}%end of proof of 393I

\leader{393J}{Lemma}\cmmnt{ ({\smc Maharam 1947})} Let $\frak A$ be a
ccc Boolean algebra with a T$_1$ topology $\frak T$ such that
(i) $\Bcup:\frak A\times\frak A\to\frak A$ is continuous at $(0,0)$ (ii)
whenever $\sequencen{a_n}$ is a non-increasing
sequence in $\frak A$ with infimum $0$, then $\sequencen{a_n}\to 0$ for
$\frak T$.   Then $\frak A$ has a strictly positive Maharam submeasure.

\proof{{\bf (a)} For any $e\in\frak A\setminus\{0\}$, there is a Maharam
submeasure $\nu$ on $\frak A$ such that $\nu e>0$.

\medskip

\Prf{\bf (i)} Choose a sequence $\sequencen{G_n}$ of neighbourhoods
of $0$, as follows.   Because $\frak T$ is T$_1$,
$G_0=\frak A\setminus\{0\}$ is a
neighbourhood of $0$ not containing $e$.   Given $G_n$, choose a
neighbourhood $G_{n+1}$ of $0$ such that $G_{n+1}\subseteq G_n$ and
$a\Bcup b\Bcup c\in G_n$ whenever $a$, $b$, $c\in G_{n+1}$.   (Take
neighbourhoods $H$, $H'$ of $0$ such that $a\Bcup b\in G_n$ for $a$,
$b\in H$, $b\Bcup c\in H$ for $b$, $c\in H'$ and set
$G_{n+1}=H\cap H'\cap G_n$.)   Define $\nu_0:\frak A\to[0,1]$ by setting

$$\eqalign{\nu_0a
&=1\text{ if }a\notin G_0,\cr
&=2^{-n}\text{ if }a\in G_n\setminus G_{n+1},\cr
&=0\text{ if }a\in\bigcap_{n\in\Bbb N}G_n.\cr}$$

\noindent Then whenever $a_0,\ldots,a_r\in\frak A$, $n\in\Bbb N$ and
$\sum_{i=0}^r\nu_0a_i<2^{-n}$, $\sup_{i\le r}a_i\in G_n$.   To see this,
induce on $r$.   If $r=0$ then we have $\nu_0a_0<2^{-n}$ so
$a_0\in G_{n+1}\subseteq G_n$.   For the inductive step to $r\ge 1$,
there must
be a $k\le r$ such that $\sum_{i<k}\nu_0a_i<2^{-n-1}$ and
$\sum_{k<i\le n}\nu_0a_i<2^{-n-1}$ (allowing $k=0$ or $k=n$, in which
case one of the
sums will be zero).  (If $\sum_{i=0}^r\nu_0a_i<2^{-n-1}$, take $k=n$;
otherwise, take $k$ to be the least number such that
$\sum_{i=0}^k\nu_0a_i\ge 2^{-n-1}$.)   By the inductive hypothesis, and
because $0$ certainly belongs to $G_{n+1}$, $b=\sup_{i<k}a_i$ and
$c=\sup_{k<i\le r}a_i$ both belong to $G_{n+1}$;  but also
$\nu_0a_k<2^{-n}$ so $a_k\in G_{n+1}$.   Accordingly, by the choice of
$G_{n+1}$,

\Centerline{$\sup_{i\le r}a_i=b\Bcup a_k\Bcup c$}

\noindent belongs to $G_n$, and the induction continues.

\medskip

\quad{\bf (ii)} Set

\Centerline{$\nu_1a
=\inf\{\sum_{i=0}^r\nu_0a_i:a_0,\ldots,a_r\in\frak A$,
$a=\sup_{i\le r}a_i\}$}

\noindent for every $a\in\frak A$.   It is easy to see that
$\nu_1(a\Bcup b)\le\nu_1a+\nu_1b$ for all $a$, $b\in\frak A$;  also
$a\in G_n$ whenever $\nu_1a<2^{-n}$, so, in particular, $\nu_1e\ge 1$,
because $e\notin G_0$.

Set

\Centerline{$\nu a=\inf\{\nu_1b:a\Bcap e\Bsubseteq b\Bsubseteq e\}$}

\noindent for every $a\in\frak A$.   Then of course $0\le\nu a\le\nu b$
whenever $a\Bsubseteq b$, and

\Centerline{$\nu 0\le\nu_10\le\nu_00=0$,}

\noindent so $\nu 0=0$.   If $a$, $b\in\frak A$ and $\epsilon>0$, there
are $a'$, $b'$ such that $a\Bcap e\Bsubseteq a'\Bsubseteq e$,
$b\Bcap e\Bsubseteq b'\Bsubseteq e$, $\nu_1a'\le\nu a+\epsilon$ and
$\nu_1 b'\le\nu b+\epsilon$;  so that
$(a\Bcup b)\Bcap e\Bsubseteq a'\Bcup b'\Bsubseteq e$ and

\Centerline{$\nu(a\Bcup b)\le\nu_1(a'\Bcup b')\le\nu_1a'+\nu_1b'
\le\nu a+\nu b+2\epsilon$.}

\noindent As $\epsilon$, $a$ and $b$ are arbitrary, $\nu$ is a
submeasure.   Next, if $\sequence{i}{a_i}$ is any non-increasing
sequence in $\frak A$ with infimum $0$, $\sequence{i}{a_i\Bcap e}$ is
another, so converges to $0$ for $\frak T$.   If $n\in\Bbb N$ there is
an $m$ such that $a_i\Bcap e\in G_n$ for every $i\ge m$, so that

\Centerline{$\nu a_i\le\nu_1(a_i\Bcap e)\le\nu_0(a_i\Bcap e)\le 2^{-n}$}

\noindent for every $i\ge m$.   As $n$ is arbitrary,
$\lim_{i\to\infty}\nu a_i=0$;  as $\sequence{i}{a_i}$ is arbitrary,
$\nu$ is a Maharam submeasure.   Finally,

\Centerline{$\nu e=\nu_1e\ge 1$,}

\noindent so $\nu e\ne 0$.\ \Qed

\medskip

{\bf (b)} Write $C$ for the set of those $c\in\frak A$ such that
there is a strictly positive Maharam submeasure on the principal ideal
$\frak A_c$.   Then $C$ is order-dense in $\frak A$.  \Prf\ Take any
$e\in\frak A\setminus\{0\}$.   By (a), there is a
Maharam submeasure $\nu$ such that $\nu e>0$.   Set
$A=\{e\Bsetminus a:\nu a=0\}$.   Because $\nu$ is a submeasure, $A$ is
downwards-directed.
\Quer\ If $\inf A=0$ then, because $\frak A$ is ccc, there is a
non-increasing sequence
$\sequencen{a_n}$ in $A$ with infimum $0$;  because
$\nu$ is a Maharam submeasure,

\Centerline{$\nu e\le\inf_{n\in\Bbb N}\nu a_n+\nu(e\Bsetminus a_n)
=\inf_{n\in\Bbb N}\nu a_n=0$.  \Bang}

\noindent Thus $A$ has a non-zero lower bound $c$, and
$\nu\restrp\frak A_c$ is a strictly positive Maharam submeasure, while
$c\Bsubseteq e$.\ \Qed

\medskip

{\bf (c)} Because $\frak A$ is ccc, there is a sequence
$\sequencen{c_n}$ in $C$ with supremum $1$.   For each $n$, let $\nu_n$
be a strictly positive Maharam submeasure on $\frak A_{c_n}$;
multiplying by a scalar if necessary, we may suppose that
$\nu_nc_n\le 2^{-n}$.   We can therefore define $\nu:\frak A\to[0,2]$ by
setting $\nu a=\sum_{n=0}^{\infty}\nu_n(a\Bcap c_n)$ for every
$a\in\frak A$, and it is easy to check that $\nu$ is a strictly positive
Maharam submeasure on $\frak A$.
}%end of proof of 393J

\leader{*393K}{Theorem}
Let $\frak A$ be a ccc Dedekind complete Boolean algebra.
Then $\frak A$ is a Maharam algebra iff there is a Hausdorff linear
space topology $\frak T$ on $L^0(\frak A)$ such that for every
neighbourhood $G$ of $0$ there is a neighbourhood $H$ of $0$ such that
$u\in G$ whenever $v\in H$ and $|u|\le|v|$.

\proof{{\bf (a)} Suppose that $\frak A$ is a Maharam algebra;  let $\nu$ be
a strictly positive Maharam submeasure on $\frak A$.

\medskip

\quad{\bf (i)} For $u\in L^0=L^0(\frak A)$ set

\Centerline{$\tau(u)
=\inf\{\alpha:\alpha\ge 0,\,\nu\Bvalue{|u|>\alpha}\le\alpha\}$.}

Then $\tau$ is an F-seminorm (definition:  2A5B\Latereditions).
\Prf\ (i) It will save a moment
if we observe that whenever $\beta>\tau(u)$ there is an $\alpha\le\beta$
such that $\nu\Bvalue{|u|>\alpha}\le\alpha$, so that

\Centerline{$\nu\Bvalue{|u|>\beta}\le\nu\Bvalue{|u|>\alpha}
\le\alpha\le\beta$.}

\noindent Also, because $\nu$ is sequentially order-continuous,

\Centerline{$\nu\Bvalue{|u|>\tau(u)}
=\lim_{n\to\infty}\nu\Bvalue{|u|>\tau(u)+2^{-n}}
\le\lim_{n\to\infty}\tau(u)+2^{-n}
=\tau(u)$.}

\noindent (ii) So

$$\eqalignno{\nu\Bvalue{|u+v|>\tau(u)+\tau(v)}
&\le\nu\Bvalue{|u|+|v|>\tau(u)+\tau(v)})\cr
&\le\nu(\Bvalue{|u|>\tau(u)}\Bcup\Bvalue{|v|>\tau(v)})\cr
\noalign{\noindent (364Ea)}
&\le\nu\Bvalue{|u|>\tau(u)}+\nu\Bvalue{|v|>\tau(v)}
\le\tau(u)+\tau(v),\cr}$$

\noindent and $\tau(u+v)\le\tau(u)+\tau(v)$.   (iii) If $|\alpha|\le 1$
then

\Centerline{$\nu\Bvalue{|\alpha u|>\tau(u)}
\le\nu\Bvalue{|u|>\tau(u)}
\le\tau(u)$,}

\noindent and $\tau(\alpha u)\le\tau(u)$.   (iv)
$\lim_{n\to\infty}\nu\Bvalue{|u|>n}=0$ because
$\sequencen{\Bvalue{|u|>n}}$ is a non-increasing sequence with infimum
$0$.   So if $\epsilon>0$, there is an $n\ge 1$ such that
$\nu\Bvalue{|u|>n\epsilon}\le\epsilon$, in which case
$\nu\Bvalue{|\alpha u|>\epsilon}\le\epsilon$ whenever
$|\alpha|\le\bover1n$, so that $\tau(\alpha u)\le\epsilon$ whenever
$|\alpha|\le\bover1n$.   As $\epsilon$ is arbitrary, 
$\lim_{\alpha\to 0}\tau(\alpha u)=0$.   Thus all the conditions of
2A5B are satisfied and $\tau$ is an F-seminorm.\ \Qed

\medskip

\quad{\bf (ii)} Accordingly we have a 
pseudometric $(u,v)\mapsto\tau(u-v)$ which
defines a linear space topology $\frak T$ on $L^0$ (2A5B).   
In fact this is a metric, because if $\tau(u-v)=0$ then
$\nu\Bvalue{|u-v|>0}=0$ and (since $\nu$ is strictly positive) $u=v$.
So $\frak T$ is Hausdorff.   Now let $G$
be an open set containing $0$.   Then there is an $\epsilon>0$ such that
$H=\{u:\tau(u)<\epsilon\}$ is included in $G$.   If $v\in H$ and
$|u|\le|v|$, then

\Centerline{$\nu\Bvalue{|u|>\tau(v)}
\le\nu\Bvalue{|v|>\tau(v)}\le\tau(v)$,}

\noindent so $\tau(u)\le\tau(v)$ and $u\in H\subseteq G$.   So $\frak T$
satisfies all the conditions.

\medskip

{\bf (b)} Given such a topology $\frak T$ on $L^0$,
let $\frak S$ be the topology on $\frak A$ induced
by $\frak T$ and the function
$\chi:\frak A\to L^0$;  that is, $\frak S=\{\chi^{-1}[G]:G\in\frak T\}$.
Then $\frak S$ satisfies the conditions of 393J.   \Prf\ (i) Because
$\frak T$ is Hausdorff and $\chi$ is injective, $\frak S$ is Hausdorff,
therefore T$_1$.
(ii) If $0\in G\in\frak S$, there is an $H\in\frak T$ such that
$G=\chi^{-1}[H]$.   Now $0$ (the zero of $L^0$) belongs to $H$, so there
is an open set $H_1$ containing $0$ such that $u\in H$ whenever
$v\in H_1$ and $|u|\le|v|$.
Next, addition on $L^0$ is continuous for $\frak T$,
so there is an open set $H_2$ containing $0$ such that $u+v\in H_1$
whenever $u$, $v\in H_2$.   Consider $G'=\chi^{-1}[H_2]$.   This is an
open set in $\frak A$ containing $0_{\frak A}$, and if $a$, $b\in G'$ then

\Centerline{$|\chi(a\Bcup b)|\le\chi a+\chi b\in H_2+H_2\subseteq H_1$,}

\noindent so $\chi(a\Bcup b)\in H$ and $a\Bcup b\in G$.   As $G$ is
arbitrary, $\Bcup$ is continuous at $(0,0)$.   (iii) If
$\sequencen{a_n}$ is
a non-increasing sequence in $\frak A$ with infimum $0$,
$u_0=\sup_{n\in\Bbb N}n\chi a_n$ is defined in $L^0$ (use the criterion
of 364L(a-i):

\Centerline{$\inf_{m\in\Bbb N}\sup_{n\in\Bbb N}\Bvalue{n\chi a_n>m}
=\inf_{m\in\Bbb N}a_{m+1}=0$.)}

\noindent If $0\in G\in\frak S$, take $H\in\frak T$ such that
$G=\chi^{-1}[H]$, and $H_1\in\frak T$ such that $0\in H_1$ and $u\in H$
whenever $v\in H_1$ and $|u|\le|v|$.   Because scalar multiplication is
continuous for $\frak T$, there is a $k\ge 1$ such that
$\bover1ku_0\in H_1$.   For any $n\ge k$,
$\chi a_n\le\bover1ku_0$ so $\chi a_n\in H$
and $a_n\in G$.   As $G$ is arbitrary, $\sequencen{a_n}\to 0$ for
$\frak S$.   As $\sequencen{a_n}$ is arbitrary, condition (ii) in
the statement of 393J is satisfied.\ \Qed

So 393J tells us that $\frak A$ has a strictly positive Maharam
submeasure, and is a Maharam algebra.
}%end of proof of 393K

\leader{393L}{}\dvAformerly{3{}92K}\cmmnt{ I now turn to some very
remarkable
ideas relating the order*-convergence of \S367 to the questions here.

\medskip

\noindent}{\bf Definition} Let $P$ be a lattice, and consider the
relation `$\sequencen{p_n}$ order*-converges to $p$' as a relation
between $P^{\Bbb N}$ and $P$.
\dvro{There}{By 367Bc, this satisfies the hypothesis
of 3A3Pa, so there} is a unique topology on $P$ for which a set
$F\subseteq P$ is closed iff $a\in F$ whenever $\sequencen{a_n}$ is a
sequence in $F$ which order*-converges to $a$ in $P$.
I will call this topology the {\bf order-sequential topology} of $P$.

\cmmnt{{\bf Warning!} For the next few paragraphs I shall be closely
following the papers {\smc Balcar G{\l}owczy\'nski \& Jech 98}
and {\smc Balcar Jech \& Paz\'ak 05}.   I should therefore note
explicitly that if $\frak A$ is a Boolean algebra which is
neither Dedekind $\sigma$-complete nor ccc, my `order-sequential
topology' on $\frak A$ may not be identical to theirs.}

\leader{393M}{Lemma} Let $\frak A$ be a Boolean algebra.

(a) A sequence $\sequencen{a_n}$ order*-converges to $a\in\frak A$ iff
there is a partition $B$ of unity in $\frak A$ such that
$\{n:n\in\Bbb N$, $(a_n\Bsymmdiff a)\Bcap b\ne 0\}$ is finite for every
$b\in B$.

(b) If $\sequencen{a_n}$ order*-converges to $a$ and $c\in\frak A$, then
$\sequencen{a_n\Bcup c}$,
$\sequencen{a_n\Bcap c}$ and $\sequencen{a_n\Bsymmdiff c}$
order*-converge to
$a\Bcup c$, $a\Bcap c$ and $a\Bsymmdiff c$ respectively.

(c) The operations $\Bcap$, $\Bcup$ and $\Bsymmdiff$ are separately
continuous for the order-sequential topology.

(d) Every disjoint sequence in $\frak A$ is order*-convergent to $0$.

\proof{{\bf (a)} Let $\sequencen{a_n}$ be a sequence in $\frak A$ and
$a\in\frak A$;  set

\Centerline{$C=\{c:\,\Exists n\in\Bbb N$,
$c\Bsubseteq a_i$ for every $i\ge n\}$,}

\Centerline{$D=\{d:\,\Exists n\in\Bbb N$,
$a_i\Bsubseteq d$ for every $i\ge n\}$.}

\medskip

\quad{\bf (i)} If $\sequencen{a_n}$ order*-converges to $a$, then
$a=\sup C=\inf D$ (367Be).   Since

\Centerline{$\inf\{d\Bsetminus a:d\in D\}
=\inf\{a\Bsetminus c:c\in C\}=0$,}

\Centerline{$E=\{(d\Bsetminus a)\Bcup(a\Bsetminus c):c\in C$, $d\in D\}$}

\noindent also has infimum $0$ (313A, 313B).   So there is a partition $B$ of unity
such that for every $b\in B$ there is an $e\in E$ such that $b\Bcap e=0$.
Now, given $b\in B$, there are $c\in C$ and $d\in D$ such that
$b\Bcap(d\setminus c)=0$;  there are $n_1$, $n_2\in\Bbb N$ such that
$c\Bsubseteq a_n$ for $n\ge n_1$ and $a_n\Bsubseteq d$ for $n\ge n_2$;
so that $\{n:(a_n\Bsymmdiff a)\Bcap b\ne 0\}$
is bounded above by $\max(n_1,n_2)$ and is
finite.   So $B$ witnesses that the condition is satisfied.

\medskip

\quad{\bf (ii)} Now suppose that $B$ is a partition of unity such that
$\{n:(a_n\Bsymmdiff a)\Bcap b\ne 0\}$ is finite for every $b\in B$.
Then $a\Bcup(1\Bsetminus b)\in D$ for every $b\in B$, because
$\{n:a_n\notBsubseteq a\Bcup(1\Bsetminus b)\}
\subseteq\{n:(a_n\Bsymmdiff a)\Bcap b\ne 0\}$ is finite.   So any lower
bound for $D$ is also a lower bound for $\{a\Bcup(1\Bsetminus b):b\in B\}$
and is included in $a$.   Similarly, any upper bound for $C$ includes $a$;
as $c\Bsubseteq d$ whenever $c\in C$ and $d\in D$, $a=\sup C=\inf D$ and
$\sequencen{a_n}$ order*-converges to $a$.

\medskip

{\bf (b)} These are all immediate from (a), because

\Centerline{$(a_n\Bcup c)\Bsymmdiff(a\Bcup c)
\Bsubseteq a_n\Bsymmdiff a$,
\quad$(a_n\Bcap c)\Bsymmdiff(a\Bcap c)
\Bsubseteq a_n\Bsymmdiff a$,}

\Centerline{$(a_n\Bsymmdiff c)\Bsymmdiff(a\Bsymmdiff c)
=a_n\Bsymmdiff a$}

\noindent for every $n$.

\medskip

{\bf (c)} By (b), we can apply 3A3Pb to each of the functions
$a\mapsto a\Bcap b=b\Bcap a$,
$a\mapsto a\Bcup b=b\Bcup a$ and
$a\mapsto a\Bsymmdiff b=b\Bsymmdiff a$
to see that these are all continuous
for every $b\in\frak A$.

\medskip

{\bf (d)} If $\sequencen{a_n}$ is a disjoint sequence in $\frak A$, there
is a partition $B$ of unity in $\frak A$ containing every $a_n$ (311Gd);
now $B$ witnesses that the condition of (a) is satisfied.
}%end of proof of 393M

\leader{393N}{Proposition}\dvAnew{2008}
Let $\frak A$ be a Maharam algebra.   Then the
Maharam-algebra topology on $\frak A$ is the
order-sequential topology.

\proof{ Let $\frak T_o$ be the order-sequential topology
on $\frak A$, $\nu$ a strictly
positive Maharam submeasure on $\frak A$, $\rho$ the metric defined from
$\nu$ (392H) and $\frak T_M$ the Maharam-algebra
topology induced by $\rho$ (393G).

\medskip

{\bf (a)} If $\sequencen{a_n}$ is a sequence in $\frak A$ such that
$\sequencen{a_n}\to^*a$ in $\frak A$, then
$\lim_{n\to\infty}\rho(a_n,a)=0$.   \Prf\ By 393Mb,
$\sequencen{a_n\Bsymmdiff a}\to^*0$;  by 367Bf,
$0=\inf_{n\in\Bbb N}\sup_{i\ge n}(a_i\Bsymmdiff a)$, so

\Centerline{$\rho(a_n,a)\le\nu(\sup_{i\ge n}a_i\Bsymmdiff a)\to 0$}

\noindent as $n\to\infty$.\ \Qed

It follows that every $\frak T_M$-closed set is $\frak T_o$-closed,
and $\frak T_o\subseteq\frak T_M$.

\medskip

{\bf (b)} Conversely, suppose that $\sequencen{a_n}$ is a sequence in
$\frak A$ converging for $\frak T_M$ to $a\in\frak A$.   Then
$\sequencen{a_n}$ has a subsequence $\sequencen{a'_n}$ such that
$\rho(a'_n,a)\le 2^{-n}$ for every $n\in\Bbb N$.   In this case, setting
$b_m=\sup_{n\ge m}a'_n\Bsymmdiff a$ for each $m$,
$\nu b_m\le 2^{-m+1}$ for every $m$ (393Bb),
so $\inf_{m\in\Bbb N}b_m=0$, and
$\sequencen{a'_n\Bsymmdiff a}\to^*0$, that is, $\sequencen{a'_n}\to^*a$.

Thus every $\frak T_M$-convergent sequence has an order*-convergent
subsequence
with the same limit;  it follows that every $\frak T_o$-closed set is
$\frak T_M$-closed, that is, $\frak T_M\subseteq\frak T_o$.
}%end of proof of 393N

\leader{393O}{Proposition} Let $\frak A$ be a ccc Dedekind
$\sigma$-complete Boolean
algebra, with its order-sequential topology, and $\frak B$ a
subalgebra of $\frak A$.   Then the topological closure of $\frak B$ is the
smallest order-closed set including $\frak B$;
\cmmnt{in particular, }$\frak B$ is
order-closed iff it is topologically closed.

\proof{{\bf (a)} Let $\overline{\frak B}$ be the topological closure of
$\frak B$, and $\frak B^{\sim}$ the smallest order-closed set including
$\frak B$.

\medskip

\quad{\bf (i)}
Suppose that $\sequencen{b_n}$ is a non-decreasing sequence in
$\overline{\frak B}$ with supremum $b$ in $\frak A$;  then
$\sequencen{b_n}\to^*b$, by 367Bf or 367Xa.
So $b\in\overline{\frak B}$.  Similarly,
$\inf_{n\in\Bbb N}b_n\in\overline{\frak B}$ for every non-increasing
sequence in $\overline{\frak B}$.   Thus $\overline{\frak B}$ is
sequentially order-closed.   But this means that it is order-closed, by
316Fb.   So $\overline{\frak B}\supseteq\frak B^{\sim}$.

\medskip

\quad{\bf (ii)} By 313Fc, $\frak B^{\sim}$ is a subalgebra of $\frak A$.
Now suppose that $\sequencen{b_n}$ is a sequence in $\frak B^{\sim}$
which order*-converges to $a\in\frak A$.   Then
$c_{mn}=\sup_{m\le i\le n}b_i$ belongs to $\frak B^{\sim}$ whenever
$m\le n$;  as $\langle c_{mn}\rangle_{n\ge m}$ is non-decreasing,
$c_m=\sup_{i\ge m}b_i=\sup_{n\ge m}c_{mn}$ belongs to $\frak B^{\sim}$
for every $m\in\Bbb N$;  as $\sequence{m}{c_m}$ is non-increasing,
$\inf_{m\in\Bbb N}c_m\in\frak B^{\sim}$.   But $c=b$ (367Bf).
As $\sequencen{b_n}$ is arbitrary, $\frak B^{\sim}$ is closed for the
order-sequential topology, and must include $\overline{\frak B}$.

Thus $\overline{\frak B}=\frak B^{\sim}$, as claimed.

\medskip

{\bf (b)} Now

\Centerline{$\frak B\text{ is order-closed}
\iff\frak B=\frak B^{\sim}
\iff\frak B=\overline{\frak B}
\iff\frak B$ is topologically closed.}
}%end of proof of 393O

\leader{393P}{Lemma} Let $\frak A$ be a ccc \wsid\ Boolean algebra,
endowed with its order-sequential topology.

(a) If $\langle a_{mn}\rangle_{m,n\in\Bbb N}$, $\sequence{m}{a_m}$ and
$a$ are such that $\sequencen{a_{mn}}$ order*-converges to $a_m$ for
each $m$, while $\sequence{m}{a_m}$ order*-converges
to $a$, then there is a sequence $\sequence{m}{k(m)}$ in $\Bbb N$ such
that $\sequence{m}{a_{m,k(m)}}$
order*-converges to $a$.

(b) If $A\subseteq\frak A$ and $a\in\overline{A}$, there is a sequence
in $A$ which order*-converges to $a$.

(c) If $G$ is a neighbourhood of $0$ in $\frak A$ then there is an open
neighbourhood $H$ of $0$, included in $G$, such that
$[0,a]\subseteq H$ for every $a\in H$.

(d) For $A\subseteq\frak A$, set $\bigvee_0(A)=\{0\}$ and
$\bigvee_{n+1}(A)=\{a\Bcup b:a\in\bigvee_n(A)$, $b\in A\}$ for
$n\in\Bbb N$.

\quad(i) If $A\subseteq\frak A$ is such that $[0,a]\subseteq A$ for every
$a\in A$, and $n\in\Bbb N$, then $[0,a]\subseteq\bigvee_n(A)$ for every
$a\in\bigvee_n(A)$.

\quad(ii) If $H\subseteq\frak A$ is an open set containing $0$ such that
$[0,a]\subseteq H$ for every $a\in H$, then $\bigvee_{n+1}(H)$ is open
and $\overline{\bigvee_n(H)}\subseteq\bigvee_{n+1}(H)$ for every
$n\in\Bbb N$.

(e) Suppose that $\frak A$ is Dedekind $\sigma$-complete.
Then for every open set $G$ containing $0$ there is an open set $H$
containing $0$ such that $\bigvee_3(H)\subseteq\bigvee_2(G)$.

\proof{{\bf (a)} Let $C_m$, for $m\in\Bbb N$, be partitions of unity in
$\frak A$ such that

\Centerline{$\{m:(a_m\Bsymmdiff a)\Bcap c\ne 0\}$ is finite for every
$c\in C_0$,}

\Centerline{$\{n:(a_{mn}\Bsymmdiff a_m)\Bcap c\ne 0\}$ is finite whenever
$m\in\Bbb N$ and $c\in C_{m+1}$}

\noindent (393Ma).   Because $\frak A$ is \wsid, there is a partition $B$
of unity such that $\{c:c\in C_m$, $c\Bcap b\ne 0\}$ is finite whenever
$m\in\Bbb N$ and $b\in B$ (316H(ii)).
Because $\frak A$ is ccc, there is a sequence
$\sequencen{b_n}$ running over $B\cup\{0\}$.   Now, for each $m$, any
sufficiently large $k(m)$ will be such that
$(a_{m,k(m)}\Bsymmdiff a_m)\Bcap b_i=0$ for every $i\le m$.   In this case,
for any $i$,

\Centerline{$\{m:(a_{m,k(m)}\Bsymmdiff a)\Bcap b_i\ne 0\}
\subseteq\{m:m<i\}\cup\{m:(a_m\Bsymmdiff a)\cap b_i\ne 0\}$}

\noindent is finite, so $B$ witnesses that
$\sequence{m}{a_{m,k(m)}}\to^*a$ (393Ma, in the other direction).

\medskip

{\bf (b)} Let $A^{\sim}$ be the set of order*-limits of sequences in
$A$.   Of course $A^{\sim}$ must be
included in $\overline{A}$.   But from (a) we see that the limit
of any order*-convergent sequence in $A^{\sim}$ belongs to $A^{\sim}$.
So $A^{\sim}$ is closed and is equal to
$\overline{A}$.   Turning this round, we see that $\overline{A}$ is just
the set of order*-limits of sequences in $A$,
as claimed.

\medskip

{\bf (c)} Set $D=\{d:d\in\frak A$, $[0,d]\not\subseteq G\}$,
$H=\frak A\setminus\overline{D}$.   Since
$D\supseteq\frak A\setminus G$, $H$ is an open subset of $G$.

\Quer\ If $0\in\overline{D}$, then (b) tells us that there is a sequence
$\sequencen{d_n}$ in $D$ order*-converging to
$0$.   Now there is for each $n\in\Bbb N$ a $c_n\Bsubseteq d_n$ such
that $c_n\notin G$.   By 367Be or 393Ma,
$\sequencen{c_n}$ order*-converges to $0$, and
$0\in\overline{\frak A\setminus G}$;  but $G$ is supposed to be
a neighbourhood of $0$.\ \BanG\   Thus $0\in H$ and $H$ is a
neighbourhood of $0$.

\Quer\ If $a\in H$ and $b\in[0,a]\setminus H$, then $b\in\overline{D}$,
so there is a sequence $\sequencen{d_n}$ in
$D$ order*-converging to $b$.   In this case, $\sequencen{d_n\Bcup a}$
order*-converges to $b\Bcup a=a$, by
393Mb.   But also $[0,d_n\Bcup a]\supseteq[0,d_n]$ is not included in
$G$, so $d_n\Bcup a\in D$ for each $n$,
and $a\in\overline{D}$;  which is impossible.\ \BanG\   Thus
$[0,a]\subseteq H$ for every $a\in H$, and $H$ has the
properties declared.

\medskip

{\bf (d)(i)} This is an elementary induction on $n$.

\medskip

\quad{\bf (ii)} The point is that
$\bigvee_{n+1}(H)=\{a\Bsymmdiff b:a\in\bigvee_n(H)$, $b\in H\}$.
\Prf\ If $a\in\bigvee_n(H)$ and $b\in H$, then
$a\Bsetminus b\in\bigvee_n(H)$, by (i), and $b\setminus a\in H$, so
$a\Bsymmdiff b\in\bigvee_{n+1}(H)$.   On the other hand, if
$c\in\bigvee_{n+1}(H)$, it is expressible as
$a\Bcup b=a\Bsymmdiff(b\Bsetminus a)$ where $a\in\bigvee_n(H)$ and $b$ and
$b\Bsetminus a$ belong to $H$.\ \Qed

Since $\Bsymmdiff$ is separately continuous, it follows at once that

\Centerline{$\bigvee_{n+1}(H)
=\bigcup_{a\in\bigvee_n(H)}\{a\Bsymmdiff b:b\in H\}
=\bigcup_{a\in\bigvee_n(H)}\{b:a\Bsymmdiff b\in H\}$}

\noindent is open, because $\Bsymmdiff$ is separately continuous (393Mc).
Next, if $d\in\overline{\bigvee_n(H)}$, then
there is a sequence $\sequencen{d_n}$ in
$\bigvee_n(H)$ order*-converging to $d$, by (b).   Now
$\sequencen{d_n\Bsymmdiff d}\to^* 0$, by 393Mb, so
$\sequencen{d_n\Bsymmdiff d}$ converges topologically to $0$, by 3A3Pa, and
there is an $n\in\Bbb N$ such that $d_n\Bsymmdiff d\in H$;  in which case
$d=d_n\Bsymmdiff(d_n\Bsymmdiff d)$ belongs to $\bigvee_{n+1}(H)$.
As $d$ is arbitrary, $\overline{\bigvee_n(H)}\subseteq\bigvee_{n+1}(H)$.

\medskip

{\bf (e)} \Quer\ Suppose, if possible, otherwise.

\medskip

\quad{\bf (i)} Choose $H_n$, $a_n$, $b_n$ and $c_n$
inductively, as follows.   $H_0\subseteq G$ is to be an open neighbourhood
of $0$ such that $[0,a]\subseteq H_0$ whenever $a\in H_0$
((c) above).   Given that $H_n$ is an open set
containing $0$ and including $[0,a]$ whenever it contains $a$,
we are supposing that
$\bigvee_3(H_n)\not\subseteq\bigvee_2(G)$;
choose $a_n$, $b_n$, $c_n\in H_n$ such that
$a_n\Bcup b_n\Bcup c_n\notin\bigvee_2(G)$, and set

\Centerline{$H_{n+1}=\{a:a$, $a\Bcup a_n$, $a\Bcup b_n$ and
$a\Bcup c_n$ all belong to $H_n\}$,}

\noindent so that $H_{n+1}$ is an open set containing $0$,
and $[0,a]\subseteq H_{n+1}$ for every $a\in H_{n+1}$.
Continue.

\medskip

\quad{\bf (ii)} At the end of the induction, set
$F=\bigcap_{n\in\Bbb N}\overline{H}_n$ and
$a^*=\inf_{n\in\Bbb N}\sup_{i\ge n}a_i$.   Then $a^*\Bcup d\in F$ for every
$d\in F$.   \Prf\ For $m\le n\in\Bbb N$,
$\sup_{m\le i\le n}a_i\Bcup d\in H_m$ for every $d\in H_{n+1}$ (induce
downwards on $m$).   Because $\Bcup$ is separately continuous,
$\sup_{m\le i\le n}a_i\Bcup d\in\overline{H}_m$ for
every $d\in F$.   Letting $n\to\infty$,
$d\Bcup\sup_{i\ge m}a_i\in\overline{H}_m$ whenever $d\in F$ and
$m\in\Bbb N$.
Next, for any $b\in\frak A$, $\{a:a\Bcap b\in\overline{H}_m\}$ is a closed
set including $H_m$, so $a\Bcap b\in\overline{H}_m$ for every
$a\in\overline{H}_m$;  that is, $[0,a]\Bsubseteq\overline{H}_m$ for every
$a\in\overline{H}_m$.   As $a^*\Bsubseteq\sup_{i\ge m}a_i$,
$d\Bcup a^*\in\overline{H}_m$ for every
$d\in F$.   As $m$ is arbitrary, $d\Bcup a^*\in F$ for
every $d\in F$.\ \Qed

Similarly, setting $b^*=\inf_{n\in\Bbb N}\sup_{i\ge n}b_i$ and
$c^*=\inf_{n\in\Bbb N}\sup_{i\ge n}c_i$, $d\Bcup b^*$ and $d\Bcup c^*$
belong to $F$ for every $d\in F$;  and of course $0\in F$.
So $e=a^*\Bcup b^*\Bcup c^*$ belongs to
$F$.   For each $n\in\Bbb N$, $a_n\Bcup b_n\Bcup c_n\notin\bigvee_2(H_0)$;
but $[0,a]\subseteq\bigvee_2(H_0)$ for every $a\in\bigvee_2(H_0)$,
by (d-i),
so $\sup_{i\ge n}a_i\Bcup b_i\Bcup c_i\notin\bigvee_2(H_0)$.
Accordingly $e=\inf_{n\in\Bbb N}\sup_{i\ge n}a_i\Bcup b_i\Bcup c_i$
does not belong to the open set $\bigvee_2(H_0)$, and
$e\notin\overline{H}_0$, by (d-ii).
So $e\in F\setminus\overline{H}_0$;  which is impossible.\ \Bang
}%end of proof of 393P

\leader{393Q}{Theorem}\cmmnt{ ({\smc Balcar G{\l}owczy\'nski \& Jech
98}, {\smc Balcar Jech \& Paz\'ak 05})} Let
$\frak A$ be a Dedekind $\sigma$-complete Boolean algebra.
Then the following are equiveridical:

(i) $\frak A$ is a Maharam algebra;

(ii) $\frak A$ is ccc and the order-sequential topology is Hausdorff;

(iii) $\frak A$ is \wsid\ and
$\{0\}$ is a G$_{\delta}$ set for the order-sequential
topology of $\frak A$;

(iv) $\frak A$ is ccc and there is a T$_1$ topology on $\frak A$
such that ($\alpha$) $\Bcup:\frak A\times\frak A\to\frak A$ is
continuous at $(0,0)$ ($\beta$) whenever $\sequencen{a_n}$ is a
non-decreasing
sequence in $\frak A$ with infimum $0$, then $\sequencen{a_n}\to 0$.

\proof{{\bf (a)(i)$\Rightarrow$(ii)}
By 393Eb, $\frak A$ is ccc.
By 393N, the order-sequential topology is metrizable, therefore Hausdorff.

\medskip

{\bf (b)(ii)$\Rightarrow$(iii)} Suppose that the conditions of (ii) are
satisfied.   In the following argument,
all topological terms will refer to the
order-sequential topology on $\frak A$.

\medskip

\quad\grheada\ $\frak A$ is \wsid.   \Prf\ Let $\sequencen{A_n}$
be a sequence of partitions of unity in $\frak A$, and set

\Centerline{$D=\{d:d\in\frak A$, $\{a:a\in A_n$, $a\Bcap d\ne 0\}$ is
finite for every $n\in\Bbb N\}$.}

\noindent Take any
$c\in\frak A^+$.   Let $G$, $H$ be disjoint open sets
containing $0$, $c$ respectively.   Choose $\sequencen{c_n}$ inductively,
as follows.   $c_0=c$.   Given $c_n\in H$, let $\sequence{i}{a_{ni}}$ be a
sequence running over $A_n$, and set $c_{nj}=\sup_{i\le j}c_n\Bcap a_{ni}$;
then $\sequence{j}{c_{nj}}$ order*-converges to $c_n$ (367Bf/367Xa),
so there is a
$j_n$ such that $c_{nj_n}\in H$;  set $c_{n+1}=c_{nj_n}$, and continue.

This gives us a non-increasing sequence $\sequencen{c_n}$ in $H$.   Set
$d=\inf_{n\in\Bbb N}c_n$;  then $d\notin G$ so $d\ne 0$, while
$d\Bsubseteq\sup_{i\le j_n}a_{ni}$ for each $n$, so $d\in D$.

As $c$ is arbitrary, $D$ is order-dense in $\frak A$ and includes a
partition of unity.   As $\sequencen{A_n}$ is arbitrary, $\frak A$ is
\wsid\ (316H).\ \Qed

\medskip

\quad\grheadb\ For any $a\in\frak A^+$ there is a sequence
$\sequencen{H_n}$ of neighbourhoods of $0$ such that
$a\notBsubseteq\sup(\bigcap_{n\in\Bbb N}H_n)$.   \Prf\ For
$A\subseteq\frak A$ and $n\in\Bbb N$, define $\bigvee_n(A)$ as in
393Pd.   Let $G$, $G'$ be disjoint neighbourhoods of $0$ and $a$
respectively, and set $G_0=G\cap\{a\Bsymmdiff b:b\in G'\}$;  then
$G_0$ is a neighbourhood of $0$ (393Mc).
By 393Pc, we can find a neighbourhood $H_0$ of $0$ such that
$H_0\subseteq G_0$ and $[0,b]\subseteq H_0$ for every $b\in H_0$,
in which case $[0,b]\subseteq\bigvee_2(H_0)$ for every
$b\in\bigvee_2(H_0)$, while $a\notin\bigvee_2(H_0)$.
By 393Pe, we can choose neighbourhoods $H_n$ of $0$
such that $H_n\subseteq H_{n-1}$ and
$\bigvee_3(H_n)\subseteq\bigvee_2(H_{n-1})$ for every $n\ge 1$;
by 393Pc, we can suppose that $[0,b]\subseteq H_n$ whenever $b\in H_n$.
But this will ensure that $\bigvee_4(H_{n+2})\subseteq\bigvee_2(H_n)$ for
every $n$, so that $\bigvee_{2^k}(H_{2k})\subseteq\bigvee_2(H_2)$ for
every $k\ge 1$.   Set $F=\bigcap_{n\in\Bbb N}H_n$.   Then

\Centerline{$\bigvee_{2^k}(F)
\subseteq\bigvee_{2^k}(H_{2k})
\subseteq\bigvee_2(H_2)$}

\noindent for every $k\ge 1$.   Since $\sup F$ is the limit of a sequence
in $\bigcup_{k\ge 1}\bigvee_{2^k}(F)$,

\Centerline{$\sup F\in\overline{\bigvee_2(H_2)}\subseteq\bigvee_3(H_2)
\subseteq\bigvee_2(H_0)$}

\noindent (using 393P(d-ii) for the first inequality)
and cannot include $a$.\ \Qed

\medskip

\quad\grheadc\ Now consider the set $D$ of those $d\in\frak A$ such that
there is a sequence $\sequencen{H_n}$ of neighbourhoods of $0$ such that
$d\Bcap\sup(\bigcap_{n\in\Bbb N}H_n)=0$.
By ($\beta$), $D$ is order-dense, so
includes a partition of unity $A$.   $A$ is countable, so there is a
sequence $\sequencen{H_n}$ of neighbourhoods of $0$ such that
$d\Bcap\sup(\bigcap_{n\in\Bbb N}H_n)=0$ for every $d\in A$;  but this means
that $\bigcap_{n\in\Bbb N}H_n=\{0\}$.   So (iii) is true.

\medskip

{\bf (c)(iii)$\Rightarrow$(iv)} Now suppose that the conditions in (iii)
are satisfied.   As in (b),
all topological terms will refer to the order-sequential
topology on $\frak A$.

\medskip

\quad\grheada\ There is a non-increasing sequence $\sequencen{G_n}$ of open
neighbourhoods of $0$ such that
$\bigcap_{n\in\Bbb N}\overline{G}_n=\{0\}$.
\Prf\ Let $\sequencen{U_n}$
be a sequence of open sets with intersection $\{0\}$.   Set $G_0=\frak A$,
and for $n\in\Bbb N$ choose an open neighbourhood
$G_{n+1}$ of $0$, included in $U_n\cap G_n$, such that
$[0,a]\subseteq G_{n+1}$ for
every $a\in G_{n+1}$ (393P).   \Quer\ If
$0\ne d\in\bigcap_{n\in\Bbb N}\overline{G}_n$, then for each
$n\in\Bbb N$ we can find a sequence
$\sequence{i}{a_{ni}}$ in $G_n$ order*-converging to $d$ (393Pc).   By
393Pa, there is a sequence $\sequencen{k(n)}$
in $\Bbb N$ such that $\sequencen{a_{n,k(n)}}$ order*-converges to $d$.
Now $d=\sup_{n\in\Bbb N}\inf_{i\ge n}a_{i,k(i)}$ (367Bf), so there is an
$n\in\Bbb N$ such that
$c=\inf_{i\ge n}a_{i,k(i)}$ is non-zero.   But in this case we must have
$c\le a_{i,k(i)}\in G_i$ and $c\in G_i\subseteq U_j$ whenever
$i\ge\max(n,j+1)$, so $c=0$.\ \BanG\
Thus $\bigcap_{n\in\Bbb N}\overline{G}_n=\{0\}$, as
required.\ \Qed

\medskip

\quad\grheadb\ For every neighbourhood $G$ of $0$ there is a
neighbourhood $H$ of $0$ such that $a\Bcup b\in G$ for all
$a$, $b\in H$.   \Prf\Quer\ Otherwise, choose $\sequencen{H_n}$,
$\sequencen{a_n}$ and $\sequencen{b_n}$ inductively,
as follows.   Start with an open neighbourhood $H_0$ of $0$ such
that $H_0\subseteq G$ and $[0,a]\subseteq H_0$ for every $a\in H_0$.
Given that $H_n$ is an open neighbourhood of $0$, let $a_n$,
$b_n\in H_n$ be such that $a_n\Bcup b_n\notin G$.   Because
the maps $a\mapsto a\Bcup a_n$ and $a\mapsto a\Bcup b_n$ are continuous,
there is an open neighbourhood $H_{n+1}$ of $0$ such that $a\Bcup a_n$
and $a\Bcup b_n$ belong to $H_n$ for every
$a\in H_{n+1}$;  and we may suppose that $H_{n+1}\Bsubseteq G_n$.
Continue.

An easy induction on $k$ shows that $a\Bcup\sup_{n\le i\le n+k}a_i$ and
$a\Bcup\sup_{n\le i\le n+k}b_i$ belong to $H_n$
whenever $k\in\Bbb N$ and $a\in H_{n+k+1}$.   In particular,
$\sup_{n\le i\le n+k}a_i\in H_n$ for every $k$;
since $\sequence{k}{\sup_{n\le i\le n+k}a_i}$ is
order*-convergent to $\sup_{i\ge n}a_i$,
$\sup_{i\ge n}a_i\in\overline{H}_n\subseteq\overline{G}_n$ for every
$n$.   Set $a^*=\inf_{n\in\Bbb N}\sup_{i\ge n}a_i$.
Then $\sequencen{\sup_{i\ge n}a_i}\to^* a^*$, and
$\sup_{i\ge n}a_i\in\overline{G}_m$ whenever $n\ge m$, so
$a^*\in\overline{G}_m$ for every $m$, and $a^*=0$.

In the same way, $\inf_{n\in\Bbb N}\sup_{i\ge n}b_i=0$.   It follows
that $\inf_{n\in\Bbb N}c_n=0$, where
$c_n=\sup_{i\ge n}a_i\Bcup\sup_{i\ge n}b_i$ for each $n$.   But now
$\sequencen{c_n}$ is a non-increasing sequence
with infimum $0$, so order*-converges to $0$, and there must be an $n$
such that $c_n\in H_0$.   Since
$a_n\Bcup b_n\Bsubseteq c_n$, $a_n\Bcup b_n\in H_0\subseteq G$, contrary
to the choice of $a_n$ and $b_n$.\ \Bang\Qed

\medskip

\quad\grheadc\ $\frak A$ is ccc.   \Prf\ Let $\sequencen{U_n}$ be a
sequence of open sets with intersection $\{0\}$, and
$A\subseteq\frak A\setminus\{0\}$ a partition of unity.
If $\sequence{i}{a_i}$ is a sequence
of distinct elements of $A$, then $\sequence{i}{a_i}\to^*0$ (393Md);  so
$A\setminus U_n$ is finite for every $n$, and $A$ is countable.\ \Qed

\medskip

\quad\grheadd\ ($\beta$) means just that $\Bcup$ is continuous at
$(0,0)$.   Also a non-increasing sequence
with infimum $0$ order*-converges to $0$, so converges topologically to
$0$ (3A3Pa);  and the topology is certainly T$_1$.   So all the
conditions of (iv) are satisfied by the order-sequential topology.

\medskip

{\bf (d)(iv)$\Rightarrow$(i)} By 393J, there is a strictly positive
Maharam submeasure on $\frak A$;  as $\frak A$ is Dedekind
$\sigma$-complete, it is a Maharam algebra.
}%end of proof of 393Q

\leader{393R}{Definition}\dvAnew{2008}
Let $\frak A$ be a Boolean algebra.   Then
$\frak A$ is {\bf\sfcc} if $\frak A$ can be expressed as
$\bigcup_{n\in\Bbb N}A_n$ where no $A_n$ includes any infinite disjoint
set.

\leader{393S}{Theorem}\dvAnew{2008}\cmmnt{ ({\smc Todor\v{c}evi\'c 04})}
Let $\frak A$ be a
Boolean algebra.   Then $\frak A$ is a Maharam algebra iff it is
\sfcc, \wsid\ and Dedekind $\sigma$-complete.

\proof{ (B.Balcar){\bf (a)} If $\frak A$ is a Maharam algebra, then of
course it is Dedekind $\sigma$-complete, and we have known since 393C that
it is \wsid.   Also it carries a strictly positive exhaustive submeasure,
so is \sfcc.

Of course $\{0\}$ is a Maharam algebra.
For the rest of the proof, therefore, I suppose that $\frak A$ is a
non-trivial algebra satisfying the conditions, and seek to show that it is
a Maharam algebra.

\medskip

{\bf (b)(i)} Let $\sequencen{A_n}$ be a
sequence of sets, with union $\frak A^+$, such that no $A_n$
includes any infinite disjoint set.   For each $n$, set
$B_n=\bigcup_{m\le n}\bigcup_{a\in A_m}[a,1]$,  so that $B_n$ includes no
infinite
disjoint subset.   Now there is an $n$ such that $1$ is in the interior
of $B_n$ for the order-sequential topology.   \Prf\Quer\
Otherwise, of course $\frak A$ is ccc, so
there is for each $n\in\Bbb N$ a sequence $\sequence{i}{b_{ni}}$
in $\frak A\setminus B_n$ which is order*-convergent to $1$
(393Pb).  By 393Pa, there is a
sequence $\sequencen{k(n)}$ in $\Bbb N$ such that $\sequencen{b_{n,k(n)}}$
order*-converges to $1$.   As $1\ne 0$, there must be an $m\in\Bbb N$ such that
$c=\inf_{i\ge m}b_{i,k(i)}\ne 0$.   There is an $n$ such that
$c\in A_n$, in which case $b_{i,k(i)}\in B_n\subseteq B_i$ for
every $i\ge\max(m,n)$.\ \Bang\Qed

\medskip

\quad{\bf (ii)} Set $H=\interior B_n$.   Then there is a $c\in H$ such that for
every $d\in\frak A$ one of $c\Bcap d$, $c\Bsetminus d\notin H$.
\Prf\Quer\
Otherwise, we can choose a sequence $\sequence{i}{c_i}$ in $H$ such that
$c_0=1$ and, for each $i\in\Bbb N$, $c_{i+1}\Bsubseteq c_i$ and
$c_i\Bsetminus c_{i+1}\in H$.   But in this case
$\sequence{i}{c_i\Bsetminus c_{i+1}}$ is a disjoint sequence in $B_n$,
which is impossible.\ \Bang\Qed

\medskip

\quad{\bf (iii)} $0$ and $1$ can be separated by open sets.   \Prf\ Take
$H$ and $c$ from (ii).   Then $G_0=\{d:c\Bsetminus d\in H\}$ and
$G_1=\{d:c\Bcap d\in H\}$ are disjoint open sets containing $0$ and $1$
respectively.\ \Qed

\medskip

{\bf (b)} It follows that $\frak A$ is actually Hausdorff in the
order-sequential topology.   \Prf\ Let $a_0$, $a_1\in\frak A$ be such that
$b=a_1\Bsetminus a_0$ is non-zero.
Consider the principal ideal $\frak A_b$.   Like $\frak A$, this is \sfcc,
\wsid\ and Dedekind $\sigma$-complete.
By (a), there are disjoint subsets $U$, $V$
of $\frak A_b$, open for the order-sequential topology of $\frak A_b$, such
that $0\in U$ and $b\in V$.   The function
$a\mapsto a\Bcap b:\frak A\to\frak A_b$ is continuous for the
order-sequential topologies (3A3Pb), so
$G=\{a:a\Bcap b\in U\}$ and $H=\{a:a\Bcap b\in V\}$ are open.   Now $G$ and
$H$ are
disjoint open sets in $\frak A$ containing $a_0$, $a_1$ respectively.
As $a_0$ and $a_1$ are arbitrary, $\frak A$ is Hausdorff.\ \Qed

By 393Q, $\frak A$ is a Maharam algebra.
}%end of proof of 393S


\exercises{
\leader{393X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 393Xa
Let $\frak A$ be the finite-cofinite algebra on an
uncountable set (316Yl).  (i) Set $\nu_10=0$, $\nu_1a=1$ for
$a\in\frak A\setminus\{0\}$.   Show that $\nu_1$ is a strictly positive
Maharam
submeasure but is not exhaustive.   (ii) Set $\nu_2a=0$ for finite $a$,
$1$ for cofinite $a$.   Show that $\nu_2$ is a uniformly exhaustive
Maharam submeasure but is not order-continuous.
%393B

\sqheader 393Xb Let $\frak A$ be a Boolean algebra and $\nu$ a
submeasure on $\frak A$.   Set $I=\{a:\nu a=0\}$.   Show that
(i) $I$ is an ideal of $\frak A$ (ii) there is a submeasure $\bar\nu$ on
$\frak A/I$ defined by setting $\bar\nu a^{\ssbullet}=\nu a$ for every
$a\in\frak A$ (iii) if $\nu$ is exhaustive, so is $\bar\nu$ (iv) if
$\nu$ is uniformly exhaustive, so is $\bar\nu$ (v) if $\nu$ is a Maharam
submeasure, $I$ is a $\sigma$-ideal (vi) if $\nu$ is a Maharam
submeasure and $\frak A$ is Dedekind $\sigma$-complete, $\bar\nu$ is a
Maharam submeasure.
%393B

\spheader 393Xc Let $\frak A$ be a Dedekind complete Boolean algebra and
$\nu$ an order-continuous submeasure on $\frak A$.   Show that $\nu$ has
a unique support $a\in\frak A$ such that $\nu\restrp\frak A_a$ is
strictly positive and $\nu\restrp\frak A_{1\Bsetminus a}$ is identically
zero.
%393B

\spheader 393Xd
Let $\frak A$ be a Boolean algebra and $\nu$ an exhaustive submeasure on
$\frak A$ such that
$\nu a=\lim_{n\to\infty}\nu a_n$ whenever $\sequencen{a_n}$ is a
non-decreasing sequence in $\frak A$ with supremum $a$.
Show that $\nu$ is a Maharam submeasure.
%393C

\spheader 393Xe Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\nu$ a uniformly exhaustive Maharam submeasure on
$\frak A$.   Show that there is a non-negative countably additive
functional $\mu$ on $\frak A$ such that $\{a:\mu a=0\}=\{a:\nu a=0\}$.
\Hint{393Xb(vi).}
%393D, 393Xb

\spheader 393Xf Let $\frak A$ be a Maharam algebra with its Maharam-algebra
topology and uniformity.   (i) Let $B\subseteq\frak A$ be a non-empty
upwards-directed set.   For
$b\in B$ set $F_b=\{c:b\Bsubseteq c\in B\}$.
Show that $\{F_b:b\in B\}$ generates a Cauchy filter
$\Cal F(B\closeuparrow)$ on $\frak A$ which converges to $\sup B$.
(ii) Show that closed subsets of $\frak A$ are order-closed.
(iii) Show that an order-dense subalgebra of $\frak A$ must be dense in the
topological sense.
%393G

\spheader 393Xg Let $\frak A$ be a Maharam algebra.   Show that it is a
measurable algebra iff for every $A\subseteq\frak A$ including antichains
of all finite sizes there is a sequence in $A$ which is order*-convergent
to $0$.
%393G

\spheader 393Xh Let $\frak A$ be a Boolean algebra.   Suppose that
$\sequencen{a_n}$, $\sequencen{b_n}$ are sequences in $\frak A$
order*-converging to $a$, $b$ respectively.   Show that
$\sequencen{a_n\frmedcirc b_n}\to^*a\frmedcirc b$ when $\frmedcirc$ 
is any of the operations $\Bcup$, $\Bcap$, $\Bsymmdiff$ or $\Bsetminus$.
%393M

\spheader 393Xi
Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra.   Write
$\frak T_{\text{os}}$ for the order-sequential topology on $\frak A$ and
$\frak T_{\text{ma}}$ for the measure-algebra topology.
Show that $\frak T_{\text{os}}\supseteq\frak T_{\text{ma}}$,
with equality iff $(\frak A,\bar\mu)$ is $\sigma$-finite.
%393N

\spheader 393Xj ({\smc Jech 08}) Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra and $\sequencen{A_n}$ a sequence of
subsets of $\frak A$ such that ($\alpha$) for
every $n\in\Bbb N$, any antichain in $A_n$ has at most $n$ elements
($\beta$)
a sequence $\sequence{k}{a_k}$ in $\frak A$ is order*-convergent to $0$ iff
$\{k:a_k\in A_n\}$ is finite for every $n\in\Bbb N$.
(i) Show that $\frak A$ is ccc.   (ii) Show that $\frak A$ is \wsid.
\Hint{if $C_n$ is non-empty and
downwards-directed with infimum $0$ for each $n$, show that there is a
sequence $\sequencen{a_n}\to^*0$ such that $a_n\in C_n$ for every $n$.}
(iii) Show that $\frak A$ is a Maharam algebra.  \Hint{393S.}
(iv) Show that any Maharam submeasure on $\frak A$ is uniformly exhaustive.
(v) Show that $\frak A$ is a measurable algebra.
%393S

\leader{393Y}{Further exercises (a)}
%\spheader 393Ya
Let $\frak A$ be any Boolean algebra with a strictly positive Maharam
submeasure.   Show that $\frak A$ is weakly $\sigma$-distributive.
%393C

\spheader 393Yb Let $U$ be a Riesz space, with its order-sequential
topology.   (i) Show that addition and
subtraction are separately continuous.   (ii) Show that $U$ is
Archimedean iff scalar multiplication is separately continuous
as a function from $\Bbb R\times U$ to $U$, and that in
this case scalar multiplication is actually continuous.
%393M

\spheader 393Yc Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra, and give $L^0=L^0(\frak A)$ its order-sequential topology.
Suppose $h:\Bbb R\to\Bbb R$ is continuous, and let $\bar h:L^0\to L^0$ be
the corresponding function as defined in 364H.   Show that $\bar h$ is
continuous.
%393M

\spheader 393Yd Let $\frak A$
be a Maharam algebra.   Show that a topology $\frak T$ on $L^0(\frak A)$
defined by the method of 393K must be the order-sequential topology on
$L^0(\frak A)$.
%393N

\spheader 393Ye Let $U$ be a \wsid\ Riesz space with the countable sup
property, with its
order-sequential topology, and $A$ a subset of $U$.
Show that $\overline{A}$ is the set of
order*-limits of sequences in $A$.
%393P

\spheader 393Yf Let $U$ be a \wsid\ Dedekind complete Riesz space with the
countable sup property, endowed with its order-sequential topology,
and $\frak A$ its band algebra.   Show that the following are
equiveridical:   (i) $\frak A$ is a Maharam algebra;
(ii) $U$ is Hausdorff;
(iii) addition on $U$ is continuous at $(0,0)$;
(iv) $\vee:U\times U\to U$ is continuous at $(0,0)$.
%393Q

\spheader 393Yg Let $\frak G$ be the regular open algebra of
$\Bbb R$, with its order-sequential topology.   (i) Show that if $U$, $V$
are open sets in $\frak G$ containing $0_{\frak G}=\emptyset$ and
$1_{\frak G}=\Bbb R$ respectively, then $U\cap V\ne\emptyset$.
(ii) Show that if $U$ is an open set in $\frak G$ containing
$\emptyset$ then there are $G$, $H\in U$ such that
$H=\Bbb R\setminus\overline{G}$.
(iii) Show that $\{\emptyset\}$ is a G$_{\delta}$ set in $\frak G$.
(iv) Show that there is no non-zero Maharam submeasure on $\frak G$.
(v) Show that there is no non-zero countably additive functional on
$\frak G$.
%393Q

\spheader 393Yh In 393Xj, show that each of the sets $A_n$ must have
non-zero intersection number.
%393Xj 393S

\spheader 393Yi Let $\frak A$ be an atomless Boolean algebra with countable
Maharam type.   Show that there is a submeasure $\mu$ on
$\frak A$, order-continuous on the left, such that whenever
$a\in\frak A\setminus\{0\}$ there is a $b\Bsubseteq a$ such that
$\mu b<\mu a$.
%G\lowcy\'nski 08, 3.6  mt39bits
}%end of exercises

\cmmnt{\Notesheader{393}
For many years it was not known whether there were any Maharam algebras
which were not measurable algebras;  this was the famous `control measure
problem', eventually solved by M.Talagrand.   I will present his
example in the next section.   We now know that we have a larger class, but
it remains very poorly understood, and the material presented here must be
regarded as work in progress.   As in \S\S391-392, the stimulus for these
ideas has been the attempt to characterize measurable algebras in more or
less algebraic terms.   If we are prepared to allow
order*-convergence of sequences to be an `algebraic' notion, then 393Xj is
such a characterization;  but it shares with Kelley's criterion 391K the
need for a sequence $\sequencen{A_n}$, covering $\frak A^+$, with defined
properties.   The advance, if any, is that the properties ($\alpha$) and
($\beta$) of 393Xj are a good deal farther from any formula for a measure.

The first few results of this section, down to 393G, are concerned with
checking that Maharam algebras share properties with measurable algebras,
and the proofs use the same ideas, with occasional minor modifications.
In 393H we have to think a little, since exhaustivity is
less familiar, and harder to apply, than additivity.   From this
proposition we see that exhaustive submeasures are to uniformly exhaustive
submeasures something like
what Maharam algebras are to measurable algebras.
393K is a further example of a well-known construction -- this time,
convergence in measure -- which has a version based on Maharam algebras.

In \S367 I examined order*-convergence in Riesz spaces, without
explicitly discussing the associated topology, and in
393L-393Q %393L 393M 393O 393P 393N 393Q
here I look at
Boolean algebras.   In both cases the usefulness of the idea starts with
the fact that the algebraic operations are separately
continuous (367Ca, 393M), which is itself a consequence of the strong
distributive laws in 313A-313B and 352E.
It is easy to see that in a Maharam algebra the order-sequential topology
is the Maharam-algebra topology (393N).   What is remarkable is that
natural questions about the order-sequential topology lead to
characterizations of Maharam algebras (393Q).
This leads directly to an astonishing algebraic characterization of Maharam
algebras (393S).   (But once again we
need to hypothesize the existence of a
suitable sequence of sets covering $\frak A^+$.)
}

\discrpage

\frfilename{mt394.tex}
\versiondate{13.6.11/30.8.18}
\copyrightdate{2017}

\def\chaptername{Measurable algebras}
\def\sectionname{Talagrand's example}

\def\sbcc{$\sigma$-bounded-cc}
\def\sfcc{$\sigma$-finite-cc}

\newsection{394\dvAnew{2018}}

I rewrite the construction in {\smc Talagrand 08} of an
exhaustive submeasure which is not uniformly exhaustive, generalized as in
{\smc Perovi\'c \& Veli\v{c}kovi\'c 18}.

\leader{394A}{PV norms (a)} I will say that a {\bf PV norm} is a function
$\|\,\|:[\Bbb N]^{<\omega}\to\Bbb N$ such that

\inset{----- $\|\emptyset\|=0$,\quad$\|I\|=1$ if $\#(I)=1$,

----- $\|I\cup J\|\le\|I\|+\|J\|$ for all $I$, $J\in[\Bbb N]^{<\omega}$,

%----- $\|I\|\le\|J\|$ whenever $I\subseteq J\in[\Bbb N]^{<\omega}$,

----- $\|I\|\le\|J\|$ whenever $I$, $J\in[\Bbb N]^{<\omega}$ and
$\#(I\cap n)\le\#(J\cap n)$ for every $n\in\Bbb N$,

----- $\lim_{n\to\infty}\|A\cap n\|=\infty$ for every infinite
$A\subseteq\Bbb N$}

\noindent({\smc Perovi\'c \& Veli\v{c}kovi\'c 18}, 2.2).

\medskip

{\bf (b)} Note that if $\|\,\|$ is a PV norm then
$\|I\|\le\|J\|\le\#(J)$ whenever $I\subseteq J\in[\Bbb N]^{<\omega}$.
We see also that if $I\in[\Bbb N]^{<\omega}$ and $k<\|I\|$ there is an
$n\in I$ such that $\|I\cap n\|=k$.

\medskip

{\bf (c)} The version of Talagrand's example in the 2012 edition of Volume 3
corresponds to the case in which $\|I\|=\#(I)$ for every
$I\in[\Bbb N]^{<\omega}$.   For the work of this section there is no need to
consider any other, and some of the formulae in 394D become more readable
if you make this simplification;  but it makes no real difference to the ideas
required.

\leader{394B}{Definitions} \cmmnt{We are ready to begin work.
The construction is complex and demands a large volume of special notation.

\medskip

}{\bf (a)}  %\Tal{2.1}
I shall work throughout with $X=\prod_{n\in\Bbb N}T_n$ where
$\sequencen{T_n}$ is a sequence of non-empty finite sets
and $\sup_{n\in\Bbb N}\#(T_n)$ is infinite.
 %394Ib
$X$ may be
regarded as a compact Hausdorff space with the product of the discrete
topologies on the $T_n$.   For each $n\in\Bbb N$,
$\frak B_n$ will be the algebra of subsets of $X$ determined by
coordinates less than $n$  %T(9.1)4/27
and $\Cal A_n$ the set of its atoms\cmmnt{, that is,
the family of sets of the
form $\{x:z\subseteq x\in X\}$ for some $z\in\prod_{i<n}T_i$}.
 %T(9.1)4/31
$\frak B\cmmnt{\mskip5mu=\bigcup_{n\in\Bbb N}\frak B_n}$ will be the
algebra of open-and-closed subsets of $X$.  %T(9.1)4/28
For $I\subseteq\Bbb N$ and $z\in\prod_{n\in I}T_n$, $Y_z$ will be
$\{x:z\subseteq x\in X\}$.   Finally, $\|\,\|$ will be a PV norm on
$[\Bbb N]^{<\omega}$.

\spheader 394Bb We shall need
a sequence $\sequence{k}{\alpha_k}$ in $\Bbb R$
and a sequence $\sequence{k}{N_k}$ in $\Bbb N$.   \cmmnt{It is easy
enough to give
appropriate formulae but perhaps the ideas will be clearer if instead} I
declare the properties they must have.

\medskip

\quad{\bf (i)}  %\Tal{3.3}
$\alpha_k>0$ and
$(2^{k+4})^{\alpha_k}\le 2$
 %394Fc
for every $k\in\Bbb N$, $\sequence{k}{\alpha_k}$ is non-increasing,
 %394E
and $\sum_{k=0}^{\infty}\alpha_k\le\bover12$.
 %394Cg
 %\alpha_k\le 1 394K
 %4^{\alpha_k}\le 2 394K

\medskip

\quad{\bf (ii)}  %\Tal{4.2, 5.2}
$N_k\in\Bbb N$ and
 %394M
 %394E
$2^{-k}(2^{-2k-12}N_k)^{\alpha_k}\ge 2^4$
 %394Gb
for every $k\in\Bbb N$.

\spheader 394Bc %\Tal{3.1}
\cmmnt{Now we come to some of the key ideas.}   For a set
$\Cal I\subseteq\Cal PX\times\Cal P\Bbb N\times\coint{0,\infty}$,
define\cmmnt{ its `spread'} $\spread\Cal I$ to be
$\bigcup_{(E,I,w)\in\Cal I}E$
and\cmmnt{ its `weight'} $\weight\Cal I$ to be $\sum_{(E,I,w)\in\Cal I}w$.

\spheader 394Bd %\Tal{3.2}
For any family
$\Cal E\subseteq\Cal PX\times\Cal P\Bbb N\times\coint{0,\infty}$ define
$\phi_{\Cal E}:\frak B\to[0,\infty]$ by setting

\Centerline{$\phi_{\Cal E}E
=\inf\{\weight\Cal I:\Cal I\subseteq\Cal E$ is finite,
$E\subseteq\spread\Cal I\}$,}

\noindent counting $\inf\emptyset$ as $\infty$.   \cmmnt{So
$\phi_{\emptyset}\emptyset=0$ and $\phi_{\emptyset}E=\infty$ for
$E\in\frak B\setminus\{\emptyset\}$.}

\spheader 394Be %\Tal{2.8}
For $D\subseteq X$ and $I\subseteq\Bbb N$ set

\Centerline{$\theta_I(D)
=\{y:y\in X,\,y\restr I=x\restr I\text{ for some }x\in D\}$.}

\spheader 394Bf{\bf (i)} %\Tal{\S3.4}
If $m<n$ in $\Bbb N$, $\phi:\frak B\to[0,\infty]$ is
a function and $E\in\frak B$, then $E$ is
{\bf $\phi$-thin between $m$ and $n$} if
$\phi(X\setminus\theta_{n\setminus m}(A\cap E))\ge 1$ for
every $A\in\Cal A_m$.
 %T(9.1)11/10

\medskip

\quad{\bf (ii)} If $I\subseteq\Bbb N$,
$\phi:\frak B\to[0,\infty]$ is a function and $E\in\frak B$, then
$E$ is {\bf $\phi$-thin along $I$} if it is $\phi$-thin between $m$ and $n$
whenever $m$, $n\in I$ and $m<n$.
 %T(9.1)11/17

\spheader 394Bg %\Tal{\S4.0}
For $k\le p\in\Bbb N$ define $\Cal C_{kp}$
and $\nu_{kp}$ by downwards induction on $k$\cmmnt{, as follows}.
\cmmnt{Start with }$\Cal C_{pp}=\emptyset$ for every $p$.
\cmmnt{Given $\Cal C_{kp}$, set }$\nu_{kp}=\phi_{\Cal C_{kp}}$.
Given that $k<p$ and
$\Cal C_{k+1,p}$
and $\nu_{k+1,p}=\phi_{\Cal C_{k+1,p}}$ have been defined, set

$$\eqalign{\Cal E_{kp}
&=\{(E,I,w):E\in\frak B,\,I\in[\Bbb N]^{<\omega},\,
1\le\|I\|\le N_k,\cr
&\mskip150mu
w\ge 2^{-k}\bigl(\Bover{N_k}{\|I\|}\bigr)^{\alpha_k},\,
E\text{ is }\nu_{k+1,p}\text{-thin along }I\},\cr
\Cal C_{kp}&=\Cal E_{kp}\cup\Cal C_{k+1,p}\cr}$$

\noindent and continue.

\spheader 394Bh %\Tal{\S5.0}
Define $\sequence{k}{c_k}$ by setting
$c_0=8$, $c_{k+1}=2^{2\alpha_k}c_k$ for every $k$.


\leader{394C}{Very elementary facts} \cmmnt{In the hope of aiding
digestion of
the definitions here, of which 394Bf and 394Bg are likely to be wholly
obscure to anyone who has not worked through this proof before, I run
over some obvious facts which will be used below.

\medskip

}{\bf (a)}  %\Tal{3.2}
$\phi_{\Cal E}:\frak B\to[0,\infty]$ is a submeasure
for any $\Cal E\subseteq\Cal PX\times\Cal P\Bbb N\times\coint{0,\infty}$.
\cmmnt{(Subadditivity and monotonicity are written into the definition.)}

\spheader 394Cb
If $I$, $J\subseteq\Bbb N$ then $\theta_I\theta_J=\theta_{I\cap J}$.
 %394E
If $I\subseteq J\subseteq\Bbb N$ then
$\theta_I(D)=\theta_I\theta_J(D)\supseteq\theta_J(D)$ for all
$D\subseteq X$.
If $I\subseteq\Bbb N$ then
$\theta_I(D\cap\theta_I(E))=\theta_I(E\cap\theta_I(D))$ for all $D$,
$E\subseteq X$.
 %394E
For any $I\subseteq\Bbb N$ and any family $\Cal D$ of subsets of $X$,
$\theta_I(\bigcup\Cal D)=\bigcup_{D\in\Cal D}\theta_I(D)$.
 %394Fc

For $n\in\Bbb N$ and $D\subseteq X$, $D\in\frak B_n$ iff $\theta_n(D)=D$.
If $E\in\frak B$ and $I\subseteq\Bbb N$, $\theta_I(E)\in\frak B$.
 %394L
If $m\le n$ in $\Bbb N$, $A\in\Cal A_m$ and $A_1\in\Cal A_n$, then
$A\cap\theta_{n\setminus m}(A_1)\in\Cal A_n$.
 %394Gd
If $m\in\Bbb N$ and $A\in\Cal A_m$ then
$E\mapsto\theta_{\Bbb N\setminus m}(A\cap E):\frak B\to\frak B$ is a
Boolean homomorphism.
 %394L

\spheader 394Cc If $m<n$, $\phi:\frak B\to[0,\infty]$ is a
non-decreasing function,
$E\in\frak B$ is $\phi$-thin between $m$ and $n$
and $E'\in\frak B$ is included in $E$,
then $E'$ is $\phi$-thin between $m$ and $n'$ for
every $n'\ge n$.
 %394Gf

\spheader 394Cd\cmmnt{ All the classes} $\Cal E_{kp}$,
$\Cal C_{kp}$ are closed under increases in the scalar variable and
decreases in the first variable, that is,

----- if $k<p$, $(E,I,w)\in\Cal E_{kp}$, $E'\in\frak B$, $E'\subseteq E$
and $w'\ge w$ then $(E',I,w')\in\Cal E_{kp}$,

----- if $k\le p$, $(E,I,w)\in\Cal C_{kp}$, $E'\in\frak B$, $E'\subseteq E$
and $w'\ge w$ then $(E',I,w')\in\Cal C_{kp}$.

\spheader 394Ce If $k\le p$\cmmnt{ in $\Bbb N$},
$\Cal C_{kp}=\bigcup_{k\le l<p}\Cal E_{lp}$.

\spheader 394Cf If $k<p$\cmmnt{ in $\Bbb N$},
$\nu_{kp}\le\nu_{k+1,p}$\cmmnt{,
because $\Cal C_{kp}\supseteq\Cal C_{k+1,p}$}.

\spheader 394Cg $8\le c_k\le 16$ for every
$k\in\Bbb N$\prooflet{, because
$\sum_{k=0}^{\infty}2\alpha_k\le 1$}.

\spheader 394Ch If $k<p$\cmmnt{ in $\Bbb N$},
then\cmmnt{ $(X,\{0\},2^{-k}N_k^{\alpha_k})\in\Cal E_{kp}$
so} $\nu_{kp}X\le 2^{-k}N_k^{\alpha_k}$ and $\nu_{kp}$ is totally finite.

\leader{394D}{}\cmmnt{ Moving up a gear, we have the following.

\medskip

\noindent}{\bf Lemma}
Suppose that $\Cal K$ is a non-empty finite family of subsets
of $\Bbb N$ and $r\in\Bbb N$ is such that $\|K\|\ge r\#(\Cal K)$ for every
$K\in\Cal K$.   Then we have an enumeration $\ofamily{i}{s}{K_i}$ of
$\Cal K$ and a
non-decreasing family $\langle n_i\rangle_{i\le s}$ such that
$\|K_i\cap n_{i+1}\setminus n_i\|=r$ for every $i<s$.
 %T(9.1)2.3

(b) Suppose that $\ofamily{i}{s}{K_i}$ is a family of finite subsets of
$\Bbb N$ such that $\|K_i\|\ge n\ge 3$ for every $i<s$ and
$\max K_i<\min K_{i+1}$ for $i\le s-2$, and that $A\in[\Bbb N]^{<\omega}$ is
such that $\|A\|\le 1$.   Let $\Cal J$ be a finite subset of
$\Cal PX\times([\Bbb N]^{<\omega}\setminus\{\emptyset\})\times\coint{0,\infty}$.
Then we can find
$\ofamily{i}{s}{u_i}$ and $\ofamily{i}{s}{v_i}$ such that
$u_i$, $v_i\in K_i$ and $u_i<v_i$ for each $i<s$ and, setting
$W=\bigcup_{i<s}v_i\setminus u_i$, $A\cap W=\emptyset$ and

\Centerline{$\weight\{(E,I,w):(E,I,W)\in\Cal J$,
$\|I\setminus W\|<\bover12\|I\|\}
\le\Bover1{n-2}\weight\Cal J$.}

% PV 3.9

\proof{{\bf (a)} If $r=0$ we can take any enumeration of $\Cal K$
and set $n_i=0$ for every $i$.   Otherwise, write
$s$ for $\#(\Cal K)$ and choose $n_i$, $K_i$ inductively, as follows.   Start
with $n_0=0$.   Given $j<s$, $n_j\in\Bbb N$ and $\ofamily{i}{j}{K_i}$ such that
$\|K\setminus n_j\|\ge r(s-j)$ for every
$K\in \Cal K_j=\Cal K\setminus\{K_i:i<j\}$, set

\Centerline{$n_{j+1}=\min\{n:\|K\cap n\setminus n_j\|\ge r$ for some
$K\in \Cal K_j\}$}

\noindent and choose $K_j\in\Cal K_j$ such that
$\|K_j\cap n_{j+1}\setminus n_j\|\ge r$.   Observe that

\Centerline{$\|K\cap n_{j+1}\setminus n_j\|
\le\|K\cap(n_{j+1}-1)\setminus n_j\|+\|\{n_{j+1}-1\}\|
\le(r-1)+1=r$}

\noindent for every $K\in\Cal K_j$, so in fact
$\|K_j\cap n_{j+1}\setminus n_j\|=r$ and also

\Centerline{$\|K\setminus n_{j+1}\|
\ge\|K\setminus n_j\|-\|K\cap n_{j+1}\setminus n_j\|
\ge r(s-j)-r=r(s-(j+1))$}

\noindent for every $K\in\Cal K_j$, so the induction will continue.

\medskip

{\bf (b)}
For $i<s$ and $k\in K_i\setminus\{\min K_i\}$ write $k^-$ for the greatest
member of $K_i$ less than $k$.   Set

\Centerline{$K'_i=\{k:k\in K_i\setminus\{\min K_i\}$,
$A\cap k\setminus k^-\ne\emptyset\}$,
\quad$K''_i=K_i\setminus(K'_i\cup\{\min K_i\})$.}

\noindent Then $\|K'_i\|\le 1$.   \Prf\
For $k\in K'_i$ set
$g(k)=\min(A\cap k\setminus k^-)$.   Then
$g:K'_i\to A$ is injective and
$g(k)\le k$ for every $k\in K'_i$.   So
$\#(K'_i\setminus m)\le\#(g[K'_i]\setminus m)\le\#(A\setminus m)$ for every
$m\in\Bbb N$, and $\|K'_i\|\le\|A\|\le 1$.\ \Qed

Consequently

\Centerline{$\#(K''_i)\ge\|K''_i\|\ge\|K_i\|-\|K'_i\|-\|\{\min K_i\}\|
\ge n-2$}

\noindent and we have a family $\ofamily{j}{n-2}{k_{ij}}$ of distinct elements
of $K''_i$, so that $A\cap k_{ij}\setminus k_{ij}^-=0$ for every $j<n-2$.

For $j<n-2$, set

\Centerline{$W_j=\bigcup_{i<s}k_{ij}\setminus k_{ij}^-$,
\quad$\Cal J_j=\{(E,I,w):(E,I,w)\in\Cal J$,
$\|I\setminus W_j\|<\bover12\|I\|\}$.}

\noindent Then $W_0,\ldots,W_{n-3}$ are disjoint and none of them meet $A$.
Since $\|\,\|$ is subadditive and $I=(I\setminus W_j)\cup(I\setminus W_{j'})$
whenever $j$, $j'$ are distinct,
$\Cal J_0,\ldots,\Cal J_{n-3}$ are disjoint.   So
$\weight\Cal J\ge\sum_{j=0}^{n-3}\weight\Cal J_j$
and there is a $j<n-2$ such that
$\weight\Cal J_j\le\Bover1{n-2}\weight\Cal J$.
Take $u_i=k_{ij}^-$ and $v_i=k_{ij}$ for each
$i$, so that $W=\bigcup_{i<s}v_i\setminus u_i$ is $W_j$, and we have an
appropriate pair of sequences.
}%end of proof of 394D

\leader{394E}{Lemma}  %\Tal{\S5.2}
Suppose that $k\le p$, $m<n$,
$A\in\Cal A_m$, $(E,I,w)\in\Cal C_{kp}$ and $I'=I\cap n\setminus m$
is non-empty.   If $E'=\theta_{n\setminus m}(E\cap A)$ and
$w'\ge\bigl(\Bover{\|I\|}{\|I'\|}\bigr)^{\alpha_k}w$,
then $(E',I',w')\in\Cal C_{kp}$.
 %T(9.1)5.2

\proof{ There is an $l$ such that $k\le l<p$ and
$(E,I,w)\in\Cal E_{lp}$.   Now
$E'$ is $\nu_{l+1,p}$-thin along $I'$.   \Prf\
Suppose that $i$, $j\in I'$ and $i<j$, so that $m\le i<j<n$.
Take any $A_1\in\Cal A_i$, and set $A_2=A\cap\theta_{n\setminus m}(A_1)$,
so that $A_2$ also belongs to $\Cal A_i$.   Then, using the list in 394Cb,

$$\eqalignno{\theta_{j\setminus i}(E'\cap A_1)
&=\theta_{j\setminus i}(A_1\cap\theta_{n\setminus m}(E\cap A))\cr
&=\theta_{j\setminus i}(\theta_{n\setminus m}
 (A_1\cap\theta_{n\setminus m}(E\cap A)))\cr
&=\theta_{j\setminus i}(\theta_{n\setminus m}
 (E\cap A\cap\theta_{n\setminus m}(A_1)))\cr
&=\theta_{j\setminus i}(\theta_{n\setminus m}(E\cap A_2))
=\theta_{j\setminus i}(E\cap A_2).\cr}$$

\noindent So

\Centerline{$\nu_{l+1,p}(X\setminus\theta_{j\setminus i}(E'\cap A_1))
=\nu_{l+1,p}(X\setminus\theta_{j\setminus i}(E\cap A_2))
\ge 1$}

\noindent because $E$ is $\nu_{l+1,p}$-thin between $i$ and $j$.
As $i$, $j$ and $A_1$ are arbitrary, $E'$ is
$\nu_{l+1,p}$-thin along $I'$.\ \Qed

Of course $\|I'\|\le\|I\|\le N_l$.   Finally, because
$\alpha_l\le\alpha_k$, we have

\Centerline{$w'\ge\bigl(\Bover{\|I\|}{\|I'\|}\bigr)^{\alpha_k}w
\ge\bigl(\Bover{\|I\|}{\|I'\|}\bigr)^{\alpha_l}
   \cdot 2^{-l}\bigl(\Bover{N_l}{\|I\|})^{\alpha_l}
=2^{-l}\bigl(\Bover{N_l}{\|I'\|}\bigr)^{\alpha_l}$,}

\noindent and $(E',I',w')\in\Cal E_{lp}\subseteq\Cal C_{kp}$.
}  %end of proof of 394E

\leader{394F}{Corollary} (a)
Suppose that $n\in\Bbb N$ and $k\le p$ and that
$\Cal I\subseteq\Cal C_{kp}$
is a finite set such that $\|I\cap n\|\ge\bover14\|I\|$
whenever $(E,I,w)\in\Cal I$.
Then $\nu_{kp}(\theta_n(\spread\Cal I))\le 2\weight\Cal I$.

(b) Suppose that $m\in\Bbb N$, $k\le p$ and $A\in\Cal A_m$.   Let
$\Cal I$ be a finite subset of $\Cal C_{kp}$ such that
$\|I\setminus m\|\ge\bover14\|I\|$ whenever $(E,I,w)\in\Cal I$.   Then
$\nu_{kp}(\theta_{\Bbb N\setminus m}(A\cap\spread\Cal I))
\le 2\weight\Cal I$.

(c) Suppose that $m<n$ in $\Bbb N$, $k\le p$ and $A\in\Cal A_m$.   Let
$\Cal I$ be a finite subset of $\Cal C_{kp}$ such that
$\|I\cap n\setminus m\|\ge 2^{-k-4}\|I\|$ whenever $(E,I,w)\in\Cal I$.
Then $\nu_{kp}(\theta_{n\setminus m}(A\cap\spread\Cal I))
\le 2\weight\Cal I$.

\proof{{\bf (a)} For each $(E,I,w)\in\Cal I$ set
$E'=\theta_n(E)\in\Cal B_n$, $I'=I\cap n$ and

\Centerline{$w'
=\bigl(\Bover{\|I\|}{\|I'\|}\bigr)^{\alpha_k}w\le 4^{\alpha_k}w
\le 2w$.}

\noindent By 394E, with $m=0$ and $A=X$, $(E',I',w')\in\Cal C_{kp}$.
Set $\Cal J=\{(E',I',w'):(E,I,w)\in\Cal I\}$ and $B=\spread\Cal J$.
Then

\Centerline{$B=\bigcup_{(E,I,w)\in\Cal I}\theta_n(E)
=\theta_n(\spread\Cal I)$}

\noindent and

\Centerline{$\nu_{kp}B\le\weight\Cal J
=\sum_{(E,I,w)\in\Cal I}w'
\le 2\weight\Cal I$,}

\noindent as required.

\medskip

{\bf (b)} This time, take $n>m$ so large that $I\subseteq n$ whenever
$(E,I,w)\in\Cal I$.   For $(E,I,w)\in\Cal I$, set

\Centerline{$E'=\theta_{n\setminus m}(A\cap E)$,
\quad$I'=I\setminus m=I\cap n\setminus m$,
\quad
$w'=\bigl(\Bover{\|I\|}{\|I'\|}\bigr)^{\alpha_k}w\le 2w$.}

\noindent Then 394E tells us that $(E',I',w')\in\Cal C_{kp}$.
Setting $\Cal J=\{(E',I',w'):(E,I,w)\in\Cal I\}$,

\Centerline{$\theta_{\Bbb N\setminus m}(A\cap\spread\Cal I)
=\bigcup_{(E,I,w)\in\Cal I}\theta_{\Bbb N\setminus m}(A\cap E)
\subseteq\bigcup_{(E,I,w)\in\Cal I}E'
=\spread\Cal J$,}

\noindent so

\Centerline{$\nu_{kp}(\theta_{N\setminus m}(A\cap\spread\Cal I))
\le\weight\Cal J
\le 2\weight\Cal I$.}

\medskip

{\bf (c)} For $(E,I,w)\in\Cal I$ set

\Centerline{$E'=\theta_{n\setminus m}(A\cap E)$,
\quad$I'=I\cap n\setminus m$,
\quad$w'=\bigl(\Bover{\|I\|}{\|I'\|}\bigr)^{\alpha_k}w
\le(2^{k+4})^{\alpha_k}w\le 2w$.}

\noindent Then $(E',I',w')\in\Cal C_{kp}$.
Setting $\Cal J=\{(E',I',w'):(E,I,w)\in\Cal I\}$,

\Centerline{$\theta_{n\setminus m}(A\cap\spread\Cal I)
=\bigcup_{(E,I,w)\in\Cal I}\theta_{n\setminus m}(A\cap E)
=\bigcup_{(E,I,w)\in\Cal I}E'=\spread\Cal J$,}

\noindent so

\Centerline{$\nu_{kp}(\theta_{n\setminus m}(A\cap\spread\Cal I))
\le\weight\Cal J
\le 2\weight\Cal I$.}
}  %end of proof of 394F

\leader{394G}{}\cmmnt{ We are at the centre of the argument.

\medskip

\noindent}{\bf Lemma} Suppose that $L\in[\Bbb N]^{<\omega}$ is such that
$\|L\|\le 1$, and $z\in\prod_{r\in L}T_r$.    Then
$\nu_{kp}Y_z\ge c_k$ whenever $k\le p$ in $\Bbb N$.
 %T(9.1)5.1

\proof{ Induce on $p-k$.

\medskip

{\bf (a)} If $k=p$ then

\Centerline{$\Cal C_{pp}=\emptyset$,
\quad$\nu_{pp}Y_z=\infty$.}

\noindent For the downwards step to $k<p$,
given that $\nu_{k+1,p}Y_z\ge c_{k+1}$, take a
finite set $\Cal I\subseteq\Cal C_{kp}$ such that $\weight\Cal I<c_k$.
The rest of this proof is devoted to showing that
$Y_z\not\subseteq\spread\Cal I$.

\medskip

{\bf (b)} It will help to get a trivial case out of the way.   If
$\Cal I\subseteq\Cal C_{k+1,p}$, then we have

\Centerline{$\weight\Cal I<c_k\le c_{k+1}\le\nu_{k+1,p}Y_z$}

\noindent by the inductive hypothesis,
so certainly $Y_z\not\subseteq\spread\Cal I$.   Accordingly we may
suppose henceforth that $\Cal I\not\subseteq\Cal C_{k+1,p}$.

A second elementary point is that $\|I\|\ge 2^{2k+12}$ whenever
$(E,I,w)\in\Cal I$.   \Prf\ We have an $l$ such that $k\le l<p$ and
$(E,I,w)\in\Cal E_{kp}$, so

\Centerline{$2^{-l}\bigl(\Bover{N_l}{\|I\|}\bigr)^{\alpha_l}
\le w\le c_k\le 2^4$}

\noindent and
$\|I\|\ge 2^{2l+12}\ge 2^{2k+12}$, by the choice of $N_l$.\ \Qed

\medskip

{\bf (c)} Express $\Cal I$ as
$\Cal J\cup\Cal K$ where $\Cal J\subseteq\Cal C_{k+1,p}$ and
$\Cal K\subseteq\Cal E_{kp}$.
Set $s=\#(\Cal K)>0$.   For $(E,I,w)\in\Cal K$ we have
$w\ge 2^{-k}$, so $s\le 2^kc_k\le 2^{k+4}$ (394Cg).
Consequently $\|I\|\ge 2^{2k+12}\ge 2^{k+8}s$ whenever $(E,I,w)\in\Cal K$.
By 394Da, we can find $m_0<m_1<\ldots<m_s$
and an enumeration $\ofamily{i}{s}{(E_i,K_i,w_i)}$ of $\Cal K$ such
that $\|K_i\cap m_{i+1}\setminus m_i\|=2^{k+8}$ for $i<s$.
By 394Db we can find members $u_i$, $v_i$ of $K_i\cap m_{i+1}\setminus m_i$
such that $u_i<v_i$, for $i<s$, and
setting $W=\bigcup_{i<s}v_i\setminus u_i$, $L\cap W=\emptyset$ and

$$\eqalign{S&
=\weight\{(E,I,w):(E,I,w)\in\Cal J,\,\|I\setminus W\|<\Bover12\|I\|\}\cr&
\le\Bover1{2^{k+8}-2}\weight\Cal J
\le 2^{-k-6}c_k
\le 2^{-k-2}.\cr}$$

\medskip

{\bf (d)} Set

\Centerline{$\Cal J_1=\{(E,I,w):(E,I,w)\in\Cal J$,
$\|I\setminus W\|<\Bover12\|I\|\}$,
\quad$\Cal J_2=\Cal J\setminus\Cal J_1$;}

\noindent then

\Centerline{$\weight\Cal J_1=S\le 2^{-k-2}\le\Bover14$}

\noindent and $\|I\cap W\|>\bover12\|I\|$ whenever $(E,I,w)\in\Cal J_1$.
For $i<s$ set

\Centerline{$\Cal J_{1i}=\{(E,I,w):(E,I,w)\in\Cal J_1$,
$\|I\cap v_i\setminus u_i\|\ge 2^{-k-5}\|I\|\}$.}

\noindent Since $s\le 2^{k+4}$, $\Cal J_1=\bigcup_{i<s}\Cal J_{1i}$.

\medskip

{\bf (e)} Suppose that $i<s$ and $A\in\Cal A_{u_i}$.
Then there is an $A'\in\Cal A_{v_i}$ such that
$A'\subseteq A\setminus(E_i\cup\spread\Cal J_{1i})$.   \Prf\   Set
$C=\theta_{v_i\setminus u_i}(A\cap\spread\Cal J_{1i})\in\frak B_{v_i}$.
By 394Fc, applied in $\Cal C_{k+1,p}$,

\Centerline{$\nu_{k+1,p}C\le 2\weight\Cal J_{1i}\le 2\weight\Cal J_1<1$.}

\noindent As $(E_i,K_i,w_i)\in\Cal E_{kp}$,
$E_i$ is $\nu_{k+1,p}$-thin between $u_i$ and $v_i$,
$\nu_{k+1,p}(X\setminus\theta_{v_i\setminus u_i}(A\cap E_i))\ge 1$ and
$C$ does not include $X\setminus\theta_{v_i\setminus u_i}(A\cap E_i)$.
Since these sets both belong to $\frak B_{v_i}$
there is an $A_1\in\Cal A_{v_i}$ disjoint from both $C$ and
$\theta_{v_i\setminus u_i}(A\cap E_i)$, that is, disjoint from
$\theta_{v_i\setminus u_i}(A\cap(E_i\cup\spread\Cal J_{1i}))$.
Now $A'=A\cap\theta_{v_i\setminus u_i}(A_1)$
belongs to $\Cal A_{v_i}$, is included in $A$
and is disjoint from $E_i\cup\spread\Cal J_{1i}$.\ \Qed

\medskip

{\bf (f)} We can therefore find a function $\Gamma:X\to X$ such that
$\Gamma[X]$ is disjoint from $\spread(\Cal K\cup\Cal J_1)$, while
$\Gamma(x)\restr m$ is determined by $x\restr m$ for every $m\in\Bbb N$.
\Prf\ By (e) just above, we have for each $i<s$
a function $q_i:\Cal A_{u_i}\to\Cal A_{v_i}$
such that $q_i(A)\subseteq A\setminus(E_i\cup\spread\Cal J_{1i})$ for every
$A\in\Cal A_{u_i}$.   We can re-interpret $q_i$ as a function
$h_i:\prod_{n<u_i}T_n\to\prod_{n<v_i}T_n$ defined by saying that if
$A=\{x:x\restr u_i=y\}$ then $q_i(A)=\{x:x\restr v_i=h_i(y)\}$;  note that
$y=h_i(y)\restr u_i$ for every $y\in\prod_{n<u_i}T_n$.
Now, for $x\in X$, define $\Gamma(x)(n)$ inductively by saying that

$$\eqalign{\Gamma(x)(n)
&=x(n)\text{ if }n\in\Bbb N\setminus W,\cr
&=h_i(\Gamma(x)\restr u_i)(n)\text{ if }i<s\text{ and }u_i\le n<v_i.\cr}$$

\noindent Of course this ensures that $\Gamma(x)\restr m$ is determined by
$x\restr m$ for every $m$.   If $i<s$, $x\in X$, and $A\in\Cal A_{u_i}$ is
such that
$\Gamma(x)\in A$, then $\Gamma(x)\in q_i(A)$, which is disjoint from
$E_i\cup\spread\Cal J_{1i}$.   Thus $\Gamma[X]$ is disjoint from
$\bigcup_{i<s}E_i\cup\spread\Cal J_{1i}=\spread(\Cal K\cup\Cal J_1)$.\ \Qed

\medskip

{\bf (g)} Take $(E,I,w)\in\Cal J_2$ and consider
$\nu_{k+1,p}(\Gamma^{-1}[E])$.

\medskip

\quad{\bf (i)} There is an $l$ such that $k\le l<p$
and $(E,I,w)\in\Cal E_{lp}$.
Now if $m$, $n\in I$ are such that $m<n$ and
$n\setminus m$ is disjoint from $W$, $\Gamma^{-1}[E]$ is
$\nu_{l+1,p}$-thin between $m$ and $n$.   \Prf\ Take any $A\in\Cal A_m$.
Because $\Gamma(x)\restr m$ is determined by $x\restr m$,
we can find an $A'\in\Cal A_m$ such that $\Gamma[A]\subseteq A'$.
In this case,

\Centerline{$A\cap\Gamma^{-1}[E]
\subseteq\Gamma^{-1}[\Gamma[A]\cap E]
\subseteq\Gamma^{-1}[A'\cap E]
\subseteq\theta_{n\setminus m}(A'\cap E)$}

\noindent because $\Gamma(x)(i)=x(i)$ whenever $x\in X$ and
$i\in n\setminus m$.   So
$\theta_{n\setminus m}(A\cap\Gamma^{-1}[E])
\subseteq\theta_{n\setminus m}(A'\cap E)$ and

\Centerline{$\nu_{l+1,p}
  (X\setminus\theta_{n\setminus m}(A\cap\Gamma^{-1}[E]))
\ge\nu_{l+1,p}(X\setminus\theta_{n\setminus m}(A'\cap E))
\ge 1$}

\noindent because $E$ is $\nu_{l+1,p}$-thin between $m$ and $n$.\ \Qed

\medskip

\quad{\bf (ii)} As noted in (b), $\|I\|\ge 2^{2k+12}\ge 4s$.
For each $i<s$ such that $\min I\le u_i$,
let $u^-_i$ be the largest element of
$I$ which is less than or equal to $u_i$.
Set $I'=I\setminus(W\cup\{u^-_i:i<s$, $\min I\le u_i\})$.   Then

\Centerline{$\|I'\|\ge\Bover{\|I\|}2-s\ge\Bover{\|I\|}4$.}

\noindent Now $\Gamma^{-1}[E]$ is $\nu_{l+1,p}$-thin along $I'$.   \Prf\
Suppose
that $m$, $n\in I'$ and $m<n$.   Let $m^+$ be the least element of $I$ such
that $m<m^+$.   Then $m^+\le n$.   \Quer\ If
$W\cap m^+\setminus m\ne\emptyset$,
there is an $i<s$ such that $m^+\setminus m$ meets $v_i\setminus u_i$, that
is, $m<v_i$ and $u_i<m^+$.
Since $m\in I'\subseteq\Bbb N\setminus W$, $m\le u_i$ and $u^-_i$ is
defined;   now $m\ne u^-_i$ so $m<u^-_i\in I$ and $m^+\le u^-_i\le u_i$.\
\BanG\ Thus
$W\cap m^+\setminus m$ is empty and (i) tells us that $\Gamma^{-1}[E]$ is
$\nu_{l+1,p}$-thin between $m$ and $m^+$, therefore
$\nu_{l+1,p}$-thin between $m$ and $n$ (394Cc).   As $m$
and $n$ are arbitrary, $\Gamma^{-1}[E]$ is $\nu_{l+1,p}$-thin
along $I'$.\ \Qed

\medskip

\quad{\bf (iii)} If we now set $w'=4^{\alpha_l}w$, we see that

\Centerline{$1\le\|I'\|\le\|I\|\le N_l$,
\quad$w'\ge 2^{-l}4^{\alpha_l}\bigl(\Bover{N_l}{\|I\|}\bigr)^{\alpha_l}
\ge 2^{-l}\bigl(\Bover{N_l}{\|I'\|}\bigr)^{\alpha_l}$,}

\noindent so

\Centerline{$(\Gamma^{-1}[E],I',w')\in\Cal E_{lp}\subseteq\Cal C_{k+1,p}$}

\noindent and

\Centerline{$\nu_{k+1,p}(\Gamma^{-1}[E])\le w'
=4^{\alpha_l}w\le 4^{\alpha_k}w$.}

\medskip

{\bf (h)} We are nearly done.   Applying (g) to each member of $\Cal J_2$,

\Centerline{$\nu_{k+1,p}(\Gamma^{-1}[\spread\Cal J_2])
\le 4^{\alpha_k}\weight\Cal J_2
\le 4^{\alpha_k}\weight\Cal I
<4^{\alpha_k}c_k=c_{k+1}\le\nu_{k+1,p}Y_z$}

\noindent by the inductive hypothesis in its full strength.
So there is a
$y\in Y_z\setminus\Gamma^{-1}[\spread\Cal J_2]$.
With (f), this means that $\Gamma(y)$ does not belong to

\Centerline{$\spread(\Cal K\cup\Cal J_1)\cup\spread(\Cal J_2)
=\spread\Cal I$.}

\noindent On the other hand, $\Gamma(y)\in Y_z$ because $L\cap W=\emptyset$.
As $\Cal I$ was arbitrary, $\nu_{kp}Y_z$ must be at least
$c_k$, which is what we need to know to proceed with the induction.
}  %end of proof of 394G

\leader{394H}{Definitions}  %\Tal{4.1}
\cmmnt{ I present the last two
definitions required.}
Fix on a non-principal ultrafilter $\Cal F$ on $\Bbb N$.
For $k\in\Bbb N$, set

\Centerline{$\nu_kE=\lim_{p\to\Cal F}\nu_{kp}E\in[0,\infty]$}

\noindent for every $E\in\frak B$;  \cmmnt{finally, }write $\nu$ for
$\nu_0$.

\leader{394I}{Proposition} (a) %\Tal{\S6.2}
For every $k\in\Bbb N$,
$\nu_k$ is a totally finite submeasure and $\nu_kX\ge 8$.

(b) %\Tal{\S4.1}
$\nu$ is not uniformly exhaustive.

\proof{{\bf (a)} It follows directly from the definition in
392A that $\nu_k$, being a limit of
submeasures, is a submeasure.   By 394Ch,
$\nu_kX\le 2^{-k}N_k^{\alpha_k}$ is finite.   By 394G and 394Cg,

\Centerline{$\nu_kX=\lim_{p\to\Cal F}\nu_{kp}X\ge c_k\ge 8$.}

\medskip

{\bf (b)} For any $n\in\Bbb N$ and $t\in T_n$,

\Centerline{$\nu Y_{nt}=\lim_{p\to\Cal F}\nu_{0p}Y_{nt}\ge 8$}

\noindent by 394G.    As $\sup_{n\in\Bbb N}\#(T_n)$ is infinite,
and $\family{t}{T_n}{Y_{nt}}$ is disjoint for every $n$, $\nu$ is not
uniformly exhaustive.
}  %end of proof of 394I

\leader{394J}{Lemma} Suppose that $k\in\Bbb N$, $E\in\frak B$,
$I\in[\Bbb N]^{<\omega}$ and
$E$ is $\bover12\nu_k$-thin along $I$.   Then

\Centerline{$\{p:p\ge k$, $E$ is $\nu_{kp}$-thin along $I\}\in\Cal F$.}

\noindent If $k\ge 1$ and $\|I\|=N_{k-1}$, then
$\nu_{k-1}E\le 2^{-k+1}$.

\proof{ If $m$, $n\in I$, $m<n$ and $A\in\Cal A_m$, then
$\nu_k(X\setminus\theta_{n\setminus m}(A\cap E))\ge 2$.   So

\Centerline{$U_{An}
=\{p:p\ge k$, $\nu_{kp}(X\setminus\theta_{n\setminus m}(A\cap E))\ge 1\}$}

\noindent belongs to
$\Cal F$.   Setting $U=\bigcap_{m<n\text{ in }I,A\in\Cal A_m}U_{An}$,
$U\in\Cal F$ and $E$ is $\nu_{kp}$-thin along $I$ for every $p\in U$.

If $k\ge 1$ and $\|I\|=N_{k-1}$, then
$(E,I,2^{-k+1})\in\Cal E_{k-1,p}$ for every
$p\in U$, so $\nu_{k-1,p}E\le 2^{-k+1}$ for every $p\in U$ and
$\nu_{k-1}E\le 2^{-k+1}$.
}%end of proof of 394J

\vleader{72pt}{394K}{Lemma} Let $m$, $k\in\Bbb N$ and
let $\sequence{i}{E_i}$ be a sequence in $\frak B$ such that

\inset{every $E_i$ is determined by coordinates in $\Bbb N\setminus m$,

$\nu_k(\bigcup_{i\le n}E_i)<2$ for every $n\in\Bbb N$.}

\noindent Then for every $\eta>0$ there is a $C\in\frak B$, determined by
coordinates in $\Bbb N\setminus m$, such that $\nu_kC\le 4$ and
$\nu_k(E_i\setminus C)\le\eta$ for each $i$.
 %T(9.1)6.5

\proof{{\bf (a)} For each $n>m$, set

\Centerline{$\tilde E_n=\bigcup\{E_i:i\le n$, $E_i\in\frak B_n\}$,}

\noindent so that $\tilde E_n$ is determined by coordinates in
$n\setminus m$ and $\nu_k\tilde E_n<2$.   Set

\Centerline{$U_n=\{p:p\ge k$, $\nu_{kp}\tilde E_n<2\}
\in\Cal F$.}

\noindent For $p\in U_n$ we can find a finite
$\Cal I_{np}\subseteq\Cal C_{kp}$ such that
$\tilde E_n\subseteq\spread\Cal I_{np}$ and
$\weight\Cal I_{np}\le 2$.   For $r>m$ set

$$\eqalign{\Cal I_{npr}
&=\{(E,I,w):(E,I,w)\in\Cal I_{np},\cr
&\mskip140mu\|I\cap(r-1)\setminus m\|<\Bover12\|I\|
  \le\|I\cap r\setminus m\|\},\cr}$$

\noindent and set

\Centerline{$\Cal I'_{np}=\{(E,I,w):(E,I,w)\in\Cal I_{np}$,
$\|I\cap m\|\ge\Bover14\|I\|\}$.}

\noindent Set $B_{np}=\theta_m(\spread\Cal I'_{np})$;  then

\Centerline{$\nu_{kp}B_{np}\le 2\weight\Cal I'_{np}\le 4$,}

\noindent by 394Fa.   Since $\nu_{kp}X\ge c_k\ge 8$
(394G, 394Cg again), $B_{np}\ne X$ and there is an
$A_{np}\in\Cal A_m$ disjoint from $\spread\Cal I'_{np}$.   Next, for
$m<r\le n$ and $p\in U_n$ set

\Centerline{$\Cal J_{npr}
=\{(\theta_{r\setminus m}(A_{np}\cap E),I\cap r\setminus m,2w):
(E,I,w)\in\Cal I_{npr}\}$,
\quad$F_{npr}=\spread\Cal J_{npr}$.}

\noindent By 394E, $\Cal J_{npr}\subseteq\Cal C_{kp}$, so
$\nu_{kp}F_{npr}\le\weight\Cal J_{npr}\le 2\weight\Cal I_{npr}$.
Note that $F_{npr}$ is determined by coordinates in
$r\setminus m$ and includes $A_{np}\cap\spread\Cal I_{npr}$.
Now if $m<j\le n$ and $p\in U_n$,
$\tilde E_j\subseteq\bigcup_{m<r\le j}F_{npr}$.   \Prf\Quer\
Otherwise, since both sets are determined by coordinates in $j\setminus m$,
and since $A_{np}\in\Cal A_m$, there is an $A\in\Cal A_j$ with

\Centerline{$A\subseteq A_{np}\cap\tilde E_j
   \setminus\bigcup_{m<r\le j}F_{npr}
\subseteq A_{np}\cap\tilde E_j
   \setminus\bigcup_{m<r\le j}\spread\Cal I_{npr}$.}

\noindent Since $A$ is also disjoint from $\spread\Cal I'_{np}$ and
$A\subseteq\tilde E_j\subseteq\spread\Cal I_{np}$,
$A\subseteq\spread\Cal I$, where

$$\eqalign{\Cal I
&=\Cal I_{np}
  \setminus(\Cal I'_{np}\cup\bigcup_{m<r\le j}\Cal I_{npr})\cr
&\subseteq\{(E,I,w):(E,I,w)\in\Cal I_{np},\,
  \|I\setminus j\|\ge\Bover14\|I\|\}.\cr}$$

\noindent Since $\Cal I\subseteq\Cal C_{kp}$,

$$\eqalignno{8
&\le\nu_{kp}X
=\nu_{kp}(\theta_{\Bbb N\setminus j}(A))
=\nu_{kp}(\theta_{\Bbb N\setminus j}(A\cap\spread\Cal I))
\le 2\weight\Cal I\cr
\displaycause{394Fb}
&\le 4.  \text{ \Bang\Qed}\cr}$$

 %T26/5

\medskip

{\bf (b)} For $r>m$ we can find $F_r\in\frak B$ such that

\inset{$\sum_{r=m+1}^{\infty}\nu_kF_r\le 4$,

$\tilde E_j\subseteq\bigcup_{m<r\le j}F_r$ for every $j>m$,

$F_r$ is determined by coordinates in $r\setminus m$.}

\noindent\Prf\ If $n\ge r>m$, then, because $\frak B_r$ is finite,
there is a set $F_{nr}\in\frak B_r$ such that
$\{p:p\in U_n,\,F_{npr}=F_{nr}\}$ belongs to $\Cal F$.
Next, if $r>m$ there is an $F_r\in\frak B_r$ such that
$\{n:n\ge r$, $F_{nr}=F_r\}$ belongs to $\Cal F$.   Now

$$\eqalignno{\nu_kF_r
&=\lim_{n\to\Cal F}\nu_kF_{nr}\cr
\displaycause{because
$\{n:\nu_kF_{nr}=\nu_kF_r\}\supseteq\{n:F_{nr}=F_r\}\in\Cal F$}
&=\lim_{n\to\Cal F}\lim_{p\to\Cal F}\nu_{kp}F_{nr}
=\lim_{n\to\Cal F}\lim_{p\to\Cal F}\nu_{kp}F_{npr}
\le 2\lim_{n\to\Cal F}\lim_{p\to\Cal F}\weight\Cal I_{npr}.\cr}$$

\noindent So, for $s>m$,

$$\eqalign{\sum_{r=m+1}^s\nu_kF_r
&\le 2\sum_{r=m+1}^s\lim_{n\to\Cal F}\lim_{p\to\Cal F}
   \weight\Cal I_{npr}\cr
&=2\lim_{n\to\Cal F}\lim_{p\to\Cal F}
   \sum_{r=m+1}^s\weight\Cal I_{npr}
\le 2\lim_{n\to\Cal F}\lim_{p\to\Cal F}
   \weight\Cal I_{np}
\le 4.\cr}$$

\noindent As $s$ is arbitrary, $\sum_{r=m+1}^{\infty}\nu_kF_r\le 4$.

If $n\ge j>m$, then we saw in (a) that
$\tilde E_j\subseteq\bigcup_{m<r\le j}F_{npr}$ for every
$p\in U_n$.   Since there are many $p$ such that $F_{nr}=F_{npr}$ whenever
$m<r\le j$, $\tilde E_j\subseteq\bigcup_{m<r\le j}F_{nr}$.   Now, given
$j>m$, there are many $n$ such that $F_{nr}=F_r$ whenever $m<r\le j$, so
$\tilde E_j\subseteq\bigcup_{m<r\le j}F_r$.

Finally, take any $r>m$.   Since $F_{npr}$ is determined by coordinates in
$r\setminus m$ whenever $n\ge r$ and $p\in U_n$, $F_{nr}$ is determined by
coordinates in $r\setminus m$ whenever $n\ge r$, and $F_r$ also is
determined by coordinates in $r\setminus m$.\ \Qed

\medskip

{\bf (c)} Let $r_0\ge m$ be such that
$\sum_{r=r_0+1}^{\infty}\nu_kF_r\le\eta$.   Set
$C=\bigcup_{m<r\le r_0}F_r$.   Then
$C$ is determined by coordinates in $\Bbb N\setminus m$ and

\Centerline{$\nu_kC\le\sum_{r=m+1}^{r_0}\nu_kF_r\le 4$.}

\noindent For any $i\in\Bbb N$, there is some $j>r_0$ such that
$E_i\subseteq\tilde E_j$, in which case

\Centerline{$E_i\setminus C\subseteq\bigcup_{r_0<r\le j}F_r$}

\noindent and

\Centerline{$\nu_k(E_i\setminus C)
\le\sum_{r=r_0+1}^j\nu_kF_r\le\eta$,}

\noindent as required.
}  %end of proof of 394K

\leader{394L}{Lemma}  %\Tal{\S6.6}
Suppose that $k\in\Bbb N$, $\epsilon>0$, $m\in\Bbb N$,
$B\in\frak B_m$ and that
$\sequence{i}{E_i}$ is a disjoint sequence in $\frak B$.   Then there
are $n>m$ and $B'\in\frak B_n$ such that $B'\subseteq B$, $B'$ is
$\bover12\nu_k$-thin between $m$ and $n$ and
$\limsup_{i\to\infty}\nu_k(E_i\cap B\setminus B')\le\epsilon$.

\proof{ Set $\eta=\Bover{\epsilon}{\#(\Cal A_m)}$.   (This is where we need to
know that all the $T_n$ are finite.)   For
those $A\in\Cal A_m$ included in $B$ define $C'_A\subseteq A$ as follows.

\medskip

\quad{\bf case 1} If there is some $r$ such that
$\nu_k(\theta_{\Bbb N\setminus m}(A\cap\bigcup_{i\le r}E_i))\ge 2$, set
$C'_A=A\setminus\bigcup_{i\le r}E_i$, so that
$\bover12\nu_k(\theta_{\Bbb N\setminus m}(A\setminus C'_A))\ge 1$ and
$E_i\cap A\setminus C'_A=\emptyset$ for $i>r$.

\medskip

\quad{\bf case 2} If
$\nu_k(\theta_{\Bbb N\setminus m}(A\cap\bigcup_{i\le r}E_i))<2$
for every $r$, then by 394K, applied to the sequence
$\sequence{i}{\theta_{\Bbb N\setminus m}(A\cap E_i)}$,
 we can find a $C\in\frak B$,
determined by coordinates in $\Bbb N\setminus m$,
such that $\nu_kC\le 4$ and
$\nu_k(\theta_{\Bbb N\setminus m}(A\cap E_i)\setminus C)\le\eta$
for every $i$.
Set $C'_A=C\cap A$.   Because $C$ is determined by coordinates in
$\Bbb N\setminus m$ and $A\in\Cal A_m$,
$\nu_k(\theta_{\Bbb N\setminus m}(C'_A))=\nu_kC\le 4$.
Also $E_i\cap A\setminus C'_A
\subseteq\theta_{\Bbb N\setminus m}(A\cap E_i)\setminus C$
so $\nu_k(E_i\cap A\setminus C'_A)\le\eta$ for every $i$.

\medskip

Set

\Centerline{$B'=\bigcup\{C'_A:A\in\Cal A_m$, $A\subseteq B\}$.}

\noindent Then $B'\in\frak B$, $B'\subseteq B$ and

$$\eqalign{\limsup_{i\to\infty}\nu_k(E_i\cap B\setminus B')
&\le\sum_{A\in\Cal A_m,A\subseteq B}\limsup_{i\to\infty}
  \nu_k(E_i\cap A\setminus C'_A)\cr
&\le\sum_{A\in\Cal A_m,A\subseteq B}\eta
\le\epsilon.\cr}$$

Let $n>m$ be such that $C'_A\in\frak B_n$ whenever $A\in\Cal A_m$ and
$A\subseteq B$.   Then $B'$ is $\bover12\nu_k$-thin between $m$ and $n$.
\Prf\ Take any $A\in\Cal A_m$ and set
$\tilde C=\theta_{n\setminus m}(A\cap B')$.
If $A\not\subseteq B$ then $A\cap B'$ and $\tilde C$
are empty and $\nu_k(X\setminus\tilde C)\ge 8$ (394Ia).
Otherwise, $A\cap B'=C'_A\in\frak B_n$ so
$\tilde C=\theta_{\Bbb N\setminus m}(C'_A)$ is disjoint from
$\theta_{\Bbb N\setminus m}(A\setminus C'_A)$ (see the last remark in
394Cb).   If $C'_A$ was chosen as in case 1 above,

\Centerline{$\nu_k(X\setminus\tilde C)
\ge\nu_k(\theta_{\Bbb N\setminus m}(A\setminus C'_A))\ge 2$.}

\noindent
If $C'_A$ was chosen as in case 2,

\Centerline{$\nu_k(X\setminus\tilde C)
=\nu_k(X\setminus\theta_{\Bbb N\setminus m}(C'_A))
\ge\nu_kX-\nu_k(\theta_{\Bbb N\setminus m}(C'_A))
\ge 8-4$.}

\noindent So in all three cases we
have $\bover12\nu_k(X\setminus\tilde C)\ge 1$, as required.\ \Qed

Thus we have an appropriate $B'$.
}  %end of proof of 394L

\leader{394M}{Theorem} $\nu$ is exhaustive.

\proof{ Let $\sequence{i}{E_i}$ be a disjoint sequence in $\frak B$.
Take any $k\in\Bbb N$ and $\epsilon>0$, and choose $\sequence{j}{B_j}$ and
$\sequence{j}{n_j}$ inductively, as
follows.    $B_0=X$ and $n_0=0$.   Given that $B_j\in\frak B_{n_j}$,
take $n_{j+1}>n_j$ and $B_{j+1}\in\frak B_{n_{j+1}}$ such that
$B_{j+1}\subseteq B_j$, $B_{j+1}$ is
$\bover12\nu_{k+1}$-thin between $n_j$ and $n_{j+1}$, and
$\limsup_{i\to\infty}
  \nu_{k+1}(E_i\cap B_j\setminus B_{j+1})\le 2^{-j}\epsilon$
(394L).   Continue.   Note that
$\limsup_{i\to\infty}\nu_{k+1}(E_i\setminus B_j)\le 2\epsilon$
for every $j$.

Let $l$ be so large that $I=\{n_j:j<l\}$ has $\|I\|=N_k$.   (This is where we
need to know that $\lim_{l\to\infty}\|A\cap l\|=\infty$ for every infinite
$A\subseteq\Bbb N$.)   Set $B=B_{l-1}$.    Then $B$ is
$\bover12\nu_{k+1}$-thin along $I$ (use 394Cc).   By 394J,
$\nu_kB\le 2^{-k}$.

Of course $\nu\le\nu_k\le\nu_{k+1}$ (394Cf).   So

\Centerline{$\limsup_{i\to\infty}\nu E_i
\le\nu_kB+\limsup_{i\to\infty}\nu_{k+1}(E_i\setminus B)
\le 2^{-k}+2\epsilon$.}

\noindent As $k$, $\epsilon$ and $\sequence{i}{E_i}$ are arbitrary,
$\nu$ is exhaustive.
}%end of proof of 394M

\leader{394N}{Remarks (a)}
Note that the whole construction is invariant under
the action of the group $\prod_{n\in\Bbb N}G_n$ where $G_n$ is the group of
all permutations of $T_n$ for each $n$.   In particular, if we
give each $T_n$ a group structure and
$X$ the product group structure, then $\nu$ is translation-invariant.

\spheader 394Nb\cmmnt{ It follows that} $\nu$ is strictly positive.
\prooflet{\Prf\ For each $n\in\Bbb N$, $\nu$ is constant on $\Cal A_n$, so
$\nu E\ge\nu X/\#(\Cal A_n)>0$ for every non-empty $E\in\frak B_n$.\ \QeD}

\spheader 394Nc We can\cmmnt{ therefore} form the metric completion
$\widehat{\frak B}$ of $\frak B$\cmmnt{, as in 392H},
and $\widehat{\frak B}$ will
be a Maharam algebra, with a strictly positive Maharam submeasure $\hat\nu$
continuously extending $\nu$\cmmnt{ (393H)}.
\cmmnt{Now} $\widehat{\frak B}$ is not
measurable.   \prooflet{\Prf\Quer\ Otherwise, let $\bar\mu$ be such that
$(\widehat{\frak B},\bar\mu)$ is a probability algebra.   Then $\bar\mu$
and $\hat\nu$ are strictly positive Maharam submeasures on
$\widehat{\frak B}$, so $\hat\nu$ is absolutely continuous with respect to
$\bar\mu$ (393F).   Let $n\ge 1$ be such that $\hat\nu b<8$ whenever
$\bar\mu b\le 1/\#(T_n)$.   Then there must be a $t\in T_n$ such that
$\bar\mu Y_{nt}\le 1/\#(T_n)$;
but $\hat\nu Y_{nt}=\nu Y_{nt}\ge 8$
(see the proof of 394Ib).\ \Bang\Qed}

\cmmnt{In fact, $\widehat{\frak B}$ is nowhere measurable (394Ya).}

\leader{*394O}{Control measures}\dvAformerly{3{}93O}\cmmnt{ One
of the original
reasons for studying Maharam submeasures was their connexion with the
following notion.}
Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra and $U$ a Hausdorff linear topological
space.   \cmmnt{ (The idea is intended to apply, in particular, when
$\frak A$ is a $\sigma$-algebra of subsets of a set.)}   A function
$\theta:\frak A\to U$ is a {\bf vector measure} if
$\sum_{n=0}^{\infty}\theta a_n=\lim_{n\to\infty}\sum_{i=0}^n\theta a_i$
is defined in $U$ and equal to $\theta(\sup_{n\in\Bbb N}a_n)$ for every
disjoint sequence $\sequencen{a_n}$ in $\frak A$.
In this case, a non-negative countably additive
functional $\mu:\frak A\to\coint{0,\infty}$ is a {\bf control measure}
for $\theta$ if $\theta a=0$ whenever $\mu a=0$.

%what about an absolute-continuity definition?

\leader{*394P}{Example} There are a metrizable linear topological space
$U$ and a vector measure $\theta:\Sigma\to U$, where $\Sigma$ is a
$\sigma$-algebra of sets, such that $\theta$ has no control measure.

\proof{ As in 394Nc, let $\widehat{\frak B}$ be the metric completion of
$\frak B$, and $\hat\nu$ the
continuous extension of $\nu$ to $\widehat{\frak B}$.   Give
$L^0=L^0(\widehat{\frak B})$ the topology defined from $\hat\nu$ as in 393K, so that
$L^0$ is a metrizable linear topological space.   By 314M, we can
identify $\widehat{\frak B}$ with a quotient algebra $\Sigma/\Cal N$ where $\Sigma$
is a $\sigma$-algebra of subsets of a set $\Omega$
and $\Cal N$ is a $\sigma$-ideal in $\Sigma$.
Set $\theta E=\chi E^{\ssbullet}\in L^0$ for $E\in\Sigma$.   Then $\theta$
is a vector measure.
\Prf\ If $\sequencen{E_n}$ is a disjoint sequence in $\Sigma$
with union $E$, set $F_n=\bigcup_{i\le n}E_i$, so that
$\chi F_n^{\ssbullet}=\sum_{i=0}^n\chi E_i^{\ssbullet}$ for each $n$.
We have $\hat\nu(E^{\ssbullet}\Bsetminus F_n^{\ssbullet})\to 0$, so that

\Centerline{$\tau(\theta E-\theta F_n)
=\tau(\chi E^{\ssbullet}-\chi F_n^{\ssbullet}))
=\min(1,\hat\nu(E^{\ssbullet}\Bsetminus F_n^{\ssbullet}))\to 0$,}

\noindent where $\tau$ is the functional of the proof of 393K, and
$\theta E=\sum_{i=0}^{\infty}\theta E_i$ in $L^0$.\ \Qed

If $\mu$ is a totally finite measure with domain $\Sigma$, set

\Centerline{$\lambda a=\inf\{\mu E:E\in\Sigma$, $E^{\ssbullet}=a\}$}

\noindent for every $a\in\widehat{\frak B}$.   Note that the infimum is always
attained.   \Prf\ If $\sequencen{E_n}$ is a sequence in $\Sigma$ such that
$E_n^{\ssbullet}=a$ for every $n\in\Bbb N$ and
$\lambda a=\lim_{n\to\infty}\mu E_n$, set $E=\bigcap_{n\in\Bbb N}E_n$;
then $E^{\ssbullet}=a$ and $\mu E=\lambda a$.\ \QeD\  Next, $\lambda$ is
countably additive.   \Prf\ If $\sequencen{a_n}$ is a disjoint sequence in
$\widehat{\frak B}$ with supremum $a$,
take $E_n\in\Sigma$ such that $E_n^{\ssbullet}=a_n$ and
$\mu E_n=\lambda a_n$ for each $n$, and $E\in\Sigma$ such that
$E^{\ssbullet}=a$ and $\mu E=\lambda a$.   Set
$F_n=E\cap E_n\setminus\bigcup_{i<n}E_i$ for each $n$, and
$F=\bigcup_{n\in\Bbb N}F_n$.   Then $F_n^{\ssbullet}=a_n$ and
$F_n\subseteq E_n$, so $\mu F_n=\lambda a_n$ for each $n$;  similarly,
$F^{\ssbullet}=a$ and $F\subseteq E$, so $\mu F=\lambda a$.   Also
$\sequencen{F_n}$ is disjoint and has union $F$.   Accordingly

\Centerline{$\lambda a=\mu F=\sum_{n=0}^{\infty}\mu F_n
=\sum_{n=0}^{\infty}\lambda a_n$.  \Qed}

Since $\widehat{\frak B}$ is not a measurable algebra, $\lambda$ cannot be strictly
positive, and there is a non-zero $a\in\widehat{\frak B}$ such that $\lambda a=0$.
Let $E\in\Sigma$ be such that $E^{\ssbullet}=a$ and $\mu E=0$;  then
$\theta E=\chi a\ne 0$.   So $\mu$ is not a control measure for $\theta$.
}  %end of proof of 394P

\leader{*394Q}{}\cmmnt{ This is not a book about vector measures, but
having gone so far I ought to note
that the generality of the phrase `metrizable linear topological space' in
394P is essential.   If we look only at normed spaces the situation is very
different.

\medskip

\noindent}{\bf Theorem} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra, $U$ a normed space and $\theta:\frak A\to U$ a vector
measure.   Then $\theta$ has a control measure.

\proof{{\bf (a)} Since $U$ can certainly be embedded in a Banach space
$\hat U$ (3A5Jb), and as $\theta$ will still be a vector measure when
regarded as a map from $\frak A$ to $\hat U$, we may assume from the
beginning that $U$ itself is complete.

\medskip

{\bf (b)} $\theta$ is bounded (that is, $\sup_{a\in\frak A}\|\theta a\|$
is finite).   \Prf\Quer\ (Cf.\ 326M.)
Suppose, if possible, otherwise.   Choose
$\sequencen{a_n}$ inductively, as follows.   $a_0=1$.   Given that
$\sup_{a\Bsubseteq a_n}\|\theta a\|=\infty$, choose $b\Bsubseteq a_n$
such that $\|\theta b\|\ge\|\theta a_n\|+1$.   Then
$\|\theta(a_n\Bsetminus b)\|\ge 1$.   Also

\Centerline{$\sup_{a\Bsubseteq a_n}\|\theta a\|
\le\sup_{a\Bsubseteq a_n}\|\theta(a\Bcap b)\|+\|\theta(a\Bsetminus
b)\|$,}

\noindent so at least one of $\sup_{a\Bsubseteq b}\|\theta a\|$,
$\sup_{a\Bsubseteq a_n\Bsetminus b}\|\theta a\|$ must be infinite.   We
may therefore take $a_{n+1}$ to be either $b$ or $a_n\Bsetminus b$ and
such that $\sup_{a\Bsubseteq a_{n+1}}\|\theta a\|=\infty$.   Observe
that in either case we shall have $\|\theta(a_n\Bsetminus a_{n+1})\|\ge
1$.   Continue.

At the end of the induction we shall have a disjoint sequence
$\sequencen{a_n\Bsetminus a_{n+1}}$ such that
$\|\theta(a_n\Bsetminus a_{n+1})\|\ge 1$ for every $n$, so that
$\sum_{n=0}^{\infty}\theta(a_n\Bsetminus a_{n+1})$ cannot be defined in
$U$;  which is impossible.\ \Bang\Qed

\medskip

{\bf (c)} Accordingly we have a bounded linear operator
$T:L^{\infty}\to U$, where $L^{\infty}=L^{\infty}(\frak A)$, such that
$T\chi=\theta$ (363Ea).

Now the key to the proof is the following fact:  if $\sequencen{u_n}$ is
a disjoint order-bounded sequence in $(L^{\infty})^+$,
$\sequencen{Tu_n}\to 0$ in $U$.   \Prf\ Let $\gamma$ be such that
$u_n\le\gamma\chi 1$ for every $n$.   Let $\epsilon>0$, and let $k$ be
the integer part of $\gamma/\epsilon$.   For $n\in\Bbb N$, $i\le k$ set
$a_{ni}=\Bvalue{u_n>\epsilon(i+1)}$;   then $\sequencen{a_{ni}}$ is
disjoint for each $i$, and if we set $v_n=\epsilon\sum_{i=0}^k\chi
a_{ni}$, we get $v_n\le u_n\le v_n+\epsilon\chi 1$, so
$\|u_n-v_n\|_{\infty}\le\epsilon$.

Because $\sequencen{a_{ni}}$ is disjoint,
$\sum_{n=0}^{\infty}\theta a_{ni}$ is defined in $U$,
and $\sequencen{\theta a_{ni}}\to 0$, for each $i\le k$.   Consequently

\Centerline{$Tv_n=\epsilon\sum_{i=0}^k\theta a_{ni}\to 0$}

\noindent as $n\to\infty$.  But

\Centerline{$\|Tu_n-Tv_n\|\le\|T\|\|u_n-v_n\|_{\infty}\le\epsilon\|T\|$}

\noindent for each $n$, so
$\limsup_{n\to\infty}\|Tu_n\|\le\epsilon\|T\|$.   As $\epsilon$ is
arbitrary, $\lim_{n\to\infty}\|Tu_n\|=0$.\ \Qed

\medskip

{\bf (d)} Consider the adjoint operator $T':U^*\to (L^{\infty})^*$.
Recall that $L^{\infty}$ is an $M$-space (363Ba) so that its dual is an
$L$-space (356N).   Write

\Centerline{$A=\{T'g:g\in U^*,\,\|g\|\le 1\}\subseteq(L^{\infty})^*
=(L^{\infty})^{\sim}$.}

\noindent If $u\in L^{\infty}$, then

\Centerline{$\sup_{f\in A}|f(u)|=\sup_{\|g\|\le 1}|(T^*g)(u)|
=\sup_{\|g\|\le 1}|g(Tu)|=\|Tu\|$.}

\noindent Now $A$ is uniformly integrable.   \Prf\ I use the criterion
of 356O.   Of course $\|f\|\le\|T'\|$ for every $f\in A$, so $A$ is
norm-bounded.   If $\sequencen{u_n}$ is an order-bounded disjoint
sequence in $(L^{\infty})^+$, then

\Centerline{$\sup_{f\in A}|f(u_n)|=\|Tu_n\|\to 0$}

\noindent as $n\to\infty$.   So $A$ is uniformly integrable.\ \Qed

\medskip

{\bf (e)} Next, $A\subseteq(L^{\infty})^{\sim}_c$.   \Prf\ If $f\in A$,
it is of the form $T'g$ for some $g\in U^*$, that is,

\Centerline{$f(\chi a)=(T'g)(\chi a)=gT(\chi a)=g(\theta a)$}

\noindent for every $a\in\frak A$.   If now $\sequencen{a_n}$ is a
disjoint sequence in $\frak A$ with supremum $a$,

\Centerline{$f(\chi a)
=g(\theta(\sup_{n\in\Bbb N}a_n))
=g(\sum_{n=0}^{\infty}\theta a_n)
=\sum_{n=0}^{\infty}g(\theta a_n)
=\sum_{n=0}^{\infty}f(\chi a_n)$.}

\noindent So $f\chi$ is countably additive.   By 363K,
$f\in(L^{\infty})^{\sim}_c$.\ \Qed

\medskip

{\bf (f)} Because $A$ is uniformly integrable, there is for each
$m\in\Bbb N$ an $f_m\ge 0$ in $(L^{\infty})^*$ such that
$\|(|f|-f_m)^+\|\le 2^{-m}$ for every $f\in A$;  moreover, we can
suppose that $f_m$ is of the form $\sup_{i\le k_m}|f_{mi}|$ where every
$f_{mi}$ belongs to $A$ (354R(b-iii)), so that
$f_m\in(L^{\infty})^{\sim}_c$ and $\mu_m=f_m\chi$ is countably additive.
Set

\Centerline{$\mu=\sum_{m=0}^{\infty}\Bover1{2^m(1+\mu_m1)}\mu_m$;}

\noindent  then $\mu:\frak A\to\coint{0,\infty}$ is a non-negative
countably additive functional.

Now $\mu$ is a control measure for $\theta$.   \Prf\ If $\mu a=0$, then
$\mu_ma=0$, that is, $f_m(\chi a)=0$, for every $m\in\Bbb N$.   But this
means that if $g\in U^*$ and $\|g\|\le 1$,

\Centerline{$|g(\theta a)|=|(T'g)(\chi a)|
\le f_m(\chi a)+\|(|T'g|-f_m)^+\|\le 2^{-m}$}

\noindent for every $m$, by the choice of $f_m$;  so that $g(\theta
a)=0$.   As $g$ is arbitrary, $\theta a=0$;  as $a$ is arbitrary, $\mu$
is a control measure for $\theta$.\ \Qed
}  %end of proof of 394Q

%\leader{394S}{}
%\leader{394T}{}
%\leader{394U}{}
%\leader{394V}{}
%\leader{394W}{}

\exercises{
\leader{394X}{Basic exercises (a)}
%\spheader 394Xa
Show that the metric completion
$\widehat{\frak B}$ of $\frak B$, as defined in
394N, always has many involutions (definition:  382O).
%394N

\leader{394Y}{Further exercises (a)}%
%\spheader 394Ya
(i)
Show that if $r\in\Bbb N$, $k\le p$ and $E\in\frak B_{r+1}$ are such that
$\nu_{kp}E<c_k$, then $\nu_{kp}(\theta_r(E))\le\Bover{32}{c_k}\,\nu_{kp}E$.
%mt39bits
(ii) Show that if $E\in\Cal B_r$ then
$\nu(E\cap Y_{rt})\ge\min(8,\bover14\nu E)$ for every $t\in T_r$.
(iii) Let $\widehat{\frak B}$ be the metric completion of $\frak B$ and
$\hat\nu$ the continuous extension of $\nu$ to $\widehat{\frak B}$.
Show that for every $a\in\frak B$ and
$n\in\Bbb N$ there is a disjoint family
$\langle c_i\rangle_{i\le n}$ such that $c_i\Bsubseteq a$ and
$\hat\nu c_i\ge\min(7,\bover15\hat\nu a)$ for every $i\le n$.
(iv) Show that the
only countably additive real-valued functional on $\widehat{\frak B}$ is
the zero functional.
(v) Show that $\widehat{\frak B}$ is nowhere measurable.
(vi) Show that if $\nu'$ is a uniformly exhaustive submeasure on
$\frak B$ which is absolutely continuous with respect to $\nu$, then
$\nu'=0$.

\query
%394G 394N
}  %end of exercises

\leader{394Z}{Problems}  Suppose that
$\|\,\|$, $\sequencen{T_n}$, $\frak B$,
$\sequence{k}{\alpha_k}$ and $\sequence{k}{N_k}$ satisfy the conditions of
394Ba-394Bb.  Let $\nu$ be the exhaustive submeasure on $\frak B$
constructed by the method of 394B and 394H, and $\widehat{\frak B}$ the
corresponding Maharam algebra.

\spheader 394Za
Does $\widehat{\frak B}$ have an order-closed subalgebra isomorphic to
the measure algebra of Lebesgue measure?   In particular, if we take
$\frak C\subseteq\frak B$ to be the algebra of sets generated by sets of
the form $\{x:x\in X$, $x(n)=0\}$ for $n\in\Bbb N$, is $\nu\restr\frak C$
uniformly exhaustive?

\spheader 394Zb Suppose that instead of taking large sets $T_n$, we simply
set $T_n=\{0,1\}$ for every $n$, but otherwise used the same construction.
Should we then find that $\nu$ was uniformly exhaustive?
\cmmnt{(This might be relevant to (a) above.)}

\spheader 394Zc Is the Boolean algebra $\widehat{\frak B}$ homogeneous?

\endnotes{
\Notesheader{394} `Maharam's problem', or the `control measure problem',
was for fifty years one of the most vexing questions in abstract measure
theory.   To begin with, there were reasonable hopes that there was a
positive answer -- in the language of this book, that every Maharam algebra
was a measurable algebra.   If this had been the case, there would have
been consequences all over the theories of topological Boolean algebras,
topological Riesz spaces and vector measures.   In the 1970s, it began to
seem too much to ask for.   In 1983 the Kalton-Roberts theorem gave new
life to the conjecture for a moment, but {\smc Roberts 93} demonstrated a
major obstacle, which Talagrand (building on some further ideas of I.Farah)
eventually developed into the construction above.   The ideas which for a
generation were collected together by their association with the control
measure problem no longer have this as a unifying principle, and (as after
any successful revolution) are now more naturally grouped in other ways.
There is a relic of this era in 394P.

Now that we know for sure that there are non-measurable Maharam algebras,
it becomes possible to ask questions about their structure.
Frustratingly, practically none of these questions has yet been answered
even for the examples constructed by Talagrand's original method, in which
$\|I\|=\#(I)$ for every $I$.   (Of course this allows variations in the
parameters $\sequencen{T_n}$,
$\sequence{k}{\alpha_k}$ and $\sequence{k}{N_k}$ and the filter
$\Cal F$, and there is every reason to suppose that $\frak c$
non-isomorphic examples can be constructed by the formulae set out above.)
I will return briefly to
such questions in Volumes 4 and 5, as I come to further properties of
measure algebras which can be interpreted in Maharam algebras.   In particular,
following {\smc Perovi\'c \& Veli\v{c}kovi\'c 18}, I will show in \S539
how different PV norms can give rise to distinguishable Maharam
algebras.
}  %end of notes

\discrpage

\frfilename{mt395.tex}
\versiondate{15.6.08}
\copyrightdate{1996}
\def\Gte{$G$-\vthsp$\tau$-\vthsp equidecomposable}
\def\IMPLY#1#2{{\bf (#1)$\Rightarrow$(#2)}}
\def\ptG{\preccurlyeq^{\tau}_G}

\def\chaptername{Measurable algebras}
\def\sectionname{Kawada's theorem}

\newsection{395\dvAformerly{\S3{}94}}

I now describe a completely different characterization
of (homogeneous) measurable algebras, based on the special nature of
their automorphism groups.   The argument depends on the notion of
`non-paradoxical' group of automorphisms;  this is an idea of great
importance in other contexts, and I therefore aim at a fairly thorough
development, with proofs which are adaptable to other circumstances.

\leader{395A}{Definitions} Let $\frak A$ be a Dedekind complete Boolean
algebra, and $G$ a subgroup of $\Aut\frak A$.
For $a$, $b\in\frak A$ I will say that an isomorphism
$\phi:\frak A_a\to\frak A_b$ between the corresponding principal ideals
belongs to
the {\bf full local semigroup generated by $G$} if there are a partition
of unity $\langle a_i\rangle_{i\in I}$ in $\frak A_a$ and a family
$\langle \pi_i\rangle_{i\in I}$ in $G$ such that $\phi c=\pi_ic$
whenever $i\in I$ and $c\Bsubseteq a_i$.   If such an isomorphism exists
I will say that $a$ and $b$ are {\bf $G$-$\tau$-equidecomposable}.

I will write $a\ptG b$ to mean that there is a $b'\Bsubseteq b$ such
that $a$ and $b'$ are \Gte.

For any function $f$ with domain $\frak A$, I will say that $f$ is {\bf
$G$-invariant} if $f(\pi a)=f(a)$ whenever $a\in\frak A$ and $\pi\in G$.

\leader{395B}{}\cmmnt{ The notion of `full local semigroup' is of
course an extension of the idea of `full subgroup' (381Be;  see also
381Yb).   The
word `semigroup' is justified by (c) of the following lemma, and the
word `full' by (e).

\medskip

\noindent}{\bf Lemma} Let $\frak A$ be a Dedekind complete Boolean
algebra and $G$ a subgroup of $\Aut\frak A$.   Write $G^*_{\tau}$ for
the full local semigroup generated by $G$.

(a) Suppose that $a$, $b\in\frak A$ and that
$\phi:\frak A_a\to\frak A_b$ is an isomorphism.   Then the following are
equiveridical:

\quad (i) $\phi\in G^*_{\tau}$;

\quad (ii) for every non-zero
$c_0\Bsubseteq a$ there are a non-zero $c_1\Bsubseteq c_0$ and a
$\pi\in G$ such that $\phi c=\pi c$ for every $c\Bsubseteq c_1$;

\quad (iii) for every non-zero
$c_0\Bsubseteq a$ there are a non-zero $c_1\Bsubseteq c_0$ and a
$\psi\in G^*_{\tau}$ such that $\phi c=\psi c$ for every
$c\Bsubseteq c_1$.

(b) If $a$, $b\in\frak A$ and $\phi:\frak A_a\to\frak A_b$ belongs to
$G^*_{\tau}$, then $\phi^{-1}:\frak A_b\to\frak A_a$ also belongs to
$G^*_{\tau}$.

(c) Suppose that $a$, $b$, $a'$, $b'\in\frak A$ and that $\phi:\frak
A_a\to\frak A_{a'}$, $\psi:\frak A_{b}\to\frak A_{b'}$ belong to
$G^*_{\tau}$.
Then $\psi\phi\in G^*_{\tau}$;
its domain is $\frak A_c$ where $c=\phi^{-1}(b\Bcap a')$, and its
set of values is $\frak A_{c'}$ where $c'=\psi(b\Bcap a')$.

(d) If $a$, $b\in\frak A$ and $\phi:\frak A_a\to\frak A_b$ belongs to
$G^*_{\tau}$, then $\phi\restrp\frak A_c\in G^*_{\tau}$ for any $c\Bsubseteq a$.

(e) Suppose that $a$, $b\in\frak A$ and that
$\psi:\frak A_a\to\frak A_b$ is an isomorphism such that there are a partition of unity $\langle a_i\rangle_{i\in I}$ in $\frak A_a$ and a family $\langle\phi_i\rangle_{i\in I}$ in $G^*_{\tau}$ such that
$\psi c=\phi_ic$ whenever $i\in I$ and $c\Bsubseteq a_i$.   Then
$\psi\in G^*_{\tau}$.

\proof{{\bf (a)} (Compare 381I.)

\medskip

\quad{\bf (i)$\Rightarrow$(iii)} is trivial, since of course
$G\subseteq G^*_{\tau}$.

\medskip

\quad{\bf (iii)$\Rightarrow$(ii)} Suppose that $\phi$ satisfies (iii),
and that $0\ne c_0\Bsubseteq a$.   Then we can find a $\psi\in
G^*_{\tau}$ and a non-zero $c_1\Bsubseteq c_0$ such that $\phi$ agrees
with $\psi$ on $\frak A_{c_1}$.   Suppose that $\dom\psi=\frak A_d$,
where necessarily $d\Bsupseteq c_1$.   Then there are a partition of
unity $\langle d_i\rangle_{i\in I}$ in $\frak A_d$ and a family
$\langle\pi_i\rangle_{i\in I}$ such that $\psi c=\pi_ic$ whenever
$c\Bsubseteq d_i$.   There is some $i\in I$ such that
$c_2=c_1\Bcap d_i\ne 0$, and we see that $\phi c=\psi c=\pi_ic$ for
every $c\Bsubseteq c_2$.   As $c_0$ is arbitrary, $\phi$ satisfies (ii).

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} If $\phi$ satisfies (ii), set

\Centerline{$D=\{d:d\Bsubseteq a$, there is some $\pi\in G$ such that
$\pi c=\phi c$ for every $c\Bsubseteq d\}$.}

\noindent The hypothesis is that $D$ is order-dense in $\frak A$, so
there is a partition of unity $\langle a_i\rangle_{i\in I}$ of
$\frak A_a$ lying within $D$ (313K);  for each $i\in I$ take
$\pi_i\in G$ such that $\phi c=\pi_ic$ for $c\Bsubseteq a_i$;  then
$\langle a_i\rangle_{i\in I}$ and $\langle\pi_i\rangle_{i\in I}$ witness
that $\phi\in G^*_{\tau}$.

\medskip

{\bf (b)} This is elementary;  if $\langle a_i\rangle_{i\in I}$,
$\langle\pi_i\rangle_{i\in I}$ witness that $\phi\in G^*_{\tau}$, then
$\langle\phi a_i\rangle_{i\in I}=\langle\pi_ia_i\rangle_{i\in I}$,
$\langle\pi_i^{-1}\rangle_{i\in I}$ witness that
$\phi^{-1}\in G^*_{\tau}$.

\medskip

{\bf (c)} I ought to start by computing the domain of $\psi\phi$:

$$\eqalign{d\in\dom(\psi\phi)
&\iff d\in\dom\phi,\,\phi d\in\dom\psi\cr
&\iff d\Bsubseteq a,\,\phi d\Bsubseteq b
\iff d\Bsubseteq\phi^{-1}(a'\Bcap b)=c.\cr}$$

\noindent So the domain of $\psi\phi$ is indeed $\frak A_c$;  now
$\phi\restrp\frak A_c$ is an isomorphism between $\frak A_c$ and $\frak
A_{\phi c}$, where $\phi c=a'\Bcap b\in\frak A_b$, so $\psi\phi$ is an
isomorphism between $\frak A_c$ and $\frak A_{\psi\phi c}=\frak A_{c'}$.
Let $\langle a_i\rangle_{i\in I}$, $\langle b_j\rangle_{j\in J}$ be
partitions of unity in $\frak A_a$, $\frak A_b$ respectively, and
$\langle\pi_i\rangle_{i\in I}$, $\langle\theta_j\rangle_{j\in J}$
families in $G$ such that $\phi d=\pi_id$ for $d\Bsubseteq a_i$,
$\psi e=\theta_je$ for $e\Bsubseteq b_j$.   Set
$c_{ij}=a_i\Bcap\pi_i^{-1}b_j$;  then
$\langle c_{ij}\rangle_{i\in I,j\in J}$ is a partition of
unity in $\frak A_c$ and $\psi\phi d=\theta_j\pi_id$ for
$d\Bsubseteq c_{ij}$, so $\psi\phi\in G^*_{\tau}$ (because all the
$\theta_j\pi_i$ belong to $G$).

\medskip

{\bf (d)} This is nearly trivial;  use the definition of $G^*_{\tau}$ or
the criteria of (a), or apply (c) with the identity map on $\frak A_c$
as one of the factors.

\medskip

{\bf (e)} This follows at once from the criterion (a-iii) above, or
otherwise.
}%end of proof of 395B

\leader{395C}{Lemma} Let $\frak A$ be a Dedekind complete Boolean
algebra and $G$ a subgroup of $\Aut\frak A$.   Write $G^*_{\tau}$ for
the full local semigroup generated by $G$.

(a) For $a$, $b\in\frak A$, $a\ptG b$ iff there is a
$\phi\in G^*_{\tau}$ such that $a\in\dom\phi$ and $\phi a\Bsubseteq b$.

(b)(i) $\ptG$ is transitive and reflexive;

\quad(ii) if $a\ptG b$ and $b\ptG a$ then $a$ and $b$ are \Gte.

(c) $G$-$\tau$-equidecomposability is an equivalence relation on
$\frak A$.

(d) If $\langle a_i\rangle_{i\in I}$ and $\langle b_i\rangle_{i\in I}$
are families in $\frak A$, of which $\langle b_i\rangle_{i\in I}$ is
disjoint, and $a_i\ptG b_i$ for every $i\in I$, then
$\sup_{i\in I}a_i\ptG\sup_{i\in I}b_i$.

\proof{{\bf (a)} This is immediate from the definition of `\Gte' and
395Bd.

\medskip

{\bf (b)(i)} $a\ptG a$ because the identity homomorphism belongs to
$G^*_{\tau}$.   If $a\ptG b\ptG c$ there are $\phi$, $\psi\in
G^*_{\tau}$ such that $\phi a\Bsubseteq b$, $\psi b\Bsubseteq c$ so that
$\psi\phi a\Bsubseteq c$;  as $\psi\phi\in G^*_{\tau}$ (395Bc),
$a\ptG c$.

\medskip

\quad{\bf (ii)} (This is of course a Schr\"oder-Bernstein theorem, and
the proof is the usual one.) Take $\phi$, $\psi\in G^*_{\tau}$ such that
$\phi a\Bsubseteq b$, $\psi b\Bsubseteq a$.   Set $a_0=a$, $b_0=b$,
$a_{n+1}=\psi b_n$ and $b_{n+1}=\phi a_n$ for each $n$.   Then
$\sequencen{a_n}$, $\sequencen{b_n}$ are non-increasing sequences;  set
$a_{\infty}=\inf_{n\in\Bbb N}a_n$, $b_{\infty}=\inf_{n\in\Bbb N}b_n$.
For each $n$,

\Centerline{$\phi\restrp\frak A_{a_{2n}\Bsetminus a_{2n+1}}:
\frak A_{a_{2n}\Bsetminus a_{2n+1}}\to\frak A_{b_{2n+1}\Bsetminus
b_{2n+2}}$,}

\Centerline{$\psi\restrp\frak A_{b_{2n}\Bsetminus b_{2n+1}}:
\frak A_{b_{2n}\Bsetminus b_{2n+1}}\to\frak A_{a_{2n+1}\Bsetminus
a_{2n+2}}$}

\noindent are isomorphisms, while

\Centerline{$\phi\restrp\frak A_{a_{\infty}}:\frak A_{a_{\infty}}
\to\frak A_{b_{\infty}}$}

\noindent is another.   So we can define an isomorphism $\theta:\frak
A_a\to\frak A_b$ by setting

$$\eqalign{\theta c&=\phi c\text{ if }c\Bsubseteq
a_{\infty}\Bcup\sup_{n\in\Bbb N}a_{2n}\Bsetminus a_{2n+1},\cr
&=\psi^{-1}c\text{ if }c\Bsubseteq
  \sup_{n\in\Bbb N}a_{2n+1}\Bsetminus a_{2n+2}.\cr}$$

\noindent By 395Be, $\theta\in G^*_{\tau}$, so $a$ and $b$ are \Gte.

\medskip

{\bf (c)} This is easy to prove directly from the results in 395B, but
also follows at once from (b);  any transitive reflexive relation
gives rise to an equivalence relation.

\medskip

{\bf (d)} We may suppose that $I$ is well-ordered by a relation $\le$.
For $i\in I$, set $a'_i=a_i\Bsetminus\sup_{j<i}a_j$.   Set $a=\sup_{i\in
I}a_i=\sup_{i\in I}a'_i$, $b=\sup_{i\in I}b_i$.   For each $i\in I$, we
have a $b'_i\Bsubseteq b_i$ and a $\phi_i\in G^*_{\tau}$ such that
$\phi_ia'_i=b'_i$.   Set $b'=\sup_{i\in I}b'_i\Bsubseteq b$;  then we
have an isomorphism $\psi:\frak A_a\to\frak A_{b'}$ defined by setting
$\psi d=\phi_id$ if $d\Bsubseteq a'_i$, and $\psi\in G^*_{\tau}$, so $a$
and $b'$ are \Gte\ and $a\ptG b$.
}%end of proof of 395C

\leader{395D}{Theorem} Let $\frak A$ be a Dedekind complete Boolean
algebra and $G$ a subgroup of $\Aut\frak A$.   Then the following are
equiveridical:

(i) there is an $a\ne 1$ such that $a$ is $G$-$\tau$-equidecomposable
with $1$;

(ii) there is a disjoint sequence $\sequencen{a_n}$ of non-zero elements
of $\frak A$ which are all $G$-$\tau$-equidecomposable;

(iii) there are non-zero \Gte\ $a$, $b$, $c\in\frak A$ such that $a\Bcap
b=0$ and $a\Bcup b\Bsubseteq c$;

(iv) there are $G$-$\tau$-equidecomposable $a$, $b\in\frak A$ such that
$a\Bsubset b$.

\proof{ Write $G^*_{\tau}$ for the full local semigroup generated by
$G$.

\medskip

\IMPLY{i}{ii} Assume (i).   There is a $\phi\in G^*_{\tau}$ such that
$\phi 1=a$.   Set $a_n=\phi^n(1\Bsetminus a)$ for each $n\in\Bbb N$;
because every $\phi^n$ belongs to $G^*_{\tau}$ (counting $\phi^0$ as the
identity operator on $\frak A$, and using 395Bc), with
$\dom\phi^n=\frak A$, $a_n$ is \Gte\ with $a_0=1\Bsetminus a$ for every
$n$.   Also $a_n=\phi^n1\Bsetminus\phi^{n+1}1$ for each $n$, while
$\sequencen{\phi^n1}$ is non-increasing, so $\sequencen{a_n}$ is
disjoint.   Thus (ii) is true.

\medskip

\IMPLY{ii}{iii} Assume (ii).   Set $a=\sup_{n\in\Bbb N}a_{2n}$,
$b=\sup_{n\in\Bbb N}a_{2n+1}$, $c=\sup_{n\in\Bbb N}a_n$, so that
$a\Bcap b=0$ and $a\Bcup b=c$.   For each $n$ we have a
$\phi_n\in G^*_{\tau}$ such that $\phi_na_0=a_n$.   So if we set

\Centerline{$\psi d
=\sup_{n\in\Bbb N}\phi_n\phi_{2n}^{-1}(d\Bcap a_{2n})$ for
$d\Bsubseteq a$,}

\noindent $\psi$ belongs to $G^*_{\tau}$ (using 395B) and witnesses that
$a$ and $c$ are \Gte.   Similarly, $b$ and $c$ are \Gte, so (iii) is
true.

\medskip

\IMPLY{iii}{iv} is trivial.

\medskip

\IMPLY{iv}{i} Take $\phi\in G^*_{\tau}$ such that $\phi b=a$.   Set

\Centerline{$\psi d=\phi(d\Bcap b)\Bcup(d\Bsetminus b)$}

\noindent for $d\in\frak A$;  then $\psi\in G^*_{\tau}$ witnesses
that $1$ is \Gte\ with $a\Bcup(1\Bsetminus b)\ne 1$.
}%end of proof of 395D

\leader{395E}{Definition} Let $\frak A$ be a Dedekind complete Boolean
algebra and $G$ a subgroup of $\Aut\frak A$.   I will say that $G$ is
{\bf fully non-paradoxical} if the statements of 395D are false;  that
is, if one of the following equiveridical statements is true:

\inset{(i)
if $a$ is $G$-$\tau$-equidecomposable with $1$ then $a=1$;}

\inset{(ii) there is no disjoint sequence $\sequencen{a_n}$ of
non-zero elements of $\frak A$ which are all \Gte;}

\inset{(iii) there are no non-zero \Gte\ $a$, $b$, $c\in\frak A$ such
that $a\Bcap b=0$ and $a\Bcup b\Bsubseteq c$;}

\inset{(iv) if $a\Bsubseteq b\in\frak A$ and $a$, $b$ are \Gte\ then
$a=b$.}

%add: there are $a\in\frak A$, $\phi$ in the full subgroup generated
%by $\pi$ which is not recurrent on $a$.

\noindent Note that if $G$ is fully non-paradoxical, and $H$ is a
subgroup of $G$, then $H$ also is
fully non-paradoxical\prooflet{, because if $a\preccurlyeq^{\tau}_Hb$ then $a\ptG b$, so that $a$ and $b$ are \Gte\ whenever they are
$H$-\vthsp$\tau$-\vthsp equidecomposable}.

\leader{395F}{Proposition} Let $(\frak A,\bar\mu)$ be a totally finite
measure algebra, and $G=\AmuA$ the group of all
measure-preserving automorphisms of $\frak A$.   Then $G$ is fully
non-paradoxical.

\proof{ If $\phi:\frak A\to\frak A_a$ belongs to the full local semigroup
generated by $G$, then we have a partition of unity $\langle
a_i\rangle_{i\in I}$ and a family $\langle\pi_i\rangle_{i\in I}$ in $G$
such that $\phi a_i=\pi_ia_i$ for every $i$;  but this means that

\Centerline{$\bar\mu a=\sum_{i\in I}\bar\mu\phi_ia_i=\sum_{i\in
I}\bar\mu\pi_ia_i
=\sum_{i\in I}\bar\mu a_i=\bar\mu 1$.}

\noindent As $\bar\mu 1<\infty$, we can conclude that $a=1$, so that $G$
satisfies the condition (i) of 395E.
}%end of proof of 395F

\leader{395G}{The fixed-point subalgebra of a group} Let $\frak A$ be a
Boolean algebra and $G$ a subgroup of $\Aut\frak A$.

\spheader 395Ga By the {\bf fixed-point subalgebra} of $G$ I mean

\Centerline{$\frak C=\{c:c\in\frak A,\,\pi c=c$ for every $\pi\in G\}$.}

\noindent\cmmnt{(I looked briefly at this construction in 333R, and
in the special case of a group generated by a single
element it appeared at various points in Chapter 38.)}
This is a subalgebra of $\frak A$, and is
order-closed\cmmnt{, because every $\pi\in G$ is order-continuous}.

\spheader 395Gb Now suppose that $\frak A$ is Dedekind complete.   In
this case $\frak C$ is Dedekind complete\cmmnt{ (314Ea)}, and we have,
for any $a\in\frak A$, an upper envelope $\upr(a,\frak C)$ of $\frak C$,
defined by setting

\Centerline{$\upr(a,\frak C)
=\inf\{c:a\Bsubseteq c\in\frak C\}$\dvro{.}{}}

\noindent\cmmnt{(313S).   }Now
$\upr(a,\frak C)=\sup\{\pi a:\pi\in G\}$.
\prooflet{\Prf\ Set $c_1=\upr(a,\frak C)$, $c_2=\sup\{\pi a:\pi\in G\}$.
(i) Because $a\Bsubseteq c_1\in\frak C$, $\pi a\Bsubseteq \pi c_1=c_1$
for every $\pi\in G$, and $c_2\Bsubseteq c_1$.   (ii) For any
$\phi\in G$,

\Centerline{$\phi c_2=\sup_{\pi\in G}\phi\pi a
=\sup_{\pi\in G}\pi a=c_2$}

\noindent because $G=\{\phi\pi:\pi\in G\}$.   So $c_2\in\frak C$;  since
also $a\Bsubseteq c_2$, $c_1\Bsubseteq c_2$, and $c_1=c_2$, as
claimed.\ \Qed}

\spheader 395Gc Again supposing that $\frak A$ is Dedekind complete,
write $G^*_{\tau}$ for the full local semigroup generated by
$G$.   Then $\phi(a\Bcap c)=\phi a\Bcap c$ whenever $\phi\in
G^*_{\tau}$, $a\in\dom\phi$ and $c\in\frak C$.
\prooflet{\Prf\ We have $\phi a=\sup_{i\in I}\pi_ia_i$, where
$a=\sup_{i\in I}a_i$ and $\pi_i\in G$ for every $i$.   Now

\Centerline{$\phi(a\Bcap c)=\sup_{i\in I}\pi_i(a_i\Bcap c)
=\sup_{i\in I}\pi_ia_i\Bcap c=\phi a\Bcap c$.  \Qed}}

\cmmnt{\noindent Consequently} $\upr(\phi a,\frak C)=\upr(a,\frak C)$
whenever $\phi\in G^*_{\tau}$ and $a\in\dom\phi$.   \prooflet{\Prf\ For
$c\in\frak C$,

\Centerline{$a\Bsubseteq c
\iff a\Bcap c=a
\iff\phi(a\Bcap c)=\phi a
\iff\phi a\Bcap c=\phi a
\iff\phi a\Bsubseteq c$.  \Qed}}

\cmmnt{\noindent It follows
that} $\upr(a,\frak C)\Bsubseteq\upr(b,\frak C)$ whenever $a\ptG b$.

\spheader 395Gd Still supposing that $\frak A$ is Dedekind complete,
we also find that if
$a\ptG b$ and $c\in\frak C$ then $a\Bcap c\ptG b\Bcap c$.
\prooflet{\Prf\ There is a $\phi\in G^*_{\tau}$ such that $\phi
a\Bsubseteq b$;  now $\phi(a\Bcap c)=\phi a\Bcap c\Bsubseteq b\Bcap c$.\
\QeD}   Hence\cmmnt{, or
otherwise,} $a\Bcap c$ and $b\Bcap c$ are \Gte\ whenever $a$ and $b$ are
\Gte\ and $c\in\frak C$.

\spheader 395Ge\dvArevised{2010}\cmmnt{ By analogy with the notion of
`ergodic automorphism',}
I will say that $G$ is {\bf ergodic} if $\sup_{\pi\in G}\pi a=1$
for every non-zero $a\in\frak A$.
\cmmnt{Thus an automorphism $\pi$ is
ergodic in the sense of 372Oa iff the group $\{\pi^n:n\in\Bbb Z\}$ it
generates is ergodic (372Pb).}

\spheader 395Gf\dvAnew{2010} If $G$ is ergodic, then $\frak C=\{0,1\}$.
\prooflet{(If $c\in\frak C\setminus\{0\}$, then
$1=\sup_{\pi\in G}\pi c=c$.)}
If $\frak A$ is Dedekind complete and $\frak C=\{0,1\}$
then $G$ is ergodic.   \prooflet{(If $a\in\frak A\setminus\{0\}$,
then $1=\upr(a,\frak C)=\sup_{\pi\in G}\pi a$, by (b) above.)}
\cmmnt{(Cf.\ 392Sa, 392Sc.)}

\leader{395H}{}\cmmnt{ I now embark on a series of lemmas leading to
the main theorem (395N).

\medskip

\noindent}{\bf Lemma} Let $\frak A$ be a Dedekind complete Boolean
algebra and $G$ a fully non-paradoxical subgroup of $\Aut\frak A$.
Write $\frak C$ for the
fixed-point subalgebra of $G$.   Take any $a$, $b\in\frak A$.   Set
$c_0=\sup\{c:c\in\frak C,\,a\Bcap c\ptG b\}$;  then $a\Bcap c_0\ptG b$
and $b\Bsetminus c_0\ptG a$.

\proof{ Enumerate $G$ as $\langle\pi_{\xi}\rangle_{\xi<\kappa}$, where
$\kappa=\#(G)$.   Define $\langle a_{\xi}\rangle_{\xi<\kappa}$,
$\langle b_{\xi}\rangle_{\xi<\kappa}$ inductively,  setting

\Centerline{$a_{\xi}=(a\Bsetminus\sup_{\eta<\xi}a_{\eta})
\Bcap\pi_{\xi}^{-1}(b\Bsetminus\sup_{\eta<\xi}b_{\eta})$,
\quad$b_{\xi}=\pi_{\xi}a_{\xi}$.}

\noindent Then $\langle a_{\xi}\rangle_{\xi<\kappa}$ is a disjoint
family in $\frak A_a$ and $\langle b_{\xi}\rangle_{\xi<\kappa}$ is a
disjoint family in $\frak A_b$, and $\sup_{\xi<\kappa}a_{\xi}$ is \Gte\
with $\sup_{\xi<\kappa}b_{\xi}$.
Set $a'=a\Bsetminus\sup_{\xi<\kappa}a_{\xi}$,
$b'=b\Bsetminus\sup_{\xi<\kappa}b_{\xi}$,

\Centerline{$\tilde c_0=1\Bsetminus\upr(a',\frak C)
=\sup\{c:c\in\frak C,\,c\Bcap a'=0\}$.}

\noindent Then

\Centerline{$a\Bcap\tilde c_0\Bsubseteq\sup_{\xi<\kappa}a_{\xi}\ptG b$,}

\noindent so $\tilde c_0\Bsubseteq c_0$.

Now $b'\Bsubseteq\tilde c_0$.   \Prf\Quer\ Otherwise, because
$\tilde c_0=1\Bsetminus\sup_{\xi<\kappa}\pi_{\xi}a'$ (395Gb), there must
be a $\xi<\kappa$ such that $\pi_{\xi}a'\Bcap b'\ne 0$.   But in this
case $d=a'\Bcap\pi_{\xi}^{-1}b'\ne 0$, and we have

\Centerline{$d\Bsubseteq(a\Bsetminus\sup_{\eta<\xi}a_{\eta})
\Bcap\pi_{\xi}^{-1}(b\Bsetminus\sup_{\eta<\xi}b_{\eta})$,}

\noindent so that $d\Bsubseteq a_{\xi}$, which is absurd.\ \Bang\QeD\
Consequently

\Centerline{$b\Bsetminus\tilde c_0
\Bsubseteq\sup_{\xi<\kappa}b_{\xi}\ptG a$.}

Now take any $c\in\frak C$ such that $a\Bcap c\ptG b$, and consider
$c'=c\Bsetminus\tilde c_0$.   Then $b'\Bcap c'=0$, that is,
$b\Bcap c'=\sup_{\xi<\kappa}b_{\xi}\Bcap c'$, which is \Gte\ with
$\sup_{\xi<\kappa}a_{\xi}\Bcap c'=(a\Bsetminus a')\Bcap c'$ (395Gd).
But now

\Centerline{$a\Bcap c'=a\Bcap c\Bcap c'
\ptG b\Bcap c'\ptG(a\Bcap c')\Bsetminus(a'\Bcap c')$;}

\noindent because $G$ is fully non-paradoxical, $a'\Bcap c'$ must be
$0$, that is, $c'\Bsubseteq\tilde c_0$ and $c'=0$.   As $c'$ is
arbitrary, $c_0\Bsubseteq\tilde c_0$ and $c_0=\tilde c_0$.
So $c_0$ has the required properties.
}%end of proof of 395H

\cmmnt{\medskip

\noindent{\bf Remark} By analogy with the notation I used in discussing
the Hahn decomposition of countably additive functionals (326S-326T), we
might denote $c_0$ as `$\Bvalue{a\ptG b}$', or perhaps
`$\Bvalue{a\ptG b}_{\frak C}$', `the region (in $\frak C$) where
$a\ptG b$'.   The same
notation would write $\upr(a,\frak C)$ as `$\Bvalue{a\ne 0}_{\frak C}$'.
}%end of comment

\leader{395I}{}\cmmnt{ The construction I wish to use depends
essentially on $L^0$ spaces as described in \S364.   The next step is
the following.

\medskip

\noindent}{\bf Lemma} Let $\frak A$ be a Dedekind complete Boolean
algebra, not $\{0\}$, and $G$ a fully non-paradoxical subgroup of
$\Aut\frak A$.   Let $\frak C$ be the fixed-point subalgebra of $G$.
Suppose that $a$, $b\in\frak A$ and that $\upr(a,\frak C)=1$.   Then
there are $u$, $v\in L^0=L^0(\frak C)$ such that

$$\eqalign{\Bvalue{u\ge n}
&=\max\{c:c\in\frak C,
  \text{ there is a disjoint family }\langle d_i\rangle_{i<n}\cr
&\qquad\qquad\qquad\text{such that }c\Bcap a\ptG d_i\Bsubseteq b
   \text{ for every }i<n\},\cr
\Bvalue{v\le n}
&=\max\{c:c\in\frak C,\text{ there is a family }\langle
d_i\rangle_{i<n}\cr
&\qquad\qquad\qquad\text{such that }d_i\ptG a\text{ for every }i<n
   \text{ and }b\Bcap c\Bsubseteq\sup_{i<n}d_i\}\cr}$$

\noindent for every $n\in\Bbb N$.   Moreover, we can arrange that

\quad(i) $\Bvalue{u\in\Bbb N}=\Bvalue{v\in\Bbb N}=1$,

\quad(ii) $\Bvalue{v>0}=\upr(b,\frak C)$,

\quad(iii) $u\le v\le u+\chi 1$.

\cmmnt{\medskip

\noindent{\bf Remark} By writing `max' in the formulae above, I mean to
imply that the elements $\Bvalue{u\ge n}$, $\Bvalue{v\le n}$ belong to
the sets described.}

\proof{{\bf (a)} Choose $\sequencen{c_n}$, $\sequencen{b_n}$ as follows.
Given $\langle b_i\rangle_{i<n}$, set $b'_n=b\Bsetminus\sup_{i<n}b_i$,

\Centerline{$c_n=\sup\{c:c\in\frak C$, $a\Bcap c\ptG b'_n\}$,}

\noindent so that $a\Bcap c_n\ptG b'_n$ (395H);  choose
$b_n\Bsubseteq b'_n$ such that $a\Bcap c_n$ is \Gte\ with $b_n$, and continue.   Then
$\sequencen{b_n}$ is a disjoint sequence in $\frak A_b$ and
$\sequencen{c_n}$ is a
non-increasing sequence in $\frak C$.

For each $n$, we have
$b'_n\Bsetminus c_n\ptG a$, by 395H;  while $a\Bcap c\not\ptG b'_n$
whenever $c\in\frak C$ and $c\notBsubseteq c_n$.   Note also that,
because $\upr(a,\frak C)=1$,

\Centerline{$c_n=\upr(a\Bcap c_n,\frak C)=\upr(b_n,\frak C)
\Bsubseteq\upr(b'_n,\frak C)$}

\noindent(using 395Gc for the second equality).
\medskip

{\bf (b)} Now $\inf_{n\in\Bbb N}c_n=0$.   \Prf\ Setting
$c_{\infty}=\inf_{n\in\Bbb N}c_n$,
$\sequencen{b_n\Bcap c_{\infty}}$ is a disjoint sequence, all \Gte\ with
$a\Bcap c_{\infty}$, so $a\Bcap c_{\infty}=0$, because $G$ is fully
non-paradoxical;  because
$\upr(a,\frak C)=1$, it follows that $c_{\infty}=0$.\ \Qed\
Accordingly, if we set
$u=\sup_{n\in\Bbb N}(n+1)\chi c_n$, $u\in L^0$ and
$\Bvalue{u\ge n}=c_{n-1}$ for $n\ge 1$.   The construction ensures that
$\Bvalue{u\in\Bbb N}$, as defined in 364G, is equal to $1$.

\medskip

{\bf (c)} Consider next $c'_0=\upr(b,\frak C)$,
$c'_n=c_{n-1}\Bcap\upr(b'_n,\frak C)$ for $n\ge 1$.   Then
$\sequencen{c'_n}$ is a non-increasing sequence with zero infimum, so
again we can define $v\in L^0$ by setting
$v=\sup_{n\in\Bbb N}(n+1)\chi c'_n$.   Once again,
$\Bvalue{v\in\Bbb N}=1$, and $\Bvalue{v\le n}=1\Bsetminus c'_n$ for each
$n$.

Of course $\Bvalue{v>0}=c'_0=\upr(b,\frak C)$.   Because
$c_n\Bsubseteq c'_n\Bsubseteq c_{n-1}$,

\Centerline{$(n+1)\chi c_n\le(n+1)\chi c'_n\le
n\chi c_{n-1}+\chi 1$}

\noindent for each $n\ge 1$, and $u\le v\le u+\chi 1$.

\medskip

{\bf (d)} Now set

$$\eqalign{C_n
&=\{c:c\in\frak C,\text{ there is a disjoint family }
  \langle d_i\rangle_{i<n}\cr
&\qquad\qquad\text{such that }c\Bcap a\ptG d_i\Bsubseteq b
   \text{ for every }i<n\}.\cr}$$

\noindent Then $c_n=\max C_{n+1}$.

\medskip

\Prf\grheada\ Because
$c_n\Bsubseteq c_{n-1}\Bsubseteq\ldots\Bsubseteq c_0$,
$a\Bcap c_n\ptG b_i$ for every $i\le n$, so that
$\langle b_i\rangle_{i\le n}$ witnesses that $c_n\in C_{n+1}$.

\medskip

\quad\grheadb\ Suppose that $c\in C_{n+1}$;  let
$\langle d_i\rangle_{i\le n}$ be a disjoint family such that
$c\Bcap a\ptG d_i\Bsubseteq b$ for every $i$.   Set
$c'=c\Bsetminus c_n$.   For each $i<n$, $b_i\ptG a$, so

\Centerline{$b_i\Bcap c'\ptG a\Bcap c'\ptG d_i\Bcap c'$,}

\noindent while also

\Centerline{$b'_n\Bcap c'\ptG a\Bcap c'\ptG d_n\Bcap c'$.}

\noindent Take $d\Bsubseteq d_n\Bcap c'$ such that $b'_n\Bcap c'$ is
\Gte\ with $d$.   Then

\Centerline{$b\Bcap c'=(b'_n\Bcap c')\Bcup\sup_{i<n}(b_i\Bcap c')
\ptG d\Bcup\sup_{i<n}(d_i\Bcap c')\Bsubseteq b\Bcap c'$.}

\noindent Because $G$ is fully non-paradoxical,
$d\Bcup\sup_{i<n}(d_i\Bcap c')$ must be exactly $b\Bcap c'$, so $d$ must
be the whole of $d_n\Bcap c'$, and

\Centerline{$a\Bcap c'\ptG d_n\Bcap c'=d\ptG b'_n$.}

\noindent But this means that $c'\Bsubseteq c_n$.   Thus $c'=0$ and
$c\Bsubseteq c_n$.   So $c_n=\sup C_{n+1}=\max C_{n+1}$.\ \Qed

Accordingly

\Centerline{$\Bvalue{u\ge n}=c_{n-1}=\max C_n$}

\noindent for $n\ge 1$.   For $n=0$ we have
$\Bvalue{u\ge 0}=1=\max C_0$.   So $\Bvalue{u\ge n}=\max C_n$ for every
$n$, as required.

\medskip

{\bf (e)} Similarly, if we set

$$\eqalign{C'_n
&=\{c:c\in\frak C,\text{ there is a family }\langle d_i\rangle_{i<n}\cr
&\qquad\qquad\text{such that }d_i\ptG a\text{ for every }i<n
   \text{ and }b\Bcap c\Bsubseteq\sup_{i<n}d_i\}\cr}$$

\noindent then $1\Bsetminus c'_n=\max C'_n$ for every $n$.

\medskip

\Prf\grheada\ If $n=0$, then of course (interpreting $\sup\emptyset$ as
$0$) $1\Bsetminus c'_0\in C'_0$ because $b\Bsubseteq c'_0$.
For each $n\in\Bbb N$, set

\Centerline{$\tilde b_n=b_n\Bcup(b'_n\Bsetminus c_n)
=(b_n\Bcap c_n)\Bcup(b'_n\Bsetminus c_n)$.}

\noindent Because $b_n\ptG a$ and $b'_n\Bsetminus c_n\ptG a$, we have
$b_n\Bcap c_n\ptG a\Bcap c_n$ and
$b'_n\Bsetminus c_n\ptG a\Bsetminus c_n$, so $\tilde b_n\ptG a$
(395Cd).   If we look at

\Centerline{$\sup_{i<n}\tilde b_i
\Bsupseteq\sup_{i<n}b_i\Bcup(b'_{n-1}\Bsetminus c_{n-1})$,}

\noindent we see that, for $n\ge 1$,

\Centerline{$b\Bsetminus\sup_{i<n}\tilde b_i
\Bsubseteq b'_n\Bcap c_{n-1}\Bsubseteq c'_n$,}

\noindent so that $b\Bsetminus c'_n\Bsubseteq\sup_{i<n}\tilde b_i$ and
$\{\tilde b_i:i<n\}$ witnesses that $1\Bsetminus c'_n\in C'_n$.

\medskip

\quad\grheadb\ Now take any $c\in C'_n$ and a corresponding family
$\langle d_i\rangle_{i<n}$ such that $d_i\ptG a$ for every $i<n$ and
$b\Bcap c\Bsubseteq\sup_{i<n}d_i$.

Set $c'=c\Bcap c'_n$.   For each $i<n$,

\Centerline{$c'\Bcap d_i\ptG c'\Bcap a\ptG b_i$}

\noindent because $c'\Bsubseteq c_i$.   So (by 395Cd, as usual)

\Centerline{$c'\Bcap b\ptG c'\Bcap\sup_{i<n}b_i\Bsubseteq c'\Bcap b$,}

\noindent and (again because $G$ is fully non-paradoxical)
$c'\Bcap b=c'\Bcap\sup_{i<n}b_i$, that is, $c'\Bcap b'_n=0$.   But
$c'\Bsubseteq c'_n\Bsubseteq\penalty-100\discretionary{}{}{}
\upr(b'_n,\frak C)$, so $c'$ must be $0$,
which means that $c\Bsubseteq 1\Bsetminus c'_n$.   As $c$ is arbitrary,
$1\Bsetminus c'_n=\sup C'_n=\max C'_n$.\ \Qed

Thus $\Bvalue{v\le n}=\max C'_n$, as declared.
}%end of proof of 395I

\leader{395J}{Notation} Observe that the specification of $\Bvalue{u\ge n}$
and $\Bvalue{v\le n}$, together with the declaration that
$\Bvalue{u\in\Bbb N}=\Bvalue{v\in\Bbb N}=1$, determine $u$ and $v$
uniquely\cmmnt{, because $\sequencen{\Bvalue{u=n}}$ and
$\sequencen{\Bvalue{v=n}}$
must be partitions of unity}.   So\cmmnt{, in the context of 395I,} we
can write $\low{b:a}$ for $u$ and $\high{b:a}$ for $v$.

\leader{395K}{Lemma} Let $\frak A$ be a Dedekind complete Boolean
algebra, not $\{0\}$, and $G$ a fully non-paradoxical subgroup of
$\Aut\frak A$ with fixed-point subalgebra $\frak C$.   Suppose that $a$,
$b$, $b_1$, $b_2\in\frak A$ and that $\upr(a,\frak C)=1$.

(a) $\low{0:a}=\high{0:a}=0$, $\low{1:a}\ge\chi 1$ and
$\low{1:1}=\chi 1$.

(b) If $b_1\ptG b_2$ then $\low{b_1:a}\le\low{b_2:a}$ and
$\high{b_1:a}\le\high{b_2:a}$.

(c) $\high{b_1\Bcup b_2:a}\le\high{b_1:a}+\high{b_2:a}$.

(d) If $b_1\Bcap b_2=0$,
$\low{b_1:a}+\low{b_2:a}\le\low{b_1\Bcup b_2:a}$.

(e) If $c\in\frak C$ is such that $a\Bcap c$ is a relative atom over
$\frak C$\cmmnt{ (definition:  331A)}, then
$c\Bsubseteq\Bvalue{\high{b:a}-\low{b:a}=0}$.

\proof{{\bf (a)-(b)} are immediate from the definitions and the basic
properties of $\ptG$, $\high{\ldots}$ and $\low{\ldots}$, as listed in
395C and 395I.

\medskip

{\bf (c)} For $j$, $k\in\Bbb N$, set
$c_{jk}=\Bvalue{\high{b_1:a}=j}\Bcap\Bvalue{\high{b_2:a}=k}$.   Then

\Centerline{$c_{jk}
\Bsubseteq\Bvalue{\high{b_1\Bcup b_2:a}\le j+k}
  \Bcap\Bvalue{\high{b_1:a}+\high{b_2:a}=j+k}$.}

\noindent\Prf\ We may suppose that $c_{jk}\ne 0$.   Of course

\Centerline{$c_{jk}\Bsubseteq\Bvalue{\high{b_1:a}+\high{b_2:a}=j+k}$.}

\noindent Next, there are sets $J$, $J'\subseteq\frak A$ such that
$d\ptG a$ for every $d\in J\cup J'$, $\#(J)\le j$, $\#(J')\le k$,
$\sup J\Bsupseteq b_1\Bcap c_{jk}$ and
$\sup J'\Bsupseteq b_2\Bcap c_{jk}$.
So $\sup(J\cup J')\Bsupseteq(b_1\Bcup b_2)\Bcap c_{jk}$ and $J\cup J'$
witnesses that
$c_{jk}\Bsubseteq\Bvalue{\high{b_1\Bcup b_2:a}\le j+k}$.\ \Qed

Accordingly

\Centerline{$c_{jk}\Bsubseteq
\Bvalue{\high{b_1:a}+\high{b_2:a}-\high{b_1\Bcup b_2:a}\ge 0}$.}

\noindent Now as $\sup_{j,k\in\Bbb N}c_{jk}=1$, we must have
$\high{b_1\Bcup b_2:a}\le\high{b_1:a}+\high{b_2:a}$.

\medskip

{\bf (d)} This time,
set $c_{jk}=\Bvalue{\low{b_1:a}=j}\Bcap\Bvalue{\low{b_2:a}=k}$ for $j$,
$k\in\Bbb N$.   Then

\Centerline{$c_{jk}\Bsubseteq\Bvalue{\low{b_1\Bcup b_2:a}\ge
j+k}\Bcap\Bvalue{\low{b_1:a}+\low{b_2:a}=j+k}$}

\noindent for every $j$, $k\in\Bbb N$.   \Prf\ Once again, we surely
have

\Centerline{$c_{jk}\Bsubseteq\Bvalue{\low{b_1:a}+\low{b_2:a}=j+k}$.}

\noindent Next, we can find a family $\langle d_i\rangle_{i<j+k}$ such
that

\Centerline{$\langle d_i\rangle_{i<j}$ is disjoint, $a\Bcap c_{jk}\ptG
d_i\Bsubseteq b_1$ for every $i<k$,}

\Centerline{$\langle d_i\rangle_{j\le i<j+k}$ is disjoint,
$a\Bcap c_{jk}\ptG d_i\Bsubseteq b_2$ for $j\le i<j+k$.}

\noindent As $b_1\Bcap b_2=0$, the whole family
$\langle d_i\rangle_{i<j+k}$ is disjoint and witnesses that
$c_{jk}\Bsubseteq\Bvalue{\low{b_1\Bcup b_2:a}\ge j+k}$.\ \Qed

So

\Centerline{$c_{jk}
\Bsubseteq\Bvalue{\low{b_1\Bcup b_2:a}-\low{b_1:a}-\low{b_2:a}\ge 0}$}

\noindent Since $\sup_{j,k\in\Bbb N}c_{jk}=1$, as before, we must have
$\low{b_1\Bcup b_2:a}\ge\low{b_1:a}+\low{b_2:a}$.

\medskip

{\bf (e)} \Quer\ Otherwise, there must be some $k\in\Bbb N$ such that

\Centerline{$c_0=c\Bcap\Bvalue{\low{b:a}=k}\Bcap\Bvalue{\high{b:a}>k}
\ne 0$.}

\noindent Let $\ofamily{i}{k}{d_i}$ be a disjoint family in $\frak A_b$
such that $a\Bcap c_0\ptG d_i$ for each $i$;  cutting the $d_i$ down if
necessary, we may suppose that $a\Bcap c_0$ is \Gte\ with $d_i$ for each
$i$.   As $c_0\notBsubseteq\Bvalue{\high{b:a}\le k}$,
$b\Bcap c_0\notBsubseteq\sup_{i<k}d_i$;  set
$d=b\Bcap c_0\Bsetminus\sup_{i<k}d_i\ne 0$.
By 395H, there is a $c_1\in\frak C$ such that $d\Bcap c_1\ptG a$ and
$a\Bsetminus c_1\ptG d$.   Setting $d_k=d$, $\langle d_i\rangle_{i\le k}$
witnesses that $c_0\Bsubseteq c_1\Bsubseteq\Bvalue{\low{b:a}\ge k+1}$,
so $c_0\Bsubseteq c_1$ must be $0$ and $d\Bcap c_0\ptG a$.
There is therefore a non-zero
$\tilde a\Bsubseteq a\Bcap c_0$ such that $\tilde a\ptG d$.   But now
remember that $a\Bcap c$ is supposed to be a relative atom over
$\frak C$, so $\tilde a=a\Bcap\tilde c$ for some $\tilde c\in\frak C$
such that $\tilde c\Bsubseteq c_0$.   In this case,
$a\Bcap\tilde c\ptG d_i$ for every $i<k$ and also
$a\Bcap\tilde c\ptG d$, so
$0\ne\tilde c\Bsubseteq\Bvalue{\low{b:a}\ge k+1}$, which is
absurd.\ \Bang
}%end of proof of 395K

\leader{395L}{Lemma} Let $\frak A$ be a Dedekind complete Boolean
algebra, not $\{0\}$, and $G$ a fully non-paradoxical subgroup of
$\Aut\frak A$ with fixed-point subalgebra $\frak C$.   Suppose that
$a_1$, $a_2$, $b\in\frak A$ and that
$\upr(a_1,\frak C)=\upr(a_2,\frak C)=1$.   Then

\Centerline{$\low{b:a_2}\ge\low{b:a_1}\times\low{a_1:a_2}$,
\quad$\high{b:a_2}\le\high{b:a_1}\times\high{a_1:a_2}$.}

\proof{ I use the same method as in 395K.   As usual, write $G^*_{\tau}$
for the full local semigroup generated by $G$.

\medskip

{\bf (a)} For $j$, $k\in\Bbb N$ set

\Centerline{$c_{j,k}=\Bvalue{\low{b:a_1}=j}
\Bcap\Bvalue{\low{a_1:a_2}=k}$.}

\noindent Then

\Centerline{$c_{j,k}\Bsubseteq\Bvalue{\low{b:a_1}\times\low{a_1:a_2}=jk}
\Bcap\Bvalue{\low{b:a_2}\ge jk}$.}

\noindent\Prf\ Write $c$ for $c_{j,k}$.   As in parts (c) and (d) of the
proof of 395K, the fact that
$c\Bsubseteq\Bvalue{\low{b:a_1}\times\low{a_1:a_2}=jk}$ is elementary;
what we need to check is that $c\Bsubseteq\Bvalue{\low{b:a_2}\ge jk}$.
Again, we may suppose that $c\ne 0$.   There are families
$\langle d_i\rangle_{i<j}$, $\langle d^*_l\rangle_{l<k}$ such that

\Centerline{$\langle d_i\rangle_{i<j}$ is disjoint,
$a_1\Bcap c\ptG d_i\Bsubseteq b$ for every $i<j$,}

\Centerline{$\langle d^*_l\rangle_{l<k}$ is disjoint,
$a_2\Bcap c\ptG d^*_l\Bsubseteq a_1$ for every $l<k$.}

\noindent For each $i<j$, let $\phi_i\in G^*_{\tau}$ be such that
$\phi_i(a_1\Bcap c)\Bsubseteq d_i$.   If $i<j$ and $l<k$, then

\Centerline{$a_2\Bcap c\ptG d^*_l\Bcap c\ptG\phi_i(d^*_l\Bcap c)
\Bsubseteq\phi_i(a_1\Bcap c)\Bsubseteq d_i\Bsubseteq b$.}

\noindent Also $\langle\phi_i(d^*_l\Bcap c)\rangle_{i<j,l<k}$ is
disjoint because $\langle\phi_i(a_1\Bcap c)\rangle_{i<j}$ and
$\langle d^*_l\rangle_{l<k}$ are, so witnesses that
$c\Bsubseteq\Bvalue{\low{b:a_2}\ge jk}$.\ \Qed

Now, just as in 395K, it follows from the fact that
$\sup_{j,k\in\Bbb N}c_{j,k}=1$ that
$\low{b:a_1}\times\low{a_1:a_2}\le\low{b:a_2}$.

\medskip

{\bf (b)} For $j$, $k\in\Bbb N$ set

\Centerline{$c_{j,k}=\Bvalue{\high{b:a_1}=j}
\Bcap\Bvalue{\high{a_1:a_2}=k}$.}

\noindent Then

\Centerline{$c_{j,k}\Bsubseteq
\Bvalue{\high{b:a_1}\times\high{a_1:a_2}=jk}
\Bcap\Bvalue{\high{b:a_2}\le jk}$.}

\noindent\Prf\ Write $c$ for $c_{j,k}$.   Then
$c\Bsubseteq\Bvalue{\high{b:a_1}\times\high{a_1:a_2}=jk}$.    There are
families $\langle d_i\rangle_{i<j}$, $\langle d_l^*\rangle_{l<k}$ such
that $d_i\ptG a_1$ for every $i<j$, $d^*_l\ptG a_2$ for every
$l<k$, $b\Bcap c\Bsubseteq\sup_{i<j}d_i$ and
$a_1\Bcap c\Bsubseteq\sup_{l<k}d^*_l$.
For each $i<j$, let $d_i'\Bsubseteq a_1$ be \Gte\ with $d_i$, and take
$\phi_i\in G^*_{\tau}$ such that $\phi_id_i'=d_i$.   Then

\Centerline{$\phi_i(d'_i\Bcap d^*_l)\ptG d^*_l\ptG a_2$ for every
$i<j$, $l<k$,}

$$\eqalign{\sup_{i<j,l<k}\phi_i(d'_i\Bcap d^*_l)
&=\sup_{i<j}\phi_i(d'_i\Bcap\sup_{l<k}d^*_l)
\Bsupseteq\sup_{i<j}\phi_i(d'_i\Bcap c)\cr
&=\sup_{i<j}d_i\Bcap c
\Bsupseteq b\Bcap c.\cr}$$

\noindent So $\langle\phi_i(d'_i\Bcap d^*_l)\rangle_{i<j,l<k}$ witnesses
that $c\Bsubseteq\Bvalue{\high{b:a_2}\le jk}$.\ \Qed

Once again, it follows easily that
$\high{b:a_1}\times\high{a_1:a_2}\ge\high{b:a_2}$.
}%end of proof of 395L

\vleader{60pt}{395M}{Lemma} Let $\frak A$ be a Dedekind complete Boolean
algebra, not $\{0\}$, and $G$ a subgroup of
$\Aut\frak A$ with fixed-point subalgebra $\frak C$.

(a) For any $a\in\frak A$, there is a $b\Bsubseteq a$ such that
$b\ptG a\Bsetminus b$ and $a\Bsetminus\upr(b,\frak C)$ is a either $0$
or a relative atom over $\frak C$.

(b) Now suppose that $G$ is fully non-paradoxical.   Then for any
$\epsilon>0$ there is an $a\in\frak A$ such that
$\upr(a,\frak C)=1$ and $\high{b:a}\le\low{b:a}+\epsilon\low{1:a}$ for
every $b\in\frak A$.

\proof{{\bf (a)} Set $B=\{d:d\Bsubseteq a,\,d\ptG a\Bsetminus d\}$ and
let $D\subseteq B$ be a maximal subset such that
$\upr(d,\frak C)\Bcap\upr(d',\frak C)\penalty-100\discretionary{}{}{}=0$ for all distinct $d$, $d'\in\frak D$.   Set
$b=\sup D$.   For any $d\in D$, $d\ptG a\Bsetminus d$, so

$$\eqalign{b\Bcap\upr(d,\frak C)
&=\sup_{d'\in D}d'\Bcap\upr(d,\frak C)
=\sup_{d'\in D}d'\Bcap\upr(d',\frak C)\Bcap\upr(d,\frak C)
=d\Bcap\upr(d,\frak C)\cr
&\ptG(a\Bsetminus d)\Bcap\upr(d,\frak C)
=(a\Bsetminus b)\Bcap\upr(d,\frak C)
\Bsubseteq a\Bsetminus b\cr}$$

\noindent by 395Gc.   By 395H,

\Centerline{$b=\sup_{d\in D}b\Bcap\upr(d,\frak C)\ptG a\Bsetminus b$.}

\Quer\ Suppose, if possible, that $a'=a\Bsetminus\upr(b,\frak C)$ is
neither $0$ nor
a relative atom over $\frak C$.   Let $d_0\Bsubseteq a'$ be an element
not expressible as $a'\Bcap c$ for any $c\in\frak C$;  then
$d_0\ne a\Bcap\upr(d_0,\frak C)$ and there must be a $\pi\in G$ such that
$d_1=\pi d_0\Bcap a\Bsetminus d_0$ is non-zero (395Gb).    In this case

\Centerline{$d_1\ptG \pi^{-1}d_1\Bsubseteq d_0
\Bsubseteq a\Bsetminus d_1$,}

\noindent so $d_1\in B$;  but also

\Centerline{$d_1\Bcap\upr(d,\frak C)
\Bsubseteq d_1\Bcap\upr(b,\frak C)=0$,}

\noindent so $\upr(d_1,\frak C)\Bcap\upr(d,\frak C)=0$, for every
$d\in D$, and we ought to have put $d_1$ into $D$.\ \Bang

Thus $b$ has the required properties.

\medskip

{\bf (b)(i)} For every $n\in\Bbb N$ we can find $a_n\in\frak A$ and
$c_n\in\frak C$ such that $\upr(a_n,\frak C)=1$, $a_n\Bsetminus c_n$ is
either $0$ or a
relative atom over $\frak C$, and $\low{1:a_n}\ge 2^n\chi c_n$.   \Prf\
Induce on $n$.   The induction starts with $a_0=c_0=1$, because
$\low{1:1}=\chi 1$.   For the inductive step, having found $a_n$ and
$c_n$, let $d\Bsubseteq a_n\Bcap c_n$ be such that
$d\ptG a_n\Bcap c_n\Bsetminus d$ and
$a_n\Bcap c_n\Bsetminus\upr(d,\frak C)$ is either $0$ or a
relative atom over $\frak C$, as in (a).   Set
$c_{n+1}=\upr(d,\frak C)$, $a_{n+1}=(a_n\Bsetminus c_{n+1})\Bcup d$;
then

$$\eqalign{\upr(a_{n+1},\frak C)
&=\upr(a_n\Bsetminus c_{n+1},\frak C)\Bcup\upr(d,\frak C)\cr
&=(\upr(a_n,\frak C)\Bsetminus c_{n+1})\Bcup c_{n+1}
=(1\Bsetminus c_{n+1})\Bcup c_{n+1}
=1\cr}$$

\noindent by 313Sb-313Sc and the inductive hypothesis.

We have $c_{n+1}\Bcap d\ptG c_{n+1}\Bcap a_n\Bsetminus d$, so

\Centerline{$c_{n+1}\Bcap a_{n+1}=d\Bsubseteq a_n$,
\quad$c_{n+1}\Bcap a_{n+1}\ptG a_n\Bsetminus d$,}

\noindent and $\low{a_n:a_{n+1}}\ge 2\chi c_{n+1}$;  by 395L,

\Centerline{$\low{1:a_{n+1}}\ge\low{1:a_n}\times\low{a_n:a_{n+1}}
\ge 2^n\chi c_n\times 2\chi c_{n+1}=2^{n+1}\chi c_{n+1}$.}

If

\Centerline{$b\Bsubseteq a_{n+1}\Bsetminus c_{n+1}
=(a_n\Bsetminus c_n)\Bcup(a_n\Bcap c_n\Bsetminus c_{n+1})$,}

\noindent then, because both terms on the right are either $0$ or
relative atoms over
$\frak C$, there are $c'$, $c''\in\frak C$ such that

$$\eqalign{b
&=(b\Bcap a_n\Bsetminus c_n)
  \Bcup(b\Bcap a_n\Bcap c_n\Bsetminus c_{n+1})\cr
&=(c'\Bcap a_n\Bsetminus c_n)
   \Bcup(c''\Bcap a_n\Bcap c_n\Bsetminus c_{n+1})
=c\Bcap a_{n+1}\Bsetminus c_{n+1}\cr}$$

\noindent where $c=(c'\Bsetminus c_n)\Bcup(c''\Bcap c_n)$ belongs to
$\frak C$.   So $a_{n+1}\Bsetminus c_{n+1}$ is either $0$ or
a relative atom over $\frak C$.

Thus the induction continues.\ \Qed

\medskip

\quad{\bf (ii)} Now suppose that $\epsilon>0$.   Take $n$ such that
$2^{-n}\le\epsilon$, and consider $a_n$, $c_n$ taken from (i) above.

Let $b\in\frak A$.   Set

\Centerline{$c
  =\Bvalue{\high{b:a_n}-\low{b:a_n}-\epsilon\low{1:a_n}>0}
  \in\frak C$.}

\noindent Since we know that

\Centerline{$\epsilon\low{1:a_n}\ge 2^{-n}2^n\chi c_n=\chi c_n$,
\quad$\high{b:a_n}\le\low{b:a_n}+\chi 1$,}

\noindent we must have $c\Bcap c_n=0$.   But this means that
$a_n\Bcap c$ is either $0$ or a relative atom over $\frak C$.   By 395Ke,
$c$ is included in $\Bvalue{\high{b:a_n}-\low{b:a_n}=0}$;  as also
$\low{1:a_n}\ge\chi 1$ (395Ka), $c$ must be zero, that is,
$\high{b:a_n}\le\low{b:a_n}+\epsilon\low{1:a_n}$.
}%end of proof of 395M

\leader{395N}{}\cmmnt{ We are at last ready for the theorem.

\medskip

\noindent}{\bf Theorem} Let $\frak A$ be a Dedekind complete Boolean
algebra and $G$ a fully non-paradoxical subgroup of $\Aut\frak A$ with
fixed-point subalgebra $\frak C$.   Then there is a unique function
$\theta:\frak A\to L^{\infty}(\frak C)$ such that

(i) $\theta$ is additive, non-negative and order-continuous;

(ii) $\Bvalue{\theta a>0}=\upr(a,\frak C)$ for every $a\in\frak A$;  in
particular, $\theta a=0$ iff $a=0$;

(iii) $\theta 1=\chi 1$;

(iv) $\theta(a\Bcap c)=\theta a\times\chi c$ for every $a\in\frak A$,
$c\in\frak C$;  in particular, $\theta c=\chi c$ for every
$c\in\frak C$;

(v) If $a$, $b\in\frak A$ are \Gte, then $\theta a=\theta b$;  in
particular, $\theta$ is $G$-invariant.

\proof{ If $\frak A=\{0\}$ this is trivial;  so I suppose henceforth
that $\frak A\ne\{0\}$.

\wheader{395N}{4}{2}{2}{36pt}

{\bf (a)} Set $A^*=\{a:a\in\frak A,\,\upr(a,\frak C)=1\}$ and for
$a\in A^*$, $b\in\frak A$ set

$$\theta_a(b)=\bover{\high{b:a}}{\low{1:a}}\in L^0=L^0(\frak C);$$

\noindent the first thing to note is that because $\low{1:a}\ge\chi 1$,
we can always do the divisions to obtain elements $\theta_a(b)$ of
$L^0(\frak A)$ (364N).    Set

\Centerline{$\theta b=\inf_{a\in A^*}\theta_ab$}

\noindent for $b\in\frak A$.
(Note that $L^0(\frak C)$ is Dedekind complete, by 364M, so the
infimum is defined.)

\medskip

{\bf (b)} The formulae of 395K tell us that, for $a\in A^*$ and $b_1$,
$b_2\in\frak A$,

\Centerline{$\theta_a0=0$,
\quad$\theta_ab_1\le\theta_ab_2$ if $b_1\Bsubseteq b_2$,}

\Centerline{$\theta_a(b_1\Bcup b_2)\le\theta_ab_1+\theta_ab_2$,}

\Centerline{$\theta_a1\ge\chi 1$.}

\noindent It follows at once that

\Centerline{$\theta 0=0$,
\quad$\theta b_1\le\theta b_2$ if $b_1\Bsubseteq b_2$,}

\Centerline{$\theta 1\ge\chi 1$.}

\medskip

{\bf (c)} For each $n\in\Bbb N$ there is an $e_n\in A^*$ such that
$\high{b:e_n}\le\low{b:e_n}+2^{-n}\low{1:e_n}$ for every $b\in\frak A$
(395Mb).   Now $\theta_{e_n}b\le\theta_ab+2^{-n}\high{b:a}$ for every
$a\in A^*$, $b\in\frak A$.   \Prf
$\high{a:e_n}\le\low{a:e_n}+2^{-n}\low{1:e_n}$, so

$$\eqalign{\high{a:e_n}\times\low{1:a}
&\le\low{a:e_n}\times\low{1:a}+2^{-n}\low{1:e_n}\times\low{1:a}\cr
&\le\low{1:e_n}+2^{-n}\low{1:e_n}\times\low{1:a}\cr}$$

\noindent (by 395L);  accordingly

$$\eqalignno{\high{b:e_n}\times\low{1:a}
&\le\high{b:a}\times\high{a:e_n}\times\low{1:a}\cr
\noalign{\noindent (by the other half of 395L)}
&\le\high{b:a}\times\low{1:e_n}
    +2^{-n}\high{b:a}\times\low{1:e_n}\times\low{1:a}\cr}$$

\noindent and, dividing by $\low{1:a}\times\low{1:e_n}$, we get
$\theta_{e_n}b\le\theta_ab+2^{-n}\high{b:a}$.\ \Qed

\medskip

{\bf (d)} Now $\theta$ is additive.   \Prf\ Taking $\sequencen{e_n}$
from (c), observe first that

\Centerline{$\inf_{n\in\Bbb N}\theta_{e_n}b
\le\theta_ab+\inf_{n\in\Bbb N}2^{-n}\high{b:a}=\theta_ab$}

\noindent for every $a\in A^*$, $b\in\frak A$, so that $\theta
b=\inf_{n\in\Bbb N}\theta_{e_n}b$ for every $b$.   Now suppose that
$b_1$, $b_2\in\frak A$ and $b_1\Bcap b_2=0$.   Then, for any
$n\in\Bbb N$,

$$\eqalignno{\high{b_1:e_n}+\high{b_2:e_n}
&\le\low{b_1:e_n}+\low{b_2:e_n}+2^{-n+1}\low{1:e_n}\cr
&\le\low{b_1\Bcup b_2:e_n}+2^{-n+1}\low{1:e_n}\cr
\noalign{\noindent (by 395Kd)}
&\le\high{b_1\Bcup b_2:e_n}+2^{-n+1}\low{1:e_n}.\cr}$$

\noindent Dividing by $\low{1:e_n}$, we have

\Centerline{$\theta b_1+\theta b_2
\le\theta_{e_n}b_1+\theta_{e_n}b_2
\le\theta_{e_n}(b_1\Bcup b_2)+2^{-n+1}\chi 1$.}

\noindent Taking the infimum over $n$, we get

\Centerline{$\theta b_1+\theta b_2
\le\theta(b_1\Bcup b_2)$.}

In the other direction, if $a$, $a'\in A^*$ and $n\in\Bbb N$,

$$\eqalign{\theta(b_1\Bcup b_2)
&\le\theta_{e_n}(b_1\Bcup b_2)
\le\theta_{e_n}(b_1)+\theta_{e_n}(b_2)\cr
&\le\theta_a(b_1)+2^{-n}\high{b_1:a}
  +\theta_{a'}(b_2)+2^{-n}\high{b_2:a'}.\cr}$$

\noindent As $n$ is arbitrary,
$\theta(b_1\Bcup b_2)\le\theta_a(b_1)+\theta_{a'}(b_2)$;  as $a$ and
$a'$ are arbitrary, $\theta(b_1\Bcup b_2)\le\theta b_1+\theta b_2$
(using 351Dc).

As $b_1$ and $b_2$ are arbitrary, $\theta$ is additive.\ \Qed

We see also that $\high{1:e_n}\le(1+2^{-n})\low{1:e_n}$, so that
$\theta_{e_n}1\le(1+2^{-n})\chi 1$ for each $n$;  since we already know
that $\theta 1\ge\chi 1$, we have $\theta 1=\chi 1$ exactly.

\medskip

{\bf (e)} If $c\in\frak C$ then

\Centerline{$\Bvalue{\theta c>0}\Bsubseteq\Bvalue{\theta_1 c>0}
\Bsubseteq\Bvalue{\high{c:1}>0}=\upr(c,\frak C)=c$}

\noindent (395I(ii)).   It follows that

\Centerline{$\theta(b\Bcap c)\le\theta b\wedge\theta c\le
\theta b\times\chi c$}

\noindent for any $b\in\frak A$, $c\in\frak C$.   Similarly,
$\theta(b\Bsetminus c)\le\theta b\times\chi(1\Bsetminus c)$;  adding, we
must have equality in both, and $\theta(b\Bcap c)=\theta b\times\chi c$.

Rather late, I point out that

\Centerline{$0\le\theta a\le\theta 1=\chi 1\in
L^{\infty}=L^{\infty}(\frak C)$}

\noindent for every $a\in\frak A$, so that $\theta a\in L^{\infty}$ for
every $a$.

\medskip

{\bf (f)}
If $b\in\frak A\Bsetminus\{0\}$, then

\Centerline{$\Bvalue{\theta b>0}\Bsubseteq\Bvalue{\theta_1b>0}
\Bsubseteq\Bvalue{\high{b:1}>0}=\upr(b,\frak C)$}

\noindent by 395I(ii) again.   \Quer\ Suppose, if possible, that
$\Bvalue{\theta b>0}\ne\upr(b,\frak C)$.   Set
$c_0=\upr(b,\frak C)\Bsetminus\Bvalue{\theta b>0}$,
$a_0=b\Bcup(1\Bsetminus\upr(b,\frak C))\discretionary{}{}{}\in A^*$.
Let $k\ge 1$ be such
that $c_1=c_0\Bcap\Bvalue{\high{1:a_0}\le k}\ne 0$.
Then $a_0\Bcap c_1=b\Bcap c_1$, so

\Centerline{$\theta a_0\times\chi c_1
=\theta(a_0\Bcap c_1)
=\theta(b\Bcap c_1)
=\theta b\times\chi c_1
=0$.}

\noindent By 364L(b-ii), there is an $a\in A^*$ such that
$c_1\notBsubseteq\Bvalue{\theta_aa_0\times\chi c_1\ge\bover1k}$, that
is, $c_2=c_1\Bcap\Bvalue{\theta_aa_0<\bover1k}\ne 0$.   Now

\Centerline{$c_2\Bsubseteq\Bvalue{\low{1:a}-k\high{a_0:a}>0}
\Bsubseteq\Bvalue{\high{1:a_0}\times\high{a_0:a}-k\high{a_0:a}>0}
\Bsubseteq\Bvalue{\high{1:a_0}>k}$,}

\noindent which is impossible, as $c_2\Bsubseteq c_1$.\ \Bang

Thus $\Bvalue{\theta b>0}=\upr(b,\frak C)$.   In particular, $\theta
b=0$ iff $b=0$.

\medskip

{\bf (g)} If $b$, $b'\in\frak A$ and $b\ptG b'$, then
$\theta b\le\theta b'$.   \Prf\ For every $a\in A^*$,
$\high{b:a}\le\high{b':a}$ (395Kb) so
$\theta_ab\le\theta_ab'$.\ \Qed\   So if $b$, $b'\in\frak A$ and
$c=\Bvalue{\theta b-\theta b'>0}$, $b'\Bcap c\ptG b$.
\Prf\Quer\ Otherwise, by 395H, there is a non-zero $c'\Bsubseteq c$ such
that $b\Bcap c'\ptG b'$.   But in this case
$\theta b\times\chi c'=\theta(b\Bcap c')\le\theta b'$ and
$c'\Bsubseteq\Bvalue{\theta b'-\theta b\ge 0}$.\ \Bang\Qed

\medskip

{\bf (h)} If $\langle a_i\rangle_{i\in I}$ is any disjoint family in
$\frak A$ with supremum $a$,
$\theta a=\sum_{i\in I}\theta a_i$, where the sum is to be interpreted
as $\sup_{J\subseteq I\text{ is finite}}\sum_{i\in J}\theta a_i$.
\Prf\ Induce on $\#(I)$.   If $\#(I)$ is finite, this is just finite
additivity ((d) above).   For the inductive step to
$\#(I)=\kappa\ge\omega$, we may suppose that $I$ is actually equal to
the cardinal $\kappa$.   Of course

\Centerline{$\theta a\ge\theta(\sup_{\xi\in J}a_{\xi})
=\sum_{\xi\in J}\theta a_{\xi}$}

\noindent for every finite $J\subseteq\kappa$, so (because
$L^{\infty}(\frak C)$ is Dedekind complete)
$u=\sum_{\xi<\kappa}\theta a_{\xi}$ is defined, and $u\le\theta a$.

For $\zeta<\kappa$, set $b_{\zeta}=\sup_{\xi<\zeta}a_{\xi}$.   By the
inductive hypothesis,

\Centerline{$\theta b_{\zeta}=\sum_{\xi<\zeta}\theta a_{\xi}
=\sup_{J\subseteq\zeta\text{ is finite}}\sum_{\xi\in J}\theta a_{\xi}
\le u$.}

\noindent At the same time, if $J\subseteq\kappa$ is finite, there is
some $\zeta<\kappa$ such that $J\subseteq\zeta$, so that
$\sum_{\xi\in J}\theta a_{\xi}\le\theta b_{\zeta}$;  accordingly
$\sup_{\zeta<\kappa}\theta b_{\zeta}=u$.

\Quer\ Suppose, if possible, that $u<\theta a$;  set $v=\theta a-u$.
Take $\delta>0$ such that $c_0=\Bvalue{v>\delta}\ne 0$.   Let
$\zeta<\kappa$ be such that
$c_1=c_0\Bsetminus\Bvalue{u-\theta b_{\zeta}>\delta}$ is
non-zero (cf.\ 364L(b-ii)).   Now
$v=\theta a-u\le\theta(a\Bsetminus b_{\zeta})$, so

\Centerline{$c_1\Bsubseteq\Bvalue{v>\delta}
\Bsubseteq\Bvalue{\theta(a\Bsetminus b_{\zeta})>0}
=\upr(a\Bsetminus b_{\zeta},\frak C)$,}

\noindent and $c_1\Bcap(a\Bsetminus b_{\zeta})\ne 0$;  there is
therefore an $\eta'\ge\zeta$ such that $d=c_1\Bcap a_{\eta'}\ne 0$.
Since $\theta d\le u-\theta b_{\zeta}$ and
$c_1$ is included in
$\Bvalue{u-\theta b_{\zeta}\le\delta}\Bcap\Bvalue{v>\delta}$,
$\Bvalue{v-\theta d>0}\Bsupseteq c_1$.

Choose $\langle d_{\xi}\rangle_{\xi<\kappa}$ inductively, as follows.
Given that $\langle d_{\eta}\rangle_{\eta<\xi}$ is a disjoint family in
$\frak A_{a\Bsetminus d}$ such that $d_{\eta}$ is \Gte\ with
$a_{\eta}\Bcap c_1$ for every $\eta<\xi$, then
$e_{\xi}=\sup_{\eta<\xi}d_{\eta}$ is \Gte\ with $b_{\xi}\Bcap c_1$, so
that $\theta e_{\xi}\le\theta b_{\xi}$, and

$$\eqalign{\Bvalue{\theta(a\Bsetminus(d\Bcup e_{\xi}))-\theta a_{\xi}>0}
&=\Bvalue{\theta a-\theta d-\theta e_{\xi}-\theta a_{\xi}>0}
\Bsupseteq\Bvalue{\theta a-\theta d-\theta b_{\xi}-\theta a_{\xi}>0}\cr
&=\Bvalue{\theta a-\theta d-\theta b_{\xi+1}>0}
\Bsupseteq\Bvalue{v-\theta d>0}
\Bsupseteq c_1.\cr}$$

\noindent By (g), $a_{\xi}\Bcap c_1\ptG a\Bsetminus(d\Bcup e_{\xi})$;
take $d_{\xi}\Bsubseteq a\Bsetminus(d\Bcup e_{\xi})$ \Gte\ with
$a_{\xi}\Bcap c_1$, and continue.

At the end of this induction, we have a disjoint family $\langle
d_{\xi}\rangle_{\xi<\kappa}$ in $\frak A_{a\Bsetminus d}$ such that
$d_{\xi}$ is \Gte\ with $a_{\xi}\Bcap c_1$ for every $\xi$.   But this
means that $a'=\sup_{\xi<\kappa}d_{\xi}$ is \Gte\ with $a\Bcap c_1$,
while $a'\Bsubseteq (a\Bsetminus d)\Bcap c_1$;  since
$d\Bcap a\Bcap c_1\ne 0$,
$G$ cannot be fully non-paradoxical.\ \Bang

Thus $\theta a=u=\sum_{\xi<\kappa}\theta a_{\xi}$ and the induction
continues.\ \Qed

\medskip

{\bf (i)} It follows that $\theta$ is order-continuous.   \Prf\
($\alpha$) If $B\subseteq\frak A$ is non-empty and upwards-directed and
has supremum $e$, then $\bigcup_{b\in B}\frak A_b$ is order-dense in
$\frak A_e$, so includes a partition of unity $A$ of $\frak A_e$;  now
(h) tells us that

\Centerline{$\theta e=\sum_{a\in A}\theta a\le\sup_{b\in B}\theta b$.}

\noindent   Since of course $\theta b\le\theta e$ for every $b\in B$,
$\theta e=\sup_{b\in B}\theta b$.   ($\beta$) If $B\subseteq\frak A$ is
non-empty and downwards-directed and has infimum $e$, then, using
($\alpha$), we see that

\Centerline{$\theta 1-\theta e
=\theta(1\Bsetminus e)=\sup_{b\in B}\theta(1\Bsetminus b)
=\sup_{b\in B}\theta 1-\theta b$,}

\noindent so that $\theta e=\inf_{b\in B}\theta b$.\ \Qed

\medskip

{\bf (j)} I still have to show that $\theta$ is unique.   Let
$\theta':\frak A\to L^{\infty}$ be any non-negative order-continuous
$G$-invariant additive function such that $\theta'c=\chi c$ for every
$c\in\frak C$.

\medskip

\quad{\bf (i)} Just as in (e) of this proof, but more easily, we
see that $\theta'(b\Bcap c)=\theta'b\times\chi c$ whenever
$b\in\frak A$ and $c\in\frak C$.

\medskip

\quad{\bf (ii)} If $\langle a_i\rangle_{i\in I}$ is a disjoint family in
$\frak A$ with supremum $a$, then
$\langle\sup_{i\in J}a_i\rangle_{J\subseteq I\text{ is finite}}$ is an
upwards-directed family with supremum $a$, so that

\Centerline{$\theta'a
=\sup_{J\subseteq I\text{ is finite}}\theta'
(\sup_{i\in J}a_i)
=\sup_{J\subseteq I\text{ is finite}}\sum_{i\in J}\theta'a_i
=\sum_{i\in I}\theta'a_i$.}

\medskip

\quad{\bf (iii)} $\theta'a=\theta'b$ whenever $a$ and $b$ are \Gte.
\Prf\ Take a partition $\langle a_i\rangle_{i\in I}$ of $a$ and a family
$\langle\pi_i\rangle_{i\in I}$ in $G$ such that
$\langle\pi_ia_i\rangle_{i\in I}$ is a partition of $b$.   Then

\Centerline{$\theta'a=\sum_{i\in I}\theta'a_i
=\sum_{i\in I}\theta'\pi_ia_i=\theta'b$.  \Qed}

\noindent Consequently $\theta'a\le\theta'b$ whenever $a\ptG b$.

\medskip

\quad{\bf (iv)} Take $a\in A^*$, $b\in\frak A$ and for $j$, $k\in\Bbb N$
set $c_{jk}=\Bvalue{\low{1:a}=j}\Bcap\Bvalue{\high{b:a}=k}$.  Then

\Centerline{$\high{b:a}\times\chi
c_{jk}\ge\theta'b\times\low{1:a}\times\chi
c_{jk}$.}

\noindent\Prf\ If $c_{jk}=0$ this is trivial;  suppose $c_{jk}\ne 0$.
Now we have sets $I$, $J$ such that $\#(I)=j$, $\#(J)\le k$, $a\Bcap
c_{jk}\ptG d$ for every $d\in I$, $e\ptG a$ for every $e\in J$, $I$ is
disjoint, and $b\Bcap c_{jk}\Bsubseteq\sup J$.   So

$$\eqalign{\theta'b\times\low{1:a}\times\chi c_{jk}
&=j\theta'b\times\chi c_{jk}
=j\theta'(b\Bcap c_{jk})
\le j\sum_{e\in J}\theta'(e\Bcap c_{jk})\cr
&\le jk\theta'(a\Bcap c_{jk})
\le k\sum_{d\in I}\theta'(d\Bcap c_{jk})
\le k\theta'c_{jk}\cr
&=k\chi c_{jk}
=\high{b:a}\times\chi c_{jk}. \text{ \Qed}\cr}$$

\noindent Summing over $j$ and $k$,
$\high{b:a}\ge\theta'b\times\low{1:a}$, that is, $\theta_ab\ge\theta'b$.
Taking the infimum over $a$, $\theta b\ge\theta'b$.   But also

\Centerline{$\theta b=\chi 1-\theta(1\Bsetminus b)
\le\chi 1-\theta'(1\Bsetminus b)=\theta'b$,}

\noindent so $\theta b=\theta'b$.   As $b$ is arbitrary,
$\theta=\theta'$.   This completes the proof.
}%end of proof of 395N

\leader{395O}{}\cmmnt{ We have reached the summit.   The rest of the
section is a list of easy corollaries.

\medskip

\noindent}{\bf Theorem} Let $\frak A$ be a Dedekind complete Boolean
algebra, not $\{0\}$, and $G$ a fully non-paradoxical subgroup of
$\Aut\frak A$.   Then there is a $G$-invariant additive functional
$\nu:\frak A\to[0,1]$ such that $\nu 1=1$.

\proof{ Let $\frak C$ be the fixed-point subalgebra of $G$, and
$\theta:\frak A\to L^{\infty}(\frak C)$ the function of 395N.   By
311D, there is a ring homomorphism $\nu_0:\frak C\to\{0,1\}$ such
that $\nu_01=1$;  now $\nu_0$ can also be regarded as an additive
functional from $\frak C$ to $\Bbb R$.   Let
$f_0:L^{\infty}(\frak C)\to\Bbb R$ be the
corresponding positive linear functional (363K).  Set $\nu=f_0\theta$.
Then $\nu$ is order-preserving and additive because $f_0$ and $\theta$
are, $\nu 1=f_0(\chi 1)=\nu_01>0$, and $\nu$ is $G$-invariant because
$\theta$ is.

}%end of proof of 395O

\vleader{60pt}{395P}{Theorem} Let $\frak A$ be a Dedekind complete Boolean
algebra and $G$ a fully non-paradoxical subgroup of $\Aut\frak A$ with
fixed-point subalgebra $\frak C$.   Then the following are
equiveridical:

(i) $\frak A$ is a measurable algebra;

(ii) $\frak C$ is a measurable algebra;

(iii) there is a strictly positive $G$-invariant countably additive
real-valued functional on $\frak A$.

\proof{ (iii)$\Rightarrow$(i)$\Rightarrow$(ii) are trivial.   For
(ii)$\Rightarrow$(iii), let $\theta:\frak A\to L^{\infty}(\frak C)$ be
the function of 395N, and $\bar\nu:\frak C\to\Bbb R$ a strictly positive
countably additive functional.
Let $f:L^{\infty}(\frak C)\to\Bbb R$ be the corresponding linear
operator;  then $f$ is sequentially order-continuous (363K again).   Set
$\bar\mu=f\theta$.   Then $\bar\mu$ is additive and order-preserving and
sequentially order-continuous because $f$ and $\theta$ are.   It is also
strictly positive, because if $a\in\frak A\setminus\{0\}$ then
$\theta a>0$ (395N(ii)), that is, there is some $\delta>0$ such that
$\Bvalue{\theta a>\delta}\ne 0$, so that

\Centerline{$\bar\mu a\ge\delta\bar\nu\Bvalue{\theta a>\delta}>0$.}

\noindent Finally, $\bar\mu$ is $G$-invariant because $\theta$ is.
}%end of proof of 395P

\leader{395Q}{Corollary:  Kawada's theorem} Let $\frak A$ be a Dedekind
complete Boolean algebra such that $\Aut\frak A$ has a subgroup which
is ergodic and fully non-paradoxical.   Then $\frak A$ is measurable.

\proof{ By 395Gf, this is the case $\frak C=\{0,1\}$ of 395P.
}%end of proof of 395Q

\leader{395R}{}\cmmnt{ Thus the existence of an ergodic fully
non-paradoxical subgroup is a sufficient condition for a Dedekind
complete Boolean algebra to be measurable.   It is not quite necessary,
because if a measure algebra $\frak A$ is not homogeneous then its
automorphism group is not ergodic.
But for homogeneous algebras the condition is necessary as well as
sufficient, by the following result.

\medskip

\noindent}{\bf Proposition} If $(\frak A,\bar\mu)$ is a homogeneous
totally finite measure algebra, $\AmuA$ is ergodic.

\proof{ If $\frak A=\{0,1\}$ this is trivial.   Otherwise, $\frak A$ is
atomless.   If $a$, $b\in\frak A\setminus\{0,1\}$, set
$\gamma=\min(\bar\mu a,\bar\mu b)$;  then there are $a'\Bsubseteq a$ and
$b'\Bsubseteq b$ such that $\bar\mu a'=\bar\mu b'=\gamma$.
By 383Fb, there is a $\pi\in G$ such
that $\pi a'=b'$, so that $\pi a\Bcap a\ne 0$.   As $b$ is arbitrary,
$\sup_{\pi\in G}\pi a=1$;  as $a$ is arbitrary, $G$ is ergodic.
}%end of proof of 395R

\exercises{
\leader{395X}{Basic exercises (a)}
%\spheader 395Xa
Re-write the section on the assumption that every group $G$ is ergodic,
so that $L^0(\frak C)$ may be
identified with $\Bbb R$, the functions $\high{\ldots}$ and
$\low{\ldots}$ become real-valued, the functionals $\theta_a$ (395N)
become submeasures and $\theta$ becomes a measure.
%-

\spheader 395Xb Let $\frak A$ be a Dedekind complete Boolean algebra and
$G$ a subgroup of $\Aut\frak A$ with fixed-point subalgebra $\frak C$.
Suppose that $\familyiI{c_i}$ is a partition of unity in $\frak C$ and
that $a$, $b\in\frak A$ are such that $a\Bcap c_i\ptG b$ for every
$i\in I$.   Show that $a\ptG b$.
%395G

\spheader 395Xc\dvAformerly{3{}94Xc}
Let $\frak A$ be a Dedekind complete Boolean algebra and
$G$ a subgroup of $\Aut\frak A$ with fixed-point subalgebra $\frak C$.
Show that $\frak A$ is relatively atomless over $\frak C$ iff the full
subgroup generated by $G$ has many involutions (definition:  382O).
%395H

\spheader 395Xd
Let $\frak A$ be a Dedekind complete Boolean algebra and $G$ a fully
non-paradoxical subgroup of $\Aut\frak A$ with fixed-point subalgebra
$\frak C$.   Show that the following are equiveridical:
(i) $\frak A$ is chargeable (definition:  391Bb);
(ii) $\frak C$ is chargeable;  (iii) there is a strictly positive
$G$-invariant real-valued additive functional on $\frak A$.
%395N

\spheader 395Xe
Let $\frak A$ be a Dedekind complete Boolean algebra and $G$ a fully
non-paradoxical subgroup of $\Aut\frak A$ with fixed-point subalgebra
$\frak C$.   Show that the following are equiveridical:
(i) there is a non-zero completely additive functional on $\frak A$;
(ii) there is a non-zero completely additive functional on $\frak C$;
(iii) there is a non-zero $G$-invariant completely additive functional
on $\frak A$.
%395N

\spheader 395Xf Let $\frak A$ be a ccc Dedekind complete Boolean
algebra.   Show that it is a measurable algebra iff there is a fully
non-paradoxical subgroup $G$ of $\Aut\frak A$ such that the fixed-point
subalgebra of $G$ is purely atomic.
%395R

\spheader 395Xg Let $(\frak A,\bar\mu)$ be a localizable measure
algebra.    Show that the following are equiveridical:
(i) $\AmuA$ is ergodic;  (ii) $\frak A$ is quasi-homogeneous
in the sense of 374G.
%395R

\spheader 395Xh Let $(\frak A,\bar\mu)$ be a localizable measure
algebra.   Show that $\AmuA$ is fully non-paradoxical
iff (i) for every infinite cardinal $\kappa$, the Maharam-type-$\kappa$
component of $\frak A$ (definition:  332Gb) has finite measure (ii) for
every $\gamma\in\ooint{0,\infty}$ there are only finitely many atoms of
measure $\gamma$.
%395R

\spheader 395Xi Let $\frak A$ be a Boolean algebra, $G$ a subgroup of
$\Aut\frak A$, and $G^*$ the full subgroup of $\Aut\frak A$ generated by
$G$.   Show that $G^*$ is ergodic iff $G$ is ergodic.
%395G out of order query

\leader{395Y}{Further exercises (a)}
%\spheader 395Ya
Let $\frak A$ be a Dedekind complete Boolean algebra, $G$ a subgroup of
$\Aut\frak A$, and $G^*_{\tau}$ the full local semigroup generated by
$G$.   For $\phi$, $\psi\in G^*_{\tau}$, say that $\phi\le\psi$ if
$\psi$ extends $\phi$.   (i) Show that every member of $G^*_{\tau}$ can
be extended to a maximal member of $G^*_{\tau}$.   (ii) Show that $G$ is
fully non-paradoxical iff every maximal member of $G^*_{\tau}$ is
actually a Boolean automorphism of $\frak A$.
%395E

\spheader 395Yb Let $\frak A$ be a ccc Dedekind complete Boolean algebra
and $G$
a subgroup of $\Aut\frak A$.   Show that $G$ is fully non-paradoxical iff
$\sequencen{\pi_na_n}$ order*-converges to $0$ whenever $\sequencen{a_n}$
is a disjoint sequence in $\frak A$ and $\sequencen{\pi_n}$
is a sequence in $G$.
%395E  easy from 395D(ii) but we don't use order*-cgt much

\spheader 395Yc Let $\frak A$ be a Dedekind complete Boolean algebra and
$G$ a fully non-paradoxical subgroup of $\Aut\frak A$ with fixed-point
subalgebra $\frak C$.   (i)
Show that $\frak A$ is ccc iff $\frak C$ is ccc.
\Hint{if $\frak C$ is ccc, $L^{\infty}(\frak C)$ has the countable sup
property.}
(ii) Show that $\frak A$ is \wsid\ iff $\frak C$ is.
(iii) Show that $\frak A$ is a Maharam algebra iff $\frak C$ is.
%395N   %mt39bits

\spheader 395Yd Let $\frak A$ be a Dedekind complete Boolean algebra,
$G$ an ergodic subgroup of $\Aut\frak A$, and $G^*_{\tau}$ the full local
semigroup generated by $G$.   Suppose that there is a non-zero
$a\in\frak A$ for which there is no $\phi\in G^*_{\tau}$ such that
$\phi a\Bsubset a$.   Show that there is a measure $\bar\mu$ such that
$(\frak A,\bar\mu)$ is a localizable measure algebra.   \Hint{show that
$\frak A_a$ is a measurable algebra.}
%395Q

\spheader 395Ye Show that there are a
semi-finite measure algebra $(\frak A,\bar\mu)$
and a subgroup $G$ of $\AmuA$ such that $G$ is not ergodic but
has fixed-point algebra $\{0,1\}$.
%395G mt39bits out of order query
}%end of exercises

\leader{395Z}{Problem} Suppose that $\frak A$ is a Dedekind complete
Boolean algebra, not $\{0\}$, and $G$ a subgroup of $\Aut\frak A$ such
that whenever $\langle a_i\rangle_{i\le n}$ is a finite partition of
unity in $\frak A$ and we are given $\pi_i$, $\pi'_i\in G$ for every
$i\le n$, then the elements $\pi_0a_0$, $\pi'_0a_0$, $\pi_1a_1$,
$\pi'_1a_1,\ldots,\pi'_na_n$ are not all disjoint.   Must there be a
non-zero non-negative $G$-invariant finitely additive functional
$\theta$ on $\frak A$?

\cmmnt{(See `Tarski's theorem' in the notes below.)}

\endnotes{\Notesheader{395} Regarded as a sufficient condition for
measurability, Kawada's theorem suffers from the obvious defect that it
is going to be rather rarely that we can verify the existence of an
ergodic fully non-paradoxical group of automorphisms without having some
quite different reason for supposing that our algebra is measurable.
If we think of it as a criterion for the existence of a
$G$-invariant measure, rather than as a criterion for measurability in
the abstract, it seems to make better sense.   But if we know from the
start that the algebra $\frak A$ is measurable, the argument
short-circuits, as we shall see in \S396.

I take the trouble to include the `$\tau$' in every `\Gte',
`$G^*_{\tau}$' and `$\ptG$' because there are important variations on
the concept, in which the partitions $\langle a_i\rangle_{i\in I}$ of
395A are required to be finite or countable.   Indeed {\bf Tarski's
theorem} relies on one of these.   I spell it out because it is close to
Kawada's in spirit, though there are significant differences in the
ideas needed in the proof:

\inset{Let $X$ be a set and $G$ a subgroup of
$\Aut\Cal PX$.   Then the following are equiveridical:  (i) there is a
$G$-invariant additive functional $\theta:\Cal PX\to[0,1]$ such that
$\theta A=1$;   (ii) there are no
$A_0,\ldots,A_n$, $\pi_0,\ldots,\pi_n$, $\pi'_0,\ldots,\pi'_n$ such that
$A_0,\ldots,A_n$ are subsets of $X$ covering $X$,
$\pi_0,\ldots,\pi'_n$ belong to $G$, and $\pi_0[A_0]$, $\pi'_0[A_0]$,
$\pi_1[A_1]$, $\pi'_1[A_1],\ldots,\pi'_n[A_n]$ are all disjoint.}

\noindent For a proof, see 449L in Volume 4;  for an illuminating
discussion of this
theorem, see {\smc Wagon 85}, Chapter 9.   But it seems to be unknown
whether the natural translation of this result is valid in all Dedekind
complete Boolean algebras (395Z).   Note that we are looking for
theorems which do not depend on any special properties of the group $G$
or the Boolean algebra $\frak A$.   For abelian or `amenable' groups, or
\wsid\ algebras, for instance, much more can be done, as described in
396Ya and \S449.

The methods of this section can, however, be used to prove similar
results for {\it countable} groups of automorphisms on Dedekind
$\sigma$-complete Boolean algebras;  I will return to such questions in
\S448.   The
presentation here owes a good deal to {\smc Nadkarni 90} and something
to {\smc Becker \& Kechris 96}.

As noted, Kawada ({\smc Kawada 1944}) treated the case in which the group
$G$ of automorphisms is ergodic, that is, the fixed-point subalgebra
$\frak C$ is trivial.   Under this hypothesis the proof is of course
very much
simpler.   (You may find it useful to reconstruct the original version,
as suggested in 395Xa.)   I give the more general argument partly for
the sake of 395O, partly to separate out the steps which really need
ergodicity from those which depend only on
non-paradoxicality, partly to prepare the ground for the countable
version in the next volume, partly to show off the power of the
construction in \S364, and
partly to get you used to `Boolean-valued' arguments.   A bolder use of
language could indeed simplify some formulae slightly by writing (for
instance) $\Bvalue{k\high{a_0:a}<\low{1:a}}$ in place of
$\Bvalue{\low{1:a}-k\high{a_0:a}>0}$ (see part (f) of the proof of
395N).   As in \S388, the differences involved in the extension to
non-ergodic groups are, in a sense, just a matter of
technique;  but this time the technique is more obtrusive.   In \S556 of
Volume 5 I will try to explain a general approach to questions of this
kind, using metamathematical ideas.
}%end of notes

\discrpage

\frfilename{mt396.tex}
\versiondate{15.8.08}
\copyrightdate{1997}

\def\chaptername{Measurable algebras}
\def\sectionname{The Hajian-Ito theorem}

\newsection{396}

In the notes to the last section, I said that the argument there
short-circuits if we are told that we are dealing with a measurable
algebra.   The point is that in this case there is a much simpler
criterion for the existence of a $G$-invariant measure (396B(ii)), with
a proof which is independent of \S395 in all its non-trivial parts,
which makes it easy to prove that non-paradoxicality is sufficient as
well as necessary.

\leader{396A}{Lemma} Let $(\frak A,\bar\mu)$ be a localizable measure
algebra.

(a) Let $\pi\in\Aut\frak A$ be a Boolean automorphism\cmmnt{ (not
necessarily measure-preserving)}, and $T_{\pi}$ the corresponding
Riesz homomorphism from $L^0=L^0(\frak A)$ to itself\cmmnt{ (364P)}.
Then there is a unique $w_{\pi}\in(L^0)^+$ such that $\int w_{\pi}\times
v=\int T_{\pi}v$ for every $v\in(L^0)^+$.

(b) If $\phi$, $\pi\in\Aut\frak A$ then $w_{\pi\phi}=w_{\phi}\times
T_{\phi^{-1}}w_{\pi}$.

(c) For each $\pi\in\Aut\frak A$ we have a norm-preserving isomorphism
$U_{\pi}$ from $L^2=L^2(\frak A,\bar\mu)$ to itself defined by
setting

\Centerline{$U_{\pi}v=T_{\pi}v\times\sqrt{w_{\pi^{-1}}}$}

\noindent for every $v\in L^2$, and $U_{\pi\phi}=U_{\pi}U_{\phi}$ for
all $\pi$, $\phi\in\Aut\frak A$.

\proof{{\bf (a)} Set $\bar\nu a=\bar\mu(\pi a)$ for $a\in\frak A$.
Then $(\frak A,\bar\nu)$ is a semi-finite measure algebra.   \Prf\
$\bar\nu 0=\bar\mu 0=0$;  if $\sequencen{a_n}$ is a disjoint sequence in
$\frak A$ with supremum $a$, then $\sequencen{\pi a_n}$ is disjoint and
(because $\pi$ is sequentially order-continuous)
$a=\sup_{n\in\Bbb N}\pi a_n$, so
$\bar\nu a=\sum_{n=0}^{\infty}\bar\nu a_n$;   if $a\ne 0$ then
$\pi a\ne 0$ so $\bar\nu a>0$.    Thus $(\frak A,\bar\nu)$ is a measure
algebra.   If $a\ne 0$ there is a $b\Bsubseteq\pi a$ such that
$0<\bar\mu b<\infty$, and now $\pi^{-1}b\Bsubseteq a$ and
$0<\bar\nu(\pi^{-1}b)<\infty$;  thus $\bar\nu$ is semi-finite.\ \Qed

By 365S, there is a unique $w_{\pi}\in(L^0)^+$ such that
$\int_aw_{\pi}=\bar\mu(\pi a)$ for every $a\in\frak A$.   If we look at

\Centerline{$W=\{v:v\in (L^0)^+$, $\int v\times w_{\pi}
=\int T_{\pi}v\}$,}

\noindent we see that $W$ contains $\chi a$ for every $a\in\frak A$,
that $v+v'\in W$ and $\alpha v\in W$ whenever $v$, $v'\in W$ and
$\alpha\ge 0$, and that $\sup_{n\in\Bbb N}v_n\in W$ whenever
$\sequencen{v_n}$ is a non-decreasing sequence in $W$ which is bounded
above in $L^0$.   By 364Jd, $W=(L^0)^+$, as required.

\medskip

{\bf (b)} For any $v\in(L^0)^+$,

$$\eqalignno{\int w_{\pi\phi}\times v
&=\int T_{\pi\phi}v
=\int T_{\pi}T_{\phi}v\cr
\noalign{\noindent (364Pe)}
&=\int w_{\pi}\times T_{\phi}v
=\int T_{\phi}(T_{\phi^{-1}}w_{\pi}\times v)\cr
\noalign{\noindent (recalling that $T_{\phi}$ is multiplicative)}
&=\int w_{\phi}\times T_{\phi^{-1}}w_{\pi}\times v.\cr}$$

\noindent As $v$ is arbitrary (and $(\frak A,\bar\mu)$ is semi-finite),
$w_{\pi\phi}=w_{\phi}\times T_{\phi^{-1}}w_{\pi}$.

\woddheader{396A}{6}{2}{2}{36pt}

{\bf (c)(i)} For any $v\in L^0$,

\Centerline{$\int(T_{\pi}v\times\sqrt{w_{\pi^{-1}}})^2
=\int T_{\pi}v^2\times w_{\pi^{-1}}
=\int T_{\pi^{-1}}T_{\pi}v^2
=\int v^2$.}

\noindent So $U_{\pi}v\in L^2$
and $\|U_{\pi}v\|_2=\|v\|_2$ whenever $v\in L^2$, and $U_{\pi}$ is a
norm-preserving operator on $L^2$.

\medskip

\quad{\bf (ii)} Now consider $U_{\pi\phi}$.   For any $v\in L^2$, we
have

$$\eqalignno{U_{\pi}U_{\phi}v
&=T_{\pi}(T_{\phi}v\times\sqrt{w_{\phi^{-1}}})
      \times\sqrt{w_{\pi^{-1}}}\cr
&=T_{\pi}T_{\phi}v
  \times\sqrt{T_{\pi}w_{\phi^{-1}}\times w_{\pi^{-1}}}\cr
\noalign{\noindent (using 364Pd)}
&=T_{\pi\phi}v\times\sqrt{w_{\phi^{-1}\pi^{-1}}}\cr
\noalign{\noindent (by (b) above)}
&=U_{\pi\phi}v.\cr}$$

\noindent So $U_{\pi\phi}=U_{\pi}U_{\phi}$.

\medskip

\quad{\bf (iii)} Writing $\iota$ for the identity operator on $\frak A$,
we see that $T_{\iota}$ is the identity operator on $L^0$,
$w_{\iota}=\chi 1$ and $U_{\iota}$ is the identity operator on $L^2$.
Since $U_{\pi^{-1}}U_{\pi}=U_{\pi}U_{\pi^{-1}}=U_{\iota}$,
$U_{\pi}:L^2\to L^2$ is an isomorphism, with inverse $U_{\pi^{-1}}$, for
every $\pi\in\Aut\frak A$.
}%end of proof of 396A

\leader{396B}{Theorem}\cmmnt{ ({\smc Hajian \& Ito 69})} Let $\frak A$
be a measurable algebra and $G$ a subgroup of $\Aut\frak A$.   Then the
following are equiveridical:

(i) there is a $G$-invariant functional $\bar\nu$ such that
$(\frak A,\bar\nu)$ is a totally finite measure algebra;

(ii) whenever $a\in\frak A\setminus\{0\}$ and $\sequencen{\pi_n}$ is a
sequence in $G$, $\sequencen{\pi_na}$ is not disjoint;

(iii) $G$ is fully non-paradoxical\cmmnt{ (definition:  395E)}.

\proof{{\bf (a)} (i)$\Rightarrow$(iii) by the argument of 395F, and
(iii)$\Rightarrow$(ii) by the criterion (ii) of 395E.   So for the rest
of the proof I assume that (ii) is true and seek to prove (i).

\medskip

{\bf (b)} Let $\bar\mu$ be such that $(\frak A,\bar\mu)$ is a totally
finite measure algebra.   If $a\in\frak A\setminus\{0\}$, then
$\inf_{\pi\in G}\bar\mu(\pi a)>0$.   \Prf\Quer\ Otherwise, let
$\sequencen{\pi_n}$ be a sequence in $G$ such that
$\bar\mu\pi_na\le 2^{-n}$ for each $n\in\Bbb N$.   Set
$b_n=\sup_{k\ge n}\pi_ka$ for each
$n$;  then $\inf_{n\in\Bbb N}b_n=0$, so that

\Centerline{$\inf_{n\in\Bbb N}\pi b_n=0$,
\quad$\lim_{n\to\infty}\bar\mu(\pi\pi_na)=0$}

\noindent for every $\pi\in\Aut\frak A$.   Choose $\sequence{i}{n_i}$
inductively so that

\Centerline{$\bar\mu(\pi_{n_i}^{-1}\pi_{n_j}a)\le 2^{-j-2}\bar\mu a$}

\noindent whenever $i<j$.   Set

\Centerline{$c=a\Bsetminus\sup_{i<j}\pi_{n_i}^{-1}\pi_{n_j}a$.}

\noindent Because

\Centerline{$\sum_{j=1}^{\infty}\sum_{i=0}^{j-1}
  \bar\mu(\pi_{n_i}^{-1}\pi_{n_j}a)
<\bar\mu a$,}

\noindent $c\ne 0$, while $\pi_{n_i}c\Bcap\pi_{n_j}c=0$ whenever $i<j$,
contrary to the hypothesis (ii).\ \Bang\Qed

\medskip

{\bf (c)} For each $\pi\in G$, define $w_{\pi}\in L^0=L^0(\frak A)$ and
$U_{\pi}:L^2\to L^2$ as in 396A, where $L^2=L^2(\frak A,\bar\mu)$.   If
$a\in\frak A\setminus\{0\}$, then
$\inf_{\pi\in G}\int_a\sqrt{w_{\pi}}>0$.   \Prf\Quer\ Otherwise, there
is a sequence $\sequencen{\pi_n}$ in $G$ such that
$\int_av_n\le 4^{-n-2}\bar\mu a$ for every $n$, where
$v_n=\sqrt{w_{\pi_n}}$.   In this case,
$\bar\mu(a\Bcap\Bvalue{v_n\ge 2^{-n}})\le 2^{-n-2}\bar\mu a$ for every
$n$, so that $b=a\Bsetminus\sup_{n\in\Bbb N}\Bvalue{v_n\ge 2^{-n}}$ is
non-zero.   But now

\Centerline{$\bar\mu(\pi_nb)=\int_bw_{\pi_n}
=\int_bv_n^2\le 4^{-n}\bar\mu b\to 0$}

\noindent as $n\to\infty$, contradicting (b) above.\ \Bang\Qed

\medskip

{\bf (d)} Write $e=\chi 1$ for the standard weak order unit of $L^0$ or
$L^2$.   Let $C\subseteq L^2$ be the convex hull of $\{U_{\pi}e:\pi\in
G\}$.   Then $C$ and its norm closure $\overline{C}$ are $G$-invariant
in the sense that $U_{\pi}v\in C$, $U_{\pi}v'\in\overline{C}$ whenever
$v\in C$, $v'\in\overline{C}$ and $\pi\in G$.   By 3A5Md, there is a
unique $u_0\in\overline{C}$ such that $\|u_0\|_2\le\|u\|_2$ for every
$u\in\overline{C}$.   Now if $\pi\in G$, $U_{\pi}u_0\in\overline{C}$,
while $\|U_{\pi}u_0\|_2=\|u_0\|_2$; so $U_{\pi}u_0=u_0$.   Also, if
$a\in\frak A\setminus\{0\}$,

$$\eqalignno{\int_au_0
&\ge\inf_{u\in\overline{C}}\int_au
=\inf_{u\in C}\int_au\cr
\noalign{\noindent (because $u\mapsto\int_au$ is $\|\,\|_2$-continuous)}
&=\inf_{\pi\in G}\int_aU_{\pi}e
=\inf_{\pi\in G}\int_aT_{\pi}e\times\sqrt{w_{\pi^{-1}}}\cr
&=\inf_{\pi\in G}\int_a\sqrt{w_{\pi^{-1}}}
=\inf_{\pi\in G}\int_a\sqrt{w_{\pi}}
>0\cr}$$

\noindent by (c).   So $\Bvalue{u_0>0}=1$.

\medskip

{\bf (e)} For $a\in\frak A$, set $\bar\nu a=\int_au_0^2$.   Because
$u_0\in L^2$, $\bar\nu$ is a non-negative countably additive functional
on $\frak A$;  because $\Bvalue{u_0^2>0}=\Bvalue{u_0>0}=1$, $\bar\nu$ is
strictly positive, and $(\frak A,\bar\nu)$ is a totally finite measure
algebra.   Finally, $\bar\nu$ is $G$-invariant.   \Prf\ If $a\in\frak A$
and $\pi\in G$, then

$$\eqalignno{\bar\nu(\pi a)
&=\int_{\pi a}u_0^2
=\int u_0^2\times\chi(\pi a)
=\int T_{\pi}(T_{\pi^{-1}}u_0^2\times\chi a)\cr
&=\int w_{\pi}\times T_{\pi^{-1}}u_0^2\times\chi a
=\int_a(T_{\pi^{-1}}u_0\times\sqrt{w_{\pi}})^2\cr
&=\int_a(U_{\pi^{-1}}u_0)^2
=\int_au_0^2
=\bar\nu a.\text{ \Qed}\cr}$$

\noindent So (i) is true.
}%end of proof of 396B

\cmmnt{
\leader{396C}{Remark} If $\frak A$ is a Boolean algebra and $G$
a subgroup of $\Aut\frak A$, a non-zero element $a$ of $\frak A$ is
called {\bf weakly wandering} if there is a sequence $\sequencen{\pi_n}$
in $G$ such that $\sequencen{\pi_na}$ is disjoint.   Thus condition (ii)
of 396B may be read as `there is no weakly wandering element of
$\frak A$'.
}%end of comment

\exercises{\leader{396X}{Basic exercises (a)}
%\spheader 396Xa
Let $(\frak A,\bar\mu)$ be a totally finite measure algebra, and
$\pi:\frak A\to\frak A$ an order-continuous Boolean homomorphism.   Let
$T_{\pi}:L^0(\frak A)\to L^0(\frak A)$ be the corresponding Riesz
homomorphism.   Show that there is a unique $w_{\pi}\in L^1(\frak
A,\bar\mu)$ such that $\int T_{\pi}v=\int v\times w_{\pi}$ for every
$v\in L^0(\frak A)^+$.
%396A

\spheader 396Xb In 396A, show that the map
$\pi\mapsto U_{\pi}:\Aut\frak A\to\eurm B(L^2;L^2)$ is injective.
%396A

\spheader 396Xc Let $\frak A$ be a measurable algebra and $G$ a subgroup
of $\Aut\frak A$.   Suppose that there is a strictly positive
$G$-invariant finitely additive functional on $\frak A$.   Show that
there is a $G$-invariant $\bar\mu$ such that  $(\frak A,\bar\mu)$ is a
totally finite measure algebra.
%396B

\leader{396Y}{Further exercises (a)}
Let $\frak A$ be a \wsid\ Dedekind complete Boolean algebra and $G$ a
subgroup of $\Aut\frak A$.   For $a$, $b\in\frak A$, say that $a$ and
$b$ are {\bf $G$-equidecomposable} if there are {\it finite} partitions
of unity $\familyiI{a_i}$ in $\frak A_a$ and $\familyiI{b_i}$ in
$\frak A_b$, and a family $\familyiI{\pi_i}$ in $G$, such that
$\pi_ia_i=b_i$
for every $i\in I$.   Show that the following are equiveridical:  (i)
$G$ is fully non-paradoxical in the sense of 395E;  (ii) if
$\sequencen{a_n}$ is a disjoint sequence of mutually
$G$-equidecomposable elements of $\frak A$, they must all be $0$.
%396B
}%end of exercises

\endnotes{
\Notesheader{396}
I have separated these few pages from \S395 partly because \S395 was
already up to full weight and partly in order that the ideas here should
not be entirely overshadowed by those of the earlier section.   It will
be evident that the construction of the $U_{\pi}$ in 396A, providing us
with a faithful representation, acting on a Hilbert space, of the whole
group $\Aut\frak A$, is a basic tool for the study of that group.
}%end of notes

\frnewpage


%\ifresultsonly\else
\frfilename{mt39conc.tex}
\versiondate{21.10.07}
\copyrightdate{2007}

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}

\def\chaptername{Measurable algebras}
\def\sectionname{Concordance}

\newsection{}

I list here the section and paragraph numbers which have (to my
knowledge) appeared in print in references to this chapter, and which
have since been changed.

\medskip

\leader{}{3{}93B} The association of a metric with a strictly positive
submeasure, used in the 2003 and 2006 editions of Volume 4, is now in
392H and 393H.

\leader{}{3{}93C} 
The result that a non-negative additive functional on a
Boolean algebra can be factored through a measure algebra, used in the 2003
and 2006 editions of Volume 4, is now in 392I.

\leader{}{3{}93O} The note on control measures for
vector measures, referred to in the
2003 and 2006 editions of Volume 4, is now in 394Q.

\leader{}{\S3{}94} Kawada's theorem, referred to in the 2003 and 2006 
editions of Volume 4, is now in \S395.

\frnewpage
%\fi

\references{
\frfilename{mt3r.tex}
\versiondate{25.8.17}
\copyrightdate{1999}

% except for references in ...X., ...Y., ...Z. give locations in form
%  342L  rather than  342Lb
%  \S511 {\it notes\/}, \S512 {\it intro.}]

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
\gdef\chaptername{References}
\gdef\sectionname{References}
\gdef\headlinesectionname{References}

\centerline{\bf References for Volume 3}

\medskip

%A

Anderson I.\ [87] {\it Combinatorics of Finite Sets.}  Oxford U.P., 1987.
\cmmnt{[332Xk, 3A1K.]}

% Anzai H.\ [51] `Ergodic skew product transformations of the torus',
% % Osaka Math.\ J.\ 3 (1951) 83-99.

Anzai H.\ [51] `On an example of a measure-preserving transformation
which is not conjugate to its inverse',
Proc.\ Japanese Acad.\ Sci.\ 27 (1951) 517-522.
\cmmnt{[\S382 {\it notes\/}.]}

\medskip
%B

Balcar B., G\l\'owczy\'nski W. \& Jech T.\ [98] `The sequential
topology on complete Boolean algebras', Fundamenta Math.\ 155 (1998)
59-78.
\cmmnt{[393L, 393Q.]}

Balcar B., Jech T.\ \& Paz\'ak T.\ [05]
`Complete ccc Boolean algebras, the order sequential topology,
and a problem of von Neumann',
Bull.\ London Math.\ Soc.\ 37 (2005) 885-898.
\cmmnt{[393L, 393Q.]}
% arXiv.math.LO/0312473

Becker H.\ \& Kechris A.S.\ [96] {\it The descriptive set theory of
Polish group actions.}  Cambridge U.P., 1996
(London Math.\ Soc.\ Lecture Note Series 232).
\cmmnt{[\S395 {\it notes\/}.]}

Bekkali M.\ \& Bonnet R.\ [89] `Rigid Boolean algebras', pp.\ 637-678 in
{\smc Monk 89}.
\cmmnt{[384L.]}

Bellow A.\ \& K\"olzow D.\ [76] {\it Measure Theory, Oberwolfach 1975.}
Springer, 1976 (Lecture Notes in Mathematics 541).

Bhaskara Rao, K.P.S.\ \& Bhaskara Rao, M.\ {\it Theory of Charges.}
Academic, 1983.
%326H

Billingsley P.\ [65] {\it Ergodic Theory and Information.}  Wiley, 1965.
\cmmnt{[\S372 {\it notes\/}.]}

Bollob\'as B.\ [79] {\it Graph Theory.}   Springer, 1979.
\cmmnt{[332Xk, 3A1K.]}

Bose R.C.\ \& Manvel B.\ [84] {\it Introduction to Combinatorial Theory.}   Wiley, 1984.
\cmmnt{[3A1K.]}

Bourbaki N.\ [66] {\it General Topology.}  Hermann/Addison-Wesley, 1968.
\cmmnt{[\S3A3, \S3A4.]}

Bourbaki N. [68] {\it Theory of Sets.}   Hermann/Addison-Wesley, 1968.
\cmmnt{[\S315 {\it notes\/}.]}

Bourbaki N. [87] {\it Topological Vector Spaces.}  Springer, 1987.
\cmmnt{[\S3A5.]}

Bukhvalov A.V.\ [95] `Optimization without compactness, and its
applications', pp.\ 95-112 in
{\smc Huijsmans Kaashoek Luxemburg \& Pagter 95}.
\cmmnt{[367U.]}

Burke M.R.\ [93] `Liftings for Lebesgue measure', pp.\ 119-150 in
{\smc Judah 93}.   \cmmnt{[341L, 345F.]}

Burke M.R.\ [n95] `Consistent liftings', privately circulated, 1995.
\cmmnt{[346Ya.]}

Burnside W.\ [1911] {\it Theory of Groups of Finite Order.}   Cambridge
U.P., 1911 (reprinted by Dover, 1955).
\cmmnt{[\S384 {\it notes\/}.]}

\medskip
%C

Chacon R.V. [69]
`Weakly mixing transformations which are not strongly mixing',
Proc.\ Amer.\ Math.\ Soc.\ 22 (1969) 559-562.
\cmmnt{[372R.]}

Chacon R.V.\ \& Krengel U.\ [64] `Linear modulus of a linear
operator', Proc.\ Amer.\ Math.\ Soc.\ 15 (1964) 553-559.
\cmmnt{[\S371 {\it notes\/}.]}

Choksi J.R.\ \& Prasad V.S.\ [82] `Ergodic theory of homogeneous measure
algebras', pp.\ 367-408 of {\smc K\"olzow \& Maharam-Stone 82}.
\cmmnt{[383I.]}

%Christenson C.O.\ \& Voxmann L.\ [77] {\it Aspects of Topology.}
%Dekker, 1977.

Cohn H.\ [06] `A short proof of the simple continued fraction expansion of
$e$', Amer.\ Math.\ Monthly 113 (2006) 57-62;
{\tt arXiv:math.NT/0601660}.
\cmmnt{[372L.]}

Coleman A.J.\ \& Ribenboim P.\ [67] (eds.) {\it Proceedings of the Symposium in Analysis, Queen's University, June 1967.}  Queen's University, Kingston, Ontario, 1967.

Comfort W.W.\ \& Negrepontis S.\ [82] {\it Chain Conditions in
Topology.}   Cambridge U.P., 1982.   \cmmnt{[\S391 {\it notes\/}.]}

\leaveitout{
Cornfeld I.P., Fomin S.V.\ \& Sina\v\i Ya.G.\ [82] {\it Ergodic Theory.}
Springer, 1982 (Grundlehren der mathematischen Wissenschaften 245)
}

Czisz\'ar I.\ [67] `Information-type measures of difference of
probability distributions and indirect observations',
Studia Scientiarum Math.\ Hungarica 2 (1967) 299-318.   \cmmnt{[386G.]}

\medskip
%D

Davey B.A. \& Priestley H.A. [90] `Introduction to Lattices and Order',
Cambridge U.P., 1990.  \cmmnt{[3A6C.]}

%Dow A.\ \& Stepr\=ans J.\ [93] `The $\sigma$-linkedness of the measure algebra', Canad.\ Math.\ Bulletin 37 (1993) 42-45.

Dugundji J.\ [66] {\it Topology.}  Allyn \& Bacon, 1966.
\cmmnt{[\S3A3, 3A4Bb.]}

Dunford N.\ [1936] `Integration and linear operators',
Trans.\ Amer.\ Math.\ Soc.\ 40 (1936) 474-494.
\cmmnt{[\S376 {\it notes\/}.]}

Dunford N.\ \& Schwartz J.T.\ [57] {\it Linear Operators I.}  Wiley,
1957 (reprinted 1988).
\cmmnt{[\S356 {\it notes\/}, \S371 {\it notes\/}, \S376 {\it notes\/},
\S3A5.]}

Dye H.A.\ [59] `On groups of measure preserving transformations I',
Amer.\ J.\ Math.\ 81 (1959) 119-159.   \cmmnt{[\S388 {\it notes\/}.]}

\medskip
%E

Eigen S.J.\ [82] `The group of measure-preserving transformations of
$[0,1]$ has no outer automorphisms', Math.\ Ann.\ 259 (1982) 259-270.
\cmmnt{[384D.]}

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).  \cmmnt{[\S3A3, \S3A4.]}

Enderton H.B.\ [77] {\it Elements of Set Theory.}   Academic, 1977.
\cmmnt{[\S3A1.]}
%QA248.E6  7850 2996 6

Erd\H{o}s P.\ \& Oxtoby J.C.\ [55] `Partitions of the plane into sets
having positive measure in every non-null product set', Trans.\ Amer.\
Math.\ Soc.\ 79 (1955) 91-102.   \cmmnt{[\S325 {\it notes\/}.]}

\medskip
%F

%Farah I.\ [04] `Examples of $\epsilon$-exhaustive pathological
%submeasures', Fundamenta Math.\ 181 (2004) 257-272.

Fathi A.\ [78] `Le groupe des transformations de $[0,1]$ qui
pr\'eservent la mesure de Lebesgue est un groupe simple', Israel J.\
Math.\ 29 (1978) 302-308.  \cmmnt{[\S382 {\it notes\/}, 383I.]}

Fremlin D.H.\ [74a] {\it Topological Riesz Spaces and
Measure Theory.}   Cambridge U.P., 1974.
\cmmnt{[Chap.\ 35 {\it intro\/}., 354Yb, \S355 {\it notes\/},
\S356 {\it notes\/}, \S363 {\it notes\/}, \S365 {\it notes\/},
\S371 {\it notes\/}, \S376 {\it notes\/}.]}

Fremlin D.H.\ [74b] `A characterization of $L$-spaces', Indag.\ Math.\
36 (1974) 270-275.  \cmmnt{[\S371 {\it notes\/}.]}

%Fremlin D.H.\ [84] {\it Consequences of Martin's Axiom.}
%Cambridge U.P., 1984.

\leaveitout{Fremlin D.H.\ [89] `Measure algebras', pp.\ 876-980 in
{\smc Monk 89}.}

Fremlin D.H., de Pagter B.\ \&  Ricker W.J.\ [05]
`Sequential closedness of Boolean algebras of projections in Banach
spaces', Studia Math.\ 167 (2005) 45-62.
\cmmnt{[\S323 {\it notes\/}.]}

Frol\'\i k Z.\ [68] `Fixed points of maps of extremally disconnected
spaces and complete Boolean algebras', Bull.\ Acad.\ Polon.\ Sci.\ 16
(1968) 269-275.  \cmmnt{[382E.]}

\medskip
%G

Gaal S.A.\ [64] {\it Point Set Topology.}  Academic, 1964.
\cmmnt{[\S3A3, \S3A4.]}

Gaifman H.\ [64] `Concerning measures on Boolean algebras', Pacific J.\
Math.\ 14 (1964) 61-73.  \cmmnt{[\S391 {\it notes\/}.]}

Gale D. [60] {\it The theory of linear economic models.}  McGraw-Hill,
1960.  \cmmnt{[3A5D.]}

Gnedenko B.V.\ \& Kolmogorov A.N.\ [54] {\it Limit Distributions for
Sums of Independent Random Variables.}  Addison-Wesley, 1954.
\cmmnt{[\S342 {\it notes\/}.]}

Graf S.\ \& Weizs\"acker H.von [76] `On the existence of lower densities in non-complete measure spaces',
pp.\ 155-158 in {\smc Bellow \& K\"olzow 76}.  \cmmnt{[341L.]}

\medskip%H

Hajian A.\ \& Ito Y.\ [69]
`Weakly wandering sets and invariant measures for a group of
transformations', J.\ of Math.\ and Mech.\ 18 (1969) 1203-1216.
\cmmnt{[396B.]}

Hajian A., Ito Y.\ \& Kakutani S.\ [75] `Full groups and a theorem of
Dye', Advances in Math.\ 17 (1975) 48-59.
\cmmnt{[\S388 {\it notes\/}.]}

Halmos P.R.\ [1948] `The range of a vector measure', Bull.\ Amer.\ Math.\
Soc.\ 54 (1948) 416-421.
\cmmnt{[326Yk.]}

Halmos P.R.\ [60] {\it Naive Set Theory.} Van Nostrand, 1960.
\cmmnt{[3A1D.]}

\leaveitout{
Hopf E.\ [1932] `Theory of measures and invariant integrals', Trans.\
Amer.\ Math.\ Soc.\ 34 (1932) 372-393.
}

Huijsmans C.B., Kaashoek M.A., Luxemburg W.A.J.\ \& de Pagter B.\ [95]
(eds.) {\it Operator Theory in Function Spaces and Banach Lattices.}
Birkh\"auser, 1995.

\medskip %I

Ionescu Tulcea C.\ \& Ionescu Tulcea A.\ [69] {\it Topics in the Theory
of Lifting.}  Springer, 1969.  \cmmnt{[\S341 {\it notes\/}.]}

\medskip %J

James I.M.\ [87] {\it Topological and Uniform Spaces.}  Springer, 1987.
\cmmnt{[\S3A3, \S3A4.]}

Jech T.\ [03] {\it Set Theory, Millennium Edition.}  Springer, 2002.
\cmmnt{[\S3A1.]}

Jech T.\ [08] `Algebraic characterizations of measure algebras',
Proc.\ Amer.\ Math.\ Soc.\ 136 (2008) 1285-1294.
%{\tt http://www.math.psu.edu/jech/preprints/papers.html}
\cmmnt{[393Xj.]}

Johnson R.A.\ [80] `Strong liftings which are not Borel liftings',
Proc.\ Amer.\ Math.\ Soc.\ 80 (1980) 234-236.   \cmmnt{[345F.]}
%Russell A. Johnson

Judah H.\ [93] (ed.) {\it Proceedings of the Bar-Ilan Conference on Set
Theory and the Reals, 1991.} Amer.\ Math.\ Soc.\ (Israel Mathematical
Conference Proceedings 6), 1993.

\medskip %K

Kakutani S.\ [1941] `Concrete representation of abstract $L$-spaces
and the mean ergodic theorem', Annals of Math. 42 (1941) 523-537.
\cmmnt{[369E.]}

Kalton N.J., Peck N.T.\ \& Roberts J.W.\ [84] `An F-space sampler',
Cambridge U.P., 1984.  \cmmnt{[\S375 {\it notes\/}.]}

Kalton N.J.\ \& Roberts J.W.\ [83] `Uniformly exhaustive submeasures and
nearly additive set functions', Trans.\ Amer.\ Math.\ Soc.\ 278 (1983)
803-816.  \cmmnt{[392D, \S392 {\it notes\/}.]}

Kantorovich L.V., Vulikh B.Z. \& Pinsker A.G.\ [50]
{\it Functional Analysis in
Partially Ordered Spaces}, Gostekhizdat, 1950.
\cmmnt{[391D.]}

Kawada Y.\ [1944] `\"Uber die Existenz der invarianten Integrale', Jap.\
J.\ Math.\ 19 (1944) 81-95.  \cmmnt{[\S395 {\it notes\/}.]}

Kelley J.L.\ [59] `Measures on Boolean algebras', Pacific J.\ Math.\ 9
(1959) 1165-1177.  \cmmnt{[\S391 {\it notes\/}.]}

\leaveitout{
Kelley J.L. \& Namioka I.\ [76] {\it Linear Topological Spaces.}
Springer, 1976.
}

\leaveitout{
King J.L.F.\ [00] `The generic transformation has roots of all orders',
Colloquium Mathematicum 84/85  (2000) 521-547.
  % if  \frak A  is the Lebesgue measure algebra
  % a comeager set in the Polish group
  % \Aut_{\bar\mu}\frak A   has roots of all orders
}

Kolmogorov A.N. [58] `New metric invariants of transitive dynamical
systems and automorphisms of Lebesgue spaces', Dokl.\ Akad.\ Nauk SSSR
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%Wendt S IN5 EL24 T IT

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\medskip %L

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\medskip %M

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\medskip %N

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\medskip %O

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\medskip%P

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\medskip %Q%R

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\medskip %S

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%Store A	QA 611.S3    6640 0159 9

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\medskip %T

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\medskip %U%V

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\medskip %W

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\medskip %X%Y%Z

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}%end of references


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