% 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{mt53.tex}
\versiondate{30.8.14}
\copyrightdate{2007}

\def\chaptername{Topologies and measures III}

\newchapter{53}

In this chapter I return to the concerns of earlier volumes, looking for
results which can be expressed in the language so far developed in this
volume.
In Chapter 43 I examined relationships between measure-theoretic and
topological properties.   The concepts we now have available (in
particular, the notion of `precaliber') make it possible to extend this
work in a new direction, seeking to understand the possible Maharam
types of measures on a given topological space.
\S531 deals with general Radon measures;  new patterns arise if we
restrict ourselves to completion regular Radon measures (\S532).
In \S533 I give a brief account of some further
results depending on assumptions
concerning the cardinals examined in Chapter 52, including
notes on uniformly regular measures and a description
of the cardinals $\kappa$ for which $\BbbR^{\kappa}$ is measure-compact
(533J).

In \S534 I set out the elementary theory of `strong measure zero' ideals in
uniform spaces, concentrating on aspects which can be studied in terms of
concepts already introduced.   Here there are some very natural questions
which have not as far as I know been answered (534Z).   
In the same section I run through elementary
properties of Hausdorff measures when examined in the light of
the concepts in Chapter 52.   In \S535 I look at
liftings and strong liftings, extending the results of \S\S341 and 453;
in particular, asking which
non-complete probability spaces have liftings.   In \S536 I run over
what is known about Alexandra Bellow's problem concerning pointwise
compact sets of continuous functions, mentioned in \S463.   With a little
help from special axioms, there are some striking possibilities concerning
repeated integrals, which I examine in \S537.
Moving into new territory, I devote a section (\S538) to a study of special
types of filter on $\Bbb N$ associated with measure-theoretic phenomena,
and to medial limits.
In \S539, I complete my account of the result of B.Balcar, T.Jech and
T.Paz\'ak that it is consistent to suppose that every Dedekind complete ccc
\wsid\ Boolean algebra is a Maharam algebra, and work through applications
of the methods of Chapter 52 to Maharam submeasures and algebras.

\discrpage

\frfilename{mt531.tex}
\versiondate{27.2.24}
\copyrightdate{2003}

\def\PR{P_{\text{R}}}

\def\chaptername{Topologies and measures III}
\def\sectionname{Maharam types of Radon measures}

\newsection{531}

In the introduction to \S434 I asked

\Centerline{\it What kinds of measures can arise on what kinds of
topological space?}

\noindent In \S\S434-435, and again in \S438, I considered a variety of
topological properties and their relations with measure-theoretic
properties of Borel and Baire measures.   I passed over, however, some
natural questions concerning possible Maharam types, to which I now
return.   For a given Hausdorff space $X$, the possible measure algebras
of totally finite Radon measures on $X$ can be described in terms of the
set $\MahR(X)$ of Maharam types of \Mth\ Radon probability measures on
$X$ (531F).   For $X\ne\emptyset$, $\MahR(X)$ is of the form
$\{0\}\cup\coint{\omega,\kappa^*}$ for some infinite cardinal $\kappa^*$
(531Ef).   In 531E and 531G I give basic results from which $\MahR(X)$
can often be determined;  for obvious reasons we are primarily concerned
with compact spaces $X$.   In more abstract contexts, there are striking
relationships between precalibers of measure algebras, the sets
$\MahR(X)$ and continuous surjections onto powers of $\{0,1\}$, which I
examine in 531L-531M, 531T and 531V.
Intertwined with these, we have results relating the
character of $X$ to $\MahR(X)$ (531N-531O).   The arguments here depend
on an analysis of the structure of homogeneous measure algebras (531J,
531K, 531R).

%531A%531B easy basics
%531C M type of product
%531D defn of MahR
%531E%531F%531G%531H basic props of MahR
%531I notation
%531J%531K lemmas
%531L%531M%531V%531N%531O precalibers and Haydon's property
%531Q HS HL type omega_1
   %531R lemma
   %531S%531T with m_K

\leader{531A}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon
measure space with measure algebra $(\frak A,\bar\mu)$.

(a)\cmmnt{ The Maharam type} $\tau(\frak A)$\cmmnt{ of $\frak A$} is
at most\cmmnt{ the weight} $w(X)$\cmmnt{ of $X$}.

(b)\cmmnt{ The cellularity} $c(\frak A)$\cmmnt{ of $\frak A$} is at
most\cmmnt{ the hereditary Lindel\"of number} $\hL(X)$\cmmnt{ of
$X$}.   If $\mu$ is locally finite, $c(\frak A)$ is at most\cmmnt{ the
Lindel\"of number} $L(X)$\cmmnt{ of $X$}.

(c) $\#(\{a:a\in\frak A$, $\bar\mu a<\infty\})
\le\max(1,w(X)^{\omega})$\cmmnt{, where $w(X)^{\omega}$ is the
cardinal power}.

(d) If $X$ is Hausdorff and $\mu$ is a Radon measure,
then\cmmnt{ the Maharam type} $\tau(\frak A)$\cmmnt{ of $\frak A$} is at
most\cmmnt{ the network weight} $\nw(X)$\cmmnt{ of $X$}.

\proof{{\bf (a)} Let $\Cal U$ be a base for $\frak T$ with
$\#(\Cal U)=w(X)$.   Set $B=\{U^{\ssbullet}:U\in\Cal U\}$ and let
$\frak B$ be the order-closed subalgebra of $\frak A$ generated by $B$;
set $\Tau=\{E:E\in\Sigma$, $E^{\ssbullet}\in\frak B\}$.   Then $\Tau$ is
a $\sigma$-subalgebra of $\Sigma$ containing every negligible set.

If $G\subseteq X$ is open, then $G\in\Tau$.   \Prf\ By 414Aa,
$G^{\ssbullet}=\sup\{U^{\ssbullet}:U\in\Cal U$, $U\subseteq G\}$ belongs
to $\frak B$.\ \QeD\   So every Borel set belongs to $\Tau$.   If
$E\in\Sigma$ and $\mu E<\infty$, then, because $\mu$ is inner regular
with respect to the Borel sets, there is a Borel subset $F$ of $E$ with
the same measure, so $F$, $E\setminus F$ and $E$ belong to $\Tau$.
Thus $\{a:a\in\frak A$, $\bar\mu a<\infty\}\subseteq\frak B$;  because
$\mu$ is semi-finite, $\frak B=\frak A$ and
$\tau(\frak A)\le\#(B)\le\#(\Cal U)=w(X)$.

\medskip

{\bf (b)(i)} If $L(X)=n$ is finite, and $F_0,\ldots,F_n\subseteq X$ are
disjoint closed sets, then at least one of them is empty.   \Prf\ For
$i\le n$, set $G_i=X\setminus\bigcup_{j\le n,j\ne i}F_j$;  then
$\bigcup_{i\le n}G_i=X$, so there is some $k\le n$ such that
$\bigcup_{i\ne k}G_i=X$, and now $F_k=\emptyset$.\ \QeD\   As $\mu$ is
inner regular with respect to the closed sets,
$c(\frak A)\le n=L(X)\le\hL(X)$.

\medskip

\quad{\bf (ii)} Suppose that $\omega\le L(X)\le\hL(X)$.   Let $\Cal G$
be the family of open subsets of $X$ of finite measure.   Then there is
a set $\Cal H\subseteq\Cal G$, with cardinal at most $\hL(X)$, such that
$\bigcup\Cal H=\bigcup\Cal G$ (5A4Bf).
Now $\sup_{H\in\Cal H}H^{\ssbullet}=1$,
because $\mu$ is effectively locally finite.

If $D\subseteq\frak A\setminus\{0\}$ is disjoint, then for each
$d\in D$ take $H_d\in\Cal H$ such that $d\Bcap H_d^{\ssbullet}\ne 0$.
If $H\in\Cal H$, then $\{d:H_d=H\}$ must be countable, since
$\mu H<\infty$.   So $\#(D)\le\max(\omega,\#(\Cal H))$;  as $D$ is
arbitrary, $c(\frak A)\le\max(\omega,\hL(X))=\hL(X)$.

\medskip

\quad{\bf (iii)} Finally, if $\omega\le L(X)$ and $\mu$ is locally
finite, then in (ii) above we have $X=\bigcup\Cal G$, so we can take
$\Cal H$ to have size at most $L(X)$, and continue as before, ending
with $c(\frak A)\le\max(\omega,\#(\Cal H))=L(X)$.

\medskip

{\bf (c)} Again let $\Cal U$ be a base for the topology of $X$ with
cardinal $w(X)$.   Let $\Tau$ be the $\sigma$-subalgebra of $\Sigma$
generated by $\Cal U$.   If $E\in\Sigma$ and $\mu E<\infty$, then for
each $n\in\Bbb N$ we can find an open set $G_n$ such that
$\mu(G_n\symmdiff E)\le 2^{-n}$;  now there is an
open set $H_n$, a finite union of members of $\Cal U$, such that
$H_n\subseteq G_n$ and $\mu(G_n\setminus H_n)\le 2^{-n}$.   Setting
$F=\bigcup_{m\in\Bbb N}\bigcap_{n\ge m}H_n$, we see that $F\in\Tau$
and $E\symmdiff F$ is negligible.   Thus
$\{F^{\ssbullet}:F\in\Tau\}\supseteq\{a:\bar\mu a<\infty\}$ and

\Centerline{$\#(\{a:\bar\mu a<\infty\})\le\#(\Tau)
\le\max(1,\#(\Cal U)^{\omega})=\max(1,w(X)^{\omega})$.}

\medskip

{\bf (d)} If $a\in\frak A\setminus\{0\}$ and the principal ideal
$\frak A_a$ is \Mth, then $\tau(\frak A_a)\le\nw(X)$.   \Prf\ There is a
compact set $K\subseteq X$ such that
$0\ne K^{\ssbullet}\subseteq\frak A_a$;  let $\mu_K$ be the subspace
measure on $K$.   Then

$$\eqalignno{\tau(\frak A_a)
&=\tau(\mu_K)\le w(K)\cr
\displaycause{by (a)}
&=\nw(K)\cr
\displaycause{5A4C(a-i)}
&\le\nw(X)\cr}$$

\noindent (5A4Bb).\ \Qed

By (b), $c(\frak A)\le\#(\frak T)\le 2^{\nw(X)}$ (5A4Ba);  so 332S tells
us that $\tau(\frak A)\le\nw(X)$.
}%end of proof of 531A

\leader{531B}{}\cmmnt{ For strictly positive measures we have some
easy inequalities in the other direction.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space,
with measure algebra $\frak A$, and $\frak T$ a topology on $X$ such
that $\Sigma$ includes a base for $\frak T$ and $\mu$ is strictly
positive.

(a) If $X$ is regular, then $w(X)\le\#(\frak A)$.

(b) If $X$ is Hausdorff, then $\#(X)\le 2^{\#(\frak A)}$.

\proof{ Set $\Cal V=\Sigma\cap\frak T$, so that $\Cal V$ is a base for
$\frak T$.  If $V$, $W\in\Cal V$ and  $V^{\ssbullet}=W^{\ssbullet}$ in
$\frak A$, then $\interior\overline{V}=\interior\overline{W}$.    \Prf\
$\mu^*(V\setminus\overline{W})\le\mu(V\setminus W)=0$, so (because $\mu$
is strictly positive) $V\subseteq\overline{W}$ and
$\overline{V}\subseteq\overline{W}$ and
$\interior\overline{V}\subseteq\interior\overline{W}$.   Similarly,
$\interior\overline{W}\subseteq\interior\overline{V}$.\ \QeD\  So if we
set $\Cal W=\{\interior\overline{V}:V\in\Cal V\}$,
$\#(\Cal W)\le\#(\frak A)$.

\medskip

{\bf (a)} If $\frak T$ is regular, $\Cal W$ is a base for $\frak T$, so
$w(X)\le\#(\Cal W)\le\#(\frak A)$.

\medskip

{\bf (b)} If $\frak T$ is Hausdorff, then for any distinct $x$,
$y\in X$, there is a $W\in\Cal W$ containing $x$ but not $y$.   \Prf\
Let $G$, $H$ be disjoint open sets containing $x$, $y$ respectively.
Take $V\in\Cal V$ such that $x\in V\subseteq G$, and set
$W=\interior\overline{V}$.\ \QeD\  So
$\#(X)\le 2^{\#(\Cal W)}\le 2^{\#(\frak A)}$.
}%end of proof of 531B

\leader{531C}{Lemma} Let $\familyiI{X_i}$ be a family of topological
spaces with product $X$, and $\mu$ a totally finite quasi-Radon measure
on $X$ with Maharam type $\kappa$.   For each $i\in I$, let $\mu_i$ be
the marginal measure on $X_i$, and $\kappa_i$ its Maharam
type.   Then $\kappa$ is at most the cardinal sum
$\sum_{i\in I}\kappa_i$.

\proof{ For each $i\in I$, let $\ofamily{\xi}{\kappa_i}{E_{i\xi}}$ be a
family in $\dom\mu_i$ such that
$\{E_{i\xi}^{\ssbullet}:\xi<\kappa_i\}\,\,\tau$-generates the measure
algebra of $\mu_i$.   Consider
$\Cal W=\{\pi_i^{-1}[E_{i\xi}]:i\in I$, $\xi<\kappa_i\}$, so that
$\Cal W\subseteq\dom\mu$ and
$\#(\Cal W)\le\sum_{i\in I}\kappa_i$.   Let $\frak B$ be the closed
subalgebra of the measure algebra $\frak A$ of $\mu$ generated by
$\{W^{\ssbullet}:W\in\Cal W\}$.

For each $i\in I$, the canonical map $\pi_i:X\to X_i$
induces a measure-preserving homomorphism
$\phi_i$ from the measure algebra $\frak A_i$ of $\mu_i$ to $\frak A$
(324M).   Now $\phi_i^{-1}[\frak B]$ is a closed subalgebra of
$\frak A_i$ containing $E_{i\xi}^{\ssbullet}$ for every $\xi<\kappa_i$,
so is the whole of $\frak A_i$, that is,
$\phi_i[\frak A_i]\subseteq\frak B$.   In particular, if
$G\subseteq X_i$ is open,
$\pi_i^{-1}[G]^{\ssbullet}=\phi_i(G^{\ssbullet})$ belongs to $\frak B$.
Now the family $\Cal V$ of open sets $V\subseteq X$ such that
$V^{\ssbullet}\in\frak B$ is closed under finite intersections and
contains $\pi_i^{-1}[G]$ whenever $i\in I$ and $G\subseteq X_i$ is open,
so $\Cal V$ is a base for the topology of $X$.   But also $\Cal V$ is
closed under arbitrary unions, because $\frak B$ is closed and $\mu$ is
$\tau$-additive (414Aa again).   So $V^{\ssbullet}\in\frak B$ for every open
set $V\subseteq X$, and therefore for every Borel set $V\subseteq X$;
as $\mu$ is inner regular with respect to the Borel sets,
$\frak B=\frak A$.

Thus $\{W^{\ssbullet}:W\in\Cal W\}$ witnesses that the Maharam type
$\tau(\frak A)$ of $\mu$ is at most $\sum_{i\in I}\kappa_i$, as claimed.
}%end of proof of 531C

\leader{531D}{Definition} If $X$ is a Hausdorff space, I write
$\MahR(X)$ for the set of Maharam types of \Mth\ Radon probability
measures on $X$.   \cmmnt{Note that} $0\in\MahR(X)$ iff $X$ is
non-empty, and\cmmnt{ that} any member of $\MahR(X)$ is either $0$ or
an infinite cardinal.

\leader{531E}{Proposition} Let $X$ be a Hausdorff space.

(a) $\kappa\le w(X)$ for every $\kappa\in\MahR(X)$.

(b) $\MahR(Y)\subseteq\MahR(X)$ for every $Y\subseteq X$.

(c) $\MahR(X)=\bigcup\{\MahR(K):K\subseteq X$ is compact$\}$.

(d)\dvArevised{2015}
If $X$ is K-analytic (in particular, if $X$ is compact)
and $Y$ is a continuous image of $X$, $\MahR(Y)\subseteq\MahR(X)$.

(e) $\omega\in\MahR(X)$ iff $X$ has a compact subset which is not
scattered.

(f)\cmmnt{ ({\smc Haydon 77})} If
$\omega\le\kappa'\le\kappa\in\MahR(X)$ then $\kappa'\in\MahR(X)$.

(g) If $Y$ is another Hausdorff space, and neither $X$ nor $Y$ is empty,
then $\MahR(X\times Y)=\MahR(X)\cup\MahR(Y)$;  generally, for any
non-empty finite family $\familyiI{X_i}$ of non-empty Hausdorff spaces,
$\MahR(\prod_{i\in I}X_i)\penalty-100=\bigcup_{i\in I}\MahR(X_i)$.

\proof{{\bf (a)} This is immediate from 531Aa.

\medskip

{\bf (b)} If $\kappa\in\MahR(Y)$, there is a \Mth\ Radon probability
measure $\mu$ on $Y$ with Maharam type $\kappa$.   Set

\Centerline{$\Sigma'=\{E:E\subseteq X$, $\mu$ measures $Y\cap E\}$,}

\Centerline{$\mu'E=\mu(Y\cap E)$ for $E\in\Sigma'\}$.}

\noindent It is easy to check that $\mu'$ is a Radon probability measure on $X$
(see 416Xc and 418I), and that $\mu'$ and $\mu$ have isomorphic measure
algebras (cf.\ 322J).   So $\mu'$ is \Mth\ and has Maharam type $\kappa$, and
$\kappa\in\MahR(X)$.

\medskip

{\bf (c)} By (b), $\MahR(K)\subseteq\MahR(X)$ for every compact set
$K\subseteq X$.   In the other direction, if $\kappa\in\MahR(X)$, there
is a \Mth\ Radon probability measure $\mu$ on $X$ with Maharam type
$\kappa$.   Let $K\subseteq X$ be a compact set with $\mu K>0$.   Then
the normalized subspace measure $\mu'=(\mu K)^{-1}\mu_K$ is a Radon
probability measure on $K$, and its measure algebra is isomorphic to a
principal ideal of the measure algebra of $\mu$, so is \Mth\ with
Maharam type $\kappa$.   Accordingly $\kappa\in\MahR(K)$.

\medskip

{\bf (d)} Take $\kappa\in\MahR(Y)$.   Then there is a \Mth\ Radon
probability measure $\nu$ on $Y$ with Maharam type $\kappa$.   Let
$f:X\to Y$ be a continuous surjection.   By 432G, there is a Radon
measure $\mu$ on $X$ such that $f$ is \imp\ for $\mu$ and $\nu$.
Let $K\subseteq X$ be a compact set such that $\mu K>0$.   Then
$f[K]\subseteq Y$ is compact and

\Centerline{$\nu f[K]=\mu[f^{-1}[f[K]]\ge\mu K>0$.}

Let $\nu_1=\Bover1{\nu f[K]}\nu_{f[K]}$ be the normalized subspace measure
on $f[K]$.   Then $\nu_1$ is a \Mth\ Radon probability measure on $f[K]$
with Maharam type $\kappa$.   By 418L, there is a Radon measure $\mu_1$ on
$K$ such that $f\restr K$ is \imp\ for $\mu_1$ and $\nu_1$ and induces an
isomorphism of their measure algebras.   So $\mu_1$ witnesses that
$\kappa\in\MahR(K)$;  by (b), $\kappa\in\MahR(X)$.

\medskip

{\bf (e)(i)} If $X$ has a compact subset $K$ which is not scattered,
then there is a continuous
surjection from $K$ onto $[0,1]$ (4A2G(j-iv)).   Of course Lebesgue
measure witnesses that $\omega\in\MahR([0,1])$, so (d) and (b)
tell us that $\omega\in\MahR(K)\subseteq\MahR(X)$.

\medskip

\quad{\bf (ii)} If every compact subset of $X$ is scattered and $\mu$ is
a \Mth\ Radon
probability measure on $X$, let $K$ be a compact set of non-zero measure
and $Z\subseteq K$ a closed self-supporting set.   Then $Z$
has an isolated point $z$ say;  in this case, $\mu\{z\}>0$ so $\{z\}$ is
an atom for $\mu$ and (because $\mu$ is \Mth) the Maharam type of $\mu$
is $0$.   As $\mu$ is arbitrary, $\omega\notin\MahR(X)$.

\medskip

{\bf (f)(i)} Suppose first that $X$ is compact.   Let $\mu$ be a \Mth\
Radon probability measure on $X$ with Maharam type $\kappa$.   Let
$\ofamily{\xi}{\kappa}{E_{\xi}}$ be a stochastically independent family
in $\dom\mu$ with
$\mu E_{\xi}=\bover12$ for every $\xi$.   For each $\xi<\kappa'$ and
$n\in\Bbb N$, let $f_{\xi n}\in C(X)$ be such that
$\int|f_{\xi n}-\chi E_{\xi}|\le 2^{-n}$ (416I).   Define
$f:X\to\BbbR^{\kappa'\times\Bbb N}$ by setting
$f(x)(\xi,n)=f_{\xi n}(x)$ for $x\in X$, $\xi<\kappa'$ and $n\in\Bbb N$.
Then $f$ is continuous, so by 418I the image measure $\nu=\mu f^{-1}$ on
the compact set $f[X]$ is a Radon measure.   For each $\xi<\kappa'$, the
set

\Centerline{$F_{\xi}=\{w:w\in f[X]$, $\lim_{n\to\infty}w(\xi,n)=1\}$}

\noindent is a Borel set, and $f^{-1}[F_{\xi}]\symmdiff E_{\xi}$ is
$\mu$-negligible;  so $\ofamily{\xi}{\kappa'}{F_{\xi}}$ is a
stochastically independent family of subsets of $f[X]$ with measure
$\bover12$.   If $\frak B$ is the measure algebra of $\nu$, and
$\frak C$ the closed subalgebra of $\frak B$ generated by
$\{F_{\xi}^{\ssbullet}:\xi<\kappa'\}$, then $\frak C$ is \Mth, with
Maharam type $\kappa'$;  at the same time,

\Centerline{$\tau(\frak B)\le w(f[X])
\le w(\BbbR^{\kappa'\times\Bbb N})=\kappa'$.}

\noindent By 332N, $\frak B$ can be embedded in $\frak C$;  by 332Q,
$\frak B$ and $\frak C$ are isomorphic, that is, $\frak B$ is \Mth\ with
Maharam type $\kappa'$, and $\nu$ witnesses that
$\kappa'\in\MahR(f[X])$.   By (d), $\kappa'\in\MahR(X)$.

\medskip

\quad{\bf (ii)} In general, (c) tells us that there is a compact set
$K\subseteq X$ such that $\kappa\in\MahR(K)$, so
$\kappa'\in\MahR(K)\subseteq\MahR(X)$.

\medskip

{\bf (g)} Because neither $Y$ nor $X$ is empty, both $X$ and $Y$ are
homeomorphic to subspaces of $X\times Y$, so (b) tells us that
$\MahR(X\times Y)\supseteq\MahR(X)\cup\MahR(Y)$.   In the other
direction, given a \Mth\ Radon probability measure $\mu$ on $X\times Y$,
let $\mu_1$, $\mu_2$ be the marginal measures on $X$ and $Y$
respectively, so that each $\mu_k$ is a Radon probability measure
(418I again).
Let $\familyiI{E_i}$, $\family{j}{J}{F_j}$ be countable partitions of
$X$, $Y$ into Borel sets such that all the subspace measures
$(\mu_1)_{E_i}$ and $(\mu_2)_{F_j}$ are \Mth.   Then
there must be $i\in J$, $j\in J$ such that $\mu(E_i\times F_j)>0$.   Let
$\mu'$ be the subspace measure $\mu_{E_i\times F_j}$;  then the Maharam
type of $\mu'$ is $\kappa$, because $\mu$ is \Mth.   Let $\mu'_1$,
$\mu'_2$ be the marginal measures of $\mu'$ on $E_i$ and $F_j$
respectively.   Then $\mu'_1$ is an indefinite-integral measure over
$(\mu_1)_{E_i}$ (415Oa), so its measure algebra is isomorphic to a
principal ideal of the measure algebra of $(\mu_1)_{E_i}$
(322K), and has the same Maharam type $\kappa_1$ say.
As in (b) above, $\kappa_1\in\MahR(X)$.   Similarly, the Maharam type
$\kappa_2$ of $\mu'_2$ belongs to $\MahR(Y)$.   Now 531C tells us that
$\kappa\le\kappa_1+\kappa_2$.   Since $\kappa$ is either zero or
infinite, it must be less than or equal to at least one of them, and
belongs to $\MahR(X)\cup\MahR(Y)$ by (f) above.

The result for general finite products now follows easily by induction
on $\#(I)$.
}%end of proof of 531E

\leader{531F}{Proposition} Let $X$ be a Hausdorff space.   Then a
totally finite measure algebra $(\frak A,\bar\mu)$ is isomorphic to the
measure algebra of a Radon measure on $X$ iff ($\alpha$) whenever
$\frak A_a$ is a non-trivial homogeneous principal ideal of $\frak A$
then $\tau(\frak A_a)\in\MahR(X)$ ($\beta$) $c(\frak A)\le\#(X)$.

\proof{{\bf (a)} If $\mu$ is a totally finite Radon measure on $X$ with
measure algebra $\frak A$ and the principal ideal $\frak A_a$ generated
by $a\in\frak A\setminus\{0\}$ is homogeneous, then there is an
$E\in\dom\mu$ such that $E^{\ssbullet}=a$.   Let $\nu$ be the probability
measure
$(\mu F)^{-1}\mu\LLcorner F$, that is, $\nu H=\mu(H\cap F)/\mu F$
whenever $H\subseteq X$ is such that $\mu$ measures $H\cap F$ (234M).
Then $\nu$ is a Radon measure (416Sa),
the measure algebra of $\nu$ is isomorphic to a principal ideal of
$\frak A_a$ (322K) so is homogeneous with the same Maharam type, and
$\nu$ witnesses that $\tau(\frak A_a)\in\MahR(X)$.   Thus $\frak A$
satisfies ($\alpha$).   As for ($\beta$), if $X$ is infinite this is
trivial (because $(\frak A,\bar\mu)$ is totally finite, so $\frak A$ is
ccc), and otherwise $\frak A$ is finite, with

\Centerline{$c(\frak A)=\#(\{a:a\in\frak A$ is an atom$\})
=\#(\{x:x\in X$, $\mu\{x\}>0\})\le\#(X)$.}

\medskip

{\bf (b)} Now suppose that $(\frak A,\bar\mu)$ is a  totally finite
measure algebra satisfying the conditions.   Express it as the simple
product of a countable family $\familyiI{(\frak A_i,\bar\mu'_i)}$ of
non-zero homogeneous measure algebras (332B);  we may suppose that
$I\subseteq\Bbb N$.   For $n\in I$, set
$\kappa_n=\tau(\frak A_n)$ and $\gamma_n=\bar\mu'_n1_{\frak A_n}$.
$(\beta$) tells us that $\#(I)\le\#(X)$;  let $\family{n}I{x_n}$ be a
family of distinct elements of $X$.

Set $J=\{n:n\in I$, $\kappa_n\ge\omega\}$.   For each $n\in J$,
($\alpha$) tells us that there is a \Mth\ Radon probability measure
$\mu_n$ on $X$ with Maharam type $\kappa_n$.   Now there is a disjoint
family $\family{n}{\Bbb N}{E_n}$ of Borel subsets of $X\setminus\{x_n:n\in I\}$
such that $\mu_nE_n>0$ for every $n\in J$.   \Prf\
Choose $\sequencen{E_n}$, $\sequencen{F_n}$ inductively, as follows.
$F_0=X\setminus\{x_n:n\in I\}$.   Given that $F_n$ is a Borel set and
$\mu_jF_n>0$ for every $j\in J\setminus n$, then if $n\notin J$ set
$E_n=\emptyset$
and $F_{n+1}=F_n$.   Otherwise, for each $j\in J$ such that $j>n$, we can
partition $F_n$ into finitely many Borel sets of $\mu_n$-measure less than
$2^{-j}\mu_nF_n$, because $\mu_n$ is atomless;   take one of these,
$G_{nj}$ say, such that $\mu_jG_{nj}>0$;  now set
$F_{n+1}=\bigcup_{j\in J,j>n}G_{nj}$ and $E_n=F_n\setminus F_{n+1}$.
Continue.\ \QeD\   Now set

\Centerline{$\mu E
=\sum_{n\in I\setminus J,x_n\in E}\gamma_n
  +\sum_{n\in J}(\mu_nE_n)^{-1}\gamma_n\mu_n(E\cap E_n)$}

\noindent whenever $E\subseteq X$ is such that $\mu_n$ measures
$E\cap E_n$ for every $n\in J$.   Of course $\mu$ is a measure.
Because every $\mu_n$ is a topological measure, so is $\mu$;  because
every $\mu_n$ is inner regular with respect to the compact sets, so is
$\mu$;  because every $\mu_n$ is complete, so is $\mu$;  thus $\mu$ is a
Radon measure.   Because every subspace measure
$(\mu_n)_{E_n}$ is \Mth\ with Maharam type $\kappa_n$, the measure
algebra of $\mu$ is isomorphic to $(\frak A,\bar\mu)$.
}%end of proof of 531F

\leader{531G}{Proposition} Let $\familyiI{X_i}$ be a family of non-empty
Hausdorff spaces with product $X$.   Then an infinite cardinal $\kappa$
belongs to $\MahR(X)$ iff {\it either}
$\kappa\le\#(\{i:i\in I$, $\#(X_i)\ge 2\})$ {\it or} $\kappa$ is
expressible as $\sup_{i\in I}\kappa_i$ where $\kappa_i\in\MahR(X_i)$ for
every $i\in I$.

\proof{{\bf (a)(i)} Suppose that $\kappa=\sup_{i\in I}\kappa_i$ where
$\kappa_i\in\MahR(X_i)$ for each $i\in I$.   For each $i$, let $\mu_i$
be a \Mth\ Radon probability measure on $X_i$ with Maharam type
$\kappa_i$ and compact support (see the proof of 531Ec).
Let $\lambda$ be the ordinary product of the measures
$\mu_i$.   By 325I, the measure algebra of $\lambda$ can be identified with
the probability
algebra free product of the measure algebras of the $\mu_i$.   It is
therefore isomorphic to the measure algebra of the usual measure on
$\{0,1\}^{\kappa'}$, where $\kappa'$ is the cardinal sum
$\sum_{i\in I}\kappa_i$;  in particular, it is homogeneous with
Maharam type $\kappa'$ (since we are supposing that $\kappa\ge\omega$).
By 417E(b-i)\formerly{4{}17E(ii)},
the measure algebra of the $\tau$-additive product $\mu$ of
$\familyiI{\mu_i}$ can be identified with the measure algebra of
$\lambda$, while $\mu$ is a Radon measure (417Q).
So $\mu$ witnesses that $\kappa'\in\MahR(X)$;  by
531Ef, $\kappa\in\MahR(X)$.

\medskip

\quad{\bf (ii)} Suppose that $\omega\le\kappa\le\#(I')$ where
$I'=\{i:i\in I$, $\#(X_i)\ge 2\}$.   For $i\in I'$, let $x_i$, $y_i$ be
distinct points of $X_i$ and $\mu_i$ the
point-supported probability measure on $X_i$ such that
$\mu_i\{x_i\}=\mu_i\{y_i\}=\bover12$;  for $i\in I\setminus I'$, let
$\mu_i$ be the unique Radon probability measure on $X_i$.   As in (i)
above, the Radon measure product of $\familyiI{\mu_i}$ is \Mth, with
Maharam type $\#(I')$, so $\#(I')\in\MahR(X)$;  by
531Ef again, $\kappa\in\MahR(X)$.

\medskip

{\bf (b)} Now suppose that $\omega\le\kappa\in\MahR(X)$ and that
$\kappa>\#(I')$.   For each $i\in I$, let $\theta_i$ be the least
cardinal greater than every member of $\MahR(X_i)$.   Note that
$\kappa'\in\MahR(X_i)$ whenever $\kappa'$ is a cardinal and
$\omega\le\kappa'<\theta_i$.   Set

\Centerline{$I_1=\{i:i\in I$, $\kappa<\theta_i\}$,
\quad$Z_1=\prod_{i\in I_1}X_i$,}

\Centerline{$I_2=\{i:i\in I$, $\theta_i\le\kappa$,
$\cf\theta_i>\omega\}$,
\quad$Z_2=\prod_{i\in I_2}X_i$,}

\Centerline{$I_3=\{i:i\in I$, $\theta_i=\kappa$, $\cf\theta_i=\omega\}$,
\quad$Z_3=\prod_{i\in I_3}X_i$,}

\Centerline{$I_4=\{i:i\in I$, $\theta_i<\kappa$, $\cf\theta_i=\omega\}$,
\quad$Z_4=\prod_{i\in I_4}X_i$,}

\Centerline{$I_5=\{i:i\in I$, $\theta_i=1$, $\#(X_i)>1\}$,
\quad$Z_5=\prod_{i\in I_5}X_i$,}

\Centerline{$I_6=\{i:i\in I$, $\#(X_i)=1\}$,
\quad$Z_6=\prod_{i\in I_6}X_i$.}

\noindent Then $X$ can be identified with $\prod_{1\le k\le 6}Z_k$, so
531Eg tells us that $\kappa\in\MahR(Z_k)$ for some $k$.   As $Z_6$ is a
singleton, we actually have $\kappa\in\MahR(Z_k)$ for some $k\le 5$.

\medskip

\quad{\bf case 1} Suppose $\kappa\in\MahR(Z_1)$.   Then, in particular,
$I_1\ne\emptyset$ and there is a $j\in I$ such that $\kappa<\theta_j$.
In this case, $\kappa\in\MahR(X_j)$, and we can set $\kappa_j=\kappa$,
$\kappa_i=0$ for $i\ne j$ to find a family in $\prod_{i\in I}\MahR(X_i)$
with supremum $\kappa$.

\medskip

\quad{\bf case 2} Suppose that $\kappa\in\MahR(Z_2)$.   Let $\mu$ be a
Radon probability measure on $Z_2$ with Maharam type $\kappa$.   For
each $i\in Z_2$, let $\mu'_i$ be the marginal measure on $X_i$, and
$\kappa'_i$ its Maharam type.   By 531C,

\Centerline{$\kappa\le\sum_{i\in I_2}\kappa'_i
\le\max(\omega,\#(I_2),\sup_{i\in I}\kappa'_i)$}

\noindent (5A4F(b-i));  since $\emptyset\ne I_2\subseteq I'$,
$\#(I_2)<\kappa\le\sup_{i\in I_2}\max(\omega,\kappa'_i)$;
since $\kappa$ is infinite, it must be less than or
equal to $\sup_{i\in I_2}\max(\omega,\kappa'_i)$.   On the other hand,
by 531F, each $\kappa'_i$ is either finite or the supremum of some
countable subset of $\MahR(X_i)$;  because $\cf\theta_i>\omega$,
$\kappa'_i<\theta_i$ and $\max(\omega,\kappa'_i)\in\MahR(X_i)$.
Setting

$$\eqalign{\kappa_i
&=\med(\kappa'_i,\omega,\kappa)\text{ for }i\in I_2,\cr
&=0\text{ for }i\in I\setminus I_2,\cr}$$

\noindent we have $\kappa_i\in\MahR(X_i)$ for every $i\in I$ and
$\kappa=\sup_{i\in I}\kappa_i$.

\medskip

\quad{\bf case 3} Suppose that $\kappa\in\MahR(Z_3)$.   Because
$\kappa=\theta_i\notin\MahR(X_i)$ for $i\in I_3$, 531Eg tells us that
$I_3$ must be infinite.   Let $\sequencen{i_n}$ be a sequence of
distinct elements of $I_3$.   Of course $\kappa$ itself is uncountable
and has countable cofinality, so we can find a sequence $\kappa'_n$ of
infinite cardinals less than $\kappa$ with supremum $\kappa$.   Setting
$\kappa_{i_n}=\kappa'_n$, $\kappa_i=0$ for
$i\in I\setminus\{i_n:n\in\Bbb N\}$, we have $\kappa_i\in\MahR(X_i)$ for
every $i$ and $\kappa=\sup_{i\in I}\kappa_i$.

\medskip

\quad{\bf case 4} Suppose that $\kappa\in\MahR(Z_4)$.  Following the
scheme of
case 2 above, let $\mu$ be a Radon probability measure on $Z_4$ with
Maharam type $\kappa$, and for each $i\in I_4$ let $\mu'_i$ be the
marginal measure on $X_i$ and $\kappa'_i$ its Maharam type.   Then, as
before, $\kappa\le\sup_{i\in I_4}\max(\omega,\kappa'_i)$.
At the same time,
$\kappa'_i\le\theta_i<\kappa$ for every $i$, so we must have
$\kappa=\sup_{i\in I_4}\theta_i$.   Set $\delta=\cf\kappa$.   Then we
can choose $\ofamily{\xi}{\delta}{i_{\xi}}$ inductively in $I_4$ so that
$\theta_{i_{\eta}}<\theta_{i_{\xi}}$ whenever $\eta<\xi<\delta$ and
$\sup_{\xi<\delta}\theta_{i_{\xi}}=\kappa$.   Now define
$\familyiI{\kappa_i}$ by saying

$$\eqalign{\kappa_{i_{\xi+1}}
&=\theta_{i_{\xi}}\text{ whenever }\xi<\delta,\cr
\kappa_i&=0\text{ if }i\in I\setminus\{i_{\xi+1}:\xi<\delta\}.\cr}$$

\noindent This gives $\kappa_i\in\MahR(X_i)$ for every $i$ and
$\kappa=\sup_{i\in I}\kappa_i$.

\medskip

\quad{\bf case 5} \Quer\ Suppose, if possible, that
$\kappa\in\MahR(Z_5)$.
Once again, we can find a Radon probability measure $\mu$ on $Z_5$ with
Maharam type $\kappa$, and look at its marginal measures $\mu'_i$ for
$i\in I_5$.   This time, however, every $\mu'_i$ must be purely atomic
and has Maharam type $\kappa'_i\le\omega$;  also $\#(I_5)<\kappa$.   So
our formula $\kappa\le\sum_{i\in I_5}\kappa'_i$ becomes $\kappa\le\omega$.
In this case $I_5$ must be finite and
$\kappa\in\bigcup_{i\in I_5}\MahR(X_i)=\{0\}$, which is absurd.\ \Bang

Thus this case evaporates and the proof is complete.
}%end of proof of 531G

\cmmnt{
\leader{531H}{Remarks} The results above already enable us to
calculate $\MahR(X)$ for many spaces.   Of course we begin with compact
spaces (531Ec).   If $X$ is compact and Hausdorff, and $\{0,1\}^{\kappa}$
is a continuous image of a closed subset of $X$, where $\kappa$ is
an infinite cardinal, then $\kappa\in\MahR(X)$ (531Ed);  so if
$\{0,1\}^{w(X)}$ is a continuous image of a closed subset of
$X$, then $\MahR(X)$ is completely specified, being
$\{0\}\cup\{\kappa:\omega\le\kappa\le w(X)\}$ (531Ea, 531Ef).   Of
course it is not generally true that $w(X)\in\MahR(X)$ (531Xc).   But it
is quite often the case that $\{0,1\}^{\kappa}$ is a continuous image of
a closed subset of
$X$ for every $\kappa\in\MahR(X)$, and I will now investigate this
phenomenon.
}%end of comment

\leader{531I}{Notation}\cmmnt{ For the rest of the section, I will use
the following notation, mostly familiar from earlier chapters of this
volume.}   For
any set $I$, let $\nu_I$ be the usual measure on
$\{0,1\}^I$, $\Tau_I$ its domain, $\Cal N_I$ its
null ideal and $(\frak B_I,\bar\nu_I)$ its measure
algebra.   \cmmnt{In this context,} I will write
$\familyiI{e_i}$ for the standard generating family in $\frak B_I$
(525A).
For $J\subseteq I$ let $\frak C_J$ be the closed subalgebra of
$\frak B_I$ generated by $\{e_i:i\in J\}$.\cmmnt{   Now for a new idea.}
For each $i\in I$, let
$\phi_i:\frak B_I\to\frak B_I$ be the
measure-preserving involution corresponding to reversal of the $i$th
coordinate in $\{0,1\}^I$, that is,
$\phi_i(e_i)=1\Bsetminus e_i$ and
$\phi_i(e_j)=e_j$ for $j\ne i$.

\leader{531J}{Lemma} Let $I$ be a set, and take
$\frak B_I$, $\frak C_J$, for $J\subseteq I$, and
$\phi_i$, for $i\in I$, as in 531I.

(a) $\bigcup\{\frak C_J:J\in[I]^{<\omega}\}$ is dense in
$\frak B_I$ for the measure-algebra topology of
$\frak B_I$.

(b) For every $a\in\frak B_I$, there is a (unique) countable
$J^*(a)\subseteq I$ such that, for $J\subseteq I$,
$a\in\frak C_J$ iff $J\supseteq J^*(a)$.

(c) $J^*(1\Bsetminus a)=J^*(a)$ for every $a\in\frak B_I$.

(d) $\phi_i\phi_j=\phi_j\phi_i$ for all $i$, $j\in I$.

(e) If $J\subseteq I$, $a\in\frak C_J$ and $i\in I$, then
$a\Bcap\phi_ia$, $a\Bcup\phi_ia$ belong to $\frak C_{J\setminus\{i\}}$.

(f) For $a\in\frak B_I$ and $i\in I$ we have $\phi_ia=a$
iff $i\notin J^*(a)$.

(g) $\phi_ia\in\frak C_J$ whenever $J\subseteq I$, $i\in I$
and $a\in\frak C_J$.

\proof{{\bf (a)} See 254Fe.

\medskip

{\bf (b)} See 254Rd or 325Mb.

\medskip

{\bf (c)} For $J\subseteq I$, $1\Bsetminus a\in\frak C_J$ iff $a\in\frak C_J$.

\medskip

{\bf (d)} Because $\{e_k:k\in I\}\,\,\tau$-generates
$\frak B_I$, it is enough to check that
$\phi_i\phi_je_k=\phi_j\phi_ie_k$ for all
$i$, $j$, $k\in I$, and this is easy.

\medskip

{\bf (e)} The subalgebra
$\{(c\Bcap e_i)\Bcup(c'\Bsetminus e_i):c$, $c'\in\frak C_{J\setminus\{i\}}\}$
generated by
$\frak C_{J\setminus\{i\}}\cup\{e_i\}$ is closed (323K), so includes
$\frak C_J$ and contains $a$.   If $c$,
$c'\in\frak C_{J\setminus\{i\}}$ are such that
$a=(c\Bcap e_i)\Bcup(c'\Bsetminus e_i)$, then
$\phi_ia=(c\Bsetminus e_i)\Bcup(c'\Bcap e_i)$ and
$a\Bcap\phi_ia=c\Bcap c'$, $a\Bcup\phi_ia=c\Bcup c'$
belong to $\frak C_{J\setminus\{i\}}$.

\medskip

{\bf (f)} If $i\notin J^*(a)$ then $\phi_ia=a$ because
$\phi_i(e_j)=e_j$ for every $j\ne i$.   If
$\phi_ia=a$ then
$a=a\Bcap\phi_ia\in\frak C_{I\setminus\{i\}}$, by (e), and
$J^*(a)\subseteq I\setminus\{i\}$, that is, $i\notin J^*(a)$.

\medskip

{\bf (g)} $\frak C_J$ is the closed subalgebra of $\frak B_I$ generated by
$\{e_j:j\in J\}$, so $\phi_i[\frak C_J]$ is the closed
subalgebra generated by
$\{\phi_ie_j:j\in J\}\subseteq\frak C_J$ (324L).
}%end of proof of 531J

\leader{531K}{Lemma} Let $\kappa\ge\omega_2$ be a cardinal, and
$\ofamily{\xi}{\kappa}{e_{\xi}}$ the standard generating family in
$\frak B_{\kappa}$.   Suppose that we are given a family
$\ofamily{\xi}{\kappa}{a_{\xi}}$ in $\frak B_{\kappa}$.
Then there are a set $\Gamma\in[\kappa]^{\kappa}$ and a family
$\ofamily{\xi}{\kappa}{c_{\xi}}$ in $\frak B_{\kappa}$ such that

\Centerline{$c_{\xi}\subseteq a_{\xi}$,
\quad$\bar\nu_{\kappa}c_{\xi}\ge 2\bar\nu_{\kappa}a_{\xi}-1$}

\noindent for every $\xi$, and

\Centerline{$\bar\nu_{\kappa}(\inf_{\xi\in I}(c_{\xi}\Bcap e_{\xi})
\Bcap\inf_{\eta\in J}(c_{\eta}\Bsetminus e_{\eta}))
=\Bover1{2^{\#(I\cup J)}}
  \bar\nu_{\kappa}(\inf_{\xi\in I\cup J}c_{\xi})$}

\noindent whenever $I$, $J\subseteq\Gamma$ are disjoint finite sets.
%for 531L

\proof{ Let $e_{\xi}$, $\phi_{\xi}$, for $\xi<\kappa$, $\frak C_L$, for
$L\subseteq\kappa$, and $J^*(a)$, for $a\in\frak B_{\kappa}$, be as in
531I-531J.   Set $L_{\xi}=J^*(a_{\xi})$ and
$c_{\xi}=a_{\xi}\Bcap\phi_{\xi}a_{\xi}$ for each $\xi$;  then

\Centerline{$\bar\nu_{\kappa}c_{\xi}
=\bar\nu_{\kappa}a_{\xi}+\bar\nu_{\kappa}(\phi_{\xi}a_{\xi})
  -\bar\nu_{\kappa}(a_{\xi}\Bcup\phi_{\xi}a_{\xi})
\ge 2\bar\nu_{\kappa}a_{\xi}-1$}

\noindent and $c_{\xi}\in\frak C_{L_{\xi}\setminus\{\xi\}}$ (531Je).
By Hajnal's Free Set Theorem (5A1J(a-iii)), there is a set
$\Gamma\in[\kappa]^{\kappa}$ such that $\xi\notin L_{\eta}$ whenever $\xi$,
$\eta$ are distinct members of $\Gamma$.   (This is where we use the
hypothesis that $\kappa\ge\omega_2$.)   Now suppose that $I$,
$J\subseteq\Gamma$ are finite and disjoint.   Then
$(L_{\xi}\setminus\{\xi\})\cap(I\cup J)=\emptyset$,  so
$c_{\xi}\in\frak C_{\kappa\setminus(I\cup J)}$, for every
$\xi\in I\cup J$.   Accordingly $c=\inf_{\xi\in I\cup J}c_{\xi}$ belongs
to $\frak C_{\kappa\setminus(I\cup J)}$.   This means that $c$ and the
$e_{\xi}$, for $\xi\in I\cup J$, are stochastically independent, and

\Centerline{$\bar\nu_{\kappa}(c\Bcap\inf_{\xi\in I}e_{\xi}
  \Bcap\inf_{\eta\in J}(1\Bsetminus e_{\eta}))
=\bar\nu_{\kappa}c\cdot\prod_{\xi\in I}\bar\nu_{\kappa}e_{\xi}
  \cdot\prod_{\eta\in J}\bar\nu_{\kappa}(1\Bsetminus e_{\eta})
=\Bover1{2^{\#(I\cup J)}}\bar\nu_{\kappa}c$,}

\noindent as claimed.
}%end of proof of 531K

\leader{531L}{Theorem}\dvArevised{2015} Let $X$ be a Hausdorff space.

(a)\cmmnt{ ({\smc Haydon 77})} If $\omega\in\MahR(X)$ then
$\{0,1\}^{\omega}$ is a continuous image of a compact subset of $X$.

(b)\cmmnt{ ({\smc Haydon 77}, {\smc Plebanek 97})} If
$\kappa\ge\omega_2$ belongs to $\MahR(X)$ and
$\lambda\le\kappa$ is an infinite cardinal such that $(\kappa,\lambda)$
is a measure-precaliber pair of every probability algebra, then
$\{0,1\}^{\lambda}$ is a continuous image of a compact subset of $X$.

\proof{{\bf (a)} If $\omega\in\MahR(X)$ then $X$ has a compact subset
$K$ which is not scattered (531Ee) and
there is a continuous surjection from $K$ onto $[0,1]$ (4A2G(j-iv) again).
As there is a continuous surjection from $[0,1]$ onto $[0,1]^{\omega}$
(5A4I(b-ii)), there is a continuous surjection $f:K\to[0,1]^{\omega}$.
Setting $K'=f^{-1}[\{0,1\}^{\omega}]$, $K'$ is a compact subset of $X$ and
$\{0,1\}^{\omega}$ is a continuous image of $K'$.

\medskip

{\bf (b)} Let $\mu$ be a \Mth\ Radon probability measure on $X$ with
Maharam type $\kappa$, $\Sigma$ its domain, and $(\frak A,\bar\mu)$ its
measure algebra, so that $(\frak A,\bar\mu)$ is isomorphic to the
measure algebra
$(\frak B_{\kappa},\bar\nu_{\kappa})$ as discussed in 531I-531K.   Let
$\ofamily{\xi}{\kappa}{e_{\xi}}$ be a stochastically independent
$\tau$-generating set of elements of measure $\bover12$ in $\frak A$, so
that $(\frak A,\ofamily{\xi}{\kappa}{e_{\xi}})$ is isomorphic to
$\frak B_{\kappa}$ with its standard generating family.
For each $\xi<\kappa$, let $E_{\xi}\in\Sigma$ be such that
$E_{\xi}^{\ssbullet}=e_{\xi}$ in $\frak A$.   Let
$K'_{\xi}\subseteq E_{\xi}$, $K''_{\xi}\subseteq X\setminus E_{\xi}$ be
compact sets of measure at least $\bover13$, and set
$K_{\xi}=K'_{\xi}\cup K''_{\xi}$, $a_{\xi}=K_{\xi}^{\ssbullet}$ for
$\xi<\kappa$.   By 531K, copied into $\frak A$,
there are $\ofamily{\xi}{\kappa}{c_{\xi}}$ and
$\Gamma_0\in[\kappa]^{\kappa}$ such that $c_{\xi}\Bsubseteq a_{\xi}$ and
$\bar\mu c_{\xi}\ge\bover13$ for each $\xi$, and

\Centerline{$\bar\mu(\inf_{\xi\in I}(c_{\xi}\Bcap e_{\xi})
  \Bcap\inf_{\eta\in J}(c_{\eta}\Bsetminus e_{\eta}))
=\Bover1{2^{\#(I\cup J)}}
  \bar\mu(\inf_{\xi\in I\cup J}c_{\xi})$}

\noindent whenever $I$, $J\subseteq\Gamma_0$ are disjoint finite sets.

At this point, recall that $(\kappa,\lambda)$ is supposed to be a
measure-precaliber pair of every probability algebra.   So there is a
$\Gamma\in[\Gamma_0]^{\lambda}$ such that $\inf_{\xi\in I}c_{\xi}\ne 0$ for
every finite $I\subseteq\Gamma$.   It follows at once that
$\inf_{\xi\in I}(a_{\xi}\Bcap e_{\xi})
\Bcap\inf_{\eta\in J}(a_{\eta}\Bsetminus e_{\eta})$ is non-zero
for all disjoint finite sets $I$, $J\subseteq\Gamma$.   But this means that
$X\cap\bigcap_{\xi\in I}K'_{\xi}\cap\bigcap_{\eta\in J}K''_{\eta}$ is
non-negligible, therefore non-empty, for all disjoint finite $I$,
$J\subseteq\Gamma$.

Set $K=\bigcap_{\xi\in\Gamma}K_{\xi}$, so that $K\subseteq X$ is
compact.   Then we have a continuous function $f:K\to\{0,1\}^{\Gamma}$
defined by setting

$$\eqalign{f(x)(\xi)&=1\text{ if }x\in K\cap E_{\xi}=K\cap K'_{\xi},\cr
&=0\text{ if }x\in K\setminus E_{\xi}=K\cap K''_{\xi}.\cr}$$

\noindent Now $f$ is surjective.   \Prf\ If $w\in\{0,1\}^{\Gamma}$ and
$L\subseteq\Gamma$ is finite, then

$$\eqalign{F_L
&=\{x:x\in X,\,x\in K'_{\xi}\text{ whenever }\xi\in L
  \text{ and }w(\xi)=1,\cr
&\mskip200mu x\in K''_{\xi}\text{ whenever }\xi\in L
  \text{ and }w(\xi)=0\}\cr}$$

\noindent is a non-empty closed set.   The family
$\{F_L:L\in[\Gamma]^{<\omega}\}$ is downwards-directed, so has non-empty
intersection;  and if $x$ is any point of the intersection, $x\in K$ and
$f(x)=w$.\ \Qed

As $\#(\Gamma)=\lambda$, $\{0,1\}^{\lambda}$ is a continuous image of a
compact subset of $X$.
}%end of proof of 531L

\leader{531M}{Proposition}\dvArevised{2015}\cmmnt{ ({\smc Plebanek 97})}
If $\kappa$ is
an infinite cardinal and $\{0,1\}^{\kappa}$ is a continuous image
of a closed subset of $X$ whenever $X$ is a compact Hausdorff space such
that $\kappa\in\MahR(X)$, then $\kappa$ is a measure-precaliber of every
probability algebra.

\proof{ It will be enough to show that $\kappa$ is a measure-precaliber
of $(\frak B_{\kappa},\bar\nu_{\kappa})$ (525I(a-i)).   Let
$\ofamily{\xi}{\kappa}{a_{\xi}}$ be a family in $\frak B_{\kappa}$
such that
$\inf_{\xi<\kappa}\bar\nu_{\kappa}a_{\xi}=\alpha>0$.   Choose
$\ofamily{\xi}{\kappa}{b_{\xi}}$ in
$\frak B_{\kappa}$ inductively, as follows.   Given
$\ofamily{\eta}{\xi}{b_{\eta}}$, let $\frak D_{\xi}$ be the closed
subalgebra of $\frak B_{\kappa}$ generated by
$\{b_{\eta}:\eta<\xi\}\cup\{a_{\xi}\}$.   Because
$\frak B_{\kappa}$ is homogeneous with Maharam type
$\kappa>\tau(\frak D_{\xi})$, it is relatively atomless over
$\frak D_{\xi}$, and there is a $b\in\frak B_{\kappa}$ such that
$\bar\nu_{\kappa}(b\Bcap c)=\bover12\bar\nu_{\kappa}c$ for every
$c\in\frak D_{\xi}$ (331B).   Set $b_{\xi}=b\Bcap a_{\xi}$;  then for
any $\eta<\xi$ we have

$$\eqalign{\bar\nu_{\kappa}(b_{\xi}\Bsymmdiff b_{\eta})&
=\bar\nu_{\kappa}b_{\xi}+\bar\nu_{\kappa}b_{\eta}
  -2\bar\nu_{\kappa}(b_{\xi}\Bcap b_{\eta})\cr&
=\Bover12\bar\nu_{\kappa}a_{\xi}+\bar\nu_{\kappa}b_{\eta}
  -\bar\nu_{\kappa}(a_{\xi}\Bcap b_{\eta})
\ge\Bover12\bar\nu_{\kappa}a_{\xi}\ge\Bover{\alpha}2.\cr}$$

\noindent Continue.

Let $\frak C$ be the subalgebra of $\frak B_{\kappa}$ generated by
$\{b_{\xi}:\xi<\kappa\}$, and $X$ its Stone space.   Then $\frak C$ is
isomorphic to the algebra of open-and-closed subsets of $X$, so we have
a Radon measure $\mu$ on $X$ defined by saying that
$\mu\widehat{c}=\bar\nu_{\kappa}c$ for every $c\in\frak C$, writing
$\widehat{c}$ for the open-and-closed subset of $X$ corresponding to $c$
(416Qa).   Now $\mu$ is strictly positive and we can identify $\frak C$
with a topologically dense subalgebra of the measure algebra of $\mu$.
It follows that $\mu$ has a \Mth\ component of type at least $\kappa$.
\Prf\Quer\ Otherwise, there would be a set $E\subseteq X$, of measure at
least $1-\bover14\alpha$, such that the Maharam type of the subspace
measure $\mu_E$ was less than $\kappa$.   But

\Centerline{$\mu(E\cap\widehat{b}_{\xi}\symmdiff\widehat{b}_{\eta})
\ge\bar\nu_{\kappa}(b_{\xi}\Bsymmdiff b_{\eta})-\Bover{\alpha}4
\ge\Bover{\alpha}4$}

\noindent whenever $\eta<\xi<\kappa$, so the topological density of the
measure algebra of $\mu_E$ is at least $\kappa$ (5A4B(h-ii))
and the Maharam
type of $\mu_E$ is at least $\kappa$ (521E(a-ii)).\ \Bang\QeD\  Thus
$\kappa\in\MahR(X)$.

Accordingly $\{0,1\}^{\kappa}$ is a continuous image
of a closed subset of $X$.   By 5A4C(d-iii),
there is a non-empty closed subset $K$ of $X$ such
that $\chi(x,K)\ge\kappa$ for every $x\in K$.   Let $D\subseteq\kappa$
be a maximal set such that $\{K\}\cup\{\widehat{b}_{\xi}:\xi\in D\}$ has
the finite intersection property.   Set
$Z=K\cap\bigcap_{\xi\in D}\widehat{b}_{\xi}$;  then $Z$ contains a point
$z$ say.   Because $\{b_{\xi}:\xi\in D\}$ is centered, so is
$\{a_{\xi}:\xi\in D\}$.

If $x\in X\setminus\{z\}$, then there is a $c\in\frak C$ such that
$x\in\widehat{c}$ and $z\notin\widehat{c}$;  accordingly there is a
$\zeta<\kappa$ such that one of $x$, $z$ belongs to
$\widehat{b}_{\zeta}$ and the other does not.   If
$\zeta\in D$ then $z\in\widehat{b}_{\zeta}$ and
$x\notin\widehat{b}_{\zeta}$, so $x\notin Z$.   If $\zeta\notin D$ then,
by the maximality of $D$, $Z\cap\widehat{b}_{\zeta}=\emptyset$, so that
$z\notin\widehat{b}_{\zeta}$, $x\in\widehat{b}_{\zeta}$ and again
$x\notin Z$.

Thus $Z=\{z\}$, and $\{z\}$ can be expressed as the intersection of
$\#(D)$ relatively open sets in $K$.   By 4A2Gd, it follows that
$\#(D)\ge\chi(z,K)\ge\kappa$, and we have already seen that
$\{a_{\xi}:\xi\in D\}$ is centered.   As $\ofamily{\xi}{\kappa}{a_{\xi}}$
is arbitrary, $\kappa$ is a measure-precaliber of $\frak B_{\kappa}$, as
required.
}%end of proof of 531M

\leader{531N}{}\cmmnt{ In 531M we have a space $X$ out of which there is
no surjection onto $\{0,1\}^{\kappa}$ because every non-empty closed set
has a point of character less than $\kappa$.   From stronger properties
of $\kappa$ we can get compact spaces with stronger topological
properties, as in the next two results.

\medskip

\noindent}{\bf Proposition} Let $\kappa$, $\kappa'$ and $\lambda$ be
infinite cardinals such that $(\kappa,\kappa')$ is not a
measure-precaliber pair of
$(\frak B_{\lambda},\bar\nu_{\lambda})$.   Then there is a compact
Hausdorff space $X$ such that $\kappa\in\MahR(X)$ and
$\chi(x,X)<\max(\kappa',\lambda^+)$ for every $x\in X$.

\proof{ Let
$\ofamily{\xi}{\kappa}{a_{\xi}}$ be a family in $\frak B_{\lambda}$,
with no centered subfamily with cardinal $\kappa'$, such that
$\inf_{\xi<\kappa}\bar\mu a_{\xi}=\alpha>0$.   Let
$\psi:\frak B_{\lambda}\to\Tau_{\lambda}$ be a lifting;  for each
$\xi<\kappa$, let $K_{\xi}\subseteq\psi a_{\xi}$ be a compact set of
measure at least $\bover12\alpha$.   If $D\subseteq\kappa$ and
$\#(D)=\kappa'$, then there is a finite set $I\subseteq D$ such that
$\inf_{\xi\in I}a_{\xi}=0$, in which case
$\bigcap_{\xi\in I}K_{\xi}
\subseteq\bigcap_{\xi\in I}\psi a_{\xi}=\emptyset$.   Thus
$\{\xi:x\in K_{\xi}\}$ has cardinal less than $\kappa'$ for every
$x\in\{0,1\}^{\lambda}$.

Set

\Centerline{$X=\bigcap_{\xi<\kappa'}\{(x,y):x\in\{0,1\}^{\lambda}$,
$y\in\{0,1\}^{\kappa}$, $x\in K_{\xi}$ or $y(\xi)=0\}$,}

\noindent so that $X$ is a compact subset of
$\{0,1\}^{\lambda}\times\{0,1\}^{\kappa}$.   Now
$\chi((x,y),X)<\max(\kappa',\lambda^+)$ for every $(x,y)\in X$.   \Prf\
Set $D=\{\xi:\xi<\kappa$, $x\in K_{\xi}\}$, so that $\#(D)<\kappa'$.
For $I\in[\lambda]^{<\omega}$ and $J\in[D]^{<\omega}$ set

\Centerline{$V_{IJ}
=\{(x',y'):(x',y')\in X$, $x'\restr I=x\restr I$,
$y'\restr J=y\restr J\}$,}

\noindent so that
$\Cal V=\{V_{IJ}:I\in[\lambda]^{<\omega}$, $J\in[D]^{<\omega}\}$ is a
downwards-directed family of closed neighbourhoods of $(x,y)$.   If
$(x',y')\in\bigcap\Cal V$, then $x'=x$, so $x'\notin K_{\xi}$ for
$\xi\in\kappa\setminus D$, and $y'(\xi)=y(\xi)=0$ for $\xi\notin D$;
also $y'\restr D=y\restr D$, so $(x',y')=(x,y)$.   Thus
$\bigcap\Cal V=\{(x,y)\}$;  by 4A2Gd again, $\Cal V$ is a base of
neighbourhoods of $(x,y)$, and

\Centerline{$\chi((x,y),X)\le\#(\Cal V)\le\max(\#(D),\lambda)
<\max(\kappa',\lambda^+)$. \Qed}

Define $g:\{0,1\}^{\lambda}\times\{0,1\}^{\kappa}\to\{0,1\}^{\kappa}$
and $h:\{0,1\}^{\lambda}\times\{0,1\}^{\kappa}\to X$ by setting

$$\eqalign{g(x,y)(\xi)&=y(\xi)\text{ if }x\in K_{\xi},\cr
&=0\text{ otherwise},\cr
h(x,y)&=(x,g(x,y)),\cr}$$

\noindent for $\xi<\kappa$, $x\in\{0,1\}^{\lambda}$ and
$y\in\{0,1\}^{\kappa}$.   Write $\Sigma$ for the domain of the product
measure $\nu=\nu_{\lambda}\times\nu_{\kappa}$ on
$\{0,1\}^{\lambda}\times\{0,1\}^{\kappa}$.   Then the
$\sigma$-algebra
$\{F:F\subseteq X$, $h^{-1}[F]\in\Sigma\}$ contains all sets of the form
$\{(x,y):x(\eta)=1\}$ and $\{(x,y):y(\xi)=1\}$, so includes a base for
the topology of $X$ and therefore contains every open-and-closed set.
Accordingly we have an additive functional
$U\mapsto\nu h^{-1}[U]$ on the algebra of open-and-closed subsets of
$X$, which extends to a Radon probability measure $\mu$ on $X$
(416Qa again).
Set $F_{\xi}=\{(x,y):(x,y)\in X$, $y(\xi)=1\}$ for each $\xi<\kappa$;
then for any $\eta<\xi<\kappa$,

$$\eqalign{\mu(F_{\xi}\setminus F_{\eta})
&=\nu h^{-1}[F_{\xi}\setminus F_{\eta}]\cr
&\ge\nu\{(x,y):x\in K_{\xi},\,y(\xi)=1,\,y(\eta)=0\}
=\Bover14\nu_{\lambda}K_{\xi}
\ge\Bover18\alpha.\cr}$$

\noindent As in the proof of 531M,
this shows that the measure algebra of $\mu$ must
have a homogeneous principal ideal with Maharam type at least $\kappa$,
and $\kappa\in\Mah(X)$.
}%end of proof of 531N

\leader{531O}{}\cmmnt{ Putting these ideas together with 531L, we come to
the following.

\medskip

\noindent}{\bf Proposition}\cmmnt{ ({\smc Kunen \& Mill 95},
{\smc Plebanek 95})}
Let $\kappa$ be a regular
%\query:  is this needed? OK if \cf\kappa\ge\omega_2
%the problem seems to be in (b)not(i)=>not(ii)
%what about  \kappa=\omega_{\omega} ?  using measure-precaliber, of course
infinite cardinal.   Then the following are equiveridical:

(i) $\kappa$ is a measure-precaliber of every measurable algebra;

(ii) if $X$ is a compact Hausdorff space such that $\kappa\in\MahR(X)$,
then $\chi(x,X)\ge\kappa$ for some $x\in X$.

\proof{{\bf (a)} Consider first the case $\kappa\ge\omega_2$.

\medskip

\quad{\bf (i)$\Rightarrow$(ii)} If $\kappa\in\MahR(X)$, then
$\{0,1\}^{\kappa}$ is a continuous image of a compact subset of
$X$, by 531Lb.   By 5A4C(d-iii) again and 5A4Bb, it follows at once that
$\chi(x,X)\ge\kappa$ for many points $x\in X$.

\medskip

\quad{\bf not-(i)$\Rightarrow$not-(ii)} By
525Ib there is a $\lambda<\kappa$ such that $\kappa$ is not a precaliber
of $\frak B_{\lambda}$, and therefore not a measure-precaliber of
$(\frak B_{\lambda},\bar\nu_{\lambda})$ (525Db).
Now 531N tells us that there is a compact
Hausdorff space $X$ such that $\kappa\in\MahR(X)$ and
$\chi(x,X)<\max(\kappa,\lambda^+)=\kappa$ for every $x\in X$.

\medskip

{\bf (b)} Now suppose that $\kappa=\omega_1$.

\medskip

\quad{\bf (i)$\Rightarrow$(ii)}
\Quer\ Suppose, if possible, that $\omega_1$ is a precaliber of every
probability algebra, but that there is a first-countable compact
Hausdorff space $X$ with $\omega_1\in\MahR(X)$.   Let $\mu$ be a \Mth\
Radon probability measure on $X$ with Maharam type $\omega_1$, and
$(\frak A,\bar\mu)$ its measure algebra;  let
$\ofamily{\xi}{\omega_1}{c_{\xi}}$ be a $\tau$-generating stochastically
independent family of elements of measure $\bover12$ in $\frak A$.   As
in 531J, there is for each $a\in\frak A$ a countable
$J^*(a)\subseteq\omega_1$ such that $a$ belongs to the closed subalgebra
of $\frak A$ generated by $\{c_{\xi}:\xi\in J^*(a)\}$.

For each $x\in X$, let $\Cal U_x$ be a countable base of open
neighbourhoods of $x$, and set
$A_x=\{U^{\ssbullet}:U\in\Cal U_x\}$,
$J^{\dagger}(x)=\bigcup_{a\in A_x}J^*(a)$.   Then $J^{\dagger}(x)$ is
countable.   For $\xi<\omega_1$, set
$D_{\xi}=\{x:J^{\dagger}(x)\subseteq\xi\}$;  then
$\ofamily{\xi}{\omega_1}{D_{\xi}}$ is a non-decreasing family with union
$X$.   Now $\omega_1$ is supposed to be a
precaliber of $\frak A$, so there must be a $\xi<\omega_1$ such that
$D_{\xi}$ has full outer measure (525Cc).

Let $G\subseteq X$ be open.  Then $G^{\ssbullet}$ belongs to the closed
subalgebra $\frak C_{\xi}$ of $\frak A$ generated by
$\{c_{\eta}:\eta<\xi\}$.   \Prf\ For each $x\in G\cap D_{\xi}$, there is
a $U_x\in\Cal U_x$ such that $U_x\subseteq G$.   Set
$H=\bigcup\{U_x:x\in G\cap D_{\xi}\}$, so that $H\subseteq G$ is open and
$G\cap D_{\xi}=H\cap D_{\xi}$;  as $D_{\xi}$ has full outer measure,
$G\setminus H$ is negligible and $H^{\ssbullet}=G^{\ssbullet}$.   But
414Aa once more tells us that
$H^{\ssbullet}=\sup_{x\in D_{\xi}}U_x^{\ssbullet}$,
and this belongs to $\frak C_{\xi}$, because
$J^{\dagger}(x)\subseteq\xi$ for every $x\in D_{\xi}$.\ \Qed

It follows at once that $F^{\ssbullet}\in\frak C_{\xi}$ for every closed
$F\subseteq X$.   Because $\mu$ is inner regular with respect to the
closed sets, $\frak C_{\xi}$ is order-dense in $\frak A$ and
$\frak A=\frak C_{\xi}$ has Maharam type $\#(\xi)<\omega_1$.\ \Bang

Thus (i)$\Rightarrow$(ii).

\medskip

\quad{\bf not-(i)$\Rightarrow$not-(ii)} Suppose that (i) is false.

\medskip

\qquad\grheada\ By 525J,
$\cov\Cal N_{\omega_1}=\omega_1$ and there is
a family $\ofamily{\xi}{\omega_1}{A_{\xi}}$ of negligible
subsets of $\{0,1\}^{\omega_1}$ covering $\{0,1\}^{\omega_1}$.
For each $\xi<\omega_1$, let $A'_{\xi}\supseteq A_{\xi}$ be a
negligible set determined by
coordinates in a countable set $J_{\xi}\subseteq\omega_1$;  set
$\tilde A_{\xi}=\bigcup\{A'_{\eta}:\eta<\xi$, $J_{\eta}\subseteq\xi\}$;
then $\tilde A_{\xi}$ is determined by coordinates less than $\xi$.
Set $H_{\xi}=\{y\restr\xi:y\in\tilde A_{\xi}\}$, so that
$H_{\xi}$ is a $\nu_{\xi}$-negligible subset of $\{0,1\}^{\xi}$.

We see that
$\ofamily{\xi}{\omega_1}{\tilde A_{\xi}}$ is non-decreasing, and

\Centerline{$\bigcup_{\xi<\omega_1}\tilde A_{\xi}
=\bigcup_{\xi<\omega_1}A'_{\xi}=\{0,1\}^{\omega_1}$.}

\noindent Consequently $y\restr\xi\in H_{\xi}$ whenever
$\eta\le\xi<\omega_1$, $y\in\{0,1\}^{\omega_1}$ and
$y\restr\eta\in H_{\eta}$, while for every
$y\in\{0,1\}^{\omega_1}$ there is a $\xi<\omega_1$ such that
$y\restr\xi\in H_{\xi}$.

\medskip

\qquad\grheadb\ Set $Y=\{0\}\cup\{2^{-n}:n\in\Bbb N\}\subseteq[0,1]$.
For $\xi\le\omega_1$ define $\phi_{\xi}:Y^{\xi}\to\{0,1\}^{\xi}$ by setting

$$\eqalign{\phi_{\xi}(x)(\eta)
&=0\text{ if }\eta<\xi\text{ and }x(\eta)=0,\cr
&=1\text{ for other }\eta<\xi.\cr}$$

\noindent Observe that $\phi_{\xi}$ is Borel measurable for every
$\xi<\omega_1$.   Choose $\langle X_{\xi}\rangle_{\xi<\omega_1}$,
and $\langle K_{\xi n}\rangle_{\xi<\omega_1,n\in\Bbb N}$
inductively, as follows.   The inductive hypothesis will be that $X_{\xi}$
is a compact subset of $Y^{\xi}$, $\phi_{\xi}[X_{\xi}]$ is
$\nu_{\xi}$-conegligible
in $\{0,1\}^{\xi}$, $\phi_{\xi}\restr X_{\xi}$ is injective
and $x\restr\eta\in X_{\eta}$ whenever $x\in X_{\xi}$
and $\eta\le\xi<\omega_1$.

Start with
$X_0=Y^0=\{\emptyset\}$ and $\phi_0:X_0\to\{0,1\}^0$ the identity map.

Given $\xi<\omega_1$ and $X_{\xi}\subseteq Y^{\xi}$,
then 433D tells us that there is a Radon measure $\mu_{\xi}$ on
$X_{\xi}$ such that $\nu_{\xi}$ is the image measure
$\mu_{\xi}\phi_{\xi}^{-1}$.   Let $\sequencen{K_{\xi n}}$ be a
disjoint sequence of compact subsets of
$X_{\xi}\setminus\phi_{\xi}^{-1}[H_{\xi}]$ with $\mu_{\xi}$-conegligible
union.   Set

$$\eqalign{X_{\xi+1}
&=\{x:x\in Y^{\xi+1},\,x\restr\xi\in X_{\xi},\,x(\xi)=0\}\cr
&\mskip150mu
\cup\bigcup_{n\in\Bbb N}\{x:x\in Y^{\xi+1},\,x\restr\xi\in K_{\xi n},\,
x(\xi)=2^{-n}\}.\cr}$$

\noindent It is easy to see that $X_{\xi+1}$ is compact and
$\phi_{\xi+1}\restr X_{\xi+1}$ is injective, while surely
$x\restr\eta\in X_{\eta}$ whenever $x\in X_{\xi+1}$ and $\eta\le\xi+1$,
just because $x\restr\xi\in X_{\xi}$.   Also

\Centerline{$\phi_{\xi+1}[X_{\xi+1}]
\supseteq\{y:y\in\{0,1\}^{\xi+1}$,
  $y\restr\xi\in\bigcup_{n\in\Bbb N}\phi_{\xi}[K_{\xi n}]\}$}

\noindent is $\nu_{\xi+1}$-conegligible because
$\phi_{\xi}[K_{\xi n}]$ must be analytic for every $n$ and

\Centerline{$\nu_{\xi}(\bigcup_{n\in\Bbb N}\phi_{\xi}[K_{\xi n}])
=\mu_{\xi}(\bigcup_{n\in\Bbb N}K_{\xi n})=1$}

\noindent because $\phi_{\xi}\restr X_{\xi}$ is injective.

Given that $X_{\eta}$ has been defined for $\eta<\xi$,
where $\xi<\omega_1$ is a non-zero limit ordinal, set

\Centerline{$X_{\xi}=\{x:x\in Y^{\xi}$, $x\restr\eta\in X_{\eta}$ for
every $\eta<\xi\}$.}

\noindent Of course $X_{\xi}$ is compact and $\phi_{\xi}\restr X_{\xi}$ is
injective.    To see that $\phi_{\xi}[X_{\xi}]$ is conegligible, observe
that

\Centerline{$W
=\bigcap_{\eta<\xi}\{y:y\in\{0,1\}^{\xi}$,
$y\restr\eta\in\phi_{\eta}[X_{\eta}]\}$}

\noindent is conegligible.   But if $y\in W$ and we choose
$x_{\eta}\in X_{\eta}$ such that $\phi_{\eta}(x_{\eta})=y\restr\eta$ for
each $\eta<\xi$, then we must have $x_{\zeta}=x_{\eta}\restr\zeta$ whenever
$\zeta\le\eta<\xi$, because $\phi_{\zeta}\restr X_{\zeta}$ is injective;
so there is an $x\in Y^{\xi}$ such that $x_{\eta}=x\restr\eta$ for every
$\eta<\xi$, in which case $x\in X_{\xi}$ and $\phi_{\xi}(x)=y$.   Thus
$\phi_{\xi}[X_{\xi}]\supseteq W$ is conegligible.

\medskip

\qquad\grheadc\ At the end of the induction, set

\Centerline{$X=\{x:x\in Y^{\omega_1}$,
$x\restr\xi\in X_{\xi}$ for every $\xi<\omega_1\}$,
\quad$\phi=\phi_{\omega_1}\restr X$.}

\noindent As in the limit stage of the construction in ($\beta$), we see
that $X$ is a closed subset of $Y^{\omega_1}$, so with the subspace
topology is a zero-dimensional compact Hausdorff space.
This time, we do not
expect that $\phi[X]$ should be conegligible in $\{0,1\}^{\omega_1}$, but
we find that it has full outer measure.   \Prf\ If
$K\subseteq\{0,1\}^{\omega_1}$ is a
non-negligible closed G$_{\delta}$ set, there is a $\xi<\omega_1$
such that $K$ is determined by coordinates less than $\xi$.   Set
$K'=\{y\restr\xi:y\in K\}$;  then $\nu_{\xi}K'=\nu_{\omega_1}K>0$,
so there is an $x_0\in X_{\xi}$ such that $\phi_{\xi}(x_0)\in K'$.
Extending $x_0$ to $x\in Y^{\omega_1}$ by setting $x(\eta)=0$ for
$\xi\le\eta<\omega_1$, we see by induction on $\zeta$ that
$x\restr\zeta\in X_{\zeta}$ for $\xi\le\zeta<\omega_1$, so
$x\in X$;  also $\phi(x)\restr\xi=\phi_{\xi}(x_0)\in K'$, so
$\phi(x)\in K$ and $K$ meets $\phi[X]$.   As $\nu_{\omega_1}$
is completion regular, $\phi[X]$ has full outer measure.\ \Qed

\medskip

\qquad\grheadd\ $X$ is first-countable.   \Prf\ If $x\in X$, $\xi<\omega_1$
and $x(\xi)\ne 0$, then $x\restr(\xi+1)$ belongs to $X_{\xi+1}$, and there
must be some $n\in\Bbb N$ such that $x(\xi)=2^{-n}$ and
$x\restr\xi\in K_{\xi n}$;  in which case
$\phi_{\xi}(x\restr\xi)\notin H_{\xi}$.   Now take any $x\in X$.
Then there is a $\xi<\omega_1$ such that $\phi(x)\in\tilde A_{\xi}$
and $\phi_{\xi}(x)=\phi(x)\restr\xi$ belongs to $H_{\xi}$.   In this case,
$V=\{x':x'\in X$, $x'\restr\xi=x\restr\xi\}$ is a G$_{\delta}$ subset of
$X$ containing $x$.   But if $x'\in V$ then, for any $\eta\ge\xi$,
$\phi_{\eta}(x'\restr\eta)\in H_{\eta}$ and $x'(\eta)=0$.   Thus $V=\{x\}$.
By 4A2Gd once more, $x$ has a countable base of neighbourhoods in $X$;
as $x$ is arbitrary, $X$ is first-countable.\ \Qed

\medskip

\qquad\grheade\  By 234F, there is a measure $\lambda$ on
$X$ such that $\phi$ is \imp\ for $\lambda$ and $\nu_{\omega_1}$.
Of course $\lambda$ is a probability measure.   Now for
any $\xi<\omega_1$ and $n\in\Bbb N$,

\Centerline{$\{x:x\in X$, $x(\xi)=0\}=\{x:\phi(x)(\xi)=0\}$,}

$$\eqalign{\{x:x\in X,\,x(\xi)=2^{-n}\}
&=\{x:\phi(x)(\xi)=1,\,x\restr\xi\in K_{\xi n}\}\cr
&=\{x:\phi(x)(\xi)=1,\,\phi_{\xi}(x\restr\xi)\in\phi_{\xi}[K_{\xi n}]\}\cr
&=\{x:\phi(x)(\xi)=1,\,\phi(x)\restr\xi\in\phi_{\xi}[K_{\xi n}]\}\cr}$$

\noindent are measured by $\lambda$.   So the domain of $\lambda$ includes
a base for the topology of the zero-dimensional compact Hausdorff space
$X$.   By 416Qa once more,
there is a Radon measure $\mu$ on $X$ agreeing with $\lambda$ on the
open-and-closed subsets of $X$;  by the Monotone Class Theorem
(136C), $\mu$ and
$\lambda$ agree on the $\sigma$-algebra generated by the open-and-closed
sets, that is, the Baire $\sigma$-algebra of $X$ (4A3Od).
In particular, setting
$E_{\xi}=\{x:x\in X$, $x(\xi)=0\}$ for $\xi<\omega_1$,

$$\eqalign{\mu(E_{\xi}\cap E_{\eta})
=\lambda(E_{\xi}\cap E_{\eta})
&=\nu_{\omega_1}\{y:y\in\{0,1\}^{\omega_1},\,y(\xi)=y(\eta)=0\}\cr
&=\Bover12\text{ if }\xi=\eta<\omega_1,\cr
&=\Bover14\text{ if }\xi,\,\eta<\omega_1\text{ are different.}\cr}$$

\noindent It follows that $\mu(E_{\xi}\symmdiff E_{\eta})=\bover12$ for all
distinct $\xi$, $\eta<\omega_1$, so $\mu$ has uncountable Maharam type and
$\omega_1\in\MahR(X)$.   Thus $X$ and $\mu$ witness that (ii) is false.

\medskip

{\bf (c)} Finally, if $\kappa=\omega$, both (i) and (ii) are true for
elementary reasons (525Fa).
}%end of proof of 531O

\leader{531P}{}\cmmnt{ In 531O we saw that if $\omega_1$ is not a
precaliber of every measurable algebra then there is a first-countable
compact Hausdorff space with a Radon measure with Maharam type $\omega_1$.
With a sharper hypothesis, and rather more work, we can get a
stronger version, as follows.

\medskip

\noindent}{\bf Lemma}\dvAnew{2024} Let $Y$ be a zero-dimensional compact
metrizable space, $\mu$ an atomless
Radon probability measure on $Y$, $A\subseteq Y$ a $\mu$-negligible set
and $\Cal Q$ a countable family of closed subsets of $Y$.
Then there are closed sets $K$, $L\subseteq Y$, with union $Y$, such that

\inset{$K\cup L=Y$,\quad $K\cap L\cap A=\emptyset$,
\quad$\mu(K\cap L)\ge\bover12$,

$K\cap Q=\overline{Q\setminus L}$
and $L\cap Q=\overline{Q\setminus K}$ for every $Q\in\Cal Q$.}

\proof{ We can of course suppose that $\emptyset\in\Cal Q$.
For each $Q\in\Cal Q$ let $D_Q$ be a countable dense subset of $Q$;  let
$S\subseteq Y\setminus(A\cup\bigcup_{Q\in\Cal Q}D_Q)$ be a closed set of
measure at least $\bover12$.
(This is where we need to know that $\mu$ is atomless, so that every
$D_Q$ is negligible.)
Let $\Cal U$ be a countable base for the topology of $Y$ consisting of
open-and-closed sets and
let $\sequencen{(U_n,Q_n)}$ run over $\Cal U\times\Cal Q$.   Choose
inductively sequences
$\sequencen{G_n}$, $\sequencen{H_n}$ of open-and-closed subsets of
$Y\setminus S$, as follows.
Start with $G_0=H_0=\emptyset$.   Given that $G_n$ and $H_n$ are
disjoint from each other and from
$S$, then

\inset{----- if $U_n\cap S=\emptyset$, take
$G_{n+1}=G_n\cup(U_n\setminus H_n)$ and
$H_{n+1}=H_n$;

----- if $U_n\cap S\cap Q_n\ne\emptyset$, $U=U_n\setminus(G_n\cup H_n)$
is open and includes $U_n\cap S\cap Q_n$;  as $D_{Q_n}$ is dense in $Q_n$,
$\overline{U\cap D_{Q_n}}$ includes $U\cap Q_n$ which meets $S$ so cannot be
included in $D_{Q_n}$, and $U\cap D_{Q_n}$ must be infinite;
take two of its points $y$, $y'$ say;  neither belongs to $S$ so
we can enlarge $G_n$ and $H_n$ to disjoint open-and-closed subsets
$G_{n+1}$, $H_{n+1}$ of $Y\setminus S$
containing $y$, $y'$ respectively, and therefore both meeting
$U_n\cap Q_n$;

----- otherwise, take $G_{n+1}=G_n$ and $H_{n+1}=H_n$.}

At the end of the induction, set $G=\bigcup_{n\in\Bbb N}G_n$ and
$H=\bigcup_{n\in\Bbb N}H_n$, so
that $G$ and $H$ are disjoint open subsets of $Y\setminus S$.   Now if
$y$ is any point of $Y\setminus S$, there must be some $n$ such that
$y\in U_n\subseteq Y\setminus S$, so that
$y\in G_{n+1}\cup H_n$;  thus $Y=G\cup H\cup S$.   Set $K=G\cup S=Y\setminus H$,
$L=H\cup S=Y\setminus G$;  then $K$ and $L$ are closed sets with union
$Y$, and $K\cap L=S$ has measure at least $\bover12$ and
is disjoint from $A$.

If $Q\in\Cal Q$, $y\in S\cap Q$ and $U$ is any neighbourhood
of $y$, there is an$\pi[Z\setminus W]\ne Q$
$n\in\Bbb N$ such that $Q_n=Q$ and $y\in U_n\subseteq U$.   In this
case, $U_n\cap S\cap Q_n\ne\emptyset$ and $G\cap Q\supseteq G_{n+1}\cap Q_n$,
$H\cap Q\supseteq H_{n+1}\cap Q$ both meet
$U_n\cap Q$.   As $U$ and $y$ are arbitrary,

\Centerline{$K\cap L\cap Q=S\cap Q
\subseteq\overline{G\cap Q}\cap\overline{H\cap Q}$,}

\Centerline{$K\cap Q\subseteq(S\cap Q)\cup(G\cap Q)
\subseteq\overline{G\cap Q}=\overline{Q\setminus L}$}

\noindent and similarly $L\cap Q\subseteq\overline{Q\setminus K}$.
At the same time,
$K\supseteq Q\setminus L$ and $L\supseteq Q\setminus K$,
so $K\cap Q=\overline{Q\setminus L}$ and $L\cap Q=\overline{Q\setminus K}$.
Thus $K$ and $L$ fulfil all the specifications.
}%end of proof of 531P

\leader{531Q}{Proposition}\dvAformerly{5{}31P} Suppose that
$\cf\Cal N_{\omega}=\omega_1$.   Then there is a hereditarily separable
perfectly normal compact Hausdorff space $X$, of weight $\omega_1$,
with a Radon probability
measure of Maharam type $\omega_1$ such that every negligible
set is metrizable.

\proof{ For $\eta\le\xi\le\omega_1$ and $x\in\{0,1\}^{\xi}$, set
$\pi_{\eta\xi}(x)=x\restr\eta$;  write $\pi_{\eta}$ for
$\pi_{\eta\omega_1}:\{0,1\}^{\omega_1}\to\{0,1\}^{\eta}$.
As in 531I, $\nu_{\xi}$ is to be the usual measure on $\{0,1\}^{\xi}$.

\medskip

{\bf (a)} Choose

\Centerline{$\langle f_{\xi}\rangle_{\omega\le\xi\le\omega_1}$,
$\langle X_{\xi}\rangle_{\omega\le\xi\le\omega_1}$,
$\langle\mu_{\xi}\rangle_{\omega\le\xi<\omega_1}$,
$\langle K_{\xi}\rangle_{\omega\le\xi<\omega_1}$,
$\langle L_{\xi}\rangle_{\omega\le\xi<\omega_1}$,}

\Centerline{$\langle
Q'_{\xi\theta}\rangle_{\omega\le\xi\le\theta<\omega_1}$,
$\langle Q_{\delta\xi}\rangle_{\omega\le\delta\le\xi<\omega_1}$,
$\langle
Q_{\eta\delta\xi}\rangle_{\omega\le\eta\le\delta\le\xi<\omega_1}$,
$\langle A_{\xi\theta}\rangle_{\omega\le\xi\le\theta<\omega_1}$,
$\langle A_{\xi}\rangle_{\omega\le\xi<\omega_1}$}

\noindent inductively,
as follows.   Every $X_{\xi}$, $K_{\xi}$, $L_{\xi}$, $Q'_{\xi\theta}$,
$Q_{\delta\xi}$ and $Q_{\eta\delta\xi}$ is to be a closed subset of
$\{0,1\}^{\xi}$, every $f_{\xi}$ will be a Baire measurable
surjection from $\{0,1\}^{\xi}$ onto $X_{\xi}$, $\mu_{\xi}$ will always be the
Radon probability measure $\nu_{\xi}f_{\xi}^{-1}$ on $\{0,1\}^{\xi}$, and
$A_{\xi}$ and $A_{\xi\theta}$ will always be $\mu_{\xi}$-negligible subsets of
$\{0,1\}^{\xi}$.

Given that $\omega\le\xi\le\omega_1$ and that $K_{\eta}$, $L_{\eta}$ are closed
subsets of $\{0,1\}^{\eta}$ covering $\{0,1\}^{\eta}$ whenever
$\omega\le\eta<\xi$, then define $f_{\xi}(x)(\eta)$, for
$x\in\{0,1\}^{\xi}$ and $\eta<\xi$, by setting

$$\eqalign{f_{\xi}(x)(\eta)
&=1\text{ if }\eta\ge\omega\text{ and }x\restr\eta\notin L_{\eta},\cr
&=0\text{ if }\eta\ge\omega\text{ and }x\restr\eta\notin K_{\eta},\cr
&=x(\eta)\text{ otherwise.}\cr}$$

\noindent (Thus the induction starts with $f_{\omega}(x)=x$ for
$x\in\{0,1\}^{\omega}$.)   Then $f_{\xi}:\{0,1\}^{\xi}\to\{0,1\}^{\xi}$ is Baire
measurable (4A3Ne).   Set

\Centerline{$X_{\xi}=\bigcap_{\omega\le\eta<\xi}\{x:x\in\{0,1\}^{\xi}$,
$x(\eta)=1$ or $x\restr\eta\in L_{\eta}$, $x(\eta)=0$ or
$x\restr\eta\in K_{\eta}\}$;}

\noindent then $X_{\xi}\subseteq\{0,1\}^{\xi}$ is compact,
$f_{\xi}(x)\in X_{\xi}$ for every
$x\in\{0,1\}^{\xi}$, and $f_{\xi}(x)=x$ for every $x\in X_{\xi}$.
So $f_{\xi}[\{0,1\}^{\xi}]=X_{\xi}$.

If now $\xi<\omega_1$, $f_{\xi}:\{0,1\}^{\xi}\to\{0,1\}^{\xi}$ is Borel
measurable;   by 433E, $f_{\xi}$ is $\nu_{\xi}$-almost-continuous, and the
image measure $\mu_{\xi}=\nu_{\xi}f_{\xi}^{-1}$ is a Radon measure on the
compact metrizable space $\{0,1\}^{\xi}$ (418I).   Of course
$\mu_{\xi}X_{\xi}=1$.   Because $\{0,1\}^{\xi}$ has countable weight, or
otherwise, $\mu_{\xi}$ has countable Maharam type (531Ad);   by 524Pb,
$\mu_{\xi}$ is inner regular with respect to a family with cardinal at most
$\cf\Cal N_{\omega}=\omega_1$, which we may suppose to consist of closed sets;
let $\langle Q'_{\xi\theta}\rangle_{\xi\le\theta<\omega_1}$ run over such a
family.   Similarly, there is a family
$\langle A_{\xi\theta}\rangle_{\xi\le\theta<\omega_1}$ running over a
cofinal subset of the null ideal of $\mu_{\xi}$ (524Pf).
Next, for $\omega\le\delta\le\xi$, let
$Q_{\delta\xi}\subseteq\pi_{\delta\xi}^{-1}[Q'_{\delta\xi}]$ be the
compact $\mu_{\xi}$-self-supporting set of the same $\mu_{\xi}$-measure as
$\pi_{\delta\xi}^{-1}[Q'_{\delta\xi}]$ (414F).
Note that $Q_{\delta\xi}$ will always be
included in $X_{\xi}$, because $\mu_{\xi}X_{\xi}=1$.   Set
$Q_{\eta\delta\xi}=X_{\xi}\cap\pi_{\delta\xi}^{-1}[Q_{\eta\delta}]$
for $\omega\le\eta\le\delta\le\xi$, and

\Centerline{$\Cal A_{\xi}
=\{\pi_{\eta\xi}^{-1}[A_{\eta\delta}]:
   \omega\le\eta\le\delta\le\xi\}$,
\quad$A_{\xi}=\bigcup\{A:A\in\Cal A_{\xi}$, $\mu_{\xi}A=0\}$;}

\noindent because $\Cal A_{\xi}$ is countable, $A_{\xi}$ is
$\mu_{\xi}$-negligible.   By 531P, we can find
closed sets $K_{\xi}$, $L_{\xi}$ covering $\{0,1\}^{\xi}$ such that
$\mu_{\xi}(K_{\xi}\cap L_{\xi})\ge\bover12$,
$K_{\xi}\cap L_{\xi}\cap A_{\xi}=\emptyset$,
$K_{\xi}\cap Q_{\eta\delta\xi}=\overline{Q_{\eta\delta\xi}\setminus L_{\xi}}$
and
$L_{\xi}\cap Q_{\eta\delta\xi}=\overline{Q_{\eta\delta\xi}\setminus K_{\xi}}$
whenever $\omega\le\eta\le\delta\le\xi$.

This deals with the inductive step to a successor ordinal $\xi+1$ when
$\omega\le\xi<\omega_1$.   For limit ordinals $\xi\in\ooint{\omega,\omega_1}$,
$f_{\xi}$ is defined by $\langle(K_{\eta},L_{\eta})\rangle_{\omega\le\eta<\xi}$,
so the induction proceeds directly to $\xi$.

\medskip

{\bf (b)} At the end of the induction, write $f$ for $f_{\omega_1}$ and
$X$ for $X_{\omega_1}=f[\{0,1\}^{\omega_1}]$.
If $z\in\{0,1\}^{\omega_1}$ and $\xi<\omega_1$, the formula for
$f_{\xi}$ in (a) shows that
$f(z)(\eta)=f_{\xi}(z\restr\xi)(\eta)$ for every $\eta<\xi$, that is,
that $f(z)\restr\xi=f_{\xi}(z\restr\xi)$.   Next, $\nu_{\omega_1}f^{-1}$
measures every Baire subset of $\{0,1\}^{\omega_1}$ (use 4A3Na), so we have
a Radon measure $\mu$ on $\{0,1\}^{\omega_1}$ defined by saying that
$\mu V=\nu_{\omega_1}f^{-1}[V]$ for every Baire set
$V\subseteq\{0,1\}^{\omega_1}$ (432F);  of course $\mu V=0$ for every
open-and-closed set $V$ disjoint from $X$, so $\mu X=1$.

At the same time, it will be helpful to fill in the definition of
$\langle Q_{\eta\delta\xi}\rangle_{\omega\le\eta\le\delta\le\xi<\omega_1}$
by taking
$Q_{\eta\delta\omega_1}=X\cap\pi_{\delta}^{-1}[Q_{\eta\delta}]$ when
$\omega\le\eta\le\delta<\omega_1$.

\medskip

{\bf (c)} Some simple facts.

\medskip

\quad{\bf (i)} I have already observed that
$\pi_{\xi}f=f_{\xi}\pi_{\xi}$ for $\omega\le\xi<\omega_1$;  consequently

\Centerline{$X_{\xi}=f_{\xi}[\{0,1\}^{\xi}]
=f_{\xi}[\pi_{\xi}[\{0,1\}^{\omega_1}]]
=\pi_{\xi}[f[\{0,1\}^{\omega_1}]]=\pi_{\xi}[X]$}

\noindent and

$$\eqalign{\mu_{\xi}V
&=\nu_{\xi}f_{\xi}^{-1}[V]
=(\nu_{\omega_1}\pi_{\xi}^{-1})f_{\xi}^{-1}[V]
=\nu_{\omega_1}(f_{\xi}\pi_{\xi})^{-1}[V]\cr
&=\nu_{\omega_1}(\pi_{\xi}f)^{-1}[V]
=(\nu_{\omega_1}f^{-1})\pi_{\xi}^{-1}[V]
=\mu\pi_{\xi}^{-1}[V]\cr}$$

\noindent for every open-and-closed set $V\subseteq\{0,1\}^{\xi}$.
Thus the Radon measures
$\mu\pi_{\xi}^{-1}$ and $\mu_{\xi}$ are identical.

\medskip

\quad{\bf (ii)} Equally, if
$\omega\le\eta\le\xi<\omega_1$,

\Centerline{$X_{\eta}=\pi_{\eta}[X]=\pi_{\eta\xi}[\pi_{\xi}[X]]
=\pi_{\eta\xi}[X_{\xi}]$}

\noindent and

\Centerline{$\mu_{\eta}
=\mu\pi_{\eta}^{-1}=\mu(\pi_{\eta\xi}\pi_{\xi})^{-1}
=(\mu\pi_{\xi}^{-1})\pi_{\eta\xi}^{-1}=\mu_{\xi}\pi_{\eta\xi}^{-1}$.}

\noindent Accordingly, if $\omega\le\eta\le\delta\le\xi<\omega_1$,

\Centerline{$\mu_{\xi}\pi_{\eta\xi}^{-1}[A_{\eta\delta}]
=\mu_{\eta}A_{\eta\delta}=0$.}

\noindent Thus in fact $A_{\xi}=\bigcup\Cal A_{\xi}$ and $K_{\xi}\cap
L_{\xi}$ is disjoint from
$\pi_{\eta\xi}^{-1}[A_{\eta\delta}]$ whenever
$\omega\le\eta\le\delta\le\xi<\omega_1$.

\medskip

\quad{\bf (iii)} If $\omega\le\eta\le\delta\le\zeta\le\xi\le\omega_1$ and
$\delta<\omega_1$, then

$$\eqalignno{\pi_{\zeta\xi}[Q_{\eta\delta\xi}]&
=\pi_{\zeta\xi}[X_{\xi}\cap\pi_{\delta\xi}^{-1}[Q_{\eta\delta}]]
=\pi_{\zeta\xi}[X_{\xi}
  \cap\pi_{\zeta\xi}^{-1}[\pi_{\delta\zeta}^{-1}[Q_{\eta\delta}]]]
=X_{\zeta}\cap\pi_{\delta\zeta}^{-1}[Q_{\eta\delta}]
\Displaycause{because $\pi_{\zeta\xi}[X_{\xi}]=X_{\zeta}$}
=Q_{\eta\delta\zeta}.\cr}$$

\medskip

\quad{\bf (iv)} If $z$, $z'\in X$,
$\omega\le\eta\le\delta<\omega_1$, $z\restr\delta=z'\restr\delta$ and
$z\restr\eta\in A_{\eta\delta}$,
then $z'=z$.  \Prf\ Suppose that $\delta\le\xi<\omega_1$ and
$z\restr\xi=z'\restr\xi$.   Then
$K_{\xi}\cap L_{\xi}$ does not meet
$\pi_{\eta\xi}^{-1}[A_{\eta\delta}]$, so does not contain
$z\restr\xi$.   Accordingly

\Centerline{$z(\xi)=1
\Longrightarrow z\restr\xi\in K_{\xi}
\Longrightarrow z\restr\xi\notin L_{\xi}
\Longrightarrow z'\restr\xi\notin L_{\xi}
\Longrightarrow z'(\xi)=1$,}

\noindent and similarly $z(\xi)=0\Rightarrow z'(\xi)=0$.   So an easy
induction on $\xi$ shows that $z(\xi)=z'(\xi)$ whenever
$\delta\le\xi<\omega_1$, and $z=z'$.\ \Qed


\medskip

{\bf (d)} We come now to the first key idea of this construction.
If $\omega\le\eta\le\delta\le\xi\le\omega_1$ and $\delta<\omega_1$, then
$g_{\eta\delta\xi}=\pi_{\delta\xi}\restr Q_{\eta\delta\xi}$ is an
irreducible continuous surjection onto $Q_{\eta\delta}$.   \Prf\
Of course $g_{\eta\delta\xi}$ is continuous, just because
$\pi_{\delta\xi}:\{0,1\}^{\xi}\to\{0,1\}^{\delta}$ is continuous, and
it is a surjection onto $Q_{\eta\delta}$ by (c-iii) just above.   To see that it
is irreducible, induce on $\xi$.   At the start,
$Q_{\eta\delta\delta}=Q_{\eta\delta}$ and
$g_{\eta\delta\delta}$ is an identity function, so is certainly
irreducible.

For the inductive step to $\xi+1$, given that
$\eta\le\delta\le\xi<\omega_1$ and $g_{\eta\delta\xi}$ is irreducible, consider
$h=\pi_{\xi,\xi+1}\restr Q_{\eta,\delta,\xi+1}$.   By (c-iii),
$h[Q_{\eta,\delta,\xi+1}]=Q_{\eta\delta\xi}$.   Note that $X_{\xi+1}$
can be identified with

\Centerline{$\{(x,1):
   x\in X_{\xi}\cap K_{\xi}\}\cup\{(x,0):x\in X_{\xi}\cap L_{\xi}\}
\subseteq\{0,1\}^{\xi}\times\{0,1\}$,}

\noindent so that $Q_{\eta,\delta,\xi+1}$ is identified with

\Centerline{$\{(x,1):x\in Q_{\eta\delta\xi}\cap K_{\xi}\}
     \cup\{(x,0):x\in Q_{\eta\delta\xi}\cap L_{\xi}\}$,}

\noindent and that with this identification $h$ becomes the first-coordinate
projection from $Q_{\eta,\delta,\xi+1}$ onto $Q_{\eta\delta\xi}$, while
$Q_{\eta\delta\xi}
=(Q_{\eta\delta\xi}\cap K_{\xi})\cup(Q_{\eta\delta\xi}\cap L_{\xi})$.
So 5A4L tells us that $h$ is irreducible.   But this means that
$g_{\eta,\delta,\xi+1}=hg_{\eta\delta\xi}$ is a composition of irreducible
continuous surjections and is irreducible, by 5A4C(d-iv)).

For the inductive step to a limit ordinal $\xi$ such that
$\delta<\xi\le\omega_1$, take a
cylinder set $V\subseteq\{0,1\}^{\xi}$ meeting $Q_{\eta\delta\xi}$.
This time, because $\xi$ is a limit ordinal, there is a $\zeta$ such
that $\delta\le\zeta<\xi$ and $V$ is determined by coordinates less than
$\zeta$.   Set $V'=\pi_{\zeta\xi}[V]$;  then $V'$ is a cylinder set in
$\{0,1\}^{\zeta}$ meeting
$\pi_{\zeta\xi}[Q_{\eta\delta\xi}]=Q_{\eta\delta\zeta}$.   Now

$$\eqalign{\pi_{\delta\xi}[Q_{\eta\delta\xi}\setminus V]
&=\pi_{\delta\zeta}[\pi_{\zeta\xi}
  [Q_{\eta\delta\xi}\setminus\pi_{\zeta\xi}^{-1}[V']]]\cr
&=\pi_{\delta\zeta}[\pi_{\zeta\xi}[Q_{\eta\delta\xi}]\setminus V']
=\pi_{\delta\zeta}[Q_{\eta\delta\zeta}\setminus V']
\subset Q_{\eta\delta}\cr}$$

\noindent because
$g_{\eta\delta\zeta}:Q_{\eta\delta\zeta}\to Q_{\eta\delta}$ is irreducible.
Thus the induction continues.\ \Qed

\medskip

{\bf (e)} It follows that if $H\subseteq X$ is closed, there is a
$\xi<\omega_1$ such that
$H=X\cap\pi_{\xi}^{-1}[\pi_{\xi}[H]]$ and $\pi_{\xi}\restr H$ is
irreducible.
\Prf\ For $\omega\le\eta\le\xi<\omega_1$,

\Centerline{$\mu_{\eta}\pi_{\eta}[H]
=\mu_{\xi}\pi_{\eta\xi}^{-1}[\pi_{\eta}[H]]
=\mu_{\xi}\pi_{\eta\xi}^{-1}[\pi_{\eta\xi}[\pi_{\xi}[H]]]
\ge\mu_{\xi}\pi_{\xi}[H]$.}

\noindent So we have an $\eta<\omega_1$ such that
$\mu_{\eta}\pi_{\eta}[H]=\mu_{\xi}\pi_{\xi}[H]$ whenever
$\eta\le\xi<\omega_1$.
Now recall that $\mu_{\eta}$ is inner regular with respect to
$\{Q'_{\eta\delta}:\eta\le\delta<\omega_1\}$.   So there is a countable
set
$I\subseteq\omega_1\setminus\eta$ such that
$\family{\delta}I{Q'_{\eta\delta}}$ is disjoint,
$Q'_{\eta\delta}\subseteq\pi_{\eta}[H]$ for every $\delta\in I$ and
$\sum_{\delta\in I}\mu_{\eta}Q'_{\eta\delta}=\mu_{\eta}\pi_{\eta}[H]$.

For each $\delta\in I$,

$$\eqalignno{\mu_{\delta}(Q_{\eta\delta}\setminus\pi_{\delta}[H])
&\le\mu_{\delta}
  (\pi_{\eta\delta}^{-1}[Q'_{\eta\delta}]\setminus\pi_{\delta}[H])\cr
&\le\mu_{\delta}
  (\pi_{\eta\delta}^{-1}[\pi_{\eta}[H]]\setminus\pi_{\delta}[H])\cr
&=\mu_{\delta}(\pi_{\eta\delta}^{-1}[\pi_{\eta}[H]])
 -\mu_{\delta}\pi_{\delta}[H]\cr
\displaycause{because surely
$\pi_{\delta}[H]\subseteq\pi_{\eta\delta}^{-1}[\pi_{\eta}[H]]$}
&=\mu_{\eta}\pi_{\eta}[H]-\mu_{\delta}\pi_{\delta}[H]
=0\cr}$$

\noindent by (c-ii) and the choice of $\eta$.   Because $Q_{\eta\delta}$ was
$\mu_{\delta}$-self-supporting,
and $\pi_{\delta}[H]$ is closed,
$Q_{\eta\delta}\subseteq\pi_{\delta}[H]$.   Because
$\pi_{\delta}\restr X\cap\pi_{\delta}^{-1}[Q_{\eta\delta}]$ is
irreducible,
$X\cap\pi_{\delta}^{-1}[Q_{\eta\delta}]\subseteq H$.

Set $\zeta=\sup(I\cup\{\eta\})<\omega_1$.   Since
$Q_{\eta\delta}\subseteq\pi^{-1}_{\eta\delta}[Q'_{\eta\delta}]$,
$\pi_{\delta\zeta}^{-1}[Q_{\eta\delta}]
\subseteq\pi_{\eta\zeta}^{-1}[Q'_{\eta\delta}]$ for
each $\delta\in I$;   as $\family{\delta}I{Q'_{\eta\delta}}$ is
disjoint, so is
$\family{\delta}I{\pi_{\delta\zeta}^{-1}[Q_{\eta\delta}]}$;  and

$$\eqalign{\sum_{\delta\in I}
  \mu_{\zeta}\pi_{\delta\zeta}^{-1}[Q_{\eta\delta}]
&=\sum_{\delta\in I}\mu_{\delta}Q_{\eta\delta}
=\sum_{\delta\in I}\mu_{\delta}\pi_{\eta\delta}^{-1}[Q'_{\eta\delta}]\cr
&=\sum_{\delta\in I}\mu_{\eta}Q'_{\eta\delta}
=\mu_{\eta}\pi_{\eta}[H]
=\mu_{\zeta}\pi_{\zeta}[H].\cr}$$

\noindent Because $X\cap\pi_{\delta}^{-1}[Q_{\eta\delta}]\subseteq H$,
$\pi_{\zeta}[H]\supseteq X_{\zeta}
  \cap\pi_{\delta\zeta}^{-1}[Q_{\eta\delta}]$
%$=Q_{\eta\delta\zeta}$
for every  $\delta\in I$.   So
$\pi_{\zeta}[H]\setminus\bigcup_{\delta\in I}
  \pi_{\delta\zeta}^{-1}[Q_{\eta\delta}]$
is $\mu_{\zeta}$-negligible and is included in $A_{\zeta\xi}$ for some
$\xi\ge\zeta$.
Repeating the arguments of the last two sentences at the new level, we
see that

\Centerline{$X_{\xi}\cap\bigcup_{\delta\in I}
  \pi_{\delta\xi}^{-1}[Q_{\eta\delta}]
\subseteq\pi_{\xi}[H]
\subseteq\bigcup_{\delta\in I}\pi_{\delta\xi}^{-1}[Q_{\eta\delta}]
  \cup\pi_{\zeta\xi}^{-1}A_{\zeta\xi}$.}

Now suppose that $V\subseteq\{0,1\}^{\omega_1}$ is an open set meeting
$H$.   Take $z\in V\cap H$.   If $z\in\pi_{\zeta}^{-1}[A_{\zeta\xi}]$, then
$z\restr\xi\ne z'\restr\xi$ for any other
$z'\in X$, by (c-iv);  so $z\restr\xi\notin\pi_{\xi}[H\setminus V]$ and
$\pi_{\xi}[H\setminus V]\ne\pi_{\xi}[H]$.
Otherwise, there is a $\delta\in I$ such that
$z\restr\xi\in\pi_{\delta\xi}^{-1}[Q_{\eta\delta}]$,
$z\in\pi_{\delta}^{-1}[Q_{\eta\delta}]$ and
$V\cap\pi_{\delta}^{-1}[Q_{\eta\delta}]$ is not empty.   Because
$\pi_{\delta}\restr X\cap\pi_{\delta}^{-1}[Q_{\eta\delta}]$ is
irreducible, $\pi_{\delta}[H\setminus V]$ cannot cover
$Q_{\eta\delta}\subseteq\pi_{\delta}[H]$.
Thus $\pi_{\delta}[H\setminus V]\ne\pi_{\delta}[H]$;  it follows at once
that $\pi_{\xi}[H\setminus V]\ne\pi_{\xi}[H]$ in this case also.
As $V$ is arbitrary, $\pi_{\xi}\restr H$ is irreducible.

I have still to check that $H=X\cap\pi_{\xi}^{-1}[\pi_{\xi}[H]]$.   If
$z$, $z'\in X$, $z\in H$ and $z'\restr\xi=z\restr\xi$, then if
$z\in\pi_{\zeta}^{-1}[A_{\zeta\xi}]$ we have
$z'=z\in H$.   Otherwise, there is some $\delta\in I$ such that
$z\in\pi_{\delta}^{-1}[Q_{\eta\delta}]$.   In this case,
$z'\restr\delta=z\restr\delta\in Q_{\eta\delta}$;   but
$X\cap\pi_{\delta}^{-1}[Q_{\eta\delta}]\subseteq H$, so again $z'\in H$.
So we have the
result.\ \Qed

\medskip

{\bf (f)} We are within sight of the end.   From (e) we see, first, that
if $H\subseteq X$ is closed then it is of the form
$X\cap\pi_{\xi}^{-1}[\pi_{\xi}[H]]$ for some $\xi<\omega_1$, so is a
zero set in $X$;  accordingly $X$ is perfectly normal, therefore
first-countable (5A4Cb).   Second, for any closed $H\subseteq X$,
there is an irreducible continuous surjection from $H$ onto a compact
metrizable space
$\pi_{\xi}[H]$;  because $\pi_{\xi}[H]$ is separable, so is $H$
(5A4C(d-i)).   It follows that
$X$ is hereditarily separable.   \Prf\ If $A\subseteq X$, then
$\overline{A}$ is separable; let
$D\subseteq\overline{A}$ be a countable dense set.   Because $X$ is
first-countable, each member of $\overline{A}$ is in the closure of a
countable subset of $A$, and there is a countable
$C\subseteq A$ such that $D\subseteq\overline{C}$.   Now $C$ is a
countable dense subset of $A$.\ \Qed

\medskip

{\bf (g)} We need to check that $\omega_1\in\MahR(X)$.   For
$\omega\le\xi<\omega_1$, set
$U_{\xi}=\{z:z\in\{0,1\}^{\omega_1}$,
$z\restr\xi\in K_{\xi}\cap L_{\xi}$, $z(\xi)=1\}$.   Then
$\mu(U_{\xi}\symmdiff E)\ge\bover14$ whenever
$E\subseteq\{0,1\}^{\omega_1}$ is a Baire set determined by
coordinates less than $\xi$.   \Prf\ Set $E'=\pi_{\xi}[E]$, so that
$E=\pi_{\xi}^{-1}[E']$ and $E'$ is a Baire set.   Then

$$\eqalign{\mu(U_{\xi}\setminus E)
&=\nu_{\omega_1}f^{-1}[U_{\xi}\setminus E]
=\nu_{\omega_1}\{z:f(z)\restr\xi\in K_{\xi}\cap L_{\xi}\setminus
E',\,f(z)(\xi)=1\}\cr
&=\nu_{\omega_1}\{z:f_{\xi}(z\restr\xi)\in K_{\xi}\cap L_{\xi}\setminus
E',\,z(\xi)=1\}\cr
&=\Bover12\nu_{\omega_1}\{z:f_{\xi}(z\restr\xi)\in K_{\xi}\cap
L_{\xi}\setminus E'\}
=\Bover12\mu_{\xi}(K_{\xi}\cap L_{\xi}\setminus E'),\cr}$$

\noindent while

$$\eqalign{\mu(E\setminus U_{\xi})
&=\nu_{\omega_1}f^{-1}[E\setminus U_{\xi}]
\ge\nu_{\omega_1}f^{-1}[E\cap\pi_{\xi}^{-1}[K_{\xi}\cap
L_{\xi}]\setminus U_{\xi}]\cr
&=\nu_{\omega_1}\{z:f(z)\restr\xi\in K_{\xi}\cap L_{\xi}\cap
E',\,f(z)(\xi)=0\}\cr
&=\nu_{\omega_1}\{z:f_{\xi}(z\restr\xi)\in K_{\xi}\cap L_{\xi}\cap
E',\,z(\xi)=0\}\cr
&=\Bover12\nu_{\omega_1}\{z:f_{\xi}(z\restr\xi)\in K_{\xi}\cap
L_{\xi}\cap E'\}
=\Bover12\mu_{\xi}(K_{\xi}\cap L_{\xi}\cap E').\cr}$$

\noindent Putting these together,

\Centerline{$\mu(E\symmdiff U_{\xi})\ge\Bover12\mu_{\xi}(K_{\xi}\cap
L_{\xi})\ge\Bover14$.
\Qed}

In particular, $\mu(U_{\eta}\symmdiff U_{\xi})\ge\bover14$ whenever
$\omega\le\eta<\xi<\omega_1$.
So $\mu$ has uncountable Maharam type.   As $\mu X=1$, the
subspace measure $\mu_X$ on $X$ also has uncountable Maharam type,
and $\omega_1\in\MahR(X)$ (531Ef).   Now we have

$$\eqalignno{\omega_1
&\le\tau(\mu_X)\le w(X)
\Displaycause{531Aa}
\le w(\{0,1\}^{\omega_1})
=\omega_1,
\cr}$$

\noindent so $\tau(\mu_X)=w(X)=\omega_1$.

\medskip

{\bf (h)} Finally, I come to the metrizability of negligible subsets of $X$.
Suppose that $A\subseteq X$ and $\mu_XA=0$.   Then we have a sequence
$\sequencen{H_n}$ of closed subsets of $X\setminus A$ such
that $\lim_{n\to\infty}\mu H_n=1$.   For each $n\in\Bbb N$ there is an
$\eta_n<\omega_1$ such that
$H_n=X\cap\pi_{\eta_n}^{-1}[\pi_{\eta_n}[H_n]]$, by (e);  setting
$\eta=\max(\omega,\sup_{n\in\Bbb N}\eta_n)$,
$H_n=X\cap\pi_{\eta}^{-1}[\pi_{\eta}[H_n]]$ for every $n$, so
$\pi_{\eta}^{-1}[\pi_{\eta}[A]]$ is disjoint from every $H_n$
and is $\mu$-negligible.   As $\mu_{\eta}=\mu\pi_{\eta}^{-1}$,
$\mu_{\eta}\pi_{\eta}[A]=0$.
There must therefore be a $\delta\ge\eta$ such that
$\pi_{\eta}[A]\subseteq A_{\eta\delta}$.   By (c-iv), $\pi_{\delta}\restr A$
is injective.

Write $B$ for $\pi_{\delta}[A]$ and $h$ for the inverse function
$(\pi_{\delta}\restr A)^{-1}:B\to A$.   Then
$x\mapsto h(x)(\xi):B\to\{0,1\}$ is continuous for every
$\xi<\omega_1$.   \Prf\ Induce on $\xi$.
For $\xi<\delta$ the result is trivial, as
$x=h(x)\restr\delta$ for every $x\in B$.   For the inductive
step to $\xi\ge\delta$, we have

$$\eqalign{\{x:x\in B,\,h(x)(\xi)=1\}
&=\{x:x\in B,\,h(x)\restr\xi\notin L_{\xi}\}\cr&
=\{x:x\in B,\,h(x)\restr\xi\in K_{\xi}\},
\cr}$$

\noindent but $x\mapsto h(x)\restr\xi$ is continuous,
by the inductive hypothesis, so this is relatively open-and-closed in $B$.
Thus $x\mapsto h(x)(\xi)$ is continuous and the induction continues.\ \Qed

Accordingly
$\pi_{\delta}\restr A$ and $h$ are the two parts of a homeomorphism
between $A$ and $B\subseteq\{0,1\}^{\delta}$, and $A$ is metrizable.
So $X$ and $\mu_X$ have all the properties claimed.
}%end of proof of 531Q

\leader{531R}{}\dvAformerly{5{}31Q}\cmmnt{ Returning to the ideas of 531K, we have the
following construction.

\wheader{531R}{4}{2}{2}{108pt}

\noindent}{\bf Lemma} Let $I$ be a set, and let
$\frak B_{I}$, $\familyiI{e_i}$,
$\familyiI{\phi_i}$,
$\langle\frak C_K\rangle_{K\subseteq I}$ and
$J^*:\frak B_{I}\to[I]^{\le\omega}$ be as in 531I-531K.   For
$a\in\frak B_{I}$ and $K\subseteq I$, set
$S_K(a)=\upr(a,\frak C_K)=\min\{c:a\Bsubseteq c\in\frak C_K\}$\cmmnt{,
the upper envelope of $a$ in $\frak C_K$ (313S)}.

(a) For all $a\in\frak B_{I}$, $i\in I$ and $K$,
$L\subseteq I$,

\quad(i) $S_{I}(a)=a$,

\quad(ii) $S_L(a)\Bsubseteq S_K(a)$ if $K\subseteq L$,

\quad(iii) $J^*S_K(a)\subseteq J^*(a)\cap K$,

\quad(iv) $S_{I\setminus\{i\}}(a)=a\Bcup\phi_ia$,

\quad(v) $S_KS_L(a)=S_{K\cap L}(a)$.

(b) Whenever $a\in\frak B_{I}$, $\epsilon>0$ and $m\in\Bbb N$,
there is a finite $L\subseteq I$ such that
$\bar\nu_{I}(S_K(a)\Bsetminus a)\le\epsilon$ whenever
$L\subseteq K\subseteq I$ and $\#(I\setminus K)\le m$.

\proof{{\bf (a)(i)} $\frak C_{I}=\frak B_{I}$ contains $a$.

\medskip

\quad{\bf (ii)} If $K\subseteq L$ then
$\frak C_L\supseteq\frak C_K$ contains $S_K(a)$.

\medskip

\quad{\bf (iii)} If $i\in I\setminus(J^*(a)\cap K)$ then
$S_K(a)\in\frak C_K$ so $\phi_iS_K(a)\in\frak C_K$, by 531Kg.
Also $\phi_iS_K(a)\Bsupseteq a$.  \Prf\ If $ i\notin K$ then
$\phi_iS_K(a)=S_K(a)\Bsupseteq a$, by 531Kf.
If $ i\notin J^*(a)$ then
$\phi_iS_K(a)\Bsupseteq\phi_ia=a$.\ \QeD\
So $\phi_iS_K(a)\Bsupseteq S_K(a)$;  but they have the same measure,
so $\phi_iS_K(a)=S_K(a)$.
As $ i$ is arbitrary, $J^*S_K(a)\subseteq J^*(a)\cap K$,
by 531Kf in the other direction.

\medskip

\quad{\bf (iv)} By 531Kf again, $S_{I\setminus\{i\}}(a)
=\phi_iS_{I\setminus\{i\}}(a)\Bsupseteq\phi_ia$;  so
$S_{I\setminus\{i\}}(a)\Bsupseteq a\Bcup\phi_ia$.
On the other hand, by
531Ke, $a\Bcup\phi_i(a)$ belongs to
$\frak C_{I\setminus\{i\}}$,
so includes $S_{I\setminus\{i\}}(a)$.

\medskip

\quad{\bf (v)} By (iii),

\Centerline{$J^*S_KS_L(a)\subseteq J^*S_L(a)\cap K
\subseteq J^*(a)\cap L\cap K$,}

\noindent and $S_KS_L(a)\in\frak C_{K\cap L}$;  since
also $S_KS_L(a)\Bsupseteq S_L(a)\Bsupseteq a$,
$S_KS_L(a)\Bsupseteq S_{K\cap L}(a)$.   On the other hand,
$S_{K\cap L}(a)$ belongs to $\frak C_K$ and includes
$S_L(a)$, so includes $S_KS_L(a)$.

\medskip

{\bf (b)} Induce on $m$.   For $m=0$ the result is immediate from
(a-i).   For the
inductive step to $m+1$, take $L_0\in[I]^{<\omega}$ such that
$\bar\nu_{I}(S_K(a)\Bsetminus a)\le\bover13\epsilon$ whenever
$L_0\subseteq K$ and $\#(I\setminus K)\le m$.
By 531Ja, there are a finite
set $L_1\subseteq I$ and a $b\in\frak C_{L_1}$ such that
$\bar\nu_{I}(a\Bsymmdiff b)\le\bover13\epsilon$;  set
$L=L_0\cup L_1$.   Suppose $L\subseteq J$ and
$\#(I\setminus J)=m+1$;  take
$ i\in I\setminus J$ and set $K=J\cup\{i\}$.   Then

\Centerline{$S_J(a)=S_{I\setminus\{i\}}S_K(a)
=S_K(a)\Bcup\phi_iS_K(a)$}

\noindent by (a-v) and (a-iv).   So

$$\eqalignno{\bar\nu_{I}(S_J(a)\Bsetminus a)
&\le\bar\nu_{I}(S_K(a)\Bsetminus a)
  +\bar\nu_{I}(\phi_iS_K(a)\Bsetminus a)\cr
&\le\Bover{\epsilon}3
  +\bar\nu_{I}\phi_i(S_K(a)\Bsetminus a)
  +\bar\nu_{I}(\phi_ia\Bsetminus a)\cr
&\le\Bover{\epsilon}3
  +\bar\nu_{I}(S_K(a)\Bsetminus a)
  +\bar\nu_{I}\phi_i(a\Bsetminus b)
  +\bar\nu_{I}(\phi_ib\Bsetminus b)
  +\bar\nu_{I}(b\Bsetminus a)\cr
&\le\Bover{2\epsilon}3
  +\bar\nu_{I}(a\Bsetminus b)
  +\bar\nu_{I}(b\Bsetminus a)
\le\epsilon\cr}$$

\noindent because $\phi_i$ is a measure-preserving Boolean
homomorphism and $\phi_ib=b$.   Thus the induction continues.
}%end of proof of 531R

\leader{531S}{}\dvAformerly{5{}31R}\cmmnt{ Moving on from hypotheses expressible as
statements about measure algebras, we have a further result which can be
used when Martin's axiom is true.

\medskip

\noindent}{\bf Lemma} Suppose that
$\omega_1<\frak m_{\text{K}}$\cmmnt{ (definition:  517O)}.
Let $\ofamily{\xi}{\omega_1}{e_{\xi}}$ be the standard generating family in
$\frak B_{\omega_1}$, and
$\ofamily{\xi}{\omega_1}{a_{\xi}}$ a family of elements of
$\frak B_{\omega_1}$ of measure greater than $\bover12$.
Then there is an uncountable set $\Gamma\subseteq\omega_1$
such that $\inf_{\xi\in I}a_{\xi}\Bcap e_{\xi}$ meets
$\inf_{\eta\in J}a_{\eta}\Bsetminus e_{\eta}$
whenever $I$, $J\subseteq\Gamma$ are finite and disjoint.

\proof{{\bf (a)} Define $J^*(a)$, for $a\in\frak B_{\omega_1}$, and
$S_I(a)$, for $a\in\frak B_{\omega_1}$ and $I\subseteq\omega_1$, as in
531J and 531R.
Let $P$ be the set of pairs $(c,I)$ where $I\subseteq\omega_1$ is finite,
$0\ne c\Bsubseteq\inf_{\xi\in I}a_{\xi}$ and $I\cap J^*(c)=\emptyset$.
Order $P$ by saying that $(c,I)\le(c',I')$ if $I\subseteq I'$ and
$c'\Bsubseteq c$.   Then $P$ is a partially ordered set.   For each
$\xi<\omega_1$, $a_{\xi}\Bcap\phi_{\xi}a_{\xi}$ belongs to
$\frak C_{\kappa\setminus\{\xi\}}$ (531Je) and is non-zero, so
$p_{\xi}=(a_{\xi}\Bcap\phi_{\xi}a_{\xi},\{\xi\})$ belongs to
$P$.   The point of the proof is the following fact.

\medskip

{\bf (b)} $P$ satisfies Knaster's condition upwards.   \Prf\ Let
$\ofamily{\xi}{\omega_1}{(c_{\xi},I_{\xi})}$ be a family in $P$.   Then
there are an $\alpha>0$ and an uncountable $A_0\subseteq\omega_1$ such
that $\bar\nu_{\omega_1}(c_{\xi}\Bcap c_{\eta})\ge\alpha$ for all $\xi$,
$\eta\in A_0$ (525Tc).   Next, there is an uncountable
$A_1\subseteq A_0$ such that $\family{\xi}{A_1}{I_{\xi}}$ is a
$\Delta$-system with root $I$ say (4A1Db);  let $m\in\Bbb N$ be such
that $A_2=\{\xi:\xi\in A_1$, $\#(I_{\xi}\setminus I)=m\}$ is
uncountable.
Finally, because $J^*(c_{\eta})$ is countable for each $\eta$, and
$\family{\xi}{A_2}{I_{\xi}\setminus I}$ is disjoint, we can find an
uncountable $A_3\subseteq A_2$ such that
$J^*(c_{\eta})\cap I_{\xi}\setminus I=\emptyset$ whenever $\eta$,
$\xi\in A_3$ and $\eta<\xi$.

Take a strictly increasing sequence $\sequence{k}{\eta_k}$ in $A_3$ and
a $\zeta\in A_3$ greater than every $\eta_k$.   By 531Rb, there is a
finite set $K\subseteq\omega_1$ such that
$\bar\nu_{\omega_1}
(S_J(1\Bsetminus c_{\zeta})\Bsetminus(1\Bsetminus c_{\zeta}))<\alpha$
whenever
$K\subseteq J\subseteq\omega_1$ and $\#(\omega_1\setminus J)=m$.
Let $k\in\Bbb N$ be
such that $I_{\eta_k}\setminus I$ does not meet $K$.   Set
$c'_{\zeta}
=S_{\kappa\setminus(I_{\eta_k}\setminus I)}(1\Bsetminus c_{\zeta})$.
Then

\Centerline{$\bar\nu_{\omega_1}(c'_{\zeta}\Bcap c_{\zeta})
\le\alpha
<\bar\nu_{\omega_1}(c_{\zeta}\Bcap c_{\eta_k})$,}

\noindent so
$c=c_{\eta_k}\Bsetminus c'_{\zeta}$ is non-zero;  as
$c'_{\zeta}\Bsupseteq 1\Bsetminus c_{\zeta}$, $c\Bsubseteq c_{\zeta}$.
Set $L=I_{\eta_k}\cup I_{\zeta}$.
Then $J^*(c_{\eta_k})$ is disjoint from $I_{\eta_k}$ and from
$I_{\zeta}\setminus I$, by the choice of $A_3$, while

\Centerline{$J^*(c'_{\zeta})
\subseteq J^*(1\Bsetminus c_{\zeta})\setminus(I_{\eta_k}\setminus I)
=J^*(c_{\zeta})\setminus(I_{\eta_k}\setminus I)$}

\noindent (531R(a-iii)) is also disjoint from $L$;  so
$J^*(c)\subseteq J^*(c_{\eta_k})\cup J^*(c'_{\zeta})$ is disjoint from
$L$.   Finally,

\Centerline{$c\Bsubseteq c_{\eta_k}\Bcap c_{\zeta}
\Bsubseteq\inf_{\xi\in I_{\eta_k}}a_{\xi}
  \Bcap\inf_{\xi\in I_{\zeta}}a_{\xi}=\inf_{\xi\in L}a_{\xi}$,}

\noindent so $(c,L)\in P$.   Now $(c,L)$ dominates both
$(c_{\eta_k},I_{\eta_k})$ and $(c_{\zeta},I_{\zeta})$.

What this shows is that if we write $Q$ for

\Centerline{$\{\{\eta,\zeta\}:\eta$, $\zeta\in A_3$,
$(c_{\eta},I_{\eta})$ and $(c_{\zeta},I_{\zeta})$ are compatible upwards
in $P\}$,}

\noindent then whenever $\zeta\in A_3$ and $M\subseteq A_3\cap\zeta$ is
infinite there is an $\eta\in M$ such that $\{\eta,\zeta\}\in Q$.   By
5A1Hb, there is an uncountable
$A_4\subseteq A_3$ such that $[A_4]^2\subseteq Q$, that is,
$\family{\xi}{A_4}{(c_{\xi},I_{\xi})}$ is upwards-linked.   As
$\ofamily{\xi}{\omega_1}{(c_{\xi},I_{\xi})}$ is arbitrary, $P$ satisfies
Knaster's condition upwards.\ \Qed

\medskip

{\bf (c)} By 517S, there is a sequence $\sequencen{R_n}$ of
upwards-directed subsets of $P$ covering $\{p_{\xi}:\xi<\omega_1\}$;  as $\omega_1$
is uncountable, there must be some $n$ such that
$\Gamma=\{\xi:p_{\xi}\in R_n\}$ has cardinal $\omega_1$.   In this case,
$\{p_{\xi}:\xi\in\Gamma\}$ is upwards-centered in $P$.   If $I$,
$J\subseteq\Gamma$ are finite and disjoint, then there must be a
$(c,K)\in P$ which
is an upper bound for $\{p_{\xi}:\xi\in I\cup J\}$;  now
$I\cup J\subseteq K$ does not meet $J^*(c)$, while
$c\Bsubseteq\inf_{\xi\in I\cup J}a_{\xi}$.   But this means that

\Centerline{$\bar\nu_{\omega_1}(c\Bcap\inf_{\xi\in I}e_{\xi}
  \Bcap\inf_{\eta\in J}(1\Bsetminus e_{\eta}))
=2^{-\#(I\cup J)}\bar\nu_{\omega_1}c>0$,}

\noindent and

\Centerline{$0\ne c\Bcap\inf_{\xi\in I}e_{\xi}
  \Bcap\inf_{\eta\in J}(1\Bsetminus e_{\eta})
\Bsubseteq\inf_{\xi\in I}(a_{\xi}\Bcap e_{\xi})
  \Bcap\inf_{\eta\in J}(a_{\eta}\Bsetminus e_{\eta})$.}

\noindent So we have a set $\Gamma$ of the kind required.
}%end of proof of 531S

\leader{531T}{Theorem}\dvAformerly{5{}31S, revised 2015}
\cmmnt{ ({\smc Fremlin 97})}
Suppose that
$\omega\le\kappa<\frak m_{\text{K}}$.   If $X$ is a Hausdorff
space and $\kappa\in\MahR(X)$, then $\{0,1\}^{\kappa}$ is a continuous
image of a compact subset of $X$.

\proof{{\bf (a)} Because $\kappa<\frak m_{\text{K}}\le\frak m(\frak A)$
for every
probability algebra $\frak A$ (525Tb), $\kappa$ is a measure-precaliber of all
probability algebras (525Fb).

\medskip

{\bf (b)} If $\kappa=\omega$, $X$ has a compact subset $K$
which is not scattered (531E(e-ii), $[0,1]$ is a continuous image of $K$
(4A2G(j-iv) again)
and $\{0,1\}^{\Bbb N}$ is a continuous image of a closed subset of $K$
(using 4A2Uc).

\medskip

{\bf (c)} If $\kappa=\omega_1$, let $K$ be a compact subset of $X$ such that
$\omega_1\in\MahR(K)$, $\mu$ a
\Mth\ Radon probability measure on $K$ with Maharam type $\omega_1$, and
$(\frak A,\bar\mu)$ its measure algebra.   Let
$\ofamily{\xi}{\omega_1}{d_{\xi}}$ be a $\tau$-generating
stochastically independent
family of elements of $\frak A$ of measure $\bover12$.    For
$\xi<\omega_1$ let
$E_{\xi}\in\dom\mu$ be such that $E_{\xi}^{\ssbullet}=d_{\xi}$, and
$K'_{\xi}\subseteq E_{\xi}$,
$K''_{\xi}\subseteq K\setminus E_{\xi}$ compact sets of measure greater
than $\bover14$;  set $K_{\xi}=K'_{\xi}\cup K''_{\xi}$ and
$a_{\xi}=K_{\xi}^{\ssbullet}$ in $\frak A$.   Because
$(\frak A,\bar\mu,\ofamily{\xi}{\omega_1}{d_{\xi}})$ is isomorphic
to $(\frak B_{\omega_1},
\bar\nu_{\omega_1},\ofamily{\xi}{\omega_1}{e_{\xi}})$,
531S tells us that there is an
uncountable set $\Gamma\subseteq\omega_1$ such that

\Centerline{$0\ne\inf_{\xi\in I}(a_{\xi}\Bcap d_{\xi})
  \Bcap\inf_{\eta\in J}(a_{\eta}\Bsetminus d_{\eta})
=(K\cap\bigcap_{\xi\in I}
  K'_{\xi}\cap\bigcap_{\eta\in J}K''_{\eta})^{\ssbullet}$}

\noindent whenever $I$, $J\subseteq\Gamma$ are finite.   Just as in part
(b) of the proof of 531L, it follows that there is a continuous
surjection from $\bigcap_{\xi\in\Gamma}K_{\xi}$ onto
$\{0,1\}^{\Gamma}\cong\{0,1\}^{\omega_1}$.

\medskip

{\bf (d)} If $\kappa\ge\omega_2$, then 531Lb gives the result.
}%end of proof of 531T

\leader{531U}{}\dvAformerly{5{}31T}\cmmnt{ If we are willing to settle
for weaker conclusions, there are
versions of 531L which do not call for any information on precalibers.

\medskip

\noindent}{\bf Proposition} Let $X$ be a Hausdorff space.

(a)\dvAnew{2015} Give the space $\PR(X)$ of Radon probability measures on $X$ 
its narrow topology\cmmnt{ (437J)}.   If $\kappa\ge\omega_2$ belongs to
$\MahR(X)$, then $\{0,1\}^{\kappa}$ is a continuous image of a compact
subset of $\PR(X)$.

(b)\dvAnew{2023} Give the space $\PR(X\times X)$ its
narrow topology.   Then its tightness $t(\PR(X\times X))$ is at least
$\sup\MahR(X)$.

\proof{{\bf (a)}({\smc Plebanek 02}){\bf (i)}
The argument begins by copying part of the proof of 531Lb.   By 531Ec, there
is a compact set $K\subseteq X$ such that $\kappa\in\MahR(K)$.
Let $\mu$ be a \Mth\ Radon probability measure on $K$ with Maharam
type $\kappa$, $\Sigma$ its domain, and $(\frak A,\bar\mu)$ its measure
algebra.   Let
$\ofamily{\xi}{\kappa}{e_{\xi}}$ be a stochastically independent
$\tau$-generating set of elements of measure $\bover12$ in $\frak A$.
For each $\xi<\kappa$, let $E_{\xi}\in\Sigma$ be such that
$E_{\xi}^{\ssbullet}=e_{\xi}$ in $\frak A$;   let
$K'_{\xi}\subseteq E_{\xi}$, $K''_{\xi}\subseteq K\setminus E_{\xi}$ be
compact sets of measure at least $\bover5{12}$.
By 531K, copied into $\frak A$,
there are $\ofamily{\xi}{\kappa}{c_{\xi}}$ and
$\Gamma\in[\kappa]^{\kappa}$ such that
$c_{\xi}\Bsubseteq(K'_{\xi}\cup K''_{\xi})^{\ssbullet}$ and
$\bar\mu c_{\xi}\ge\bover23$ for each $\xi$, and

\Centerline{$\bar\mu(\inf_{\xi\in I}c_{\xi}\Bcap e_{\xi}
  \Bcap\inf_{\eta\in J}c_{\eta}\Bsetminus e_{\eta})
=\Bover1{2^{\#(I\cup J)}}
  \bar\mu(\inf_{\xi\in I\cup J}c_{\xi})$}

\noindent whenever $I$, $J\subseteq\Gamma$ are disjoint finite sets.
In particular, $\inf_{\xi\in I}c_{\xi}\Bcap e_{\xi}$ meets
$\inf_{\eta\in J}c_{\eta}\Bsetminus e_{\eta}$
whenever $I$, $J\subseteq\Gamma$ are disjoint finite sets and
$\inf_{\xi\in I\cup J}c_{\xi}\ne 0$.
But as $c_{\xi}\Bcap e_{\xi}\Bsubseteq(K'_{\xi})^{\ssbullet}$ and
$c_{\xi}\Bsetminus e_{\xi}\Bsubseteq(K''_{\xi})^{\ssbullet}$ for every
$\xi$, we see that
$\bigcap_{\xi\in I}K'_{\xi}\cap\bigcap_{\xi\in J}K''_{\xi}\ne\emptyset$
whenever $I$, $J\subseteq\Gamma$ are disjoint finite sets and
$\inf_{\xi\in I\cup J}c_{\xi}\ne 0$.

\medskip

\quad{\bf (ii)} Now for the new idea.   For each $I\subseteq\Gamma$, set

\Centerline{$L_I
=\{\nu:\nu\in\PR(K)$, $\nu K'_{\xi}\ge\Bover23$ for $\xi\in I$ and
$\nu K''_{\xi}\ge\Bover23$ for $\xi\in\Gamma\setminus I\}$.}

\noindent Then $L_I\ne\emptyset$.   \Prf\ Consider the families
$\{c_{\xi}:\xi\in\Gamma\}\subseteq\frak A$ and
$\{K'_{\xi}:\xi\in I\}\cup\{K''_{\xi}:\xi\in\Gamma\setminus I\}
\subseteq\Cal PX$.   Because $\bar\mu:\frak A\to[0,1]$ is additive and
$\bar\mu c_{\xi}\ge\bover23$ for every $\xi\in\Gamma$, the intersection
number of $\{c_{\xi}:\xi\in\Gamma\}$ must be at least $\bover23$
(391I).   So if $\xi_0,\ldots,\xi_n\in\Gamma$ there is a set
$J\subseteq n+1$ such that $\#(J)\ge\bover23(n+1)$ and
$\inf_{j\in J}c_{\xi_j}\ne 0$.   In this case,
setting $J'=\{j:j\in J$, $\xi_j\in I\}$ and
$J''=J\setminus J'$ we have
$\bigcap_{j\in J'}K'_{\xi_j}\cap\bigcap_{j\in J''}K''_{\xi_j}\ne\emptyset$.
As $\xi_0,\ldots,\xi_n$ are arbitrary,
$\{K'_{\xi}:\xi\in I\}\cup\{K''_{\xi}:\xi\in\Gamma\setminus I\}$ has
intersection number at least $\bover23$.

By 391I in the other direction,
there is an additive functional $\tilde\nu:\Cal PK\to[0,1]$ such
that $\tilde\nu K=1$, $\tilde\nu K'_{\xi}\ge\bover23$ for every
$\xi\in I$ and $\tilde\nu K''_{\xi}\ge\bover23$ for every
$\xi\in\Gamma\setminus I$.   By 416K, there is a Radon measure
$\nu'$ on $K$ such that $\nu'K'_{\xi}\ge\bover23$ for every
$\xi\in I$, $\nu'K''_{\xi}\ge\bover23$ for every
$\xi\in\Gamma\setminus I$, and $\nu'K\le 1$.   Setting
$\nu=\Bover1{\nu'K}\nu'$, we see that $\nu'\in L_I$.\ \Qed

\medskip

\quad{\bf (iii)} Set

$$\eqalign{L
=\bigcup_{I\subseteq\Gamma}L_I
&=\bigcap_{\xi\in\Gamma}(\{\nu:\nu\in\PR(K),\,\nu K'_{\xi}\ge\Bover23\}
  \cup\{\nu:\nu\in\PR(K),\,\nu K''_{\xi}\ge\Bover23\})\cr&
=\PR(K)\setminus\bigcup_{\xi\in\Gamma}
  (\{\nu:\nu(K\setminus K'_{\xi})>\Bover13\}
     \cap\{\nu:\nu(K\setminus K''_{\xi})>\Bover13\}).
\cr}$$

\noindent Then $L$ is closed in $\PR(K)$ for the narrow topology.
Since all the sets
$\{\nu:\nu K'_{\xi}\ge\bover23\}$ and $\{\nu:\nu K''_{\xi}\ge\bover23\}$
are closed in $\PR(K)$, we have a continuous function
$f:L\to\{0,1\}^{\Gamma}$ defined by saying that

$$\eqalign{f(\nu)(\xi)
&=1\text{ if }\nu K'_{\xi}\ge\Bover23\cr
&=0\text{ if }\nu K''_{\xi}\ge\Bover23,\cr}$$

\noindent and (ii) tells us that this is surjective.

\medskip

\quad{\bf (iv)} Recall now that the compact space $\PR(K)$ can be identified with
the subspace $\{\mu:\mu(X\setminus K)=0\}$ of $\PR(X)$ (use 437Nb).   So
$\{0,1\}^{\kappa}\cong\{0,1\}^{\Gamma}$ is a continuous image of a compact
subset of $\PR(X)$.

\medskip

{\bf (b)} Take $\kappa\in\MahR(X)$, and a Maharam homogeneous probability
measure $\mu$ on $X$ with Maharam type $\kappa$;  write $\Sigma$ for the domain
of $\mu$.   I need to show that $\kappa\le t(\PR(X\times X))$.

\medskip

\quad{\bf (i)}\cmmnt{({\smc Plebanek \& Sobota 15})}
To begin with (down to the end of (iii) below),
suppose that $X$ is compact and that $\kappa=\omega_1$.
In $\PR(X\times X)$ let $L$ be the set of measures with both marginals
equal to $\mu$.   By 437N(a-i), $L$ is compact.   If $E$, $F\in\Sigma$,
$\nu\mapsto\nu(E\times F):L\to\Bbb R$ is continuous.   \Prf\
Take $\nu_0\in L$ and $\epsilon>0$.   Then there are open $G\supseteq E$,
$H\supseteq F$ such that

$$\eqalign{\epsilon&
\ge\mu(G\setminus E)+\mu(H\setminus F)\cr&
=\nu((G\setminus E)\times X)+\nu(X\times(H\setminus F))
\ge\nu((G\times H)\setminus(E\times F))\cr}$$

\noindent for every $\nu\in L$.   Now
$U=\{\nu:\nu\in\PR(X\times X)$, $\nu(G\times H)>\nu_0(G\times H)-\epsilon\}$
is a neighbourhood of $\nu_0$ (437Kd), and if $\nu\in U\cap L$ then

\Centerline{$\nu(E\times F)\ge\nu(G\times H)-\epsilon
\ge\nu_0(G\times H)-2\epsilon\ge\nu_0(E\times F)-2\epsilon$.}

\noindent Similarly, there are neighbourhoods $V$, $W$ of $\nu_0$ such that

\Centerline{$\nu((X\setminus E)\times X)\ge\nu_0((X\setminus E)\times X)
   -2\epsilon$
for every $\nu\in V\cap L$,}

\Centerline{$\nu(E\times(X\setminus F))\ge\nu_0(E\times(X\setminus F))
   -2\epsilon$
for every $\nu\in W\cap L$,}

\noindent But now we see that

\Centerline{$\nu(E\times F)\le\nu_0(E\times F)+4\epsilon$
for every $\nu\in V\cap W\cap L$,}

\Centerline{$|\nu(E\times F)-\nu_0(E\times F)|\le 4\epsilon$
for every $\nu\in U\cap V\cap W\cap L$.}

\noindent As $\nu_0$ and $\epsilon$ are arbitrary, $\nu\mapsto\nu(E\times F)$ is
continuous on $L$.\ \Qed

\medskip

\quad{\bf (ii)} Choose
$\ofamily{\xi}{\omega_1}{E_{\xi}}$, $\ofamily{\xi}{\omega_1}{G_{\xi}}$
inductively in such a way that for each $\xi<\omega_1$

\Centerline{$E_{\xi}\in\Sigma$,\quad$\mu E_{\xi}=\bover12$,}

\Centerline{$E_{\xi}$ is independent of the $\sigma$-algebra generated by
$\{G_{\eta}:\eta<\xi\}\cup\{E_{\eta}:\eta<\xi\}$,}

\Centerline{$G_{\xi}\subseteq X$ is open,
\quad$E_{\xi}\subseteq G_{\xi}$,\quad$\mu G_{\xi}\le\bover35$.}

\noindent Now whenever $\xi<\omega_1$
and $I\in[\omega_1\setminus\xi]^{<\omega}$ there is a $\nu_{\xi I}\in L$
such that
$\nu_{\xi I}(G_{\eta}\times G_{\eta})\le\bover9{25}$ for every
$\eta<\xi$ and $\nu_{\xi I}(G_{\zeta}\times G_{\zeta})\ge\bover12$ for every
$\zeta\in I$.   \Prf\ Set $n=\#(I)$;  let $\Cal A$ be the set of atoms of the
algebra generated by $\{E_{\eta}:\eta\in I\}$,
$V=\bigcup_{A\in\Cal A}A\times A$ and $\nu_{\xi I}=2^n\mu^2\LLcorner V$ (234M)
where $\mu^2$ is the Radon product measure $\mu\times\mu$,
so that $\mu A=2^{-n}$ for $A\in\Cal A$, $\mu^2V=2^{-n}$
and $\nu_{\xi I}$ is a Radon probability measure on $X\times X$ (416Sa).
If $F\in\dom\mu$ then

$$\eqalign{\nu_{\xi I}(F\times X)
&=2^n\sum_{A\in\Cal A}\mu^2(A^2\cap(F\times X))\cr&
=2^n\sum_{A\in\Cal A}\mu(A\cap F)\cdot\mu A
=\sum_{A\in\Cal A}\mu(A\cap F)
=\mu F
\cr}$$

\noindent and similarly $\nu_{\xi I}(X\times F)=\mu F$;  thus $\nu_{\xi I}\in L$.   If
$\eta<\xi$ then

$$\eqalignno{\nu_{\xi I} G_{\eta}^2
&=2^n\sum_{A\in\Cal A}(\mu(A\cap G_{\eta}))^2
=2^n\sum_{A\in\Cal A}(\mu A\cdot\mu G_{\eta})^2
\Displaycause{because $G_{\eta}$ and $\family{\zeta}{I}{E_{\zeta}}$ are
independent}
=(\mu G_{\eta})^2
\le\Bover9{25}.
\cr}$$

\noindent If $\zeta\in I$ then

$$\eqalign{\nu_{\xi I}G_{\zeta}^2&
\ge\nu_{\xi I}E_{\zeta}^2
=2^n\sum_{A\in\Cal A}(\mu(A\cap E_{\zeta}))^2
=2^n\sum_{A\in\Cal A,A\subseteq E_{\zeta}}(\mu A)^2\cr&
=2^{-n}\#(\{A: A\in\Cal A,A\subseteq E_{\zeta}\})
=\Bover12.
\cr}$$

\noindent So $\nu_{\xi I}$ works.\ \Qed

\medskip

\quad{\bf (iii)} Now, for each $\xi<\omega_1$, we can choose
$\nu_{\xi}\in\bigcap_{\xi\le\zeta<\omega_1}
  \overline{\{\nu_{\xi I}:\zeta\in I\in[\omega_1\setminus\xi]^{<\omega}\}}$.
Because $L$ is closed, $\nu_{\xi}\in L$;  because
$\nu\mapsto\nu E^2:L\to\Bbb R$ is continuous whenever $E\in\Sigma$, by (i),
$\nu_{\xi}G_{\eta}^2\le\bover{9}{25}$ for $\eta<\xi$ and
$\nu_{\xi}G_{\zeta}^2\ge\bover12$ for $\zeta\ge\xi$.   Next, there is a
$\nu\in\bigcap_{\eta<\omega_1}\overline{\{\nu_{\xi}:\eta<\xi<\omega_1\}}$,
and now $\nu G_{\eta}^2\le\bover{9}{25}$ for every $\eta<\omega_1$.
Writing $D=\{\nu_{\xi}:\xi<\omega_1\}\subseteq L$, $\nu\in\overline{D}$;
but any countable $C\subseteq D$ is included in $\{\nu_{\xi}:\xi\le\zeta\}$ for
some $\zeta<\omega_1$, so that

\Centerline{$\sup\{\nu'G_{\zeta}^2:\nu'\in C\}\le\Bover{9}{25}<\Bover12
\le\inf\{\nu_{\xi}G_{\zeta}^2:\xi\le\zeta\}\le\nu G_{\zeta^2}$}

\noindent and $\nu\notin\overline{C}$.   So $D$ and $\nu$ witness that
$t(\PR(X\times X))\ge\omega_1$.

\medskip

\quad{\bf (iv)} Still supposing that $X$ is compact, consider other
possibilities for $\kappa$.   If $\kappa=0$ then of course
$\kappa\le t(\PR(X\times X))$.   If $\kappa=\omega$, then $X$ has a compact
subset which is not scattered (531Ee), so $X$ has a point
$x$ which is not isolated;  setting $A=X\setminus\{x\}$,
$x\in\overline A\setminus\overline B$ whenever $B\in[A]^{<\omega}$, so
$\omega\le t(X)\le t(X\times X)\le t(\PR(X\times X))$
by 5A4Bb and 437Jd.   If $\kappa\ge\omega_2$, then

$$\eqalignno{\kappa&
=t(\{0,1\}^{\kappa})
\Displaycause{by 5A4I(b-iii)}
\le t(\PR(X))
\Displaycause{by (a) above and 5A4Bb}
\le t(\PR(X\times X))\cr}$$

\noindent because $\PR(X)$ is homeomorphic to a subspace of $\PR(X\times X)$,
by 437Nb.

\medskip

\quad{\bf (v)} This deals with the case in which $X$ is compact.
For the general case, we see from 531Ec that there is a compact set
$K\subseteq X$ such that $\kappa\in\MahR(K)$, so (i)-(iv) tell us that
$t(\PR(K\times K))\ge\kappa$.   But $\PR(K\times K)$ can be identified with a
subset of $\PR(X\times X)$ (437Nb) and
$t(\PR(X\times X))\ge t(\PR(K\times K))\ge\kappa$.
}%end of proof of 531U

\leader{531V}{}\dvAformerly{5{}31U}\cmmnt{ 531Lb and 531U both depend on
Hajnal's Free Set Theorem (5A1Ic), which here is useful
when dealing with cardinals greater than or equal
to $\omega_2$.   More elementary arguments, as in 531La, give similar
results for $\omega$, leaving $\omega_1$ exposed as a special case.
In fact it really is different in this context,
as is shown by the following.

\medskip

\noindent}{\bf Proposition} (a) Suppose that the continuum hypothesis is
true.   Then there is a
compact Hausdorff space $X$ such that $\omega_1\in\MahR(X)$ but
$\{0,1\}^{\omega_1}$ is not a continuous image of a closed subset of
$\PR(X)$.

(b)\cmmnt{({\smc Plebanek 97})} Suppose that
there is a family $\ofamily{\xi}{\omega_1}{W_{\xi}}$ in
$\Cal N_{\omega_1}$ such that every closed subset of
$\{0,1\}^{\omega_1}\setminus\bigcup_{\xi<\omega_1}W_{\xi}$ is scattered.
Then there is a compact Hausdorff space $X$ such that
$\omega_1\in\MahR(X)$ but $\{0,1\}^{\omega_1}$ is not a continuous image
of a closed subset of $X$.

\proof{{\bf (a)(i)} In fact this is witnessed by the space $X$ described in 531Q.
(Since we are assuming that $\frak c=\omega_1$, we certainly have
$\cf\Cal N_{\omega}=\omega_1$, so we can perform the construction in 531Q.)
For the present argument, all we need to know is that
$X$ is a compact Hausdorff space of weight at most $\omega_1$ carrying a
Radon probability measure with uncountable Maharam type
for which every negligible subset is separable and metrizable.

Let $\hat\mu$ be such a measure.   Then

\Centerline{$\omega_1\le\tau(\hat\mu)\le w(X)\le\omega_1$}

\noindent by 511Gc and 531Aa.   Next,
by 524Pf (or directly from the construction in 531Q), the cofinality of
the null ideal $\Cal N(\hat\mu)$ is
$\max(\cf\Cal N_{\omega},\cf[\omega_1]^{\le\omega})=\omega_1$;
let $\ofamily{\xi}{\omega_1}{H_{\xi}}$ be a cofinal family in
$\Cal N(\hat\mu)$ consisting of Borel sets.

\medskip

\quad{\bf (ii)} Write $\Cal B$ for the Borel $\sigma$-algebra of $X$,
$M$ for the set of totally finite Borel measures on $X$
which are absolutely continuous with respect to
$\mu=\hat\mu\restr\Cal B$, and for
$\xi<\omega_1$ write $M_{\xi}$ for the set of totally finite
Borel measures on $X$ for
which $X\setminus H_{\xi}$ is negligible.

\medskip

\qquad\grheada\ $\#(M)\le\frak c$.   \Prf\ For a non-negative
$\mu$-integrable real-valued function $f$ write $\nu_f$ for the
corresponding indefinite-integral measure, so that
$\nu_fE=\int f\times\chi E\,d\mu$ for $E\in\Cal B$.
By the Radon-Nikod\'ym theorem (232F), every
member of $M$ is expressible as $\nu_f$ for some $f$;  moreover, if
$f\eae g$ then $\nu_f=\nu_g$.   So $\#(M)\le\#(L^1(\mu))$.   Now
$L^1(\mu)$ can be identified with $L^1(\frak A,\bar\mu)$, where
$(\frak A,\bar\mu)$ is the measure algebra of $\mu$ (365B);  since
$\hat\mu$ is the completion of $\mu$, we can identify
$(\frak A,\bar\mu)$ with the measure algebra of $\hat\mu$ (322Da) and apply
529Ba to see that the topological density of $L^1(\frak A,\bar\mu)$ is
$\omega_1$.   Since $L^1(\frak A,\bar\mu)$ is metrizable,

\Centerline{$\#(M)\le\#(L^1(\mu))=\#(L^1(\frak A,\bar\mu))
\le d(L^1(\frak A,\bar\mu))^{\omega}=\omega_1^{\omega}=\frak c$.
\Qed}

\medskip

\qquad\grheadb\ For every $\xi<\omega_1$, $\#(M_{\xi})\le\frak c$.
\Prf\ A member of $M_{\xi}$ is determined by its restriction to the
Borel $\sigma$-algebra of
$H_{\xi}$.   Now $H_{\xi}$ is separable and metrizable,
therefore second-countable, and its topology has a countable base
$\Cal U_{\xi}$ containing $H_{\xi}$ and
closed under finite intersections.   If $\nu$, $\nu'$
are different totally finite Borel measures on $H_{\xi}$, then
$\nu\restr\Cal U_{\xi}\ne\nu'\restr\Cal U_{\xi}$ by the Monotone
Class Theorem (136C again), so the same
is true if $\nu$, $\nu'$ are different members of $M_{\xi}$, and

\Centerline{$\#(M_{\xi})\le\#(\BbbR^{\Cal U_{\xi}})=\frak c$.  \Qed}

\medskip

\qquad\grheadc\ Every totally finite Borel measure $\nu$ on $X$
can be expressed as a sum $\nu'+\nu''$
where $\nu'\in M$ and $\nu''\in\bigcup_{\xi<\omega_1}M_{\xi}$.   \Prf\
By 232Ia, we can express $\nu$ as $\nu'+\nu''$ where $\nu'$, $\nu''$ are
countably additive, $\nu'\in M$
and $\nu''$ is singular with respect to $\mu$.   There is a
a Borel set $F$ such that $\mu F=\nu''(X\setminus F)=0$;  and now there
is a $\xi<\omega_1$ such that $F\subseteq H_{\xi}$, so that
$\nu''\in M_{\xi}$ and we have found a suitable expression for $\nu$.
\Qed

\medskip

\quad{\bf (iii)} Of course every member of $\PR(X)$ is determined by its
restriction to $\Cal B$.   We therefore have

$$\eqalignno{\#(\PR(X))
&\le\#(\bigcup_{\xi<\omega_1}M\times M_{\xi})
\le\max(\omega_1,\sup_{\xi<\omega_1}\#(M)\cdot\#(M_{\xi}))
\Displaycause{taking the cardinal products}
=\frak c
=\omega_1
<2^{\omega_1}
=\#(\{0,1\}^{\omega_1}).\cr}$$

\noindent So there cannot possibly be a continuous surjection from
any subset of $\PR(X)$ onto $\{0,1\}^{\omega_1}$.

\medskip

{\bf (b)} For the second example
we can use a variation in the construction in 531M.

\medskip

\quad{\bf (i)} Set $E_{\xi}=\{z:z\in\{0,1\}^{\omega_1}$, $z(\xi)=1\}$
for each $\xi<\omega_1$.   Choose a family
$\langle K_{\xi n}\rangle_{\xi<\omega_1,n\in\Bbb N}$ of compact sets in
$\{0,1\}^{\omega_1}$ as follows.   Given
$\langle K_{\eta n}\rangle_{\eta<\xi,n\in\Bbb N}$, where $\xi<\omega_1$,
such that $\bigcup_{n\in\Bbb N}K_{\eta n}$ is conegligible for every
$\eta<\xi$, then for each $j\in\Bbb N$ we can find a family
$\ofamily{\eta}{\xi}{n(\xi,\eta,j)}$ in $\Bbb N$ such that
$L_{\xi j}=\bigcap_{\eta<\xi}\bigcup_{i\le n(\xi,\eta,j)}K_{\eta i}$ has
measure at least $1-2^{-j-4}$.   For $j\in\Bbb N$
choose a compact set $K'_{\xi j}\subseteq L_{\xi j}
\setminus(W_{\xi}\cup\bigcup_{i<j}K'_{\xi i})$ of measure at least
$1-2^{-j-3}-\nu_{\omega_1}(\bigcup_{i<j}K'_{\xi i})$.   Set

\Centerline{$K_{\xi,2i}=K'_{\xi i}\cap E_{\xi}$,
\quad$K_{\xi,2i+1}=K'_{\xi i}\setminus E_{\xi}$}

\noindent for each $i\in\Bbb N$, and continue.   Note that if $\xi<\omega_1$,
$i\in\Bbb N$ and $z$, $z'\in K_{\xi i}$ then $z(\xi)=z'(\xi)$.

\medskip

\quad{\bf (ii)} At the end of the induction, let $\frak C$ be the algebra of
subsets of $\{0,1\}^{\omega_1}$ generated by
$\{K_{\xi i}:\xi<\omega_1$, $i\in\Bbb N\}$, and $X$ its Stone space.
Then we have a Radon probability measure $\mu$ on $X$ defined by setting
$\mu\widehat{C}=\nu_{\omega_1}C$ for every $C\in\frak C$, where
$\widehat{C}$ is the open-and-closed subset of $X$ corresponding to $C$.   For
$\eta<\xi<\omega_1$, we have

$$\eqalign{\mu(\widehat{K}_{\eta 0}\symmdiff\widehat{K}_{\xi 0})
&=\nu_{\omega_1}(K_{\eta 0}\symmdiff K_{\xi 0})\cr
&=\nu_{\omega_1}((E_{\eta}\cap K'_{\eta 0})\symmdiff(E_{\xi}\cap K'_{\xi
0}))\cr
&\ge\nu_{\omega_1}(E_{\eta}\symmdiff E_{\xi})
  -\nu_{\omega_1}(E_{\eta}\setminus K'_{\eta 0})
  -\nu_{\omega_1}(E_{\xi}\setminus K'_{\xi 0})\cr
&\ge\Bover12-\Bover18-\Bover18
=\Bover14,\cr}$$

\noindent so the Maharam type of $\mu$ is at least $\omega_1$ and
$\omega_1\in\MahR(X)$.

\medskip

\quad{\bf (iii)} Let $F\subseteq X$ be a non-scattered closed set.
Then there is a $\zeta<\omega_1$ such that
$F\not\subseteq\bigcup_{i\in\Bbb N}\widehat{K}_{\zeta i}$.   \Prf\Quer\
Otherwise, set

\Centerline{$R
=\bigcap_{\xi<\omega_1}\bigcup_{i\in\Bbb N}
  (F\cap\widehat{K}_{\xi i})\times K_{\xi i}
\subseteq X\times\{0,1\}^{\omega_1}$.}

\noindent Note that for each $\xi<\omega_1$ the $\widehat{K}_{\xi i}$
are disjoint open-and-compact sets covering the compact set $F$, so
$\{i:F\cap\widehat{K}_{\xi i}\ne\emptyset\}$ is finite and
$\bigcup_{i\in\Bbb N}(F\cap\widehat{K}_{\xi i})\times K_{\xi i}$ is
compact;  thus $R$ is compact.   If $(x,z)$ and $(x',z')\in R$ and
$x\ne x'$, there must be some $C\in\frak C$ such that $x\in\widehat{C}$
and $x'\notin\widehat{C}$, so there must be some $\xi<\omega_1$ and
$i\in\Bbb N$ such that
just one of $x$, $x'$ belongs to $\widehat{K}_{\xi i}$;  in this case,
only the corresponding one of $z$, $z'$ can belong to $K_{\xi i}$, and
$z\ne z'$.

Conversely, if $(x,z)$ and $(x',z')\in R$ and $z\ne z'$, there is some
$\xi$ such that $z(\xi)\ne z'(\xi)$.   In this case, if $i$,
$j\in\Bbb N$ are such that $(x,z)\in\widehat{K}_{\xi i}\times K_{\xi i}$
and $(x',z')\in\widehat{K}_{\xi j}\times K_{\xi j}$, $i\ne j$ and
$x\ne x'$.

This shows that $R$ is the graph of a bijection from $F$ to $R[F]$.
Because $R$ is a compact subset of $F\times R[F]$, it is a
homeomorphism, and $R[F]$ is not scattered.   But, for each
$\xi<\omega_1$,
$R[F]\subseteq\bigcup_{i\in\Bbb N}K_{\xi i}$ is disjoint from $W_{\xi}$;
and all compact subsets of
$\{0,1\}^{\omega_1}\setminus\bigcup_{\xi<\omega_1}W_{\xi}$ are supposed
to be scattered.\ \Bang\Qed

\medskip

\quad{\bf (iv)} Take
$x\in F\setminus\bigcup_{i\in\Bbb N}\widehat{K}_{\zeta i}$.   Then
$\chi(x,X)\le\omega$.   \Prf\ Consider the set

\Centerline{$V=\bigcap_{\eta\le\zeta,i\in\Bbb N}\{x':x'\in X$,
$x'\in\widehat{K}_{\eta i}\iff x\in\widehat{K}_{\eta i}\}$.}

\noindent This is a G$_{\delta}$ set containing $x$.   \Quer\ If there
is an $x'\in V\setminus\{x\}$, there must be some $\xi<\omega_1$ and
$j\in\Bbb N$ such that just one of $x$, $x'$ belongs to
$\widehat{K}_{\xi j}$.   In this case, $\xi>\zeta$, so
$K_{\xi j}\subseteq\bigcup_{i\le k}K_{\zeta i}$ and
$\widehat{K}_{\xi j}\subseteq\bigcup_{i\le k}\widehat{K}_{\zeta i}$ for
some $k\in\Bbb N$.   But neither $x$ nor $x'$ belongs to
$\bigcup_{i\le k}\widehat{K}_{\zeta i}$.\ \BanG\   Thus $V=\{x\}$;  by
4A2Gd as usual, $\chi(x,X)\le\omega$.\ \Qed

\medskip

\quad{\bf (v)} Thus we see that whenever $F\subseteq X$ is a non-scattered
closed set, there is an $x\in F$ such that $\chi(x,X)$ is countable.
By 5A4C(d-iii), $\{0,1\}^{\omega_1}$ is not a continuous image of
a closed subset of $X$.
}%end of proof of 531V




\exercises{\leader{531X}{Basic exercises (a)}
%\spheader 531Xa
Show that there is a Hausdorff completely regular
quasi-Radon probability space $(X,\frak T,\Sigma,\mu)$ with Maharam type
greater than $\#(X)$.   \Hint{523Ib.}
%531A

\spheader 531Xb Give an example of a separable
Radon measure space with magnitude $2^{\frakc}$.   \Hint{4A2B(e-ii).}
%531A

\spheader 531Xc Let $I^{\|}$ be the split interval (343J, 419L).   Show
that $\MahR(I^{\/})=\{0,\omega\}$.
%531E

\spheader 531Xd Let $I$ be an infinite set, and $\beta I$ the
Stone-\v{C}ech compactification of the discrete space $I$.   Show that
$2^{\#(I)}$ is the greatest member of $\MahR(\beta I)$.
\Hint{5A4Ia,  515I.}
%531E

\spheader 531Xe For a topological space $X$, write $\MahqR(X)$ for the
set of Maharam types of \Mth\ quasi-Radon probability measures on $X$.
(i) Show that $\kappa\le w(X)$ for every $\kappa\in\MahqR(X)$.   (ii)
Show that $\MahqR(Y)\subseteq\MahqR(X)$ for every $Y\subseteq X$.
(iii) Show that if $Y$ is another topological space, and neither $X$ nor
$Y$ is empty, then $\MahqR(X\times Y)=\MahqR(X)\cup\MahqR(Y)$.
%531E

\sqheader 531Xf Let $X$ be a Hausdorff topological group carrying Haar
measures, and $\frak A$ its Haar measure algebra (442H, 443A).   Show that
$w(X)=\max(c(\frak A),\tau(\frak A))$.
\Hint{443Gf, 529Ba.}
Show that if $X$ is $\sigma$-compact, locally
compact, Hausdorff and not discrete then $w(X)\in\MahR(X)$.
%531E

\spheader 531Xg\dvAnew{2016} Let $X$ be a Hausdorff space such that
$\MahR(X)\subseteq\{0,\omega\}$, and $\Cal N$ the null ideal of Lebesgue measure
on $\Bbb R$.   Show that the union of fewer than $\add\Cal N$ universally
Radon-measurable subsets of $X$ is universally Radon-measurable.
%434E 521Aa 524I 531E out of order query

\spheader 531Xh Let $X$ be a completely regular Hausdorff space and $\kappa$ an
infinite cardinal.   Suppose that whenever $Y$ is a Hausdorff continuous
image of $X$ of weight $\kappa$ then $\MahR(Y)\subseteq\kappa$.   Show
that $\MahR(X)\subseteq\kappa$.
%531F mt53bits 4A2F(h-iii)

\spheader 531Xi Let $X$ be a Hausdorff space, and $\familyiI{E_i}$ a
family of universally Radon-measurable subsets of $X$ such that
$\#(I)<\cov\Cal N_{\kappa}$ for every $\kappa$.   Show that
$\MahR(\bigcup_{i\in I}E_i)=\bigcup_{i\in I}\MahR(E_i)$.
%531F

\spheader 531Xj Let $K$ be an Eberlein compactum.   Show that
$\MahR(K)\subseteq\{0,\omega\}$.   \Hint{467Xj.}
%531F

\spheader 531Xk\dvAnew{2015}
Let $X$ be a Hausdorff space and $\kappa$ a cardinal.   Show that there is
a Radon probability measure on $X$ with Maharam type $\kappa$ iff
{\it either} $\kappa$ is finite and $2^{\kappa}\le 2\#(X)$ {\it or}
$\kappa=\omega\le\#(X)$ {\it or} $\kappa\in\MahR(X)$ {\it or}
$\cf\kappa=\omega$ and $\kappa=\sup\MahR(X)$.
%531F out of order query

\sqheader 531Xl\dvAnew{2015} Let $X$ be a Hausdorff space and $\kappa$ an
infinite cardinal.   (i) Show that $\{0,1\}^{\kappa}$ is a continuous image
of a compact subset of $X$ iff $[0,1]^{\kappa}$ is a continuous image of a
compact subset of $X$, and that in this case $\{0,1\}^{\kappa}$ is a
continuous image of a compact subset of $\PR(X)$.   \Hint{437Xt.}
(ii) Show that if $X$ is normal and
$\{0,1\}^{\kappa}$ is a continuous image
of a closed subset of $X$ then $[0,1]^{\kappa}$ is a continuous image of
$X$.   \Hint{4A2F(d-ix).}
(iii) Show that if $X$ is completely regular and
$\{0,1\}^{\kappa}$ is a continuous image
of a compact subset of $X$ then $[0,1]^{\kappa}$ is a continuous image of
$X$.  \Hint{4A2F(h-iii).}
%531L

\spheader 531Xm Let $(X,\frak T,\Sigma,\mu)$ be a Hausdorff quasi-Radon
probability space.   Show that the Maharam type of $\mu$ is at most
$\max(\omega,2^{\chi(X)})$.   \Hint{5A4Ba, 5A4Bg.}
%531N

\spheader 531Xn In the language of 531R, show that if $a$,
$b\in\frak B_I$ and $K\subseteq I\setminus J^*(b)$ is finite,
then $\bar\nu_I(a\Bsymmdiff S_{I\setminus K}(a))
\le 2^{\#(I)}\bar\nu_I(a\Bsymmdiff b)$.
%531R

\spheader 531Xo Show that if $\frak m_{\text{K}}>\omega_1$ and $X$ is a
countably tight compact Hausdorff space, then $\omega_1\notin\MahR(X)$.
%531T 5A4Bc
%is it enough if\omega_1 a precaliber of all prob algs?
%\v{S}apirovski\v\i's theorem:  a compact countably compact Hausdorff
%space has a point-countable \pi-base (proof by Stevo).
%In 531Q we have a countably tight X such that\omega_1\in\MahR(X)

\spheader 531Xp Let $X$ be an infinite
compact Hausdorff space with a strictly
positive Radon measure $\mu$.   Show that the
topological density of $\PR(X)$, with its narrow topology, is at most
the Maharam type of $\mu$.   \Hint{the indefinite-integral measures over
$\mu$ are dense in $\PR(X)$.}
%531U
%- but can d(X)  be greater than  \tau(\mu) ?
%(\query:  compact needed?)

\spheader 531Xq Let $\Cal W\subseteq\Cal N_{\omega}$ be such that every
compact subset of $\{0,1\}^{\omega}\setminus\bigcup\Cal W$ is scattered.
Show that there is a family $\Cal W'\subseteq\Cal N_{\omega_1}$
such that $\#(\Cal W')=\#(\Cal W)$ and every compact subset of
$\{0,1\}^{\omega_1}\setminus\bigcup\Cal W'$ is scattered.
%531V out of order query

\leader{531Y}{Further exercises (a)}
%\spheader 531Ya
Let $\kappa$ be an infinite cardinal such that $\kappa=\kappa^{\frakc}$.
Show that there is a set $X\subseteq\{0,1\}^{\kappa}$, of full outer
measure for $\nu_{\kappa}$, such that every subset of $X$ with cardinal
$\frak c$ is discrete.   Show that $\MahqR(X)$ (531Xe) contains $\kappa$
but not $\omega$.
%531Xe 531E 521Dc

\spheader 531Yb
Let $X$ and $Y$ be infinite compact Hausdorff spaces, and suppose that
there is a norm-preserving linear isomorphism between the dual spaces
$C(X)^*$ and $C(Y)^*$.   Show that $\MahR(X)=\MahR(Y)$.
%531F 529B in an L-space, u & v disjoint iff \|\alpha u+\beta v\| is
%  what it should be
%Suppose that  C(X)^*  and  C(Y)^*  are isomorphic as linear topological
%  spaces?

\spheader 531Yc Let $\mu$ be a
$\tau$-additive Borel probability measure on a topological
space $X$, and $\kappa$ a cardinal of uncountable cofinality such
that (i) $\chi(x,X)<\cf\kappa$ for every $x\in X$ (ii) no non-negligible
measurable set can be covered by $\cf\kappa$ negligible sets.   Show
that the Maharam type of $\mu$ cannot be $\kappa$.
%531O Query:  could the Maharam type be greater than $\kappa$? mt53bits

\spheader 531Yd Let $X$ be a completely regular Hausdorff space and
$\kappa\ge\omega_2$ a cardinal.   Show that if $\kappa\in\MahR(X)$ then
the Banach space $\ell^1(\kappa)$ is isomorphic, as linear topological
space, to a subspace of the Banach space $C_b(X)$.
%242Xa for \ell^1(\kappa) 354Hb for C_b(X)
\leaveitout{\Hint{take $K'_{\xi}$, $K''_{\xi}$ as in the proof of 531Ua;
let $g_{\xi}\in C_b(X)$ be such that
$2\chi K'_{\xi}-\chi X\le g_{\xi}\le 2\chi(X\setminus K''_{\xi})-\chi X$;
set $Tu=\sum_{\xi<\kappa}u(\xi)g(\xi)$ for $u\in\ell^1(\kappa)$;  show that
$\|Tu\|_{\infty}\le\|u\|_1\le\bover32\|Tu\|_{\infty}$.}}
%531U

\spheader 531Ye Let $X$ be a locally compact Hausdorff space and $\kappa$
an infinite cardinal such that $\ell^1(\kappa)$ is isomorphic, as linear
topological space, to a subspace of $C_0(X)$ (definition:  436I).   Show
that $\kappa\in\MahR(X)$.   \Hint{Reduce to the case in which $X$ is
compact.   Show that if $\sequence{i}{e_i}$ is the standard generating
family in $\ell^1$, $n\in\Bbb N$ and
$\langle\alpha_{ij}\rangle_{i<j\le n}$ is a family in $\coint{0,\infty}$,
then there is a family $\langle\epsilon_{ij}\rangle_{i<j\le n}$ in
$\{-1,1\}$ such that
$\|\sum_{i<j\le n}\epsilon_{ij}\alpha_{ij}(e_i-e_j)\|_1
\ge\sum_{i<j\le n}\alpha_{ij}$.   See {\smc Pe{\l}czy\'nski 68}.}
%531Yd 531U mt53bits
}%end of exercises

\leader{531Z}{Problems (a)} Can there be a perfectly normal compact
Hausdorff space $X$ such that $\omega_2\in\MahR(X)$?   (See 531Q, 554Xd.)

\spheader 531Zb Can there be a hereditarily separable compact
Hausdorff space $X$ such that $\omega_2\in\MahR(X)$?
%531Q
%Recall that a countably tight compact Hausdorff space has a
%point-countable \pi_base.
%What about hereditary \pi-weight-countable?

\leaveitout{The JKR space is hereditarily separable and carries a
non-trivial Borel
measure (a copy of Lebesgue measure).}

\leaveitout{Find a ZFC example of a topological space $X$ and a Borel
probability measure on $X$ with Maharam type greater than $w(X)$.
(Note that if $\kappa$ is atomlessly-measurable then $X=\kappa$ with
its discrete topology serves.   Also there is then no non-trivial Borel
probability measure on $X$ of low Maharam type.} %531A

\leaveitout{\spheader 531Z? Let $X$ be a topological space, and suppose
that there is a quasi-Radon probability measure on $X$ with Maharam type
$\omega_1$.   Does it follow that there is an atomless quasi-Radon
measure on $X$ with Maharam type $\omega$?} %531E

\leaveitout{If $X$ is compact Hausdorff and $P(X)$ is countably tight,
can $\omega_1\in\MahR(X)$?   see {\smc Plebanek \& Sobota p14}.
}

\leaveitout{If $\sequencen{\kappa_n}$ is a sequence of cardinals, each
a measure-precaliber of every probability algebra, what about
$\sup_{n\in\Bbb N}\kappa_n$?}

\endnotes{
\Notesheader{531}
This section is directed to Radon measures, studying $\MahR(X)$;  of
course we can look at Maharam types of quasi-Radon measures (531Xe,
531Ya), or Borel or Baire measures for that matter.   In the next
section I shall have something to say about completion regular measures.
The function $X\mapsto\MahR(X)$ has a much more satisfying list of basic
properties (531E, 531G) than the others.

From 531L and 531T we see that there are many cardinals $\kappa$ such
that whenever $X$ is a Hausdorff space and $\kappa\in\MahR(X)$,
then there is a continuous function from a closed subset of $X$ onto
$\{0,1\}^{\kappa}$.
Such cardinals are said to have {\bf Haydon's property}.   From 531L,
531M and 531T we see that

\inset{$\omega$ has Haydon's property (531La);

if $\kappa\ge\omega_2$ and $\kappa$ is a measure-precaliber of
$\frak B_{\kappa}$ then $\kappa$ has Haydon's property (531Lb);

$\frak c^{(+n)}$ has Haydon's property for $n\ge 1$ (525K);

if $\kappa\ge\omega$ is not a measure-precaliber of $\frak B_{\kappa}$
then $\kappa$ does not have Haydon's property (531M);

if $\omega_1<\frak m_{\text{K}}$ then $\omega_1$ has Haydon's property
(531T).}

\noindent (See also 544D.)
Thus if $\frak m_{\text{K}}>\omega_1$, an infinite cardinal
$\kappa$ has Haydon's property iff it is a measure-precaliber of every
probability algebra.   $\omega_1$ really is different;  it is possible
that $\omega_1$ is a precaliber of every probability algebra but does
not have Haydon's property.   To check this, it is enough to find a
model of set theory in which $\cov\Cal N_{\omega_1}>\omega_1$ (525Gc) but there
is a family $\ofamily{\xi}{\omega_1}{W_{\xi}}$ as in 531Vb;  one is
described in 553F.

You will observe that the key arguments of this section all depend on
analysis of the measure algebras $\frak B_{\kappa}$.   We have already
seen in \S524 that many properties of a Radon measure can be determined
from its measure algebra.   Here we find that some important topological
properties of compact Hausdorff spaces can be determined by the measure
algebras of the Radon measures they carry.   The results here largely
depend for their applications on knowing enough about precalibers;  I
remind you that it seems to be still unknown whether it is possible that
every infinite cardinal should be a measure-precaliber of every probability
algebra.

The remarks above have concerned the existence of continuous surjections
onto $\{0,1\}^{\kappa}$;  a natural place to start, because measures of
Maharam type $\kappa$ arise immediately from such surjections.   In
531N-531Q %531N 531O 531Q
I look at different measures of the richness of a compact space $X$.
Concerning characters, 531N-531O give us quite a lot of information,
slightly irregular at the edges.   I ought to offer a remark on the
context of 531Q.
In some set theories (for instance, when $\frak m>\omega_1$), we find
not only that $\omega_1$ is a precaliber of every measurable algebra,
but also that a compact Hausdorff space is hereditarily separable iff it is
hereditarily Lindel\"of ({\smc Fremlin 84a}, 44H);  so that, for
instance, a hereditarily separable compact Hausdorff space must be
first-countable, so cannot carry
a Radon measure of uncountable Maharam type.   Typically, the situation
is very different if the continuum hypothesis or
Jensen's $\diamondsuit$ is true,
and 531Q is a descendant of the construction in {\smc Kunen 81} of a
non-separable hereditarily Lindel\"of compact Hausdorff space.   See
{\smc D\v{z}amonja \& Kunen 93} for further
exploration of these questions.

Following the lead of {\smc Haydon 77}, more than half of
this section is devoted
to investigating properties of compact Hausdorff spaces carrying
Radon measures of particular Maharam types.   Most of the topological properties
considered are very natural ones in this context.   But in 531U I add an
interesting pair of results concerning topological properties of $\PR(X)$ or
$\PR(X\times X)$, less obviously connected to individual Radon measures on $X$.

}%end of notes


\frfilename{mt532.tex}
\versiondate{1.6.13}
\copyrightdate{2004}

\def\chaptername{Topologies and measures III}
\def\sectionname{Completion regular measures on $\{0,1\}^I$}


\newsection{532}
\def\headlinesectionname{Completion regular measures on
$\{0,1{\delimiter"5267309}^I$}

As I remarked in the introduction to \S434, the trouble with topological
measure theory is that there are too many questions to ask.   In \S531 I
looked at the problem of determining the possible Maharam types of Radon
measures on a Hausdorff space $X$.   But we can ask the same
question for any of the other classes of topological measures listed in
\S411.   It turns out that the very narrowly focused topic of completion
regular Radon measures on powers of $\{0,1\}$ already leads us to some
interesting arguments.

I define the classes $\MahcrR(X)$, corresponding to the $\MahR(X)$
examined in \S531, in 532A.   They are less accessible, and I almost
immediately specialize to the relation
$\lambda\in\MahcrR(\{0,1\}^{\kappa})$.   This at least is more or less
convex (532G, 532K), and can be characterized in terms of the measure
algebras $\frak B_{\lambda}$ (532I).   On the way it is helpful to extend
the treatment of completion regular measures given in \S434 (532D, 532E,
532H).   For fixed infinite $\lambda$, there is a critical cardinal
$\kappa_0\le(2^{\lambda})^+$ such that
$\lambda\in\MahcrR(\{0,1\}^{\kappa})$ iff $\lambda\le\kappa<\kappa_0$;
under certain conditions, when $\lambda=\omega$,
we can locate $\kappa_0$ in terms of the
cardinals of Cicho\'n's diagram (532P, 532Q).   This depends on facts
about the Lebesgue measure algebra (532M, 532O) which are of independent
interest.    Finally, for other $\lambda$ of countable cofinality, the
square principle and Chang's transfer principle are relevant (532R-532S).

\leader{532A}{Definition} If $X$ is a topological space, I
write $\Mahcr(X)$ for the set of Maharam types of \Mth\ completion
regular topological probability measures on $X$.
If $X$ is a Hausdorff space, I write
$\MahcrR(X)$ for the set of Maharam types of \Mth\ completion regular
Radon probability measures on $X$.

\leader{532B}{Proposition} Let $X$ be a Hausdorff space.   Then a
probability algebra $(\frak A,\bar\mu)$ is isomorphic to the measure
algebra of a completion regular Radon probability measure on $X$ iff
($\alpha$) $\tau(\frak A_a)\in\MahcrR(X)$ whenever $\frak A_a$ is a
non-zero homogeneous principal ideal of $\frak A$ ($\beta$) the number
of atoms of $\frak A$ is not greater than the number of points $x\in X$
such that $\{x\}$ is a zero set.

\proof{{\bf (a)} Suppose that $\mu$ is a completion regular Radon
probability measure on $X$ and $\frak A_a$ is a non-zero homogeneous
principal ideal of its measure algebra $\frak A$.   Let $F$ be such that
$F^{\ssbullet}=a$ and $\nu$ the indefinite-integral measure over $\mu$
defined by the function $\Bover1{\mu F}\chi F$.   Then $\nu$ is a Radon
measure (416Sa), inner regular with respect to the zero sets (412Q);  and
its measure algebra is isomorphic, up to a scalar multiple, to
$\frak A_a$, so is homogeneous with Maharam type
$\tau(\frak A_a)$.   So $\nu$ witnesses that
$\tau(\frak A_a)\in\MahcrR(X)$.    This shows that $\frak A$ satisfies
condition ($\alpha$).

As for condition ($\beta$), each atom of $\frak A$ is of the form
$\{x\}^{\ssbullet}$ for some $x\in X$ such that $\mu\{x\}>0$ (414G, or
otherwise).   In this case, because $\mu$ is completion regular, $\{x\}$
must be a zero set.   So we have at least as many singleton zero sets as
we have atoms in $\frak A$.

\medskip

{\bf (b)} Now suppose that $(\frak A,\bar\mu)$ satisfies the conditions.
I copy the argment of 531F.   Express $(\frak A,\bar\mu)$ as the simple
product of a countable family $\familyiI{(\frak A_i,\bar\mu'_i)}$ of
non-zero homogeneous measure algebras.   For $i\in I$, set
$\kappa_i=\tau(\frak A_i)$ and $\gamma_i=\bar\mu'_i1_{\frak A_i}$.   Set
$J=\{i:i\in I$, $\kappa_i\ge\omega\}$.   $(\beta$) tells us that
$\#(I\setminus J)$ is less than or equal to the number of singleton zero
sets in $X$;  let $\family{i}{I\setminus J}{x_i}$ be a
family of distinct elements of $X$ such that every $\{x_i\}$ is a zero
set.

For each $i\in J$,
($\alpha$) tells us that there is a completion regular \Mth\ Radon
probability measure
$\mu_i$ on $X$ with Maharam type $\kappa_i$.   Now there is a disjoint
family $\family{i}{J}{E_i}$ of Baire subsets of $X$ such that $\mu_iE_i>0$
for every $i\in J$.   \Prf\ We may suppose that $J\subseteq\Bbb N$.
Choose $\sequence{i}{E_i}$, $\sequence{i}{F_i}$ inductively, as follows.
$F_0=X\setminus\{x_i:i\in I\setminus J\}$.   Given that $F_i$ is a Baire
set and $\mu_jF_i>0$ for every $j\in J\setminus i$, then if $i\notin J$
set $E_i=\emptyset$
and $F_{i+1}=F_i$;  otherwise, because $\mu_i$ is atomless and
completion regular, we can find,
for each $j\in J$ such that $j>i$, a Baire set $G_{ij}\subseteq F_i$
such that $\mu_iG_{ij}<2^{-j}\mu_iF_i$ and $\mu_jG_{ij}>0$;  set
$F_{i+1}=\bigcup_{j\in J,j>i}G_{ij}$ and $E_i=F_i\setminus F_{i+1}$;
continue.\ \QeD\   Now set

\Centerline{$\mu E
=\sum_{i\in I\setminus J,x_i\in E}\gamma_i
  +\sum_{i\in J}(\mu_iE_i)^{-1}\gamma_i\mu_i(E\cap E_i)$}

\noindent whenever $E\subseteq X$ is such that $\mu_i$ measures
$E\cap E_i$ for every $i\in J$.   Of course $\mu$ is a probability
measure.   Because every $\mu_i$ is a topological measure, so is $\mu$;
because every $\mu_i$ is inner regular with respect to the compact sets,
so is $\mu$;  because every $\mu_i$ is complete, so is $\mu$;  so $\mu$
is a Radon measure.   Because every subspace measure
$(\mu_i)_{E_i}$ is \Mth\ with Maharam type $\kappa_i$, the measure
algebra of $\mu$ is isomorphic to $(\frak A,\bar\mu)$.
Because all the $\{x_i\}$ are zero sets and all the $\mu_i$ are
completion regular, $\mu$ is completion regular.
}%end of proof of 532B

\leader{532C}{Remarks}\cmmnt{ Nearly the whole of this section will be
devoted to the usual measures on powers of $\{0,1\}$.   Accordingly the
following notation will be useful, as previously in this volume.}
If $I$ is any set, $\nu_I$ will be the usual measure on $\{0,1\}^I$,
$\frak B_I$ its measure algebra and $\Cal N_I$ its null ideal.
\cmmnt{In this context,
$\familyiI{e_i}$ will be the standard generating family in
$\frak B_I$ (525A), and for $J\subseteq I$, $\frak C_J$ will be the closed
subalgebra of $\frak B_I$ generated by $\{e_i:i\in J\}$.}

\cmmnt{If $X$
is a topological space, $\Cal B(X)$ will be its Borel $\sigma$-algebra.}

Let $\kappa$ be an infinite cardinal.   Then\cmmnt{ $\nu_{\kappa}$ is
a completion regular Radon probability measure (416U), and
$\frak B_{\kappa}$ is homogeneous with Maharam type $\kappa$.   So}
$\kappa\in\MahcrR(\{0,1\}^{\kappa})$.
\cmmnt{Next, any Radon measure on $\{0,1\}^{\kappa}$ can have Maharam
type at most $w(\{0,1\}^{\kappa})$ (531Aa), so }$\lambda\le\kappa$ for
every $\lambda\in\MahcrR(\{0,1\}^{\kappa})$.   \cmmnt{At the bottom
end,} $0\in\MahcrR(\{0,1\}^{\kappa})$ iff\cmmnt{ $\{0,1\}^{\kappa}$
has a singleton G$_{\delta}$ set, that is, iff} $\kappa=\omega$.

\cmmnt{From this we see already that we do not have direct
equivalents of any of the results
531Eb-531Ef. %531Eb 531Ec 531Ed 531Ee 531Ef
However the class
$\{(\lambda,\kappa):\lambda\in\MahcrR(\{0,1\}^{\kappa})\}$ is convex in
two senses (532G, 532K).   For the first of these, it will be useful to
have a result left over from \S434.}

\leader{532D}{Theorem}\cmmnt{ ({\smc Fremlin \& Grekas 95})} Let
$(X,\mu_1)$ and $(Y,\mu_2)$ be
effectively locally finite topological measure spaces of which
$X$ is quasi-dyadic\cmmnt{ (definition:  434O)}, $\mu_1$ is completion
regular and $\mu_2$ is $\tau$-additive.
Let $\mu$ be the c.l.d.\ product measure on
$X\times Y$\cmmnt{ as defined in \S251}.   Then $\mu$ is a
$\tau$-additive topological measure.

\proof{{\bf (a)} To begin with (down to the end of (e)) let us suppose
that $\mu_1$ and $\mu_2$ are complete and totally finite and
inner regular with respect to
the Borel sets.   Let $\familyiI{X_i}$ be a family of separable
metrizable spaces such that there is a continuous surjection
$f:\prod_{i\in I}X_i\to X$.   For each $i\in I$, let $\Cal U_i$ be a
countable base for the topology of $X_i$ not containing $\emptyset$;
for $J\subseteq I$, let
$\Cal C_J$ be the family of cylinder sets expressible in the form
$\{z:z\in\prod_{i\in I}X_i$, $z(i)\in U_i$ for every $i\in K\}$ where
$K\subseteq J$ is finite and $U_i\in\Cal U_i$ for each $i\in K$.

\medskip

{\bf (b)} \Quer\ Suppose, if possible, that $\mu$ is not a topological
measure.   Let $W\subseteq X\times Y$ be a closed set which is not
measured by $\mu$.   By 434Q, $\mu_1$ is $\tau$-additive;
by 417C, there is a $\tau$-additive topological
measure $\tilde\mu$ extending $\mu$, and $\mu^*W=\tilde\mu W$ (apply
417C(b-v-$\alpha$) to the complement of $W$).

\medskip

{\bf (c)} If $J\subseteq I$ is countable, there are sets $H$, $V$, $V'$
such that $H\subseteq Y$ is open, $V\in\Cal C_J$,
$V'\in\Cal C_{I\setminus J}$, $f[V\cap V']\times H$ is disjoint from
$W$, and $\mu^*(W\cap(f[V]\times H))>0$.   \Prf\ For $V\in\Cal C_J$, set

\Centerline{$\Cal H_V
=\bigcup_{V'\in\Cal C_{I\setminus J}}
  \{H:H\subseteq Y$ is open, $W\cap(f[V\cap V']\times H)=\emptyset\}$,}

\Centerline{$H_V=\bigcup\Cal H_V$,}

\noindent and choose a measurable envelope $F_V$ of $f[V]$.   As
$\Cal C_J$ is countable,

\Centerline{$W_1
=(X\times Y)\setminus\bigcup_{V\in\Cal C_J}F_V\times H_V$}

\noindent is measured by $\mu$;  also $W_1\subseteq W$ because

\Centerline{$\{f[V\cap V']\times H:
V\in\Cal C_J$, $V'\in\Cal C_{I\setminus J}$, $H\subseteq Y$ is open$\}$}

\noindent is a network for the topology of $X\times Y$.   So

\Centerline{$\tilde\mu W_1=\mu W_1\le\mu_*W<\mu^*W=\tilde\mu W$}

\noindent and $\tilde\mu(W\setminus W_1)>0$.   There must therefore be a
$V\in\Cal C_J$ such that $\tilde\mu(W\cap(F_V\times H_V))>0$.   Next,
because $\mu_2$ is $\tau$-additive, there is a countable
$\Cal H\subseteq\Cal H_V$ such that
$\mu_2(H_V\setminus\bigcup\Cal H)=0$, and now
$\tilde\mu(W\cap(F_V\times\bigcup\Cal H))
=\tilde\mu(W\cap(F_V\times H_V))$ is non-zero.   Accordingly there is an
$H\in\Cal H$ such that $\tilde\mu(W\cap(F_V\times H))>0$.   By
417G\formerly{4{}17H},

\Centerline{$\int_{F_V}\mu_2(W[\{x\}]\cap H)\mu_1(dx)
=\tilde\mu(W\cap(F_V\times H))$}

\noindent is greater than $0$.   But this means that
$\mu_1\{x:x\in F_V$, $\mu_2(W[\{x\}]\cap H)>0\}>0$.   (Recall that we
are supposing that $\mu_1$ is complete.)   So
$\{x:x\in f[V]$, $\mu_2(W[\{x\}]\cap H)>0\}$ is not $\mu_1$-negligible,
and $W\cap(f[V]\times H)$ is not $\mu$-negligible.
Finally, because $H\in\Cal H_V$, there is a $V'\in\Cal C_{I\setminus J}$
such that $W\cap(f[V\cap V']\times H)=\emptyset$.\ \Qed

\medskip

{\bf (d)} We may therefore choose inductively families
$\ofamily{\xi}{\omega_1}{J_{\xi}}$,
$\ofamily{\xi}{\omega_1}{H_{\xi}}$,
$\ofamily{\xi}{\omega_1}{V_{\xi}}$,
$\ofamily{\xi}{\omega_1}{V'_{\xi}}$ in such a way that, for every
$\xi<\omega_1$,

\inset{$J_{\xi}$ is a countable subset of $I$,

$H_{\xi}$ is an open subset of $Y$,

$V_{\xi}\in\Cal C_{J_{\xi}}$, $V'_{\xi}\in\Cal C_{I\setminus J_{\xi}}$,

$W\cap(f[V_{\xi}\cap V'_{\xi}]\times H_{\xi})=\emptyset$,

$\mu^*(W\cap(f[V_{\xi}]\times H_{\xi}))>0$,

$\bigcup_{\eta<\xi}J_{\eta}\subseteq J_{\xi}$,

$V_{\xi}$, $V'_{\xi}\in\Cal C_{J_{\xi+1}}$.}

\noindent For each $\xi<\omega_1$, let $K_{\xi}$ be a finite subset of
$J_{\xi+1}$ such that $V_{\xi}$ and $V'_{\xi}$ are determined by
coordinates in $K_{\xi}$.   By the $\Delta$-system Lemma (4A1Db), there
is an uncountable set $A\subseteq\omega_1$ such that
$\family{\xi}{A}{K_{\xi}}$ is a $\Delta$-system with root $K$ say.   Set
$\zeta_0=\min A$.    Express each $V_{\xi}$ as
$\tilde V_{\xi}\cap\hat V_{\xi}$ where  $\tilde V_{\xi}\in\Cal C_K$ and
$\hat V_{\xi}\in\Cal C_{K_{\xi}\setminus K}$;   because $\Cal C_K$ is
countable, there is a $\tilde V$ such that
$B=\{\xi:\xi\in A$, $\xi>\zeta_0$, $\tilde V_{\xi}=\tilde V\}$ is
uncountable.   Note that $\mu_1^*f[\tilde V]>0$, because
$\mu_1^*f[\tilde V]\ge\mu^*(W\cap(f[V_{\xi}]\times H_{\xi}))$ for any
$\xi\in B$.   Also

\Centerline{$K\subseteq K_{\zeta_0}\subseteq J_{\zeta_0+1}
\subseteq J_{\xi}$,}

\noindent so $V'_{\xi}$ is determined by coordinates in
$K_{\xi}\setminus J_{\xi}\subseteq K_{\xi}\setminus K$, for every
$\xi\in B$.

\medskip

{\bf (e)} Set $H'_{\xi}=\bigcup_{\eta\in B\setminus\xi}H_{\eta}$ for
each $\xi<\omega_1$.   Then $\ofamily{\xi}{\omega_1}{H'_{\xi}}$ is
non-increasing, so there is a $\zeta<\omega_1$ such that
$\mu_2H'_{\xi}=\mu_2H'_{\zeta}$ whenever $\xi\ge\zeta$.   Now consider
$F=\{x:\mu_2(W[\{x\}]\cap H'_{\zeta})>0\}$.   Applying 417G to
the indicator function of $W\cap(X\times H'_{\zeta})$, and
recalling once more that $\mu_1$ is complete, we see that $\mu_1$
measures $F$.   Also $\mu_1^*(F\cap f[\tilde V])>0$.   \Prf\ Take any
$\xi\in B\setminus\zeta$.   Then

\Centerline{$F\cap f[\tilde V]
\supseteq\{x:x\in f[\tilde V_{\xi}]$, $\mu_2(W[\{x\}]\cap H_{\xi})>0\}$}

\noindent must be non-$\mu_1$-negligible because
$W\cap(f[\tilde V_{\xi}]\times H_{\xi})$ is not $\mu$-negligible.\ \Qed

At this point, recall that we are supposing that $\mu_1$ is completion
regular.   So there is a zero set $Z\subseteq F$ such that
$\mu_1Z>\mu_1F-\mu_1^*(F\cap f[\tilde V])$, and
$Z\cap f[\tilde V]\ne\emptyset$, that is, $\tilde V\cap f^{-1}[Z]$ is
not empty.   $f^{-1}[Z]$ is a zero set (4A2C(b-iv)), so there is a
countable set $J\subseteq I$ such that $f^{-1}[Z]$ is determined by
coordinates in $J$ (4A3Nc);  we may suppose that $K\subseteq J$.
Because $\family{\eta}{A}{K_{\eta}\setminus K}$ is disjoint, there is a
$\xi\ge\zeta$ such that $J\cap K_{\eta}=K$ for every
$\eta\in A\setminus\xi$.

Take any $w\in\tilde V\cap f^{-1}[Z]$ and modify it to produce
$w'\in\prod_{i\in I}X_i$ such that $w'\restr J=w\restr J$ and
$w'\in\hat V_{\eta}\cap V'_{\eta}$ for every $\eta\in B\setminus\xi$;
this is possible because $\hat V_{\eta}\cap V'_{\eta}$ is determined by
coordinates in $K_{\eta}\setminus K$ for each $\eta$, and $J$ and the
$K_{\eta}\setminus K$ are disjoint.
Set $x=f(w')$;  then $x\in Z\subseteq F$, so
$\mu_2(W[\{x\}]\cap H'_{\zeta})>0$.

$w'\in\tilde V$, because $w\in\tilde V$
and $\tilde V$ is determined by coordinates in $K\subseteq J$;  so
$w'\in\tilde V\cap\hat V_{\eta}\cap V'_{\eta}=V_{\eta}\cap V'_{\eta}$ for
every $\eta\in B\setminus\xi$.
Accordingly $x\in f[V_{\eta}\cap V'_{\eta}]$;  as
$W\cap(f[V_{\eta}\cap V'_{\eta}]\times H_{\eta})=\emptyset$,
$W[\{x\}]$ does not meet $H_{\eta}$.   As $\eta$ is arbitrary,
$W[\{x\}]$ does not meet $H'_{\xi}$ and $W[\{x\}]\cap H'_{\zeta}$ is
$\mu_2$-negligible.   But this is impossible.\ \Bang

\medskip

{\bf (f)} This contradiction shows that $\mu$ will be a topological
measure, at least if $\mu_1$ and $\mu_2$ are complete, totally finite
and inner regular
with respect to the Borel sets.   Now suppose just that $\mu_1$ and
$\mu_2$ are totally finite.   Let $\mu'_1$ and $\mu'_2$ be the completions
of the Borel measures $\mu_1\restr\Cal B(X)$ and $\mu_2\restr\Cal B(Y)$,
and $\mu'$ their c.l.d.\
product.   Then $\mu_1\restr\Cal B(X)$ and $\mu'_1$ are
completion regular topological measures, while $\mu_2\restr\Cal B(Y)$ and
$\mu'_2$ are $\tau$-additive.   So (a)-(e) tell us that $\mu'$ measures
every open set.   Now the completions $\hat\mu_1$, $\hat\mu_2$ extend
$\mu'_1$ and $\mu'_2$, and $\mu$ is the c.l.d.\ product of $\hat\mu_1$ and
$\hat\mu_2$ (251T), so $\mu$ extends $\mu'$
(251L).   Thus we again have a topological product measure $\mu$.

\medskip

{\bf (g)} In the general case, let $W\subseteq X\times Y$ be an open set,
$E\subseteq X$ a zero set of finite measure, and $F\subseteq Y$ any set of
finite measure.   Then $\mu$ measures $W\cap(E\times F)$.   \Prf\ Let
$(\mu_1)_E$ and $(\mu_2)_F$ be the subspace measures.   Then both are
totally finite topological measures, $(\mu_1)_E$ is inner regular with
respect to the zero sets (412Pd), $E$ is
quasi-dyadic (434Pc), and $(\mu_2)_F$ is $\tau$-additive (414K).   So
the product $(\mu_1)_E\times(\mu_2)_F$ is a topological measure and
measures $W\cap(E\times F)$.    By 251Q, $\mu$ measures
$W\cap(E\times F)$.\ \Qed

Let $\Cal K$ be the family of zero sets of finite measure in $X$, $\Cal L$
the family of Borel sets of finite measure in $Y$, and $\Cal M$ the family
of sets $M\subseteq X\times Y$ such that $\mu$ measures $W\cap M$.
Because $\mu_1$ is inner regular with respect to $\Cal K$,
$\mu_2$ is inner regular with respect to $\Cal L$,
$E\times F\in\Cal M$ for every $E\in\Cal K$ and $F\in\Cal L$, and
$\Cal M$ is a $\sigma$-algebra of sets, 412R tells us
that $\mu$ is inner regular with respect to $\Cal M$.   As $\mu$ is
complete and locally determined, it must measure $W$ (412Ja).   As $W$ is
arbitrary, $\mu$ is a topological measure.

\medskip

{\bf (h)} Finally, as noted in (b), $\mu_1$ is $\tau$-additive and
there is a $\tau$-additive topological measure $\tilde\mu$ on $X\times Y$
extending $\mu$.   (434Q and 417C still apply.)
So $\mu$ too must be $\tau$-additive.
}%end of proof of 532D

\leader{532E}{Corollary} Let $\familyiI{X_i}$ be a family of regular
spaces with countable networks, and $Y$ any topological space.   Suppose
that we are given a strictly positive topological probability measure
$\mu_i$ on each $X_i$, and a $\tau$-additive topological probability
measure $\nu$ on $Y$.   Let $\mu$ be the ordinary product measure on
$Z=\prod_{i\in I}X_i\times Y$.

(a) $\mu$ is a topological measure.

(b) $\mu$ is $\tau$-additive.

(c) If $\nu$ is completion regular, and every $\mu_i$ is inner regular with
respect to the Borel sets, then $\mu$ is completion regular.

\proof{{\bf (a)} For each $i$, $X_i$ is hereditarily Lindel\"of (4A2Nb), so
$\mu_i$ is $\tau$-additive (414O);  let $\mu'_i$ be the completion of the
Borel measure $\mu_i\restr\Cal B(X_i)$.   Then $\mu'_i$ is a
quasi-Radon measure (415C).
By 4A2Nb, $X_i$ is perfectly normal, so $\mu'_i$ is completion regular.
By 434Pb-434Pc, $\prod_{i\in I}X_i$ is
quasi-dyadic.   The product $\nu_1$ of the $\mu'_i$ is a topological
measure (453I) and inner regular with respect to the zero sets (412Ub);
so the product $\mu'$ of $\nu_1$ and $\nu$ is a topological measure, by
532D.   Now $\mu'$ is also the product of the measures
$\mu_i\restr\Cal B(X_i)$ and $\nu$ (254I, 254N),
so $\mu$ extends $\mu'$ (254H) and $\mu$ also is a topological measure.

\medskip

{\bf (b)} Because every $\mu_i$ is $\tau$-additive, as is $\nu$,
417E tells us that
there is a $\tau$-additive measure extending $\mu$, so $\mu$ itself must
be $\tau$-additive.

\medskip

{\bf (c)} For any $i\in I$, we know from (a) that
$\mu'_i$ is inner regular with respect to
the zero sets.   Now every non-$\mu_i$-negligible set includes a
non-$\mu_i$-negligible Borel set, which includes a non-$\mu_i$-negligible
zero set;  accordingly $\mu_i$ is completion regular.
By 412Ub again, $\mu$ is inner regular with respect to the
zero sets, so is completion regular.
}%end of proof of 532E

\leader{532F}{Corollary} Let $\familyiI{(X_i,\mu_i)}$ be a family of
quasi-dyadic compact Hausdorff spaces with strictly positive completion
regular Radon measures.   Then the ordinary product measure $\mu$ on
$\prod_{i\in I}X_i$ is a completion regular Radon measure.

\proof{ By 532D, the ordinary product measure on
$\prod_{i\in J}X_i$ is a topological measure, for every finite
$J\subseteq I$.   By 417Sc, $\mu$ is the $\tau$-additive product
measure on $\prod_{i\in I}X_i$,
which by 417Q is a Radon measure.   By 412Ub once more, $\mu$
is completion regular.
}%end of proof of 532F

\leader{532G}{Proposition} Suppose that $\lambda$, $\lambda'$ and
$\kappa$ are cardinals such that
$\max(\omega,\lambda)\le\lambda'\le\kappa$
and $\lambda\in\MahcrR(\{0,1\}^{\kappa})$.   Then
$\lambda'\in\MahcrR(\{0,1\}^{\kappa})$.

\proof{ Let $\nu$ be a completion regular \Mth\ Radon probability
measure on $\{0,1\}^{\kappa}$ with Maharam type $\lambda$, and consider
the ordinary product measure $\mu$ of $\nu_{\lambda'}$ and $\nu$ on
$X=\{0,1\}^{\lambda'}\times\{0,1\}^{\kappa}$.   Applying 532E with
$Y=\{0,1\}^{\kappa}$ and $X_{\xi}=\{0,1\}$ for $\xi<\lambda'$, we see
that $\mu$ is a completion regular topological probability measure on a
compact Hausdorff space, therefore (being complete) a
Radon measure.   By 334A, the Maharam type of $\mu$ is at most
$\max(\omega,\lambda',\lambda)=\lambda'$, so the measure algebra
$(\frak A,\bar\mu)$ of $\mu$ can be embedded in $\frak B_{\lambda'}$.
At the same time, the \imp\ projection from $X$ onto
$\{0,1\}^{\lambda'}$ induces a measure-preserving embedding of
$\frak B_{\lambda'}$ into $\frak A$.   By 332Q, $(\frak A,\bar\mu)$ and
$(\frak B_{\lambda'},\bar\nu_{\lambda'})$ are isomorphic, that is, $\mu$
is \Mth\ with Maharam type $\lambda'$.   So $\mu$ witnesses that
$\lambda'\in\MahcrR(X)=\MahcrR(\{0,1\}^{\kappa})$.
}%end of proof of 532G

\vleader{48pt}{532H}{Lemma} Let $\familyiI{X_i}$ be a family of separable
metrizable spaces, and $\mu$ a totally finite completion regular
topological measure on $X=\prod_{i\in I}X_i$.   Then

(a) the support of $\mu$ is a zero set;

(b) $\mu$ is inner regular with respect to the self-supporting zero
sets.

\proof{{\bf (a)} Recall from 434Q that $\mu$ is $\tau$-additive, so has
a support $Z$.   Let $\sequencen{K_n}$ be a sequence of zero sets such
that $K_n\subseteq Z$ and $\mu K_n\ge\mu X-2^{-n}$ for each $n$.   Then
there is a countable set $J\subseteq I$ such that every $K_n$ is
determined by coordinates in $J$ (4A3Nc again).
So $\bigcup_{n\in\Bbb N}K_n$ and
$Z'=\overline{\bigcup_{n\in\Bbb N}K_n}$ are determined by coordinates in
$J$ (4A2B(g-i)), and $Z'$ is a zero set, by 4A3Nc in the other direction.
But $Z'\subseteq Z$ and $\mu Z'=\mu Z$ so
$Z=Z'$ is a zero set.

\medskip

{\bf (b)} If $\mu E>\gamma$ then there is a zero set $K\subseteq E$ such
that $\mu K\ge\gamma$.   Now $\mu\LLcorner K$ (234M) is a totally finite
topological measure on $X$ which is completion regular (412Q),
so its support $Z$ is a
zero set, by (a);  and $Z\subseteq K\subseteq E$
is self-supporting for $\mu$ with $\mu Z\ge\gamma$.
}%end of proof of 532H

\leader{532I}{}\cmmnt{ There is a useful general characterization of
the sets $\MahcrR(\{0,1\}^{\kappa})$ in terms of the measure algebras
$\frak B_{\lambda}$.   At the same time, we can check that other
products of separable metrizable spaces follow powers of $\{0,1\}$, as
follows.

\medskip

\noindent}{\bf Theorem}\cmmnt{ ({\smc Choksi \& Fremlin 79})} Let
$\lambda\le\kappa$ be infinite cardinals.   Then the following are
equiveridical:

(i) $\lambda\in\MahcrR(\{0,1\}^{\kappa})$;

(ii) there is a family $\ofamily{\xi}{\kappa}{X_{\xi}}$ of non-singleton
separable metrizable spaces such that
$\lambda\in\Mahcr(\prod_{\xi<\kappa}X_{\xi})$;

(iii) there is a Boolean-independent family
$\ofamily{\xi}{\kappa}{b_{\xi}}$ in
$\frak B_{\lambda}$ with the following property:   for every
$a\in\frak B_{\lambda}$
there is a countable set $J\subseteq\kappa$ such that the subalgebras
generated by $\{a\}\Bcup\{b_{\xi}:\xi\in J\}$ and
$\{b_{\eta}:\eta\in\kappa\setminus J\}$ are Boolean-independent.

\proof{ If $\kappa=\omega$ then $\lambda=\omega$ and (i)-(iii) are all
true.   So we may assume that $\kappa$ is uncountable.

\medskip

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

\medskip

{\bf (ii)$\Rightarrow$(iii)}\grheada\
If $\lambda\in\Mahcr(X)$, where every $X_{\xi}$ is a
non-trivial separable metrizable space and
$X=\prod_{\xi<\kappa}X_{\xi}$, let $\mu$ be a \Mth\ completion regular
topological probability measure on $X$ with Maharam type $\lambda$.   By
532Ha and 4A3Nc,
the support $Z$ of $\mu$ is determined by coordinates in a
countable subset $L$ of $\kappa$.

\medskip

\quad\grheadb\
Let $\frak A$ be the measure algebra of $\mu$.   For each $\xi<\kappa$,
let $f_{\xi}:X_{\xi}\to[0,1]$ be a continuous function taking both values
$0$ and $1$;  let $t_{\xi}\in\ooint{0,1}$ be such that
$\mu\{x:x\in X$, $f_{\xi}(x(\xi))=t_{\xi}\}=0$.   Set
$U_{\xi}=\{x:f_{\xi}(x(\xi))<t_{\xi}\}$,
$V_{\xi}=\{x:f_{\xi}(x)>t_{\xi}\}$;  then $U_{\xi}$ and $V_{\xi}$ are
disjoint non-empty open sets in $X$, both determined by coordinates in
$\{\xi\}$, and $\mu(U_{\xi}\cup V_{\xi})=1$.   Set
$b_{\xi}=U_{\xi}^{\ssbullet}$ in $\frak A$.   Then
$\ofamily{\xi}{\kappa\setminus L}{b_{\xi}}$ is Boolean-independent.
\Prf\ If $I$, $I'\subseteq\kappa\setminus L$ are disjoint finite sets,
then $H=X\cap\bigcap_{\xi\in I}U_{\xi}\cap\bigcap_{\xi\in I'}V_{\xi}$ is
a non-empty open set in $X$.   As $H$ is determined by coordinates in
$I\cup I'$ and $Z$ is determined by coordinates in $L$, $Z\cap H$ is
non-empty and therefore non-negligible;  so $\mu H>0$ and
$\inf_{\xi\in I}b_{\xi}\Bsetminus\sup_{\xi\in I'}b_{\xi}$ is non-zero in
$\frak A$.\ \Qed

\medskip

\quad\grheadc\
If $a\in\frak A$ let $E$ be such that $E^{\ssbullet}=a$.   By 532Hb, we
can choose for each $n\in\Bbb N$ self-supporting zero sets
$K_n\subseteq E$, $\tilde K_n\subseteq X\setminus E$ such that
$\mu K_n+\mu\tilde K_n\ge 1-2^{-n}$.   Let $J\subseteq\kappa\setminus L$
be a countable set such that every $K_n$ and every $\tilde K_n$ is
determined by coordinates in $J\cup L$.
Now the subalgebras $\frak D_1$, $\frak D_2$ generated by
$\{a\}\cup\{b_{\xi}:\xi\in J\}$ and
$\{b_{\xi}:\xi\in(\kappa\setminus L)\setminus J\}$ are
Boolean-independent.   \Prf\ Take non-zero $d_1\in\frak D_1$ and
$d_2\in\frak D_2$.   Suppose for the moment that
$d_1\Bcap a\ne 0$.   As in ($\beta$),
there is an open set $G$, determined by coordinates in
$J$, such that $0\ne a\Bcap G^{\ssbullet}\Bsubseteq d_1$.
There is also an open set $H$, determined by coordinates in
$\kappa\setminus(J\cup L)$, such that
$0\ne H^{\ssbullet}\Bsubseteq d_2$.   Next, as
$a=\sup_{n\in\Bbb N}K_n^{\ssbullet}$, there is an $n\in\Bbb N$ such that
$0\ne K_n^{\ssbullet}\Bcap G^{\ssbullet}$, that is,
$K_n\cap G\ne\emptyset$.   As $K_n\cap G$ is determined by coordinates
in $J\cup L$ and $H$ is determined by coordinates in
$\kappa\setminus(J\cup L)$,
$K_n\cap G\cap V\ne\emptyset$;  as $K_n$ is self-supporting,

\Centerline{$0\ne(K_n\cap G\cap H)^{\ssbullet}\Bsubseteq d_1\Bcap d_2$.}

\noindent In the same way, using $K'_n$ in place of $K_n$, we see that
$d_1\Bcap d_2\ne 0$ if $d_1\Bsetminus a\ne 0$.   As $d_1$ and $d_2$ are
arbitrary, $\frak D_1$ and $\frak D_2$ are Boolean-independent.\ \Qed

\medskip

\quad\grheadd\
As $\#(\kappa\setminus L)=\kappa$ and $\frak A\cong\frak B_{\lambda}$,
$\family{\xi}{\kappa\setminus L}{b_{\xi}}$, suitably reinterpreted,
witnesses that (iii) is satisfied.

\medskip

{\bf (iii)$\Rightarrow$(i)} Now suppose that the conditions of (iii) are
satisfied.   Let $(Z,\nu)$ be the Stone space of
$(\frak B_{\lambda},\bar\nu_{\lambda})$.   (See 411P
for a summary of the properties of these spaces.)   For
$b\in\frak B_{\lambda}$ write $\widehat{b}$ for the corresponding
open-and-closed subset of $Z$.    Define $\phi:Z\to\{0,1\}^{\kappa}$ by
setting $\phi(z)=\ofamily{\xi}{\kappa}{\chi\widehat{b}_{\xi}(z)}$ for
$z\in Z$.   Then $\phi$ is continuous;  let $\mu=\nu\phi^{-1}$ be the
image Radon measure on $\{0,1\}^{\kappa}$ (418I).   Now $\mu$ is
completion regular.   \Prf\ Suppose that $K\subseteq\{0,1\}^{\kappa}$ is
compact and self-supporting.   Identifying
$\frak B_{\lambda}$ with the measure algebra of $\nu$, we have a Boolean
homomorphism $\psi:\dom\mu\to\frak B_{\lambda}$ defined by setting
$\psi E=(\phi^{-1}[E])^{\ssbullet}$ whenever $\mu$ measures $E$, and
$\bar\nu_{\lambda}\psi E=\nu\phi^{-1}[E]=\mu E$ for every $E$;  setting
$E_{\xi}=\{x:x\in\{0,1\}^{\kappa}$, $x(\xi)=1\}$,
$\psi E_{\xi}=b_{\xi}$.   Set $a=\psi K$.   Let $J\subseteq\kappa$ be a
countable set such that
the subalgebras $\frak D_1$, $\frak D_2$ generated by
$\{a\}\Bcup\{b_{\xi}:\xi\in J\}$ and
$\{b_{\eta}:\eta\in\kappa\setminus J\}$ are Boolean-independent.
\Quer\ If $x\in K$, $y\in\{0,1\}^{\kappa}\setminus K$ and
$x\restr J=y\restr J$, let $U$ be an open cylinder containing $y$
and disjoint from $K$.   Express $U$ as $U'\cap U''$ where $U'$ is
determined by coordinates in $J$ and $U''$ by coordinates in
$\kappa\setminus J$.   Then $\psi U'\in\frak D_1$ and
$\psi U''\in\frak D_2$.   As $\ofamily{\xi}{\kappa}{b_{\xi}}$ is
Boolean-independent,
$\psi U''\ne 0$.   Now $K$ is self-supporting and $x\in K\cap U'$, so
$\mu(K\cap U')>0$ and $\psi(K\cap U')=a\Bcap\psi U'$ is non-zero;  also
$a\Bcap\psi U'\in\frak D_1$;  because $\frak D_1$ and $\frak D_2$ are
Boolean-independent, $\psi(K\cap U)=a\Bcap\psi U'\Bcap\psi U''\ne 0$ and
$K\cap U$ cannot be empty, contrary to the choice of $U$.\ \Bang

This shows that $K$ is determined by coordinates in $J$ and is a zero
set (4A3Nc, in the other direction).
As $K$ is arbitrary, we see that all self-supporting compact sets
are zero sets.   But as $\mu$ is a Radon measure, it is inner regular
with respect to the self-supporting compact sets, therefore with respect
to the zero sets, and is completion regular.\ \Qed

The \imp\ function $\phi$ (and, of course, the Boolean homomorphism
$\psi$) correspond to an embedding of the measure algebra
of $\mu$ into
$\frak B_{\lambda}$.   So the Maharam type of $\mu$ is at most
$\lambda$.
There is therefore a $\lambda'\in\MahcrR(\{0,1\}^{\kappa})$ such
that $\lambda'\le\lambda$ (532B).   By 532G,
$\lambda\in\MahcrR(\{0,1\}^{\kappa})$.
}%end of proof of 532I

\leader{532J}{Corollary}
(a) Suppose that $\lambda$, $\kappa$ are infinite cardinals and
$\lambda\in\MahcrR(\{0,1\}^{\kappa})$.   Then
$\kappa$ is at most the cardinal power $\lambda^{\omega}$.

(b) If $\kappa$ is an infinite cardinal such that
$\lambda^{\omega}<\kappa$ for every $\lambda<\kappa$\cmmnt{ (e.g.,
$\kappa=\frak c^+$)}, then $\MahcrR(\{0,1\}^{\kappa})=\{\kappa\}$.

\proof{{\bf (a)} By 532I, $\kappa\le\#(\Cal B_{\lambda})$;  by
524Ma, $\#(\Cal B_{\lambda})\le\lambda^{\omega}$.

\medskip

{\bf (b)} By (a), no infinite cardinal less than $\kappa$ can belong to
$\MahcrR(\{0,1\}^{\kappa})$.   Also $\kappa$ is uncountable, so the
remarks in 532C tell us the rest of what we need.
}%end of proof of 532J

\leader{532K}{Corollary} If $\omega\le\lambda\le\kappa'\le\kappa$ and
$\lambda\in\MahcrR(\{0,1\}^{\kappa})$ then
$\lambda\in\MahcrR(\{0,1\}^{\kappa'})$.

\proof{ If $\ofamily{\xi}{\kappa}{b_{\xi}}$ witnesses the truth of
532I(iii) for $\lambda$ and $\kappa$, then its subfamily
$\ofamily{\xi}{\kappa'}{b_{\xi}}$ witnesses the truth of 532I(iii) for
$\lambda$ and $\kappa'$.   \Prf\ Of course
$\ofamily{\xi}{\kappa'}{b_{\xi}}$ is Boolean-independent.   If
$a\in\frak B_{\lambda}$, there is a countable set $J\subseteq\kappa$ such
that the subalgebras generated by $\{a\}\cup\{b_{\xi}:\xi\in J\}$ and
$\{b_{\eta}:\eta\in\kappa\setminus J\}$ are Boolean-independent.
Now $J'=J\cap\kappa'$ is a countable subset of $\kappa'$ and
the subalgebras generated by $\{a\}\cup\{b_{\xi}:\xi\in J'\}$ and
$\{b_{\eta}:\eta\in\kappa'\setminus J'\}$ are Boolean-independent.\ \Qed
}%end of proof of 532K

\leader{532L}{Corollary} If $\omega\le\lambda\le\lambda'$ and
$\cff[\lambda']^{\le\lambda}<\cf\kappa$ and
$\lambda'\in\MahcrR(\{0,1\}^{\kappa})$, then
$\lambda\in\MahcrR(\{0,1\}^{\kappa})$.

\proof{ Let $\ofamily{\xi}{\kappa}{b_{\xi}}$ be a family in
$\frak B_{\lambda'}$ satisfying (iii) of 532I.   Let
$\ofamily{\eta}{\lambda'}{e_{\eta}}$ be the standard generating
family in $\frak B_{\lambda'}$, and
$\Cal J$ a cofinal subset of $[\lambda']^{\lambda}$ with cardinal less
than $\cf\kappa$.   For each $\xi<\kappa$, there are a countable set
$L\subseteq\lambda'$ such that $b_{\xi}$ belongs to the closed
subalgebra $\frak C_L$ of $\frak B_{\lambda'}$ generated by
$\{e_{\eta}:\eta\in L\}$, and a $J_{\xi}\in\Cal J$ such that
$L\subseteq J_{\xi}$.   Because
$\#(J)<\cf\kappa$, there is a $J\in\Cal J$ such that
$A=\{\xi:\xi<\kappa$, $J_{\xi}=J\}$ has cardinal $\kappa$.   Now the
closed subalgebra $\frak C_J$ of $\frak B_{\lambda'}$ generated by
$\{e_{\eta}:\eta\in J\}$ is isomorphic to $\frak B_{\lambda}$, and the
Boolean-independent $\family{\xi}{A}{b_{\xi}}$ in $\frak C_J$
witnesses that 532I(iii) is true of $\lambda$
and $\kappa$, as in the proof of 532K.
}%end of proof of 532L

\leader{532M}{}\cmmnt{ I turn now to the question of identifying those
$\kappa$ for which $\omega\in\MahcrR(\{0,1\}^{\kappa})$.   We know from
532C and 532Ja that
they all lie between $\omega$ and $\frak c$.   To go farther we need to
look at some of the cardinals from \S522.

\medskip

\noindent}{\bf Proposition} If
$A\subseteq\frak B_{\omega}\setminus\{0\}$ and
$\#(A)<\frak d\cmmnt{\mskip5mu=\cf(\BbbN^{\Bbb N})}$, then there is
a $c\in\frak B_{\omega}$ such that neither $c$ nor $1\Bsetminus c$
includes any member of $A$.

\proof{ Let $\sequencen{e_n}$ be the standard generating family in
$\frak B_{\omega}=\frak B_{\Bbb N}$.
For $a\in\frak A$ and $n\in\Bbb N$ let
$f_a(n)\in\Bbb N$ be such that there is a $b$ in the subalgebra
$\frak C_{f_a(n)^2}$ generated by $\{e_i:i<f_a(n)^2\}$ such that
$\bar\nu_{\omega}(b\Bsymmdiff a)<2^{-n-3}\bar\mu a$.   Because
$\#(A)<\frak d$, there is an $f\in\BbbN^{\Bbb N}$ such that
$f\not\le f_a$ for every $a\in\frak A$;  we may suppose that $f$ is
strictly increasing and $f(0)>0$.   Note that

\Centerline{$f(n)^2+n+1<f(n)^2+2f(n)+1\le f(n+1)^2$}

\noindent for every $n$.   For
each $n\in\Bbb N$, set

\Centerline{$I_n=f(n)^2\subseteq\Bbb N$,
\quad$I'_n=I_{n+1}\setminus I_n$,}

\Centerline{$c'_n=\inf_{f(n)^2\le i\le f(n)^2+n+1}e_i
\in\frak C_{I'_n}$;}

\noindent then $\bar\nu_{\omega}c'_n=2^{-n-2}$ for each
$n$.   Define $c_n\in\frak C_{I_n}$,
for $n\in\Bbb N$, by setting $c_0=0$ and
$c_{n+1}=c_n\Bsymmdiff c'_n$ for each $n$.   Then
$\bar\nu_{\omega}(c_{n+1}\Bsymmdiff c_n)=2^{-n-2}$ for every $n$, so
$\sequencen{c_n}$ is a Cauchy sequence for the measure metric on
$\frak B_{\omega}$, and has a limit $c$.   Note that

\Centerline{$\sum_{i=n}^{m-1}2^{-i-3}
\le\bar\nu_{\omega}(c_m\Bsymmdiff c_n)
\le\sum_{i=n}^{m-1}2^{-i-2}$}

\noindent whenever $m\ge n$.   \Prf\ Induce on $m$.   For $m=n$ the
result is trivial (interpreting $\sum_{i=n}^{n-1}$ as zero).   For the
inductive step to $m+1$, $c'_m\in\frak C_{I'_m}$ is stochastically
independent of $c_m\Bsymmdiff c_n\in\frak C_{I_m}$, so

$$\eqalignno{\bar\nu_{\omega}(c_{m+1}\Bsymmdiff c_n)
&=\bar\nu_{\omega}(c'_m\Bsymmdiff c_m\Bsymmdiff c_n)\cr
&=\bar\nu_{\omega}c'_m+\bar\nu_{\omega}(c_m\Bsymmdiff c_n)
   -2\bar\nu_{\omega}(c'_m\Bcap(c_m\Bsymmdiff c_n))\cr
&=2^{-m-2}+(1-2^{-m-1})\bar\nu_{\omega}(c_m\Bsymmdiff c_n)\cr
&\ge 2^{-m-2}+(1-2^{-m-1})\sum_{i=n}^{m-1}2^{-i-3}\cr
\displaycause{by the inductive hypothesis}
&=\sum_{i=n}^{m-1}2^{-i-3}+2^{-m-3}(2-4\sum_{i=n}^{m-1}2^{-i-3})
\ge\sum_{i=n}^m2^{-i-3};\cr}$$

\noindent on the other hand,

\Centerline{$\bar\nu_{\omega}(c_{m+1}\Bsymmdiff c_n)
\le 2^{-m-2}+\bar\nu_{\omega}(c_m\Bsymmdiff c_n)
\le\sum_{i=n}^m2^{-i-2}$.}

\noindent So the induction proceeds.\ \QeD\  Taking the limit
as $m\to\infty$, we see
that $2^{-n-2}\le\bar\nu_{\omega}(c\Bsymmdiff c_n)\le 2^{-n-1}$ for
every $n\in\Bbb N$.

Take any $a\in A$.   Let $n\in\Bbb N$ be such that $f_a(n)<f(n)$.   Then
there is a $b\in\frak C_{I_n}$ such that
$\bar\nu_{\omega}(a\Bsymmdiff b)<2^{-n-3}\bar\mu a$.   Now
$c\Bsymmdiff c_n\in\frak C_{\Bbb N\setminus I_n}$
is stochastically independent of both
$b\Bsetminus c_n$ and $b\Bcap c_n$, so

$$\eqalign{\bar\nu_{\omega}(b\Bsetminus c)
&=\bar\nu_{\omega}(((b\Bsetminus c_n)\Bsetminus(c\Bsymmdiff c_n))
 \Bcup((b\Bcap c_n)\Bcap(c\Bsymmdiff c_n)))
\cr
&=\bar\nu_{\omega}(b\Bsetminus c_n)
  (1-\bar\nu_{\omega}(c\Bsymmdiff c_n))
 +\bar\nu_{\omega}(b\Bcap c_n)\cdot\bar\nu_{\omega}(c\Bsymmdiff c_n)
\cr
&\ge\bar\nu_{\omega}(b\Bsetminus c_n)(1-2^{-n-1})
  +2^{-n-2}\bar\nu_{\omega}(b\Bcap c_n)
\ge 2^{-n-2}\bar\nu_{\omega}b
\ge 2^{-n-3}\bar\nu_{\omega}a.\cr}$$

\noindent So

\Centerline{$\bar\nu_{\omega}(a\Bsetminus c)
\ge 2^{n-3}\bar\nu_{\omega}a-\bar\nu_{\omega}(b\Bsetminus a)
>0$,}

\noindent and $a\notBsubseteq c$.   Similarly,

$$\eqalign{\bar\nu_{\omega}(b\Bcap c)
&=\bar\nu_{\omega}(b\cap c_n)
  (1-\bar\nu_{\omega}(c\Bsymmdiff c_n))
 +\bar\nu_{\omega}(b\Bsetminus c_n)
  \cdot\bar\nu_{\omega}(c\Bsymmdiff c_n)\cr
&\ge\bar\nu_{\omega}(b\Bcap c_n)(1-2^{-n-1})
  +2^{-n-2}\bar\nu_{\omega}(b\Bsetminus c_n)
\ge 2^{-n-2}\bar\nu_{\omega}b,\cr}$$

\noindent and $\bar\nu_{\omega}(a\Bcap c)>0$.

As $a$ is arbitrary, we have found an appropriate $c$.
}%end of proof of 532M

\leader{532N}{}\cmmnt{ It will be useful to have a classic example
relevant to a question already examined in 325F.

\medskip

\noindent}{\bf Lemma} There is a Borel set
$W\subseteq\{0,1\}^{\Bbb N}\times\{0,1\}^{\Bbb N}$ such that whenever
$E$, $F\subseteq\{0,1\}^{\Bbb N}$ have positive measure for
$\nu_{\omega}$ then neither $(E\times F)\cap W$ nor
$(E\times F)\setminus W$ is negligible for the product measure
$\nu_{\omega}^2$ on  $\{0,1\}^{\Bbb N}\times\{0,1\}^{\Bbb N}$.

\proof{{\bf (a)} (Cf.\ 134Jb.)   There is a Borel set
$H\subseteq\{0,1\}^{\Bbb N}$ such that both $H$ and its complement meet
every non-empty open set in a set of non-zero measure.   \Prf\ For
$x\in\{0,1\}^{\Bbb N}$ set $I_x=\{n:x(i)=0$ for $2^n\le i<2^{n+1}\}$.   Set
$H=\{x:I_x$ is finite and not empty and $\max I_x$ is even$\}$.\ \Qed

\medskip

{\bf (b)} Let $+$ be the usual group operation on
$\{0,1\}^{\Bbb N}\cong\Bbb Z_2^{\Bbb N}$.   In this group, addition and
subtraction are identical, as $x+x=0$ for every $x$;  but the formulae
may be easier to read if I use the symbol $-$ when it seems appropriate.
Set $W=\{(x,y):x$, $y\in\{0,1\}^{\Bbb N}$, $x-y\in H\}$.

Let $E$, $F\subseteq\{0,1\}^{\Bbb N}$ be sets of positive measure.   Then
$\{z:z\in\{0,1\}^{\Bbb N}$, $\nu_{\omega}(E\cap(F+z))>0\}$ is open
(443C) and not empty (443Da), so meets $H$ in a set of positive measure.
Now

$$\eqalignno{\nu_{\omega}^2((E\times F)\cap W)
&=\nu_{\omega}^2\{(x,y):x\in E,\,y\in F,\,x-y\in H\}\cr
&=\nu_{\omega}^2\{(x,z):x\in E,\,x-z\in F,\,z\in H\}\cr
\displaycause{because $(x,y)\mapsto(x,x-y)$ is a measure space automorphism
for $\nu_{\omega}^2$, as in 255Ae or 443Xa}
&=\nu_{\omega}^2\{(x,z):x\in E,\,x\in F+z,\,z\in H\}\cr
&=\int_H\nu_{\omega}(E\cap(F+z))\nu_{\omega}(dz)
>0.\cr}$$

\noindent Applying the same argument with $\{0,1\}^{\Bbb N}\setminus H$
in the
place of $H$, we see that the same is true of $(E\times F)\setminus W$.
}%end of proof of 532N

\leader{532O}{Proposition} If
$A\subseteq\frak B_{\omega}\setminus\{0\}$ and
$\#(A)<\cov\Cal N_{\omega}$, then there is a $c\in\frak B_{\omega}$
such that neither $c$ nor $1\Bsetminus c$ includes any member of $A$.

\proof{ Take $W\subseteq\{0,1\}^{\Bbb N}\times\{0,1\}^{\Bbb N}$ as in
532N.   For $x\in\{0,1\}^{\Bbb N}$, set $c_x=W[\{x\}]^{\ssbullet}$ in
$\frak B_{\omega}$.
If $a\in A$, then $\{x:a\Bsubseteq c_x\}\in\Cal N_{\omega}$.   \Prf\ Let
$F\in\Tau_{\omega}$ be such that $F^{\ssbullet}=a$, and set
$E=\{x:a\Bsubseteq c_x\}$.   Because
$x\mapsto c_x$ is measurable when $\frak B_{\omega}$ is given its
measure-algebra topology (418Ta), $E\in\Tau_{\omega}$.   For every
$x\in E$, $F\setminus W[\{x\}]$ is negligible, so
$(E\times F)\setminus W$ is negligible, by Fubini's theorem (252D).
But this means that at least
one of $E$, $F$ must be negligible;  since $F^{\ssbullet}=a\ne 0$,
$\nu_{\omega}E=0$, as required.\ \Qed

Similarly, $\{x:a\Bcap c_x=0\}$ is negligible.   Since
$\{0,1\}^{\Bbb N}$ cannot be covered by $\#(A)$ negligible sets, there
is an $x\in\{0,1\}^{\Bbb N}$ such that $c_x$ neither includes, nor is
disjoint from, any member of $A$.
}%end of proof of 532O

\leader{532P}{Proposition} Set
$\kappa=\max(\frak d,\cov\Cal N_{\omega})$.   If
$\FN(\Cal P\Bbb N)=\omega_1$, then
$\omega\in\MahcrR(\{0,1\}^{\kappa})$.   In particular, if
$\frak c=\omega_1$ then $\omega\in\MahcrR(\{0,1\}^{\omega_1})$.

\proof{{\bf (a)} By 524O(b-ii), $\FN(\frak B_{\omega})=\omega_1$;  let
$f:\frak B_{\omega}\to[\frak B_{\omega}]^{\le\omega}$ be a
Freese-Nation function.   By 532M (if $\kappa=\frak d$) or 532O (if
$\kappa=\cov\Cal N_{\omega}$), we can choose inductively a family
$\ofamily{\xi}{\kappa}{b_{\xi}}$ in $\frak B_{\omega}$ such that neither
$b_{\xi}$ nor $1\Bsetminus b_{\xi}$ includes any nonzero member of
$\frak D_{\xi}$, where
$\frak D_{\xi}$ is the smallest subalgebra of $\frak B_{\omega}$
including $\{b_{\eta}:\eta<\xi\}$ and such that
$f(d)\subseteq\frak D_{\xi}$ for every $d\in\frak D_{\xi}$.
Of course this
implies that $\ofamily{\xi}{\kappa}{b_{\xi}}$ is Boolean-independent.

\medskip

{\bf (b)} For $K$, $L\subseteq\kappa$ set
$d_{KL}=\inf_{\xi\in K}b_{\xi}\Bsetminus\sup_{\xi\in L}b_{\xi}$.
For $a\in\frak B_{\omega}$, set
$Q_a=\{(K,L):K$, $L\in[\kappa]^{<\omega}$ are disjoint,
$d_{KL}\Bsubseteq a\}$, and let $Q'_a$ be the set of minimal members of
$Q_a$, taking $(K,L)\le(K',L')$ if $K\subseteq K'$ and $L\subseteq L'$.
Of course $Q_a$ is well-founded so $Q'_a$ is coinitial with $Q_a$.
Now $R_{an}=\{(K,L):(K,L)\in Q'_a$, $\#(K\cup L)=n\}$ is countable for
every $n\in\Bbb N$ and $a\in\frak B_{\omega}$.   \Prf\ Induce on $n$.
If $n=0$ this is trivial.   For the inductive step to $n+1$, set
$R'_{\zeta}
=\{(K,L):K\cup L\subseteq\zeta$, $(K\cup\{\zeta\},L)\in R_{a,n+1}\}$ for
each $\zeta<\kappa$.   For $(K,L)\in R'_{\zeta}$,
$b_{\zeta}\cap d_{KL}=d_{K\cup\{\zeta\},L}$ is included in $a$,
so there is a
$c_{KL\zeta}\in f(d_{K\cup\{\zeta\},L})\cap f(a)$ such that
$d_{K\cup\{\zeta\},L}\Bsubseteq c_{KL\zeta}\Bsubseteq a$, in which case
$b_{\zeta}\Bsubseteq c_{KL\zeta}\Bcup(1\Bsetminus d_{KL})$.   If
$\zeta<\zeta'<\kappa$, $(K,L)\in R'_{\zeta}$ and
$(K',L')\in R'_{\zeta'}$, then $d_{K'L'}\notBsubseteq a$ (because
$(K'\cup\{\zeta'\},L')$ is a minimal member of $Q_a$), so
$c_{KL\zeta}\Bcup(1\Bsetminus d_{K'L'})\ne 1$;  as $c_{KL\zeta}$ and
$d_{K'L'}$ both belong to $\frak D_{\zeta'}$,
$b_{\zeta'}\notBsubseteq c_{KL\zeta}\Bcup(1\Bsetminus d_{K'L'})$ and
$c_{KL\zeta}\ne c_{K'L'\zeta'}$.   As $f(a)$ is countable,
$A=\{\zeta:R'_{\zeta}\ne\emptyset\}$ is countable.   Next, for any
$\zeta\in A$ and $(K,L)\in R'_{\zeta}$, we see that
$d_{KL}\subseteq a\Bcup(1\Bsetminus b_{\zeta})$, and indeed that
$(K,L)\in Q'_{a\Bcup(1\Bsetminus b_{\zeta})}$, so that
$(K,L)\in R_{a\Bcup(1\Bsetminus b_{\zeta}),n}$.   By the inductive
hypothesis, $R'_{\zeta}$ is countable.

This shows that
$\{(K,L,\zeta):K\cup L\subseteq\zeta$,
$(K\cup\{\zeta\},L)\in R_{a,n+1}\}$ is countable.    In the same way,
applying the ideas above to
$1\Bsetminus b_{\zeta}$ in place of $b_{\zeta}$,
$\{(K,L,\zeta):K\cup L\subseteq\zeta$,
$(K,L\cup\{\zeta\})\in R_{a,n+1}\}$ is countable;  so $R_{a,n+1}$ is
countable and the induction proceeds.\ \Qed

It follows that $Q'_a$ is countable for every $a\in\frak B_{\omega}$.

\medskip

{\bf (c)} Now take any $a\in\frak B_{\omega}$ and let
$J\subseteq\kappa$ be a countable set such that $K\cup L\subseteq J$
whenever $(K,L)\in Q'_a\cup Q'_{1\Bsetminus a}$.   \Quer\ Suppose, if
possible, that the algebras $\frak E_1$, $\frak E_2$ generated by
$\{a\}\Bcup\{b_{\xi}:\xi\in J\}$ and
$\{b_{\eta}:\eta\in\kappa\setminus J\}$ are not Boolean-independent.
Then there must be finite subsets $K$, $L$, $K'$ and $L'$ of $\kappa$
such that $K\cup L\subseteq J$,
$K'\cup L'\subseteq\kappa\setminus J$, $d_{K'L'}\ne 0$, and either

\Centerline{$d_{KL}\Bcap a\ne 0$,
$d_{K'L'}\Bcap d_{KL}\Bcap a=0$}

\noindent or

\Centerline{$d_{KL}\Bsetminus a\ne 0$,
$d_{K'L'}\Bcap d_{KL}\Bsetminus a=0$.}

\noindent Suppose the former.   Then
$(K\cup K',L\cup L')\in Q_{1\Bsetminus a}$ so there is a
$(K'',L'')\in Q'_{1\Bsetminus a}$ such that $K''\subseteq K\cup K'$ and
$L''\subseteq L\cup L'$;  in which case $K''\cup L''\subseteq J$ so in
fact $K''\subseteq K$, $L''\subseteq L$ and
$d_{KL}\Bcap a\Bsubseteq d_{K''L''}\Bcap a=0$, which is impossible.
Replacing $a$ by $1\Bsetminus a$ we get a similar contradiction in the
second case.\ \BanG\  So $\frak E_1$ and $\frak E_2$ are
Boolean-independent.

\medskip

{\bf (d)} As $a$ is arbitrary, (c) shows that
$\ofamily{\xi}{\kappa}{b_{\xi}}$ satisfies the conditions of 532I(iii),
so that $\omega$ belongs to $\MahcrR(\{0,1\}^{\kappa})$, as claimed.
}%end of proof of 532P

\leader{532Q}{Proposition} Suppose that $\add\Cal N_{\omega}>\omega_1$.

(a) $\lambda\notin\MahcrR(\{0,1\}^{\kappa})$ whenever $\lambda\ge\omega$
and $\max(\omega,\cff[\lambda]^{\le\omega})<\kappa$.

(b) If $\omega_1\le\kappa\le\omega_{\omega}$ then
$\MahcrR(\{0,1\}^{\kappa})=\{\kappa\}$.

\proof{{\bf (a)} \Quer\ If $\lambda\in\MahcrR(\{0,1\}^{\kappa})$, set
$\kappa'=(\max(\omega,\cff[\lambda]^{\le\omega}))^+$;  then
$\lambda<\kappa'$ so
$\lambda\in\MahcrR(\{0,1\}^{\kappa'})$ (532K).   As
$\cff[\lambda]^{\le\omega}<\cf\kappa'$, $\omega$ belongs to
$\MahcrR(\{0,1\}^{\lambda^+})$ (532L) and therefore to
$\MahcrR(\{0,1\}^{\omega_1})$ (532K again).

Let $\ofamily{\xi}{\omega_1}{b_{\xi}}$ be a
family in $\frak B_{\omega}$ satisfying the conditions of 532I(iii).
By 524Mb, $\omega_1<\wdistr(\frak B_{\omega})$;  by 514K, there is a
countable $C\subseteq\frak B_{\omega}\setminus\{0\}$ such that for every
$\xi<\omega_1$ there is a $c\in C$ such that
$c\Bsubseteq b_{\xi}$.   Let $a\in C$ be such that
$\{\xi:\xi<\omega_1$, $a\Bsubseteq b_{\xi}\}$ is uncountable.
There is
supposed to be a countable $J\subseteq\omega_1$ such that the
subalgebras generated by $\{a\}$ and
$\{b_{\xi}:\xi\in\omega_1\setminus J\}$ are
Boolean-independent;  but then $\{\xi:a\Bsubseteq b_{\xi}\}\subseteq J$,
which is impossible.\ \Bang

This shows that (a) is true.

\medskip

{\bf (b)} If $\omega\le\lambda<\kappa\le\omega_{\omega}$, then
$\cff[\lambda]^{\le\omega}\le\lambda<\kappa$ (5A1F(e-iv)),
so (a) tells us that
$\lambda\notin\MahcrR(\{0,1\}^{\kappa})$.   From 532C we see that
$\MahcrR(\{0,1\}^{\kappa})$ must be $\{\kappa\}$ exactly.
}%end of proof of 532Q

\leader{532R}{}\cmmnt{ Two combinatorial principles already used in
524O are relevant to the questions treated here.

\medskip

\noindent}{\bf Proposition} Suppose that $\lambda$ is an uncountable
cardinal with countable cofinality such that
$\square_{\lambda}$\cmmnt{ (definition:  5A6D(a-ii))} is
true.   Set $\kappa=\lambda^+$.   Then
$\lambda\in\MahcrR(\{0,1\}^{\kappa})$.

\proof{{\bf (a)} Let $\ofamily{\xi}{\kappa}{I_{\xi}}$ be a family of
countably infinite subsets of $\lambda$ as in 5A6E.   For each
$\xi<\kappa$, let $\sequencen{I_{\xi n}}$, $\sequencen{\alpha_{\xi n}}$
be such that $\sequencen{I_{\xi n}}$ is a disjoint sequence of subsets
of $I_{\xi}$ with $\#(I_{\xi n})=n$ for each $n$ and
$\sequencen{\alpha_{\xi n}}$ is a sequence of distinct points in
$I_{\xi}\setminus\bigcup_{n\in\Bbb N}I_{\xi n}$.   Set

\Centerline{$U_{\xi n}
=\{x:x\in\{0,1\}^{\lambda}$, $x(\eta)=0$ for every $\eta\in I_{\xi n}\}$,}

\Centerline{$V_{\xi n}
=\{x:x\in U_{\xi n}\setminus\bigcup_{m>n}U_{\xi m}$,
$x(\alpha_{\xi n})=1\}$,}

\Centerline{$\tilde V_{\xi n}
=\{x:x\in U_{\xi n}\setminus\bigcup_{m>n}U_{\xi m}$,
$x(\alpha_{\xi n})=0\}$}

\noindent for $n\in\Bbb N$.   Note that as
$\nu_{\kappa}U_{\xi m}=2^{-m}$ for each $n$, $V_{\xi n}$ and
$\tilde V_{\xi n}$ are non-negligible, while both are determined by
coordinates in
$\{\alpha_{\xi n}\}\cup\bigcup_{m\ge n}I_{\xi m}\subseteq I_{\xi}$.   Set

\Centerline{$F_{\xi}=\bigcup_{n\in\Bbb N}V_{\xi n}$,
\quad$b_{\xi}=F_{\xi}^{\ssbullet}\in\frak B_{\lambda}$.}

\noindent Note that $F_{\xi}\cap\tilde V_{\xi n}=\emptyset$ for every $n$.

\medskip

{\bf (b)} Take any $a\in\frak B_{\lambda}$.   Then we can express $a$ as
$E^{\ssbullet}$ where $E\subseteq\{0,1\}^{\lambda}$ is a Baire set;  let
$I\subseteq\lambda$ be a countable set such that $E$ is determined by
coordinates in $I$.   By the choice of $\ofamily{\xi}{\kappa}{I_{\xi}}$
there is a countable set $J\subseteq\kappa$ such that $I\cap I_{\xi}$ is
finite for every $\xi\in\kappa\setminus J$.   Let $\frak D_1$,
$\frak D_2$ be the subalgebras of $\frak B_{\lambda}$ generated by
$\{a\}\cup\{b_{\xi}:\xi\in J\}$ and
$\{b_{\xi}:\xi\in\kappa\setminus J\}$ respectively.   Then $\frak D_1$
and $\frak D_2$ are Boolean-independent.   \Prf\ If $d_1\in\frak D_1$
and $d_2\in\frak D_2$ are non-zero, we can express $d_1$ as
$H_1^{\ssbullet}$ where $H_1\subseteq\{0,1\}^{\lambda}$ is a Baire set
determined by coordinates in $L=I\cup\bigcup_{\xi\in K}I_{\xi}$ for some
finite $K\subseteq J$.   Next, we can find disjoint finite sets $K'$,
$K''\subseteq\kappa\setminus J$ such that
$d_2\Bsupseteq\inf_{\xi\in K'}b_{\xi}\Bsetminus\sup_{\xi\in K''}b_{\xi}$.
Because all the sets $I_{\xi}\cap I_{\eta}$, for distinct $\xi$,
$\eta<\kappa$, and also the sets $I\cap I_{\xi}$, for
$\xi\in\kappa\setminus J$, are finite, there is an $m\in\Bbb N$ such
that all the sets
$J_{\xi}=\{\alpha_{\xi m}\}\cup\bigcup_{n\ge m}I_{\xi n}$, for
$\xi\in K'\cup K''$, are disjoint from each other and from $I$.   Look
at the sets $V_{\xi m}$, for $\xi\in K'$, and $\tilde V_{\xi m}$,
for $\xi\in K''$.   Set
$H_2=\{0,1\}^{\lambda}\cap\bigcap_{\xi\in K'}V_{\xi m}
\cap\bigcap_{\xi\in K''}\tilde V_{\xi m}$.   Then
$H_2^{\ssbullet}\Bsubseteq d_2$.   But observe now that all the
$V_{\xi m}$ and $\tilde V_{\xi m}$ are non-negligible and that $V_{\xi m}$,
$\tilde V_{\xi m}$ are determined by coordinates in $J_{\xi}$ for each
$\xi\in K'\cup K''$.   So the sets $H_1$, $V_{\xi m}$ (for $\xi\in K'$)
and $\tilde V_{\xi m}$ (for $\xi\in K''$) are stochastically
independent, and

\Centerline{$\bar\nu_{\lambda}(d_1\Bcap d_2)
\ge\nu_{\lambda}(H_1\cap H_2)
=\nu_{\lambda} H_1\cdot\prod_{\xi\in K'}\nu_{\lambda} V_{\xi m}
  \cdot\prod_{\xi\in K''}\nu_{\lambda}\tilde V_{\xi m}
>0$.}

\noindent Thus $d_1\Bcap d_2\ne 0$;  as $d_1$ and $d_2$ are arbitrary,
$\frak D_1$ and $\frak D_2$ are stochastically independent.\ \Qed

\medskip

{\bf (c)} The argument of (b) works equally well with $I=\emptyset$
and $J$ an arbitrary
finite subset of $\kappa$ to show that $\ofamily{\xi}{\kappa}{b_{\xi}}$
is Boolean-independent.   So the conditions of 532I(iii) are satisfied
and $\kappa\in\MahcrR(\lambda)$, as claimed.
}%end of proof of 532R

\leader{532S}{Proposition} Suppose that $\add\Cal N_{\omega}>\omega_1$
and that $\lambda$ is an infinite cardinal such that
CTP$(\lambda^+,\lambda)$\cmmnt{ (definition:  5A6Fa)}
is true.   Then $\lambda\notin\MahcrR(\{0,1\}^{\kappa})$ for any
$\kappa>\lambda$.

\proof{ By 532K, it is enough to consider the case $\kappa=\lambda^+$.
\Quer\ Suppose, if possible, that there is a family
$\ofamily{\xi}{\kappa}{b_{\xi}}$ in $\frak B_{\lambda}$ satisfying the
conditions of 532I(iii).   Let
$\ofamily{\eta}{\lambda}{e_{\eta}}$ be the standard generating
family in $\frak B_{\lambda}$.   Then for
each $\xi<\kappa$ we have a countable set $I_{\xi}\subseteq\lambda$ such
that $b_{\xi}$ belongs to the closed subalgebra of $\frak B_{\lambda}$
generated by $\{e_{\eta}:\eta\in I_{\xi}\}$.   Because
CTP$(\kappa,\lambda)$ is true, there is an uncountable set
$A\subseteq\kappa$ such that $J=\bigcup_{\xi\in A}I_{\xi}$ is
countable (5A6F(b-ii)).   Now the closed subalgebra $\frak C_J$ generated
by $\{e_{\eta}:\eta\in J\}$ is isomorphic to $\frak B_{\omega}$, so
$\family{\xi}{A}{b_{\xi}}$ witnesses that
$\omega\in\MahcrR(\{0,1\}^{\omega_1})$;  but this contradicts 532Qa.\ \Bang
}%end of proof of 532S

\exercises{
\leader{532X}{Basic exercises (a)}
%\spheader 532Xa
Let $X$ be a normal Hausdorff space and $Y\subseteq X$ a
zero set.   Show that $\MahcrR(Y)\subseteq\MahcrR(X)$.
%532A

\spheader 532Xb Let $\beta\Bbb N$ be the Stone-\v{C}ech
compactification of $\Bbb N$.   (i) Show that
$\MahcrR(\beta\Bbb N)=\{0\}$.   \Hint{non-empty zero sets in
$\beta\Bbb N\setminus\Bbb N$ are never ccc.}
(ii) Give an example of a non-empty
compact Hausdorff space $X$ such that $\MahcrR(X)=\emptyset$.
%532A

\spheader 532Xc Let $X$ and $Y$ be compact Hausdorff spaces.   Show that
$\MahcrR(X\times Y)\subseteq\MahcrR(X)\cup\MahcrR(Y)$.
\Hint{434U.}
%532B

\spheader 532Xd Let $\lambda$ and $\kappa$ be infinite cardinals such that
$\lambda\in\MahcrR(\{0,1\}^{\kappa})$.   (i) Show that there is a
strictly positive \Mth\ completion regular Radon probability measure on
$\{0,1\}^{\kappa}$ with Maharam type $\lambda$.   (ii) Suppose that
$\lambda$ is uncountable and that $H\subseteq\{0,1\}^{\kappa}$ is a
non-empty G$_{\delta}$ set.   Show that $\lambda\in\MahcrR(H)$.
%532E

\spheader 532Xe Find a proof of 532E which does not rely on 532D.
\Hint{415E.}
%532E

\spheader 532Xf Let $\familyiI{(X_i,\mu_i)}$ be a family of
quasi-dyadic
spaces with strictly positive completion regular topological probability
measures.   Show that the ordinary product measure on
$\prod_{i\in I}X_i$ is a strictly positive completion regular
$\tau$-additive topological probability measure.
%532F
%each \mu_i \tau-additive 434Q
%so finite products \tau-add topological measures 532D
%so product topological measure \tau-add top measure 417Sc
%completion regular by 412Ub
%strictly positive 411Xk

\leader{532Y}{Further exercises (a)}
%\spheader 532Ya
Let $Z$ be the Stone space of $\frak B_{\lambda}$, where
$\lambda\ge\omega$.   (i) Show that if $F\subseteq Z$ is a
non-empty nowhere dense zero set then it is not ccc.   (ii) Show that
$\MahcrR(Z)=\{\lambda\}$.
(iii) Show that $\MahcrR(Z\times Z)=\emptyset$.
%532Xb 532Xc 532B mt53bits

\spheader 532Yb Let $\familyiI{X_i}$ be a family of topological spaces
with countable networks, and $Y$ any topological space.   Suppose that
we are given a strictly positive topological probability measure $\mu_i$
on each $X_i$, and a $\tau$-additive topological probability measure
$\nu$ on $Y$.   Show that the ordinary product measure on
$\prod_{i\in I}X_i\times Y$ is a topological measure.
%532E
%reduce to case \supp\nu=Y , use 417Sc

\spheader 532Yc Suppose that $\FN(\Cal P\Bbb N)=\omega_1$.   Show that
there are a Hausdorff space $X$ and a completion regular Radon measure
$\mu$ on $X$ such that the Maharam type of $\mu$ is $\omega$, but the
Maharam type of $\mu\restr\Cal B(X)$ is $\omega_1$.   \Hint{419C.}
%532P
%\omega\in\MahcrR(\{0,1\}^{\omega_1\setminus\xi}) 532G
% construct  X  as in 419C
% witnessed by \mu_{\xi} , \mu  direct sum of  \mu_{\xi}
% if we have open sets G_n\subseteq X, there is a cofinal closed set
% V  such that  \mu_{\xi}(G_n\cap X_{\xi})\in\{0,1\}  for every
% \xi\in V ;  so  \mu\restr\Cal B  has Maharam type > \omega
}%end of exercises

\leader{532Z}{Problems (a)} In 532P, can we take
$\kappa=\cf\Cal N_{\omega}$?
%532P

\spheader 532Zb We have $\omega\in\MahcrR(\{0,1\}^{\omega_1})$ if
$\FN(\Cal P\Bbb N)=\omega_1$\cmmnt{ (532P, 532K)} and not if
$\add\Cal N_{\omega}>\omega_1$\cmmnt{ (532Q)}.   Can we narrow the
gap?
%532Q

\spheader 532Zc For a Hausdorff space $X$ let $\MahspcrR(X)$ be the set of
Maharam types of strictly positive Maharam homogeneous completion regular
Radon measures on $X$.
Describe the sets $\Gamma$ of cardinals for which there
are compact Hausdorff spaces $X$ such that $\MahspcrR(X)=\Gamma$.

\endnotes{
\Notesheader{532}
I have spent a good many pages on a rather specialized topic.   But I
think the patterns here are instructive.   When looking at $\MahR(X)$,
as in \S531, we quickly come to feel that it is a measure of a
certain kind of complexity;  the richer the space $X$, the larger
$\MahR(X)$ will be.   531Eb and 531Ed are direct manifestations of this,
and 531G develops the theme.   $\MahcrR(X)$ can sometimes tell us more
about $X$;  knowing $\MahcrR(X)$ we may have a lower bound on the
complexity
of $X$ as well as an upper bound.   (On the other hand, $\MahcrR(X)$ can
evaporate for non-trivial reasons, as in 532Xb and 532Ya, and leave us
with very
little idea of what $X$ might be like.)   In place of the
straightforward facts in 531E, we have the relatively complex and
partial results in 532G and 532K.   As soon as we leave the constrained
context of powers of $\{0,1\}$, the most natural questions seem to be
obscure (532Zc).

However, if we follow the paths which are open, rather than those we
might otherwise have chosen, we come to some interesting ideas, starting
with 532I.   Here, as happened in \S531, we see that a proper
understanding of the
measure algebras $\frak B_{\lambda}$ will take us a long way;  and once
again we find that this understanding has to be conditional on the model
of set theory we are working in.   Even to decide which powers of
$\{0,1\}$ carry completion regular Radon measures with countable Maharam
type we need to examine some new aspects of the Lebesgue measure algebra
(532M-532O).  %532M 532N 532O
Moreover, as well as the familiar cardinals of Cicho\'n's diagram, we
have to look at the Freese-Nation number of $\Cal P\Bbb N$ (532P).
For larger Maharam types, in a way that we are becoming accustomed to,
other combinatorial principles become relevant (532R, 532S).
}%end of notes

\discrpage

\frfilename{mt533.tex}
\versiondate{4.1.14}
\copyrightdate{2007}

\def\chaptername{Topologies and measures III}
\def\sectionname{Special topics}

\def\ureg{\mathop{\text{ureg}}}

\newsection{533}

I present notes on certain questions which can be answered if we make
particular assumptions concerning values of the cardinals considered in
\S\S523-524.   The first cluster (533A-533E) %533A 533B 533C 533D 533E
looks at Radon and quasi-Radon measures in contexts in which the additivity
of Lebesgue measure is large compared with other cardinals of the
structures considered.   Developing ideas which arose in the course of
\S531, I discuss `uniform regularity' in perfectly normal and
first-countable spaces (533H).   We also have a complete description of
the cardinals $\kappa$ for which $\BbbR^{\kappa}$ is measure-compact
(533J).

As previously, I write $\Cal N(\mu)$ for the null ideal of a measure
$\mu$;   $\nu_{\kappa}$ will be the usual measure on $\{0,1\}^{\kappa}$
and $\Cal N_{\kappa}=\Cal N(\nu_{\kappa})$ its null ideal.

\leader{533A}{Lemma} Let $(X,\Sigma,\mu)$ be a semi-finite measure space
with measure algebra $(\frak A,\bar\mu)$.   If
$\ofamily{\xi}{\kappa}{\Cal K_{\xi}}$ is a family of ideals in $\Sigma$
such that $\mu$ is inner regular with respect to every
$\Cal K_{\xi}$ and
$\kappa<\min(\add\Cal N(\mu),\wdistr(\frak A))$, then $\mu$ is inner
regular with respect to $\bigcap_{\xi<\kappa}\Cal K_{\xi}$.

\proof{ Take $E\in\Sigma$ and $\gamma<\mu E$.   Then there is an
$E_1\in\Sigma$ such that $E_1\subseteq E$ and $\gamma<\mu E_1<\infty$.
For $\xi<\kappa$, $D_{\xi}=\{K^{\ssbullet}:K\in\Cal K_{\xi}\}$ is closed
under finite unions and is order-dense in $\frak A$, so includes a
partition of unity $A_{\xi}$.   Now there is a partition $B$ of unity in
$\frak A$ such that $\{a:a\in A_{\xi}$, $a\Bcap b\ne 0\}$ is finite for
every $b\in B$ and $\xi<\kappa$.   Let $B'\subseteq B$ be a finite set
such that $\bar\mu(E_1^{\ssbullet}\Bcap\sup B')\ge\gamma$, and let
$E_2\subseteq E_1$ be such that
$E_2^{\ssbullet}=E_1^{\ssbullet}\Bcap\sup B'$.   For any $\xi<\kappa$,

\Centerline{$A'_{\xi}
=\{a:a\in A_{\xi}$, $a\Bcap E_2^{\ssbullet}\ne 0\}
\subseteq\bigcup_{b\in B'}\{a:a\in A_{\xi}$, $a\Bcap b\ne 0\}$}

\noindent is finite, so $\sup A'_{\xi}$ belongs to $D_{\xi}$
and can be expressed
as $K_{\xi}^{\ssbullet}$ for some $K_{\xi}\in\Cal K_{\xi}$.   Now
$E_2^{\ssbullet}\Bsubseteq\sup A'_{\xi}$ so $E_2\setminus K_{\xi}$ is
negligible.   As $\kappa<\add\Cal N(\mu)$, we have a negligible
$H\in\Sigma$ including $\bigcup_{\xi<\kappa}E_2\setminus K_{\xi}$;  now
$E'=E_2\setminus H\subseteq E$, $\mu E'\ge\gamma$ and
$E'\in\bigcap_{\xi<\kappa}\Cal K_{\xi}$.   As $E$ and $\gamma$ are
arbitrary, $\mu$ is inner regular with respect to
$\bigcap_{\xi<\kappa}\Cal K_{\xi}$.
}%end of proof of 533A

\cmmnt{\medskip

\noindent{\bf Remark} Of course this result is covered by 412Ac unless
$\wdistr(\frak A)>\omega_1$, which nearly forces $\frak A$ to have
countable Maharam type (524Mb).
}

\leader{533B}{Corollary} Let $(X,\Sigma,\mu)$ be a totally finite
measure space with countable Maharam type.   If $\Cal E\subseteq\Sigma$,
$\#(\Cal E)<\min(\add\Cal N_{\omega},\add\Cal N(\mu))$ and $\epsilon>0$,
there is a set $F\in\Sigma$ such that $\mu(X\setminus F)\le\epsilon$ and
$\{E\cap F:E\in\Cal E\}$ is countable.

\proof{ Let $(\frak A,\bar\mu)$ be the measure algebra of $\mu$.   Then
$\frak A$ is separable in its measure-algebra topology (521Ea).   Let
$\Cal H\subseteq\Sigma$ be a countable set
such that $\{H^{\ssbullet}:H\in\Cal H\}$ is
dense in $\frak A$.   For $E\in\Cal E$ and $n\in\Bbb N$ choose
$H_{En}\in\Cal H$ such that
$\mu(E\symmdiff H_{En})\le 2^{-n}$;  let $\Cal K_E$ be the family of
measurable sets $K$ such that $K$ is disjoint from
$\bigcup_{i\ge n}E\symmdiff H_{Ei}$ for some $n$.   Then $\mu$ is inner
regular with respect to $\Cal K_E$.   Because
$\#(\Cal E)<\min(\wdistr(\frak A),\add\Cal N(\mu))$ (524Mb), $\mu$ is
inner regular with respect to $\bigcap_{E\in\Cal E}\Cal K_E$ (533A) and
there is an $F\in\bigcap\Cal K_E$ such that $\mu F\ge\mu X-\epsilon$.
If $E\in\Cal E$, there is an $n\in\Bbb N$ such that
$F\cap(E\symmdiff H_{En})=\emptyset$, that is, $F\cap E=F\cap H_{En}$;
so $\{F\cap E:E\in\Cal E\}\subseteq\{F\cap H:H\in\Cal H\}$ is countable.
}%end of proof of 533B

\leader{533C}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon
measure space with countable Maharam type.

(a) If $w(X)<\add\Cal N_{\omega}$, then $\mu$ is inner regular with
respect to the second-countable subsets of $X$;  if moreover $\frak T$
is regular and Hausdorff, then $\mu$ is inner regular with respect to
the metrizable subsets of $X$.

(b) If $Y$ is a topological space of weight less than
$\add\Cal N_{\omega}$, then any measurable function $f:X\to Y$ is almost
continuous.

(c) If $\familyiI{Y_i}$ is a family of topological spaces, with
$\#(I)<\add\Cal N_{\omega}$, and
$f_i:X\to Y_i$ is almost continuous for every $i$, then
$x\mapsto\cmmnt{ f(x)=}\familyiI{f_i(x)}:X\to\prod_{i\in I}Y_i$ is
almost continuous.

\proof{ Note first that $\add\Cal N(\mu)\ge\add\Cal N_{\omega}$, by
524Ta.

\medskip

{\bf (a)} Let
$\Cal U$ be a base for $\frak T$ with $\#(\Cal U)<\add\Cal N_{\omega}$.
Set

\Centerline{$\Cal F=\{F:F\subseteq X$, $\{F\cap U:U\in\Cal U\}$ is
countable$\}$.}

\noindent Then $\mu$ is inner regular with respect to $\Cal F$.   \Prf\
If $E\in\Sigma$ and $\gamma<\mu E$, let $H\in\Sigma$ be such that
$H\subseteq E$ and $\gamma<\mu H<\infty$.   Then the subspace measure
$\mu_H$ still has countable Maharam type (use 322I and 514Ed) and

\Centerline{$\add\Cal N(\mu_H)\ge\add\Cal N(\mu)\ge\add\Cal N_{\omega}
>\#(\{H\cap U:U\in\Cal U\})$.}

\noindent By 533B, there is an $F\in\dom\mu_H$ such that
$\mu_HF\ge\gamma$ and $\{F\cap H\cap U:U\in\Cal U\}$ is countable;  now
$F\in\Cal F$, $F\subseteq E$ and $\mu F\ge\gamma$.\ \QeD\   But every
member of $\Cal F$ is second-countable (use 4A2B(a-vi)).
If $\frak T$ is regular and Hausdorff, then every member of $\Cal F$ is
separable and metrizable (4A2Pb).

\medskip

{\bf (b)} If $f:X\to Y$ is measurable, let $\Cal V$ be a base for the
topology of $Y$ with $\#(\Cal V)<\add\Cal N_{\omega}$.   Suppose that
$E\in\Sigma$ and $\gamma<\mu E$.   By 533B, there is an $F\in\Sigma$
such that $F\subseteq E$, $\gamma<\mu F<\infty$ and
$\{F\cap f^{-1}[V]:V\in\Cal V\}$ is countable.
It follows that $\{f[F]\cap V:V\in\Cal V\}$ is countable, so that the
subspace topology on $f[F]$ is second-countable (4A2B(a-vi) again).
Giving $F$ its subspace topology $\frak T_F$ and measure $\mu_F$, $\mu_F$
is inner regular with respect to the closed sets (412Pc).   If
$H\subseteq f[F]$ is relatively open in $f[F]$, it is of the form
$G\cap f[F]$ where $G$ is an open subset of $Y$, so that
$(f\restr F)^{-1}[H]=F\cap f^{-1}[G]$ is measured by $\mu_F$;  thus
$f\restr F:F\to f[F]$ is measurable.   By 418J, $f\restr F$ is
almost continuous, and there is a $K\in\Sigma$ such that
$K\subseteq F$, $\mu K\ge\gamma$ and $f\restr K$ is continuous.

As $E$ and $\gamma$ are arbitrary, $f$ is almost continuous.

\medskip

{\bf (c)} For each $i\in I$, set
$\Cal K_i=\{K:K\in\Sigma$, $f_i\restr K$ is continuous$\}$.
Then $\Cal K_i$ is an ideal in $\Sigma$ and $\mu$ is inner regular with
respect to $\Cal K_i$.    Also, as in 533B, $\#(I)<\wdistr(\frak A)$,
where $\frak A$ is the measure algebra of $\mu$.
So $\mu$ is inner regular with respect to
$\Cal K=\bigcap_{i\in I}\Cal K_i$, by 533A.   But $f\restr K$ is
continuous for every $K\in\Cal K$, so $f$ is almost continuous.
}%end of proof of 533C

\leader{533D}{Proposition} Let $(X,\frak T)$ be a
first-countable compact Hausdorff space such that
$\cff[w(X)]^{\le\omega}<\add\Cal N_{\omega}$, and $\mu$ a Radon measure
on $X$ with countable Maharam type.   Then $\mu$ is inner regular with
respect to the metrizable zero sets.

\proof{ Set $\kappa=w(X)$.   Then there is an injective continuous
function $f:X\to[0,1]^{\kappa}$ (5A4Cc).   Let $\Cal I$ be a cofinal
subset of $[\kappa]^{\le\omega}$ with $\#(\Cal I)<\add\Cal N_{\omega}$.
By 524Pa, $\add\mu\ge\add\Cal N_{\omega}$.

For $I\in\Cal I$ and $x\in X$ set $f_I(x)=f(x)\restr I$.   We need to know
that for every $x\in X$ there is an $I\in\Cal I$ such that
$\{x\}=f_I^{-1}[f_I[\{x\}]]$.   \Prf\ Set $F_I=f_I^{-1}[f_I[\{x\}]]$ for
each $I$.   Because $\Cal I$ is upwards-directed,
$\family{I}{\Cal I}{F_I}$ is downwards-directed.   Because $f$ is
injective and $\bigcup\Cal I=\kappa$, $\bigcap_{I\in\Cal I}F_I=\{x\}$.
Let $\Cal V$ be a countable base of open neighbourhoods of $x$.   For
each $V\in\Cal V$ there is an $I_V\in\Cal I$ such that
$F_{I_V}\cap(X\setminus V)=\emptyset$.   Let $I\in\Cal I$ be such that
$\bigcup_{V\in\Cal V}I_V\subseteq I$;  then $F_I=\{x\}$.\ \Qed

For $I\in\Cal I$, let $\lambda_I$ be the image measure $\mu f_I^{-1}$ on
$[0,1]^I$;  note that $\lambda_I$ is a Radon measure (418I).   Of course
$\add\lambda_I$ is also at least
$\add\Cal N_{\omega}$, and in particular is greater than $\kappa$.   If
$G\subseteq X$ is open, then $G$ and $f_I[G]$ are expressible as unions
of at most $\kappa$ compact sets, so $\lambda_I$ measures $f_I[G]$.

There is an $I\in\Cal I$ such that $\mu f_I^{-1}[f_I[G]]=\mu G$ for
every open set $G\subseteq X$.   \Prf\Quer\ Suppose, if possible,
otherwise.   For each $I\in\Cal I$ choose an open set $G_I\subseteq X$
such that $E_I=f_I^{-1}[f_I[G_I]]\setminus G_I$ is non-negligible;
because $\lambda_I$ measures $f_I[G_I]$, $\mu$ measures $E_I$.   Set
$E'_I=\bigcup_{J\in\Cal I,J\supseteq I}E_J$ for each $I\in\Cal I$;
because $\#(\Cal I)<\add\mu$, $\mu$ measures $E'_I$.   Note that
$E'_I\subseteq E'_J$ whenever $J\subseteq I$ in $\Cal I$;  moreover, any
sequence in $\Cal I$ has an upper bound in $\Cal I$.   There is
therefore an $M\in\Cal I$ such that $E'_M\setminus E'_I$ is negligible
for every $I\in\Cal I$.   Again because $\#(\Cal I)<\add\mu$,
$E'_M\setminus\bigcap_{I\in\Cal I}E'_I$ is negligible;  as $E'_M$ is not
negligible, there is an $x\in\bigcap_{I\in\Cal I}E'_I$.   But there is
an $I\in\Cal I$ such that $\{x\}=f_I^{-1}[f_I[\{x\}]]$, so
$x\notin E_J$ for any $J\supseteq I$.\ \Bang\Qed

Let $\Cal U$ be a base for the topology of $X$ with $\#(\Cal U)=\kappa$.
Then $\bigcup_{U\in\Cal U}f_I^{-1}[f_I[U]]\setminus U$ is
$\mu$-negligible;  let $Y$ be its complement.   If $x\in X$ and $y\in Y$
and $x\ne y$, there is a $U\in\Cal U$ containing $x$ but not $y$, so
$f_I^{-1}[f_I[U]]$ contains $x$ and not $y$ and $f(x)\ne f(y)$.   If
$F\subseteq Y$ is compact, then $F$ is homeomorphic to the metrizable
$f_I[F]$, so is metrizable, and $F=f_I^{-1}[f_I[F]]$ is a zero set.   As
$\mu$ is surely inner regular with respect to the compact subsets of the
conegligible set $Y$, it is inner regular with respect to the metrizable
zero sets.
}%end of proof of 533D

\leader{533E}{Corollary} Suppose that
$\cov\Cal N_{\omega_1}>\omega_1$.   Let $(X,\frak T)$ be a
first-countable K-analytic Hausdorff space such that
$\cff[w(X)]^{\le\omega}<\add\Cal N_{\omega}$.   Then $X$ is a Radon
space.

\proof{ Let $\mu$ be a totally finite Borel measure on $X$,
$E\subseteq X$ a Borel set and $\gamma<\mu E$.   Because $X$ is
K-analytic, there is a compact set $K\subseteq X$ such that
$\mu(E\cap K)>\gamma$ (apply 432B to the measure $\mu\LLcorner E$).
Let $\lambda$ be the Radon measure on
$K$ defined by saying that $\int fd\lambda=\int_Kfd\mu$ for every
$f\in C(K)$ (using the Riesz Representation Theorem, 436J/436K).
Because $\cov\Cal N_{\omega_1}>\omega_1$, $\omega_1$ is a precaliber of
every measurable algebra (525J);  as $K$ is first-countable,
$\omega_1\notin\MahR(K)$ (531O) and $\lambda$ must have countable
Maharam type (531Ef).   By 533D, $\lambda$ is completion regular.   But if
$F\subseteq K$ is a zero set (for the subspace topology of $K$), there
is a non-increasing sequence $\sequencen{f_n}$ in $C(K)$ with infimum
$\chi F$, so

\Centerline{$\lambda F=\lim_{n\to\infty}\int f_nd\lambda
=\lim_{n\to\infty}\int_Kf_nd\mu=\mu F$.}

\noindent Accordingly

\Centerline{$\lambda H
=\sup\{\lambda F:F\subseteq H$ is a zero set$\}
=\sup\{\mu F:F\subseteq H$ is a zero set$\}\le\mu H$}

\noindent for every Borel set $H\subseteq K$.   As $\lambda K=\mu K$,
$\lambda$ agrees with $\mu$ on the Borel subsets of $K$.   In
particular, $\lambda(E\cap K)>\gamma$;  now there is a compact set
$L\subseteq E\cap K$ such that $\gamma\le\lambda L=\mu L$.

As $E$ and $\gamma$ are arbitrary, $\mu$ is tight;  as $\mu$ is
arbitrary, $X$ is a Radon space.
}%end of proof of 533E

\leader{533F}{Definition} Let $X$ be a topological space and $\mu$ a
topological measure on $X$.   I will say that $\mu$ is
{\bf uniformly regular} if there is a countable family $\Cal V$ of open
sets in $X$ such that $G\setminus\bigcup\{V:V\in\Cal V$, $V\subseteq G\}$
is negligible for every open set $G\subseteq X$.

\vleader{72pt}{533G}{Lemma} Let $(X,\frak T,\Sigma,\mu)$ be a compact Radon
measure space.

(a) The following are equiveridical:

\quad(i) $\mu$ is uniformly regular;

\quad(ii) there are a metrizable space $Z$ and a continuous function
$f:X\to Z$ such that $\mu f^{-1}[f[F]]=\mu F$ for every closed
$F\subseteq X$;

\quad(iii) there is a countable family $\Cal H$ of cozero sets in $X$
such that $\mu G=\sup\{\mu H:H\in\Cal H$, $H\subseteq G\}$ for every
open set $G\subseteq X$;

\quad(iv) there is a countable family $\Cal E$ of zero sets in $X$ such
that $\mu G=\sup\{\mu E:E\in\Cal E$, $E\subseteq G\}$ for every open set
$G\subseteq X$.

(b) If $\frak T$ is perfectly normal, the following are equiveridical:

\quad(i) $\mu$ is uniformly regular;

\quad(ii) there are a metrizable space $Z$ and a continuous function
$f:X\to Z$ such that $\mu f^{-1}[f[E]]=\mu E$ for every $E\in\Sigma$;

\quad(iii) there are a metrizable space $Z$ and a continuous function
$f:X\to Z$ such that $f[G]\ne f[X]$ whenever $G\subseteq X$ is open and
$\mu G<\mu X$;

\quad(iv) there is a countable family $\Cal E$ of closed sets in $X$
such that $\mu G=\sup\{\mu E:E\in\Cal E$, $E\subseteq G\}$ for every
open set $G\subseteq X$.

\proof{{\bf (a)(i)$\Rightarrow$(iii)} Given $\Cal V$ as in 533F, then for
each $V\in\Cal V$ there is a cozero set $H_V\subseteq V$ of the same
measure.   \Prf\ $\frak T$ is completely regular, so
$\Cal H_V=\{H:H\subseteq V$ is a cozero set$\}$ has union $V$;  $\mu$ is
$\tau$-additive, so there is a sequence $\sequencen{H_n}$ in $\Cal H_V$
such that $\mu V=\mu(\bigcup_{n\in\Bbb N}H_n)$;  set
$H_V=\bigcup_{n\in\Bbb N}H_n$;  by 4A2C(b-iii), $H_V$ is a cozero
set.\ \QeD\  Now $\Cal H=\{H_V:V\in\Cal V\}$ witnesses that (iii) is
true.

\medskip

\quad{\bf (iii)$\Rightarrow$(iv)} Given $\Cal H$ as in (iii), then for
each $H\in\Cal H$ let $\sequencen{F_n(H)}$ be a non-decreasing sequence
of zero sets with union $H$ (4A2C(b-vi)).   Set
$\Cal E=\{F_n(H):H\in\Cal H$, $n\in\Bbb N\}$, so that $\Cal E$ is a
countable family of zero sets.   If $G\subseteq X$ is open,

\Centerline{$\mu G=\sup_{H\in\Cal H,H\subseteq G}\mu H
=\sup_{H\in\Cal H,H\subseteq G,n\in\Bbb N}\mu F_n(H)
\le\sup_{E\in\Cal E,E\subseteq G}\mu E\le\mu G$,}

\noindent so $\Cal E$ witnesses that (iv) is true.

\medskip

\quad{\bf (iv)$\Rightarrow$(ii)} Given $\Cal E$ as in (iv),
then for each $E\in\Cal E$ choose a continuous $f_E:X\to\Bbb R$ such
that $E=f_E^{-1}[\{0\}]$, and set $f(x)=\family{E}{\Cal E}{f_E(x)}$ for
$x\in X$.   Then
$f:X\to Z=\BbbR^{\Cal E}$ is continuous and $Z$ is metrizable and
$f^{-1}[f[E]]=E$ for every $E\in\Cal E$.   If $F\subseteq X$ is closed,
set $\Cal E_0=\{E:E\in\Cal E$, $E\cap F=\emptyset\}$.   Then
$\bigcup\Cal E_0$ has the same measure as $X\setminus F$ and does not
meet $f^{-1}[f[F]]$, so $\mu f^{-1}[f[F]]=\mu F$.   As $F$ is arbitrary,
$f$ and $Z$ witness that $\mu$ satisfies (ii).

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} Take $Z$ and
$f:X\to Z$ as in (ii).   Replacing $Z$ by $f[X]$ if
necessary, we may suppose that $f$ is surjective, so that $Z$ is
compact, therefore second-countable (4A2P(a-ii)).   Let $\Cal U$ be a
countable base for the topology of $Z$ closed under finite unions, and
set $\Cal V=\{f^{-1}[U]:U\in\Cal U\}$, so that $\Cal V$ is a countable
family of open sets in $X$.   If $G\subseteq X$ is
open, set $F=X\setminus G$,
$\Cal U_0=\{U:U\in\Cal U$, $U\cap f[F]=\emptyset\}$,
$\Cal V_0=\{f^{-1}[U]:U\in\Cal U_0\}$.   Then
$Z\setminus f[F]=\bigcup\Cal U_0$ so
$X\setminus f^{-1}[f[F]]=\bigcup\Cal V_0$ and (because $\Cal U_0$ and
$\Cal V_0$ are closed under finite unions)

$$\eqalign{\sup\{\mu V:V\in\Cal V,\,V\subseteq G\}
&\ge\sup_{V\in\Cal V_0}\mu V
=\mu(X\setminus f^{-1}[f[F]])\cr
&=\mu X-\mu f^{-1}[f[F]]
=\mu X-\mu F
=\mu G.\cr}$$

\noindent Thus $\Cal V$ witnesses that $\mu$ is uniformly regular.

\medskip

{\bf (b)(i)$\Rightarrow$(iii)} If $\mu$ is uniformly regular, then by
(a-ii) there are
a metrizable space $Z$ and a continuous function $f:X\to Z$ such that
$\mu f^{-1}[f[F]]=\mu F$ for every closed $F\subseteq X$.   If now
$G\subseteq X$ is open and $\mu G<\mu X$, there is a sequence
$\sequencen{F_n}$ of closed sets with union $G$, because $\frak T$ is
perfectly normal.   In this case
$f^{-1}[f[G]]=\bigcup_{n\in\Bbb N}f^{-1}[f[F_n]]$ has the same measure
as $G$, so is not the whole of $X$, and $f[G]\ne f[X]$.   Thus $f$ and
$Z$ witness that (iii) is true.

\medskip

\quad{\bf (iii)$\Rightarrow$(ii)} Take $Z$ and $f$ from (iii).
Let $\nu$ be the image measure $\mu f^{-1}$ on $Z$;  then $\mu$ is a
Radon measure (418I again).   \Quer\ If $E\in\Sigma$ and
$\mu^*f^{-1}[f[E]]>\mu E$, let $E'\supseteq E$ be a Borel set such that
$\mu E'=\mu E$.   Because $X$ is perfectly normal, $E'$ belongs to the
Baire $\sigma$-algebra of $X$ (4A3Kb), so is Souslin-F (421L), therefore
K-analytic (422Hb);  consequently $f[E']$ is K-analytic (422Gd)
therefore measured by $\nu$ (432A).   This means that
$f^{-1}[f[E']]\in\Sigma$, and of course

\Centerline{$\mu f^{-1}[f[E']]\ge\mu^*f^{-1}[f[E]]>\mu E=\mu E'$.}

\noindent We can therefore find open sets $G\supseteq E'$ and
$G'\supseteq X\setminus f^{-1}[f[E']]$ such that $\mu G+\mu G'<\mu X$.
But now $G\cup G'$ is an open set of measure less than $\mu X$ and
$f[G\cup G']=f[X]$, which is supposed to be impossible.\ \Bang

Thus, for any $E\in\Sigma$, we have $\mu^*f^{-1}[f[E]]=\mu E$;  of
course it follows at once that $f^{-1}[f[E]]$ is measurable, with the
same measure as $E$, as required by (ii).

\medskip

\quad{\bf (ii)$\Rightarrow$(i)$\Leftrightarrow$(iv)}
These follow immediately from (a), because all closed sets in $X$ are zero
sets.
}%end of proof of 533G

\leader{533H}{Theorem} (a) Suppose that
$\cov\Cal N_{\omega_1}>\omega_1$.   Let
$X$ be a perfectly normal compact Hausdorff space.   Then every Radon
measure on $X$ is uniformly regular.

(b)\cmmnt{ ({\smc Plebanek 00})} Suppose that
$\cov\Cal N_{\omega_1}>\omega_1=\non\Cal N_{\omega}$.   Let
$X$ be a first-countable compact Hausdorff space.   Then every Radon
measure on $X$ is uniformly regular.

\proof{{\bf (a)} Let $\mu$ be a Radon measure on $X$.   \Quer\ If $\mu$
is not uniformly regular, then we can choose
$\ofamily{\xi}{\omega_1}{g_{\xi}}$ and
$\ofamily{\xi}{\omega_1}{G_{\xi}}$
inductively, as follows.   Given that $g_{\eta}:X\to\Bbb R$ is
continuous for every $\eta<\xi$, set
$f_{\xi}(x)=\ofamily{\eta}{\xi}{g_{\eta}(x)}$ for $x\in X$, so that
$f_{\xi}:X\to\BbbR^{\xi}$ is continuous.   By 533G(b-iii), there is an
open set $G_{\xi}$ such that $\mu G_{\xi}<\mu X$ and
$f_{\xi}[G_{\xi}]=f_{\xi}[X]$;  now $G_{\xi}$ is a cozero set and there
is a continuous function $g_{\xi}:X\to\Bbb R$ such that
$G_{\xi}=\{x:g_{\xi}(x)\ne 0\}$.   Continue.

At the end of the induction, we have a continuous function
$f_{\omega_1}:X\to\BbbR^{\omega_1}$, setting
$f_{\omega_1}(x)=\ofamily{\xi}{\omega_1}{g_{\xi}(x)}$ for each $x$.
Now $\omega_1$ is a precaliber of every measurable algebra (525J again),
and
$\mu(X\setminus G_{\xi})>0$ for each $\xi$, so there is an $x\in X$ such
that $A=\{\xi:x\notin G_{\xi}\}$ is uncountable (525Ca).   Set
$H=\{y:f_{\omega_1}(y)\ne f_{\omega_1}(x)\}$;  then $H$ is an open set,
so expressible as $\bigcup_{n\in\Bbb N}K_n$ where each $K_n$ is compact.
For each $\xi\in A$ there is an
$x_{\xi}\in G_{\xi}$ such that $f_{\xi}(x_{\xi})=f_{\xi}(x)$.   As
$g_{\xi}(x_{\xi})\ne 0=g_{\xi}(x)$, $x_{\xi}\in H$.   Let $n\in\Bbb N$
be such that $A'=\{\xi:\xi\in A$, $x_{\xi}\in K_n\}$ is uncountable.
Then

\Centerline{$f_{\omega_1}(x)
\in\overline{\{f_{\omega_1}(x_{\xi}):\xi\in A'\}}
\subseteq f_{\omega_1}[K_n]$;}

\noindent but this is impossible, because $K_n\subseteq H$.\ \Bang

So $\mu$ must be uniformly regular, as required.

\medskip

{\bf (b)} Let $\mu$ be a Radon measure on $X$.   If $\mu X=0$ then of
course $\mu$ is uniformly regular;  suppose $\mu X>0$.   As in (a)
and the proof of 533E, the Maharam type of $\mu$ is
countable.   Let $\frak A$ be the measure algebra of $\mu$;  then
$d(\frak A)\le\non\Cal N_{\omega}$ (524Me), so there is a set
$A\subseteq X$, of full outer measure, with $\#(A)\le\omega_1$ (521Lc).
For each
$x\in X$, let $\sequencen{V_{xn}}$ run over a base of neighbourhoods of
$x$.   Let $\Cal H$ be the family of sets
expressible as finite unions of $V_{xn}$ for $x\in A$ and $n\in\Bbb N$,
so that $\Cal H$ is a family of open sets in $X$ and
$\#(\Cal H)\le\omega_1$.

For any open $G\subseteq X$,
$\mu G=\sup\{\mu H:H\in\Cal H$, $H\subseteq G\}$.   \Prf\ Set
$H^*=\bigcup\{H:H\in\Cal H$, $H\subseteq G\}$.   For any $x\in A\cap G$,
there is an $n\in\Bbb N$ such that $V_{xn}\subseteq G$, and now
$V_{xn}\in\Cal H$, so $x\in H^*$.   Thus $G\setminus H^*$ does not meet
$A$;  as $A$ has full outer measure,

\Centerline{$\mu G=\mu H^*=\sup\{\mu H:H\in\Cal H$, $H\subseteq G\}$}

\noindent because $\{H:H\in\Cal H$, $H\subseteq G\}$ is closed under
finite unions.\ \QeD\  So there is a countable
$\Cal H'\subseteq\{H:H\in\Cal H$, $H\subseteq G\}$ such that
$\mu G=\sup_{H\in\Cal H'}\mu H$.

Let $\ofamily{\xi}{\omega_1}{H_{\xi}}$ run over $\Cal H$.   For
$\xi<\omega_1$, set

\Centerline{$\Cal G_{\xi}
=\{G:G\subseteq X$ is open, $\mu G=\sup\{\mu H_{\eta}:\eta\le\xi$,
$H_{\eta}\subseteq G\}\}$.}

\noindent Then $\bigcup_{\xi<\omega_1}\Cal G_{\xi}=\frak T$.   For each
$\xi<\omega_1$, set

\Centerline{$Y_{\xi}
=\{y:y\in X$, $V_{yn}\in\Cal G_{\xi}$ for every $n\in\Bbb N\}$;}

\noindent then $X=\bigcup_{\xi<\omega_1}Y_{\xi}$.   Now there is a
$\xi<\omega_1$ such that $Y_{\xi}$ has full outer measure.   \Prf\ Let
$\xi$ be such that $\mu^*Y_{\xi}=\mu^*Y_{\eta}$ for every $\eta\ge\xi$.
\Quer\ If $\mu^*Y_{\xi}<\mu X$, let $K\subseteq X\setminus Y_{\xi}$ be a
non-negligible measurable set.   Then the subspace measure $\mu_K$ is a
Radon measure with countable Maharam type, so

\Centerline{$\cov\Cal N(\mu_K)\ge\cov\Cal N_{\omega}
\ge\cov\Cal N_{\omega_1}>\omega_1$.}

\noindent Since $K\subseteq\bigcup_{\eta<\omega_1}Y_{\eta}$, there must
be some $\eta<\omega_1$ such that $\mu^*_K(K\cap Y_{\eta})>0$;   but now
$\mu^*(K\cap Y_{\eta})>0$ and $\eta>\xi$ and

\Centerline{$\mu^*Y_{\eta}
=\mu^*(Y_{\eta}\setminus K)+\mu^*(Y_{\eta}\cap K)>\mu^*Y_{\xi}$.  \Bang}

\noindent So $Y_{\xi}$ has full outer measure.\ \Qed

Set $\Cal H_{\xi}=\{H_{\eta}:\eta\le\xi\}$.   If $G\subseteq X$ is open,
and $H^*=\bigcup\{H:H\in\Cal H_{\xi}$, $H\subseteq G\}$, then
$G\setminus H^*$ is negligible.   \Prf\ Set
$\Cal V=\{V_{yn}:y\in Y_{\xi}$, $n\in\Bbb N$, $V_{yn}\subseteq G\}$,
$H_1^*=\bigcup\Cal V$.   Then $Y_{\xi}$ does not meet
$G\setminus H_1^*$, so $\mu H_1^*=\mu G$.   Let
$\Cal V_0\subseteq\Cal V$ be a countable set such that
$\mu(\bigcup\Cal V_0)=\mu G$.   If $V\in\Cal V_0$, then
$V\in\Cal G_{\xi}$ and $V\subseteq G$ so $V\setminus H^*$ is negligible.
Accordingly

\Centerline{$G\setminus H^*
\subseteq(G\setminus\bigcup\Cal V_0)
  \cup\bigcup_{V\in\Cal V_0}(V\setminus H^*)$}

\noindent is negligible.\ \QeD\   So if we take $\Cal H'$ to be the set
of finite unions of members of $\Cal H_{\xi}$, $\Cal H'$ will be a
countable family of open sets and $\mu G=\sup\{\mu H:H\in\Cal H'$,
$H\subseteq G\}$ for every open $G\subseteq X$.   Thus $\mu$ is
uniformly regular.
}%end of proof of 533H

%{\bf Problem} For (a), is it enough if $\add\Cal N_{\omega}>\omega_1$?

\leader{533I}{}\cmmnt{ We know from 435Fb/435H and 439P
that $\BbbR^{\Bbb N}$ is measure-compact and $\BbbR^{\frak c}$ is not.
It turns out that we already have a language in which to express a
necessary and sufficient condition for $\BbbR^{\kappa}$ to be
measure-compact.   To give the result in its full strength I repeat a
definition from 435Xk.

\medskip

\noindent}{\bf Definition} A completely regular space $X$ is {\bf
strongly measure-compact} if
$\mu X=\sup\{\mu^*K:K\subseteq X$ is compact$\}$ for every totally
finite Baire measure $\mu$ on $X$.

\cmmnt{\medskip

\noindent{\bf Remark} For the elementary properties of these spaces, see
435Xk.   I repeat one here:  a completely regular space $X$ is strongly
measure-compact iff it is measure-compact and pre-Radon.
\prooflet{\Prf{\bf (i)}
Suppose that $X$ is measure-compact and pre-Radon and
that $\mu$ is a totally finite Baire measure on $X$.   Because $X$ is
measure-compact, $\mu$ has an extension to a quasi-Radon measure
$\tilde\mu$ (435D);  because $X$ is pre-Radon, $\tilde\mu$ is Radon
(434Jb) and

$$\eqalign{\mu X
&=\tilde\mu X
=\sup_{K\subseteq X\text{ is compact}}\tilde\mu K\cr
&=\sup_{K\subseteq X\text{ is compact}}\tilde\mu^*K
\le\sup_{K\subseteq X\text{ is compact}}\mu^*K
\le\mu X.\cr}$$

\noindent As $\mu$ is arbitrary, $X$ is strongly measure-compact.
{\bf (ii)}
Suppose that $X$ is strongly measure-compact.   ($\alpha$) Let $\mu$ be
a Baire probability measure on $X$.   Then there is a
non-negligible compact set, so $X$ cannot be covered by the negligible
open sets;  by 435Fa, this is enough to ensure that $X$ is
measure-compact.   ($\beta$) Now let $\mu$ be a totally finite
$\tau$-additive Borel measure on $X$.   Write $\nu$ for the restriction
of $\mu$ to the Baire $\sigma$-algebra of $X$.   Then there is a compact
set $K\subseteq X$ which is not $\nu$-negligible.   \Quer\ If
$\mu(X\setminus K)=\mu X$, then, because $\mu$ is $\tau$-additive and
$X$ is regular, there is a closed set
$F\subseteq X\setminus K$ such that $\mu F+\nu^*K>\mu X$.   Because $X$
is completely regular, there is a zero set $G$ including $K$ and
disjoint from $F$, in which case $\nu^*K>\mu G=\nu G$, which is
impossible.\ \BanG\   So $\mu K>0$;  by 434J(a-iii), this tells us that
$X$ is pre-Radon.\ \Qed
}}

\vleader{72pt}{533J}{Theorem}\cmmnt{ (see {\smc Fremlin 77})} Let $\kappa$
be a cardinal.   Then the following are equiveridical:

(i) $\BbbR^{\kappa}$ is measure-compact;

(ii) if $\ofamily{\xi}{\kappa}{X_{\xi}}$ is a family of strongly
measure-compact completely regular Hausdorff spaces then
$\prod_{\xi<\kappa}X_{\xi}$ is measure-compact;

(iii) whenever $X$ is a compact Hausdorff space and
$\ofamily{\xi}{\kappa}{G_{\xi}}$ is a family of cozero sets in $X$, then
$X\cap\bigcap_{\xi<\kappa}G_{\xi}$ is measure-compact;

(iv) for any Radon measure, the union of $\kappa$ or fewer closed
negligible sets has inner measure zero;

(v) for any Radon measure, the union of $\kappa$ or fewer
negligible sets has inner measure zero;

(vi) $\kappa<\cov\Cal N(\mu)$ for any Radon measure $\mu$;

(vii) $\kappa<\cov\Cal N_{\kappa}$;

(viii) $\kappa<\frak m(\frak A)$ for every measurable algebra $\frak A$.

\proof{{\bf not-(iv)$\Rightarrow$not-(i)} Suppose that $X$ is a
Hausdorff space, $\mu$ is a Radon measure on $X$ and
$\ofamily{\xi}{\kappa}{F_{\xi}}$ is a family of closed $\mu$-negligible
subsets of $X$ such that $\mu_*(\bigcup_{\xi<\kappa}F_{\xi})>0$.   Then
there is a compact set $K\subseteq\bigcup_{\xi<\kappa}F_{\xi}$ such that
$\mu K>0$.

For each $\xi<\kappa$, there is a continuous $g_{\xi}:K\to\coint{0,1}$
such that $g_{\xi}(z)=0$ for $z\in K\cap F_{\xi}$ and
$g_{\xi}^{-1}[\{0\}]$ is negligible.   \Prf\ For each $n\in\Bbb N$,
there is a compact set $L_n\subseteq K\setminus F_{\xi}$ such that
$\mu L_n\ge\mu K-2^{-n}$;  there is a continuous $f_n:K\to[0,1]$ such that
$f_n(z)=0$ for $z\in K\cap F_{\xi}$, $1$ for $z\in L_n$;  set
$g_{\xi}=\sum_{n=0}^{\infty}2^{-n-2}f_n$.\ \QeD\   Set
$g(z)=\ofamily{\xi}{\kappa}{g_{\xi}(z)}$ for $z\in K$, so that
$g:K\to\coint{0,1}^{\kappa}$ is continuous.

Let $\nu$ be the Baire measure on $[0,1]^{\kappa}$ defined by
setting $\nu H=\mu g^{-1}[H]$ for every Baire set
$H\subseteq[0,1]^{\kappa}$.   Then $\ooint{0,1}^{\kappa}$ has full outer
measure for $\nu$.   \Prf\ If $H\subseteq[0,1]^{\kappa}$ is a Baire
set including $\ooint{0,1}^{\kappa}$, then $H$ is determined by
coordinates in some countable subset $I$ of $\kappa$ (4A3Mb).   If
$z\in K$ and $g_{\xi}(z)>0$ for every $\xi\in I$, then
$g(z)\restr I\in\ooint{0,1}^I$ is equal to $w\restr I$ for some
$w\in H$, so $g(z)\in H$.   Thus $g^{-1}[H]$ includes
$\{z:z\in K$, $g_{\xi}(z)>0$ for every $\xi\in I\}$ and

\Centerline{$\nu H=\mu g^{-1}[H]
\ge\mu\{z:g_{\xi}(z)>0$ for every $\xi\in I\}
=\mu K=\nu[0,1]^{\kappa}$. \Qed}

On the other hand, every point $y$ of $\ooint{0,1}^{\kappa}$ belongs to
a $\nu$-negligible cozero set.   \Prf\ $g[K]$ is a compact set not
containing $y$, so there is a cozero set $W$ containing $y$ and disjoint
from $g[K]$, and now $\nu W=0$.\ \Qed

Let $\nu_0$ be the subspace measure on $\ooint{0,1}^{\kappa}$.   By
4A3Nd, $\nu_0$ is a Baire measure on $\ooint{0,1}^{\kappa}$.   If
$y\in\ooint{0,1}^{\kappa}$ it belongs to a $\nu$-negligible cozero
set $W\subseteq[0,1]^{\kappa}$, and now $W\cap\ooint{0,1}^{\kappa}$ is a
$\nu_0$-negligible cozero set in $\ooint{0,1}^{\kappa}$ containing
$y$.   At the same time,

\Centerline{$\nu_0\ooint{0,1}^{\kappa}=\nu[0,1]^{\kappa}
=\mu K>0$.}

\noindent So $\nu_0$ witnesses that $\ooint{0,1}^{\kappa}$ is
not measure-compact;  as $\BbbR^{\kappa}$ is homeomorphic to
$\ooint{0,1}^{\kappa}$, it also is not measure-compact.

\medskip

{\bf (iv)$\Rightarrow$(iii)} Suppose that (iv) is true and that we have
$X$ and $\ofamily{\xi}{\kappa}{G_{\xi}}$, as in (iii), with a Baire
probability measure $\mu$ on $Y=X\cap\bigcap_{\xi<\kappa}G_{\xi}$.   Let
$\nu$ be the Radon probability measure on $X$ defined by saying that
$\int fd\nu=\int(f\restr Y)d\mu$ for every $f\in C(X)$ (436J/436K again).
Then $\nu G_{\xi}=1$ for each $\xi<\kappa$.   \Prf\ Let $f:X\to\Bbb R$
be a continuous function such that $G_{\xi}=\{x:x\in X$, $f(x)\ne 0\}$.
Set $f_n=n|f|\wedge\chi X$ for each $n$.   Then
$\lim_{n\to\infty}f_n=\chi G_{\xi}$, so

\Centerline{$\nu G_{\xi}
=\lim_{n\to\infty}\int f_nd\nu
=\lim_{n\to\infty}\int(f_n\restr Y)d\mu=\mu Y=1$.\ \Qed}

\noindent By (iv), $\nu_*(\bigcup_{\xi<\kappa}(X\setminus G_{\xi}))=0$,
that is, $Y$ has full outer measure.   In particular, $Y$ must meet the
support of $\nu$;  take any $z$ in the intersection.   If $U$ is a
cozero set in $Y$ containing $z$, there is an open set $G\subseteq X$
such that $U=G\cap Y$;  now there is a continuous $f:X\to[0,1]$ such
that $f(z)=1$ and $f(x)=0$ for $x\in X\setminus G$;  in this case

\Centerline{$\mu U\ge\int(f\restr Y)d\mu=\int fd\nu>0$}

\noindent because $\{x:f(x)>0\}$ is an open set meeting the support of
$\nu$.   This shows that $Y$ is not covered by the $\mu$-negligible
relatively cozero sets;  as $\mu$ is arbitrary, $Y$ is measure-compact
(435Fa).

\medskip

{\bf (iii)$\Rightarrow$(i)} We can express $\BbbR^{\kappa}$ in the form of
(iii) by taking $X=[-\infty,\infty]^{\kappa}$ and
$G_{\xi}=\{x:x(\xi)$ is finite$\}$ for each $\xi$.

\medskip

{\bf (iv)$\Rightarrow$(vii)} Let $Z$ be the Stone space of the measure
algebra of $\nu_{\kappa}$, and $\lambda$ its usual measure.   If
$\ofamily{\xi}{\kappa}{E_{\xi}}$ is a family of $\lambda$-negligible
sets, then, because $\lambda$ is inner regular with respect to the
open-and-closed sets, we can find negligible zero sets
$F_{\xi}\supseteq E_{\xi}$ for each $\xi$.   By (iv),
$\{F_{\xi}:\xi<\kappa\}$ cannot cover $Z$, so the same is true of
$\{E_{\xi}:\xi<\kappa\}$.   Thus $\cov\Cal N(\lambda)>\kappa$.   By
524Jb, $\cov\Cal N_{\kappa}>\kappa$.

\medskip

{\bf (vii)$\Rightarrow$(vi)} Let $\theta$ be
$\min\{\cov\Cal N(\nu):\nu$ is a non-zero Radon measure$\}$.   By 524Pc,
there is an infinite cardinal $\kappa'$ such that
$\theta=\cov\Cal N_{\kappa'}$;  by 523F, $\theta=\cov\Cal N_{\theta}$.
\Quer\ If $\theta\le\kappa$, then 523B tells us that

\Centerline{$\kappa<\cov\Cal N_{\kappa}
\le\cov\Cal N_{\theta}=\theta$.  \Bang}

\noindent So $\theta>\kappa$, as required.

\medskip

{\bf (vi)$\Rightarrow$(v)} If (vi) is true, $(X,\mu)$
is a Radon measure space, $\ofamily{\xi}{\kappa}{F_{\xi}}$ is a family
of negligible sets, and $E\subseteq\bigcup_{\xi<\kappa}F_{\xi}$
is a measurable set, then the subspace measure $\mu_E$ is a Radon
measure (416Rb), while $E$ can be covered by $\kappa$ negligible sets;
by (vi), $\mu E=0$;  as $E$ is arbitrary,
$\mu_*(\bigcup_{\xi<\kappa}F_{\xi})=0$.

\medskip

{\bf (v)$\Rightarrow$(ii)} Suppose that (v) is true, that
$\ofamily{\xi}{\kappa}{X_{\xi}}$ is a family of strongly
measure-compact completely regular Hausdorff spaces with product $X$,
and that $\mu$ is a Baire probability measure on $X$.   For each
$\xi<\kappa$ let $Z_{\xi}$ be the Stone-\v{C}ech compactification of
$X_{\xi}$;  set $Z=\prod_{\xi<\kappa}Z_{\xi}$, and $\pi_{\xi}(z)=z(\xi)$
for $z\in Z$, $\xi<\kappa$.   Then we have a Radon probability measure
$\lambda$ on $Z$ defined by saying that
$\int g\,d\lambda=\int_X(g\restr X)d\mu$ for every $g\in C(Z)$.   Note
that if $W\subseteq Z$ is a zero set, there is a non-increasing sequence
$\sequencen{g_n}$ in $C(Z)$ with infimum $\chi W$, so that

\Centerline{$\lambda W=\inf_{n\in\Bbb N}\int g_nd\lambda
=\inf_{n\in\Bbb N}\int_X(g_n\restr X)d\mu=\mu(W\cap X)$.}

Now $\lambda\pi_{\xi}^{-1}[X_{\xi}]=1$ for each $\xi$.   \Prf\ Let
$\epsilon>0$.   We have a Baire probability measure $\mu_{\xi}$ on
$X_{\xi}$ defined by setting $\mu_{\xi}E=\mu(X\cap\pi_{\xi}^{-1}[E])$
for every Baire set $E\subseteq X_{\xi}$, and a Radon measure
$\lambda_{\xi}=\lambda\pi_{\xi}^{-1}$ on $Z_{\xi}$.   Because $X_{\xi}$
is strongly measure-compact, there is a compact set $K\subseteq X_{\xi}$
such that
$\mu_{\xi}^*K\ge 1-\epsilon$.   Now $K$ is still compact when regarded
as a subset of $Z_{\xi}$, so there is a zero set $F\subseteq Z_{\xi}$,
including $K$, such that $\lambda_{\xi}F=\lambda_{\xi}K$.   In this
case, $F\cap X_{\xi}$ is a zero set in $X_{\xi}$ including $K$, so

$$\eqalign{\lambda_*\pi_{\xi}^{-1}[X_{\xi}]
&\ge\lambda\pi_{\xi}^{-1}[K]
=\lambda_{\xi}K
=\lambda_{\xi}F
=\lambda\pi_{\xi}^{-1}[F]\cr
&=\mu(X\cap\pi_{\xi}^{-1}[F])
=\mu_{\xi}(F\cap X_{\xi})
\ge\mu_{\xi}^*K
\ge 1-\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, we have the result.\ \Qed

By (v), $X=Z\cap\bigcap_{\xi<\kappa}\pi_{\xi}^{-1}[X_{\xi}]$ has full
outer measure for $\lambda$.   Let $\Cal G$ be the family of
$\mu$-negligible cozero sets in $X$ and $\Cal H$ the family of
$\lambda$-negligible open sets in $Z$.   If $x\in G\in\Cal G$, then
there is a continuous function $g:Z\to[0,1]$ such that $g(x)=1$ and
$H=\{y:y\in X$, $g(y)>0\}$ is included in $G$;  now
$\int g\,d\lambda=\int(g\restr X)d\mu=0$, so $\lambda H=0$.   This shows
that $\bigcup\Cal G\subseteq\bigcup\Cal H$ is $\lambda$-negligible, and,
in particular, is not the whole of $X$.   By 435Fa as usual,
this is enough to show that $X$ is measure-compact, as required.

\medskip

{\bf (ii)$\Rightarrow$(i)} is elementary, because $\Bbb R$ is certainly
strongly measure-compact.

\medskip

{\bf (vi)$\Rightarrow$(viii)$\Rightarrow$(vii)} are immediate from 524Md.
}%end of proof of 533J

\exercises{\leader{533X}{Basic exercises (a)}
%\spheader 533Xa
Describe a family $\family{t}{\Bbb R}{\Cal K_t}$ such that every $\Cal K_t$
consists of compact sets,
Lebesgue measure on $\Bbb R$ is inner regular with respect to every
$\Cal K_t$, but $\bigcap_{t\in\Bbb R}\Cal K_t=\emptyset$.
%533A

\spheader 533Xb Let $\mu$ be a uniformly regular topological
measure on a topological space $X$.   (i) Show that if $A\subseteq X$
then the subspace measure on $A$ is uniformly regular.   (ii) Show that any
indefinite-integral measure over $\mu$ is uniformly regular.   (iii) Show
that if $Y$ is another topological space and $f:X\to Y$ is a continuous
open map, then the image measure $\mu f^{-1}$ is uniformly regular.
%533F

\spheader 533Xc Show that any Radon measure on the split interval is
uniformly regular.   \Hint{419L.}
%533F

\spheader 533Xd ({\smc Babiker 76}) Let $X$ and $Y$ be compact Hausdorff
spaces, $\mu$ a Radon measure on $X$, $f:X\to Y$ a continuous surjection
and $\nu=\mu f^{-1}$ the image measure on $Y$.   Show that the following
are equiveridical:  (i) $\nu f[F]=\mu F$ for every closed
$F\subseteq X$;  (ii)
$\int g\,d\mu=\inf\{\int h\,d\nu:h\in C(Y)$, $hf\ge g\}$ for every
$g\in C(X)$;  (iii) for every $g\in C(X)$,
$\{y:g$ is constant on $f^{-1}[\{y\}]\}$ is $\nu$-conegligible.
%533F

\spheader 533Xe Show that any uniformly regular Borel measure has
countable Maharam type.
%533G

\spheader 533Xf Let $\familyiI{X_i}$ be a countable family of topological
spaces with product $X$, and $\mu$ a $\tau$-additive topological
measure on $X$.   Suppose that the marginal measure of $\mu$ on $X_i$ is
uniformly regular for every $i\in I$.
Show that $\mu$ is uniformly regular.
%533G mt53bits

\spheader 533Xg Let $X$ be $[0,1]\times\{0,1\}$ with the topology
generated by

$$\eqalign{\{G\times\{0,1\}:G\subseteq[0,1]
&\text{ is relatively open for the usual topology}\}\cr
&\cup\{\{(t,1)\}:t\in[0,1]\}
\cup\{X\setminus\{(t,1)\}:t\in[0,1]\}.\cr}$$

\noindent Show that $X$ is compact and Hausdorff.   Let $\mu$ be the
Radon measure on $X$ which is the image of Lebesgue measure on $[0,1]$
under the map $t\mapsto(t,0)$.   Show that $\mu$ is uniformly regular
but not completion regular.
%533G

\spheader 533Xh Let $X$ be a topological space and $\mu$ a
uniformly regular topological probability measure on $X$.
Show that there is an equidistributed sequence in $X$.
%533G

\spheader 533Xi Show that there is a first-countable compact Hausdorff
space with a uniformly regular topological probability measure,
inner regular with respect to the closed sets, which is not
$\tau$-additive.   \Hint{439K.}
%533G

\leader{533Y}{Further exercises (a)}
%\spheader 533Ya
({\smc Pol 82}) Let $X$ be a compact Hausdorff space and
$\mu$ a uniformly regular Radon measure on $X$.   Show that if we give
the space $M^+_{\text{R}}$ of Radon measures on $X$ its narrow topology
(437Jd) then $\chi(\mu,M^+_{\text{R}})\le\omega$.
%533G

\spheader 533Yb For a topological measure $\mu$ on a space
$X$, write $\ureg(\mu)$ for the smallest size of any family $\Cal V$ of
open subsets of $X$ such that
$G\setminus\bigcup\{V:V\in\Cal V$, $V\subseteq G\}$ is negligible for every
open $G\subseteq X$.   (i) Show that if $\mu$ is inner regular with
respect to the Borel sets then the Maharam type $\tau(\mu)$ of $\mu$
is at most $\ureg(\mu)$.   (ii) Show that if $X$ is compact and Hausdorff
and $\mu$ is a Radon measure, then
$\ureg(\mu)\le\max(\non\Cal N_{\tau(\mu)},\chi(X))$.
(iii) Show that if $X$ is compact and Hausdorff,
$\mu$ is a Radon probability measure and
$\cov\Cal N_{\tau(\mu)}>\ureg(\mu)$, then $\mu$ has an equidistributed
sequence.
%533H

\spheader 533Yc ({\smc Plebanek 00})  Suppose that $\kappa$ is a regular
infinite cardinal such that
$\non\Cal N_{\kappa}<\cov\Cal N_{\kappa}=\kappa$.   Let $(X,\mu)$ be a
Radon probability space such that $\chi(X)<\kappa$.   Show that $\mu$
has an equidistributed sequence.
%533Yb
%By 531O, $\kappa>\tau(\mu)$ so $\ureg(\mu)<\kappa$ if $X$ compact

\spheader 533Yd Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space with
countable Maharam type, $\Cal A\subseteq\Sigma$ a set with cardinal less
than $\add\Cal N_{\omega}$, and $\frak S$ the topology on $X$ generated by
$\frak T\cup\Cal A$.   Show that $\mu$ is $\frak S$-Radon.
%533B out of order query
}%end of exercises

\leader{533Z}{Problem} For which cardinals $\kappa$ is $\BbbR^{\kappa}$
Borel-measure-compact?

\discrversionA{If $\kappa<\cov\Cal N_{\kappa}$, yes, by 435Fd.   What if
$\kappa=\omega_1=\cov\Cal N_{\omega}<\frak p$?
$\BbbR^{\omega_1}$ is never Borel-measure-complete, by 434Id, because it
includes $\omega_1$.}{}

\endnotes{
\Notesheader{533} I suppose that from the standpoint of measure theory
the most fundamental of all the properties of $\omega$ is the fact that
the union of countably many Lebesgue negligible sets is again Lebesgue
negligible;  this is of course shared by every
$\kappa<\add\Cal N_{\omega}$ (which is in effect the definition of
$\add\Cal N_{\omega}$).   In 533A-533E %533A 533B 533C 533D 533E
and 533J we have results showing that uncountable cardinals can be
`almost countable' in other ways.   In each case the fact that $\omega$
has the property examined is either trivial (as in 533B) or a basic
result from Volume 4 (as in 533Cb, 533Cc and 533E).   Similarly, the
fact that $\frak c$ does {\it not} have any of these properties is
attested by classical examples.
If you are familiar with Martin's axiom you will not be surprised to
observe that everything here is sorted out if we assume that
$\frak m=\frak c$.

533H does not quite fit this pattern, and the hypothesis in 533Hb
definitely contradicts Martin's axiom.   `Uniformly regular' measures
got squeezed out of \S434 by shortage of space;  in the exercises
533Xb-533Xi % 533Xb 533Xc 533Xd 533Xe 533Xf 533Xg 533Xh 533Xi
I sketch some of what was missed.   Here I mention them
just to show that
there is more to say on the subject of first-countable and perfectly
normal spaces than I put into 531O and 531Q.   Another phenomenon of
interest is the occurrence of measures which are inner regular with respect
to a family of compact metrizable sets (462J, 533Ca, 533D).
}%end of notes

\discrpage

\frfilename{mt534.tex}
\versiondate{27.6.22}
\copyrightdate{2003}

\def\chaptername{Topology and measure III}
\def\sectionname{Hausdorff measures, strong measure zero and Rothberger's
property}
\def\czz{ CHECK\query\ }

\newsection{534}

In this section I look at constructions which are primarily metric
rather than topological.   I start with a note on Hausdorff
measures, spelling out connexions between Hausdorff $r$-dimensional
measure on a
separable metric space and the basic $\sigma$-ideal $\Cal N$ (534B).

The main part of the section section is a brief introduction to a
class of ideals which are of
great interest in set-theoretic analysis.   While the most important
ones are based on separable metric spaces, some of the ideas can be
expressed in more general contexts, and I give a definition of `strong measure
zero' in terms of uniformities (534Ca).   An
associated topological notion is what I call `Rothberger's property'
(534Cb).   A famous
characterization of sets of strong measure zero in $\Bbb R$ in terms
of translations of meager sets can also be represented as a theorem about
$\sigma$-compact groups (534K).   There are few elementary results
describing the cardinal functions of strong measure zero ideals, but I
give some information on their additivities (534M) and uniformities (534Q).
There seem to be some interesting questions concerning
spaces with isomorphic strong measure zero ideals, which I consider in
534N-534P.  %534N 534O 534P
A particularly important question, from the very beginning of the
topic in {\smc Borel 1919}, concerns the possible cardinals of sets of
strong measure zero;  in 534Q-534S %534Q 534R 534S
I give some sample facts and
illustrative examples.

\leader{534A}{}\dvAformerly{5{}34A}\cmmnt{ An elementary lemma will be useful.

\medskip

\noindent}{\bf Lemma} Let $(X,\rho)$ be a separable metric space.
Then there is a countable family $\Cal C$ of subsets of $X$ such that
whenever $A\subseteq X$ has finite diameter and $\eta>0$ then there
is a $C\in\Cal C$ such that $A\subseteq C$ and $\diam C\le\eta+2\diam A$.

\proof{ Let $D$ be a countable dense subset of $X$ and set
$\Cal C=\{\emptyset\}\cup\{B(x,q):x\in D$, $q\in\Bbb Q$, $q\ge 0\}$.
If $A\subseteq X$ has finite diameter and $\eta>0$, then if
$A=\emptyset$ we can take $C=\emptyset$.   Otherwise, take $y\in A$
and $q\in\Bbb Q$ such that
$\diam A+\bover14\eta\le q\le\diam A+\bover12\eta$.   Let $x\in D$ be
such that $\rho(x,y)\le\bover14\eta$;  then $C=B(x,q)\in\Cal C$,
$A\subseteq B(y,\diam A)\subseteq C$ and $\diam C\le 2q\le\eta+2\diam A$.
}%end of proof of 534A

\leader{534B}{Hausdorff \dvrocolon{measures}}\dvAformerly{5{}34B}
\cmmnt{ There are
difficult questions concerning the cardinals associated with
even the most
familiar Hausdorff measures.   However we do have some easy results.

\medskip

\noindent}{\bf Theorem} Let $(X,\rho)$ be a metric space and
$r>0$.   Write $\mu_{Hr}$ for $r$-dimensional Hausdorff measure on
$X$, $\Cal N(\mu_{Hr})$ for its null ideal,
$\Cal N$ for the null ideal of Lebesgue measure on $\Bbb R$ and
$\Cal M$ for the ideal of meager subsets of $\Bbb R$.

(a) $\add\mu_{Hr}=\add\Cal N(\mu_{Hr})$.

(b) If $X$ is separable,
$\Cal N(\mu_{Hr})\prT\Cal N$, so that
$\add\mu_{Hr}\ge\add\Cal N$ and $\cf\Cal N(\mu_{Hr})\le\cf\Cal N$.

(c) If $X$ is separable,
$(X,\in,\Cal N(\mu_{Hr}))\prGT(\Cal M,\not\ni,\Bbb R)$, so that
$\cov\Cal N(\mu_{Hr})\le\non\Cal M$ and
$\frakmctbl\le\non\Cal N(\mu_{Hr})$.

(d) If $X$ is analytic and $\mu_{Hr}X>0$, then
$\add\mu_{Hr}=\add\Cal N$, $\cf\Cal N(\mu_{Hr})=\cf\Cal N$,
$\non\Cal N(\mu_{Hr})\le\non\Cal N$ and
$\cov\Cal N(\mu_{Hr})\ge\cov\Cal N$.
%see also 534Ya

\proof{{\bf (a)} 521Ac.

\medskip

{\bf (b)(i)} Let $\Cal C$ be a countable family of subsets of $X$ such
that whenever $A\subseteq X$ has finite diameter and $\eta>0$ there
is a $C\in\Cal C$
such that $A\subseteq C$ and $\diam C\le\eta+2\diam A$ (534A).

If $A\subseteq X$, then $A\in\Cal N(\mu_{Hr})$ iff for
every $\epsilon>0$ there is a sequence $\sequencen{C_n}$ in $\Cal C$
such that $A\subseteq\bigcup_{n\in\Bbb N}C_n$ and
$\sum_{n=0}^{\infty}(\diam C_n)^r\le\epsilon$.   \Prf\ If $A$ is
negligible and $\epsilon>0$, then (by the definition in 471A) there
must be a sequence $\sequencen{A_n}$ of subsets of $X$ such that
$A\subseteq\bigcup_{n\in\Bbb N}A_n$ and
$\sum_{n=0}^{\infty}(\diam A_n)^r<2^{-r}\epsilon$.   Let
$\sequencen{\eta_n}$ be a sequence of strictly positive real numbers
such that $\sum_{n=0}^{\infty}(\eta_n+2\diam A_n)^r\le\epsilon$.
For each $n$ we
can find $C_n\in\Cal C_n$ such that $A_n\subseteq C_n$ and
$\diam C_n\le\eta_n+2\diam A_n$, so that
$\sum_{n=0}^{\infty}(\diam C_n)^r\le\epsilon$, while
$A\subseteq\bigcup_{n\in\Bbb N}C_n$.

On the other hand, if $A$ satisfies the condition, then for every
$\epsilon$, $\delta>0$ there is a sequence $\sequencen{C_n}$ of
subsets of $X$ such that $A\subseteq\bigcup_{n\in\Bbb N}C_n$ and
$\sum_{n=0}^{\infty}(\diam C_n)^r\le\min(\epsilon,\delta^r)$.   In
this case, $\diam C_n\le\delta$ for every $n$, so $\theta_{r\delta}A$, as
defined in 471A, is at most $\epsilon$.   As $\epsilon$ is arbitrary,
$\theta_{r\delta}A=0$;  as $\delta$ is arbitrary, $A$ is
$\mu_{Hr}$-negligible.\ \Qed

\medskip

\quad{\bf (ii)} It follows that
$(\Cal N(\mu_{Hr}),\subseteq,\Cal N(\mu_{Hr}))
\prGT(\BbbN^{\Bbb N},\subseteq^*,\Cal S)$, where
$(\BbbN^{\Bbb N},\subseteq^*,\Cal S)$ is the $\Bbb N$-localization relation
(522K).

\medskip

\Prf\grheada\ For each $n\in\Bbb N$, let $\Cal I_n$ be the family of
finite subsets $I$ of $\Cal C$ such that
$\sum_{C\in I}(\diam C)^r\le 4^{-n}$.   Let $\sequence{j}{I_{nj}}$ be
a sequence running over $\Cal I_n$.   Now, given $A\in\Cal N(\mu_{Hr})$,
then for each $n\in\Bbb N$ let $\sequence{i}{C_{ni}}$ be a sequence in
$\Cal C$, covering $A$, such that
$\sum_{i=0}^{\infty}(\diam C_{ni})^r\le 2^{-n-1}$.   Let
$\sequence{i}{C_i}$ be a re-indexing of the family
$\langle C_{ni}\rangle_{n,i\in\Bbb N}$, so that $\sequence{i}{C_i}$
is a sequence in $\Cal C$, $\sum_{i=0}^{\infty}(\diam C_i)^r\le 1$, and
$A\subseteq\bigcap_{m\in\Bbb N}\bigcup_{i\ge m}C_i$.   Let
$\sequencen{k(n)}$ be a strictly increasing sequence in $\Bbb N$ such
that $k(0)=0$ and $\sum_{i=k(n)}^{\infty}(\diam C_i)^r\le 4^{-n}$ for
every $n$.   Now, for $n\in\Bbb N$, let $\phi(A)(n)$ be such that
$\{C_i:k(n)\le i<k(n+1)\}=I_{n,\phi(A)(n)}$.

This process defines a function $\phi:\Cal N(\mu_{Hr})\to\BbbN^{\Bbb N}$
such that

\Centerline{$A
\subseteq\bigcap_{m\in\Bbb N}\bigcup_{n\ge m}\bigcup I_{n,\phi(A)(n)}$}

\noindent for every $A\in\Cal N(\mu_{Hr})$.

\medskip

\qquad\grheadb\ For $S\in\Cal S$, set

\Centerline{$\psi(S)
=\bigcap_{m\in\Bbb N}\bigcup_{n\ge m}\bigcup_{i\in S[\{n\}]}\bigcup I_{ni}
\subseteq X$.}

\noindent If $n\in\Bbb N$, then

\Centerline{$\sum\{(\diam C)^r:C\in\bigcup_{i\in S[\{n\}]}I_{ni}\}
\le 2^n\cdot 4^{-n}=2^{-n}$,}

\noindent because $\#(S[\{n\}])\le 2^n$ and
$\sum\{(\diam C)^r:C\in I_{ni}\}\le 4^{-n}$ for every $i$.   But this
means that, for any $m\in\Bbb N$,

\Centerline{$\sum\{(\diam C)^r:C\in\bigcup_{n\ge m}
  \bigcup_{i\in S[\{n\}]}I_{ni}\}
\le 2^{-m+1}$,}

\noindent while

\Centerline{$\psi(S)
\subseteq\bigcup(\bigcup_{n\ge m}\bigcup_{i\in S[\{n\}]}I_{ni})$.}

\noindent So $\psi(S)\in\Cal N(\mu_{Hr})$ for every $S\in\Cal S$.

\medskip

\qquad\grheadc\ Suppose that $A\in\Cal N(\mu_{Hr})$ and
$\phi(A)\subseteq^*S\in\Cal S$.   Then there is some $m_0\in\Bbb N$
such
that $(n,\phi(A)(n))\in S$ for every $n\ge m_0$.   Now, for any
$m\in\Bbb N$, we have

$$\eqalign{A
&\subseteq\bigcup_{n\ge\max(m,m_0)}\bigcup I_{n,\phi(A)(n)}\cr
&\subseteq\bigcup_{n\ge\max(m,m_0)}\bigcup_{i\in S[\{n\}]}\bigcup
I_{ni}
\subseteq\bigcup_{n\ge m}\bigcup_{i\in S[\{n\}]}\bigcup I_{ni},\cr}$$

\noindent so $A\subseteq\psi(S)$.   This shows that $(\phi,\psi)$ is a
Galois-Tukey connection from
$(\Cal N(\mu_{Hr}),\subseteq\nobreak,\penalty-100\Cal N(\mu_{Hr}))$
to $(\BbbN^{\Bbb N},\subseteq^*,\Cal S)$, and
$(\Cal N(\mu_{Hr}),\subseteq,\Cal N(\mu_{Hr}))
\prGT(\BbbN^{\Bbb N},\subseteq^*,\Cal S)$.\ \Qed

\medskip

\quad{\bf (iii)} Since $(\Cal N,\subseteq,\Cal N)
\equivGT(\NN,\subseteq^*,\Cal S)$ (522M),
$(\Cal N(\mu_{Hr}),\subseteq,\Cal N(\mu_{Hr}))
\prGT(\Cal N,\subseteq\nobreak,\penalty-100\Cal N)$,
that is, $\Cal N(\mu_{Hr})\penalty-100\prT\Cal N$.

\medskip

\quad{\bf (iv)} By 513Ee, as usual, we can conclude that
$\add\Cal N(\mu_{Hr})\ge\add\Cal N$ and
$\cf\Cal N(\mu_{Hr})\le\cf\Cal N$.

\medskip

{\bf (c)(i)} If $\mu_{Hr}X=0$, the result is trivial.   \Prf\
Set $\phi(x)=\emptyset$ for $x\in X$, $\psi(t)=X$ for
$t\in\Bbb R$;  then $(\phi,\psi)$ is a Galois-Tukey connection from
$(X,\in,\Cal N(\mu_{Hr}))$ to $(\Cal M,\not\ni,\Bbb R)$.\ \QeD\  So
let
us suppose that $X$ is infinite.

\medskip

\quad{\bf (ii)} Let $F$ be the set of $1$-Lipschitz functions
$f:X\to[0,1]$.   Define $T:X\to\ell^{\infty}(F)$ by setting
$(Tx)(f)=f(x)$ for $f\in F$ and $x\in X$.   Then

\Centerline{$\|Tx-Ty\|_{\infty}
=\sup_{f\in F}|f(x)-f(y)|=\min(1,\rho(x,y))$}

\noindent for all $x$, $y\in X$.   \Prf\ Of course
$\sup_{f\in F}|f(x)-f(y)|\le\min(1,\rho(x,y))$, by the definition of
$F$.   On the other hand, we can set
$f(z)=\min(1,\rho(z,x))$ for every $z\in X$;  then $f\in F$ and
$|f(x)-f(y)|=\min(1,\rho(x,y))$.   So we have equality.\ \QeD\   Thus
$T$ is $1$-Lipschitz for $\rho$ and the usual metric on
$\ell^{\infty}(F)$, and $T[X]$ is a separable subset of
$\ell^{\infty}(F)$ (4A2B(e-iii)).   Let $V$ be the closed linear
subspace of $\ell^{\infty}(F)$ generated by $T[X]$;  then $V$ is
separable (4A4Bg).   Being a closed subset of the complete metric
space
$\ell^{\infty}(F)$, $V$ is a Polish space.   Since $X$ has more than
one
point, and $T$ is injective, $V$ is non-empty and has no isolated
points.

Let $\sequencen{v_n}$ enumerate a dense subset of $V$.   Set

\Centerline{$E
=\bigcap_{n\in\Bbb N}\bigcup_{i\ge n}U(v_i,2^{-i-1})$}

\noindent where $U(v,\delta)=\{u:u\in V$, $\|u-v\|_{\infty}<\delta\}$
for $v\in V$, $\delta>0$.   Then $E$ is the intersection of a sequence
of dense open sets in $V$, so is comeager, and $M=V\setminus E$
belongs to the ideal $\Cal M(V)$ of meager subsets of $V$.
For any $v\in V$, the map $u\mapsto u-v:V\to V$ is a
homeomorphism, so $M-v\in\Cal M(V)$.   Define $\phi:X\to\Cal M(V)$ by
setting $\phi(x)=M-Tx$ for $x\in X$.

In the other direction, define $\psi:V\to\Cal PX$ by setting
$\psi(v)=T^{-1}[E-v]$ for $v\in V$.   Then $\psi(v)\in\Cal
N(\mu_{Hr})$
for every $v\in V$.   \Prf\ If $v\in V$ and $\delta\le\bover12$, then
$\|u-u'\|_{\infty}<1$ for all $u$, $u'\in U(v,\delta)$, so
$\rho(x,x')\le\|Tx-Tx'\|_{\infty}$ whenever $x$,
$x'\in T^{-1}[U(v,\delta)]$.   Accordingly
$\diam T^{-1}[U(v_i-v,2^{-i-1})]\le 2^{-i}$ for every $i\in\Bbb N$.
This means that

$$\eqalign{\mu_{Hr}^*T^{-1}[E-v]
&=\mu_{Hr}^*(\bigcap_{n\in\Bbb N}\bigcup_{i\ge n}
  T^{-1}[U(v_i-v,2^{-i-1})]\cr
&\le\inf_{n\in\Bbb N}\sum_{i=n}^{\infty}(2^{-i})^r
=0.  \text{ \Qed}\cr}$$

\noindent So $\psi$ is a function from $V$ to $\Cal N(\mu_{Hr})$.   We
now see that

$$\eqalign{\phi(x)\not\ni v
&\Longrightarrow v\notin M-Tx
\Longrightarrow Tx\notin M-v\cr
&\Longrightarrow Tx\in E-v
\Longrightarrow x\in\psi(v).\cr}$$

\noindent Thus $(\phi,\psi)$ is a Galois-Tukey connection from
$(X,\in,\Cal N(\mu_{Hr}))$ to $(\Cal M(V),\not\ni,V)$ and

\Centerline{$(X,\in,\Cal N(\mu_{Hr}))\prGT(\Cal M(V),\not\ni,V)
\cong(\Cal M,\not\ni,\Bbb R)$}

\noindent (522Wb).

\medskip

\quad{\bf (iii)} Now

\Centerline{$\cov\Cal N(\mu_{Hr})=\cov(X,\in,\Cal N(\mu_{Hr}))
\le\cov(\Cal M,\not\ni,\Bbb R)=\non\Cal M$,}

\Centerline{$\non\Cal N(\mu_{Hr})=\add(X,\in,\Cal N(\mu_{Hr}))
\ge\add(\Cal M,\not\ni,\Bbb R)=\cov\Cal M=\frakmctbl$}

\noindent (512D, 512Ed, 522Sa).

\medskip

{\bf (d)} If $X$ is analytic and $\mu_{Hr}X>0$, then by Howroyd's
theorem (471S) there is a compact set $K\subseteq X$ such that
$0<\mu_{Hr}K<\infty$.   Now the subspace measure $\mu_{Hr}^{(K)}$ on $K$
is an atomless Radon measure (471E, 471Dg, 471F)
on a compact metric space, so

\Centerline{$\add\Cal N\le\add\Cal N(\mu_{Hr})
\le\add\Cal N(\mu_{Hr}^{(K)})=\add\Cal N$,}

\Centerline{$\cf\Cal N\ge\cf\Cal N(\mu_{Hr})
\ge\cf\Cal N(\mu_{Hr}^{(K)})=\cf\Cal N$,}

\Centerline{$\non\Cal N(\mu_{Hr})
\le\non\Cal N(\mu_{Hr}^{(K)})=\non\Cal N$,}

\Centerline{$\cov(X,\Cal N(\mu_{Hr}))
\ge\cov(K,\Cal N(\mu_{Hr}^{(K)}))=\cov\Cal N$}

\noindent by (b) above, 521F and 522Wa.
}%end of proof of 534B

\leader{534C}{Strong measure zero and Rothberger's property}
\dvAformerly{5{}34C, 5{}34E}
{\bf (a)} Let $(X,\Cal W)$ be a uniform space and $A\subseteq X$.
I say that $A$ has {\bf strong measure zero} or {\bf property C} in $X$ if
for any sequence $\sequencen{W_n}$ in $\Cal W$ there is a
cover $\sequencen{A_n}$ of $A$ such that $A_n\times A_n\subseteq W_n$
for every $n\in\Bbb N$.   If $(X,\rho)$
is a metric space, a subset $A$ of $X$ has strong measure zero in $X$
if it has strong measure
zero for the uniformity defined by the metric\cmmnt{ (3A4B),
that is, for any
sequence $\sequencen{\epsilon_n}$ of strictly positive real numbers
there is a
cover $\sequencen{A_n}$ of $X$ such that $\diam A_n\le\epsilon_n$ for
every $n\in\Bbb N$}.

I will write $\CalSmz(X,\Cal W)$ or $\CalSmz(X,\rho)$ for the family of
sets of strong measure zero in a uniform space $(X,\Cal W)$ or a metric
space $(X,\rho)$.

\medskip

{\bf (b)} If $X$ is a topological space and $A$ is a
subset of $X$, I will say that $A$ has {\bf Rothberger's property in}
$X$ if for every sequence
$\sequencen{\Cal G_n}$ of non-empty open covers of $X$ there is a
sequence $\sequencen{G_n}$ such that $G_n\in\Cal G_n$ for every
$n\in\Bbb N$ and $A\subseteq\bigcup_{n\in\Bbb N}G_n$.
I will write $\CalRbg(X)$ for the family
of subsets of $X$ with Rothberger's property in $X$.

%534Xh, 534Xm

\leader{534D}{Proposition}\dvAformerly{5{}34D, 5{}34F}
(a)(i) If $(X,\Cal W)$ is a uniform space and $A\subseteq X$, then $A$ has
strong measure zero in $X$ iff it has strong measure zero in itself when it is
given its subspace uniformity.

\quad(ii) If $(X,\Cal W)$ is a uniform space,
then $\CalSmz(X,\Cal W)$ is a $\sigma$-ideal
containing all the countable subsets of $X$.

\quad(iii) If $(X,\Cal W)$ and $(Y,\Cal V)$ are uniform spaces and
$f:X\to Y$ is uniformly continuous, then $f[A]\in\CalSmz(Y,\Cal V)$ whenever
$A\in\CalSmz(X,\Cal W)$.

\quad(iv) Let $(X,\Cal W)$ be a uniform space and $A\subseteq X$.   Then
$A\in\CalSmz(X,\Cal W)$ iff $f[A]\in\CalSmz(Y,\rho)$ whenever $(Y,\rho)$ is
a metric space and $f:X\to Y$ is uniformly continuous.

\quad(v) Let $(X,\Cal W)$ be a uniform space and $A\in\CalSmz(X,\Cal W)$.
If $B\subseteq X$ is such that $B\setminus G\in\CalSmz(X,\Cal W)$
whenever $G$ is an open set including $A$, then
$B\in\CalSmz(X,\Cal W)$.

(b) Let $X$ be a topological space.

\quad (i) $\CalRbg(X)$ is
a $\sigma$-ideal containing all the countable subsets of $X$.

\quad (ii) If $Y$ is another topological space, $f:X\to Y$ is continuous and
$A\in\CalRbg(X)$, then $f[A]\in\CalRbg(Y)$.

\quad(iii)\dvAformerly{5{}34Xh} If $A\in\CalRbg(X)$
and $B\subseteq X$ is such that $B\setminus G\in\CalRbg(X)$
whenever $G$ is an open set including $A$, then $B\in\CalRbg(X)$.

\quad(iv)\dvAformerly{5{}34Xf} If $F\subseteq X$ is closed, then
$\CalRbg(F)=\{A:A\in\CalRbg(X)$, $A\subseteq F\}$.

\proof{{\bf (a)(i)} Recall that the subspace uniformity on $A$ is just
$\Cal W_A=\{W\cap(A\times A):W\in\Cal W\}$ (3A4D).   If
$A\in\CalSmz(A,\Cal W_A)$ and $\sequencen{W_n}$ is a sequence in $\Cal W$, then
$\sequencen{W_n\cap(A\times A)}$ is a sequence in $\Cal W_A$, so we have
a sequence $\sequencen{A_n}$ of sets covering $A$ with
$A_n\times A_n\subseteq W_n\cap(A\times A)\subseteq W_n$ for every $n$;  as
$\sequencen{W_n}$ is arbitrary, $A\in\CalSmz(X,\Cal W)$.   If
$A\in\CalSmz(X,\Cal W)$ and $\sequencen{V_n}$ is a sequence in $\Cal W_A$, we
can choose for each $n$ a $W_n\in\Cal W$ such that $V_n=W_n\cap(A\times A)$;
now we have a sequence $\sequencen{A_n}$ of sets covering $A$ with
$A_n\times A_n\subseteq W_n$ for every $n$, in which case
$\sequencen{A_n\cap A}$ covers $A$ and
$(A_n\cap A)\times(A_n\cap A)\subseteq V_n$ for every $n$;   as
$\sequencen{V_n}$ is arbitrary, $A\in\CalSmz(A,\Cal A)$.

\medskip

\quad{\bf (ii)} It is immediate from the definition that any subset of
a set in $\CalSmz(X,\Cal W)$ belongs to $\CalSmz(X,\Cal W)$,
and so does any countable set.   Now suppose that
$\sequencen{A_n}$ is a sequence in $\CalSmz(X,\Cal W)$.
Let $\sequencen{W_n}$ be any sequence in
$\Cal W$.   For each $k\in\Bbb N$, $\sequence{i}{W_{2^k(2i+1)}}$ is a
sequence in $\Cal W$, so there is a sequence $\sequence{i}{A_{ki}}$,
covering $A_k$, such that $A_{ki}\times A_{ki}\subseteq W_{2^k(2i+1)}$
for every $i$.   Set $B_0=\emptyset$ and $B_n=A_{ki}$ if $n=2^k(2i+1)$
where $k$, $i\in\Bbb N$;  then $A\subseteq\bigcup_{n\in\Bbb N}B_n$ and
$B_n\times B_n\subseteq W_n$ for every $n$.   As $\sequencen{W_n}$ is
arbitrary, $A$ has strong measure zero;  as $\sequencen{A_n}$ is
arbitrary, $\CalSmz(X,\Cal W)$ is a $\sigma$-ideal.

\medskip

\quad{\bf (iii)} Let $\sequencen{V_n}$ be a sequence in $\Cal V$.   For each
$n\in\Bbb N$, there is a $W_n\in\Cal W$
such that $(f(x),f(x'))\in V_n$ whenever
$(x,x')\in W_n$.   Because $A\in\CalSmz(X,\Cal W)$, there is
a cover $\sequencen{A_n}$ of $A$ such that
$A_n\times A_n\subseteq W_n$
for every $n$;  now $f[A_n]\times f[A_n]\subseteq V_n$ for every $n$
and $\bigcup_{n\in\Bbb N}f[A_n]=f[A]$.   As $\sequencen{V_n}$ is
arbitrary, $f[A]\in\CalSmz(Y,\Cal V)$.

\medskip

\quad{\bf (iv)} If $A$ has strong measure zero, then of course $f[A]$ has
strong measure zero for any uniformly continuous function $f$ from $X$
to a metric space, by (iii).   Now suppose that
$A$ satisfies the
condition, and that $\sequencen{W_n}$ is a sequence in $\Cal W$.   Then
there is a pseudometric $\rho$ on $X$, compatible with the uniformity in
the sense that $\{(x,y):\rho(x,y)\le\epsilon\}\in\Cal W$ for every
$\epsilon>0$, such that $\{(x,y):\rho(x,y)<2^{-n}\}\subseteq W_n$ for
every $n$ (4A2Ja).   Set $\sim\mskip5mu=\{(x,y):\rho(x,y)=0\}$.   Then
$\sim$ is
an equivalence relation on $X$.   If $Y$ is the set of equivalence
classes, we have a metric $\tilde\rho$ on $Y$ defined by setting
$\tilde\rho(x^{\ssbullet},y^{\ssbullet})=\rho(x,y)$ for all $x$,
$y\in X$.   Setting $f(x)=x^{\ssbullet}$ for $x\in X$, $f:X\to Y$ is
uniformly continuous.   So $f[A]\in\CalSmz(Y,\tilde\rho)$.   Let
$\sequencen{B_n}$ be a cover of $f[A]$ such that $\diam B_n\le 2^{-n-1}$
for every $n$, and set $A_n=f^{-1}[B_n]$ for each $n$.   Then
$\sequencen{A_n}$ is a cover of $A$.   If $n\in\Bbb N$ and $x$,
$y\in A_n$, then $\rho(x,y)=\tilde\rho(f(x),f(y))\le 2^{-n-1}$, so
$(x,y)\in W_n$.   Thus $A_n\times A_n\subseteq W_n$.   As
$\sequencen{W_n}$ is arbitrary, $A\in\CalSmz(X,\Cal W)$.

\medskip

\quad{\bf (v)} Let $\sequencen{W_n}$ be any sequence in $\Cal W$.   For each
$n\in\Bbb N$, let $V_n\in\Cal W$ be such that
$V_n\frsmallcirc V_n\frsmallcirc V_n^{-1}\subseteq W_{2n}$.   Then there
is a sequence $\sequencen{A_n}$, covering $A$, such that
$A_n\times A_n\subseteq V_n$ for every $n$.   Set
$B_{2n}=\interior V_n[A_n]$ for
each $n$, and $G=\bigcup_{n\in\Bbb N}B_{2n}$;  then
$B_{2n}\times B_{2n}\subseteq W_{2n}$ for every $n$ and $G$ is an open
set including $A$.   Accordingly $B\setminus G\in\CalSmz(X,\Cal W)$
and there is a sequence $\sequencen{B_{2n+1}}$, covering $B\setminus G$,
such that $B_{2n+1}\times B_{2n+1}\subseteq W_{2n+1}$ for every $n$.
Now $\sequencen{B_n}$ covers $B$ and $B_n\times B_n\subseteq W_n$ for
every $n$.   As $\sequencen{W_n}$ is arbitrary, $B\in\CalSmz(X,\Cal W)$.

\medskip

{\bf (b)(i)} We can copy the argument of (a-ii).
As before, it is immediate from the definition that any subset
of a set in $\CalRbg(X)$, and any countable subset of $X$, belong to
$\CalRbg(X)$.   Now
suppose that $\sequencen{A_n}$ is a sequence in $\CalRbg(X)$,
with union $A$.   Let $\sequencen{\Cal G_n}$ be any sequence of
non-empty open covers of $X$.   For each $k\in\Bbb N$,
$\sequence{i}{\Cal G_{2^k(2i+1)}}$ is a
sequence of open covers of $X$, so there is a sequence
$\sequence{i}{G_{ki}}$,
covering $A_k$, such that $G_{ki}\in\Cal G_{2^k(2i+1)}$
for every $i$.   Take $G_0$ to be any member of $\Cal G_0$, and set
$G_n=G_{ki}$ if $n=2^k(2i+1)$
where $k$, $i\in\Bbb N$;  then $A\subseteq\bigcup_{n\in\Bbb N}G_n$ and
$G_n\in\Cal G_n$ for every $n$.   As $\sequencen{\Cal G_n}$ is
arbitrary, $A$ has Rothberger's property in $X$.

\medskip

\quad{\bf (ii)} This uses the idea of (a-iii).
Let $\sequencen{\Cal H_n}$ be a sequence of non-empty open
covers of $Y$.   For each $n\in\Bbb N$, set
$\Cal G_n=\{f^{-1}[H]:H\in\Cal H_n\}$;  then $\Cal G_n$ is a non-empty
open cover of $X$.   Because $A\in\CalRbg(X)$, there
is a cover $\sequencen{G_n}$ of $A$ such that $G_n\in\Cal G_n$ for
every $n\in\Bbb N$.   Expressing $G_n$ as $f^{-1}[H_n]$
where $H_n\in\Cal H_n$
for each $n\in\Bbb N$,  $f[A]\subseteq\bigcup_{n\in\Bbb N}H_n$.
As $\sequencen{\Cal H_n}$ is
arbitrary, $f[A]$ has Rothberger's property in $Y$.

\medskip

\quad{\bf (iii)} And here we can copy from (a-v).
Let $\sequencen{\Cal G_n}$ be any sequence of open covers of $X$.
Then there is a sequence $\sequencen{G_{2n}}$, covering $A$, such that
$G_{2n}\in\Cal G_{2n}$ for every $n$.   Set $H=\bigcup_{n\in\Bbb N}G_{2n}$;
then there is a sequence $\sequencen{G_{2n+1}}$, covering $B\setminus H$, such
that $G_{2n+1}\in\Cal G_{2n+1}$ for each $n$.   Putting these together, we
have a sequence $\sequencen{G_n}$ covering $B$ such
that $G_n\in\Cal G_n$ for every $n$.   As $\sequencen{\Cal G_n}$ is arbitrary,
$B$ has Rothberger's property in $X$.

\medskip

\quad{\bf (iv)} If $A\in\CalRbg(F)$ then $A\in\CalRbg(X)$ by (ii), because the
identity map from $F$ to $X$ is continuous.   Conversely, if $A\subseteq F$ and
$A\in\CalRbg(X)$, let $\sequencen{\Cal G_n}$ be a sequence of
non-empty relatively open covers of $F$.   For $n\in\Bbb N$ set

\Centerline{$\Cal H_n=\{H:H\subseteq X$ is open and there is a
$G\in\Cal G_n$ such that $H\cap F\subseteq G\}$.}

\noindent Then $\Cal H_n$ is a non-empty
open cover of $X$ because $\Cal G_n$ covers
$F$ and $X\setminus F\in\Cal H_n$.   Because $A\in\CalRbg(X)$, there is a
sequence $\sequencen{H_n}$ such that $H_k\in\Cal H_n$
for every $n\in\Bbb N$ and $A\subseteq\bigcup_{n\in\Bbb N}H_n$.
For $n\in\Bbb N$ choose $G_n\in\Cal G_n$ such that
$H_n\cap F\subseteq G_n$;  then $A\subseteq\bigcup_{n\in\Bbb N}G_n$.  As
$\sequencen{\Cal G_n}$ is arbitrary, $A\in\CalRbg(F)$.
}%end of proof of 534D

\vleader{48pt}{534E}{Proposition}\dvAformerly{5{}34F}
Let $(X,\Cal W)$ be a uniform space, and give $X$ the topology induced by
$\Cal W$.

(a) $\CalRbg(X)\subseteq\CalSmz(X,\Cal W)$.

(b) If $X$ is $\sigma$-compact, $\CalRbg(X)=\CalSmz(X,\Cal W)$.

\proof{{\bf (a)} Suppose that $A\in\CalRbg(X)$,
and that $\sequencen{W_n}$ is any sequence in $\Cal W$.   For each
$n\in\Bbb N$, set
$\Cal G_n=\{G:G\subseteq X$ is open, $G\times G\subseteq W_n\}$;  then
$\Cal G_n$ is a non-empty open cover of $X$.   So we can find a cover
$\sequencen{G_n}$ of $A$ such that $G_n\in\Cal G_n$, that is,
$G_n\times G_n\subseteq W_n$,  for each $n$.   As $\sequencen{W_n}$ is
arbitrary, $A\in\CalSmz(X,\Cal W)$.

\medskip

{\bf (b)(i)} Let $K\subseteq X$ be compact and $\Cal G$ an open cover
of $X$.   Then there is a $W\in\Cal W$ such that whenever $x\in K$ there
is a $G\in\Cal G$ such that $W[\{x\}]\subseteq G$.   \Prf\ (Cf.\ 2A2Ed.)
Set

\Centerline{$Q
=\{(x,V):x\in X$, $V\in\Cal W$, $V[V[\{x\}]]\subseteq G$ for some
$G\in\Cal G\}$.}

\noindent Then for every $x\in X$ there are a $G\in\Cal G$ such that
$x\in G$ and a $V\in\Cal W$ such that $V[V[\{x\}]]\subseteq G$, and in
this case $(x,V)\in Q$ and $x\in\interior V[\{x\}]$.   So
$\{\interior V[\{x\}]:(x,V)\in Q\}$ is an open cover of $X$ and
there is a finite set $Q_0\subseteq Q$ such that
$K\subseteq\bigcup\{\interior V[\{x\}]:(x,V)\in Q_0\}$.
Let $W\in\Cal W$ be such
that $W\subseteq V$ whenever $(x,V)\in Q_0$.   If $x\in K$, there
is an $(x',V)\in Q_0$ such that $x\in V[\{x'\}]$;  and now there is a
$G\in\Cal G$ including $V[V[\{x'\}]]\supseteq W[\{x\}]$.\ \Qed

\medskip

\quad{\bf (ii)} Suppose that $K\subseteq X$ is compact and
$A\in\CalSmz(X,\Cal W)$.   Then $A\cap K\in\CalRbg(X)$.
\Prf\ Let $\sequencen{\Cal G_n}$ be a sequence of non-empty open
covers of $X$.   For each $n\in\Bbb N$ let $W_n\in\Cal W$ be such that
$\{W_n[\{x\}]:x\in K\}$ refines $\Cal G_n$.   Let $\sequencen{A_n}$ be a
cover of $A$ such that $A_n\times A_n\subseteq W_n$ for every $n$.
If $n\in\Bbb N$ and $A_n\cap K=\emptyset$, take any $G_n\in\Cal G_n$.
Otherwise, take $x_n\in A_n\cap K$ and $G_n\in\Cal G_n$ such that
$W_n[\{x_n\}]\subseteq G_n$.   If $x\in A\cap K$, there is an
$n\in\Bbb N$ such that $x\in A_n$;  now
$(x_n,x)\in A_n\times A_n\subseteq W_n$ and
$x\in W_n[\{x_n\}]\subseteq G_n$.  As $x$ is arbitrary,
$A\cap K\subseteq\bigcup_{n\in\Bbb N}G_n$.   As $\sequencen{\Cal G_n}$
is arbitrary, $A\cap K$ has Rothberger's property in $X$.\ \Qed

\medskip

\quad{\bf (iii)} $\CalSmz(X,\Cal W)\subseteq\CalRbg(X)$.
\Prf\ If $A\in\CalSmz(X,\Cal W)$, let
$\sequencen{K_n}$ be a sequence of compact subsets of $X$ covering $X$.
By (ii) here, $A\cap K_n\in\CalRbg(X)$ for each $n$;
by 534D(b-i), $A\in\CalRbg(X)$.\ \QeD\   Putting this together with (a), we
see that $\CalRbg(X)=\CalSmz(X,\Cal W)$.
}%end of proof of 534E

\leader{534F}{}\cmmnt{ Another case in which Rothberger's property and strong
measure zero coincide is the following.

\medskip

\noindent}{\bf Proposition}\dvAformerly{5{}34Xi}
Let $X$ be a regular paracompact space, and
$\Cal W$ the uniformity on $X$ defined by the family of all continuous
pseudometrics on $X$.   Then

$$\eqalign{\CalRbg(X)&
=\CalSmz(X,\Cal W)\cr&
=\{A:A\subseteq X,\,f[A]\in\CalSmz(Y,\rho)\text{ whenever }(Y,\rho)
\text{ is a metric space}\cr&
\mskip 200mu
\text{and }f:X\to Y\text{ is continuous}\}.\cr}$$
%3A4Ba

\proof{ $X$ is normal, therefore completely regular (4A2Ge),
so $\Cal W$ induces its topology (4A2J(g-i)),
and $\CalRbg(X)\subseteq\CalSmz(X,\Cal W)$ by 534Eb.
If $A\in\CalSmz(X,\Cal W)$, $(Y,\rho)$ is a metric space and $f:X\to Y$ is
continuous, then $(x,y)\mapsto\rho(f(x),f(y))$ is a continuous pseudometric
on $X$ so is one of the pseudometrics defining $\Cal W$, and $f$ is uniformly
continuous;  now $f[A]\in\CalSmz(Y,\rho)$ by 534D(b-ii).

Now suppose that $A\subseteq X$ is such that
$f[A]\in\CalSmz(Y,\rho)$ whenever $(Y,\rho)$ is a metric
space and $f:X\to Y$ is continuous, and let $\sequencen{\Cal G_n}$ be a sequence
of open covers of $X$.   By 5A4Fb there is for each $n\in\Bbb N$ a continuous
pseudometric $\sigma_n$ on $X$ such that every subset of $X$ of
$\sigma_n$-diameter at most $1$ is included in a member of $\Cal G_n$.   Set

\Centerline{$\sigma(x,y)=\sum_{n=0}^{\infty}2^{-n}\min(2,\sigma_n(x,y))$}

\noindent for $x$, $y\in X$.   Then $\sigma$ is a continuous pseudometric on
$X$.   Let $\sim$ be the corresponding equivalence relation
$\{(x,y):\sigma(x,y)=0\}$ and $Y=X/\sim$ the set of equivalence classes;  then
$Y$ has a metric $\rho$ defined by saying that
$\rho(x^{\ssbullet},y^{\ssbullet})=\sigma(x,y)$ for $x$, $y\in X$, and
$x\mapsto x^{\ssbullet}:X\to Y$ is continuous.   Accordingly
$f[A]=\{x^{\ssbullet}:x\in A\}$ belongs to $\CalSmz(Y,\rho)$, and there is a
sequence $\sequencen{B_n}$ of subsets of $Y$, covering $f[A]$, such that the
$\rho$-diameter of $B_n$ is at most $2^{-n}$ for each $n$.
Setting $A_n=f^{-1}[B_n]$ for each $n$, $A\subseteq\bigcup_{n\in\Bbb N}A_n$.
If $n\in\Bbb N$ and $x$, $y\in A_n$, then $\sigma(x,y)\le 2^{-n}$ so
$\sigma_n(x,y)\le 1$;  by the choice of $\sigma_n$, there is a set
$G_n\in\Cal G_n$ including $A_n$.   But now we see that
$A\subseteq\bigcup_{n\in\Bbb N}\Cal G_n$.   As $\sequencen{\Cal G_n}$ is
arbitrary, $A\in\CalRbg(X)$.
}%end of proof of 534F

\leader{534G}{Remarks}\dvAformerly{5{}34G}
We see from 534E that in Euclidean space, the context of the original
investigation of these
ideas, what I call Rothberger's property and strong measure
zero coincide;  and as the latter phrase is more commonly used and has a more
generally accepted meaning, it is tempting to prefer it.   But in the
framework of this treatise, devoted as it is to maximal convenient generality,
the concepts diverge.   Strong measure zero has an obvious interpretation in
any metric space, and can readily be applied in general uniform spaces;  while
Rothberger's property is a topological notion.
They have very different natures as soon as we leave the area
of $\sigma$-compact spaces.   In particular, the Polish space $\NN$,
topologically identifiable with $\Bbb R\setminus\Bbb Q$, has a wide variety of
compatible uniformities, giving rise to potentially
very different strong measure
zero ideals.   So we find ourselves with the possibility that
$\CalRbg(\Bbb R\setminus\Bbb Q)$ may be much smaller than the trace of
$\CalRbg(\Bbb R)$ on the subset $\Bbb R\setminus\Bbb Q$, even though
$\Bbb Q\in\CalRbg(\Bbb R)$ (534Sb).
Strong measure zero, of course, is much more manageable on subsets (534D(a-i)).

\leader{534H}{}\cmmnt{ Of course sets with strong measure zero or Rothberger's
property are necessarily small in other ways.

\medskip

\noindent}{\bf Proposition}
If $(X,\rho)$ is a metric space and $A\in\CalSmz(X,\rho)$, then
$A$ is separable, zero-dimensional and universally negligible, and all compact
subsets of $A$ are countable.

\proof{{\bf (a)} \Quer\ If $A$ is not separable, there is an uncountable
$B\subseteq A$ such that $\epsilon=\inf_{x,y\in B,x\ne y}\rho(x,y)$ is
greater than $0$ (5A4B(h-iii)).
Now there can be no cover $\sequencen{A_n}$ of $B$ by sets of
diameter less than $\epsilon$.\ \BanG\ Thus $A$ is separable.

\medskip

{\bf (b)} Now suppose that $\mu$ is a Borel probability measure on $A$.   Then
there is a $\delta>0$ such that for every $n\in\Bbb N$ there is a
relatively Borel set $E_n\subseteq A$
with $\diam E_n\le 2^{-n}$ and $\mu E_n\ge\delta$.
\Prf\Quer\ Otherwise, we can find for each $n\in\Bbb N$ an
$\epsilon_n>0$ such that $\mu E\le 2^{-n-2}$ whenever $E\subseteq A$ is a
relatively Borel
set and $\diam E\le\epsilon_n$.   Let $\sequencen{A_n}$ be a cover of $A$
such that $\diam A_n\le\epsilon_n$ for every $n$;  then
$\diam\overline{A}_n\le\epsilon_n$, so $\mu(A\cap\overline{A}_n)\le 2^{-n-2}$
for every $n$, and

\Centerline{$\mu A\le\sum_{n=0}^{\infty}\mu(A\cap\overline{A}_n)<1$.
\Bang\Qed}

Now consider $E=\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}E_m$.   Since
$\mu E\ge\delta>0$, there is an $x\in E$.   For any $n\in\Bbb N$,
there is an $m\ge n$ such that

\Centerline{$x\in E_m\subseteq B(x,2^{-m})\subseteq B(x,2^{-n})$,}

\noindent so

\Centerline{$\mu\{x\}=\inf_{n\in\Bbb N}\mu(A\cap B(x,2^{-n}))\ge\delta>0$.}

\noindent As $\mu$ is arbitrary, this shows that $A$ is universally
negligible.

\medskip

{\bf (c)} In particular, $[0,1]$, with its usual metric, is not of strong
measure zero.   Now if $G\subseteq X$ is open and $x\in G$, let $\delta>0$ be
such that $B(x,\delta)\subseteq G$, and set
$f(y)=\max(0,1-\bover1{\delta}\rho(y,x))$ for $y\in X$;  then $f:X\to[0,1]$ is
uniformly continuous, so $f[X]$ has strong measure zero (534D(a-iii)) and cannot
be the whole of $[0,1]$.   As $f(x)=1$, there is an
$\alpha\in\coint{0,1}\setminus f[X]$, and
$f^{-1}[\mskip1mu[\alpha,1]\mskip1mu]
=f^{-1}[\mskip2mu\ocint{\alpha,1}\mskip1mu]$ is an open-and-closed
neighbourhood of $x$ included in $G$.   As $x$ and $G$ are arbitrary, $X$ is
zero-dimensional.

\medskip

{\bf (d)} If $K\subseteq X$ is compact, it must be scattered (439C(a-v));
because it is first-countable, it must be countable (4A2G(j-vi)).
}%end of proof of 534H

\leader{534I}{}\dvAnew{2022} Let $X$ be a regular topological space.   Then $X$
has Rothberger's property in itself iff it is Lindel\"of and zero-dimensional
and $f[X]\in\CalRbg(\Bbb R\setminus\Bbb Q)$ whenever
$f:X\to\Bbb R\setminus\Bbb Q$ is continuous.

\proof{{\bf (a)} Suppose that $X$ has Rothberger's property in itself.
Let $\Cal G$ be an open cover of $X$.   If $X$ is empty then $\emptyset$ is a
countable subset of $\Cal G$ covering $X$.   Otherwise, setting
$\Cal G_n=\Cal G$ for each $n$, we have a sequence
$\sequencen{G_n}$ covering $X$ such that $G_n\in\Cal G_n$ for every $n$,
and $\{G_n:n\in\Bbb N\}$ is a countable subcover of $\Cal G$.   So $X$ is
Lindel\"of.

Thus $X$ is Lindel\"of and regular, therefore normal and completely regular
(4A2H(b-i)).   If $G\subseteq X$ is open and $x\in G$, there is a continuous
$f:X\to[0,1]$ such that $f(x)=1$ and $f(y)=0$ for every $y\in X\setminus G$.
Since $f[X]\in\CalRbg([0,1])$ (534D(b-ii)), $f[X]\ne[0,1]$;  taking
$\alpha\in[0,1]\setminus X$, $f^{-1}[\ocint{\alpha,1}]=f^{-1}[[\alpha,1]]$ is an
open-and-closed subset of $G$ containing $x$.   Thus $X$ is zero-dimensional.
And of course $f[X]\in\CalRbg(\Bbb R\setminus\Bbb Q)$ for every continuous
$f:X\to\Bbb R\setminus\Bbb Q$, by by 534D(b-ii) again.

\medskip

{\bf (b)} Suppose that $X$ has the given properties.

\medskip

\quad{\bf (i)} If $Z$ is a zero-dimensional
Polish space and $f:X\to Z$ is continuous then
$f[X]\in\CalRbg(Z)$.   \Prf\ By 5A4If, we can suppose that $Z$ is a closed
subspace of $\NN$.   In this case, $f[X]\in\CalRbg(\NN)$, by hypothesis;
by 534D(b-iv), $f[X]\in\CalRbg(Z)$.

\medskip

\quad{\bf (ii)} Now take a sequence $\sequencen{\Cal G_n}$ of non-empty open
covers of $X$.   For each $n\in\Bbb N$, let $\Cal G'_n$ be the family of
open-and-closed subsets of $X$ included in members of $\Cal G_n$;  as $X$ is
zero-dimensional, each $\Cal G'_n$ is a non-empty open cover of $X$.
As $X$ is Lindel\"of, there is for each $n\in\Bbb N$ a sequence
$\sequence{i}{G_{ni}}$ in $\Cal G'_n$ such that $X=\bigcup_{i\in\Bbb N}G_{in}$.
Define $f:X\to Z=\{0,1\}^{\Bbb N\times\Bbb N}$ by setting
$f(x)(n,i)=\chi G_{ni}$ for $x\in X$ and $n$, $i\in\Bbb N$;  as every $G_{ni}$
is open-and-closed, $f$ is continuous.   For $n$, $i\in\Bbb N$, set
$H_{ni}=\{z:z\in Z$, $z(n,i)=1\}$, so that
$G_{ni}=f^{-1}[H_{ni}$.   Set
$E=\bigcap_{n\in\Bbb N}\bigcup_{i\in\Bbb N}H_{ni}$, so that $E$ is a
G$_{\delta}$ subset of $Z$ including $f[X]$.   By 4A2Qd, $E$ is Polish, so
$f[X]\in\CalRbg(E)$, by (i) above.

For each $n\in\Bbb N$, $\{E\cap H_{ni}:i\in\Bbb N\}$ is a
relatively open cover of $E$.
There is therefore a sequence $\sequencen{i_n}$ in $\Bbb N$ such that
$f[X]\subseteq\bigcup_{n\in\Bbb N}H_{ni_n}$ and
$X=\bigcup_{n\in\Bbb N}G_{ni_n}$.   Finally there is for each
$n\in\Bbb N$ a $G_n\in\Cal G_n$ such that $G_{ni_n}\subseteq G_n$, so that
$X=\bigcup_{n\in\Bbb N}G_n$.   As $\sequencen{\Cal G_n}$ is arbitrary,
$X$ has Rothberger's property in itself.
}%end of proof of 534I

\leader{534J}{Proposition} Let $X$ be a Hausdorff space, and $K$ a compact
subset of $X$.   Then $K$ belongs to $\CalRbg(X)$ iff it is scattered.

\proof{{\bf (a)} Set

\Centerline{$\Cal F=\{F:F\subseteq K$ is closed, $L\in\CalRbg(X)$ for every
closed $L\subseteq K\setminus F\}$.}

\medskip

{\bf (b)} $F_1\cap F_2\in\Cal F$ whenever $F_1$, $F_2\in\Cal F$.
\Prf\ If $L\subseteq K\setminus(F_1\cap F_2)$ is closed then $L\cap F_1$,
$L\cap F_2$ are disjoint compact subsets of the Hausdorff space $X$, so there
are disjoint open subsets $G_1$, $G_2$ of $X$ such that $L\cap F_1\subseteq G_1$
and $L\cap F_2\subseteq G_2$ (4A2Fh).   Now $L\setminus G_2$ is a closed subset
of $X$ disjoint from
$F_1$, so belongs to $\CalRbg(X)$, and similarly $L\setminus G_1\in\CalRbg(X)$,
so $L=(L\setminus G_1)\cup(L\setminus G_2)$ belongs to $\CalRbg(X)$.   As $L$ is
arbitrary, $F_1\cap F_2\in\Cal F$.\ \Qed

\medskip

{\bf (c)} $K^*=\bigcap\Cal F$ belongs to $\Cal F$.   \Prf\ Since $K\in\Cal F$,
$K^*\subseteq K$ and $K^*$ is closed.   If
$L\subseteq K\setminus K^*$ is closed, therefore compact, there must be a finite
subset $\Cal F_0$ of $\Cal F$ such that $L\cap\bigcap\Cal F_0$ is empty;  we can
take it that $K\in\Cal F_0$, and now (b) assures us that
$\bigcap\Cal F_0\in\Cal F$ so $L\in\CalRbg(X)$.   As $L$ is arbitrary,
$K^*\in\Cal F$.\ \Qed

\medskip

{\bf (d)} $K^*$ has no isolated point.   \Prf\Quer\ If $x\in K^*$ is an isolated
point of $K^*$, set $F=K^*\setminus\{x\}$.   Then $F$ is a closed subset of
$K$ not belonging to $\Cal F$, so there is a closed set
$L\subseteq K\setminus F$ which does not belong to $\CalRbg(X)$.
Now $\{x\}$ certainly belongs to $\CalRbg(X)$, so by 534D(b-iii) there is an
open set $H$ containing $x$ such that $L\setminus H\notin\CalRbg(X)$.   But
$L\setminus H$ is a closed subset of $K\setminus K^*$ and $K^*\in\Cal F$,
by (c).\ \Bang\Qed

\medskip

{\bf (e)} If $K$ is scattered, then $K^*$ must be empty,
$K\subseteq K\setminus K^*$ and $K\in\CalRbg(X)$.

\medskip

{\bf (f)} Finally, if $K$ is not scattered then there is a continuous surjection
from $K$ to $[0,1]$ (4A2G(j-iv));  now $[0,1]\notin\CalSmz(\Bbb R,\rho)$, where
$\rho$ is the usual metric on $\Bbb R$, by 534H, so
$[0,1]\notin\CalRbg(\Bbb R)$ (534Sa), $K\notin\CalRbg(K)$ (534D(b-ii))
and $K\notin\CalRbg(X$ (534D(b-iv)).
}%end of proof of 534J

\vleader{48pt}{534K}{Theorem}\dvAformerly{5{}34H}
Let $X$ be a $\sigma$-compact locally compact
Hausdorff topological group and $A$ a subset of $X$.   Then the
following are equiveridical:

(i) $A\in\CalRbg(X)$;

(ii) for any sequence $\sequencen{U_n}$ of neighbourhoods of the
identity $e$ of $X$, there is a sequence $\sequencen{x_n}$ in $X$ such
that $A\subseteq\bigcup_{n\in\Bbb N}U_nx_n$;

(iii) $FA\ne X$ for any nowhere dense set $F\subseteq X$;

(iv) $EA\ne X$ for any meager set $E\subseteq X$;

(v) $AF\ne X$ for any nowhere dense set $F\subseteq X$;

(vi) $AE\ne X$ for any meager set $E\subseteq X$.

\cmmnt{\medskip

\noindent{\bf Remark} For the general theory of topological groups see
\S4A5 and Chapter 44.   Readers unfamiliar with this theory, or
impatient with the extra discipline needed to deal with non-commutative
groups, may prefer to start by assuming that $X=\BbbR^2$, so that
every $xU$ becomes $x+U$, every $V^{-1}V$ becomes $V-V$, and the right
uniformity is the Euclidean metric uniformity.
}%end of comment

\proof{{\bf (i)$\Rightarrow$(ii)} Suppose that (i) is true, and that
$\sequencen{U_n}$ is any sequence of neighbourhoods of $e$.   Then
$\{\interior U_nx:x\in X\}$ is an open cover of $X$ for each $n$, so there
is a sequence $\sequencen{x_n}$ such that
$A\subseteq\bigcup_{n\in\Bbb N}U_nx_n$.

\medskip

{\bf (ii)$\Rightarrow$(i)} Suppose that (ii) is true, and that
$\sequencen{W_n}$ is any sequence in the right uniformity
$\Cal W$ of $X$ (4A5Ha).   Then
for each $n\in\Bbb N$ there is a neighbourhood $U_n$ of $e$ such that
$W_n\supseteq\{(x,y):xy^{-1}\in U_n\}$;  let $V_n$ be a neighbourhood
of $e$ such that $V_nV_n^{-1}\subseteq U_n$.   By (ii), there is a
sequence $\sequencen{x_n}$ such
that $A\subseteq\bigcup_{n\in\Bbb N}V_nx_n$.   Set $A_n=V_nx_n$ for
each $n$.   Then $A_nA_n^{-1}=V_nV_n^{-1}\subseteq U_n$, so
$A_n\times A_n\subseteq W_n$, for each $n$, while $\sequencen{A_n}$
covers $A$.   As $\sequencen{W_n}$ is arbitrary, $A\in\CalSmz(X,\Cal W)$.
By 534Eb, $A\in\CalRbg(X)$.

\medskip

{\bf (ii)$\Rightarrow$(iv)} Suppose that $A$ satisfies (ii), and that
$E\subseteq X$ is meager.

\medskip

\quad\grheada\
If $K\subseteq X$ is compact and nowhere dense, then
there is a sequence $\sequencen{U_n}$ of neighbourhoods of $e$ such
that $K'=\bigcap_{n\in\Bbb N}U_nK$ is still compact and nowhere dense.
\Prf\ By 443N(ii), there is a nowhere dense zero set $F\supseteq K$.
Now $F$ is a G$_{\delta}$ set;  suppose that $F=\bigcap_{n\in\Bbb N}G_n$
where $G_n$ is open for each $n$.   As $K\subseteq G_n$,
the open set $U'_n=\{x:xK\subseteq G_n\}$ (4A5Ei) contains $e$;  let
$U_n$ be a compact
neighbourhood of $e$ included in $U'_n$.   Then $U_nK\subseteq G_n$
for every $n$, so $K'=\bigcap_{n\in\Bbb N}U_nK\subseteq F$ is nowhere
dense, while $K'$ is compact (use 4A5Ef).\ \Qed

\medskip

\quad\grheadb\ Let $K\subseteq X$ be compact and nowhere dense and
$U$ a neighbourhood of $e$.   Then there is a neighbourhood $V$ of $e$ such
that for every $x\in X$ there is an $x'\in Ux$ such that
$Vx'\cap K=\emptyset$.   \Prf\ Let $\sequencen{U_n}$ be a sequence of
neighbourhoods of $e$ such that $K'=\bigcap_{n\in\Bbb N}U_nK$ is
compact and nowhere dense (($\alpha$) above).   Choose a sequence
$\sequencen{V_n}$ of compact neighbourhoods of $e$ such that
$V_0\subseteq U$ and $V_{n+1}V_{n+1}^{-1}\subseteq U_n\cap V_n$ for
each $n\in\Bbb N$.   Then $Y=\bigcap_{n\in\Bbb N}V_n$ is a compact subgroup
of $X$ (see the proof of 4A5S), and $YK=\bigcap_{n\in\Bbb N}V_nK$
(4A5Eh).   \Quer\ If for every $n\in\Bbb N$ there is an
$x_n\in X$ such that $V_n^{-1}x'\cap K\ne\emptyset$ for every
$x'\in Ux_n$, then, in particular, $V_n^{-1}x_n\cap K\ne\emptyset$, so
$x_n\in V_nK$.   Since $\sequencen{V_nK}$ is a non-increasing sequence
of compact sets, $\sequencen{x_n}$ has a cluster point

\Centerline{$x^*\in\bigcap_{n\in\Bbb N}V_nK=YK\subseteq K'$.}

Because $K'$ is nowhere dense, $V_1x^*\not\subseteq K'$;  take
$x\in V_1x^*\setminus K'$.   Let $W$ be an open
neighbourhood of $e$ such that $Wx\cap K'=\emptyset$.   Then $Wx$ is
disjoint from $YK=Y^{-1}YK$ so $YWx\cap YK=\emptyset$.   Now
$YW$ is an open set including $Y=\bigcap_{n\in\Bbb N}V_n$, and
all the $V_n$ are compact, so there is an $m\ge 1$ such that
$V_m\subseteq Y W$ and $V_mx\cap YK=\emptyset$.

But observe that there is an $n>m$ such that
$x_n\in V_1x^*$, so that

\Centerline{$x\in V_1V_1^{-1}x_n\subseteq V_0x_n\subseteq Ux_n$,}

\noindent while $V_n^{-1}x\cap K\subseteq V_mx\cap YK$ is empty.\
\Bang

Thus we can take $V=V_n^{-1}$ for some $n$.\ \Qed

\medskip

\quad\grheadc\ Because $X$ is
$\sigma$-compact, any F$_{\sigma}$ set in $X$ is actually K$_{\sigma}$,
and there is
a sequence $\sequencen{K_n}$ of nowhere dense compact sets covering $E$;
we can suppose that $\sequencen{K_n}$ is non-decreasing.   Choose
inductively sequences $\sequencen{U_n}$, $\sequencen{V_n}$,
$\sequencen{V'_n}$ and $\sequencen{V''_n}$ of neighbourhoods of $e$
such that

\inset{$U_0$ is any compact neighbourhood of $e$,

given $U_n$, $V_n$ is to be a neighbourhood of $e$ such that
$V_nV_n\subseteq U_n$,

given $V_n$, $V'_n$ is to be a neighbourhood of $e$ such that for
every $y\in X$ there is a $z\in V_ny$ such that $V'_nz\cap
K_{n+1}=\emptyset$}

\noindent (using ($\beta$)),

\inset{given $V'_n$, $V''_n$ is to be an open neighbourhood of $e$
such that $(V''_n)^{-1}V''_n\subseteq V'_n$,

given $V''_n$, $U_{n+1}$ is to be a compact neighbourhood of $e$,
included in $V_n\cap V''_n$, such that
$K_{n+1}U_{n+1}\subseteq V''_nK_{n+1}$.}

\noindent (This last is possible by 4A5Ei, because $V''_nK_{n+1}$ is
an open set including $K_{n+1}$,
so $\{x:K_{n+1}x\subseteq V''_nK_{n+1}\}$ is
an open set containing $e$.)

\medskip

\quad\grheadd\ For each $k\in\Bbb N$, $\sequence{i}{U_{2^k(2i+1)}}$
is a sequence of neighbourhoods of $e$, so there must be a sequence
$\sequence{i}{x_{ki}}$ such that
$A\subseteq\bigcup_{i\in\Bbb N}U_{2^k(2i+1)}x_{ki}$.
Set $x_0=e$ and $x_n=x_{ki}$ if $n=2^k(2i+1)$.   For any $k\in\Bbb N$,

\Centerline{$A\subseteq\bigcup_{i\in\Bbb N}A_{ki}
\subseteq\bigcup_{i\in\Bbb N}U_{2^k(2i+1)}x_{ki}
\subseteq\bigcup_{n\ge 2^k}U_nx_n\subseteq\bigcup_{n\ge k}U_nx_n$.}

\noindent This means that $EA\subseteq\bigcup_{n\ge 1}K_nU_nx_n$.
\Prf\ If $z\in EA$, we can express it as $xy$ where $x\in E$ and
$y\in A$.   There are a $k\ge 1$ such that $x\in K_k$ and an
$n\ge k$ such that $y\in U_nx_n$, in which case $z\in K_nU_nx_n$.\
\Qed

\medskip

\quad\grheade\ Now choose $\sequencen{y_n}$, $\sequencen{z_n}$ as
follows.   Start from $y_0=e$.   Given $y_n$, let
$z_n\in V_ny_nx_{n+1}^{-1}$ be such that
$V'_nz_n\cap K_{n+1}=\emptyset$;  this is possible by the choice of
$V'_n$.   Now set $y_{n+1}=z_nx_{n+1}$, and continue.

For each $n$,

$$\eqalignno{U_{n+1}y_{n+1}
&\subseteq V_ny_{n+1}\cr
\displaycause{by the choice of $U_{n+1}$}
&=V_nz_nx_{n+1}
\subseteq V_nV_ny_nx_{n+1}^{-1}x_{n+1}\cr
\displaycause{by the choice of $z_n$}
&\subseteq U_ny_n\cr}$$

\noindent by the choice of $V_n$.   Consequently,
$U_{n+1}y_{n+1}\cap K_{n+1}U_{n+1}x_{n+1}=\emptyset$.   \Prf\ We chose
$z_n$ such that $V'_nz_n\cap K_{n+1}=\emptyset$.   Because
$(V''_n)^{-1}V''_n\subseteq V'_n$,
$V''_nz_n\cap V''_nK_{n+1}=\emptyset$.   Because
$K_{n+1}U_{n+1}\subseteq V''_nK_{n+1}$ and $U_{n+1}\subseteq V''_n$,
$U_{n+1}z_n\cap K_{n+1}U_{n+1}=\emptyset$, that is,
$U_{n+1}y_{n+1}\cap K_{n+1}U_{n+1}x_{n+1}=\emptyset$.\ \Qed

\medskip

\quad{\bf ($\pmb{\zeta}$)} From ($\epsilon$) we see that
$\sequencen{U_ny_n}$ is a non-increasing sequence of compact sets, so
has non-empty intersection.   Take any $x\in\bigcap_{n\in\Bbb
N}U_ny_n$.
Then $x\notin K_{n+1}U_{n+1}x_{n+1}$ for any $n$, so
$x\notin\bigcup_{n\ge 1}K_nU_nx_n\supseteq EA$.   Thus $EA\ne X$.   As
$E$ is arbitrary, (iv) is true.

\medskip

{\bf (iv)$\Rightarrow$(iii)} is trivial.

\medskip

{\bf (iii)$\Rightarrow$(ii)} Suppose that (iii) is true.   Let
$\sequencen{U_n}$ be any sequence of open neighbourhoods of $e$.
Then there is a sequence $\sequencen{x_n}$ in $X$ such that
$G=\bigcup_{n\in\Bbb N}x_nU_n^{-1}$ is dense.   \Prf\ Let
$\sequencen{V_n}$ be a sequence of neighbourhoods of $e$ such that
$V_{n+1}V_{n+1}^{-1}\subseteq V_n\cap U_n^{-1}$ for every $n\in\Bbb N$.
Then there is a compact normal subgroup $Y$ of $X$ such that
$Y\subseteq\bigcap_{n\in\Bbb N}V_n$ and $X/Y$ is metrizable (4A5S).
The canonical map $x\mapsto x^{\ssbullet}:X\to X/Y$ is continuous, so
$X/Y$ is $\sigma$-compact, therefore separable (4A2P(a-ii)).   Let
$\sequencen{x_n}$ be a sequence in $X$ such that
$\{x_n^{\ssbullet}:n\in\Bbb N\}$ is dense in $X/Y$.   Set
$G_0=\bigcup_{n\in\Bbb N}x_nV_{n+1}Y$.   \Quer\ If
$H=X\setminus\overline{G}_0$ is non-empty, then
$\{x^{\ssbullet}:x\in H\}$
is open (4A5Ja) so contains $x_n^{\ssbullet}$ for some $n$.   But
$x_nY\subseteq x_nV_{n+1}Y\subseteq G_0$, so there can be no $x\in H$
such that $x^{\ssbullet}=x_n^{\ssbullet}$.\ \BanG\  Thus $G_0$ is
dense. But, for any $n\in\Bbb N$,
$Y\subseteq V_{n+1}^{-1}$ so $V_{n+1}Y\subseteq U_n^{-1}$, and
$G=\bigcup_{n\in\Bbb N}x_nU_n^{-1}$ includes $G_0$.   Thus $G$ is dense,
as required.\ \Qed

Accordingly $F=X\setminus G$ is nowhere dense, and
$FA\ne X$;  suppose $x\in X\setminus FA$.   Then
$F\cap xA^{-1}=\emptyset$, that is,
$xA^{-1}\subseteq\bigcup_{n\in\Bbb N}x_nU_n^{-1}$, that is,
$A^{-1}\subseteq\bigcup_{n\in\Bbb N}x^{-1}x_nU_n^{-1}$, that is,
$A\subseteq\bigcup_{n\in\Bbb N}U_nx_n^{-1}x$.   As $\sequencen{U_n}$ is
arbitrary, (ii) is true.

\medskip

{\bf (i)$\Leftrightarrow$(v)$\Leftrightarrow$(vi)}
Because $x\mapsto x^{-1}$ is a homeomorphism,

$$\eqalignno{A\in\CalRbg(X)&
\Longrightarrow A^{-1}\in\CalRbg(X)\cr&
\Longrightarrow EA^{-1}\ne X
  \text{ whenever }E\subseteq X\text{ is meager}\cr
& \Longrightarrow E^{-1}A^{-1}\ne X
  \text{ whenever }E\subseteq X\text{ is meager}\cr
\displaycause{because $E^{-1}$ is meager if $E$ is}
&\mskip-6mu\iff AE\ne X\text{ whenever }E\subseteq X
  \text{ is meager}\cr
& \Longrightarrow AF^{-1}\ne X\text{ whenever }F\subseteq X
  \text{ is nowhere dense}\cr
\displaycause{because $F^{-1}$ is nowhere dense if $F$ is}
& \Longrightarrow FA^{-1}\ne X\text{ whenever }F\subseteq X
  \text{ is nowhere dense}\cr
& \Longrightarrow A^{-1}\in\CalRbg(X)\cr&
\Longrightarrow A\in\CalRbg(X).\cr}$$
}%end of proof of 534K

\cmmnt{\medskip

\noindent{\bf Remark} The case $X=\Bbb R$ is due to {\smc Galvin
Mycielski \& Solovay 79}.
}%end of comment

\leader{534L}{Proposition}\dvAformerly{5{}34J}
\cmmnt{ ({\smc Fremlin 91})} Let
$(X,\rho)$ be a separable metric space.
Then $\CalSmz(X,\rho)\prT\Cal N^{\frak d}$\cmmnt{,
where $\Cal N$ is the null ideal
of Lebesgue measure on $\Bbb R$ and $\frak d$ is the dominating
number (522A)}.

\proof{{\bf (a)} By 534A, there is a countable family $\Cal C$ of
subsets of $X$ such that whenever $A\subseteq X$ has finite diameter
and $\eta>0$, there is a $C\in\Cal C$ such that $A\subseteq C$ and
$\diam C\le\eta+2\diam A$.   For each $i\in\Bbb N$, let
$\sequence{j}{C_{ij}}$ be a sequence running over $\{C:C\in\Cal C$,
$\diam C\le 2^{-i}\}$.   Let $(\NN,\subseteq^*,\Cal S)$ be the
$\Bbb N$-localization relation.

\medskip

{\bf (b)} Let $D\subseteq\NN$ be a cofinal set with cardinal $\frak d$.
For each $d\in D$ we can find a function $\phi_d:\CalSmz(X,\rho)\to\NN$ such
that $A\subseteq\bigcap_{n\in\Bbb N}\bigcup_{i\ge n}
C_{d(i),\phi_d(A)(i)}$ for every $A\in\CalSmz(X,\rho)$.   \Prf\ For
$A\in\CalSmz(X,\rho)$ and $k\in\Bbb N$, choose a sequence
$\sequence{i}{A_{ki}}$
of sets covering $A$ such that $2\diam A_{ki}<2^{-d(2^k(2i+1))}$ for
every $i\in\Bbb N$.   For $n=2^k(2i+1)$, let $A_n\in\Cal C$ be such
that $A_{ki}\subseteq A_n$ and $\diam A_n\le 2^{-d(n)}$;  choose
$\phi_d(A)(n)$ such that $A_n=C_{d(n),\phi_d(A)(n)}$.\ \QeD\  Define
$\phi:\CalSmz(X,\rho)\to(\NN)^D$ by setting
$\phi(A)=\family{d}{D}{\phi_d(A)}$
for $A\in\CalSmz(X,\rho)$.

\medskip

{\bf (c)} For $S\in\Cal S$ and $d\in D$, define

\Centerline{$\psi_d(S)
=\bigcap_{n\in\Bbb N}\bigcup_{i\ge n}\bigcup_{j\in S[\{i\}]}C_{d(i),j}
\subseteq X$.}

\noindent For $\family{d}{D}{S_d}\in\Cal S^D$ set
$\psi(\family{d}{D}{S_d})=\bigcap_{d\in D}\psi_d(S_d)$.   Then
$A=\psi(\family{d}{D}{S_d})$ has strong measure zero.   \Prf\ Let
$\sequence{i}{\epsilon_i}$ be any family of strictly positive real
numbers.   Let $d\in D$ be such that $2^{-d(k)}\le\epsilon_i$ whenever
$k\in\Bbb N$ and $i<2^{k+1}$.   For each $k\in\Bbb N$,
$\#(S_d[\{k\}])\le 2^k$, so we can find a sequence $\sequence{i}{A_i}$
such that
$\langle A_i\rangle_{2^k\le i<2^{k+1}}$ is a re-enumeration of
$\family{j}{S[\{k\}]}{C_{d(k),j}}$ supplemented by empty sets if
necessary.   This will ensure that if $2^k\le i<2^{k+1}$ then $\diam
A_i\le 2^{-d(k)}\le\epsilon_i$, while

\Centerline{$A\subseteq\psi_d(S_d)
\subseteq\bigcup_{k\in\Bbb N}\bigcup_{j\in S_d[\{k\}]}C_{d(k),j}
=\bigcup_{(k,j)\in S_d}C_{d(k),j}
=\bigcup_{i\in\Bbb N}A_i$.}

\noindent As $\sequence{i}{\epsilon_i}$ is arbitrary, $A\in\CalSmz(X,\rho)$.\
\Qed

\medskip

{\bf (d)} Taking
$(\BbbN^{\Bbb N},\subseteq^*,\Cal S)$ to be the $\Bbb N$-localization relation,
as in the proof of 534B, $(\phi,\psi)$ is a Galois-Tukey connection from
$(\CalSmz(X,\rho),\subseteq,\CalSmz(X,\rho))$ to
$(\NN,\penalty-100\subseteq^*\nobreak,\penalty-100\Cal S)^D$, that is,
$((\NN)^D,T,\penalty-100\Cal S^D)$,
where $T$ is the simple product relation as defined in 512H.   \Prf\
$\phi:\CalSmz(X,\rho)\to(\NN)^D$
and $\psi:\Cal S^D\to\CalSmz(X,\rho)$ are functions.   Suppose that
$A\in\CalSmz(X,\rho)$ and $\family{d}{D}{S_d}$ are such that
$(\phi(A),\family{d}{D}{S_d})\in T$, that is,
$\phi_d(A)\subseteq^*S_d$
for every $d$.   Fix $d\in D$ for the moment.   Then there is an
$n\in\Bbb N$ such that $(i,\phi_d(A)(i))\in S_d$ for every $i\ge n$.
Now, for any $m\ge n$,

\Centerline{$A\subseteq\bigcup_{i\ge m}C_{d(i),\phi_d(A)(i)}
\subseteq\bigcup_{i\ge m}\bigcup_{j\in S_d[\{i\}]}C_{d(i),j}$.}

\noindent Thus

\Centerline{$A\subseteq
\bigcap_{m\in\Bbb N}\bigcup_{i\ge m}\bigcup_{j\in
S_d[\{i\}]}C_{d(i),j)}
=\psi_d(S_d)$.}

\noindent This is true for every $d$, so
$A\subseteq\psi(\family{d}{D}{S_d})$.   As $A$ and
$\family{d}{D}{S_d}$
are arbitrary, $(\phi,\psi)$ is a Galois-Tukey connection.\ \Qed

\medskip

{\bf (e)} Thus
$(\CalSmz(X,\rho),\subseteq,\CalSmz(X,\rho))
\prGT(\NN,\subseteq^*,\Cal S)^D$.   But
$(\NN,\subseteq^*,\Cal S)
\equivGT(\Cal N,\penalty-100\subseteq\nobreak,\penalty-100\Cal N)$
(522M), so $(\NN,\subseteq^*,\penalty-100\Cal S)^D
\equivGT(\Cal N,\subseteq,\Cal N)^D$ (512Hb) and

\Centerline{$(\CalSmz(X,\rho),\subseteq,\CalSmz(X,\rho))
\prGT(\Cal N,\subseteq,\Cal N)^D=(\Cal N^D,\le,\Cal N^D)$}

\noindent where $\le$ is the natural partial order of the product
partially ordered set $\Cal N^D$.   Accordingly
$\CalSmz(X,\rho)\prT\Cal N^D\cong\Cal N^{\frak d}$, as claimed.
}%end of proof of 534L

\leader{534M}{Corollary}\dvAformerly{5{}34K}
(a) If $(X,\Cal W)$ is a Lindel\"of uniform
space, then $\add\CalSmz(X,\Cal W)\ge\add\Cal N$,
where $\Cal N$ is the null ideal of Lebesgue measure on $\Bbb R$.

(b) If $X$ is a Lindel\"of regular topological space, then
$\add\CalRbg(X)\ge\add\Cal N$.

\proof{{\bf (a)(i)} If $(X,\rho)$ is a separable metric space, then
534L tells us that $\CalSmz(X,\rho)\prT\Cal N^{\frak d}$, so
$\add\CalSmz(X,\rho)\ge\add\Cal N^{\frak d}=\add\Cal N$ (513E(e-ii), 511Hg).

\medskip

\quad{\bf (ii)} In general, if $\Cal A\subseteq\CalSmz(X,\Cal W)$ and
$\#(\Cal A)<\add\Cal N$, take any metric space $(Y,\rho)$ and uniformly
continuous $f:X\to Y$.   Then $f[X]$ is Lindel\"of (5A4Bc), therefore separable
(4A2Pc), and $f[A]$ has strong measure zero in
$f[X]$ for every $A\in\Cal A$ (534D(a-iii)), so
$f[\bigcup\Cal A]=\bigcup_{A\in\Cal A}f[A]$ has strong measure zero, by (i).
As $f$ is arbitrary, $\bigcup\Cal A$ has strong measure zero, by 534D(a-iv);
as $\Cal A$ is arbitrary, $\add\CalSmz(X,\Cal W)\ge\add\Cal N$.

\medskip

{\bf (b)} Being Lindel\"of and regular, $X$ is paracompact and normal
(4A2H(b-i)), so there is a uniformity $\Cal W$ on $X$, inducing its topology,
with $\CalRbg(X)=\CalSmz(X,\Cal W)$ (534F);  so
$\add\CalRbg(X)=\add\CalSmz(X,\Cal W)\ge\add\Cal N$, by (a).
}%end of proof of 534M

\leader{534N}{}{\bf $\CalSmz$-equivalence (a)}\dvAformerly{5{}34L}
 If $(X,\Cal V)$ and
$(Y,\Cal W)$ are uniform spaces, \cmmnt{I say that} they are
{\bf $\CalSmz$-equivalent} if there is a bijection $f:X\to Y$ such
that
a set $A\subseteq X$ has strong measure zero in $X$ iff $f[A]$ has
strong measure zero in $Y$.   \cmmnt{Of course this is an
equivalence relation on the class of uniform spaces.}

\spheader 534Nb If $(X,\Cal V)$ and $(Y,\Cal W)$ are uniform
spaces,\cmmnt{ I
say that} $X$ is {\bf $\CalSmz$-embeddable} in $Y$ if it is
$\CalSmz$-equivalent to a subspace of $Y$\cmmnt{ (with the subspace
uniformity, of course)}.   Evidently this is transitive in the sense
that if $X$ is $\CalSmz$-embeddable in $Y$ and $Y$ is
$\CalSmz$-embeddable in $Z$ then $X$ is $\CalSmz$-embeddable in $Z$.

\leaveitout{\spheader 534Nc When $X$ and $Y$ are topological spaces,\cmmnt{ I
will say that} they are {\bf $\CalRbg$-equivalent} if there is a
bijection $f:X\to Y$ such that
$A\subseteq X$ has Rothberger's property in $X$ iff $f[A]$ has
Rothberger's property in $Y$.
}

\leader{534O}{Lemma}\dvAformerly{5{}34M}
(a) Suppose that $(X,\Cal W$) and $(Y,\Cal V)$
are uniform spaces and that $\sequencen{X_n}$, $\sequencen{Y_n}$ are
partitions of $X$, $Y$ respectively such that $X_n$ is
$\CalSmz$-equivalent to $Y_n$ for every $n$.   Then $X$ is
$\CalSmz$-equivalent to $Y$.

(b) Suppose that $(X,\Cal W)$ and $(Y,\Cal V)$ are uniform spaces such that
$X$ is $\CalSmz$-embeddable in $Y$ and $Y$ is $\CalSmz$-embeddable
in $X$.   Then $(X,\Cal W)$ and $(Y,\Cal V)$ are $\CalSmz$-equivalent.

\proof{{\bf (a)} For each $n\in\Bbb N$, let $f_n:X_n\to Y_n$ be a
bijection identifying the ideals of sets with strong measure zero.
Then $f=\bigcup_{n\in\Bbb N}f_n$ is a bijection identifying
$\CalSmz(X,\Cal W)$ and $\CalSmz(Y,\Cal V)$.

\medskip

{\bf (b)} (Compare 344D.) Let $X_1\subseteq X$ and $Y_1\subseteq Y$ be
$\CalSmz$-equivalent to $Y$, $X$ respectively;  let $f:X\to Y_1$ and
$g:Y\to X_1$ be bijections identifying the ideals of strong measure
zero in each pair.   Set $X_0=X$, $Y_0=Y$, $X_{n+1}=g[Y_n]$ and
$Y_{n+1}=f[X_n]$ for
each $n\ge 1$;  then $\sequencen{X_n}$ is a non-increasing sequence
of subsets of $X$ and $\sequencen{Y_n}$ is a non-increasing sequence of
subsets of $Y$.   Set $X_{\infty}=\bigcap_{n\in\Bbb N}X_n$,
$Y_{\infty}=\bigcap_{n\in\Bbb N}Y_n$.   Then
$f\restr X_{2k}\setminus X_{2k+1}$ is an $\CalSmz$-equivalence between
$X_{2k}\setminus X_{2k+1}$ and $Y_{2k+1}\setminus Y_{2k+2}$, while
$g\restrp Y_{2k}\setminus Y_{2k+1}$ is an $\CalSmz$-equivalence
between
$Y_{2k}\setminus Y_{2k+1}$ and $X_{2k+1}\setminus X_{2k+2}$;  and
$g\restrp Y_{\infty}$ is an $\CalSmz$-equivalence between $Y_{\infty}$
and $X_{\infty}$.    So (a) gives the required $\CalSmz$-equivalence
between $X$ and $Y$.
}%end of proof of 534O

\leader{534P}{Proposition}\dvAformerly{5{}34N}
 $\BbbR^r$, $\ooint{0,1}^r$, $[0,1]^r$ and $\{0,1\}^{\Bbb N}$
are $\CalSmz$-equivalent for every integer $r\ge 1$.

\proof{ As these spaces are $\sigma$-compact and completely regular, we do not
have to specify the uniformities we are thinking of, by 534Eb;  in each case,
the sets with strong measure zero are the sets with Rothberger's property.

\medskip

{\bf (a)} Give $\Bbb R$ its usual metric $\rho$.   Of course the
identity maps are $\CalSmz$-embeddings of
$\ooint{0,1}$ in $[0,1]$ and $[0,1]$ in $\Bbb R$.
To complete the circuit, use 534Eb;
any homeomorphism between $\Bbb R$ and $\ooint{0,1}$ matches
$\CalRbg(\Bbb R)=\CalSmz(\Bbb R,\rho)$ with
$\CalRbg(\ooint{0,1})=\CalSmz(\ooint{0,1},\rho)$.
By 534Ob, $\Bbb R$ and $[0,1]$ and $\ooint{0,1}$ are
$\CalSmz$-equivalent.

\medskip

{\bf (b)} Give $\{0,1\}^{\Bbb N}$ the metric $\rho$ defined by saying
that

\Centerline{$\rho(x,y)
=\inf\{2^{-n}:n\in\Bbb N$, $x\restr n=y\restr n\}$}

\noindent for $x$, $y\in\{0,1\}^{\Bbb N}$.   Define
$f:\{0,1\}^{\Bbb N}\to[0,1]$ by setting
$f(x)=\sum_{n=0}^{\infty}2^{-n-1}x(n)$ for $x\in\{0,1\}^{\Bbb N}$.
Then $f$ is continuous, therefore uniformly continuous, so $f[A]$ has
strong measure zero in $[0,1]$ whenever $A\subseteq\{0,1\}^{\Bbb N}$
has strong measure zero in $\{0,1\}^{\Bbb N}$.   It is also the case that
$f^{-1}[B]$ has strong measure zero whenever $B\subseteq[0,1]$ does.
\Prf\ Let $\sequencen{\epsilon_n}$ be any sequence of strictly
positive
numbers.   Then there is a sequence $\sequencen{B_n}$, covering $B$,
such that $\diam B_n<\bover12\min(1,\epsilon_{2n},\epsilon_{2n+1})$
for every $n$.   Fix $n$ for the moment and consider $f^{-1}[B_n]$.   If
$k$ is such that $2^{-k-1}\le\diam B_n<2^{-k}$, then $B_n$ can meet at
most two intervals of the type $I_{ki}=[2^{-k}i,2^{-k}(i+1)]$.   So
$f^{-1}[B_n]$
can meet at most two sets of the type $\{x:x\restr k=z\}$, and we can
express it as $A_{2n}\cup A_{2n+1}$ where

\Centerline{$\max(\diam A_{2n},\diam A_{2n+1})
\le 2^{-k}\le 2\diam B_n\le\min(\epsilon_{2n},\epsilon_{2n+1})$.}

\noindent Putting these together, we have a cover $\sequencen{A_n}$ of
$\bigcup_{n\in\Bbb N}f^{-1}[B_n]\supseteq f^{-1}[B]$ such that
$\diam A_n\le\epsilon_n$ for every $n$;  as $\sequencen{\epsilon_n}$
is arbitrary, $f^{-1}[B]$ has strong measure zero.\ \Qed

Of course $f$ is not a bijection, so it is not in itself an
$\CalSmz$-equivalence.   But if we set

\Centerline{$D_1
=\{x:x\in\{0,1\}^{\Bbb N}$, $x$ is eventually constant$\}$,}

\Centerline{$D_2=\{2^{-k}i:k\in\Bbb N$, $i\le 2^k\}$,}

\noindent then $D_1\subseteq\{0,1\}^{\Bbb N}$ and $D_2\subseteq[0,1]$
are countably infinite, and $f\restr\{0,1\}^{\Bbb N}\setminus D_1$ is
an $\CalSmz$-equivalence between $\{0,1\}^{\Bbb N}\setminus D_1$ and
$[0,1]\setminus D_2$.   Putting this together with any bijection
between $D_1$ and $D_2$, we have an $\CalSmz$-equivalence between
$\{0,1\}^{\Bbb N}$ and $[0,1]$.

\medskip

{\bf (c)(i)} I show by induction on $r$ that $[0,1]^r$ is
$\CalSmz$-equivalent to $\Bbb R$ and therefore to $[0,1]$.   The case
$r=1$ is covered by (a).   For the inductive step to $r\ge 2$, I adapt
the method of (b).   Give $\{0,1\}^{\Bbb N\times r}$ the metric $\rho$
defined by setting

\Centerline{$\rho(x,y)
=\inf\{2^{-n}:n\in\Bbb N$, $x\restr(n\times r)=y\restr(n\times r)\}$}

\noindent for $x$, $y\in\{0,1\}^{\Bbb N\times r}$.   Define
$f:\{0,1\}^{\Bbb N\times r}\to[0,1]^r$ by setting

\Centerline{$f(x)=\ofamily{j}{r}{\sum_{i=0}^{\infty}2^{-i-1}x(i,j)}$}

\noindent for $x\in \{0,1\}^{\Bbb N\times r}$.   Then $f$ is uniformly
continuous, so $f[A]$ has strong measure zero in $[0,1]^r$ whenever
$A$
has strong measure zero in $\{0,1\}^{\Bbb N\times r}$.   Moreover, we
find once again that $f^{-1}[B]$ has strong measure zero whenever
$B\subseteq[0,1]^r$ has strong measure zero.   \Prf\ Let
$\sequencen{\epsilon_n}$ be a sequence of strictly positive real
numbers.   This time, set $m=2^r$ and let $\sequence{n}{B_n}$ be a
cover of $B$ such that
$\diam B_n<\bover12\min(1,\inf_{mn\le i<mn+m}\epsilon_i)$ for every
$n$.   (For definiteness, let me say that I am giving $[0,1]^r$ its
Euclidean
metric.)   In this case, if $2^{-k-1}\le\diam B_n<2^{-k}$, $B_n$ can
meet at most $2^r$ intervals of the form
$[2^{-k}\pmb{n},2^{-k}(\pmb{n}+\pmb{1})]$ where $\pmb{n}\in\BbbN^r$ and
$\pmb{1}=(1,\ldots,1)$.   So $f^{-1}[B_n]$ can meet at most $2^r=m$ sets
of the form $\{x:x\restr(k\times r)=z\}$, and can be covered by $m$
sets $\langle A_j\rangle_{mn\le j<mn+m}$ where

\Centerline{$\diam A_j\le 2^{-k}\le 2\diam B_n\le\epsilon_j$}

\noindent for every $j$.   Putting these together, we have a cover
$\sequence{j}{A_j}$ of $f^{-1}[B]$ such that $\diam A_j\le\epsilon_j$
for every $j$;  as $\sequencen{\epsilon_n}$ is arbitrary, $f^{-1}[B]$
has strong measure zero.\ \Qed

The function $f$ here is very far from being one-to-one.   But if we
set

\Centerline{$D_1^*=\bigcup_{j<r}\{x:x\in\{0,1\}^{\Bbb N\times r}$,
$\sequence{i}{x(i,j)}\in D_1\}$,}

\Centerline{$D_2^*=\bigcup_{j<r}\{z:z\in[0,1]^r$, $z(j)\in D_2\}$,}

\noindent where $D_1\subseteq\{0,1\}^{\Bbb N}$, $D_2\subseteq[0,1]$
are defined as in the proof of (b), then $f$ is a bijection between
$\{0,1\}^{\Bbb N\times r}\setminus D^*_1$ and
$[0,1]^r\setminus D^*_2$,
so is an $\CalSmz$-equivalence between these.   Accordingly
$[0,1]^r\setminus D^*_2$ is $\CalSmz$-embeddable in
$\{0,1\}^{\Bbb N\times r}$, which is homeomorphic, therefore uniformly
equivalent, to $\{0,1\}^{\Bbb N}$, which is in turn
$\CalSmz$-equivalent
to $\ooint{0,1}$;  so  $[0,1]^r\setminus D^*_2$ is
$\CalSmz$-embeddable in $\ooint{0,1}$.

Now consider $D_2^*$.   This is a countable union of sets which are
isometric, therefore $\CalSmz$-equivalent, to $[0,1]^{r-1}$ and
therefore to $\ooint{0,1}$, by the inductive hypothesis.   We can
therefore express $D_2^*$ as $\bigcup_{n\in\Bbb N}X_n$ where
$\sequencen{X_n}$ is disjoint and every $X_n$ is $\CalSmz$-embeddable
in $\ooint{0,1}$ and therefore in $\ooint{n+1,n+2}$.   Assembling these
with the $\CalSmz$-equivalence between $[0,1]^r\setminus D_2^*$ and
$\ooint{0,1}$ we have already found, we have an $\CalSmz$-embedding
from $[0,1]^r$ to $\Bbb R$.   In the other direction, we certainly have an
isometric embedding of $[0,1]$ in $[0,1]^r$ and therefore a
$\CalSmz$-embedding of $\Bbb R$ in $[0,1]^r$;  so $\Bbb R$ and
$[0,1]^r$ are $\CalSmz$-equivalent.   Thus the induction proceeds.

\medskip

\quad{\bf (ii)} As for $\BbbR^r$, we have a homeomorphism between
$\BbbR^r$ and $\ooint{0,1}^r$, which (because these again are
$\sigma$-compact) is an $\CalSmz$-equivalence and therefore an
$\CalSmz$-embedding of $\BbbR^r$ in $[0,1]^r$.   So 534Ob, once more,
tells us that $\BbbR^r$ and $[0,1]^r$ and $[0,1]$ are
$\CalSmz$-equivalent.

\medskip

{\bf (d)} Thus $\BbbR^r$, $\ooint{0,1}^r$, $[0,1]^r$ and
$\{0,1\}^{\Bbb N}$ are $\CalSmz$-equivalent, for any uniformities inducing their
usual topologies.
}%end of proof of 534P

\leader{534Q}{Large sets with strong measure
\dvrocolon{zero}}\cmmnt{ It is a remarkable fact that it is relatively
consistent with ZFC to
suppose that the only subsets of $\Bbb R$ with strong measure zero are
the countable sets ({\smc Laver 76}, {\smc Ihoda 88} or
{\smc Bartoszy\'nski \& Judah 95}, \S8.3).   We therefore find
ourselves
investigating constructions of non-trivial sets with strong measure
zero under special axioms.

\medskip

\noindent}{\bf Proposition}\dvAformerly{5{}34N}
(a) Let $X$ be a Lindel\"of space.   Then $\non\CalRbg(X)\ge\frakmctbl$.

(b)\cmmnt{ (see {\smc Fremlin \& Miller 88})}
Give $\NN$ the metric $\rho$ defined by setting
$\rho(x,y)
=\inf\{2^{-n}:n\in\Bbb N$, $x\restr n=y\restr n\}$ for $x$, $y\in\NN$.
Then $\non\CalSmz(\NN,\rho)=\non\CalRbg(\NN)=\frakmctbl$.

\proof{{\bf (a)} Suppose that $A\subseteq X$ and $\#(A)<\frakmctbl$.
Let $\sequencen{\Cal G_n}$ be a sequence of non-empty open covers of
$X$.   Because $X$ is Lindel\"of, we can choose for each $n$ a
non-empty
countable $\Cal G'_n\subseteq\Cal G_n$ covering $X$.
Let $P$ be the set of finite sequences $p=\ofamily{i}{n}{p(i)}$ such
that $p(i)\in\Cal G'_i$ for every $i<n$;  say
that $p\le q$ in $P$ if $q$ extends $P$.   Then $P$ is a countable
partially ordered set.   For each $x\in A$, the set
$Q_x=\{p:x\in p(i)$
for some $i<\#(p)\}$ is cofinal with $P$.   \Prf\ Given $p\in P$, set
$n=\#(p)$;  let $G\in\Cal G'_n$ be such that $x\in G$;  set
$q=p\cup\{(n,G)\}$;  then $p\le q\in Q_x$.\ \Qed

Because $\#(A)<\frakmctbl\le\frak m^{\uparrow}(P)$ (517Pc), there is
an upwards-directed family $R\subseteq P$ meeting every $Q_x$ (517B(iv)).
Now $p^*=\bigcup R$ is a function;  $A\subseteq\bigcup_{i\in\dom p^*}p^*(i)$
and $p^*(i)\in\Cal G_i$ for every $i\in\dom p^*$.   As
$\sequencen{\Cal G_n}$ is arbitrary, $A$ has Rothberger's property in
$X$;  as $A$ is arbitrary, $\non\CalRbg(X)\ge\frakmctbl$.

\medskip

{\bf (b)} By 522Sb, there is a set
$A\subseteq\NN$, with cardinal $\frakmctbl$, such that for every $y\in\NN$
there is an $x\in A$ such that $x(n)\ne y(n)$ for every $n$.   \Quer\
If $A\in\CalSmz(\NN,\rho)$, take a sequence $\sequencen{y_n}$ in $\NN$ such that
$A\subseteq\bigcup_{n\in\Bbb N}B(y_n,2^{-n-1})$.   Set $y(n)=y_n(n)$
for every $n$.   Then there is an $x\in A$ such that $x(n)\ne y(n)$ for
every $n$.   But in this case $x(n)\ne y_n(n)$ and
$x\restr n+1\ne y_n\restr n+1$ and $x\notin B(y_n,2^{-n-1})$ for every
$n$.\ \Bang

Thus $A$ witnesses that $\non\CalSmz(\NN,\rho)\le\frakmctbl$.   But we know from
(a) that $\non\CalRbg(\NN)\ge\frakmctbl$ and from 534Ea that
$\non\CalSmz(\NN,\rho)\ge\non\CalRbg(\NN)$, so the three cardinals are equal.
}%end of proof of 534Q
%re  cov(\CalSmz) see Barto & Judah 8.4.7

\leader{534R}{Proposition}\dvAformerly{5{}34O}
(a) If $(X,\rho)$ is a separable metric space and
$A\subseteq X$ has cardinal less than $\frak c$, there is a Lipschitz
function $f:X\to\Bbb R$ such that $f\restr A$ is injective.

(b)\cmmnt{ ({\smc Carlson 93})} If $\kappa<\frak c$ is a cardinal
and there is any separable metric space with a set with cardinal $\kappa$ which
is of strong measure zero, then there is a subset of $\Bbb R$ with cardinal
$\kappa$ which has Rothberger's property in $\Bbb R$.

(c)(i) If $\cf(\frakmctbl)=\frak b$ there is a
subset of $\Bbb R$ with cardinal $\frakmctbl$ which has Rothberger's property
in itself.

\quad(ii)\cmmnt{ ({\smc Rothberger 1941})}
If $\frak b=\omega_1$ there is a subset of $\Bbb R$ with cardinal
$\omega_1$ which has Rothberger's property in itself.

\quad(iii) If $\frakmctbl=\frak d$ there is a
subset of $\Bbb R$ with cardinal $\frakmctbl$ which has Rothberger's property
in itself.

\proof{{\bf (a)} If $X=\emptyset$ this is trivial.   Otherwise, let
$\sequencen{x_n}$ run over a dense sequence in $X$, and for $x\in X$
define $g_x:\Bbb R\to\Bbb R$ by setting

\Centerline{$g_x(t)
=\sum_{n=0}^{\infty}\Bover{\min(1,\rho(x,x_n))}{n!}t^n$}

\noindent for $t\in\Bbb R$.   Then $g_x$ is a real-entire function
(5A5A).   If $x$, $y\in X$ are distinct, then there must be some $n$
such that $\min(1,\rho(x,x_n))\ne\min(1,\rho(y,x_n))$, so that one of
the coefficients of $g_x-g_y$ is non-zero, and $\{t:g_x(t)=g_y(t)\}$
is countable (5A5A).   So if $A\subseteq X$ and $\#(A)<\frak c$, we can
find a $t\ge 0$ such that $g_x(t)\ne g_y(t)$ for all distinct $x$,
$y\in A$.   Set $f(x)=g_x(t)$ for $x\in X$;  then $f:X\to\Bbb R$ is a
function such that $f\restr A$ is injective.   If $x$, $y\in X$ then

$$\eqalign{|f(x)-f(y)|
&=|\sum_{n=0}^{\infty}(\min(1,\rho(x,x_n))-\min(1,\rho(y,y_n))
  \Bover{t^n}{n!}|\cr
&\le e^t\sup_{n\in\Bbb N}|\rho(x,x_n)-\rho(y,y_n)|
\le e^t\rho(x,y),\cr}$$

\noindent so that $f$ is Lipschitz.

\medskip

{\bf (b)} Let $(X,\rho)$ be a separable metric space with a set
$A\in[X]^{\kappa}$ of strong measure zero.   Then (a) tells us that we
have a uniformly continuous function $f:X\to\Bbb R$ which is injective
on $A$, so that $f[A]\in[\Bbb R]^{\kappa}$ has strong measure zero
in $\Bbb R$ (534D(a-iii)).

\medskip

{\bf (c)(i)} Let $\ofamily{\xi}{\frak b}{x_{\xi}}$ be a family
in $\NN$ which is increasing and
unbounded for the pre-order $\le^*$ of 522C(i).   Let
$C\subseteq\frakmctbl$ be a closed cofinal set with cardinal $\frak b$
(5A1Ae), and
$\ofamily{\xi}{\frak b}{\zeta_{\xi}}$ the increasing enumeration of $C$;
let $\ofamily{\eta}{\frakmctbl}{y_{\eta}}$ be a family of distinct elements
of $\NN$ such that $y_{\eta}\ge x_{\xi}$ whenever $\xi<\frak b$ and
$\zeta_{\xi}\le\eta<\zeta_{\xi+1}$.

If $K\subseteq\NN$ is compact, then $\{\eta:y_{\eta}\in K\}$ has cardinal
strictly less than $\frakmctbl$.   \Prf\ Set $x(n)=\sup_{y\in K}y(n)$ for
each $n\in\Bbb N$ (I pass over the trivial case $K=\emptyset$).   Then
there is a $\xi<\frak b$ such that $x_{\xi}\not\le^*x$.   If
$\zeta_{\xi}\le\eta<\frakmctbl$, there is a $\xi'\ge\xi$ such that
$\zeta_{\xi'}\le\eta<\zeta_{\xi'+1}$ (this is where we need to know that
$C$ is closed), and now

\Centerline{$y_{\eta}\ge x_{\xi'}\ge^*x_{\xi}$,
\quad$y_{\eta}\not\le x$,
\quad$y_{\eta}\notin K$.}

\noindent So $\{\eta:y_{\eta}\in K\}\subseteq\zeta_{\xi}$ has cardinal less
than $\frakmctbl$.\ \Qed

Let $f:\NN\to[0,1]\setminus\Bbb Q$ be any homeomorphism (4A2Ub), and
consider $A=\{f(y_{\eta}):\eta<\frakmctbl\}\cup\Bbb Q$.   Then $\#(A)=\frakmctbl$.
Also $A$ has Rothberger's property in $A$.   \Prf\ Of course $\Bbb Q$, being
countable, has Rothberger's property in $A$.   Let $G\subseteq\Bbb R$ be an
open set including $\Bbb Q$.   Then $[0,1]\setminus G$ and
$K=f^{-1}[\,[0,1]\setminus G]$ are compact.   Now

\Centerline{$\#(A\setminus G)
=\#(\{\eta:y_{\eta}\in K\})<\frakmctbl$}

\noindent so $A\setminus G$ has Rothberger's property in $A$, by 534Qa.   By
534D(b-iii), this is enough to show that $A$ has Rothberger's property
in itself.\ \Qed

Thus we have a set of the required kind.

\medskip

\quad{\bf (ii)} This follows immediately if $\frakmctbl=\omega_1$, and otherwise
we can take any subset of $\Bbb R$ of cardinal $\omega_1$.

\medskip

\quad{\bf (iii)} The argument is similar to that in (i).   This time, let
$\ofamily{\xi}{\frak d}{x_{\xi}}$ be a cofinal family in $\NN$.   For each
$\xi<\frak d$, let $y_{\xi}\in\NN$ be such that $y_{\xi}\ge x_{\xi}$ and
$y_{\xi}\not\le x_{\eta}$ for any $\eta<\xi$.   Again, if
$K\subseteq\NN$ is compact, then $\{\eta:y_{\eta}\in K\}$ has cardinal
strictly less than $\frakmctbl$.   \Prf\ Taking $x=\sup K$ as before,
there is a $\xi<\frak d=\frakmctbl$
such that $x\le x_{\xi}$;  now for any $\eta>\xi$
we know that $y_{\eta}\not\le x_{\xi}$ so $y_{\eta}\notin K$.\ \QeD\
The rest of the proof proceeds as before.   (The set
$\{y_{\eta}:\eta<\frak d\}$ has cardinal $\frak d$ because it is cofinal
with $\NN$.)
}%end of proof of 534R

\leader{534S}{}\dvAformerly{5{}34P}
\cmmnt{ Subject to the continuum hypothesis we have many
ways of building sets with strong measure zero, in addition to those in
the proof of 534R.   I give one example to
suggest what can be done with a weak form of Martin's axiom.

\medskip

\noindent}{\bf Example} Suppose that $\frakmctbl=\frak c$.

(a) There is a set $A\subseteq[0,1]\setminus\Bbb Q$ such that

\quad($\alpha$) $\#(A\cap K)<\frak c$ for every compact
$K\subseteq[0,1]\setminus\Bbb Q$,

\quad($\beta$) there is a continuous function $f:[0,1]\setminus\Bbb Q\to[0,1]$
such that $f[A]=[0,1]$,

\quad($\gamma$) $A+A\supseteq\ocint{0,1}$.

(b) Now $A\cup\Bbb Q$ has Rothberger's property in itself,
$A\in\CalRbg(\Bbb R)$, $A$ is not meager,
$A\notin\CalRbg(\Bbb R\setminus\Bbb Q)$ and
$A\times A\notin\CalRbg(\Bbb R^2)$.

\proof{{\bf (a)(i)} For $x\in\NN$, define $\psi(x)\in\{0,1\}^{\Bbb N}$ by
setting $\psi(x)(n)=0$ if $x(n)$ is even, $1$ if $x(n)$ is odd.
Then $\psi:\NN\to\{0,1\}^{\Bbb N}$ is a continuous surjection.   Let
$\phi:\NN\to[0,1]\setminus\Bbb Q$ be a homeomorphism (4A2Ub again).
Enumerate $\NN$ as
$\ofamily{\xi}{\frak c}{x_{\xi}}$ and $\ocint{0,1}$ as
$\ofamily{\xi}{\frak c}{t_{\xi}}$.
For $\xi\le\frak c$, set $K_{\xi}=\{x:x\in\NN$, $x\le x_{\xi}\}$, so that
$K_{\xi}$ is compact and $\phi[K_{\xi}]$ is a compact subset of
$[0,1]\setminus\Bbb Q$, therefore nowhere dense in $\Bbb R$.
Write $\Cal M$ for the ideal of meager subsets of $\Bbb R$, as in
\S522.

Choose $\ofamily{\xi}{\frak c}{a_{\xi}}$,
$\ofamily{\xi}{\frak c}{b_{\xi}}$ and
$\ofamily{\xi}{\frak c}{c_{\xi}}$
as follows.   For each $\xi<\frak c$, $\{x_{\eta}:\eta<\xi\}$ is not
cofinal with $\NN$, because

\Centerline{$\cf\NN=\frak d\ge\cov\Cal M=\frakmctbl=\frak c$}

\noindent (522I, 522Sa again), so we can find a $y_{\xi}\in\NN$ such that
$y_{\xi}\not\le x_{\eta}$ for any $\eta<\xi$;   raising $y_{\xi}$ if
need be, we can arrange that
$\psi(y_{\xi})=\psi(x_{\xi})$.   Set $a_{\xi}=\phi(y_{\xi})$.   Consider

\Centerline{$\Cal E_{\xi}
=\{\phi[K_{\eta}]:\eta<\xi\}
 \cup\{t_{\xi}-\phi[K_{\eta}]:\eta<\xi\}
 \cup\{\Bbb Q\}\cup\{t_{\xi}-\Bbb Q\}$.}

\noindent This is a family of fewer than $\frak c=\frakmctbl$ meager
subsets of $\Bbb R$, so does not cover $\ooint{0,t_{\xi}}$ (522Sa once more).
Take any $b_{\xi}\in\ooint{0,t_{\xi}}\setminus\bigcup\Cal E_{\xi}$;  then
neither $b_{\xi}$ nor $c_{\xi}=t_{\xi}-b_{\xi}$ belongs to
$\Bbb Q\cup\bigcup_{\eta<\xi}\phi[K_{\eta}]$.

At the end of the process, set

\Centerline{$A=\{a_{\xi}:\xi<\frak c\}\cup\{b_{\xi}:\xi<\frak c\}
\cup\{c_{\xi}:\xi<\frak c\}$.}

\medskip

\quad{\bf (ii)}\grheada\ If $K\subseteq[0,1]\setminus\Bbb Q$ is compact, then
$\phi^{-1}[K]\subseteq\NN$ is compact, so there is an $\eta<\frak c$ such that
$\phi^{-1}[K]\subseteq K_{\eta}$ and $K\subseteq\phi[K_{\eta}]$.   If
$\eta<\xi<\frak c$, $y_{\xi}\notin K_{\eta}$ so $a_{\xi}\notin K$, while
neither $b_{\xi}$ nor $c_{\xi}$ belongs to $\phi[K_{\eta}]\supseteq K$.
So $A\cap K\subseteq\{a_{\xi}:\xi\le\eta\}\cup\{b_{\xi}:\xi\le\eta\}
\cup\{c_{\xi}:\xi\le\eta\}$ has cardinal less than $\frak c$.

\medskip

\qquad\grheadb\ For $x\in\{0,1\}^{\Bbb N}$ set
$h(x)=\Bover12\sum_{i=0}^{\infty}2^{-i}x(i)$, so that
$h:\{0,1\}^{\Bbb N}\to[0,1]$ is a continuous surjection.   Set
$f=h\psi\phi^{-1}:[0,1]\setminus\Bbb Q\to[0,1]$.   Then $f$ is continuous.
Since
$\psi\phi^{-1}(a_{\xi})=\psi(y_{\xi})=\psi(x_{\xi})$ for every $\xi<\frak c$,
$\psi\phi^{-1}[A]=\{0,1\}^{\Bbb N}$ and $f[A]=[0,1]$.   So $f\restr A$ is a
surjection from $A$ onto $[0,1]$.

\medskip

\qquad\grheadc\ Since $t_{\xi}=b_{\xi}+c_{\xi}\in A+A$ for every $\xi<\frak c$,
$A+A\supseteq\ocint{0,1}$.

\medskip

{\bf (b)(i)} Let $H\subseteq A\cup\Bbb Q$ be a relatively open set including
$\Bbb Q$, and take an open set $G\subseteq\Bbb R$ such that
$H=G\cap(A\cup\Bbb Q)$.   Then
$K=[0,1]\setminus G$ is a compact subset of $[0,1]\setminus\Bbb Q$ and
$\phi^{-1}[K]$ is a compact subset of $\NN$.   There is therefore an
$\eta<\frak c$ such that $\phi^{-1}[K]$ is bounded above by $x_{\eta}$,
that is, $\phi^{-1}[K]\subseteq K_{\eta}$ and $K\subseteq\phi[K_{\eta}]$.
So neither $a_{\xi}$ nor $b_{\xi}$ nor $c_{\xi}$ can belong to $K$ for any
$\xi>\eta$, and $\#(A\cap K)<\frak c=\frakmctbl$.   By 534Qa,
$(A\cup\Bbb Q)\setminus H=(A\cup\Bbb Q)\setminus G=A\cap K$
belongs to $\CalRbg(A\cup\Bbb Q)$;  as
$\Bbb Q\in\CalRbg(A\cup\Bbb Q)$ and
$H$ is arbitrary, $A\in\CalRbg(A\cup\Bbb Q)$ (534D(b-iii)).

\medskip

\quad{\bf (ii)} As the embedding $A\cup\Bbb Q\embedsinto\Bbb R$ is continuous,
$A\cup\Bbb Q\in\CalRbg(\Bbb R)$ (534D(b-ii)) and $A\in\CalRbg(\Bbb R)$
(534D(b-i)).

\medskip

\quad{\bf (iii)} By 534Ea, $A$ is of strong measure zero for the usual metric on
$\Bbb R$.   Setting $B=A+\Bbb Z$, $B$ is the
union of a sequence of sets isometric
to $A$, so is of strong measure zero.   As $A+A\supseteq\ocint{0,1}$,
$B+A=\Bbb R$;  by 534K, $A$ is not meager.

\medskip

\quad{\bf (iv)} Of course $[0,1]$ is not of strong measure zero for its usual
metric (534H) so does not belong to $\CalRbg([0,1])$ (534Ea);  now
(a-$\beta$) here and 534D(b-ii) tell us that $A$ cannot belong to
$\CalRbg([0,1]\setminus\Bbb Q)$.   But $[0,1]\setminus\Bbb Q$ is relatively
closed in $\Bbb R\setminus\Bbb Q$, so $A$ cannot belong to
$\CalRbg(\Bbb R\setminus\Bbb Q)$, by 534D(b-iv).

\medskip

\quad{\bf (v)} If we give $\Bbb R$ and $\Bbb R^2$ their usual metrics,
addition is a uniformly continuous function from $\BbbR^2$ to $\Bbb R$, while
$A+A\supseteq\ocint{0,1}$ is not of strong measure zero.  So $A\times A$ is not
of strong measure zero (534D(a-iii)) and cannot belong to $\CalRbg(\BbbR^2)$.
}%end of proof of 534S


\exercises{\leader{534X}{Basic exercises (a)}%
%\spheader 534Xa
(i) Let $(X,\rho)$ be a metric space, $r>0$ and $A\subseteq X$ a set
with strong measure zero.   Show that $A$ has zero Hausdorff
$r$-dimensional measure.   (ii) Find a subset of $\BbbR^2$ which is
universally
negligible but does not have strong measure zero (for the usual metric
on $\BbbR^2$).   \Hint{439G.}   (iii)
Find a subset of $\{0,1\}^{\Bbb N}$ which is universally negligible
but does not have strong measure zero for the metric of 534Qb.
%534D

\spheader 534Xb Let $r$, $s\ge 1$ be integers.   Let $A\subseteq\BbbR^r$
be a set with strong measure zero, and $f:A\to\BbbR^s$ a function which
is differentiable relative to its domain at every point of $A$.   Show
that $f[A]$ has strong measure zero.  \Hint{262N.}
%534D

\spheader 534Xc Let $(X,\Cal W)$ and $(Y,\Cal V)$ be uniform spaces and
$f:X\to Y$ a continuous function.   Suppose that $A\in\CalSmz(X,\Cal W)$ is
covered by a sequence of compact subsets of $X$.   Show that
$f[A]\in\CalSmz(Y,\Cal V)$.
%534E

\spheader 534Xd Let $X$ be a $\sigma$-compact topological space
which is either Hausdorff or regular, and $A\subseteq X$.   Show that
$A\in\CalRbg(X)$ iff for every sequence
$\sequencen{\Cal G_n}$ of finite open
covers of $X$, there is a sequence $\sequencen{G_n}$, covering $A$,
such that $G_n\in\Cal G_n$ for every $n$.
%534F

\spheader 534Xe Let $(X,\Cal W)$ be a Hausdorff uniform space with
strong measure zero.   Show that $X$ is universally negligible iff it
is a Radon space.
%534H

\spheader 534Xf(i) Let $(X,\Cal W)$ be a
Hausdorff uniform space.   Show that if $X$ has strong measure zero
then it is universally $\tau$-negligible.
(ii) Let $X$ be a Hausdorff topological space.   Show
that if $A\in\CalRbg(X)$ then $A$ is
universally $\tau$-negligible (definition: 439Xh).
%534H

\spheader 534Xg Give $\omega_1+1$ its order topology.   Show that it
has Rothberger's property in itself but is not universally negligible.
%534I

\spheader 534Xh Give $\omega_1+1$ its order topology.   Show that $\omega_1$
has Rothberger's property in $\omega_1+1$ but not in itself.
%534I

\spheader 534Xi Let $X$ be a locally compact Hausdorff topological
group.   Show that a subset of $X$ has Rothberger's property in $X$
iff it has strong measure zero for the right uniformity of $X$
iff it has strong measure zero for the bilateral uniformity of $X$.
%534K 534E

\spheader 534Xj(i) Let $(X,\Cal W)$ be a Lindel\"of uniform space.   Show
that there is some $\kappa$ such that
$\CalSmz(X,\Cal W)\prT\Cal N^{\kappa}$,
where $\Cal N$ is the null ideal of Lebesgue measure on $\Bbb R$.
%534M   idea of "uniform weight" would specify \kappa
(ii) Let $X$ be a regular Lindel\"of space.   Show
that there is some $\kappa$ such that
$\CalRbg(X)\prT\Cal N^{\kappa}$.
%534M

\spheader 534Xk Show that every separable metric space $(X,\rho)$ is
uniformly equivalent to a subspace of $[0,1]^{\Bbb N}$ and is
therefore $\CalSmz$-embeddable in $[0,1]^{\Bbb N}$.
%534N

\spheader 534Xl Let $\sequencen{I_n}$ be a non-decreasing sequence of
finite sets covering $\Bbb Z$.   For $x$, $y\in\BbbZ^{\Bbb N}$ set
$\rho(x,y)=\inf\{2^{-n}:n\in\Bbb N$, $x\restr I_n=y\restr I_n\}$.   Show that
$\rho$ is a metric on $\BbbZ^{\Bbb N}$ inducing its topological group uniformity
(4A5He), and that $\non\CalSmz(\BbbZ^{\BbbN},\rho)=\frakmctbl$.
%534Q

\spheader 534Xm(i) Show that no cofinal subset of $\NN$ has strong
measure zero for the metric $\rho$ of 534Qb.
(ii) Suppose that $\frakmctbl=\frak d$.   Show that there are a
subset $A$ of $\Bbb R\setminus\Bbb Q$ and a metric $\rho'$ on
$\Bbb R\setminus\Bbb Q$ inducing the usual
topology of $\Bbb R\setminus\Bbb Q$ such that $A$ has strong measure zero for
the usual metric on $\Bbb R$ but not for $\rho'$.
%534R(c-iii) 534Qb

\spheader 534Xn\dvArevised{2022}
Let $A$ be the set constructed in 534Sa on the assumption
that $\frakmctbl=\frak c$.   Show that $A$ has strong measure zero for the usual
metric of $\Bbb R$, and describe a metric on $[0,1]\setminus\Bbb Q$, inducing
the usual topology on $[0,1]\setminus\Bbb Q$, for which $A$ does not have
strong measure zero.   (See also 534Ye.)
%534S

\spheader 534Xo [In this exercise, I will say that a topological space
which has Rothberger's property in itself has `property C$'$'.]  
(i) Show that any Lindel\"of space with cardinal 
less than $\frakmctbl$ has property C$'$.   (ii) Show that if $X$ is a
topological space expressible as the union of a sequence of subspaces with
property C$'$, then $X$ has property C$'$.   (iii) Show that if $X$ is a regular
Lindel\"of space expressible as the union of fewer than $\add\Cal N$ 
subspaces with property C$'$, then $X$ has property C$'$.   (iv) Show that a
continuous image of a space with property C$'$ has property C$'$.   
(v) Show that a closed subset of a space with property C$'$ has property C$'$.   
(vi) Show that if $X$ is a topological space, $A\subseteq X$ has property C$'$
and every closed subset of $X\setminus A$ has property C$'$, then $X$ has
property C$'$.   
%534D 534M 534Q

\leader{534Y}{Further exercises (a)}
%\spheader 534Ya
Let $(X,\rho)$ be an analytic metric space and $\mu_{Hr}$ Hausdorff
$r$-dimensional measure on $X$, where $r>0$;  suppose that
$\mu_{Hr}X>0$.   Let $\Cal I$ be the
$\sigma$-ideal of subsets of $X$ generated by
$\{A:\mu_{Hr}^*A<\infty\}$.   Show that

$$\eqalign{\non\Cal N(\mu_{Hr})
=\min(\non\Cal N,\non\Cal I)
&=\non\Cal N\text{ if }\mu_{Hr}\text{ is }\sigma\text{-finite},\cr
&=\non\Cal I\text{ otherwise}.\cr}$$
%\noindent\Hint{if $\mu_{Hr}$ is not $\sigma$-finite, then by 471S there
%is an uncountable disjoint family of Borel sets of non-zero finite
%measure.}
%534B mt53bits

\spheader 534Yb(i) Set
$\Cal I=\{\coint{4^{-m}i,4^{-m}(i+1)}:m\in\Bbb N$, $i\in\Bbb Z\}$.   For
$A\subseteq\Bbb R$ set
$\theta(A)=\inf\{\sum_{I\in\Cal I'}\sqrt{\diam I}:
\Cal I'\subseteq\Cal I$ covers $A\}$.
Show that if $\mu^{(1)}_{H,1/2}$ is Hausdorff $\bover12$-dimensional
measure on $\Bbb R$, then $\mu^{(1)}_{H,1/2}(A)=0$ iff $\theta(A)=0$.
(ii) Set $\Cal J
=\{\coint{2^{-m}i,2^{-m}(i+1)}\times\coint{2^{-m}j,2^{-m}(j+1)}:
  m\in\Bbb N$, $i$, $j\in\Bbb Z\}$, and for $A\subseteq\BbbR^2$ set
$\theta'(A)=\inf\{\sum_{J\in\Cal J'}\diam J:\Cal J'\subseteq\Cal J$
covers $A\}$.
Show that if $\mu^{(2)}_{H1}$ is Hausdorff $1$-dimensional measure on
$\BbbR^2$, then $\mu^{(2)}_{H1}(A)=0$ iff $\theta'(A)=0$.
(iii) Show that the null ideals $\Cal N(\mu^{(1)}_{H,1/2})$ and
$\Cal N(\mu^{(2)}_{H1})$ are isomorphic.
%534B

\spheader 534Yc Show that if {\it either}
$\non\Cal N=\cf\Cal N$ {\it or} $\non\Cal N<\cov\Cal N$, where $\Cal N$ is
the null ideal of Lebesgue measure on $\Bbb R$,
then Hausdorff one-dimensional measure on $\BbbR^2$ does not have the
measurable envelope property.
%Problem:  is it consistent to suppose that \mu_{H1} *does* have the
%measurable envelope property?
%534B mt53bits

\spheader 534Ye Suppose that $\frakmctbl=\frak c$.   Let $X$ be the
group of all permutations of $\Bbb N$, regarded as the isometry group of
$\Bbb N$ with its $\{0,1\}$-valued metric, so that $X$ is a Polish group
(441Xq).   Show that there is a subset $A$ of $X$ such that $A$
has strong measure zero for the right uniformity of $X$ but $A^{-1}$
does not.
%534K mt53bits

\spheader 534Yd
Let $\frak G$ be a collection of families of sets.   Let us say that a
set $A$ has the
{\bf $\frak G$-Rothberger property} if for every sequence
$\sequencen{\Cal G_n}$ in $\frak G$ there is a cover $\sequencen{G_n}$
of $A$ such that $G_n\in\Cal G_n$ for every $n\in\Bbb N$.   (i) Show
that the family $\Cal I$ of sets with the $\frak G$-Rothberger
property
is a $\sigma$-ideal of sets containing every countable subset of
$\bigcap_{\Cal G\in\frak G}\bigcup\Cal G$.
(ii) Show that if $\frak H$ is another collection of families of sets,
and $f$ is a function such that for every $\Cal H\in\frak H$ there is
a member of $\frak G$ refining $\{f^{-1}[H]:H\in\Cal H\}$, then $f[A]$
has the $\frak H$-Rothberger property whenever $A\in\Cal I$.   (iii)
Suppose that $\frak G$ is a collection of families of open subsets of a
topological space  $X$, that $A\in\Cal I$ has the
$\frak G$-Rothberger property, and that $B\subseteq X$ is such that
$B\setminus G\in\Cal I$ for every open set $G\supseteq A$.   Show that
$B\in\Cal I$.   (iv) Suppose that $X=\bigcup\Cal G$ for every
$\Cal G\in\frak G$, and that every member of $\frak G$ is countable.
Show that $\non(\Cal I,X)\ge\frakmctbl$.
%534Qa

\spheader 534Ye Suppose that $\frakmctbl=\frak d$.
Show that there are two complete metrics $\rho$, $\rho'$ on
$\NN$, both inducing the usual topology of $\NN$, such that
$\CalSmz(\NN,\rho)\ne\CalSmz(\NN,\rho')$.
%534E 534S 534Ye 534Q 534Xm mt53bits 534R

\leaveitout{
\spheader 534Y? Let $(X,\Cal W)$ be a uniform space.
Consider the following infinite game for two players, `Cover' and
`Noncover'.   (Compare 451V.)
Noncover chooses $W_0\in\Cal W$;  Cover chooses
$A_0$ such that $A_0\times A_0\subseteq W_0$;
Noncover chooses $W_1\in\Cal W$;  Cover chooses
$A_1$ such that $A_1\times A_1\subseteq W_1$;  and so on.   At the end of
the game, Cover wins iff $X\subseteq\bigcup_{n\in\Bbb N}A_n$.   $X$ is {\bf
strategically of strong measure zero} if Cover has a winning strategy in
this game.   (i) Show that $\omega_1+1$, with its usual uniformity, is
strategically of strong measure zero.   (ii) Show that a metric space is
strategically of strong measure zero iff it is countable.
}%end of leaveitout
}%end of exercises

\leader{534Z}{Problems (a)}
%\spheader 534Za
Let $\mu^{(2)}_{H1}$ be one-dimensional
Hausdorff measure on $\BbbR^2$.
Is the covering number $\cov\Cal N(\mu^{(2)}_{H1})$ necessarily equal
to $\cov\Cal N$?
\cmmnt{As observed in 534Bc-534Bd, we have
$\cov\Cal N\le\cov\Cal N(\mu^{(2)}_{H1})\le\non\Cal M$.
We can ask the same question for $r$-dimensional Hausdorff
measure on $\BbbR^n$ whenever $0<r<n$;
in particular, for $r$-dimensional Hausdorff measure on $[0,1]$, where
$0<r<1$, and these questions are strongly connected (534Yb).
{\smc Shelah \& Stepr\=ans 05} show that
$\non\Cal N(\mu^{(2)}_{H1})$ can be less than $\non\Cal N$;
of course this is possible only because $\mu^{(2)}_{H1}$ is not
semi-finite (439H, 521Xg).
}%end of comment

\spheader 534Zb Can $\cf\CalRbg(\Bbb R)$ be $\omega_1$?
%534G

\spheader 534Zc How many types of complete separable metric spaces under
$\CalSmz$-equivalence can there be?   If we give $\NN$ the metric of
534Qb, can it fail to be $\CalSmz$-equivalent to $[0,1]^{\Bbb N}$ with the
metric $(x,y)\mapsto\sup_{n\in\Bbb N}2^{-n}|x(n)-y(n)|$?
%534Yd 534P

\spheader 534Zd Suppose that there is a separable metric space
with cardinal $\frak c$ with strong measure zero.   Must
there be a subset of $\Bbb R$ with cardinal $\frak c$ with
Rothberger's property in $\Bbb R$?
%534R

\spheader 534Ze\dvAnew{2022} On $\Bbb R$, let $\frak T$ be the usual topology
and $\frak S$ the right-facing Sorgenfrey topology (415Xc).   Must 
$\CalRbg(\Bbb R,\frak S)$ and $\CalRbg(\Bbb R,\frak T)$ be the same?
%534J

\endnotes{
\Notesheader{534}
I have very little to say about Hausdorff measures, and 534B is here
only because it would seem even lonelier in a section by itself.   All I
have tried to do is to run through the obvious questions connecting \S471
with Chapter 52.   But at the next level there is surely much more to be
done (534Za).

`Strong measure zero' has attracted a great deal of attention, starting
with the work of E.Borel, who suggested that every subset of
$\Bbb R$ with strong measure zero must be countable;  this is the
{\bf Borel conjecture}.   It turns out that this is undecidable in
ZFC (see the preamble to 534Q),
and that if the Borel conjecture is true then there are no uncountable
sets of strong measure zero in any separable metric space (534Rb).
So we have some questions of a new kind:  in the ideals
$\CalSmz(X,\Cal W)$ of sets
of strong measure zero, in addition to the standard cardinals $\add$,
$\non$, $\cov$ and $\cf$, we find ourselves asking for the possible
cardinals of sets belonging to the ideal.

The next point is that strong measure zero is not (or rather, not
always) either a topological property or a metric property;  it is a
property of uniform spaces.   We must therefore be prepared to examine
uniformities, even if we are happy to stay with metrizable ones.   In
534Xm we see that we can have a set which has strong measure
zero for one of two equivalent metrics and not for the other.
{\smc Goldstern Judah \& Shelah 93} describe a model in which
$\frakmctbl=\omega_1$, $\add\CalRbg(\Bbb R)=\frak c=\omega_2$ and
there is a subset of $\Bbb R$ of cardinal
$\omega_2$ with strong measure zero.    So in this case $\NN$,
with the metric described in 534Qb, is not even $\CalSmz$-embeddable in
$\Bbb R$ with its usual metric.
Of course in models of set theory in which the Borel
conjecture is true we do have a topologically determined structure on any
separable metrizable space.

Note that for any uncountable complete separable metric space $(X,\rho)$, there
is a subset of $X$ homeomorphic to $\{0,1\}^{\Bbb N}$
(423Ba, 423K\formerly{4{}23J}), and the
homeomorphism must be a uniform equivalence;  so that $\{0,1\}^{\Bbb N}$
and its companions $[0,1]^r$, $\BbbR^r$ (534P) must be $\CalSmz$-embeddable
in $X$.   In this sense they are the `simplest' uncountableomplete
metric spaces.   In the same sense,
$[0,1]^{\Bbb N}$ is the most complex separable metric space (534Xk).

For $\sigma$-compact spaces, strong measure zero becomes a topological
property (534Eb), corresponding to what I call `Rothberger's property'
(534Cb).   {\smc Rothberger 1938b} investigated subsets of $\Bbb R$
which have Rothberger's property in themselves, under the name `property
C$'$'.   The ideas of 534Da and 534L-534Ma can be re-presented as
theorems about Rothberger's property (534Db,
534Mb, 534Xj	);  the machinery of 534Yd is supposed to
suggest a reason for this.   It is natural to be attracted to a
topological concept, but there is a difficulty in that Rothberger's
property is not hereditary in the usual way (534Xh, 534Xm, 534Xn).
I note that while 534P can be stated in terms of
$\CalRbg$-equivalence, isomorphism of the ideals of sets with the
appropriate Rothberger's property, the concept of strong measure zero
seems to be necessary in the Schr\"oder-Bernstein arguments based on 534O.
Of course the spaces here are paracompact and normal, so 534F gives us an
alternative approach to this issue.

For a fuller discussion of strong measure zero in $\Bbb R$, see
{\smc Bartoszy\'nski \& Judah 95}, chap.\ 8, from which many of the
ideas of this section are taken.
}%end of notes

\discrpage

\discrversionA{Any hope of a version of 534K for Polish groups?

\medskip

what about $\shr(\CalSmz)$?

\medskip

In the model of {\smc Goldstern Judah \& Shelah 93},
there is a set $X\subseteq\NN$ such that $X\notin\CalRbg(X)$ but
$f[X]\in\CalRbg(\Bbb R)$ whenever $f:X\to\Bbb R$ is continuous.   So in
534I we cannot use $\Bbb R$ in place of $\Bbb R\setminus\Bbb Q$.

}{}
\frfilename{mt535.tex}
\versiondate{1.6.11}
\copyrightdate{2003}

\def\chaptername{Topologies and measures III}
\def\sectionname{Liftings}

\newsection{535}

I introduced the Lifting Theorem\cmmnt{ (\S341)} as one of the
fundamental facts
about complete strictly localizable measure spaces.   Of course we can
always complete a measure space and thereby in effect obtain a lifting
for any $\sigma$-finite measure.   For the applications of the Lifting
Theorem in \S\S452-453 this procedure is natural and effective;  and
generally in this treatise I have taken the view that one should work
with completed measures unless there is some strong reason not to.   But
I have also embraced the principle of maximal convenient generality,
seeking formulations which will exhibit the full force of each idea in
the context appropriate to that idea, uncluttered by the special
features of intended applications.   So the question of when, and why,
liftings for incomplete measures can be found is one which automatically
arises.   It turns out to be a fruitful question, in the sense that it
leads us to new arguments, even though the answers so far available are
unsatisfying.

As usual, much of what we want to know depends on the behaviour of the
usual measures on powers of $\{0,1\}$ (535B).   An old argument relying
on the continuum hypothesis shows that Lebesgue measure can have a Borel
lifting;  this has been usefully refined, and I give a strong version in
535D-535E.
We know that we cannot expect to have translation-invariant Borel
liftings\cmmnt{ (345F)}, but strong Borel liftings are possible
(535H-535I), and in some cases can be built from Borel liftings
(535J-535N).   %535J 535N

For certain applications in functional analysis, we are more interested
in liftings for $L^{\infty}$ spaces than in liftings for measure algebras;
and it is sometimes sufficient to have a `linear lifting', not
necessarily corresponding to a lifting in the strict sense (535O,
535P).   I give a couple of paragraphs to linear liftings because in
some ways they are easier to handle and it is conceivable that they are
relevant to the main outstanding problem (535Zf).

\leader{535A}{Notation (a)}\cmmnt{ The most interesting questions to
be examined in this section can be phrased in the following language.}
If $(X,\Sigma,\mu)$ is a measure space and $\frak T$ a topology on $X$,
\cmmnt{I will say that} a {\bf Borel lifting} of $\mu$ is a lifting which takes
values in the Borel $\sigma$-algebra $\Cal B(X)$ of $X$.   \cmmnt{(As
usual, I will use the word `lifting' indifferently for homomorphisms
from $\Sigma$ to itself, or from $\frak A$ to $\Sigma$, where $\frak A$
is the measure algebra of $\mu$.   Of course a homomorphism
$\theta:\frak A\to\Sigma$ is a Borel lifting iff the corresponding
homomorphism $E\mapsto\theta E^{\ssbullet}:\Sigma\to\Sigma$ is a Borel
lifting.)}   Similarly, a {\bf Baire lifting} of $\mu$ is a lifting
which takes values in the Baire $\sigma$-algebra $\CalBa(X)$ of $X$.

\spheader 535Ab I remark at once that if $(X,\frak T,\Sigma,\mu)$ is a
topological measure space and $\phi:\Sigma\to\Cal B(X)$ is a Borel
lifting for $\mu$, then $\phi\restr\Cal B(X)$ is a lifting for the Borel
measure $\mu\restr\Cal B(X)$.   Conversely, if
$\phi':\Cal B(X)\to\Cal B(X)$ is a lifting for $\mu\restr\Cal B(X)$, and
if for every $E\in\Sigma$ there is a Borel set $E'$ such that
$E\symmdiff E'$ is negligible, then $\phi'$ extends uniquely to a Borel
lifting $\phi$ of $\mu$.

In the same way, any Baire lifting for a measure $\mu$ which measures
every zero set will give\cmmnt{ us} a lifting for $\mu\restr\CalBa(X)$;  and a lifting for $\mu\restr\CalBa(X)$ will correspond to a Baire lifting for $\mu$ if\cmmnt{, for instance,} $\mu$ is completion regular\cmmnt{, as in 535B below}.

\spheader 535Ac \cmmnt{As in Chapter 52,} I will say that, for any set
$I$, $\nu_I$ is the usual measure on $\{0,1\}^I$ and $\frak B_I$ its
measure algebra.

\leader{535B}{Proposition} Let $(X,\Sigma,\mu)$ be a strictly
localizable measure space with non-zero measure.   Suppose that
$\nu_{\kappa}$ has a Baire lifting\cmmnt{ (that is,
$\nu_{\kappa}\restr\CalBa(\{0,1\}^{\kappa})$ has a lifting)} for every
infinite cardinal $\kappa$
such that the Maharam-type-$\kappa$ component of the measure algebra of
$\mu$ is non-zero.   Then $\mu$ has a lifting.

\proof{ Write $(\frak A,\bar\mu)$ for the measure algebra of $\mu$.

\medskip

{\bf (a)} Suppose first that $\mu$ is a Maharam-type-homogeneous
probability measure.   In this case $\frak A$ is either $\{0,1\}$ or
isomorphic to $\frak B_{\kappa}$ for some infinite $\kappa$.   The case
$\frak A=\{0,1\}$ is trivial, as we can set $\phi E=\emptyset$ if
$E\in\Sigma$ is negligible, $\phi E=X$ if $E\in\Sigma$ is conegligible.
Otherwise, $\frak A$ is
$\tau$-generated by a stochastically independent family
$\ofamily{\xi}{\kappa}{e_{\xi}}$ of elements of measure $\bover12$.
For each $\xi<\kappa$, choose $E_{\xi}\in\Sigma$ such that
$E_{\xi}^{\ssbullet}=e_{\xi}$, and define $f:X\to\{0,1\}^{\kappa}$ by
setting $f(x)(\xi)=\chi E_{\xi}(x)$ for $x\in X$ and $\xi<\kappa$.
Then $\{F:F\subseteq\{0,1\}^{\kappa}$, $\nu F$ and $\mu f^{-1}[F]$ are
defined and equal$\}$ is a Dynkin class containing all the measurable
cylinders in $\{0,1\}^{\kappa}$, so includes
$\CalBa_{\kappa}=\CalBa(\{0,1\}^{\kappa})$,
and $f$ is \imp\ for $\mu$ and
$\nuprime_{\kappa}=\nu_{\kappa}\restr\CalBa_{\kappa}$.   Note that
$\frak B_{\kappa}$ can be identified with the measure algebra of
$\nuprime_{\kappa}$ (put 415E and
322Da together, or see 415Xs\formerly{4{}15Xp}).   So we have an induced
measure-preserving Boolean homomorphism $\pi:\frak B_{\kappa}\to\frak A$
defined by setting $\pi F^{\ssbullet}=f^{-1}[F]^{\ssbullet}$ for every
$F\in\CalBa_{\kappa}$.   Since $\pi[\frak B_{\kappa}]$ is an
order-closed subalgebra of $\frak A$ (324Kb) containing every $e_{\xi}$,
it is the whole of $\frak A$.

We are supposing that there is a lifting
$\theta:\frak B_{\kappa}\to\CalBa_{\kappa}$ of
$\nu_{\kappa}$.   Define
$\theta_1:\frak A\to\Sigma$ by setting
$\theta_1a=f^{-1}[\theta\pi^{-1}a]$ for every $a\in\frak A$;  then
$\theta_1$ is a Boolean homomorphism because $\theta$ and $\pi^{-1}$
are, and

\Centerline{$(\theta_1a)^{\ssbullet}=\pi((\theta\pi^{-1}a)^{\ssbullet})
=\pi\pi^{-1}a=a$}

\noindent for every $a\in\frak A$, so $\theta_1$ is a lifting for $\mu$.

\medskip

{\bf (b)} It follows at once that if $\mu$ is any non-zero totally
finite Maharam-type-homogeneous measure, then it will have a lifting, as
we can apply (a) to a scalar multiple of $\mu$.   Now consider the
general case.   Let $\Cal K$ be the family of measurable subsets $K$ of
$X$ such that the subspace measure $\mu_K$ is non-zero, totally finite
and Maharam-type-homogeneous.   Then $\mu$ is inner regular with respect
to $\Cal K$, by Maharam's theorem (332B).   By 412Ia, there is a
decomposition $\familyiI{X_i}$ of $X$ such that at most
one $X_i$ does not belong to $\Cal K$, and that exceptional one, if any,
is negligible;  adding a trivial element $X_k=\emptyset$ if necessary,
we may suppose that there is exactly one $k\in I$ such that
$\mu X_k=\emptyset$.   For each $i\in I\setminus\{k\}$, let $\mu_i$ be
the subspace measure on $X_i$, and $\Sigma_i$ its domain;  then $\mu_i$
has a lifting $\phi_i:\Sigma_i\to\Sigma_i$.   (The point is that if the
Maharam type $\kappa$ of $\mu_i$ is infinite, then the
Maharam-type-$\kappa$ component of $\frak A$ includes $X_i^{\ssbullet}$
and is non-zero, so our hypothesis tells us that $\nu_{\kappa}$ has a
Baire lifting.)   At this point, recall that we are also supposing that
$\mu X>0$, so there is some $j\in I\setminus\{k\}$;  fix $z\in X_j$, and
define $\phi:\Sigma\to\Cal PX$ by setting

$$\eqalign{\phi E&=\bigcup_{i\in I\setminus\{k\}}\phi_i(E\cap X_i)
  \text{ if }z\notin\phi_j(E\cap X_j),\cr
&=X_k\cup\bigcup_{i\in I\setminus\{k\}}\phi_i(E\cap X_i)
  \text{ if }z\in\phi_j(E\cap X_j).\cr}$$

\noindent Then $\phi$ is a lifting for $\mu$.   \Prf\ It is a Boolean
homomorphism because every $\phi_i$ is.   If $E\in\Sigma$, then
$X_i\cap\phi E=\phi_i(E\cap X_i)$ if $i\in I\setminus\{k\}$, and is
either $X_k$ or $\emptyset$ if $i=k$;  in any case, it belongs to
$\Sigma_i$;  as $\familyiI{X_i}$ is a decomposition for $\mu$,
$\phi E\in\Sigma$.   Also

\Centerline{$\mu(E\symmdiff\phi E)
\le\mu X_k+\sum_{i\in I\setminus\{k\}}
  \mu_i((E\cap X_i)\symmdiff\phi_i(E\cap X_i))
=0$.}

\noindent Finally, if $\mu E=0$, then $\mu_i(E\cap X_i)=0$ and
$\phi_i(E\cap X_i)=\emptyset$ for every $i\in I\setminus\{k\}$, so
$\phi E=\emptyset$.\ \Qed
}%end of proof of 535B

\leader{535C}{Proposition} If $\lambda$ and $\kappa$ are cardinals with
$\lambda=\lambda^{\omega}\le\kappa$, and $\nu_{\kappa}$ has a Baire
lifting, then $\nu_{\lambda}$ has a Baire lifting.

\proof{ If $\lambda$ is finite, the result is trivial, so we may suppose
that $\lambda\ge\omega$ (and therefore that $\lambda\ge\frak c$).   For
$I\subseteq\kappa$, write $\CalBa_I$ for the Baire
$\sigma$-algebra of $\{0,1\}^I$ and $\Tau_I$ for the family of those
$E\in\CalBa_{\kappa}$ which are determined by coordinates in $I$.
Set $\pi_I(x)=x\restr I$ for every $x\in\{0,1\}^{\kappa}$;  then
$H\mapsto\pi_I^{-1}[H]$ is a Boolean isomorphism between $\CalBa_I$ and
$\Tau_I$, with inverse $E\mapsto\pi_I[E]$.   \Prf\ Because $\pi_I$ is
continuous,
$\pi_I^{-1}[H]\in\CalBa_{\kappa}$ for every $H\in\CalBa_I$.   Of course
$H\mapsto\pi_I^{-1}[H]$ is a Boolean homomorphism, and it is injective
because $\pi_I$ is surjective.   Identifying $\{0,1\}^{\kappa}$ with
$\{0,1\}^I\times\{0,1\}^{\kappa\setminus I}$, we have a function
$h:\{0,1\}^I\to\{0,1\}^{\kappa}$ defined by setting $h(v)=(v,\tbf{0})$
for $v\in\{0,1\}^I$.   This is continuous, therefore
$(\CalBa_I,\CalBa_{\kappa})$-measurable.   If $E\in\Tau_I$, then
$E=\pi_I^{-1}[\pi_I[E]]=\pi_I^{-1}[h^{-1}[E]]$;   so
$H\mapsto\pi_I^{-1}[H]$ is surjective and is an isomorphism.\ \QeD\

Consequently $\#(\Tau_I)\le\frak c$ for every countable
$I\subseteq\kappa$
(4A1O, because $\CalBa_I$ is $\sigma$-generated by the cylinder sets, by
4A3Na).   For any $I$, $\Tau_I=\bigcup_{J\in[I]^{\le\omega}}\Tau_J$,
because every member of $\CalBa_I$ is determined by coordinates in a
countable set (4A3Nb).   So
$\#(\Tau_I)\le\max(\frak c,\#(I)^{\omega})=\lambda$ whenever
$I\subseteq\kappa$ and $\#(I)=\lambda$.

Let $\phi$ be a Baire lifting for $\nu_{\kappa}$.
Choose a non-decreasing family $\ofamily{\xi}{\omega_1}{J_{\xi}}$ in
$[\kappa]^{\lambda}$ such that $J_0=\lambda$ and
$\phi E\in\Tau_{J_{\xi+1}}$ whenever $\xi<\omega_1$ and
$E\in\Tau_{J_{\xi}}$.   Set $J=\bigcup_{\xi<\omega_1}J_{\xi}$;  then
$\Tau_J=\bigcup_{\xi<\omega_1}\Tau_{J_{\xi}}$, so $\phi E\in\Tau_J$ for
every $E\in\Tau_J$.

We therefore have a Boolean homomorphism $\phi_1:\CalBa_J\to\CalBa_J$
defined by setting $\phi_1H=\pi_J[\phi(\pi_J^{-1}[H])]$ for every
$H\in\CalBa_J$.   If $\nu_JH=0$, then $\nu_{\kappa}\pi_J^{-1}[H]=0$ and
$\phi_1H=\phi(\pi_J^{-1}[H])=0$.   For any $H\in\CalBa_J$,

\Centerline{$\pi_J^{-1}[H\symmdiff\phi_1H]
=\pi_J^{-1}[H]\symmdiff\phi(\pi_J^{-1}[H])$}

\noindent is $\nu_{\kappa}$-negligible, so $H\symmdiff\phi_1H$ is
$\nu_J$-negligible.   Thus $\phi_1$ is a lifting for
$\nu_J\restr\CalBa_J$.   As $\nu_J\restr\CalBa_J$ is isomorphic to
$\nu_{\lambda}\restr\CalBa_{\lambda}$, the latter also has a lifting.
As $\nu_{\lambda}$ is completion regular (416U), the measure algebra of
$\nu_{\lambda}\restr\CalBa_{\lambda}$ can be identified with
$\frak B_{\lambda}$, and we can interpret a lifting for
$\nu_{\lambda}\restr\CalBa_{\lambda}$ as a Baire lifting for its
completion $\nu_{\lambda}$.
}%end of proof of 535C

\leader{535D}{}\cmmnt{ The following result covers most of the cases
in which non-complete probability measures are known to have liftings.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ be a measure space such
that $\mu X>0$, and suppose that its measure algebra is tightly
$\omega_1$-filtered\cmmnt{ (definition:  511Di)}.   Then $\mu$ has a
lifting.

\proof{ This is a special case of 518L.
}%end of proof of 535D

\leader{535E}{Proposition} Suppose that $\frak c\le\omega_2$ and
the Freese-Nation number $\FN(\Cal P\Bbb N)$ is $\omega_1$.

(a) If $\frak A$ is a measurable
algebra with cardinal at most $\omega_2$, it is tightly
$\omega_1$-filtered.

(b) ({\smc Mokobodzki 7?})
% Corollary 19
Let $(X,\Sigma,\mu)$ be a
$\sigma$-finite measure space with non-zero measure and
Maharam type at most $\omega_2$.

\quad(i) $\mu$ has a lifting.

\quad(ii) If $\frak T$ is a topology on $X$ such that $\mu$ is inner
regular with respect to the Borel sets, then $\mu$ has a Borel lifting.

\quad(iii) If $\frak T$ is a topology on $X$ such that $\mu$ is inner
regular with respect to the zero sets, then $\mu$ has a Baire lifting.

\proof{{\bf (a)} By 524O(b-iii), $\FN(\frak A)\le\omega_1$, so 518M gives
the result.

\medskip

{\bf (b)(i)} By 514De, the measure algebra of $\mu$ has
cardinal at most

\Centerline{$\omega_2^{\omega}=\max(\frak c,\omega_2)\le\omega_2$}

\noindent (5A1F(e-iii)).   So we can put (a) and 535D together.

\medskip

\quad{\bf (ii)} Because $\mu$ is $\sigma$-finite and inner regular with
respect to the Borel sets, every measurable set can be expressed as the
union of a Borel set and a negligible set.   By (i),
$\mu\restr\Cal B(X)$ has a lifting, which can be interpreted as a Borel
lifting for $\mu$, as in 535Ab.

\medskip

\quad{\bf (iii)} As (ii), but with $\CalBa(X)$ in place of $\Cal B(X)$.
}%end of proof of 535E

\leader{535F}{}\cmmnt{ Using the continuum hypothesis, we can go a little
farther with ideas from 341J.

\wheader{535F}{6}{2}{2}{48pt}

\noindent}{\bf Proposition}
Let $(X,\Sigma,\mu)$ be a measure space such that
$\mu X>0$ and $\#(\frak A)\le\omega_1$,
where $\frak A$ is the measure algebra of $\mu$, and suppose that
$\undtheta:\frak A\to\Sigma$ is such that

\Centerline{$\undtheta 0=\emptyset$,
\quad$\undtheta(a\Bcap b)=\undtheta a\cap\undtheta b$ for all $a$, $b\in\frak A$,
\quad$(\undtheta a)^{\ssbullet}\Bsubseteq a$ for every $a\in\frak A$.}

\noindent Then $\mu$ has a lifting $\theta:\frak A\to\Sigma$ such that
$\theta E^{\ssbullet}\supseteq\undtheta E$ for every $E\in\Sigma$.

\proof{{\bf (a)}
Adjusting $\undtheta 1$ if necessary, we can suppose that $\undtheta 1=X$.
Note that $\undtheta a\subseteq\undtheta b$ whenever $a\Bsubseteq b$ in $\frak A$.
Let $\ofamily{\xi}{\omega_1}{a_{\xi}}$ be a family running over $\frak A$,
and for $\alpha\le\omega_1$ let $\frak C_{\alpha}$ be the subalgebra of
$\frak A$ generated by $\{a_{\xi}:\xi<\alpha\}$.   Define Boolean
homomorphisms $\theta_{\alpha}:\frak C_{\alpha}\to\Sigma$ inductively, as
follows.   The inductive hypothesis will be that
$(\theta_{\alpha}c)^{\ssbullet}=c$ and $\theta_{\alpha}c\supseteq\undtheta c$
for every $c\in\frak C_{\alpha}$, while $\theta_{\alpha}$ extends
$\theta_{\beta}$ for every $\beta\le\alpha$.
Start with $\theta_00=\emptyset$, $\theta_01=X$.

\medskip

{\bf (b)} Given $\theta_{\alpha}$, where $\alpha<\omega_1$, set

\Centerline{$F
=\bigcup\{\undtheta(c\Bcup a_{\alpha})\Bsetminus\theta_{\alpha}c:
  c\in\frak C_{\alpha}\}$,}

\Centerline{$G
=\bigcup\{\undtheta(c\Bcup(1\Bsetminus a_{\alpha}))\Bsetminus\theta_{\alpha}c:
  c\in\frak C_{\alpha}\}$.}

\noindent Because $\frak C_{\alpha}$ is countable, $F$ and
$G$ belong to $\Sigma$.   If $c\in\frak C_{\alpha}$, then

\Centerline{$(\undtheta(c\Bcup a_{\alpha})\setminus\theta_{\alpha}c)^{\ssbullet}
=\undtheta(c\Bcup a_{\alpha})^{\ssbullet}\Bsetminus c
\Bsubseteq(c\Bcup a_{\alpha})\Bsetminus c
\Bsubseteq a_{\alpha}$,}

\noindent so $F^{\ssbullet}\Bsubseteq a_{\alpha}$;  similarly,
$G^{\ssbullet}\Bsubseteq 1\Bsetminus a_{\alpha}$.
Next, $F\cap G=\emptyset$.   \Prf\ If $b$,
$c\in\frak C_{\alpha}$, then

$$\eqalign{(\undtheta(b\Bcup a_{\alpha})\setminus\theta_{\alpha}b)
   \cap(\undtheta(c\Bcup(1\Bsetminus a_{\alpha}))\setminus\theta_{\alpha}c)
&=\undtheta((b\Bcup a_{\alpha})\Bcap(c\Bcup(1\Bsetminus a_{\alpha})))
   \setminus(\theta_{\alpha}b\Bcup\theta_{\alpha}c)\cr
&\subseteq\undtheta(b\Bcup c)
   \setminus\theta_{\alpha}(b\Bcup c)
=\emptyset.  \text{ \Qed}\cr}$$

\noindent Choose any $E\in\Sigma$ such that $E^{\ssbullet}=a_{\alpha}$ and
set $E_{\alpha}=(E\cup F)\setminus G$;  then
$E_{\alpha}^{\ssbullet}=a_{\alpha}$, $F\subseteq E_{\alpha}$ and
$G\cap E_{\alpha}=\emptyset$.

If $c\in\frak C_{\alpha}$ and $c\Bsubseteq a_{\alpha}$, then
$\undtheta((1\Bsetminus c)\Bcup a_{\alpha})=\undtheta 1=X$, so

\Centerline{$\theta_{\alpha}c
=\undtheta((1\Bsetminus c)\Bcup a_{\alpha})
  \setminus\theta_{\alpha}(1\Bsetminus c)
\subseteq F\subseteq E_{\alpha}$.}

\noindent Similarly, if $c\in\frak C_{\alpha}$ and
$c\Bcap a_{\alpha}=0$, then

\Centerline{$\theta_{\alpha}c
=\undtheta((1\Bsetminus c)\Bcup(1\Bsetminus
a_{\alpha}))\setminus\theta_{\alpha}(1\Bsetminus c)
\subseteq G$}

\noindent is disjoint from $E_{\alpha}$.   We can therefore define a
Boolean homomorphism $\theta_{\alpha+1}:\frak C_{\alpha+1}\to\Sigma$ by
setting

\Centerline{$\theta_{\alpha+1}((b\Bcap a_{\alpha})
  \Bcup(c\Bsetminus a_{\alpha}))
=(\theta_{\alpha}b\cap E_{\alpha})
  \cup(\theta_{\alpha}c\setminus E_{\alpha})$}

\noindent for all $b$, $c\in\frak C_{\alpha}$ (312O), and
$\theta_{\alpha+1}$ will extend $\theta_{\beta}$ for every
$\beta\le\alpha+1$.
Because
$(\theta_{\alpha+1}a_{\alpha})^{\ssbullet}=E_{\alpha}^{\ssbullet}
=a_{\alpha}$ and
$\theta_{\alpha+1}c=\theta_{\alpha}c$ for every $c\in\frak C_{\alpha}$,
$(\theta_{\alpha+1}a)^{\ssbullet}=a$ for every $a\in\frak C_{\alpha+1}$.

I have still to check the other part of the inductive hypothesis.   If $b$,
$c\in\frak C_{\alpha}$, then

$$\eqalign{
\undtheta((b\Bcap a_{\alpha})\Bcup(c\Bsetminus a_{\alpha}))
&=\undtheta((b\Bcup c)\Bcap(c\Bcup a_{\alpha})
   \Bcap(b\Bcup(1\Bsetminus a_{\alpha})))\cr
&=\undtheta(b\Bcup c)\cap\undtheta(c\Bcup a_{\alpha})
   \cap\undtheta(b\Bcup(1\Bsetminus a_{\alpha}))\cr
&\subseteq\theta_{\alpha}(b\Bcup c)\cap(F\cup\theta_{\alpha}c)
   \cap(G\cup\theta_{\alpha}b)\cr
&\subseteq\theta_{\alpha+1}(b\Bcup c)
  \cap(\theta_{\alpha+1}a_{\alpha}\cup\theta_{\alpha+1}c)
  \cap(\theta_{\alpha+1}(1\Bsetminus a_{\alpha})\cup\theta_{\alpha+1}b)\cr
&=\theta_{\alpha+1}((b\Bcap a_{\alpha})\Bcup(c\Bsetminus a_{\alpha})),
\cr}$$

\noindent which is what we need to know.

\medskip

{\bf (c)} For non-zero limit ordinals $\alpha\le\omega_1$, we have
$\frak C_{\alpha}=\bigcup_{\beta<\alpha}\frak C_{\beta}$ so we can, and
must, take $\theta_{\alpha}=\bigcup_{\beta<\alpha}\theta_{\beta}$.

At the end of the induction, $\theta_{\omega_1}:\frak A\to\Sigma$ is an
appropriate lifting.
}%end of proof of 535F

\leader{535G}{Corollary}\cmmnt{ (see {\smc Neumann 1931})} % p. 111
Suppose that $\frak c=\omega_1$.   Then for any integer $r\ge 1$ there is
a Borel lifting $\theta$ of Lebesgue measure  on $\Bbb R^r$ such that
$x\in\theta E^{\ssbullet}$ whenever $E\subseteq\Bbb R^r$ is a Borel set and
$x$ is a density point of $E$.

\proof{ In 535F, let $\undtheta$ be lower Lebesgue density (341E),
interpreted
as a function from the Lebesgue measure algebra to the Borel
$\sigma$-algebra.   We need to check that $\undtheta E^{\ssbullet}$ is indeed
always a Borel set;  this is because

\Centerline{$\undtheta E^{\ssbullet}=\intstar E
=\{x:\lim_{n\to\infty}\Bover{\mu(E\cap B(x,2^{-n}))}{\mu B(x,2^{-n})}=1\}$}

\noindent and the functions $x\mapsto\mu(E\cap B(x,2^{-n}))$ are all
continuous (use 443B).
}%end of proof of 535G

\leader{535H}{}\cmmnt{ Again using the continuum hypothesis, we have
some results on `strong' liftings, as described in \S453.

\medskip

\noindent}{\bf Theorem} Let
$(X,\frak T,\Sigma,\mu)$ be an effectively
locally finite $\tau$-additive topological measure space with measure
algebra $\frak A$.   If $\#(\frak A)\le\add\mu$ and
$\mu$ is strictly positive, then $\mu$ has a strong lifting.

\wheader{535H}{0}{0}{0}{36pt}

\proof{{\bf (a)} For each $a\in\frak A$, set

\Centerline{$\overline{a}=\bigcap\{F:F\subseteq X$ is closed,
$F^{\ssbullet}\supseteq a\}$.}

\noindent Then $\overline{a}$ is closed and
$\overline{a}^{\ssbullet}\Bsupseteq a$ (414Ac).   If $a$, $b\in\frak A$,
then $\overline{a\Bcup b}=\overline{a}\cup\overline{b}$.   \Prf\ Of
course $\overline{a\Bcup b}\supseteq\overline{a}\cup\overline{b}$,
because the operation $\overline{\phantom{a}}$ is order-preserving.   On
the other hand, $\overline{a}\cup\overline{b}$ is a closed set and
$(\overline{a}\cup\overline{b})^{\ssbullet}\Bsupseteq a\Bcup b$, so
$\overline{a}\cup\overline{b}\supseteq\overline{a\Bcup b}$.\ \Qed

For a subalgebra $\frak B$ of $\frak A$, say that a function
$\theta:\frak B\to\Sigma$ is `potentially a strong lifting' if it is a
Boolean homomorphism and $(\theta b)^{\ssbullet}=b$ and
$\theta b\subseteq\overline{b}$ for every $b\in\frak B$.

\medskip

{\bf (b)} (The key.)  Suppose that $\frak B$ is a subalgebra of
$\frak A$, with cardinal less than $\add\mu$, and $c\in\frak A$;  let
$\frak B_1$ be the subalgebra of $\frak A$ generated by
$\frak B\cup\{c\}$.   If $\theta:\frak B\to\Sigma$ is potentially a
strong lifting, then it has an extension $\theta_1:\frak B_1\to\Sigma$
which is also potentially a strong lifting.

\Prf\ Set

\Centerline{$C_0=\bigcup\{\theta a:a\in\frak B,\,a\Bsubseteq c\}$,}

\Centerline{$D_0=\bigcap\{\theta b:b\in\frak B,\,c\Bsubseteq b\}$,}

\Centerline{$C_1
=\bigcup\{\theta a\setminus\overline{a\Bsetminus c}:a\in\frak B\}$,}

\Centerline{$D_1
=\bigcap\{(X\setminus\theta b)\cup\overline{b\Bcap c}:b\in\frak B\}$.}

\noindent Fix $E_0\in\Sigma$ such that $E_0^{\ssbullet}=c$.

If $a$, $a'$, $b$, $b'\in\frak B$ and $a'\Bsubseteq c\Bsubseteq b'$,
then

\Centerline{$a'\Bsubseteq b'$, so $\theta a'\subseteq\theta b'$;}

\Centerline{$\theta a'\Bcap\theta b=\theta(a'\Bcap b)
\subseteq\overline{a'\Bcap b}\subseteq\overline{b\Bcap c}$,
so
$\theta a'\subseteq(X\setminus\theta b)\cup\overline{b\Bcap c}$;}

\Centerline{$\theta a\setminus\theta b'=\theta(a\Bsetminus b')
\subseteq\overline{a\Bsetminus b'}\subseteq\overline{a\Bsetminus c}$,
so
$\theta a\setminus\overline{a\Bsetminus c}\subseteq\theta b'$;}

\Centerline{$\theta a\cap\theta b
=\theta(a\Bcap b)\subseteq\overline{a\Bcap b}
=\overline{a\Bcap b\Bcap c}\cup\overline{a\Bcap b\Bsetminus c}
\subseteq\overline{a\Bsetminus c}\cup\overline{b\Bcap c}$,}

\noindent so

\Centerline{$\theta a\setminus\overline{a\Bsetminus c}
\subseteq(X\setminus\theta b)\cup\overline{b\Bcap c}$.}

\noindent This shows that $C_0\cup C_1\subseteq D_0\cap D_1$.   At the
same time,

\Centerline{$E_0^{\ssbullet}=c\Bsupseteq a'$, so
$\theta a'\setminus E_0$ is negligible;}

\Centerline{$E_0^{\ssbullet}=c\Bsubseteq b'$, so $E_0\setminus\theta b'$
is negligible;}

\Centerline{$(E_0\cup\overline{a\Bsetminus c})^{\ssbullet}
\Bsupseteq c\Bcup(a\Bsetminus c)\Bsupseteq a=(\theta a)^{\ssbullet}$}

\noindent so
$(\theta a\setminus\overline{a\Bcap c})\setminus E_0$ is negligible;

\Centerline{$E_0^{\ssbullet}=c\Bsubseteq(1\Bsetminus b)\Bcup(b\Bcap c)
\Bsubseteq(X\Bsetminus\theta b)^{\ssbullet}
  \Bcup\overline{b\Bcap c}^{\ssbullet}$,}

\noindent so $E_0\setminus((X\setminus\theta b)\cup\overline{b\Bcap c})$
is negligible.   Because $\#(\frak B)<\add\mu$,
$(C_0\cup C_1)\setminus E_0$ and $E_0\setminus(D_0\cap D_1)$ are
measurable and negligible.

If we set

\Centerline{$E=(E_0\cup C_0\cup C_1)\cap(D_0\cap D_1)$,}

\noindent then $E\in\Sigma$, $E^{\ssbullet}=c$ and
$C_0\cup C_1\subseteq E_0\subseteq D_0\cap D_1$.   So we can set
$\theta_1c=E$ to define a homomorphism from $\frak B_1$ to $\Sigma$
(312O again), and we shall have $(\theta_1d)^{\ssbullet}=d$ for every
$d\in\frak B_1$.

We must check that $\theta_1d\subseteq\overline{d}$ for every
$d\in\frak B_1$.   Now $d$ is expressible as
$(b\Bcap c)\Bcup(a\Bsetminus c)$ for some $a$, $b\in\frak B$, and in this
case

\Centerline{$\theta b\cap E
\subseteq\theta b\cap((X\setminus\theta b)\cup\overline{b\Bcap c})
\subseteq\overline{b\Bcap c}$,}

\Centerline{$\theta a\setminus E
\subseteq\theta a\setminus(\theta a\setminus\overline{a\Bsetminus c})
\subseteq\overline{a\Bsetminus c}$,}

\noindent so

\Centerline{$\theta_1d=(\theta b\cap E)\cup(\theta a\setminus E)
\subseteq\overline{b\Bcap c}\cup\overline{a\Bsetminus c}
=\overline{d}$.}

\noindent So $\theta_1$ is a potential strong lifting, as required.\
\Qed

\wheader{535H}{4}{2}{2}{36pt}

{\bf (c)} Enumerate $\frak A$ as $\family{\xi}{\kappa}{a_{\xi}}$ where
$\kappa\le\add\mu$, and for $\alpha\le\kappa$ let $\frak B_{\alpha}$ be
the subalgebra of $\frak A$ generated by $\{a_{\xi}:\xi<\alpha\}$.
Then (b) tells us that we can choose inductively a family
$\ofamily{\alpha}{\kappa}{\theta_{\alpha}}$ such that
$\theta_{\alpha}:\frak B_{\alpha}\to\Sigma$ is a potential strong
lifting and $\theta_{\alpha+1}$ extends $\theta_{\alpha}$ for each
$\alpha<\kappa$.   (At non-zero limit ordinals $\alpha$,
$\frak B_{\alpha}=\bigcup_{\xi<\alpha}\frak B_{\xi}$ so we can take
$\theta_{\alpha}$ to be the common extension of
$\bigcup_{\xi<\alpha}\theta_{\xi}$.   We need to know that $\mu$ is
strictly positive in order to be sure that $\overline{1}=X$, so that we
can take $\theta_01=X$.)   In this way we obtain a lifting
$\theta=\theta_{\kappa}$ of $\mu$.   Also
$\theta a\subseteq\overline{a}$ for every $a\in\frak A$.   Looking at
this from the other side, if $F\subseteq X$ is closed then
$\overline{F^{\ssbullet}}\subseteq F$ so
$\theta(F^{\ssbullet})\subseteq F$, and $\theta$ is a strong lifting.
}%end of proof of 535H

\leader{535I}{Corollary}\cmmnt{ (see {\smc Mokobodzki 75})}
%Theorem 5
Suppose that $\frak c=\omega_1$.   Let $(X,\frak T,\Sigma,\mu)$ be a
strictly positive $\sigma$-finite quasi-Radon measure space with Maharam
type at most $\omega_1=\frak c$.   Then $\mu$ has a strong Borel
lifting.

%any hope of strong Baire lifting?  or of getting to $\omega_2$?

\proof{ Because $\mu$ is $\sigma$-finite, its measure algebra $\frak A$
is ccc, and has size at most $\frak c^{\omega}=\omega_1$;  so we can
apply 535H to $\mu\restr\Cal B(X)$.
}%end of proof of 535I

\leader{535J}{}\cmmnt{ Under certain conditions, we can deduce the
existence of a strong lifting from the existence of a lifting.   The
basic case is the following.

\medskip

\noindent}{\bf Lemma} Let $(X,\frak T,\Sigma,\mu)$ be a completely
regular totally finite topological measure space with a Borel lifting
$\phi$.   Suppose that $K\subseteq X$ is a self-supporting set of
non-zero measure, homeomorphic to $\{0,1\}^{\Bbb N}$, such that
$K\cap G\subseteq\phi G$
for every open set $G\subseteq X$.   Then the subspace measure $\mu_K$
has a strong Borel lifting.

\proof{{\bf (a)} Taking $\Cal E$ to be the algebra of
relatively open-and-closed subsets of $K$, we have a Boolean
homomorphism $\psi_0:\Cal E\to\Cal B(X)$ such that
$E\subseteq\interior\psi_0E$ for every $E\in\Cal E$.   \Prf\ We have a
Boolean-independent sequence $\sequencen{E_n}$ in $\Cal E$ which generates
$\Cal E$ and separates the points of $K$.   Because every member
of $\Cal E$ is compact, we can choose for each $n\in\Bbb N$ an open
$H_n\subseteq X$ such that $E_n=K\cap H_n=K\cap\overline{H}_n$.
Define $h:X\to K$ by saying that, for every $n\in\Bbb N$
and $x\in X$, $h(x)\in E_n$ iff $x\in H_n$.
Define $\psi_0:\Cal E\to\Cal B(X)$ by setting $\psi_0E=h^{-1}[E]$ for
$E\in\Cal E$.   Then $\psi_0$ is a Boolean homomorphism.   The set

\Centerline{$\{E:E\in\Cal E$, $E\subseteq\interior\psi_0E$,
$K\setminus E\subseteq\interior\psi_0(K\setminus E)\}$}

\noindent is a subalgebra of $\Cal E$ containing every $E_n$, so is the
whole of $\Cal E$, and $\psi_0$ has the required property.\ \Qed

\medskip

{\bf (b)} Let $\frak A$ be the measure algebra of $\mu$, and
$\theta:\frak A\to\Cal B(X)$ the lifting corresponding to $\phi$.   Set
$\psi_1E=(\psi_0E)^{\ssbullet}$ for $E\in\Cal E$, so that
$\psi_1:\Cal E\to\frak A$ is a Boolean homomorphism.   Let
$\Cal I$ be the null ideal of $\mu_K$.   Because $K$ is
self-supporting, $\Cal E\cap\Cal I=\{\emptyset\}$.   Taking
$\Cal E'=\{E\symmdiff F:E\in\Cal E$, $F\in\Cal I\}$, $\Cal E'$ is a
subalgebra of $\Cal PK$, and we have a Boolean homomorphism
$\psi':\Cal E'\to\Cal E$ defined by setting $\psi'(E\symmdiff F)=E$
whenever $E\in\Cal E$ and $F\in\Cal I$;  set $\psi_1'=\psi_1\psi'$, so
that $\psi_1':\Cal E'\to\frak A$ is a Boolean homomorphism extending
$\psi_1$, and $\psi_1'F=0$ whenever $F\in\Cal I$.   Because $\mu$ is
totally finite, $\frak A$ is Dedekind complete, and there is a Boolean
homomorphism $\tilde\psi_1:\Cal PK\to\frak A$ extending $\psi_1'$
(314K).   Now set

\Centerline{$\phi_1E
=K\cap(\phi E\cup(\theta\tilde\psi_1E\setminus\phi K))$}

\noindent for every measurable $E\subseteq K$.   Then $\phi_1$ is a
strong lifting for $\mu_K$.    \Prf\ $\phi\restr\Sigma_K$
is a Boolean homomorphism from the domain $\Sigma_K$ of $\mu_K$ to
$\Cal B(\phi K)$, while $E\mapsto\theta\tilde\psi_1E\setminus\phi K$ is
a Boolean homomorphism from $\Sigma_K$ to $\Cal B(X\setminus\phi K)$;
putting these together, $\phi_1$ is a Boolean homomorphism from
$\Sigma_K$ to $\Cal B(K)$.   If $E\in\Sigma_K$, then
$E\symmdiff(K\cap\phi E)$ and
$K\setminus\phi K$ are negligible, so
$E\symmdiff\phi_1E$ is negligible.   If $E\in\Sigma_K$ is negligible,
then $\phi E=\emptyset$, $\psi_1'E=0$ and $\phi_1E$ is empty.
Thus $\phi_1$ is a lifting for $\mu_K$.   Morover, if $E\in\Cal E$, set
$G=\interior\psi_0E$, so that $E=K\cap G$.   In this case,

\Centerline{$E\subseteq\phi G=\theta G^{\ssbullet}
\subseteq\theta(\psi_0E)^{\ssbullet}=\theta\psi_1E
=\theta\tilde\psi_1E$,}

\noindent while

\Centerline{$E\cap\phi K\subseteq\phi G\cap\phi K=\phi E$;}

\noindent so $E\subseteq\phi_1E$.   So if $V\subseteq K$ is relatively
open,

\Centerline{$V=\bigcup\{E:E\in\Cal E$, $E\subseteq V\}
\subseteq\bigcup\{\phi_1E:E\in\Cal E$, $E\subseteq V\}
\subseteq\phi_1V$.}

\noindent Thus $\phi_1$ is strong.\ \Qed
}%end of proof of 535J

\leader{535K}{Lemma} Let $X$ be a metrizable space, $\mu$ an atomless
Radon measure on $X$ and $\nu$ an atomless strictly positive Radon
measure on $\{0,1\}^{\Bbb N}$.   Let $\Cal K$ be the family of those
subsets $K$ of $X$ such that $K$, with the subspace topology and
measure, is isomorphic to $\{0,1\}^{\Bbb N}$ with its usual topology and
a scalar multiple of $\nu$.   Then $\mu$ is inner regular with respect
to $\Cal K$.

\proof{{\bf (a)} It will be helpful to note that if $E\in\dom\mu$ and
$\gamma<\mu E$ there is a compact set $K\subseteq E$ such that
$\mu K=\gamma$.   \Prf\ Let $\sequencen{\gamma_n}$ be a strictly
decreasing sequence with $\gamma_0<\mu E$ and $\inf_{n\in\Bbb
N}\gamma_n=\gamma$.   Choose $\sequencen{K_n}$, $\sequencen{E_n}$
inductively as follows.   $E_0=E$.   Given that $\mu E_n>\gamma_n$, let
$K_n\subseteq E_n$ be a compact set such that $\mu K_n\ge\gamma_n$;  now
let $E_{n+1}$ be a measurable set with measure $\gamma_n$ (215D, because
$\mu$ is atomless).   At the end of the induction, set
$K=\bigcap_{n\in\Bbb N}K_n$.\ \Qed

\medskip

{\bf (b)} Now for the main argument.   Suppose that $0\le\gamma<\mu E$.
Let $\sequencen{\gamma_n}$ be a strictly decreasing sequence with
$\gamma_0<\mu E$ and $\inf_{n\in\Bbb N}\gamma_n=\gamma$.   Set
$\gamma'_n=\bover12(\gamma_n+\gamma_{n+1})$ for each $n$.   For
$\sigma\in\bigcup_{n\in\Bbb N}\{0,1\}^n$, set
$I_{\sigma}=\{z:\sigma\subseteq z\in\{0,1\}^{\Bbb N}\}$.
Let $K_0$ be a compact subset of $E$ of measure $\gamma_0$;  because $X$
is metrizable, $K_0$ is second-countable; let $\sequencen{V_n}$ run over
a base for the topology of $K_0$.   Choose $\sequencen{m(n)}$ and
$L_{\sigma}$, for $\sigma\in\{0,1\}^{m(n)}$, as follows.   Start with
$m(0)=0$ and $L_{\emptyset}=K_0$.   Given that
$\family{\sigma}{\{0,1\}^{m(n)}}{L_{\sigma}}$ is a disjoint family of
compact subsets of $X$ with $\mu L_{\sigma}=\gamma_n\nu I_{\sigma}$ for
every $\sigma\in\{0,1\}^{m(n)}$, let $m(n+1)>m(n)$ be so large that
$\gamma_{n+1}\nu I_{\tau}<(\gamma_n-\gamma_{n+1})\nu I_{\sigma}$
whenever $\sigma\in\{0,1\}^{m(n)}$ and $\tau\in\{0,1\}^{m(n+1)}$.
(This is where we need to know that $\nu$ is atomless and strictly
positive.)    Now, for each $\sigma\in\{0,1\}^{m(n)}$, enumerate
$\{\tau:\sigma\subseteq\tau\in\{0,1\}^{m(n+1)}\}$ as
$\ofamily{i}{2^{m(n+1)-m(n)}}{\tau(\sigma,i)}$.   Choose inductively
disjoint compact sets $L_{\tau(\sigma,i)}\subseteq L_{\sigma}$, for
$i<2^{m(n+1)-m(n)}$, in such a way that
$\mu L_{\tau(\sigma,i)}=\gamma_{n+1}\nu I_{\tau(\sigma,i)}$ and
$L_{\tau(\sigma,i)}$ is always either included in $V_n$ or disjoint from
it;  this will be possible because when we come to choose
$L_{\tau(\sigma,i)}$, the measure of the set
$F=L_{\sigma}\setminus\bigcup_{j<i}L_{\tau(\sigma,j)}$ available will be

$$\eqalign{\gamma_n\nu I_{\sigma}
   -\sum_{j<i}\gamma_{n+1}\nu I_{\tau(\sigma,j)}
&\ge(\gamma_n-\gamma_{n+1})\nu I_{\sigma}
  +\gamma_{n+1}\nu I_{\tau(\sigma,i)}\cr
&>2\gamma_{n+1}\nu I_{\tau(\sigma,i)},\cr}$$

\noindent so at least one of $F\cap V_n$, $F\setminus V_n$ will be of
measure greater than $\gamma_{n+1}\nu I_{\tau(\sigma,i)}$.
Continue.

Set $K_n=\bigcup\{L_{\sigma}:\sigma\in\{0,1\}^{m(n)}\}$ for each
$n\in\Bbb N$, and $K=\bigcap_{n\in\Bbb N}K_n$.
The construction ensures that whenever $n\le k$,
$\sigma\in\{0,1\}^{m(n)}$, $\tau\in\{0,1\}^{m(k)}$ and
$\sigma\subseteq\tau$, then $L_{\tau}\subseteq L_{\sigma}$.   We
therefore have a function $f:K\to\{0,1\}^{\Bbb N}$ defined by saying
that $f(x)\restr m(n)=\sigma$ whenever $n\in\Bbb N$,
$\sigma\in\{0,1\}^{m(n)}$ and $x\in K\cap L_{\sigma}$.   Because all the
$L_{\sigma}$ are compact, $f$ is continuous.   But it is also
injective.   \Prf\ If $x$, $y\in K$ are different, there is an
$n\in\Bbb N$ such that $x\in V_n$ and $y\notin V_n$;  now
$f(x)\restr m(n+1)\ne f(y)\restr m(n+1)$.\ \Qed

For any $n\in\Bbb N$, $\sigma\in\{0,1\}^{m(n)}$ and $k\ge n$,

\Centerline{$\mu(\bigcup\{L_{\tau}:
  \sigma\subseteq\tau\in\{0,1\}^{m(k)}\})
=\sum_{\sigma\subseteq\tau\in\{0,1\}^{m(k)}}\gamma_k\nu I_{\tau}
=\gamma_k\nu I_{\sigma}$.}

\noindent So

\Centerline{$\mu(f^{-1}[I_{\sigma}])
=\inf_{k\ge n}\gamma_k\nu I_{\sigma}=\gamma\nu I_{\sigma}$.}

\noindent Thus the Radon measure $\mu f^{-1}$ on $\{0,1\}^{\Bbb N}$
agrees with the Radon measure $\gamma\nu$ on
$\{I_{\sigma}:\sigma\in\bigcup_{n\in\Bbb N}\{0,1\}^{m(n)}\}$;  as this
is a base for the topology of $\{0,1\}^{\Bbb N}$ closed under finite
intersections, $\mu f^{-1}$ and $\gamma\nu$ are identical (415H(v)).
Once again because $\nu$ is strictly positive, $f$ is surjective and is
a homeomorphism.   So $f$ witnesses that $K\in\Cal K$.
As $E$ and $\gamma$ are arbitrary, $\mu$ is inner regular with respect
to $\Cal K$.
}%end of proof of 535K

\leader{535L}{Lemma} (a) If $(X,\frak T)$ is a separable metrizable
space, there is a zero-dimensional separable metrizable topology
$\frak S$ on $X$, finer than $\frak T$, with the same Borel sets as
$\frak T$, such that $\frak T$ is a $\pi$-base for $\frak S$.

(b) If $X$ is a non-empty zero-dimensional separable metrizable space
without isolated points, it is homeomorphic to a dense subset of
$\{0,1\}^{\Bbb N}$.

(c) Any completely regular space with cardinal less than $\frak c$
is zero-dimensional.

\proof{{\bf (a)} Enumerate a countable base for $\frak T$ as
$\sequencen{U_n}$.   Define a sequence $\sequencen{\frak S_n}$ of
topologies on $X$ by saying that $\frak S_0=\frak T$ and that
$\frak S_{n+1}$ is the topology on $X$ generated by
$\frak S_n\cup\{V_n\}$, where $V_n$ is the closure of $U_n$ for
$\frak S_n$.   Inducing on $n$, we see that $\frak S_n$ is
second-countable and has the same Borel sets as $\frak T$, for every
$n$.   So taking $\frak S$ to be the topology generated by
$\bigcup_{n\in\Bbb N}\frak S_n$ (that is, the topology generated by
$\{U_n:n\in\Bbb N\}\cup\{V_n:n\in\Bbb N\}$), this also is
second-countable and has the same Borel sets as $\frak T$.   Each $V_n$
is open for $\frak S_{n+1}$ and closed for $\frak S_n$, so is
open-and-closed for $\frak S$.   Moreover, since

\Centerline{$U_n
=\bigcup\{U_m:m\in\Bbb N$, $\overline{U}^{\frak T}_m\subseteq U_n\}
=\bigcup\{V_m:m\in\Bbb N$, $V_m\subseteq U_n\}$}

\noindent for each $n$, $\{V_n:n\in\Bbb N\}$ is a base for $\frak S$
consisting of open-and-closed sets for $\frak S$, and $\frak S$ is
zero-dimensional.   Finally, observe that if $V_n$ is not empty, then
$V_n\supseteq U_n\ne\emptyset$, so $\frak T\supseteq\{U_n:n\in\Bbb N\}$
is a $\pi$-base for $\frak S$.

\medskip

{\bf (b)} The family $\Cal E_0$ of open-and-closed subsets of $X$ is a
base for the topology of $X$, so includes a countable base $\Cal U$
(4A2P(a-iii)).   Because $X$ has no isolated points, the subalgebra
$\Cal E_1$ of $\Cal E_0$ generated by $\Cal U$ is countable, atomless
and non-trivial, and must be isomorphic to the algebra $\Cal E$ of
open-and-closed subsets of $\{0,1\}^{\Bbb N}$
(316M).   Let
$\pi:\Cal E\to\Cal E_1$ be an isomorphism.   Then we have a function
$f:X\to\{0,1\}^{\Bbb N}$ defined by saying that, for $E\in\Cal E$,
$f(x)\in E$ iff $x\in\pi E$.   Because $\pi E\ne\emptyset$ for every
non-empty $E\in\Cal E$, $f[X]$ is dense in $\{0,1\}^{\Bbb N}$.
Because $\{f^{-1}[E]:E\in\Cal E\}=\Cal E_1\supseteq\Cal U$ is a base
for the topology of $X$, $f$ is a homeomorphism between $X$ and
$f[X]$.

\medskip

{\bf (c)} If $X$ is a completely regular space and $\#(X)<\frak c$,
$G\subseteq X$ is open and $x\in G$, let $f:X\to[0,1]$ be a continuous
function such that $f(x)=1$ and $f(y)=0$ for $y\in X\setminus G$.
Because $\#(X)<\frak c$, there is an $\alpha\in[0,1]\setminus f[X]$, and
now $\{y:f(x)>\alpha\}=\{y:f(x)\ge\alpha\}$ is an open-and-closed set
containing $x$ and included in $G$.   As $x$ and $G$ are arbitrary, $X$
is zero-dimensional.
}%end of proof of 535L

\leader{535M}{Lemma} Suppose that there is a Borel probability measure
on $\{0,1\}^{\Bbb N}$ with a strong lifting.   Then whenever $X$ is a
separable metrizable space and $D\subseteq X$ is a dense set, there is a
Boolean homomorphism $\phi$ from $\Cal PD$ to the Borel $\sigma$-algebra
$\Cal B(X)$ of $X$ such that
$\phi A\subseteq\overline{A}$ for every $A\subseteq D$.

\proof{{\bf case 1} Suppose that $X$ is countable.   Then it is
zero-dimensional (535Lc), so has a base $\Cal U$ consisting of
open-and-closed sets;  let $\Cal E$ be the algebra of sets generated by
$\Cal U$.   For $E\in\Cal E$ set $\pi E=E\cap D$;   then $\pi$ is an
isomorphism between $\Cal E$ and a subalgebra $\Cal E'$ of $\Cal PD$.
Because $\Cal B(X)=\Cal PX$ is Dedekind complete, the Boolean homomorphism
$\pi^{-1}:\Cal E'\to\Cal E$ extends to a Boolean homomorphism
$\phi:\Cal PD\to\Cal PX=\Cal B(X)$ (314K again).   If $A\subseteq D$ and
$x\in X\setminus\overline{A}$, then there is a $U\in\Cal U$ such that
$x\in U$ and $A\cap U=\emptyset$, in which case

\Centerline{$\phi A\subseteq\pi^{-1}(D\setminus U)=X\setminus U$}

\noindent does not contain $x$.   As $x$ is arbitrary,
$\phi A\subseteq\overline{A}$;  as $A$ is arbitrary, $\phi$ has the
required property.

\medskip

{\bf case 2} Suppose that $X$ is zero-dimensional and has no isolated
points.   If $X$ is empty the result is trivial;  otherwise, by 535Lb,
we may suppose that $X$ is a dense subset of $\{0,1\}^{\Bbb N}$.   This
time, let $\Cal E$ be the algebra of open-and-closed subsets of
$\{0,1\}^{\Bbb N}$.   For $E\in\Cal E$, set
$\pi E=E\cap D$.   Because $D$ is dense in $X$ and therefore in
$\{0,1\}^{\Bbb N}$, $\pi$ is an isomorphism between $\Cal E$ and a
subalgebra $\Cal E'$ of $\Cal PD$.    Fix a Borel probability measure
$\mu$ on $\{0,1\}^{\Bbb N}$ with a strong lifting $\theta$,
and let $\frak A$ be the measure algebra of
$\mu$.   Then $A\mapsto(\pi^{-1}A)^{\ssbullet}$ is a Boolean
homomorphism from $\Cal E'$ to $\frak A$;  because $\frak A$ is Dedekind
complete, it extends to a Boolean homomorphism $\psi:\Cal PD\to\frak A$.
For $E\subseteq\{0,1\}^{\Bbb N}$, set $\tilde\pi E=E\cap X$.
Then $\phi=\tilde\pi\theta\psi$ is a Boolean homomorphism from
$\Cal PD$ to $\Cal B(X)$.   If $A\subseteq D$ and
$x\in X\setminus\overline{A}$, then there is an $E\in\Cal E$ such that
$x\in E$ and $A\cap E=\emptyset$, in which case

\Centerline{$\phi A\subseteq\theta\psi A
\subseteq\theta\psi(D\setminus E)
=\theta(\{0,1\}^{\Bbb N}\setminus E)^{\ssbullet}
=\{0,1\}^{\Bbb N}\setminus E$,}

\noindent and $x\notin\phi A$.   As $x$ and $A$ are arbitrary, $\phi$ is
a suitable homomorphism.

\medskip

{\bf case 3} Suppose that $X$ has no isolated points.   Write $\frak T$
for the given topology on $X$.   By 535La, there is a finer
zero-dimensional separable metrizable topology $\frak S$ on $X$, with
the same Borel sets, such that $\frak T$ is a $\pi$-base for $\frak S$.
If $V\in\frak S$ is non-empty, there is a non-empty $U\in\frak T$ such
that $U\subseteq V$, and $D\cap V\supseteq D\cap U$ is non-empty;  so
$D$ is $\frak S$-dense.   By case 2, there is a Boolean homomorphism
$\phi:\Cal PD\to\Cal B(X,\frak S)$ such that
$\phi A\subseteq\overline{A}^{\frak S}$ for every $A\subseteq D$.   As
$\Cal B(X,\frak S)=\Cal B(X,\frak T)$, and
$\overline{A}^{\frak S}\subseteq\overline{A}^{\frak T}$ for every
$A\subseteq X$, this $\phi$ satisfies the conditions required.

\medskip

{\bf general case} In general, let $\Cal G$ be the family of countable
open subsets of $X$, and $G_0=\bigcup\Cal G$;  because $X$ is separable
and metrizable, therefore hereditarily Lindel\"of, $G_0$ is countable.
Set $Z=X\setminus G_0$, and let $D_0$ be a countable dense subset of
$Z$;  set $Y=D\cup G_0\cup D_0$.   By case 1, there is a Boolean
homomorphism $\phi_0:\Cal PD\to\Cal PY$ such that
$\phi_0A\subseteq\overline{A}$ for every $A\subseteq D$.   By case 3,
there is a Boolean homomorphism $\phi_1:\Cal P(Y\cap Z)\to\Cal B(Z)$
such that $\phi_1B\subseteq\overline{B}$ for every $B\subseteq Y\cap Z$.
Now set

\Centerline{$\phi A=(\phi_0A\setminus Z)\cup\phi_1(Z\cap\phi_0A)$}

\noindent for every $A\subseteq D$.   Then $\phi$ is a Boolean
homomorphism from $\Cal PD$ to $\Cal B(X)$;  and if
$A\subseteq D$, then

\Centerline{$\phi A
\subseteq\phi_0A\cup\phi_1(Z\cap\phi_0A)
\subseteq\overline{A}\cup\overline{Z\cap\overline{A}}
=\overline{A}$,}

\noindent so in this case also we have a homomorphism of the kind we
need.
}%end of proof of 535M

\leader{535N}{Theorem} Suppose there is a metrizable space $X$ with
a non-zero atomless semi-finite
tight Borel measure $\mu$ which has a lifting.
Then whenever $Y$ is a metrizable space and $\nu$ is a strictly positive
$\sigma$-finite Borel measure on $Y$, $\nu$ has a strong lifting.

\proof{{\bf (a)} Let $\phi$ be a lifting for $\mu$.   Then there is
a Borel set $E\subseteq X$, of non-zero finite measure, such that
$E\cap G\subseteq\phi G$ for every open $G\subseteq X$.
\Prf\ Let $L_0\subseteq X$ be a compact set of
non-zero measure;  then $L_0$ has a countable base $\Cal U$;  set
$E=L_0\cap\phi L_0\setminus\bigcup_{U\in\Cal U}(U\symmdiff\phi U)$,
so that $\mu E=\mu L_0\in\ooint{0,\infty}$.   If $G\subseteq X$ is open
and $x\in E\cap G$, then there is a
$U\in\Cal U$ such that $x\in U\subseteq G$.   Since
$x\in E\cap U$, $x\in\phi U\subseteq\phi G$.   As $x$ and $G$ are
arbitrary, we have an appropriate $E$.\ \Qed

\medskip

{\bf (b)} Let $\lambda$ be any strictly positive atomless Radon measure
on $\{0,1\}^{\Bbb N}$.   There is a compact set $K\subseteq E$ such that
$K$, with its induced topology and measure, is isomorphic to
$\{0,1\}^{\Bbb N}$ with its usual topology and a non-zero multiple of
$\lambda$, by 535K.   In particular, $K$ is self-supporting.   By 535J,
the subspace measure on $K$ has a strong Borel lifting.   It follows at
once that $\lambda$ has a strong Borel lifting.

\medskip

{\bf (c)} Refining (b) slightly, we see that if
$Y\subseteq\{0,1\}^{\Bbb N}$ is a dense set and $\lambda$ is a strictly
positive atomless totally finite Borel measure on $Y$, then $\lambda$
has a strong lifting.   \Prf\ There is a Radon measure $\nu$ on
$\{0,1\}^{\Bbb N}$ such that $\nu E=\lambda(Y\cap E)$ for every Borel
set $E\subseteq\{0,1\}^{\Bbb N}$ (416F);
because $\lambda$ is atomless, so is
$\nu$;  because $\lambda$ is strictly positive and $Y$ is dense, $\nu$
is strictly positive.   So $\nu$ has a strong Borel lifting $\psi_0$
say.   If $E$, $F\in\Cal B(\{0,1\}^{\Bbb N})$ and $E\cap Y=F\cap Y$,
then $\nu(E\symmdiff F)=0$ and $\psi_0E=\psi_0F$;  we therefore have a
Boolean homomorphism $\psi:\Cal B(Y)\to\Cal B(Y)$ defined by setting
$\psi(E\cap Y)=Y\cap\psi_0E$ for every Borel set
$E\subseteq\{0,1\}^{\Bbb N}$.   It is easy to check that $\psi$ is a
lifting for $\lambda$, and it is strong because if
$G\subseteq\{0,1\}^{\Bbb N}$ is open then
$\psi(Y\cap G)=Y\cap\psi_0G\subseteq Y\cap G$.\ \Qed

\medskip

{\bf (d)} If $(Y,\frak S)$ is a separable metrizable space with a
strictly positive atomless totally finite Borel measure $\nu$, then $\nu$
has a strong lifting.   \Prf\ If $Y=\emptyset$ the
result is trivial.   Otherwise, by 535La, there is a finer separable
metrizable topology $\frak S'$ on $Y$ with the same Borel sets such that
$\frak S$ is a $\pi$-base for $\frak S'$.   Because $\frak S$ and
$\frak S'$ have the same Borel sets, $\nu$ is a Borel measure for
$\frak S'$;  because every non-empty $\frak S'$-open set includes a
non-empty $\frak S$-open set, $\nu$ is strictly positive for $\frak S'$;
because $\nu$ is atomless, $Y$ has no $\frak S'$-isolated points.
By 535Lb, $(Y,\frak S')$ is homeomorphic to a dense subset of
$\{0,1\}^{\Bbb N}$;  by (c) above, $\nu$ has a lifting $\phi$ which is
strong with respect to the topology $\frak S'$.    But now $\phi$ is
still strong with respect to the coarser topology $\frak S$.\ \Qed

\medskip

{\bf (e)} Now suppose that $Y$ is a separable metrizable space with a
strictly positive totally finite Borel measure $\nu$.   Then $\nu$ has a
strong lifting.   \Prf\ The set $D=\{y:\nu\{y\}>0\}$ is countable.   If
$D$ is empty, then the result is immediate from (d) applied to a scalar
multiple of $\nu$.  (If $\nu Y=0$ then $Y=\emptyset$ and the result is
trivial.)    Otherwise, let
$\nu_{Y\setminus D}$ be the subspace measure;  then $\nu_{Y\setminus D}$
is a totally finite Borel measure on $Y\setminus D$, and is zero on
singletons, so must be atomless.   Because $Y\setminus D$ is
hereditarily Lindel\"of, $\nu_{Y\setminus D}$ is $\tau$-additive;  let
$Z$ be its support, and $\nu_Z$ the subspace measure on $Z$.   Then
$\nu_Z$ has a strong Borel lifting $\psi_0$, by (d) again.   Next, $Z$
is relatively closed
in $Y\setminus D$, so is expressible as $F\setminus D$ for some closed
set $F\subseteq Y$.   If $x\in Y\setminus F$ and $G$ is an open set
containing $x$, then $G'=G\setminus F$ is a non-empty open set, so has
non-zero measure, while
$\nu_{Y\setminus D}(G'\setminus D)=0$;  accordingly
$G'\cap D\ne\emptyset$.   This shows that
$Y\setminus F\subseteq\overline{D}$ so $D$ is dense in $Y\setminus Z$.
Now 535M (with (b) above) tells us that there is a Boolean homomorphism
$\psi_1:\Cal PD\to\Cal B(Y\setminus Z)$ such that
$\psi_1A\subseteq\overline{A}$ for every $A\subseteq D$.   Define
$\psi:\Cal B(Y)\to\Cal B(Y)$ by setting

\Centerline{$\psi E
=\psi_0(E\cap Z)\cup(E\cap D)\cup(\psi_1(E\cap D)\setminus D)$}

\noindent for every Borel set $E\subseteq Y$.   $\psi$ is a Boolean
homomorphism because $\psi_0$ and $\psi_1$ are.   If $\nu E=0$, then
$\nu_Z(E\cap Z)=0$ and $E\cap D=\emptyset$, so $\psi E=\emptyset$.
For any $E\in\Cal B(Y)$, $\psi_0(E\cap Z)\symmdiff(E\cap Z)$ and
$Y\setminus(D\cup Z)$ are negligible, so $E\symmdiff\psi E$ is
negligible.   Thus $\psi$ is a lifting for $\nu$.   Finally, for any $E$,

\Centerline{$\psi E
\subseteq\psi_0(E\cap Z)\cup(E\cap D)\cup\psi_1(E\cap D)
\subseteq\overline{E}$,}

\noindent so $\psi$ is a strong lifting.\ \Qed

\medskip

{\bf (f)} Finally, if $Y$ is a metrizable space and $\nu$ is a strictly
positive $\sigma$-finite Borel measure on $Y$, then $Y$ must be ccc,
therefore separable;  and there is a totally finite Borel measure
$\nuprime$ with the same null ideal as $\nu$, so that $\nuprime$ has a
strong lifting, by (e), which is also a strong lifting for $\nu$.
}%end of proof of 535N

%{\smc Burke 93}, 3.6 gives a version with an incorrect proof

\leader{535O}{Linear liftings} Let $(X,\Sigma,\mu)$ be a measure space,
with measure algebra $\frak A$.   Write
$\eusm L^{\infty}(\Sigma)$ for the space of bounded $\Sigma$-measurable
real-valued functions on $X$.  A {\bf linear lifting} for $\mu$ is

\inset{{\it either} a positive linear operator
$T:L^{\infty}(\mu)\to\eusm L^{\infty}(\Sigma)$ such that
$T(\chi X^{\ssbullet})=\chi X$ and
$(Tu)^{\ssbullet}=u$ for every $u\in L^{\infty}(\mu)$

{\it or} a positive linear operator
$S:\eusm L^{\infty}(\Sigma)\to\eusm L^{\infty}(\Sigma)$ such that
$S(\chi X)=\chi X$, $Sf=0$ whenever $f=0$ a.e.\ and $Sf\eae f$ for every
$f\in\eusm L^{\infty}(\Sigma)$.}

\cmmnt{\noindent As with liftings (see 341A-341B) we have a direct
correspondence between the two kinds of linear operator;  given $T$ as
in the first formulation, we can set $Sf=T(f^{\ssbullet})$ for every
$f\in\eusm L^{\infty}(\Sigma)$;  given $S$ as in the second formulation,
we can set $T(f^{\ssbullet})=Sf$ for every
$f\in\eusm L^{\infty}(\Sigma)$.
}%end of comment

If $\theta:\frak A\to\Sigma$ is a lifting for $\mu$, then we have a
corresponding Riesz homomorphism
$T:L^{\infty}(\frak A)\to\eusm L^{\infty}(\Sigma)$ such that
$T(\chi a)=\chi(\theta a)$ for every $a\in\frak A$\cmmnt{, by 363F}.
Identifying $L^{\infty}(\frak A)$ with $L^{\infty}(\mu)$\cmmnt{ as in
363I}, \cmmnt{we
see that} $T$ can be regarded as a linear lifting.   \cmmnt{Of course
the associated linear operator from
$\eusm L^{\infty}(\Sigma)$ to itself is the operator derived \cmmnt{by
the process of 363F} from the Boolean homomorphism
$E\mapsto\theta E^{\ssbullet}:\Sigma\to\Sigma$.}

\cmmnt{As in 535Aa,} I will say that a {\bf Borel linear lifting} is a
linear lifting such that all its values are Borel measurable functions;
\cmmnt{similarly,}
a {\bf Baire linear lifting} is a linear lifting such that all its
values are Baire measurable functions.

\leader{535P}{}\cmmnt{ I give a sample result to show that for some
purposes linear liftings are adequate.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a countably compact
measure space such that $\Sigma$ is countably generated, $(Y,\Tau,\nu)$
a $\sigma$-finite measure space with a linear lifting, and
$f:X\to Y$ an \imp\ function.   Then there is a disintegration
$\family{y}{Y}{\mu_y}$ of $\mu$ over $\nu$, consistent with $f$, such
that $y\mapsto\mu_yE$ is a $\Tau$-measurable function for every
$E\in\Sigma$.

\proof{ I use the method of 452H-452I.

\medskip

{\bf (a)} Suppose first that $\mu$ and $\nu$ are probability measures.
Let $S:L^{\infty}(\nu)\to\eusm L^{\infty}(\Tau)$ be a linear lifting for
$\nu$.   Let $T:L^{\infty}(\mu)\to L^{\infty}(\nu)$ be the positive
linear operator defined by saying that $\int_FTu=\int_{f^{-1}[F]}u$
whenever $u\in L^{\infty}(\mu)$ and $F\in\Tau$ (as in part (a) of the
proof of 452I).   For $y\in Y$ and $E\in\Sigma$, set

\Centerline{$\psi_yE=(ST(\chi E^{\ssbullet}))(y)$}

\noindent as in part (b) of the proof of 452H.   Because $\mu$ is
countably compact, we can use the argument of 452H to see that we have a
family $\family{y}{Y}{\mu'_y}$ of totally finite measures on $X$ such
that, for any $E\in\Sigma$, $\mu'_yE=\psi_yE$ for almost every $y\in Y$.

Let $\Cal H$ be a countable subalgebra of $\Sigma$ such that $\Sigma$ is
the $\sigma$-algebra of sets generated by $\Cal H$.   Set
$Y_0=\{y:\mu'_yH=\psi_yH$ for every $H\in\Cal H\}$, so that $Y_0$ is
conegligible;  let $Y_1\subseteq Y_0$ be a measurable conegligible set;
set $\mu_y=\mu'_y$ for $y\in Y_1$, and take $\mu_y$ to be the zero
measure on $X$ for $y\in Y\setminus Y_1$.   If $H\in\Cal H$, then

\Centerline{$\mu_yH=\psi_yH=ST(\chi H^{\ssbullet})(y)$}

\noindent for every $y\in Y_1$, so $y\mapsto\mu_yH$ is
$\Tau$-measurable;  also, of course,

\Centerline{$\int_F\mu_yH\nu(dy)
=\int_FST(\chi H^{\ssbullet})d\nu
=\int_FT(\chi H^{\ssbullet})
=\int_{f^{-1}[F]}\chi H^{\ssbullet}
=\mu(H\cap f^{-1}[F])$.}

Now consider the family $\Cal E$ of those $E\in\Sigma$ such that
$y\mapsto\mu_yE$ is $\Tau$-measurable and
$\int_F\mu_yE\nu(dy)=\mu(E\cap f^{-1}[F])$ for every $F\in\Tau$.   This
is a Dynkin class including $\Cal H$, so is the whole of
$\Sigma$;  which is what we need to know.

\medskip

{\bf (b)} In general, if $\nu Y=0$, the result is trivial.   Otherwise,
apply (a) to a suitable pair of indefinite-integral measures over $\mu$
and $\nu$, as in part (c) of the proof of 452I.
}%end of proof of 535P

\leader{535Q}{Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
probability spaces, and $\lambda$ the c.l.d.\ product measure on
$X\times Y$.   Suppose that $\lambda\restr\Sigma\tensorhat\Tau$ has a
linear lifting.   Then $\mu$ has a linear lifting.

\proof{ Let $S:\eusm L^{\infty}(\Sigma\tensorhat\Tau)
\to\eusm L^{\infty}(\Sigma\tensorhat\Tau)$ be a linear lifting for
$\lambda\restr\Sigma\tensorhat\Tau$.   For
$h\in\eusm L^{\infty}(\Sigma\tensorhat\Tau)$, set
$(Uh)(x)=\int h(x,y)\nu(dy)$ for every $x\in X$;  by 252P, $Uh$ is
well-defined and is $\Sigma$-measurable.   Now $U$ is a positive linear
operator from $\eusm L^{\infty}(\Sigma\tensorhat\Tau)$ to
$\eusm L^{\infty}(\Sigma)$, and $U(\chi(X\times Y))=\chi X$, because
$\nu Y=1$.   Note that

\Centerline{$\int|Uh|d\mu\le\int U|h|d\mu
=\iint|h(x,y)|\nu(dy)\mu(dx)=\int|h|d\lambda$}

\noindent for every $h\in\eusm L^{\infty}(\Sigma\tensorhat\Tau)$ (252P
again).
Next, for $f\in\eusm L^{\infty}(\Sigma)$ set $(Vf)(x,y)=f(x)$ for every
$x\in X$ and $y\in Y$, so that $V$ is a positive linear operator from
$\eusm L^{\infty}(\Sigma)$ to
$\eusm L^{\infty}(\Sigma\tensorhat\Tau)$.

Consider $S_1=USV:\eusm L^{\infty}(\Sigma)\to\eusm L^{\infty}(\Sigma)$.
This is a positive linear operator and $S_1(\chi X)=\chi X$.   If
$f\in\eusm L^{\infty}(\Sigma)$ and $f=0\,\,\mu$-a.e., then
$Vf=0\,\,\lambda$-a.e.\ and $SVf=0$, so $S_1f=0$.   For any
$f\in\eusm L^{\infty}(\Sigma)$,

\Centerline{$\int|f-S_1f|d\mu=\int|f-USVf|d\mu
=\int|UVf-USVf|d\mu\le\int|Vf-SVf|d\lambda=0$,}

\noindent so $f\eae S_1f$;  thus $S_1$ is a linear lifting for $\mu$.
}%end of proof of 535Q

\leader{535R}{Proposition} Write $\nu_{\omega}^2$ for the usual measure on $(\{0,1\}^{\omega})^2$, and $\Tau_{\omega}^{(2)}$ for its domain.
Suppose that $\nu_{\kappa}$ has a Baire linear lifting for some
$\kappa\ge\frak c^{++}$.   Then there is a Borel linear lifting $S$ for
$\nu_{\omega}^2$ which respects coordinates in the sense that if
$f\in\eusm L^{\infty}(\Tau_{\omega}^{(2)})$ is determined by a single
coordinate, then $Sf$ is determined by the same coordinate.

\proof{ Because $(\{0,1\}^{\kappa},\nu_{\kappa})$ is isomorphic, as
topological measure space, to
$(\{0,1\}^{\kappa\times\omega},\nu_{\kappa\times\omega})$, the latter
has a Baire linear lifting $S_0$ say.   For $I\subseteq\kappa$, let
$\Tau_I$ be the $\sigma$-algebra of Baire subsets of
$\{0,1\}^{\kappa\times\omega}$ determined by coordinates in
$I\times\omega$.   Then $\#(\Tau_I)\le\frak c$ whenever
$\#(I)\le\frak c$.   Also $\CalBa(\{0,1\}^{\kappa\times\omega})
=\bigcup\{\Tau_I:I\in[\kappa]^{\le\omega}\}$ (4A3N).   It follows that
for every $\xi<\kappa$ there is a set $I_{\xi}\subseteq\kappa$, with cardinal
at most $\frak c$, such that $\xi\in I_{\xi}$ and $S_0(\chi E)$ is
$\Tau_{I_{\xi}}$-measurable whenever $E\in\Tau_{I_{\xi}}$;  so that
$S_0f$ is $\Tau_{I_{\xi}}$-measurable whenever
$f:\{0,1\}^{\kappa\times\omega}\to\Bbb R$ is bounded and
$\Tau_{I_{\xi}}$-measurable.

Because $\kappa\ge\frak c^{++}$, there are $\xi$, $\eta<\kappa$ such
that $\xi\notin I_{\eta}$ and $\eta\notin I_{\xi}$ (5A1J(a-iii)).   Set
$J=\{\xi\}\times\omega$, $K=\{\eta\}\times\omega$ and
$L=(\kappa\times\omega)\setminus(J\cup K)$, so that
$\{0,1\}^{\kappa\times\omega}$ can be identified with
$\{0,1\}^{J\cup K}\times\{0,1\}^L$ and
$\CalBa(\{0,1\}^{\kappa\times\omega})$
with $\CalBa(\{0,1\}^{J\cup K})\tensorhat\CalBa(\{0,1\}^L)$.   Set
$(Vf)(w,z)=f(w)$ when $f:\{0,1\}^{J\cup K}\to\Bbb R$ is a function,
$w\in\{0,1\}^{J\cup K}$ and $z\in\{0,1\}^L$;   and
$(Uh)(w)=\int h(w,z)\nu_L(dz)$ when
$h:\{0,1\}^{\kappa\times\omega}\to\Bbb R$ is a bounded Baire measurable
function and $w\in\{0,1\}^{J\cup K}$.
Then $S_1=US_0V$ is a Baire linear lifting for $\nu_{J\cup K}$, just as
in 535Q.   Moreover, if $f:\{0,1\}^{J\cup K}\to\Bbb R$ is a bounded
Baire measurable function determined by coordinates in $J$, in the sense
that $f(x,y)=f(x,y')$ whenever $x\in\{0,1\}^J$ and $y$,
$y'\in\{0,1\}^K$, then $S_1f$ is determined by coordinates in $J$.
\Prf\ $Vf$ is determined by coordinates in $J$, so $S_0Vf$ is determined
by coordinates in $I_{\xi}\times\omega$;  since
$K\cap(I_{\xi}\times\omega)$ is empty, $S_0Vf(x,y,z)=S_0Vf(x,y',z)$ for
all $x\in\{0,1\}^J$, $z\in\{0,1\}^L$ and $y$, $y'\in\{0,1\}^K$.   It
follows at once that

\Centerline{$S_1f(x,y)=\int S_0Vf(x,y,z)\nu_L(dz)
=\int S_0Vf(x,y',z)\nu_L(dz)=S_1f(x,y')$}

\noindent whenever $x\in\{0,1\}^J$ and $y$, $y'\in\{0,1\}^K$.\ \QeD\
Similarly, if
$f:\{0,1\}^{J\cup K}\to\Bbb R$ is a bounded Baire measurable function
determined by coordinates in $K$, then $S_1f$ is determined by
coordinates in $K$.

Now we can
transfer $S_1$ from $\{0,1\}^{J\cup K}\cong\{0,1\}^J\times\{0,1\}^K$ to
$(\{0,1\}^{\omega})^2$, and we shall obtain a Baire (or Borel)
linear lifting $S$ for $\nu_{\omega}^2$ which respects coordinates.
}%end of proof of 535R

\exercises{\leader{535X}{Basic exercises (a)}
%\spheader 535Xa
Let $(X,\Sigma,\mu)$ be a measure space with a lifting,
and $A$ any subset of $X$.   Show that if $A$ has a measurable envelope
then the subspace measure $\mu_A$ has a lifting.   \Hint{322I.}
%-

\spheader 535Xb Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of
measure spaces, with $\mu_iX_i>0$ for every $i\in I$, and
$(X,\Sigma,\mu)$ their direct sum.   Show that $\mu$ has a lifting iff
every $\mu_i$ has a lifting.
%-

\spheader 535Xc Let $\frak A$ be a Boolean algebra and $I$ a proper
ideal of $\frak A$.   Suppose that $\sup A$ is defined in $\frak A$ and
belongs to $I$ whenever $A\subseteq I$ and $\#(A)<\#(\frak A)$.
Show that there is a Boolean homomorphism $\theta:\frak A/I\to\frak A$
such that $(\theta b)^{\ssbullet}=b$ for every $b\in\frak A/I$.
\Hint{enumerate $\frak A$ as $\{a_{\xi}:\xi<\kappa\}$;  let
$\frak C_{\xi}$ be the subalgebra of $\frak A/I$ generated by
$\{a_{\eta}^{\ssbullet}:\eta<\xi\}$;  construct
$\theta\restrp\frak C_{\xi}$ inductively by choosing
$\theta a_{\xi}^{\ssbullet}$ appropriately.}
%535D

\spheader 535Xd Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $I$ a proper ideal of $\frak A$.   Show that if the quotient
Boolean algebra $\frak A/I$ is tightly $\omega_1$-filtered, then
there is a Boolean homomorphism $\theta:\frak A/I\to\frak A$ such that
$(\theta b)^{\ssbullet}=b$ for every $b\in\frak A/I$.
%535D

\spheader 535Xe Let $\frak A$ be a tightly $\omega_1$-filtered Boolean
algebra, $\frak B$ a Dedekind $\sigma$-complete Boolean
algebra and $\frak A_0$ a countable subalgebra of $\frak A$.   Show that
every Boolean homomorphism from $\frak A_0$ to $\frak B$ extends to a
Boolean homomorphism from $\frak A$ to $\frak B$.
%535D

\spheader 535Xf Let $\frak A$, $\frak B$ be
Boolean algebras such that $\sup A$ is defined in $\frak A$ whenever
$A\subseteq\frak A$ and $\#(A)<\#(\frak B)$,
and $\pi:\frak A\to\frak B$ a surjective
Boolean homomorphism.   Suppose that
$\undtheta:\frak B\to\frak A$ is such that $\undtheta 0=0$,
$\pi\undtheta b\Bsubseteq b$ for every $b\in\frak B$
and $\undtheta(b\Bcap c)=\undtheta b\Bcap\undtheta c$ for all
$b$, $c\in\frak B$.   Show that there is a Boolean homomorphism
$\theta:\frak B\to\frak A$ such that $\undtheta b\Bsubseteq\theta b$ and
$\pi\theta b=b$ for every $b\in\frak B$.
%535F

\spheader 535Xg Suppose that $\frak c\le\omega_2$ and
$\FN(\Cal P\Bbb N)=\omega_1$.   Show that $\nu_{\kappa}$ has a strong
Baire lifting whenever $\kappa\le\omega_2$.   \Hint{let
$\ofamily{\xi}{\kappa}{e_{\xi}}$ be the standard generating family for
$\frak B_{\kappa}$.   Show that there is a tight $\omega_1$-filtration
$\ofamily{\eta}{\zeta}{a_{\eta}}$ of
$\frak B_{\kappa}$ such that for every $\xi<\kappa$ there is an
$\eta<\zeta$ such that the closed subalgebras generated by
$\{e_{\delta}:\delta<\xi\}$ and $\{a_{\delta}:\delta<\eta\}$ are the
same and $e_{\xi}=a_{\eta}$.}
%535I method of 518M does the job

\spheader 535Xh Suppose that $\frak c\le\omega_2$ and
$\FN(\Cal P\Bbb N)=\omega_1$.   Show that
whenever $X$ is a separable metrizable space and $D\subseteq X$ is a
dense set, there is a Boolean homomorphism $\phi:\Cal PD\to\Cal B(X)$
such that $\phi A\subseteq\overline{A}$ for every $A\subseteq D$.
%535M 535Xe

\spheader 535Xi Let $(X,\Sigma,\mu)$ be a measure space.   Show that a
linear lifting $S:\eusm L^{\infty}(\Sigma)\to\eusm L^{\infty}(\Sigma)$
of $\mu$ corresponds to a lifting iff it is `multiplicative', that is,
$S(f\times g)=Sf\times Sg$ for all $f$, $g\in\eusm L^{\infty}(\Sigma)$.
%535O

\spheader 535Xj Let $(X,\Sigma,\mu)$ be a strictly localizable measure
space with non-zero measure.   Suppose that $\nu_{\kappa}$ has a Baire
linear lifting for every infinite cardinal $\kappa$ such that the
Maharam-type-$\kappa$ component of the measure algebra of $\mu$ is
non-zero.   Show that $\mu$ has a linear lifting.
%535O 535B

\spheader 535Xk Let $(X,\Sigma,\mu)$ be a probability space such that
whenever $\Cal E\subseteq\Sigma$, $\#(\Cal E)\le\frak c$ and
$\bigcup\Cal E$ is negligible, then $\bigcup\Cal E\in\Sigma$.   Show
that $\mu$ has a linear lifting.   \Hint{363Yf.}
%535O

\spheader 535Xl Let $(Y,\Tau,\nu)$ be a $\sigma$-finite measure space
with a linear lifting, $Z$ a set, $\Upsilon$ a countably generated
$\sigma$-algebra of subsets of $Z$, and $\mu$ a measure with domain
$\Tau\tensorhat\Upsilon$ such that $\nu$ is the marginal measure of
$\mu$ on $Y$ and the marginal measure of $\mu$ on $Z$ is countably compact.
Show that there is a family
$\family{y}{Y}{\mu_y}$ of measures with domain $\Upsilon$ such that
$y\mapsto\mu_yH$ is a $\Tau$-measurable function for every
$H\in\Upsilon$ and $\mu W=\int\mu_yW[\{y\}]\nu(dy)$ for every
$W\in\Tau\tensorhat\Upsilon$.
%535P 452M

\spheader 535Xm Let $(X,\frak T,\Sigma,\mu)$ and $(Y,\frak S,\Tau,\nu)$
be $\tau$-additive topological probability spaces, and $\lambda$ the
$\tau$-additive product measure on $X\times Y$ (417F\formerly{4{}17G}).   
Suppose that
$\lambda$ has a Borel linear lifting and that $\mu$ is inner regular
with respect to the Borel sets.   Show that $\mu$ has a Borel linear
lifting.
%

\leader{535Y}{Further exercises (a)}
%\spheader 535Ya
Suppose that we are provided with a bijection between
$\Cal B(\Bbb R)$ and $\omega_1$, but are otherwise not permitted to use
the axiom of choice.   Show that we can construct a Borel lifting for
Lebesgue measure.
%535D

\spheader 535Yb Suppose that for every cardinal $\kappa$ there is a
Baire linear lifting for $\nu_{\kappa}$.   Show that for every
$n\in\Bbb N$ there is a Borel linear lifting $S$ for Lebesgue measure on
$[0,1]^n$ which ($\alpha$) respects coordinates in the sense that if
$f:[0,1]^n\to\Bbb R$ is a bounded measurable function determined by
coordinates in
$I\subseteq n$, then $Sf$ also is determined by coordinates in $I$
($\beta$) is symmetric in the sense that if $\rho:n\to n$ is any
permutation and
$(\hat\rho f)(x)=f(x\rho)$ for $x\in[0,1]^n$ and $f:[0,1]^n\to\Bbb R$,
then $S$ commutes with $\hat\rho$.
\Hint{5A1Jb.}
%535O

%\spheader ????b For any $n\in\Bbb N$ there is an infinite $\kappa$
%such that for $f:[\kappa]^n\to[\kappa]^{\le\frakc}$ there is an
%$I\in[\kappa]^{n+1}$ such that $I\cap f(J)\subseteq J$ for every
%$J\in[I]^n$.

\spheader 535Yc Let $(X,\Sigma,\mu)$ be a countably compact
measure space, $(Y,\Tau,\nu)$ a $\sigma$-finite measure space with a
linear lifting, and
$f:X\to Y$ an \imp\ function.   Suppose there is a family
$\Cal H\subseteq\Sigma$ such that $\Sigma$ is the $\sigma$-algebra of
sets generated by $\Cal H$ and $\#(\Cal H)<\add\nu$.   Show that there
is a disintegration
$\family{y}{Y}{\mu_y}$ of $\mu$ over $\nu$, consistent with $f$, such
that $y\mapsto\mu_yE$ is a $\Tau$-measurable function for every
$E\in\Sigma$.
%535P

\spheader 535Yd\dvAnew{2009}
({\smc T\"ornquist 11}) Let $(X,\Sigma,\mu)$ be a countably separated
perfect complete strictly localizable measure space,
$\frak A$ its measure algebra and $G$ a subgroup of $\Aut\frak A$ of
cardinal at most $\min(\add\Cal N,\frak p)$, where $\Cal N$ is the
null ideal of Lebesgue measure on $\Bbb R$.   Show that there is an action
$\action$ of $G$ on $X$ such that
$\pi\action E=\{\pi\action x:x\in E\}$ belongs to $\Sigma$ and
$(\pi\action E)^{\ssbullet}=\pi(E^{\ssbullet})$ whenever
$\pi\in G$ and $E\in\Sigma$.   \Hint{344C, 425Ya.}
%425Ba mt53bits
}%end of exercises

\leader{535Z}{Problems (a)} Can it be that every probability space has a
lifting?

\cmmnt{By 535B, it is enough to consider
$(\{0,1\}^{\kappa},\vthsp\CalBa(\{0,1\}^{\kappa}),
\vthsp\nu_{\kappa}\restr\CalBa(\{0,1\}^{\kappa}))$ where $\kappa$ is a
cardinal.
Since Mokobodzki's theorem (535Eb) deals with $\kappa\le\omega_2$ when
$\frak c=\omega_1$, the key case to consider seems to be
$\kappa=\omega_3$.
}%end of comment

\spheader 535Zb Suppose that $\frak c\ge\omega_3$.
Does $\nu_{\omega}$ have a Borel lifting?

\cmmnt{It is known to be relatively consistent with ZFC to suppose
that $\frak c=\omega_2$ and that $\FN(\Cal P\Bbb N)=\omega_1$ (554G-554H).
In this case
$\nu_{\omega}$ has a Borel lifting (535E(b-ii)).   But if
$\frak c\ge\omega_3$ then $\frak B_{\omega}$ is not tightly
$\omega_1$-filtered (518S).
}%end of comment

\spheader 535Zc (A.H.Stone) Can there be a countable ordinal $\zeta$ and
a lifting $\phi$
of $\nu_{\omega}$ such that $\phi E$ is a Borel set,
with Baire class at most $\zeta$, for every Borel set
$E\subseteq\{0,1\}^{\omega}$?

\cmmnt{The point of this question is that while, subject to the
continuum hypothesis, we can almost write down a formula for a Borel
lifting for Lebesgue measure (535Ya), the method gives no control over
the Baire classes of the sets constructed.
}%end of comment

\spheader 535Zd Can there be a strictly positive Radon probability
measure of countable Maharam type which does not have a strong lifting?
\cmmnt{(See 453G, 453N, 535I, 535Xg.)}
%535I

\discrversionA{535Zg\query\ Must there be a Boolean homomorphism
$\phi:\Cal P\Bbb Q\to\Cal B(\Bbb R)$ such that
$\phi A\subseteq\overline{A}$ for every $A\subseteq\Bbb Q$?}{}
%535M

\spheader 535Ze Is there a probability space which has a linear lifting
but no lifting?
%535O

\spheader 535Zf Can there be a Borel linear lifting for the usual measure
on $(\{0,1\}^{\omega})^2$ which respects coordinates in the sense of 535R?
%535R

\cmmnt{It seems possible that there is a proof in ZFC that there is no
such lifting;  in which case 535R shows that we should have a negative
answer to (a).
}%end of comment

\endnotes{
\Notesheader{535} For a fuller account of this topic, see {\smc Burke
93}.

{\smc Neumann \& Stone 1935} used a direct construction along the lines of
535Xc to show that if the continuum hypothesis is true then Lebesgue
measure has a Borel lifting.   The method works equally well for
$\nu_{\omega_1}$, but for $\nu_{\omega_2}$ we need a further idea from
{\smc Mokobodzki 7?}; %\query;
the version I give here is based on {\smc Geschke 02},
itself derived at some remove from
{\smc Carlson Frankiewicz \& Zbierski 94},
who showed that we could have a Borel
lifting for Lebesgue measure in a model in which the continuum hypothesis
is false (554I).

It is not a surprise that there should be a model of set theory in which
Lebesgue measure has no Borel lifting.   Nor is it a surprise that the
first such model should have been found by S.Shelah ({\smc Shelah 83}).
What does remain surprising is that in most of the vast number of models
of set theory which have been studied, we do not know whether there is
such a lifting.   Only in the familiar case $\frak c=\omega_1$, the
special combination $\frak c=\omega_2=\FN(\Cal P\Bbb N)^+$ (535E),
and in variations of Shelah's
model, do we have definite information.   It remains possible that in
any model in which $\frak m>\omega_1$ or $\frak c=\omega_3$ there is no
Borel lifting for Lebesgue measure.   When we leave the real line, the
position is even more open;  conceivably it is relatively consistent
with ZFC to suppose that every probability space has a lifting, and at
least equally believably it is a theorem of ZFC that $\nu_{\omega_3}$
does not have a Baire lifting.

From 535I we see that $\omega_2$ appears in Losert's example (453N) for
a good reason.   Once again, it seems to be unknown whether it is
consistent to suppose that there is a (completed) strictly positive
Radon probability measure with countable Maharam type which has no
strong lifting (535Zd).   When we come to look for strong Borel
liftings, we have some useful information in the separable metrizable
case (535N).   The result is natural enough.   We are used to supposing
that Polish spaces are all very much the same, and that point-supported
measures are trivial.   But because the concept of `strong' lifting is
topological, and cannot easily be reduced to the Borel structure, we
have to work a bit;  and it seems also that
point-supported measures need care (535M).
%535Zg\query

`Linear liftings' (535O-535R) %535O 535P 535Q 535R
remain poor relations.   I give them house room here partly for
completeness and partly because of a slender hope that they will lead us
to a solution of 535Za.   Of course the match between $\omega_3$
in 535Za and $\frak c^{++}$ in 535R may show only a temporarily
coincidental frontier of ignorance.
{\smc Burke \& Shelah 92} have shown that it is relatively
consistent with ZFC to suppose that $\nu_{\omega}$ has no Borel linear
lifting.

}%end of notes

\discrpage

\frfilename{mt536.tex}
\versiondate{20.2.12}
\copyrightdate{2003}

\def\chaptername{Topologies and measures III}
\def\sectionname{Alexandra Bellow's problem}

\newsection{536}

In 463Za I mentioned a curious problem concerning pointwise compact sets
of continuous functions.   This problem is known to be soluble if we are
allowed to assume the continuum hypothesis, for instance.   Here I
present the relevant arguments, with supplementary remarks on `stable' sets of
measurable functions (536E-536F).

\leader{536A}{The problem}\cmmnt{ I recall some ideas from \S463.}
Let $(X,\Sigma,\mu)$ be a measure space, and
$\eusm L^0\cmmnt{\mskip5mu =\eusm L^0(\Sigma)}$ the space of all
$\Sigma$-measurable functions from $X$ to $\Bbb R$\cmmnt{, so that
$\eusm L^0$ is a linear subspace of $\BbbR^X$}.   On $\eusm L^0$ we have
the linear space topologies $\frak T_p$ and $\frak T_m$ of pointwise
convergence and convergence in measure\cmmnt{ (462Ab, 245Ab)}.
$\frak T_p$ is Hausdorff and locally convex;  if $\mu$ is $\sigma$-finite,
$\frak T_m$ is pseudometrizable.
The question\cmmnt{, already asked in 463Za,} is this:  suppose that
$K\subseteq\eusm L^0$ is compact for $\frak T_p$,
and that $\frak T_m$ is Hausdorff on $K$.   Does it follow that
$\frak T_p$ and $\frak T_m$ agree on $K$?

\leader{536B}{Known cases} Let $(X,\Sigma,\mu)$ be a $\sigma$-finite
measure space.   Given that $K\subseteq\eusm L^0$ is compact
for $\frak T_p$, and $\frak T_m$ is Hausdorff on $K$, and

\inset{{\it either} $K$ is sequentially compact for $\frak T_p$

{\it or} $K$ is countably tight for $\frak T_p$

{\it or} $K$ is convex

{\it or} $X$ has a topology for which $K\subseteq C(X)$,
$\mu$ is a strictly positive topological measure, and every function
$h\in\BbbR^X$ which is continuous on every relatively countably compact
set is continuous

{\it or} $\mu$ is perfect

{\it or} $K$ is stable\cmmnt{, in the sense of 465A},}

\noindent then $K$ is metrizable for $\frak T_p$, and $\frak T_p$ and
$\frak T_m$ agree on $K$\cmmnt{ (463Cd, 463F, 463G, 463H,
463Lc, 465G)}.

\cmmnt{Now for the new results.}

\leader{536C}{Proposition}\dvAnew{2013}\cmmnt{ (see {\smc Talagrand 84},
9-3-3.)}   Let $(X,\Sigma,\mu)$ be a probability space
such that the $\pi$-weight $\pi(\mu)$ of $\mu$ is at most $\frak p$.   If
$K\subseteq\eusm L^0$ is $\frak T_p$-compact then it is
$\frak T_m$-compact.

\proof{{\bf (a)} For the time being (down to the end of (d) below),
suppose that $|f|\le\chi X$ for every $f\in K$.   Let $\sequence{i}{f_i}$
be any sequence in $K$.

\medskip

{\bf (b)} For $I\in[\Bbb N]^{\omega}$, write
$\limsup_{i\to I}f_i$ for $\inf_{n\in\Bbb N}\sup_{i\in I\setminus n}f_i$
and
$\liminf_{i\to I}f_i$ for $\sup_{n\in\Bbb N}\inf_{i\in I\setminus n}f_i$.
Then there is an $I\in[\Bbb N]^{\omega}$ such
that $\liminf_{i\to J}f_i\eae\liminf_{i\to I}f_i$ and
$\limsup_{i\to J}f_i\eae\limsup_{i\to I}f_i$ for every $J\in[I]^{\omega}$.
\Prf\ (See the proof of 463D.)   For $I$, $J\in[\Bbb N]^{\omega}$ set
$\Delta(I)=\int\limsup_{i\to I}f_i-\liminf_{i\to I}f_i$ and say that
$J\preceq I$ if either $J\subseteq I$ or
$J\setminus I$ is finite and $I\setminus J$
is infinite.   Then $\Delta(J)\le\Delta(I)$ whenever $J\preceq I$, and
any non-increasing sequence in $[\Bbb N]^{\omega}$ has a $\preceq$-lower
bound
in $[\Bbb N]^{\omega}$.   By 513P, inverted, there is an
$I\in[\Bbb N]^{\omega}$ such
that $\Delta(J)=\Delta(I)$ whenever $J\preceq I$, and this $I$ will serve.\
\Qed

Set $g=\liminf_{i\to I}f_i$ and $h=\limsup_{i\to I}f_i$.

\medskip

{\bf (c)} \Quer\ Suppose, if possible, that $E=\{x:g(x)<h(x)\}$ is not
negligible.   Let $\Cal H$ be a coinitial subset of
$\Sigma\setminus\Cal N(\mu)$, where $\Cal N(\mu)$ is the null ideal of
$\mu$, with cardinal $\pi(\mu)\le\frak p$, and
$\ofamily{\xi}{\frak p}{H_{\xi}}$ a family running over
$\{H:H\in\Cal H$, $H\subseteq E\}$.
Choose $\ofamily{\xi}{\frak p}{I_{\xi}}$,
$\ofamily{\xi}{\frak p}{x_{\xi}}$ and
$\ofamily{\xi}{\frak p}{y_{\xi}}$ inductively, as follows.
The inductive hypothesis will be that, for any $\xi<\frak p$,
$\ofamily{\eta}{\xi}{I_{\eta}}$ is a family of infinite subsets of
$\Bbb N$ such that $I_{\eta}\setminus I_{\zeta}$ is finite whenever
$\zeta\le\eta<\xi$.   Start with
$I_0=I$.   For the inductive step to $\xi+1$, where $\xi<\frak p$,
since $g\eae\liminf_{i\to I_{\xi}}f_i$, there must be
an $x_{\xi}\in H_{\xi}\cap E$ such that
$g(x_{\xi})=\liminf_{i\to I_{\xi}}f_i(x)$.
Let $J\in[I_{\xi}]^{\omega}$ be such that
$\lim_{i\to J}f_i(x_{\xi})=g(x_{\xi})$.
Now $\limsup_{i\to J}f_i\eae h$, so we can find a
$y_{\xi}\in E\cap H_{\xi}$ such that
$\limsup_{i\to J}f_i(y_{\xi})=h(y_{\xi})$ and an
$I_{\xi+1}\in[J]^{<\omega}$ such that
$\lim_{i\to I_{\xi+1}}f_i(y_{\xi})=h(y_{\xi})$.

For non-zero limit ordinals $\xi<\frak p$, let $I_{\xi}$ be an infinite
subset of $I$ such that $I_{\xi}\setminus I_{\eta}$ is finite for every
$\eta<\xi$.

At the end of the induction, there will be a non-principal ultrafilter
$\Cal F$ on $\Bbb N$ containing $I_{\xi}$ for every $\xi<\frak p$.
Set $f=\lim_{i\to\Cal F}f_i$.   Because $K$ is $\frak T_p$-compact,
$f\in K\subseteq\eusm L^0$.   So at least one of the measurable sets
$E'=\{x:x\in E$, $g(x)<f(x)\}$ and $E''=\{x:x\in E$, $f(x)<h(x)\}$ is
non-negligible and contains $H_{\xi}$ for some $\xi<\frak p$.   Now
$I_{\xi+1}\in\Cal F$, so
$f(x_{\xi})=\lim_{i\to I_{\xi+1}}f_i(x_{\xi})=g(x_{\xi})$ and
$f(y_{\xi})=h(y_{\xi})$.   But this means that
$x_{\xi}\in H_{\xi}\setminus E''$ and $y_{\xi}\in H_{\xi}\setminus E'$, so
$H_{\xi}$ cannot be included in either $E'$ or $E''$.\ \Bang

\medskip

{\bf (d)} So $g\eae h$ and $\{x:g(x)=\lim_{i\to I}f_i(x)\}$ includes the
conegligible set $\{x:g(x)=h(x)\}$.
We also have a $g_0\in K$ which is a $\frak T_p$-cluster point of
$\langle f_i\rangle_{i\in I}$.   Of course $g\le g_0\le h$, and
all three must be equal $\mu$-a.e.
But this means that $\langle f_i\rangle_{i\in I}$ converges
almost everywhere to $g_0$, and
therefore converges in measure to $g_0$ (245Ec).
Now recall that
$\sequence{i}{f_i}$ was an arbitrary sequence in $K$.   So we see that
every sequence in $K$ has a subsequence which is $\frak T_m$-convergent to
a point of $K$.   As $\frak T_m$ is pseudometrizable, $K$ is
$\frak T_m$-compact (4A2Le).

\medskip

{\bf (e)} This concludes the proof when $|f|\le\chi X$ for every $f\in K$.
For the general case, let $\phi:\Bbb R\to\ooint{-1,1}$ be a homeomorphism,
and consider $K'=\{\phi f:f\in K\}$.   Since
$f\mapsto\phi f$ is a $\frak T_p$-continuous function from $\eusm L^0$
to itself, $K'$ is $\frak T_p$-compact,
therefore $\frak T_m$-compact, by (a)-(c).
Next, $f\mapsto\phi^{-1}f:K'\to K$
is $\frak T_m$-continuous.  \Prf\ If $\sequencen{f_n}$ is a sequence in
$K'$ which is $\frak T_m$-convergent to $f\in K'$, and
$\sequencen{g_n}$ is a subsequence of $\sequencen{f_n}$, then
$\sequencen{g_n}$ has a sub-subsequence $\sequencen{h_n}$ converging a.e.\
to $f$ (245Ka);
now $\phi^{-1}h_n$ converges a.e.\ to $\phi^{-1}f\in K$, so
converges in measure to $\phi^{-1}f$.   As $\sequencen{g_n}$ is arbitrary,
$\sequencen{\phi^{-1}f_n}$ converges in measure to $\phi^{-1}f$.
Thus $f\mapsto\phi^{-1}f$ is sequentially continuous for $\frak T_m$,
therefore continuous (4A2Ld).\ \QeD\  So $K=\{\phi^{-1}f:f\in K'\}$ is
$\frak T_m$-compact, as claimed.
}%end of proof of 536C

\leader{536D}{Theorem}\dvAformerly{5{}36C}
Let $(X,\Sigma,\mu)$ be a probability space, and
$\eusm L^0$ the space of $\Sigma$-measurable real-valued functions on
$X$.   Write $\frak T_p$, $\frak T_m$ for the topologies of pointwise
convergence and convergence in measure on $\eusm L^0$.   Suppose that
$K\subseteq\eusm L^0$ is $\frak T_p$-compact and that
$\mu\{x:f(x)\ne g(x)\}>0$ for any distinct $f$, $g\in K$, but that $K$
is not $\frak T_p$-metrizable.

(a) Every infinite Hausdorff space which is a continuous image of a
closed subset of $K$ has a non-trivial convergent sequence.

(b) There is a continuous surjection from a closed subset of $K$ onto
$\{0,1\}^{\omega_1}$.

(c) Every infinite compact Hausdorff space of weight at most $\omega_1$
has a non-trivial convergent sequence.

(d) $\frak c>\omega_1$.

(e) The Maharam type of $\mu$ is at least $2^{\omega_1}$.

(f) There is a non-negligible measurable set in $\Sigma$ which can be
covered by $\omega_1$ negligible sets.

(g)\dvAnew{2013} $\pi(\mu)>\frak p$.

(h) $\frakmctbl=\omega_1$.

\proof{ For $f$, $g\in\eusm L^0$ set
$\rho(f,g)=\int\min(1,|f-g|)$;  then $\rho$ is a pseudometric on
$\eusm L^0$ defining $\frak T_m$, and
$\rho\restr K\times K$ is a metric on $K$.
Set $\Delta(\emptyset)=0$, and for non-empty $A\subseteq\eusm L^0$ set
$\Delta(A)=\sup\{\rho(\inf L,\sup L):\emptyset\ne L\in[A]^{<\omega}\}$.
Note that if $A\subseteq K$ has more than one member then $\Delta(A)>0$,
and that $\Delta(A)\le\Delta(B)$ whenever $A\subseteq B$.

\medskip

{\bf (a)(i)} \Quer\ Suppose, if possible, that $Z$ is an infinite
Hausdorff space, $K_0\subseteq K$ is closed, $\phi:K_0\to Z$ is a
continuous surjection and there is no non-trivial
convergent sequence in $Z$.   Write $\Cal L$ for the family of closed
subsets $L$ of $K_0$ such that $\phi[L]$ is infinite.   Then
$L=\bigcap_{n\in\Bbb N}L_n$ belongs to
$\Cal L$ for every non-increasing sequence $\sequencen{L_n}$ in
$\Cal L$.   \Prf\ $\sequencen{\phi[L_n]}$ is a non-increasing sequence of
infinite closed
subsets of $Z$;  because $Z$ is supposed to have no non-trivial
convergent sequence, $M=\bigcap_{n\in\Bbb N}\phi[L_n]$ is infinite
(4A2G(h-i)).   Since $\phi[L]=M$ (5A4Cf), $L\in\Cal L$.\ \QeD\
By 513P again, there is a
$K_1\in\Cal L$ such that $\Delta(L)=\Delta(K_1)$ for every $L\in\Cal L$
such that $L\subseteq K_1$.

\medskip

\quad{\bf (ii)} Now there is no non-trivial convergent sequence in
$\phi[K_1]$, so $\phi[K_1]$ cannot be scattered (4A2G(h-ii)), and there is
a continuous surjection $\psi:\phi[K_1]\to[0,1]$ (4A2G(j-iv)).   Let
$M\subseteq\phi[K_1]$ be a closed set such that $\psi[M]=[0,1]$ and
$\psi\restr M$ is irreducible (4A2G(i-i)).   Then $M$ is infinite,
has a countable $\pi$-base and no isolated points (4A2G(i-ii)).
Let $K_2\subseteq\phi^{-1}[M]$ be a closed set such that $\phi[K_2]=M$
and $\phi\restr K_2$ is irreducible.   Then $K_2$ has a countable
$\pi$-base, and $\phi[K_2]$ is infinite, so $\Delta[K_2]=\Delta[K_1]$.

Let $\Cal V$ be a countable $\pi$-base for the topology of $K_2$, not
containing $\emptyset$.   For each $V\in\Cal V$, choose
$h_V\in V$.    Set $g_0=\inf_{V\in\Cal V}h_V$,
$g_1=\sup_{V\in\Cal V}h_V$ in $\BbbR^X$.   Then $g_0$ and $g_1$
are measurable, and

\Centerline{$\int g_1-g_0\ge\Delta(K_2)=\Delta(K_1)>0$.}

\noindent Set $g(x)=\max(\bover12(g_0(x)+g_1(x)),g_1(x)-\bover12)$ for
$x\in X$, and

\Centerline{$E=\{x:g_0(x)<g_1(x)\}=\{x:g(x)<g_1(x)\}
=\{x:g_0(x)<g(x)\}$,}

\noindent so that
$\mu E>0$.   For $x\in E$, the set $F_x=\{f:f\in K_2,\,f(x)\le g(x)\}$
is a proper closed subset of $K_2$, so there is
some $V\in\Cal V$ such that $V\cap F_x=\emptyset$.   Because
$\Cal V$ is countable, there is a $V\in\Cal V$ such that
$D=\{x:x\in E,\,V\cap F_x=\emptyset\}$ is non-negligible.
But now observe that
$f(x)>g(x)$ whenever $f\in V$ and $x\in D$, so
$h_U(x)>g(x)$
whenever $U\in\Cal V$, $U\subseteq V$ and $x\in D$.   Set
$\Cal V'=\{U:U\in\Cal V$, $U\subseteq V\}$,
$g'_0=\inf_{U\in\Cal V'}h_U$ and
$L=\{f:f\in K_2$, $g'_0\le f\le g_1\}$.   Then $g\le g'_0$ and

\Centerline{$\{x:x\in X$, $g_1(x)-g'_0(x)<\min(1,g_1(x)-g_0(x))\}
\supseteq D$}

\noindent is non-negligible, so

\Centerline{$\Delta(L)\le\int\min(1,g_1-g'_0)
<\int\min(1,g_1-g_0)=\Delta(K_1)$.}

\noindent On the other hand, $L$ meets every member of $\Cal V'$,
so $L\cap V$ is dense in $V$ and $L$ includes $V$.   Because
$\phi\restr K_2$ is irreducible, $\phi[K_2\setminus V]\ne M$ and
$\phi[L]$ includes the non-empty open subset
$M\setminus\phi[K_2\setminus V]$ of $M$, which is infinite because
$M$ has no isolated points.   So $\Delta(L)$ ought to be equal to
$\Delta(K_1)$, by the choice of $K_1$.\ \Bang

Thus (a) is true.

\medskip

{\bf (b)} If $\sequencen{f_n}$ is a sequence in $K$ which converges
at almost every point of $X$, then any two $\frak T_p$-cluster points of
$\sequencen{f_n}$ must be equal a.e.\ and therefore equal, so
$\sequencen{f_n}$ is $\frak T_p$-convergent (5A4Ce).

\Quer\ Suppose, if possible, that there is no continuous
surjection from a closed subset of $K$ onto $\{0,1\}^{\omega_1}$.   Then
463D tells us that every sequence in $K$ has a subsequence
which is convergent almost everywhere, therefore convergent.
So $K$ is sequentially compact, which is impossible, as noted in 536B.\
\Bang

\medskip

{\bf (c)} Since $[0,1]$ is a continuous image of $\{0,1\}^{\Bbb N}$,
$[0,1]^{\omega_1}$ is a continuous image of
$\{0,1\}^{\omega_1\times\Bbb N}\cong\{0,1\}^{\omega_1}$ and therefore of
a closed subset of $K$.   If $Z$ is an infinite compact Hausdorff space
of weight at most $\omega_1$, it is homeomorphic to a closed subset of
$[0,1]^{\omega_1}$ (5A4Cc) and therefore to a continuous image of a
closed subset of $K$.   By (a), $Z$ must have a non-trivial convergent
sequence.

\medskip

{\bf (d)} Since $\beta\Bbb N$ has weight $\frak c$ (5A4Ia), is infinite,
but has no non-trivial convergent subsequence (4A2I(b-v)), we must have
$\omega_1<\frak c$.

\medskip

{\bf (e)(i)} If $F_1$, $F_2$ are disjoint non-empty $\frak T_p$-closed
subsets of $K$, then $\rho(F_1,F_2)>0$.   \Prf\Quer\ Otherwise, there
are sequences $\sequencen{f_n}$ in $F_1$, $\sequencen{g_n}$ in $F_2$
such that $\rho(f_n,g_n)\le 2^{-n}$ for every $n\in\Bbb N$.   Let
$\Cal F$ be any non-principal ultrafilter on
$\Bbb N$ and set $f=\lim_{n\to\Cal F}f_n$, $g=\lim_{n\to\Cal F}g_n$,
taking the limits in $K$ for the topology $\frak T_p$.   Then, for any
$n\in\Bbb N$,

\Centerline{$\{x:|f(x)-g(x)|>2^{-n}\}
\subseteq\bigcup_{i\ge 2n}\{x:|f_i(x)-g_i(x)|>2^{-n}\}$}

\noindent has measure at most $\sum_{i=2n}^{\infty}2^{-i+n}=2^{-n+1}$,
so $f\eae g$ and $f=g$;  but $f\in F_1$ and $g\in F_2$, so this is
impossible.\ \Bang\Qed

\medskip

\quad{\bf (ii)} By (b), there are a closed subset $K_0$ of $K$ and a
continuous surjection $\psi:K_0\to\{0,1\}^{\omega_1}$.   For
$\xi<\omega_1$, set $F_{\xi}=\{f:f\in K_0$, $\psi(f)(\xi)=0\}$,
$F'_{\xi}=\{f:f\in K_0$, $\psi(f)(\xi)=1\}$;  then
$\rho(F_{\xi},F'_{\xi})>0$.   There must therefore be a $\delta>0$ such
that $C=\{\xi:\rho(F_{\xi},F'_{\xi})\ge\delta\}$ is uncountable.   For
each $D\subseteq C$, choose $h_D\in K_0$ such that
$\psi(h_D)\restr C=\chi D$.   Then $\rho(h_D,h_{D'})\ge\delta$ for all
distinct $D$, $D'\subseteq C$.   Thus
$A=\{h_D^{\ssbullet}:D\subseteq C\}$ is a subset of $L^0=L^0(\mu)$, of
cardinal $2^{\omega_1}$, such that any two members of $A$ are distance
at least $\delta$ apart for the metric on $L^0$ corresponding to $\rho$.
Accordingly the cellularity and topological density of $L^0$ are at
least $2^{\omega_1}$;  by 529Bb, the Maharam type of $\mu$ is at least
$2^{\omega_1}$.

\medskip

{\bf (f)(i)} By (b), there is a continuous surjection
$\psi:K_0\to\{0,1\}^{\omega_1}$ where $K_0\subseteq K$ is closed.
Let $Q$ be the set of pairs $(F,C)$ such that $F\subseteq K_0$ is
closed, $C\subseteq\omega_1$ is closed and cofinal and
$\{\psi(f)\restr C:f\in F\}=\{0,1\}^C$.   If $\sequencen{(F_n,C_n)}$ is
a non-increasing sequence in $Q$, then it has a lower bound in $Q$.
\Prf\ Set $F=\bigcap_{n\in\Bbb N}F_n$ and $C=\bigcap_{n\in\Bbb N}C_n$.
Then for any $z\in\{0,1\}^C$ and $n\in\Bbb N$ there is an $f_n\in F_n$
such that $\psi(f_n)\restr C=z$;  now take a $\frak T_p$-cluster point
$f$ of $\sequencen{f_n}$, and see that $f\in F$ and that
$\psi(f)\restr C=z$.  As $z$ is arbitrary, $(F,C)\in Q$.\ \QeD\
By 513P once more, there is a member $(K_1,C^*)$ of $Q$
such that $\Delta(F)=\Delta(K_1)$ whenever $(F,C)\in Q$,
$F\subseteq K_1$ and $C\subseteq C^*$.   Now $C^*$ is order-isomorphic
to $\omega_1$ and its order topology agrees with the subspace topology
induced by the order topology of $\omega_1$ (4A2Rm).   Let
$\theta:\omega_1\to C^*$ be an order-isomorphism and set
$\psi_1(f)=\psi(f)\theta$ for $f\in K_1$.   Then
$\psi_1:K_1\to\{0,1\}^{\omega_1}$ is a continuous surjection, and if
$F\subseteq K_1$ is closed, $C\subseteq\omega_1$ is closed and cofinal
and $\{\psi_1(f)\restr C:f\in F\}=\{0,1\}^C$, then $(F,\theta[C])\in Q$
so $\Delta(F)=\Delta(K_1)$.

\medskip

\quad{\bf (ii)} Let $K_2\subseteq K_1$ be a compact set such that
$\psi_1\restr K_2$ is an irreducible surjection onto
$\{0,1\}^{\omega_1}$ (4A2G(i-i) again).
Because $\{0,1\}^{\omega_1}$ is separable
(4A2B(e-ii)), so is $K_2$ (5A4C(d-i)).   Let $\sequencen{f_n}$ enumerate a
dense subset of $K_2$.   Because $K_2$ is compact in $\Bbb R^X$,
$h_1=\sup_{n\in\Bbb N}f_n$ and $h_0=\inf_{n\in\Bbb N}f_n$ are defined in
$\BbbR^X$, and of course they belong to $\eusm L^0$.   If $f\in K_2$,
then

\Centerline{$f(x)\in\overline{\{f_n(x):n\in\Bbb N\}}
\subseteq[h_0(x),h_1(x)]$}

\noindent for every $x$, and $h_0\le f\le h_1$.   Accordingly we have

\Centerline{$\Delta(K_2)\le\rho(h_0,h_1)
=\sup_{n\in\Bbb N}\rho(\inf_{i\le n}f_i,\sup_{i\le n}f_i)
\le\Delta(K_2)$.}

Let $\Cal U$ be the family of
non-empty cylinder sets in $\{0,1\}^{\omega_1}$.   For $U\in\Cal U$ set
$I_U=\{n:n\in\Bbb N$, $\psi_1(f_n)\in U\}$ and
$g_U=\inf\{f_n:n\in I_U\}$.   Observe that
$F_U=\{f:f\in K_2$, $g_U\le f\le h_1\}$ is a closed subset of $K_1$ and
that $F_U\cap\psi_1^{-1}[U]$ is dense in $\psi_1^{-1}[U]$, so
$U\cap\psi_1[F_U]$ must be dense in $U$ and $U\subseteq\psi_1[F_U]$.
There is a finite set $I\subseteq\omega_1$ such that $U$ is determined
by coordinates in $I$;  in this case, $C=\omega_1\setminus I$ is closed
and cofinal in $\omega_1$, and $\{z\restr C:z\in U\}=\{0,1\}^C$.   By
the choice of $K_1$, $\Delta(F_U)=\Delta(K_1)$.   As
$F_U\subseteq[g_U,h_1]$ in $\eusm L^0$,
$\rho(g_U,h_1)=\Delta(K_1)=\rho(h_0,h_1)$, and
$\min(1,h_1-g_U)\eae\min(1,h_1-h_0)$.

Set $h(x)=\max(\bover12(h_0(x)+h_1(x)),h_1(x)-\bover12)$ for $x\in X$,
and $E=\{x:h_0(x)<h_1(x)\}=\{x:h(x)<h_1(x)\}$, so that $E$ is measurable
and not negligible.   If  $U\in\Cal U$, then

$$\eqalign{E_U
&=\{x:x\in E,\,h(x)\le g_U(x)\}\cr
&\subseteq\{x:x\in E,\,h_1(x)-g_U(x)<\min(1,h_1(x)-h_0(x))\}\cr}$$

\noindent is negligible.

For every $x\in E$,
$F'_x=\{f:f\in K_2$, $f(x)\le h(x)\}$ is a proper closed subset of
$K_2$, so $\psi_1[F'_x]\ne\{0,1\}^{\omega_1}$ and there is some
$U\in\Cal U$ such that $U\cap\psi_1[F'_x]=\emptyset$.   In this case
$f_n\notin F'_x$, that is, $f_n(x)>h(x)$, for every $n\in I_U$, so
$g_U(x)\ge h(x)$.   Thus $E=\bigcup_{U\in\Cal U}E_U$ is a non-negligible
measurable set covered by $\omega_1$ negligible sets.

\medskip

{\bf (g)} This is immediate from 536C, since we already know that $K$ cannot
be stable.

\medskip

{\bf (h)} Continuing the argument from (f), define
$\phi:X\to\BbbR^{\Bbb N}$ by setting $\phi(x)=\sequencen{f_n(x)}$ for
$x\in X$.   Then $\phi$ is measurable (418Bd), so we have a non-zero
totally finite Borel measure $\nu$ on $\BbbR^{\Bbb N}$ defined by
setting $\nu H=\mu(E\cap\phi^{-1}[H])$ for every Borel set
$H\subseteq\BbbR^{\Bbb N}$.   Note that $\phi[X]\subseteq\ell^{\infty}$
and that $\ell^{\infty}
=\bigcup_{n\in\Bbb N}\bigcap_{i\in\Bbb N}\{w:|w(i)|\le n\}$ is an
F$_{\sigma}$ set in $\BbbR^{\Bbb N}$.   Now set

\Centerline{$h'_1(w)=\sup_{n\in\Bbb N}w(n)$,
\quad$h'_0(w)=\inf_{n\in\Bbb N}w(n)$,}

\Centerline{$h'(w)=\max(\bover12(h'_0(w)+h'_1(w)),h'_1(w)-\bover12)$}

\noindent for $w\in\ell^{\infty}$, so that $h_1=h'_1\phi$,
$h_0=h'_0\phi$ and $h=h'\phi$;  for $U\in\Cal U$, set

\Centerline{$E'_U
=\{w:w\in\ell^{\infty}$, $h'(w)\le\inf_{n\in I_U}w(n)\}$}

\noindent so that $E'_U$ is an F$_{\sigma}$ set and
$E_U=E\cap\phi^{-1}[E'_U]$;  accordingly $\nu E'_U=0$.   Because
$E\subseteq\bigcup_{U\in\Cal U}E_U$,
$\phi[E]\subseteq\bigcup_{U\in\Cal U}E'_U$.

Thus we have a non-negligible subset of $\BbbR^{\Bbb N}$ which is
covered by $\omega_1$ negligible F$_{\sigma}$ sets and therefore by
$\omega_1$ closed negligible sets.   By 526M, $\frakmctbl=\omega_1$.
}%end of proof of 536D

\leader{536E}{}\cmmnt{ The discussion of stable sets in \S465 emphasized their
connection with pointwise compactness.   
In 465D and 465G we saw that stable sets 
are relatively pointwise compact and that on a stable set $\frak T_m$ is
coarser than $\frak T_p$.   The question of when we might be able
to be sure that a pointwise compact set is stable was 
left open (but see 465Xj and 465Xn).   We now have the concepts to
take another step in this direction, which fits fairly naturally here, 
though it is not obviously connected with the question in 536A.

\medskip

\noindent}{\bf Proposition}\dvAnew{2016}
Let $(X,\Sigma,\mu)$ be a semi-finite measure space,
with null ideal $\Cal N(\mu)$.   For $E\in\Sigma$ let $\mu_E$ be the
subspace measure on $E$.
Suppose that $\pi(\mu_E)\le\cov(E,\Cal N(\mu))$ whenever 
$E\in\Sigma$ and $0<\mu E<\infty$.   Then every $\frak T_p$-separable
$\frak T_p$-compact subset of $\eusm L^0=\eusm L^0(\Sigma)$ is stable.

\proof{{\bf (a)} \Quer\ Suppose that $K$ is a 
$\frak T_p$-separable $\frak T_p$-compact subset of $\eusm L^0$ 
which is not stable.   Let $A$ be
a countable $\frak T_p$-dense subset of $K$.   By 465C(a-ii), $A$ is not stable.
So there are a set $E\in\Sigma$ and $\alpha<\beta$ in $\Bbb R$
such that $0<\mu E<\infty$ and, in the language of 465A,
$(\mu^{2k})^*D_k(A,E,\alpha,\beta)=(\mu E)^{2k}$ for every $k\ge 1$.
Because $A$ is countable, 

$$\eqalign{D_k(A,E,\alpha,\beta)
=\bigcup_{f\in A}\{w:w\in E^{2k},\,
&f(w(2i))\le\alpha,\cr
&f(w(2i+1))\ge\beta\text{ for }i<k\}.\cr}$$

\noindent is measured by the product measure $\mu^{2k}$ for every $k$, so that
$E^{2k}\setminus D_k(A,E,\alpha,\beta)$ is $\mu^{2k}$-negligible for every $k$.

\medskip

{\bf (b)} For sets $I$, $J\subseteq E$ set

\Centerline{$A_{IJ}
=\{f:f\in A$, $f(x)\le\alpha$ for $x\in I$, $f(x)\ge\beta$ for $i\in J\}$.}

\noindent Let $Q$ be the family of pairs $(I,J)$ of finite subsets of
$E$ such that 
$E^{2k}\setminus D_k(A_{IJ},E,\alpha,\beta)$ is $\mu^{2k}$-negligible 
for every $k$.   Then whenever $(I,J)\in Q$, the set 
$\{(x,y):x$, $y\in E$, $(I\cup\{x\},J\cup\{y\})\notin Q\}$
is $\mu^2$-negligible.   \Prf\ For any $k\ge 1$, if we identify $E^{2k+2}$ with
$E^{2k}\times E^2$,

$$\eqalign{D_{k+1}(A_{IJ},E,\alpha,\beta)
&=\bigcup_{f\in A_{IJ}}\{(w,(x,y)):w\in E^{2k},\,x,\,y\in E,\,
    f(x)\le\alpha,\,f(y)\ge\beta,\cr&
\mskip100mu f(w(2i))\le\alpha,\,f(w(2i+1))\ge\beta\text{ for }i<k\}\cr&
=\{(w,(x,y)):
   x,\,y\in E,\,w\in D_k(A_{I\cup\{x\},J\cup\{y\}},E,\alpha,\beta)\}.\cr}$$

\noindent Let $F_k$ be the set of those $(x,y)\in E^2$ such that
$E^{2k}\setminus D_k(A_{I\cup\{x\},J\cup\{y\}},E,\alpha,\beta)$ is not
$\mu^{2k}$-negligible.   As $E^{2k+2}\setminus D_{k+1}(A_{IJ},E,\alpha,\beta)$ 
is $\mu^{2k+2}$-negligible, $F_k$ is $\mu^2$-negligible (252D).   
As $k$ is arbitrary. 

\Centerline{$\{(x,y):x$, $y\in E$, $(I\cup\{x\},J\cup\{y\})\notin Q\}
=\bigcup_{k\ge 1}F_k$}

\noindent is $\mu^2$-negligible.\ \Qed

\medskip

{\bf (c)} Set $\kappa=\pi(\mu_E)$;  
then $\kappa\ge\cov(E,\Cal N(\mu))$ is infinite.
Let $\ofamily{\xi}{\kappa}{H_{\xi}}$ run over a coinitial set in
$\{H:H\in\Sigma\setminus\Cal N(\mu)$, $H\subseteq E\}$.   Then we can choose
$\ofamily{\xi}{\kappa}{(x_{\xi},y_{\xi})}$ in such a way that, for
each $\xi<\kappa$,

\Centerline{$x_{\xi}$, $y_{\xi}\in H_{\xi}$,
\quad$(\{x_{\eta}:\eta\in I\},\{y_{\eta}:\eta\in I\})\in Q$
for every finite $I\subseteq\xi$,}

\noindent\Prf\ When we come to choose $(x_{\xi},y_{\xi})$ we shall need to
find a point $(x,y)$ of $H_{\xi}^2$ such that

\Centerline{$(\{x\}\cup\{x_{\eta}:\eta\in I\},
\{y\}\cup\{y_{\eta}:\eta\in I\})$}

\noindent belongs to $Q$ for every finite $I\subseteq\xi$.   By (b) and the
inductive hypothesis, the forbidden set

\Centerline{$H_{\xi}^2\cap\bigcup_{I\in[\xi]^{<\omega}} 
\{(x,y):(\{x\}\cup\{x_{\eta}:\eta\in I\},
\{y\}\cup\{y_{\eta}:\eta\in I\})\notin Q\}$}

\noindent is the union of fewer than $\kappa\,\,\mu^2$-negligible subsets of 
$H_{\xi}^2$ and cannot cover $H_{\xi}^2$, by 521Jd, since 
$\kappa\ge\cov(H_{\xi},\Cal N(\mu))$.   
We therefore have a candidate eligible to be $(x_{\xi},y_{\xi})$, and the
induction can proceed.\ \Qed

\medskip

{\bf (d)} At the end of the induction, we see that 

\Centerline{$C_I=A_{\{x_{\eta}:\eta\in I\},\{y_{\eta}:\eta\in I\}}$}

\noindent is non-empty for every finite $I\subseteq\kappa$.   Let $\Cal F$ be
the filter on $K$ generated by $\{C_I:I\in[\kappa]^{<\omega}\}$.   Because
$K$ is $\frak T_p$-compact, 
$\Cal F$ has a $\frak T_p$-cluster point $f\in K\subseteq\eusm L^0$.
Now one of $\{x:x\in E$, $f(x)<\beta\}$ and
$\{x:x\in E$, $f(x)>\alpha\}$ must belong to $\Sigma\setminus\Cal N(\mu)$ and
include some $H_{\xi}$;  but $x_{\xi}$, $y_{\xi}\in H_{\xi}$,
while $f(x_{\xi})\le\alpha$ and $f(y_{\xi})\ge\beta$.\ \Bang  

\medskip

{\bf (e)} Thus every pointwise separable-and-compact subset of $\eusm L^0$
must be stable, as claimed.
}%end of proof of 536E

\leader{536F}{Proposition} Suppose that $\cov\Cal N=\cf\Cal N$, 
where $\Cal N$ is
the null ideal of Lebesgue measure on $\Bbb R$.   Let $(X,\frak T,\Sigma,\mu)$
be a Radon measure space.   Then every $\frak T_p$-separable $\frak T_p$-compact
subset of $\eusm L^0(\mu)$ is stable.

\proof{{\bf (a)} To begin with (down to the end of (c) below), suppose that
$\mu$ is totally finite.   Let $K\subseteq\eusm L^0(\mu)$ be
$\frak T_p$-separable and $\frak T_p$-compact.   If $K$ is empty, it is surely
stable and we can stop.   Otherwise, let $\sequencen{f_n}$ be a sequence running
over a $\frak T_p$-dense subset of $K$.   Define $\phi:X\to\BbbR^{\Bbb N}$ by
setting $\phi(x)=\sequencen{f_n(x)}$ for $n\in\Bbb N$.   Then $\phi$ is 
measurable (418Bd again), therefore almost continuous (418J).   Set $Z=\phi[X]$, and
let $\nu$ be the image measure $\mu\phi^{-1}$ on $Z$;  
then $\nu$ is a Radon measure (418I).   (This is
where it helps to assume that $\mu$ is totally finite.)

\medskip

{\bf (b)} Consider the set $L=\{g:g\in\BbbR^Z$, $g\phi\in K\}$.   

\medskip

\quad{\bf (i)} $L\subseteq\eusm L^0(\nu)$.   
\Prf\ If $g\in L$ and $\alpha>0$, then

\Centerline{$\phi^{-1}[\{z:z\in Z$, $g(z)>\alpha\}]
=\{x:x\in X$, $g\phi(x)>\alpha\}$}

\noindent is measured by $\mu$ so $\{z:g(z)>\alpha\}$ is measured by
$\nu$.\ \Qed

\medskip

\quad{\bf (ii)} $L$ is $\frak T_p$-separable.   \Prf\ Set $g_n(z)=z(n)$ for
$n\in\Bbb N$ and $z\in Z$.   Then $g_n\phi=f_n\in K$ so $g_n\in L$.   If
$g\in L$, there is a filter $\Cal F$ on $\Bbb N$ such that
$g\phi$ is the $\frak T_p$-limit $\lim_{n\in\Cal F}f_n$, that is,

\Centerline{$g\phi(x)=\lim_{n\to\Cal F}f_n(x)=\lim_{n\to\Cal F}g_n\phi(x)$}

\noindent for every $x$.   But now $g(z)=\lim_{n\to\Cal F}g_n(z)$ for every
$z\in Z$ and $g=\lim_{n\to\Cal F}g_n$ belongs to the
$\frak T_p$-closure of $\{g_n:n\in\Bbb N\}$.   So the countable set
$\{g_n:n\in\Bbb N\}$ witnesses that $L$ is $\frak T_p$-separable.\ \Qed

\medskip

\quad{\bf (iii)} $L$ is $\frak T_p$-compact.  
\Prf\ Note first that if $z\in Z$ there is an $x\in\phi^{-1}[\{z\}]$, and now

\Centerline{$\sup_{g\in L}|g(z)|
=\sup_{g\in L}|g\phi(x)|\le\le\sup_{f\in K}|f(x)|$}

\noindent is finite.   As $g$ is arbitrary,
$L$ is relatively $\frak T_p$-compact in $\BbbR^Z$;
write $\overline{L}$ for its $\frak T_p$-closure.   The map 
$g\mapsto g\phi:\BbbR^Z\to\BbbR^X$ is continuous for the pointwise topologies
and $g\phi\in K$ for every $g\in L$, so $g\phi\in\overline{K}=K$ for every
$g\in\overline{L}$, and $L=\overline{L}$ is $\frak T_p$-compect.\ \Qed

\medskip

\quad{\bf (iv)} $K=\{g\phi:g\in L\}$.   \Prf\ As the function
$g\mapsto g\phi$ is continuous, $K'=\{g\phi:g\in L\}$ is $\frak T_p$-compact,
therefore $\frak T_p$-closed;  since it contains $f_n=g_n\phi$ for every
$n$, it includes $K$.   By the definition of $L$, $K'\subseteq K$ and they are
equal.\ \Qed

\medskip

{\bf (c)} Now note that $\nu$ is a Radon measure on a separable metrizable
space.   So $\tau(\nu)\le\omega$ (531Ad),
$\pi(\nu)\le\cf\Cal N$ (524Pb) and $\cov\Cal N(\nu_F)\ge\cov\Cal N$ for every
non-negligible set $F\in\dom\nu$ (524Pc).   We are supposing that 
$\cov\Cal N=\cf\Cal N$, so 563E assures us that $L$ is stable.
Since $\phi$ is \imp, $K=\{g\phi:g\in L\}$ is stable (465Cd\formerly{4{}65Xe}).

\medskip

{\bf (d)} This deals with the case of totally finite $\mu$.   For the general
case, take any $E\in\Sigma$ such that $\mu E<\infty$.   Then
$A_E=\{f\restr E:f\in A\}$ is included in $\eusm L^0(\dom\mu_E)$, and it is 
$\frak T_p$-separable and $\frak T_p$-compact because the map
$f\mapsto f\restr E$ is pointwise continuous.   Also $\mu_E$ is a Radon measure,
by 416Rb.   So $A_E$ is stable, by (a)-(c).   As $E$ is arbitrary,
$A$ is stable (456C(c-iv)).
}%end of proof of 536F

\exercises{\vleader{36pt}{536X}{Basic exercises (a)}
%\spheader 536Xa
Let $(X,\Sigma,\mu)$ be a complete measure space, with null ideal
$\Cal N(\mu)$.   Suppose that $\add\Cal N(\mu)=\cov\Cal N(\mu)$.
Show that there is a $\frak T_p$-compact
$\frak T_m$-compact $K\subseteq\eusm L^0(\Sigma)$
such that the identity map on $K$ is not
$(\frak T_p,\frak T_m)$-continuous.
%463Xh

\spheader 536Xb Let $(X,\Sigma,\mu)$ be a perfect measure space.
Suppose that
$\non(E,\Cal N(\mu))<\cov(E,\Cal N(\mu))$ for every non-negligible
measurable set $E$ of finite measure.   Show that if
$K\subseteq\eusm L^0(\Sigma)$ is $\frak T_p$-compact, then the identity
map on $K$ is $(\frak T_p,\frak T_m)$-continuous.
%536Xa
%463Lb  K  is  \frakT_m-compact

\leader{536Y}{Further exercises (a)}
%\spheader 536Ya
Suppose that the additivity and covering number of the Lebesgue
null ideal are equal.   Find a strictly
localizable perfect measure space $(X,\Sigma,\mu)$ and a
$\frak T_p$-compact $K\subseteq\eusm L^0(\Sigma)$ such that $\frak T_m$
is Hausdorff on $K$ but $K$ is not $\frak T_m$-compact.
%463L mt53bits
}%end of exercises

\endnotes{
\Notesheader{536} The methods of 536C-536D are derived from ideas of
M.Talagrand.   They seem frustratingly close to delivering an answer to
the original question.   But it seems clear that even if a positive
answer -- every $\frak T_p$-compact
$\frak T_m$-separated set is metrizable -- is true in ZFC, 
some further idea will be needed in the proof.   
On the other side, while it may well
be that in some familiar model of set theory there is a negative answer,
parts (c), (d) and (g) of 536D give simple tests to rule out many
candidates.
}%end of notes

\discrpage

\frfilename{mt537.tex}
\versiondate{12.8.13}
\copyrightdate{2005}

\def\shrplusN{\shr^+\mskip-4mu\Cal N}

\def\chaptername{Topologies and measures III}
\def\sectionname{Sierpi\'nski sets, shrinking numbers
and strong Fubini theorems}

\newsection{537}

\leaveitout{
A defn Sierp set
B elem facts
C defn entangled
D elem facts
E lemma
F Sierp > entangled
G entangled > nonproductive ccc
H scal mable fns
I,J  \iint f_y(u_x)dxdy = \iint f_y(u_x)dydx
537K repeated integration:  Shipman
537L Sierp
537M upper integrals
537N,537O,537P,537Q repeated upper/lower integration;  non/shr v cov
537R,537S  repeated integration:  shr/cov
}

W.Sierpi\'nski observed that if the continuum hypothesis is true then there
are uncountable subsets of $\Bbb R$ which have no uncountable negligible
subsets, and that such sets lead to curious phenomena;  he also observed
that, again assuming the continuum hypothesis, there would be a
(non-measurable) function $f:[0,1]^2\to\{0,1\}$ for which Fubini's theorem
failed radically, in the sense that

\Centerline{$\iint f(x,y)dxdy=0$,\quad$\iint f(x,y)dydx=1$.}

\noindent In this section I set out to explore these two insights in the
light of the concepts introduced in Chapter 52.   I start with
definitions of `Sierpi\'nski' and `strongly Sierpi\'nski' set (537A), with
elementary facts and an excursion into the theory of `entangled' sets
(537C-537G).  %537C 537D 537E 537F 537G
Turning to repeated integration, I look at three interesting cases in
which,
for different reasons, some form of separate measurability is enough
to ensure equality of repeated integrals (537I, 537L, 537S).
Working a bit harder, we find
that there can be valid non-trivial inequalities of the form
$\overline{\int}\underline{\int}dxdy
\le\overline{\int}\,\overline{\int}dydx$
(537N-537Q).  %537N 537O 537P 537Q

As elsewhere, I will write $\Cal N(\mu)$ for the null ideal of a measure
$\mu$.

\leader{537A}{Definitions (a)}
If $(X,\Sigma,\mu)$ is a measure space, a subset of $X$ is a
{\bf Sierpi\'nski set} if it is uncountable but meets every negligible set
in a countable set.

\spheader 537Ab If $(X,\Sigma,\mu)$ is a measure space, a subset $A$ of $X$
is a
{\bf strongly Sierpi\'nski set} if it is uncountable and for every $n\ge 1$
and for every set $W\subseteq X^n$ which is negligible for
the\cmmnt{ (c.l.d.)} product measure on $X^n$, the set
$\{u:u\in A^n\cap W$, $u(i)\ne u(j)$ for $i<j<n\}$ is countable.

\leader{537B}{Proposition} (a) Let $(X,\Sigma,\mu)$
be a measure space and $A\subseteq X$ a Sierpi\'nski set.

\quad(i) $\add\Cal N(\mu)=\non\Cal N(\mu)=\omega_1$
and $\cov\Cal N(\mu)\ge\#(A)$.

\quad(ii) If $\{x\}$ is negligible for every $x\in A$, then
$\cf\Cal N(\mu)\ge\cff([\#(A)]^{\le\omega})$.

(b) Suppose that $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ are measure spaces
such that singleton subsets of $Y$ are negligible.   Let
$f:X\to Y$ be an \imp\ function.

\quad(i) If $A\subseteq X$ is a Sierpi\'nski set, then $f[A]$ is a
Sierpi\'nski set in $Y$ and $\#(f[A])=\#(A)$.

\quad(ii) Now suppose that $\nu$ is $\sigma$-finite.
If $A\subseteq X$ is a strongly Sierpi\'nski set, then $f[A]$ is a
strongly Sierpi\'nski set in $Y$.

(c) Suppose that $\lambda$ and $\kappa$ are infinite cardinals and that
$(X,\Sigma,\mu)$ is a locally compact semi-finite measure space of Maharam
type at most $\lambda$ in which singletons are negligible and $\mu X>0$.
Give $\{0,1\}^{\lambda}$ its usual measure.

\quad(i) If $\{0,1\}^{\lambda}$ has a
Sierpi\'nski subset with cardinal $\kappa$, then $X$ has a
Sierpi\'nski subset with cardinal $\kappa$.

\quad(ii) If $\{0,1\}^{\lambda}$ has a
strongly Sierpi\'nski subset with cardinal $\kappa$, then $X$ has a
strongly Sierpi\'nski subset with cardinal $\kappa$.

\proof{{\bf (a)(i)} We are told that $A$ is uncountable;  now any subset of
$A$ with $\omega_1$ members witnesses that $\non\Cal N(\nu)\le\omega_1$.
On the other hand, if $\Cal E$ is a family of negligible sets covering $X$,
then $\#(A)\le\max(\omega,\#(\Cal E))$, so $\#(\Cal E)\ge\#(A)$;  as
$\Cal E$ is arbitrary, $\cov\Cal N(\mu)\ge\#(A)$.

\medskip

\quad{\bf (ii)} If $\{x\}$ is negligible for every $x\in A$, then $[A]^{\le\omega}\subseteq\Cal N(\mu)$,
and the identity function is a Tukey function
from $[A]^{\le\omega}$ to $\Cal N(\mu)$;
so $\cff[A]^{\le\omega}\le\cf\Cal N(\mu)$.

\medskip

{\bf (b)(i)} If $y\in Y$, then $\{y\}$ and
$f^{-1}[\{y\}]$ are negligible, so $A\cap f^{-1}[\{y\}]$ is countable;
consequently $\#(A)\le\max(\omega,\#(f[A]))$ and $\#(f[A])=\#(A)$.
If $F\subseteq Y$ is negligible, then $f^{-1}[F]$ is negligible so
$A\cap f^{-1}[F]$ and $f[A]\cap F$ are countable.   So $f[A]$ is a
Sierpi\'nski set.

\medskip

\quad{\bf (ii)} Let $W\subseteq Y^n$ be a negligible set
for the product measure $\lambda'$ on $Y^n$, where $n\ge 1$.   Define
$\pmb{f}:X^n\to Y^n$ by saying that
$\pmb{f}(x_0,\ldots,x_{n-1})=(f(x_0),\ldots,f(x_{n-1}))$ for
$x_0,\ldots,x_{n-1}\in X$.   Because $\nu$ is $\sigma$-finite,
$\pmb{f}$ is \imp\ for $\lambda$ and $\lambda'$ (251Wp).
If $W$ is $\lambda'$-negligible,
then $\pmb{f}^{-1}[W]$ is $\lambda$-negligible, and
$B=\{u:u\in A^n\cap\pmb{f}^{-1}[W],\,u(i)\ne u(j)$ for $i<j<n\}$ is
countable.   Consequently

\Centerline{$\{v:v\in f[A]^n\cap W,\,v(i)\ne v(j)\text{ for }i<j<n\}
\subseteq\pmb{f}[B]$}

\noindent is countable.

\medskip

{\bf (c)} Take any set $E\subseteq X$ of non-zero finite measure, and give
$E$ its normalized subspace measure $\mu'_E=(\mu E)^{-1}\mu_E$.
Then there is an
$f:\{0,1\}^{\lambda}\to E$ which is \imp\ for $\nu_{\lambda}$ and
$\mu'_E$ (343Cd).
So (b) tells us that $E$ has a subset $A$ with cardinal
$\kappa$ which is Sierpi\'nski or strongly Sierpi\'nski for $\mu'_E$.
But now $A$ is still Sierpi\'nski or strongly Sierpi\'nski for $\mu$.
}%end of proof of 537B

\leader{537C}{Entangled sets (a) Definition}
If $X$ is a totally ordered set, then $X$ is {\bf $\omega_1$-entangled}
if whenever $n\ge 1$, $I\subseteq n$ and
$\langle x_{\xi i}\rangle_{\xi<\omega_1,i<n}$
is a family of distinct elements of $X$, then
there are distinct $\xi$, $\eta<\omega_1$ such that
$I=\{i:i<n$, $x_{\xi i}\le x_{\eta i}\}$.

\medskip

{\bf (b)} Give $\{0,1\}^{\Bbb N}$ its lexicographic ordering, that is,

\Centerline{$x\le y$ iff either $x=y$ or there is an $n\in\Bbb N$ such that
$x\restr n=y\restr n$ and $x(n)<y(n)$.}

\noindent Then\cmmnt{ the map
$x\mapsto\bover23\sum_{n=0}^{\infty}3^{-n}x(n):\{0,1\}^{\Bbb N}\to\Bbb R$
is an order-isomorphism between $\{0,1\}^{\Bbb N}$ and the Cantor set, so}
any
$\omega_1$-entangled subset of $\{0,1\}^{\Bbb N}$ can be transferred to an
$\omega_1$-entangled subset of $\Bbb R$.

\leader{537D}{Lemma} Let $X$ be an $\omega_1$-entangled totally ordered set.

(a) There is a countable set $D\subseteq X$ which meets $[x,y]$ whenever
$x<y$ in $X$.

(b) Whenever $n\ge 1$, $I\subseteq n$ and
$\langle x_{\xi i}\rangle_{\xi<\omega_1,i<n}$
is a family of distinct elements of $X$,
there are $\xi<\eta<\omega_1$ such that
$I=\{i:i<n$, $x_{\xi i}\le x_{\eta i}\}$.

\proof{{\bf (a)(i)} There is a countable set $D_0\subseteq X$ which meets
$[x,z]$ whenever $x<y<z$ in $X$.
\Prf\Quer\ Otherwise, choose
$\langle x_{\xi i}\rangle_{\xi<\omega_1,i<3}$ inductively so that
$x_{\xi 0}<x_{\xi 1}<x_{\xi 2}$ and $[x_{\xi 0},x_{\xi 2}]$ does not meet
$\{x_{\eta i}:\eta<\xi$, $i<3\}$.    Now, if $\xi$, $\eta<\omega_1$ are
different, we cannot have

\Centerline{$x_{\xi 0}\le x_{\eta 0}$,
\quad$x_{\xi 1}>x_{\eta 1}$,
\quad$x_{\xi 2}\le x_{\eta 2}$.}

\noindent So $\langle x_{\xi i}\rangle_{\xi<\omega_1,i<3}$ witnesses that
$X$ is not $\omega_1$-entangled.\ \Bang\Qed

\medskip

\quad{\bf (ii)}
Set $A=\{(x,y):x<y$, $[x,y]\cap D_0=\emptyset\}$.   Note that if
$(x,y)$, $(x',y')\in A$ are distinct, then $[x,y]\cap[x',y']=\emptyset$,
since otherwise $[\min(x,x'),\max(y,y')]$ would be an interval
disjoint from $D_0$ with at least three elements.    It follows that $A$ is
countable.
\Prf\Quer\ Otherwise, let $\ofamily{\xi}{\omega_1}{(x_{\xi 0},x_{\xi 1})}$
be a family of
distinct elements of $A$.   Then all the $x_{\xi i}$ are
distinct.   But if $\xi$, $\eta<\omega_1$ are different, we cannot have

\Centerline{$x_{\xi 0}\le x_{\eta 0}$,\quad$x_{\xi 1}>x_{\eta 1}$.}

\noindent So $\langle x_{\xi i}\rangle_{\xi<\omega_1,i<2}$ witnesses that
$X$ is not entangled.\ \Bang\Qed

\medskip

\quad{\bf (iii)} So if we set $D=D_0\cup\{x:(x,y)\in A\}$ we shall have a suitable
countable set.

\medskip

{\bf (b)} For $i<n$ write $\le_i\,=\,\le$ if $i\in I$, $\le_i\,=\,\ge$ if
$i\in n\setminus I$;  we are seeking $\xi<\eta$ such that
$x_{\xi i}\le_i x_{\eta i}$ for every $i<n$.   For each family
$\pmb{d}=\ofamily{i}{n}{d_i}$ in $D$, set
$A_{\pmb{d}}=\{\xi:x_{\xi i}\le_id_i$ for each $i<n\}$.   Let
$\zeta<\omega_1$ be
such that $A_{\pmb{d}}\cap\zeta\ne\emptyset$ whenever $\pmb{d}\in D^n$ and
$A_{\pmb{d}}\ne\emptyset$.   Now there are distinct $\xi'$,
$\eta\in\omega_1\setminus\zeta$ such that $x_{\xi'i}\le_i x_{\eta i}$ for
every $i<n$.   For each $i<n$, there is a $d_i\in D$ such that
$x_{\xi'i}\le_i d_i\le_i x_{\eta i}$.   Set $\pmb{d}=\ofamily{i}{n}{d_i}$;
then $\xi'\in A_{\pmb{d}}$ so there is a $\xi\in\zeta\cap A_{\pmb{d}}$.
Now $\xi<\eta$ and $x_{\xi i}\le_i x_{\eta i}$ for every $i$, as
required.
}%end of proof of 537D

\leader{537E}{Lemma} Suppose that $n\ge 1$, $I\subseteq n$ and that
$A\subseteq(\{0,1\}^{\Bbb N})^n$ is non-negligible for the usual product
measure $\nu_{\Bbb N}^n$ on $(\{0,1\}^{\Bbb N})^n$.
Let $\le$ be the lexicographic
ordering of $\{0,1\}^{\Bbb N}$.   Then there are $v$, $w\in A$
such that $v(i)\ne w(i)$ for every $i<n$ and $\{i:i<n$, $v(i)\le w(i)\}=I$.

\proof{ For each $k\in\Bbb N$ let $\Sigma_k$ be the algebra of subsets of
$X=(\{0,1\}^{\Bbb N})^n$ generated by sets of the form
$\{v:v\in X$, $v(i)(j)=1\}$ for $i<n$ and $j<k$.   Then
$\sequence{k}{\Sigma_k}$ is a non-decreasing sequence of
finite algebras
and the $\sigma$-algebra generated by $\bigcup_{k\in\Bbb N}\Sigma_k$ is the
Borel $\sigma$-algebra $\Cal B(X)$ of $X$.   Let $E\in\Cal B(X)$ be a
measurable envelope of $A$ for $\nu_{\Bbb N}^n$.   For each $k\in\Bbb N$,
let $f_k$ be the conditional expectation of $\chi E$ on $\Sigma_k$,
that is,

\Centerline{$f_k(u)
=2^{kn}\nu_{\Bbb N}^n\{v:v\in E$, $v(i)\restr k=u(i)\restr k$
for every $i<n\}$}

\noindent for $u\in X$.   By L\'evy's martingale theorem (275I),
$\chi E\eae\lim_{k\to\infty}f_k$.   In particular, there are a $u\in A$
and a $k\in\Bbb N$ such that $f_k(u)>1-2^{-n}$.   But this means that

\Centerline{$F=\{v:v\in E$, $v(i)\restr k=u(i)\restr k$ for every $i<n\}$}

\noindent has measure greater than $2^{-kn}(1-2^{-n})$, and both
the sets

\Centerline{$F'=\{v:v\in F$, $v(i)(k)=0$ for $i\in I$, $v(i)(k)=1$ for
$i\in n\setminus I\}$,}

\Centerline{$F''=\{w:w\in F$, $w(i)(k)=1$ for $i\in I$, $w(i)(k)=0$ for
$i\in n\setminus I\}$,}

\noindent must have positive measure.   Accordingly we can find
$v\in A\cap F'$ and $w\in A\cap F''$, and these will serve.
}%end of proof of 537E

\leader{537F}{Corollary} Suppose that $A\subseteq\{0,1\}^{\Bbb N}$ is
strongly Sierpi\'nski for the usual measure on $\{0,1\}^{\Bbb N}$.
Then $A$ is $\omega_1$-entangled for the
lexicographic ordering of $\{0,1\}^{\Bbb N}$.

\proof{ Let $\langle x_{\xi i}\rangle_{\xi<\omega_1,i<n}$ be a family of
distinct points in $A$, where $n\ge 1$, and $I$ a subset of $n$.
Then $x_{\xi}=\ofamily{i}{n}{x_{\xi i}}$ belongs to $A^n$, and has no two
coordinates the same, for every $\xi<\omega_1$.   So
$D=\{x_{\xi}:\xi<\omega_1\}$ cannot be negligible.   By 537E, there are
distinct $\xi$, $\eta<\omega_1$ such that
$I=\{i:x_{\xi i}\le x_{\eta i}\}$.
}%end of proof of 537F

\leader{537G}{Theorem}\cmmnt{ ({\smc Todor\v{c}evi\'c 85})}
Suppose that there is an $\omega_1$-entangled totally ordered set $X$ of
size $\kappa\ge\omega_1$.
Then there are two upwards-ccc partially ordered
sets $P$, $Q$ such that $c^{\uparrow}(P\times Q)\ge\kappa$.

\proof{{\bf (a)} Let $Y\subseteq X$ be a set such that
$\#(Y)=\#(X\setminus Y)=\kappa$, and $f:Y\to X\setminus Y$ an injective
function.   Set

\Centerline{$P
=\{I:I\in[Y]^{<\omega}$, $f\restr I$ is order-preserving$\}$,}

\Centerline{$Q
=\{I:I\in[Y]^{<\omega}$, $f\restr I$ is order-reversing$\}$,}

\noindent both ordered by $\subseteq$.   Then
$\{(\{y\},\{y\}):y\in Y\}$ is an up-antichain in $P\times Q$, so
$c^{\uparrow}(P\times Q)\ge\kappa$.

\medskip

{\bf (b)} $P$ is upwards-ccc.   \Prf\ Let
$\ofamily{\alpha}{\omega_1}{I_{\alpha}}$ be a family in $P$.
By the $\Delta$-system Lemma (4A1Db), there is an uncountable set
$A\subseteq\omega_1$ such that $\family{\alpha}{A}{I_{\alpha}}$ is a
$\Delta$-system with root $I$ say;  now there is an $n\in\Bbb N$ such that
$B=\{\alpha:\alpha\in A$, $\#(I_{\alpha}\setminus I)=n\}$ is uncountable.
If $n=0$ then $I_{\alpha}=I_{\beta}$ are upwards-compatible for any
$\alpha$, $\beta\in B$ and we can stop.

If $n\ge 1$, enumerate $I_{\alpha}\setminus I$ in increasing order as
$\ofamily{i}{n}{x_{\alpha i}}$, for each $\alpha\in B$.
Let $D\subseteq X$ be a countable set such that $D$ meets every interval in
$X$ with more than one member (537Da).   For $i<j<n$ and
$\alpha\in B$ let $d_{\alpha i j}$, $d'_{\alpha i j}\in D$ be such that
$x_{\alpha i}\le d_{\alpha i j}\le x_{\alpha j}$ and
$f(x_{\alpha i})\le d'_{\alpha i j}\le f(x_{\alpha j})$.   (Because
$I_{\alpha}\in P$, $f\restr I_{\alpha}$ is order-preserving so
$f(x_{\alpha i})<f(x_{\alpha j})$.)   Let
$\langle d_{ij}\rangle_{i<j<n}$, $\langle d'_{ij}\rangle_{i<j<n}$
be such that

\Centerline{$C=\{\alpha:\alpha\in B$,
$d_{\alpha ij}=d_{ij}$ and $d'_{\alpha ij}=d'_{ij}$ whenever $i<j<n\}$}

\noindent is uncountable.

Consider the family $\langle y_{\alpha i}\rangle_{\alpha\in C,i<2n}$ where
$y_{\alpha i}=x_{\alpha i}$ and $y_{\alpha,i+n}=f(x_{\alpha i})$ if $i<n$.
Because $X$ is entangled, there must be distinct $\alpha$, $\beta\in C$
such that $y_{\alpha i}\le y_{\beta i}$ for every $i<2n$, that is,
$x_{\alpha i}\le x_{\beta i}$ and $f(x_{\alpha i})\le f(x_{\beta i})$ for
every $i<n$.   But now examine $I=I_{\alpha}\cup I_{\beta}$.   If $x$,
$x'\in I$ and $x\le x'$,

\inset{{\it either} both $x$ and $x'$ belong to $I_{\alpha}$ and
$f(x)\le f(x')$ because $I_{\alpha}\in P$,

{\it or} both $x$ and $x'$ belong to $I_{\beta}$ and
$f(x)\le f(x')$,

{\it or} $x=x_{\alpha i}$ and $x'=x_{\beta j}$ where $i<j<n$, so that

\Centerline{$f(x)=f(x_{\alpha i})\le d'_{ij}\le f(x_{\beta_j})=f(x')$,}

{\it or} $x=x_{\beta i}$ and $x'=x_{\alpha j}$ where $i<j<n$, so that
$f(x)\le f(x')$,

{\it or} $x=x_{\alpha i}$ and $x'=x_{\beta i}$ where $i<n$, so that
$f(x)=f(x_{\alpha i})\le f(x_{\beta i})=f(x')$.}

\noindent (Note that we cannot have $x=x_{\alpha i}$ and $x'=x_{\beta j}$
with $j<i$, because in this case $x_{\beta j}\le d_{ji}\le x_{\alpha i}$
while $x_{\beta j}\ne x_{\alpha i}$;  nor can we have
$x=x_{\beta i}<x'=x_{\alpha i}$ with $i<n$.)
So $f\restr I$ is order-preserving and
$I\in P$ witnesses that $I_{\alpha}$ and $I_{\beta}$ are upwards-compatible
in $P$.   As $\ofamily{\alpha}{\omega_1}{I_{\alpha}}$ is arbitrary,
$P$ is upwards-ccc.\ \Qed

\woddheader{537G}{4}{2}{2}{48pt}

{\bf (c)} Similarly, $Q$ is upwards-ccc.   \Prf\ The principal
changes needed in the argument above are

\inset{----- in the choice of the $d'_{\alpha ij}$, we need to write
`$f(x_{\alpha i})\ge d'_{\alpha i j}\ge f(x_{\alpha j})$';

----- in the choice of particular $\alpha$ and $\beta$ in the set $C$,
we need to write
`$y_{\alpha i}\le y_{\beta i}$ for $i<n$ and
$y_{\alpha i}\ge y_{\beta i}$ for $n\le i<2n$'.  \Qed}

\noindent So $P$ and $Q$ satisfy our requirements.
}%end of proof of 537G

\leader{537H}{Scalarly measurable functions (a) Definition}
Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of
subsets of $X$ and $U$ a linear topological space.
A function $\phi:X\to U$ is {\bf scalarly \hbox{($\Sigma$-)}measurable} if
$f\phi:X\to\Bbb R$ is ($\Sigma$-)measurable for every $f\in U^*$.

\spheader 537Hb If $\phi:X\to U$ is scalarly measurable, $V$ is another
linear topological space and $T:U\to V$ is a continuous linear operator,
then
$T\phi:X\to V$ is scalarly measurable\prooflet{, because $hT\in U^*$ for
every $h\in V^*$}.

\spheader 537Hc If $U$ is a separable metrizable locally convex space
and $\phi:X\to U$ is scalarly measurable, then it is measurable.
\prooflet{\Prf\
$\Tau=\{F:F\subseteq U$, $\phi^{-1}[F]\in\Sigma\}$ includes the
cylindrical $\sigma$-algebra of $U$ (4A3U\formerly{4{}A3T}), which is the Borel
$\sigma$-algebra (4A3W\formerly{4{}A3V}).\ \Qed}

\leader{537I}{Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
probability spaces  and $U$ a reflexive Banach space.   Suppose that
$x\mapsto u_x:X\rightarrow  U$ and $y\mapsto f_y:Y\rightarrow U^*$ are
bounded scalarly measurable functions.   Then
$\iint f_y(u_x)\mu(dx)\nu(dy)$ and
$\iint f_y(u_x)\nu(dy)\mu(dx)$ are defined and equal.

\proof{{\bf (a)(i)} Recall from 467Hc that if
$V\subseteq U$ and $W\subseteq U^*$ are closed linear subspaces, I call
them a
`projection pair' if $U=V\oplus W^{\smallcirc}$ and $\|v+v'\|\ge\|v\|$ for
all $v\in V$ and $v'\in W^{\smallcirc}$.   We need to know that this
is symmetric;  that is, that in this case

\Centerline{$U^*=W\oplus V^{\smallcirc}$,
\quad$\|g+g'\|\ge\|g\|$ for all $g\in W$, $g'\in V^{\smallcirc}$.}

\noindent\Prf\ Note first that if $g\in W\cap V^{\smallcirc}$, then
$g(u)=0$ for every $u\in W^{\smallcirc}+V$, that is, $g=0$.
Now take any $f\in U^*$.   Define $g:U\to\Bbb R$ by saying that
$g(v+v')=f(v)$ for $v\in V$, $v'\in W^{\smallcirc}$.   Then $g$ is linear
and continuous and $\|g\|\le\|f\|$.   Now $g(v')=0$ for every
$v'\in W^{\smallcirc}$, that is, $g\in W^{\smallcirc\smallcirc}$, which is
the weak*-closure of $W$ (4A4Eg);  but as $U$ and $U^*$ are reflexive, this
is just the norm-closure of $W$, which is equal to $W$.
Set $g'=f-g$.   Then $g'\in V^{\smallcirc}$.   This shows that
$f\in W+V^{\smallcirc}$;  as $f$ is arbitrary,
$U^*=W\oplus V^{\smallcirc}$.   Finally, I remarked in the course of the
argument that $\|g\|\le\|f\|$, which is what we need to know to check that
$\|g\|\le\|g+g'\|$ whenever $g\in W$ and $g'\in V^{\smallcirc}$.\ \Qed

\medskip

\quad{\bf (ii)} Because $U$ is reflexive, its unit ball is weakly compact, so $U$
is surely weakly compactly generated, therefore weakly K-countably
determined (467M).   Now turn to Lemma 467J.   This tells us that
there is a family $\Cal M$ of subsets of $U\cup U^*$ such that

\inset{for every $B\subseteq X\cup X^*$ there is an $M\in\Cal M$ such that
$B\subseteq M$ and $\#(M)\le\max(\omega,\#(B))$;

whenever $\Cal M'\subseteq\Cal M$ is upwards-directed, then
$\bigcup\Cal M'\in\Cal M$;

whenever $M\in\Cal M$ then $(V_M,W_M)$
is a projection pair of subspaces of $U$ and $U^*$,}

\noindent where I write $V_M=\overline{M\cap U}$ and
$W_M=\overline{M\cap U^*}$.   For $M\in\Cal M$,

\Centerline{$U=V_M\oplus W_M^{\smallcirc}$,
\quad$U^*=W_M\oplus V_M^{\smallcirc}$;}

\noindent let $P_M:U\to V_M$ and $Q_M:U^*\to W_M$ be the corresponding
projections.   Since $\|v\|\le\|v+v'\|$ whenever $v\in V_M$ and
$v'\in W_M^{\smallcirc}$, $\|P_M\|\le 1$;  similarly, $\|Q_M\|\le 1$.

If $u\in U$, $f\in U^*$ and $M\in\Cal M$, then

\Centerline{$f(P_Mu)=(Q_Mf)(u)=(Q_Mf)(P_Mu)$.}

\noindent\Prf\ Express $u$ as $v+v'$ and $f$ as $g+g'$, where $v\in V_M$,
$v'\in W_M^{\smallcirc}$, $g\in W_M$ and $g'\in V_M^{\smallcirc}$.   Then

\Centerline{$f(v)=g(v)=g(u)$,}

\noindent that is,

\Centerline{$f(P_Mu)=(Q_Mf)(P_Mu)=(Q_Mf)(u)$.  \Qed}

\medskip

\quad{\bf (iii)} If $M_0$, $M_1\in\Cal M$ and $M_0\subseteq M_1$ then
$P_{M_0}=P_{M_0}P_{M_1}=P_{M_1}P_{M_0}$.   \Prf\ If $u\in U$, express it as
$v_0+v'_0$ where $v_0\in V_{M_0}$ and $v'_0\in W_{M_0}^{\smallcirc}$;
now express $v'_0$ as $v_1+v'_1$ where $v_1\in V_{M_1}$ and
$v'_1\in W_{M_1}^{\smallcirc}$.   Then

\Centerline{$P_{M_0}u=v_0\in V_{M_1}$,}

\noindent so $P_{M_1}P_{M_0}u=P_{M_0}u$.   On the other hand,
$u=v_0+v_1+v'_1$ where $v_0+v_1\in V_{M_1}$ and
$v'_1\in W_{M_1}^{\smallcirc}$,  so $P_{M_1}u=v_0+v_1$;  and as
$v_1=v'_0-v'_1$ belongs to $W_{M_0}^{\smallcirc}$,
$P_{M_0}P_{M_1}u=v_0=P_{M_0}u$.\ \Qed

\medskip

\quad{\bf (iv)} If $\ofamily{\xi}{\zeta}{M_{\xi}}$ is a non-decreasing
family in $\Cal M$,
where $\zeta$ is a non-zero limit ordinal, then we know
that $M=\bigcup_{\xi<\zeta}M_{\xi}$ belongs to $\Cal M$.   Now

\Centerline{$P_Mu=\lim_{\xi\uparrow\zeta}P_{M_{\xi}}u$}

\noindent for every $u\in U$, the limit being for the norm topology on $U$.
\Prf\ Let $\epsilon>0$.   We know that $P_Mu\in V_M=\overline{M\cap U}$,
so there is a $u'\in M\cap U$ such that $\|u'-P_Mu\|\le\bover12\epsilon$.
Let $\xi<\zeta$ be such that $u'\in M_{\xi}$.   If $\xi\le\eta<\zeta$, then

$$\eqalignno{\|P_{M_{\eta}}u-P_Mu\|
&=\|P_{M_{\eta}}(P_Mu-u')+P_M(u'-P_Mu)\|\cr
\displaycause{because $u'\in V_{M_{\eta}}$, so $P_Mu'=P_{M_{\eta}}u'=u'$}
&\le 2\|P_Mu-u'\|
\le\epsilon.  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (v)} Similarly,

\Centerline{$Q_{M_0}=Q_{M_0}Q_{M_1}=Q_{M_1}Q_{M_0}$}

\noindent whenever $M_0$, $M_1\in\Cal M$ and $M_0\subseteq M_1$, and

\Centerline{$Q_Mf=\lim_{\xi\uparrow\zeta}Q_{M_{\xi}}f$}

\noindent whenever $f\in U^*$ and $\zeta$ is a non-zero limit ordinal
and $\ofamily{\xi}{\zeta}{M_{\xi}}$ is a non-decreasing family in
$\Cal M$ with union $M$.

\medskip

{\bf (b)} Now let $\Cal M_0$ be $\{M:M\in\Cal M$, $\#(M)\le\omega\}$.
Then there is an $M_0\in\Cal M_0$ such that

\Centerline{$P_{M_0}(u_x)=P_{M}(u_x)\,\,\mu$-a.e.($x$)}

\noindent whenever $M_0\subseteq M\in\Cal M_0$.

\Prf\Quer\ Suppose, if possible,
otherwise.   Then we can choose inductively an increasing
family $\langle M_{\xi}\rangle_{\xi<\omega_1}$ in $\Cal M_0$
such that

\Centerline{$\mu\{x:P_{M_{\xi+1}}(u_x)\ne P_{M_{\xi}}(u_x)\}
>0$ for every $\xi<\omega_1$,}

\Centerline{$M_{\xi}=\bigcup_{\eta<\xi}M_{\eta}$ whenever
$\xi<\omega_1$ is a non-zero countable limit ordinal.}

\noindent (The set of $x$ for which
$P_{M_{\xi+1}}(u_x)\ne P_{M_{\xi}}(u_x)$ is necessarily measurable
because $x\mapsto P_{M_{\xi+1}}u_x-P_{M_{\xi}}u_x$ is scalarly measurable,
by 537Hb, therefore measurable for the norm topology, by 537Hc, since
$V_{M_{\xi+1}}$ is separable.)   Now there must be a $\delta>0$ such that

\Centerline{$A=\{\xi:\xi<\omega_1$, $\mu E_{\xi}\ge\delta\}$}

\noindent is infinite, where

\Centerline{$E_{\xi}
=\{x:\|P_{M_{\xi+1}}(u_x)-P_{M_{\xi}}(u_x)\|\ge\delta\}$}

\noindent for each $\xi<\omega_1$.   But in this case
there must be an $x\in X$ such that

\Centerline{$A'=\{\xi:\xi\in A$, $x\in E_{\xi}\}$}

\noindent is infinite.   (Take a sequence $\sequencen{\xi_n}$ of distinct
points in $A$, and $x\in\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}E_{\xi_m}$.)
Let $\zeta$ be any cluster point of $A'$ in $\omega_1$.   Then

\Centerline{$P_{M_{\zeta}}(u_x)
=\lim_{\xi\uparrow\zeta}P_{M_{\xi}}(u_x)$}

\noindent ((a-iv) above), which is impossible.\ \Bang\Qed

\medskip


{\bf (c)} Similarly, there is an $M_1\in\Cal M_0$ such that
$M_1\supseteq M_0$ and

\Centerline{$P_{M_1}(f_y)=P_M(f_y)\,\,\nu$-a.e.($y$)}

\noindent whenever $M_1\subseteq M\in\Cal M_0$.
Because $x\mapsto P_{M_1}(u_x)$ and $y\mapsto Q_{M_1}(f_y)$
are scalarly measurable maps to norm-separable spaces, they are
norm-measurable;  again because $V_{M_1}$ and $W_{M_1}$ are separable,
$(x,y)\to(P_{M_1}u_x,Q_{M_1}f_y):X\times Y\to V_{M_1}\times W_{M_1}$ is
$\Sigma\tensorhat\Tau$-measurable (418Bd).   Because
$(f,x)\mapsto f(x):U^*\times U\to\Bbb R$ is norm-continuous,
$(x,y)\mapsto(Q_{M_1}f_y)(P_{M_1}u_x)$ is
$\Sigma\tensorhat\Tau$-measurable, and

\Centerline{$\iint(Q_{M_1}f_y)(P_{M_1}u_x)\mu(dx)\nu(dy)
=\iint(Q_{M_1}f_y)(P_{M_1}u_x)\nu(dy)\mu(dx)$}

\noindent by Fubini's theorem (252C).

Now observe that if $y\in Y$ there is an $M\in\Cal M_0$ such that
$M_1\subseteq M$ and $f_y\in M$.   So

$$\eqalign{\int f_y(u_x)\mu(dx)
&=\int (Q_Mf_y)(u_x)\mu(dx)
=\int f_y(P_Mu_x)\mu(dx)\cr
&=\int f_y(P_{M_1}u_x)\mu(dx)
=\int(Q_{M_1}f_y)(P_{M_1}u_x)\mu(dx).\cr}$$

\noindent This is true for every $y$.   So
$\iint f_y(u_x)\mu(dx)\nu(dy)$ is defined and equal to
$\iint(Q_{M_1}f_y)(P_{M_1}u_x)\mu(dx)\nu(dy)$.
Similarly,

\Centerline{$\iint f_y(u_x)\nu(dy)\mu(dx)
= \iint(Q_{M_1}f_y)(P_{M_1}u_x)\nu(dy)\mu(dx)$.}

\noindent Putting these together, we have the result.
}%end of proof of 537I

\leader{537J}{Corollary} Let $(X,\Sigma,\mu)$, $(Y,\Tau,\nu)$ and
$(Z,\Lambda,\sigma)$ be
probability spaces.   Let $x\mapsto U_x:X\rightarrow\Lambda$
and $y\mapsto V_y:Y\rightarrow\Lambda$ be functions such that

\Centerline{$x\mapsto\sigma(U_x\cap W)$,
\quad$y\mapsto\sigma(V_y\cap W)$}

\noindent are measurable for every $W\in\Lambda$.   Then
$\iint\sigma(U_x\cap V_y)\mu(dx)\nu(dy)$ and
$\iint\sigma(U_x\cap V_y)\nu(dy)\mu(dx)$ are defined and equal.

\proof{{\bf (a)} For $x\in X$ set $u_x=(\chi U_x)^{\ssbullet}$ in
$L^2(\sigma)$.   Then $x\mapsto u_x$ is scalarly measurable.   \Prf\
If $f\in U^*$, there is a $v\in L^2(\sigma)$ such that
$f(u)=\int u\times v$ for every $u\in L^2(\sigma)$ (244K).
Let $\epsilon>0$.   Then there are $W_0,\ldots,W_n\in\Lambda$ and
$\alpha_0,\ldots,\alpha_n\in\Bbb R$ such that
$\|v-\sum_{i=0}^n\alpha_i(\chi W_i)^{\ssbullet}\|_2\le\epsilon$ (244Ha),
so that

\Centerline{$|f(u_x)-\sum_{i=0}^n\alpha_i\sigma(U_x\cap W_i)|
=|\int u_x\times v-\int u_x\times\sum_{i=0}^n\alpha_i(\chi W_i)^{\ssbullet}|
\le\epsilon\|u_x\|_2\le\epsilon$}

\noindent for every $x\in X$.   Now the function
$x\mapsto\sum_{i=0}^n\alpha_i\sigma(U_x\cap W_i)$ is $\Sigma$-measurable.
So we see that the function $x\mapsto f(u_x)$ is uniformly approximated by
$\Sigma$-measurable functions and is itself $\Sigma$-measurable.   As $f$
is arbitrary, $x\mapsto u_x$ is scalarly measurable.\ \Qed

\medskip

{\bf (b)} Similarly, setting $v_y=(\chi V_y)^{\ssbullet}$ for $y\in Y$,
$y\mapsto v_y:Y\to L^2(\sigma)$ is scalarly measurable.   Identifying
$L^2(\sigma)$ with its dual, 537I tells us that


\Centerline{$\iint (u_x|v_y)\mu(dx)\nu(dy)
= \iint (u_x|v_y)\nu(dy)\mu(dx)$,}

\noindent that is, that

\Centerline{$\iint\sigma(U_x\cap V_y)\mu(dx)\nu(dy)
= \iint\sigma(U_x\cap V_y)\nu(dy)\mu(dx)$.}
}%end of proof of 537J

\leader{537K}{}\cmmnt{ The next few paragraphs
will be concerned with upper and lower integrals.   For the basic theory of
these, see \S133 and 214J.

\medskip

\noindent}{\bf Theorem}\cmmnt{ ({\smc Freiling 86}, {\smc Shipman 90})}
Let $\langle(X_j,\Sigma_j,\mu_j)\rangle_{j\le m}$ be a
finite sequence of probability spaces and $\langle\kappa_j\rangle_{j\le m}$
a sequence of cardinals such that $X_j^{\Bbb N}$, with its
product measure $\mu_j^{\Bbb N}$, has a subset with cardinal $\kappa_j$
which is not covered by $\kappa_{j-1}$ negligible sets (if $j\ge 1$) and is
not negligible (if $j=0$).    Let $f:\prod_{j\le m}X_j\to\Bbb R$ be a
bounded function, and suppose that $\sigma:m+1\to m+1$ and
$\tau:m+1\to m+1$ are permutations.   Set

\Centerline{$I=\underlineint\ldots\underlineint f(x_0,\ldots,x_m)
   dx_{\sigma(m)}\ldots dx_{\sigma(0)}$,}

\Centerline{$I'=\overlineint\ldots\overlineint f(x_0,\ldots,x_m)
   dx_{\tau(m)}\ldots dx_{\tau(0)}$.}

\noindent Then $I\le I'$.

\proof{ Let $M\ge 0$ be such that $|f(x_0,\ldots,x_m)|\le M$ for
all $x_0,\ldots,x_m$.

\medskip

{\bf (a)} Set $Z=\prod_{j\le m}X_j^{\Bbb N}$.
The key fact is that we can find negligible sets
$W(\pmb{u})\subseteq X_k^{\Bbb N}$, for $k\le m$ and
$\pmb{u}\in\prod_{j\le m,j\ne k}X_j^{\Bbb N}$, such that

\Centerline{$I
\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(t_{0i},\ldots,t_{mi})$}

\noindent whenever
$\langle t_j\rangle_{j\le m}=\langle\sequence{i}{t_{ji}}\rangle_{j\le m}$
is such that
$t_k\notin W(t_0,\ldots,t_{k-1},t_{k+1},\ldots,t_m)$ for every $k$.
\Prf\ Because the formula

\Centerline{$\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^n
  f(t_{0i},\ldots,t_{mi})$}

\noindent is tolerant of permutations of the coordinates $0,\ldots,m$, it
is enough to consider the case $\sigma(j)=j$ for $j\le m$, so that

\Centerline{$I=\underlineint\ldots\underlineint f(x_0,\ldots,x_m)
   dx_m\ldots dx_0$.}

\medskip

\quad{\bf (i)} Define $D_0,\ldots,D_{m+1}$ as follows.
$D_0=\{\emptyset\}=\prod_{j<0}X_j^{\Bbb N}$.   For $0<k\le m$ let
$D_k$ be the set of those $(t_0,\ldots,t_{k-1})\in\prod_{j<k}X_j^{\Bbb N}$
such that

\Centerline{$I\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^n
  \underlineint\ldots\underlineint f(t_{0i},\ldots,t_{k-1,i},x_k,\ldots,x_m)
     dx_m\ldots dx_k$,}

\noindent where $t_j=\sequence{i}{t_{ji}}$ for
$j<k$.   For $k<m$ and
$\pmb{u}=(u_0,\ldots,u_{k-1},u_{k+1},\ldots,u_m)$ in
$\prod_{j\le m,j\ne k}X_j^{\Bbb N}$, set

$$\eqalign{W(\pmb{u})&=\emptyset\text{ if }
  (u_0,\ldots,u_{k-1})\notin D_k,\cr
&=\{t:t\in X_k^{\Bbb N},\,(u_0,\ldots,u_{k-1},t)\notin D_{k+1}\}
  \text{ otherwise}.\cr}$$

\medskip

\quad{\bf (ii)} $W(\pmb{u})\subseteq X_k^{\Bbb N}$ is negligible.
To see this, we need consider only
the case in which $(u_0,\ldots,u_{k-1})$ belongs to $D_k$.
Express $u_j$ as
$\sequence{i}{u_{ji}}$ for $j<k$, and for $i\in\Bbb N$ define
$h_i:X_k\to\Bbb R$ by setting

\Centerline{$h_i(x)=\underlineint\ldots\underlineint
f(u_{0i},\ldots,u_{k-1,i},x,x_{k+1},\ldots,x_m)
  dx_m\ldots dx_{k+1}$}

\noindent for $x\in X_k$.    Now the definition of $D_k$ tells us just that

\Centerline{$I\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^n
  \underlineint\ldots\underlineint
     f(u_{0i},\ldots,u_{k-1,i},x_k,\ldots,x_m)dx_m\ldots dx_k$,}

\noindent that is, that

\Centerline{$I\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^n
   \underlineint h_i(x)dx$.}

\noindent For each $i\in\Bbb N$ let $g_i:X_k\to[-M,M]$ be a measurable
function such that $g_i(x)\le h_i(x)$ for every $x$ and
$\int g_id\mu_k=\underline{\int}h_id\mu_k$.
Now consider the functions $\tilde g_i:X_k^{\Bbb N}\to\Bbb R$ defined by
setting $\tilde g_i(t)=g_i(t_i)$ for $t=\sequence{i}{t_i}\in X_k^{\Bbb N}$.
We have $\int\tilde g_id\mu_k^{\Bbb N}=\underline{\int}h_id\mu_k$
for each $i$, while $\sequence{i}{\tilde g_i}$ is a uniformly bounded
independent sequence of random variables.   By the strong law of large
numbers in the form 273H,

\Centerline{$\lim_{n\to\infty}\Bover1{n+1}
  \sum_{i=0}^n(\tilde g_i(t)-\int\tilde g_id\mu_k^{\Bbb N})
=0$}

\noindent for almost every $t\in X_k^{\Bbb N}$.   Since

\Centerline{$\liminf_{n\to\infty}\Bover1{n+1}
  \sum_{i=0}^n\int\tilde g_id\mu_k^{\Bbb N}
=\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^n\underlineint h_id\mu_k
\ge I,$}

\noindent we have

$$\eqalign{I
&\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^n\tilde g_i(t)
\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nh_i(t_i)\cr
&=\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^n
  \underline{\int}\ldots\underline{\int}
  f(u_{0i},\ldots,u_{k-1,i},t_i,x_{k+1},\ldots,x_m)
    dx_m\ldots dx_{k+1}\cr}$$

\noindent for almost every $t=\sequence{i}{t_i}\in X_k^{\Bbb N}$, that is,
$(u_0,\ldots,u_{k-1},t)\in D_{k+1}$ for almost every $t\in X_k^{\Bbb N}$,
that is, $W(\pmb{u})$ is negligible, as required.

\medskip

\quad{\bf (iii)} Suppose that
$\pmb{t}=(t_0,\ldots,t_m)\in Z$ is such that
$t_k\notin W(t_0,\ldots,t_{k-1},t_{k+1},\ldots,t_m)$ for every $k<m$.
Then $(t_0,\ldots,t_k)\in D_{k+1}$ for every $k$;  in particular,
$\pmb{t}\in D_{m+1}$ and, writing $t_j=\sequence{i}{t_{ji}}$ for each $j$,

\Centerline{$I
\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(t_{0i},\ldots,t_{mi})$.
\Qed}

\medskip

{\bf (b)} Similarly, or applying the argument above to $-f$,
we have negligible sets
$W'(\pmb{u})\subseteq X_k^{\Bbb N}$, for $k\le m$ and
$\pmb{u}\in\prod_{j\le m,j\ne k}X_j^{\Bbb N}$, such that

\Centerline{$I'
\ge\limsup_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(t_{0i},\ldots,t_{mi})$}

\noindent whenever
$\langle t_j\rangle_{j\le m}=\langle\sequence{i}{t_{ji}}\rangle_{j\le m}$
is such that
$t_k\notin W'(t_0,\ldots,t_{k-1},t_{k+1},\ldots,t_m)$ for every $k$.
Enlarging the
$W'(\pmb{u})$ if necessary, we may suppose that
$W'(\pmb{u})\supseteq W(\pmb{u})$ for every $\pmb{u}$.

\medskip

{\bf (c)} Now the point of the construction is that
we can find a $\pmb{t}=(t_0,\ldots,t_m)\in Z$ such that
$t_k\notin W'(t_0,\ldots,t_{k-1},t_{k+1},\penalty-100\ldots,t_m)$ for
every $k$.   \Prf\ For each $k\le m$ let $A_k\subseteq X_k^{\Bbb N}$
be a non-negligible set
with cardinal $\kappa_k$ which (if $k\ge 1$)
cannot be covered by $\kappa_{k-1}$ negligible sets.
Choose $t_m$, $t_{m-1},\ldots,t_0$ in such a way that

\Centerline{$t_k\in A_k$,
\quad$t_k\notin W(\pmb{u})$ whenever
$\pmb{u}\in\prod_{j<k}A_j\times\prod_{k<j\le m}\{t_j\}$;}

\noindent this is always possible because
$\#(A_0\times\ldots\times A_{k-1})=\kappa_{k-1}$ if $k\ge 1$.\ \Qed

So we get

$$\eqalign{I
&\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(t_{0i},\ldots,t_{mi})\cr
&\le\limsup_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(t_{0i},\ldots,t_{mi})
\le I',\cr}$$

\noindent as claimed.
}%end of proof of 537K

\leader{537L}{Corollary}
Let $\langle(X_j,\Sigma_j,\mu_j)\rangle_{j\le m}$ be a
finite sequence of probability spaces such that $X_j^{\Bbb N}$, with its
product measure $\mu_j^{\Bbb N}$, has a Sierpi\'nski set with cardinal
$\omega_{j+1}$ for each $j\le m$.    Let $f:\prod_{j\le m}X_j\to\Bbb R$ be a
bounded function, and suppose that $\sigma:m+1\to m+1$ and
$\tau:m+1\to m+1$ are permutations such that the two repeated integrals

\Centerline{$I=\int\ldots\int f(x_0,\ldots,x_m)
   dx_{\sigma(m)}\ldots dx_{\sigma(0)}$,}

\Centerline{$I'=\int\ldots\int f(x_0,\ldots,x_m)
   dx_{\tau(m)}\ldots dx_{\tau(0)}$,}

\noindent are both defined.   Then $I=I'$.

\proof{ Apply 537K in both directions.
}%end of proof of 537L

\leader{537M}{}\cmmnt{ A pair of simple facts which I never got round to
spelling out will be useful below.

\medskip

\noindent}{\bf Lemma} Suppose that $(X,\Sigma,\mu)$ is a totally
finite measure space and $f$ is a $[0,\infty]$-valued function defined
almost everywhere in $X$.

(a) If
$\gamma<\overline{\int}f$, then there is a measurable integrable function
$g:X\to\coint{0,\infty}$ such that $\int g\ge\gamma$ and
$\{x:x\in\dom f$, $g(x)\le f(x)\}$ has full outer measure in $X$.

(b) If $\underline{\int}f<\gamma$, then there is a measurable integrable
function $g:X\to\coint{0,\infty}$ such that $\int g\le\gamma$ and
$\{x:x\in\dom f$, $f(x)\le g(x)\}$ has full outer measure in $X$.

\proof{{\bf (a)} By 135H(b-i),

\Centerline{$\overlineint f
=\sup_{k\in\Bbb N}\overlineint\min(f(x),k)\mu(dx)$;}

\noindent let $k\in\Bbb N$ be such that
$\overline{\int}f_k>\gamma$, where $f_k(x)=\min(f(x),k)$ for
$x\in\dom f$.   Because $\mu X<\infty$, $\overline{\int}f_k$ is finite.
By 133J(a-i), there is an integrable $h$ such that
$\int h=\overline{\int}f_k$ and $f_k\leae h$;  adjusting $h$ on a
negligible set if necessary, we can arrange that $h$ is defined (and
finite) everywhere
on $X$ and is measurable.   Set $\epsilon=(\int h-\gamma)/(1+\mu X)$, and
$g=h-\epsilon\chi X$;  then by the last part of 133J(a-i),

\Centerline{$\{x:x\in\dom f$, $g(x)\le f(x)\}
=\{x:x\in\dom f$, $h(x)\le f(x)+\epsilon\}$}

\noindent has full outer measure in $X$, while $\int g\ge\gamma$.

\medskip

{\bf (b)} By 135Ha,
there is a measurable $h:X\to[0,\infty]$
such that $h\leae f$ and $\int h=\underline{\int}f$;  as $\int h$ is
finite, $h$ is finite a.e.\ and can be adjusted to be finite
everywhere.   Set
$\epsilon=(\gamma-\int h)/(1+\mu X)$, and
$g=h+\epsilon\chi X$;  then $\int g\le\gamma$ and $\{x:f(x)\le g(x)\}$ has
full outer measure.
}%end of proof of 537M

\leader{537N}{}\cmmnt{ For ordinary two-variable repeated integrals we
can squeeze a little bit more out than is given by 537K.

\medskip

\noindent}{\bf Proposition}
Let $(X,\Sigma,\mu)$ be a semi-finite measure space,
$(Y,\Tau,\nu)$ a probability space, and
$\nu^{\Bbb N}$ the product measure on $Y^{\Bbb N}$.
If $\non(E,\Cal N(\mu))<\cov\Cal N(\nu^{\Bbb N})$ for every
$E\in\Sigma\setminus\Cal N(\mu)$, then

\Centerline{$\underlineint\,\underlineint f(x,y)\nu(dy)\mu(dx)
\le\overlineint\,\overlineint f(x,y)\mu(dx)\nu(dy)$}

\noindent for every function $f:X\times Y\to[0,\infty]$.

\proof{{\bf (a)} To begin with, suppose that $\mu X<\infty$ and
$\#(X)<\cov\Cal N(\nu^{\Bbb N})$.    For each $y\in Y$, let
$h_y:X\to[0,\infty]$ be a measurable function
such that $f(x,y)\le h_y(x)$ for every
$x\in X$ and $\int h_yd\mu=\overline{\int}f(x,y)\mu(dx)$;
let $v:Y\to[0,\infty]$
be a measurable function such that $\int h_yd\mu\le v(y)$ for every
$y\in Y$ and
$\int v\,d\nu=\overline{\int}\,\overline{\int}f(x,y)\mu(dx)\nu(dy)$.
If this is infinite, we can stop.   Otherwise, for each $x\in X$ let
$g_x:Y\to[0,\infty]$ be a measurable function such that
$g_x(y)\le f(x,y)$ for every $y\in Y$ and
$\int g_xd\nu=\underline{\int}f(x,y)\nu(dy)$, and let
$u:X\to[0,\infty]$ be a measurable function such that
$u(x)\le\int g_x\,d\nu$ for every $x$ and
$\int u\,d\mu=\underline{\int}\,\underline{\int}f(x,y)\nu(dy)\mu(dx)$.

As $\#(X)<\cov\Cal N(\nu^{\Bbb N})$,
we can find a sequence $\sequence{i}{y_i}$ in $Y$ such that

\Centerline{$\int v\,d\nu
=\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nv(y_i)$}

\noindent and

\Centerline{$\int g_xd\nu
=\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^ng_x(y_i)$}

\noindent for every $x\in X$.   (For by 273J, the set of such sequences
is the
intersection of fewer than $\cov\Cal N(\nu^{\Bbb N})$ conegligible sets
in $Y^{\Bbb N}$, and cannot be empty.)   If $x\in X$, then

\Centerline{$u(x)
\le\int g_xd\nu
=\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^ng_x(y_i)
\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nh_{y_i}(x)$.}

\noindent So

$$\eqalignno{\underline{\int}\,\underline{\int}f(x,y)\nu(dy)\mu(dx)
&=\int u\,d\mu
\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^n\int h_{y_i}d\mu\cr
\displaycause{by Fatou's Lemma}
&\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nv(y_i)
=\int v\,d\nu\cr
&=\overline{\int}\,\overline{\int}f(x,y)\mu(dx)\nu(dy),\cr}$$

\noindent as required.

\medskip

{\bf (b)} Now suppose that $\mu$ is totally finite and that
$X$ has a subset $A$ of full outer measure with
$\#(A)<\cov\Cal N(\nu^{\Bbb N})$.   Let $\mu_A$ be the subspace measure on
$A$.   Then for any $q:X\to[0,\infty]$ we have

\Centerline{$\underlineint q\,d\mu
\le\underlineint(q\restr A)d\mu_A
\le\overlineint(q\restr A)d\mu_A
\le\overlineint q\,d\mu$}

\noindent (214J).   So, writing
$f_A$ for the restriction of $f$ to $A\times Y$,

$$\eqalignno{\underline{\int}\,\underline{\int}f(x,y)\nu(dy)\mu(dx)
&\le\underline{\int}\,\underline{\int}f_A(x,y)\nu(dy)\mu_A(dx)\cr
&\le\overline{\int}\,\overline{\int}f_A(x,y)\mu_A(dx)\nu(dy)\cr
\displaycause{by (a)}
&\le\overline{\int}\,\overline{\int}f(x,y)\mu(dx)\nu(dy).\cr}$$

\medskip

{\bf (c)} For the general case, let $u:X\to[0,\infty]$ be a
measurable function such that $u(x)\le\underline{\int}f(x,y)\nu(dy)$ for
every $x\in X$ and
$\int u\,d\mu=\underline{\int}\,\underline{\int}f(x,y)\nu(dy)\mu(dx)$.
Take any $\gamma<\int u\,d\mu$.   Because $\mu$ is semi-finite, there is
a non-empty set $F\in\Sigma$ of finite measure such that
$\int_Fu\,d\mu>\gamma$.
Now let $\Cal E$ be the family of measurable
sets $E\subseteq F$ of finite measure for which there is a non-empty
set $A\subseteq E$, with cardinal
less than $\cov\Cal N(\nu^{\Bbb N})$, such that $\mu^*A=\mu E$, that is,
$A$ has full outer measure for the subspace measure $\mu_E$, that is,
$E$ is a measurable envelope of $A$.   Then $\Cal E$ is closed under finite
unions and every non-empty member of $\Sigma$ includes a member of
$\Cal E$.   So there
is a non-decreasing sequence $\sequence{k}{E_k}$ in $\Cal E$ such that
$\bigcup_{k\in\Bbb N}E_k\subseteq F$ and
$F\setminus\bigcup_{k\in\Bbb N}E_k$ is negligible.   In this case,
$\gamma<\int_Fu\,d\mu=\lim_{k\to\infty}\int_{E_k}u\,d\mu$, so there is a
$k\in\Bbb N$ such that $\gamma\le\int_{E_k}u\,d\mu$.   Set $E=E_k$.

Consider the restriction $f_E$ of $f$ to $E\times Y$ and the subspace
measure $\mu_E$ on $E$.   We have

$$\eqalignno{\gamma
&\le\int_Eu\,d\mu
=\int(u\restr E)d\mu_E
\le\underline{\int}\,\underline{\int}f_E(x,y)\nu(dy)\mu_E(dx)\cr
&\le\overline{\int}\,\overline{\int}f_E(x,y)\mu_E(dx)\nu(dy)\cr
\displaycause{because $E\in\Cal E$, so we can use (b)}
&\le\overline{\int}\,\overline{\int}f(x,y)\mu(dx)\nu(dy)\cr}$$

\noindent because $\overline{\int}f_E(x,y)\mu_E(dx)
\le\overline{\int}f(x,y)\mu(dx)$ for every $y$, by 214Ja or otherwise.
Since $\gamma$ is arbitrary,

\Centerline{$\underlineint\,\underlineint f(x,y)\nu(dy)\mu(dx)
\le\overlineint\,\overlineint f(x,y)\mu(dx)\nu(dy)$}

\noindent in this case also.
}%end of proof of 537N

\leader{537O}{Corollary}
Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be probability spaces, and
$\nu^{\Bbb N}$ the product measure on $Y^{\Bbb N}$.
If $\shrplusN(\mu)\le\cov\Cal N(\nu^{\Bbb N})$ then

\Centerline{$\overlineint\underlineint f(x,y)\nu(dy)\mu(dx)
\le\overlineint\,\overlineint f(x,y)\mu(dx)\nu(dy)$}

\noindent for every
function $f:X\times Y\to\coint{0,\infty}$.

\proof{ Take any
$\gamma<\overline{\int}\underline{\int}f(x,y)\nu(dy)\mu(dx)$.
By 537Ma, there are a measurable function $u:X\to\coint{0,\infty}$ and a
set $A$ of full outer measure in $X$ such that $\int u\,d\mu\ge\gamma$ and
$u(x)\le\underline{\int}f(x,y)\nu(dy)\mu(dx)$ for every $x\in A$.
Let $\mu_A$ be the subspace measure on $A$, and $f_A$ the restriction
of $f$ to $A\times Y$.   If $B\subseteq A$ is any non-negligible relatively
measurable set, there is a non-negligible $D\subseteq B$ such that
$\#(D)<\shrplusN(\mu)$, so

\Centerline{$\non(B,\Cal N(\mu_A))
=\non(B,\Cal N(\mu))\le\#(D)<\cov\Cal N(\nu^{\Bbb N})$.}

\noindent So

$$\eqalignno{\gamma
&\le\int u\,d\mu
=\int(u\restr A)d\mu_A
\le\underline{\int}\,\underline{\int}f_A(x,y)\nu(dy)\mu_A(dx)\cr
\displaycause{because $u\restr A$ is measurable and
$(u\restr A)(x)\le\underline{\int}f_A(x,y)\nu(dy)$ for every $x\in A$}
&\le\overline{\int}\,\overline{\int}f_A(x,y)\mu_A(dx)\nu(dy)\cr
\displaycause{by 537N}
&\le\overline{\int}\,\overline{\int}f(x,y)\mu(dx)\nu(dy)\cr}$$

\noindent because $\overline{\int}f_A(x,y)\mu_A(dx)
\le\overline{\int}f(x,y)\mu(dx)$ for every $y$, by 214J again.
As $\gamma$ is arbitrary, we have the result.
}%end of proof of 537O

\cmmnt{\medskip

\noindent{\bf Remark} There is a similar inequality, under different
hypotheses, in 543C below.
}

\leader{537P}{Corollary}
Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be probability spaces, and
$\nu^{\Bbb N}$ the product measure on $Y^{\Bbb N}$;  suppose that
$\shrplusN(\mu)\le\cov\Cal N(\nu^{\Bbb N})$, and that
$f:X\times Y\to\Bbb R$ is bounded.

(a)

\Centerline{$\overlineint\underlineint f(x,y)\nu(dy)\mu(dx)
\le\overlineint\,\overlineint f(x,y)\mu(dx)\nu(dy)$,}

\Centerline{$\underlineint \,\underlineint f(x,y)\mu(dx)\nu(dy)
\le\underlineint\overlineint f(x,y)\nu(dy)\mu(dx)$.}

(b) If $\iint f(x,y)\mu(dx)\nu(dy)$ is defined, and $\int f(x,y)\nu(dy)$
is defined for almost every $x$, then the other repeated integral
$\iint f(x,y)\nu(dy)\mu(dx)$ is
defined and equal to $\iint f(x,y)\mu(dx)\nu(dy)$.

\proof{{\bf (a)} Apply 537O to the functions $(x,y)\mapsto M+f(x,y)$,
$(x,y)\mapsto M-f(x,y)$ for suitable $M$.

\wheader{537P}{6}{2}{2}{60pt}

{\bf (b)} By (a),

$$\eqalign{\iint f(x,y)\mu(dx)\nu(dy)
&\le\underline{\int}\int f(x,y)\nu(dy)\mu(dx)\cr
&\le\overline{\int}\int f(x,y)\nu(dy)\mu(dx)
\le\iint f(x,y)\mu(dx)\nu(dy).\cr}$$
}%end of proof of 537P

\leader{537Q}{}\cmmnt{ We can extend the second part of 537Pa, as well as
the first, to unbounded functions, if we strengthen the set-theoretic
hypothesis.

\medskip

\noindent}{\bf Proposition}\cmmnt{ ({\smc Humke \& Laczkovich 05})}
Let $(X,\Sigma,\nu)$ and $(Y,\Tau,\mu)$ be probability spaces, and
$\mu^{\Bbb N}$, $\nu^{\Bbb N}$ the product measures on $X^{\Bbb N}$,
$Y^{\Bbb N}$ respectively.
If $\shrplusN(\mu^{\Bbb N})\le\cov\Cal N(\nu^{\Bbb N})$ then
$\underline{\int}\,\underline{\int}f(x,y)\mu(dx)\nu(dy)
\le\underline{\int}\overline{\int}f(x,y)\nu(dy)\mu(dx)$
for every function $f:X\times Y\to\coint{0,\infty}$.

\proof{ \Quer\ Suppose, if possible, otherwise.

\medskip

{\bf (a)} There is a measurable function $u:Y\to\coint{0,\infty}$ such that

\Centerline{$u(y)\le\underlineint f(x,y)\mu(dx)$ for every $y$,
\quad$\underlineint\overlineint f(x,y)\nu(dy)\mu(dx)
<\int u\,d\nu$.}

\noindent
Since $\int u\,d\nu$ is the supremum of the integrals of the non-negative
simple functions dominated by $u$, we may suppose that $u$ itself is a
simple function;  express it as $\sum_{j=0}^m\alpha_j\chi F_j$ where
$\alpha_j\ge 0$ for each $i$ and $(F_0,\ldots,F_m)$ is a partition of $Y$
into measurable sets.   Now

$$\eqalignno{\sum_{j=0}^m\underline{\int}\overline{\int}
  f(x,y)\chi F_j(y)\nu(dy)\mu(dx)
&\le\underline{\int}\sum_{j=0}^m
  \overline{\int}f(x,y)\chi F_j(y)\nu(dy)\mu(dx)\cr
\displaycause{133J(b-v)}
&\le\underline{\int}\overline{\int}f(x,y)\nu(dy)\mu(dx)\cr
\displaycause{because if $x\in X$ and $q:Y\to[0,\infty]$ is measurable and
$f(x,y)\le q(y)$ for every $y$, then
\ifnum\stylenumber=12\else the sum \fi
$\sum_{j=0}^m\overline{\int}f(x,y)\chi F_j(y)\nu(dy)$ is at most
$\sum_{j=0}^m\int q\times\chi F_jd\nu=\int q\,d\nu$}
&<\int u\,d\nu
=\sum_{j=0}^m\alpha_j\nu F_j.\cr}$$

\noindent There are therefore a $j\le m$ and a $\gamma<1$ such that

\Centerline{$\underlineint\overlineint
  f(x,y)\chi F_j(y)\nu(dy)\mu(dx)
<\gamma\alpha_j\nu F_j$.}

\noindent Now there is a measurable function
$v:X\to\coint{0,\infty}$ such that
$\int v\,d\mu\le\gamma\alpha_j\nu F_j$ and

\Centerline{$D
=\{x:x\in X$, $\overlineint f(x,y)\chi F_j(y)\nu(dy)\le v(x)\}$}

\noindent has full outer measure in $X$, by 537Mb.

\medskip

{\bf (b)} For $y\in Y$ and $\pmb{x}=\sequence{i}{x_i}\in X^{\Bbb N}$, set
$h(\pmb{x},y)=\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(x_i,y)$.
If $y\in Y$, then $\underline{\int}f(x,y)\mu(dx)\le h(\pmb{x},y)$ for
$\mu^{\Bbb N}$-almost every $\pmb{x}$.   \Prf\ We have a measurable
function $q:X\to\coint{0,\infty}$ such that $q(x)\le f(x,y)$ for every $x$
and

$$\eqalign{\underline{\int}f(x,y)\mu(dx)
&=\int q\,d\mu
\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nq(x_i)\cr
&\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(x_i,y)
=h(\pmb{x},y)\cr}$$

\noindent for almost every $\pmb{x}=\sequence{i}{x_i}$.\ \QeD\   At the
same time,

\Centerline{$V
=\{\sequence{i}{x_i}:\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nv(x_i)
   \le\gamma\alpha_j\nu F_j\}$}

\noindent is conegligible in $X^{\Bbb N}$, because
$\int v\,d\mu\le\gamma\alpha_j\nu F_j$.

\wheader{537Q}{6}{2}{2}{36pt}

{\bf (c)} Set

\Centerline{$W=\{(\pmb{x},y):\pmb{x}\in X^{\Bbb N}$, $y\in F_j$,
$h(\pmb{x},y)\ge\alpha_j\}$}

\noindent and consider the function $\chi W:X^{\Bbb N}\times Y\to\{0,1\}$.
If $y\in F_j$ then $\underline{\int}f(x,y)\mu(dx)\ge\alpha_j$
so $W^{-1}[\{y\}]$ is conegligible in $X^{\Bbb N}$.
On the other hand, if $\pmb{x}=\sequence{i}{x_i}$ belongs to
$V\cap D^{\Bbb N}$,

$$\eqalignno{\overline{\int}\alpha_j\chi W(\pmb{x},y)\nu(dy)
&\le\overline{\int}h(\pmb{x},y)\chi F_j(y)\nu(dy)\cr
&=\overline{\int}\liminf_{n\to\infty}\Bover1{n+1}
  \sum_{i=0}^nf(x_i,y)\chi F_j(y)\nu(dy)\cr
&\le\liminf_{n\to\infty}\overline{\int}\Bover1{n+1}
  \sum_{i=0}^nf(x_i,y)\chi F_j(y)\nu(dy)\cr
\displaycause{133Kb}
&\le\liminf_{n\to\infty}\Bover1{n+1}
  \sum_{i=0}^n\overline{\int}f(x_i,y)\chi F_j(y)\nu(dy)\cr
\displaycause{133J(b-ii)}
&\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nv(x_i)
\le\gamma\alpha_j\nu F_j.\cr}$$

\medskip

{\bf (d)}
As $V$ is conegligible and $D^{\Bbb N}$ has full outer measure (254Lb),

$$\eqalign{\underline{\int}\overline{\int}
  \chi W(\pmb{x},y)\nu(dy)\mu^{\Bbb N}(d\pmb{x})
&\le\gamma\nu F_j
<\nu F_j
=\iint\chi W(\pmb{x},y)\mu^{\Bbb N}(d\pmb{x})\nu(dy)\cr
&=\underline{\int}\,\underline{\int}
  \chi W(\pmb{x},y)\mu^{\Bbb N}(d\pmb{x})\nu(dy).\cr}$$

\noindent But we are supposing that
$\shrplusN(\mu^{\Bbb N})\le\cov\Cal N(\nu^{\Bbb N})$, so this contradicts
537P.\ \Bang

So we have the result.
}%end of proof of 537Q

\leader{537R}{Lemma} Let $(X,\Sigma,\mu)$ be a complete probability space
and $(Y,\Tau,\nu)$ a
probability space such that
$\shrplusN(\mu)\penalty-50\le\cov\Cal N(\nu^{\Bbb N})$,
where $\nu^{\Bbb N}$ is the product measure on $Y^{\Bbb N}$.   Let
$f:X\times Y\to\Bbb R$ be a bounded function which is measurable in
each variable separately, and set
$u(x)=\int f(x,y)\nu(dy)$ for $x\in X$.  Then $u:X\to\Bbb R$ is measurable.

\proof{ \Quer\ Otherwise, there are a non-negligible measurable set
$E\subseteq X$ and $\alpha$, $\beta\in\Bbb R$ such that
$\alpha<\beta$ and

\Centerline{$\mu^*\{x:x\in E$, $u(x)\le\alpha\}
=\mu^*\{x:x\in E$, $u(x)\ge\beta\}=\mu E$}

\noindent (413G).   Let $A\subseteq\{x:x\in E$, $u(x)\le\alpha\}$ and
$B\subseteq\{x:x\in E$, $u(x)\ge\beta\}$ be sets with cardinal less than
$\shrplusN(\mu)$ and outer measure greater than $\bover12\mu E$
(521Ca).   Let $\sequence{i}{y_i}$ be a sequence in $Y$ such that

\Centerline{$u(x)=\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(x,y_i)$}

\noindent for every $x\in A\cup B$.   Because $x\mapsto f(x,y_i)$ is
measurable for each $i$, $u\restr A\cup B$ is measurable;   but this means
that $A$ and $B$ can be separated by measurable sets, which is impossible,
because $\mu^*A+\mu^*B>\mu E$.\ \Bang
}%end of proof of 537R

\leader{537S}{Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
probability spaces such that

\Centerline{$\shrplusN(\mu)\le\cov\Cal N(\nu^{\Bbb N})$,}

\noindent where $\nu^{\Bbb N}$ is the product measure on $Y^{\Bbb N}$,
and

\Centerline{$\cff([\tau(\nu)]^{\le\omega})<\cov(E,\Cal N(\mu))$ for every
$E\in\Sigma\setminus\Cal N(\mu)$,}

\noindent where $\tau(\nu)$ is the Maharam type of $\nu$.
Let $f:X\times Y\to\coint{0,\infty}$ be a
function which is measurable in each variable separately.  Then
$\iint f(x,y)\mu(dx)\nu(dy)$ and $\iint f(x,y)\nu(dy)\mu(dx)$ exist and
are equal.

\proof{{\bf (a)}
Let $\tilde\Lambda\supseteq\Sigma\tensorhat\Tau$ be the
$\sigma$-algebra of sets
$W\subseteq X\times Y$ such that all the vertical and horizontal
sections of $W$ are measurable.   If $W\in\tilde\Lambda$, then
$x\mapsto\nu W[\{x\}]:X\to[0,1]$ is measurable,
by 537R. If $W\in\tilde\Lambda$ and almost every horizontal section of
$W$ is negligible, then

$$\eqalign{\overline{\int}\nu W[\{x\}]\mu(dx)
&=\overline{\int}\underline{\int}\chi W(x,y)\nu(dy)\mu(dx)\cr
&\le\overline{\int}\,\overline{\int}\chi W(x,y)\mu(dx)\nu(dy)
=0\cr}$$

\noindent by 537Pa, so almost every vertical section of $W$ is
negligible.

\medskip

{\bf (b)} Let $(\frak B,\bar\nu)$ be the measure algebra of $(Y,\Tau,\nu)$.
If $W\in\tilde\Lambda$ and there is a metrically separable subalgebra
$\frak C$ of
$\frak B$ containing $W[\{x\}]^{\ssbullet}$ for every $x\in X$, then
there is a $W'\in\Sigma\tensorhat\Tau$ such that
$W[\{x\}]\symmdiff W'[\{x\}]$ is negligible for almost every $x$.
\Prf\ Note first that for every $F\in\Tau$ the map

\Centerline{$x\mapsto\nu(W[\{x\}]\symmdiff F)
=\nu((W\symmdiff(X\times F))[\{x\}]$}

\noindent is measurable, by (a).   So
$x\mapsto W[\{x\}]^{\ssbullet}:X\to\frak C$ is measurable, by 418Bc.   By
418T(b-ii), there is a $W'\in\Sigma\tensorhat\Tau$ such that
$W[\{x\}]^{\ssbullet}=W'[\{x\}]^{\ssbullet}$ for almost every $x$.\ \Qed

\medskip

{\bf (c)} In fact we find that for any $W\in\tilde\Lambda$ there is a
$W'\in\Sigma\tensorhat\Tau$ such that
$W[\{x\}]\symmdiff W'[\{x\}]$ is negligible for almost every $x$.
\Prf\ Set $\kappa=\tau(\nu)=\tau(\frak B)$, and let
$\ofamily{\xi}{\kappa}{e_{\xi}}$ generate $\frak B$.   Let
$\Cal K\subseteq[\kappa]^{\le\omega}$ be a cofinal set with cardinal
$\cff[\kappa]^{\le\omega}$.   For $K\in\Cal K$, let $\frak B_K$
be the closed subalgebra of $\frak B$ generated by
$\{e_{\xi}:\xi\in K\}$ and $A_K$ the set
$\{x:x\in X$, $W[\{x\}]^{\ssbullet}\in\frak B_K\}$.
Note that $K\mapsto A_K$ is non-decreasing and that the union of any
sequence in $\Cal K$ is included in a member of $\Cal K$.
So there is a $K_0\in\Cal K$
such that $\mu^*A_{K_0}=\sup_{K\in\Cal K}\mu^*A_K$.

If $E$ is a measurable envelope of $A_{K_0}$, then
$\{A_K\setminus E:K\in\Cal K\}$ is a cover of $X\setminus E$ by negligible
sets.
So $\cov(X\setminus E,\penalty-100\Cal N(\mu))\le\cff[\kappa]^{\le\omega}$ and
$X\setminus E$ must be negligible, that is, $A_{K_0}$ has full outer
measure.

Taking a sequence
$\sequencen{F_n}$ in $\Tau$ such that $\{F_n^{\ssbullet}:n\in\Bbb N\}$
is dense in $\frak B_{K_0}$, we see from (a) that
\ifnum\stylenumber=12 the function \else\fi
$x\mapsto\inf_{n\in\Bbb N}\nu(W[\{x\}]\symmdiff F_n)$ is measurable, while
it is zero on $A_{K_0}$.   So
$W[\{x\}]^{\ssbullet}\in\frak B_{K_0}$ for almost every $x\in X$,
that is, $A_{K_0}$ is actually conegligible.   Taking a measurable
conegligible set $E'\subseteq A_{K_0}$ and applying (b) to
$W\cap(E'\times Y)$, we see that there is
a $W'\in\Sigma\tensorhat\Tau$ such that $W[\{x\}]\symmdiff W'[\{x\}]$ is
negligible for almost every $x\in X$.\ \Qed

\medskip

{\bf (d)} Now turn to the function $f$ under consideration.
For $q\in\Bbb Q$ set
$W_q=\{(x,y):f(x,y)\ge q\}\in\tilde\Lambda$.   By (c), we have
$V_q\in\Sigma\tensorhat\Tau$ such that
$V_q[\{x\}]\symmdiff W_q[\{x\}]$ is $\nu$-negligible for $\mu$-almost
every $x$, and therefore $W_q^{-1}[\{y\}]\symmdiff V_q^{-1}[\{y\}]$ is
$\mu$-negligible for $\nu$-almost every $y$, by (a).   If $q\le q'$ then
$W_{q'}\setminus W_q$ is empty, so $V_{q'}[\{x\}]\setminus V_q[\{x\}]$ is
$\nu$-negligible for $\mu$-almost every $x$, and
$V_{q'}\setminus V_q$ is $(\mu\times\nu)$-negligible, where $\mu\times\nu$
is the product measure on
$X\times Y$.   Similarly, $\bigcap_{q'<q}V_{q'}\setminus V_q$ is
negligible for every $q$.   Moreover, writing $V_{\infty}$ for
$\bigcap_{q\in\Bbb Q}V_q$, $V_{\infty}[\{x\}]$ is $\nu$-negligible for
$\mu$-almost every $x$, so $(\mu\times\nu)V_{\infty}=0$;  similarly,
$(\mu\times\nu)V_0=1$.   There is therefore a
$\Sigma\tensorhat\Tau$-measurable $g:X\times Y\to\coint{0,\infty}$
such that
$V_q\symmdiff\{(x,y):g(x,y)\ge q\}$ is $(\mu\times\nu)$-negligible for
every $q\in\Bbb Q$.   In this case,

\Centerline{$\{x:f(x,y)\ne g(x,y)\}$ is $\mu$-negligible for
$\nu$-almost every $y$,}

\Centerline{$\{y:f(x,y)\ne g(x,y)\}$ is $\nu$-negligible for
$\mu$-almost every $x$,}

\noindent and

$$\eqalign{\iint f(x,y)\mu(dx)\nu(dy)
&=\iint g(x,y)\mu(dx)\nu(dy)\cr
&=\iint g(x,y)\nu(dy)\mu(dx)
=\iint f(x,y)\nu(dy)\mu(dx)\cr}$$

\noindent by 252H.

\medskip

{\bf (e)} Finally, if $f$ is unbounded, set $f_k(x,y)=\min(f(x,y),k)$ for
each $k\in\Bbb N$.   Then

$$\eqalign{\iint f(x,y)\mu(dx)\nu(dy)
&=\lim_{k\to\infty}\iint f_k(x,y)\mu(dx)\nu(dy)\cr
&=\lim_{k\to\infty}\iint f_k(x,y)\nu(dy)\mu(dx)
=\iint f(x,y)\nu(dy)\mu(dx).\cr}$$
}%end of proof of 537S

\exercises{\leader{537X}{Basic exercises (a)}(i)
%\sqheader 537Xa
Let $(X,\Sigma,\mu)$ be a measure space such that
singletons are negligible and $\cf\Cal N(\mu)=\omega_1$.   Show that there
is a Sierpi\'nski subset of $X$.   (ii) Show that if $\mu$ is Lebesgue
measure on $\Bbb R$ and
$\cf\Cal N(\mu)=\omega_1$, then there is a strongly Sierpi\'nski subset of
$\Bbb R$.
%537A

\spheader 537Xb Show that for any uncountable cardinal $\kappa$
there is a purely atomic probability space with a
strongly Sierpi\'nski set with cardinal $\kappa$.
%537A

\spheader 537Xc Let $(X,\Sigma,\mu)$ be a measure space.   Show that the
union of any sequence of Sierpi\'nski sets in $X$ is again a Sierpi\'nski
set in $X$.
%537A

\spheader 537Xd Let $(X,\Sigma,\mu)$ be a measure space and $Y$ any
subspace of $X$.   Show that a subset of $Y$ is a Sierpi\'nski set for
the subspace measure on $Y$ iff it is a Sierpi\'nski set for $\mu$.
%537B

\spheader 537Xe Suppose that  $\lambda$ is an infinite cardinal
and the usual measure $\nu_{\lambda}$ on $\{0,1\}^{\lambda}$ has a
Sierpi\'nski set with cardinal $\kappa$.
Show that $\nu_{\lambda}$ has a Sierpi\'nski set $A$ such that
$\#(A\cap E)=\kappa$ whenever $\nu_{\lambda}E>0$.
%537B

\spheader 537Xf Let $(X,\rho)$ be a non-separable metric space with
$r$-dimensional Hausdorff measure, where $r>0$.   Show that $X$ has a
Sierpi\'nski subset with cardinal equal to the topological density of
$X$.
%537B

\sqheader 537Xg Suppose that $\non\Cal N<\cov\Cal N$, where $\Cal N$ is the
null ideal of Lebesgue measure on $\Bbb R$.   Let $(X,\frak T,\Sigma,\mu)$
and $(Y,\frak S,\Tau,\nu)$ be Radon probability spaces of countable Maharam
type, and $f:X\times Y\to\coint{0,\infty}$ a function such that
$I=\iint f(x,y)\mu(dx)\nu(dy)$ and $I'=\iint f(x,y)\nu(dy)\mu(dx)$ are both
defined.   Show that $I=I'$.
%537N

\sqheader 537Xh Let $(X,\Sigma,\mu)$ be a probability space in which
there is a well-ordered family in $\Cal N(\mu)$ with union $X$;  e.g.,
because
$\non\Cal N(\mu)=\#(X)$ or $\add\Cal N(\mu)=\cov\Cal N(\mu)$.
Show that there is a function $f:X\times X\to[0,1]$ such that
$\int f(x,y)\mu(dx)=0$ for every $y\in X$ and $\int f(x,y)\mu(dy)=1$
for every $x\in X$.

\sqheader 537Xi (In this exercise, all integrals are to be taken with
respect to one-dimensional Lebesgue measure $\mu$.)
(i) Find a function $f:[0,1]^2\to\{0,1\}$
such that $\int\overline{\int}f(x,y)dxdy=1$ but $\iint f(x,y)dydx=0$.
\Hint{there is a disjoint family $\family{y}{[0,1]}{A_y}$ of sets of full
outer measure.}  (ii) Find a function $f:[0,1]^2\to\{0,1\}$
such that $\iint f(x,y)dxdy=1$ but $\int\underline{\int}f(x,y)dydx=0$.
(iii) Find a function $f:[0,1]^2\to\{0,1\}$ such that
$\overline{\int}\int f(x,y)dxdy=1$ but
$\underline{\int}\int f(x,y)dydx=0$.   \Hint{enumerate $[0,1]$ as
$\ofamily{\xi}{\frak c}{x_{\xi}}$ in such a way that
$\{x_{\xi}:\xi<\non\Cal N(\mu)\}$ has full outer measure;  set
$f(x_{\xi},x_{\eta})=1$ if $\eta<\xi$.}

\leader{537Y}{Further exercises (a)}\dvAnew{2016}
%\spheader 537Ya
Let $\langle(X_j,\Sigma_j,\mu_j)\rangle_{j\le m}$ be a
finite sequence of probability spaces and $\langle\kappa_j\rangle_{j\le m}$
a sequence of cardinals such that $X_j$ has a subset with cardinal $\kappa_j$
which is not covered by $\kappa_{j-1}$ negligible sets (if $j\ge 1$) and is
not negligible (if $j=0$).   Set $X=\prod_{j\le m}X_j$, and for $k\le m$
write $Z_k$ for $\prod_{j\le m,j\ne k}X_j$.
Suppose that for each $k\le m$ we have a set
$A_k\subseteq X$ such that, identifying $X$ with $X_k\times Z_k$,
$\{z:(x,z)\in A_k\}\subseteq Z_k$ is negligible for the product measure on
$Z_k$ whenever $x\in X_k$.   Show that $\bigcup_{k\le m}A_k\ne X$.   \query
%537K mt53bits
}%end of exercises

\leader{537Z}{Problems (a)} Is it relatively consistent with ZFC to suppose
that $\Bbb R$, with Lebesgue measure, has a Sierpi\'nski subset but no
strongly Sierpi\'nski subset?

\spheader 537Zb Is it relatively consistent with ZFC to suppose that there
is a probability space $(X,\mu)$ such that $(X,\mu)$ has a Sierpi\'nski set
but its power $(X^{\Bbb N},\mu^{\Bbb N})$ does not?

\endnotes{
\Notesheader{537}
It is easy to see that if $\frak c=\omega_1$ then there is a
strongly Sierpi\'nski
set with cardinal $\omega_1$ for Lebesgue measure (537Xa).
Countable-cocountable measures have strongly Sierpi\'nski sets for trivial
reasons.   To eliminate all Sierpi\'nski sets (on the definition of 537A)
from atomless complete locally determined measure spaces, it is enough to
ensure that the uniformity of Lebesgue measure is greater than $\omega_1$
(537Bb).   For the simplest models with
non-trivial Sierpi\'nski sets with cardinal greater than $\omega_1$, see 552E
below.

The `entangled sets' of 537C-537G %537C 537D 537E 537F 537G
belong rather to combinatorics than to measure theory;  I go as far as I do
into this theory because
it is interesting in view of 552E.   But it includes a
proof that if the continuum hypothesis is true then there are two ccc
partially ordered sets whose product is not ccc, which in its own context
is of great importance.

Fubini's theorem is so important in measure theory that exploration of its
boundaries has been a perennial challenge.   I gave elementary examples in
252Xf-252Xg to show that as soon as we abandon the requirement that
$\iint|f(x,y)|dxdy<\infty$ our repeated integrals can be expected to be
unreliable.   But for non-negative functions $f$ on $\sigma$-finite spaces,
measurability is enough to ensure that repeated integrals are equal
(252H).   In this section I look for results which will be valid for
non-measurable functions.   In 537I-537J we have a rather esoteric example
-- or, some would say, an example from a topic which I have neglected in
this book -- which is unusual in that it is a theorem of ZFC;  for a note
on its ancestry see {\smc Fremlin 93}, 5L.   In 537K-537L we see that,
in the presence of a sufficient supply of Sierpi\'nski sets,  for
instance, we
must have $\iint f(x,y)dxdy=\iint f(x,y)dydx$  for ordinary
bounded real-valued functions on the product of probability spaces,
as long as both repeated integrals are defined.   The argument here depends
on using the strong law of large numbers to replace an integral
$\int f(x,y)dx$ by the limit of a sequence of averages of values
$f(x_i,y)$.   This is why the Sierpi\'nski sets must be available not in
the original probability spaces $X_0,\ldots,X_m$ but in their powers
$X_j^{\Bbb N}$.   Of course for our favourite spaces, starting with
$[0,1]$, $(X^{\Bbb N},\mu^{\Bbb N})$ is isomorphic to $(X,\mu)$, so this
does not seem too large a step;  but it begs an obvious question (537Zb).
For any result of this kind we certainly need some special axiom (537Xh).

In 537L the hypothesis includes strong `separate measurability' conditions;
we need not only separate measurability, but
measurability of the functions $x\mapsto\int f(x,y)dy$ and
$y\mapsto\int f(y,x)dx$.   With a different set-theoretic hypothesis we can
relax these (537S).   I approach this form through ideas
from {\smc Humke \& Laczkovich 05}, where there is a careful analysis of
repeated integrals of the form $\underline{\int}\overline{\int}$, etc.
My own version is in 537N-537Q.   %537N 537O 537P 537Q
At every step there are ZFC examples to show that we cannot change the
formulae involving
$\underline{\int}$, $\overline{\int}$ without disaster (537Xi);
but it is not so clear that the set-theoretic hypotheses offered are
unimprovable.

}%end of notes

\discrpage

\frfilename{mt538.tex}
\versiondate{18.2.14}
\copyrightdate{2009}

\def\chaptername{Topologies and Measures III}
\def\sectionname{Filters and limits}

\def\FS{\mathop{\text{FS}}\nolimits}

\newsection{538}

A great many special types of filter have been studied.   In this section I
look at some which are particularly interesting from the point of view of
measure theory:  Ramsey ultrafilters, measure-converging filters and
filters with the Fatou property.   About half the section is directed
towards Benedikt's theorem (538M) on extensions of perfect probability
measures;  on the way we need to look at
measure-centering ultrafilters (538G-538K) %538G 538H 538I 538J 538K
and iterated
products of filters (538E, 538L).   The second major topic here is a study
of `medial limits' (538P-538S); %538P 538Q 538R 538S
these are Banach limits of a very special type.
In between, the measure-converging property (538N)
and the Fatou property (538O) offer some intriguing patterns.

%538A definitions - filter types
%538B R-K ordering
%538C ultrafilters and R-K
%538D \Cal F\ltimes\Cal G
%538E iteration
%538F Ramsey ufilters
%538H measure-centering ultrafilters
%538G characterizations
%538I extending perfect probabilities
%538J   properties of construction
%538K   more
%538L iterated product of Ramsey ufilters is measure-centering
%538M universal extension of perfect probability
%538N measure-converging filters.
%538O Fatou property
%538P medial functionals:  equivalent conditions
%538Q    definition
%538R    properties
%538S    existence

\leader{538A}{\dvrocolon{Filters}}\cmmnt{ For ease of reference, I begin
the section with a list of the special types of filter on $\Bbb N$
which we shall be looking at later.

\medskip

\noindent}{\bf  Definitions}
Let $\Cal F$ be a filter on $\Bbb N$.

\spheader 538Aa $\Cal F$ is {\bf free} if it\cmmnt{ contains every cofinite
subset of $\Bbb N$, that is,} includes the Fr\'echet filter.

\spheader 538Ab $\Cal F$ is a {\bf $p$-point filter} if it is free and
for every sequence
$\sequencen{A_n}$ in $\Cal F$ there is an $A\in\Cal F$ such that
$A\setminus A_n$ is finite for every $n\in\Bbb N$.
\cmmnt{(Compare 5A6Ga.)}

\spheader 538Ac $\Cal F$ is {\bf Ramsey} or {\bf selective}
if it is free and for every
$f:[\Bbb N]^2\to\{0,1\}$ there is an $A\in\Cal F$ such that $f$ is constant
on $[A]^2$.
%C&N p 212 gives defn for uniform ultrafilters on other cardinals

\spheader 538Ad $\Cal F$ is {\bf rapid} if it is free and for
every sequence $\sequencen{t_n}$
of real numbers which converges to $0$, there is an $A\in\Cal F$ such that
$\sum_{n\in A}|t_n|$ is finite.   Note that a free filter $\Cal F$ on
$\Bbb N$ is rapid iff for every $f\in\BbbN^{\Bbb N}$ there is an
$A\in\Cal F$ such that $\#(A\cap f(k))\le k$ for every $k\in\Bbb N$.
\prooflet{\Prf\ (i) If $\Cal F$ is rapid and $f\in\BbbN^{\Bbb N}$, let
$g\in\BbbN^{\Bbb N}$ be a strictly increasing sequence such that $f\le g$.
Set $t_i=2$ if $i<g(0)$, $\Bover1{k+1}$ if $g(k)\le i<g(k+1)$;  then there
is an $A\in\Cal F$ such that $\sum_{i\in A}t_i$ is finite;  as $\Cal F$ is
free, there is an $A\in\Cal F$ such that $\sum_{i\in A}t_i\le 1$, in which
case $\#(A\cap f(k))\le\#(A\cap g(k))\le k$ for every $k\in\Bbb N$.
(ii) If $\Cal F$ satisfies the condition and $\sequence{i}{t_i}\to 0$, take
a strictly increasing
$f\in\BbbN^{\Bbb N}$ such that $|t_i|\le 2^{-k}$ whenever $k\in\Bbb N$ and
$i\ge f(k)$;  let $A\in\Cal F$ be such that $\#(A\cap f(k))\le k$ for every
$k$;  then $\sum_{i\in A}|t_i|
\le\sum_{k=0}^{\infty}2^{-k}\#(A\cap f(k+1)\setminus f(k))$ is finite.\
\Qed}

\spheader 538Ae $\Cal F$ is {\bf nowhere dense} if for every sequence
$\sequencen{t_n}$ in $\Bbb R$ there is an $A\in\Cal F$ such that
$\{t_n:n\in A\}$ is nowhere dense.

\spheader 538Af $\Cal F$ is {\bf measure-centering} or has {\bf property M}
if whenever $\frak A$ is a Boolean algebra,
$\nu:\frak A\to\coint{0,\infty}$ is an additive functional, and
$\sequencen{a_n}$ is a sequence in $\frak A$ such that
$\inf_{n\in\Bbb N}\nu a_n>0$, there is an $A\in\Cal F$ such that
$\{a_n:n\in A\}$ is centered.

\spheader 538Ag $\Cal F$ is {\bf measure-converging} if
whenever $(X,\Sigma,\mu)$ is a probability space,
$\sequencen{E_n}$ is a sequence in
$\Sigma$, and $\lim_{n\to\infty}\mu E_n=1$, then
$\bigcup_{A\in\Cal F}\bigcap_{n\in A}E_n$ is conegligible.
%this is "measure-converging over the Frechet filter"

\spheader 538Ah $\Cal F$ has the {\bf Fatou property} if whenever
$(X,\Sigma,\mu)$ is a probability space, $\sequencen{E_n}$ is a sequence in
$\Sigma$, and $X=\bigcup_{A\in\Cal F}\bigcap_{n\in A}E_n$, then
$\lim_{n\to\Cal F}\mu E_n$ is defined and equal to $1$.

\spheader 538Ai For any countably infinite set $I$, I will say that a
filter $\Cal F$ on $I$ is free, or a $p$-point filter, or Ramsey, etc., if
it is isomorphic to such a filter on $\Bbb N$.   \cmmnt{Of course this
usage is
possible only because every property here is invariant under permutations
of $\Bbb N$.}   For `rapid' and `measure-converging'
filters, we need an appropriate translation of
`sequence converging to $0$';  the corresponding notion on an arbitrary
index set $I$ is a function $u\in\pmb{c}_0(I)$, that is,
a real-valued function $u$ on $I$ such that
$\{i:i\in I$, $|u(i)|\ge\epsilon\}$ is finite for every
$\epsilon>0$\cmmnt{;  if we give $I$
its discrete topology, $c_0(I)$ is $C_0(I)$ as defined in 436I}.

%rare=Q-point:  for any partition of \Bbb N into finite sets,
%\exists A\in\Cal F  meeting each member of the partition in at most
%one point.   Rare => rapid;  Ramsey <=> rare p-point ultrafilter.

\leader{538B}{}\cmmnt{ We need a number of basic ideas which can
profitably be examined in a rather more general context.   I start with a
fundamental pre-order on the class of all filters.

\medskip

\noindent}{\bf The Rudin-Keisler ordering} If
$\Cal F$, $\Cal G$ are filters on sets $I$, $J$ respectively, I will
say that
$\Cal F\leRK\Cal G$ if there is a function $f:J\to I$ such that

\Centerline{$\Cal F=f[[\Cal G]]=\{A:A\subseteq I$, $f^{-1}[A]\in\Cal G\}$,}

\noindent the filter on $I$ generated by
$\{f[B]:B\in\Cal G\}$.\cmmnt{  (I ought to remark that while this is a
standard idea for ultrafilters, in the case of general filters the
terminology is not well established.)}\cmmnt{   Of course}
$\leRK$ is reflexive and transitive.   If $\Cal F\leRK\Cal G$ and
$\Cal G$ is an ultrafilter, then $\Cal F$ is an
ultrafilter\cmmnt{ (2A1N)}.   If
$\Cal F$ is a principal ultrafilter then $\Cal F\leRK\Cal G$ for every
filter $\Cal G$.
%check Mildenberger
%C&N p 206 for ultrafilters

\leader{538C}{Lemma} (a) If $I$ is a set, $\Cal F$ is an ultrafilter on
$I$ and $f:I\to I$ is a function such that $f[[\Cal F]]=\Cal F$, then
$\{i:f(i)=i\}\in\Cal F$.
%C&N 9.2a

(b) If $I$ is a set, $\Cal F$ and $\Cal G$ are ultrafilters on $I$,
$\Cal F\leRK\Cal G$ and $\Cal G\leRK\Cal F$, then there is a permutation
$h:I\to I$ such that $h[[\Cal F]]=\Cal G$;  that is, $\Cal F$ and
$\Cal G$ are isomorphic.
%C&N 9.2b/9.3

\proof{{\bf (a)} It is enough to consider the case in which $I=\kappa$ is a
cardinal.

\medskip

\quad{\bf (i)} $\{\xi:\xi<\kappa$, $\xi\le f(\xi)\}\in\Cal F$.
\Prf\ Define $\sequencen{D_n}$, $\sequencen{E_n}$ by saying that

\Centerline{$D_0=\kappa$,
\quad$D_{n+1}=\{\xi:\xi\in D_n$, $f(\xi)\in D_n$, $f(\xi)<\xi\}$,
\quad$E_n=D_n\setminus D_{n+1}$}

\noindent for $n\in\Bbb N$.   If $\xi\in D_n$ then
$\xi>f(\xi)>\ldots>f^n(\xi)$, so $\bigcap_{n\in\Bbb N}D_n=\emptyset$ and
$\sequencen{E_n}$ is a partition of $\kappa$.   If $\xi\in E_{n+1}$ then
$f^{n+1}(\xi)<f^n(\xi)<\ldots<\xi$, $f^{n+1}(\xi)\le f^{n+2}(\xi)$, so
$f(\xi)\in E_n$.   Set $E=\bigcup_{n\ge 1}E_{2n}$,
$E'=\bigcup_{n\in\Bbb N}E_{2n+1}$;  then $f[E]\subseteq E'$ is disjoint
from $E$, so $E\notin\Cal F$.   Also $f[E']\subseteq E\cup E_0$ is disjoint
from $E'$, so $E'\notin\Cal F$.   Because $\Cal F$ is an ultrafilter,
$E_0\in\Cal F$, as claimed.\ \Qed

\medskip

\quad{\bf (ii)} If $A\subseteq I$ and $A\notin\Cal F$ then
$B=\bigcup_{n\in\Bbb N}(f^n)^{-1}[A]$ does not belong to $\Cal F$.   \Prf\
For $\xi\in B$ set $m(\xi)=\min\{n:n\in\Bbb N$, $f^n(\xi)\in A\}$.
If $m(\xi)>0$ then $m(f(\xi))=m(\xi)-1$.   So setting
$C=\{\xi:m(\xi)$ is even and not $0\}$, $C'=\{\xi:m(\xi)$ is odd$\}$ we
have $f[C]\cap C=\emptyset$, $f[C']\cap C'=\emptyset$ and
$B\subseteq A\cup C\cup C'$;  so $B\notin\Cal F$.\ \Qed

Turning this round, if $A\in\Cal F$ then
$\bigcup_{n\in\Bbb N}(f^n)^{-1}[\kappa\setminus A]\notin\Cal F$ and
$\bigcap_{n\in\Bbb N}(f^n)^{-1}[A]\in\Cal F$.

\medskip

\quad{\bf (iii)} For $\xi<\kappa$ set

\Centerline{$g(\xi)=\min\{\zeta:$ there is some $n\in\Bbb N$ such that
$f^n(\zeta)=\xi\}$.}

\noindent Then $g[[\Cal F]]=\Cal F$.   \Prf\ If $A\in\Cal F$ then
$\Cal F$ contains $\bigcap_{n\in\Bbb N}(f^n)^{-1}[A]\subseteq g^{-1}[A]$,
so $g^{-1}[A]\in\Cal F$.   Thus $\Cal F\subseteq g[[\Cal F]]$;  as $\Cal F$
is an ultrafilter, $\Cal F=g[[\Cal F]]$.\ \Qed

Now $g(\xi)\le\xi$ for every $\xi<\kappa$;  applying (i) to $g$,
we see that $G=\{\xi:g(\xi)=\xi\}\in\Cal F$.   But consider
$H=\{\xi:\xi<f(\xi)\}$.   Then $g(\eta)<\eta$ for every $\eta\in f[H]$, so
$f[H]\notin\Cal F$ and $H\notin\Cal F$.   Since we already know that
$\{\xi:\xi\le f(\xi)\}\in\Cal F$, we see that $\{\xi:f(\xi)=\xi\}$ belongs
to $\Cal F$, as claimed.

\medskip

{\bf (b)} Let $f$, $g:I\to I$ be such that $f[[\Cal F]]=\Cal G$ and
$g[[\Cal G]]=\Cal F$.   Then $(gf)[[\Cal F]]=g[[f[[\Cal F]]\,]]=\Cal F$, so
$J_0=\{i:g(f(i))=i\}\in\Cal F$, by (a).   Similarly, $J_1=\{i:f(g(i))=i\}$
belongs to $\Cal G$.   Set $J=J_0\cap f^{-1}[J_1]\in\Cal F$;
then $g(f(i))=i$ for
every $i\in J$ and $f(g(j))=j$ for every $j\in f[J]$, so $f\restr J$ and
$g\restr f[J]$ are inverse bijections between $J\in\Cal F$ and
$f[J]\in\Cal G$.   If $J$ is finite, then certainly
$\#(I\setminus J)=\#(I\setminus f[J])$ and there is an extension of
$f\restr J$ to a permutation of $I$.   If $J$ is infinite, let
$J'\subseteq J$ be a set such that $\#(J')=\#(J\setminus J')=\#(J)$ and
$J'\in\Cal F$;  then $\#(I\setminus J')=\#(I\setminus f[J'])=\#(I)$
so there is an extension of $f\restr J'$ to a permutation of $I$.

Thus in either case we have a permutation
$h:I\to I$ and a $K\in\Cal F$ such
that $K\subseteq J$ and $h\restr K=f\restr K$.   But now
$h[[\Cal F]]=\Cal G$ and $h$ is an isomorphism between $(I,\Cal F)$ and
$(I,\Cal G)$.
}%end of proof of 538C

\leader{538D}{Finite products of filters (a)}
Suppose that $\Cal F$, $\Cal G$ are filters on sets $I$, $J$ respectively.
I will write $\Cal F\ltimes\Cal G$ for

\Centerline{$\{A:A\subseteq I\times J$,
$\{i:A[\{i\}]\in\Cal G\}\in\Cal F\}$.}

\noindent\cmmnt{It is easy to check that }$\Cal F\ltimes\Cal G$
is a filter.
\cmmnt{ (Compare the skew product $\Cal I\ltimes\Cal J$ of ideals
defined in 527Ba.)}

\spheader 538Db
If $\Cal F$ and $\Cal G$ are ultrafilters, so is $\Cal F\ltimes\Cal G$.
\prooflet{\Prf\ If $A\subseteq I\times J$ and $A\notin\Cal F\ltimes\Cal G$,
then $\{i:A[\{i\}]\in\Cal G\}\notin\Cal F\}$ and

\Centerline{$\{i:((I\times J)\setminus A)[\{i\}]\in\Cal G\}
=\{i:i\in I,\,J\setminus A[\{i\}]\in\Cal G\}
=I\setminus\{i:A[\{i\}]\in\Cal G\}
\in\Cal F$,}

\noindent so $(I\times J)\setminus A\in\Cal F\ltimes\Cal G$.\ \Qed}

\spheader 538Dc If $\Cal F$, $\Cal G$ and $\Cal H$ are filters on $I$,
$J$, $K$ respectively, then the natural bijection between
$(I\times J)\times K$ and $I\times(J\times K)$ is an isomorphism between
$(\Cal F\ltimes\Cal G)\ltimes\Cal H$ and
$\Cal F\ltimes(\Cal G\ltimes\Cal H)$.   \prooflet{\Prf\ If
$A\subseteq I\times(J\times K)$ and $B=\{(i,j),k):(i,(j,k))\in A\}$, then

$$\eqalign{A\in\Cal F\ltimes(\Cal G\ltimes\Cal H)
&\iff\{i:A[\{i\}]\in\Cal G\times\Cal H\}\in\Cal F\cr
&\iff\{i:\{j:(A[\{i\}])[\{j\}]\in\Cal H\}\in\Cal G\}\in\Cal F\cr
&\iff\{(i,j):(A[\{i\}])[\{j\}]\in\Cal H\}\in\Cal F\ltimes\Cal G\cr
&\iff\{(i,j):B[\{(i,j)\}]\in\Cal H\}\in\Cal F\ltimes\Cal G\cr
&\iff B\in(\Cal F\ltimes\Cal G)\ltimes\Cal H.  \text{ \Qed}\cr}$$}

\spheader 538Dd It follows that we can define a product
$\Cal F_0\ltimes\ldots\ltimes\Cal F_n$ of any finite string
$\Cal F_0,\ldots,\Cal F_n$ of filters, and under the natural
identifications of the base sets we shall have
$(\Cal F_0\ltimes\ldots\ltimes\Cal F_n)
\ltimes(\Cal F_{n+1}\ltimes\ldots\ltimes\Cal F_m)$ identified with
$\Cal F_0\ltimes\ldots\ltimes\Cal F_m$ whenever
$\Cal F_0,\ldots,\Cal F_n,\ldots,\Cal F_m$ are filters.

\spheader 538De\cmmnt{ For any filters $\Cal F$ and $\Cal G$,
$\Cal F\leRK\Cal F\ltimes\Cal G$ and $\Cal G\leRK\Cal F\ltimes\Cal G$.
\prooflet{\Prf\ Taking the base sets to be $I$, $J$ respectively and
$f(i,j)=i$, $g(i,j)=j$ for $i\in I$ and $j\in J$, we have
$\Cal F=f[[\Cal F\ltimes\Cal G]]$ and $\Cal G=g[[\Cal F\ltimes\Cal G]]$.\
\Qed}

Inducing on $n$, we see that
$\Cal F_n\leRK\Cal F_0\ltimes\ldots\ltimes\Cal F_n$ whenever
$\Cal F_0,\ldots,\Cal F_n$ are filters;  consequently}
$\Cal F_m\leRK\Cal F_0\ltimes\ldots\ltimes\Cal F_n$ whenever
$\Cal F_0,\ldots,\Cal F_n$ are filters and $m\le n$.

\spheader 538Df If $\Cal F$, $\Cal F'$, $\Cal G$ and $\Cal G'$ are
filters, with $\Cal F\leRK\Cal F'$ and $\Cal G\leRK\Cal G'$, then
$\Cal F\ltimes\Cal G\leRK\Cal F'\ltimes\Cal G'$.   \prooflet{\Prf\ Let the
base sets of the filters be $I$, $I'$, $J$ and $J'$, and let $f:I'\to I$
and $g:J'\to J$ be such that $\Cal F=f[[\Cal F']]$ and
$\Cal G=g[[\Cal G']]$.   Set $h(i,j)=(f(i),g(j))$ for $i\in I$ and
$j\in J$.   If $A\subseteq I\times J$, then

$$\eqalign{h^{-1}[A]\in\Cal F'\ltimes\Cal G'
&\iff\{i:(h^{-1}[A])[\{i\}]\in\Cal G'\}\in\Cal F'\cr
&\iff\{i:g^{-1}[A[\{f(i)\}]]\in\Cal G'\}\in\Cal F'\cr
&\iff\{i:A[\{f(i)\}]\in\Cal G\}\in\Cal F'\cr
&\iff\{i:A[\{i\}]\in\Cal G\}\in\Cal F
\iff A\in\Cal F\ltimes\Cal G.\cr}$$

\noindent So $\Cal F\ltimes\Cal G=h[[\Cal F'\ltimes\Cal G']]$ and
$\Cal F\ltimes\Cal G\leRK\Cal F'\ltimes\Cal G'$.\ \Qed}

\cmmnt{Accordingly} $\Cal F_0\ltimes\ldots\ltimes\Cal F_n
\leRK\Cal G_0\ltimes\ldots\ltimes\Cal G_n$ whenever
$\Cal F_i\leRK\Cal G_i$ for every $i\le n$.

\spheader 538Dg It follows that if $\Cal F_0,\ldots,\Cal F_n$ are filters
and $k_0<\ldots<k_m\le n$, then
$\Cal F_{k_0}\ltimes\ldots\ltimes\Cal F_{k_m}
\leRK\Cal F_0\ltimes\ldots\ltimes\Cal F_n$.   \prooflet{\Prf\ Induce on $m$
to see that $\Cal F_{k_0}\ltimes\ldots\ltimes\Cal F_{k_m}
\leRK\Cal F_0\ltimes\ldots\ltimes\Cal F_{k_m}$.\ \Qed}

\vleader{72pt}{538E}{}\cmmnt{ There are many variations on the construction
here.   A fairly elaborate extension will be needed in 538L below.

\medskip

\noindent}{\bf Iterated products of filters (a)}\cmmnt{ First,
a scrap of notation for the rest of the first half of this section (down to
538M).}   Set $S=\bigcup_{i\in\Bbb N}\BbbN^i$.
Fix on a family
$\langle\theta(\xi,k)\rangle_{1\le\xi<\omega_1,k\in\Bbb N}$
such that each $\sequence{k}{\theta(\xi,k)}$ is a non-decreasing sequence
running over a cofinal subset of $\xi$.
\cmmnt{(You will
probably prefer to suppose that when $\xi=\eta+1$ is a successor ordinal,
then $\theta(\xi,k)=\eta$ for every $k\in\Bbb N$.)}

\spheader 538Eb Now suppose that $\zeta$ is a non-zero
countable ordinal.
Let $\langle\Cal F_{\xi}\rangle_{1\le\xi\le\zeta}$ be a family of
filters on
$\Bbb N$.   For $\xi\le\zeta$, define $\Cal G_{\xi}\subseteq\Cal PS$ as
follows.   Start by taking $\Cal G_0$ to be
the principal filter generated by
$\{\emptyset\}$.   For $1\le\xi\le\zeta$, set

\Centerline{$\Cal G_{\xi}
=\{A:A\subseteq S,\,\{k:k\in\Bbb N,\,
    \{\tau:\fraction{k}^{\smallfrown}\tau\in A\}
    \in\Cal G_{\theta(\xi,k)}\}\in\Cal F_{\xi}\}$.}

\noindent (See 5A1C for the notation here.)
It is elementary to check that every $\Cal G_{\xi}$ is a
filter, and that if every $\Cal F_{\xi}$ is free, so is every
$\Cal G_{\xi}$.
Moreover, if every $\Cal F_{\xi}$ is an ultrafilter, so is every
$\Cal G_{\xi}$.

\spheader 538Ec \cmmnt{Continuing from (b), we find that}
$\Cal F_{\xi}\leRK\Cal G_{\xi}$ whenever $1\le\xi\le\zeta$
and $\Cal G_{\eta}\leRK\Cal G_{\xi}$
whenever $0\le\eta\le\xi\le\zeta$.   \prooflet{\Prf\ Induce on $\xi$.
(i) If $\xi\ge 1$, define $f:S\to\Bbb N$ by setting $f(\tau)=\tau(0)$ if
$\tau\ne\emptyset$,
$f(\emptyset)=0$.   Then, for $B\subseteq\Bbb N$,

$$\eqalign{f^{-1}[B]\in\Cal G_{\xi}
&\iff\{k:\{\tau:\fraction{k}^{\smallfrown}\tau\in f^{-1}[B]\}
   \in\Cal G_{\theta(\xi,k)}\}\in\Cal F_{\xi}\cr
&\iff B\in\Cal F_{\xi},\cr}$$

\noindent so $\Cal F_{\xi}=f[[\Cal G_{\xi}]]\leRK\Cal G_{\xi}$.
(ii) If $\eta=\xi\le\zeta$ then of course $\Cal G_{\eta}\leRK\Cal G_{\xi}$.
(iii) If $0\le\eta<\xi$ then there is a $k_0$ such that
$\eta\le\theta(\xi,k)$
for $k\ge k_0$.   For $k\ge k_0$,
$\Cal G_{\eta}\leRK\Cal G_{\theta(\xi,k)}$ by the inductive hypothesis;
let $g_k:S\to S$ be such that
$\Cal G_{\eta}=g_k[[\Cal G_{\theta(\xi,k)}]]$.   Now define $g:S\to S$
by setting

$$\eqalign{g(\tau)
&=g_k(\sigma)\text{ if }k\ge k_0
    \text{ and }\tau=\fraction{k}^{\smallfrown}\sigma,\cr
&=\emptyset\text{ otherwise}.\cr}$$

\noindent For $B\subseteq S$,

$$\eqalignno{g^{-1}[B]\in\Cal G_{\xi}
&\iff\{k:\{\sigma:\fraction{k}^{\smallfrown}\sigma\in g^{-1}[B]\}
  \in\Cal G_{\theta(\xi,k)}\}\in\Cal F_{\xi}\cr
&\iff\{k:k\ge k_0,\,
  \{\sigma:g(\fraction{k}^{\smallfrown}\sigma)\in B\}
  \in\Cal G_{\theta(\xi,k)}\}\in\Cal F_{\xi}\cr
\displaycause{because $\Cal F_{\xi}$ is free}
&\iff\{k:k\ge k_0,\,
  \{\sigma:g_k(\sigma)\in B\}
  \in\Cal G_{\theta(\xi,k)}\}\in\Cal F_{\xi}\cr
&\iff\{k:k\ge k_0,\,
  B\in\Cal G_{\eta}\}\in\Cal F_{\xi}
\iff B\in\Cal G_{\eta},\cr}$$

\noindent so $\Cal G_{\eta}=g[[\Cal G_{\xi}]]\leRK\Cal G_{\xi}$.\ \Qed}

\spheader 538Ed\cmmnt{ It follows that if
$1\le\xi_0<\ldots<\xi_n\le\zeta$ then
$\Cal F_{\xi_n}\ltimes\ldots\ltimes\Cal F_{\xi_0}\leRK\Cal G_{\xi_n}$.
\prooflet{\Prf\ Induce on the pair $(\xi_n,n)$.   If $\xi_n=1$ then $n=0$
and we just have $\Cal F_1\leRK\Cal G_1$, as in part (i) of the proof of
(c).   For the inductive step to
$\xi_n=\xi>1$, if $n=0$ then again we need only note that
$\Cal F_{\xi_0}=\Cal F_{\xi}\leRK\Cal G_{\xi}$.    If $n>0$, let
$k_0\ge 1$ be such that $\xi_{n-1}\le\theta(\xi,k)$ for every
$k\ge k_0$.   For $k\ge k_0$,

\Centerline{$\Cal F_{\xi_{n-1}}\ltimes\ldots\ltimes\Cal F_{\xi_0}
\leRK\Cal G_{\xi_{n-1}}\le\Cal G_{\theta(\xi,k)}$}

\noindent by the inductive hypothesis, so
we have a function $g_k:S\to\BbbN^n$ such that
$\Cal F_{\xi_{n-1}}\ltimes\ldots\ltimes\Cal F_{\xi_0}
=g_k[[\Cal G_{\theta(\xi,k)}]]$.   Define $g:S\to\BbbN^{n+1}$ by setting

$$\eqalign{g(\tau)
&=\fraction{k}^{\smallfrown}g_k(\sigma)\text{ if }k\ge k_0
    \text{ and }\tau=\fraction{k}^{\smallfrown}\sigma,\cr
&=\text{the constant function with value }0\text{ otherwise}.\cr}$$

\noindent Then, for $B\subseteq\BbbN^{n+1}$,

$$\eqalignno{g^{-1}[B]\in\Cal G_{\xi}
&\iff\{k:\{\sigma:g(\fraction{k}^{\smallfrown}\sigma)\in B\}
  \in\Cal G_{\theta(\xi,k)}\}\in\Cal F_{\xi}\cr
&\iff\{k:k\ge k_0,\,\{\sigma:\fraction{k}^{\smallfrown}g_k(\sigma)\in B\}
  \in\Cal G_{\theta(\xi,k)}\}\in\Cal F_{\xi}\cr
&\iff\{k:k\ge k_0,\,\{\sigma:g_k(\sigma)\in B_k\}
  \in\Cal G_{\theta(\xi,k)}\}\in\Cal F_{\xi}\cr
\displaycause{writing
$B_k=\{\sigma:\fraction{k}^{\smallfrown}\sigma\in B\}\subseteq\BbbN^n$
for $k\in\Bbb N$}
&\iff\{k:k\ge k_0,\,
  B_k\in\Cal F_{\xi_{n-1}}\ltimes\ldots\ltimes\Cal F_{\xi_0}\}
  \in\Cal F_{\xi}\cr
&\iff\{k:k\in\Bbb N,\,
  B_k\in\Cal F_{\xi_{n-1}}\ltimes\ldots\ltimes\Cal F_{\xi_0}\}
  \in\Cal F_{\xi}\cr
&\iff B\in\Cal F_{\xi_n}\ltimes\ldots\ltimes\Cal F_{\xi_0}.\cr}$$

\noindent Thus $g$ witnesses that
$\Cal F_{\xi_n}\ltimes\ldots\ltimes\Cal F_{\xi_0}\leRK\Cal G_{\xi_n}$, and
the induction proceeds.\ \Qed}

Consequently}
$\Cal F_{\xi_n}\ltimes\ldots\ltimes\Cal F_{\xi_0}\leRK\Cal G_{\zeta}$
whenever $1\le\xi_0<\ldots<\xi_n\le\zeta$.

\spheader 538Ee \cmmnt{The following special remark will be useful in
Theorem 538L.}
Suppose that we are given $A_{\xi}\in\Cal F_{\xi}$ for each
$\xi\in[1,\zeta]$.   Define $T\subseteq S$ and $\alpha:T\to[0,\zeta]$ as
follows.   Start by saying that $\emptyset\in T$ and
$\alpha(\emptyset)=\zeta$.   Having determined $T\cap\BbbN^n$ and
$\alpha\restr T\cap\BbbN^n$, where $n\in\Bbb N$, then for
$\tau\in\BbbN^{n+1}$ say that $\tau\in T$ iff $\tau$ is of the form
$\sigma^{\smallfrown}\fraction{k}$ where

\Centerline{$\sigma\in T\cap\BbbN^n$,
\quad$\alpha(\sigma)>0$,
\quad$k\in A_{\alpha(\sigma)}$,
\quad$\sigma(i)<k$ for every $i<n$,}

\noindent and in this case set $\alpha(\tau)=\theta(\alpha(\sigma),k)$.
Continue.   Observe that $\alpha(\tau)<\alpha(\sigma)$ whenever $\sigma$,
$\tau\in T$ and $\tau$ properly extends $\sigma$.

Suppose that $D\in\bigcap_{1\le\xi\le\zeta}\Cal F_{\xi}$.   Then
$T^*_D=\{\tau:\tau\in T\cap\bigcup_{n\in\Bbb N}D^n$, $\alpha(\tau)=0\}$
belongs to $\Cal G_{\zeta}$.
\prooflet{\Prf\ I aim to show by induction on
$\xi$ that if $\tau\in T\cap\bigcup_{n\in\Bbb N}D^n$ and
$\alpha(\tau)=\xi$ then
$\{\sigma:\tau^{\smallfrown}\sigma\in T_D^*\}$ belongs to $\Cal G_{\xi}$.
If $\xi=0$ then of course
$\{\sigma:\tau^{\smallfrown}\sigma\in T_D^*\}=\{\emptyset\}\in\Cal G_0$.
For the inductive step to $\xi>0$,

$$\eqalignno{\{k:
  \{\sigma:\tau^{\smallfrown}\fraction{k}^{\smallfrown}&\sigma\in T_D^*\}
       \in\Cal G_{\theta(\xi,k)}\}\cr
&\supseteq\{k:k\in D,\,\tau^{\smallfrown}\fraction{k}\in T,\,
    \alpha(\tau^{\smallfrown}\fraction{k})=\theta(\xi,k)\}\cr
\displaycause{by the inductive hypothesis}
&\supseteq\{k:k\in A_{\xi}\cap D,\,
    \tau(i)<k\text{ for every }i<\dom\tau\}\cr
&\in\Cal F_{\xi},\cr}$$

\noindent so $\{\sigma:\tau^{\smallfrown}\sigma\in T_D^*\}\in\Cal G_{\xi}$.
At the end of the induction, we can apply this to $\tau=\emptyset$
and $\xi=\zeta$.\ \Qed}

\leader{538F}{Ramsey \dvrocolon{filters}}\cmmnt{ There is an extensive
and fascinating theory of Ramsey filters;
see, for instance, {\smc Comfort
\& Negrepontis 74}.   Here, however, I will give only those fragments which
are directly relevant to the other work of this section.

\medskip

\noindent}{\bf Proposition} (a) A Ramsey filter on $\Bbb N$ is a
rapid $p$-point ultrafilter.

(b) If $\Cal F$ is a Ramsey ultrafilter on $\Bbb N$, $\Cal G$ is a
non-principal ultrafilter on $\Bbb N$, and $\Cal G\leRK\Cal F$,
then $\Cal F$ and $\Cal G$ are isomorphic and $\Cal G$
is a Ramsey ultrafilter.

(c) Let $\Cal F$ be a Ramsey filter on $\Bbb N$.
Suppose that $\sequencen{A_n}$ is any sequence in $\Cal F$.   Then
there is an $A\in\Cal F$ such that $n\in A_m$ whenever $m$, $n\in A$ and
$m<n$.

(d) Let $\Cal F$ be a Ramsey filter on $\Bbb N$.   Let
$\Cal S\subseteq[\Bbb N]^{<\omega}$ be such that $\emptyset\in\Cal S$ and
$\{n:I\cup\{n\}\in\Cal S\}\in\Cal F$ for every $I\in\Cal S$.   Then there
is an $A\in\Cal F$ such that $[A]^{<\omega}\subseteq\Cal S$.

(e) If $\frak F$ is a countable family of distinct Ramsey filters on
$\Bbb N$, there is a disjoint family $\family{\Cal F}{\frak F}{A_{\Cal F}}$
of subsets of $\Bbb N$ such that $A_{\Cal F}\in\Cal F$ for every
$\Cal F\in\frak F$.

(f) Let $\frak F$ be a countable
family of non-isomorphic Ramsey ultrafilters on $\Bbb N$, and
$\frak h:\Bbb N\to[\frak F]^{<\omega}$ a function.
Suppose that we are given
an $A_{\Cal F}\in\Cal F$ for each $\Cal F\in\frak F$.
Then there is an $A\in\bigcap\frak F$ such that whenever $i$, $j\in A$,
$\Cal F\in\frak h(i)$ and $i<j$, there is a $k\in A_{\Cal F}$ such that
$i<k<j$.

(g) If $\frakmctbl=\frak c$, there is a Ramsey ultrafilter on $\Bbb N$.

\woddheader{538F}{0}{0}{0}{54pt}

\proof{{\bf (a)} Let $\Cal F$ be a Ramsey filter on $\Bbb N$.

\medskip

\quad{\bf (i)} $\Cal F$ is an ultrafilter.   \Prf\ Let $A$ be any subset of
$\Bbb N$.   Define $f:[\Bbb N]^2\to\{0,1\}$ by setting $f(I)=1$ if
$\#(I\cap A)=1$, $0$ otherwise.   Then we have an $I\in\Cal F$ such that
$f$ is constant on $[I]^2$.   As $\Cal F$ is free, $\#(I)\ge 3$ and
the constant value of $f$ cannot be $1$.   So either $I\subseteq A$ and
$A\in\Cal F$, or $I\cap A=\emptyset$ and $\Bbb N\setminus A\in\Cal F$.
As $A$ is arbitrary, $\Cal F$ is an ultrafilter.\ \Qed

\medskip

\quad{\bf (ii)} $\Cal F$ is a $p$-point filter.   \Prf\ Let
$\sequencen{I_n}$ be a sequence in $\Cal F$.   Set
$K_n=(\Bbb N\setminus n)\cap\bigcap_{i<n}I_i$,
$J_n=K_n\setminus K_{n+1}$ for each $n$;
then $\sequencen{J_n}$ is a partition of $\Bbb N$.
Define $f:[\Bbb N]^2\to\{0,1\}$ by
setting $f(a)=0$ if there is an $n\in\Bbb N$ such that $a\subseteq J_n$,
$1$ otherwise.   Let $I\in\Cal F$ be such that $f$ is constant on $[I]^2$.

Since $\Bbb N\setminus J_n\in\Cal F$ for every $n$, there must be two
points in $I$ belonging to different $J_n$;  so that the constant value of
$f$ must be $1$, and no two points of $I$ belong to the same $J_n$.
In particular, $I\cap J_n$ is always finite, and
$I\setminus I_n\subseteq\bigcup_{i\le n}I\cap J_i$ is always finite.
As $\sequencen{I_n}$ is arbitrary, $\Cal F$ is a $p$-point filter.\ \Qed

\medskip

\quad{\bf (iii)} $\Cal F$ is rapid.   \Prf\ Let $\sequencen{t_n}$ be a
sequence converging to $0$.   For each $n$, set
$I_n=\{i:|t_i|\le 2^{-n}\}$;  as $\Cal F$ is free, $I_n\in\Cal F$.
Looking again at the proof of (ii) above, we see that the construction
there gives us an $I\in\Cal F$ such that $\#(I\setminus I_n)\le n+1$ for
every $n$.   We can therefore enumerate $I$ as $\sequencen{k_n}$ in such a
way that $k_{n+1}\in I_n$ for every $n$.   But this means that

\Centerline{$\sum_{i\in I}|t_i|=\sum_{n=0}^{\infty}|t_{k_n}|
\le|t_{k_0}|+\sum_{n=1}^{\infty}2^{-n+1}<\infty$.}

\noindent As $\sequencen{t_n}$ is arbitrary, $\Cal F$ is rapid.\ \Qed

\medskip

{\bf (b)} Let $f:\Bbb N\to\Bbb N$ be
such that $f[[\Cal F]]=\Cal G$.   For $K\in[\Bbb N]^2$, set
$h(K)=0$ if $f\restr K$ is constant, $1$ otherwise.
Then there is an $A\in\Cal F$ such that $h$ is constant on $[A]^2$,
that is, $f$ is either
constant or injective on $A$.   Since $f[A]\in\Cal G$, $f[A]$ is infinite,
so $f$ is injective on $A$.   Let $g:\Bbb N\to\Bbb N$ be any function
extending $(f\restr A)^{-1}$;  then $gf(n)=n$ for every $n\in A$, so

\Centerline{$(gf)[[\Cal F]]
=\{I:(gf)^{-1}[I]\in\Cal F\}
=\{I:A\cap(gf)^{-1}[I]\in\Cal F\}
=\{I:A\cap I\in\Cal F\}
=\Cal F$.}

\noindent But this means that $g[[\Cal G]]=\Cal F$ and $\Cal F\leRK\Cal G$.

By 538Cb, $\Cal F$ and $\Cal G$ are isomorphic, so $\Cal G$ also must be a
Ramsey ultrafilter.

\medskip

{\bf (c)} For $m<n$ in $\Bbb N$, set $h(\{m,n\})=1$ if $n\in A_m$, $0$
otherwise.   Then there is an $A\in\Cal F$ such that $h\restr[A]^2$ is
constant.   Setting $k=\min A$, $A$ meets $A_k\setminus(k+1)$, so $h$ takes
the value $1$ on $[A]^2$;   consequently
$n\in A_m$ whenever $m$, $n\in A$ and $m<n$.

\medskip

{\bf (d)} For $n\in\Bbb N$, set

\Centerline{$A_n
=\{i:I\cup\{i\}\in\Cal S$ whenever $I\subseteq n+1$ and $I\in\Cal S\}
\in\Cal F$.}

\noindent By (c), there is an $A\in\Cal F$ such that $n\in A_m$ whenever
$m$, $n\in A$ and $m<n$;  and we can suppose that $A\subseteq A_0$, so that
$\{n\}\in\Cal S$ for every $n\in A$.   Now an easy induction on $n$ shows
that $\Cal P(A\cap n)\subseteq\Cal S$ for every $n$, so
$[A]^{<\omega}\subseteq\Cal S$.

\medskip

{\bf (e)} Enumerate $\frak F$ as $\ofamily{n}{\#(\frak F)}{\Cal F_n}$.
For distinct $m$, $n<\#(\frak F)$ there is a
$B_{mn}\in\Cal F_m\setminus\Cal F_n$.
\Prf\ We know that there is a set in $\Cal F_m\symmdiff\Cal F_n$;  now
either this set or its complement will serve for $B_{mn}$.\ \QeD\  Because
every member of $\frak F$ is a $p$-point filter ((a) above), we can find
for each $n<\#(\frak F)$ a set $C_n\in\Cal F_n$ such that
$C_n\setminus(B_{nm}\setminus B_{mn})$ is finite for every $m<\#(\frak F)$
such that $m\ne n$.   Set $A_{\Cal F_n}=C_n\setminus\bigcup_{m<n}C_m$ for
$n<\#(\frak F)$;  then $\family{\Cal F}{\frak F}{A_{\Cal F}}$ is disjoint.
Since

\Centerline{$C_m\cap C_n
\subseteq(C_m\setminus B_{mn})\cup(C_n\cap B_{mn})$}

\noindent is finite whenever $m\ne n$, $C_n\setminus A_{\Cal F_n}$ is
finite and  $A_{\Cal F_n}\in\Cal F_n$ for each $n<\#(\frak F)$.

\medskip

{\bf (f)(i)} We can suppose that $\frak h(i)\subseteq\frak h(j)$ whenever
$i\le j$, and that $\frak F=\bigcup_{i\in\Bbb N}\frak h(i)$.
Let $g:\Bbb N\to\Bbb N$ be a strictly increasing function
such that $g(0)>0$ and whenever $i\in\Bbb N$ and
$\Cal F\in \frak h(i)$, there is a $k\in A_{\Cal F}$ such that $i<k<g(i)$.
Set $l_m=g^m(0)$ and
$J_m=l_{m+1}\setminus l_m$ for each $m$, so that $\sequence{m}{J_m}$ is a
partition of $\Bbb N$.   Let $\ofamily{\xi}{\omega_1}{a_{\xi}}$ be a family
of infinite subsets of $\Bbb N$, all containing $0$,
such that $a_{\xi}\cap a_{\eta}$ is finite
for all distinct $\xi$, $\eta<\omega_1$ (5A1Ga), and set
$M_{\xi}=\bigcup_{m\in a_{\xi}}J_m$ for each $\xi$;  then
$M_{\xi}\cap M_{\eta}$ is finite for all distinct $\xi$, $\eta<\omega_1$.
It follows that each member of $\frak F$ can contain at most one $M_{\xi}$,
and there is a $\xi<\omega_1$ such that $M_{\xi}$ does not belong to any
member of $\frak F$, that is, $M=\Bbb N\setminus M_{\xi}$ belongs to
$\bigcap\frak F$.

\medskip

\quad{\bf (ii)} Define $f:\Bbb N\to\Bbb N$ by setting
$f(n)=\max\{m:m\in a_{\xi}$, $l_m\le n\}$ for $n\in\Bbb N$.   For each
$\Cal F\in\frak F$, $f[[\Cal F]]$ is isomorphic to $\Cal F$, by (b).
It follows that if $\Cal F$, $\Cal F'$ are distinct
members of $\frak F$, $f[[\Cal F]]\ne f[[\Cal F']]$.   Because $\frak F$ is
countable,
there is a disjoint family $\family{\Cal F}{\frak F}{K_{\Cal F}}$ of sets
such that $K_{\Cal F}\in f[[\Cal F]]$ for every $\Cal F\in\frak F$
((e) above).   Set $L_{\Cal F}=f^{-1}[K_{\Cal F}]\in\Cal F$ for each
$\Cal F\in\frak F$.

\medskip

\quad{\bf (iii)} For $i<j$ in $\Bbb N$, set $h(\{i,j\})=1$ if
$j<g(i)$, $0$ otherwise.
$\Cal F\in\frak F$, there is an $L'_{\Cal F}\in\Cal F$ such that
$L'_{\Cal F}\subseteq L_{\Cal F}$ and $h$ is constant on
$[L'_{\Cal F}]^2$.   As $L'_{\Cal F}$ is infinite, the constant value
cannot be $1$ and must be $0$, that is, $g(i)\le j$ whenever $i$,
$j\in L'_{\Cal F}$ and $i<j$.

\medskip

\quad{\bf (iv)} Consider
$A=\bigcup_{\Cal F\in\frak F}L'_{\Cal F}\cap M$.   Then
$A\in\bigcap\frak F$.   Suppose that $i$, $j\in A$ and $i<j$;
then $g(i)\le j$.   \Prf\ Let
$\Cal F$, $\Cal F'\in\frak F$ be such that
$i\in L'_{\Cal F}$ and $j\in L'_{\Cal F'}$.

\medskip

\qquad{\bf case 1} If $\Cal F=\Cal F'$, then both $i$ and $j$ belong to
$L'_{\Cal F}$, so $g(i)\le j$ by (iii).

\medskip

\qquad{\bf case 2} If $\Cal F\ne\Cal F'$, then $i\in L_{\Cal F}$ and
$j\in L_{\Cal F'}$, so $f(i)\in K_{\Cal F}$ and $f(j)\in K_{\Cal F'}$ and
$f(i)\ne f(j)$.   Let $m$, $n\in\Bbb N$ be such that $i\in J_m$ and
$j\in J_n$;  since $j\notin M_{\xi}$,
$n\notin a_{\xi}$ and $f(j)<n$.   As $K_{\Cal F}$ and
$K_{\Cal F'}$ are disjoint, $f(i)<f(j)$.   It follows that $m<f(j)<n$, so

\Centerline{$g(i)\le g(l_{m+1})\le g(l_{f(j)})\le l_n\le j$}

\noindent and $g(i)\le j$ in this case also.\ \Qed

By the choice of $g$,
this means that if $\Cal F\in\frak h(i)$ there must be a
$k\in A_{\Cal F}$ such that $i<k<j$, as required.

\medskip

{\bf (g)(i)} Suppose that $\Cal E\subseteq\Cal P\Bbb N$ is a filter base,
containing $\Bbb N\setminus n$ for every $n\in\Bbb N$, and with cardinal less
than $\frakmctbl$.   Let $f:[\Bbb N]^2\to\{0,1\}$ be a function.
Then there is an $F\subseteq\Bbb N$ such that $f$ is constant on $[F]^2$
and $F$ meets every member of $\Cal E$.   \Prf\ Set

\Centerline{$\Cal E^+=\{J:J\subseteq\Bbb N$, $J\cap E\ne\emptyset$ for
every $E\in\Cal E\}$,}

\Centerline{$S_n
=\{n\}\cup\{i:i\in\Bbb N\setminus\{n\}$, $f(\{i,n\})=1\}$,}

\Centerline{$S'_n
=\{n\}\cup\{i:i\in\Bbb N\setminus\{n\}$, $f(\{i,n\})=0\}$}

\noindent for $n\in\Bbb N$.

\medskip

\qquad{\bf case 1} Suppose that $\{n:n\in J$, $J\cap S_n\in\Cal E^+\}$
belongs to $\Cal E^+$ for every $J\in\Cal E^+$.   Set

\Centerline{$\Cal I
=\{I:I\in[\Bbb N]^{<\omega}$, $f(K)=1$ for every $K\in[I]^2$,
$\Bbb N\cap\bigcap_{i\in I}S_i\in\Cal E^+\}$.}

\noindent
If $I\in\Cal I$, $J=\Bbb N\cap\bigcap_{i\in I}S_i$ and $E\in\Cal E$, then
$J\in\Cal E^+$;  because $\Cal E$ is a filter base, $J\cap E\in\Cal E^+$;
by hypothesis, $\{n:n\in J\cap E$, $J\cap E\cap S_n\in\Cal E^+\}$ belongs
to $\Cal E^+$ and is not empty.   There is therefore some $n\in J\cap E$
such that $J\cap S_n\in\Cal E^+$, in which case
$I\cup\{n\}\in\Cal I$.

In particular, there is some $k\in\Bbb N$ such that $\{k\}\in\Cal I$.   Set

\Centerline{$C
=\{\alpha:\alpha\in\NN$, $\{\alpha(i):i<m\}\in\Cal I$ for every
$m\in\Bbb N\}$.}

\noindent Then $C$ is compact, and it is non-empty because the constant
function with value $k$ belongs to $C$.   Moreover, if $\alpha\in C$ and
$m\in\Bbb N$ and $E\in\Cal E$, there is an $n\in E$ such that
$\{\alpha(i):i<m\}\cup\{n\}\in\Cal I$, so there is a $\beta\in C$ such that
$\beta(i)=\alpha(i)$ for $i<m$ and $\beta(m)=n$.   Thus
$\{\beta:\beta\in C$, $E\cap\beta[\Bbb N]\ne\emptyset\}$ is a dense open
subset of $C$.   Writing $\Cal M(C)$ for the ideal of meager subsets of
$C$, $\cov\Cal M(C)$ is either $\infty$ (if $C$ has an isolated point)
or $\cov\Cal M(\Bbb R)=\frakmctbl$, by 522Wb and 522Sa;
in either case, it is greater than
$\#(\Cal E)$.   There is therefore some $\alpha\in C$ such that
$F=\alpha[\Bbb N]$ meets every member of $\Cal E$;  in this case,
$f$ is equal to $1$ everywhere in $[F]^2$, so we have an appropriate $F$.

\medskip

\qquad{\bf case 2} Otherwise, there is a $K\in\Cal E^+$ such that
$\{n:n\in K$, $K\cap S_n\in\Cal E^+\}$ does not belong to $\Cal E^+$.
Let $E_0\in\Cal E$ be disjoint from
$\{n:n\in K$, $K\cap S_n\in\Cal E^+\}$.
Set $\Cal G=\Cal E\cup\{K\cap E:E\in\Cal E\}$,
so that $\Cal G$ is a filter base and $\#(\Cal G)<\frakmctbl$.
If $n\in E_0$ then there is an
$E'_n\in\Cal E$ disjoint from $K\cap S_n$.
So if $J\in\Cal G^+$,
$J\cap S'_n\supseteq(J\cap K\cap E'_n)\setminus\{n\}$ belongs to
$\Cal G^+$ for every $n\in E_0$;  accordingly
$\{n:n\in J$, $J\cap S'_n\in\Cal G^+\}\supseteq J\cap E_0$
belongs to $\Cal G^+$.

We can therefore apply the argument of case 1 to $\Cal G$ and
the function $1-f$ to see that there is an $F\subseteq\Bbb N$, meeting
every member of $\Cal G\supseteq\Cal E$, such that $f=0$ on
$[F]^2$.\ \Qed

\medskip

\quad{\bf (ii)}
Enumerate the set of functions from $[\Bbb N]^2$ to $\{0,1\}$ as
$\ofamily{\xi}{\frak c}{f_{\xi}}$.   Choose a non-decreasing family
$\langle\Cal E_{\xi}\rangle_{\xi\le\frak c}$ inductively, as follows;
the inductive
hypothesis will be that $\Cal E_{\xi}\subseteq\Cal P\Bbb N$ is a
filter base with cardinal at most $\max(\omega,\#(\xi))$.
Start with $\Cal E_0=\{\Bbb N\setminus n:n\in\Bbb N\}$.
Given $\Cal E_{\xi}$, where $\xi<\frak c=\frakmctbl$,
use (i) to find a set $F_{\xi}$, meeting every member of $\Cal E_{\xi}$,
such that $f_{\xi}$ is constant on $[F_{\xi}]^2$;  take
$\Cal E_{\xi+1}=\Cal E_{\xi}\cup\{E\cap F_{\xi}:E\in\Cal E_{\xi}\}$.
Given $\ofamily{\eta}{\xi}{\Cal E_{\eta}}$, where $\xi\le\frak c$ is a
non-zero limit ordinal, set $\Cal E_{\xi}=\bigcup_{\eta<\xi}\Cal E_{\eta}$.

At the end of the induction, let $\Cal F$ be the filter generated by
$\Cal E_{\frakc}$;  then $\Cal F$ is a Ramsey filter.
}%end of proof of 538F

\vleader{60pt}{538G}{Measure-centering filters:  Theorem}
Let $\Cal F$ be a free filter on $\Bbb N$.   Write
$\nu_{\omega}$ for the usual measure on $\{0,1\}^{\Bbb N}$,
$\Tau_{\omega}$ for its domain and $(\frak B_{\omega},\bar\nu_{\omega})$
for its measure algebra.   Then the following are equiveridical:

(i) $\Cal F$ is measure-centering;

(ii) whenever $\sequencen{a_n}$ is a sequence in $\frak B_{\omega}$ such
that $\inf_{n\in\Bbb N}\bar\nu_{\omega}a_n>0$, there is an $A\in\Cal F$
such that $\{a_n:n\in A\}$ is centered;

(iii) whenever $\sequencen{F_n}$ is a sequence in $\Tau_{\omega}$ such that
$\inf_{n\in\Bbb N}\nu_{\omega}F_n>0$, there is an $A\in\Cal F$ such that
$\bigcap_{n\in A}F_n\ne\emptyset$;

\allowmorestretch{468}{
(iv) whenever $(X,\Sigma,\mu)$ is a perfect totally finite measure space
and $\sequencen{F_n}$ is a sequence in $\Sigma$,
$\mu^*(\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n)
\ge\liminf_{n\to\Cal F}\mu F_n$;
}

(v) whenever $\mu$ is a Radon probability measure on $\Cal P\Bbb N$, then
$\mu^*\Cal F\ge\liminf_{n\to\Cal F}\mu E_n$, where
$E_n=\{a:n\in a\subseteq\Bbb N\}$ for each $n$.

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

\medskip

\allowmorestretch{468}{
{\bf not-(iv)$\Rightarrow$not-(ii)} Suppose there are
a perfect totally finite measure space $(X,\Sigma,\mu)$ and a sequence
$\sequencen{F_n}$ in $\Sigma$ such that
$\liminf_{n\in\Bbb N}\mu F_n
>\mu^*(\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n)$.
Let $F$ be a measurable envelope of
$\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n$.
Let $\Tau$ be the $\sigma$-subalgebra of
$\Sigma$ generated by $\{F\}\cup\{F_n:n\in\Bbb N\}$;
then $\mu\restrp\Tau$ is a compact measure (451F).   Let
$\nu$ be its normalization $\Bover1{\mu X}\mu\restrp\Tau$;
then $\nu$ is a compact probability measure.   We see that
$\liminf_{n\to\Cal F}\nu F_n>\nu F$;  take $\gamma$ such that
$\nu F<\gamma<\liminf_{n\to\Cal F}\nu F_n$, and set
$C=\{n:\nu F_n>\gamma\}$, so that $C\in\Cal F$.
}%end of allowmorestretch

Let $\Cal K$ be a compact class such that
$\nu$ is inner regular with respect to $\Cal K$.   For $n\in C$, let
$K_n\in\Cal K\cap\Tau$
be such that $K_n\subseteq F_n\setminus F$ and
$\nu K_n\ge\gamma-\nu F$;  for $n\in\Bbb N\setminus C$ set $K_n=X$.

The measure algebra $(\frak B,\bar\nu)$ of $\nu$ is a probability algebra
with countable Maharam type, so there is a measure-preserving Boolean
homomorphism $\pi:\frak B\to\frak B_{\omega}$ (332P or 333D).   Set
$a_n=\pi K_n^{\ssbullet}$ for each $n$.   Then

\Centerline{$\bar\nu_{\omega}a_n=\nu K_n\ge\gamma-\nu F>0$}

\noindent for every $n$.   On the other hand, if $A\in\Cal F$, then
$A\cap C\in\Cal F$ so
$\bigcap_{n\in A\cap C}K_n\subseteq\bigcap_{n\in A\cap C}F_n\setminus F$ is
empty.   As $K_n$ belongs to the compact class $\Cal K$ for every
$n\in A\cap C$, there must be a finite set $I\subseteq A\cap C$ such that
$\bigcap_{n\in I}K_n=\emptyset$, in which case
$\inf_{n\in I}a_n=\pi(\bigcap_{n\in I}K_n)^{\ssbullet}=0$.   This shows
that $\{a_n:n\in A\}$ is not centered.   So $\sequencen{a_n}$ witnesses
that (ii) is false.

\medskip

{\bf (iv)$\Rightarrow$(i)} Suppose that (iv) is true.   Take a Boolean
algebra $\frak A$, an additive functional $\nu:\frak A\to\coint{0,\infty}$
and a sequence $\sequencen{a_n}$ in $\frak A$ such that
$\inf_{n\in\Bbb N}\nu a_n>0$.   By 311E and 311H, we can suppose that
$\frak A$ is the algebra of open-and-closed subsets of a compact
zero-dimensional space $Z$.   In this case, there is a Radon measure
$\mu$ on $Z$ extending $\nu$ (416Qa).   Of course $\mu$ is perfect
(416Wa), and $\liminf_{n\to\Cal F}\mu a_n\ge\inf_{n\in\Bbb N}\nu a_n>0$,
so (iv) tells us that there is an $A\in\Cal F$ such that
$\bigcap_{n\in A}a_n\ne\emptyset$, in which case $\{a_n:n\in\Bbb N\}$ is
centered in $\frak A$.   As
$\frak A$, $\nu$ and $\sequencen{a_n}$ are arbitrary, $\Cal F$ is
measure-centering.

\medskip

{\bf (iv)$\Rightarrow$(v)} The point is simply that $\mu$ is perfect
(416Wa again) and that

\Centerline{$\bigcup_{A\in\Cal F}\bigcap_{n\in A}E_n
=\bigcup_{A\in\Cal F}\{a:A\subseteq a\subseteq\Bbb N\}
=\Cal F$.}

\medskip

{\bf (v)$\Rightarrow$(iii)} Suppose that (v) is true, and that
$\sequencen{F_n}$ is a sequence in $\Tau_{\omega}$ such that
$\inf_{n\in\Bbb N}\nu_{\omega}F_n>0$.   Define
$\phi:\{0,1\}^{\Bbb N}\to\Cal P\Bbb N$ by setting
$\phi(x)=\{n:x\in F_n\}$ for each $n$.   Then $\phi$ is almost continuous
(418J), so the image measure $\mu=\nu_{\omega}\phi^{-1}$ is a Radon
probability measure on $\Cal P\Bbb N$ (418I).
Defining $E_n$ as in (v), we have

\Centerline{$\mu E_n=\nu_{\omega}\phi^{-1}[E_n]=\nu_{\omega}F_n$}

\noindent for every $n\in\Bbb N$, so

\Centerline{$0<\inf_{n\in\Bbb N}\bar\nu_{\omega}a_n
\le\liminf_{n\to\Cal F}\mu E_n\le\mu^*\Cal F
=\nu_{\omega}^*\phi^{-1}[\Cal F]$}

\noindent (451Pc).   In particular, there must be an
$x\in\phi^{-1}[\Cal F]$, so that $A=\{n:x\in F_n\}$ belongs to $\Cal F$,
and $\bigcap_{n\in A}F_n$ is non-empty.

\medskip

{\bf (iii)$\Rightarrow$(ii)} Assume (iii).   Let $\sequencen{a_n}$ be a
sequence in $\frak B_{\omega}$ such that
$\inf_{n\in\Bbb N}\bar\nu_{\omega}a_n>0$.   Let
$\theta:\frak B_{\omega}\to\Tau_{\omega}$ be a lifting (341K), and set
$F_n=\theta a_n$ for each $n$.   Then $\nu_{\omega}F_n=\bar\nu_{\omega}a_n$
for every $n$, so (iii) tells us that there is an $A\in\Cal F$ such that
$\bigcap_{n\in A}F_n\ne\emptyset$.   In this case,
$\theta(\inf_{n\in I}a_n)=\bigcap_{n\in I}F_n\ne\emptyset$
for every non-empty finite $I\subseteq A$, so $\{a_n:n\in A\}$ is centered.
}%end of proof of 538G

\leader{538H}{Proposition}
(a) Any measure-centering filter on $\Bbb N$ is an ultrafilter.

(b) If $\Cal F$ is a measure-centering
ultrafilter on $\Bbb N$ and $\Cal G$ is a filter on
$\Bbb N$ such that $\Cal G\leRK\Cal F$, then $\Cal G$ is measure-centering.

(c) Every Ramsey ultrafilter on $\Bbb N$ is measure-centering.

(d)\cmmnt{ ({\smc Shelah 98b})} Every
measure-centering ultrafilter on $\Bbb N$ is a nowhere dense ultrafilter.

(e)\cmmnt{ ({\smc Benedikt 99})} If $\cov\Cal N=\frak c$, where $\Cal N$
is the Lebesgue null ideal, then there is a measure-centering ultrafilter
on $\Bbb N$.
%would  \cov\Cal N=\cf\Cal N  be enough?

\proof{{\bf (a)} Let $a$, $b$ be disjoint non-zero elements of
$\frak B_{\omega}$, where $(\frak B_{\omega},\bar\nu_{\omega})$ is the
measure algebra of the usual measure on $\{0,1\}^{\Bbb N}$, as in 538G.
Given $I\subseteq\Bbb N$, set $a_n=a$ if $n\in I$, $b$ if
$n\in\Bbb N\setminus I$.   Then $\inf_{n\in\Bbb N}\bar\nu_{\omega}a_n>0$,
so there is a $J\in\Cal F$ such that $\{a_n:n\in J\}$ is
centered, in which case either $J\subseteq I$ or $J\cap I=\emptyset$;  so
that one of $I$, $\Bbb N\setminus I$ must belong to $\Cal F$.

\medskip

{\bf (b)} Let $f:\Bbb N\to\Bbb N$ be such that $f[[\Cal F]]=\Cal G$.
Let $(\frak A,\bar\mu)$ be a totally finite measure algebra and
$\sequencen{a_n}$ a sequence in
$\frak A$ with $\inf_{n\in\Bbb N}\bar\mu a_n>0$.   Then
$\sequencen{a_{f(n)}}$ has the same property, so there is an
$A\in\Cal F$ such that $\{a_{f(n)}:n\in A\}$ is centered.   Now
$f[A]\in\Cal G$ and $\{a_m:m\in f[A]\}$ is centered.

\medskip

{\bf (c)} Let $\Cal F$ be a Ramsey ultrafilter
and $\sequencen{b_n}$ a sequence in $\frak B_{\omega}$
such that $\gamma=\inf_{n\in\Bbb N}\bar\nu_{\omega}b_n$ is greater than
$0$.
Set $b=\inf_{A\in\Cal F}\sup_{n\in A}b_n$;  then
$\bar\nu_{\omega}b\ge\gamma$.
Set $\Cal S=\{I:I\in[\Bbb N]^{<\omega}$, $b\Bcap\inf_{n\in I}b_n\ne 0\}$.
Then $\emptyset\in\Cal S$.   If $I\in\Cal S$, set
$c=b\Bcap\inf_{n\in I}b_n$ and $C=\{n:c\Bcap b_n=0\}$.   Then
$\sup_{n\in C}b_n$ does not meet $c$ so does not include $b$, and
$C\notin\Cal F$.   Accordingly

\Centerline{$\{n:I\cup\{n\}\in\Cal S\}=\Bbb N\setminus C\in\Cal F$.}

\noindent By 538Fd, there is an $A\in\Cal F$ such that
$[A]^{<\omega}\subseteq\Cal S$, in which case $\{b_n:n\in A\}$ is centered.
As $\sequencen{b_n}$ is arbitrary, $\Cal F$ is measure-centering.

\medskip

{\bf (d)} Let $\Cal F$ be a measure-centering ultrafilter, and
$\sequencen{t_n}$ a sequence in $\Bbb R$.   Let $F\subseteq\coint{0,1}$ be
a nowhere dense set with non-zero Lebesgue measure, and set
$H=\bigcup_{k\in\Bbb Z}F+k$, so that $H$ is nowhere dense in $\Bbb R$;  let
$\mu$ be Lebesgue measure on $[0,1]$.
For $n\in\Bbb N$ set

\Centerline{$E_n=\{x:x\in[0,1]$, $x+t_n\in H\}
=[0,1]\cap\bigcup_{k\in\Bbb Z}F-t_n+k$,}

\noindent so that $\mu E_n=\mu F>0$.   By 538G(iv), there is an
$A\in\Cal F$ such that $\bigcap_{n\in A}E_n$ is non-empty;  take
$x\in\bigcap_{n\in A}E_n$, so that $t_n\in H-x$ for every $n\in A$, and
$\{t_n:n\in A\}$ is nowhere dense.
As $\sequencen{t_n}$ is arbitrary, $\Cal F$ is a nowhere dense filter.

\medskip

{\bf (e)(i)} Let $\sequencen{a_n}$ be a sequence in
$\frak B_{\omega}$ such that $\inf_{n\in\Bbb N}\bar\nu_{\omega}a_n>0$, and
$\Cal C\subseteq\Cal P\Bbb N$ a filter base such that
$\#(\Cal C)<\cov\Cal N$.   Then there is an $A\subseteq\Bbb N$ such that
$A$ meets every member of $C$ and
$\{a_n:n\in A\}$ is centered.   \Prf\ Set
$\epsilon=\inf_{n\in\Bbb N}\bar\nu_{\omega}a_n$.
For $C\in\Cal C$ set $b_C=\sup_{n\in C}a_n$;  because $C\ne\emptyset$,
$\bar\nu_{\omega}b_C\ge\epsilon$.
Set $b=\inf_{C\in\Cal C}b_C$;  because $\Cal C$ is downwards-directed,
$\bar\nu_{\omega}b\ge\epsilon$ (321F) and $b\ne 0$.

Let $\theta:\frak B_{\omega}\to\Tau_{\omega}$ be a lifting (341K).   For
$C\in\Cal C$, set $F_C=\bigcup_{n\in C}\theta a_n$;  then

\Centerline{$F_C^{\ssbullet}=b_C\Bsupseteq b$,}

\noindent so $\theta b\setminus F_C$ is negligible.   Because
$b\ne 0$, $\theta b$ is not negligible;  because
$\#(\Cal C)<\cov\Cal N$, $\theta b\cap\bigcap_{C\in\Cal C}F_C$ is non-empty
(apply 522Wa to the subspace measure on $\theta b$).   Take any $x$ in the
intersection, and set $A=\{n:x\in\theta a_n\}$.   For every $C\in\Cal C$,
there is an $n\in C$ such that $x\in\theta a_n$, so $A\cap C\ne\emptyset$.
If $I\subseteq A$ is finite and not empty, then
$\theta(\inf_{n\in I}a_n)=\bigcap_{n\in I}\theta a_n$ contains $x$, so
$\inf_{n\in I}a_n\ne 0$;  thus $\{a_n:n\in A\}$ is centered.\ \Qed

\medskip

\quad{\bf (ii)} Since $\#(\frak B_{\omega})=\frak c$ (524Ma), we can
enumerate as $\ofamily{\xi}{\frak c}{\sequencen{a_{\xi n}}}$ the family of
all sequences $\sequencen{a_n}$ in $\frak B_{\omega}$ such that
$\inf_{n\in\Bbb N}\bar\nu_{\omega}a_n>0$.   Choose
$\ofamily{\xi}{\frak c}{\Cal C_{\xi}}$ inductively, as follows.
The inductive hypothesis will be that $\Cal C_{\xi}\subseteq\Cal P\Bbb N$
is a filter base and $\#(\Cal C_{\xi})\le\max(\omega,\#(\xi))$.
Start with $\Cal C_0=\{\Bbb N\setminus n:n\in\Bbb N\}$.
Given $\Cal C_{\xi}$, where $\xi<\frak c$, such that

\Centerline{$\#(\Cal C_{\xi})\le\max(\omega,\#(\xi))<\frak c=\cov\Cal N$,}

\noindent (i) tells us that there is an $A_{\xi}\subseteq\Bbb N$, meeting
every member of $\Cal C_{\xi}$, such
that $\{a_{\xi n}:n\in A_{\xi}\}$ is centered;  set

\Centerline{$\Cal C_{\xi+1}
=\Cal C_{\xi}\cup\{C\cap A_{\xi}:C\in\Cal C_{\xi}\}$.}

\noindent For a non-zero limit ordinal $\xi\le\frak c$, set
$\Cal C_{\xi}=\bigcup_{\eta<\xi}\Cal C_{\eta}$.
Let $\Cal F$ be
the filter generated by $\Cal C_{\frak c}$;  then $\Cal F$ is a free filter
satisfying 538G(ii), so is measure-centering.
}%end of proof of 538H

\leader{538I}{Theorem} Suppose that $\Cal F$ is a measure-centering
ultrafilter on $\Bbb N$, and that $(X,\Sigma,\mu)$ is a perfect probability
space.   Let $\Cal A$ be the family of all sets of the form
$\lim_{n\to\Cal F}E_n$ where $\sequencen{E_n}$ is a sequence in $\Sigma$.
Then there is a unique complete measure $\lambda$ on $X$ such that
$\lambda$ is inner regular with respect to $\Cal A$ and
$\lambda(\lim_{n\to\Cal F}E_n)=\lim_{n\to\Cal F}\mu E_n$
for every sequence $\sequencen{E_n}$ in $\Sigma$;  and $\lambda$ extends
$\mu$.

\medskip

\noindent{\bf Remark} By `$\lim_{n\to\Cal F}E_n$' I mean the limit in the
compact Hausdorff space $\Cal PX$, that is,

\Centerline{$\{x:\{n:x\in E_n\}\in\Cal F\}
=\bigcup_{A\in\Cal F}\bigcap_{n\in A}E_n
=\bigcap_{A\in\Cal F}\bigcup_{n\in A}E_n$.}

\proof{{\bf (a)}
$\Cal A$ is an algebra of subsets of $X$.   \Prf\ If $\sequencen{E_n}$,
$\sequencen{F_n}$ are sequences in $\Sigma$, then

\Centerline{$\lim_{n\to\Cal F}(E_n\cap F_n)
=(\lim_{n\to\Cal F}E_n)\cap(\lim_{n\to\Cal F}F_n)$,}

\Centerline{$\lim_{n\to\Cal F}(E_n\symmdiff F_n)
=(\lim_{n\to\Cal F}E_n)\symmdiff(\lim_{n\to\Cal F}F_n)$}

\noindent because $\Cal F$ is an ultrafilter.\ \QeD\   Of course
$\Sigma\subseteq\Cal A$, because if $E_n=E$ for every $n$ then
$\lim_{n\to\Cal F}E_n=E$.

\wheader{538I}{6}{2}{2}{72pt}

{\bf (b)} If $\sequencen{E_n}$ and $\sequencen{F_n}$ are sequences in
$\Sigma$ and $\lim_{n\to\Cal F}E_n=\lim_{n\to\Cal F}F_n$, then
$\lim_{n\to\Cal F}\mu E_n=\lim_{n\to\Cal F}\mu F_n$.   \Prf\

$$\eqalignno{|\lim_{n\to\Cal F}\mu E_n-\lim_{n\to\Cal F}\mu F_n|
&=\lim_{n\to\Cal F}|\mu E_n-\mu F_n|
\le\lim_{n\to\Cal F}\mu(E_n\symmdiff F_n)
\le\mu^*(\lim_{n\to\Cal F}E_n\symmdiff F_n)\cr
\displaycause{538G(iv)}
&=\mu^*(\lim_{n\to\Cal F}E_n\symmdiff\lim_{n\to\Cal F}F_n)
=\mu^*\emptyset
=0.  \text{ \Qed}\cr}$$

\medskip

{\bf (c)} We therefore have a functional $\phi:\Cal A\to[0,1]$ defined by
setting $\phi(\lim_{n\to\Cal F}E_n)=\lim_{n\to\Cal F}\mu E_n$ for every
sequence $\sequencen{E_n}$ in $\Sigma$.   Clearly $\phi$ extends $\mu$.
Also $\phi$ is additive.   \Prf\ If $\sequencen{E_n}$, $\sequencen{F_n}$
are sequences in $\Sigma$ such that $\lim_{n\to\Cal F}E_n$ and
$\lim_{n\to\Cal F}F_n$ are disjoint, then

$$\eqalign{\phi(\lim_{n\to\Cal F}E_n\cup\lim_{n\to\Cal F}F_n)
&=\phi(\lim_{n\to\Cal F}E_n\cup F_n)
=\lim_{n\to\Cal F}\mu(E_n\cup F_n)\cr
&=\lim_{n\to\Cal F}\mu E_n+\mu F_n-\mu(E_n\cap F_n)\cr
&=\lim_{n\to\Cal F}\mu E_n
  +\lim_{n\to\Cal F}\mu F_n
  -\lim_{n\to\Cal F}\mu(E_n\cap F_n)\cr
&=\phi(\lim_{n\to\Cal F}E_n)
  +\phi(\lim_{n\to\Cal F}F_n)
  -\phi(\lim_{n\to\Cal F}E_n\cap F_n)\cr
&=\phi(\lim_{n\to\Cal F}E_n)
  +\phi(\lim_{n\to\Cal F}F_n).  \text{ \Qed}\cr}$$

\medskip

{\bf (d)} Next, if
$\sequence{m}{A_m}$ is a non-increasing sequence in $\Cal A$, and
$0\le\gamma<\inf_{m\in\Bbb N}\phi A_m$,
there is an $A\in\Cal A$ such that
$A\subseteq\bigcap_{m\in\Bbb N}A_m$ and $\phi A\ge\gamma$.
\Prf\ We can suppose that $A_0=X$.   For each
$m\in\Bbb N$, let $\sequencen{E_{mn}}$ be a sequence in $\Sigma$ such that
$A_m=\lim_{n\to\Cal F}E_{mn}$, starting with $E_{0n}=X$ for every $n$.
For $m\in\Bbb N$, set $E'_{mn}=\bigcap_{i\le m}E_{in}$ for
$n\in\Bbb N$;  then

\Centerline{$A_m=\bigcap_{i\le m}A_i=\lim_{n\to\Cal F}E'_{mn}$;}

\noindent set $I_m=\{n:n\in\Bbb N$, $\mu E'_{mn}\ge\gamma\}$.  Since
$\lim_{n\to\Cal F}\mu E'_{mn}=\phi A_m>\gamma$, $I_m\in\Cal F$.
For $n\in\Bbb N$, set
$F_n=\bigcap\{E'_{mn}:m\in\Bbb N$, $\mu E'_{mn}\ge\gamma\}$;
set $A=\lim_{n\to\Cal F}F_n$.   Then $\mu F_n\ge\gamma$ for every $n$,
so $\phi A\ge\gamma$.   Also, for $m\in\Bbb N$,
$F_n\subseteq E'_{mn}$ whenever $n\in I_m$, so
$A\subseteq\lim_{n\to\Cal F}E'_{mn}=A_m$.\ \Qed

\medskip

{\bf (e)} In particular, $\inf_{m\in\Bbb N}\phi A_m$ must be $0$ whenever
$\sequence{m}{A_m}$ is a non-increasing sequence in $\Cal A$ with empty
intersection.   By 413K, there is a complete measure $\lambda$ on $X$
extending $\phi$ and inner regular with respect to $\Cal A_{\delta}$, the
set of intersections of sequences in $\Cal A$.   But
$\lambda C=\sup\{\lambda A:A\in\Cal A$, $A\subseteq C\}$ for every
$C\in\Cal A_{\delta}$.   \Prf\ Suppose that $0\le\gamma<\lambda C$.
There is a sequence $\sequence{m}{A_m}$ in
$\Cal A$ with intersection $C$;  because $\Cal A$ is an algebra of sets, we
can suppose that $\sequence{m}{A_m}$ is non-increasing.   Now

\Centerline{$\gamma<\lambda C=\inf_{m\in\Bbb N}\lambda A_m
=\inf_{m\in\Bbb N}\phi A_m$,}

\noindent so (d) tells us that there is an $A\in\Cal A$ such that
$A\subseteq C$ and $\gamma\le\phi A=\lambda A$.\ \QeD\  It follows at once
that $\lambda$ is inner regular with respect to $\Cal A$.

\medskip

{\bf (f)} If $E\in\Sigma$ and we set $E_n=E$ for every $n\in\Bbb N$, then
$E=\lim_{n\to\Cal F}E_n$ belongs to $\Cal A$ and

\Centerline{$\lambda E=\phi E=\lim_{n\to\Cal F}\mu E_n=\mu E$.}

\noindent So $\lambda$ extends $\mu$.   Finally, we see from 412Mb, as
usual, that $\lambda$ is uniquely defined.
}%end of proof of 538I

\medskip

\noindent{\bf Notation} In this context, I will call $\lambda$ the
{\bf $\Cal F$-extension} of $\mu$.

\leader{538J}{Proposition} Let $\Cal F$ be a measure-centering ultrafilter
on $\Bbb N$ and $(X,\Sigma,\mu)$ a
perfect probability space;  let $\lambda$ be the
$\Cal F$-extension of $\mu$ as defined in 538I.

(a) Let $(\frak A,\bar\mu)$ be the measure algebra of $\mu$,
$(\frak B,\bar\lambda)$ the measure algebra of $\lambda$, and
$(\frak C,\bar\nu)$ the probability algebra reduced power
$(\frak A,\bar\mu)^{\Bbb N}|\Cal F$
(328C).  Then we have a
measure-preserving isomorphism $\pi:\frak B\to\frak C$ defined by saying
that

\Centerline{$\pi((\lim_{n\to\Cal F}E_n)^{\ssbullet})
=\sequencen{E_n^{\ssbullet}}^{\ssbullet}$}

\noindent for every sequence $\sequencen{E_n}$ in $\Sigma$.

(b) Let $(X',\Sigma',\mu')$ be another
perfect probability space, and $\phi:X\to X'$ an \imp\ function.
Let $\lambda'$ be the $\Cal F$-extension of $\mu'$.
Then $\phi$ is \imp\ for $\lambda$ and $\lambda'$.

(c) Let $\Cal F'$ be a filter on $\Bbb N$ such that $\Cal F'\leRK\Cal F$,
and $\lambda'$ the
$\Cal F'$-extension of $\mu$.   Then $\lambda$ extends $\lambda'$.

\proof{{\bf (a)(i)} I had better check first that the formula for $\pi$
defines a function.   If $\sequencen{E_n}$, $\sequencen{F_n}$ are sequences
in $\Sigma$ such that
$(\lim_{n\to\Cal F}E_n)^{\ssbullet}=(\lim_{n\to\Cal F}F_n)^{\ssbullet}$ in
$\frak B$, then

$$\eqalign{0
&=\lambda(\lim_{n\to\Cal F}E_n\symmdiff\lim_{n\to\Cal F}F_n)
=\lim_{n\to\Cal F}\mu(E_n\symmdiff F_n)\cr
&=\lim_{n\to\Cal F}\bar\mu(E^{\ssbullet}_n\Bsymmdiff F_n^{\ssbullet})
=\bar\nu(\sequencen{E_n^{\ssbullet}}^{\ssbullet}
  \Bsymmdiff\sequencen{F_n^{\ssbullet}}^{\ssbullet}),\cr}$$

\noindent so $\sequencen{E_n^{\ssbullet}}^{\ssbullet}
=\sequencen{F_n^{\ssbullet}}^{\ssbullet}$ in $\frak C$.

\medskip

\quad{\bf (ii)} Setting $\frak B_0=\{E^{\ssbullet}:E\in\Cal A\}$, where
$\Cal A$ is as in 538I, it is now routine to check that
$\pi:\frak B_0\to\frak C$ is a surjective
measure-preserving Boolean homomorphism.   (Recall that $\frak C$ is, by
definition, the quotient of $\frak A^{\Bbb N}$ by the ideal
$\{\sequencen{a_n}:\lim_{n\to\Cal F}\bar\mu a_n=0\}$.)
But of course this means that
$\frak B_0$ is isomorphic to $\frak C$, therefore Dedekind complete.
Since $\lambda$ is inner regular with respect to $\Cal A$ (538I),
$\frak B_0$ is order-dense in $\frak B$, and must be the whole of
$\frak B$.

\wheader{538J}{6}{2}{2}{36pt}

{\bf (b)} Setting

\Centerline{$\Cal A
=\{\lim_{n\to\Cal F}E_n:E_n\in\Sigma\Forall n\in\Bbb N\}$,
\quad$\Cal A'
=\{\lim_{n\to\Cal F}F_n:F_n\in\Sigma'\Forall n\in\Bbb N\}$}

\noindent as in 538I, $\phi^{-1}[C]\in\Cal A$ and
$\lambda\phi^{-1}[C]=\lambda'C$ for every $C\in\Cal A'$.   \Prf\ Let
$\sequencen{F_n}$ be a sequence in $\Sigma'$ such that
$C=\lim_{n\to\Cal F}F_n$;  then

$$\eqalign{\lambda\phi^{-1}[C]
&=\lambda\phi^{-1}[\lim_{n\to\Cal F}F_n]
=\lambda(\lim_{n\to\Cal F}\phi^{-1}[F_n])\cr
&=\lim_{n\to\Cal F}\mu\phi^{-1}[F_n]
=\lim_{n\to\Cal F}\mu'F_n
=\lambda'C.  \text{ \Qed}\cr}$$

\noindent By 412K, $\phi$ is \imp\ for $\lambda$ and $\lambda'$.

\medskip

{\bf (c)} By 538Hb, $\Cal F'$ is measure-centering.
Let $f:\Bbb N\to\Bbb N$ be such that $\Cal F'=f[[\Cal F]]$.
Setting

\Centerline{$\Cal A
=\{\lim_{n\to\Cal F}E_n:E_n\in\Sigma\Forall n\in\Bbb N\}$,
\quad$\Cal A'
=\{\lim_{n\to\Cal F'}E_n:E_n\in\Sigma\Forall n\in\Bbb N\}$,}

\noindent $\Cal A'\subseteq\Cal A$ and $\lambda A=\lambda'A$ for every
$A\in\Cal A'$.   \Prf\ Let
$\sequencen{E_n}$ be a sequence in $\Sigma$ such that
$A=\lim_{n\to\Cal F'}E_n$;  then $A=\lim_{n\to\Cal F}E_{f(n)}$, so

\Centerline{$\lambda A=\lim_{n\to\Cal F}\mu E_{f(n)}
=\lim_{n\to\Cal F'}\mu E_n=\lambda'A$.  \Qed}

\noindent By 412K again, the identity map from $X$ to itself is
\imp\ for $\lambda$ and $\lambda'$, that is, $\lambda$ extends $\lambda'$.
}%end of proof of 538J

\leader{538K}{}\cmmnt{ Having identified the measure algebra of a
measure-centering-ultrafilter extension $\lambda$
as a probability algebra reduced
product (538Ja), we are in a position to apply the results of \S377.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a perfect probability
space, $\Cal F$ a measure-centering ultrafilter on $\Bbb N$ and $\lambda$
the $\Cal F$-extension of $\mu$\cmmnt{ as constructed in 538I}.

(a)(i) Let $\sequencen{f_n}$ be a sequence in $\eusm L^0(\mu)$ such that
$\{f_n^{\ssbullet}:n\in\Bbb N\}$ is bounded in the linear topological space
$L^0(\mu)$.   Then

\qquad($\alpha$) $f(x)=\lim_{n\to\Cal F}f_n(x)$ is defined in $\Bbb R$
for $\lambda$-almost every $x\in X$;

\qquad($\beta$) $f\in\eusm L^0(\lambda)$.

\quad(ii) For every $f\in\eusm L^0(\lambda)$ there is a sequence
$\sequencen{f_n}$ in $\eusm L^0(\mu)$, bounded in the sense of (i), such
that $f=\lim_{n\to\Cal F}f_n\,\,\lambda$-a.e.

(b) Suppose that $1\le p\le\infty$, and
$\sequencen{f_n}$ is a sequence in
$\eusm L^p(\mu)$ such that $\sup_{n\in\Bbb N}\|f_n\|_p$ is finite.
Set $f(x)=\lim_{n\to\Cal F}f_n(x)$ whenever this is defined in $\Bbb R$.

\quad(i)($\alpha$) $f\in\eusm L^p(\lambda)$;

\qquad($\beta$) $\|f\|_p\le\lim_{n\to\Cal F}\|f_n\|_p$.
%Fatou's Lemma

\quad(ii) Let $g$ be a conditional expectation of $f$ on $\Sigma$.

\qquad($\alpha$) If $p=1$ and $\{f_n:n\in\Bbb N\}$ is uniformly integrable, then
then $\|f\|_1=\lim_{n\to\Cal F}\|f_n\|_1$ and
$g^{\ssbullet}=\lim_{n\to\Cal F}f_n^{\ssbullet}$ for the weak topology
of $L^1(\mu)$.

\qquad($\beta$) If $1<p<\infty$, then
$g^{\ssbullet}=\lim_{n\to\Cal F}f_n^{\ssbullet}$ for the weak topology of
$L^p(\mu)$.

\woddheader{538K}{0}{0}{0}{36pt}

(c) Suppose that $1\le p\le\infty$ and $f\in\eusm L^p(\lambda)$.

\quad(i) There is a
sequence $\sequencen{f_n}$ in $\eusm L^p(\mu)$ such that
$f=\lim_{n\to\Cal F}f_n\,\,\lambda$-a.e.\
and $\|f\|_p=\sup_{n\in\Bbb N}\|f_n\|_p$.

\quad(ii) If $p=1$, we can arrange that $\sequencen{f_n}$ should
be uniformly integrable.

(d) Let $(X',\Sigma',\mu')$ be another perfect measure space, and
$\lambda'$ the $\Cal F$-extension of $\mu'$.
Let $S:L^1(\mu)\to L^1(\mu')$ be a bounded linear operator.

\allowmorestretch{468}{
\quad(i) There is a unique bounded linear operator
$\hat S:L^1(\lambda)\to L^1(\lambda')$ such that
$\hat Sf^{\ssbullet}=g^{\ssbullet}$
whenever $\sequencen{f_n}$, $\sequencen{g_n}$ are uniformly integrable
sequences in $\eusm L^1(\mu)$, $\eusm L^1(\nu)$ respectively,
$f=\lim_{n\to\Cal F}f_n\,\,\lambda$-a.e.,
$g=\lim_{n\to\Cal F}g_n\,\,\lambda'$-a.e.,
and $g_n^{\ssbullet}=Sf_n^{\ssbullet}$ for every $n\in\Bbb N$.
}

\quad(ii) The map
$S\mapsto\hat S:\eurm B(L^1(\mu);L^1(\mu'))
\to\eurm B(L^1(\lambda);L^1(\lambda'))$
is a norm-preserving Riesz homomorphism.

\proof{ We shall find that most of the work for this result has been done
in \S377.   The only new step is in (a)(i), but we shall have some checking
to do.

\medskip

{\bf (a)(i)} Let $\sequencen{\tilde f_n}$ be a sequence of
$\Sigma$-measurable functions from $X$ to $\Bbb R$ such that
$\tilde f_n=f_n\,\,\mu$-a.e.\ for every $n\in\Bbb N$.

\medskip

\qquad\grheada\ Let $\epsilon>0$.   Applying 367Rd\Latereditions\ to
$\{\tilde f_n^{\ssbullet}:n\in\Bbb N\}=\{f_n^{\ssbullet}:n\in\Bbb N\}$,
there is a $\gamma>0$ such that
$\mu E_n\le\epsilon$ for every
$n\in\Bbb N$, where $E_n=\{x:|\tilde f_n(x)|\ge\gamma\}$.
Set $E=\lim_{n\to\Cal F}E_n$,
so that $\lambda E\le\epsilon$.   For $x\in X\setminus E$,
$\{n:|\tilde f_n(x)|\le\gamma\}\in\Cal F$, so
$\lim_{n\to\Cal F}\tilde f_n(x)$ is defined in $\Bbb R$.   As $\epsilon$ is
arbitrary, $\lim_{n\to\Cal F}\tilde f_n(x)$ is defined in $\Bbb R$ for
$\lambda$-almost every $x$.   Since

\Centerline{$\{x:x\in\dom f_n$ and
$f_n(x)=\tilde f_n(x)$ for every $n\in\Bbb N\}$}

\noindent is $\mu$-conegligible, therefore $\lambda$-conegligible,
$\lim_{n\to\Cal F}f_n$ is defined in $\Bbb R\,\,\lambda$-a.e.

\medskip

\qquad\grheadb\ For any $\alpha\in\Bbb R$,

\Centerline{$\{x:\lim_{n\to\Cal F}\tilde f_n(x)>\alpha\}
=\bigcup_{k\in\Bbb N}\lim_{n\to\Cal F}\{x:f_n(x)\ge\alpha+2^{-k}\}
\in\dom\lambda$.}

\noindent So $f\eae\lim_{n\to\Cal F}\tilde f_n$ belongs to
$\eusm L^0(\lambda)$.

\medskip

\quad{\bf (ii)} At this point I seek to import the machinery of \S377.

\medskip

\qquad\grheada\
Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\lambda)$ be the measure algebras
of $\mu$, $\lambda$ respectively;  recall that we can identify
$L^0(\mu)$ and $L^0(\lambda)$ with $L^0(\frak A)$ and $L^0(\frak B)$
(364Ic).   Write
$(\frak C,\bar\nu)$ for the probability algebra reduced power
$(\frak A,\bar\mu)^{\Bbb N}|\Cal F$;  let
$\phi:\frak A^{\Bbb N}\to\frak C$ be the canonical surjection, and
$\pi:\frak B\to\frak C$ the isomorphism of 538Ja;  set
$\psi=\pi^{-1}\phi:\frak A^{\Bbb N}\to\frak B$.   Then
$\bar\lambda\psi(\sequencen{a_n})=\lim_{n\to\Cal F}\bar\mu a_n$ for every
sequence $\sequencen{a_n}$ in $\frak A$, and $\psi$ is surjective.

\medskip

\qquad\grheadb\
Let $W_0\subseteq L^0(\frak A)^{\Bbb N}$ be the set of sequences bounded
for the topology of convergence in measure, and
$\eusm W_0\subseteq\eusm L^0(\mu)^{\Bbb N}$ the set of sequences
$\sequencen{f_n}$ such that $\sequencen{f_n^{\ssbullet}}\in W_0$.
Then we have a Riesz
homomorphism $T:W^0\to L^0(\frak B)$ defined by saying that
$T(\sequencen{f_n^{\ssbullet}})=(\lim_{n\to\Cal F}f_n)^{\ssbullet}$
whenever $\sequencen{f_n}\in\eusm W_0$.   \Prf\ We know from (i) that
$(\lim_{n\to\Cal F}f_n)^{\ssbullet}$ is defined in
$L^0(\lambda)\cong L^0(\frak B)$ whenever $\sequencen{f_n}\in\eusm W_0$.
(I am taking the domain of $\lim_{n\to\Cal F}f_n$ to be
$\{x:\lim_{n\to\Cal F}f_n(x)$ is defined in $\Bbb R\}$.)   Since

\Centerline{$\lim_{n\to\Cal F}f_n\eae\lim_{n\to\Cal F}g_n$}

\noindent whenever $f_n\eae g_n$ for every $n$, $T$ is
well-defined.   Since

\Centerline{$\lim_{n\to\Cal F}f_n+g_n
\eae\lim_{n\to\Cal F}f_n+\lim_{n\to\Cal F}g_n$,}

\Centerline{$\lim_{n\to\Cal F}\alpha f_n
\eae\alpha\lim_{n\to\Cal F}f_n$,
\quad$\lim_{n\to\Cal F}|f_n|
\eae|\lim_{n\to\Cal F}f_n|$}

\noindent whenever $\sequencen{f_n}$, $\sequencen{g_n}\in\eusm W_0$ and
$\alpha\in\Bbb R$, $T$ is a Riesz homomorphism.\ \Qed

\medskip

\qquad\grheadc\ If $\sequencen{a_n}$ is any sequence in $\frak A$,
$T(\sequencen{\chi a_n})=\chi\psi(\sequencen{a_n})$.   \Prf\
Express each $a_n$ as $E_n^{\ssbullet}$, where $E_n\in\Sigma$, and set
$F=\lim_{n\to\Cal F}E_n$.   In the language of 538Ja,

\Centerline{$\psi(\sequencen{a_n})
=\pi^{-1}\phi(\sequencen{a_n})
=\pi^{-1}(\sequencen{a_n}^{\ssbullet})
=F^{\ssbullet}$,}

\noindent so

\Centerline{$T(\sequencen{\chi a_n})
=(\lim_{n\to\Cal F}\chi E_n)^{\ssbullet}
=(\chi F)^{\ssbullet}
=\chi(F^{\ssbullet})
=\chi\psi(\sequencen{a_n})$.  \Qed}

\medskip

\qquad\grheadd\ Recalling that $W_0$ is just the family of sequences
$\sequencen{u_n}$ in $L^0$ such that
$\inf_{k\in\Bbb N}\sup_{n\in\Bbb N}\bar\mu\Bvalue{|u_n|>k}=0$
(367Rd again),
($\gamma$) means that we can identify $T:W_0\to L^0(\frak B)$
with the Riesz homomorphism described in 377B.   By 377D(d-i),
$T[W_0]=L^0(\frak B)$, which is what we need to prove the immediate result
here.

\medskip

{\bf (b)(i)} As in part (a) of the proof of
377C, we see that a $\|\,\|_p$-bounded
sequence in $\eusm L^p(\mu)$ will belong to $\eusm W_0$.   So we can use
377Db.

\medskip

\quad{\bf (ii)} Use 377Ec.

\medskip

{\bf (c)} Use 377Dd.

\medskip

{\bf (d)} Use 377F.
}%end of proof of 538K

\leader{538L}{Theorem} Suppose that $\zeta$ is a non-zero
countable ordinal and
$\langle\Cal F_{\xi}\rangle_{1\le\xi\le\zeta}$ is a family of Ramsey
ultrafilters on $\Bbb N$, no two isomorphic.   Let
$\langle\Cal G_{\xi}\rangle_{\xi\le\zeta}$ be the corresponding iterated
product system, as described in 538E.   Then
$\Cal G_{\zeta}$ is a measure-centering ultrafilter.

\proof{{\bf (a)}
Define $\langle(\frak A_{\xi},\bar\mu_{\xi})\rangle_{\xi\le\zeta}$
inductively, as follows.
$(\frak A_0,\bar\mu_0)=(\frak B_{\omega},\bar\nu_{\omega})$ is to be
the measure
algebra of the usual measure on $\{0,1\}^{\Bbb N}$.   Given
$\ofamily{\eta}{\xi}{(\frak A_{\eta},\bar\mu_{\eta})}$, where
$0<\xi\le\zeta$,
let $(\frak A_{\xi},\bar\mu_{\xi})$ be the probability algebra reduced
product
$\prod_{k\in\Bbb N}(\frak A_{\theta(\xi,k)},\bar\mu_{\theta(\xi,k)})
|\Cal F_{\xi}$, as described in 328A-328C.  %328A 328B 328C
At the end of the induction, write
$(\frak C,\bar\nu)$ for $(\frak A_{\zeta},\bar\mu_{\zeta})$.

\medskip

{\bf (b)} We have a family
$\langle\phi_{\xi\eta}\rangle_{\eta\le\xi\le\zeta}$
defined by induction on $\xi$, as follows.
The inductive hypothesis will be that $\phi_{\eta'\eta}$ is a
measure-preserving Boolean homomorphism from $\frak A_{\eta}$ to
$\frak A_{\eta'}$, and that
$\phi_{\eta''\eta}=\phi_{\eta''\eta'}\phi_{\eta'\eta}$ whenever
$\eta\le\eta'\le\eta''<\xi$.   For the inductive step to $\xi$,
take $\phi_{\xi\xi}$ to be the identity map on $\frak A_{\xi}$.
If $\xi>0$, set $\tilde\phi_{kj}=\phi_{\theta(\xi,k),\theta(\xi,j)}$ for
$j\le k$ in $\Bbb N$;  then 328Ea
tells us that we have measure-preserving Boolean homomorphisms
$\tilde\phi_k:\frak A_{\theta(\xi,k)}\to\frak A_{\xi}$ such that
$\tilde\phi_j=\tilde\phi_k\tilde\phi_{kj}$ for $j\le k$.   If $j\le k$ and
$\eta\le\theta(\xi,j)$, then

\Centerline{$\tilde\phi_k\phi_{\theta(\xi,k),\eta}
=\tilde\phi_k\tilde\phi_{kj}\phi_{\theta(\xi,j),\eta}
=\tilde\phi_j\phi_{\theta(\xi,j),\eta}$}

\noindent whenever $k\ge j$;  so we can take this common value for
$\phi_{\xi\eta}$.   If $\eta\le\eta'<\xi$, then take $k$ such that
$\eta'\le\theta(\xi,k)$, and see that

\Centerline{$\phi_{\xi\eta'}\phi_{\eta'\eta}
=\tilde\phi_k\phi_{\theta(\xi,k),\eta'}\phi_{\eta'\eta}
=\tilde\phi_k\phi_{\theta(\xi,k),\eta}
=\phi_{\xi\eta}$,}

\noindent so the induction proceeds.

For each $\xi\le\zeta$,
write $\pi_{\xi}$ for $\phi_{\zeta\xi}:\frak A_{\xi}\to\frak C$, and
$\frak C_{\xi}$ for the subalgebra $\pi_{\xi}[\frak A_{\xi}]$.   Of
course $\pi_{\xi}\phi_{\xi\eta}=\pi_{\eta}$,
so that $\frak C_{\eta}\subseteq\frak C_{\xi}$,
whenever $\eta\le\xi\le\zeta$.

\medskip

{\bf (c)} For each $\xi>0$, we have a canonical map
$\sequence{k}{a_k}\to\sequence{k}{a_k}^{\ssbullet}:
\prod_{k\in\Bbb N}\frak A_{\theta(\xi,k)}\to\frak A_{\xi}$.   Since every
$\pi_{\xi}:\frak A_{\xi}\to\frak C_{\xi}$ is a measure-preserving
isomorphism, we have a corresponding map
$\psi_{\xi}:\prod_{k\in\Bbb N}\frak C_{\theta(\xi,k)}\to\frak C_{\xi}$.
Reading off the basic facts of 328Ab and 328Eb, we see that

\inset{----- $\bar\nu\psi_{\xi}(\sequence{k}{c_k})
=\lim_{k\to\Cal F_{\xi}}\bar\nu c_k$ for every sequence
$\sequence{k}{c_k}\in\prod_{k\in\Bbb N}\frak C_{\theta(\xi,k)}$,

----- $\psi_{\xi}(\sequence{k}{c_k})\Bsubseteq\sup_{k\in A}c_k$ whenever
$\sequence{k}{c_k}\in\prod_{k\in\Bbb N}\frak C_{\theta(\xi,k)}$ and
$A\in\Cal F_{\xi}$}

\noindent (we can take the supremum in $\frak C$ because $\frak C_{\xi}$ is
regularly embedded in $\frak C$, as noted in 314Ga).

\medskip

{\bf (d)} Let $\family{\tau}{S}{a_{\tau}}$ be a family in
$\frak A_0=\frak B_{\omega}$ such that
$\gamma=\inf_{\tau\in S}\bar\mu a_{\tau}$ is non-zero.
By 538Fe, we can find a
disjoint family $\langle A_{\xi}\rangle_{1\le\xi\le\zeta}$ of subsets of
$\Bbb N$ such that $A_{\xi}\in\Cal F_{\xi}$ for every $\xi$.
Use these to define $T\subseteq S$ and $\alpha:T\to[0,\zeta]$ as in
538Ee.   Set $c_{\tau}=0$ for $\tau\in S\setminus T$.
For $\tau\in T$ define $c_{\tau}\in\frak C_{\alpha(\tau)}$
by induction on $\alpha(\tau)$, as
follows.   If $\alpha(\tau)=0$, set $c_{\tau}=\pi_0a_{\tau}$.   For the
inductive step to $\alpha(\tau)=\xi>0$, we know that
$\tau^{\smallfrown}\fraction{k}\in T$ and
$\alpha(\tau^{\smallfrown}\fraction{k})=\theta(\xi,k)$ whenever
$k\in A_{\xi}$ and $\tau(i)<k$ for every $i<\dom\tau$;  for other
$k$, $\tau^{\smallfrown}\fraction{k}\notin T$ so
$c_{\tau^{\smallfrown}\fraction{k}}=0\in\frak C_{\theta(\xi,k)}$.
Thus $c_{\tau^{\smallfrown}\fraction{k}}\in\frak C_{\theta(\xi,k)}$ for
every $k$, and
$\psi_{\xi}(\sequence{k}{c_{\tau^{\smallfrown}\fraction{k}}})
\in\frak C_{\xi}$;  take this for $c_{\tau}$.   Note that

\Centerline{$\bar\nu c_{\tau}
=\lim_{k\to\Cal F_{\xi}}\bar\nu c_{\tau^{\smallfrown}\fraction{k}}
\ge\inf\{\bar\nu c_{\tau^{\smallfrown}\fraction{k}}:k\in\Bbb N$,
   $\tau^{\smallfrown}\fraction{k}\in T\}$.}

\noindent
Inducing on $\alpha(\tau)$, we see that $\bar\nu c_{\tau}\ge\gamma$ for
every $\tau\in T$.   In particular, $\bar\nu c_{\emptyset}\ge\gamma$.

\medskip

{\bf (e)} For $I\subseteq\Bbb N$, set $T_I=T\cap\bigcup_{n\in\Bbb N}I^n$
and $e_I=\inf_{\tau\in T_I}c_{\tau}$;  let $\Cal S$ be the family of those
finite subsets $I$ of $\Bbb N$ such that $e_I\ne\emptyset$.   Then
$T_{\emptyset}=\{\emptyset\}$, $e_{\emptyset}=c_{\emptyset}$ and
$\emptyset\in\Cal S$.   Moreover, if $I\in\Cal S$ and $1\le\xi\le\zeta$,
then $\{k:I\cup\{k\}\in\Cal S\}\in\Cal F_{\xi}$.   \Prf\ Set
$k_0=\sup I+1$.  If $k\in A_{\xi}$ and $k\ge k_0$, set

\Centerline{$d_k
=\inf\{c_{\tau^{\smallfrown}\fraction{k}}:
  \tau\in T_I$, $\alpha(\tau)=\xi\}$.}

\noindent Set $B=\{k:k\in A_{\xi},\,k\ge k_0,\,d_k\Bcap e_I\ne 0\}$.
If $k\in B$, then

\Centerline{$T_{I\cup\{k\}}
=T_I\cup\{\tau^{\smallfrown}\fraction{k}:\tau\in T_I$,
  $\alpha(\tau)=\xi\}$,}

\noindent because every member of $T$ is strictly increasing
and $\tau^{\smallfrown}\fraction{k}$ can belong to $T$ only when
$k\in A_{\alpha(\tau)}$, that is, $\alpha(\tau)=\xi$.   So
$e_{I\cup\{k\}}=d_k\Bcap e_I\ne 0$ and $I\cup\{k\}\in\Cal S$.

\Quer\ If $B\notin\Cal F_{\xi}$, then
$B'=\{k:k\in A_{\xi}$, $k\ge k_0$, $d_k\Bcap e_I=0\}$ belongs to
$\Cal F_{\xi}$.   So

$$\eqalignno{e_I
&\Bsubseteq\inf\{c_{\tau}:\tau\in T_I,\,\alpha(\tau)=\xi\}\cr
&=\inf_{\Atop{\tau\in T_I}{\alpha(\tau)=\xi}}
   \psi_{\xi}(\sequence{k}{c_{\tau^{\smallfrown}\fraction{k}}})
=\psi_{\xi}(\sequence{k}{\inf_{\Atop{\tau\in T_I}{\alpha(\tau)=\xi}}
   c_{\tau^{\smallfrown}\fraction{k}}})\cr
\displaycause{because $\psi_{\xi}$ is a Boolean homomorphism and $T_I$ is
finite}
&\Bsubseteq\sup_{k\in B'}\inf_{\Atop{\tau\in T_I}{\alpha(\tau)=\xi}}
   c_{\tau^{\smallfrown}\fraction{k}}\cr
\displaycause{by (c)}
&=\sup_{k\in B'}d_k.\cr}$$

\noindent But $e_I\Bcap d_k=0$ for every $k\in B'$ and $e_I\ne 0$.\ \Bang

Thus $\{k:I\cup\{k\}\in\Cal S\}\supseteq B\in\Cal F_{\xi}$.\ \Qed

\medskip

{\bf (f)} For $i\in\Bbb N$ set

\Centerline{$C_i=\{k:I\cup\{k\}\in\Cal S$ whenever $I\in\Cal S$ and
  $I\subseteq i\}$,}

\noindent so that $C_i\in\Cal F_{\xi}$ for every $\xi\in[1,\zeta]$.
At this point, recall that every $\Cal F_{\xi}$ is supposed to be a Ramsey
ultrafilter.   So for each $\xi\in[1,\zeta]$ we have an
$A'_{\xi}\in\Cal F_{\xi}$ such that $A'_{\xi}\subseteq A_{\xi}\cap C_0$ and
$j\in C_i$ whenever $i$, $j\in A'_{\xi}$ and $i<j$ (538Fc).
Next, for $i\in\Bbb N$ set
$M_i=\{\alpha(\tau):\tau\in T$, $\tau(j)\le i$ whenever $j<\dom\tau\}$;
then $M_i$ is finite, so there is a
$D\in\bigcap_{1\le\xi\le\zeta}\Cal F_{\xi}$ such that whenever
$i$, $j\in D$, $i<j$ and $\xi\in M_i$, there is a $k\in A'_{\xi}$ such that
$i<k<j$ (538Ff).   Of course we can suppose that
$D\subseteq\bigcup_{1\le\xi\le\zeta}A'_{\xi}$, so that
$D\cap A_{\xi}=D\cap A'_{\xi}$ for every $\xi$.

\medskip

{\bf (g)} $J\in\Cal S$ for every finite subset $J$ of $D$.   \Prf\ Induce
on $\#(J)$.   We know that $\emptyset\in\Cal S$.   If
$i\in D$, then $\{i\}\in\Cal S$ because $D\subseteq C_0$.   For the
inductive step to $\#(J)\ge 2$, set $j=\max J$, $I=J\setminus\{j\}$ and
$i=\max I$.   Then $I\in\Cal S$, by the inductive hypothesis;  so if
$T_J=T_I$, we certainly have $J\in\Cal S$.   Otherwise, there is a
member of $T_J\setminus T_I$, and this must be of the form
$\tau^{\smallfrown}\fraction{j}$ where $\tau\in T_I$ and
$j\in A_{\alpha(\tau)}$;  as $j\in D$, $j\in A'_{\alpha(\tau)}$.
But this means that $\alpha(\tau)\in M_i$ and
there is a $k\in A'_{\alpha(\tau)}$ such that $i<k<j$.   In this case,
$j\in C_k$ and $I\subseteq k$, so $J=I\cup\{k\}$ belongs to $\Cal S$, and
the induction proceeds.\ \Qed

\medskip

{\bf (h)} Thus $\{c_{\tau}:\tau\in T_D\}$ is centered;  setting
$T^*_D=\{\tau:\tau\in T_D$, $\alpha(\tau)=0\}$,
$\{c_{\tau}:\tau\in T^*_D\}$ and therefore
$\{a_{\tau}:\tau\in T^*_D\}$ are centered.   But $T^*_D$ belongs to
$\Cal G_{\zeta}$, by 538Ee.

Since $\family{\tau}{S}{a_{\tau}}$ was chosen arbitrarily in (d) above,
$\Cal G_{\zeta}$ satisfies the condition of 538G(ii), translated to the
countably infinite set $S$, and is measure-centering.
}%end of proof of 538L

\leader{538M}{Benedikt's theorem}\cmmnt{ ({\smc Benedikt 98})}
Let $(X,\Sigma,\mu)$
be a perfect probability space.   Then there is a measure
$\lambda$ on $X$, extending $\mu$, such that
$\lambda(\lim_{n\to\Cal F}E_n)$ is defined and equal to
$\lim_{n\to\Cal F}\mu E_n$ for every sequence $\sequencen{E_n}$ in
$\Sigma$ and every Ramsey filter $\Cal F$ on $\Bbb N$.

%what is the Maharam type of $\lambda$?

\proof{{\bf (a)} If there are no Ramsey filters, we can take $\lambda=\mu$ and
stop;  so let us suppose that there is at least one Ramsey filter.
Let $\frak F$ be a family of Ramsey filters consisting of just one member
of each isomorphism class, so that every Ramsey filter is isomorphic to
some member of $\frak F$, and no two members of $\frak F$ are isomorphic.
Fix a well-ordering $\preccurlyeq$ of $\frak F$ with a greatest member
$\Cal F^*$ and a family
$\langle\theta(\xi,k)\rangle_{1\le\xi<\omega_1,k\in\Bbb N}$ such that
$\sequence{k}{\theta(\xi,k)}$ is always a non-decreasing sequence of
ordinals less than $\xi$ such that $\{\theta(\xi,k):k\in\Bbb N\}$ is
cofinal with $\xi$.

\medskip

{\bf (b)(i)} For any
non-empty countable set $D\subseteq\frak F$ containing $\Cal F^*$,
enumerate it in $\preccurlyeq$-increasing order as
$\langle\Cal F_{\xi}\rangle_{1\le\xi\le\zeta}$, and let $\Cal G_D$ be the
measure-centering ultrafilter constructed from
$\langle\Cal F_{\xi}\rangle_{1\le\xi\le\zeta}$ and
$\langle\theta(\xi,k)\rangle_{1\le\xi\le\zeta,k\in\Bbb N}$ by the method of
538E-538L;  let $\lambda_D$ be the $\Cal G_D$-extension of $\mu$ as
defined in 538I.

\medskip

\quad{\bf (ii)} For any non-empty finite set $I\subseteq\frak F$,
list it in $\preccurlyeq$-increasing order as $\Cal F_0,\ldots,\Cal F_n$,
and set $\Cal H_I=\Cal F_n\ltimes\ldots\ltimes\Cal F_0$ as defined in 538D.
By 538Ed, or otherwise, $\Cal H_I\leRK\Cal G_I$, so $\Cal H_I$ is
measure-centering (538Hb);
let $\lambda'_I$ be the $\Cal H_I$-extension of $\mu$.

\medskip

{\bf (c)} If $\emptyset\ne I\subseteq J\in[\frak F]^{<\omega}$, then
$\Cal H_I\leRK\Cal H_J$, by 538Dg, and $\lambda'_J$ extends $\lambda'_I$,
by 538Jc.   Thus
$\langle\lambda'_I\rangle_{\emptyset\ne I\in[\frak F]^{<\omega}}$ is an
upwards-directed family of probability measures on $X$.

If $\Cal I\subseteq[\frak F]^{<\omega}\setminus\{\emptyset\}$ is
countable, we have a non-empty
countable set $D\subseteq\frak F$
including $\{\Cal F^*\}\cup\bigcup\Cal I$.   Now 538Ed
tells us that $\Cal H_I\leRK\Cal G_D$ for every $I\in\Cal I$, so that
$\lambda_D$ extends $\lambda'_I$ for every $I\in\Cal I$
(538Jc again).   Thus for every countable subset of
$\{\lambda'_I:I\in[\frak F]^{<\omega}\setminus\{\emptyset\}\}$ there is a
measure on $X$ extending them all.   By 457G, there is a measure $\lambda$
on $X$ extending every $\lambda'_I$.

\medskip

{\bf (d)} If $\Cal F$ is a Ramsey ultrafilter and $\sequencen{E_n}$ is a
sequence in $\Sigma$, there is an $\Cal F'\in\frak F$ such that
$\Cal F$ and $\Cal F'$ are isomorphic.   In particular,
$\Cal F\leRK\Cal F'$, so $\tilde\lambda_{\Cal F'}$ extends
$\tilde\lambda_{\Cal F}$, where
$\tilde\lambda_{\Cal F}$, $\tilde\lambda_{\Cal F'}$ are the
$\Cal F$-extension and $\Cal F'$-extension of
$\mu$.   But $\lambda$ extends
$\lambda'_{\{\Cal F'\}}=\tilde\lambda_{\Cal F'}$ and therefore extends
$\tilde\lambda_{\Cal F}$.   Accordingly $\lambda(\lim_{n\to\Cal F}E_n)$ is
defined and equal to
$\tilde\lambda_{\Cal F}(\lim_{n\to\Cal F}E_n)=\lim_{n\to\Cal F}\mu E_n$, as
required.
}%end of proof of 538M

\leader{538N}{Measure-converging filters:  Proposition} (a) Let $\Cal F$
be a free filter on $\Bbb N$.
Let $\nu_{\omega}$ be the usual measure on $\{0,1\}^{\Bbb N}$, and
$\Tau_{\omega}$ its domain.   Then the following are equiveridical:

\inset{(i) $\Cal F$ is measure-converging;

(ii) whenever $\sequencen{F_n}$ is a sequence in $\Tau_{\omega}$
and $\lim_{n\to\infty}\nu_{\omega}F_n=1$, then
$\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n$ is conegligible;

(iii) whenever $(X,\Sigma,\mu)$ is a measure space with locally determined
negligible sets\cmmnt{ (definition:  213I)}, and $\sequencen{f_n}$ is a
sequence in $\eusm L^0\cmmnt{\mskip5mu=\eusm L^0(\mu)}$
which converges in measure to
$f\in\eusm L^0$, then $\lim_{n\to\Cal F}f_n\eae f$;

(iv) whenever $\mu$ is a Radon measure on $\Cal P\Bbb N$ such that
$\lim_{n\to\infty}\mu E_n=1$, where $E_n=\{a:n\in a\subseteq\Bbb N\}$ for
each $n$, then $\mu\Cal F=1$.}

(b)\dvAnew{2012} Every measure-converging filter is free.

(c)\dvAnew{2012} Suppose that $\Cal F$ is a measure-converging filter.

\quad(i) If $\Cal G$ is a filter on $\Bbb N$ including $\Cal F$, then
$\Cal G$ is measure-converging.

\quad(ii) If $\Cal G$ is a filter on $\Bbb N$ and
$\Cal G\leRB\Cal F$\cmmnt{ (definition:  5A6Ic)}, then $\Cal G$ is
measure-converging.

(d)\cmmnt{ (M.Foreman)} Every rapid filter is measure-converging.

(e)\cmmnt{ (M.Talagrand)} If there is a measure-converging filter,
there is a measure-converging filter which is not rapid.

(f) Let $\Cal F$ be a measure-converging filter on $\Bbb N$ and
$\Cal G$ any filter on $\Bbb N$.   Then $\Cal G\ltimes\Cal F$ is
measure-converging.

(g) If $\frakmctbl=\frak d$, there is a rapid filter.

\proof{{\bf (a)(i)$\Rightarrow$(iii)} Suppose that $\Cal F$ is
measure-converging, and that $(X,\Sigma,\mu)$, $\sequencen{f_n}$ and $f$
are as in (iii).   Let $H\in\Sigma$ be a conegligible
set such that $H\subseteq\dom f\cap\dom f_n$ and $f\restr H$ and
$f_n\restr H$ are measurable for
every $n\in\Bbb N$.   Let $k\in\Bbb N$;  set
$H_k=\{x:x\in H$, $\limsup_{n\to\Cal F}|f_n(x)-f(x)|>2^{-k}\}$.   Then
$H_k\cap F$ is negligible whenever $F\in\Sigma$ and $\mu F<\infty$.   \Prf\
If $\mu F=0$ this is trivial.   Otherwise, let $\nu=\Bover1{\mu F}\mu_F$ be
the normalized subspace measure on $F$.   For each $n\in\Bbb N$, set
$F_n=\{x:x\in F\cap H$, $|f_n(x)-f(x)|\le 2^{-k}\}$.   Then

\Centerline{$\lim_{n\to\infty}\nu(F\setminus F_n)
\le\Bover{2^k}{\mu F}\lim_{n\to\infty}\int\min(|f_n-f|,\chi F)d\mu
=0$}

\noindent because $\sequencen{f_n}\to f$ in measure.   So
$\lim_{n\to\infty}\nu F_n=1$ and
$H'=\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n$ is $\nu$-conegligible.
But $H'\cap H_k=\emptyset$, so $\mu^*(H_k\cap F)=\nu^*(H_k\cap F)=0$.\ \Qed

Since $\mu$ has locally determined negligible sets, $H_k$ is negligible.
This is true for every $k\in\Bbb N$, so $H\setminus\bigcup_{k\in\Bbb N}H_k$
is conegligible;  and $\lim_{n\to\Cal F}f_n(x)=f(x)$ for every
$x\in H\setminus\bigcup_{k\in\Bbb N}H_k$, so $\lim_{n\to\Cal F}f_n=f$ a.e.,
as required.

\medskip

\quad{\bf (iii)$\Rightarrow$(iv)} Assuming (iii), let $\mu$ and
$\sequencen{E_n}$ be as in (iv).   Set
$f_n=\chi(\Cal P\Bbb N\setminus E_n)$ for each $n$;  then
$\lim_{n\to\infty}\int f_nd\mu=0$, so $\sequencen{f_n}\to 0$ in measure,
and $H=\{a:\lim_{n\to\Cal F}f_n(a)=0\}$ is conegligible.
But for any $a\in H$,

\Centerline{$a=\{n:a\in E_n\}=\{n:f_n(x)\le\Bover12\}$}

\noindent belongs to $\Cal F$, so $H\subseteq\Cal F$ and $\mu\Cal F=1$.

\medskip

\quad{\bf (iv)$\Rightarrow$(ii)} Assume (iv), and let $\sequencen{F_n}$ be
as in (ii).   Define $\phi:\{0,1\}^{\Bbb N}\to\Cal P\Bbb N$ by setting
$\phi(x)=\{n:x\in F_n\}$ for $x\in\{0,1\}^{\Bbb N}$.   Then $\phi$ is
almost continuous (418J), so the image measure
$\mu=\nu_{\omega}\phi^{-1}$ on $\Cal P\Bbb N$ is a Radon measure (418I).
Since $F_n=\phi^{-1}[E_n]$ for each $n$, $\lim_{n\to\infty}\mu E_n=1$ and
$1=\mu\Cal F=\nu_{\omega}\phi^{-1}[\Cal F]$.   But now

\Centerline{$\bigcup_{a\in\Cal F}\bigcap_{n\in a}F_n
=\bigcup_{a\in\Cal F}\{x:a\subseteq\phi(x)\}
=\phi^{-1}[\Cal F]$}

\noindent is $\nu_{\omega}$-conegligible, as required.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} Assume (ii), and take a probability space
$(X,\Sigma,\mu)$ and a sequence $\sequencen{H_n}$ in $\Sigma$ such that
$\lim_{n\to\infty}\mu H_n\penalty-50=1$;  set
$G=\bigcup_{A\in\Cal F}\bigcap_{n\in A}H_n$.

Let $\lambda$ be the c.l.d. product
measure on $X\times\{0,1\}^{\Bbb N}$, and set

\Centerline{$W_n=H_n\times\{0,1\}^{\Bbb N}$,
\quad$V_n=\{(x,y):x\in X$, $y\in\{0,1\}^{\Bbb N}$, $y(n)=1\}$}

\noindent for $n\in\Bbb N$.   Let $\Lambda_1$ be the $\sigma$-algebra of
subsets of $X\times\{0,1\}^{\Bbb N}$ generated by
$\{W_n:n\in\Bbb N\}\cup\{V_n:n\in\Bbb N\}$, and $\lambda_1$ the completion
of the restriction $\lambda\restr\Lambda_1$.   Note that as the identity
map from $X\times\{0,1\}^{\Bbb N}$ is \imp\ for $\lambda$ and
$\lambda\restr\Lambda_1$, it is \imp\ for their completions
(234Ba);  but
$\lambda$ is complete, so this just means that $\lambda$ extends
$\lambda_1$.   Then $\lambda_1$ is a complete
atomless probability measure with countable Maharam type.   Its measure
algebra $\frak C$ is therefore isomorphic, as measure algebra, to the
measure algebra $\frak B_{\omega}$ of $\nu_{\omega}$;  let
$\pi:\frak B_{\omega}\to\frak C$ be a measure-preserving isomorphism.
By 343B, or otherwise, there is a realization
$\phi:X\times\{0,1\}^{\Bbb N}\to\{0,1\}^{\Bbb N}$,
\imp\ for $\lambda_1$ and $\nu_{\omega}$, such that
$\phi^{-1}[F]^{\ssbullet}=\pi F^{\ssbullet}$ in $\frak C$ for every
$F\in\Tau_{\omega}$.   Because $\pi$ is surjective, there is for each
$n\in\Bbb N$ an $F_n\in\Tau_{\omega}$ such that
$\phi^{-1}[F_n]\symmdiff W_n$ is $\lambda_1$-negligible.

Since

\Centerline{$\lim_{n\to\infty}\nu_{\omega}F_n
=\lim_{n\to\infty}\lambda_1W_n
=\lim_{n\to\infty}\lambda W_n
=\lim_{n\to\infty}\mu H_n=1$,}

\noindent $F=\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n$ is
$\nu_{\omega}$-conegligible, and
$\phi^{-1}[F]=\bigcup_{A\in\Cal F}\bigcap_{n\in A}\phi^{-1}[F_n]$ is
$\lambda_1$-conegligible.   We have
$G\times\{0,1\}^{\Bbb N}=\bigcup_{A\in\Cal F}\bigcap_{n\in A}W_n$,
so

\Centerline{$(G\times\{0,1\}^{\Bbb N})\symmdiff\phi^{-1}[F]
\subseteq\bigcup_{n\in\Bbb N}W_n\symmdiff\phi^{-1}[F_n]$}

\noindent is $\lambda_1$-negligible.
Thus $G\times\{0,1\}^{\Bbb N}$ is $\lambda_1$-conegligible, therefore
$\lambda$-conegligible.   But this means that $G$ is $\mu$-conegligible, by
252D applied to $G\times\{0,1\}^{\Bbb N}$;  and this is what we
needed to know.

\medskip

{\bf (b)} Let $\Cal F$ be a measure-converging filter and $m\in\Bbb N$.
Take a singleton set $X=\{x\}$ and the probability measure $\mu$ on $X$;
set $E_i=\emptyset$ for $i<n$, $X$ for $i\ge n$.   Then
$\lim_{i\to\infty}\mu E_i=1$, so there is an $A\in\Cal F$ such that
$\bigcap_{i\in A}E_i$ is non-empty.   Now $\Bbb N\setminus n\supseteq A$
belongs to $\Cal F$;  as $n$ is arbitrary, $\Cal F$ is free.

\medskip

{\bf (c)(i)} Immediate from the definition in 538Ag.

\medskip

\quad{\bf (ii)} Let $f:\Bbb N\to\Bbb N$ be a finite-to-one function such
that $\Cal G=f[[\Cal F]]$.
Let $(X,\Sigma,\mu)$ be a probability space and
$\sequencen{E_n}$ a sequence in $\Sigma$ such that
$\lim_{n\to\infty}\mu E_n=1$.   Set $F_n=E_{f(n)}$ for $n\in\Bbb N$;
because $f$ is finite-to-one, $\lim_{n\to\infty}\mu F_n=1$.   So
$H=\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n$ is conegligible.   If
$x\in H$, set $A_x=\{n:x\in E_n\}$;  then

\Centerline{$f^{-1}[A_x]=\{n:x\in E_{f(n)}\}=\{n:x\in F_n\}$}

\noindent belongs to $\Cal F$ so $A_x\in f[[\Cal F]]$ and
$x\in\bigcup_{B\in f[[\Cal F]]}\bigcap_{n\in B}E_n$.   Thus
$\bigcup_{B\in f[[\Cal F]]}\bigcap_{n\in B}E_n\supseteq H$ is conegligible.
As $(X,\Sigma,\mu)$ and $\sequencen{E_n}$ are arbitrary, $f[[\Cal F]]$ is
measure-converging.

\medskip

{\bf (d)} Let $\Cal F$ be a rapid filter on $\Bbb N$,
and $\sequencen{H_n}$ a sequence in $\Tau_{\omega}$ such that
$\lim_{n\to\infty}\nu_{\omega}H_n=1$.   Set
$G=\bigcup_{A\in\Cal F}\bigcap_{n\in A}H_n$.   Since
$\lim_{n\to\infty}(1-\nu_{\omega}H_n)=0$, there is an $A\in\Cal F$ such
that $\sum_{n\in A}1-\nu_{\omega}H_n<\infty$;   set
$H=\bigcup_{m\in\Bbb N}\bigcap_{n\in A\setminus m}H_n\subseteq G$.
Then

\Centerline{$\nu_{\omega}H
\ge\sup_{m\in\Bbb N}1-\sum_{n\in A\setminus m}(1-\nu_{\omega}H_n)
=1$,}

\noindent so $G$ is conegligible.   Thus $\Cal F$ satisfies (a-ii) and is
measure-converging.

\medskip

{\bf (e)} Let $\Cal F$ be a measure-converging filter.
Let $\sequencen{I_n}$ be a
sequence of non-empty finite subsets of $\Bbb N$ such that
$\lim_{n\to\infty}\#(I_n)=\infty$.   Let $\Cal G$ be

\Centerline{$\{A:A\subseteq\Bbb N$,
$\lim_{n\to\Cal F}\Bover1{\#(I_n)}\#(A\cap I_n)=1\}$.}

\noindent Then $\Cal G$ is a filter on $\Bbb N$.

\medskip

\quad{\bf (i)} $\Cal G$ is measure-converging.  \Prf\ Let
$\sequence{i}{H_i}$ be a sequence in $\Tau_{\omega}$ such that
$\lim_{i\to\infty}\nu_{\omega}H_i=1$, and set
$G=\bigcup_{A\in\Cal G}\bigcap_{i\in A}H_i$.   Set
$g_n=\Bover1{\#(I_n)}\sum_{i\in I_n}\chi H_i$ for each $n$;  then

\Centerline{$\lim_{n\to\infty}\int g_n
=\lim_{n\to\infty}\Bover1{\#(I_n)}\sum_{i\in I_n}\nu_{\omega}H_i
=1$}

\noindent because $\lim_{n\to\infty}\#(I_n)=\infty$ and
$\lim_{i\to\infty}\nu_{\omega}H_i=1$.   Since
$0\le g_n\le\chi\{0,1\}^{\Bbb N}$ for every $n$,
$\sequencen{g_n}\to\chi\{0,1\}^{\Bbb N}$ in measure.   By (a-iii) above,
$H=\{x:\lim_{n\to\Cal F}g_n(x)=1\}$ is conegligible.

For $x\in H$, set $A_x=\{i:x\in H_i\}$.   Then

\Centerline{$\Bover1{\#(I_n)}\#(I_n\cap A_x)
=g_n(x)\to 1$}

\noindent as $n\to\Cal F$, so $A_x\in\Cal G$ and $x\in G$.   Accordingly
$G\supseteq H$ is conegligible.   As $\sequence{i}{H_i}$ is arbitrary,
$\Cal G$ is measure-converging.\ \Qed

\medskip

\quad{\bf (ii)} $\Cal G$ is not rapid.   \Prf\
Define $\sequence{i}{t_i}$ by saying that

\Centerline{$t_i=\sup\{\Bover1{\#(I_n)}:n\in\Bbb N$, $i\in I_n\}$}

\noindent for $i\in\Bbb N$, counting $\sup\emptyset$ as $0$.
Then $\lim_{i\to\infty}t_i=0$.   If $A\in\Cal G$ and $m\in\Bbb N$, then
$B=\{n:\#(A\cap I_n)\ge\bover23\#(I_n)\}$
belongs to $\Cal F$, and must be infinite, by (b) above.
So there is an $n\in B$ such that $\#(I_n)\ge 3m$, and now

\Centerline{$\sum_{i\in A\setminus m}t_i
\ge\#(A\cap I_n\setminus m)\cdot\Bover1{\#(I_n)}\ge\Bover13$.}

\noindent As $m$ is arbitrary, $\sum_{i\in A}t_i=\infty$;  as $A$ is
arbitrary, $\Cal G$ is not rapid.\ \Qed

\medskip

{\bf (f)} Let
$\langle E_{ij}\rangle_{i,j\in\Bbb N}$ be a family in $\Tau_{\omega}$
such that $\sequence{n}{\nu_{\omega}E_{i_nj_n}}\to 1$ for some, or
any, enumeration $\sequencen{(i_n,j_n)}$ of $\Bbb N\times\Bbb N$.
Set $G=\bigcup_{C\in\Cal G\ltimes\Cal F}\bigcap_{(i,j)\in C}E_{ij}$.
For each $i\in\Bbb N$, $\lim_{j\to\infty}\nu_{\omega}E_{ij}=1$, so
$G_i=\bigcup_{A\in\Cal F}\bigcap_{j\in A}E_{ij}$ is conegligible;  set
$H=\bigcap_{i\in\Bbb N}G_i$.   For $x\in H$, set
$A_x=\{(i,j):x\in E_{ij}\}$.   As $x\in G_i$, $A_x[\{i\}]\in\Cal F$ for
every $i\in\Bbb N$;  thus $A_x\in\Cal G\ltimes\Cal F$ and $x\in G$.
So $G$ includes the conegligible set $H$, and is itself conegligible.
As $\langle E_{ij}\rangle_{i,j\in\Bbb N}$ is arbitrary, $\Cal G$ is
measure-converging.

\medskip

{\bf (g)(i)} Suppose that $\Cal E\subseteq[\Bbb N]^{\omega}$ is a
family with $\#(\Cal E)<\frakmctbl$, and that $f\in\NN$
is non-decreasing.   Then there
is an $A\subseteq\Bbb N$, meeting every member of $\Cal E$,
such that $\#(A\cap f(n))\le n$ for every $n\in\Bbb N$.   \Prf\
Consider
$X=\prod_{n\in\Bbb N}\Bbb N\setminus f(n)$.   Then $X$ is a closed subset
of $\NN$, homeomorphic to $\NN$.   For $E\in\Cal E$, set

\Centerline{$G_E=\{x:x\in X$, $E\cap x[\Bbb N]\ne\emptyset\}$;}

\noindent then $G_E$ is a dense open subset of $X$.   Writing $\Cal M(X)$
for the ideal of meager subsets of $X$,
$\#(\Cal E)<\frakmctbl=\cov\Cal M(X)$, so there is an
$x\in X\cap\bigcap_{E\in\Cal E}G_E$;  set $A=x[\Bbb N]$.\ \Qed

\medskip

\quad{\bf (ii)} Let $\ofamily{\xi}{\frak d}{f_{\xi}}$ be a cofinal family
in $\BbbN^{\Bbb N}$;  we may suppose that every $f_{\xi}$ is
strictly increasing.    Choose a non-decreasing family
$\langle\Cal E_{\xi}\rangle_{\xi\le\frak d}$
inductively, as follows.
$\Cal E_0=\{\Bbb N\setminus n:n\in\Bbb N\}$.   Given that
$\xi<\frak d=\frakmctbl$ and that
$\Cal E_{\xi}\subseteq[\Bbb N]^{<\omega}$ is a filter base with cardinal at
most $\max(\omega,\#(\xi))$, use (i) to find a set
$A_{\xi}\subseteq\Bbb N$, meeting every member of $\Cal E_{\xi}$, such that
$\#(A_{\xi}\cap f_{\xi}(n))\le n$ for every $n$;  set

\Centerline{$\Cal E_{\xi+1}
=\Cal E_{\xi}\cup\{A_{\xi}\cap E:E\in\Cal E_{\xi}\}$.}

\noindent For non-zero limit ordinals $\xi\le\frak d$ set
$\Cal E_{\xi}=\bigcup_{\eta<\xi}\Cal E_{\eta}$.

At the end of the induction, let $\Cal F$ be the filter on $\Bbb N$
generated by $\Cal E_{\frak d}$.   Then $\Cal F$ is rapid.   \Prf\ It is
free because $\Cal E_0\subseteq\Cal F$.   If $\sequencen{t_n}$ is a
sequence in $\Bbb R$ converging to $0$, let $f\in\NN$ be such that
$|t_i|\le 2^{-n}$ whenever $n\in\Bbb N$ and $i\ge f(n)$, and let
$\xi<\frak d$ be such that $f\le f_{\xi}$.   Then $A_{\xi}\in\Cal F$ and

\Centerline{$\sum_{i\in A_{\xi}}|t_i|
\le\sum_{n=0}^{\infty}
   2^{-n}\#(A_{\xi}\cap f_{\xi}(n+1)\setminus f_{\xi}(n))
\le\sum_{n=0}^{\infty}2^{-n}(n+1)$}

\noindent is finite.\ \Qed
}%end of proof of 538N

\leader{538O}{The Fatou property:  Proposition} (a)
Let $\Cal F$ be a filter on $\Bbb N$.   Let $\nu_{\omega}$ be the usual
measure on $\{0,1\}^{\Bbb N}$, and $\Tau_{\omega}$ its domain.   Then the
following are equiveridical:

\inset{(i) $\Cal F$ has the Fatou property;

(ii) whenever $\sequencen{F_n}$ is a sequence in $\Tau_{\omega}$ and
$\nu_{\omega}^*(\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n)=1$, then
$\lim_{n\to\Cal F}\nu_{\omega}F_n=1$;

(iii) whenever $(X,\Sigma,\mu)$ is a measure space and
$\sequencen{f_n}$ is a sequence of non-negative functions in
$\eusm L^0(\mu)$, then $\overline{\int}\liminf_{n\to\Cal F}f_nd\mu
\le\liminf_{n\to\Cal F}\int f_nd\mu$;

(iv) whenever $\mu$ is a Radon
probability measure on $\Cal P\Bbb N$, then
$\mu^*\Cal F\le\liminf_{n\to\Cal F}\mu E_n$, where
$E_n=\{a:n\in a\subseteq\Bbb N\}$ for each
$n\in\Bbb N$.}

(b) If $\Cal F$ and $\Cal G$ are filters on $\Bbb N$, $\Cal G\leRK\Cal F$
and $\Cal F$ has the Fatou property, then $\Cal G$ has the Fatou property.

(c) If $\Cal F$ and $\Cal G$ are filters on $\Bbb N$ with
the Fatou property, then $\Cal F\ltimes\Cal G$ has the Fatou property.

\medskip

\proof{{\bf (a) not-(iii)$\Rightarrow$not-(i)}
Suppose that $(X,\Sigma,\mu)$ is a measure space and
$\sequencen{f_n}$ a sequence of non-negative functions
in $\eusm L^0$ such that
$\overline{\int}\liminf_{n\to\Cal F}f_nd\mu
>\liminf_{n\to\Cal F}\int f_nd\mu$.   Changing the $f_n$ on negligible sets
does not change either $\overline{\int}\liminf_{n\to\Cal F}f_nd\mu$ or
$\overline{\int}\liminf_{n\to\Cal F}f_nd\mu$, so we may
assume that every $f_n$ is defined everywhere in $X$ and is
$\Sigma$-measurable.   Take $\alpha$ such that
$\overline{\int}\liminf_{n\to\Cal F}f_nd\mu
>\alpha>\liminf_{n\to\Cal F}\int f_nd\mu$;  set
$A=\{n:\int f_nd\mu\le\alpha\}$;  then $A$ meets every member of
$\Cal F$.   Since $f_n$ is integrable for every $n\in A$,
the set $G=\{x:\sup_{n\in A}f_n(x)>0\}$ is a countable
union of sets of finite measure.

Let $\lambda$ be the c.l.d.\ product measure on $G\times\Bbb R$, and
consider the ordinate sets
$W_n=\{(x,\beta):x\in G$, $0\le\beta<f_n(x)\}$ for $n\in A$.   Set
$W=\bigcup_{C\in\Cal F}\bigcap_{n\in C\cap A}W_n$;
writing $g$ for $\liminf_{n\to\Cal F}f_n$,

\Centerline{$\{(x,\beta):x\in G,\,0\le\beta<g(x)\}
\subseteq W$.}

\noindent Since $\lambda$ is a product of two $\sigma$-finite measures it
is $\sigma$-finite, and $W$ has a measurable envelope $\tilde W$ say.
Now $\lambda^*W>\alpha$.   \Prf\Quer\ Otherwise,
$\lambda\tilde W\le\alpha$.   Writing $\mu_L$ for Lebesgue measure on
$\Bbb R$,

$$\eqalignno{\alpha
\ge\lambda\tilde W
&=\int_G\mu_L\tilde W[\{x\}]\mu(dx)\cr
\displaycause{252D}
&\ge\overline{\int}_Gg\,d\mu
>\alpha.  \text{ \Bang\Qed}\cr}$$

There is therefore a set $V\subseteq\tilde W$ such that
$\alpha<\lambda V<\infty$, and now $\lambda^*(V\cap W)>\alpha$.
Let $\nu$ be the subspace measure on $V\cap W$.   Set

$$\eqalign{V_n
&=V\cap W\cap W_n\text{ if }n\in A,\cr
&=V\cap W\text{ if }n\in\Bbb N\setminus A.\cr}$$

\noindent Then

$$\eqalign{\liminf_{n\to\Cal F}\nu V_n
&=\sup_{C\in\Cal F}\inf_{n\in C}\nu V_n
\le\sup_{n\in A}\nu V_n\cr
&\le\sup_{n\in A}\lambda W_n
=\sup_{n\in A}\int f_nd\mu
\le\alpha.\cr}$$

\noindent On the other hand,

\Centerline{$\bigcup_{C\in\Cal F}\bigcap_{n\in C}V_n
=\bigcup_{C\in\Cal F}\bigcap_{n\in C\cap A}V\cap W\cap W_n
=V\cap W$}

\noindent and $\nu(V\cap W)=\lambda^*(V\cap W)>\alpha$.
Moving to a normalization of $\nu$, we see that (i) is false.

\medskip

\quad{\bf (iii)$\Rightarrow$(iv)} If $\Cal F$ satisfies (iii) and
$\mu$ is a Radon probability measure on $\Cal P\Bbb N$, set
$g=\liminf_{n\to\Cal F}\chi E_n$.   If $a\in\Cal F$, then
$\{n:\chi E_n(a)=1\}=a\in\Cal F$, so $g(a)=1$;  thus

$$\eqalignno{\mu^*\Cal F
&=\overline{\int}\chi\Cal Fd\mu\cr
\displaycause{133Je}
&\le\overline{\int}g\,d\mu
\le\liminf_{n\to\Cal F}\int\chi E_n
=\liminf_{n\to\Cal F}\mu E_n,\cr}$$

\noindent as required.

\medskip

\quad{\bf (iv)$\Rightarrow$(ii)} Given (iv),
suppose that $\sequencen{F_n}$ is a sequence in $\Tau_{\omega}$
and $\nu_{\omega}^*(\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n)=1$.
As in the corresponding part of the argument for 538Na, define
$\phi:\{0,1\}^{\Bbb N}\to\Cal P\Bbb N$ by setting
$\phi(x)=\{n:x\in F_n\}$, and let $\mu$ be the Radon measure
$\nu_{\omega}\phi^{-1}$.   Then

\Centerline{$\mu^*\Cal F=\nu_{\omega}^*\phi^{-1}[\Cal F]
=\nu_{\omega}^*(\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n)=1$}

\noindent (using 451Pc again for the first equality),
so $\lim_{n\to\Cal F}\nu_{\omega}F_n=\lim_{n\to\Cal F}\mu E_n=1$.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)}
Assume (ii), and take a probability space
$(X,\Sigma,\mu)$ and a sequence $\sequencen{H_n}$ in $\Sigma$ such that
$X=\bigcup_{A\in\Cal F}\bigcap_{n\in A}H_n$.

As in the corresponding part of the argument for 538Na,
let $\lambda$ be the c.l.d. product
measure on $X\times\{0,1\}^{\Bbb N}$, and set

\Centerline{$W_n=H_n\times\{0,1\}^{\Bbb N}$,
\quad$V_n=\{(x,y):x\in X$, $y\in\{0,1\}^{\Bbb N}$, $y(n)=1\}$}

\noindent for $n\in\Bbb N$.   Let $\Lambda_1$ be the $\sigma$-algebra of
subsets of $X\times\{0,1\}^{\Bbb N}$ generated by
$\{W_n:n\in\Bbb N\}\cup\{V_n:n\in\Bbb N\}$, and $\lambda_1$ the completion
of the restriction $\lambda\restr\Lambda_1$.   As before, there is a
function
$\phi:X\times\{0,1\}^{\Bbb N}\to\{0,1\}^{\Bbb N}$, \imp\ for
$\lambda_1$ and $\nu_{\omega}$, such that there is for each
$n\in\Bbb N$ an $F_n\in\Tau_{\omega}$ such that
$\phi^{-1}[F_n]\symmdiff W_n$ is $\lambda_1$-negligible.
Set $G=\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n$.

Since $X=\bigcup_{A\in\Cal F}\bigcap_{n\in A}H_n$,
$X\times\{0,1\}^{\Bbb N}=\bigcup_{A\in\Cal F}\bigcap_{n\in A}W_n$ and

\Centerline{$(X\times\{0,1\}^{\Bbb N})\setminus\phi^{-1}[G]
\subseteq\bigcup_{n\in\Bbb N}\phi^{-1}[F_n]\symmdiff W_n$}

\noindent is $\lambda_1$-negligible.   By 413Eh,

\Centerline{$\nu^*_{\omega}G\ge\lambda_1\phi^{-1}[G]=1$.}

\noindent By (ii), $\lim_{n\to\Cal F}\nu_{\omega}F_n=1$.   But

\Centerline{$\nu_{\omega}F_n
=\lambda_1\phi^{-1}[F_n]=\lambda_1W_n=\lambda W_n=\mu H_n$}

\noindent for each $n$, so $\lim_{n\to\Cal F}\mu H_n=1$.
As $(X,\Sigma,\mu)$ and $\sequencen{H_n}$ are arbitrary, $\Cal F$ has the
Fatou property.

\medskip

{\bf (b)} Let $h:\Bbb N\to\Bbb N$ be such that $\Cal G=h[[\Cal F]]$.
Let $(X,\Sigma,\mu)$ be a probability space and $\sequencen{H_n}$ a
sequence in $\Sigma$ such that

$$\eqalign{X
&=\bigcup_{A\in\Cal G}\bigcap_{n\in A}H_n\cr
&=\bigcup_{A\subseteq\Bbb N,h^{-1}[A]\in\Cal F}\bigcap_{n\in A}H_n
=\bigcup_{A\in\Cal F}\bigcap_{n\in A}H_{h(n)}.\cr}$$

\noindent Then

$$\eqalign{1
&=\liminf_{n\to\Cal F}\mu H_{h(n)}
=\sup_{A\in\Cal F}\inf_{n\in A}\mu H_{h(n)}\cr
&\le\sup_{A\in\Cal G}\inf_{n\in A}\mu H_n
=\liminf_{n\to\Cal G}\mu H_n.\cr}$$

\noindent As $(X,\Sigma,\mu)$ and $\sequencen{H_n}$ are arbitrary,
$\Cal G$ has the Fatou property.

\medskip

{\bf (c)} Let $(X,\Sigma,\mu)$ be a probability space and
$\langle E_{ij}\rangle_{i,j\in\Bbb N}$ a family in $\Sigma$ such that
$X=\bigcup_{C\in\Cal F\ltimes\Cal G}\bigcap_{(i,j)\in C}E_{ij}$.
For each $i\in\Bbb N$, set
$F_i=\bigcup_{B\in\Cal G}\bigcap_{j\in B}E_{ij}$, and let $G_i\in\Sigma$ be
a measurable envelope of $F_i$.   Then
$\bigcup_{A\in\Cal F}\bigcap_{i\in A}G_i=X$.   \Prf\ If $x\in X$, there is
a $C\in\Cal F\ltimes\Cal G$ such that $x\in E_{ij}$ whenever
$(i,j)\in C$.   Set $A=\{i:C[\{i\}]\in\Cal G\}\in\Cal F$.
If $i\in A$, then

\Centerline{$x\in\bigcap_{j\in C[\{i\}]}E_{ij}\subseteq F_i
\subseteq G_i$,}

\noindent so $x\in\bigcap_{i\in A}G_i$.\ \Qed

Accordingly $\lim_{i\to\Cal F}\mu G_i=1$.   Take $\epsilon>0$;  then
$A=\{i:\mu G_i\ge 1-\epsilon\}$ belongs to $\Cal F$.   For each
$i\in A$,

$$\eqalignno{1-\epsilon
&\le\mu G_i
=\mu^*F_i
=\overline{\int}\chi F_i
=\overline{\int}\liminf_{j\to\Cal G}\chi E_{ij}
\le\liminf_{j\to\Cal G}\int\chi E_{ij}\cr
\displaycause{by (a-iii) above}
&=\liminf_{j\to\Cal G}\mu E_{ij},\cr}$$

\noindent so $\{j:\mu E_{ij}\ge 1-2\epsilon\}\in\Cal G$.   But this means
that $\{(i,j):\mu E_{ij}\ge 1-2\epsilon\}\in\Cal F\ltimes\Cal G$.
As $\epsilon$ is arbitrary,
$\lim_{(i,j)\to\Cal F\ltimes\Cal G}\mu E_{ij}=1$.   As $(X,\Sigma,\mu)$ and
$\langle E_{ij}\rangle_{i,j\in\Bbb N}$ are arbitrary, $\Cal F\ltimes\Cal G$
has the Fatou property.
}%end of proof of 538O

\leader{538P}{Theorem} Let $\nu:\Cal P\Bbb N\to\Bbb R$ be a bounded
finitely additive functional.   Write $\dashint\ldots d\nu$ for the
associated linear functional on $\ell^{\infty}$\cmmnt{ (see 363L)},
and set $E_n=\{a:n\in a\subseteq\Bbb N\}$ for each
$n\in\Bbb N$.   Then the following are equiveridical:

(i) whenever $\mu$ is a Radon probability measure on $\Cal P\Bbb N$,
$\int\nu(a)\mu(da)$ is defined and equal to $\dashint\mu E_n\nu(dn)$;

\allowmorestretch{468}{
(ii) whenever $\mu$ is a Radon probability measure on $[0,1]^{\Bbb N}$,
$\idashint x\,d\nu\,\mu(dx)$ is defined and equal to
\penalty-100$\dashiint x(n)\mu(dx)\nu(dn)$;
}

(iii) whenever $(X,\Sigma,\mu)$ is a probability space and
$\sequencen{f_n}$ is a uniformly bounded sequence of measurable
real-valued functions on $X$, then $\idashint f_n(x)\nu(dn)\mu(dx)$ is
defined and equal to $\dashiint f_nd\mu\,\nu(dn)$;

(iv) whenever $\sequencen{F_n}$ is a sequence of Borel subsets of
$\{0,1\}^{\Bbb N}$, $\idashint\chi F_n(x)\nu(dn)\nu_{\omega}(dx)$ is defined
and equal to $\dashint\nu_{\omega}F_n\nu(dn)$, where $\nu_{\omega}$ is the
usual measure on $\{0,1\}^{\Bbb N}$.

\proof{{\bf (i)$\Rightarrow$(ii)}\grheada\
For $t\in[0,1]$ define $h_t:[0,1]^{\Bbb N}\to\Cal P\Bbb N$ by setting
$h_t(x)=\{n:x(n)\ge t\}$ for $x\in[0,1]^{\Bbb N}$, and let
$\mu_t=\mu h_t^{-1}$ be the image measure on $\Cal P\Bbb N$.   Then
$\mu_t$ is a Radon measure for each $t$.   \Prf\
Because $h_t$ is Borel measurable and
$\Cal P\Bbb N$ is metrizable, $h_t$ is almost continuous (418J),
so $\mu_t$ is a Radon measure (418I).\ \Qed

\medskip

\quad\grheadb\
For $m\in\Bbb N$ define $v_m\in[0,1]^{\Bbb N}$ by setting

\Centerline{$v_m(n)=2^{-m}\sum_{k=1}^{2^m}\mu\{x:x(n)\ge 2^{-m}k\}$.}

\noindent Then $\|v_{m+1}-v_m\|_{\infty}\le 2^{-m-1}$.   \Prf\ For any
$n\in\Bbb N$,

$$\eqalign{v_m(n)-v_{m+1}(n)
&=2^{-m}\sum_{k=1}^{2^m}\mu\{x:x(n)\ge 2^{-m}k\}
  -2^{-m-1}\sum_{k=1}^{2^{m+1}}\mu\{x:x(n)\ge 2^{-m-1}k\}\cr
&=2^{-m-1}\sum_{k=1}^{2^m}(2\mu\{x:x(n)\ge 2^{-m}k\}
  -\mu\{x:x(n)\ge 2^{-m}k\}\cr
&\mskip150mu-\mu\{x:x(n)\ge 2^{-m-1}(2k+1)\})\cr
&=2^{-m-1}\sum_{k=1}^{2^m}\mu\{x:2^{-m}k\le x(n)<2^{-m-1}(2k+1)\}
\le 2^{-m-1}.  \text{ \Qed}\cr}$$

\noindent So $v=\lim_{m\to\infty}v_m$ is defined in $\ell^{\infty}$ and
$\dashint v\,d\nu=\lim_{m\to\infty}\dashint v_md\nu$.
Also $v(n)=\int x(n)\mu(dx)$ for every $n\in\Bbb N$, so
$\dashiint x(n)\mu(dx)\nu(dn)=\dashint v\,d\nu$.

\medskip

\quad\grheadc\ Set

\Centerline{$f(t)=\dashint\mu_tE_n\nu(dn)=\int\nu(a)\mu_t(da)$}

\noindent for each $t\in[0,1]$ (using (i)).
Then, for any $m\in\Bbb N$,

$$\eqalignno{\dashint v_md\nu
&=2^{-m}\sum_{k=1}^{2^m}\dashint\mu\{x:x(n)\ge 2^{-m}k\}\nu(dn)\cr
&=2^{-m}\sum_{k=1}^{2^m}\dashint\mu\{x:h_{2^{-m}k}(x)\in E_n\}\nu(dn)\cr
&=2^{-m}\sum_{k=1}^{2^m}\dashint\mu_{2^{-m}k}E_n\nu(dn)
=2^{-m}\sum_{k=1}^{2^m}f(2^{-m}k).\cr}$$

\medskip

\quad\grheadd\ Next, for $m\in\Bbb N$ and $x\in[0,1]^{\Bbb N}$, set
$q_m(x)=2^{-m}\sum_{k=1}^{2^m}\chi h_{2^{-m}k}(x)$, so that
$\sequence{m}{q_m(x)}$ is non-decreasing and
$\|x-q_m(x)\|_{\infty}\le 2^{-m}$ for each $m$, while
$q_m:[0,1]^{\Bbb N}\to[0,1]^{\Bbb N}$ is Borel measurable.   Now

$$\eqalign{\dashiint q_m(x)d\nu\,\mu(dx)
&=2^{-m}\sum_{k=1}^{2^m}\int\nu(h_{2^{-m}k}(x))\mu(dx)\cr
&=2^{-m}\sum_{k=1}^{2^m}\int\nu(a)\mu_{2^{-m}k}(da)
=2^{-m}\sum_{k=1}^{2^m}f(2^{-m}k).\cr}$$

\noindent Also $\sequence{m}{\dashint q_m(x)d\nu}\to\dashint x\,d\nu$
uniformly for
$x\in[0,1]^{\Bbb N}$, so $\idashint x\,d\nu\,\mu(dx)$ is defined and equal to

$$\eqalign{\lim_{m\to\infty}\idashint q_m(x)d\nu\,\mu(dx)
&=\lim_{m\to\infty}2^{-m}\sum_{k=1}^{2^m}f(2^{-m}k)
=\lim_{m\to\infty}\dashint v_md\nu\cr
&=\dashint v\,d\nu
=\dashiint x(n)\mu(dx)\nu(dn).\cr}$$

\noindent As $\mu$ is arbitrary, (ii) is true.

\medskip

{\bf (ii)$\Rightarrow$(iii)} Assume (ii), and let
$(X,\Sigma,\mu)$ be a probability space and $\sequencen{f_n}$ a
uniformly bounded sequence of measurable
real-valued functions on $X$.   As completing $\mu$ does not affect the
integral $\int\ldots d\mu$ (212Fb),
we may suppose that $\mu$ is complete.   Let
$\gamma>0$ be such that $|f_n(x)|\le\gamma$ for every $n\in\Bbb N$ and
$x\in X$,
and set $q(x)(n)=\Bover1{2\gamma}(\gamma+f_n(x))$ for all $n$ and $x$.
Then
$q:X\to[0,1]^{\Bbb N}$ is measurable, so there is a Radon probability
measure $\lambda$ on $[0,1]^{\Bbb N}$ such that $q$ is \imp\ for $\mu$
and $\lambda$.   \Prf\ Taking $\lambda_0E=\mu q^{-1}[E]$ for Borel sets
$E\subseteq[0,1]^{\Bbb N}$, $q$ is \imp\ for $\mu$ and $\lambda_0$;  taking
$\lambda$ to be the completion of $\lambda_0$, $q$ is \imp\ for $\mu$ and
$\lambda$, by 234Ba;  and $\lambda$ is a Radon measure by 433Cb.\ \QeD\
Now

$$\eqalignno{\dashiint f_nd\mu\,\nu(dn)
&=2\gamma\dashiint q(x)(n)\mu(dx)\nu(dn)-\gamma\cr
&=2\gamma\dashiint z(n)\lambda(dz)\nu(dn)-\gamma\cr
\displaycause{235Gc}
&=2\gamma\idashint z(n)\nu(dn)\lambda(dz)-\gamma\cr
\displaycause{by (ii)}
&=2\gamma\idashint q(x)(n)\nu(dn)\mu(dx)-\gamma
=\idashint f_n(x)\nu(dn)\mu(dx).\cr}$$

\noindent As $\mu$ and $\sequencen{f_n}$ are arbitrary, (iii) is true.

\medskip

{\bf (iii)$\Rightarrow$(iv)} is elementary, taking $f_n=\chi F_n$ and
$\mu=\nu_{\omega}$.

\medskip

{\bf (iv)$\Rightarrow$(i)} If (iv) is true and $\mu$ is a Radon
probability measure on $\Cal P\Bbb N$, there is an \imp\ function $\phi$
from $(\{0,1\}^{\Bbb N},\nu_{\omega})$ to $(\Cal P\Bbb N,\mu)$
(343Cd).    For each $n\in\Bbb N$, set $F_n=\phi^{-1}[E_n]$ for each $n$
and choose a Borel set $F'_n\subseteq\{0,1\}^{\Bbb N}$ such that
$\nu_{\omega}(F'_n\symmdiff F_n)=0$.   Then
$\idashint\chi F'_n(x)\nu(dn)\nu_{\omega}(dx)$ is defined and equal to

\Centerline{$\dashint\nu_{\omega}F'_n\nu(dn)
=\dashint\nu_{\omega}F_n\nu(dn)=\dashint\mu E_n\nu(dn)$.}

\noindent Now

$$\eqalignno{\dashint\mu E_n\nu(dn)
&=\idashint\chi F'_n(x)\nu(dn)\nu_{\omega}(dx)
=\idashint\chi F_n(x)\nu(dn)\nu_{\omega}(dx)\cr
\displaycause{because for almost every $x$, $\chi F_n(x)=\chi F'_n(x)$ for
every $n$}
&=\idashint\chi E_n(\phi(x))\nu(dn)\nu_{\omega}(dx)
=\idashint\chi E_n(a)\nu(dn)\mu(da)\cr
\displaycause{235Gc again}
&=\idashint\chi a(n)\nu(dn)\mu(da)
=\int\nu(a)\,\mu(da).\cr}$$

\noindent As $\mu$ is arbitrary, (i) is true.
}%end of proof of 538P

\leader{538Q}{Definition} I will say that a bounded
finitely additive functional
$\nu$ satisfying (i)-(iv) of 538P is a {\bf medial functional};
if, in addition, $\nu$ is non-negative, $\nu a=0$ for every finite set
$a\subseteq\Bbb N$ and $\nu\Bbb N=1$, I will call $\nu$
a {\bf medial limit}.
\cmmnt{I should remark that the term `medial limit' is
normally used for the associated linear
functional $\dashint\ldots d\nu$ on $\ell^{\infty}$, rather than
the additive functional $\nu$ on $\Cal P\Bbb N$;
thus $h\in(\ell^{\infty})^*$
is a medial limit if $h\ge 0$, $h(w)=\lim_{n\to\infty}w(n)$ for every
convergent sequence $w\in\ell^{\infty}$ and
$\int h(\sequencen{f_n(x)})\mu(dx)$ is defined and equal to
$h(\sequencen{\int f_nd\mu})$ whenever $(X,\Sigma,\mu)$ is a
probability space and $\sequencen{f_n}$ is a uniformly bounded sequence
of measurable real-valued functions on $X$.}

\cmmnt{Note that }538P(i) tells us that a medial
limit\cmmnt{ $\nu:\Cal P\Bbb N\to\Bbb R$} is universally
Radon-measurable\cmmnt{ (definition:  434Ec)}, therefore universally
measurable\cmmnt{ (434Fc)}.

\vleader{48pt}{538R}{Proposition}
Let $M\cmmnt{\mskip5mu\cong(\ell^{\infty})^*}$
be the $L$-space of bounded finitely additive
functionals on $\Cal P\Bbb N$, and $M_{\text{med}}\subseteq M$ the set of
medial functionals.

(a) $M_{\text{med}}$ is a band in $M$, and if
$T\in L^{\times}(\ell^{\infty};\ell^{\infty})$\cmmnt{ (definition:
355G)} and $T':M\to M$ is its
adjoint, then $T'\nu\in M_{\text{med}}$ for every $\nu\in M_{\text{med}}$.

(b) Taking $M_{\tau}$ to be the band of completely additive functionals
on $\Cal P\Bbb N$ and $M_{\text{m}}$ the band of measurable functionals,
as described in \S464,
$M_{\tau}\subseteq M_{\text{med}}\subseteq M_{\text{m}}$.

(c) Suppose that $\sequence{k}{\nu_k}$ is a norm-bounded sequence in
$M_{\text{med}}$, and that $\nu\in M_{\text{med}}$.   Set
$\tilde\nu(a)=\dashint\nu_k(a)\nu(dk)$ for $a\subseteq\Bbb N$.   Then
$\tilde\nu\in M_{\text{med}}$.

(d) Suppose that $\nu\in M$ is a medial limit, and set
$\Cal F=\{a:a\subseteq\Bbb N$, $\nu(a)=1\}$.   Then $\Cal F$ is a
measure-converging filter with the Fatou property.
%any hope of `rapid'?  did Matt know?

(e) Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\lambda)$ be probability spaces, and
$T\in L^{\times}(L^{\infty}(\mu);L^{\infty}(\lambda))$.
Let $\sequencen{f_n}$, $\sequencen{g_n}$ be sequences in
$\eusm L^{\infty}(\mu)$, $\eusm L^{\infty}(\nu)$ respectively such that
$Tf_n^{\ssbullet}=g_n^{\ssbullet}$ for every $n$ and
$\sequencen{f_n^{\ssbullet}}$ is norm-bounded in $L^{\infty}(\mu)$.
Let $\nu\in M$ be a medial functional.   Then
$f(x)=\dashint f_n(x)\nu(dn)$ and
$g(y)=\dashint g_n(y)\nu(dn)$ are defined for almost every $x\in X$ and
$y\in Y$;  moreover,
$f\in\eusm L^{\infty}(\mu)$, $g\in\eusm L^{\infty}(\lambda)$ and
$Tf^{\ssbullet}=g^{\ssbullet}$.

\proof{{\bf (a)(i)}
Any of the four conditions of 538P makes it clear that
$M_{\text{med}}$ is a linear subspace of $M$.

We see also that $M_{\text{med}}$ is norm-closed in $M$.   \Prf\ Let
$\sequencen{\nu_n}$ be a sequence in $M_{\text{med}}$ which is
norm-convergent to $\nu\in M$.   If $\mu$ is a Radon probability measure on
$[0,1]^{\Bbb N}$, then
$\sequencen{\dashint x\,d\nu_n}\to\dashint x\,d\nu$ uniformly for
$x\in[0,1]^{\Bbb N}$, so

$$\eqalign{\idashint x\,d\nu\,\mu(dx)
&=\lim_{n\to\infty}\idashint x\,d\nu_n\mu(dx)\cr
&=\lim_{n\to\infty}\dashiint x(i)\mu(dx)\nu_n(di)
=\dashiint x(i)\,\mu(dx)\nu(di).\cr}$$

\noindent As $\mu$ is arbitrary, $\nu\in M_{\text{med}}$.\ \Qed

\medskip

\quad{\bf (ii)} Before completing the proof that $M_{\text{med}}$ is a
band, I deal with the second clause of (a).

\medskip

\qquad\grheada\ Recall from \S355 that
$L^{\times}(\ell^{\infty};\ell^{\infty})$ is the set of differences of
order-continuous positive linear operators from $\ell^{\infty}$ to itself.
Since $M$ can be identified with $(\ell^{\infty})^*$, any
$T\in L^{\times}(\ell^{\infty};\ell^{\infty})$ has an adjoint
$T':M\to M$ defined by saying that
$(T'\nu)(a)=\dashint T(\chi a)d\nu$ for every $a\subseteq\Bbb N$.
Since $x\mapsto\dashint Tx\,d\nu$ and $x\mapsto\dashint x\,d(T'\nu)$
both belong to $(\ell^{\infty})^*$ and agree on
$\{\chi a:a\subseteq\Bbb N\}$, they are equal, that is,
$\dashint Tx\,d\nu=\dashint x\,d(T'\nu)$ for every $x\in\ell^{\infty}$.

\medskip

\qquad\grheadb\
If $T:\ell^{\infty}\to\ell^{\infty}$ is an order-continuous positive linear
operator, it is a norm-bounded linear operator (355C), and all the functionals
$x\mapsto(Tx)(n)$ are order-continuous, therefore represented by members of
$\ell^1$;  that is, we have a family
$\langle\alpha_{ni}\rangle_{n,i\in\Bbb N}$ in $\coint{0,\infty}$ such that

\Centerline{$(Tx)(n)
=\sum_{i=0}^{\infty}\alpha_{ni}x(i)$ whenever $x\in\ell^{\infty}$
and $n\in\Bbb N$,}

\Centerline{$\sup_{n\in\Bbb N}\sum_{i=0}^{\infty}\alpha_{ni}
=\|T\|$ is finite.}

\noindent In this case, if $\nu\in M$ and $\nu'=T'\nu$ in $M$,

\Centerline{$\dashint x\,d\nu'
=\dashint(Tx)(n)\nu(dn)
=\dashint\sum_{i=0}^{\infty}\alpha_{ni}x(i)\nu(dn)$}

\noindent for every $x\in\ell^{\infty}$.

Now suppose that that $\|T\|\le 1$, so that
$\sum_{i=0}^{\infty}\alpha_{ni}\le 1$ for every $n$.   Consider the
function $\phi=T\restr[0,1]^{\Bbb N}$.   This is a function from
$[0,1]^{\Bbb N}$ to itself, and it is continuous for the product topology
on $\Bbb N$.

Take any $\nu\in M$ and Radon probability measure $\mu$ on
$[0,1]^{\Bbb N}$;  then the image measure $\mu_1=\mu\phi^{-1}$ on
$[0,1]^{\Bbb N}$ is a Radon probability measure (418I), and
$\int f(\phi(x))\mu(dx)=\int f(x)\mu_1(dx)$ for any $\mu_1$-integrable
function $f$.   In particular, setting $f(x)=\dashint x\,d\nu$,

\Centerline{$\idashint\phi(x)\,d\nu\,\mu(dx)
=\idashint x\,d\nu\,\mu_1(dx)=\dashiint x(n)\mu_1(dx)\nu(dn)$}

\noindent because $\nu\in M_{\text{med}}$.

Set $\nu'=T'\nu$.   Then we can calculate

$$\eqalignno{\dashiint x(n)\mu(dx)\nu'(dn)
&=\dashint\sum_{i=0}^{\infty}\alpha_{ni}\int x(i)\mu(dx)\nu(dn)
=\dashiint\sum_{i=0}^{\infty}\alpha_{ni}x(i)\mu(dx)\nu(dn)\cr
\displaycause{the inner integral is with respect to a genuine
$\sigma$-additive measure, so we have B.Levi's theorem}
&=\dashiint\phi(x)(n)\mu(dx)\nu(dn)
=\dashiint x(n)\mu_1(dx)\nu(dn)\cr
&=\idashint\phi(x)\,d\nu\,\mu(dx)
=\idashint Tx\,d\nu\,\mu(dx)
=\idashint x\,d\nu'\,\mu(dx).\cr}$$

\noindent As $\mu$ is arbitrary, $\nu'$ satisfies 538P(ii), and is a medial
functional.

\medskip

\qquad\grheadc\ Thus $T'\nu\in M_{\text{med}}$ whenever
$\nu\in M_{\text{med}}$ and $T:\ell^{\infty}\to\ell^{\infty}$ is positive,
order-continuous and of norm at most $1$.   As $M_{\text{med}}$ is a linear
subspace of $M$, the same is true for every positive order-continuous $T$
and for differences of such operators, that is, for every
$T\in L^{\times}(\ell^{\infty};\ell^{\infty})$, as claimed.

\medskip

\quad{\bf (iii)} I now return to the question of showing that
$M_{\text{med}}$ is a band.   The point is that if
$\nu$ is a medial functional and $b\subseteq\Bbb N$, then $\nu_b$ is a
medial functional, where $\nu_b(a)=\nu(a\cap b)$ for every
$a\subseteq\Bbb N$.   \Prf\ Define $T:\ell^{\infty}\to\ell^{\infty}$ by
setting $Tx=x\times\chi b$ for $x\in\ell^{\infty}$.   Then $T$ is a
positive order-continuous operator, and $T'\nu\in M_{\text{med}}$, by
(iii) above.   But

\Centerline{$(T'\nu)(a)=\dashint T(\chi a)d\nu=\dashint\chi(a\cap b)d\nu
=\nu(a\cap b)=\nu_b(a)$}

\noindent for every $a\subseteq\Bbb N$, so $\nu_b=T'\nu$ is a medial
functional.\ \Qed

By 436M, this is enough to ensure that $M_{\text{med}}$ is a band in $M$.

\medskip

{\bf (b)(i)} Recall that an additive functional on $\Cal P\Bbb N$
is completely additive iff it corresponds to an element of $\ell^1$,
that is,
belongs to the band generated by the elementary functionals $\delta_k$
where $\delta_k(a)=\chi a(k)$ for $k\in\Bbb N$ and $a\subseteq\Bbb N$.
To see that $\delta_k$ belongs to $M_{\text{med}}$, all we have to do is
to note that $\delta_k=\chi E_k$ where $E_k$ is defined as in 538P;
so if $\mu$ is a  Radon probability measure on $\Cal P\Bbb N$, we shall
have

\Centerline{$\int\delta_kd\mu=\mu E_k=\dashint\mu E_n\delta_k(dn)$.}

\noindent Since $M_{\text{med}}$ is a band, it must include $M_{\tau}$.

\medskip

\quad{\bf (ii)} On the other side, 538P(i) tells us that every member of
$M_{\text{med}}$ is universally measurable, and therefore belongs to
$M_{\text{m}}$, which is just the set of bounded additive
functionals which are $\Sigma$-measurable, where $\Sigma$ is the domain of
the usual measure on $\Cal P\Bbb N$.

\medskip

{\bf (c)(i)} Because $\sequence{k}{\nu_k}$ is norm-bounded, $\tilde\nu$ is
well-defined and additive;  also it is bounded.
\Prf\ If $\gamma$ is such that
$\|\nu\|\le\gamma$ and $\|\nu_k\|\le\gamma$ for every $k$, then

\Centerline{$|\tilde\nu(a)|
\le\gamma\sup_{k\in\Bbb N}|\nu_k(a)|\le\gamma^2$}

\noindent for every $a\subseteq\Bbb N$.\ \Qed

Note that

\Centerline{$\dashint\chi a\,d\tilde\nu=\tilde\nu(a)
=\dashint\nu_k(a)\nu(dk)=\dashidashint\chi a\,d\nu_k\nu(dk)$}

\noindent for every $a\subseteq\Bbb N$, so that

\Centerline{$\dashint x\,d\tilde\nu
=\dashint x(n)\,\tilde\nu(dn)
=\dashidashint x(n)\,\nu_k(dn)\nu(dk)
=\dashidashint x\,d\nu_k\nu(dk)$}

\noindent whenever $x\in\ell^{\infty}$ is a linear combination of
indicator functions, and therefore for every $x\in\ell^{\infty}$.

\medskip

\quad{\bf (ii)}
Now suppose that $(X,\Sigma,\mu)$ is a probability space and that
$\sequencen{f_n}$ is a uniformly bounded sequence of measurable real-valued
functions on $X$.   Let $(X,\hat\Sigma,\hat\mu)$ be the completion of
$(X,\Sigma,\mu)$.   For $k\in\Bbb N$ and $x\in X$ set
$g_k(x)=\dashint f_n(x)\nu_k(dn)$;  because $\nu_k$ is a medial functional,
we know that
$\int g_kd\mu=\dashiint f_n(x)\mu(dx)\nu_k(dn)$ is defined, so $g_k$ is
$\hat\Sigma$-measurable.   Consequently $\idashint g_k(x)\nu(dk)\hat\mu(dx)$
is defined and equal to $\dashiint g_k(x)\hat\mu(dx)\nu(dk)$.   It follows that

$$\eqalign{\dashiint f_n(x)\mu(dx)\tilde\nu(dn)
&=\dashidashiint f_n(x)\mu(dx)\nu_k(dn)\nu(dk)\cr
&=\dashiidashint f_n(x)\nu_k(dn)\mu(dx)\nu(dk)
=\dashiint g_k(x)\hat\mu(dx)\nu(dk)\cr
&=\idashint g_k(x)\nu(dk)\hat\mu(dx)
=\idashidashint f_n(x)\nu_k(dn)\nu(dk)\hat\mu(dx)\cr
&=\idashint f_n(x)\tilde\nu(dn)\hat\mu(dx)
=\idashint f_n(x)\tilde\nu(dn)\mu(dx).\cr}$$

\noindent (Recall that $\mu$ and $\hat\mu$ give rise to the same integrals,
by 212Fb again.)
As $(X,\Sigma,\mu)$ and $\sequencen{f_n}$ are arbitrary,
$\tilde\nu\in M_{\text{med}}$.

\medskip

{\bf (d)} Of course $\Cal F=\{\Bbb N\setminus a:\nu(a)=0\}$ is a filter.

\medskip

\quad{\bf (i)}
If $(X,\Sigma,\mu)$ is a probability space, $\sequencen{E_n}$ is a sequence
in $\Sigma$, and $\lim_{n\to\infty}\mu E_n=1$, then

\Centerline{$\idashint\chi E_n(x)\nu(dn)\mu(dx)
=\dashiint\chi E_nd\mu\,\nu(dn)=\dashint\mu E_n\nu(dn)=1$.}

\noindent So $E=\{x:\dashint\chi E_n(x)\nu(dn)=1\}$ is $\mu$-conegligible.
But if $x\in E$ and $a=\{n:x\in E_n\}$, then
$\nu a=\dashint\chi E_n(x)\nu(dn)=1$ and $a\in\Cal F$ and
$x\in\bigcap_{n\in a}E_n$.   Thus
$\bigcup_{a\in\Cal F}\bigcap_{n\in a}E_n\supseteq E$ is conegligible.
As $(X,\Sigma,\mu)$ and $\sequencen{E_n}$ are arbitrary,
$\Cal F$ is measure-converging.

\medskip

\quad{\bf (ii)} If $(X,\Sigma,\mu)$ is a probability space,
$\sequencen{E_n}$ is a sequence in $\Sigma$, and
$X=\bigcup_{A\in\Cal F}\bigcap_{n\in A}E_n$, then $\{n:x\in E_n\}\in\Cal F$
for every $x\in X$, and

\Centerline{$\dashint\mu E_n\nu(dn)
=\idashint\chi E_n(x)\nu(dn)\mu(dx)
=\int\nu\{n:x\in E_n\}\mu(dx)=1$.}

\noindent So for any $\epsilon>0$, $\nu\{n:\mu E_n\le 1-\epsilon\}=0$
and $\{n:\mu E_n\ge 1-\epsilon\}\in\Cal F$;  accordingly
$\lim_{n\to\Cal F}\mu E_n=1$.   As $(X,\Sigma,\mu)$ and $\sequencen{E_n}$
are arbitrary, $\Cal F$ has the Fatou property.

\medskip

{\bf (e)(i)}
For each $n\in\Bbb N$, we can find a $\Sigma$-measurable function
$f'_n:X\to\Bbb R$, equal almost everywhere to $f_n$, and such that
$\sup_{x\in X}|f'_n(x)|=\esssup|f_n|$.   Now $\sequencen{f'_n}$ is
uniformly bounded, so $f'(x)=\dashint f'_n(x)\nu(dn)$ is defined for every
$x\in X$;  and $f(x)$ is defined and equal to $f'(x)$ for
$\mu$-almost every $x$.   Since $f'$ is integrable, $f'$ and $f$ are
$\mu$-virtually measurable and essentially bounded, and
$f\in\eusm L^{\infty}(\mu)$.
Similarly, $g\in\eusm L^{\infty}(\lambda)$.

\medskip

\quad{\bf (ii)} If $h\in\eusm L^1(\mu)$, then
$\int f\times h\,d\mu=\dashiint f_n\times h\,d\mu\,\nu(dn)$.   \Prf\ ($\alpha$)
If $h$ is defined everywhere, measurable and bounded, then, taking $f'_n$
and $f'$ as in (i),
$(f'\times h)(x)=\dashint f'_n(x)h(x)\nu(dn)$ for every $x\in X$, so

$$\eqalign{\int f\times h\,d\mu
&=\int f'\times h\,d\mu
=\idashint(f'_n\times h)(x)\nu(dn)\mu(dx)\cr
&=\dashiint f'_n\times h\,d\mu\,\nu(dn)
=\dashiint f_n\times h\,d\mu\,\nu(dn).\cr}$$

\noindent($\beta$) In general, set $\gamma=\sup_{n\in\Bbb N}\esssup f_n$.
Given $\epsilon>0$, there is a simple
function $h'$ such that $\|h-h'\|_1\le\epsilon$, and now

$$\eqalign{|\int f\times h\,d\mu&-\dashiint f_n\times h\,d\mu\,\nu(dn)|\cr
&\le|\int f\times h\,d\mu-\int f\times h'\,d\mu|
+|\int f\times h'\,d\mu-\dashiint f_n\times h'\,d\mu\,\nu(dn)|\cr
&\mskip200mu
+|\dashiint f_n\times h'\,d\mu\,\nu(dn)-\dashiint f_n\times h\,d\mu\,\nu(dn)|\cr
&\le\|f\|_{\infty}\|h-h'\|_1
+\sup_{n\in\Bbb N}|\int f_n\times h'\,d\mu-\int f_n\times h\,d\mu|
\le 2\epsilon\gamma.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\int f\times h\,d\mu=\dashiint f_n\times h\,d\mu\,\nu(dn)$.\ \Qed

Similarly, $\int g\times h\,d\lambda=\dashiint g_n\times h\,d\lambda\,\nu(dn)$
for every $\lambda$-integrable $h$.

\medskip

\quad{\bf (iii)} If $h\in\eusm L^1(\lambda)$ there is an
$\tilde h\in\eusm L^1(\mu)$ such that
$\int\tilde h^{\ssbullet}\times v=\int h^{\ssbullet}\times Tv$ for every
$v\in L^{\infty}(\mu)$.   \Prf\ Recall that $L^1(\mu)$, $L^1(\lambda)$ can
be identified with $L^{\infty}(\mu)^{\times}$ and
$L^{\infty}(\nu)^{\times}$ (365Lb\formerly{3{}65Mb});
perhaps I should remark that the
formulae $\int\tilde h^{\ssbullet}\times v$,
$\int h^{\ssbullet}\times Tv$ represent abstract integrals taken in
$L^1(\mu)$, $L^1(\lambda)$ respectively (242B).   Setting
$\phi(w)=\int h^{\ssbullet}\times w$ for $w\in L^{\infty}(\lambda)$,
$\phi\in L^{\infty}(\lambda)^{\times}$, so
$\phi T\in L^{\infty}(\mu)^{\times}$ (355G) and there is an
$\tilde h\in\eusm L^1(\mu)$ such that

\Centerline{$\int\tilde h^{\ssbullet}\times v=\phi(Tv)
=\int h^{\ssbullet}\times Tv$}

\noindent for every $v\in L^{\infty}(\mu)$.\ \Qed

\medskip

\quad{\bf (iv)} Take $h$ and $\tilde h$ as in (iii), and consider

$$\eqalignno{\int h^{\ssbullet}\times g^{\ssbullet}
&=\int h\times g\,d\lambda
=\dashiint h\times g_n\,d\lambda\,\nu(dn)\cr
\displaycause{by (ii)}
&=\dashint(\int h^{\ssbullet}\times g_n^{\ssbullet})\nu(dn)
=\dashint(\int h^{\ssbullet}\times Tf_n^{\ssbullet})\nu(dn)\cr
&=\dashint(\int\tilde h^{\ssbullet}\times f_n^{\ssbullet})\nu(dn)
=\dashiint\tilde h\times f_nd\mu\,\nu(dn)
=\int\tilde h\times fd\mu\cr
\displaycause{by (ii) again}
&=\int\tilde h^{\ssbullet}\times f^{\ssbullet}
=\int h^{\ssbullet}\times Tf^{\ssbullet}.\cr}$$

\noindent As $h$ is arbitrary, and the duality between $L^{\infty}(\mu)$
and $L^1(\lambda)$ is separating, $Tf^{\ssbullet}=g^{\ssbullet}$, as
required.
}%end of proof of 538R

\leader{538S}{Theorem} (a) If
$\frakmctbl=\frak c$, there is a medial limit.
%Foreman thinks enough if $\frakmctbl=\frak d$, email 5.9.07

(b)\dvAnew{2011}\cmmnt{ ({\smc Larson 09})} Suppose that the filter
dichotomy\cmmnt{ (5A6Id)} is true.   If $I$ is any set and $\nu$ is a
finitely additive real-valued functional on $\Cal PI$ which is universally
measurable for the usual topology on $\Cal PI$, then $\nu$ is completely
additive.\cmmnt{\footnote{The result developed into this form in the
course of correspondence with J.Pachl.}} %6.11
Consequently there is no medial limit.
%universally Radon-measurable will do I suppose, but demands a little
%more from Chap 43

\proof{{\bf (a)(i)}
Let $M$ be the $L$-space of bounded additive functionals
on $\Cal P\Bbb N$.   Let us say that a subset $C$ of
$M$ is {\bf rationally convex} if $\alpha\nu+(1-\alpha)\nu'\in C$
whenever $\nu$, $\nu'\in C$ and $\alpha\in[0,1]\cap\Bbb Q$;  for
$A\subseteq M$, write $\Gamma_{\Bbb Q}(A)$ for the smallest
rationally convex set including $A$.   Set
$Q=\Gamma_{\Bbb Q}(\{\delta_n:n\in\Bbb N\})$ where
$\delta_n(a)=\chi a(n)$ for $a\subseteq\Bbb N$ and $n\in\Bbb N$.
In the language of 538Rb, $Q\subseteq M_{\tau}\subseteq M_{\text{med}}$,
so 538P(i) tells us that $\int\nu\,d\mu=\dashint\mu E_n\nu(dn)$ for every
$\nu\in Q$, where $E_n=\{a:n\in a\subseteq\Bbb N\}$ as usual.

\medskip

\quad{\bf (ii)}
Suppose that $\Cal F$ is a filter base on $Q$, consisting of rationally
convex sets, with cardinal less than $\frakmctbl$.   Let
$\mu$ be a Radon probability measure on $\Cal P\Bbb N$.
Then there is a sequence
$\sequence{k}{\nu_k}$ in $Q$ such that

\inset{$\sum_{k=0}^{\infty}\int|\nu_{k+1}(a)-\nu_k(a)|\mu(da)<\infty$,

$\{k:k\in\Bbb N$, $\nu_k\in F\}$ is infinite for every $F\in\Cal F$.}

\noindent\Prf\ Each $\nu\in Q$ is a
bounded Borel measurable real-valued function
on $\Cal P\Bbb N$;  let $u\in L^2=L^2(\mu)$ be a
$\frak T_s(L^2,L^2)$-cluster point
of $\family{\nu}{Q}{\nu^{\ssbullet}}$
along the filter generated by $\Cal F$.   For
any $F\in\Cal F$, the $\|\,\|_2$-closure of the
rationally convex set $\{\nu^{\ssbullet}:\nu\in F\}\subseteq L^2$
is convex, so
includes the weak closure of $\{\nu^{\ssbullet}:\nu\in F\}$ and
therefore contains $u$.   So $\{\nu^{\ssbullet}:\nu\in F\}$ meets
$\{v:v\in L^2$, $\|v-u\|_2\le\epsilon\}$ for every $\epsilon>0$.

Set $H_k=\{\nu:\nu\in Q$, $\|\nu^{\ssbullet}-u\|_2\le 2^{-k}\}$ for
each $k\in\Bbb N$;  then every $H_k$ meets every member of $\Cal F$.
If we give each $H_k$ its discrete topology, and take $H$ to be the product
$\prod_{k\in\Bbb N}H_k$, then $H$ is homeomorphic to $\NN$.   Writing
$\Cal M(H)$ for the ideal of meager subsets of $H$,
$\cov\Cal M(H)=\frakmctbl>\#(\Cal F)$, while

\Centerline{$\bigcup_{k\ge n}\{\alpha:\alpha\in H$, $\alpha(k)\in F\}$}

\noindent is a dense open subset of $H$ for every $F\in\Cal F$
and $n\in\Bbb N$.   There is
therefore an $\alpha\in H$ such that $\{k:\alpha(k)\in F\}$ is infinite for
every $F\in\Cal F$;  take $\nu_k=\alpha(k)$ for each $k$.   Since
$\mu$ is a probability measure,

$$\eqalignno{\int|\nu_{k+1}-\nu_k|d\mu
&\le\|\nu_{k+1}^{\ssbullet}-\nu_k^{\ssbullet}\|_2\cr
\displaycause{244E;  see 244Xd}
&\le 2^{-k-1}+2^{-k}\cr}$$

\noindent for every $k$, and
$\sum_{k=0}^{\infty}\int|\nu_{k+1}-\nu_k|d\mu$
is finite.\ \Qed

\medskip

\quad{\bf (iii)} Because a Radon
probability measure on $\Cal P\Bbb N$ is defined by its
values on the countable algebra $\frak B$ of open-and-closed sets, the
number of such measures is at most $\#(\BbbR^{\frak B})=\frak c$.
Enumerate them as $\ofamily{\xi}{\frakc}{\mu_{\xi}}$.   Choose a
non-decreasing family
$\langle\Cal F_{\xi}\rangle_{\xi\le\frakc}$ of filter bases on $Q$, as
follows.   The inductive hypothesis will be that $\Cal F_{\xi}$ has
cardinal at most $\max(\omega,\#(\xi))$ and consists of rationally convex
sets.   Start with $\Cal F_0=\{F_n:n\in\Bbb N\}$ where
$F_n=\Gamma_{\Bbb Q}(\{\delta_i:i\ge n\})$ for each $n$.
Given $\Cal F_{\xi}$ where $\xi<\frak c$, apply (ii) with
$\mu=\mu_{\xi}$ to see that there is
a sequence $\sequence{k}{\nu_{\xi k}}$ in $Q$ such that

\inset{$\sum_{k\in\Bbb N}\int|\nu_{\xi,k+1}-\nu_{\xi k}|d\mu_{\xi}<\infty$,

$\{k:\nu_{\xi k}\in F\}$ is infinite for every $F\in\Cal F_{\xi}$.}

\noindent Let $\Cal F_{\xi+1}$ be

\Centerline{$\Cal F_{\xi}
\cup\{F\cap\Gamma_{\Bbb Q}(\{\nu_{\xi k}:k\ge l\}):
F\in\Cal F_{\xi}$, $l\in\Bbb N\}$.}

\noindent For non-zero limit ordinals $\xi\le\frak c$, set
$\Cal F_{\xi}=\bigcup_{\eta<\xi}\Cal F_{\eta}$.

\medskip

\quad{\bf (iv)} At the end of the induction, let $\Cal F$ be the filter on
$M\cong(\ell^{\infty})^*$
generated by $\Cal F_{\frakc}$, and let $\theta$ be a cluster point of
$\Cal F$ for the
weak* topology of $(\ell^{\infty})^*$.   Then $\theta$ is a medial limit.
\Prf\ If $\mu$ is a Radon probability measure on $\Cal P\Bbb N$,
take $\xi<\frak c$ such that $\mu=\mu_{\xi}$.   Because
$\Gamma_{\Bbb Q}(\{\nu_{\xi k}:k\ge l\})$ belongs to $\Cal F$ for every
$l\in\Bbb N$,
$\dashint u(n)\theta(dn)=\lim_{k\to\infty}\dashint u(n)\nu_{\xi k}(dn)$ for every
$u\in\ell^{\infty}$ for which the limit is defined.   In particular,
$\theta(a)=\lim_{k\to\infty}\nu_{\xi k}(a)$ whenever $a\subseteq\Bbb N$ is
such that the limit is defined.   Because
$\sum_{k\in\Bbb N}\int|\nu_{\xi,k+1}-\nu_{\xi k}|d\mu$ is finite,
this is the case for $\mu$-almost every $a$, so

\Centerline{$\int\theta(a)\mu(da)
=\lim_{k\to\infty}\int\nu_{\xi k}(a)\mu(da)
=\lim_{k\to\infty}\dashint\mu E_n\nu_{\xi k}(dn)$;}

\noindent and because the latter limit is defined it is equal to
$\dashint\mu E_n\theta(dn)$.   As $\mu$ is arbitrary, $\theta$ satisfies
condition (i) of 538P, and is a medial functional;  because $Q\in\Cal F$,
$\theta\Bbb N=1$;  and because $\Cal F_0\subseteq\Cal F$, $\theta(a)=0$ for
every finite $a\subseteq\Bbb N$, so $\theta$ is a medial limit.\ \Qed

\medskip

{\bf (b)(i)} The key is the following.   Suppose that
$\nu:\Cal PI\to\Bbb R$ is a universally measurable additive
functional.

\medskip

\qquad\grheada\ For every set $J$ and function
$\phi:I\to J$, $\nu\phi^{-1}$ is
universally measurable, where $(\nu\phi^{-1})(b)=\nu(\phi^{-1}[b])$ for
every $b\subseteq J$.   \Prf\ We have only to observe that
$b\mapsto\phi^{-1}[b]:\Cal PJ\to\Cal PI$ is continuous, and
apply 434Df.\ \Qed

\medskip

\qquad\grheadb\ $\nu$ is bounded.   \Prf\Quer\
Otherwise, there is a disjoint sequence
$\sequence{k}{c_k}$ of subsets of $I$ such that
$\lim_{k\to\infty}|\nu c_k|=\infty$ (326D(ii)).   Enlarging $c_0$ if
necessary, we can suppose that $\bigcup_{k\in\Bbb N}c_k=I$.
Set $\phi(i)=k$ for $k\in\Bbb N$ and $i\in c_k$.   Then
$\nu\phi^{-1}[\{k\}]\to\infty$ as $k\to\infty$.   But $\nu'=\nu\phi^{-1}$
is universally measurable, therefore
$\Tau_{\Bbb N}$-measurable, where $\Tau_{\Bbb N}$ is the
domain of the usual measure $\lambda_{\Bbb N}$ on $\Cal P\Bbb N$.
Let $M$ be such
that $\lambda_{\Bbb N}E>0$ where $E=\{a:|\nu'a|\le M\}$.   Then there are
an $n\in\Bbb N$ such that for every $k\ge n$ there are $a$, $b\in E$ such
that $a\symmdiff b=\{k\}$ (345E;  recall that the natural
bijection $a\to\chi a:\Cal P\Bbb N\to\{0,1\}^{\Bbb N}$ identifies
$\lambda_{\Bbb N}$ with the usual measure on $\{0,1\}^{\Bbb N}$).
In this case, $k$ belongs to
exactly one of $a$, $b$;  suppose that $k\in a\setminus b$;  then
$|\nu'\{k\}|=|\nu a-\nu'b|\le 2M$.   This is supposed to be true for
every $k\ge n$, so $\limsup_{k\to\infty}|\nu'\{k\}|\le 2M$.\ \Bang\Qed

\medskip

\qquad\grheadc\ $|\nu|$ is universally measurable.
\Prf\ As in part (b-i) of the proof of 464K, there is a sequence
$\sequencen{c_n}$ in $\Cal PI$ such that
$\nu^+a=\lim_{n\to\infty}\nu(a\cap c_n)$ for every $a\subseteq I$.
Since all the functions $a\mapsto a\cap c_n$ are continuous,
$a\mapsto\nu(a\cap c_n)$ is universally measurable for every $n$, and
$\nu^+$ is universally measurable (use 418C).
Consequently $|\nu|=2\nu^+-\nu$ is universally measurable.\ \Qed

\medskip

\quad{\bf (ii)} If $\nu:\Cal P\Bbb N\to\coint{0,\infty}$
is a universally measurable additive functional and $\nu\{n\}=0$ for every
$n\in\Bbb N$, then $\nu=0$.   \Prf\Quer\ Otherwise, consider
$\Cal F=\{a:\nu a=\nu\Bbb N\}$.   This is a filter on $\Bbb N$
containing every cofinite set.   Let $\phi:\Bbb N\to\Bbb N$ be
finite-to-one, and write $\nu'$ for $\nu\phi^{-1}$.
Setting $\Cal I=\{a:\nu'a=0\}$, we have
a strictly positive additive functional on the quotient algebra
$\Cal P\Bbb N/\Cal I$, so $\Cal P\Bbb N/\Cal I$ is ccc and $\Cal I$ cannot
be $[\Bbb N]^{<\omega}$, that is, $\phi[[\Cal F]]$ is not
the Fr\'echet filter.   On the other hand,
$\nu'$ is universally measurable, by (i-$\alpha$), so

\Centerline{$\phi[[\Cal F]]=\{a:\phi^{-1}[a]\in\Cal F\}
=\{a:\nu'a=\nu'\Bbb N\}$}

\noindent is a universally measurable subset of $\Cal P\Bbb N$, and
cannot be an ultrafilter (464Ca).   Thus $\Cal F$ witnesses that the filter
dichotomy is false.\ \Bang\Qed

\medskip

\quad{\bf (iii)} Returning to the general case of a universally measurable
additive functional $\nu:\Cal PI\to\Bbb R$, set $\gamma_i=\nu\{i\}$ for
$i\in I$.   By (i-$\beta$),
$\sup_{J\in[I]^{<\omega}}|\sum_{j\in J}\gamma_j|
=\sup_{J\in[I]^{<\omega}}|\nu J|$ is finite, so
$\sum_{i\in I}|\gamma_i|<\infty$, and we have a functional
$\nu_1:\Cal PI\to\Bbb R$ defined by setting $\nu_1a=\sum_{i\in a}\gamma_i$
for every $a\subseteq I$.   $\nu_1$ is continuous for the topology of
$\Cal PI$, so $\nu_2=\nu-\nu_1$ is universally measurable, and
$\nu'=|\nu_2|$ is universally measurable, by (i-$\gamma$).

$\nu'J=0$ for every countable set $J\subseteq I$.   \Prf\ If $J$ is finite,
this is trivial, because

\Centerline{$|\nu_2|\{i\}
=|\nu_2\{i\}|=|\nu\{i\}-\nu_1\{i\}|=|\gamma_i-\gamma_i|=0$}

\noindent for every $i\in I$.   If $J$ is countably infinite, then
the embedding $\Cal PJ\embedsinto\Cal PI$ is continuous, so
$\nu'\restr\Cal PJ$ is universally measurable for the usual topology on
$\Cal PJ$;  also it is still zero on singletons, so (ii) tells us that it
is zero on the whole of $\Cal PJ$.\ \Qed

It follows that
$\nu'$ is zero everywhere.   \Prf\ Take $c\subseteq I$ and $\epsilon>0$.
$\nu'$ must be $\Tau_I$-measurable, where $\Tau_I$ is the domain of
the usual measure $\lambda_I$ on $\Cal PI$.   Since $\lambda_I$ is a
completion regular Radon measure (416U), there must be a non-negligible
zero set $K\subseteq\Cal PI$ such that $|\nu'a-\nu'b|\le\epsilon$ for all
$a$, $b\in K$;  and there is a countable set $J\subseteq I$ such that $K$
is determined by coordinates in $J$ (4A3Nc, applied to
$\{0,1\}^I\cong\Cal PI$).   Take any $a\in K$.   Then
$c_1=(c\setminus J)\cup(a\cap J)$ and $a\cap J$ both belong to $K$.
But as $\nu'(c\cap J)=0$,

\Centerline{$|\nu'c|=|\nu'c_1-\nu'(a\cap J)|\le\epsilon$.}

\noindent As $c$ and $\epsilon$ are arbitrary, $\nu'=0$.\ \Qed

Accordingly $\nu_2=0$ and $\nu=\nu_1$.   But of course $\nu_1$ is
completely additive.

\medskip

\quad{\bf (iv)} Finally, a medial limit would be a non-zero additive
functional from $\Cal P\Bbb N$ to $[0,1]$
which was universally measurable, as noted in
538Q, and zero on singletons;  and this has already been ruled out by (ii).
}%end of proof of 538S

\cmmnt{\medskip

\noindent{\bf Remark} It is possible to have medial limits when
$\frakmctbl\ll\frak c$;  see 553N.}

\exercises{\leader{538X}{Basic exercises (a)}
%\spheader 538Xa
Let $\Cal F$ be a filter on $\Bbb N$, and $I$ an infinite
subset of $\Bbb N$ such that $\Bbb N\setminus I\notin\Cal F$;  write
$\Cal F\lceil I$ for the filter $\{A\cap I:A\in\Cal F\}$.
Show that if
$\Cal F$ is free, or a $p$-point filter, or Ramsey, or rapid, or nowhere
dense, or measure-centering,
or measure-converging, or with the Fatou property, then so is
$\Cal F\lceil I$.
%538A

\spheader 538Xb
For $A\in[\Bbb N]^{\omega}$ let $f_A:\Bbb N\to A$ be the
increasing enumeration of $A$.   Let $\Cal F$ be a free filter on $\Bbb N$.
Show that $\Cal F$ is rapid iff $\{f_A:A\in\Cal F\}$ is cofinal with $\NN$.
%538A

\spheader 538Xc\dvAnew{2009}
Let $\Cal F$ be a filter which is universally
measurable (regarded as a subset
of $\Cal P(\bigcup\Cal F)$ with its usual topology), and
$\Cal G$ another filter such that $\Cal G\leRK\Cal F$.   Show that $\Cal G$
is universally measurable.
%538B Also works for universally Radon-measurable.

\spheader 538Xd
Let $\CalFr$ be the Fr\'echet filter and $\Cal F_d$ the
asymptotic density filter, the filter
of subsets of $\Bbb N$ with asymptotic density $1$.   (i) Show that
$\CalFr$ and $\Cal F_d$ are $p$-point filters.   (ii) Show that
$\CalFr\leRB\Cal F_d$ but that $\CalFr\ltimes\CalFr\not\leRK\Cal F_d$.
%538D

\spheader 538Xe(i) Let $\sequencen{\Cal F_n}$ be a sequence of
filters on $\Bbb N$, and $\Cal F$ a filter on $\Bbb N$.   Write
$\lim_{n\to\Cal F}\Cal F_n$ for the filter
$\{A:A\subseteq\Bbb N$, $\{n:n\in\Bbb N$, $A\in\Cal F_n\}\in\Cal F\}$.
Show that if every $\Cal F_n$ is rapid, then
$\lim_{n\to\Cal F}\Cal F_n$ is rapid.
(ii) Let $\Cal F$ be a rapid filter, and $\Cal G$ any filter on $\Bbb N$.
Show that $\Cal G\ltimes\Cal F$ is
rapid.   (iii) In 538E, suppose that $\Cal F_1$ is rapid.   Show that
$\Cal G_{\xi}$ is rapid for every $\xi\ge 1$.
%538E

\spheader 538Xf(i) Let $\Cal F$ be a nowhere dense filter, and
$\Cal G$ a filter on $\Bbb N$ such that $\Cal G\leRK\Cal F$.   Show that
$\Cal G$ is nowhere dense.
(ii)\dvAnew{2009} Show that a $p$-point ultrafilter is nowhere dense.
(iii)\dvAnew{2009}
In 538E, show that if every $\Cal F_{\xi}$ is a nowhere dense ultrafilter,
then $\Cal G_{\zeta}$ is a nowhere dense ultrafilter.
%538E 538Xe

\sqheader 538Xg
Let $\Cal F$ be a free filter on $\Bbb N$.   Show that the
following are equiveridical:  (i) $\Cal F$ is a Ramsey filter;  (ii)
whenever $K$ is finite, $k\in\Bbb N$ and $f:[\Bbb N]^k\to K$ is a function,
there is an $F\in\Cal F$ such that $f$ is constant on $[F]^k$;
(iii) $\Cal F$ is a $p$-point filter and whenever $\sequencen{E_n}$ is a
disjoint sequence in $[\Bbb N]^{<\omega}$, there is an $F\in\Cal F$
such that $\#(F\cap E_n)\le 1$ for every $n$;  (iv) whenever
$\sequencen{E_n}$ is a disjoint sequence in $\Cal P\Bbb N\setminus\Cal F$,
there is an $F\in\Cal F$ such that $\#(F\cap E_n)\le 1$ for every $n$.
%538F

\spheader 538Xh Let $\frak F$ be a countable family of distinct $p$-point
ultrafilters on $\Bbb N$.   Show that there is a disjoint family
$\family{\Cal F}{\frak F}{A_{\Cal F}}$ of subsets of $\Bbb N$ such that
$A_{\Cal F}\in\Cal F$ for every $\Cal F\in\frak F$.
%538F

\spheader 538Xi\dvAnew{2010}
Let $(X,\Sigma,\mu)$ be a complete
perfect probability space, $(Y,\frak S)$ a
perfectly normal compact Hausdorff space,
$\sequencen{f_n}$ a sequence of measurable functions from $X$ to $Y$,
$\Cal F$ a measure-centering ultrafilter on $\Bbb N$ and $\lambda$ the
$\Cal F$-extension of $\mu$.   (i) Setting
$f(x)=\lim_{n\to\Cal F}f_n(x)$ for $x\in X$, show that $f$ is
$\dom\lambda$-measurable.   (ii) For each $n\in\Bbb N$, show that there is
a unique Radon measure $\nu_n$ on $Y$ such that $f_n$ is \imp\ for $\mu$
and $\nu_n$.
(iii) Let $\nu$ be the limit $\lim_{n\to\Cal F}\nu_n$ for the
narrow topology on the space of Radon probability measures on $Y$ (437Jd).
Show that $f$ is \imp\ for $\lambda$ and $\nu$.   \Hint{look at the Radon
measure associated with the image measure $\lambda f^{-1}$.   You may
prefer to begin with metrizable $Y$.}
%538I

\spheader 538Xj Let $\familyiI{(\frak A_i,\bar\mu_i)}$ be a family of
probability algebras, $\Cal F$ an ultrafilter on $I$, and
$(\frak A,\bar\mu)$ the probability algebra reduced product of
$\prod_{i\in I}(\frak A_i,\bar\mu_i)|\Cal F$.   For each $i\in I$,
let $\Bsubseteqshort_i$ be the order relation on $\frak A_i$;  set
$P=\prod_{i\in I}\frak A_i$ and let $P|\Cal F$ be the
partial order reduced product of $\familyiI{(\frak A_i,\Bsubseteqshort_i)}$
modulo $\Cal F$ as defined in 5A2A.   Describe a canonical order-preserving
map from $P|\Cal F$ to $\frak A$.
%538J

\spheader 538Xk(i) Let $(\frak A,\bar\mu)$ be a homogeneous
probability algebra with Maharam type $\kappa$,
$I$ a non-empty set, $\Cal F$ an ultrafilter on $I$ and
$(\frak C,\bar\nu)$ the probability algebra reduced power
$(\frak A,\bar\mu)^I|\Cal F$.   Show that $\frak C$ is homogeneous, with
Maharam type the transversal number $\Tr_{\Cal I}(I;\kappa)$
(definition:  5A1M), where $\Cal I=\{I\setminus A:A\in\Cal F\}$.
\Hint{5A1Nd, 521Eb.} %\Tr attained, M type=top density
(ii)\dvAnew{2009} % {\smc Farah Hart \& Sherman p09}
Show that if $(\frak A,\bar\mu)$ is any probability algebra
and $\Cal F$ and $\Cal G$ are non-principal
ultrafilters on $\Bbb N$, then the probability algebra reduced powers
$(\frak A,\bar\mu)^{\Bbb N}|\Cal F$ and
$(\frak A,\bar\mu)^{\Bbb N}|\Cal G$ are isomorphic.
%538J

\spheader 538Xl Let $(X,\Sigma,\mu)$ be a perfect probability space and
$\mu'$ an indefinite-integral measure over $\mu$ which is also a
probability measure.   Let $\Cal F$ be a measure-centering ultrafilter on
$\Bbb N$ and $\lambda$, $\lambda'$ the $\Cal F$-extensions of $\mu$ and
$\mu'$.   Show that $\lambda'$ is an
indefinite-integral measure over $\lambda$.
%538K  take care over  \dom(\lambda')

\sqheader 538Xm  ({\smc Benedikt 98})
(i) Let $\Cal F$ be any free filter on $\Bbb N$.   Show that
$\Cal F\ltimes\Cal F$ is not measure-centering.   \Hint{let
$\sequencen{e_n}$ be the standard generating family in $\frak B_{\omega}$,
and consider $a_{mn}=e_m\Bsetminus e_n$ if $m<n$, $1$ otherwise.}
(ii) Let $\Cal F$ be a measure-centering ultrafilter on $\Bbb N$.
Show that if $f$, $g\in\NN$ and $\{n:f(n)\ne g(n)\}\in\Cal F$, then
$f[[\Cal F]]\ne g[[\Cal F]]$.   \Hint{consider
$a_n=e_{f(n)}\Bsetminus e_{g(n)}$ if $f(n)\ne g(n)$.}
%538L "Hausdorff filter"

\spheader 538Xn\dvAnew{2009}
Let $X$ be a locally compact Hausdorff topological group,
and $\mu$ a left Haar measure on $X$.   Show that there is a
complete locally determined \lti\ measure
$\lambda$ on $X$ such that
$\lambda(\lim_{n\to\Cal F}E_n)$ is defined and equal to
$\sup_{K\subseteq X\text{ is compact}}\lim_{n\to\Cal F}\mu(E_n\cap K)$
whenever $\Cal F$ is a Ramsey ultrafilter on
$\Bbb N$ and $\sequencen{E_n}$ is a sequence of Haar measurable subsets of
$X$.
%538M

\spheader 538Xo(i) Let $\sequencen{\Cal F_n}$ be a sequence of
measure-converging filters on $\Bbb N$.
Show that $\bigcap_{n\in\Bbb N}\Cal F_n$ is measure-converging, so that
$\lim_{n\to\Cal F}\Cal F_n$ (538Xe) is measure-converging for any
filter $\Cal F$ on $\Bbb N$.
(ii) In 538E, suppose that $\Cal F_1$ is measure-converging.
Show that $\Cal G_{\xi}$ is measure-converging for every $\xi\in[1,\zeta]$.
%538N 538Xg

\spheader 538Xp\dvAnew{2012} Suppose that
$\ofamily{\xi}{\kappa}{\Cal F_{\xi}}$ is a family of measure-converging
filters, where $\kappa$ is non-zero and
less than the additivity $\add\Cal N$ of
Lebesgue measure.   Show that $\bigcap_{\xi<\kappa}\Cal F_{\xi}$ is
measure-converging.
%538N(iv) 538Xo

\spheader 538Xq(i)\dvAnew{2014}
Let $\Cal F$ be a filter on $\Bbb N$.   Show that
$\Cal F$ has the Fatou property iff $\int fd\mu$ and
$\lim_{n\to\Cal F}\int f_nd\mu$ are defined and
equal whenever $(X,\Sigma,\mu)$ is a
measure space, $g:X\to\coint{0,\infty}$ is an integrable function and
$\sequencen{f_n}$ is a sequence of measurable functions on $X$ such that
$|f_n|\leae g$ for every $n$ and $\lim_{n\to\Cal F}f_n\eae f$.
(ii) Show that a non-principal ultrafilter on $\Bbb N$ cannot
have the Fatou property.   \Hint{464Ca.}
%538O

\spheader 538Xr
Show that the asymptotic density filter (538Xd) has the Fatou property.
%538O 538Xd 538Ym

\spheader 538Xs(i)
Let $\sequencen{\Cal F_n}$ be a sequence of filters with
the Fatou property, and $\Cal F$ a filter with the Fatou property.   Show
that $\lim_{n\to\Cal F}\Cal F_n$ (538Xe) has the Fatou property.
(ii) In 538E, suppose that $\Cal F_{\xi}$ has the Fatou property for every
$\xi\in[1,\zeta]$.
Show that $\Cal G_{\xi}$ has the Fatou property for every $\xi\le\zeta$.
%538O

\spheader 538Xt Let $\nu:\Cal P\Bbb N\to\Bbb R$ be a bounded additive
functional.   (i) Show that $\nu$ is a medial functional iff
$\int\nu\{n:x\in E_n\}\mu(dx)$ is defined and equal to
$\dashint\mu E_n\nu(dn)$ whenever
$(X,\Sigma,\mu)$ is a probability space and $\sequencen{E_n}$ is a sequence
in $\Sigma$.   (ii) Show that in this case $a\mapsto\nu\phi^{-1}[a]$ is a
medial functional for any $\phi:\Bbb N\to\Bbb N$.
%538Q

\sqheader 538Xu Let $(X,\Sigma,\mu)$ be a probability space, and $\Tau$ a
$\sigma$-subalgebra of $\Sigma$.   Let $\sequencen{f_n}$ be a sequence in
$\eusm L^{\infty}(\mu)$ such that $\sup_{n\in\Bbb N}\esssup|f_n|$ is
finite, and for each $n\in\Bbb N$ let $g_n$ be a conditional expectation of
$f_n$ on $\Tau$.   Suppose that $\nu$ is a medial functional.   Show that
$f(x)=\dashint f_n(x)\nu(dn)$ and
$g(x)=\dashint g_n(x)\nu(dn)$ are defined for
almost every $x$, that $f\in\eusm L^{\infty}(\mu)$, and that $g$ is a
conditional expectation of $f$ on $\Tau$.
%538R

\spheader 538Xv (V.Bergelson) Show that there are a probability algebra
$(\frak A,\bar\mu)$ and a sequence $\sequencen{a_n}$ in $\frak A$ such that
$\inf_{n\in\Bbb N}\bar\mu a_n\penalty-100>0$
but $a_m\Bcap a_n\Bcap a_{m+n}=0$ whenever
$m$, $n>0$.   \Hint{for $n\ge 1$, set
$E_n=\{x:x\in[0,1]$, $\lfloor 3nx\rfloor\equiv 1\mod 3\}$.}
%538+

\leader{538Y}{Further exercises (a)}\dvAnew{2014}
%\spheader 538Ya
Show that if $\Cal F$ and $\Cal G$ are
filters and $\Cal F\leRK\Cal G$, then, in the language of
512A, $(\Cal F,\supseteq,\Cal F)\prGT(\Cal G,\supseteq,\Cal G)$, so that
$\ci\Cal F\le\ci\Cal G$ and $\Cal F$ is $\kappa$-complete whenever
$\kappa$ is a cardinal and $\Cal G$ is $\kappa$-complete.
%538B

\spheader 538Yb
Let $\Cal F$ be a free ultrafilter on $\Bbb N$, and suppose
that whenever $\Cal G$ is a free filter on $\Bbb N$ and
$\Cal G\leRK\Cal F$, then $\Cal F\leRK\Cal G$.   Show that $\Cal F$ is a
Ramsey ultrafilter.   \Hint{{\smc Comfort \& Negrepontis 74}.} %Th 9.6
%538D

\spheader 538Yc Show that if $\frak p=\frak c$ then there are
$2^{\frakc}$
Ramsey ultrafilters on $\Bbb N$, and therefore $2^{\frakc}$ isomorphism
classes of Ramsey ultrafilters.
%538F

\spheader 538Yd Let $\Cal F$ be an ultrafilter on $\Bbb N$.   Show that
$\Cal F$ is measure-centering iff whenever $\frak A$ is a Boolean algebra,
$D\subseteq\frak A\setminus\{0\}$ has intersection number greater than
$0$ (definition: 391H) and $\sequencen{a_n}$ is a sequence in $D$, then
there is an $A\in\Cal F$ such that $\{a_n:n\in A\}$ is centered.
%538G out of order query

\spheader 538Ye(i) Show that if $\cov\Cal N=\frak c$, there is a
measure-centering ultrafilter on $\Bbb N$
including the asymptotic density filter (538Xd).   (ii) Show that an
ultrafilter on
$\Bbb N$ including the asymptotic density filter cannot be a
$p$-point filter.
(iii) Show that a filter on $\Bbb N$ including the asymptotic density
filter cannot be a rapid filter.
%538H out of order query 53bits

\spheader 538Yf(i) Let $\Cal F$, $\Cal G$ be free filters on $\Bbb N$ such
that $\Cal F\ltimes\Cal G$ is measure-centering.   Show that there is no
free filter $\Cal H$ such that $\Cal H\leRK\Cal F$ and $\Cal H\leRK\Cal G$.
(ii) Show that if there are two non-isomorphic Ramsey
ultrafilters on $\Bbb N$, then there are  two
non-isomorphic measure-centering ultrafilters $\Cal F$, $\Cal G$ on
$\Bbb N$ such that $\Cal F\ltimes\Cal G$ is not measure-centering.
% either  \CalH_1 , \CalH_1\ltimes\CalH_2  or
%  \CalH_1\ltimes\CalH_2, \CalH_2
%538Xm 538L

\spheader 538Yg For an uncountable set $I$,
let us say that a filter $\Cal F$
on $I$ is {\bf uniform and measure-centering} if $\#(A)=\#(I)$ for every
$A\in\Cal F$ and whenever $\frak A$ is a Boolean algebra,
$\nu:\frak A\to\coint{0,\infty}$ is an additive functional, and
$\familyiI{a_i}$ is a family in $\frak A$ with
$\inf_{i\in I}\nu a_i>0$, there is an $A\in\Cal F$ such that
$\{a_i:i\in A\}$ is centered.   (i) State and prove a result corresponding
to 538G for such filters.   \Hint{in the part corresponding to 538G(iv),
use `compact' measures rather than `perfect' measures.}   (ii) State and
prove a result corresponding to 538H.   \Hint{set $\kappa=\#(I)$.
In the part corresponding to
538Hc, suppose that you have a $\kappa$-complete ultrafilter on $I$,
rather than a
Ramsey ultrafilter;  see 4A1L.   In the part corresponding to 538He,
suppose that $\kappa$ is regular and that
$\cov\Cal N_{\kappa}=2^{\kappa}$, where $\Cal N_{\kappa}$ is
the null ideal of the usual measure on $\{0,1\}^{\kappa}$.}
(iii) State and prove results corresponding to 538I-538K. %538I 538J 538K
(iv) State and prove results corresponding to 538L-538M, but with
`normal ultrafilters' in place of `Ramsey ultrafilters'.
%538M
%question:  can there be a uniform measure-centering ultrafilter on
%\frak c ?

\spheader 538Yh\dvAnew{2009} Show that if $\Cal F$ and $\Cal G$ are filters
on $\Bbb N$, $\Cal F$ is rapid and $\Cal G\leRB\Cal F$, then $\Cal G$ is
rapid.
%538N

\spheader 538Yi Give an example of filters $\Cal F$, $\Cal G$ on $\Bbb N$
such that $\Cal F$ has the Fatou property, $\Cal G\subseteq\Cal F$ and
$\Cal G$ does not have the Fatou property.
%538O  \Cal G=\Cal F_d\cap\Cal H  where  \Cal H  ultrafilter containing
%set of density  0

\spheader 538Yj(i) Let $\Cal F$ be a nowhere dense filter on $\Bbb N$,
and $\Cal I$ the
ideal $\{\Bbb N\setminus A:A\in\Cal F\}$.   Show that
$\Cal P\Bbb N/\Cal I$ is finite.
(ii) Show that a free filter with the Fatou property cannot be nowhere
dense.
%538Xq 538O

\spheader 538Yk\dvAnew{2009} Let $(X,\Sigma,\mu)$ be a probability space
and $\sequence{m}{f_m}$, $\sequencen{g_n}$ two uniformly bounded
sequences of real-valued measurable functions defined on $X$.
Let $\nu$,
$\nu':\Cal P\Bbb N\to\Bbb R$ be bounded additive functionals.   Show that
$\dashidashiint f_m\times g_nd\mu\,\nu(dm)\nu'(dn)
=\dashidashiint f_m\times g_nd\mu\,\nu'(dn)\nu(dm)$.
%538P
%both equal to $\int f\times g$.

\spheader 538Yl ({\smc Meyer 73})
Let $\nu$ be a medial limit.   Write $U$ for the set of
sequences $u\in\BbbR^{\Bbb N}$ such that
$\sup\{\dashint v\,d\nu:v\in\ell^{\infty}$, $v\le|u|\}$ is
finite;  for $u\in U$, write $\dashint u\,d\nu$ for
$\lim_{m\to\infty}\dashint\med(-m,u(n),m)\nu(dn)$
(see 364Xj).
Suppose that $(X,\Sigma,\mu)$ is a probability space and
$\sequencen{f_n}$ a sequence of $\mu$-integrable real-valued functions
on $X$ such that $\sequencen{\int|f_n|d\mu}\in U$.
(i) Show that $\sequencen{f_n(x)}\in U$ for $\mu$-almost every
$x\in X$.   Set $f(x)=\dashint f_n(x)\nu(dn)$ whenever
$\sequencen{f_n(x)}\in U$.
(ii) Show that if every $f_n$ is non-negative then
$\int fd\mu\le\dashiint f_nd\mu\,\nu(dn)$.
(iii) Show that if $\{f_n:n\in\Bbb N\}$ is uniformly integrable then
$\int fd\mu=\dashiint f_nd\mu\,\nu(dn)$.
(iv)\dvAnew{2012}
Show that if $\sequencen{f_n^{\ssbullet}}$ is weakly convergent to $0$
in $L^1(\mu)$, then $f\eae 0$.   (v)\dvAnew{2012}
Suppose that $\sequencen{f_n}$ is uniformly integrable.
Let $\Tau$ be a $\sigma$-subalgebra of $\Sigma$, and for
each $n\in\Bbb N$ let $g_n$ be a conditional expectation of $f_n$ on
$\Tau$;  set
$g(x)=\dashint g_n(x)\nu(dn)$ whenever $\sequencen{g_n(x)}\in U$.
Show that
$g$ is a conditional expectation of $f$ on $\Tau$.
%538R 53bits 538Xu

\spheader 538Ym\dvAnew{2014}
Suppose that $\Cal F$ is a filter on $\Bbb N$ with the Fatou property, and
$\sequence{n}{\nu_n}$ a sequence of medial limits.   Set
$\Cal G=\{A:A\subseteq\Bbb N$, $\lim_{n\to\Cal F}\nu_nA=1\}$.   Show that
$\Cal G$ is a filter with the Fatou property.
%538Xr 538R mt53bits

\spheader 538Yn\dvAnew{2009} Show that
$\frak u\ge\frak r(\omega,\omega)\ge\max(\cov\Cal N,\frakmctbl)$
(definitions:  5A6Ia, 529G).
%538S 529Xg

\spheader 538Yo\dvAnew{2009}(i) Show that if $\Cal F$ is a rapid filter
on $\Bbb N$, then
$\ci\Cal F\ge\frak d$.   (ii) Show that $\frak d\ge\frak g$ (definition:
5A6I(b-ii)).   (iii) Show that if $\frak u<\frak g$ there are no rapid
filters on $\Bbb N$, and if there is a measure-converging filter
there is a measure-converging ultrafilter with coinitiality $\frak u$.
%538S 5A6J 538Ya 53bits

\spheader 538Yp\dvAnew{2011} Suppose that the filter dichotomy is true.
(i) Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra.   Show that if $\nu:\frak A\to\Bbb R$ is an additive
functional which is universally measurable for the order-sequential
topology of $\frak A$, then
$\nu$ is countably additive.   (ii) Let $(\frak A,\bar\mu)$ be a
localizable measure algebra.   Show that if $\nu:\frak A\to\Bbb R$ is an
additive functional which is universally measurable for the measure-algebra
topology on $\frak A$, then it is continuous.
%538S

\spheader 538Yq(i) Show that there is a semigroup operation $\dot+$ on
the set $\beta\Bbb N$ of ultrafilters on $\Bbb N$ defined by saying that
$\Cal F\dot+\Cal G=+[[\Cal F\ltimes\Cal G]]$ for all $\Cal F$,
$\Cal G\in\beta\Bbb N$, where $+:\Bbb N\times\Bbb N\to\Bbb N$ is addition.
(ii) Show that if we identify $\beta\Bbb N$ with the Stone-\v{C}ech
compactification of $\Bbb N$ (4A2I(b-i)), then $\dot+$ is
continuous in the first variable.   (iii) Show that there is a
non-principal ultrafilter $\Cal F$ on $\Bbb N$ which is {\bf idempotent},
that is, $\Cal F\dot+\Cal F=\Cal F$.
\Hint{consider a minimal closed sub-semigroup of the set of non-principal
ultrafilters.}
(iv) For any function $f\in\NN$, write $\FS(f)$ for
$\{\sum_{n\in K}f(n):K\in[\Bbb N]^{<\omega}\}$;  say a {\bf finite sum set}
is a set of the form $\FS(f)$ for some strictly increasing function
$f\in\NN$.   Show that if $\Cal F$ is a non-principal idempotent
ultrafilter on $\Bbb N$ and $I\in\Cal F$, then $I$ includes a finite sum
set.   (This is a version of {\bf Hindman's theorem}.)
(v) Show that if $I\subseteq\Bbb N$ is a finite sum set
there is an idempotent ultrafilter containing $I$.
(vi) Suppose that $(\frak A,\bar\mu)$ is a probability algebra and
$\pi:\frak A\to\frak A$ is a measure-preserving Boolean homomorphism.
($\alpha$) Show that if $\Cal F$ is an idempotent ultrafilter on $\Bbb N$,
then
$\lim_{n\to\Cal F}\mu(a\Bcap\pi^na)\ge(\mu a)^2$ for every $a\in\frak A$
($\beta$) Show that there is a finite sum set $I\subseteq\Bbb N$ such that
$\{\pi^na:n\in I\}$ is centered.
(vii) Show that no idempotent ultrafilter is measure-centering.
\Hint{538Xv.}
(viii) Show that if $\Cal F$ is a $p$-point ultrafilter then
$\Cal F\dot+\Cal F$ is isomorphic to $\Cal F\ltimes\Cal F$ and is
not measure-centering.
%mt53bits
(ix) Repeat, as far as possible, for semigroups other than $(\Bbb N,+)$.
%538+

\spheader 538Yr (V.Bergelson-M.Talagrand) Show that there are a probability
algebra $(\frak A,\bar\mu)$ and a sequence $\sequencen{a_n}$ in $\frak A$
such that $\bar\mu a_n=\bover12$ for every $n\in\Bbb N$ but
$\inf_{m,n\in I}\bar\mu(a_m\Bcap a_n)=0$ whenever
$I\subseteq\Bbb N$ does not have asymptotic density $0$.
%538+ n08707.tex
}%end of exercises

\leader{538Z}{Problem}
Show that it is relatively consistent with ZFC to suppose that there are
no measure-converging filters on $\Bbb N$.
%What happens if we add Cohen reals to a model of GCH?
%What if  \frak u < \frak g ? 538Ys

\leaveitout{For a function $f:\Bbb N\to\Bbb N$, write $\FS(f)$ for
$\{\sum_{n\in I}f(n):I\in[\Bbb N]^{<\omega}\}$.   A {\bf finite sum set} is
a set of the form $\FS(f)$ for a strictly increasing sequence $f$.
Suppose that $(\frak A,\bar\mu)$ is a probability algebra and
$\sequencen{a_n}$ is a sequence in $\frak A$ such that
$\inf_{n\in\Bbb N}\mu a_n>0$.   Must there be a finite sum set
$F\subseteq\Bbb N$ such that $\{a_n:n\in F\}$ is centered?

In 538Yq(vi-$\beta$), if $\Cal F$ is idempotent,
do we actually have a set $I\in\Cal F$ such that $\{\pi^na:n\in I\}$ is
centered?
}

\endnotes{
\Notesheader{538} This is a long section, and rather a lot of ideas are
crowded into it, starting with the list in 538A.   If you have looked at
ultrafilters on $\Bbb N$ at all, you are likely to have encountered
`$p$-point', `rapid' and `Ramsey' ultrafilters, and most of
538B-538D %538B 538C 538D
and 538F will probably be familiar.   The `iterated products' of 538E will
also be a matter of adapting known concepts to my particular formulation.

Some of the slightly contorted language of 538Fe and 538Ff
(with references to
`$\#(\frak F)$') is there because we do not know how many isomorphism
classes of Ramsey filters there are.   If there are none (as in random real
models, see 553H), or one ({\smc Shelah 82}, \S VI.5),
then things are very simple.   If there are infinitely many then we could
rephrase 538Ff in terms of sequences of non-isomorphic filters.
But it is possible that there should be two, or seventeen ({\smc Shelah
98a}, p.\ 335).
%e-mail to S.S., 9.7.07;  reply 4.5.14

In 538H-538M %538I 538L 538M
I try to set out, and expand, some of the principal ideas of
{\smc Benedikt 98}.   The starting point is the observation that a Ramsey
ultrafilter gives us an extension of Lebesgue measure on $[0,1]$,
indeed of any perfect probability measure.   Observing that this property
is preserved by iterations, we are led to `measure-centering' ultrafilters.
Once we have the idea of measure-centering-ultrafilter extension
of a perfect
probability measure, we can set out to look at its properties in terms of
the (by now very extensive) general theory of this treatise.   The first
step has to be the identification of its measure algebra (538Ja, 538Xk),
followed, if
possible, by the identification of the corresponding Banach function
spaces.   It turns out that these can be reached by an alternative route
{\it not} involving special properties of the ultrafilter or the
probability space, which I have expressed in general forms in
\S\S328 and 377.   This gives a
long list of facts, which I have written out in 538Ja and 538K.
Minor variations of the measure and the filter are straightforward
(538Jb, 538Jc, 538Xl).   For iterated products of filters we have more
work to do (538L), especially if we are
to express them in a form adequate for
the objective, the universal-extension result of 538M.

You will have noticed that in the statement of 538G I speak of
`$\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n$' and
`$\liminf_{n\to\Cal F}\mu F_n$'.   Something of the sort is necessary since
in that theorem I do not insist from the outset
that $\Cal F$ should be an ultrafilter.   Of course only ultrafilters are
of interest in this context, by 538Ha, and for these we have
$\bigcup_{A\in\Cal F}\bigcap_{n\in A}F_n=\lim_{n\to\Cal F}F_n$ and
$\liminf_{n\to\Cal F}\mu F_n=\lim_{n\to\Cal F}\mu F_n$, as in 538I.

For most of this section I have kept firmly to the study of filters on
$\Bbb N$.   For measure-centering filters, at least, there are interesting
extensions to filters on uncountable sets, which I mention in 538Yg.   We
can do a good deal with the ideas of 538G-538K on cardinals less than
$\frak c$ in the presence of (for instance) Martin's axiom;  but for
anything corresponding to 538L-538M it seems that we must use a \2vm\
cardinal (541M below).

Measure-converging filters (538N) and filters with the Fatou property
(538O) form an oddly complementary pair.   I have tried to emphasize the
correspondence in the
characterizations 538Na and 538Oa
(compare 538G(v), 538Na(iv) and 538Oa(iv)),
but after this they seem to diverge.   The phrase `Fatou
property' comes from 538O(a-iii);  if you like, Fatou's Lemma says that the
Fr\'echet filter has the Fatou property.   From 538Xq(i) I see that I could
just as well have called it the `Lebesgue property'.
Note that any filter
larger than a measure-converging filter is again measure-converging, so
that if there is a measure-converging filter there is a measure-converging
ultrafilter;  but that no non-principal ultrafilter can have the Fatou
property (538Xq(ii)).   On the other hand, there are many free filters with the
Fatou property, but it is not known for sure whether there have to be
measure-converging filters.   It is possible for a
measure-converging filter to have the Fatou property (538Rd).

In the last part of the section I look at a different kind of limit.   A
`Banach limit' is an extension to $\ell^{\infty}$
of the ordinary limit regarded as a linear functional on the closed
subspace of convergent sequences;  a `medial limit' is a Banach limit which
commutes with integration in appropriate settings.   To study these
I use the formulae of repeated integration to do some
surprising things.   In 363L I tried to explain what I meant by the formula
`$\dashint\ldots d\nu$' for a {\it finitely} additive functional $\nu$.
This defines linear functionals which are positive for non-negative $\nu$.
In `repeated integrals' like $\dashiint f_n(x)\mu(dx)\nu(dn)$ (538P(iii)),
we must interpret the formula
as $\dashint\bigl(\int f_n(x)\mu(dx)\bigr)\nu(dn)$;  the `inner
integral' is an ordinary integral with respect to the countably additive
measure $\mu$, and the `outer integral' is a name for a linear functional.
In the integral $\dashint\ldots d\nu$
we have no problem with measurability,
though we must check that the integrand $n\mapsto\int f_nd\mu$ is bounded
(or, at least, satisfies the condition in 538Yl);
but when we look at the other repeated integrals, $\int\nu(a)\mu(da)$ or
$\idashint x\,d\nu\,\mu(dx)$ or $\idashint f_n(x)\nu(dn)\mu(dx)$,
the conditions of 538P must explicitly assert
that the outer integrals are defined.

Because we don't need to consider measurability, the `finitely additive
integrals' here are in some ways easy to deal with;  `disintegrations' like
$\tilde\nu=\dashint\nu_k\,\nu(dk)$
(538Rc) slide past all the usual questions.
However we must always be vigilant against the temptations of limiting
processes.   As with the Riemann integral, of course, we can integrate
the limit of a uniformly convergent sequence of functions.   But see the
manoeuvres of part (a-iii) of the proof of 538R, where
the sums $\sum_{i=0}^{\infty}\alpha_{ni}...$ demand different treatments at
different points.   And Fubini's theorem nearly disappears;  the point of
`medial functionals' is that something extraordinary has to happen before
we can expect to change the order of integration.

I have used the language of Volume 3 to express 538Re in a general form.
Of course by far the most important example is when the operator $T$ is a
conditional expectation operator (538Xu).   For more examples of operators
in $L^{\times}(L^{\infty};L^{\infty})$, see \S\S373-374.

For most of the classes of filter here, there is a question concerning
their existence.   Subject to the continuum hypothesis, there are many
Ramsey ultrafilters, and refining the argument we find that the same is
true if $\frak p=\frak c$ (538Yc).
There are many ways of forcing non-existence
of Ramsey ultrafilters, of which one of
the simplest is in 553H below.   With more difficulty, we can eliminate
$p$-point ultrafilters ({\smc Wimmers 82})
or rapid filters ({\smc Miller 80}) or nowhere dense
filters and therefore measure-centering ultrafilters
(538Hd, {\smc Shelah 98b}).   It is not known for sure that we can
eliminate measure-converging filters (538Z).

}%end of notes

\discrpage

\frfilename{mt539.tex}
\versiondate{24.5.14/30.8.18}
\copyrightdate{2007}

\def\chaptername{Topologies and measures III}
\def\sectionname{Maharam submeasures}

\newsection{539}

Continuing the work of \S\S392-394 %392 393 394
and 496, I return to Maharam
submeasures and the forms taken by the ideas of the present volume in this
context.   At least for countably generated algebras, and in some cases
more generally, many of the methods of Chapter 52 can be applied
(539B-539K). %539B 539C 539D 539G 539H 539J 539K
In 539L-539N  %539L 539M 539N
I give the main result of {\smc Balcar Jech \& Pazak 05} and
{\smc Veli\v{c}kovi\'c 05}:  it is consistent
to suppose that every Dedekind complete ccc \wsid\ Boolean algebra is a
Maharam algebra.   In 539R-539U %539R 539S 539T 539W
I introduce the idea of `exhaustivity rank' of an exhaustive submeasure.

\cmmnt{\leader{539A}{The story so far} As submeasures have hardly
appeared before in this volume, I begin by repeating some of
the essential ideas.

\spheader 539Aa If $\frak A$ is a Boolean algebra, a
{\bf submeasure} on $\frak A$ is a functional $\nu:\frak A\to[0,\infty]$
such that $\nu 0=0$, $\nu a\le\nu b$ whenever $a\Bsubseteq b$, and
$\nu(a\Bcup b)\le\nu a+\nu b$ for all $a$, $b\in\frak B$
(392A);  it is {\bf totally finite} if $\nu 1<\infty$.
If $\nu$ is a submeasure defined on an algebra of subsets of a set $X$, I
say that the {\bf null ideal} of $\nu$ is the ideal
$\Cal N(\nu)$ of subsets of $X$
generated by $\{E:\nu E=0\}$ (496Bc).
A submeasure $\nu$ on a Boolean algebra $\frak A$
is {\bf exhaustive} if $\lim_{n\to\infty}\nu a_n=0$ for every
disjoint sequence $\sequencen{a_n}$ in $\frak A$;
it 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 $i\le n$ (392Bc).
A {\bf Maharam submeasure} is a totally finite
sequentially order-continuous submeasure (393A);
a Maharam submeasure on a Dedekind
$\sigma$-complete Boolean algebra is exhaustive (393Bc).

\spheader 539Ab
A {\bf Maharam algebra} is a Dedekind $\sigma$-complete Boolean algebra
with a
strictly positive Maharam submeasure.   Any Maharam algebra is ccc and
\wsid\ (393Eb).   A Maharam algebra is measurable
iff it carries a strictly positive uniformly exhaustive submeasure (393D).
If $\nu$ is any Maharam submeasure on a Dedekind $\sigma$-complete Boolean
algebra $\frak A$, its Maharam algebra is the quotient
$\frak A/\{a:\nu a=0\}$ (496Ba).

\spheader 539Ac If $\nu$ is any
strictly positive totally finite submeasure on a Boolean algebra
$\frak A$, there is an associated metric $(a,b)\mapsto\nu(a\Bsymmdiff b)$
on $\frak A$;
the completion $\widehat{\frak A}$
of $\frak A$ under this metric is a Boolean algebra (392Hc).
If $\nu$ is exhaustive, then
$\widehat{\frak A}$ is a Maharam algebra (393H).   If $\nu$ and $\nuprime$ are
both strictly positive Maharam submeasures on the same Maharam algebra
$\frak A$, $\nu$ is absolutely continuous with respect to $\nuprime$ (393F).
Consequently the associated metrics are uniformly equivalent, and $\frak A$
has a canonical topology and uniformity, its {\bf Maharam-algebra topology}
and {\bf Maharam-algebra uniformity} (393G).

\spheader 539Ad Let $\frak A$ be a Boolean algebra.

\medskip

\quad{\bf (i)} A sequence $\sequencen{a_n}$ in $\frak A$
order*-converges to $a\in\frak A$ (definition:  367A)
iff there is a partition $B$ of unity in
$\frak A$ such that $\{n:b\Bcap(a_n\Bsymmdiff a)\ne 0\}$ is finite for
every $b\in B$ (393Ma).

\medskip

\quad{\bf (ii)} The {\bf order-sequential topology} on $\frak A$
is the topology for which the closed sets are just the sets closed under
order*-convergence (393L).

\medskip

\quad{\bf (iii)}
If $\frak A$ is ccc and Dedekind $\sigma$-complete, a subalgebra of
$\frak A$
is order-closed iff it is closed for the order-sequential topology (393O).

\medskip

\quad{\bf (iv)} If $\frak A$ is ccc and \wsid, then the closure of a
set $A\subseteq\frak A$ for the order-sequential topology is the set of
order*-limits of sequences in $A$ (393Pb).

\medskip

\quad{\bf (v)}
If $\frak A$ is a Maharam algebra, then its Maharam-algebra topology
is its order-sequential topology (393N).

\medskip

\quad{\bf (vi)}
If $\frak A$ is a Dedekind $\sigma$-complete ccc \wsid\ Boolean algebra,
and $\{0\}$
is a G$_{\delta}$ set for the order-sequential topology, then $\frak A$ is
a Maharam algebra (393Q).

\spheader 539Ae
It was a long-outstanding problem (the `Control Measure Problem') whether
every Maharam algebra is in fact a measurable algebra;  this was solved by
a counterexample in {\smc Talagrand 08}, described in \S394.

\spheader 539Af
If $X$ is a Hausdorff space, a {\bf totally finite Radon submeasure} on
$X$ is a
totally finite submeasure $\nu$ defined on a $\sigma$-algebra $\Sigma$ of
subsets of $X$ such that (i) if $E\subseteq F\in\Sigma$ and $\nu F=0$ then
$E\in\Sigma$ (ii) every open set belongs to $\Sigma$ (iii) if $E\in\Sigma$
and $\epsilon>0$ there is a compact set $K\subseteq E$ such that
$\nu(E\setminus K)\le\epsilon$ (496C).
Every totally finite Radon submeasure is a Maharam submeasure (496Da).
If $X$ is a Hausdorff space and $\nu$
is a totally finite Radon submeasure on $X$,
a set $E\in\dom\nu$ is {\bf self-supporting} if $\nu(E\cap G)>0$ whenever
$G\subseteq X$ is an open set meeting $E$.   If
$E\in\dom\nu$ and $\epsilon>0$, there is a compact self-supporting
$K\subseteq E$ such that $\nu(E\setminus K)\le\epsilon$ (496Dd).

Let $\nu$ be a strictly positive Maharam submeasure on a
Dedekind $\sigma$-complete Boolean algebra $\frak A$.
Let $Z$ be the Stone space
of $\frak A$, and write $\widehat{a}$ for the open-and-closed subset of $Z$
corresponding to each $a\in\frak A$.
Then there is a unique totally finite Radon
submeasure $\nuprime$ on $Z$ such that
$\nuprime\widehat{a}=\nu a$ for every $a\in\frak A$; the null ideal of $\nuprime$
is the nowhere dense ideal of $Z$ (496G).

\spheader 539Ag For a cardinal $\kappa$, I write $\Cal N_{\kappa}$ for the
null ideal of the usual measure on $\{0,1\}^{\kappa}$;
$\Cal N\cong\Cal N_{\omega}$ will be the null ideal of Lebesgue measure on
$\Bbb R$, and $\Cal M$ the meager ideal of $\Bbb R$.
}%end of comment

\leader{539B}{Proposition} Let $\frak A$ be a Maharam algebra,
$\tau(\frak A)$ its Maharam type and $d_{\frak T}(\frak A)$ its
topological density for its Maharam-algebra topology.
Then $\tau(\frak A)\le d_{\frak T}(\frak A)\le\max(\omega,\tau(\frak A))$.

\proof{ Recall that the Maharam-algebra topology is the order-sequential
topology (539A(d-v)).   $\frak A$ is ccc and \wsid\ (539Ab), so if
$D\subseteq\frak A$ is topologically dense, then
every element of $\frak A$ is expressible as the order*-limit
$\inf_{n\in\Bbb N}\sup_{m\ge n}a_m$ of some sequence $\sequencen{a_n}$ in
$D$ (539A(d-iv)).   In this case
$D\,\,\tau$-generates $\frak A$ and $\tau(\frak A)\le\#(D)$;
accordingly $\tau(\frak A)\le d_{\frak T}(\frak A)$.   If
$D\subseteq\frak A\,\,\tau$-generates $\frak A$, let $\frak B$ be the
subalgebra of $\frak A$ generated by $D$ and $\overline{\frak B}$ its
topological closure.   Then $\overline{\frak B}$ is order-closed (because
$\frak A$ is ccc), so is the whole of $\frak A$, and
$d_{\frak T}(\frak A)\le\#(\frak B)\le\max(\omega,\#(D))$;  accordingly
$d_{\frak T}(\frak A)\le\max(\omega,\tau(\frak A))$.
}%end of proof of 539B

\leader{539C}{Theorem} Let $\frak A$ be a Maharam algebra.

(a)

\Centerline{$(\frak A^+,\Bsupseteqshort\strprime,
[\frak A^+]^{\le\max(\omega,\tau(\frak A))})
\prGT(\Pou(\frak A),\sqsubseteq^*,\Pou(\frak A))$\dvro{.}{,}}

\cmmnt{\noindent where $\frak A^+=\frak A\setminus\{0\}$,
$(\frak A^+,\Bsupseteqshort\strprime,[\frak A^+]^{\le\kappa})$ is
defined as in 512F, and $(\Pou(\frak A),\sqsubseteq^*)$
as in 512Ee.}

(b) $\Pou(\frak A)\prT\Cal N_{\tau(\frak A)}$.

\proof{ If $\frak A=\{0\}$ these are both trivial;  suppose otherwise.
Fix a strictly positive Maharam submeasure $\nu$ on $\frak A$ such that
$\nu 1=1$.   Let $\frak B$ be a subalgebra of $\frak A$ which is
dense in $\frak A$ for the metric $(a,b)\mapsto\nu(a\Bsymmdiff b)$ and has
cardinal at most $\kappa=\max(\omega,\tau(\frak A))$ (539B).

\medskip

{\bf (a)(i)}
For $a\in\frak A^+$ choose $\phi(a)\in\Pou(\frak A)$ as follows.   Start by
taking $d_n\in\frak B$, for $n\in\Bbb N$, such that
$\nu(d_n\Bsymmdiff(1\Bsetminus a))\le 2^{-n-2}\nu a$ for each $n$;  set
$b_n=d_n\Bsetminus\sup_{i<n}b_i$ for $n\in\Bbb N$,
$a'=1\Bsetminus\sup_{n\in\Bbb N}b_n=1\Bsetminus\sup_{n\in\Bbb N}d_n$;
then every $b_n$ belongs to $\frak B$,

\Centerline{$\nu(a'\Bsetminus a)
\le\inf_{n\in\Bbb N}\nu((1\Bsetminus d_n)\setminus a)
\le\inf_{n\in\Bbb N}\nu(d_n\Bsymmdiff(1\Bsetminus a))
=0$,}

\Centerline{$\nu(a\Bsetminus a')\le\sum_{n=0}^{\infty}\nu(a\Bcap d_n)
<\nu a$,}

\noindent so $0\ne a'\Bsubseteq a$.   Now set
$\phi(a)=\{a'\}\Bcup\{b_n:n\in\Bbb N\}$.

\medskip

\quad{\bf (ii)} For $C\in\Pou(\frak A)$, set

\Centerline{$\psi(C)=\{c\Bcap b:c\in C$, $b\in\frak B\}\setminus\{0\}
\in[\frak A^+]^{\le\kappa}$.}

\medskip

\quad{\bf (iii)} Suppose that $a\in\frak A^+$, $C\in\Pou(\frak A)$ and
$\phi(a)\sqsubseteq^*C$.   Then there is a $b\in\psi(C)$ such that
$b\Bsubseteq a$.   \Prf\ Let $c\in C$ be such that $c\Bcap a'\ne 0$, where
$a'$ is defined as in (i) above.   Then
$B=\{b:b\in\phi(a)\setminus\{a'\}$, $c\Bcap b\ne 0\}$ is a finite subset of
$\frak B$, so $\sup B\in\frak B$ and $c\Bsetminus\sup B\in\psi(C)$.
But $c\Bsetminus\sup B=c\Bcap a'\Bsubseteq a$.\ \QeD\  Thus
$a\mskip5mu\Bsupseteqshort\strprime\mskip5mu\psi(C)$.

As $a$ is arbitrary, $(\phi,\psi)$ is a Galois-Tukey connection from
$(\frak A^+,\Bsupseteqshort\strprime,
[\frak A^+]^{\le\kappa})$ to
$(\Pou(\frak A),\penalty-100\sqsubseteq^*\penalty+100,\Pou(\frak A)$, and
$(\frak A^+,\Bsupseteqshort\strprime,
\penalty-100[\frak A^+]^{\le\kappa})
\prGT(\Pou(\frak A),\sqsubseteq^*,\Pou(\frak A)$.

\medskip

{\bf (b)(i)} If $\tau(\frak A)$ is finite,
then $\frak A$ is purely atomic and
$\Pou(\frak A)$ has an upper bound in itself,
as does $\Cal N_{\kappa}$;  so the
result is trivial.   Accordingly we may suppose henceforth that
$\tau(\frak A)=\kappa$ is infinite.

\medskip

\quad{\bf (ii)} If $C\in\Pou(\frak A)$,
there is a sequence $\sequencen{b_n}$ in $\frak B$ such
that $\nu b_n\le 4^{-n}$ for every $n\in\Bbb N$ and
$\{c:c\in C$, $c\notBsubseteq\sup_{i\ge n}b_i\}$ is finite for every
$n\in\Bbb N$.   \Prf\ If $C$ is finite this is trivial.   Otherwise,
set $\epsilon_n=4^{-n}/(n+2)$ for each $n\in\Bbb N$, and
enumerate $C$ as $\sequencen{c_n}$.   Let $\sequencen{k(n)}$ be a strictly
increasing sequence such that $\nu c'_n\le\epsilon_n$ for every $n$, where
$c'_n=\sup_{i\ge k(n)}c_i$;  choose $\sequencen{b_n}$ in $\frak B$
inductively so that

\Centerline{$\nu(b_n\symmdiff\sup_{j\le n}
   (c'_j\Bsetminus\sup_{j\le i<n}b_i))
\le\epsilon_{n+1}$}

\noindent for each $n\in\Bbb N$.   Then we see by induction on $n$ that

\Centerline{$\nu(c'_j\setminus\sup_{j\le i<n}b_i)\le\epsilon_n$}

\noindent whenever $j\le n$ in $\Bbb N$, and therefore that

\Centerline{$\nu b_n\le\epsilon_{n+1}+(n+1)\epsilon_n\le 4^{-n}$}

\noindent for each $n$;  while $c'_j\Bsubseteq\sup_{i\ge j}b_i$ for every
$j$, so

\Centerline{$1\Bsetminus\sup_{i\ge n}b_i\Bsubseteq 1\Bsetminus c'_n
=\sup_{i<k(n)}c_i$}

\noindent meets only finitely many members of $C$, for every $n$.\ \Qed

\medskip

\quad{\bf (iii)} Now fix on an enumeration $\ofamily{\xi}{\kappa}{b_{\xi}}$
of $\frak B$.   Consider the $\kappa$-localization relation
$(\kappa^{\Bbb N},\subseteq^*,\Cal S_{\kappa})$ (522K).
We see from (ii) that we can find a function
$\phi:\Pou(\frak A)\to\kappa^{\Bbb N}$ such that

\Centerline{$\nu b_{\phi(C)(n)}\le 4^{-n}$ for every $n\in\Bbb N$,}

\Centerline{$1\Bsetminus\sup_{i\ge n}b_{\phi(C)(i)}$ meets only finitely
many members of $C$, for every $n\in\Bbb N$.}

\noindent Next, define $\psi:\Cal S_{\kappa}\to\Pou(\frak A)$ as follows.
Given $S\in\Cal S_{\kappa}$, set $a_0(S)=1$,

\Centerline{$a_{n+1}(S)
=\sup_{m\ge n}\sup\{b_{\xi}:(m,\xi)\in S$, $\nu b_{\xi}\le 4^{-m}\}$}

\noindent for each $n$;  then
$\nu a_{n+1}(S)\le\sum_{m=n}^{\infty}2^{-m}=2^{-n+1}$ for every $n$, so
$\psi(S)=\{a_n(S)\Bsetminus a_{n+1}(S):n\in\Bbb N\}$
is a partition of unity in $\frak A$.

\medskip

\quad{\bf (iv)} Suppose that $C\in\Pou(\frak A)$ and $S\in\Cal S_{\kappa}$ are such
that $\phi(C)\subseteq^*S$.   In this case there is an $m\in\Bbb N$ such
that $(n,\phi(C)(n))\in S$ for every $n\ge m$.   Since
$\nu b_{\phi(C)(n)}\le 4^{-n}$ for every $n$,
$\sup_{i\ge n}b_{\phi(C)(i)}\Bsubseteq a_{n+1}(S)$
and $1\Bsetminus a_{n+1}(S)$ meets only finitely many members of
$C$, for every $n\ge m$.   Thus every member of $\psi(S)$ meets only
finitely many members of $C$, and $C\sqsubseteq^*\psi(S)$.

This shows that $(\phi,\psi)$ is a Galois-Tukey connection from
$(\Pou(\frak A),\sqsubseteq^*,\Pou(\frak A))$ to
$(\kappa^{\Bbb N},\subseteq^*,\Cal S_{\kappa})$, and
$(\Pou(\frak A),\sqsubseteq^*,\Pou(\frak A))
\prGT(\kappa^{\Bbb N},\subseteq^*,\Cal S_{\kappa})$.
On the other side, we know already that
$(\kappa^{\Bbb N},\subseteq^*,\Cal S_{\kappa})
\prGT(\Cal N_{\kappa},\subseteq,\penalty-50\Cal N_{\kappa})$ (524G);  so
$(\Pou(\frak A),\sqsubseteq^*,\Pou(\frak A))
\prGT(\Cal N_{\kappa},\penalty-100\subseteq\penalty+100,\Cal N_{\kappa})$,
that is, $\Pou(\frak A)\prT\Cal N_{\kappa}$.
}%end of proof of 539C

\leader{539D}{Corollary} Let $\frak A$ be a Maharam algebra.

(a) $\pi(\frak A)\le\max(\cff[\tau(\frak A)]^{\le\omega},\cf\Cal N)$.

(b) If $\tau(\frak A)\le\omega$, then $\wdistr(\frak A)\ge\add\Cal N$.

\proof{ Set $\kappa=\tau(\frak A)$.

\medskip

{\bf (a)} If $\pi(\frak A)$ is countable, or
$\pi(\frak A)\le\cff[\kappa]^{\le\omega}$, we can stop.   Otherwise,
$\kappa$ is infinite and

$$\eqalign{\max(\omega,\kappa)
&\le\max(\omega,\cff[\kappa]^{\le\omega})
<\pi(\frak A)\cr
&=\cov(\frak A^+,\Bsupseteqshort,\frak A^+)
\le\max(\omega,\kappa,
   \cov(\frak A^+,\Bsupseteqshort\strprime,[\frak A^+]^{\le\kappa}))
   \cr}$$

\noindent (512Gf), so

$$\eqalignno{\pi(\frak A)
&\le\cov(\frak A^+,\Bsupseteqshort\strprime,[\frak A^+]^{\le\kappa})
\le\cov(\Pou(\frak A),\sqsubseteq^*,\Pou(\frak A))\cr
\displaycause{539Ca, 512Da}
&=\cf\,\Pou(\frak A)
\le\cf\Cal N_{\kappa}\cr
\displaycause{539Cb, 513E(e-i)}
&=\max(\cff[\kappa]^{\le\omega},\cf\Cal N)\cr}$$

\noindent (523N).

\medskip

{\bf (b)} If $\kappa$ is finite, $\wdistr(\frak A)=\infty$ and we can stop.
Otherwise, $\kappa=\omega$ and

$$\eqalignno{\wdistr(\frak A)
&=\add\Pou(\frak A)
\Displaycause{512Ee}
\ge\add\Cal N_{\kappa}
\Displaycause{539Cb, 513E(e-ii)}
=\add\Cal N.
\cr}$$
}%end of proof of 539D

\leader{539E}{Proposition}\cmmnt{ ({\smc Veli\v{c}kovi\'c 05},
{\smc Balcar Jech \& Paz\'ak 05})} If
$\frak A$ is an atomless Maharam algebra, not $\{0\}$, there is a sequence
$\sequencen{a_n}$ in $\frak A$ such that $\sup_{n\in I}a_n=1$ and
$\inf_{n\in I}a_n=0$ for every infinite $I\subseteq\Bbb N$.

\proof{ Fix a strictly positive Maharam submeasure
$\nu$ on $\frak A$.

\medskip

{\bf (a)} If $\sequencen{a_n}$ is a sequence in $\frak A$ such that
$\delta=\inf_{n\in\Bbb N}\nu a_n$ is greater than $0$, there
are a non-zero $d\in\frak A$ and an infinite $I\subseteq\Bbb N$ such
that $d\Bsubseteq\sup_{i\in J}a_i$ for every infinite
$J\subseteq I$.   \Prf\Quer\ Otherwise, set $b_J=\sup_{i\in J}a_i$ for
$J\subseteq\Bbb N$.
Choose $\ofamily{\xi}{\omega_1}{I_{\xi}}$,
$\ofamily{\xi}{\omega_1}{c_{\xi}}$ and
$\ofamily{\xi}{\omega_1}{d_{\xi}}$ inductively, as follows.
$I_0=\Bbb N$.   The inductive hypothesis will be that $I_{\xi}$ is an
infinite subset of $\Bbb N$,
$I_{\xi}\setminus I_{\eta}$ is finite whenever $\eta\le\xi$, and
$c_{\xi}\Bcap b_{I_{\xi+1}}=0$ for every $\xi<\omega_1$.
Given $\langle I_{\eta}\rangle_{\eta\le\xi}$ where $\xi<\omega_1$,
set $d_{\xi}=\inf_{n\in\Bbb N}b_{I_{\xi}\setminus n}$.   Since
$\nu b_J\ge\delta$ for every non-empty $J\subseteq\Bbb N$,
$\nu d_{\xi}\ge\delta$
and $d_{\xi}\ne 0$.   By hypothesis, there is an infinite
$I_{\xi+1}\subseteq I_{\xi}$ such that
$c_{\xi}=d_{\xi}\Bsetminus b_{I_{\xi+1}}$
is non-zero.
Given $\ofamily{\eta}{\xi}{I_{\eta}}$ where $\xi<\omega_1$ is a non-zero
limit ordinal, let $I_{\xi}$ be an infinite set such that
$I_{\xi}\setminus I_{\eta}$ is finite for every $\eta<\xi$, and
continue.

Now observe that if $\eta<\xi<\omega_1$, $I_{\xi}\setminus I_{\eta}$ is
finite, so that there is an $n\in\Bbb N$ such that
$I_{\xi}\setminus n\subseteq I_{\eta+1}$, and

\Centerline{$c_{\xi}\Bsubseteq d_{\xi}
\Bsubseteq b_{I_{\xi}\setminus n}\Bsubseteq b_{I_{\eta+1}}$}

\noindent is disjoint from $c_{\eta}$.   But this means that
$\ofamily{\xi}{\omega_1}{c_{\xi}}$ is disjoint, which is impossible,
because $\frak A$ is ccc.\ \Bang\Qed

\medskip

{\bf (b)} Let us say that a Boolean algebra $\frak B$ {\bf splits
reals} if there is a sequence $\sequencen{b_n}$ in $\frak B$ such
that $\sup_{n\in I}b_n=1$ and $\inf_{n\in I}b_n=0$ for every infinite
$I\subseteq\Bbb N$.   Now the set of those
$d\in\frak A$ such that the principal ideal $\frak A_d$ generated by
$d$ splits reals is order-dense in $\frak A$.  \Prf\
Let $a\in\frak A^+$.

\medskip

\quad{\bf case 1} If $\nu\restrp\frak A_a$ is uniformly exhaustive, then
$\frak A_a$ is measurable (539Ab).
Let $\bar\mu$ be a probability measure on
$\frak A_a$;  because $\frak A_a$, like $\frak A$, is atomless, there
is a stochastically independent family $\sequencen{a_n}$ in
$\frak A_a$ with $\bar\mu a_n=\bover12$ for every $n$, and now
$\sequencen{a_n}$ witnesses that $\frak A_a$ splits reals.

\medskip

\quad{\bf case 2} If $\nu\restrp\frak A_a$ is not uniformly exhaustive,
let $\langle b_{ni}\rangle_{i\le n\in\Bbb N}$ be a family of elements
of $\frak A_a$ such that $\langle b_{ni}\rangle_{i\le n}$ is disjoint
for each $n$ and
$\epsilon=\inf_{i\le n\in\Bbb N}\nu b_{ni}$ is greater than $0$.   There is
a family $\ofamily{\xi}{\omega_1}{f_{\xi}}$ in
$\prod_{n\in\Bbb N}\{0,\ldots,n\}$ such that
$\{n:f_{\xi}(n)=f_{\eta}(n)\}$ is finite whenever
$\eta<\xi<\omega_1$.   (For each $\xi<\omega_1$ let
$\theta_{\xi}:\xi\to\Bbb N$ be injective.   Now define
$\ofamily{\xi}{\omega_1}{f_{\xi}}$ inductively by saying that

\Centerline{$f_{\xi}(n)
=\min(\Bbb N\setminus\{f_{\eta}(n):\eta<\xi$, $\theta_{\xi}(\eta)<n\})$}

\noindent for every $\xi<\omega_1$ and $n\in\Bbb N$.)

\Quer\ If for every
$\xi<\omega_1$ and $I\in[\Bbb N]^{\omega}$ there is a
$J\in[I]^{\omega}$ such that
$\inf_{i\in J}b_{i,f_{\xi}(i)}\ne 0$, choose
$\ofamily{\xi}{\omega_1}{I_{\xi}}$ inductively so that
$I_{\xi}\in[\Bbb N]^{\omega}$, $I_{\xi}\setminus I_{\eta}$ is finite
for every $\eta<\xi$, and
$c_{\xi}=\inf_{i\in I_{\xi}}b_{i,f_{\xi}(i)}$ is non-zero for every
$\xi<\omega_1$.   Then
whenever $\eta<\xi$ the set $I_{\xi}\cap I_{\eta}$ is infinite, so
there is an
$i\in I_{\xi}\cap I_{\eta}$ such that $f_{\xi}(i)\ne f_{\eta}(i)$;
now $c_{\xi}\Bcap c_{\eta}
\Bsubseteq b_{i,f_{\xi}(i)}\Bcap b_{i,f_{\eta}(i)}=0$.
But this means that we have an uncountable
disjoint family in $\frak A_a$, which is impossible, because
$\frak A$ is ccc.\ \Bang

Thus we have a $\xi<\omega_1$ and an infinite $I\subseteq\Bbb N$ such
that $\inf_{i\in J}d_i=0$ for every infinite $J\subseteq I$, where
$d_i=b_{i,f_{\xi}(i)}$ for $i\in I$.   Next, applying (a) to
$\familyiI{d_i}$, we have an infinite $K\subseteq I$ and a $d\ne 0$
such that $d=\sup_{i\in J}d_i$ for every infinite $J\subseteq K$.
But this means that
$\family{i}{K}{d\Bcap d_i}$ witnesses that $\frak A_d$ splits reals;
while $d\Bsubseteq a$.

As $a$ is arbitrary, we have the result.\ \Qed

\medskip

{\bf (c)} By 313K, there is a partition $D$ of unity in $\frak A$
such that
$\frak A_d$ splits reals for every $d\in D$;  choose a sequence
$\sequencen{a_{dn}}$ in $\frak A_d$ witnessing this for each $d\in D$.
Set $a_n=\sup_{d\in D}a_{dn}$ for each $n$.   If $I\subseteq\Bbb N$ is
infinite, then

\Centerline{$\sup_{n\in I}a_n=\sup_{d\in D}\sup_{n\in I}a_{dn}
=\sup D=1$,}

\noindent while

\Centerline{$d\Bcap\inf_{n\in I}a_n=\inf_{n\in I}a_{dn}=0$}

\noindent for every $d\in D$, so $\inf_{n\in I}a_n=0$.   Thus
$\sequencen{a_n}$ witnesses that $\frak A$ splits reals, as claimed.
}%end of proof of 539E

\leader{539F}{Definition}\cmmnt{ For the next result I need a name for
one more cardinal between $\omega_1$ and $\frak c$.}   The {\bf
splitting number} $\frak s$ is the least cardinal of any family
$\Cal A\subseteq\Cal P\Bbb N$ such that for every infinite
$I\subseteq\Bbb N$ there is an $A\in\Cal A$ such that $I\cap A$ and
$I\setminus A$ are both infinite.

\leader{539G}{Proposition} Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of
subsets of $X$, and $\nu$ an atomless Maharam submeasure on $\Sigma$.
Let $\Cal M$ be the ideal of meager subsets of $\Bbb R$.

(a) $\non\Cal N(\nu)\ge\max(\frak s,\frakmctbl)$.

(b) $\cov\Cal N(\nu)\le\non\Cal M$.

\proof{ If $\nu X=0$, these are both trivial;  suppose otherwise.

\medskip

{\bf (a)(i)} Suppose that $D\subseteq X$ and $\#(D)<\frakmctbl$.
For any $\epsilon>0$, there is an $F\in\Sigma$ such that
$D\subseteq F$ and $\nu F\le\epsilon$.   \Prf\ By 393I,
there is for each
$n\in\Bbb N$ a finite partition $\Cal E_n$ of $X$ into members of $\Sigma$
such that $\nu E\le 2^{-n-1}\epsilon$ for each $E\in\Cal E_n$.
Express each $\Cal E_n$ as
$\{E_{ni}:i<k(n)\}$.   For $x\in D$,
let $f_x\in\prod_{n\in\Bbb N}k(n)$ be such that $x\in E_{n,f_x(n)}$ for
every $n$.   Because $\#(D)<\frakmctbl$, there is an
$f\in\NN$ such that $f\cap f_x\ne\emptyset$ for every $x\in D$
(522Sb);  we may suppose that $f(n)<k(n)$ for every
$n$.   Set $F=\bigcup_{n\in\Bbb N}E_{n,f(n)}$;  this works.\ \Qed

Applying this repeatedly, we get a sequence
$\sequencen{F_n}$ in $\Sigma$ such that $D\subseteq F_n$ and
$\nu F_n\le 2^{-n}$ for every $n$;  now $F=\bigcap_{n\in\Bbb N}F_n$
includes $D$ and belongs to $\Cal N(\nu)$.   As $D$ is arbitrary,
$\non\Cal N(\nu)\ge\frakmctbl$.

\medskip

\quad{\bf (ii)} Set $\frak A=\Sigma/\Sigma\cap\Cal N(\nu)$, and
define $\bar\nu:\frak A\to\coint{0,\infty}$ by setting
$\bar\nu E^{\ssbullet}=\nu E$ for every $E\in\Sigma$.   Then $\bar\nu$
is a strictly positive atomless Maharam submeasure on $\frak A$.
By 539E, there is a sequence $\sequencen{a_n}$ in $\frak A$ such that
$\sup_{n\in I}a_n=1$ and $\inf_{n\in I}a_n=0$ for every infinite
$I\subseteq\Bbb N$.   For each $n\in\Bbb N$, let $E_n\in\Sigma$ be such
that $E_n^{\ssbullet}=a_n$.

Suppose that $D\subseteq X$ and $\#(D)<\frak s$.   For $x\in D$, set
$A_x=\{n:x\in E_n\}$.   Because $\#(D)<\frak s$, there is an infinite
$I\subseteq\Bbb N$ such that one of $I\cap A_x$, $I\setminus A_x$ is
finite for every $x\in D$.   Set

\Centerline{$F
=\bigcup_{m\in\Bbb N}\bigl((X\setminus\bigcup_{n\in I\setminus m}E_n)
  \cup(\bigcap_{n\in I\setminus m}E_n)\bigr)$;}

\noindent then

\Centerline{$F^{\ssbullet}
=\sup_{m\in\Bbb N}\bigl((1\Bsetminus\sup_{n\in I\setminus m}a_n)
  \cup(\inf_{n\in I\setminus m}a_n)\bigr)=0$,}

\noindent so $F\in\Cal N(\nu)$, while $D\subseteq F$.   As $D$ is
arbitrary, $\non\Cal N(\nu)\ge\frak s$.

\medskip

{\bf (b)} Let $\sequencen{k(n)}$, $\langle E_{ni}\rangle_{i<k(n)}$ and
$\family{x}{X}{f_x}$ be as in (a-i) above, with $\epsilon=1$.   Give
$Z=\prod_{n\in\Bbb N}k(n)$ its compact metrizable product topology.
By 522Wb, there is a family $\ofamily{\xi}{\non\Cal M}{g_{\xi}}$ in
$Z$ such that $\{g_{\xi}:\xi<\non\Cal M\}$ is non-meager.   For each
$f\in Z$, the set

\Centerline{$H(f)=\bigcap_{m\in\Bbb N}\bigcup_{n\ge m}
  \{g:g\in Z$, $g(n)=f(n)\}$}

\noindent is comeager in $Z$, so contains some $g_{\xi}$;  turning this
round, $Z=\bigcup_{\xi<\non\Cal M}H(g_{\xi})$.   Consider the sets
$F_{\xi}=\{x:x\in X$, $f_x\in H(g_{\xi})\}$;  then
$X=\bigcup_{\xi<\non\Cal M}F_{\xi}$, while

\Centerline{$\nu F_{\xi}
\le\inf_{m\in\Bbb N}\sum_{n=m}^{\infty}\nu E_{n,g_{\xi}(n)}
=0$}

\noindent for every $\xi$.   So $\cov\Cal N(\nu)\le\non\Cal M$.
}%end of proof of 539G

\leader{539H}{Corollary} Let $\frak A$ be an atomless Maharam
algebra, not $\{0\}$.   Then $d(\frak A)\ge\max(\frak s,\frakmctbl)$.

\proof{ Let $Z$ be the Stone space of $\frak A$ and $\nuprime$ the totally
finite Radon submeasure on $Z$ corresponding to a strictly positive
Maharam submeasure $\nu$ on $\frak A$ (539Af), so that $\Cal N(\nuprime)$ is
the ideal of meager subsets of $Z$.
Note that the meager sets of $Z$ are all nowhere dense, because $\frak A$
is \wsid\ (316I).   Because $\frak A$ is atomless, so are $\nu$ and $\nuprime$.
As every meager subset of
$Z$ is nowhere dense (and $Z\ne\emptyset$), no dense set can be meager, and

$$\eqalignno{d(\frak A)
&=d(Z)
\Displaycause{514Bd}
\ge\non\Cal N(\nuprime)\ge\max(\frak s,\frakmctbl)
\cr}$$

\noindent by 539Ga.
}%end of proof of 539H

\leader{539I}{Corollary}\dvAformerly{5{}55L} Suppose that
$\#(X)<\max(\frak s,\frakmctbl)$, where $\frak s$ is the splitting number.
Let $\Sigma$ be a $\sigma$-algebra of subsets of $X$ such that
$(X,\Sigma)$ is countably separated\cmmnt{, in the sense that there is a
sequence in $\Sigma$ separating the points of $X$},
and $\Cal I$ a $\sigma$-ideal of
$\Sigma$ containing singletons.   Then there is no non-zero Maharam
submeasure on $\Sigma/\Cal I$.

\proof{{\bf (a)}
Let $\mu$ be a Maharam submeasure on $\Sigma/\Cal I$.   Then we
have a Maharam submeasure $\nu$ on $\Sigma$ defined by setting
$\nu E=\mu E^{\ssbullet}$ for every $E\in\Sigma$, and $\nu\{x\}=0$ for
every $x\in X$.

\medskip

{\bf (b)} $\nu$ is atomless.   \Prf\ Let $\sequencen{E_n}$ be a sequence in
$\Sigma$ separating the points of $X$, and $F\in\Sigma$ such that
$\nu F>0$.
Choose $\sequencen{F_n}$ inductively so that $F_0=F$ and, given that
$\nu F_n>0$, $F_{n+1}$ is either $F_n\cap E_n$ or $F_n\setminus E_n$ and
$\nu F_{n+1}>0$.   Then $\bigcap_{n\in\Bbb N}F_n$ has at most one member,
so $\lim_{n\to\infty}\nu F_n=0$, and there is an $n$ such that
$\nu F_n=\nu(F\cap F_n)$ and $\nu(F\setminus F_n)$ are non-zero.\ \Qed

\medskip

{\bf (c)} By 539Ga,

\Centerline{$\non\Cal N(\nu)\ge\max(\frak s,\frakmctbl)>\#(X)$}

\noindent and $\nu X=0$, so $\mu$ is identically $0$.
}%end of proof of 539I

\leader{539J}{Theorem} (a) Let $\nu$ be a totally finite Radon submeasure
on a Hausdorff
space $X$\cmmnt{ (539Af)} and $\frak A$ its Maharam algebra.
Then $\Cal N(\nu)\prT\Pou(\frak A)$.

(b) Let $\nu$ be a totally finite Radon submeasure on a Hausdorff
space $X$ and $\frak A$ its Maharam algebra.

\quad(i) $\wdistr(\frak A)\le\add\Cal N(\nu)$.

\quad(ii) $\tau(\frak A)\le w(X)$.

\quad(iii)
$\cf\Cal N(\nu)\le\max(\cff[\tau(\frak A)]^{\le\omega},\cf\Cal N)$.

\quad(iv) If $\tau(\frak A)\le\omega$\cmmnt{ (e.g., because $X$ is
second-countable)}, then
$\add\Cal N(\nu)\ge\add\Cal N$ and $\cf\Cal N(\nu)\le\cf\Cal N$.

\proof{{\bf (a)} For $E\in\Cal N(\nu)$, let $\Cal K_E$ be a maximal
disjoint family of compact sets of non-zero submeasure disjoint from $E$,
and set $C_E=\{K^{\ssbullet}:K\in\Cal K_E\}$.   Because $\nu$ is inner
regular with respect to the compact sets, $C_E\in\Pou(\frak A)$.
Now $E\mapsto C_E:\Cal N(\nu)\to\Pou(\frak A)$ is a Tukey function.   \Prf\
Suppose that $\Cal E\subseteq\Cal N(\nu)$ and $D\in\Pou(\frak A)$ are such
that $C_E\sqsubseteq^*D$ for every $E\in\Cal E$;  take any $\epsilon>0$.
Because $D$ is
countable, we have a countable partition $\Cal H$ of $X$ into measurable
sets such that $D=\{H^{\ssbullet}:H\in\Cal H\}$.   Because $\nu$ is inner
regular with respect to the self-supporting compact sets (539Af),
we can find a self-supporting compact set $K\subseteq X$ such that
$\nu(X\setminus K)\le\epsilon$ and $K$ is covered by finitely many members
of $\Cal H$;  consequently $K^{\ssbullet}$ meets only finitely many members
of $D$.

If $E\in\Cal E$, then $K^{\ssbullet}$ meets only finitely many members of
$C_E$, so there is a finite set $\Cal K'_E\subseteq\Cal K_E$ such that
$K\setminus K_E$ is negligible, where $K_E=\bigcup\Cal K'_E$.   But $K_E$
is compact and $K$ is self-supporting, so $K\subseteq K_E$ and
$K\cap E=\emptyset$.

This means that $\bigcup\Cal E\subseteq X\setminus K$ is included in an
open set of submeasure at most $\epsilon$.   This is true for every
$\epsilon>0$, so $\bigcup\Cal E$ is included in a negligible G$_{\delta}$
set and belongs to $\Cal N(\nu)$;  that is, $\Cal E$ is bounded above in
$\Cal N(\nu)$.
As $\Cal E$ is arbitrary, $E\mapsto C_E$ is a Tukey function.\ \Qed

\medskip

{\bf (b)(i)} Putting (a) and 513E(e-ii) together,

\Centerline{$\wdistr(\frak A)=\add\Pou(\frak A)\le\add\Cal N(\nu)$.}

\medskip

\quad{\bf (ii)} If $\Cal U$ is a base for the topology of $X$ with
$\#(\Cal U)=w(X)$, consider $D=\{U^{\ssbullet}:U\in\Cal U\}$ and the
order-closed subalgebra $\frak B$ of $\frak A$ generated by $D$;  note that
$\frak B$ is closed for the order-sequential (or Maharam-algebra)
topology of $\frak A$ (539Ad).
Let $\Cal E$ be the algebra of sets generated by $\Cal U$.
If $F\in\dom\nu$ and $\epsilon>0$, there are compact sets
$K\subseteq F$, $L\subseteq X\setminus F$ such that
$\nu(X\setminus(K\cup L))\le\epsilon$.   There is an $E\in\Cal E$ such
that $K\subseteq E\subseteq X\setminus L$, so
$\nu(E\symmdiff F)\le\epsilon$.   Now $E^{\ssbullet}\in\frak B$ and
$\bar\nu(F^{\ssbullet}\Bsymmdiff E^{\ssbullet})\le\epsilon$;  as $\epsilon$
is arbitrary, $F^{\ssbullet}\in\frak B$;  as $F$ is arbitrary,
$\frak B=\frak A$ and $\frak A$ is $\tau$-generated by $D$.   This means
that $\tau(\frak A)\le\#(D)\le w(X)$, as required.

\medskip

\quad{\bf (iii)} Setting $\kappa=\tau(\frak A)$, (a) and 539Cb tell
us that
$\Cal N(\nu)\prT\Cal N_{\kappa}$, where $\Cal N_{\kappa}$ is the null ideal
of the usual measure on $\{0,1\}^{\kappa}$.
So $\add\Cal N(\nu)\ge\add\Cal N_{\kappa}$ and

\Centerline{$\cf\Cal N(\nu)\le\cf\Cal N_{\kappa}
\le\max(\cff[\kappa]^{\le\omega},\cf\Cal N)$}

\noindent (513E(e-i), 523N).

\medskip

\quad{\bf (iv)} If $\kappa\le\omega$ then
$\Cal N_{\kappa}\prT\Cal N$ so $\add\Cal N(\nu)\ge\add\Cal N$ and
$\cf\Cal N(\nu)\le\cf\Cal N$.
}%end of proof of 539J

\leader{539K}{}\cmmnt{ We can approach precalibers by some of the same
combinatorial methods as before.

\wheader{539K}{6}{2}{2}{72pt}

\noindent}{\bf Proposition} Let $\frak A$ be a Boolean algebra and $\nu$ an
exhaustive submeasure on $\frak A$.

(a) Let $\sequence{i}{a_i}$ be a
sequence in $\frak A$ such that $\inf_{i\in\Bbb N}\nu a_i>0$.

\quad (i) There is an infinite $I\subseteq\Bbb N$ such that
$\{a_i:i\in I\}$ is centered.

\quad(ii) For every $k\in\Bbb N$ there are an $S\in[\Bbb N]^{\omega}$ and
a $\delta>0$ such that
$\nu(\inf_{i\in J}a_i)\ge\delta$ for every $J\in[S]^k$.

(b) Suppose that $\ofamily{\xi}{\kappa}{a_{\xi}}$ is a
family in $\frak A$ such that $\inf_{\xi<\kappa}\nu a_{\xi}>0$,
where $\kappa$ is a regular uncountable cardinal.
Then for every $k\in\Bbb N$ there are a stationary set
$S\subseteq\kappa$ and a $\delta>0$ such that
$\nu(\inf_{i\in J}a_i)\ge\delta$ for every $J\in[S]^k$.

(c) If $\nu$ is strictly positive, then $(\kappa,\kappa,k)$ is
a precaliber triple of $\frak A$ for every regular uncountable
cardinal $\kappa$ and every $k\in\Bbb N$;  in particular, $\frak A$
satisfies Knaster's condition.

\proof{{\bf (a)(i)} This is 392J.

\medskip

\quad{\bf (ii)} Induce on $k$.   The cases $k=0$, $k=1$ are
trivial.   For the inductive step to $k+1$, let
$M\in[\Bbb N]^{\omega}$ and $\delta>0$ be such that
$\nu(\inf_{i\in J}a_i)\ge\delta$ for every $J\in[M]^k$.   \Quer\
Suppose, if possible, that
for every $S\in[M]^{\omega}$ and $\gamma>0$ there is a
$J\in[S]^{k+1}$ such that
$\nu(\inf_{i\in J}a_i)<\gamma$.   Using Ramsey's theorem (4A1G) repeatedly,
we can find
$\sequencen{I_n}$ such that $I_0\in[M]^{\omega}$,
$I_{n+1}\in[I_n]^{\omega}$, $r_n=\min I_n\notin I_{n+1}$ and
$\nu(\inf_{i\in J}a_i)\le 2^{-n-2}\delta$ for every $n\in\Bbb N$ and
$J\in[I_n]^{k+1}$.   Set $S=\{r_n:n\in\Bbb N\}$.
If $J\in[S]^k$ and $\min J=r_n$, then $J\cup\{r_m\}\in[I_m]^{k+1}$, so
$\nu(\inf_{i\in J}a_i\Bcap a_{r_m})\le 2^{-m-2}\delta$, for every
$m<n$.   It follows that
$\nu(\inf_{i\in J}a_i\Bcap\sup_{m<n}a_{r_m})\le\bover12\delta$ and
$\nu(\inf_{i\in J}a_i\Bsetminus\sup_{m<n}a_{r_m})\ge\bover12\delta$.
But this means that
$\nu c_n\ge\bover12\delta$ where
$c_n=a_{r_n}\Bsetminus\sup_{m<n}a_{r_m}$ for each $n$.   As
$\sequencen{c_n}$ is disjoint, this is impossible.\ \Bang

Thus we can find $\gamma>0$ and $S\in[M]^{\omega}$ such that
$\nu(\inf_{i\in J}a_i)\ge\gamma$ for every $J\in[S]^{k+1}$, and the
induction continues.

\medskip

{\bf (b)} Again induce on $k$.   The cases $k=0$, $k=1$ are
trivial.   For the inductive step to $k+1\ge 2$, write
$c_J=\inf_{i\in J}a_i$ for $J\in[\kappa]^{<\omega}$.   We know from
the inductive hypothesis that there are a stationary set
$S\subseteq\kappa$ and a $\delta>0$ such that $\nu c_J\ge 3\delta$
for every $J\in[S]^k$.   For each $\xi\in S$, choose
$m(\xi)\in\Bbb N$ and $\ofamily{i}{m(\xi)}{J_{\xi i}}$ as follows.   Given
$\ofamily{i}{j}{J_{\xi i}}$, where $j\in\Bbb N$, choose, if possible,
$J_{\xi j}\in[S\cap\xi]^k$
such that $\nu(c_{J_{\xi j}}\Bcap c_{J_{\xi i}})\le 2^{-i}\delta$ for
every $i<j$ and
$\nu(a_{\xi}\Bcap c_{J_{\xi j}})\le 2^{-j}\delta$;  if this is not
possible, set $m(\xi)=j$ and stop.   Now the point is that we always
do have to stop.   \Prf\Quer\ Otherwise, set
$d_i=c_{J_{\xi i}}$ for each $i\in\Bbb N$.   Because
$J_{\xi i}\in[S]^k$, $\nu d_i\ge 3\delta$ for each $i$;  also
$\nu(d_i\Bcap d_j)\le 2^{-i}\delta$ for $i<j$;  so $\nu d'_j\ge\delta$,
where $d'_j=d_j\Bsetminus\sup_{i<j}d_i$ for each $j$.   But now
$\sequence{j}{d'_j}$ is disjoint and $\nu$ is not exhaustive.\
\Bang\Qed

At the end of the process, we have $m(\xi)$ and
$\ofamily{i}{m(\xi)}{J_{\xi i}}$ for each
$\xi\in S$.   By the Pressing-Down Lemma (4A1Cc), there are $\tilde m$ and
$\ofamily{i}{\tilde m}{\tilde J_i}$ such that
$S'=\{\xi:\xi\in S$, $m(\xi)=\tilde m$, $J_{\xi i}=\tilde J_i$ for
every $i<\tilde m\}$ is
stationary in $\kappa$.   \Quer\ Suppose, if possible, that
$I\in[S']^{k+1}$ and
$\nu c_I\le 2^{-\tilde m}\delta$.   Set $\xi=\max I$,
$J=I\setminus\{\xi\}$, $\eta=\min I\in J$.
Then $J\in[S\cap\xi]^k$.   For each $i<\tilde m=m(\xi)$,

\Centerline{$\nu(c_J\Bcap c_{J_{\xi i}})
\le\nu(a_{\eta}\Bcap c_{J_{\xi i}})
=\nu(a_{\eta}\Bcap c_{J_{\eta i}})\le 2^{-i}\delta$,}

\noindent while

\Centerline{$\nu(a_{\xi}\Bcap c_J)=\nu c_I\le 2^{-\tilde m}\delta$.}

\noindent  But this means that we could have extended the sequence
$\ofamily{i}{\tilde m}{J_{\xi i}}$ by setting $J_{\xi\tilde m}=J$.\
\Bang

So $S'$ and $2^{-\tilde m}\delta$ provide the next step in the
induction.

\medskip

{\bf (c)} This is now immediate from (b).
}%end of proof of 539K

\leader{539L}{}\cmmnt{ I come now to the work of
{\smc Balcar Jech \& Paz\'ak 05}, based on the characterizations of Maharam
algebras set out in \S393.

\woddheader{539L}{4}{2}{2}{60pt}

\noindent}{\bf Lemma}\cmmnt{ ({\smc Quickert 02})} Let
$\frak A$ be a Boolean
algebra, and $\Cal I$ the family of countable subsets $I$ of
$\frak A$ for which there is a partition $C$ of unity
such that $\{a:a\in I$, $a\Bcap c\ne 0\}$ is finite for every $c\in C$.

(a) $\Cal I$ is an ideal of $\Cal P\frak A$ including
$[\frak A]^{<\omega}$.

(b) If $A\subseteq\frak A^+$ is such that $A\cap I$ is
finite for every $I\in\Cal I$, and
$B=\{b:b\Bsupseteq a$ for some $a\in A\}$, then $B\cap I$ is finite
for every $I\in\Cal I$.

(c) If $\frak A$ is ccc, then there is no uncountable
$B\subseteq\frak A$ such that $[B]^{\le\omega}\subseteq\Cal I$.

(d) If $\frak A$ is ccc and \wsid, $\Cal I$ is a
$p$-ideal\cmmnt{ (definition:  5A6Ga)}.

\proof{{\bf (a)} Of course every finite subset of $\frak A$
belongs to $\Cal I$.   If $I_0$, $I_1\in\Cal I$ and
$J\subseteq I_0\cup I_1$, then
$J\in[\frak A]^{\le\omega}$.   For each $j$, we have a partition $C_j$
of unity in $\frak A$ such that $\{a:a\in I_j$, $a\Bcap c\ne 0\}$ is
finite for every $c\in C_j$.   Set
$C=\{c_0\Bcap c_1:c_0\in C_0$, $c_1\in C_1\}$;  then $C$ is a
partition
of unity in $\frak A$ and $\{a:a\in J$, $a\Bcap c\ne 0\}$ is finite
for every $c\in C$.

\medskip

{\bf (b)} Take $I\in\Cal I$.   Set $J=B\cap I$.   For each
$b\in J$, let $a_b\in A$ be such that $a_b\Bsubseteq b$.   Let $C$ be
a partition of unity such that $\{b:b\in I$, $b\Bcap c\ne 0\}$ is finite
for every $c\in C$;  then $\{a_b:b\in J$, $a_b\Bcap c\ne 0\}$ is
finite
for every $c\in C$, so $\{a_b:b\in J\}$ belongs to $\Cal I$ and must be
finite.   \Quer\ If $J$ is infinite, there is an $a\in A$ such that
$K=\{b:b\in J$, $a=a_b\}$ is infinite;  but in this case there is a
$c\in C$ such that $a\Bcap c\ne 0$ and $b\Bcap c\ne 0$ for every
$b\in K$.\ \BanG\  So $J$ is finite, as claimed.

\medskip

{\bf (c)} Let $\widehat{\frak A}$ be the Dedekind completion of
$\frak A$ (314U).   Let $B\subseteq\frak A$ be an uncountable set, and
$\ofamily{\xi}{\omega_1}{b_{\xi}}$ a family of distinct elements of
$B$.   Set
$d=\inf_{\xi<\omega_1}\sup_{\xi\le\eta<\omega_1}b_{\eta}$, taken in
$\widehat{\frak A}$.   Then (because $\widehat{\frak A}$ is ccc, by
514Ee) $d=\sup_{\xi\le\eta<\omega_1}b_{\eta}$ for
some $\xi$ (316E);  in particular, $d\ne 0$.   Next, we can find a strictly
increasing sequence $\sequencen{\xi_n}$ in $\omega_1$ such that
$d\Bsubseteq\sup_{\xi_n\le\eta<\xi_{n+1}}b_{\eta}$ for every
$n\in\Bbb N$.   Set
$I=\{b_{\eta}:\eta<\sup_{n\in\Bbb N}\xi_n\}\in[B]^{\le\omega}$.   If
$C$ is any partition of unity in $\frak A$, there must be some $c\in C$
such that $c\Bcap d\ne 0$, and now $\{a:a\in I$, $a\Bcap c\ne 0\}$ is
infinite.   So $I\notin\Cal I$.

\medskip

{\bf (d)} Let $\sequencen{I_n}$ be a sequence in $\Cal I$.
For each $n\in\Bbb N$, let $C_n$ be a partition of unity such
that $\{a:a\in I_n$, $a\Bcap c\ne 0\}$ is finite for every $c\in C_n$.
Let $D$ be a partition of unity such that
$\{c:c\in C_n$, $c\Bcap d\ne 0\}$ is finite for every $d\in D$ and
$n\in\Bbb N$.   Then

\Centerline{$\{a:a\in I_n$, $a\Bcap d\ne 0\}
\subseteq\bigcup_{c\in C_n,c\Bcap d\ne 0}
  \{a:a\in I_n$, $a\Bcap c\ne 0\}$}

\noindent is finite for every $d\in D$ and $n\in\Bbb N$.   Let
$\sequencen{d_n}$ be a sequence running over $D\cup\{\emptyset\}$ and
set $I=\bigcup_{n\in\Bbb N}\{a:a\in I_n$, $a\Bcap d_i=0$ for every
$i\le n\}$.   Then

\Centerline{$I_n\setminus I\subseteq\bigcup_{i\le n}\{a:a\in I_n$,
$a\Bcap d_i\ne\emptyset\}$}

\noindent is finite for each $n$.   Also

\Centerline{$\{a:a\in I$, $a\Bcap d_n\ne 0\}
\subseteq\bigcup_{i<n}\{a:a\in I_i$, $a\Bcap d_n\ne 0\}$}

\noindent is finite for each $n$, so $I\in\Cal I$.
}%end of proof of 539L

\medskip

\noindent{\bf Remark}\cmmnt{ In this context,} $\Cal I$ is called
{\bf Quickert's ideal}.

\leader{539M}{Lemma} Let $\frak A$
be a \wsid\ ccc Dedekind $\sigma$-complete Boolean algebra, and suppose
that $\frak A^+$ is expressible as $\bigcup_{k\in\Bbb N}D_k$
where no infinite subset of any $D_k$ belongs to Quickert's ideal
$\Cal I$.   Then $\frak A$ is a Maharam algebra.

\proof{ The point is that if $\sequencen{a_n}$ is a
sequence in $\frak A$ which order*-converges to $0$, then
$\{a_n:n\in\Bbb N\}\in\Cal I$ (539A(d-i)).   So no
sequence in any $D_k$ can order*-converge to $0$.   Because $\frak A$
is \wsid\ and ccc, $0$ does not belong to the closure $\overline{D}_k$ of
$D_k$ for the order-sequential topology on $\frak A$
(539A(d-iv)).   So $\frak A^+=\bigcup_{k\in\Bbb N}\overline{D}_k$
is F$_{\sigma}$ and $\{0\}$ is G$_{\delta}$ for the order-sequential
topology.   It follows that $\frak A$ is a Maharam algebra (539A(d-vi)).
}%end of proof of 539M

\leader{539N}{Theorem} ({\smc Balcar Jech \& Paz\'ak 05},
{\smc Veli\v{c}kovi\'c 05}) Suppose that
Todor\v{c}evi\'c's $p$-ideal dichotomy\cmmnt{ (5A6Gb)} is true.
Then every Dedekind
$\sigma$-complete ccc \wsid\ Boolean algebra is a Maharam algebra.

\proof{ Let $\frak A$ be a Dedekind $\sigma$-complete ccc \wsid\
Boolean algebra.   Let $\Cal I$ be Quickert's ideal on $\frak A$;
then $\Cal I$ is a $p$-ideal (539Ld).
By 539Lc, there is no $B\in[\frak A]^{\omega_1}$ such that
$[B]^{\le\omega}\subseteq\Cal I$.
We are assuming that Todor\v{c}evi\'c's $p$-ideal dichotomy is true;  so
$\frak A$ must be expressible as $\bigcup_{n\in\Bbb N}D_n$ where no
infinite subset of any $D_n$ belongs to
$\Cal I$.   By 539M, $\frak A$ is a Maharam algebra.
}%end of proof of 539N

\leader{539O}{Corollary} Suppose that
Todor\v{c}evi\'c's $p$-ideal dichotomy is true.   Let $\frak A$ be a
Dedekind complete Boolean algebra such
that every countably generated order-closed subalgebra of $\frak A$ is a
measurable algebra.   Then $\frak A$ is a measurable algebra.

\proof{{\bf (a)} $\frak A$ is ccc.   \Prf\Quer\ Otherwise, let
$\ofamily{\xi}{\omega_1}{a_{\xi}}$ be a disjoint family of
non-zero elements of $\frak A$.
Let $f:\omega_1\to\{0,1\}^{\Bbb N}$ be an injective function, and set
$b_n=\sup\{a_{\xi}:\xi<\omega_1$, $f_{\xi}(n)=1\}$ for each $n$;  let
$\frak B$ be the order-closed subalgebra of $\frak A$ generated by
$\{b_n:n\in\Bbb N\}\cup\{\sup_{\xi<\omega_1}a_{\xi}\}$.
Then $a_{\xi}\in\frak B$ for every $\xi<\omega_1$, so $\frak B$ is not ccc;
but $\frak B$ is supposed to be measurable.\ \Bang\Qed

\medskip

{\bf (b)} $\frak A$ is \wsid.   \Prf\ Let $\sequencen{C_n}$ be a sequence
of partitions of unity in $\frak A$.   As $\frak A$ is ccc, every $C_n$ is
countable;  let $\frak B$ be the order-closed subalgebra of $\frak A$
generated by $\bigcup_{n\in\Bbb N}C_n$.   Then $\frak B$ is measurable,
therefore \wsid, and there is a partition $D$ of unity in $\frak B$ such
that $\{c:c\in C_n$, $c\Bcap d\ne 0\}$ is finite for every $n\in\Bbb N$ and
$d\in D$.   As $\frak B$ is order-closed, $D$ is still a partition of unity
in $\frak A$.   As $\sequencen{C_n}$ is arbitrary, $\frak A$ is \wsid.\
\Qed

\medskip

{\bf (c)} By 539N, $\frak A$ is a Maharam algebra;  let $\nu$ be a strictly
positive Maharam submeasure on $\frak A$.   Now $\nu$ is uniformly
exhaustive.   \Prf\Quer\ Otherwise, there are $\epsilon>0$ and a family
$\langle a_{ni}\rangle_{i\le n\in\Bbb N}$ in $\frak A$ such that
$\langle a_{ni}\rangle_{i\le n}$ is disjoint for every $n\in\Bbb N$ and
$\nu a_{ni}\ge\epsilon$ whenever $i\le n\in\Bbb N$.   Let $\frak B$ be the
order-closed subalgebra of $\frak A$ generated by
$\{a_{ni}:i\le n\in\Bbb N\}$.   Then $\frak B$ is a measurable algebra;
let $\bar\mu$ be a functional such that $(\frak B,\bar\mu)$ is a totally
finite measure algebra.   Since $\bar\mu$ and $\nu\restr B$ are both
strictly positive Maharam submeasures on $\frak B$, $\nu$ is absolutely
continuous with respect to $\bar\mu$ (539Ac).
But $\nu a_{ni}\ge\epsilon$ for every $n$ and
$i$, while $\inf_{i\le n\in\Bbb N}\bar\mu a_{ni}$ must be zero.\ \Bang\Qed

\medskip

{\bf (d)} So $\frak A$ is a Dedekind $\sigma$-complete Boolean algebra
with a strictly positive uniformly exhaustive Maharam submeasure, and is a
measurable algebra (539Ab).
}%end of proof of 539O

\leader{539P}{}\cmmnt{ I should say at once that 539N-539O really do need
some special axiom.   In fact the following example was found at the very
beginning of the study of Maharam algebras.

\medskip

\noindent}{\bf Souslin algebras:  Proposition}
Suppose that $T$ is a well-pruned
Souslin tree\cmmnt{ (554Yc, 5A1Ed)}, and set $\frak A=\RO^{\uparrow}(T)$.

(a) $\frak A$ is Dedekind complete, ccc and \wsid.

(b) If $\frak B$ is an order-closed subalgebra of $\frak A$ and
$\tau(\frak B)\le\omega$, then $\frak B\cong\Cal PI$ for some countable set
$I$;  in particular, $\frak B$ is a measurable algebra.

(c)\cmmnt{ ({\smc Maharam 1947})} The only Maharam submeasure on
$\frak A$ is identically zero.

\proof{{\bf (a)(i)} $\frak A$ is Dedekind complete just because it is a
regular open algebra.

\medskip

\quad{\bf (ii)} $T$ is upwards-ccc, so $\frak A$ is ccc, by 514Nc.

\medskip

\quad{\bf (iii)} For $t\in T$, set
$\widehat{t}=\interior\overline{\coint{t,\infty}}\in\frak A$;  then
$\{\widehat{t}:t\in T\}$ is order-dense in $\frak A$.   Let
$r:T\to\On$ be the rank function of $T$ (5A1Ea).   For each
$\xi<\omega_1$, $A_{\xi}=\{\widehat{t}:t\in T$, $r(t)=\xi\}$
is a partition of
unity in $\frak A$.   \Prf\ If $r(t)=r(t')$ and $t\ne t'$ then
$\coint{t,\infty}\cap\coint{t',\infty}=\emptyset$ so
$\widehat{t}\cap\widehat{t'}=0$ in $\frak A$;  thus $A_{\xi}$ is disjoint.
If $a\in\frak A\setminus\{0\}$, there is an $s\in T$ such that
$\widehat{s}\Bsubseteq a$;   if $r(s)\ge\xi$, there is a $t\le s$ such that
$r(t)=\xi$, and $a\Bcap\widehat{t}\ne 0$;  if $r(s)<\xi$, there is a
$t\ge s$ such that $r(t)=\xi$ (because $T$ is well-pruned),
and $\widehat{t}\Bsubseteq a$.   Thus $\sup A_{\xi}=1$ in $\frak A$.\ \Qed

If $A\subseteq\frak A$ is a partition of unity, there is a $\xi<\omega_1$
such that $A_{\xi}$ refines $A$ in the sense that every member of $A_{\xi}$
is included in some member of $A$ (see 311Ge).  \Prf\
$B=\{\widehat{t}:t\in T$, $\widehat{t}\Bsubseteq a$ for some $a\in A\}$ is
order-dense in $\frak A$, so there is a partition $C$ of unity included in
$B$;  $C$ is countable;  let $D\subseteq T$ be a countable set such that
$C=\{\widehat{t}:t\in D\}$;  set $\xi=\sup_{t\in D}r(t)$.\ \Qed

Of course $A_{\eta}$ refines $A_{\xi}$ whenever $\xi\le\eta<\omega_1$.
So if $\sequencen{C_n}$ is a sequence of partitions of unity in $\frak A$,
there is a $\xi<\omega_1$ such that $A_{\xi}$ refines $C_n$ for every
$n\in\Bbb N$, and then $\{c:c\in C_n$, $a\Bcap c\ne 0\}$ has just one
member for
every $a\in A_{\xi}$ and $n\in\Bbb N$.   As $\sequencen{C_n}$ is
arbitrary, $\frak A$ is \wsid.

\medskip

{\bf (b)} If $B\subseteq\frak A$ is a countable set $\tau$-generating
$\frak B$, there is a countable set $D\subseteq T$ such that
$b=\sup\{\widehat{t}:t\in D$, $\widehat{t}\Bsubseteq b\}$ for every
$b\in B$.   Now $\xi=\sup\{r(t):t\in D\}$ is countable, and
$b=\sup\{a:a\in A_{\xi}$, $a\Bsubseteq b\}$ for every
$b\in B$, so $\frak B$ is included in the order-closed subalgebra $\frak C$
of $\frak A$ generated by $A_{\xi}$.   Of course $A_{\xi}$ is order-dense
in $\frak C$.   For $a\in A_{\xi}$, set
$b_a=\inf\{b:b\in\frak B$, $b\Bsupseteq a\}$;  then every $b_a$ is an atom
in $\frak B$ and $\{b_a:a\in A_{\xi}\}$ is order-dense in $\frak B$, so
$\frak B$ is purely atomic.   As $\frak B$ is ccc, the set $I$ of its atoms
is countable;  being Dedekind complete, $\frak B$ is isomophic to
$\Cal PI$.  %316Xf

\medskip

{\bf (c)} Let $\nu$ be a Maharam submeasure on $\frak A$.   Then for every
$\epsilon>0$ there is a $\xi<\omega_1$ such that $\nu a\le\epsilon$ for
every $a\in A_{\xi}$.   \Prf\ Set

\Centerline{$T'=\{t:\nu\widehat{t}\ge\epsilon\}$.}

\noindent Then $T'$ is a subtree of $T$ and $\{t:t\in T'$, $r(t)=\xi\}$ is
finite for every $\xi<\omega_1$, because $\nu$ is exhaustive.   Also $T'$,
like $T$, can have no uncountable branches.   It follows that the height of
$T'$ is countable (5A1E(b-i)),
that is, that there is a $\xi<\omega_1$ such
that $r(t)<\xi$ for every $t\in T'$ and $\nu a\le\epsilon$ for every
$a\in A_{\xi}$.\ \Qed

As this is true for every $\epsilon>0$, there is actually a $\xi<\omega_1$
such that $\nu a=0$ for every $a\in A_{\xi}$.   But as $A_{\xi}$ is a
countable partition of unity and $\nu$ is a Maharam submeasure, $\nu 1=0$
and $\nu$ is identically zero.
}%end of proof of 539P

\leader{539Q}{Reflection principles }\cmmnt{In 539O, we have a theorem
of the type `if every small subalgebra of $\frak A$ is $\ldots$,
then $\frak A$ is $\ldots$'.   There was a similar result in 518I, and we
shall have another in 545G.   Here I collect some simple facts which are
relevant to the present discussion.

\medskip

}{\bf (a)} If $\frak A$ is a Boolean algebra and every subset of
$\frak A$ of
cardinal $\omega_1$ is included in a ccc subalgebra of $\frak A$, then
$\frak A$ is ccc.   \prooflet{(For there can be no disjoint set with cardinal
$\omega_1$.)}

\spheader 539Qb If $\frak A$ is ccc and every countable subset of
$\frak A$ is included
in a \wsid\ subalgebra of $\frak A$, then $\frak A$ is \wsid.
\prooflet{\Prf\ If $C_n$ is a partition of unity in $\frak A$
for every $n$, set

\Centerline{$D=\{d:\{c:c\in C_n$, $c\Bcap d\ne 0\}$ is finite
for every $n\in\Bbb N\}$.}

\noindent\Quer\ If $D$ is not order-dense in $\frak A$, take
$a\in\frak A^+$
such that $d\notBsubseteq a$ for every $d\in D$.   Let $\frak B$ be a
\wsid\ subalgebra of $\frak A$ including
$\{a\}\cup\bigcup_{n\in\Bbb N}C_n$.   Then every $C_n$ is a partition of
unity in $\frak B$, so there is a partition $B$ of unity in $\frak B$ such
that $B\subseteq D$.   But now $a\in\frak B^+$ so there is a $b\in B$ such
that $a\Bcap b\ne 0$ and $a\Bcap b\in D$.\ \Bang

So $D$ is order-dense in $\frak A$ and includes a partition of unity
in $\frak A$.   As $\sequencen{C_n}$ is arbitrary, $\frak A$ is \wsid.\
\Qed}

\spheader 539Qc\dvArevised{2014} If every countable subset of
$\frak A$ is included in a subalgebra of $\frak A$ with the
$\sigma$-interpolation property, then $\frak A$ has the
$\sigma$-interpolation property.
\prooflet{\Prf\ If $A$, $B\subseteq\frak A$ are countable and
$a\Bsubseteq b$ whenever $a\in A$ and $b\in B$, let $\frak B$ be a
subalgebra of $\frak A$, including $A\cup B$,
with the $\sigma$-interpolation property;  then there is a $c\in\frak B$
such that $a\Bsubseteq c\Bsubseteq b$ for every $a\in A$ and $b\in B$.\
\Qed}

\spheader 539Qd If $\frak A$ is a Maharam algebra and every countably
generated closed subalgebra of $\frak A$ is a measurable algebra, then
$\frak A$ is measurable.
\prooflet{(This is part (c) of the proof of 539O.)}

\spheader 539Qe
Suppose that Todor\v{c}evi\'c's $p$-ideal dichotomy is true.   Let
$\frak A$ be a Boolean algebra such that every subset of $\frak A$ of
cardinal at most $\omega_1$ is included in a subalgebra of $\frak A$ which
is a Maharam algebra.   Then $\frak A$ is a Maharam algebra.
\prooflet{\Prf\ By (a),
$\frak A$ is ccc;  by (c), $\frak A$ is Dedekind complete;  by (b),
$\frak A$ is \wsid;  by 539N, $\frak A$ is a Maharam algebra.\ \Qed}

\spheader 539Qf
Suppose that Todor\v{c}evi\'c's $p$-ideal dichotomy is true.   Let
$\frak A$ be a Boolean algebra such that every subset of $\frak A$ of
cardinal at most $\frak c$ is included in a subalgebra of $\frak A$ which
is a measurable algebra.   Then $\frak A$ is measurable.
\prooflet{\Prf\
By (a), $\frak A$ is ccc.   So if $\frak B$ is a countably generated
order-closed subalgebra, it has cardinal $\frak c$, and is included in a
measurable subalgebra $\frak C$ of $\frak A$.   Now $\frak B$ is
order-closed in $\frak C$, so is itself a measurable algebra.   By 539O,
$\frak A$ also is measurable.\ \Qed}

\spheader 539Qg
On the other hand,\cmmnt{ {\smc Farah \& Veli\v{c}kovi\'c 06} show
that} if
$\kappa$ is an infinite cardinal such that $2^{\kappa}=\kappa^+$,
$\square_{\kappa}$\cmmnt{ (5A6D)}
is true and the cardinal power $\kappa^{\omega}$ is
equal to $\kappa$, then there is a Dedekind complete Boolean algebra
$\frak A$, with cardinal $\kappa^+$, such that every order-closed subalgebra
of $\frak A$ with cardinal at most $\kappa$ is a measurable algebra, but
$\frak A$ is not a measurable
algebra (and therefore is not a Maharam algebra\cmmnt{,
by (d) above}). %539Qd
In particular, this can\cmmnt{ easily} be the case with $\kappa=\frak c$.

\leader{539R}{Exhaustivity \dvrocolon{rank}}\cmmnt{ While we now know
that there are
non-measurable Maharam algebras, we know practically nothing about their
structure.   The following idea is one tool for investigation.

\medskip

\noindent}{\bf Definitions} Suppose that $\frak A$ is a
Boolean algebra and $\nu$ an exhaustive
submeasure on $\frak A$.   For $\epsilon>0$, say that
$a\preccurlyeq_{\epsilon}b$ if either $a=b$ or $a\Bsubseteq b$ and
$\nu(b\Bsetminus a)>\epsilon$.
Then $\preccurlyeq_{\epsilon}$ is a well-founded partial order on
$\frak A$\cmmnt{ (use 5A1Dc;
if $\sequencen{a_n}$ were strictly decreasing for $\preccurlyeq_{\epsilon}$, then
$\sequencen{a_n\Bsetminus a_{n+1}}$ would be disjoint, with
$\nu(a_n\Bsetminus a_{n+1})\ge\epsilon$ for every $n$)}.   Let
$r_{\nu\epsilon}:\frak A\to\On$
be the corresponding rank function\cmmnt{, so that

\Centerline{$r_{\nu\epsilon}(a)=\sup\{r_{\nu\epsilon}(b)+1:b\Bsubseteq a$,
$\nu(a\Bsetminus b)>\epsilon\}$}

\noindent for every $a\in\frak A$ (5A1Db)}.   Now the
{\bf exhaustivity rank} of $\nu$ is $\sup_{\epsilon>0}r_{\nu\epsilon}(1)$.

\leader{539S}{Elementary facts} Let $\frak A$ be a Boolean algebra with
an exhaustive submeasure $\nu$ and associated rank functions
$r_{\nu\epsilon}$ for $\epsilon>0$.

\spheader 539Sa $r_{\nu\delta}(a)\le r_{\nu\epsilon}(b)$ whenever
$\nu(a\Bsetminus b)\le\delta-\epsilon$.
\prooflet{\Prf\ Induce on $r_{\nu\epsilon}(b)$.   If $r_{\nu\epsilon}(b)=0$,
then $\nu b\le\epsilon$ so $\nu a\le\delta$ and $r_{\nu\delta}(a)=0$.
For the inductive step to $r_{\nu\epsilon}(b)=\xi$,
if $c\Bsubseteq a$ and $\nu(a\Bsetminus c)>\delta$ then
$\nu(b\Bsetminus c)>\epsilon$ and
$r_{\nu\epsilon}(b\Bcap c)<\xi$.   Also
$\nu(c\Bsetminus b)\le\delta-\epsilon$ so, by the inductive
hypothesis, $r_{\nu\delta}(c)\le r_{\nu\delta}(b\Bcap c)<\xi$;  as $c$ is
arbitrary,
$r_{\nu\delta}(a)\le\xi$ and the induction continues.\ \QeD\ }\cmmnt{ In
particular,}

\Centerline{$r_{\nu\epsilon}(a)\le r_{\nu\epsilon}(b)$ if $a\Bsubseteq b$,
\quad$r_{\nu\delta}(a)\le r_{\nu\epsilon}(a)$ if $\epsilon\le\delta$.}

\spheader 539Sb If $a$, $b\in\frak A$ are disjoint and $\epsilon>0$, then
$r_{\nu\epsilon}(a\Bcup b)$ is at least the ordinal sum
$r_{\nu\epsilon}(a)+r_{\nu\epsilon}(b)$.
\prooflet{\Prf\ Induce on $r_{\nu\epsilon}(b)$.   If $r_{\nu\epsilon}(b)=0$, the
result is immediate from (a) above.   For the inductive step to
$r_{\nu\epsilon}(b)=\xi$, we have for any $\eta<\xi$ a $c\Bsubseteq b$
such that $\nu(b\Bsetminus c)>\epsilon$ and
$\eta\le r_{\nu\epsilon}(c)<\xi$.   Now
$r_{\nu\epsilon}(a\Bcup c)\ge r_{\nu\epsilon}(a)+\eta$,
by the inductive hypothesis, and
$\nu((a\Bcup b)\Bsetminus(a\Bcup c))>\epsilon$, so
$r_{\nu\epsilon}(a\Bcup b)>r_{\nu\epsilon}(a)+\eta$;  as $\eta$ is arbitrary,
$r_{\nu\epsilon}(a\Bcup b)\ge r_{\nu\epsilon}(a)+\xi$ and the induction
continues.\ \Qed}

\leaveitout{\spheader 539Sc
If $\nuprime$ is another exhaustive submeasure on $\frak A$
with rank functions $r'_{\epsilon}$, and $\nu a\le\alpha\nuprime a$
for every $a\in\frak A$, where $\alpha>0$, then
$r_{\alpha\epsilon}(a)\ge r'_{\epsilon}(a)$ for every $a\in\frak A$
and $\epsilon>0$ (induce on $r'_{\epsilon}(a)$, as usual).
%do we want this?
}

\leader{539T}{The rank of a Maharam algebra (a)} Note that the rank
function $r_{\nu\epsilon}$ associated with an exhaustive submeasure
$\nu$ depends only on the set $\{a:\nu a>\epsilon\}$.   In
particular, if $\nu$ and $\nuprime$ are exhaustive submeasures on a
Boolean algebra $\frak A$ and
$\nu a\le\epsilon$ whenever $\nuprime a\le\delta$, then
$r_{\nu\epsilon}(a)\le r_{\nu'\delta}(a)$ for every
$a\in\frak A$.   If $\frak A$ is a Maharam algebra, then any two
Maharam submeasures on $\frak A$ are mutually absolutely continuous
\cmmnt{(539Ac)}, so have the same exhaustivity rank;
I will call this the
{\bf Maharam submeasure rank} of $\frak A$, $\Mhsr(\frak A)$.
Note that if $a\in\frak A$ then $\Mhsr(\frak A_a)\le\Mhsr(\frak A)$.

\medskip

{\bf (b)} If $\frak A$ is a measurable algebra,
$\Mhsr(\frak A)\le\omega$\cmmnt{,
because if $\mu$ is an additive functional and $\epsilon>0$,
then $\mu a>\epsilon r_{\mu\epsilon}(a)$ for every $a\in\frak A$}.
More generally,
for any uniformly exhaustive submeasure $\nu$ and $\epsilon>0$,
$r_{\nu\epsilon}(a)$ is finite, being
the maximal size of any disjoint set
consisting of elements, included in $a$,
of submeasure greater than $\epsilon$.

\medskip

{\bf (c)}\dvAnew{2017}{\bf (i)} Suppose that $\frak A$ is a Maharam algebra with a strictly
positive Maharam submeasure $\nu$, and that $\frak B$ is a subalgebra of
$\frak A$ which is dense for the Maharam-algebra topology of $\frak A$.
For $\epsilon>0$, write $r_{\epsilon}=r_{\nu\epsilon}$ for the
corresponding rank function on $\frak A$, and
$r'_{\epsilon}=r_{\nu\restr\frak B,\epsilon}$ for the rank function on
$\frak B$ corresponding to the exhaustive submeasure $\nu\restr\frak B$.
If $0<\delta<\epsilon$, $a\in\frak A$, $b\in\frak B$, $\xi\in\On$,
$\nu(a\Bsymmdiff b)<\epsilon-\delta$ and $r_{\epsilon}(a)\ge\xi$,
then $r'_{\delta}(b)\ge\xi$.
\prooflet{\Prf\ Induce on $\xi$.   If $\xi=0$ the result is trivial.
For the inductive step to
$\xi>0$, take any $\eta<\xi$.   Then we have an $a'\Bsubseteq a$ such that
$\nu(a\Bsetminus a')>\epsilon$ and
$r_{\epsilon}(a')>\eta$.   Let
$b'\in\frak B$ be such that
$\nu(a'\Bsymmdiff b')<\epsilon-\delta-\nu(a\Bsymmdiff b)$ and consider
$b\Bcap b'$.   We have

\Centerline{$\nu(a'\Bsymmdiff(b\Bcap b'))
=\nu((a\Bcap a')\Bsymmdiff(b\Bcap b'))
\le\nu(a\Bsymmdiff b)+\nu(a'\Bsymmdiff b')
<\epsilon-\delta$}

\noindent so $r'_{\delta}(b\Bcap b')>\eta$, by the inductive hypothesis.
Moreover,

$$\eqalign{\nu(b\Bsetminus(b\Bcap b'))&
=\nu(b\Bsetminus b')
\ge\nu(a\Bsetminus a')-\nu(a\Bsetminus b)-\nu(b'\Bsetminus b)\cr&
>\epsilon-\nu(a\Bsymmdiff b)-\nu(a'\Bsymmdiff b')
>\delta\cr}$$

\noindent so $r'_{\delta}(b)\ge\eta+1$.   This is true for every $\eta<\xi$,
so $r'_{\delta}(b)\ge\xi$.\ \Qed}

\medskip

\quad{\bf (ii)} It follows that if $\frak A$ is an infinite Maharam algebra,
then $\Mhsr(\frak A)<\tau(\frak A)^+$.
\prooflet{\Prf\ $\frak A$ has a dense subalgebra $\frak B$ with cardinal
$\tau=\tau(\frak A)$ (539B).
If $\nu$ is a strictly positive Maharam submeasure on
$\frak A$, then $\nu\restr\frak B$ is an exhaustive submeasure on $\frak B$, so
$r_{\nu\restr\frak B,\delta}(1)<\tau^+$ for every
$\delta>0$, by 5A1Dd.   By (i) here,
$r_{\nu\epsilon}(1)<\tau^+$ for every $\epsilon>0$.   Since
$\cf\tau^+>\omega$, $\Mhsr(\frak A)=\sup_{n\in\Bbb N}r_{\nu,2^{-n}}(1)$
is less than $\tau^+$.\ \Qed}

\medskip

{\bf (d)}\dvAnew{2017}
The Maharam algebras described in \S394 are all defined from exhaustive
submeasures with domain the
countable algebra $\frak B$ of open-and-closed subsets of a compact metrizable
space.   By (c),
such algebras must have Maharam submeasure rank less than $\omega_1$.

\leader{539U}{Theorem} Suppose that $\frak A$ is a non-measurable
Maharam algebra.   Then
$\Mhsr(\frak A)$ is at least the ordinal power $\omega^{\omega}$.

\proof{ Let $\nu$ be a strictly positive Maharam submeasure on $\frak A$.

\medskip

{\bf (a)} For the time being (down to the end of (d) below), assume that
$\frak A$ is nowhere measurable (definition:  391Bc).
For $a\in\frak A$, set

\Centerline{$\check\nu a
=\inf_{n\in\Bbb N}\sup\{\min_{i\le n}\nu a_i:
    a_0,\ldots,a_n\Bsubseteq a$ are disjoint$\}$.}

\noindent Then $\check\nu$ is a Maharam submeasure.   \Prf\ Of course
$\check\nu 0=0$ and $\check\nu a\le\check\nu b$ whenever
$a\Bsubseteq b$.   If $a$, $b\in\frak A$ and $\epsilon>0$, then there
are $n_0$, $n_1\in\Bbb N$ such that whenever $\familyiI{c_i}$ is a
disjoint family in $\frak A$, then
$\#(\{i:\nu(c_i\Bcap a)\ge\check\nu a+\epsilon\})\le n_0$ and
$\#(\{i:\nu(c_i\Bcap b)\ge\check\nu b+\epsilon\})\le n_1$.   So

\Centerline{$\#(\{i:\nu(c_i\Bcap(a\Bcup b))
\ge\check\nu a+\check\nu b+2\epsilon\})\le n_0+n_1$.}

\noindent It follows that
$\check\nu(a\Bcup b)\le\check\nu a+\check\nu b+2\epsilon$;  as
$\epsilon$, $a$ and $b$ are arbitrary, $\check\nu$ is a submeasure.
Because $\check\nu\le\nu$, $\check\nu$ is a Maharam submeasure.\ \Qed

\medskip

{\bf (b)} Because $\frak A$ is nowhere measurable, $\check\nu$ is strictly
positive.   \Prf\ If $a\in\frak A\setminus\{0\}$, the principal ideal
$\frak A_a$ is not measurable, so the Maharam submeasure
$\nu\restrp\frak A_a$ cannot be uniformly exhaustive;  that is, there is an
$\epsilon>0$ such that there are arbitrarily long disjoint strings
$\langle a_i\rangle_{i\le n}$ in $\frak A_a$ with $\nu a_i\ge\epsilon$ for
every $i\le n$.   But this means that $\check\nu a\ge\epsilon>0$.\ \Qed

\medskip

{\bf (c)} Let $r_{\nu\epsilon}$,
$r_{\check\nu\epsilon}$ be the rank functions associated with
$\nu$ and $\check\nu$.   Then
$r_{\nu\epsilon}(a)$ is at least the ordinal product
$\omega\cdot r_{\check\nu\epsilon}(a)$ whenever $a\in\frak A$ and
$\epsilon>0$.
\Prf\ Induce on $r_{\check\nu\epsilon}(a)$.   If
$r_{\check\nu\epsilon}(a)=0$, the result is trivial.   For the
inductive step to $r_{\check\nu\epsilon}(a)=\xi+1$, take
$b\Bsubseteq a$ such that $\check\nu b>\epsilon$ and
$r_{\check\nu\epsilon}(a\Bsetminus b)=\xi$.
Then for every $n\in\Bbb N$ there are disjoint
$b_0,\ldots,b_n\Bsubseteq b$ such that $\nu b_i>\epsilon$ for every
$i$, and $r_{\nu\epsilon}(b)\ge\omega$;  by the inductive hypothesis,
$r_{\nu\epsilon}(a\Bsetminus b)\ge\omega\cdot\xi$;  by 539Sb,
$r_{\nu\epsilon}(a)\ge\omega\cdot\xi+\omega=\omega\cdot(\xi+1)$, and the
induction proceeds.   The inductive step to non-zero limit $\xi$ is
elementary.\ \Qed

\medskip

{\bf (d)} Now

$$\eqalignno{\Mhsr(\frak A)
&=\sup_{\epsilon>0}r_{\nu\epsilon}(1)
\ge\sup_{\epsilon>0}\omega\cdot r_{\check\nu\epsilon}(1)
=\omega\cdot\sup_{\epsilon>0}r_{\check\nu\epsilon}(1)\cr
\displaycause{5A1Bb}
&=\omega\cdot\Mhsr(\frak A);\cr}$$

\noindent as $\Mhsr(\frak A)>0$, $\Mhsr(\frak A)\ge\omega^{\omega}$
(5A1Bc).

\medskip

{\bf (e)} For the general case,
let $a\in\frak A^+$ be such that the
principal ideal $\frak A_a$ is nowhere measurable.   Then
$\Mhsr(\frak A)\ge\Mhsr(\frak A_a)\ge\omega^{\omega}$.
}%end of proof of 539U

\leader{539V}{PV norms and exhaustivity}\dvAnew{2017} {\bf (a)}
If we construct a submeasure $\nu$ on an algebra $\frak B$
from a PV norm $\|\,\|$ on $[\Bbb N]^{<\omega}$ and sequences
$\sequencen{T_n}$, $\sequence{k}{\alpha_k}$ and $\sequence{k}{N_k}$
as in 394B and 394H, we can relate the exhaustivity rank of $\nu$
to $\|\,\|$, as follows.   Note first that the set
$\Cal L=\{L:L\in[\Bbb N]^{<\omega}$, $\nu L\le 1\}$,
ordered by $\subseteq$,
is a tree with no infinite branches, by the last clause of 394Aa.
For $\Cal K\subseteq[\Bbb N]^{<\omega}$, set
$\partial\Cal K=\{K\setminus\{\max K\}:\emptyset\ne K\in\Cal K\}$;
iterating as in 421N, set

\Centerline{$\partial^0\Cal L=\Cal L$,
\quad$\partial^{\xi}\Cal L
=\partial(\bigcap_{\eta<\xi}\partial^{\eta}\Cal L)$}

\noindent for ordinals $\xi>0$.   Now observe that if
$L\subset L'\in\Cal L$
and $z\in\prod_{r\in L'}T_r$, then (at least if every $T_r$ has at least two members) $Y_{z\restr L}\setminus Y_z$ includes
some $Y_{z'}$ where $z'\in\prod_{r\in L'}T_r$, so
$\nu(Y_{z\restr L}\setminus Y_z)\ge 8$ (394G) and
$r_{\nu 1}(Y_{z\restr L})>r_{\nu 1}(Y_z)$.
An easy induction now shows that
$r_{\nu 1}(Y_z)\ge\xi$ whenever $L\in\partial^{\xi}\Cal L$ and
$z\in\prod_{r\in L}T_r$.   So if $\emptyset\in\partial^{\xi}\Cal L$ then
$r_{\nu 1}(X)\ge\xi$.

\spheader 539Vb Moving to the Maharam algebra $\frak A=\widehat{\frak B}$
defined from $\nu$, as in
394Nc, we see that $\frak A$ has a strictly positive Maharam submeasure
$\hat\nu$ extending $\nu$, so that the same formulae, interpreted in $\frak A$,
tell us that $\Mhsr(\frak A)\ge\xi$ whenever $\emptyset\in\partial^{\xi}\Cal L$.

\spheader 539Vc The next step is to understand which families
$\Cal L\subseteq[\Bbb N]^{<\omega}$ can be expressed as
$\{L:\|L\|\le 1\}$ for some PV norm $\|\,\|$.   Looking through the
definition in 394Aa, we see that we shall need, at least,

\inset{----- $\{n\}\in\Cal L$ for every $n\in\Bbb N$,

----- $I\in\Cal L$ whenever $J\in\Cal L$ and $\#(I\cap n)\le\#(J\cap n)$
for every $n$,

----- for every infinite $A\subseteq\Bbb N$ there is an $n\in\Bbb N$ such that
$A\cap n\notin\Cal L$.}

\noindent Following {\smc Perovi\'c \& Veli\v{c}kovi\'c 18},
I will say that a family satisfying these three conditions is
{\bf admissible}.
The point is that they are sufficient as well as necessary.   \prooflet{\Prf\
Given an admissible family $\Cal L\subseteq[\Bbb N]^{<\omega}$, set

\Centerline{$\|I\|
=\min\{\#(\Cal L_0):\Cal L_0\subseteq\Cal L$, $I\subseteq\bigcup\Cal L_0\}$}

\noindent for $I\in[\Bbb N]^{<\omega}$.   Because $\Cal L$ contains
all singletons, $\|I\|\le\#(I)$ is always finite.   $\|I\|=0$ iff
$I\subseteq\bigcup\emptyset$ iff $I=\emptyset$.   If $\#(I)=1$ then
$I\in\Cal L$ so $\|I\|\le 1$.   Of course $\|\,\|$ is subadditive.   If
$I$, $J\in[\Bbb N]^{<\omega}$ and $\#(I\cap n)\le\#(J\cap n)$ for every
$n$, there is an injective function $f:I\to J$ such that
$f(i)\le i$ for every $i\in I$ (set $f(i)=\min(J\setminus f[I\cap i])$ for
$i\in I$);  now if $\Cal L_0\subseteq\Cal L$ and
$J\subseteq\bigcup\Cal L_0$ then $f^{-1}[L]\in\Cal L$ for every $L\in\Cal L_0$
and $I\subseteq\bigcup_{L\in\Cal L_0}f^{-1}[L]$.   So $\|I\|\le\|J\|$.
Finally, if $A\subseteq\Bbb N$ and $\|A\cap n\|\le m$ for every $n\in\Bbb N$,
let $\langle L_{ni}\rangle_{n\in\Bbb N,i<m}$ be a family in $\Cal L$ such that
$A\cap n\subseteq\bigcup_{i<m}L_{ni}$ for every $n$.   For $j\in A$ let
$g_j:\Bbb N\to m$ be such that $j\in L_{n,g_j(n)}$ whenever $j<n$.
Let $h:\Bbb N\to m$ be such that for every $k\in\Bbb N$ there is an
$n\in\Bbb N$ such that $g_j(n)=h(j)$ for every $j\in A\cap k$, so that
$j\in L_{n,h(j)}$ for $j\in A\cap k$, and $h^{-1}[\{l\}]\cap A\cap k$ is
included in $L_{nl}$ and belongs to $\Cal L$.   As $k$ is arbitrary,
$h^{-1}[\{l\}]\cap A$ must be finite;  as $l$ is arbitrary, $A$ is finite.

Now we see that $\|I\|\le 1$ iff there is an $L\in\Cal L$ including $I$, that
is, iff $I\in\Cal L$.   So we have expressed $\Cal L$ in the required form.\
\Qed}

\spheader 539Vd For every $\xi<\omega_1$ there is an admissible family
$\Cal L_{\xi}\subseteq[\Bbb N]^{<\omega}$ such that
$\emptyset\in\partial^{\eta}\Cal L_{\xi}$ for every $\eta<\xi$.
\prooflet{\Prf\
Recall from 5A1Tb that there is a sequence $\sequencen{\le_n}$ of
partial orders on $\omega_1$ such that

\inset{
$\sequencen{\le_n}$ is non-decreasing and $\bigcup_{n\in\Bbb N}\le_n$ is the
usual ordering of $\omega_1$,

if $\xi<\omega_1$ and $n\in\Bbb N$ then $\{\eta:\eta\le_n\xi\}$ is finite.}

\noindent Define $\ofamily{\xi}{\omega_1}{\Cal L_{\xi}}$ inductively by saying
that $\Cal L_0=\{I:I\subseteq\Bbb N$, $\#(I)\le 1\}$ and

\Centerline{$\Cal L_{\xi}
=\Cal L_0\cup\bigcup_{\eta<\xi}\{I:I\in[\Bbb N]^{<\omega}$, $\#(I)\ge 2$,
$\eta\le_{\min I}\xi$, $I\setminus\{\min I\}\in\Cal L_{\eta}\}$}

\noindent for $0<\xi<\omega_1$.   We see at once that $\Cal L_0$ is admissible.
Supposing that $\Cal L_{\eta}$ is admissible
and $\emptyset\in\partial^{\zeta}\Cal L_{\eta}$ whenever
$\zeta<\eta<\xi$, we need to check the following.

\medskip

\quad{\bf (i)} $\{n\}\in\Cal L_{\xi}$ for every $n\in\Bbb N$, because
$\{n\}\in\Cal L_0$.

\medskip

\quad{\bf (ii)} If $J\in\Cal L_{\xi}$ and $\#(I\cap n)\le\#(J\cap n)$ for every
$n\in\Bbb N$, either $\#(I)\le 1$ and certainly $I\in\Cal L_{\xi}$, or
$\#(I)>1$, $\#(J)>1$ and $\min J\le\min I$.   In this case,
$\#((I\setminus\{\min I\})\cap n)\le\#((J\setminus\{\min J\})\cap n)$ for every
$n\in\Bbb N$.   Now there is an $\eta<\xi$ such that $\eta\le_{\min J}\xi$ and
$J\setminus\{\min J\}\in\Cal L_{\eta}$.   Because $\Cal L_{\eta}$ is admissible,
$I\setminus\{\min I\}\in\Cal L_{\eta}$;  because $\min J\le\min I$,
$\eta\le_{\min I}\xi$ and $I\in\Cal L_{\xi}$.

\medskip

\quad{\bf (iii)} If $A\subseteq\Bbb N$ is infinite, then
$D=\{\eta:\eta\le_{\min A}\xi\}$ is finite.   For each $\eta\in D$ there is an
$n_{\eta}\in\Bbb N$ such that
$(A\setminus\{\min A\})\cap n_{\eta}\notin\Cal L_{\eta}$.   Setting
$n=\max(\{1+\min(A\setminus\{\min A\})\}\cup\{n_{\eta}:\eta\in D\}$,
we see that $\#(A\cap n)\ge 2$ and
$(A\cap n)\setminus\{\min(A\cap n)\}\notin\Cal L_{\eta}$ for any $\eta\in D$, so
$A\cap n\notin\Cal L_{\xi}$.

\medskip

\quad{\bf (iv)} Thus $\Cal L_{\xi}$ is admissible.   Now suppose that
$\eta<\xi$.

\medskip

\qquad\grheada\ If $\eta\le 1$, we have
$\{\emptyset,\{0\}\}\subseteq\Cal L_{\xi}$ so
$\emptyset\in\partial\Cal L_{\xi}\subseteq\partial^{\eta}\Cal L_{\xi}$.

\medskip

\qquad\grheadb\ If $\eta\ge 2$ let $n$ be such that
$\eta\le_n\xi$ and consider
$\Cal L=\{\{n\}\cup(I+n+1):I\in\Cal L_{\eta}\}$, where I
write $I+n+1$ for $\{i+n+1:i\in I\}$.   Then $\Cal L\subseteq\Cal L_{\xi}$.
An easy induction on $\zeta$ shows that

\Centerline{$\partial^{\zeta}\Cal L_{\xi}
\supseteq\partial^{\zeta}\Cal L
\supseteq\{\{n\}\cup(I+n+1):I\in\partial^{\zeta}\Cal L_{\eta}\}$}

\noindent for every $\zeta$ such that
$\emptyset\in\partial^{\zeta}\Cal L_{\eta}$, and in particular for every
$\zeta<\eta$.   So
$\{n\}\in\bigcap_{\zeta<\eta}\partial^{\zeta}\Cal L_{\xi}$ and
$\emptyset\in\partial^{\eta}\Cal L_{\xi}$.

\medskip

\quad{\bf (v)} Inducing on $\xi$, we see that
$\emptyset\in\partial^{\eta}\Cal L_{\xi}$ whenever $\eta<\xi<\omega_1$.\ \Qed}

\spheader 539Ve Putting these together, we see that if $\xi<\omega_1$ we
have an admissible family $\Cal L_{\xi+1}$ such that we can
define a PV norm $\|\,\|_{\xi}$ from $\Cal L_{\xi+1}$ as in
(c), a submeasure $\nu_{\xi}$ on a countable atomless algebra $\frak B$ from $\|\,\|_{\xi}$ as in 394H, and a Maharam algebra
$\frak A_{\xi}$ from $\nu_{\xi}$ as in 394Nc, in such a way that the exhaustivity rank of
$\nu_{\xi}$ is at least $\xi$ and $\Mhsr(\frak A_{\xi})\ge\xi$.

\leader{539W}{The set of exhaustive submeasures:  Theorem}\dvAnew{2017}
Let $\frak C$ be a countable atomless Boolean algebra, not $\{0\}$.   Write
$M_{\text{sm}}$ for the set of totally finite submeasures on $\frak C$,
regarded as a subset of $\coint{0,\infty}^{\frak C}$, and
$M_{\text{esm}}$ for the set of exhaustive totally finite submeasures on
$\frak C$.   Then $M_{\text{sm}}$ is Polish, and
$M_{\text{esm}}\subseteq M_{\text{sm}}$ is coanalytic and not Borel.   Setting

\Centerline{$F_{\xi}
=\{\nu:\nu\in M_{\text{esm}}$ has exhaustivity rank at most $\xi\}$}

\noindent for $\xi<\omega_1$, every $F_{\xi}$ is a Borel subset of
$M_{\text{sm}}$ and every analytic subset of
$M_{\text{esm}}$ is included in some $F_{\xi}$.

\proof{{\bf (a)} Directly from the definition in
539Aa, we see that $M_{\text{sm}}$ is a closed subset of the Polish space
$\coint{0,\infty}^{\frak C}$, and is itself Polish.
Writing $D\subseteq\frak C^{\Bbb N}$ for the
set of infinite disjoint sequences in $\frak C$, we see that

\Centerline{$\{(\nu,d):\nu d(n)\ge\epsilon\text{ for every }n\in\Bbb N\}$}

\noindent is closed in $M_{\text{sm}}\times\frak C^{\Bbb N}$ (if we give
$\frak C$ its discrete topology) for every $\epsilon$, so that

\Centerline{$\{\nu:\text{ there is some }d\in D\text{ such that }
\nu(d(n))\ge\epsilon\text{ for every }n\in\Bbb N\}$}

\noindent is analytic in $M_{\text{sm}}$ for every $\epsilon$ (423B), and

$$\eqalign{&\{\nu:\nu\in M_{\text{sm}},\,\nu\text{ is not exhaustive}\}\cr&
\mskip20mu
=\bigcup_{k\in\Bbb N}\{\nu:\text{ there is some }d\in D\text{ such that }
\nu(d(n))\ge 2^{-k}\text{ for every }n\in\Bbb N\}\cr}$$

\noindent is analytic (423B, 423E).   Accordingly its complement in
$M_{\text{sm}}$, the set of $M_{\text{esm}}$ of
exhaustive totally finite submeasures, is coanalytic.

\medskip

{\bf (b)}
Define $\langle E_{a\epsilon\xi}\rangle_{a\in\frak C,\epsilon>0,\xi<\omega_1}$
by saying that

\Centerline{$E_{a\epsilon 0}=\{\nu:\nu\in M_{\text{sm}}$, $\nu a\le\epsilon\}$,}

\Centerline{$E_{a\epsilon\xi}
=\{\nu:\nu\in M_{\text{sm}}$, $\nu\in\bigcup_{\eta<\xi}E_{b\epsilon\eta}$
whenever $b\Bsubseteq a$ and $\nu(a\Bsetminus b)>\epsilon\}$}

\noindent for $a\in\frak C$, $\epsilon>0$ and $0<\xi<\omega_1$.
Then every $E_{a\epsilon\xi}$ is a Borel subset of $M_{\text{sm}}$.
\Prf\ For $\xi=0$ this is just because
$\nu\mapsto\nu a:M_{\text{sm}}\to\coint{0,\infty}$ is continuous.   For
$\xi>0$ we have

$$\eqalign{E_{a\epsilon\xi}
=\bigcap_{\Atop{b\in\frak C}{b\Bsubseteq a}}
   \bigcup_{\eta<\xi}\{\nu:\nu(a\Bsetminus b)\le\epsilon\text{ or }
        \nu\in E_{b\epsilon\eta}\}\cr}$$

\noindent which is Borel because $\frak C$ is countable.\ \QeD\   Observe also
that $E_{a\delta\xi}\subseteq E_{a\epsilon\xi}$ whenever $a\in\frak C$,
$0<\delta\le\epsilon$ and $\xi<\omega_1$.

\medskip

{\bf (c)(i)} If $\nu\in M_{\text{esm}}$, $a\in\frak C$, $\epsilon>0$ and
$\xi<\omega_1$, then
$r_{\nu\epsilon}(a)\le\xi$ iff $\nu\in E_{a\epsilon\xi}$.
\Prf\ Induce on $\xi$.   For $\xi=0$ we have

\Centerline{$r_{\nu\epsilon}(a)=0
\iff\nu a\le\epsilon\iff\nu\in E_{a\epsilon 0}$.}

\noindent For the inductive step to $\xi>0$,

$$\eqalign{r_{\nu\epsilon}(a)\le\xi&
\iff r_{\nu\epsilon}(b)<\xi\text{ whenever }b\Bsubseteq a\text{ and }
     \nu(a\Bsetminus b)>\epsilon\cr&
\iff \nu\in\bigcup_{\eta<\xi}E_{b\epsilon\eta}
     \text{ whenever }b\Bsubseteq a\text{ and }
     \nu(a\Bsetminus b)>\epsilon\cr&
\iff\nu\in E_{a\epsilon\xi}.  \text{ \Qed}\cr}$$

\noindent So for $\nu\in M_{\text{esm}}$ and $\xi\le\omega_1$,

\Centerline{$\nu\text{ has exhaustivity rank at most }\xi
\iff\nu\in E_{1\epsilon\xi}\text{ for every }\epsilon>0$.}

\medskip

\quad{\bf (ii)} Next, if $\nu\in M_{\text{sm}}$ and $\xi<\omega_1$ are such
that $\nu\in E_{1\epsilon\xi}$ for every $\epsilon>0$, then $\nu$ is exhaustive.
\Prf\Quer\ Otherwise, there are an $\epsilon>0$ and a
non-increasing sequence $\sequence{i}{a_i}$
such that $\nu(a_i\Bsetminus a_{i+1})>\epsilon$ for every $i\in\Bbb N$.
Of course we can suppose that $a_0=0$.   But now we find, inducing on $\eta$,
that
$\nu\notin E_{a_i\epsilon\eta}$ for every $i\in\Bbb N$ and $\eta<\omega_1$,
which is impossible.\ \Bang\Qed

\medskip

\quad{\bf (iii)} Setting

\Centerline{$F_{\xi}=\bigcap_{\epsilon>0}E_{1\epsilon\xi}
=\bigcap_{k\in\Bbb N}E_{1,2^{-k},\xi}$}

\noindent for $\xi<\omega_1$, we see that $F_{\xi}$ is a Borel subset of
$M_{\text{sm}}$ and is precisely
the set of exhaustive submeasures on $\frak C$ with exhaustivity rank at most
$\xi$.

\medskip

{\bf (d)(i)} For $\epsilon>0$, write $W_{\epsilon}$ for the set of triples
$(\nu,\nu',H)$ such that

\inset{$\nu$, $\nu'\in M_{\text{sm}}$ and $H\subseteq\frak C^2$.

$(1,1)\in H$,

whenever $(a,b)\in H$ there is a $b'\Bsubseteq b$ such that
$\nu'(b\Bsetminus b')>\epsilon$ and $(a',b')\in H$ whenever $a'\Bsubseteq a$ and
$\nu(a\Bsetminus a')>\epsilon$.}

\noindent Then $W_{\epsilon}$ is a Borel subset of
$M_{\text{sm}}\times M_{\text{sm}}\times\Cal P(\frak C^2)$, where the power set
$\Cal P(\frak C^2)$ is given its usual compact metrizable topology (4A2Ud).
So $V_{\epsilon}=\{(\nu,\nu'):$ there is an $H$ such that
$(\nu,\nu',H)\in W_{\epsilon}\}$ is an analytic subset of $M_{\text{sm}}^2$.

\medskip

\quad{\bf (ii)} If $\epsilon>0$, $\nu$, $\nu'\in M_{\text{esm}}$ and
$(\nu,\nu',H)\in W_{\epsilon}$ then
$r_{\nu\epsilon}(a)<r_{\nu'\epsilon}(b)$ whenever $(a,b)\in H$.
\Prf\ I show by induction on $\xi$ that if $(a,b)\in H$ and
$r_{\nu\epsilon}(a)\ge\xi$ then $r_{\nu'\epsilon}(b)>\xi$.   \Prf\
Induce on $\xi$.   If $\xi=0$ we know that there is a $b'\Bsubseteq b$ such that
$\nu'(b\Bsetminus b)>\epsilon$ so $r_{\nu'\epsilon}(b)>0$.   For the inductive
step to $\xi>0$, we know that there is a $b'\Bsubseteq b$ such that
$\nu'(b\Bsetminus b')>\epsilon$ and $(a',b')\in H$ whenever $a'\Bsubseteq a$ and
$\nu(a\Bsetminus a')>\epsilon$.   If $\eta<\xi$ then there is an
$a'\Bsubseteq a$ such that $\nu(a\Bsetminus a')>\epsilon$ and
$r_{\nu\epsilon}(a')\ge\eta$;  now $(a',b')\in H$ so
$r_{\nu'\epsilon}(b')>\eta$, by the inductive hypothesis.   As $\eta$ is
arbitrary, $\xi\le r_{\nu'\epsilon}(b')<r_{\nu'\epsilon}(b)$.    Thus the
induction continues.\ \Qed

\medskip

\quad{\bf (iii)} If $\epsilon>0$ and $\nu$, $\nu'\in M_{\text{esm}}$ then
$(\nu,\nu')\in V_{\epsilon}$ iff $r_{\nu\epsilon}(1)<r_{\nu'\epsilon}(1)$.
\Prf\ If $(\nu,\nu')\in V_{\epsilon}$ there is an $H$ such that
$(\nu,\nu',H)\in W_{\epsilon}$;  now $(1,1)\in H$ so (ii) tells us that
$r_{\nu\epsilon}(1)<r_{\nu'\epsilon}(1)$.   If
$r_{\nu\epsilon}(1)<r_{nu'\epsilon}(1)$ set
$H=\{(a,b):a$, $b\in\frak C$, $r_{\nu\epsilon}(a)<r_{\nu'\epsilon}(b)\}$;
then it is easy to check that $(\nu,\nu',H)\in W_{\epsilon}$, so
$(\nu,\nu')\in V_{\epsilon}$.\ \Qed

\medskip

{\bf (e)} Now suppose that $A\subseteq M_{\text{esm}}$ is an analytic set, and
that $\epsilon>0$.
Consider the relation $\preccurlyeq_{\epsilon}$ on $A$ defined by saying that
$\nu\preccurlyeq_{\epsilon}\nu'$ if either $\nu=\nu'$ or
$r_{\nu\epsilon}(1)<r_{\nu'\epsilon}(1)$.   This is a partial ordering, and it
is well-founded because if $B\subseteq A$ is well-founded and
$\min_{\nu\in B}r_{\nu\epsilon}(1)=\xi$ then any $\nu\in B$ such that
$r_{\nu\epsilon}(1)=\xi$ is minimal in $B$.   Now
$\{(\nu,\nu'):\nu\prec_{\epsilon}\nu'\}=A^2\cap V_{\epsilon}$, so by the
Kunen-Martin theorem (5A1De) $\preccurlyeq_{\epsilon}$ has countable height.
Since $\nu\prec_{\epsilon}\nu'$ whenever $\nu$, $\nu'\in A$ and
$r_{\nu\epsilon}(1)<r_{\nu'\epsilon}(1)$, $\{r_{\nu\epsilon}(1):\nu\in A\}$
must be countable.

This is true for every $\epsilon>0$, so
$\zeta=\sup_{\nu\in A,k\in\Bbb N}r_{\nu,2^{-k}}(1)$ is less than $\omega_1$.
But now $A\subseteq F_{\zeta}$.

\medskip

{\bf (f)} Finally, no $F_{\zeta}$ can be the whole of
$M_{\text{esm}}$.   \Prf\ We know from 539Ve that there are a countable atomless Boolean algebra $\frak B$ and a totally finite exhaustive submeasure on
$\frak B$ with exhaustivity rank at least $\zeta+1$.   But $\frak C$ and
$\frak B$ are isomorphic (316M) so the same is true of $\frak A$, that is,
$M_{\text{esm}}\setminus F_{\zeta}\ne\emptyset$.\ \QeD\
By (e) here, $M_{\text{esm}}$ cannot be analytic, so cannot be a Borel subset of the Polish space $M_{\text{sm}}$.
}%end of proof of 539W

\cmmnt{\medskip

\noindent{\bf Remark} In the language of 423S,
$\ofamily{\xi}{\omega_1}{F_{\xi}}$ is a family of Borel constituents of
$M_{\text{ens}}$.
}

\exercises{\leader{539X}{Basic exercises (a)}
%\spheader 539Xa
Let $\frak A$ be a Maharam algebra.    Show that
$\link_n(\frak A)\le\max(\omega,\tau(\frak A))$ for every $n\ge 2$.
%539B

\spheader 539Xb Show that, in the language of \S522,
$\frak p\le\frak s\le\min(\non\Cal N,\non\Cal M,\frak d)$.
%539F 53bits

\spheader 539Xc Let $\frak A$ be a Maharam algebra.   (i) Show that if

\inset{($\alpha$) $\cff[\lambda]^{\le\omega}\le\lambda^+$ for every
cardinal $\lambda\le\tau(\frak A)$,

($\beta$) $\square_{\lambda}$ is true for every uncountable cardinal
$\lambda\le\tau(\frak A)$ of countable cofinality,}

\noindent then $\FN(\frak A)\le\FN(\Cal P\Bbb N)$, with equality unless
$\frak A$ is finite.   \Hint{518D, 518I.}   (ii) Show that if
$\#(\frak A)\le\omega_2$ and $\FN(\Cal P\Bbb N)=\omega_1$, then
$\frak A$ is tightly $\omega_1$-filtered.   \Hint{518M.}
%539H

\spheader 539Xd Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of subsets of
$X$, and $\nu:\Sigma\to\coint{0,\infty}$ a non-zero Maharam submeasure;
set $\Cal I=\{E:E\in\Sigma$, $\nu E=0\}$ and $\frak A=\Sigma/\Cal I$.
Suppose that $\#(\frak A)\le\omega_2$ and $\FN(\Cal P\Bbb N)=\omega_1$.
Show that there is a lifting for $\nu$, that is, a Boolean homomorphism
$\theta:\frak A\to\Sigma$ such that $(\theta a)^{\ssbullet}=a$ for every
$a\in\frak A$.   \Hint{518L.}
%539Xc 539H out of order query

\spheader 539Xe Let $\frak A$ be a Boolean algebra, $\nu$ an exhaustive
submeasure on $\frak A$, and $\sequence{i}{a_i}$ a sequence in $\frak A$
such that $\inf_{i\in\Bbb N}\nu a_i>0$.   Let $\Cal F$ be a Ramsey
ultrafilter on $\Bbb N$.   (i) Show that there is an $I\in\Cal F$ such
$\inf_{i,j\in I}\nu(a_i\Bcap a_j)>0$.   (ii) Show that for every
$k\in\Bbb N$ there is an $I\in\Cal F$ such that
$\inf\{\nu(\inf_{i\in K}a_i):K\in[I]^k\}>0$.
(iii) Show that there is an $I\in\Cal F$ such that
$\{a_i:i\in I\}$ is centered.  \Hint{538Hc.}
%539K for (iii), work in $\widehat{\frak A}$

\spheader 539Xf Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of
subsets of $X$, and $\Cal I\normalsubgroup\Cal PX$ a $\sigma$-ideal;
suppose that $\Sigma/\Sigma\cap\Cal I$ is ccc.
Let $Y$ be a set, $\Tau$ a $\sigma$-algebra of subsets of $Y$, and
$\nu:\Tau\to\coint{0,\infty}$ a Maharam submeasure;
let $\Cal I\ltimes\Cal N(\nu)$ be the skew product.   
Show that $(\Sigma\tensorhat\Tau)
/(\Sigma\tensorhat\Tau)\cap(\Cal I\ltimes\Cal N(\nu))$ is ccc.
\Hint{527L.}
%527L

\leader{539Y}{Further exercises (a)}
%\spheader 539Ya
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra with a countable $\sigma$-generating set (331E), and $\nu$
a Maharam submeasure on $\frak A$.   Set $\Cal I=\{a:\nu a=0\}$.
Show that $\Cal I\prT\Cal N$.
%539J 341Yf 496K 539Ja 539Cb

\spheader 539Yb\dvArevised{2014}
Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of subsets of
$X$, and $\Cal I$ a proper $\sigma$-ideal of subsets of $X$ generated by
$\Sigma\cap\Cal I$;  let $\Sigma_L$ be the algebra of Lebesgue measurable
subsets of $\Bbb R$.   Write $\frak A$ for $\Sigma/\Sigma\cap\Cal I$,
$\Cal L$ for $(\Sigma\tensorhat\Sigma_L)\cap(\Cal I\ltimes\Cal N)$ and
$\frak C$ for $\Sigma\tensorhat\Sigma_L/\Cal L$.
(i) Show that $c(\frak C)=\max(\omega,c(\frak A))$ and
$\tau(\frak C)=\max(\omega,\tau(\frak A))$.
(ii) Show that $\frak C$ is \wsid\ iff $\frak A$ is.
% what about $\wdistr(\frak C)=\min(\cf\Cal N,\wdistr(\frak A))$?.
(iii) Show that $\frak C$ is measurable iff $\frak A$ is.
(iv) Show that $\frak C$ is a Maharam algebra iff $\frak A$ is.
% is $\Mhsr(\frak C)=\omega\cdot\Mhsr(\frak A)$?.\query
% not if \frak A measurable!  or \max(\omega,\Mhsr(\frak A)) ??
%539P 53bits

\spheader 539Yc Let $\frak A$ be a Boolean algebra with a
strictly positive Maharam submeasure $\hat\nu$, and $\frak B$ a
subalgebra of $\frak A$ which is dense for the associated metric (539Ac);
set $\nu=\hat\nu\restrp\frak B$, so
that $\nu$ is an exhaustive submeasure on $\frak B$.   For $\epsilon>0$ let
$r_{\nu\epsilon}:\frak B\to\On$ and
$r_{\hat\nu\epsilon}:\frak A\to\On$ be the rank functions
associated with $\nu$ and $\hat\nu$ respectively.   Show that

\Centerline{$r_{\nu\delta}(b)\le r_{\hat\nu\delta}(b)
\le r_{\nu\epsilon}(b)$}

\noindent whenever $b\in\frak B$ and $0<\epsilon<\delta$.
%539S

\spheader 539Ye (J.Kupka) Let $\nu$ be a
totally finite submeasure on a Boolean algebra $\frak A$, and set

\Centerline{$\check\nu a
=\inf_{n\in\Bbb N}\sup\{\min_{i\le n}\nu a_i:
    a_0,\ldots,a_n\Bsubseteq a$ are disjoint$\}$.}

\noindent for $a\in\frak A$, as in the proof of 539U.   Show that {\it
either} $\check\nu\ge\bover13\nu$ {\it or} there is a non-zero additive
$\mu:\frak A\to\coint{0,\infty}$ such that $\mu a\le\nu a$ for every
$a\in\frak A$.  \Hint{392D.}
%539U

\spheader 539Yf Show that the exhaustive submeasures constructed by
Talagrand's original method, as described in \S394 with
$\|I\|=\#(I)$ for $I\in[\Bbb N]^{<\omega}$,
have exhaustivity rank at most the ordinal power $\omega^{\omega^2}$.
%n06204 539U P&V

\spheader 539Yg\dvAnew{2014}
Suppose that $\frak A$ is a non-measurable Maharam algebra.
Show that $\Mhsr(\frak A)=\omega\cdot\Mhsr(\frak A)$.
%539U
}%end of exercises

\leader{539Z}{Problems (a)}
%\spheader 539Za
Let $\nu$ be a non-zero totally finite Radon submeasure on a Hausdorff
space $X$.   Must there be a lifting for $\nu$?
\cmmnt{that is, writing $\Sigma$ for the domain of $\nu$, must there be a
Boolean homomorphism $\phi:\Sigma\to\Sigma$ such that
$\nu(E\symmdiff\phi E)=0$ for every $E\in\Sigma$ and $\phi E=\emptyset$
whenever $\nu E=0$?}

\spheader 539Zb\dvAnew{2017} Is there a Maharam algebra with uncountable
Maharam submeasure rank?

\endnotes{
\Notesheader{539} During the growth of this treatise, the
sections on Maharam submeasures were twice transformed by
new discoveries, and I naturally hope that the work I have just presented
will be similarly outdated before too long.
In the pages above I have tried in
the first place to show how the cardinal functions of chapters 51 and 52
can be applied in this more general context.   With minor refinements of
technique, we can go a fair way.   Because we know we have at least two
non-trivial atomless Maharam algebras of countable type, we are led to a
more detailed analysis, as in 539Ca and 539J.

Equally instructive are the
apparent limits to what the methods can achieve, which mostly point to
remaining areas of obscurity.   I say `remaining';  but what is most
conspicuous about the present situation is our nearly total ignorance
concerning the structure of non-measurable Maharam algebras.   The
Talagrand-Perovi\'c-Veli\v{c}kovi\'c
construction, as described in \S394, gives us a family of such algebras,
but so far we can answer hardly any of the most elementary questions
about them (see 394Z).

The message of {\smc Balcar Jech \& Paz\'ak 05} is that a Dedekind
complete, ccc, \wsid\ Boolean algebra is `nearly' a Maharam algebra.   Any
further condition (e.g., the $\sigma$-finite chain condition, as in
393S) is likely to render it a Maharam algebra;
and with a little help from an extra axiom of set theory, it is
already necessarily a Maharam algebra
(539N).   Similarly, much of the work of the last sixty years on
submeasures suggests that exhaustive submeasures are `nearly'
uniformly exhaustive, and that an extra condition (e.g., sub- or
super-modularity) is enough to tip the balance (413Yh).
At both boundaries, there are few examples to limit
conjectures about further conditions on which such results might be based.
Besides 539P and Talagrand's examples, we have a further important
possibility of a not-quite-Maharam algebra in 555K below.
}%end of notes

\discrpage

\references{
\frfilename{mt5r.tex}
\versiondate{28.4.23}
\copyrightdate{2008}

% 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 5}

%use subparagraph details (5A1Aa) only for X. Y. Z.
%  \S511 {\it notes\/}, \S512 {\it intro.}]

\medskip%A

%Antonovskii M.Ja.\ \& Chudnovsky D.V.\ [76] `Some questions of general
%topology and Tikhonov semifields II', Russian Math.\ Surveys 31 (1976)
%69-128.

Argyros S.A.\ \& Kalamidas N.D.\ [82] `The $K_{\alpha n}$ property on
spaces with strictly positive measure', Canadian J.\ Math.\ 34 (1982)
1047-1058.
\cmmnt{[525T.]}

Argyros S.A.\ \& Tsarpalias A.\ [82] `Calibers of compact spaces',
Trans.\ Amer.\ Math.\ Soc.\ 270 (1982) 149-162.
\cmmnt{[525N.]}

\medskip%B

Babiker A.G.\ [76] `On uniformly regular topological measure spaces',
Duke Math.\ J.\ 43 (1976) 775-789.
\cmmnt{[533Xd.]}

Balcar B.\ \& Fran\v{e}k F.\ [82] `Independent families in complete Boolean
algebras', Trans.\ Amer.\ Math.\ Soc.\ 274 (1982) 607-618.
\cmmnt{[515H.]}

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{[\S539 {\it intro.}, 539E, 539L, 539N, \S539 {\it notes\/}, 555K.]}
 % arXiv.math.LO/0312473

Balcar B., Jech T.\ \& Zapletal J.\ [97] `Semi-Cohen Boolean algebras',
Ann.\ Pure \& Applied Logic 87 (1997) 187-208.
\cmmnt{[\S515 {\it notes\/}, \S547 {\it notes\/}.]}

Balcar B.\ \& Vojt\'a\v{s} P.\ [77] `Refining systems on Boolean algebras',
pp.\ 45-58 in {\smc Lachlan Srebny \& Zarach 77}.
\cmmnt{[515E.]}

Bar-Hillel Y.\ [70] {\it Mathematical Logic and Foundations of Set Theory.}
North-Holland, 1970.

Bartoszy\'nski T.\ [84] `Additivity of measure implies additivity of
category', Trans.\ Amer.\ Math.\ Soc.\ 281 (1984) 209-213.
\cmmnt{[Chap.\ 52 {\it intro.}, 522Q, \S524 {\it notes\/}.]}

Bartoszy\'nski T.\ [87] `Combinatorial aspects of measure and category',
Fund.\ Math.\ 127 (1987) 225-239.
\cmmnt{[522S.]}
% http://matwbn.icm.edu.pl/ksiazki/fm/fm127/fm127118.pdf

% Bartoszy\'nski T.\ [88] `On covering of real line by null sets',
% Pacific J.\ Math.\ 131 (1988) 1-12.

% Bartoszy\'nski T.\ [p97] `Invariants of measure and category', in
% preparation for {\smc Foreman Kanamori \& Magidor ??}

% Bartoszy\'nski T.\ [00] `A note on duality between measure and category',
% Proc.\ Amer.\ Math.\ Soc.\ 128 (2000) 2745-2748.
   % (2^N,+,meager) not iso to (2^N,+,negligible)

Bartoszy\'nski T.\ \& Judah H.\ [89] `On the cofinality of the smallest
covering of the real line by meager sets', J.\ Symbolic Logic 54 (1989) 828-832.
\cmmnt{[522V.]}

Bartoszy\'nski T.\ \& Judah H.\ [95] {\it Set Theory:  on the structure
of the real line}.  A.K.Peters, 1995.
\cmmnt{[522R, \S522 {\it notes\/},
526M, \S528 {\it notes\/}, 534Q, \S534 {\it notes\/}, 546I,
553C, 554Ya, 5A1Q.]}

%Bartoszy\'nski T., Judah H.\ \& Shelah S.\ [93] `The Cicho\'n diagram',
%J.\ Symbolic Logic 58 (1993) 401-423.

%Bartoszy\'nski T.\ \& Scheepers M.\ [96] (eds.) {\it Set Theory (BEST, 1992-1994).}
%Amer.\ Math.\ Soc., 1996 (Contemporary Mathematics 192).
% ISBN 0821803069

Bartoszy\'nski T.\ \& Shelah S.\ [92] `Closed measure zero sets', Ann.\
Pure \& Applied Logic 58 (1992) 93-110.
\cmmnt{[526M.]}

Baumgartner J.E., Taylor A.D.\ \& Wagon S.\ [77] `On splitting
stationary subsets of large cardinals',
J.\ Symbolic Logic 42 (1977) 203-214.
\cmmnt{[\S541 {\it notes\/}, \S544 {\it notes\/}.]}

Bell M.G.\ [81] `On the combinatorial principle $P(c)$',
Fundamenta Math.\ 114 (1981) 149-157.
\cmmnt{[\S517 {\it notes\/}.]}

Benedikt M.\ [98] `Ultrafilters which extend measures',
J.\ Symbolic Logic 63 (1998) 638-662.
\cmmnt{[538M, 538Xm, \S538 {\it notes\/}.]}

Benedikt M.\ [99] `Hierarchies of measure-theoretic ultrafilters',
Annals of Pure and Applied Logic 97 (1999) 203-219.
\cmmnt{[538H.]}

Blair C.E.\ [77] `The Baire category theorem implies the principle of
dependent choice', Bull.\ Acad.\ Polon.\ Sci.\ (Math.\ Astron.\ Phys.) 25
(1977) 933-934.
\cmmnt{[\S566 {\it notes\/}.]}

%Blass A.\ [88] `Selective ultrafilters and homogeneity', Ann.\ Pure and
%Applied Logic 38 (1988) 252-285.

%Blass A.\ [p01b] `Combinatorial cardinal characteristics of the continuum',
%pp. ?? in {\smc Foreman Kanamori \& Magidor ??}.

Blass A.\ \& Laflamme C.\ [89] `Consistency results about filters and the
number of inequivalent growth types', J.\ Symbolic Logic 54 (1989) 50-56.
% u<g  filter dichotomy
\cmmnt{[5A6Ib, 5A6J.]}

Booth D.\ [70] `Ultrafilters on a countable set',
Ann.\ Math.\ Logic 2 (1970) 1-24.
\cmmnt{[517R.]}

Borel E.\ [1919] `Sur la classification des ensembles de mesure nulle', Bull.\ Soc.\ Math.\ France 47 (1919) 97-125.
\cmmnt{[\S534 {\it intro.}]}

Brendle J.\ [00] {\it Inner model theory and large cardinals.}
Research report, March 2000.
\cmmnt{[528N, 529F-529H.]} %529F 529G 529H

%Brendle J.\ [01] `Cardinal invariants of the continuum --- a survey',
%J.\ Research Institute for Mathematical Sciences, Kyoto University
%(JRIMS Kokyuroku) 1202 (2001) 7-32.

Brendle J.\ [06] `Cardinal invariants of the continuum and combinatorics
on uncountable cardinals',
Annals of Pure and Applied Logic 144 (2006) 43-72.
\cmmnt{[529F, 529H.]}

Burke M.R.\ [93] `Liftings for Lebesgue measure', pp.\ 119-150 in
{\smc Judah 93}.
\cmmnt{[\S535 {\it notes\/}.]}

%Burke M.R.\ [n95] `Consistent liftings', privately circulated, 1995.

%Burke M.R.\ [n96] `A theorem of Gitik and Shelah on disjoint refinements
%of sequences of subsets of the real line', note of 23.8.96.
%\cmmnt{[5x46J, \S547 {\it intro.}]}

Burke M.R.\ [n05] `Non-null sets of minimal cofinality', note of 2.3.05.
\cmmnt{[523K.]}

Burke M.R.\ \& Magidor M.\ [90] `Shelah's pcf theory and
its applications', Ann.\ Pure and Applied Logic 50 (1990) 207-254.
\cmmnt{[513J, \S5A2 {\it intro.}.]}

Burke M.R.\ \& Shelah S.\ [92]
`Linear liftings for non-complete probability spaces',
Israel J.\ Math.\ 79 (1992) 289-296.
\cmmnt{[\S535 {\it notes\/}.]}

\medskip%C

Carlson T.J.\ [84] `Extending Lebesgue measure to infinitely many sets',
Pacific J.\ Math.\ 115 (1984) 33-45.
\cmmnt{[552N, \S552 {\it notes.}]}

Carlson T.J.\ [93] `Strong measure zero and strongly meager sets',
Proc.\ Amer.\ Math.\ Soc.\ 118 (1993) 577-586.
\cmmnt{[534R.]}

Carlson T., Frankiewicz R.\ \& Zbierski P.\ [94]
`Borel liftings of the measure
algebra and the failure of the continuum hypothesis', Proc.\ Amer.\ Math.\ Soc.\
120 (1994) 1247-1250.
\cmmnt{[\S535 {\it notes\/}, 554I.]}

Choksi J.R.\ \& Fremlin D.H.\ [79] `Completion regular measures on
product spaces', Math.\ Ann.\ 241 (1979) 113-128.
\cmmnt{[532I.]}

%Cicho\'n J., Kamburelis T.\ \& Pawlikowski J.\ [85] `On dense subsets of
%the measure algebra', Proc.\ Amer.\ Math.\ Soc.\ 94 (1985) 142-146.

Cicho\'n J.\ \& Pawlikowski J.\ [86]
`On ideals of subsets of the plane and on Cohen
reals', J.\ Symbolic Logic 51 (1986) 560-569.
\cmmnt{[527F.]}

%Cohen P.\ [63] `The independence of the continuum hypothesis',
% Proc.\ Nat.\ Acad.\ Sci.\ U.S.A.\ 50 (1963) 1143-1148.

Comfort W.W.\ \& Negrepontis S.\ [74] {\it The Theory of Ultrafilters.}
Springer, 1974.
\cmmnt{[538F, 538Yb.]}

Comfort W.W.\ \& Negrepontis S.\ [82] {\it Chain Conditions in
Topology.}   Cambridge U.P., 1982.
\cmmnt{[516 {\it notes\/}.]}

Cummings J.\ [92] `A model in which GCH holds at successors but fails at
limits',
Trans.\ Amer.\ Math.\ Soc.\ 329 (1992) 115-142.
\cmmnt{[525Z.]}

\medskip%D

Dellacherie C., Meyer P.A.\ \& Weil M.\ (eds.) [73]
{\it S\'eminaire de Probabilit\'es VII.}   Springer, 1973
(Springer Lecture Notes in Mathematics 321).

Devlin K.J.\ [84] {\it Constructibility.} Springer, 1984.
\cmmnt{[5A6D.]}

%Dow A.\ \& Hart K.P.\ [p98] `The measure algebra does not always embed',
%preprint, 1998

Dow A.\ \& Stepr\=ans J.\ [94] `The $\sigma$-linkedness of the measure
algebra', Canad.\ Math.\ Bulletin 37 (1994) 42-45.
\cmmnt{[524L.]}

%Drake F.\ [74] {\it Set Theory.}  North-Holland, 1974.

%Drake F.R.\ \& Singh D.\ [96] {\it Intermediate Set Theory.}
%Wiley, 1996.  \cmmnt{[\S5A1.]}
% ISBN 0 471 96496 4

D\v{z}amonja M.\ \& Kunen K.\ [93] `Measures on compact HS spaces',
Fundamenta Math.\ 143 (1993) 41-54.
\cmmnt{[\S531 {\it notes\/}.]}

%D\v{z}amonja M.\ \& Kunen K.\ [95] `Properties of the class of
%measure-separable compact spaces', Fundamenta Math.\ 147 (1995) 261-277.

D\v{z}amonja M.\ \& Plebanek G.\ [04]
`Precalibre pairs of measure algebras',
Topology Appl.\ 144 (2004) 67-94.
\cmmnt{[525L.]}

\medskip%E

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).
\cmmnt{[\S5A4.]}

%Engelking R.\ \& Kar{\l}owicz M.\ [65]
%`Some theorems of set theory and their topological consequences',
% Fund.\ Math.\ 57 (1965) 275-285.

Erd\H{o}s P., Hajnal A., M\'at\'e A.\ \& Rado R.\ [84] {\it Combinatorial
Set Theory:  Partition Relations for Cardinals.}   Akad\'emiai Kiad\'o,
1984 (Disquisitiones Math.\ Hung.\ 13).
\cmmnt{[5A1F, 5A1H, 5A6F.]}

\medskip%F

Farah I.\ [00] {\it Analytic Quotients.}
Mem.\ Amer.\ Math.\ Soc.\ 148 (2000).
\cmmnt{[5A6H.]}

Farah I.\ [03] `How many algebras $\Cal P(\Bbb N)/\Cal I$ are there?',
Illinois J.\ Math.\ 46 (2003) 999-1033.
\cmmnt{[\S556 {\it notes\/}.]}

Farah I.\ [06] `Analytic Hausdorff gaps II: the density zero ideal',
Israel J.\ Math.\ 154 (2006) 235-246.
\cmmnt{[556S.]}

Farah I.\ \& Veli\v{c}kovi\'c B.\ [06]
`Von Neumann's problem and large cardinals',
Bull.\ London Math.\ Soc.\ 38 (2006) 907-912.
\cmmnt{[539Q.]}

Feferman S. \& L\'evy A. [63] `Independence results in set theory by Cohen's
method', Notices Amer.\ Math.\ Soc.\ 10 (1963) 53.
\cmmnt{[561A.]}

Fleissner W.G.\ [91]
`Normal measure axiom and Balogh's theorems', Topology and its Appl.\
39 (1991) 123-143.
\cmmnt{[555N.]}

% Fleissner W.G., Hansell R.W.\ \& Junnila H.J.K.\ [82] `PMEA implies
% Proposition P', Topology and its Appl.\ 13 (1982) 255-262.

% Foreman M., Kanamori A.\ \& Magidor M.\ [??] {\it Handbook of Set
% Theory}.   Kluwer, to appear.

%Foreman M., Magidor M.\ \& Shelah S.\ [88] `Martin's maximum, saturated ideals, and non-regular ultrafilters', Annals of Math.\ (2) 127 (1988) 1-47.

Foreman M.\ [10] `Ideals and generic elementary embeddings',  pp.\ 885-1147
in {\smc Foreman \& Kanamori 10}, vol.\ 2.
\cmmnt{[547Z.]}

Foreman M.\ \& Kanamori A.\  [10] {\it Handbook of Set Theory.}
Springer, 2010.

Foreman M.\ \& Wehrung F.\ [91] `The Hahn-Banach theorem implies the
existence of a non-Lebesgue measurable set', Fundamenta Math.\ 138
(1991) 13-19.
\cmmnt{[563A.]}

Foreman M.\ \& Woodin W.\ [91] `The generalized continuum hypothesis can
fail everywhere', Annals of Math.\ 133 (1991) 1-35.
\cmmnt{[525Z.]}

Fossy J.\ \& Morillon M.\ [98] `The Baire category property and some
notions of compactness', J.\ London Math.\ Soc. (2) 57 (1998) 1-19.
\cmmnt{[\S566 {\it notes\/}.]}

% Frankiewicz R.\ \& Plebanek G.\ [94] `Nonaccessible filters in
% measure algebras and functionals on $L^{\infty}(\lambda)^*$', Studia
% Math.\ 108 (1994) 191-200.

%Frankiewicz R.\ \& Plebanek G.\ [96] `On asymptotic density and
%uniformly distributed sequences', Studia Math.\ 119 (1996) 17-26.

Freese R. \& Nation J.B.\ [78] `Projective lattices',
Pacific J.\ Math.\ 75 (1978) 93-106.
\cmmnt{[\S518 {\it notes\/}.]}

Freiling C.\ [86] `Axioms of symmetry;  throwing darts at the real number
line', J.\ Symbolic Logic 51 (1986) 190-200.
\cmmnt{[537K.]}

Fremlin D.H.\ [74] {\it Topological Riesz Spaces and Measure Theory.}
Cambridge U.P., 1974.
\cmmnt{[564Xc.]}

Fremlin D.H.\ [77]
`Uncountable powers of {\bf R} can be almost Lindel\"of',
Manuscripta Math.\ 22 (1977) 77-85.
\cmmnt{[533J.]}

Fremlin D.H.\ [84a] {\it Consequences of Martin's Axiom.}
Cambridge U.P., 1984.
\cmmnt{[\S511 {\it notes\/}, \S517 {\it notes\/}, \S531 {\it notes\/}.]}

Fremlin D.H.\ [84b] `On the additivity and cofinality of Radon measures',
Mathematika 31 (1984) 323-335.
\cmmnt{[\S524 {\it notes\/}.]}

Fremlin D.H.\ [87] {\it Measure-additive Coverings and
Measurable Selectors.} Dissertationes Math.\ 260 (1987).
\cmmnt{[551A.]}

Fremlin D.H.\ [88] `Large correlated families of positive
random variables',
Math.\ Proc.\ Cambridge Phil.\ Soc.\ 103 (1988) 147-162.
\cmmnt{[525S.]}

Fremlin D.H.\ [91] `The partially ordered sets of measure
theory and Tukey's ordering,' Note di Matematica  11 (1991) 177-214.
[523Yc, \S524 {\it notes\/}, 526B, 526I, 527J, 529C, 529D, 534L.]

Fremlin D.H.\ [93] `Real-valued-measurable cardinals', pp.\ 151-304
in {\smc Judah 93}.
\cmmnt{[\S537 {\it notes\/}, \S541 {\it notes\/}, \S544 {\it notes\/},
\S545 {\it notes\/}, \S5A2 {\it intro.}.]}

Fremlin D.H.\ [97] `On compact spaces carrying Radon measures of
uncountable Maharam type', Fundamenta Math.\  154 (1997) 295-304.
\cmmnt{[531T.]}

%Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces',
%Fundamenta Math.\ 165 (2000) 67-94.

Fremlin D.H.\ [03] `Skew products of ideals',
J.\ Applied Analysis 9 (2003) 1-18.
\cmmnt{[\S527 {\it notes\/}.]}
%n01910.tex

%Fremlin D.H.\ [n97] `A theorem of Gitik and Shelah on disjoint
%refinements of sequences of subsets of the real line', note of 4.4.97.
%n96j01.tex

%Fremlin D.H.\ [n99] `Embedding the Lebesgue measure algebra into
%$\Cal P\Bbb N/\Cal Z$', note of 25.5.99.
%n99525.tex

%Fremlin D.H.\ [n99] `Notes on {\smc Farah p99}', note of 26.7.99.
%n99612.tex

%Fremlin D.H.\ [n99] `Notes on {\smc Solecki p96}', note of 17.6.99.
%n99617.tex

%Fremlin D.H.\ [n02] `Games and the Tukey classification', note of 29.9.02.
%n02929.tex

Fremlin D.H.\ \& Grekas S.\ [95] `Products of completion regular
measures', Fundamenta Math.\ 147 (1995) 27-37.
\cmmnt{[532D.]}

Fremlin D.H.\ \& Miller A.W.\ [88]
`On some properties of Hurewicz, Menger and Rothberger',
Fund.\ Math.\ 129 (1988) 17-33.
\cmmnt{[534Q.]}

Fremlin D.H., Natkaniec T.\ \& Rec{\l}aw I.\ [00]
`Universally Kuratowski-Ulam spaces', Fundamenta Math.\ 165 (2000) 239-247.
\cmmnt{[\S527 {\it notes\/}.]}

%Friedman H.\ [80]
%`A consistent Fubini-Tonelli theorem for nonmeasurable functions',
% Illinois J.\ Math.\ 24 (1980) 390-395.

Friedman S.D. \& Koepke P.\ [97] `An elementary approach to the fine
structure of L', Bull.\ Symbolic Logic 3 (1997) 453-468.
\cmmnt{[5A6D.]}

Fuchino S.,  Geschke S., Shelah S.\ \& Soukup L.\ [01]
`On the weak Freese-Nation property of complete Boolean algebras',
Ann.\ Pure Appl.\ Logic 110 (2001) 89-105.
\cmmnt{[518K.]}

Fuchino S., Geschke S.\ \& Soukup L.\ [01] `On the weak Freese-Nation
property of $\Cal P(\omega)$', Arch.\ Math.\ Logic 40 (2001) 425-435.
\cmmnt{[522U, \S522 {\it notes\/}.]}

Fuchino S., Koppelberg S.\ \& Shelah S.\ [96] `Partial orderings with
the weak Freese-Nation property',
Ann.\ Pure and Applied Logic 80 (1996) 35-54.
\cmmnt{[518A, 518G, 518Yb, \S518 {\it notes\/}, 522U.]}

Fuchino S.\ \& Soukup L.\ [97] `More set theory around the weak
Freese-Nation property', Fundamenta Math.\ 154 (1997) 159-176.
\cmmnt{[518I, 518K.]}

\medskip%G

Galvin F.\ [80] `Chain conditions and products',
Fundamenta Math.\ 108 (1980) 33-48.
\cmmnt{[\S553 {\it notes\/}.]}

Galvin F., Mycielski J.\ \& Solovay R.M.\ [79] `Strong measure zero sets',
Notices Amer.\ Math.\ Soc.\ 26 (1979) A280.
%(abstract 79T-E25);  often assigned to year 1973
\cmmnt{[534K.]}

Gandy R.O.\ \& Hyland J.M.E.\ [77](eds.) {\it Logic Colloquium '76.}
North-Holland, 1977.

Geschke S.\ [02] `On tightly $\kappa$-filtered Boolean algebras',
Algebra Universalis 47 (2002) 69-93.
\cmmnt{[518P, 518S, \S535 {\it notes\/}.]}

Gitik M.\ \& Shelah S.\ [89] `Forcings with ideals and
simple forcing notions', Israel J.\ Math.\ 68 (1989) 129-160.
\cmmnt{[543E, 547F.]}

Gitik M.\ \& Shelah S.\ [93]
`More on simple forcing notions and forcings with ideals',
Annals of Pure and Applied Logic 59 (1993) 219-238.
\cmmnt{[542E, 543E, 547F.]}

Gitik M.\ \& Shelah S.\ [01]
`More on real-valued measurable cardinals and forcing with ideals',
Israel J.\ Math. 124 (2001) 221-242.
\cmmnt{[548E.]}
%GiSh582

G\l\'owczy\'nski W.\ [91] `Measures on Boolean algebras',
Proc.\ Amer.\ Math.\ Soc.\ 111 (1991) 845-849.
\cmmnt{[555K.]}

G\l\'owczy\'nski W.\ [08] `Outer measure on Boolean algebras', Acta Univ.\
Carolinae (Math.\ et Phys.) 49 (2008) 3-8.
\cmmnt{[555K.]}

Goldstern M., Judah H.\ \& Shelah S.\ [93] `Strong measure zero sets
without Cohen reals',
J.\ Symbolic Logic 58 (1993) 1323-1341.
\cmmnt{[\S534 {\it notes\/}.]}

\medskip%H

% Hart J.\ \& Kunen K.\ [98] `Weak Measure Extension Axioms',
% Topology and its Applications 85 (1998) 219-246.
% http://www.math.wisc.edu/~kunen/me.ps

Haydon R.G.\ [77] `On Banach spaces which contain $\ell^1(\tau)$ and
types of measures on compact spaces', Israel J.\ Math.\ 28 (1977) 313-324.
\cmmnt{[531E, 531L.]}

%Haydon R.G.\ [78] `On dual $L^1$-spaces and injective bidual Banach
%spaces,' Israel J.\ Math.\ 31 (1978) 142-152.

Hodges W., Hyland M., Steinhorn C.\ \& Truss J.\ [96] (eds.)
{\it Logic: from Foundations to Applications,
European Logic Colloquium, 1993.}
Clarendon, 1996.
%[ISBN 0-19-853862-6]

Humke P.D.\ \& Laczkovich M.\ [05] `Symmetrically approximately
continuous functions, consistent density theorems, and Fubini type
inequalities', Trans.\ Amer.\ Math.\ Soc.\ 357 (2005) 31-44.
\cmmnt{[537Q, \S537 {\it notes\/}.]}

\medskip%I

Ihoda J.I.\ [88] `Strong measure zero sets and rapid filters',
J.\ Symbolic Logic 53 (1988) 393-402.
%model of Borel conjecture + Ramsey filter
\cmmnt{[534Q.]}

\medskip%J

Jech T.\ [73] {\it The Axiom of Choice.}  North-Holland, 1973.
\cmmnt{[561A, 561Xc, 561Yc, 561Yi, 564Yd.]}

Jech T.\ [78] {\it Set Theory.}  Academic, 1978 (ISBN 0123819504).
\cmmnt{[{\it Intro.}, 553H,
555O, \S555 {\it notes\/}, \S562 {\it notes\/}, \S5A1, 5A6B.]}
%5A1B 5A1D 5A1E 5A1F B
%ASIN B0095LMZ9O

Jech T.\ [03] {\it Set Theory, Millennium Edition.}  Springer, 2003
(ISBN 3540440852).
\cmmnt{[\S521 {\it notes\/}, \S555 {\it notes\/},
\S562 {\it notes\/}, \S567 {\it notes\/}, \S5A1, 5A3N, 5A3P, 5A6B.]}
%5A1B 5A1D 5A1E 5A1F B

%Jin R.\ \& Shelah S.\ [p96] `Compactness of Loeb spaces', preprint, 1996.

%Jovanovi\'c A.\ [p01]
%`Real valued measurability, some set-theoretic aspects',
%pp.\ ?? in {\smc Pap 0?}

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.

Judah H.\ \& Repick\'y M.\ [95] `Amoeba reals', J.\ Symbolic Logic 60 (1995) 1168-1185.
\cmmnt{[528N.]}

%Juh\'asz I.\ [71]  {\it Cardinal Functions in Topology.}  North-Holland,
%1975 (Math.\ Centre Tracts 34).

%Juh\'asz I.\ [80]  {\it Cardinal Functions in Topology -- Ten Years Later.}  Mathematisch Centrum, Amsterdam, 1980 (Math.\ Centre Tracts 123).   ISBN 9061961963.

Just W.\ [92] `A modification of Shelah's oracle-c.c., with
applications', Trans.\ Amer.\ Math.\ Soc.\ 329 (1992) 325-356.
\cmmnt{[527M.]}

%Just W.\ \& Krawczyk A.\ [84] `On certain Boolean algebras
%$\Cal P(\omega)/I$', Trans.\ Amer.\ Math.\ Soc.\ 285 (1984) 411-429.

Just W.\ \& Weese M.\ [96] {\it Discovering Modern Set Theory I.}
Amer.\ Math.\ Soc., 1996 (Graduate Studies in Mathematics 8).
\cmmnt{[{\it Intro.}.]}

Just W.\ \& Weese M.\ [97] {\it Discovering Modern Set Theory II.}
Amer.\ Math.\ Soc., 1997 (Graduate Studies in Mathematics 18).
\cmmnt{[5A1B, 5A1F, 5A1H.]}

\medskip%K

%Kada M. \& Yuasa Y.\ [96] `Cardinal invariants about shrinkability of unbounded
%sets', Topology and its Applications 74 (1996) 215-223.

%Kamburelis A.\ [89] `Iterations of Boolean algebras with measure',
%Arch.\ Math.\ Logic 29 (1989) 21-28.
% chargeable algebras

%Kamburelis A.\ [90] `A new proof of the Gitik-Shelah theorem', Israel
%J.\ Math.\ 72 (1990) 373-380.

Kanamori A.\ [03] {\it The Higher Infinite.}   Springer, 2003.
\cmmnt{[\S541 {\it notes\/}, \S555 {\it notes\/}, \S567 {\it notes\/},
5A1H, 5A6B, 5A6F.]}

Kechris A.S.\ [95] {\it Classical Descriptive Set Theory.}  Springer,
1995.
\cmmnt{[5A1D, \S567 {\it notes\/}.]}
%QA248.K4

Keisler H.J.\ \& Tarski A.\ [64] `From accessible to
inaccessible cardinals', Fundamenta Math.\ 53 (1964) 225-308;  errata
Fundamenta Math.\ 57 (1965) 119.
\cmmnt{[\S541 {\it intro.\/}, \S541 {\it notes\/}.]}

Kelley J.L.\ [50] `The Tychonoff product theorem implies the axiom of
choice', Fundamenta Math.\ 37 (1950) 75-76.
\cmmnt{[561D.]}

Koppelberg S.\ [75]
`Homomorphic images of $\sigma$-complete Boolean algebras',
Proc.\ Amer.\ Math.\ Soc.\ 51 (1975) 171-175.
\cmmnt{[515L.]}

Koppelberg S.\ [89] {\it General Theory of Boolean
Algebras\/}, vol.\ 1 of {\smc Monk 89}.
\cmmnt{[\S515 {\it notes\/}.]}

Koppelberg S.\ \& Shelah S.\ [96]
`Subalgebras of Cohen algebras need not be Cohen',
pp.\ 261-275 in {\smc Hodges Hyland Steinhorn \& Truss 96}.
\cmmnt{[\S515 {\it notes\/}, \S547 {\it notes\/}.]}

Kraszewski J.\ [01] `Properties of ideals on the generalized Cantor spaces',
J.\ Symbolic Logic 66 (2001) 1303-1320.
\cmmnt{[523G, 523H, 523J, \S523 {\it notes\/}.]}

%Krawczyk A.\ [n85] `Dominating reals add random reals', 1985.

Kumar A.\ [13] `Avoiding rational distances', Real Analysis Exchange 38
(2012/13), 493-498.
%{\tt http://www.math.wisc.edu/\~{}akumar/ARD.pdf}.
%{\tt
%https:/{\bsp}projecteuclid.org/journals/real-analysis-exchange/volume-38%
%/issue-2/Avoiding-Rational-{\vthsp}Distances/rae/1403894908.full}
\cmmnt{[548Xb.]}

%Kumar A.\ [n16] `On a theorem of Gitik and Shelah', note of January 2016
%({\tt http:/{\bsp}www.math.huji.ac.il/\~{}akumar/otgs.pdf}).

Kumar A.\ \& Shelah S.\ [17] `A transversal of full outer measure',
Advances in Math.\ 321 (2017) 475-485.
\cmmnt{[547P, 547Q, \S548 {\it intro.\/}, 548C.]}
%http://home.iitk.ac.in/~krashu/tfom.pdf;  printed version not seen

%Kunen K.\ [75] `Some points in $\beta N$',
% Math.\ Proc.\ Cambridge Phil.\ Soc.\ 80 (1975) 385-398.

Kunen K.\ [80] {\it Set Theory.}  North-Holland, 1980.
\cmmnt{[{\it Intro.}, 511A, \S521 {\it notes\/}, \S522 {\it notes\/},
\S541 {\it notes\/}, 551Q, \S551 {\it notes\/}, 556F,
\S562 {\it notes}, Appendix {\it intro.}, \S5A1, \S5A3, 5A6B.]}

Kunen K.\ [81] `A compact $L$-space under CH', Topology and its Appl.\
12 (1981) 283-287.
\cmmnt{[\S531 {\it notes\/}.]}

Kunen K.\ [84] `Random and Cohen reals', pp.\ 887-911 in
{\smc Kunen \& Vaughan 84}.
\cmmnt{[\S554 {\it notes}.]}

Kunen K.\ [n70] `$\Pi^1_1$ reflection at the continuum', note of January
1970.
\cmmnt{[543C, 544C, 544E, 544F.]}

Kunen K.\ \& Mill J.van [95] `Measures on Corson compact spaces',
Fundamenta Math.\ 147 (1995) 61-72.
\cmmnt{[531O.]}
%http://www.math.wisc.edu/~kunen/corson.ps

Kunen K.\ \& Vaughan J.E.\ [84] (eds.) {\it Handbook of Set-Theoretic
Topology.}   North-Holland, 1984.

Kuratowski K.\ [66] {\it Topology,} vol.\ I.   Academic, 1966.
\cmmnt{[562G.]}

\medskip%L

Lachlan A., Srebny M.\ \& Zarach A.\ [77] (eds.) {\it Set Theory and Hierarchy Theory.}  Springer, 1977 (Lecture Notes in Math.\ 619).

%Laczkovich M.\ \& Miller A.W.\ [95] `Measurability of functions with
%approximately continuous vertical sections and measurable horizontal
%sections', Colloquium Math.\ 69 (1995) 299-308.

Larson P.\ [09] `The filter dichotomy and medial limits',
J.\ Math.\ Logic 9 (2009) 159-165.
\cmmnt{[538S.]}

Larson P., Neeman I.\ \& Shelah S.\ [10] `Universally measurable sets in
generic extensions', Fundamenta Math.\ 208 (2010) 173-192.
\cmmnt{[553O.]}
	
%Larson P.\ \& Shelah S.\ [p09]
%`There may be just continuum many universally measurable sets', 2009.
%LrSh:947

Laver R.\ [76] `On the consistency of Borel's conjecture', Acta Math.\
137 (1976) 151-169.
\cmmnt{[534Q.]}

Laver R.\ [87] `Random reals and Souslin trees',
Proc.\ Amer.\ Math.\ Soc.\ 100 (1987) 531-534.
\cmmnt{[553M.]}

Levinski J.-P., Magidor M.\ \& Shelah S.\ [90] `Chang's conjecture for
$\aleph_{\omega}$', Israel J.\ Math.\ 69 (1990) 161-172.
\cmmnt{[5A6F.]}

Levy A.\ [71] `The sizes of the indescribable cardinals',
pp.\ 205-218 in {\smc Scott 71}.
\cmmnt{[\S541 {\it notes\/}, \S544 {\it notes\/}.]}

Lipecki Z.\ [09]
`Semivariations of an additive function on a Boolean ring',
Colloquium Math.\ 117 (2009) 267-279.
\cmmnt{[552Xe.]}

%Loeb P.A.\ \& Talvila E.\ [p00]
%`Covering theorems and Lebesgue integration', preprint, June 2000.

%Losada M. \& Todor\v{c}evi\'c S. [p98] `Chains and antichains in Boolean
%algebras', note of 5.12.98.

%Louveau A.\ \& Veli\v{c}kovi\'c B.\ [99] `Analytic ideals and cofinal
%types', Ann.\ Pure and Applied Logic 99 (1999) 171-195.
%\ell^1\not\prT\CalNwd

\medskip%M

Maharam D.\ [1947] `An algebraic characterization of measure algebras',
Ann.\ Math.\ 48 (1947) 154-167.
\cmmnt{[539P.]}

Makowsky J.A.\ \& Ravve E.V.\ (eds.) [98]
{\it Logic Colloquium '95.}  Springer, 1998 (Lecture Notes in Logic 11).
%ISBN 3-540-63994-2

%Malliaris M.\ \& Shelah S.\ [16] `Cofinality spectrum theorems in
%model theory, set theory and general topology',
%J.\ Amer.\ Math.\ Soc.\ 29 (2016) 237-297.

Martin D.A.\ [70] `Measurable cardinals and analytic games',
Fundamenta Math.\ 66 (1970) 287-291.
\cmmnt{[567N.]}

Martin D.A.\ [75] `Borel determinacy', Ann.\ of Math.\ (2) 102 (1975)
363-371.
\cmmnt{[\S567 {\it notes\/}.]}

Martin D.A.\ \& Solovay R.M.\ [70] `Internal Cohen
extensions', Ann.\ Math.\ Logic 2 (1970) 143-178.
\cmmnt{[Chap.\ 52 {\it intro.}, \S528 {\it intro.}, \S528 {\it notes\/}.]}

%Mathias A.R.D.\ [78] `0\# and the $p$-point problem', pp.\ 375-384 in
%{\smc M\"uller \& Scott 78}.

M\'atrai T.\ [p09] `More cofinal types of definable directed orders', 2009
({\tt https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.414.3938}).
\cmmnt{[526L, \S526 {\it notes\/}.]}
% books/matrai.p09.pdf

%Mercourakis S.\ [96] `Some remarks on countably determined measures and
%uniform distribution of sequences', Monatschefte f\"ur Math.\ 121 (1996)
%79-101.

Meyer P.-A.\ [73] `Limites m\'ediales d'apr\`es Mokobodzki',
pp.\ 198-204 in {\smc Dellacherie Meyer \& Weil 73}.
\cmmnt{[538Yl.]}

Meyer P.A.\ [75] (ed.) {\it S\'eminaire de Probabilit\'es IX.}
Springer, 1975 (Lecture Notes in Mathematics 465).

Miller A.W.\ [80] `There are no Q-points in Laver's model for the Borel
conjecture', Proc.\ Amer.\ Math.\ Soc.\ 78 (1980) 103-106.
\cmmnt{[\S538 {\it notes\/}.]}
%actually, no rapid ultrafilters, therefore no rapid filters

Miller A.W.\ [81] `Some properties of measure and category',
Trans.\ Amer.\ Math.\ Soc.\ 266 (1981) 93-114.
\cmmnt{[522J.]}

Miller A.W.\ [82] `The Baire category theorem and cardinals of countable
cofinality', J.\ Symbolic Logic 47 (1982) 275-288.
\cmmnt{[552G, 552Xb.]}

%Miller A.W.\ [83] `Mapping a set of reals onto the reals',
%J.\ Symbolic Logic 48 (1983) 575-584.

Mokobodzki G.\ [75] `R\'el\`evement bor\'elien compatible avec une
classe d'ensembles n\'egligeables.  Application \grv{a} la
d\'esint\'egration des mesures', pp.\ 437-442 in {\smc Meyer 75}.
\cmmnt{[535I.]}

Mokobodzki G.\ [7?] `D\'esintegration des mesures et rel\`evements
Boreliens de sous-espaces de $L^{\infty}(X,\Cal B,\mu)$',
Lab.\ d'Analyse Fonctionnelle, Univ.\ Paris VI (?).
\cmmnt{[535E, \S535 {\it notes\/}.]\footnote{I have not been able to
locate this paper;  I believe it was a seminar report.   I took notes from
it in 1977.}}
% "On two theorems of Mokobodzki", 23.6.77.

Monk J.D.\ [89] (ed.) {\it Handbook of Boolean Algebra.}
North-Holland, 1989.

%Monk J.D.\ [96] {\it Cardinal Invariants on Boolean Algebras.}
%Birkhauser, 1996.
% QA266

Moore J.T.\ [05] `Set mapping reflection', J.\ Math.\ Logic 5 (2005) 87-98.
\cmmnt{[517O.]}

Moschovakis Y.N.\ [70] `Determinacy and prewellorderings of the continuum',
pp.\ 24-62 in {\smc Bar-Hillel 70}.
\cmmnt{[567M.]}

%M\"uller G.H.\ \& Scott D.S.\ [78] (eds.) {\it Higher Set Theory.}
%Springer, 1978 (Lecture Notes in Mathematics 669).

Mycielski J.\ [64] `On the axiom of determinateness',
Fund.\ Math.\ 53 (1964) 205-224.
\cmmnt{[567D.]}

%Mycielski J.\ \& Steinhaus H.\ [62]
%`A mathematical axiom contradicting the axiom of choice',
%Bull.\ Acad.\ Polon.\ Sci.\ (Math., Astron., Phys.) 10 (1962) 1-3.

Mycielski J.\ \& \'Swierczkowski S.\ [64]
`On the Lebesgue measurability and the axiom of determinateness',
Fund.\ Math.\ 54 (1964) 67-71.
\cmmnt{[567F.]}

\medskip%N

Naimark M.A. [70] {\it Normed Rings.}  Wolters-Noordhoff, 1970.
\cmmnt{[\S561 {\it notes\/}.]}

Neumann J.von [1931] `Algebraische Repr\"asentanten der Funktionen ``bis
auf eine Menge vom Masse Null''\nobreak', Crelle's J.\ Math.\ 165 (1931)
109-115.
\cmmnt{[535G.]}

Neumann J.von \& Stone M.H.\ [1935] `The determination of representative
elements in the residual classes of a Boolean algebra', Fundamenta
Math.\ 25 (1935) 353-378.
\cmmnt{[\S535 {\it notes\/}.]}

%Normann D.\ [76] `Martin's axiom and medial functions', Math.\
%Scandinavica 38 (1976) 167-176.

%Nyikos P.J.\ [80] `A provisional solution to the normal Moore space
%problem', Proc.\ Amer.\ Math.\ Soc.\ 78 (1980) 429-435.

\medskip%O\medskip%Pa

%Pap E.\ [0?] (ed.) {\it Handbook of Measure Theory.}  North-Holland, 200?.

Pawlikowski J.\ [86] `Why Solovay real produces Cohen real',
J.\ Symbolic Logic 51 (1986) 957-968.
\cmmnt{[552G, 552H.]}

%Pb%Pc%Pd%Pe

Pe{\l}czy'nski A.\ [68] `On Banach spaces containing $L^1(\mu)$, Studia
Math.\ 30 (1968) 231-246.
\cmmnt{[531Ye.]}

Perovi\'c \v{Z}.\ \& Veli\v{c}kovi\'c B.\ [18] `Ranks of Maharam algebras',
Advances in Math.\ 330 (2018) 253-279.
\cmmnt{[539V.]}

%Pf%Pg%Ph%Pu%Pi

% Pierce R.S.\ [58] `A note on complete Boolean algebras',
% Proc.\ Amer.\ Math.\ Soc.\ 9 (1958) 892-896.
% the Dedekind complete case of 515L

%Pj%Pk%Pl

%Plebanek G.\ [93] `Remarks on measurable Boolean algebras and
%sequential cardinals', Fundamenta Math.\ 143 (1993) 11-22.

Plebanek G.\ [95] `On Radon measures on first-countable compact spaces',
Fundamenta Math.\ 148 (1995) 159-164.
\cmmnt{[531O.]}

Plebanek G.\ [97] `Non-separable Radon measures and small compact
spaces', Fundamenta Math.\ 153 (1997) 25-40.
\cmmnt{[531L, 531M, 531V.]}

Plebanek G.\ [00]
`Approximating Radon measures on first-countable compact spaces',
Colloquium Math.\ 86 (2000) 15-23.
\cmmnt{[533H, 533Yc.]}

Plebanek G.\ [02]
`On compact spaces carrying Radon measures of large Maharam type',
Acta Univ.\ Carolinae 43 (2002) 87-99.
\cmmnt{[531U.]}

Plebanek G.\ \& Sobota D.\ [15] `Countable tightness in the spaces of
regular probability measures', Fund.\ Math.\ 229 (2015) 159-169.
\cmmnt{[531U.]}

%Pm%Pn%Po

Pol R.\ [82] `Note on the spaces $P(S)$ of regular probability measures
whose topology is determined by countable subsets', Pacific J.\ Math.\
100 (1982) 185-201.
\cmmnt{[533Ya.]}

%Pp%Pq%Pr

Prikry K.\ [75] `Ideals and powers of cardinals', Bull.\
Amer.\ Math.\ Soc.\ 81 (1975) 907-909.
\cmmnt{[555N.]}

%Ps%Pt%Pu%Pv%Pw%Px%Py%Pz
\medskip%Q

Quickert S.\ [02] `CH and the Saks property', Fundamenta Math.\ 171
(2002) 93-100.
\cmmnt{[539L.]}

\medskip%R

Raisonnier J.\ \& Stern J.\ [85] `The strength of measurability hypotheses', Israel J.\ Math.\ 50 (1985) 337-349.
\cmmnt{[Chap.\ 52 {\it intro.}, 522Q.]}

%Roitman J.\ [79] `Adding a random or a Cohen real;  topological
%consequences and the effect on Martin's axiom', Fundamenta Math.\ 103
%(1979) 47-60.

%Ros{\l}anowski A.\ \& Shelah S.\ [p00] `Measured creatures', preprint, 7.10.00

Rothberger F.\ [1938a] `Eine \"Aquivalenz zwischen der kontinuumhypothese
unter der Existenz der Lusinschen und Sierpinschischen Mengen',
Fundamenta Math.\ 30 (1938) 215-217.
\cmmnt{[522G.]}

Rothberger F.\ [1938b] `Eine Versch\"arfung der Eigenschaft C',
Fundamenta Math.\ 30 (1938) 50-55.
\cmmnt{[\S534 {\it notes\/}.]}

Rothberger F.\ [1941] `Sur les familles ind\'enombrables de suites de nombres naturels et les probl\`emes concernant la propri\'et\'e {\bf C}', Proc.\ Cambridge Phil.\ Soc.\ 37 (1941) 109-126.
\cmmnt{[534R.]}

\medskip%S

Scott D.S.\ [71] (ed.) {\it Axiomatic Set Theory.}  Amer.\
Math.\ Soc., 1971
(Proceedings of Symposia in Pure Mathematics XIII, vol.\ 1).

Shelah S.\ [82] {\it Proper Forcing.}  Springer, 1982
(Lecture Notes in Mathematics 940).
\cmmnt{[\S538 {\it notes\/}.]}

Shelah S.\ [83] `Lifting problem of the measure algebra', Israel J.\
Math.\ 45 (1983) 90-96.   % Sh185
\cmmnt{[\S535 {\it notes\/}.]}

Shelah S.\ [84] `Can you take Solovay inaccessible away?', Israel J.\ Math.\ 48 (1984) 1-47.
\cmmnt{[\S522 {\it notes\/}.]}

Shelah S.\ [92] `Cardinal arithmetic for skeptics',
Bull.\ Amer.\ Math.\ Soc.\ 26 (1992) 197-210.
% Sh400a
\cmmnt{[5A2D.]}

Shelah S.\ [94] {\it Cardinal Arithmetic.}  Oxford U.P., 1994.
% Sh400
\cmmnt{[5A2D, 5A2G.]}

Shelah S.\ [96] `Further cardinal arithmetic', Israel J.\ Math.\ 95
(1996) 61-114.
\cmmnt{[542I.]}
%Sh 430

Shelah S.\ [98a] {\it Proper and Improper Forcing.}  Springer, 1998.
\cmmnt{[\S538 {\it notes\/}.]}
%http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pl/1235419814

Shelah S.\ [98b] `There may be no nowhere dense ultrafilter', pp.\ 305-324
in {\smc Makovsky \& Ravve 98}.
\cmmnt{[538H, \S538 {\it notes\/}.]}
% Sh594;  arXiv:math.LO/9611221.

% Shelah S.\ [99] `Special subsets of $^{\cf(\mu)}\mu$, Boolean
% algebras and Maharam measure algebras',
% Topology and its Applications 99 (1999) 135-235.
 % Sh620

Shelah S.\ [00] `Covering of the null ideal may have countable
cofinality', Fundamenta Math.\ 166 (2000) 109-136.
\cmmnt{[\S522 {\it notes\/}.]}
% Sh592

Shelah S.\ [03] `The null ideal restricted to some non-null set may be
$\aleph_1$-saturated',  Fundamenta Math.\ 179 (2003) 97-129.
\cmmnt{[ \S548 {\it notes\/}.]}
% Sh619

%Shelah S.\ \& Fremlin D.H.\ [93] `Pointwise compact and stable sets of
%measurable functions', J.\ Symbolic Logic 58 (1993) 435-455.
  %FrSh 406

Shelah S.\ \& Stepr\=ans J.\ [05]
`Comparing the uniformity invariants of null sets for different measures',
Advances in Math.\ 192 (2005) 403-426.
\cmmnt{[534Za.]}
% Con(\non\Cal N(\mu_H)<\non\Cal N)

%Shelah S.\ \& Woodin H.\ [90] `Large cardinals imply that every
%reasonably definable set of reals is Lebesgue measurable', Israel J.\
%Math.\ 70 (1990) 381-394.

Shipman J.\ [90] `Cardinal conditions for strong Fubini
theorems', Trans.\ Amer.\ Math.\ Soc.\ 321 (1990) 465-481.
\cmmnt{[537K.]}

%Solecki S.\ [06] `Amenability, free subgroups, and Haar null sets in
%non-locally-compact groups',
%Proc.\ London Math.\ Soc.\ 93 (2006) 693-722.

Solecki S.\ \& Todor\v{c}evi\'c S.\ [04]
`Cofinal types of topological directed orders',
Ann.\ Inst.\ Fourier 54 (2004) 1877-1911.
\cmmnt{[513K, 513O, 513Yi.]}

Solecki S.\ \& Todor\v{c}evi\'c S.\ [10]
`Avoiding families and Tukey functions on the nowhere-dense ideal',
J.\ Inst.\ Math.\ Jussieu 9 (2010) 1-31.
\cmmnt{[526L.]}

Solovay R.M.\ [66] `New proof of a theorem of Gaifman and Hales',
Bull.\ Amer.\ Math.\ Soc.\ 72 (1966) 282-284.
\cmmnt{[514Xi.]}

Solovay R.M.\ [70] `A model of set theory in which every set of reals
is Lebesgue measurable', Annals of Math.\ 92 (1970) 1-56.
\cmmnt{[\S522 {\it notes\/}.]}

Solovay R.M.\ [71] `Real-valued measurable cardinals', pp.\ 397-428 in
{\smc Scott 71}.
\cmmnt{[541J, 541P, 541Ya, \S541 {\it notes\/}, 555D, 555O.]}

%Solovay R.M., Reinhardt W.N.\ \& Kanamori A.\ [78]
%`Strong axioms of infinity and elementary embeddings',
%Annals of Mathematical Logic 13 (1978) 73-116.

%Solovay R.M.\ \& Tennenbaum S.\ [71]
%`Iterated Cohen extensions and Souslin's problem',
%Ann.\ Math.\ (2) 94 (1971) 201-245.

%Stepr\=ans J.\ [96] `Cardinal invariants associated with Hausdorff
%capacities', pp.\ 147-184 in {\smc Bartoszy\'nski \& Scheepers 96}.

%Szyma\'nski A.\ [88] `Some remarks on real-valued
%measurable cardinals', Proc.\ Amer.\ Math.\ Soc.\ 104 (1988)
%596-602.

\medskip%T\medskip

Talagrand M.\ [84] {\it Pettis Integral and Measure Theory.}
Mem.\ Amer.\ Math.\ Soc.\  307 (1984).
\cmmnt{[536C.]}

Talagrand M.\ [08] `Maharam's problem', Annals of Math.\ 168 (2008)
981-1009.
\cmmnt{[539A.]}

Tarski A.\ [1945] `Ideale in volst\"andigen Mengenk\"orpen II', Fundamenta Math.\ 33 (1945) 51-65.
\cmmnt{[541P.]}

Todor\v{c}evi\'c S.\ [85] `Remarks on chain conditions in products',
Compositio Math.\ 55 (1985) 295-302.
\cmmnt{[537G.]}

Todor\v{c}evi\'c S.\ [87] `Partitioning pairs of countable ordinals',
Acta Math.\ 159 (1987) 261-294.
\cmmnt{[554Yc.]}

Todor\v{c}evi\'c S.\ [00] `A dichotomy for $P$-ideals of countable sets',
Fundamenta Math.\ 166 (2000) 251-267.
\cmmnt{[5A6G.]}

T\"ornquist A.\ [11] `On the pointwise implementation of near-actions',
Trans.\ Amer.\ Math.\ Soc.\ 363 (2011) 4929-4944.
%{\tt http://arxiv.org/abs/0909.3869}
\cmmnt{[535Yd.]}

Truss J.K.\ [77] `Sets having calibre $\aleph_1$',
pp.\ 595-612 in {\smc Gandy \& Hyland 77}.
\cmmnt{[522J.]}

Truss J.K.\ [88] `Connections between different amoeba algebras',
Fund.\ Math.\ 130 (1988) 137-155.
\cmmnt{[Chap.\ 52 {\it intro.}, 528A, 528Da, 528K, \S528 {\it notes\/}.]}

Tukey J.W.\ [1940] {\it Convergence and Uniformity in Topology.}
Princeton U.P., 1940 (Ann.\ Math.\ Studies 1).
\cmmnt{[513F, \S513 {\it notes}.]}

%U
\medskip

Ulam S.\ [1930] `Zur Masstheorie in der allgemeinen
Mengenlehre', Fundamenta Math.\ 16 (1930) 140-150.
\cmmnt{[\S541 {\it intro.\/}, \S541 {\it notes\/}.]}

%V
\medskip

Veli\v{c}kovi\'c B.\ [92] `Forcing axioms and stationary sets', Advances in
Math.\ 94 (1992) 256-284.
\cmmnt{[517O.]}

Veli\v{c}kovi\'c B.\ [05] `CCC forcing and splitting reals',
Israel J.\ Math.\ 147 (2005) 209-221.
\cmmnt{[\S539 {\it intro.}, 539E, 539N.]}

Vojt\'a\v{s} P.\ [93] `Generalized Galois-Tukey connections between
explicit relations on classical objects of real analysis', pp.\ 619-643
in {\smc Judah 93}.
\cmmnt{[512A.]}

%W
\medskip

Wimmers E.\ [82] `The Shelah $P$-point independence theorem',
Israel J.\ Math.\ 43 (1982) 28-48.
\cmmnt{[\S538 {\it notes\/}.]}

\medskip%X\medskip%Y\medskip%Z

Zakrzewski P.\ [92] `Strong Fubini theorems from measure extension
axioms', Comm.\ Math.\ Univ.\ Carolinae 33 (1992) 291-297.
\cmmnt{[544J.]}

%Zindulka O.\ [00] `Residual measures in locally compact spaces', Topology Appl.\ 108 (2000) 253-265.

%Zindulka O.\ [p98] `Killing residual measures', preprint, 1998.

\discrpage

}%end of references

%---------------------------------------------------------------%

\end



