% 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{mt54.tex}
\versiondate{23.10.14/17.9.16}
\copyrightdate{2005}

\def\chaptername{Real-valued-measurable cardinals}

\def\pssqa{power set $\sigma$-quotient algebra}

\newchapter{54}

Of the many questions in measure theory which involve non-trivial set
theory, perhaps the first to have been
recognised as fundamental is what I call the `Banach-Ulam problem':
is there a non-trivial measure space in which every set is measurable?
In various forms, this question has arisen repeatedly in earlier volumes
of this treatise (232Hc, 363S, 438A).   The time has now come for an
account of the developments of the last fifty years.

The measure theory of this chapter will begin in \S543;  the first two
sections deal with generalizations to wider contexts.   If $\nu$ is a
probability measure with domain $\Cal PX$, its null ideal is
$\omega_1$-additive and $\omega_1$-saturated
in $\Cal PX$.   In \S541 I look at ideals $\Cal I\normalsubgroup\Cal PX$
such that $\Cal I$ is simultaneously
$\kappa$-additive and $\kappa$-saturated for some
$\kappa$;  this is already enough to lead us to
a version of the Keisler-Tarski theorem on
normal ideals (541J), a great strengthening of Ulam's theorem on
inaccessibility of real-valued-measurable cardinals (541Lc), a form of
Ulam's dichotomy (541P), and some very striking infinitary combinatorics
(541Q-541S).   In \S542 I specialize to the case $\kappa=\omega_1$, still
without calling on the special properties of null ideals, with more
combinatorics (542E, 542I).

I have said many times in the course of this treatise that almost the first
thing to ask about any measure is, what does its measure algebra look like?
For an atomless probability measure with domain $\Cal PX$, the Gitik-Shelah
theorem (543E-543F) gives a great deal of information, associated with a
tantalizing problem (543Z).   \S544 is devoted to the measure-theoretic
consequences of assuming that there is some \am\ cardinal, with results on
repeated integration (544C, 544I, 544J),
the null ideal of a normal witnessing
probability (544E-544F) and regressive functions (544M).

I do not discuss consistency questions in this chapter (I will touch on
some of them in Chapter 55).
The ideas of \S\S541-544 would be in danger of becoming irrelevant if it
turned out that there can be no \2vm\ cardinal.   I have no real qualms
about this.   One of my reasons for
confidence is the fact that very much stronger assumptions have been
investigated without any hint of catastrophe.   Two of these, the `product
measure extension axiom' and the `normal measure axiom' are mentioned in
\S545.

One way of looking at the Gitik-Shelah theorem is to say that if $X$ is a
set and $\Cal I$ is a proper $\sigma$-ideal of subsets of $X$, then
$\Cal PX/\Cal I$ cannot be an atomless measurable algebra of small Maharam
type.   We can ask whether there are further theorems of this kind provable
in ZFC.   Two such results are in \S547:  the `Gitik-Shelah theorem for
category' (547F-547G), showing that $\Cal PX/\Cal I$ cannot be isomorphic
to $\RO(\Bbb R)$, and 547R, showing that `$\sigma$-measurable' algebras of
moderate complexity also cannot appear
as \pssqa{s}.   This leads us to a striking fact about free sets for relations
with countable equivalence classes (548C) and thence to disjoint
refinements of sequences of sets (548E).

\discrpage

\frfilename{mt541.tex}
\versiondate{10.12.12}
\copyrightdate{2004}

\def\chaptername{Real-valued-measurable cardinals}
\def\sectionname{Saturated ideals}

\def\covSh{\cov_{\text{Sh}}}
\def\Esic{$\exists$sic}
\def\Ewic{$\exists$wic}
\def\Tr{\mathop{\text{Tr}}}

\newsection{541}

If $\nu$ is a totally finite measure with domain $\Cal PX$ and null
ideal $\Cal N(\nu)$, then its measure algebra $\Cal PX/\Cal N(\nu)$ is
ccc, that is to say, $\sat(\Cal PX/\Cal N(\nu))\le\omega_1$;  while the
additivity of $\Cal N(\nu)$ is at least $\omega_1$.   It turns out that
an ideal $\Cal I$ of $\Cal PX$ such that
$\sat(\Cal PX/\Cal I)\le\add\Cal I$ is either trivial or extraordinary.
In this section I present a little of the theory of such ideals.   To
begin with, the quotient algebra has to be Dedekind complete (541B).
Further elementary ideas are in 541C (based on a method already used in
\S525) and 541D-541E.   In a less expected direction, we have a useful
fact concerning transversal numbers $\Tr_{\Cal I}(X;Y)$ (541F).

The most remarkable properties of saturated ideals arise because of
their connexions with `normal' ideals (541G).   These ideals share the
properties of non-stationary ideals (541H-541I).   If $\Cal I$ is an
$(\add\Cal I)$-saturated ideal of $\Cal PX$, we have corresponding
normal ideals on $\add\Cal I$ (541J).   Now there can be a
$\kappa$-saturated normal ideal on $\kappa$ only if there is a great
complexity of cardinals below $\kappa$ (541L).

The original expression of these ideas ({\smc Keisler \& Tarski 64})
concerned `\2vm' cardinals, on which we have normal ultrafilters (541M).
The dichotomy of {\smc Ulam 1930}\cmmnt{ (438Ce-438Cf)} reappears in the
context of $\kappa$-saturated normal ideals (541P).   For
$\kappa$-saturated ideals, `normality' implies some far-reaching
extensions (541Q).   Finally, I include a technical lemma concerning the
covering numbers $\covSh(\alpha,\beta,\gamma,\delta)$ (541S).

\leader{541A}{Definition} If $\frak A$ is a Boolean algebra, $I$ is an
ideal of $\frak A$ and $\kappa$ is a cardinal, I will say that $I$ is
{\bf $\kappa$-saturated} in $\frak A$ if
$\kappa\ge\sat(\frak A/I)$\cmmnt{;  that is, if for every family
$\ofamily{\xi}{\kappa}{a_{\xi}}$ in $\frak A\setminus I$ there are
distinct $\xi$, $\eta<\kappa$ such that
$a_{\xi}\Bcap a_{\eta}\notin I$.}

\leader{541B}{Proposition} Let $\frak A$ be a Dedekind complete Boolean
algebra and $I$ an ideal of $\frak A$ which is $(\add I)^+$-saturated in
$\frak A$.   Then the quotient algebra $\frak A/I$ is Dedekind complete.

\proof{ Take any $B\subseteq\frak A/I$.   Let $C$ be the set of those
$c\in\frak A/I$ such that {\it either} $c\Bcap b=0$ for every $b\in B$
{\it or} there is a $b\in B$ such that $c\Bsubseteq b$.   Then $C$ is
order-dense in $\frak A/I$, so there is a partition of unity
$D\subseteq C$ (313K).   Enumerate $D$ as
$\ofamily{\xi}{\kappa}{d_{\xi}}$, where

\Centerline{$\kappa=\#(D)<\sat(\frak A/I)\le(\add I)^+$.}

\noindent For each $\xi<\kappa$, choose $a_{\xi}\in\frak A$ such that
$a_{\xi}^{\ssbullet}=d_{\xi}$.   Now $\#(\xi)<\kappa\le\add I$, so
$\sup_{\eta<\xi}a_{\xi}\Bcap a_{\eta}\in I$.   Set
$\tilde a_{\xi}=a_{\xi}\Bsetminus\sup_{\eta<\xi}a_{\eta}$;  then
$\tilde a_{\xi}^{\ssbullet}=a_{\xi}^{\ssbullet}=d_{\xi}$.   Set
$L=\{\xi:\xi<\kappa$, $d_{\xi}\Bsubseteq b$ for some $b\in B\}$ and
$a=\sup_{\xi\in L}\tilde a_{\xi}$ in $\frak A$.

\Quer\ If $b\in B$ and $b\notBsubseteq a^{\ssbullet}$, there must be a
$\xi<\kappa$ such that $d_{\xi}\Bcap(b\setminus a^{\ssbullet})\ne 0$.
But $d_{\xi}\in C$, so there must be a $b'\in B$ such that
$d_{\xi}\Bsubseteq b'$;  accordingly $\xi\in L$,
$\tilde a_{\xi}\Bsubseteq a$
and $d_{\xi}\Bsubseteq a^{\ssbullet}$.\ \BanG\  Thus $a^{\ssbullet}$ is
an upper bound of $B$ in $\frak A/I$.

\Quer\ If there is an upper bound $b^*$ of $B$ such that
$a^{\ssbullet}\notBsubseteq b^*$, there must be a $\xi<\kappa$ such that
$d_{\xi}\Bcap a^{\ssbullet}\Bsetminus b^*\ne 0$.   As
$d_{\xi}\notBsubseteq b^*$, $d_{\xi}\notBsubseteq b$ for every $b\in B$,
and $\xi\notin L$.   But this means that
$\tilde a_{\xi}\Bcap\tilde a_{\eta}=0$ for every $\eta\in L$, so
$\tilde a_{\xi}\Bcap a=0$ (313Ba) and $d_{\xi}\Bcap a^{\ssbullet}=0$.\
\Bang

Thus $a^{\ssbullet}=\sup B$ in $\frak A/I$;  as $B$ is arbitrary,
$\frak A/I$ is Dedekind complete.
}%end of proof of 541B

\leader{541C}{Proposition} Let $X$ be a set, $\kappa$ a regular infinite
cardinal, $\Sigma$ an algebra of subsets of $X$ such that
$\bigcup\Cal E\in\Sigma$ whenever $\Cal E\subseteq\Sigma$ and
$\#(\Cal E)<\kappa$, and $\Cal I$ a $\kappa$-saturated $\kappa$-additive
ideal of $\Sigma$.

(a) If $\Cal E\subseteq\Sigma$ there is an
$\Cal E'\in[\Cal E]^{<\kappa}$ such that
$E\setminus\bigcup\Cal E'\in\Cal I$ for every $E\in\Cal E$.

(b) If $\ofamily{\xi}{\kappa}{E_{\xi}}$ is any family
in $\Sigma\setminus\Cal I$, and $\theta<\kappa$ is a cardinal, then
$\{x:x\in X$, $\#(\{\xi:x\in E_{\xi}\})\ge\theta\}$ includes a member of
$\Sigma\setminus\Cal I$\cmmnt{ (and, in particular, is not empty)}.

(c) Suppose that no element of $\Sigma\setminus\Cal I$ can be covered by
$\kappa$ members of $\Cal I$.   Then $\kappa$ is a precaliber of
$\Sigma/\Cal I$.

\proof{ Write $\frak A$ for $\Sigma/\Cal I$.

\medskip

{\bf (a)} Consider
$A=\{E^{\ssbullet}:E\in\Cal E\}\subseteq\frak A$.   By 514Db, there is
an $\Cal E'\in[\Cal E]^{<\sat(\frak A)}$ such that
$\{E^{\ssbullet}:E\in\Cal E'\}$ has the same upper bounds as $A$.
Now $\#(\Cal E')<\sat(\frak A)\le\kappa$, so
$F=\bigcup\Cal E'$ belongs to $\Sigma$, and $F^{\ssbullet}$ must be an
upper bound for $A$, that is, $E\setminus F\in\Cal I$ for every
$E\in\Cal E$.

\medskip

{\bf (b)} For $\alpha\le\beta<\kappa$ set
$F_{\alpha\beta}=\bigcup_{\alpha\le\xi<\beta}E_{\xi}\in\Sigma$.
Then for every $\alpha<\kappa$ we have a $g(\alpha)<\kappa$ such that
$g(\alpha)\ge\alpha$ and
$E_{\xi}\setminus F_{\alpha,g(\alpha)}\in\Cal I$ whenever
$g(\alpha)\le\xi<\kappa$, by (a) and the regularity of $\kappa$.

Define $\ofamily{\alpha}{\kappa}{h(\alpha)}$ by setting $h(0)=0$,
$h(\alpha+1)=g(h(\alpha))$ for each $\alpha<\kappa$, and
$h(\alpha)=\sup_{\beta<\alpha}h(\beta)$ for non-zero limit ordinals
$\alpha<\kappa$.   Set $G_{\alpha}=F_{h(\alpha),h(\alpha+1)}$ for each
$\alpha$.   If $\beta<\alpha$, then

\Centerline{$G_{\alpha}\setminus G_{\beta}
=\bigcup_{h(\alpha)\le\xi<h(\alpha+1)}E_{\xi}
  \setminus F_{h(\beta),g(h(\beta))}\in\Cal I$}

\noindent because $\Cal I$ is $\kappa$-additive.   Consequently
$G_{\theta}\setminus\bigcap_{\beta<\theta}G_{\beta}$ belongs to
$\Cal I$;  and $G_{\theta}\supseteq E_{h(\theta)}\notin\Cal I$, so
$G=\bigcap_{\beta<\theta}G_{\beta}\in\Sigma\setminus\Cal I$.   But if
$x\in G$ then $\{\xi:x\in E_{\xi}\}$ meets $\coint{h(\beta),h(\beta+1)}$
for every $\beta<\theta$ and has cardinal at least $\theta$.

\medskip

{\bf (c)} Let $\ofamily{\xi}{\kappa}{a_{\xi}}$ be a family of non-zero
elements in $\frak A$.   For each $\xi<\kappa$, choose
$\tilde E_{\xi}\in\Sigma$ such that
$\tilde E_{\xi}^{\ssbullet}=a_{\xi}$.   Let
$\Cal K$ be the family of all those finite subsets $K$ of $\kappa$ such
that $H_K=\bigcap_{\xi\in K}\tilde E_{\xi}$ belongs to $\Cal I$.
Now set $E_{\xi}
=\tilde E_{\xi}\setminus\bigcup\{H_K:K\in\Cal K$, $K\subseteq\xi\}$;
then $E_{\xi}\in\Sigma\setminus\Cal I$ and $E_{\xi}^{\ssbullet}=a_{\xi}$
for each $\xi<\kappa$.

Repeat the argument of (b).   Once again we get a family
$\ofamily{\alpha}{\kappa}{G_{\alpha}}$ in $\Sigma\setminus\Cal I$ such
that $G_{\alpha}\setminus G_{\beta}\in\Cal I$ whenever
$\beta\le\alpha<\kappa$.   Now, applying (a) to
$\ofamily{\alpha}{\kappa}{X\setminus G_{\alpha}}$, we have a
$\gamma<\kappa$ such that
$\bigcap_{\alpha<\gamma}G_{\alpha}\setminus G_{\beta}\in\Cal I$ for
every $\beta<\kappa$.   On the other hand,
$G_{\gamma}\setminus G_{\alpha}\in\Cal I$ for every $\alpha<\gamma$, so
in fact $G_{\gamma}\setminus G_{\beta}\in\Cal I$ for every
$\beta<\kappa$, while $G_{\gamma}\notin\Cal I$.   At this point, recall
that we are now assuming that $G_{\gamma}$ cannot be covered by
$\bigcup_{\beta<\kappa}G_{\gamma}\setminus G_{\beta}$, and there is an
$x\in\bigcap_{\beta<\kappa}G_{\beta}$.

As in (b), it follows that $\Gamma=\{\xi:\xi<\kappa$, $x\in E_{\xi}\}$
has cardinal $\kappa$.   If $K\subseteq\Gamma$ is finite and not empty,
take $\zeta\in\Gamma$ such that $K\subseteq\zeta$.   Then

\Centerline{$x\in E_{\zeta}\cap\bigcap_{\xi\in K}\tilde E_{\xi}
\subseteq E_{\zeta}\cap H_K$;}

\noindent it follows that $K\notin\Cal K$ and $H_K\notin\Cal I$,
that is, that $\inf_{\xi\in K}a_{\xi}\ne 0$ in
$\frak A$.   So $\family{\xi}{\Gamma}{a_{\xi}}$ is centered;  as
$\ofamily{\xi}{\kappa}{a_{\xi}}$ is arbitrary, $\kappa$ is a precaliber
of $\frak A$.
}%end of proof of 541C

\leader{541D}{Lemma} %RVMC 5A
Let $X$ be a set, $\Cal I$ an ideal of $\Cal PX$, $Y$ a set of
cardinal less than $\add\Cal I$ and $\kappa$ a cardinal such that
$\Cal I$ is $(\cf\kappa)$-saturated in $\Cal PX$.   Then for any
function $f:X\to[Y]^{<\kappa}$ there is an
$M\in[Y]^{<\kappa}$ such that $\{x:f(x)\not\subseteq M\}\in\Cal I$.
%do we really need quite such an elaborate version? maybe enough to have
%regular $\kappa$

\proof{ If $\kappa>\#(Y)$ this is trivial;
suppose that $\kappa\le\#(Y)<\add\Cal I$.   For cardinals
$\lambda<\kappa$ set $X_{\lambda}=\{x:\#(f(x))=\lambda\}$.   If
$A=\{\lambda:X_{\lambda}\notin\Cal I\}$ then $\Cal I$ is not
$\#(A)$-saturated in $\Cal PX$, so $\#(A)<\cf\kappa$ and
$\theta=\sup A$ is less than $\kappa$.   Set
$X'=\{x:\#(f(x))\le\theta\}$;  then
$X\setminus X'$ is the union of at most $\lambda<\add\Cal I$ members of
$\Cal I$, so belongs to $\Cal I$.

For each $x\in X'$ let
$\langle h_{\xi}(x)\rangle_{\xi<\theta}$
run over a set including $f(x)$.   For each $\xi<\theta$,

\Centerline{$Y_{\xi}=\{y:h_{\xi}^{-1}[\{y\}]\notin\Cal I\}$}

\noindent has cardinal less than $\cf\kappa$, and because
$\#(Y)<\add\Cal I$,
$h_{\xi}^{-1}[Y\setminus Y_{\xi}]\in\Cal I$.   Set
$M=\bigcup_{\xi<\theta}Y_{\xi}\in[Y]^{<\kappa}$.   (If $\kappa$ is
regular, $M$ is the union of fewer than $\kappa$ sets with cardinal less than
$\kappa$, so $\#(M)<\kappa$;  if $\kappa$ is not regular, then
$M$ is the union of fewer than $\kappa$ sets with cardinal at most
$\cf\kappa$, so again $\#(M)<\kappa$.)
Because $\theta<\add\Cal I$,

\Centerline{$\{x:f(x)\not\subseteq M\}
\subseteq(X\setminus X')
  \cup\bigcup_{\xi<\theta}h_{\xi}^{-1}[Y\setminus Y_{\xi}]
\in\Cal I$,}

\noindent as required.
}%end of proof of 541D

\leader{541E}{Corollary} Let $X$ be a set, $\Cal I$ an ideal of
$\Cal PX$, $Y$ a set of
cardinal less than $\add\Cal I$ and $\kappa$ a cardinal such that
$\Cal I$ is $(\cf\kappa)$-saturated in $\Cal PX$.   Then for any
function $g:X\to Y$ there is an
$M\in[Y]^{<\kappa}$ such that $g^{-1}[Y\setminus M]\in\Cal I$.

\proof{ Apply 541D to the function $x\mapsto\{g(x)\}$.   (In the trivial
case $\kappa=1$, $\Cal I=\Cal PX$.)
}%end of proof of 541E

\leader{541F}{Lemma} Let $X$ and $Y$ be sets, $\kappa$ a regular
uncountable cardinal, and $\Cal I$ a proper
$\kappa$-saturated $\kappa$-additive ideal of subsets of $X$.
Then $\Tr_{\Cal I}(X;Y)$\cmmnt{ (definition:  5A1Ma)} is
attained, in the sense that there is a set $G\subseteq Y^X$ such
that $\#(G)=\Tr_{\Cal I}(X;Y)$ and
$\{x:x\in X$, $g(x)=g'(x)\}\in\Cal I$ for all distinct $g$,
$g'\in G$.

\proof{ It is enough to consider the case in which
$Y=\lambda$ is a cardinal.   Set $\theta=\Tr_{\Cal I}(X;\lambda)$.

\medskip

{\bf (a)} If $\lambda^+<\kappa$ then
$\theta\le\lambda$.   \Prf\Quer\ Suppose, if possible,
that we have a family $\ofamily{\xi}{\lambda^+}{f_{\xi}}$ in $\lambda^X$
such that $\{x:f_{\xi}(x)=f_{\eta}(x)\}\in\Cal I$ whenever
$\eta<\xi<\lambda^+$.
Then $\lambda$ is surely infinite, so $\lambda^+$ is uncountable and
regular.   For each $x\in X$ there is an $\alpha_x<\lambda^+$ such that
$\{f_{\xi}(x):\xi<\alpha_x\}=\{f_{\xi}(x):\xi<\lambda^+\}$.
Setting

\Centerline{$F_{\alpha}=\{x:x\in X$, $\alpha_x=\alpha\}
\subseteq\bigcup_{\eta<\alpha}\{x:f_{\eta}(x)=f_{\alpha}(x)\}$,}

\noindent we see that $F_{\alpha}\in\Cal I$ for each $\alpha<\lambda^+$.
But $X=\bigcup_{\alpha<\lambda^+}F_{\alpha}$ and $\lambda^+<\kappa$, so
this is impossible.\ \Bang\Qed

Since we can surely find a family $\ofamily{\xi}{\lambda}{f_{\xi}}$ in
$\lambda^X$ such that $f_{\xi}(x)\ne f_{\eta}(x)$ whenever $x\in X$ and
$\eta<\xi<\lambda$, we have the result when $\lambda^+<\kappa$.

\medskip

{\bf (b)} We may therefore suppose from now on that
$\lambda^+\ge\kappa$.

If $H\subseteq\lambda^X$ is such that

\Centerline{$F=\{f:f\in\lambda^X$, $\{x:f(x)\le h(x)\}\in\Cal I$ for
every $h\in H\}\ne\emptyset$,}

\noindent then there is an $f_0\in F$ such that

\Centerline{$\{x:f(x)<f_0(x)\}\in\Cal I$ for every $f\in F$.}

\noindent\Prf\Quer\ If not, choose a family
$\langle f_{\xi} \rangle_{\xi<\kappa}$ in $F$ inductively, as follows.
$f_0$ is to be any member of $F$.   Given $f_{\xi}$, there is an
$f\in F$ such that $\{x:f(x)<f_{\xi}(x)\}\notin\Cal I$;  set
$f_{\xi+1}(x)=\min(f(x),f_{\xi}(x))$ for every $x$;  then
$f_{\xi+1}\in F$.   Given that $f_{\eta}\in F$ for every
$\eta<\xi$, where $\xi<\kappa$ is a non-zero limit
ordinal, set $f_{\xi}(x)=\min_{\eta<\xi}f_{\eta}(x)$ for each $x$;
then for any $h\in H$ we shall have

\Centerline{$\{x:f_{\xi}(x)\le h(x)\}
=\bigcup_{\eta<\xi}\{x:f_{\eta}(x)\le h(x)\}\in\Cal I$,}

\noindent so $f_{\xi}\in F$ and the induction continues.

Now consider

\Centerline{$E_{\xi}=\{x:f_{\xi+1}(x)<f_{\xi}(x)\}
\in\Cal PX\setminus\Cal I$}

\noindent for $\xi<\kappa$.   By 541Cb there is an $x\in X$ such that
$A=\{\xi:x\in E_{\xi}\}$ is infinite.   But if
$\langle\xi(n)\rangle_{n\in \Bbb N}$ is any strictly increasing sequence
in $A$, $\langle f_{\xi(n)}(x)\rangle_{n\in\Bbb N}$ is a strictly
decreasing sequence of ordinals, which is impossible.\ \Bang\Qed

\medskip

{\bf (c)} Choose a family
$\langle g_{\xi}\rangle_{\xi<\delta}$ in $\lambda^X$ as follows.   Given
$\langle g_{\eta}\rangle_{\eta<\xi}$, set

\Centerline{$F_{\xi}=\{f:f\in\lambda^X$,
$\{x:f(x)\le g_{\eta}(x)\}\in\Cal I$ for every $\eta<\xi\}$.}

\noindent If $F_{\xi}=\emptyset$, set $\delta=\xi$ and stop.
If $F_{\xi}\ne\emptyset$ use (b) to find $g_{\xi}\in F_{\xi}$ such that
$\{x:f(x)<g_{\xi}(x)\}\in\Cal I$ for every $f\in F_{\xi}$,
and continue.   Note that for $\xi<\min(\lambda,\kappa)$,
$\{x:g_{\xi}(x)\ne \xi\}\in\Cal I$.   \Prf\ Induce on $\xi$.   If
$\xi<\min(\lambda,\kappa)$ and $\{x:g_{\eta}(x)\ne\eta\}\in\Cal I$ for
every $\eta<\xi$, then the constant function with value $\xi$ belongs to
$F_{\xi}$, so $g_{\xi}$ is defined and
$\{x:g_{\xi}(x)>\xi\}\in\Cal I$.   On the other hand,
$\{x:g_{\xi}(x)=\eta\}\in\Cal I$ for $\eta<\xi$;
as $\xi<\add\Cal I$, $\{x:g_{\xi}(x)<\xi\}\in\Cal I$.\ \QeD\  Accordingly
$\delta\ge\min(\lambda,\kappa)$.

\medskip

{\bf (d)} Because $g_{\xi}\in F_{\xi}$,
$\{x:g_{\xi}(x)=g_{\eta}(x)\}\in\Cal I$ whenever $\eta<\xi<\delta$, so
$\#(\delta)\le\theta$.   On the other hand, suppose that
$F\subseteq\lambda^X$ is such that $\{x:f(x)=f'(x)\}\in\Cal I$ for all
distinct $f$, $f'\in F$.   For each $f\in F$, set

\Centerline{$\zeta'_f=\min\{\xi:\xi\le\delta$, $f\notin F_{\xi}\};$}

\noindent this must be defined because $F_{\delta}=\emptyset$.
Also $F_0=\lambda^X$ and $F_{\xi}=\bigcap_{\eta<\xi}F_{\eta}$
if $\xi\le\delta$ is a non-zero limit ordinal, so $\zeta'_f$ must
be a successor ordinal;  let $\zeta_f$ be its predecessor.
We have $f\in F_{\zeta_f}$ and

\Centerline{$\{x:f(x)<g_{\zeta_f}(x)\}\in\Cal I$,
\quad$\{x:f(x)\le g_{\zeta_f}(x)\}\notin\Cal I$,}

\noindent so that

\Centerline{$E_f=\{x:f(x)=g_{\zeta_f}(x)\}\notin\Cal I$.}

If $f$, $f'$ are distinct members of $F$ and $\zeta_f=\zeta_{f'}$,
then $E_f\cap E_{f'}\in\Cal I$.   So

\Centerline{$\{f:f\in F$, $\zeta_f=\zeta\}$}

\noindent must have cardinal less than $\kappa$ for every
$\zeta<\delta$.

If $\kappa=\lambda^+$, $\#(\{f:f\in F$, $\zeta_f=\zeta\})\le\lambda$ for
every $\zeta<\delta$, so $\#(F)\le\max(\delta,\lambda)=\delta$.   On the
other hand, if $\kappa\le\lambda$, then
$\#(F)\le\max(\delta,\kappa)=\delta$.   As $F$ is arbitrary,
$\theta=\delta$ and we may take $G=\{g_{\xi}:\xi<\delta\}$ as our
witness that $\Tr_{\Cal I}(X;\lambda)$ is attained.
}%end of proof of 541F

\leader{541G}{Definition} Let $\kappa$ be a regular uncountable
cardinal.   A {\bf normal ideal} on $\kappa$ is a proper ideal $\Cal I$
of $\Cal P\kappa$, including $[\kappa]^{<\kappa}$, such that

\Centerline{$\{\xi:\xi<\kappa$, $\xi\in\bigcup_{\eta<\xi}I_{\eta}\}$}

\noindent belongs to $\Cal I$ for every family
$\ofamily{\xi}{\kappa}{I_{\xi}}$ in $\Cal I$.
\cmmnt{It is easy to check that a proper ideal $\Cal I$ of
$\Cal P\kappa$ is
normal iff the dual filter $\{\kappa\setminus I:I\in\Cal I\}$ is normal
in the sense of 4A1Ic.}

\leader{541H}{Proposition} Let $\kappa$ be a regular uncountable
cardinal and $\Cal I$ a proper ideal of $\Cal P\kappa$ including
$[\kappa]^{<\kappa}$.   Then the following are equiveridical:

(i) $\Cal I$ is normal;

(ii) $\Cal I$ is $\kappa$-additive and whenever
$S\in\Cal P\kappa\setminus\Cal I$ and $f:S\to\kappa$ is regressive, then
there is an $\alpha<\kappa$ such that
$\{\xi:\xi\in S$, $f(\xi)\le\alpha\}$ is not in $\Cal I$;

(iii) whenever
$S\in\Cal P\kappa\setminus\Cal I$ and $f:S\to\kappa$ is regressive, then
there is a $\beta<\kappa$ such that
$\{\xi:\xi\in S$, $f(\xi)=\beta\}$ is not in $\Cal I$.

\proof{{\bf (i)$\Rightarrow$(ii)} Suppose that $\Cal I$ is normal.

\medskip

\quad\grheada\ (Cf.\ 4A1J.)  Suppose that
$\ofamily{\eta}{\alpha}{I_{\eta}}$ is a
family in $\Cal I$, where $0<\alpha<\kappa$, and
$I=\bigcup_{\eta<\alpha}I_{\eta}$.
Then $I\setminus\alpha\subseteq\{\xi:\xi\in\bigcup_{\eta<\xi}I_{\eta}\}$
belongs to $\Cal I$;  as $\alpha\in\Cal I$, $I\in\Cal I$;  as
$\ofamily{\eta}{\alpha}{I_{\eta}}$ is arbitrary, $\Cal I$ is
$\kappa$-additive.

\medskip

\quad\grheadb\ Take $S$ and $f$ as in (ii).   \Quer\ If
$I_{\alpha}=\{\xi:\xi\in S$, $f(\xi)\le\alpha\}$ belongs to $\Cal I$ for
every $\alpha$, then
$I=\{\xi:\xi<\kappa$, $\xi\in\bigcup_{\alpha<\xi}I_{\alpha}\}$ belongs
to $\Cal I$.   But if $\xi\in S$ then $f(\xi)<\xi$ and
$\xi\in I_{f(\xi)}$, so $S\subseteq I$.\ \BanG\  As $S$ and $f$ are
arbitrary, (ii) is true.

\medskip

{\bf (ii)$\Rightarrow$(iii)} Suppose (ii) is true and that $S$, $f$ are
as in (iii).   By (ii), there is an $\alpha<\kappa$ such that
$\{\xi:\xi\in S$, $f(\xi)\le\alpha\}\notin\Cal I$.   As $\Cal I$ is
$\kappa$-additive, there is a $\beta\le\alpha$ such that
$\{\xi:\xi\in S$, $f(\xi)=\beta\}\notin\Cal I$.   As $S$ and $f$ are
arbitrary, (iii) is true.

\medskip

{\bf (iii)$\Rightarrow$(i)}
Now suppose that (iii) is true, and that
$\ofamily{\xi}{\kappa}{I_{\xi}}$ is any family in $\Cal I$;  set
$S=\{\xi:\xi<\kappa$, $\xi\in\bigcup_{\eta<\xi}I_{\eta}\}$.   Then we
have a regressive function $f:S\to\kappa$ such that $\xi\in I_{f(\xi)}$
for every $\xi\in S$.   Since
$\{\xi:\xi\in S$, $f(\xi)=\beta\}\subseteq I_{\beta}\in\Cal I$ for every
$\beta<\kappa$, (iii) tells us that $S\in\Cal I$.
Since we are assuming that $\Cal I$ is a proper ideal including
$[\kappa]^{<\kappa}$, it is normal.
}%end of proof of 541H

\leader{541I}{Lemma} Let $\kappa$ be a regular uncountable cardinal.

(a) The family of non-stationary subsets of $\kappa$ is a normal ideal
on $\kappa$, and is included in every normal ideal on $\kappa$.

(b) If $\Cal I$ is a normal ideal on $\kappa$, and
$\family{K}{[\kappa]^{<\omega}}{I_K}$ is any family in $\Cal I$, then
$\{\xi:\xi<\kappa$, $\xi\in\bigcup_{K\in[\xi]^{<\omega}}I_K\}$ belongs
to $\Cal I$.

\proof{{\bf (a)} Let $\Cal I$ be the family of non-stationary subsets of
$\kappa$.

\medskip

\quad{\bf (i)} Since a subset of $\kappa$ is non-stationary iff it is
disjoint from some closed cofinal set (4A1Ca), any subset of a
non-stationary set is non-stationary.   Because the intersection of two
closed cofinal sets is again a closed cofinal set (4A1Bd), $\Cal I$ is
an ideal.   Because $\kappa\setminus\xi$ is a closed cofinal set for any
$\xi<\kappa$, and $\kappa$ is regular,
$[\kappa]^{<\kappa}\subseteq\Cal I$.

Now suppose that $\ofamily{\xi}{\kappa}{I_{\xi}}$ is any family in
$\Cal I$, and that
$I=\{\xi:\xi<\kappa$, $\xi\in\bigcup_{\eta<\xi}I_{\eta}\}$.   For each
$\xi<\kappa$ let $F_{\xi}$ be a closed cofinal subset of $\kappa$
disjoint from $I_{\xi}$, and let $F$ be the diagonal intersection of
$\ofamily{\xi}{\kappa}{F_{\xi}}$;  then $F$ is a closed cofinal set
(4A1B(c-ii)), and it is easy to check that $F$ is disjoint from $I$, so
$I\in\Cal I$.   Thus $\Cal I$ is normal.

\medskip

\quad{\bf (ii)} Let $\Cal J$ be any normal ideal on $\kappa$.   If
$F\subseteq\kappa$ is a closed cofinal set containing $0$, we have a
regressive function $f:\kappa\setminus F\to F$ defined by setting
$f(\xi)=\sup(F\cap\xi)$ for every $\xi\in\kappa\setminus F$.   If
$\alpha<\kappa$, $\{\xi:f(\xi)\le\alpha\}$ is bounded above by
$\min(F\setminus\alpha)$ so belongs to
$[\kappa]^{<\kappa}\subseteq\Cal J$;  by 541H(ii), $\kappa\setminus F$
must belong to $\Cal J$.   This works for any closed cofinal set
containing $0$;  but as $\{0\}$ surely belongs to $J$,
$\kappa\setminus F\in\Cal J$ for every closed cofinal set $F$, that is,
$\Cal I\subseteq\Cal J$.

\medskip

{\bf (b)} Set $J_{\xi}=\bigcup_{K\in[\xi+1]^{<\omega}}I_K$;  because
$\Cal I$ is $\kappa$-additive, $J_{\xi}\in\Cal I$ for each $\xi$.   Now

\Centerline{$\{\xi:\xi<\kappa$,
  $\xi\in\bigcup_{K\in[\xi]^{<\omega}}I_K\}
=\{\xi:\xi<\kappa$, $\xi\in\bigcup_{\eta<\xi}J_{\eta}\}\in\Cal I$}

\noindent because $\Cal I$ is normal.
}%end of proof of 541I

\leader{541J}{Theorem}\cmmnt{ ({\smc Solovay 71})} Let $X$ be a set
and $\Cal J$ an ideal of
subsets of $X$.   Suppose that $\add\Cal J=\kappa>\omega$ and that
$\Cal J$ is $\lambda$-saturated in $\Cal PX$, where $\lambda\le\kappa$.
Then there are $Y\subseteq X$ and $g:Y\to\kappa$ such that
$\{B:B\subseteq\kappa$, $g^{-1}[B]\in\Cal J\}$ is a $\lambda$-saturated
normal ideal on $\kappa$.

\proof{ (Cf.\ 4A1K.) Let $\ofamily{\xi}{\kappa}{J_{\xi}}$ be a family in
$\Cal J$
such that $Y=\bigcup_{\xi<\kappa}J_{\xi}\notin\Cal J$.   Let $F$ be the
set of functions $f:Y\to\kappa$ such that $f^{-1}[\alpha]\in\Cal J$ for
every $\alpha<\kappa$.   Set $f_0(y)=\min\{\xi:y\in J_{\xi}\}$ for
$y\in Y$;  then $f_0\in F$.   \Prf\ If $\alpha<\kappa$, then
$f^{-1}[\alpha]=\bigcup_{\xi<\alpha}J_{\xi}$ belongs to $\Cal J$ because
$\Cal J$ is $\kappa$-additive.\ \Qed

The point is that there is a $g\in F$ such that
$\{y:y\in Y$, $f(y)<g(y)\}\in\Cal J$ for every $f\in F$.   \Prf\Quer\
Otherwise, choose $f_{\xi}$, for $0<\xi<\kappa$, as follows.   Given
$f_{\xi}\in F$, where $\xi<\kappa$, there is an $f\in F$ such that
$A_{\xi}=\{y:f(y)<f_{\xi}(y)\}\notin\Cal J$;  set
$f_{\xi+1}(y)=\min(f(y),f_{\xi}(y))$ for every $y$.   Then

\Centerline{$f_{\xi+1}^{-1}[\alpha]
=f^{-1}[\alpha]\cup f_{\xi}^{-1}[\alpha]\in\Cal J$}

\noindent for every $\alpha<\kappa$, so $f_{\xi+1}\in F$.   Given that
$f_{\eta}\in F$ for every $\eta<\xi$, where $\xi<\kappa$ is a
non-zero limit ordinal, set $f_{\xi}(y)=\min\{f_{\eta}(y):\eta<\xi\}$
for each $y\in Y$;  then

\Centerline{$f_{\xi}^{-1}[\alpha]
=\bigcup_{\eta<\xi}f_{\eta}^{-1}[\alpha]\in\Cal J$}

\noindent for every $\alpha<\kappa$, because
$\#(\xi)<\kappa=\add\Cal J$.

This construction ensures that $\ofamily{\xi}{\kappa}{f_{\xi}(y)}$ is
non-increasing for every $y$, and that
$\{y:f_{\xi+1}(y)<f_{\xi}(y)\}=A_{\xi}\notin\Cal J$ for every
$\xi<\kappa$.   But as $\Cal J$ is $\kappa$-saturated in $\Cal PX$,
there must be a point $y$ belonging to infinitely many $A_{\xi}$ (541Cb),
so that there is a strictly decreasing sequence in
$\{f_{\xi}(y):\xi<\kappa\}$, which is impossible.\ \Bang\Qed

Now consider $\Cal I=\{B:B\subseteq\kappa$, $g^{-1}[B]\in\Cal J\}$.
Because $\Cal J$ is $\lambda$-saturated in $\Cal PX$, $\Cal I$ is
$\lambda$-saturated in $\Cal P\kappa$.   \Prf\ If
$\ofamily{\xi}{\lambda}{B_{\xi}}$ is a family in
$\Cal P\kappa\setminus\Cal I$, then
$\ofamily{\xi}{\lambda}{g^{-1}[B_{\xi}]}$  is a family in
$\Cal PX\setminus\Cal J$, so there are distinct $\xi$, $\eta<\lambda$
such that
$g^{-1}[B_{\xi}\cap B_{\eta}]=g^{-1}[B_{\xi}]\cap g^{-1}[B_{\eta}]$ does
not belong to $\Cal J$, and $B_{\xi}\cap B_{\eta}$ does not belong to
$\Cal I$.\ \QeD\   Next, $\Cal I$ is normal.   \Prf\ Of course
$\kappa=\add\Cal J$ is regular (513C(a-i)), and we are supposing that it
is uncountable.   If $S\in\Cal P\kappa\setminus\Cal I$ and
$h:S\to\kappa$
is regressive, set $f(y)=hg(y)$ if $y\in g^{-1}[S]$, $g(y)$ otherwise.
Then $\{y:f(y)<g(y)\}=g^{-1}[S]\notin\Cal J$, so $f\notin F$ and there
is an $\alpha<\kappa$ such that $f^{-1}[\alpha]\notin\Cal J$.   But

\Centerline{$f^{-1}[\alpha]
\subseteq g^{-1}[\alpha]
  \cup\bigcup_{\beta<\alpha}g^{-1}[h^{-1}[\{\beta\}]]$;}

\noindent as $\alpha<\add\Cal J$, there is a $\beta<\alpha$ such that
$g^{-1}[h^{-1}[\{\beta\}]]\notin\Cal J$ and
$h^{-1}[\{\beta\}]\notin\Cal I$.   As $h$ is arbitrary, $\Cal I$ is
normal (541H).\ \Qed
}%end of proof of 541J

\leader{541K}{Lemma} %RVMC 4Lc
Let $\kappa$ be a regular uncountable cardinal and $\Cal I$ a
normal ideal on $\kappa$ which is $\kappa'$-saturated in $\Cal P\kappa$,
where $\kappa'\le\kappa$.

(a) If $S\in\Cal P\kappa\setminus\Cal I$ and $f:S\to\kappa$ is
regressive, then there is a set $A\in[\kappa]^{<\kappa'}$ such that
$S\setminus f^{-1}[A]\in\Cal I$;  \cmmnt{consequently} there is an
$\alpha<\kappa$ such that
$\{\xi:\xi\in S$, $f(\xi)\ge\alpha\}\in\Cal I$.

(b) If $\lambda<\kappa$, then
$\{\xi:\xi<\kappa$, $\cf\xi\le\lambda\}\in\Cal I$.

(c) If for each $\xi<\kappa$ we are given a relatively closed set
$C_{\xi}\subseteq\xi$ which is cofinal with $\xi$, then

\Centerline{$C=\{\alpha:\alpha<\kappa$,
$\{\xi:\alpha\notin C_{\xi}\}\in\Cal I\}$}

\noindent is a cofinal closed set in $\kappa$.

\proof{{\bf (a)} Choose $\langle S_{\eta}\rangle_{\eta\le\gamma}$ and
$\ofamily{\eta}{\gamma}{\alpha_{\eta}}$ inductively, as follows.
$S_0=S$.   If $S_{\eta}\in\Cal I$, set $\gamma=\eta$ and stop.
Otherwise, $f\restr S_{\eta}$ is regressive, so (because $\Cal I$ is
normal) there is an $\alpha_{\eta}<\kappa$ such that
$\{\xi:\xi\in S_{\eta}$, $f(\xi)=\alpha_{\eta}\}\notin\Cal I$ (541H(iii)).
Set $S_{\eta+1}=\{\xi:\xi\in S_{\eta}$, $f(\xi)\ne\alpha_{\eta}\}$.
Given $\ofamily{\zeta}{\eta}{S_{\zeta}}$ for a non-zero limit ordinal
$\eta$, set $S_{\eta}=\bigcap_{\zeta<\eta}S_{\zeta}$.   Now
$\ofamily{\eta}{\gamma}{S_{\eta}\setminus S_{\eta+1}}$ is a disjoint
family in $\Cal P\kappa\setminus\Cal I$, so $\#(\gamma)<\kappa'$ and
$A=\{\alpha_{\eta}:\eta<\gamma\}\in[\kappa]^{<\kappa'}$, while
$S\setminus f^{-1}[A]=S_{\gamma}$ belongs to $\Cal I$.
Setting $\alpha=\sup A + 1$, $\alpha<\kappa$
(because $\kappa$ is regular) and
$\{\xi:\xi\in S$, $f(\xi)\ge\alpha\}$ belongs to $\Cal I$.

\medskip

{\bf (b)} \Quer\ Otherwise, set
$S=\{\xi:0<\xi<\kappa$, $\cf\xi\le\lambda\}$ and for $\xi\in S$ choose a
cofinal set $A_{\xi}\subseteq\xi$ with $\#(A_{\xi})\le\lambda$.   Let
$\ofamily{\eta}{\lambda}{f_{\eta}}$ be a family of functions defined on
$S$ such that $A_{\xi}=\{f_{\eta}(\xi):\eta<\lambda\}$ for each
$\xi\in S$.   By (a), we have for each $\eta<\lambda$ an
$\alpha_{\eta}<\kappa$ such that
$B_{\eta}=\{\xi:\xi\in S$, $f_{\eta}(\xi)\ge\alpha_{\eta}\}\in\Cal I$.
Set $\alpha=\sup_{\eta<\lambda}\alpha_{\eta}<\kappa$;  as
$\lambda<\kappa=\add\Cal I$ (541H), there is a
$\xi\in S\setminus\bigcup_{\eta<\lambda}B_{\eta}$ such that $\xi>\alpha$.
But now $A_{\xi}\subseteq\alpha$ is not cofinal with $\xi$.\ \Bang

\medskip

{\bf (c)} For $\alpha<\kappa$, $0<\xi<\kappa$ set

$$\eqalign{f_{\alpha}(\xi)&=\min(C_{\xi}\setminus\alpha)
\text{ if }\xi>\alpha,\cr
&=0\text{ otherwise}.\cr}$$

\noindent Then $f_{\alpha}$ is regressive, so by (a) there is a
$\zeta_{\alpha}<\kappa$ such that
$\kappa\setminus f_{\alpha}^{-1}[\zeta_{\alpha}]\in\Cal I$, that is,
$\{\xi:C_{\xi}\cap\zeta_{\alpha}\setminus\alpha=\emptyset\}\in\Cal I$.
Set $\tilde C=\{\alpha:\alpha<\kappa$, $\zeta_{\beta}<\alpha$ for every
$\beta<\alpha\}$;  then $\tilde C$ is cofinal with $\kappa$.   If
$\alpha\in\tilde C$, then

$$\eqalign{\{\xi:\xi>\alpha,\,\alpha\notin C_{\xi}\}
&\subseteq\{\xi:C_{\xi}\cap\alpha\text{ is not cofinal with }\alpha\}\cr
&\subseteq\{\xi:C_{\xi}\cap\zeta_{\beta}\setminus\beta=\emptyset
  \text{ for some }\beta<\alpha\}\cr}$$

\noindent is the union of fewer than $\kappa$ members of $\Cal I$, so
belongs to $\Cal I$, and $\alpha\in C$.   Thus $C$ is cofinal with
$\kappa$.   If $\alpha<\kappa$ and $\alpha=\sup(C\cap\alpha)$, then

\Centerline{$\{\xi:\xi>\alpha,\,\alpha\notin C_{\xi}\}
\subseteq\{\xi:\beta\notin C_{\xi}\text{ for some }\beta\in
C\cap\alpha\}$}

\noindent is again the union of fewer than $\kappa$ members of $\Cal I$,
so $\alpha\in C$.   Thus $C$ is closed.
}%end of proof of 541K

\vleader{72pt}{541L}{Theorem} Let $\kappa$ be an uncountable cardinal
such that there is a proper $\kappa$-saturated $\kappa$-additive ideal of
$\Cal P\kappa$ containing singletons.

(a) There is a $\kappa$-saturated normal ideal on $\kappa$.

(b) $\kappa$ is weakly inaccessible.

(c) The set of weakly inaccessible cardinals less than $\kappa$ is
stationary in $\kappa$.

\proof{{\bf (a)} Let $\Cal J$ be a proper
$\kappa$-saturated $\kappa$-additive ideal of $\Cal P\kappa$.
The additivity of $\Cal J$ must be exactly $\kappa$, so 541J
tells us that there is a $\kappa$-saturated normal ideal $\Cal I$
on $\kappa$.

\medskip

{\bf (b)} Of course $\kappa=\add\Cal J=\add\Cal I$ is regular.
\Quer\ Suppose, if possible, that $\kappa=\lambda^+$ is
a successor cardinal.   For each $\alpha<\kappa$ let
$\phi_{\alpha}:\alpha\to\lambda$ be an injection.   For $\beta<\kappa$ and
$\xi<\lambda$ set $A_{\beta\xi}
=\{\alpha:\beta<\alpha<\kappa$, $\phi_{\alpha}(\beta)=\xi\}$.
Then $\bigcup_{\xi<\lambda}A_{\beta\xi}=\kappa\setminus(\beta+1)
\notin\Cal I$, so there is a $\xi_{\beta}<\lambda$ such that
$A_{\beta,\xi_{\beta}}\notin\Cal I$.   Now there must be an
$\eta<\lambda$ such that $B=\{\beta:\beta<\kappa$, $\xi_{\beta}=\eta\}$
has cardinal $\kappa$.   But in this case
$\family{\beta}{B}{A_{\beta\eta}}$ is a
disjoint family in $\Cal P\kappa\setminus\Cal I$, and $\Cal I$ is not
$\kappa$-saturated in $\Cal P\kappa$.\ \Bang

Thus $\kappa$ is a regular uncountable limit cardinal, i.e., is weakly
inaccessible.

\medskip

{\bf (c)} Write $R$ for the set
of regular infinite cardinals less than $\kappa$ and $L$ for the set of
limit cardinals less than $\kappa$.

\medskip

\quad{\bf (i)} $\kappa\setminus R\in\Cal I$.  \Prf\Quer\ Otherwise,
$A=(\kappa\setminus R)\setminus\{0,1\}\notin\Cal I$.   For $\xi\in A$,
set $f(\xi)=\cf\xi$;  then $f:A\to\kappa$ is regressive.   Because
$\Cal I$ is normal, there must be a $\delta<\kappa$ such that
$B=\{\xi:\xi<\kappa$, $\cf\xi=\delta\}\notin\Cal I$.
For each $\xi\in B$, let $\ofamily{\eta}{\delta}{g_{\eta}(\xi)}$ enumerate
a cofinal subset of $\xi$.   If $\eta<\delta$, then
$g_{\eta}:B\to\kappa$ is regressive, so by 541Ka there is a
$\gamma_{\eta}<\kappa$ such that
$J_{\eta}=\{\xi:\xi\in B$, $g_{\eta}(\xi)\ge\gamma_{\eta}\}\in\Cal I$.
Set $\gamma=\sup_{\eta<\delta}\gamma_{\eta}$;  as $\kappa$ is regular,
$\gamma<\kappa$;  while
$B\setminus(\gamma+1)\subseteq\bigcup_{\eta<\delta}J_{\eta}$ belongs to
$\Cal I$, which is impossible.\ \Bang\Qed

\medskip

\quad{\bf (ii)} $R\setminus L\in\Cal I$.   \Prf\ We have a regressive
function $f:R\setminus L\to\kappa$ defined by setting
$f(\lambda^+)=\lambda$ for every infinite cardinal $\lambda<\kappa$.
Now $f^{-1}[\{\xi\}]$ is empty or a singleton for every $\xi$, so always
belongs to $\Cal I$;  because
$\Cal I$ is normal, $R\setminus L\in\Cal I$.\ \Qed

\medskip

\quad{\bf (iii)} Accordingly the set $R\cap L$ of weakly inaccessible
cardinals less than $\kappa$ cannot belong to $\Cal I$ and must be
stationary, by 541Ia.
}%end of proof of 541L

\leader{541M}{Definition (a)} A regular uncountable cardinal $\kappa$ is
{\bf \2vm}\cmmnt{ (often just {\bf measurable})} if there is a proper
$\kappa$-additive
$2$-saturated ideal of $\Cal P\kappa$ containing singletons.

\cmmnt{Of course a proper ideal $\Cal I$ of $\Cal P\kappa$ is
$2$-saturated iff it is maximal, that is, the dual filter
$\{\kappa\setminus I:I\in\Cal I\}$ is an ultrafilter;  thus $\kappa$ is
\2vm\ iff there is a non-principal $\kappa$-complete ultrafilter on
$\kappa$.   From 541J we see also that if $\kappa$ is \2vm\ then there
is a normal maximal ideal of $\Cal P\kappa$, that is, there is a normal
ultrafilter on $\kappa$, as considered in \S4A1.}

\spheader 541Mb An uncountable cardinal $\kappa$ is {\bf weakly compact}
if for every $S\subseteq[\kappa]^2$ there is a $D\in[\kappa]^{\kappa}$
such that $[D]^2$ is either included in $S$ or disjoint from $S$.

\leader{541N}{Theorem} (a) A \2vm\ cardinal is weakly compact.

(b) A weakly compact cardinal is strongly inaccessible.

\proof{{\bf (a)} If $\kappa$ is a \2vm\ cardinal, there is a
non-principal normal ultrafilter on $\kappa$, so 4A1L tells us that
$\kappa$ is weakly compact.

\medskip

{\bf (b)} Let $\kappa$ be a weakly compact cardinal.

\medskip

\quad{\bf (i)} Set $\lambda=\cf\kappa$;  let $A\in[\kappa]^{\lambda}$ be
a cofinal subset of $\kappa$, and
$\ofamily{\zeta}{\lambda}{\alpha_{\zeta}}$ the increasing enumeration of
$A$.   For $\xi<\kappa$ set $f(\xi)=\min\{\zeta:\xi<\alpha_{\zeta}\}$;
now set $S=\{I:I\in[\kappa]^2$, $f$ is constant on $I\}$.   If
$D\in[\kappa]^{\kappa}$, take any $\xi\in D$;  then there is an
$\eta\in D\setminus\alpha_{f(\xi)}$, so $f$ is not constant on
$\{\xi,\eta\}$ and $[D]^2\not\subseteq S$.   There must therefore be a
$D\in[\kappa]^2$ such that $[D]^2\cap S=\emptyset$.   But in this case
$f$ is injective on $D$, so $\lambda\ge\#(f[D])=\kappa$ and
$\cf\kappa=\kappa$.

Thus $\kappa$ is regular.

\medskip

\quad{\bf (ii)} \Quer\ Suppose, if possible, that $\kappa$ is not
strongly inaccessible.   Then there is a least cardinal $\lambda<\kappa$
such that $2^{\lambda}\ge\kappa$;  let $\phi:\kappa\to\Cal P\lambda$ be
an injective function.   Set

\Centerline{$S=\{\{\xi,\eta\}:\xi<\eta<\kappa$,
$\min(\phi(\xi)\symmdiff\phi(\eta))\in\phi(\eta)\}$.}

\noindent Because $\kappa$ is weakly compact, there is a
$D\in[\kappa]^{\kappa}$ such that either $[D]^2\subseteq S$ or
$[D]^2\cap S=\emptyset$.   Set
$B=\{\phi(\xi)\cap\gamma:\xi\in D$, $\gamma<\lambda\}$.   Then

\Centerline{$\#(B)\le\#(\bigcup_{\gamma<\lambda}\Cal P\gamma)
\le\max(\lambda,\sup_{\gamma<\lambda}2^{\gamma})<\kappa$}

\noindent because $\kappa$ is regular, $\lambda<\kappa$ and
$2^{\gamma}<\kappa$ for every $\gamma<\lambda$.   So there must be an
$\eta\in D$ such that
$B=\{\phi(\xi)\cap\gamma:\xi\in D\cap\eta$, $\gamma<\lambda\}$.   Take
$\zeta\in D$ such that $\zeta>\eta$, set
$\gamma=\min(\phi(\eta)\symmdiff\phi(\zeta))$ and take
$\xi\in D\cap\eta$ such that
$\phi(\xi)\cap(\gamma+1)=\phi(\zeta)\cap(\gamma+1)$.   Now
$\gamma=\min(\phi(\xi)\symmdiff\phi(\eta))$, so

\Centerline{$\{\xi,\eta\}\in S
\iff\gamma\in\phi(\eta)
\iff\gamma\notin\phi(\zeta)
\iff\{\eta,\zeta\}\notin S$.}

\noindent But this means that $[D]^2$ can be neither included in $S$ nor
disjoint from $S$;  contrary to the choice of $D$.\ \Bang

Thus $\kappa$ is strongly inaccessible.
}%end of proof of 541N

\leader{541O}{Lemma} Let $X$ be a set and $\Cal I$ a proper ideal of
subsets of $X$ such that $\Cal PX/\Cal I$ is atomless.   If $\Cal I$ is
$\lambda$-saturated and $\kappa$-additive, with $\lambda\le\kappa$, then
$\kappa\le\cov\Cal I\le\sup_{\theta<\lambda}2^{\theta}$.

\proof{ We may take it that $\lambda=\sat(\Cal PX/\Cal I)$.
If $\lambda>\kappa$ the result is trivial because $\Cal I$ contains
singletons.   So suppose that $\lambda\le\kappa$.
For each $A\in\Cal PX\setminus\Cal I$ choose $A'\subseteq A$ such that
neither $A'$ nor $A\setminus A'$ belongs to $\Cal I$;  this is possible because
$\Cal PX/\Cal I$ is atomless.   Define
$\ofamily{\xi}{\lambda}{\Cal A_{\xi}}$ inductively, as follows.
$\Cal A_0=\{X\}$.   Given that
$\Cal A_{\xi}\subseteq\Cal PX\setminus\Cal I$, then set
$\Cal A_{\xi+1}
=\{A':A\in\Cal A_{\xi}\}\cup\{A\setminus A':A\in\Cal A_{\xi}\}$.   For a
non-zero limit ordinal $\xi<\lambda$, set
$E_{\xi}=\bigcap_{\eta<\xi}\bigcup\Cal A_{\eta}$;  for
$x\in E_{\xi}$ set $C_{\xi x}
=\bigcap\{A:x\in A\in\bigcup_{\eta<\xi}\Cal A_{\eta}\}$;  set
$\Cal A_{\xi}=\{C_{\xi x}:x\in E_{\xi}\}\setminus\Cal I$, and
continue.   Observe that this construction ensures that each
$\Cal A_{\xi}$ is disjoint, and that if $\eta\le\xi$ and
$A\in\Cal A_{\xi}$ then there is a $B\in\Cal A_{\eta}$ such that
$A\subseteq B$.

If $x\in X$, then
$\alpha_x=\{\xi:\xi<\lambda$, $x\in\bigcup\Cal A_{\xi}\}$ is an
initial segment of $\lambda$, so is an ordinal less than or equal to
$\lambda$.   In fact $\alpha_x<\lambda$.   \Prf\ For each
$\xi<\alpha_x$ take $A_{\xi}\in\Cal A_{\xi}$ such that
$x\in A_{\xi}$, and let $B_{\xi}$ be either $A'_{\xi}$ or
$A_{\xi}\setminus A'_{\xi}$ and such that $x\notin B_{\xi}$.   Then
$\ofamily{\xi}{\alpha_x}{B_{\xi}}$ is a disjoint family in
$\Cal P\kappa\setminus\Cal I$ so has cardinal less than $\lambda$.\ \Qed

Of course each $\alpha_x$ is a non-zero limit ordinal, because
$\bigcup\Cal A_{\xi}=\bigcup\Cal A_{\xi+1}$ for each $\xi$.   Now set
$\Cal A=\bigcup_{\xi<\lambda}\Cal A_{\xi}$;  then
$\#(\Cal A)\le\lambda$.   Next, for any $x\in X$,
$\Cal B_x=\{A:A\in\Cal A$, $x\in A\}$ has cardinal less than
$\lambda$ and $C_x=\bigcap\Cal B_x$ belongs to $\Cal I$ and
contains $x$.   So $\Cal C=\{C_x:x\in X\}$ has cardinal
at most $\#([\lambda]^{<\lambda})=\sup_{\theta<\lambda}2^{\theta}$
(because $\lambda=\sat(\Cal PX/\Cal I)$ is regular, by 514Da),
and $\Cal C\subseteq\Cal I$ covers $X$, so

\Centerline{$\kappa\le\add\Cal I\le\cov\Cal I
\le\#(\Cal C)\le\sup_{\theta<\lambda}2^{\theta}$.}
}%end of proof of 541O

\vleader{48pt}{541P}{Theorem}\cmmnt{ ({\smc Tarski 1945}, {\smc Solovay 71})}
Suppose that $\kappa$ is a regular uncountable cardinal with a proper
$\lambda$-saturated $\kappa$-additive ideal
$\Cal I$ of $\Cal P\kappa$ containing singletons, where
$\lambda\le\kappa$.   Set $\frak A=\Cal P\kappa/\Cal I$.   Then

{\it either} $\kappa\le\sup_{\theta<\lambda}2^{\theta}$ and $\frak A$ is
atomless

{\it or} $\kappa$ is \2vm\ and $\frak A$ is purely atomic.

\proof{{\bf (a)} Let us begin by noting that $\Cal I$ is
$\lambda$-saturated iff $\lambda\ge\sat(\frak A)$;  so it will be enough
to prove the result when $\lambda=\sat(\frak A)$, in which case
$\lambda$ is either finite or regular and uncountable (514Da again).

\medskip

{\bf (b)} Suppose that $\frak A$ is atomless.   By 541O,
$\kappa\le\sup_{\theta<\lambda}2^{\theta}$.
So in this case we have the first alternative of the dichotomy.

\medskip

{\bf (c)} Before continuing with an analysis of atoms in $\frak A$, I
draw out some further features of the structure discussed in
the proof of 541O.   We
find that if $\frak A$ is atomless then $\kappa$ is not weakly compact.
\Prf\ Construct
$\ofamily{\xi}{\lambda}{\Cal A_{\xi}}$, $\Cal A$ and $\Cal C$
as in 541O.   Consider
$\alpha^*=\sup\{\xi:\xi<\lambda$, $\Cal A_{\xi}\ne\emptyset\}$.

\medskip

\quad{\bf case 1} If $\alpha^*<\kappa$, then $\#(\Cal A)<\kappa$,
because $\kappa$ is
regular and $\#(\Cal A_{\xi})<\lambda\le\kappa$ for every $\xi$.   In
this case $\kappa\le\#(\Cal C)\le 2^{\#(\Cal A)}$ and $\kappa$ is not
strongly inaccessible, therefore not weakly compact, by 541Nb.

\medskip

\quad{\bf case 2} If $\alpha^*=\kappa$, then $\#(\Cal A')=\kappa$, where
$\Cal A'=\bigcup_{\xi<\kappa}\Cal A_{\xi+1}$.   Note that each
$D\in\Cal A'$ has a companion $D^*\in\Cal A'$ defined by saying that if
$D\in\Cal A_{\xi+1}$ then $D^*=D_0\setminus D$ where $D_0$ is the unique
member of $\Cal A_{\xi}$ including $D$.   Consider the relation
$S=\{(D,D'):D$, $D'\in\Cal A'$, $D\cap D'=\emptyset\}$.   Take any
$\Cal D\in[\Cal A']^{\kappa}$.   Then $[\Cal D]^2\not\subseteq S$, because
$\Cal I$ is $\kappa$-saturated.   \Quer\ If
$[\Cal D]^2\cap S=\emptyset$, any two members of $\Cal D$ meet.   If $D_1$ and
$D_2$ are distinct members of $\Cal D$, then they cannot both belong to
$\Cal A_{\xi+1}$ for any $\xi$, so one must belong to $\Cal A_{\eta+1}$
and the other to $\Cal A_{\xi+1}$ where $\eta<\xi$;  say
$D_1\in\Cal A_{\eta+1}$ and $D_2\in\Cal A_{\xi+1}$.   Now
$D_2\cup D_2^*\in\Cal A_{\xi}$ meets $D_1$ and is therefore included in
$D_1$;  so $D_1^*\cap D_2^*=\emptyset$.   Thus $\{D^*:D\in\Cal D\}$ is a
disjoint family in $\Cal A$ with cardinal $\kappa$, contrary to the hypothesis
that $\Cal I$ is $\kappa$-saturated.\ \Bang

Thus if $\Cal D\in[\Cal A']^{\kappa}$, $[\Cal D]^2$ is neither included
in nor disjoint from $S$.   Since $\#(\Cal A')=\kappa$, this shows that
$\kappa$ cannot be weakly compact.\ \Qed

\medskip

{\bf (d)} Now suppose that $\frak A$ has an atom $a$.
Let $A\in\Cal P\kappa\setminus\Cal I$ be such that $A^{\ssbullet}=a$.
Set $\Cal I_A=\{I:I\subseteq\kappa$, $I\cap A\in\Cal I\}$;  then
$\Cal I_A$ is a $\kappa$-additive maximal ideal of $\kappa$ containing
singletons, so $\kappa$ is \2vm.   It follows that $\kappa$ is weakly
compact (541Na).

\Quer\ Suppose, if possible, that $\frak A$ is not purely atomic.   Then
there is a $C\in\Cal P\kappa\setminus\Cal I$ such that
$\Cal P\kappa/\Cal I_C$ is atomless, where
$\Cal I_C=\{I:I\subseteq\kappa$, $I\cap C\in\Cal I\}$.   Also $\Cal I_C$ is
$\kappa$-additive and $\lambda$-saturated.   But this is
impossible, by (c).\ \BanG\  Thus $\frak A$ is purely atomic, and we
have the second alternative of the dichotomy.
}%end of proof of 541P

\leader{541Q}{Theorem} %RVMC 5B
Let $\kappa$ be a regular uncountable cardinal and $\Cal I$ a normal
ideal on $\kappa$.   Let $\theta<\kappa$ be a cardinal of uncountable
cofinality such that
$\Cal I$ is $(\cf\theta)$-saturated in $\Cal P\kappa$, and
$f:[\kappa]^{<\omega}\to[\kappa]^{<\theta}$ any
function.   Then there are $C\in\Cal I$ and
$f^*:[\kappa\setminus C]^{<\omega}\to[\kappa]^{<\theta}$ such that
$f(I)\cap\eta\subseteq f^*(I\cap\eta)$ whenever
$I\in[\kappa\setminus C]^{<\omega}$ and $\eta<\kappa$.
%again:  do we need all this?  probably not, but it's fun

\proof{{\bf (a)} I show by induction on $n\in\Bbb N$ that if
$g:[\kappa]^{\le n}\to[\kappa]^{<\theta}$ is a
function then there are $A\in\Cal I$ and
$g^*:[\kappa\setminus A]^{\le n}\to[\kappa]^{<\theta}$ such that
$g(I)\cap\eta\subseteq g^*(I\cap\eta)$ for every
$I\in[\kappa\setminus A]^{\le n}$ and $\eta<\kappa$.

\Prf\ If $n=0$ this is trivial;  take $A=\emptyset$,
$g^*(\emptyset)=g(\emptyset)$.   For the inductive step to $n+1$, given
$g:[\kappa]^{\le n+1}\to[\kappa]^{<\theta}$, then for each $\xi<\kappa$
define $g_{\xi}:[\kappa]^{\le n}\to[\kappa]^{<\theta}$ by
setting $g_{\xi}(J)=g(J\cup\{\xi\})$ for every
$J\in[\kappa]^{\le n}$.   Set

\Centerline{$D=\{\xi:\xi<\kappa$, $\cf(\xi)\ge\theta\}$;}

\noindent then $\kappa\setminus D\in\Cal I$ (541Kb).  For $\xi\in D$ and
$J\in[\kappa]^{\le n}$ set
$\zeta_{J\xi}=\sup(\xi\cap g_{\xi}(J))<\xi$.   Then for each
$J\in[\kappa]^{\le n}$
the function $\xi\mapsto\zeta_{J\xi}:D\to\kappa$ is regressive,
so there is a $\zeta^*_J<\kappa$ such that
$\{\xi:\zeta_{J\xi}\ge\zeta^*_J\}\in\Cal I$ (541Ka).   Now
$\add\Cal I=\kappa$, by 541H, so 541D tells us
that there is an $h(J)\in[\zeta^*_J]^{<\theta}$ such that
$\{\xi:\xi\cap g_{\xi}(J)\not\subseteq h(J)\}\in\Cal I$.   By the
inductive hypothesis, there are $B\in\Cal I$ and
$h^*:[\kappa\setminus B]^{\le n}\to[\kappa]^{<\theta}$ such that
$h(J)\cap\eta\subseteq h^*(J\cap\eta)$ for every
$J\in[\kappa\setminus B]^{\le n}$ and $\eta<\kappa$.

Try setting

\Centerline{$A_J=\{\xi:\xi\cap g_{\xi}(J)\not\subseteq h(J)\}$ for
$J\in[\kappa]^{\le n}$,}

\Centerline{$A=B\cup\{\xi:\xi\in\bigcup_{J\in[\xi]^{\le n}}A_J\}$,}

$$\eqalign{g^*(I)
&=g(I)\text{ if }I\in[\kappa\setminus A]^{n+1},\cr
&=g(I)\cup h^*(I)\text{ if }I\in[\kappa\setminus A]^{\le n}.\cr}$$

\noindent Then $A_J$ always belongs to $\Cal I$, by the choice of
$h(J)$, so $A\in\Cal I$, by 541Ib,
while $g^*(I)\in[\kappa]^{<\theta}$ for every
$I\in[\kappa\setminus A]^{\le n+1}$.   Take $\eta<\kappa$ and
$I\in[\kappa\setminus A]^{\le n+1}$.   If $I\subseteq\eta$ then
$g(I)\cap\eta\subseteq g^*(I)=g^*(I\cap\eta)$.   Otherwise, set
$\xi=\max I$ and $J=I\setminus\{\xi\}$.   Then
$\eta\le\xi\in\kappa\setminus A_J$, so

\Centerline{$g(I)\cap\eta
=g_{\xi}(J)\cap\xi\cap\eta
\subseteq h(J)\cap\eta
\subseteq h^*(J\cap\eta)
=h^*(I\cap\eta)
\subseteq g^*(I\cap\eta)$.}

\noindent Thus the induction continues.\ \Qed

\medskip

{\bf (b)} Now applying (a) to $f\restr[\kappa]^{\le n}$
we obtain sets $C_n\in\Cal I$ and functions
$f^*_n:[\kappa\setminus C_n]^{\le n}\to[\kappa]^{<\theta}$ such that
$f(I)\cap\eta\subseteq f_n^*(I\cap\eta)$
whenever $I\in[\kappa\setminus C_n]^{\le n}$ and $\eta<\kappa$.   Set
$C=\bigcup_{n\in\Bbb N}C_n\in\Cal I$ and
$f^*(I)=\bigcup_{n\ge\#(I)}f^*_n(I)$ for each
$I\in[\kappa\setminus C]^{<\omega}$.   Because $\cf\theta>\omega$,
$f(I)\in[\kappa]^{<\theta}$ for every $I$.
If $I\in[\kappa\setminus C]^{<\omega}$ and
$\eta<\kappa$, set $n=\#(I)$;  then $I\in[\kappa\setminus C_n]^n$ so
$f(I)\cap\eta\subseteq f_n^*(I\cap\eta)\subseteq f^*(I\cap\eta)$, as
required.
}%end of proof of 541Q

\vleader{72pt}{541R}{Corollary} %RVMC 5C
Let $\kappa$ be a regular uncountable cardinal, $\Cal I$ a normal
ideal on $\kappa$, and $\theta<\kappa$ a cardinal of uncountable
cofinality such that
$\Cal I$ is $(\cf\theta)$-saturated in $\Cal P\kappa$.

(a) If $Y$ is a set of
cardinal less than $\kappa$ and $f:[\kappa]^{<\omega}\to[Y]^{<\theta}$ a
function, then there are $C\in\Cal I$ and $M\in[Y]^{<\theta}$ such
that $f(I)\subseteq M$ for every $I\in[\kappa\setminus C]^{<\omega}$.

(b) If $Y$ is any set and $g:\kappa\to[Y]^{<\theta}$ a function, then there
are $C\in\Cal I$ and $M\in[Y]^{<\theta}$ such that
$g(\xi)\cap g(\eta)\subseteq M$ for all distinct $\xi$,
$\eta\in\kappa\setminus C$.

\proof{{\bf (a)} We may suppose that $Y\subseteq\kappa$.   In this case, by 541Q,
we have a $C_0\in\Cal I$ and an
$f^*:[\kappa\setminus C_0]^{<\omega}\to[\kappa]^{<\theta}$ such that
$f(I)\cap\eta\subseteq f^*(I\cap\eta)$ whenever
$I\subseteq\kappa\setminus C_0$ is finite and $\eta<\kappa$.   Let
$\gamma<\kappa$ be such that $Y\subseteq\gamma$ and set
$M=Y\cap f^*(\emptyset)$, $C=C_0\cup\gamma$.   Then
$M\in[Y]^{<\theta}$, $C\in\Cal I$ and if
$I\in[\kappa\setminus C]^{<\omega}$ then

\Centerline{$f(I)=f(I)\cap\gamma
\subseteq Y\cap f^*(I\cap\gamma)\cap\gamma=M$.}

\medskip

{\bf (b)} Since $\bigcup_{\xi<\kappa}g(\xi)$ has cardinal at most $\kappa$,
we may again suppose that $Y\subseteq\kappa$.   Apply 541Q with
$f(\{\xi\})=g(\xi)$ for $\xi<\kappa$.
Taking $C$ and $f^*$ from 541Q, set $M=Y\cap f^*(\emptyset)$.
Set $F=\{\xi:\xi<\kappa$, $g(\eta)\subseteq\xi$ for every $\eta<\xi\}$;
then $F$ is a closed cofinal subset of $\kappa$ (because $\theta\le\kappa$
and $\kappa$ is regular), so $C'=C\cup(\kappa\setminus F)\in\Cal I$
(541Ia).   If $\xi$, $\eta$ belong to $\kappa\setminus C'=F\setminus C$
and $\eta<\xi$, then

\Centerline{$g(\xi)\cap g(\eta)\subseteq\xi\cap g(\xi)
=\xi\cap f(\{\xi\})\subseteq f^*(\xi\cap\{\xi\})
=f^*(\emptyset)$,}

\noindent so $g(\xi)\cap g(\eta)\subseteq M$.   Thus $C'$ serves.
}%end of proof of 541R

\vleader{36pt}{541S}{Lemma} %RVMC 7O
Let $\kappa$ be a regular uncountable cardinal and $\Cal I$ a normal
ideal on $\kappa$.   Suppose that $\gamma$ and $\delta$ are cardinals
such that
$\omega\le\gamma<\delta<\kappa\le 2^{\delta}$,
$\Cal I$ is $\delta$-saturated in
$\Cal P\kappa$, $2^{\beta}=2^{\gamma}$
for $\gamma\le\beta<\delta$, but $2^{\delta}>2^{\gamma}$.   Then
$\delta$ is regular and

\Centerline{$2^{\delta}=\covSh(2^{\gamma},\kappa,\delta^+,\delta)
=\covSh(2^{\gamma},\kappa,\delta^+,\omega_1)
=\covSh(2^{\gamma},\kappa,\delta^+,2)$.}

\proof{ By 5A1Fh, $\delta$ is regular.    Of course

\Centerline{$\covSh(2^{\gamma},\kappa,\delta^+,\delta)
\le\covSh(2^{\gamma},\kappa,\delta^+,\omega_1)
\le\covSh(2^{\gamma},\kappa,\delta^+,2)
\le\#([2^{\gamma}]^{\le\delta})\le 2^{\delta}$}

\noindent (5A2D, 5A2Ea).   For the reverse inequality, let
$\Cal E\subseteq[2^{\gamma}]^{<\kappa}$ be a set with cardinal
$\covSh(2^{\gamma},\kappa,\delta^+,\delta)$
such that every member of $[2^{\gamma}]^{\le\delta}$ is covered by
fewer than $\delta$ members of
$\Cal E$.   For each ordinal $\xi<\delta$ let
$\phi_{\xi}:\Cal P\xi\to 2^{\gamma}$ be an injective function.   For
$A\subseteq\delta$ define $f_A:\delta\to 2^{\gamma}$ by

\Centerline{$f_A(\xi)=\phi_{\xi}(A\cap\xi)$ for every $\xi<\delta$.}

\noindent Choose $E_A\in\Cal E$ such that $f^{-1}_A[E_A]$ is cofinal
with $\delta$;  such must exist because $\delta$ is regular and
$f_A[\delta]$ can be covered by fewer than $\delta$ members of $\Cal E$.

\Quer\ If $2^{\delta}>\#(\Cal E)$ then there must be an $E\in\Cal E$
and an $\Cal A\subseteq\Cal P\delta$ such that $\#(\Cal A)=\kappa$ and
$E_A=E$ for every $A\in\Cal A$.   For each pair $A$, $B$ of distinct
members of $\Cal A$ set $\xi_{AB}=\min(A\symmdiff B)<\delta$.   By 541Ra,
there is a set
$\Cal B\subseteq\Cal A$, with cardinal $\kappa$, such that
$M=\{\xi_{AB}:A$, $B\in\Cal B$, $A\ne B\}$ has cardinal less than
$\delta$.   Set $\zeta=\sup M<\delta$.   Next, for each $A\in\Cal B$,
take $\eta_A>\zeta$ such that $f_A(\eta_A)\in E$.
Let $\eta<\delta$ be such that $\Cal C=\{A:A\in\Cal B$, $\eta_A=\eta\}$
has cardinal $\kappa$.   Then we have a map

\Centerline{$A\mapsto f_A(\eta)=\phi_{\eta}(A\cap\eta):\Cal C\to E$}

\noindent which is injective, because if $A$, $B$ are distinct members
of $\Cal C$ then $\xi_{AB}\le\zeta<\eta$, so $A\cap\eta\ne B\cap\eta$.
So $\#(E)\ge\kappa$;  but $E\in\Cal E\subseteq[2^{\gamma}]^{<\kappa}$.\
\Bang

As $\Cal E$ is arbitrary,
$\covSh(2^{\gamma},\kappa,\delta^+,\delta)\ge 2^{\delta}$.
}%end of proof of 541S

\exercises{\leader{541X}{Basic exercises (a)}
%\spheader 541Xa
Let $\kappa$ be a regular infinite cardinal.   Show that
$\Cal P\kappa/[\kappa]^{<\kappa}$ is not Dedekind complete, so
$[\kappa]^{<\kappa}$ is not $\kappa^+$-saturated in $\Cal P\kappa$.
\Hint{construct a disjoint family $\ofamily{\xi}{\kappa}{A_{\xi}}$ in
$[\kappa]^{\kappa}$;  show that if $\#(A_{\xi}\setminus A)<\kappa$ for
every $\xi$ there is a $B\in[A]^{\kappa}$ such that
$\#(B\cap A_{\xi})<\kappa$ for every $\xi$.}
%541B

\spheader 541Xb
Let $\frak A$ be a Boolean algebra and $I$ an ideal of
$\frak A$.   Suppose there is a cardinal $\kappa$ such that $I$ is
$\kappa$-additive and $\kappa^+$-saturated and $\frak A$ is
Dedekind $\kappa^+$-complete in the sense that $\sup A$ is defined in
$\frak A$ whenever $A\in[\frak A]^{\le\kappa}$.   Show that $\frak A/I$
is Dedekind complete.
%541B

\spheader 541Xc Suppose that $X$ and $Y$ are sets and $\Cal I$, $\Cal J$
ideals of subsets of $X$, $Y$ respectively.   Suppose that $\kappa$ is
an infinite cardinal such that both $\Cal I$ and $\Cal J$ are
$\kappa$-saturated and $\kappa^+$-additive.   Show that
$\Cal I\ltimes\Cal J$ (definition:  527Ba) is $\kappa$-saturated and
$\kappa^+$-additive.
%541C

\spheader 541Xd Simplify the argument of 541D to give a direct proof of
541E in the case in which $\kappa$ is regular.
%541E

\sqheader 541Xe Show that there is a \2vm\ cardinal iff there are a set $I$
and a non-principal $\omega_1$-complete ultrafilter on $I$.
%541M

\sqheader 541Xf Let $\kappa$ be a \2vm\ cardinal and $\Cal I$ a normal
maximal proper ideal of $\Cal P\kappa$.   (i) Show that if
$S\subseteq[\kappa]^{<\omega}$, there is a $C\in\Cal I$ such that, for
each $n\in\Bbb N$, $[\kappa\setminus C]^n$ is either disjoint from $S$
or included in it.
(ii) Show that if $\#(Y)<\kappa$, and $f:[\kappa]^{<\omega}\to Y$ is
any function, then there is a $C\in\Cal I$ such that $f$ is constant on
$[\kappa\setminus C]^n$ for each $n\in\Bbb N$.
\Hint{4A1L.}
%541N

\spheader 541Xg Let $\kappa$ be a regular uncountable cardinal, $\Cal I$
a $\kappa$-saturated normal ideal on $\kappa$ and
$f:[\kappa]^2\to\kappa$ a function.   Show that there are a $C\in\Cal I$
and an $f^*:\kappa\to\kappa$ such that whenever
$\eta\in\kappa\setminus C$ and $\xi\in\eta\setminus C$ then either
$f(\{\xi,\eta\})\ge\eta$ or $f(\{\xi,\eta\})<f^*(\xi)$.
%541Q

\leader{541Y}{Further exercises (a)}
%\spheader 541Ya
Show that if $\kappa$ is a regular uncountable cardinal and
$S\subseteq\kappa$ is stationary, then $S$ can be partitioned into
$\kappa$ stationary sets.   ({\it Hint\/}:  reduce to the case in which there is a
$\kappa$-saturated normal ideal $\Cal I$ of $\kappa$ containing
$\kappa\setminus S$.   Define $f:S\to\kappa$ inductively by saying that

\Centerline{$f(\xi)=\min(\bigcup\{\kappa\setminus f[C]:C\subseteq\xi$ is
relatively closed and cofinal$\}$.}

\noindent Set $S_{\gamma}=f^{-1}[\{\gamma\}]$.   Apply 541Kc with
$C_{\xi}\cap S_{\gamma}=\emptyset$ for $\xi\in S_{\gamma}$ to show that
$S_{\gamma}\in\Cal I$ for every $\gamma$.   Hence show that $S_{\gamma}$
is stationary for every $\gamma$.   See
{\smc Solovay 71}.)
%541K mt54bits

\spheader 541Yb Show that if $\kappa$ is
\2vm\ and $\Cal F$ is a normal ultrafilter on $\kappa$, then the set of
weakly compact cardinals less than $\kappa$ belongs to $\Cal F$.
%541N
}%end of exercises

\endnotes{
\Notesheader{541}
The ordinary principle of exhaustion (215A) can be regarded as an
expression of $\omega_1$-saturation (compare 316E and 514Db).   In
541B-541E %541B 541C 541D 541E
we have versions of results already given in special cases;  but note
that 541B, for instance, goes a step farther than the arguments offered
in 314C and 316Fa can reach.   541Ca corresponds to 215B(iv);
541Cb-541Cc are associated with 516Q and 525C.   In all this work you
will probably find it helpful to use the words `negligible' and
`conegligible' and `almost everywhere', so that the conclusion of 541E
becomes `$g(x)\in M$ a.e.($x$)'.   I don't use this language in the
formal exposition because of the obvious danger of confusing a reader
who is skimming through without reading introductions to sections very
carefully;  but in the principal applications I have in mind, $\Cal I$
will indeed be a null ideal.   The cardinals
$\Tr_{\Cal I}(X;Y)$ will appear only occasionally in this book, but
are of great importance in infinitary combinatorics generally.   Note
that the key step in the proof of 541F (part (b) of the proof) develops
an idea from the proof of 541J.

For the special purposes of \S438 I mentioned `normal filters' in \S4A1;
I have now attempted an introduction to the general theory of normal
filters and ideals.   The central observation of {\smc Keisler \& Tarski
64} was that if $\kappa$ is uncountable then any $\kappa$-complete
non-principal ultrafilter gives rise to a normal ultrafilter on
$\kappa$.   It was noticed very quickly that something similar happens
if we have a $\kappa$-complete filter of conegligible sets in a totally
finite measure space;  the extension of the idea to general
$\kappa$-additive $\kappa$-saturated ideals is in {\smc Solovay 71}.
In this chapter I speak oftener of ideals than of filters but the ideas
are necessarily identical.   Observe that the Pressing-Down Lemma
(4A1Cc) is the special case of 541H(iii) when $\Cal I$ is the ideal of
non-stationary sets (541Ia).

541Lb here is a re-working of Ulam's theorem
(438C, {\smc Ulam 1930}).   The dramatic further step in 541Lc derives
from {\smc Keisler \& Tarski 64}.   The proof of 541Lc already makes it
plain that much more can be said;  for
extensions of these ideas, see {\smc Fremlin 93}, {\smc Levy 71} and
{\smc Baumgartner Taylor \& Wagon 77}.   In 541P we have an extension of
Ulam's dichotomy (438C, 543B).   `Weak compactness' of a cardinal
corresponds to Ramsey's theorem (4A1G);  the idea was the basis of the
proof of 451Q.   Here I treat it as a purely combinatorial concept,
but its real importance is in model theory ({\smc Kanamori 03}, \S4).

541Q is a fairly strong version of one of the typical properties of
saturated normal ideals.   The simplest not-quite-trivial case is when
we have a function $f:[\kappa]^2\to\kappa$.   In this case we find that
if we discard an appropriate
negligible set $C$ then, for the remaining doubletons
$I\in[\kappa\setminus C]^2$, $f(I)$
is either greater than or equal to $\max I$ or in a `small' set determined
by $\min I$.   In this form, with the appropriate definition of `small', it
is enough for the ideal to be
$\kappa$-saturated (541Xg).   In the intended applications of 541Q,
however, we shall be looking at functions
$f:[\kappa]^{<\omega}\to[\kappa]^{\le\omega}$ and shall need to start
from an $\omega_1$-saturated ideal to obtain the full strength of the
result.

Shelah's four-cardinal covering numbers $\covSh$ are not immediately
digestible;  in \S5A2 I give the basic pcf theory linking them to
cofinalities of products.   541S is here because it relies on a normal
ideal being saturated.

Perhaps I have not yet sufficiently emphasized that there is a good
reason why I have given no examples of normal ideals other than the
non-stationary ideals, and no discussion of the saturation of those
beyond Solovay's theorem 541Ya.   We have in fact come to an area of
mathematics which demands further acts of faith.   I will continue,
whenever possible, to express ideas as arguments in ordinary ZFC;  but
in most of the principal theorems the hypotheses will include assertions
which can be satisfied only in rather special models of set theory.
Moreover, these are special in a new sense.   By and large, the
assumptions used in the first three chapters of this volume (Martin's
axiom, the continuum hypothesis, and so on) have been proved to be
relatively consistent with ZFC (indeed, with ZF);  that is, we know how
to convert any proof in ZFC that `$\frak m=\omega_1$' into a proof in ZF
that `$0=1$'.   The formal demonstration that this can be done is of
course normally expressed in a framework reducible to Zermelo-Fraenkel
set theory;  but it is sufficiently compelling to be itself part of the
material which must be encompassed by any formal system claiming to
represent twenty-first century mathematics.   In the present chapter,
however, we are coming to results like 541P which have no content unless
there can be non-trivial $\kappa$-saturated $\kappa$-additive ideals.
And such objects are known to be strange in a different way from Souslin
lines.

To describe this difference I turn to the simplest of the new
propositions.   Write \Esic, \Ewic\ for the sentences `there is a
strongly inaccessible cardinal', `there is a weakly inaccessible
cardinal'.   Of course \Esic\ implies \Ewic, while `there is a cardinal
which is not measure-free' also implies \Ewic, by Ulam's theorem.   We
have no such implications in the other direction, but it is easy to
adapt G\"odel's argument for the relative consistency of GCH to show
that if ZFC + \Ewic\ is consistent so is ZFC + GCH + \Ewic, while
ZFC + GCH + \Ewic\ implies \Esic.   But we find also that there is a
proof in ZFC + \Esic\ that `ZFC is consistent'.   So if there were a
proof in ZFC that
`if ZFC is consistent, then ZFC + \Esic\ is consistent', there would be
a proof in ZFC + \Esic\ that `ZFC + \Esic\ is consistent';  by G\"odel's
incompleteness theorem, this would give us a proof that ZFC + \Esic\ was
in fact {\it inconsistent} (and therefore that ZFC and ZF are
inconsistent).

The last paragraph is expressed crudely, in a language which blurs some
essential distinctions;  for a more careful account see {\smc Kunen 80},
{\S}IV.10.   But what I am trying to say is that any theory involving
inaccessible cardinals -- and the theory of this chapter involves
unthinkably many such cardinals -- necessarily leads us to propositions
which are not merely unprovable, but have high `consistency strength';
we have long strings $\phi_0,\ldots,\phi_n$ of statements such that (i)
if ZFC + $\phi_{i+1}$ is consistent, so is ZFC + $\phi_i$ (ii) there can
be no proof of the reverse unless ZF is inconsistent.

We do not (and in my view cannot) know for sure that ZF is consistent.
It has now survived for a hundred years or so, which is empirical
evidence of a sort.   I do not suppose that the century of my own birth
was also the century in which the structure of formal mathematics was
determined for eternity;  I hope and trust that mathematicians will come
to look on ZFC as we now think of Euclidean geometry, as a glorious
achievement and an enduring source of inspiration but inadequate for the
expression of many of our deepest ideas.   But (arguing from the weakest
of historical analogies) I suggest that if and when a new paradox shakes
the foundations of mathematics, it will be because some new Cantor has
sought to extend apparently trustworthy methods to a totally new
context.   And I think that the mathematicians of that generation will
stretch their ingenuity to the utmost to find a resolution of the
paradox which is conservative, in that it retains as much as possible of
their predecessors' ideas, subject perhaps to re-writing a good many
proofs and tut-tutting over the naivety of essays such as this.

I think indeed (I am going a bit farther here) that they will be as
reluctant to discard measurable cardinals as our forebears were to
discard Cantor's cardinals.   There is a flourishing theory of large
cardinals in which very much stronger statements than `there is a \2vm\
cardinal' have been explored in depth without catastrophe.   (I
mention a couple of these in \S545;  another is the Axiom of Determinacy
in \S567.)   Occasionally a proof that
there are no measurable cardinals is announced;  but the last real
fright was in 1976, and most of these arguments have easily been shown
not to reach the claimed conclusion.   My best guess is that measurable
cardinals are safe.   But even if I am wrong, and they are
irreconcilable with ZFC as now formulated, it does not follow that ZFC
will be kept and measurable cardinals discarded.   It could equally
happen that one of the axioms of ZF will be modified;  or, at least,
that a modified form will become a recognized option.   This is a
partisan view from somebody who has a substantial investment to protect.
But if you wish to prove me wrong, I do not see how you can do so
without giving part of your own life to the topic.
}%end of notes

\discrpage

\frfilename{mt542.tex}
\versiondate{8.7.13}
\copyrightdate{2004}

\def\chaptername{Real-valued-measurable cardinals}
\def\sectionname{Quasi-measurable cardinals}

\newsection{542}

As is to be expected, the results of \S541 take especially dramatic
forms when we look at $\omega_1$-saturated $\sigma$-ideals.   542B-542C
spell out the application of the most important ideas from \S541 to this
special case.   In addition, we can us Shelah's pcf theory to give us
some remarkable combinatorial results concerning cardinal arithmetic
(542E) and cofinalities of partially ordered sets (542I-542J).

\leader{542A}{Definition} A cardinal $\kappa$ is {\bf quasi-measurable}
if $\kappa$ is regular and uncountable and there is an
$\omega_1$-saturated normal ideal on $\kappa$.

\leader{542B}{Proposition} If $X$ is a set and $\Cal I$ is a proper
$\omega_1$-saturated $\sigma$-ideal of $\Cal PX$ containing singletons,
then $\add\Cal I$ is quasi-measurable.

\proof{ This is immediate from the special case $\lambda=\omega_1$ of 541J.
}%end of proof of 542B

\leader{542C}{Proposition} If $\kappa$ is a \qm\ cardinal, then $\kappa$
is weakly inaccessible, the set of weakly inaccessible cardinals less
than $\kappa$ is stationary in $\kappa$, and either
$\kappa\le\frak c$ or $\kappa$ is \2vm.

\proof{ By 541L, $\kappa$ is weakly inaccessible and the set of weakly
inaccessible cardinals less than $\kappa$ is stationary in $\kappa$.
Let $\Cal I$ be an $\omega_1$-saturated normal ideal on $\kappa$
and $\frak A=\Cal P\kappa/\Cal I$.   Then 541P tells us that either
$\frak A$ is atomless and $\kappa\le\frak c$ or $\frak A$ is purely
atomic and $\kappa$ is \2vm.
}%end of proof of 542C

\leader{542D}{Proposition} %RVMC 7K
Let $\kappa$ be a quasi-measurable cardinal.

(a) Let $\langle\theta_{\zeta}\rangle_{\zeta<\lambda}$ be a family
such that $\lambda<\kappa$ is a cardinal, every $\theta_{\zeta}$
is a regular infinite cardinal and
$\lambda<\theta_{\zeta}<\kappa$ for every $\zeta<\lambda$.   Then
$\cf(\prod_{\zeta<\lambda}\theta_{\zeta})<\kappa$.

(b) If $\alpha$ and $\gamma$ are cardinals less than $\kappa$, then
$\Theta(\alpha,\gamma)$\cmmnt{ (definition: 5A2Db)} is less than
$\kappa$.

(c) If $\alpha$, $\beta$, $\gamma$ and $\delta$ are cardinals, with
$\alpha<\kappa$, $\gamma\le\beta$ and $\delta\ge\omega_1$, then
$\covSh(\alpha,\beta,\gamma,\delta)$\cmmnt{ (definition: 5A2Da)}
is less than $\kappa$.

(d) $\Theta(\kappa,\kappa)=\kappa$.

\proof{ Fix an $\omega_1$-saturated normal ideal $\Cal I$ on $\kappa$.

\medskip

{\bf (a)} \Quer\ Suppose, if possible, otherwise.   Then $\lambda$ is
surely infinite.   By 5A2Bc,
there is an ultrafilter $\Cal F$ on $\lambda$ such that
$\cf(\prod_{\zeta<\lambda}\theta_{\zeta})
=\cf(\prod_{\zeta<\lambda}\theta_{\zeta}|\Cal F)$, where the reduced
product $\prod_{\zeta<\lambda}\theta_{\zeta}|\Cal F$ is defined in 5A2A;
by 5A2C,
there is a family $\ofamily{\zeta}{\lambda}{\theta'_{\zeta}}$ of regular
cardinals such that $\lambda<\theta'_{\zeta}\le\theta_{\zeta}$ for each
$\zeta$ and $\cf(\prod_{\zeta<\lambda}\theta'_{\zeta}|\Cal F)=\kappa$.
Let $\ofamily{\xi}{\kappa}{p_{\xi}}$ be a cofinal family in
$P=\prod_{\zeta<\lambda}\theta'_{\zeta}|\Cal F$.
For each $\xi<\kappa$ we can find $q_{\xi}\in P$ such that
$q_{\xi}\not\le p_{\eta}$ for any $\eta\le\xi$;  because $P$ is
upwards-directed, we can suppose that $q_{\xi}\ge p_{\xi}$, so that
$\{q_{\xi}:\xi<\kappa\}$ also is cofinal with $P$.    Choose
$f_{\xi}\in\prod_{\zeta<\lambda}\theta'_{\zeta}$ such that
$f_{\xi}^{\ssbullet}=q_{\xi}$ for each $\xi$.

For each $\zeta<\lambda$, $\theta'_{\zeta}<\kappa$, so there is a
countable set $M_{\zeta}\subseteq\theta'_{\zeta}$ such that
$I_{\zeta}=\kappa\setminus\{\xi:f_{\xi}(\zeta)\in M_{\zeta}\}$ belongs
to $\Cal I$ (541E).   As

\Centerline{$\omega\le\lambda<\theta'_{\zeta}=\cf\theta'_{\zeta}$,}

\noindent we can find $g(\zeta)$ such that
$M_{\zeta}\subseteq g(\zeta)<\theta'_{\zeta}$.   Consider
$g^{\ssbullet}$ in $\prod_{\zeta<\lambda}\theta'_{\zeta}|\Cal F$.
There is an $\eta<\kappa$ such that $g^{\ssbullet}\le p_{\eta}$, so that
$f_{\xi}^{\ssbullet}\not\le g^{\ssbullet}$ for every $\xi\ge\eta$.
On the other hand, $\eta\cup\bigcup_{\zeta<\lambda}I_{\zeta}$ belongs to
$\Cal I$, so there is a $\xi\ge\eta$ such that
$f_{\xi}(\zeta)\in M_{\zeta}$ for every $\zeta<\lambda$;  in which case
$f_{\xi}\le g$, which is impossible.\ \Bang

\medskip

{\bf (b)} \Quer\ Suppose, if possible, otherwise.   Of course we can
suppose that $\gamma$ is infinite.   For each
$\xi<\kappa$ there must be a family
$\langle\theta_{\xi\zeta}\rangle_{\zeta<\lambda_{\xi}}$
of regular cardinals less than $\alpha$ such that
$\lambda_{\xi}<\gamma$, $\omega\le\lambda_{\xi}<\theta_{\xi\zeta}$
for every $\zeta<\lambda_{\xi}$
and $\cf(\prod_{\zeta<\lambda_{\xi}}\theta_{\xi\zeta})\ge\xi$.
Let $\lambda$ be such that
$A=\{\xi:\xi<\kappa$, $\lambda_{\xi}=\lambda\}\notin\Cal I$.
By 541Ra, applied to the function
$I\mapsto\{\theta_{\xi\zeta}:\xi\in A\cap I$, $\zeta<\lambda_{\xi}\}:
[\kappa]^{<\omega}\to[\alpha]^{<\lambda^+}$,
there are $C\in\Cal I$ and $M\in[\alpha]^{\le\lambda}$ such that
$\theta_{\xi\zeta}\in M$ whenever
$\xi\in A\setminus C$ and $\zeta<\lambda$.
Let $\langle\theta_{\zeta}\rangle_{\zeta<\lambda'}$ enumerate
$\{\theta:\theta\in M$ is a regular cardinal, $\theta>\lambda\}$.
By (a), there is a cofinal set
$F\subseteq\prod_{\zeta<\lambda'}\theta_{\zeta}$ with $\#(F)<\kappa$.
Let $\xi\in A\setminus C$ be such that $\xi>\#(F)$.
For each $f\in F$ define
$g_f\in\prod_{\zeta<\lambda}\theta_{\xi\zeta}$ by setting

\Centerline{$g_f(\zeta)=f(\zeta')$
whenever $\theta_{\xi\zeta}=\theta_{\zeta'}$.}

\noindent Then $\{g_f:f\in F\}$ is cofinal with
$\prod_{\zeta<\lambda}\theta_{\xi\zeta}$, because if
$h\in\prod_{\zeta<\lambda}\theta_{\xi\zeta}$
there is an $f\in F$ such that

\Centerline{$f(\zeta')\ge\sup\{h(\zeta):\zeta<\lambda$,
$\theta_{\xi\zeta}=\theta_{\zeta'}\}$}

\noindent for every $\zeta'<\lambda'$, and in this case $h\le g_f$.   So

\Centerline{$\#(F)<\xi\le\cf(\prod_{\zeta<\lambda}\theta_{\xi\zeta})
\le\#(F)$,}

\noindent which is absurd.\ \Bang

\medskip

{\bf (c)} This is trivial if any of the cardinals $\alpha$, $\beta$ or
$\gamma$ is finite;  let us take it that they are all infinite.   Then

\Centerline{$\covSh(\alpha,\beta,\gamma,\delta)
\le\covSh(\alpha,\gamma,\gamma,\omega_1)
\le\max(\omega,\alpha,\Theta(\alpha,\gamma))<\kappa$}

\noindent by 5A2D, 5A2G and (b) above.

\medskip

{\bf (d)} Because $\kappa$ is an uncountable limit cardinal,
$\kappa\le\Theta(\kappa,\kappa)$.   (If $\omega\le\theta<\kappa$, then
$\cf\theta^+\le\Theta(\kappa,\kappa)$.)
On the other hand, let $\lambda<\kappa$ be an infinite cardinal and
$\langle\theta_{\zeta}\rangle_{\zeta<\lambda}$ a family of infinite
regular cardinals such that $\lambda<\theta_{\zeta}<\kappa$ for every
$\zeta<\lambda$.   Then
$\alpha=\sup_{\zeta<\lambda}\theta_{\zeta}^+<\kappa$ and

\Centerline{$\cf(\prod_{\zeta<\lambda}\theta_{\zeta})
\le\Theta(\alpha,\lambda^+)<\kappa$.}

\noindent As $\langle\theta_{\zeta}\rangle_{\zeta<\lambda}$ is
arbitrary, $\Theta(\kappa,\kappa)=\kappa$.
}%end of proof of 542D

\leader{542E}{Theorem}\cmmnt{ ({\smc Gitik \& Shelah 93})}
%RVMC 7P
If $\kappa\le\frak c$ is a \qm\ cardinal, then

\Centerline{$\{2^{\gamma}:\omega\le\gamma<\kappa\}$}

\noindent is finite.

\proof{ \Quer\ Suppose, if possible, otherwise.

\medskip

{\bf (a)} Define a sequence $\langle\gamma_n\rangle_{n\in\Bbb N}$ of
cardinals by setting

\Centerline{$\gamma_0=\omega$,
\quad$\gamma_{n+1}=\min\{\gamma:2^{\gamma}>2^{\gamma_n}\}$ for
$n\in\Bbb N$.}

\noindent Then we are supposing that $\gamma_n<\kappa$ for every $n$, so
by 541S, %\gamma_n regular,
  %2^{\gamma_{n+1}}=\covSh(2^{\gamma_n},\kappa,\gamma_{n+1}^+,\omega_1)
5A2D, %\covSh(2^{\gamma_n},\kappa,\gamma_{n+1}^+,\omega_1)
  %\le\covSh(2^{\gamma_n},\gamma_{n+1}^+,\gamma_{n+1}^+,\omega_1)
5A2G %\covSh(2^{\gamma_n},\gamma_{n+1}^+,\gamma_{n+1}^+,\omega_1)
  %\le\max(2^{\gamma_n},\Theta(2^{\gamma_n},\gamma_{n+1}^+))
and 5A2F %\Theta(2^{\gamma_n},\gamma_{n+1}^+)\le 2^{\gamma_{n+1}}
$\gamma_n$ is regular and

$$\eqalign{2^{\gamma_{n+1}}
&=\covSh(2^{\gamma_n},\kappa,\gamma_{n+1}^+,\omega_1)
\le\covSh(2^{\gamma_n},\gamma_{n+1}^+,\gamma_{n+1}^+,\omega_1)\cr
&\le\max(2^{\gamma_n},\Theta(2^{\gamma_n},\gamma_{n+1}^+))
\le 2^{\gamma_{n+1}}\cr}$$

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

\medskip

{\bf (b)} Now $\Theta(2^{\gamma_n},\gamma)=\Theta(\frak c,\gamma)$
whenever $n\in\Bbb N$ and $\gamma$ is a regular cardinal with
$\gamma_n<\gamma<\kappa$.
\Prf\ Induce on $n$.   For $n=0$ we have $\frak c=2^{\gamma_0}$.   For
the inductive step to $n+1$, if $\gamma$ is regular and
$\gamma_{n+1}<\gamma<\kappa$, then
$\frak c\ge\kappa>\Theta(\gamma,\gamma)$ (542Db), so

$$\eqalignno{\Theta(2^{\gamma_{n+1}},\gamma)
&=\Theta(\Theta(2^{\gamma_n},\gamma_{n+1}^+),\gamma)\cr
\displaycause{by (a)}
&\le\Theta(\Theta(2^{\gamma_n},\gamma),\gamma)\cr
\displaycause{because $\gamma\ge\gamma_{n+1}^+$ and $\Theta$ is
order-preserving}
&=\Theta(\Theta(\frak c,\gamma),\gamma)\cr
\displaycause{by the inductive hypothesis}
&\le\Theta(\frak c,\gamma)\cr
\displaycause{5A2H}
&\le\Theta(2^{\gamma_{n+1}},\gamma)\cr}$$

\noindent (because $2^{\gamma_{n+1}}\ge\frak c$).\ \QeD\  In particular,

\Centerline{$2^{\gamma_{n+1}}=\Theta(2^{\gamma_n},\gamma_{n+1}^+)
=\Theta(\frak c,\gamma_{n+1}^+)$}

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

\medskip

{\bf (c)} For each $n\in\Bbb N$, let $\lambda_n$ be the least infinite
cardinal such that $\Theta(\lambda_n,\gamma_n^+)>\frak c$.   Then
$\langle\lambda_n\rangle_{n\in\Bbb N}$ is non-increasing;  also
$\lambda_1\le\frak c$, because

\Centerline{$\Theta(\frak c,\gamma_1^+)=2^{\gamma_1}>\frak c$,}

\noindent so there are $n\ge 1$, $\lambda\le\frak c$ such that
$\lambda_m=\lambda$
for every $m\ge n$.   Now for $m\ge n$ we have

$$\eqalignno{\frak c
<\Theta(\lambda,\gamma_m^+)
&\le\max(\lambda,
(\sup_{\lambda'<\lambda}
  \Theta(\lambda',\gamma_m^+))^{\cf\lambda})\cr
\displaycause{5A2I}
&\le\max(\lambda,\frak c^{\cf\lambda})=2^{\cf\lambda}.\cr}$$

\noindent Also we still have
$\lambda\ge\kappa>\Theta(\gamma_n^+,\gamma_n^+)$ because
$\Theta(\lambda',\gamma_n^+)<\kappa\le\frak c$
for every $\lambda'<\kappa$.   Using 5A2H again,

\Centerline{$2^{\gamma_n}
=\Theta(\frak c,\gamma_n^+)
\le\Theta(\Theta(\lambda,\gamma_n^+),\gamma_n^+)
\le\Theta(\lambda,\gamma_n^+)
\le 2^{\cf\lambda}$;}

\noindent consequently

\Centerline{$2^{\gamma_n}<2^{\gamma_{n+1}}\le 2^{\cf\lambda}$}

\noindent and $\cf\lambda>\gamma_n$.   But 5A2Ia now tells us that

\Centerline{$\Theta(\lambda,\gamma_n^+)
\le\max(\lambda,\sup_{\lambda'<\lambda}\Theta(\lambda',\gamma_n^+))
\le\frak c$,}

\noindent which is absurd.\ \Bang

This contradiction proves the theorem.
}%end of proof of 542E

\leader{542F}{Corollary} Let $\kappa\le\frak c$ be a \qm\ cardinal.

(a) There is a regular infinite cardinal $\gamma<\kappa$ such that
$2^{\gamma}=2^{\delta}$ for every cardinal $\delta$ such that
$\gamma\le\delta<\kappa$;  that is, $\#([\kappa]^{<\kappa})=2^{\gamma}$.

(b) Let $\Cal I$ be any proper $\omega_1$-saturated $\kappa$-additive
ideal of $\Cal P\kappa$ containing singletons, and
$\frak A=\Cal P\kappa/\Cal I$.
If $\gamma$ is as in (a), then the
cardinal power $\tau(\frak A)^{\gamma}$ is equal to $2^{\kappa}$.

%\quad(ii) $\tau(\frak A)^{\omega}\le 2^{\kappa}\le 2^{\tau(\frak A)}$.

\proof{{\bf (a)} By 542E, there is a first $\gamma<\kappa$ such that
$2^{\delta}=2^{\gamma}$ whenever $\gamma\le\delta<\kappa$.   Of course
$\gamma$ is infinite;  by 5A1Fh it is regular.
Because $\kappa$ is regular,

\Centerline{$[\kappa]^{<\kappa}=\bigcup_{\xi<\kappa}\Cal P\xi$,
\quad
$\#([\kappa]^{<\kappa})=\max(\kappa,\sup_{\delta<\kappa}2^{\delta})
=\sup_{\delta<\kappa}2^{\delta}=2^{\gamma}$.}

\medskip

{\bf (b)} Of course $\tau(\frak A)\le\#(\frak A)\le\#(\Cal P\kappa)$, so
$\tau(\frak A)^{\gamma}\le(2^{\kappa})^{\gamma}=2^{\kappa}$.   In the
other direction, we have an injective function
$\phi_{\xi}:\Cal P\xi\to\Cal P\gamma$ for each $\xi<\kappa$.   For
$A\subseteq\kappa$ and $\eta<\gamma$ set

\Centerline{$d_{A\eta}
=\{\xi:\xi<\kappa$, $\eta\in\phi_{\xi}(A\cap\xi)\}^{\ssbullet}
\in\frak A$.}

\noindent If $A$, $B\subseteq\kappa$ are distinct then there is a
$\zeta<\kappa$ such that $\phi_{\xi}(A\cap\xi)\ne\phi_{\xi}(B\cap\xi)$
for every $\xi\ge\zeta$, that is,

\Centerline{$\bigcup_{\eta<\gamma}
  \{\xi:\eta\in\phi_{\xi}(A\cap\xi)\symmdiff\phi_{\xi}(B\cap\xi)\}
\supseteq\kappa\setminus\zeta\notin\Cal I$.}

\noindent Because $\Cal I$ is $\kappa$-additive and $\gamma<\kappa$,
there is an $\eta<\gamma$ such that
$\{\xi:\eta\in\phi_{\xi}(A\cap\xi)\symmdiff\phi_{\xi}(B\cap\xi)\}
\notin\Cal I$, that is, $d_{A\eta}\ne d_{B\eta}$.   Thus
$A\mapsto\ofamily{\eta}{\gamma}{d_{A\eta}}:
\Cal P\kappa\to\frak A^{\gamma}$ is injective, and
$2^{\kappa}\le\#(\frak A)^{\gamma}$.   But $\frak A$ is ccc, so
$\#(\frak A)\le\max(4,\tau(\frak A)^{\omega})$ (514De) and

\Centerline{$2^{\kappa}\le(\tau(\frak A)^{\omega})^{\gamma}
=\tau(\frak A)^{\gamma}$.}
}%end of proof of 542F

\leader{542G}{Corollary} Suppose that $\kappa$ is a \qm\ cardinal.

(a) If $\kappa\le\frak c<\kappa^{(+\omega_1)}$\cmmnt{ (notation:
5A1F(a-ii))}, then
$2^{\lambda}\le\frak c$ for every cardinal $\lambda<\kappa$.

(b) Let $\Cal I$ be any proper $\omega_1$-saturated $\kappa$-additive
ideal of $\Cal P\kappa$ containing singletons, and
$\frak A=\Cal P\kappa/\Cal I$.   If $2^{\lambda}\le\frak c$ for every
cardinal $\lambda<\kappa$, then $\#(\frak A)=2^{\kappa}$.

\proof{{\bf (a)} \Quer\ Otherwise, taking $\gamma$ as in 542Fa,
$2^{\gamma}>\frak c$.   Let $\gamma_1\le\gamma$ be the first cardinal
such that $2^{\gamma_1}>\frak c$;  note that, using 542E and 5A1Fh,
we can be sure that $\gamma_1$ is regular.   Next,

\Centerline{$\covSh(\kappa,\kappa,\gamma_1^+,\gamma_1)
\le\cff[\kappa]^{<\kappa}=\kappa$,}

\noindent by 5A2Ea.   So
$\covSh(\alpha,\kappa,\gamma_1^+,\gamma_1)\le\alpha$ whenever
$\kappa\le\alpha<\kappa^{(+\gamma_1)}$ (induce on $\alpha$, using
5A2Eb).
In particular, $\covSh(\frak c,\kappa,\gamma_1^+,\gamma_1)\le\frak c$.
But 541S tells us that
$\covSh(\frak c,\kappa,\gamma_1^+,\gamma_1)=2^{\gamma_1}$.\ \Bang

\medskip

{\bf (b)} By 541B, $\frak A$ is Dedekind complete;  by 515L,
$\#(\frak A)^{\omega}=\#(\frak A)$;  so 542Fb tells us that

\Centerline{$2^{\kappa}=\tau(\frak A)^{\omega}\le\#(\frak A)
\le\#(\Cal P\kappa)=2^{\kappa}$.}
}%end of proof of 542G

\leader{542H}{Lemma} %RVMC S7A
Let $\kappa$ be a \qm\ cardinal and
$\langle\alpha_i\rangle_{i\in I}$ a countable family of ordinals less
than $\kappa$ and of cofinality at least $\omega_2$.   Then there is a
set $F\subseteq P=\prod_{i\in I}\alpha_i$ such that

(i) $F$ is cofinal with $P$;

(ii) if $\langle f_{\xi}\rangle_{\xi<\omega_1}$
is a non-decreasing family in $F$ then
$\sup_{\xi<\omega_1}f_{\xi}\in F$;

(iii) $\#(F)<\kappa$.

\proof{ Note that $\add P=\min_{i\in I}\cf\alpha_i>\omega_1$ (I am
passing over the trivial case $I=\emptyset$), so
$\sup_{\xi<\omega_1}f_{\xi}$ is defined in $P$ for every family
$\ofamily{\xi}{\omega_1}{f_{\xi}}$ in $P$.   We have

\Centerline{$\cf P
=\cf(\prod_{i\in I}\cf\alpha_i)
\le\Theta(\sup_{i\in I}(\cf\alpha_i)^+,\omega_1)
<\kappa$,}

\noindent by 542Db.   So
we can find a cofinal set $F_0\subseteq P$ of
cardinal less than $\kappa$.   Now for $0<\zeta\le\omega_2$ define
$F_{\zeta}$ by saying that

\Centerline{$F_{\zeta+1}=\{\sup_{\xi<\omega_1}f_{\xi}:
\langle f_{\xi}\rangle_{\xi<\omega_1}$ is a non-decreasing family in
$F_{\zeta}\}$,}

\Centerline{$F_{\zeta}=\bigcup_{\eta<\zeta}F_{\eta}$ for non-zero limit
ordinals $\zeta\le\omega_2$.}

\noindent Then $\#(F_{\zeta})<\kappa$ for every $\zeta$.   \Prf\ Induce
on $\zeta$.   For the inductive step to $\zeta+1$, \Quer\ suppose, if
possible, that $\#(F_{\zeta})<\kappa$ but $\#(F_{\zeta+1})\ge\kappa$.
Then there is a proper $\omega_1$-saturated $\kappa$-additive
$\Cal I\normalsubgroup\Cal PF_{\zeta+1}$ containing singletons.
For each $h\in F_{\zeta+1}$ choose a non-decreasing family
$\langle f_{h\xi}\rangle_{\xi<\omega_1}$ in $F_{\zeta}$ with supremum
$h$.   The set $h[I]$ of values of $h$ is a countable subset of
$Y=\bigcup_{i\in I}\alpha_i$, and $\#(Y)<\kappa$.   By 541D, there is a
set $H\subseteq F_{\zeta+1}$, with cardinal $\kappa$, such that
$M=\bigcup_{h\in H}h[I]$ is countable.   Now, for each $h\in H$, there
is a $\gamma(h)<\omega_1$ such that whenever $i\in I$ and $\beta\in M$
then $h(i)>\beta$ iff
$f_{h,\gamma(h)}(i)>\beta$.  If $g$, $h\in H$ and $i\in I$ and
$g(i)<h(i)$,
then $f_{g,\gamma(g)}(i)\le g(i)<f_{h,\gamma(h)}(i)$, because
$g(i)\in M$.
Thus $h\mapsto f_{h,\gamma(h)}:H\to F_{\zeta}$ is injective;  but
$\#(F_{\zeta})<\kappa=\#(H)$.\ \Bang

Thus $\#(F_{\zeta+1})<\kappa$ if $\#(F_{\zeta})<\kappa$.   At limit
ordinals $\zeta$ the induction proceeds without difficulty because
$\cf\kappa>\zeta$.\ \Qed

So $\#(F_{\omega_2})<\kappa$ and we may take $F=F_{\omega_2}$.
}%end of proof of 542H

\leader{542I}{Theorem}\cmmnt{ ({\smc Shelah 96})} %RVMC S7B
Let $\kappa$ be a \qm\ cardinal.

(a) For any cardinal $\theta$, $\cff[\kappa]^{<\theta}\le\kappa$.

(b) For any cardinal $\lambda<\kappa$, and any $\theta$,
$\cff[\lambda]^{<\theta}<\kappa$.

\proof{{\bf (a)(i)} Consider first the case $\theta=\omega_1$.
Write $G_1$ for the set of ordinals less than $\kappa$ of cofinality
less than or equal to $\omega_1$;  for $\delta\in G_1$  let
$\psi_{\delta}:\cf\delta\to\delta$ enumerate
a cofinal subset of $\delta$.   Next, write
$G_2$ for $\kappa\setminus G_1$, and for every countable set
$A\subseteq G_2$ let
$F(A)\subseteq\prod_{\alpha\in A}\alpha$ be a cofinal set, with cardinal
less than $\kappa$, closed under
suprema of non-decreasing families of length $\omega_1$;  such exists by
542H above.

\medskip

\quad{\bf (ii)} It is worth observing at this point that if
$\langle A_{\zeta}\rangle_{\zeta<\omega_1}$ is any family of countable
subsets of $G_2$, $D=\bigcup_{\zeta<\omega_1}A_{\zeta}$, and
$g\in\prod_{\alpha\in D}\alpha$, then there is an
$f\in\prod_{\alpha\in D}\alpha$ such that
$f\ge g$ and $f\restr A_{\zeta}\in F(A_{\zeta})$ for every
$\zeta<\omega_1$.   \Prf\ Let $\langle\phi(\xi)\rangle_{\xi<\omega_1}$
run over $\omega_1$ with cofinal repetitions.   Choose a
non-decreasing family $\langle f_{\xi}\rangle_{\xi<\omega_1}$ in
$\prod_{\alpha\in D}\alpha$ in such a way that $f_0=g$ and
$f_{\xi+1}\restr A_{\phi(\xi)}\in F(A_{\phi(\xi)})$ for every $\xi$;
this is possible because $\add(\prod_{\alpha\in D})\ge\omega_2$ and
$F(A)$ is cofinal with $\prod_{\alpha\in A}\alpha$ for every $A$.   Set
$f=\sup_{\xi<\omega_1}f_{\xi}$;  this
works because every $F(A)$ is closed under suprema of non-decreasing
families of length $\omega_1$.\ \Qed

\medskip

\quad{\bf (iii)} We can now find a family $\Cal A$ of countable subsets
of $\kappa$ such that

\qquad$(\alpha)$ $\{\alpha\}\in\Cal A$ for every $\alpha<\kappa$;

\qquad$(\beta)$ whenever $A$, $A'\in\Cal A$ and $\zeta<\omega_1$ then
$A\cup A'$, $A\cap G_2$ and $\{\psi_{\alpha}(\xi):
\alpha\in A\cap G_1,\,\xi<\min(\zeta,\cf\alpha)\}$ all belong to
$\Cal A$;

\qquad$(\gamma)$ whenever $A\in\Cal A\cap[G_2]^{\le\omega}$ and
$f\in F(A)$ then $f[A]\in\Cal A$;

\qquad$(\delta)$ $\#(\Cal A)\le\kappa$.

\medskip

\quad{\bf (iv)} \Quer\ Suppose, if possible, that
$\cff[\kappa]^{\le\omega}>\kappa$.   Because
$[\kappa]^{\le\omega}=\bigcup_{\lambda<\kappa}[\lambda]^{\le\omega}$,
there is a cardinal $\lambda<\kappa$ such that
$\cff[\lambda]^{\le\omega}>\kappa$.   We can therefore choose
inductively a family $\langle a_{\xi}\rangle_{\xi<\kappa}$ of countable
subsets of $\lambda$ such that

\Centerline{$a_{\xi}\not\subseteq\bigcup_{\eta\in A\cap\xi}a_{\eta}$}

\noindent whenever $\xi<\kappa$ and $A\in\Cal A$.   By 541D, there is a
set $W\subseteq\kappa$, with cardinal $\kappa$, such that
$\bigcup_{\xi\in W}a_{\xi}$ is countable.   Let $\delta<\kappa$ be such
that $W\cap\delta$ is cofinal with $\delta$ and of order type
$\omega_1$;  then $\delta\in G_1$.

\medskip

\quad{\bf (v)} I choose a family
$\langle A_{k\zeta}\rangle_{\zeta<\omega_1,k\in\Bbb N}$ in $\Cal A$ as
follows.    Start by setting
$A_{0\zeta}=\psi_{\delta}[\zeta]$ for every $\zeta<\omega_1$;  then
$A_{0\zeta}\in\Cal A$ by (iii)($\alpha$-$\beta$).   Given
$\langle A_{k\zeta}\rangle_{\zeta<\omega_1}$, set
$A'_{k\zeta}=A_{k\zeta}\cap G_2$ for each $\zeta<\omega_1$.   For
$\alpha\in D_k=\bigcup_{\zeta<\omega_1}A'_{k\zeta}$, set
$g_k(\alpha)=\sup(\alpha\cap W\cap\delta)<\alpha$;
choose $f_k\in\prod_{\alpha\in D_k}\alpha$ such that $g_k\le f_k$ and
$f_k\restr A'_{k\zeta}\in F(A'_{k\zeta})$ for every $\zeta$;  this is
possible by (ii) above.   Set

\Centerline{$A_{k+1,\zeta}
=A_{k\zeta}\cup f_k[A'_{k\zeta}]\cup\{\psi_{\alpha}(\xi):
\alpha\in A_{k\zeta}\cap G_1,\,\xi<\min(\zeta,\cf\alpha)\}
\in\Cal A$}

\noindent for each $\zeta<\omega_1$, and continue.
An easy induction on $k$ shows that
$\langle A_{k\zeta}\rangle_{\zeta<\omega_1}$ is non-decreasing for every
$k$;  also, every $A_{k\zeta}$ is a subset of $\delta$.

\medskip

\quad{\bf (vi)} Set $V_k=\bigcup_{\zeta<\omega_1}A_{k\zeta}$,
$b_k=\bigcup\{a_{\xi}:\xi\in W\cap V_k\}$;  then $b_k$ is countable and
there is a $\beta(k)<\omega_1$ such that
$b_k=\bigcup\{a_{\xi}:\xi\in W\cap A_{k,\beta(k)}\}$.   Now
$\bigcup_{k\in\Bbb N}A_{k,\beta(k)}$ is a
countable subset of $\delta$, so there is a member $\gamma$ of
$W\cap\delta$ greater than its supremum.   We have

\Centerline{$a_{\gamma}
\not\subseteq\bigcup\{a_{\eta}:\eta\in A_{k,\beta(k)}\}$,}

\noindent so $a_{\gamma}\not\subseteq b_k$ and
$\gamma\notin V_k$, for each $k$.

Set $V=\bigcup_{k\in\Bbb N}V_k$.   We have just seen that
$W\cap\delta\not\subseteq V$;  set
$\gamma_0=\min(W\cap\delta\setminus V)$.   Because
$V_0=\psi_{\delta}[\omega_1]$ is cofinal with $\delta$,
$V\setminus\gamma_0\ne\emptyset$;  let $\gamma_1$ be its least member.
Then $\gamma_1>\gamma_0$.   There must be $k\in\Bbb N$ and
$\zeta<\omega_1$ such that $\gamma_1\in A_{k\zeta}$.   Observe
that if $\alpha\in V\cap G_1$ then $V\cap\alpha$ is cofinal with
$\alpha$;  but $V\cap\gamma_1\subseteq\gamma_0$, so $\gamma_1\notin G_1$
and $\gamma_1\in A'_{k\zeta}\subseteq D_k$.   But now
$f_k(\gamma_1)\in A_{k+1,\zeta}\subseteq V$ and
$\gamma_0\le g_k(\gamma_1)\le f_k(\gamma_1)<\gamma_1$, so
$\gamma_1\ne\min(V\setminus\gamma_0)$.\ \Bang

\medskip

\quad{\bf (vii)} This contradiction shows that
$\cff[\kappa]^{\le\omega}\le\kappa$.   Now consider
$\cff[\kappa]^{\le\delta}$, where $\delta<\kappa$ is an infinite
cardinal.   Then

$$\eqalignno{\covSh(\kappa,\delta^+,\delta^+,\omega_1)
&=\max(\kappa,
  \sup_{\lambda<\kappa}\covSh(\lambda,\delta^+,\delta^+,\omega_1))\cr
\displaycause{5A2Eb again}
&\le\kappa\cr}$$

\noindent by 542Dc.   So there
is a family $\Cal B\subseteq [\kappa]^{\le\delta}$, with cardinal at most
$\kappa$,  such that every member of $[\kappa]^{\le\delta}$ is covered
by a sequence in $\Cal B$.   But now
$\cff[\Cal B]^{\le\omega}\le\kappa$, so there is a family $\frak C$ of
countable subsets of $\Cal B$ which is cofinal with
$[\Cal B]^{\le\omega}$ and with cardinal at most $\kappa$;  setting
$\Cal D=\{\bigcup\Cal C:\Cal C\in\frak C\}$, we have $\Cal D$ cofinal
with $[\kappa]^{\le\delta}$ and with cardinal at most $\kappa$.   So
$\cff[\kappa]^{\le\delta}\le\kappa$.

Finally, of course,
$[\kappa]^{<\theta}=\bigcup_{\delta<\theta}[\kappa]^{\le\delta}$, so

\Centerline{$\cff[\kappa]^{<\theta}
\le\sup_{\delta<\theta}\cff[\kappa]^{\le\delta}\le\kappa$}

\noindent whenever $\theta\le\kappa$.    For
$\theta>\kappa$ we have $\cff[\kappa]^{<\theta}=1$, so
$\cff[\kappa]^{<\theta}\le\kappa$ for every $\theta$.

\medskip

{\bf (b)} If $\Cal A$ is cofinal with $[\kappa]^{<\theta}$ then
$\{A\cap\lambda:A\in\Cal A\}$ is cofinal with $[\lambda]^{<\theta}$, so
$\cff[\lambda]^{<\theta}\le\cff[\kappa]^{\theta}
\le\kappa$, by (a).

\Quer\ If $\cff[\lambda]^{<\theta}=\kappa$, there is a cofinal family
$\ofamily{\xi}{\kappa}{A_{\xi}}$ in $[\lambda]^{<\theta}$ such that
$A_{\xi}\not\subseteq A_{\eta}$ for any $\eta<\xi<\kappa$.   Of course
$\omega<\theta\le\lambda<\kappa=\cf\kappa$, so we may suppose that every
$A_{\xi}$ is infinite.   So there is an infinite $\delta<\theta$ such
that $E=\{\xi:\xi<\kappa$, $\#(A_{\xi})=\delta\}$ has cardinal $\kappa$.
Next, by 541D applied to a suitable ideal of subsets of $E$, there is a
set $M\in[\lambda]^{\le\delta}$ such that
$F=\{\xi:\xi\in E$, $A_{\xi}\subseteq M\}$ has cardinal $\kappa$.   But
now there must be an $\eta<\kappa$ such that $M\subseteq A_{\eta}$, and
a $\xi\in F$ such that $\xi>\eta$;  which is impossible.\ \Bang

Thus $\cff[\lambda]^{<\theta}<\kappa$, as claimed.
}%end of proof of 542I

\leader{542J}{Corollary} %RVMC S7C
Let $\kappa$ be a \qm\ cardinal.   Let
$\langle P_{\zeta}\rangle_{\zeta<\lambda}$ be a family of partially
ordered sets
such that $\lambda<\add P_{\zeta}\le\cf P_{\zeta}<\kappa$ for every
$\zeta<\lambda$.   Then $\cf(\prod_{\zeta<\lambda}P_{\zeta})<\kappa$.

\proof{ For each $\zeta<\lambda$ let $Q_{\zeta}$ be a
cofinal subset of $P_{\zeta}$ with cardinal less than $\kappa$.   Set
$P=\prod_{\zeta<\kappa}P_{\zeta}$, $Z=\bigcup_{\zeta<\lambda}Q_{\zeta}$;
then $\#(Z)<\kappa$ so $\cff[Z]^{\le\lambda}<\kappa$, by 542Ib.   Let
$\Cal A$ be a cofinal subset of $[Z]^{\le\lambda}$ with
$\#(\Cal A)<\kappa$.   For each $A\in\Cal A$ choose $f_A\in P$ such that
$f_A(\zeta)$ is an upper bound for $A\cap P_{\zeta}$ for every $\zeta$;
this is possible because $\add P_{\zeta}>\#(A)$.   Set
$F=\{f_A:A\in\Cal A\}$.

If $g\in P$, then there is an $h\in\prod_{\zeta<\lambda}Q_{\zeta}$ such
that $g\le h$.   Now $h[\lambda]\in[Z]^{\le\lambda}$ so there is an
$A\in\Cal A$ such that $h[\lambda]\subseteq A$.   In this case
$h\le f_A$.   Accordingly $F$ is cofinal with $P$ and
$\cf P\le\#(F)<\kappa$, as required.
}%end of proof of 542J

\leader{542K}{}\dvAnew{2015}\cmmnt{ For an application in \S547 below,
%547R prf g
it will be useful to have the following variant of 542I.

\medskip

\noindent}{\bf Proposition} Let $\kappa$ be a \qm\ cardinal.

(a) For every cardinal $\theta<\kappa$
there is a family $\Cal D_{\theta}$ of countable sets, with cardinal
less than $\kappa$, which is stationary over $\theta$\cmmnt{ in
the sense of 5A1R}.

(b) There is a family $\Cal A$ of countable sets, with cardinal
at most $\kappa$, which is stationary over $\kappa$.
%548B

\proof{ Fix on a
proper $\kappa$-additive $\omega_1$-saturated ideal $\Cal I$ of
subsets of $\kappa$ containing all singleton subsets of $\kappa$.
Write $\Cal F$ for $\{F:F\subseteq\kappa$,
$\kappa\setminus F\in\Cal I\}$.

\medskip

{\bf (a)(i)} By 542Ib,
there is a cofinal subset $\Cal C$ of $[\theta]^{\le\omega}$ with
cardinal less than $\kappa$.   Set

\Centerline{$\Cal A=\{\bigcup_{\alpha<\omega_1}C_{\alpha}:
   \ofamily{\alpha}{\omega_1}{C_{\alpha}}$ is a non-decreasing family
   in $\Cal C\}$.}

\noindent Note that $\Cal A\subseteq[\theta]^{\le\omega_1}$.   Now
$\Cal A$ is stationary over $\theta$.   \Prf\
If $f:[\theta]^{<\omega}\to[\theta]^{\le\omega}$ is a function, choose
$\ofamily{\alpha}{\omega_1}{C_{\alpha}}$ in $\Cal C$ such that

\Centerline{$C_{\alpha}\supseteq\bigcup_{\beta<\alpha}C_{\beta}
   \cup\bigcup\{f(I):I\in[\bigcup_{\beta<\alpha}C_{\beta}]^{<\omega}\}$}

\noindent for $\alpha<\omega_1$.   Then
$A=\bigcup_{\alpha<\omega_1}C_{\alpha}$ belongs to $\Cal A$ and
$f(I)\subseteq A$ for every $I\in[A]^{<\omega}$.   As $f$ is arbitrary,
$\Cal A$ is stationary over $\theta$.\ \Qed

\medskip

\quad{\bf (ii)} \Quer\ Suppose, if possible, that $\#(\Cal A)\ge\kappa$.
Let $\ofamily{\xi}{\kappa}{A_{\xi}}$ be a family of distinct elements of
$\Cal A$.   For each $\xi<\kappa$ let
$\ofamily{\alpha}{\omega_1}{C_{\xi\alpha}}$ be a non-decreasing family
in $\Cal C$ such that $A_{\xi}=\bigcup_{\alpha<\omega_1}C_{\xi\alpha}$.
By 541D, there is a set $A\in[\theta]^{\le\omega_1}$ such that
$D=\{\xi:\xi<\kappa$, $A_{\xi}\subseteq A\}$ belongs to $\Cal F$.
Let $\ofamily{\alpha}{\omega_1}{B_{\alpha}}$ be a non-decreasing family of
countable sets with union $A$.

\medskip

\quad{\bf (iii)} For each $\alpha<\omega_1$ there is a countable set
$\Cal B_{\alpha}$ such that
$D_{\alpha}=\{\xi:\xi\in D$, $A_{\xi}\cap B_{\alpha}\in\Cal B_{\alpha}\}$
belongs to $\Cal F$.
\Prf\ For each $\xi\in D$ there is a $\gamma_{\xi}<\omega_1$ such that
$A_{\xi}\cap B_{\alpha}=C_{\xi\gamma_{\xi}}\cap B_{\alpha}$.
As $\#(\Cal C)<\kappa$, there is a countable set
$\Cal D$ such that
$D\setminus\{\xi:C_{\xi\gamma_{\xi}}\in\Cal D\}\in\Cal I$;  set
$\Cal B_{\alpha}=\{C\cap B_{\alpha}:C\in\Cal D\}$.\ \Qed

Set $D'=\bigcap_{\alpha<\omega_1}D_{\alpha}$;  then $D'\in\Cal F$.

\medskip

\quad{\bf (iv)} Whenever $\xi$, $\eta\in D'$ are distinct there is an
$\alpha<\omega_1$ such $A_{\xi}\cap B_{\alpha}\ne A_{\eta}\cap B_{\alpha}$.
There is therefore, for each $\xi\in D'$, a $\beta_{\xi}<\omega_1$ such
that $\{\eta:\eta\in D'$,
$A_{\xi}\cap B_{\beta_{\xi}}\ne A_{\eta}\cap B_{\beta_{\xi}}\}
\in\Cal F$.   Next, there is a $\beta<\omega_1$ such that
$D''=\{\xi:\xi\in D'$, $\beta_{\xi}\le\beta\}$ belongs to $\Cal I$.
However, because $\Cal B_{\beta}$ is countable, there is a
$B\in\Cal B_{\beta}$ such that $E=\{\xi:\xi\in D''$,
$A_{\xi}\cap B_{\beta}=B\}$ does not belong to $\Cal I$.   But now, if we
take any $\xi\in E$, there must be an $\eta\in E$
such that $A_{\xi}\cap B_{\beta_{\xi}}\ne A_{\eta}\cap B_{\beta_{\xi}}$, in
which case $A_{\xi}\cap B_{\beta}$ and $A_{\eta}\cap B_{\beta}$ are
distinct and cannot both be equal to $B$.\ \Bang

Thus $\#(\Cal A)<\kappa$.

\medskip

\quad{\bf (v)} Finally, observe that for each $A\in\Cal A$ there
is a family of countable sets, with cardinal at most $\omega_1$, which is
stationary over $A$.   (If $A$ is countable, we can take $\{A\}$;
otherwise, use 5A1Sc.)   Taking the union of these, we get a family of
countable sets, with cardinal less than $\kappa$, which is stationary over
$\theta$ by 5A1Sb.

\medskip

{\bf (b)} Because $\kappa$ is a limit cardinal of uncountable cofinality,
the set of cardinals less than $\kappa$ is
stationary over $\kappa$, by 5A1Sc.   Putting this together with (a) and
5A1Sb, we see that there is a
family of countable sets, with cardinal at most $\kappa$, which is stationary
over $\kappa$.
}%end of proof of 542K

\exercises{\leader{542X}{Basic exercises (a)}
%\spheader 542Xa
Let $\kappa$ be a quasi-measurable cardinal, and $\theta$ a cardinal
such that $2\le\theta\le\kappa$.   Show that
$\cff[\kappa]^{<\theta}=\kappa$.
%542I

\leader{542Y}{Further exercises (a)}
%\spheader 542Ya
Let $X$ be a \hwtr\ topological space such that there is no
quasi-measurable cardinal less than or equal to the weight of $X$, and
$\mu$ a totally finite Maharam submeasure on the Borel $\sigma$-algebra of
$X$.   (i) Show that $\mu$ is $\tau$-subadditive in the sense that if
whenever $\Cal G$ is a non-empty
upwards-directed family of open sets in $X$ with union $H$, then
$\inf_{G\in\Cal G}\mu(H\setminus G)=0$.
(ii) Show that if $X$ is Hausdorff and K-analytic, then the completion of
$\mu$ is a Radon submeasure on $X$.
%4{}38M 4{}96J

}%end of exercises

\endnotes{
\Notesheader{542} The arguments of this section have taken on a certain
density, and I ought to explain what they are for.   The cardinal
arithmetic of 542E-542G is relevant to one of the most important
questions in this chapter, to be treated in the next section:  supposing
that there is an extension of Lebesgue measure to a measure $\mu$
defined on all subsets of $\Bbb R$, what can we say about the Maharam
type of $\mu$?   And 542I-542J will tell us something about the
cofinalities of our favourite partially ordered sets under the same
circumstances.

Let me draw your attention to a useful trick, used twice above.   If
$\kappa$ is a quasi-measurable cardinal, and $X$ is any set with cardinal
at least $\kappa$, there is a non-trivial
$\omega_1$-saturated $\sigma$-ideal of subsets of $X$.   This is the
basis of the proof of 542H (taking $X=F_{\zeta+1}$ in the inductive
step) and the final step in the proof of 542I (taking $X=E$).   Exposed
like this, the idea seems obvious.   In the thickets of an argument it
sometimes demands an imaginative jump.

}%end of notes

\discrpage



\frfilename{mt543.tex}
\versiondate{11.11.13}
\copyrightdate{2004}

\def\chaptername{Real-valued-measurable cardinals}
\def\sectionname{The Gitik-Shelah theorem}

\newsection{543}

I come now to the leading case at the centre of the work of the last two
sections.   If our $\omega_1$-saturated $\sigma$-ideal of sets is the
null ideal of a measure with domain $\Cal PX$, it has some even more
striking properties than those already discussed.   I will go farther into
these later in the chapter.   But I will begin with what is known about one
of the first questions I expect a reader of this book to ask:   if
$(X,\Cal PX,\mu)$ is a probability space, what can, or must, its measure
algebra be?   There can, of course, be a purely atomic part;  the
interesting question relates to the atomless part, if any, always
remembering that we need a special act of faith to believe that there can
be an atomless case.   Here we find that the Maharam type of an atomless
probability defined on a power set must be greater than its additivity
(543F), which must itself be `large' (541L).

\leader{543A}{Definitions (a)} A {\bf\rvm\ cardinal} is an uncountable
cardinal $\kappa$ such that there is a $\kappa$-additive probability
measure $\nu$ on $\kappa$, defined on every subset of $\kappa$, for which
all singletons are negligible.   In this context I will call $\nu$ a
{\bf witnessing probability}.

\spheader 543Ab If $\kappa$ is a regular uncountable cardinal, a
probability measure $\nu$ on $\kappa$ with domain $\Cal P\kappa$ is
{\bf normal} if its null ideal $\Cal N(\nu)$ is normal.   In this
case,\cmmnt{ $\nu$ must be $\kappa$-additive (541H, 521Ad)
and zero on singletons,
so $\kappa$ is \rvm, and} I will say that $\nu$ is a {\bf normal
witnessing probability}.

\spheader 543Ac An {\bf\am\ cardinal} is a \rvm\ cardinal with an
atomless witnessing probability.

\leader{543B}{}\cmmnt{ Collecting ideas which have already
appeared, some of them more than once, we have the following.

\medskip

\noindent}{\bf Proposition} (a) Let $(X,\Cal PX,\mu)$ be a totally finite
measure space in which singletons are negligible and $\mu X>0$.
Then $\kappa=\add\mu$ is \rvm,
and there are a non-negligible $Y\subseteq X$ and a function
$g:Y\to\kappa$ such that the normalized image measure
$B\mapsto\Bover1{\mu Y}\mu g^{-1}[B]$ is a normal witnessing probability
on $\kappa$.

(b) Every \rvm\ cardinal is \qm\cmmnt{ (definition: 542A)} and has a
normal witnessing
probability;  in particular, every \rvm\ cardinal is uncountable and
regular.

(c) If $\kappa\le\frak c$ is a \rvm\ cardinal, then $\kappa$ is \am, and
every witnessing probability on $\kappa$ is atomless.

(d) If $\kappa>\frak c$ is a \rvm\ cardinal, then $\kappa$ is \2vm, and
every witnessing probability on $\kappa$ is purely atomic.

(e) A cardinal $\lambda$ is measure-free\cmmnt{ (definition: 438A)}
iff there is no \rvm\ cardinal $\kappa\le\lambda$;  $\frak c$ is
measure-free iff there is no \am\ cardinal.

(f) Again suppose that $(X,\Cal PX,\mu)$ is a totally finite
measure space.

\quad(i) If $\mu$ is purely atomic, $\add\mu$ is either $\infty$ or
a \2vm\ cardinal.

\quad(ii) If $\mu$ is not purely atomic, $\add\mu$ is \am.

\proof{{\bf (a)} By 521Ad again,
$\kappa$ is the additivity of the null ideal
$\Cal N(\mu)$ of $\mu$;  because $\mu$ is
$\sigma$-finite, $\Cal N(\mu)$ is $\omega_1$-saturated;  and of course
$\kappa\ge\omega_1$.   By 541J, there are $Y\subseteq X$ and
$g:Y\to\kappa$ such that
$\Cal I=\{B:B\subseteq\kappa$, $\mu g^{-1}[B]=0\}$
is a normal ideal on $\kappa$.   In particular, $\kappa\notin\Cal I$ and
$Y$ is non-negligible.   Set $\nu B=\Bover1{\mu Y}\mu g^{-1}[B]$ for
$B\subseteq\kappa$;  then $\nu$ is a probability measure with domain
$\Cal P\kappa$.   Its null ideal $\Cal N(\nu)=\Cal I$ is normal, so it is
a normal measure and witnesses that $\kappa$ is \rvm.

\medskip

{\bf (b)} If $\kappa$ is \rvm, then (a) tells us that any witnessing
probability on $\kappa$ can be used to define a normal witnessing
probability $\nu$ say.   Since $\kappa$ is the additivity of a
$\sigma$-ideal, it must be uncountable and regular (513C(a-i));  also
$\Cal N(\nu)$ is an $\omega_1$-saturated normal ideal, so $\kappa$ is
\qm.

\medskip

{\bf (c)-(d)} Apply 541P.   If $\kappa$ is \rvm, it is a regular
uncountable cardinal, and if $\nu$ is a witnessing probability on $\kappa$
then $\Cal N(\nu)$ is a proper $\omega_1$-saturated $\kappa$-additive
ideal of subsets of $\kappa$.    Taking $\frak A$ to be the measure
algebra of $\nu$, then 541P tells us that either $\frak A$ is atomless
and $\kappa\le\frak c$, or $\frak A$ is purely atomic and $\kappa$ is
\2vm, in which case $\kappa$ is surely greater than $\frak c$ (541N).
Turning this round, if $\kappa\le\frak c$ then $\frak A$ and $\nu$ must
be atomless and $\kappa$ is \am, while if $\kappa>\frak c$ then
$\frak A$ and $\nu$ are purely atomic and $\kappa$ is \2vm.

\medskip

{\bf (e)} If $\kappa\le\lambda$ is \rvm, then any witnessing probability
on $\kappa$ extends to a probability measure with domain $\Cal P\lambda$
which is zero on singletons, so $\lambda$ is not measure-free.   If
$\lambda$ is not measure-free, let $\mu$ be a probability measure with
domain $\Cal P\lambda$ which is zero on singletons;  then (a) tells us
that $\add\mu$ is \rvm, and $\add\mu\le\lambda$ because
$\lambda=\bigcup\Cal N(\mu)$.

If there is an \am\ cardinal $\kappa$, then $\kappa$ is \rvm\ and
there is an atomless witnessing probability on $\kappa$, and
$\kappa\le\frak c$, by (d).   So in this case $\frak c$ is not
measure-free.   On the other hand, if $\frak c$ is not measure-free, there
is a \rvm\ cardinal $\kappa\le\frak c$, which is \am, by (c).

\medskip

{\bf (f)(i)} If $\mu$ is purely atomic, and $\add\mu$ is not $\infty$,
set $\kappa=\add\mu$ and let
$\ofamily{\xi}{\kappa}{A_{\xi}}$ be a family of negligible sets in $X$ with
non-negligible union $A$.   Let $E\subseteq A$ be an atom for $\mu$.
Repeating the construction of (a),
but starting from the subspace $(E,\Cal PE,\mu_E)$,
we see that the normalized image
measure constructed on $\kappa=\add\mu_E$ can take only the two values $0$
and $1$, so that its null
ideal is $2$-saturated and witnesses that $\kappa$ is \2vm.

\medskip

\quad{\bf (ii)} If $\mu$ is not purely atomic, let $E$ be the atomless part of
$X$, so that $\mu_E$ is atomless and $\mu_{X\setminus E}$ is purely atomic.
Singletons in $E$ must be
negligible, so (a) tells us that $\add\mu_E$ is a \rvm\ cardinal;
also there is an \imp\
function from $E$ to $[0,\mu E]$ (343Cc), so $E$ can be covered by
$\frak c$ negligible sets and
$\add\mu_E\le\frak c$ is \am, by (c) here.   Now (i) tells us that
$\frak c\le\add\mu_{X\setminus E}$, so
$\add\mu=\min(\add\mu_E,\add\mu_{X\setminus E})=\add\mu_E$ is \am.
}%end of proof of 543B

\cmmnt{\medskip

\noindent{\bf Remark} 543Bc-543Bd are {\bf Ulam's dichotomy}.
}

\vleader{72pt}{543C}{Theorem}\cmmnt{ (see {\smc Kunen n70})}
Suppose that $(Y,\Cal PY,\nu)$ is a
$\sigma$-finite measure space and that $(X,\frak T,\Sigma,\mu)$ is a
$\sigma$-finite quasi-Radon measure space with $w(X)<\add\nu$.   Let
$f:X\times Y\to[0,\infty]$ be any function.   Then

\Centerline{$\overlineint\bigl(\int f(x,y)\nu(dy)\bigr)\mu(dx)
\le\int\bigl(\overline{\intop}f(x,y)\mu(dx)\bigr)\nu(dy)$.}

\proof{ Because $\mu$ is $\sigma$-finite and effectively locally finite,
there is a sequence of open sets of finite measure with conegligible
union in $X$.   Since none of the integrals are changed by deleting a
negligible subset of $X$, and the weight of any subset of $X$ is at most
the weight of $X$, we may suppose that this conegligible union is
$X$ itself, so that $\mu$ is outer regular with respect to the open sets
(412Wb).   Set $\lambda=w(X)<\add\nu$;  let
$\langle G_{\xi}\rangle_{\xi<\lambda}$ enumerate a base for the topology
of $X$.   Fix $\epsilon>0$.   Because $\nu$ is $\sigma$-finite, we have
a function $y\mapsto\epsilon_y:Y\to\ooint{0,\infty}$ such that
$\int\epsilon_y\nu(dy)\le\epsilon$.   For each $y\in Y$, let
$h_y:X\to[0,\infty]$ be a lower semi-continuous function such that
$f(x,y)\le h_y(x)$ for every $x\in X$ and

\Centerline{$\int h_y(x)\mu(dx)
\le\epsilon_y+\overline{\intop}f(x,y)\mu(dx)$}

\noindent (412Wa).   For $I\subseteq\lambda$, $x\in X$ and $y\in Y$, set

\Centerline{$f_I(x,y)
=\sup(\{0\}\cup\{s:\enskip\Exists\xi\in I$, $x\in G_{\xi}$,
  $h_y(x')\ge s\Forall x'\in G_{\xi}\})$.}

\noindent Then $f_I$ is expressible as
$\sup_{\xi\in I,s\in\Bbb Q^+}s\chi(G_{\xi}\times B_{\xi s})$,
writing $\Bbb Q^+$ for the set of non-negative rational numbers and
$B_{\xi s}$ for $\{y:h_y\ge s\chi G_{\xi}\}$.
So $f_I$ is $(\Sigma\tensorhat\Cal PY)$-measurable for all countable
$I$, and for such $I$ we shall have

\Centerline{$\iint f_I(x,y)\mu(dx)\nu(dy)
=\iint f_I(x,y)\nu(dy)\mu(dx)$,}

\noindent by Fubini's theorem (252C).   Next, for any $I\subseteq\lambda$,
$x\mapsto f_I(x,y)$ is lower semi-continuous for each $y$, and

\Centerline{$\sup_{I\in[\lambda]^{<\omega}}f_I(x,y)=h_y(x)$}

\noindent for all $x\in X$ and $y\in Y$, because each $h_y$
is lower semi-continuous.   So

\Centerline{$\sup_{I\in[\lambda]^{<\omega}}\int f_I(x,y)\mu(dx)
=\int h_y(x)\mu(dx)$}

\noindent for each $y\in Y$ (414Ba).   Because
$\lambda<\add\nu$, it follows that

\Centerline{$\sup_{I\in[\lambda]^{<\omega}}
  \iint f_I(x,y)\mu(dx)\nu(dy)
=\iint h_y(x)\mu(dx)\nu(dy)$}

\noindent (521B(d-i)).  On the other hand, if we write

\Centerline{$g_I(x)=\int f_I(x,y)\nu(dy)$}

\noindent for $x\in X$ and finite $I\subseteq\lambda$, then
$g_I$ also is lower semi-continuous.   \Prf\ If $x\in X$, set
$J=\{\xi:\xi\in I$, $x\in G_{\xi}\}$ and
$H=X\cap\bigcap_{\xi\in J}G_{\xi}$;  then $f_I(x,y)\le f_I(x',y)$ whenever
$x'\in H$ and $y\in Y$, so $g_I(x)\le g_I(x')$ for $x'\in H$, while
$x\in\interior H$.\ \QeD\   So $g=\sup_{I\in[\lambda]^{<\omega}}g_I$
is lower semi-continuous, and $\int g(x)\mu(dx)
=\sup_{I\in[\lambda]^{<\omega}}\int g_I(x)\mu(dx)$.   Also

\Centerline{$g(x)=\sup_{I\in[\lambda]^{<\omega}}\int f_I(x,y)\nu(dy)
=\int h_y(x)\nu(dy)\ge\int f(x,y)\nu(dy)$}

\noindent for every $x\in X$.   So we have

$$\eqalign{\overline{\int}\mskip-10mu\int f(x,y)\nu(dy)\mu(dx)
&\le\int g(x)\mu(dx)
=\sup_{I\in[\lambda]^{<\omega}}\int g_I(x)\mu(dx)\cr
&=\sup_{I\in[\lambda]^{<\omega}}\iint f_I(x,y)\nu(dy)\mu(dx)\cr
&=\sup_{I\in[\lambda]^{<\omega}}\iint f_I(x,y)\mu(dx)\nu(dy)\cr
&=\iint h_y(x)\mu(dx)\nu(dy)
\le\epsilon+\int\overline{\int}f(x,y)\mu(dx)\nu(dy).\cr}$$

\noindent As $\epsilon$ is arbitrary, we have the result.
}%end of proof of 543C

\cmmnt{\medskip

\noindent{\bf Remark} Compare 537O.}

\vleader{72pt}{543D}{Corollary} Let $\kappa$ be a \rvm\ cardinal, with
witnessing probability $\nu$, and $(X,\frak{T},\Sigma,\mu)$
a totally finite quasi-Radon measure space with $w(X)<\kappa$.

(a) If $C\subseteq X\times\kappa$ then

\Centerline{$\overlineint\nu C[\{x\}]\mu(dx)
\le\int\mu^*C^{-1}[\{\xi\}]\nu(d\xi)$.}

(b) If $A\subseteq X$ and $\#(A)\le\kappa$, then there is
a $B\subseteq A$ such that $\#(B)<\kappa$ and $\mu^*B=\mu^*A$.

(c) If $\langle C_{\xi}\rangle_{\xi<\kappa}$ is a
family in $\Cal PX\setminus\Cal N(\mu)$ such that
$\#(\bigcup_{\xi<\kappa}C_{\xi})<\kappa$, then there are distinct $\xi$,
$\eta<\kappa$ such that $\mu^*(C_{\xi}\cap C_{\eta})>0$.

(d) If we have a family $\langle h_{\xi}\rangle_{\xi<\kappa}$ of
functions such that $\dom h_{\xi}$ is a non-negligible subset of
$X$ for each $\xi$ and $\#(\bigcup_{\xi<\kappa}h_{\xi})<\kappa$\cmmnt{
(identifying each $h_{\xi}$ with its graph)},
then there are distinct $\xi$, $\eta<\kappa$ such that

\Centerline{$\mu^*\{x:x\in\text{dom}(h_{\xi})\cap\text{dom}(h_{\eta}),
  \,h_{\xi}(x)=h_{\eta}(x)\}>0$.}

\proof{{\bf (a)} Apply 543C to $\chi C:X\times\kappa\to\Bbb R$.

\medskip

{\bf (b)}\Quer\ Suppose, if possible, otherwise.
Then surely $\#(A)=\kappa$;  let $f:\kappa\to A$ be a bijection.   Set

\Centerline{$C=\{(f(\eta),\xi):\eta\le\xi<\kappa\}
\subseteq X\times\kappa$.}

\noindent If $x\in A$,

\Centerline{$\nu C[\{x\}]=\nu\{\xi:f^{-1}(x)\le\xi<\kappa\}=1$,}

\noindent so $\overline{\int}\nu C[\{x\}]\mu(dx)=\mu^*A$.
If $\xi<\kappa$,

\Centerline{$\mu^*C^{-1}[\{\xi\}]=\mu^*\{f(\eta):\eta\le\xi\}<\mu^*A$,}

\noindent so $\int\mu^*C^{-1}[\{\xi\}]\nu(d\xi)<\mu^*A$.
But this contradicts (a).\ \Bang

\medskip

{\bf (c)} Let $\tilde\nu$ be the probability on $\kappa\times\kappa$
defined by writing

\Centerline{$\tilde\nu A=\int\nu A[\{\xi\}]\,\nu(d\xi)$ for every
$A\subseteq\kappa\times\kappa$.}

\noindent Then $\tilde\nu$ is $\kappa$-additive, by 521B(d-ii).   Set

$$\eqalign{C
&=\{(x,(\xi,\eta)):\xi,\,\eta\text{ are distinct members of }\kappa,\,
  x\in C_{\xi}\cap C_{\eta}\}\cr
&\subseteq X\times(\kappa\times\kappa).\cr}$$

\noindent Set

\Centerline{$E=\{x:x\in X$, $\nu\{\xi:x\in C_{\xi}\}=0\}$.}

\noindent Because $\#(\bigcup_{\xi<\kappa}C_{\xi})<\kappa$,

\Centerline{$\{\xi:E\cap C_{\xi}\ne\emptyset\}
=\bigcup\{\{\xi:x\in C_{\xi}\}:x\in E\cap\bigcup_{\eta<\kappa}C_{\eta}\}$}

\noindent is $\nu$-negligible, and
there is a $\xi<\kappa$ with $C_{\xi}\cap E=\emptyset$;
thus $\mu^*(X\setminus E)>0$.   Now if $x\in X\setminus E$ then

\Centerline{$\tilde\nu\{(\xi,\eta):(x,(\xi,\eta))\in C\}
=(\nu\{\xi:x\in C_{\xi}\})^2>0$.}

\noindent So we have

$$\eqalign{0
&<\overline{\int}\tilde\nu\{(\xi,\eta):(x,(\xi,\eta))\in C\}\mu(dx)\cr
&\le\int\mu^*\{x:(x,(\xi,\eta))\in C\}\tilde\nu(d(\xi,\eta))\cr}$$

\noindent by 543C, and there must be distinct $\xi$, $\eta<\kappa$
such that $\mu^*\{x:(x,(\xi,\eta))\in C\}>0$, as required.

\medskip

{\bf (d)} Set $Y=\bigcup_{\xi<\kappa}h_{\xi}[X]$.
Give $X\times Y$ the measure $\tilde\mu$ and topology
$\frak T'$ defined as follows.   The domain of $\tilde\mu$ is to
be the family $\tilde\Sigma$ of subsets $H$ of
$X\times Y$ for which there are $E$, $E'\in\Sigma$ with
$\mu(E'\setminus E)=0$ and
$E\times Y\subseteq H\subseteq E'\times Y$;
and for such $H$, $\tilde\mu H$
is to be $\mu E=\mu E'$.   The topology $\frak{T}'$ is
to be just the family $\{G\times Y:G\in\frak{T}\}$.
It is easy to check that
$(X\times Y,\frak{T}',\tilde\Sigma,\tilde\mu)$
is a totally finite quasi-Radon measure space of weight less than
$\kappa$, and that $\tilde\mu^*h_{\xi}=\mu^*(\dom h_{\xi})>0$
for each $\xi<\kappa$.   So (c) gives the result.
}%end of proof of 543D

\leader{543M}{Lemma}\dvAnew{2020} 
Let $\kappa\le\frak c$ be a quasi-measurable cardinal and 
$\lambda<\min(\kappa^{(+\omega)},2^{\kappa})$ an infinite cardinal.
Set $\zeta=\max(\lambda^+,\kappa^+)$.   

(a) We have an infinite cardinal
$\delta<\kappa$, a stationary set $S\subseteq\zeta$, and a family
$\family{\alpha}{S}{g_{\alpha}}$ of functions from $\kappa$ to
$2^{\delta}$ such that $g_{\alpha}[\kappa]\subseteq\alpha$ for every
$\alpha\in S$ and $\#(g_{\alpha}\cap g_{\beta})<\kappa$ for distinct
$\alpha$, $\beta\in S$.   Moreover, we can arrange that

----- if $\lambda<\Tr(\kappa)$ (definition:  5A1Mb), then
$g_{\alpha}[\kappa]\subseteq\kappa$ for every $\alpha\in S$;

----- if $\lambda\ge\Tr(\kappa)$, then
$g_{\alpha}\restr\gamma=g_{\beta}\restr\gamma$ whenever $\gamma<\kappa$
is a limit ordinal and $\alpha$, $\beta\in S$ are such that
$g_{\alpha}(\gamma)=g_{\beta}(\gamma)$.

(b) Now suppose that $S_1\subseteq S$ is stationary in $\zeta$ and
$\family{\alpha}{S_1}{\theta_{\alpha}}$ is a family of limit ordinals
less than $\kappa$.
Then there are a $\theta<\kappa$ and a set
$Y\in[2^{\delta}]^{<\kappa}$ such that
$S_2=\{\alpha:\alpha\in S_1$, $\theta_{\alpha}=\theta$,
$g_{\alpha}[\theta]\subseteq Y\}$ is stationary in $\zeta$.

\proof{{\bf (a) case 1} If $\lambda<\Tr(\kappa)$, then $\zeta\le\Tr(\kappa)$
(5A1Na);  as $\zeta$ is a successor cardinal, there is a family
$\ofamily{\alpha}{\zeta}{g_{\alpha}}$ of functions from $\kappa$ to
$\kappa$ such that $\#(g_{\alpha}\cap g_{\beta})<\kappa$ for all
distinct $\alpha$, $\beta<\zeta$.   Set $S=\zeta\setminus\kappa$, so
that $S$ is a stationary set in $\zeta$ and
$g_{\alpha}[\kappa]\subseteq\alpha$ for every $\alpha\in S$.
We know that $\kappa\le\frak c$;  set $\delta=\omega$, so that
$\delta<\kappa\le 2^{\delta}$ and $g_{\alpha}$ is a function from
$\kappa$ to $2^{\delta}$ for every $\alpha$.

\medskip

\quad{\bf case 2} Suppose that $\lambda\ge\Tr(\kappa)$.   Then

\Centerline{$\kappa<\Tr(\kappa)<\lambda^+=\zeta
\le\min(2^{\kappa},\kappa^{(+\omega)})$,}

\noindent so $\sup_{\delta<\kappa}2^{\delta}\ge\zeta$, by 5A1Nb;
and as this supremum is attained (542E),
there is a cardinal $\delta<\kappa$ such that
$2^{\delta}\ge\zeta$.   Because
$\kappa<\lambda<\kappa^{(+\omega)}$,
$\lambda$ is regular, and of course $\lambda>\omega_1$.
So 5A1P gives us the functions we need.

\medskip

{\bf (b)} Because $\zeta=\cf\zeta>\kappa$, there is a $\theta<\kappa$
such that

\Centerline{$S'_1=\{\alpha:\alpha\in S_1$, $\theta_{\alpha}=\theta\}$}

\noindent is stationary in $\zeta$, by the Pressing-Down Lemma,
and of course $\theta$ is a limit ordinal.

\medskip

\quad{\bf case 1} If $\lambda<\Tr(\kappa)$, then $g_{\alpha}[\theta]$
is a subset of $\kappa$, and is therefore bounded above in $\kappa$, for
each $\alpha$.   Let $\theta'<\kappa$ be such that

\Centerline{$S_2
=\{\alpha:\alpha\in S'_1$, $g_{\alpha}[\theta]\subseteq\theta'\}$}

\noindent is stationary in $\zeta$.   As $\kappa\le\frak c\le 2^{\delta}$,
$\theta'\in[2^{\delta}]^{<\kappa}$ and we can take $Y=\theta'$.

\medskip

\quad{\bf case 2} If $\lambda\ge\Tr(\kappa)$, then
$g_{\alpha}(\theta)<\alpha$ for $\alpha\in S'_1$;  so
there is a $\theta'<\zeta$ such that

\Centerline{$S''_1
=\{\alpha:\alpha\in S_1$, $g_{\alpha}(\theta)=\theta'\}$}

\noindent is stationary in $\zeta$.   Then
$g_{\alpha}\restr\theta=g_{\beta}\restr\theta$ for all $\alpha$,
$\beta\in S''_1$;  take $Y$
to be the common value of $g_{\alpha}[\theta]$ for
$\alpha\in S''_1$.
}%end of proof of 543M


\leader{543E}{The Gitik-Shelah theorem}\dvAformerly{5{}43E}
\cmmnt{ ({\smc Gitik \& Shelah
89}, {\smc Gitik \& Shelah 93})} Let $\kappa$ be an
\am\ cardinal, with witnessing probability $\nu$.   Then the Maharam
type of $\nu$ is at least $\min(\kappa^{(+\omega)},2^{\kappa})$.

\proof{{\bf (a)} To begin with (down to the end of (g) below) let us
suppose that $\nu$ is \Mth, with Maharam type $\lambda$;
of course $\lambda$ is infinite, because $\nu$ is atomless.   Let
$(\frak A,\bar\nu)$ be the measure algebra of $\nu$,
$\nu_{\lambda}$ the usual measure of $\{0,1\}^{\lambda}$ and
$\frak B_{\lambda}$ the measure algebra of
$\nu_{\lambda}$;  then there is a
measure-preserving isomorphism $\phi:\frak B_{\lambda}\to\frak A$.
Because $\nu_{\lambda}$ is a compact measure (342Jd), there is a
function $f:\kappa\to\{0,1\}^{\lambda}$ such that
$\phi(E^{\ssbullet})=f^{-1}[E]^{\ssbullet}$ whenever $\nu_{\lambda}$
measures $E$ (343B).

\medskip

{\bf (b)}\Quer\ Suppose, if possible, that
$\lambda<\min(\kappa^{(+\omega)},2^{\kappa})$.

Of course $\kappa$ is quasi-measurable and at most $\frak c$.
So 543Ma tells us that if we set $\zeta=\max(\lambda^+,\kappa^+)$, there will
be an infinite cardinal
$\delta<\kappa$, a stationary set $S\subseteq\zeta$, and a family
$\family{\alpha}{S}{g_{\alpha}}$ of functions from $\kappa$ to
$2^{\delta}$ such that $g_{\alpha}[\kappa]\subseteq\alpha$ for every
$\alpha\in S$ and $\#(g_{\alpha}\cap g_{\beta})<\kappa$ for distinct
$\alpha$, $\beta\in S$.   Moreover, we can arrange that

----- if $\lambda<\Tr(\kappa)$ (definition:  5A1Mb), then
$g_{\alpha}[\kappa]\subseteq\kappa$ for every $\alpha\in S$;

----- if $\lambda\ge\Tr(\kappa)$, then
$g_{\alpha}\restr\gamma=g_{\beta}\restr\gamma$ whenever $\gamma<\kappa$
is a limit ordinal and $\alpha$, $\beta\in S$ are such that
$g_{\alpha}(\gamma)=g_{\beta}(\gamma)$.

\medskip

{\bf (c)} Fix an injective function $h:2^{\delta}\to\{0,1\}^{\delta}$.
For $\alpha\in S$, $\iota<\delta$ set

\Centerline{$U_{\alpha\iota}
=\{\xi:\xi<\kappa$, $(hg_{\alpha}(\xi))(\iota)=1\}$,}

\noindent and choose a Baire set
$H_{\alpha\iota}\subseteq\{0,1\}^{\lambda}$ such that
$\phi^{-1}(U^{\ssbullet}_{\alpha\iota})=H^{\ssbullet}_{\alpha\iota}$ in
$\frak B_{\lambda}$.   Define
$\tilde g_{\alpha}:\{0,1\}^{\lambda}\to\{0,1\}^{\delta}$ by setting

$$\eqalign{(\tilde g_{\alpha}(x))(\iota)
&=1\text{ if }x\in H_{\alpha\iota},\cr
&=0\text{ otherwise}.\cr}$$

\noindent Then

$$\eqalign{\{\xi:
  \xi<\kappa,\,\tilde g_{\alpha}f(\xi)\ne hg_{\alpha}(\xi)\}
&=\bigcup_{\iota<\delta}\{\xi:
  (\tilde g_{\alpha}f(\xi))(\iota)\ne(hg_{\alpha}(\xi))(\iota)\}\cr
&=\bigcup_{\iota<\delta}
  U_{\alpha\iota}\symmdiff f^{-1}[H_{\alpha\iota}]
\in\Cal N(\nu)\cr}$$

\noindent because $\delta<\kappa=\add\Cal N(\nu)$.
Set $V_{\alpha}=\{\xi:\tilde g_{\alpha}f(\xi)=hg_{\alpha}(\xi)\}$, so
that $\nu V_{\alpha}=1$, for each $\alpha\in S$.

\medskip

{\bf (d)} Because every $H_{\alpha\iota}$ is a Baire
set, there is for each $\alpha\in S$ a set $I_{\alpha}\subseteq\lambda$
such that $\#(I_{\alpha})\le\delta$ and $H_{\alpha\iota}$ is determined
by coordinates in $I_{\alpha}$ for every $\iota<\delta$, that is,
$\tilde g_{\alpha}(x)=\tilde g_{\alpha}(y)$ whenever $x$,
$y\in\{0,1\}^{\lambda}$ and $x\restr I_{\alpha}=y\restr I_{\alpha}$.
Because $\lambda<\cf\zeta$, there is an $M\subseteq\lambda$ such that

\Centerline{$S_1=\{\alpha:\alpha\in S$, $I_{\alpha}\subseteq M\}$}

\noindent is stationary in $\zeta$ and $\cf(\#(M))\le\delta$ (5A1K);
because $\lambda<\kappa^{(+\omega)}$ and $\cf(\kappa)=\kappa>\delta$,
$\#(M)<\kappa$.   Set $\pi_M(z)=z\restr M$ for $z\in\{0,1\}^{\lambda}$,
and $f_M=\pi_Mf$, so that $f_M:\kappa\to\{0,1\}^M$ is \imp\ for
$\nu$ and the usual measure $\nu_M$ of $\{0,1\}^M$.
For $w\in\{0,1\}^M$ define $\psi(w)\in\{0,1\}^{\lambda}$ by setting

$$\eqalign{\psi(w)(\xi)
&=w(\xi)\text{ if }\xi\in M,\cr
&=0\text{ otherwise}.\cr}$$

\noindent If we set

\Centerline{$g^*_{\alpha}
=\tilde g_{\alpha}\psi:\{0,1\}^M\to\{0,1\}^{\delta}$,}

\noindent then $g^*_{\alpha}$ is Baire measurable in each coordinate, while
$g^*_{\alpha}\pi_M=\tilde g_{\alpha}$ for $\alpha\in S_1$.

\medskip

{\bf (e)} For each $\alpha\in S_1$, there is a $\theta_{\alpha}<\kappa$
such that $\mu^*_M(f_M[V_{\alpha}\cap\theta_{\alpha}])=1$.
\Prf\ Apply 543Db to $f_M[V_{\alpha}]\subseteq\{0,1\}^M$.   There
must be a set $B\subseteq f_M[V_{\alpha}]$ such that $\#(B)<\kappa$ and
$\mu_M^*B=\mu_M^*(f_M[V_{\alpha}])$;  because $\kappa$ is regular,
there is a $\theta_{\alpha}<\kappa$ such that
$B\subseteq f_M[V_{\alpha}\cap\theta_{\alpha}]$.   On the other hand,
because $f_M$ is \imp,
$\mu_M^*(f_M[V_{\alpha}])\ge\nu V_{\alpha}=1$.\ \Qed

Evidently we may take it that every $\theta_{\alpha}$ is a non-zero
limit ordinal.

\medskip

{\bf (f)} Now 543Mb tells us that
there are a $\theta<\kappa$ and a $Y\in[2^{\delta}]^{<\kappa}$
such that

\Centerline{$S_2=\{\alpha:\alpha\in S_1$, $\theta_{\alpha}=\theta$,
$g_{\alpha}[\theta]\subseteq Y\}$}

\noindent is stationary in $\zeta$.

\medskip

{\bf (g)} For each $\alpha\in S_2$, set

\Centerline{$Q_{\alpha}=f_M[V_{\alpha}\cap\theta]
=f_M[V_{\alpha}\cap\theta_{\alpha}]$,}

\noindent so that $\mu^*_MQ_{\alpha}=1$.  If $y\in Q_{\alpha}$, take
$\xi\in V_{\alpha}\cap\theta$ such that $f_M(\xi)=y$;  then

\Centerline{$g_{\alpha}^*(y)=g_{\alpha}^*\pi_Mf(\xi)
=\tilde g_{\alpha}f(\xi)=hg_{\alpha}(\xi)\in h[Y]$.}

\noindent Thus
$g_{\alpha}^*\restr Q_{\alpha}\subseteq f_M[\theta]\times h[Y]$ for
every $\alpha\in S_2$, and we can apply
543Dd to $X=\{0,1\}^M$, $\mu=\mu_M$ and the family
$\langle g^*_{\alpha}\restr Q_{\alpha}\rangle_{\alpha\in S'}$,
where $S'\subseteq S_2$ is a set with cardinal $\kappa$,
to see that there are distinct $\alpha$, $\beta\in S_2$ such that
$\mu^*_M\{y:y\in Q_{\alpha}\cap Q_{\beta}$,
$g^*_{\alpha}(y)=g^*_{\beta}(y)\}>0$.
Now, however, consider

\Centerline{$E=\{y:y\in\{0,1\}^M$, $g^*_{\alpha}(y)=g^*_{\beta}(y)\}$.}

\noindent Then $E=\bigcap_{\iota<\delta}E_{\iota}$, where

\Centerline{$E_{\iota}
=\{y:y\in\{0,1\}^M$, $g^*_{\alpha}(y)(\iota)=g^*_{\beta}(y)(\iota)\}$}

\noindent is a Baire subset of $\{0,1\}^M$ for each $\iota<\delta$.
Because $\delta<\kappa$,

$$\eqalign{\nu f_M^{-1}[E]
&=\nu(\bigcap_{\iota<\delta}f_M^{-1}[E_{\iota}])
=\inf_{I\in[\delta]^{<\omega}}
  \nu(\bigcap_{\iota\in I}f_M^{-1}[E_{\iota}])\cr
&=\inf_{I\in[\delta]^{<\omega}}\mu_M(\bigcap_{\iota\in I}E_{\iota})
\ge\mu_M^*E
>0.}$$

\noindent Consequently

$$\eqalign{0
&<\nu f^{-1}_M[E]
=\nu\{\xi:g^*_{\alpha}\pi_M f(\xi)=g^*_{\beta}\pi_M f(\xi)\}\cr
&=\nu\{\xi:\tilde g_{\alpha}f(\xi)=\tilde g_{\beta}f(\xi)\}
=\nu\{\xi:\xi\in V_{\alpha}\cap V_{\beta},
  \,\tilde g_{\alpha}f(\xi)=\tilde g_{\beta}f(\xi)\}\cr
&=\nu\{\xi:hg_{\alpha}(\xi)=hg_{\beta}(\xi)\}
=\nu\{\xi:g_{\alpha}(\xi)=g_{\beta}(\xi)\}\cr}$$

\noindent (because $h$ is injective).   But this is
absurd, because in (b) above we chose
$g_{\alpha}$, $g_{\beta}$ in such a way that
$\{\xi:g_{\alpha}(\xi)=g_{\beta}(\xi)\}$ would be bounded in $\kappa$.\
\BanG\

\medskip

{\bf (h)} Thus the result is true for \Mth\ witnessing probabilities on
$\kappa$.   In general, if $\nu$ is any witnessing probability on
$\kappa$, there is a non-negligible $A\subseteq\kappa$ such that the
subspace measure on $A$ is \Mth;  setting
$\nuprime C=\Bover1{\nu A}\nu(A\cap C)$ for $C\subseteq\kappa$, we
obtain a
\Mth\ witnessing probability $\nuprime$.   Now the Maharam type of $\nu$
is at least as great as the Maharam type of $\nuprime$, so is at least
$\min(2^{\kappa},\kappa^{(+\omega)})$, as required.
}%end of proof of 543E

\leader{543F}{Theorem}\dvAformerly{5{}43F} Let $(X,\Cal PX,\mu)$ be an atomless
semi-finite measure space.   Write $\kappa=\add\mu$.   Then the
Maharam type
of $(X,\Cal PX,\mu)$ is at least $\min(\kappa^{(+\omega)},2^{\kappa})$,
and in particular is greater than $\kappa$.

\proof{ Let $\ofamily{\xi}{\kappa}{E_{\xi}}$ be a family in
$\Cal N(\mu)$ such that
$E=\bigcup_{\xi<\kappa}E_{\xi}\notin\Cal N(\mu)$.   Let $F\subseteq E$
be a set of non-zero finite measure.
Set $f(x)=\min\{\xi:x\in E_{\xi}\}$ for $x\in F$.   Let $\mu_F$ be the
subspace measure on $F$ and $\mu'=(\mu F)^{-1}\mu_F$ the corresponding
probability measure;  of course $\dom\mu'=\Cal PF$ and
$\mu'(F\cap E_{\xi})=0$ for every $\xi<\kappa$.   Note also that

\Centerline{$\add\mu'=\add\mu_F\ge\add\mu\ge\kappa$}

\noindent (521Fc).   Let $\nu$ be the image measure $\mu'f^{-1}$, so
that $\dom\nu=\Cal P\kappa$ and $\nu$ is $\kappa$-additive (521Hb).
Also $\nu\{\xi\}\le\mu'E_{\xi}=0$ for every $\xi$, so $\nu$ witnesses
that $\kappa$ is \rvm.   Next, $\mu_F$ is atomless (214Ka), so $\mu'$
also is.   There is therefore a function $g:F\to[0,1]$ which is \imp\
for $\mu'$ and Lebesgue measure (343Cb), and $F$ can be covered by
$\frak c$ negligible sets;  accordingly $\add\mu'\le\frak c$ so
$\kappa\le\frak c$ and $\nu$ must be atomless (543Bc).

Let $(\frak A,\bar\mu)$, $(\frak A',\bar\mu')$ and $(\frak B,\bar\nu)$
be the measure algebras of $\mu$, $\mu'$ and $\nu$ respectively.   Then
$\frak A'$ is isomorphic to a principal ideal of $\frak A$ (322I), so
$\tau(\frak A)\ge\tau(\frak A')$ (514Ed).   Next, $f:F\to\kappa$ induces
a measure-preserving Boolean homomorphism from $\frak B$ to $\frak A'$,
so that $\tau(\frak A')\ge\tau(\frak B)$ (332Tb).   Now 543E tells us
that

\Centerline{$\min(\kappa^{(+\omega)},2^{\kappa})
\le\tau(\frak B)\le\tau(\frak A)$,}

\noindent as required.
}%end of proof of 543F

\leader{543G}{Corollary}\dvAformerly{5{}43G} Let $(X,\Cal PX,\nu)$ be an
atomless probability space, and $\kappa=\add\nu$.   Let
$(Z,\Sigma,\mu)$ be a compact probability space with Maharam type
$\lambda\le\min(2^{\kappa},\kappa^{(+\omega)})$ (e.g.,
$Z=\{0,1\}^{\lambda}$ with its usual measure).   Then there is an \imp\
function $f:X\to Z$.

\proof{ Let $(\frak A,\bar\mu)$ and $\frak B,\bar\nu)$ be the measure
algebras of $\mu$, $\nu$ respectively.   By 543F, the Maharam type of
the subspace measure $\nu_C$ is at least $\lambda$ whenever
$C\subseteq X$ and $\nu C>0$;  that is, every non-zero principal ideal
of $\frak B$ has Maharam type at least $\lambda$.   So there is
a measure-preserving Boolean homomorphism from $\frak A$ to $\frak B$
(332P).   Because $\mu$ is compact, this is represented by an \imp\
function from $X$ to $Z$ (343B).
}%end of proof of 543G

\leader{543H}{Corollary}\dvAformerly{5{}43H} If $\kappa$ is an \am\ cardinal, and
$(Z,\mu)$ is a compact probability space with Maharam type at most
$\min(2^{\kappa},\kappa^{(+\omega)})$, then there is an extension of
$\mu$ to a $\kappa$-additive measure defined on $\Cal PZ$.

\proof{ Let $\nu$ be a witnessing probability on $\kappa$;  by 543G,
there is an \imp\ function $f:X\to Z$;  now the image measure
$\nu f^{-1}$ extends $\mu$ to $\Cal PZ$.
}%end of proof of 543H

\leader{543I}{Corollary}\dvAformerly{5{}43I} If $\kappa$ is an \am\ cardinal, with
witnessing probability $\nu$, and $2^{\kappa}\le\kappa^{(+\omega)}$,
then $(\kappa,\Cal P\kappa,\nu)$ is \Mth, with
Maharam type $2^{\kappa}$.

\proof{ If $C\in\Cal P\kappa\setminus\Cal N(\nu)$, then
the Maharam type of the subspace measure on $C$ is at least
$2^{\kappa}$, by
543F;  but also it cannot be greater than $\#(\Cal PC)=2^{\kappa}$.
}%end of proof of 543I

\leader{543J}{Proposition}\dvAformerly{5{}43J} Let $\kappa$ be an \am\ cardinal, $\nu$ a
witnessing probability on $\kappa$, and $\frak A$ the measure algebra of
$\nu$.   Then

(a) there is a $\gamma<\kappa$ such that $2^{\gamma}=2^{\delta}$ for
every cardinal $\delta$ such that $\gamma\le\delta<\kappa$;

(b) the cardinal power $\tau(\frak A)^{\gamma}$ is $2^{\kappa}$;

(c) if $\frak c<\kappa^{(+\omega_1)}$, then
$\#(\frak A)=\tau(\frak A)^{\omega}=2^{\kappa}$.

%(d) if $\kappa=\frak c$ then $\tau(\frak A)=2^{\frakc}$.

\proof{ Use 542F-542G.   Because $\kappa$ is \qm\ and
$\kappa\le\frak c$, 542Fa tells us that there is a $\gamma$ as in (a);
and now 542Fb and 515M tell us that

\Centerline{$2^{\kappa}=\tau(\frak A)^{\gamma}
=(\tau(\frak A)^{\omega})^{\gamma}=\#(\frak A)^{\gamma}$.}

\noindent If $\frak c<\kappa^{(+\omega_1)}$, then 542G tells us
that $\#(\frak A)=2^{\kappa}$.
}%end of proof of 543J

\leader{543K}{Proposition}\dvAformerly{5{}43K} Let $\kappa$ be an \am\ cardinal.
If there is a witnessing
probability on $\kappa$ with Maharam type $\lambda$, then there is a
\Mth\
normal witnessing probability $\nu$ on $\kappa$ with Maharam type
at most $\lambda$.

\proof{ Repeat the proof of 543Ba, with $X=\kappa$ and $\mu$ a witnessing
probability on $\kappa$ with Maharam type $\lambda$.   Taking a
non-negligible $Y\subseteq\kappa$ and $g:Y\to\kappa$ such that
$\nu=\Bover1{\mu Y}\mu_Yg^{-1}$ is normal, then $g$ induces an
embedding of the measure algebra of $\nu$ into a principal ideal of the
measure algebra of $\mu$, so the Maharam type of $\nu$ is at most
$\lambda$.   There is now
an $E\in\Cal P\kappa\setminus\Cal N(\nu)$ such that
the subspace measure $\nu_E$ is \Mth, and
setting $\nuprime A=\nu(A\cap E)/\nu E$ for $A\subseteq\kappa$ we
obtain a \Mth\ probability measure $\nuprime$ with Maharam
type less than or equal to $\lambda$.   Now $\nuprime$ is again normal.
\Prf\ Let $\ofamily{\xi}{\kappa}{I_{\xi}}$ be any family in
$\Cal N(\nuprime)$, and set
$I=\{\xi:\xi<\kappa$, $\xi\in\bigcup_{\eta<\xi}I_{\eta}\}$.   Then
$I_{\xi}\cap E\in\Cal N(\nu)$ for every $\xi$, so
$I\cap E=\{\xi:\xi<\kappa$, $\xi\in\bigcup_{\eta<\xi}I_{\eta}\cap E\}$
is $\nu$-negligible and $I$ is $\nuprime$-negligible.\ \QeD\   So we
have an appropriate normal witnessing probability.
}%end of proof of 543K

\leader{543L}{Proposition}\dvAformerly{5{}43L} Suppose that $\nu$ is a \Mth\
witnessing probability on an \am\ cardinal
$\kappa$ with Maharam type $\lambda$.   Then there is a
\Mth\ witnessing probability $\nuprime$ on
$\kappa$ with Maharam type at least
$\Tr_{\Cal N(\nu)}(\kappa;\lambda)$.

\proof{ Let $\nu_1$ be the $\kappa$-additive probability on
$\kappa\times\kappa$ given by

\Centerline{$\nu_1 C=\int\nu C[\{\xi\}]\nu(d\xi)$ for every
$C\subseteq\kappa\times\kappa$.}

\noindent Set $\theta=\Tr_{\Cal N(\nu)}(\kappa;\lambda)$.
By 541F there is a family $F\subseteq\lambda^{\kappa}$
such that $\#(F)=\theta$ and $\{\xi:f(\xi)=g(\xi)\}\in\Cal N(\nu)$
for all distinct $f$, $g\in F$.   Let
$\langle E_{\xi}\rangle_{\xi<\lambda}$
be a $\nu$-stochastically independent family of subsets of $\kappa$
of $\nu$-measure $\bover12$.   For each $f\in F$ set

\Centerline{$C_f=\{(\xi,\eta):\xi<\kappa$, $\eta\in E_{f(\xi)}\}$.}

\noindent Then for any non-empty finite subset $I$ of
$F$, $\nu(\bigcap_{f\in I}E_{f(\xi)})=2^{-\#(I)}$ for
$\nu$-almost every $\xi$, so that

\Centerline{$\nu_1(\bigcap_{f\in I}C_f)=2^{-\#(I)}$.}

\noindent Thus $\langle C_f\rangle_{f\in F}$ is
stochastically independent
for $\nu_1$, and the Maharam type of the subspace measure
$(\nu_1)_C$ is at least
$\#(F)=\theta$ whenever $\nu_1C>0$.   Once again, take
$\nu_2C=\nu_1(C\cap D)/\nu_1 D$ for some $D$ for which
$(\nu_1)_D$ is \Mth, to obtain a \Mth\
$\kappa$-additive probability $\nu_2$ with Maharam type at least $\theta$.
Finally, of course, $\nu_2$ can be copied onto a
probability $\nuprime$ on $\kappa$, as asked for.
}%end of proof of 543L

%\leader{543N}{}
%\leader{543O}{}
%\leader{543P}{}
%\leader{543Q}{}
%\leader{543R}{}
%\leader{543S}{}
%\leader{543T}{}
%\leader{543U}{}
%\leader{543V}{}
%\leader{543W}{}


\exercises{\leader{543X}{Basic exercises (a)}
%\spheader 543Xa
Let $\kappa$ be an \am\ cardinal.
Show that the following are equiveridical:  (i) every witnessing
probability $\nu$ on $\kappa$ is \Mth\;  (ii) any two
witnessing probabilities on $\kappa$ have the same
Maharam type.
%543K

\spheader 543Xb Let $\kappa$ be an \am\ cardinal.
Show that the following are equiveridical:  (i) every normal witnessing
probability $\nu$ on $\kappa$ is \Mth\;  (ii) any two
normal witnessing probabilities on $\kappa$ have the same
Maharam type.
%543K

\spheader 543Xc Suppose that $\frak c$ is \am.   Show that there is a
\Mth\ normal witnessing probability on $\frak c$ with
Maharam type $2^{\frakc}$.   \Hint{542Ga, 5A1Nc.}
%543L

\spheader 543Xd\dvAnew{2017}
Let $\mu$ be Lebesgue measure on $\Bbb R$, and
$\theta=\bover12(\mu^*+\mu_*)$ the outer measure described in 413Xd.   Show
that $\mu$ is the measure defined from $\theta$ by Carath\'eodory's method.
\Hint{438Ym.}
%543F out of order query
\leader{543Y}{Further exercises (a)}
%\spheader 543Ya
Let $\nu$ be a witnessing probability on an
atomlessly-measurable cardinal $\kappa$ with Maharam type $\lambda$.
Let $F$ be the set of all functions $f\subseteq\kappa\times\lambda$ such
that $\dom f\notin\Cal N(\nu)$, and let $\theta$ be

$$\eqalign{\sup\{\#(F_0):F_0&\subseteq F,\,\{\xi:\xi\in\dom f\cap\dom g,
\,f(\xi)=g(\xi)\}\in\Cal N(\nu)\cr
&\mskip250mu\text{ for all distinct }f,\,g\in F_0\}.\cr}$$

\noindent Show that there is a witnessing probability $\nuprime$ on
$\kappa$ with Maharam type at least $\theta$.
%543L mt54bits
}%end of exercises

\leader{543Z}{Problems} Let $\kappa$ be an \am\ cardinal.

\spheader 543Za Must every witnessing
probability $\nu$ on $\kappa$ be \Mth?   \cmmnt{(See 555E.)}
%543I 543J

\spheader 543Zb Must every normal witnessing probability $\nu$ on $\kappa$
be \Mth?

\leaveitout{more questions:  if $\frak c$ is \am, must any witnessing
probability have Maharam type $2^{\frakc}$?

If $\kappa$ has a witnessing probability with Maharam type $\lambda$,
does it have a normal witnessing probability with the same type?
}%end of leaveitout

\endnotes{
\Notesheader{543} The results of 543I-543J leave a tantalizingly narrow
gap;  it seems possible that the Maharam type of a witnessing probability
on an \am\ cardinal $\kappa$ is determined by $\kappa$ (543Xa, 543Za).
If so,
there is at least a chance that there is a proof depending on no ideas
more difficult than those above.   To find a counter-example, however,
we may need not only to make some strong assumptions about the
potential existence of appropriate large cardinals, but also to find a new
method of constructing models with \am\ cardinals.   Possibly we get
a different question if we look at normal witnessing
probabilities (543Zb).   A positive answer to either part of 543Z
would have implications for transversal numbers (543L, 543Ya).

}%end of notes

\discrpage



	
\frfilename{mt544.tex}
\versiondate{31.12.13}
\copyrightdate{2004}


\def\chaptername{Real-valued-measurable cardinals}
\def\sectionname{Measure theory with an \am\ cardinal}

\def\CalUn{\Cal{U}\eurm{n}}

\newsection{544}

As is to be expected, a witnessing measure on a \rvm\ cardinal has some
striking properties, especially if it is normal.   What is less obvious is
that the mere existence of such a cardinal can have implications for
apparently unrelated questions in analysis.   In 544J, for instance, we see
that if there is any \am\ cardinal then we have a version of Fubini's
theorem, $\iint f(x,y)dxdy=\iint f(x,y)dydx$, for many functions $f$
on $\BbbR^2$ which are not jointly measurable.
In this section I explore results of this
kind.   We find that, in the presence of an \am\ cardinal, the
covering number of the Lebesgue null ideal is large (544B) while
its uniformity is small (544G-544H).
There is a second inequality on repeated integrals (544C) to add to the
one already given in 543C, and which tells us
something about measure-precalibers (544D);  I add a couple of variations
(544I-544J).    Next, I give a pair of
theorems (544E-544F) on a measure-combinatorial property of the filter of
conegligible sets of a normal witnessing measure.   Revisiting the theory
of Borel measures on metrizable spaces, discussed in \S438 on the
assumption that no \rvm\ cardinal was present, we find that there are some
non-trivial arguments applicable to spaces with non-measure-free
weight (544K-544L).

In \S541 I briefly mentioned `weakly compact' cardinals.
Two-valued-measurable cardinals
are always weakly compact;  \am\ cardinals never are;  but \am\ cardinals
may or may not have a significant combinatorial property which can be
regarded as a form of weak compactness (544M, 544Yc).   Finally,
I summarise what is known about the location of an \am\ cardinal on
Cicho\'n's diagram (544N).

\leader{544A}{Notation} I repeat some of my notational conventions.
For a measure $\mu$, $\Cal N(\mu)$ will be its null ideal.
For any set $I$, $\nu_I$ will be the usual measure on $\{0,1\}^I$,
$\Cal N_I=\Cal N(\nu_I)$ its null ideal and $(\frak B_I,\bar\nu_I)$ its
measure algebra.

\leader{544B}{Proposition} Let $\kappa$ be an \am\ cardinal.
If $(X,\Sigma,\mu)$ is any locally compact\cmmnt{ (definition:  342Ad)}
semi-finite measure space with $\mu X>0$, then
$\cov\Cal N(\mu)\penalty-100\ge\kappa$.

\proof{ By 521Lb, applied to any \Mth\ subspace of $X$ with non-zero finite
measure, it is enough to show that
$\cov\Cal N_{\lambda}\ge\kappa$ for every $\lambda$;  by
523F, we need look only at the case $\lambda=\kappa$.
Fix on an atomless $\kappa$-additive
probability $\nu$ with domain $\Cal P\kappa$.   By 543G there is an
\imp\ function $f:\kappa\to\{0,1\}^{\kappa}$.   So
$\cov\Cal N_{\kappa}\ge\cov\Cal N(\nu)=\kappa$, by 521Ha.
}%end of proof of 544B

\leader{544C}{Theorem}\cmmnt{ ({\smc Kunen n70})}
Let $\kappa$ be a \rvm\ cardinal
and $\nu$ a normal witnessing probability on $\kappa$;
let $(X,\mu)$ be a compact probability space and
$f:X\times\kappa\to\coint{0,\infty}$ any function.   Then

\Centerline{$\underlineint\bigr(\int f(x,\xi)\nu(d\xi)\bigl)\mu(dx)
\le\int\bigl(\overlineint f(x,\xi)\mu(dx)\bigr)\nu(d\xi)$.}

\proof{\Quer Suppose, if possible, otherwise.

\medskip

{\bf (a)} We are supposing that there is a $\mu$-integrable function
$g:X\to\Bbb R$ such that $0\le g(x)\le\int f(x,\xi)\nu(d\xi)$
for every $x\in X$ and

\Centerline{$\int g(x)\,\mu(dx)>\int\overlineint f(x,\xi)\mu(dx)
\nu(d\xi)$.}

\noindent We can suppose that $g$ is a simple function;  express it
as $\sum_{i=0}^nt_i\chi F_i$ where $(F_0,\ldots,F_n)$ is a partition of $X$
into measurable sets.   For any $\xi<\kappa$,

\Centerline{$\overlineint f(x,\xi)\mu(dx)
=\sum_{i=0}^n\overlineint f(x,\xi)\chi F_i(x)\mu(dx)$}

\noindent (133L).    So there must be some $i\le n$ such that

\Centerline{$t_i\mu F_i
>\int\overlineint f(x,\xi)\chi F_i(x)\mu(dx)\nu(d\xi)$.}

\noindent Set $Y=F_i$, $\mu_1=(\mu F_i)^{-1}\mu\restr\Cal PF_i$,
$t=t_i$;  then  $(Y,\mu_1)$ is a compact probability space (451Da) and

$$\eqalignno{\int\overline{\int}f(y,\xi)\mu_1(dy)\nu(d\xi)
&=\Bover1{\mu F_i}\int\overline{\int}f(x,\xi)\chi F_i(x)\mu(dx)\nu(d\xi)\cr
\displaycause{135Id}
&<t
\le\inf_{y\in Y}\int f(y,\xi)\nu(d\xi).\cr}$$

\medskip

{\bf (b)} Let $(\frak A,\bar\mu_1)$ be the measure algebra of $(Y,\mu_1)$.
Then there is a cardinal $\lambda\ge\kappa$ such that $(\frak A,\bar\mu_1)$
can be embedded in $(\frak B_{\lambda},\bar\nu_{\lambda})$, the measure
algebra of $\nu_{\lambda}$.
Because $\mu_1$ is compact, there is an \imp\
function $\phi:\{0,1\}^{\lambda}\to Y$ (343B).
By 235A, $\overline{\int}f(\phi(z),\xi)\nu_{\lambda}(dz)
\le\overline{\int}f(y,\xi)\mu_1(dy)$ for every $\xi$, so
$\int\overline{\int}f(\phi(z),\xi)\nu_{\lambda}(dz)\nu(d\xi)<t$.

For each $\xi<\kappa$ choose a Baire measurable
function $h_{\xi}:\{0,1\}^{\lambda}\to\Bbb R$ such that
$f(\phi(z),\xi)\le h_{\xi}(z)$
for every $z\in\{0,1\}^{\lambda}$ and
$\int h_{\xi}(z)\nu_{\lambda}(dz)=
\overline{\int}f(\phi(z),\xi)\nu_{\lambda}(dz)$;  we can do this because
$\nu_{\lambda}$ is the completion of its restriction to the Baire
$\sigma$-algebra $\CalBa(\{0,1\}^{\lambda})$ (see 4A3Of), so we can apply
133J(a-i) to the Baire measure
$\nu_{\lambda}\restr\CalBa(\{0,1\}^{\lambda})$.
For each $\xi$, there is a countable set
$I_{\xi}\subseteq\lambda$ such that $h_{\xi}$ is determined by coordinates
in $I_{\xi}$, in the sense that $h_{\xi}(z)=h_{\xi}(z')$ whenever
$z\restr I_{\xi}=z'\restr I_{\xi}$.

By 541Rb, there are $\Gamma\subseteq\kappa$ and a countable set
$J\subseteq\lambda$ such
that $\nu\Gamma=1$ and $I_{\xi}\cap I_{\eta}\subseteq J$
whenever $\xi$, $\eta$ are distinct members of $\Gamma$.

\medskip

{\bf (c)} For $u\in\{0,1\}^J$ and $u'\in\{0,1\}^{\lambda\setminus J}$ write
$u\cup u'$ for their common extension to a member of
$\{0,1\}^\lambda$.   Set

\Centerline{$f_1(u,\xi)
=\int h_{\xi}(u\cup u')\nu_{\lambda\setminus J}(du')$}

\noindent for $u\in\{0,1\}^J$ and $\xi<\kappa$.   Then, applying
Fubini's theorem to
$\{0,1\}^{\lambda}\cong\{0,1\}^J\times\{0,1\}^{\lambda\setminus J}$,
we have

\Centerline{$\int f_1(u,\xi)\nu_J(du)
=\int h_{\xi}(z)\nu_{\lambda}(dz)$,}

\noindent  so that

\Centerline{$\iint f_1(u,\xi)\nu_J(du)\nu(d\xi)
=\int\overlineint f(\phi(z),\xi)\nu_{\lambda}(dz)\nu(d\xi)
<t,$}

\noindent and

\Centerline{$\overlineint\int f_1(u,\xi)\nu(d\xi)\nu_J(du)<t$}

\noindent by Theorem 543C.   Accordingly there is a $u\in\{0,1\}^J$
such that $\int f_1(u,\xi)\nu(d\xi)<t$.

\medskip

{\bf (d)} For each $\xi\in\Gamma$ take
$v_{\xi}\in\{0,1\}^{\lambda\setminus J}$ such that
$h_{\xi}(u\cup v_{\xi})\le f_1(u,\xi)$.
Let $w\in\{0,1\}^{\lambda}$ be such that

\Centerline{$w\restr J=u$,
\quad$w\restr I_{\xi}\setminus J=v_{\xi}\restr I_{\xi}\setminus J$
for every $\xi\in\Gamma$;}

\noindent such a $w$ exists because if $\xi$, $\eta\in\Gamma$
and $\xi\ne\eta$ then $I_{\xi}\cap I_{\eta}\subseteq J$.   Now

\Centerline{$f(\phi(w),\xi)
\le h_{\xi}(w)=h_{\xi}(u\cup v_{\xi})
\le f_1(u,\xi)$}

\noindent for every $\xi\in\Gamma$, so

\Centerline{$\int f(\phi(w),\xi)\nu(d\xi)\le\int f_1(u,\xi)\nu(d\xi)
<t$,}

\noindent contradicting the last sentence of (a) above.\ \Bang

This completes the proof.
}%end of proof of 544C

\leader{544D}{Corollary} If $\kappa$ is an \am\ cardinal and
$\omega\le\lambda\le\kappa$, then $\lambda$ is a measure-precaliber
of every probability algebra.

\proof{ If $\lambda<\kappa$ this is a corollary of 544B and 525J.
If $\lambda=\kappa$, we can use 544C and 525C.   For let
$\langle E_{\xi}\rangle_{\xi<\kappa}$ be
a non-decreasing family in $\Cal N_{\kappa}$ with union $E$.
Let $\nu$ be a normal witnessing probability on $\kappa$.   Set

\Centerline{$C=\{(x,\xi):\xi<\kappa,\,x\in E_{\xi}\}
\subseteq\{0,1\}^{\kappa}\times\kappa$.}

\noindent Then

\Centerline{$\underlineint \nu C[\{x\}]\nu_{\kappa}(dx)\ge\mu_*E$,
\quad$\int\nu_{\kappa}^*C^{-1}[\{\xi\}]\nu(d\xi)=0$,}

\noindent so 544C, applied to the indicator function of $C$, tells us
that $\mu_*E=0$;  now 525Cc tells us that
$\kappa$ is a precaliber of $\frak B_{\kappa}$, and therefore a
measure-precaliber of every
probability algebra, by 525Ia and 525Da.
}%end of proof of 544D

\leader{544E}{Theorem}\cmmnt{ ({\smc Kunen n70})} %RVMC 6D
Let $\kappa$ be a \rvm\ cardinal and $\nu$ a
normal witnessing probability on $\kappa$.   If $(X,\mu)$ is a
quasi-Radon probability space of weight strictly less than $\kappa$,
and $f:[\kappa]^{<\omega}\to\Cal N(\mu)$ is any function, then

\Centerline{$\bigcap_{V\subseteq\kappa,\nu V=1}
  \bigcup_{I\in[V]^{<\omega}}f(I)\in\Cal N(\mu)$.}

%would \Mah(\mu)<\kappa be enough?   Try $X$ the Stone space of
%[0,1].   Also in 543C.

\proof{ Let $\Cal F$ be the filter of
$\nu$-conegligible subsets of $\kappa$.

\medskip

{\bf (a)} I show by induction on $n\in\Bbb N$ that if
$g:[\kappa]^{\le n}\to\Cal N(\mu)$ is any function, then

\Centerline{$E(g)=\bigcap_{V\in\Cal F}\bigcup_{I\in[V]^{\le n}}g(I)
\in\Cal N(\mu)$.}

\noindent \Prf\ For $n=0$ this is
trivial;  $E(g)=g(\emptyset)\in\Cal N(\mu)$.   For the
inductive step to $n+1$, given
$g:[\kappa]^{\le n+1}\to\Cal N(\mu)$, then for each $\xi<\kappa$
define $g_{\xi}:[\kappa]^{\le n}\to\Cal N(\mu)$ by setting
$g_{\xi}(I)=g(I\cup\{\xi\})$ for each $I\in[\kappa]^{\le n}$.
By the inductive hypothesis,  $E(g_{\xi})\in\Cal N(\mu)$.   Set

\Centerline{$C=\{(x,\xi):x\in E(g_{\xi})\}\subseteq X\times\kappa$.}

\noindent Then

\Centerline{$\int\mu^*C^{-1}[\{\xi\}]\nu(d\xi)
=\int\mu^*E(g_{\xi})\nu(d\xi)=0$,}

\noindent so by 543C

\Centerline{$\overlineint\nu C[\{x\}]\mu(dx)=0$,}

\noindent and
$\mu D=0$, where $D=g(\emptyset)\cup\{x:\nu C[\{x\}]>0\}$.

Take any $x\in X\setminus D$ and set
$W=\kappa\setminus C[\{x\}]\in\Cal F$.
For each $\xi\in W$, $x\notin E(g_{\xi})$, so there is a
$V_{\xi}\in\Cal F$ such that $\nu V_{\xi}=1$ and $x\notin g_{\xi}(I)$
for every $I\in[V_{\xi}]^{\le n}$.   Set

\Centerline{$V=W\cap\{\xi:\xi\in V_{\eta}$ for every
$\eta<\xi\}$.}

\noindent Then $V\in\Cal F$.   If $I\in[V]^{\le n+1}$, either
$I=\emptyset$ and $x\notin g(I)$, or there is a least element $\xi$ of $I$;
in the latter case,
$\xi\in W$ and $J=I\setminus\{\xi\}\subseteq V_{\xi}$ and
$x\notin g_{\xi}(J)=g(I)$.
So $x\notin\bigcup\{g(I):I\in[V]^{\le n+1}\}$.   As $x$ is
arbitrary, $E(g)\subseteq D\in\Cal N(\mu)$ and the
induction proceeeds.\ \Qed

\medskip

{\bf (b)} Now consider

\Centerline{$G=\bigcup_{n\in\Bbb N}E(f\restr[\kappa]^{\le n})
\in\Cal N(\mu)$.}

\noindent If $x\in X\setminus G$ then for each $n\in\Bbb N$ there
is a $V_n\in\Cal F$ such that
$x\notin\bigcup\{f(I):I\in[V_n]^{\le n}\}$.   Set
$V=\bigcap_{n\in\Bbb N}V_n\in\Cal F$;
then $x\notin\bigcup\{f(I):I\in[V]^{<\omega}\}$.   As $x$
is arbitrary,

\Centerline{$\bigcap_{V\subseteq\kappa,\nu V=1}
  \bigcup_{I\in[V]^{<\omega}}f(I)\subseteq G\in\Cal N(\mu)$,}

\noindent as required.
}%end of proof of 544E

\leader{544F}{Theorem}\cmmnt{ ({\smc Kunen n70})}
Let $\kappa$ be a \rvm\ cardinal with a
normal witnessing probability $\nu$.   If $(X,\mu)$ is a
locally compact semi-finite measure space with $\mu X>0$
and $f:[\kappa]^{<\omega}\to\Cal N(\mu)$ is a function,
then there is a $\nu$-conegligible $V\subseteq\kappa$ such that
$\bigcup\{f(I):I\in[V]^{<\omega}\}\ne X$.

\proof{{\bf (a)} Consider first the case
$(X,\mu)=(\{0,1\}^{\kappa},\nu_{\kappa})$.   For any
$L\subseteq\kappa$ let $\pi_L:\{0,1\}^{\kappa}\to\{0,1\}^L$
be the restriction map.
Let $\Cal F$ be the conegligible filter on $\kappa$.

\medskip

\quad{\bf (i)}
For each $I\in[\kappa]^{<\omega}$, there is a countable set
$g(I)\subseteq\kappa$ such that
$\nu_{g(I)}(\pi_{g(I)}[f(I)])=0$ (254Od);   enlarging $f(I)$ if
necessary, we may suppose
that $f(I)=\pi_{g(I)}^{-1}[\pi_{g(I)}[f(I)]]$.
By 541Q there are a set $C\in\Cal F$ and a function
$h:[\kappa]^{<\omega}\to[\kappa]^{\le\omega}$ such that
$g(I)\cap\eta\subseteq h(I\cap\eta)$
whenever $I\in[C]^{<\omega}$ and $\eta<\kappa$.   Set

\Centerline{$\Gamma=\{\gamma:\gamma<\kappa$,
$h(I)\subseteq\gamma$ for every $I\in[\gamma]^{<\omega}\}$;}

\noindent then $\Gamma$ is a closed cofinal set in $\kappa$,
because $\cf(\kappa)>\omega$.
Let $\langle\gamma_{\eta}\rangle_{\eta\le\kappa}$ be the
increasing enumeration of $\Gamma\cup\{0,\kappa\}$.

\medskip

\quad{\bf (ii)}
For $\eta<\kappa$, set $M(\eta)=\kappa\setminus\gamma_{\eta}$
and $L(\eta)=\gamma_{\eta+1}\setminus\gamma_{\eta}$;
then $\nu_{M(\eta)}$ can be identified with the product
measure $\nu_{L(\eta)}\times\nu_{M(\eta+1)}$.   Choose
$u_{\eta}\in\{0,1\}^{\gamma_{\eta}}$, $V_{\eta}\subseteq\kappa$
inductively, as follows.   $u_0\in\{0,1\}^0$ is the empty function.
Given $u_{\eta}$, then for each $I\in[\kappa]^{<\omega}$ set

\Centerline{$f'_{\eta}(I)=\{v:v\in\{0,1\}^{L(\eta)}$,
$\nu_{M(\eta+1)}\{w:{u_{\eta}}\cup v\cup w\in f(I)\}>0\}$,}

\noindent and

$$\eqalign{f_{\eta}(I)
&=f'_{\eta}(I)\text{ if }\nu_{L(\eta)}(f'_{\eta}(I))=0,\cr
&=\emptyset\text{ otherwise}.\cr}$$

\noindent By 544E, applied to
$J\mapsto f_{\eta}(K\cup J)\in\Cal N_{L(\eta)}$,
we can find for each $K\in[\gamma_{\eta+1}]^{<\omega}$
a set $E_{\eta K}\subseteq\{0,1\}^{L(\eta)}$
such that $\nu_{L(\eta)}E_{\eta K}=1$ and for every $v\in E_{\eta K}$
there is a  set $V\in\Cal F$ such that $v\notin f_{\eta}(K\cup J)$ for
any $J\in[V]^{<\omega}$.   Choose
$v_{\eta}\in\bigcap\{E_{\eta K}: K\in[\gamma_{\eta+1}]^{<\omega}\}$
(using 544B);   for
$K\in[\gamma_{\eta+1}]^{<\omega}$ choose $V_{\eta K}\in\Cal F$
such that $v_{\eta}\notin f_{\eta}(K\cup J)$ for any
$J\in[V_{\eta K}]^{<\omega}$.   Set
$V_{\eta}=\bigcap\{V_{\eta K}:K\in[\gamma_{\eta+1}]^{<\omega}\}\in\Cal F$
and
$u_{\eta+1}={u_{\eta}}\cup v_{\eta}\in\{0,1\}^{\gamma_{\eta+1}}$.

At limit ordinals $\eta$ with $0<\eta\le\kappa$, set
$u_{\eta}=\bigcup_{\xi<\eta}u_{\xi}\in\{0,1\}^{\gamma_{\eta}}$.

\medskip

\quad{\bf (iii)}  Now consider $u=u_{\kappa}\in\{0,1\}^{\kappa}$ and

\Centerline{$V=\{\xi:\xi\in C$, $\xi\in V_{\eta}$ for every $\eta<\xi\}
\in\Cal F$.}

\noindent If $I\in[V]^{<\omega}$ then

\Centerline{$\nu_{M(\eta)}\{w:{u_{\eta}}\cup w\in f(I)\}=0$}

\noindent for every $\eta<\kappa$.  \Prf\ Induce on $\eta$.  For
$\eta=0$ this says just that
$\nu_{\kappa}f(I)=0$, which was our hypothesis on $f$.
For the inductive step to $\eta+1$, we have

\Centerline{$\nu_{M(\eta)}\{w:{u_{\eta}}\cup w\in f(I)\}=0$}

\noindent by the inductive hypothesis, so Fubini's theorem tells
us that

\Centerline{$\nu_{L(\eta)}\{v:
\nu_{M(\eta+1)}\{w:u_{\eta}\cup v\cup w\in f(I)\}>0\}=0$,}

\noindent that is, $\nu_{L(\eta)}f'_{\eta}(I)=0$, so that
$f_{\eta}(I)=f'_{\eta}(I)$.   Now setting
$K=I\cap\gamma_{\eta+1}$ and $J=I\setminus\gamma_{\eta+1}$, we see that
$J\subseteq V_{\eta}$ (because of course $\eta<\gamma_{\eta+1}$),
therefore $J\subseteq V_{\eta K}$ and
$v_{\eta}\notin f_{\eta}(K\cup J)=f'_{\eta}(I)$;  but this says just that

\Centerline{$\nu_{M(\eta+1)}\{w:
  u_{\eta}\cup v_{\eta}\cup w\in f(I)\}=0$,}

\noindent that is, that

\Centerline{$\nu_{M(\eta+1)}\{w:
  {u_{\eta+1}}\cup w\in f(I)\}=0$,}

\noindent so that the induction continues.

For the inductive step to a non-zero limit ordinal
$\eta\le\kappa$, there is a non-zero $\zeta<\eta$ such that
$I\cap\gamma_{\eta}\subseteq\gamma_{\zeta}$.   Now

\Centerline{$g(I)\cap\gamma_{\eta}
\subseteq h(I\cap\gamma_{\eta})=h(I\cap\gamma_{\zeta})
\subseteq\gamma_{\zeta}$,}

\noindent by the choice of $\Gamma$.   As $f(I)$ is determined by
coordinates in $g(I)$, this means that

\Centerline{$\{w:w\in\{0,1\}^{M(\zeta)}$,
   ${u_{\zeta}}\cup w\in f(I)\}
=\{0,1\}^{\gamma_{\eta}\setminus\gamma_{\zeta}}
  \times\{w:w\in\{0,1\}^{M(\eta)}$, ${u_{\eta}}\cup w\in f(I)\}$.}

\noindent By the inductive hypothesis,

\Centerline{$\nu_{M(\zeta)}\{w:{u_{\zeta}}\cup w\in f(I)\}=0$,}

\noindent so that

\Centerline{$\nu_{M(\eta)}\{w:{u_{\eta}}\cup w\in f(I)\}=0$}

\noindent and the induction continues.\ \Qed

\medskip

\quad{\bf (iv)}
But now, given $I\in[V]^{<\omega}$,  there is surely some $\eta<\kappa$
such that $g(I)\subseteq\gamma_{\eta}$, and in this case
$\{w:{u_{\eta}}\cup w\in f(I)\}$ is either $\emptyset$ or
$\{0,1\}^{M(\eta)}$.   As it is $\nu_{M(\eta)}$-negligible it must be
empty, and $u\notin f(I)$.

Thus we have a point $u\notin\bigcup\{f(I):I\in[V]^{<\omega}\}$,
as required.

\medskip

{\bf (b)} If $(X,\mu)$ is a compact probability space,
we have a $\lambda\ge\kappa$ and an \imp\ function
$\phi:\{0,1\}^{\lambda}\to X$.   For each $I\in[\kappa]^{<\omega}$
let $J_I\in[\lambda]^{\le\omega}$ be such that
$\nu_{J_I}\pi_{J_I}[\phi^{-1}[f(I)]]=0$, where here
$\pi_{J_I}$ is interpreted as a
map from $\{0,1\}^{\lambda}$ to $\{0,1\}^{J_I}$;  set
$J=\kappa\cup\bigcup_{I\in[\kappa]^{<\omega}}J_I$.
Let $q:\{0,1\}^J\to\{0,1\}^{\lambda}$ be any function such that
$\pi_Jq$ is the identity on $\{0,1\}^J$, and set $\psi=\phi q$.
For any $I\in[\kappa]^{<\omega}$,

\Centerline{$\psi^{-1}[f(I)]=q^{-1}[\phi^{-1}[f(I)]]
\subseteq\pi_J[\phi^{-1}[f(I)]]
\subseteq\pi_J[\pi_{J_I}^{-1}[\pi_{J_I}[\phi^{-1}[f(I)]]]]$}

\noindent is $\nu_J$-negligible because
$\pi_{J_I}^{-1}[\pi_{J_I}[\phi^{-1}[f(I)]]]$ is $\nu_{\lambda}$-neglgible
and determined by coordinates in $J$.

Because $\#(J)=\kappa$, (a) tells us that
there are $u\in\{0,1\}^J$ and a conegligible $V\subseteq\kappa$ such that
$u\notin\psi^{-1}[f(I)]$ for every
$I\in[V]^{<\omega}$;  in which case $\psi(u)\notin f(I)$ for every
$I\in[V]^{<\omega}$ and $\bigcup\{f(I):I\in[V]^{<\omega}\}\ne X$.

\medskip

{\bf (c)} For the general case, take a subset $E$ of $X$ with non-zero
finite measure, and apply (b) to the function $I\mapsto E\cap f(I)$ and
the normalized subspace measure $\Bover1{\mu E}\nu_E$.
}%end of proof of 544F

\leader{544G}{Proposition} Let $\kappa$ be an \am\ cardinal and
$\omega_1\le\lambda<\kappa$.
If $(X,\mu)$ is an atomless locally compact  semi-finite measure space of
Maharam type less than $\kappa$, and $\mu X>0$,
then there is a Sierpi\'nski set $A\subseteq X$ with cardinal $\lambda$.

%this has to work for strongly Sierpi\'nski query

\proof{{\bf (a)} To begin with, suppose that $X=\{0,1\}^{\theta}$ and
$\mu=\nu_{\theta}$ where $\theta<\kappa$.
Let $\nu$ be an atomless $\kappa$-additive
probability defined on $\Cal P\kappa$.   By 543G
there is a function $f:\kappa\to(\{0,1\}^{\theta})^{\lambda}$ which
is inverse-measure-preserving for $\nu$ and the usual measure
$\nu_{\theta}^{\lambda}$ of
$(\{0,1\}^{\theta})^{\lambda}$, which we may think of either as the
power of $\nu_{\theta}$, or as the Radon power of $\nu_{\theta}$, or as a
copy of $\nu_{\theta\times\lambda}$.   For $\xi<\kappa$, set

\Centerline{$A_{\xi}
=\{f(\xi)(\eta):\eta<\lambda\}\subseteq\{0,1\}^{\theta}$.}

\Quer Suppose, if possible, that for every
$\xi<\kappa$ there is a set $J_{\xi}\subseteq\lambda$ such that
$\#(J_{\xi})=\omega_1$ but $E_{\xi}=f(\xi)[J_{\xi}]$ is
$\nu_{\theta}$-negligible.
For each $\xi$ choose a countable set $I_{\xi}\subseteq\theta$
such that $E'_{\xi}=\pi_{I_{\xi}}^{-1}[\pi_{I_{\xi}}[E_{\xi}]$
is $\nu_{\theta}$-negligible, writing $\pi_{I_{\xi}}(x)=x\restr I_{\xi}$
for $x\in\{0,1\}^{\theta}$.   By 541D, there is a countable
$I\subseteq\theta$ such that
$V=\{\xi:I_{\xi}\subseteq I\}$ is $\nu$-conegligible.   For $\xi\in V$ set
$E^*_{\xi}=\pi_I[E_{\xi}]\subseteq\{0,1\}^I$,
so that $\nu_I E^*_{\xi}=0$.  Fix a sequence
$\sequence{m}{U_m}$ running over the open-and-closed
subsets of $\{0,1\}^I$, and for each $\xi\in V$,
$n\in\Bbb N$ choose an open set
$G_{n\xi}\subseteq\{0,1\}^I$ such that $E^*_{\xi}\subseteq G_{n\xi}$ and
$\nu_I(G_{n\xi})\le 2^{-n}$.   For $m, n\in\Bbb N$ set

\Centerline{$D_{nm}=\{\xi:\xi\in V$, $U_m\subseteq G_{n\xi}\}$.}

\noindent For each $\alpha<\lambda$, set
$f_{\alpha}(\xi)=\pi_I(f(\xi)(\alpha))\in\{0,1\}^I$ for $\xi<\kappa$;
then the functions $f_{\alpha}$ are all stochastically independent, in the
sense that the $\sigma$-algebras
$\Sigma_{\alpha}=\{f_{\alpha}^{-1}[H]:H\subseteq\{0,1\}^I$ is Borel$\}$ are
independent.   \Prf\ Suppose that
$\alpha_0,\ldots,\alpha_n<\lambda$ are distinct and
$H_0,\ldots,H_n$ are Borel subsets of $\{0,1\}^I$.   For each $i$, set

\Centerline{$W_i=\{u:u\in(\{0,1\}^{\theta})^{\lambda}$,
$u(\alpha_i)\in\pi_I^{-1}[H_i]\}$.}

\noindent Then

$$\eqalign{\nu(\bigcap_{i\le n}f_{\alpha_i}^{-1}[H_i])
&=\nu f^{-1}[\bigcap_{i\le n}W_i]
=\nu_{\theta}^{\lambda}(\bigcap_{i\le n}W_i)\cr
&=\prod_{i\le n}\nu_{\theta}^{\lambda}W_i
=\prod_{i\le n}\nu f_{\alpha_i}^{-1}[H_i].  \text{ \Qed}\cr}$$

\noindent By 272Q, there is for each $\xi<\kappa$ an
$\alpha(\xi)\in J_{\xi}$ such that $\Sigma_{\alpha(\xi)}$ is
stochastically independent of the $\sigma$-algebra $\Tau$ generated by
$\{D_{nm}:n$, $m\in\Bbb N\}$.   Because $\lambda<\kappa$ and $\nu$ is
$\kappa$-additive, there is a $\gamma<\lambda$ such that
$B=\{\xi:\alpha(\xi)=\gamma\}$ has $\nu B>0$.
Take $n\in\Bbb N$ such that $\nu(B)>2^{-n}$, and examine

\Centerline{$C=\bigcup_{m\in\Bbb N}(D_{nm}\cap f_{\gamma}^{-1}[U_{m}])$.}

\noindent Then $\nu C=(\nu\times\nu_I)(\tilde C)$ where

\Centerline{$\tilde C
=\bigcup_{m\in\Bbb N}(D_{nm}\times U_m)\subseteq\kappa\times\{0,1\}^I$}

\noindent and $\nu\times\nu_I$ is the c.l.d.\ product measure on
$\kappa\times\{0,1\}^I$.   \Prf\ Because $\Tau$ and $\Sigma_{\gamma}$ are
independent, and $\nu_I$ is the completion of its restriction to the Borel
(or the Baire) $\sigma$-algebra of $\{0,1\}^I$, the map
$\xi\mapsto(\xi,f_{\gamma}(\xi)):\kappa\to\kappa\times\{0,1\}^I$ is \imp\
for $\nu$ and $(\nu\restr\Tau)\times\nu_I$ (cf.\ 272J).
The inverse of $\tilde C$ under this map is just $C$, so

$$\eqalign{\nu C
&=((\nu\restr\Tau)\times\nu_I)(\tilde C)
=\int(\nu\restr\Tau)\tilde C^{-1}[\{u\}]\nu_I(du)\cr
&=\int\nu\tilde C^{-1}[\{u\}]\nu_I(du)
=(\nu\times\nu_I)(\tilde C). \text{ \Qed}\cr}$$

\noindent But, for each $\xi<\kappa$,
the vertical section $\tilde C[\{\xi\}]$ is just
$\bigcup\{U_m:\xi\in D_{nm}\}=G_{n\xi}$, so

\Centerline{$(\nu\times\nu_I)(\tilde C)
=\int\nu_I(G_{n\xi})\nu(d\xi)\le 2^{-n}$.}

\noindent Accordingly $\nu C\le 2^{-n}<\nu B$ and
there must be a $\xi\in B\cap V\setminus C$.
But in this case $f(\xi)(\gamma)\in E_{\xi}$, because
$\gamma=\alpha(\xi)\in J_{\xi}$, so
$f_{\gamma}(\xi)=\pi_I(f(\xi)(\gamma))\in E^*_{\xi}$.
On the other hand, $f_{\gamma}(\xi)\notin G_{n\xi}$, because
there is no $m$ such that
$f_{\gamma}(\xi)\in U_m\subseteq G_{n\xi}$;
contrary to the choice of $G_{n\xi}$.\ \Bang

So take some $\xi<\kappa$ such that
$\nu_{\theta}^*(f(\xi)[J])>0$ for every uncountable
$J\subseteq\lambda$.   Evidently $f(\xi)$
is countable-to-one, so $A_{\xi}$ must have cardinal
$\lambda$, and will serve for $A$.

\medskip

{\bf (b)} Now suppose that $(X,\mu)$ is an atomless compact probability
space with Maharam type $\theta<\kappa$.   Then we have an \imp\ map
$h:\{0,1\}^{\theta}\to X$.   Let $A\subseteq\{0,1\}^{\theta}$ be a
Sierpi\'nski set of
cardinal $\lambda$;  then $h[A]$ is a Sierpi\'nski set with cardinal
$\lambda$ in $X$, by 537B(b-i).

\medskip

{\bf (c)} Finally, for the general case as stated, we can apply (b) to a
normalized subspace measure, as usual.
}%end of proof of 544G

\leader{544H}{Corollary} Let $\kappa$ be an \am\ cardinal.

(a) $\non\Cal N_{\theta}=\omega_1$ for
$\omega\le\theta<\kappa$.

(b) $\non\Cal N_{\theta}\le\kappa$ for
$\theta\le\min(2^{\kappa},\kappa^{(+\omega)})$.

(c) $\non\Cal N_{\theta}<\theta$ for
$\kappa<\theta<\kappa^{(+\omega)}$.

\proof{{\bf (a)}  Immediate from 544G.

\medskip

{\bf (b)} If $\nu$ is any witnessing probability on $\kappa$ then
we have an \imp\ function $f:\kappa\to\{0,1\}^{\theta}$ (543G);  now
$f[\kappa]$ witnesses that $\non\Cal N_{\theta}\le\kappa$.

\medskip

{\bf (c)} Induce on $\theta$, using 523Ib.
}%end of proof of 544H

\cmmnt{\medskip

\noindent{\bf Remark} There may be more to be said;  see 544Zc.
}%end of comment

\leader{544I}{}\cmmnt{ The following is an elementary corollary of
Theorem 543C.

\medskip

\noindent}{\bf Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a totally
finite
quasi-Radon measure space and $(Y,\Cal PY,\nu)$ a probability space;
suppose that $w(X)<\add\nu$.   Let $f:X\times Y\to\Bbb R$ be a
bounded function such that all the sections
$x\mapsto f(x,y):X\to\Bbb R$ are $\Sigma$-measurable.   Then
the repeated integrals $\iint f(x,y)\nu(dy)\mu(dx)$ and
$\iint f(x,y)\mu(dx)\nu(dy)$ are defined and equal.

\proof{ If $\mu X=0$ this is trivial;  otherwise,
re-scaling $\mu$ if necessary, we may suppose that $\mu X=1$.   By 543C,

\Centerline{$\overlineint\int f(x,y)\nu(dy)\mu(dx)
\le\int\overlineint  f(x,y)\mu(dx)\nu(dy)
=\iint f(x,y)\mu(dx)\nu(dy)$.}

\noindent Similarly

\Centerline{$\overlineint\int (-f(x,y))\nu(dy)\mu(dx)
\le\iint(-f(x,y))\mu(dx)\nu(dy),$}

\noindent so that

\Centerline{$\underlineint \int f(x,y)\nu(dy)\mu(dx)
\ge\iint f(x,y)\mu(dx)\nu(dy)$.}

\noindent Putting these together we have the result.
}%end of proof of 544I

\leader{544J}{Proposition}\cmmnt{ ({\smc Zakrzewski 92})} Let
$\kappa$ be an \am\ cardinal and
$(X,\frak T,\Sigma,\penalty-100\mu)$,
$(Y,\frak S,\Tau,\nu)$  Radon probability spaces both of
weight less than $\kappa$;  let $\mu\times\nu$ be the c.l.d.\ product
measure on $X\times Y$, and $\Lambda$ its domain.
Let $f:X\times Y\to\Bbb R$ be a function
such that all its horizontal and vertical sections

\Centerline{$x\mapsto f(x,y^*):X\to\Bbb R$,
\quad$y\mapsto f(x^*,y):Y\to\Bbb R$}

\noindent are measurable.    Then

(a) if $f$ is bounded, the repeated integrals

\Centerline{$\iint f(x,y)\mu(dx)\nu(dy)$,
\quad$\iint f(x,y)\nu(dy)\mu(dx)$}

\noindent exist and are equal;

(b) in any case, there is a $\Lambda$-measurable function
$g:X\times Y\to\Bbb R$ such that all the
sections $\{x:g(x,y^*)\ne f(x,y^*)\}$, $\{y:g(x^*,y)\ne f(x^*,y)\}$ are
negligible.

\proof{{\bf (a)} By 543H there is a $\kappa$-additive measure
$\tilde\nu$ on $Y$, with domain $\Cal PY$, extending $\nu$.   Now 544I
tells us, among other things, that the function

\Centerline{$x\mapsto\int f(x,y)\nu(dy)=
\int f(x,y)\tilde\nu(dy):X\to\Bbb R$}

\noindent is $\Sigma$-measurable.   Similarly,
$y\mapsto\int f(x,y)\mu(dx)$ is $\Tau$-measurable.
So returning to 544I we get

$$\eqalign{\iint f(x,y)\mu(dx)\nu(dy)
&=\iint f(x,y)\mu(dx)\tilde\nu(dy)\cr
&=\iint f(x,y)\tilde\nu(dy)\mu(dx)
=\iint f(x,y)\nu(dy)\mu(dx).\cr}$$

\medskip

{\bf (b)(i)} Suppose first that $f$ is bounded.   By (a), we can define a
measure $\theta$ on $X\times Y$ by saying that

\Centerline{$\theta G=\int\nu G[\{x\}]\mu(dx)
=\int\mu G^{-1}[\{y\}]\nu(dy)$}

\noindent whenever $G\subseteq X\times Y$ is such that $G[\{x\}]\in\Tau$
for almost every $x\in X$ and $G^{-1}[\{y\}]\in\Sigma$ for almost every
$y\in Y$.   This $\theta$ extends $\mu\times\nu$;
so the Radon-Nikod\'ym theorem (232G) tells us that there is a
$\Lambda$-measurable function $h:X\times Y\to\Bbb R$ such
that $\int_Gf(x,y)\theta(dxdy)=\int_G h(x,y)\theta(dxdy)$ for every
$G\in\Lambda$.

Let $\Cal U$ be a base for the topology $\frak T$, closed under finite
intersections, with
$\#(\Cal U)<\kappa$.   For any $U\in\Cal U$ consider

\Centerline{$V_U=\{y:\int_Uf(x,y)\mu(dx)>\int_Uh(x,y)\mu(dx)\}$.}

\noindent The argument of (a) shows that $y\mapsto\int_Uf(x,y)\mu(dx)$
is $\Tau$-measurable, so $V_U\in\Tau$, and

$$\eqalign{\int_{V_U}\int_Uf(x,y)\mu(dx)\nu(dy)
&=\int_{U\times V_U}f(x,y)\theta(dxdy)\cr
&=\int_{U\times V_U}h(x,y)\theta(dxdy)
=\int_{V_U}\int_U h(x,y)\mu(dx)\nu(dy),\cr}$$

\noindent so $\nu V_U=0$.   Similarly

\Centerline{$\nu\{y:\int_U f(x,y)\mu(dx)<\int_U h(x,y)\mu(dx)\}=0$.}

\noindent Because $\#(\Cal U)<\kappa$, and no
non-negligible measurable set in $Y$ can be covered by fewer than
$\kappa$ negligible sets (544B), we must have

\Centerline{$\nu^*\{y:\int_U f(x,y)\mu(dx)=\int_U h(x,y)\mu(dx)$
for every $U\in\Cal U\}=1$.}

\noindent But because $\Cal U$ is a base for the topology of $X$ closed
under finite intersections, we see that

\Centerline{$\nu^*\{y:f(x,y)=h(x,y)$ for $\mu$-almost every $x\}=1$.}

\noindent (For each $y$ such that
$\int_U f(x,y)\mu(dx)=\int_U h(x,y)\mu(dx)$ for every $U\in\Cal U$,
apply 415H(v) to the indefinite-integral measures
over $\mu$ defined by the functions $x\mapsto f(x,y)$, $x\mapsto h(x,y)$;
these are quasi-Radon by 415Ob.)   Again using (a), we know that the
the repeated integral $\iint|f(x,y)-h(x,y)|\mu(dx)\nu(dy)$ exists, and
it must be $0$.   Thus

\Centerline{$\nu\{y:f(x,y)=h(x,y)$ for $\mu$-almost every $x\}=1$.}

Similarly,

\Centerline{$\mu\{x:f(x,y)=h(x,y)$ for $\nu$-almost every $y\}=1$.}

\noindent But now, changing $h$ on a set of the form
$(E\times Y)\cup(X\times F)$ where $\mu E=\nu F=0$, we can get a function
$g$, still
$\Lambda$-measurable, such that
$\{(x,y):f(x,y)\ne g(x,y)\}$ has all its horizontal  and vertical
sections negligible.

\medskip

\quad{\bf (ii)}
This deals with bounded $f$.   But for general $f$ we can look at the
truncates $(x,y)\mapsto\med(-n,f(x,y),\penalty-100n)$ for each $n$ to get a
sequence $\langle g_n\rangle_{n\in\Bbb N}$ of functions which will
converge at an adequate number of points to provide a suitable $g$.
}%end of proof of 544J

\leader{544K}{Proposition} If $X$ is a metrizable space and $\mu$
is a $\sigma$-finite Borel measure on $X$, then
$\add\Cal N(\mu)\ge\add\Cal N_{\omega}$.

\proof{{\bf (a)} If there is an
atomlessly-measurable cardinal then

\Centerline{$\add\Cal N_{\omega}\le\non\Cal N_{\omega}=\omega_1$}

\noindent (544Ha), so the result is immediate.
So let us henceforth suppose otherwise.

\medskip

{\bf (b)} Because there is a totally finite measure with the same domain
and the same null ideal as $\mu$ (215B(vii)), we can suppose that
$\mu$ itself is totally finite.   Let $(\frak A,\bar\mu)$
be the measure algebra of $\mu$ and
$\Cal U$ a $\sigma$-disjoint base for the topology of $X$
(4A2Lg);  express $\Cal U$ as $\bigcup_{n\in\Bbb N}\Cal U_n$ where each
$\Cal U_n$ is disjoint.   For $\Cal V\subseteq\Cal U_n$ set
$\nu_n\Cal V=\mu(\bigcup\Cal V)$.   Then $\nu_n$ is a
totally finite measure with domain $\Cal P\Cal U_n$.   Because there is no
\am\ cardinal, $\add\nu_n$ is either $\infty$ or a \2vm\ cardinal;  in
either case, $\nu_n$ is $\frak c$-additive and purely atomic (438Ce, 543B).

\medskip

{\bf (c)} $\mu$ has countable Maharam type.
\Prf\ Because $\nu_n$ is purely atomic, there is a sequence
$\sequence{i}{\Cal U_{ni}}$ of subsets of $\Cal U_n$ such that for every
$\Cal V\subseteq\Cal U_n$ there is a $J\subseteq\Bbb N$ such that
$\nu_n(\Cal V\symmdiff\bigcup_{i\in J}\Cal U_{ni})=0$.   Set
$W_{ni}=\bigcup\Cal U_{ni}$ for each $i$.   Let  $\frak B$ be the closed
subalgebra of $\frak A$ generated by
$\{W_{ni}^{\ssbullet}:n$, $i\in\Bbb N\}$.

If $G\subseteq X$ is open, set
$\Cal V_n=\{U:U\in\Cal U_n$, $U\subseteq G\}$ and $G_n=\bigcup\Cal V_n$
for each $n$.   Then we have $J_n\subseteq\Bbb N$ such that

\Centerline{$0
=\nu_n(\Cal V_n\symmdiff\bigcup_{i\in J_n}\Cal U_{ni})
=\mu(G_n\symmdiff\bigcup_{i\in J_n}W_{ni})$,}

\noindent so $G_n^{\ssbullet}=\sup_{i\in J_n}W_{ni}^{\ssbullet}\in\frak B$.
Now $G=\bigcup_{n\in\Bbb N}G_n$ so
$G^{\ssbullet}=\sup_{n\in\Bbb N}G_n^{\ssbullet}$ belongs to $\frak B$.

The set $\Sigma=\{E:E\subseteq X$ is Borel, $E^{\ssbullet}\in\frak B\}$
is a $\sigma$-algebra of subsets of $X$, and we have just seen that it
contains every open set;  so $\Sigma$ is the whole Borel $\sigma$-algebra
and $\frak A=\frak B$ has countable Maharam type.\ \Qed

\medskip

{\bf (d)} Next, if $\ofamily{\xi}{\kappa}{G_{\xi}}$ is a family of
open sets where $\kappa<\frak c$, and $G=\bigcup_{\xi<\kappa}G_{\xi}$,
then $G^{\ssbullet}=\sup_{\xi<\kappa}G_{\xi}^{\ssbullet}$ in $\frak A$.
\Prf\ Look at
$\Cal V_{n\xi}=\{U:U\in\Cal U_n$, $U\subseteq G_{\xi}\}$,
$\Cal V_n=\bigcup_{\xi<\kappa}\Cal V_{n\xi}$ for each $n$.
Because $\nu_n$ is $\frak c$-additive,

\Centerline{$\mu(\bigcup\Cal V_n)=\nu\Cal V_n
=\sup_{J\subseteq\kappa\text{ is finite}}
   \nu(\bigcup_{\xi\in J}\Cal V_{n\xi})$}

\noindent and there is a countable set $J_n\subseteq\kappa$ such that
$\mu(\bigcup\Cal V_n)=\mu(\bigcup_{\xi\in J_n}\bigcup\Cal V_{n\xi})$.
Now

\Centerline{$G^{\ssbullet}
=\sup_{n\in\Bbb N}(\bigcup\Cal V_n)^{\ssbullet}
=\sup_{n\in\Bbb N}\sup_{\xi\in J_n}(\bigcup\Cal V_{n\xi})^{\ssbullet}
\Bsubseteq\sup_{\xi<\kappa}G_{\xi}^{\ssbullet}
\Bsubseteq G^{\ssbullet}$.  \Qed}

\medskip

{\bf (e)} Let $\ofamily{\xi}{\kappa}{E_{\xi}}$ be a
family in $\Cal N(\mu)$ where $\kappa<\add\Cal N_{\omega}$.
Because $\mu$ is inner regular with respect to the closed sets (412D),
we can find closed sets $F_{\xi n}\subseteq X\setminus E_{\xi}$
such that $\mu F_{\xi n}\ge\mu X-2^{-n}$ for $\xi<\kappa$ and $n\in\Bbb N$.
By 524Mb and (c) above, $\wdistr(\frak A)\ge\add\Cal N_{\omega}$,
so there is a countable $C\subseteq\frak A$ such that
$F_{\xi n}^{\ssbullet}
=\sup\{c:c\in C$, $c\subseteq F_{\xi n}^{\ssbullet}\}$
for every $n\in\Bbb N$ and
$\xi<\kappa$ (514K).   Again because $\mu$ is inner
regular with respect to the closed sets, there is a
sequence$\sequence{m}{F_m}$ of closed sets such that whenever
$C'\subseteq C$ is finite then $\bar\mu(\sup C')
=\sup\{\mu F_m:m\in\Bbb N$, $F_m^{\ssbullet}\Bsubseteq\sup C'\}$.
Consequently

\Centerline{$\mu F_{\xi n}
=\sup\{\mu F_m:m\in\Bbb N$,
  $F_m^{\ssbullet}\Bsubseteq F_{\xi n}^{\ssbullet}\}$}

\noindent for every $\xi<\kappa$ and $n\in\Bbb N$.   Set

\Centerline{$H_m
=X\cap\bigcap\{F_{\xi n}:n\in\Bbb N$, $\xi<\kappa$,
$F_m^{\ssbullet}\subseteq F_{\xi n}^{\ssbullet}\}$.}

\noindent Applying (d) to
$\{X\setminus F_{\xi n}:F_m^{\ssbullet}\subseteq F_{\xi n}^{\ssbullet}\}$,
we see that $H_m^{\ssbullet}\Bsupseteq F_m^{\ssbullet}$, that is,
$F_m\setminus H_m$ is negligible, for each $m$.

Each $H_m$ is closed;  let $f_m:X\to[0,1]$ be a continuous function
such that $H_m=f_m^{-1}[\{0\}]$.   Set
$f(x)=\sequence{m}{f_m(x)}\in[0,1]^{\Bbb N}$ for $x\in X$, and let $\nu$ be
the restriction of the image measure $\mu f^{-1}$ to the Borel
$\sigma$-algebra of $[0,1]^{\Bbb N}$.   Then
$\add\Cal N(\nu)\ge\add\Cal N_{\omega}$ (apply 522W(a-i) to the atomless
part of $\nu$).   For each $\xi<\kappa$ and $n\in\Bbb N$, there is an
$m\in\Bbb N$ such that $F_m^{\ssbullet}\Bsubseteq F_{\xi n}^{\ssbullet}$
and $\mu F_m\ge\mu F_{\xi n}-2^{-n}\ge\mu X-2^{-n+1}$;  now
$H_m\subseteq F_{\xi n}$ is disjoint from $E_{\xi}$ and
$\mu H_m\ge\mu X-2^{-n+1}$.
So $\{z:z\in[0,1]^{\Bbb N}$, $z(m)>0\}$ includes
$f[E_{\xi}]$ and has measure at most $2^{-n+1}$.   Accordingly $f[E_{\xi}]$
is $\nu$-negligible.

As $\add\Cal N(\nu)>\kappa$, $\bigcup_{\xi<\kappa}f[E_{\xi}]$ is
$\nu$-negligible;  as $f$ is \imp, $\bigcup_{\xi<\kappa}E_{\xi}$ is
$\mu$-negligible;  as $\ofamily{\xi}{\kappa}{E_{\xi}}$ is arbitrary,
$\add\Cal N(\mu)\ge\add\Cal N_{\omega}$.
}%end of proof of 544K

\leader{544L}{Corollary}  Let $X$ be a metrizable space.

(a) If $\CalUn$ is the $\sigma$-ideal of universally negligible
subsets of $X$, then $\add\CalUn\ge\add\Cal N_{\omega}$.

(b) If $\Sigma_{\text{um}}$ is the $\sigma$-algebra of universally
measurable subsets of $X$, then $\bigcup\Cal E\in\Sigma_{\text{um}}$
whenever $\Cal E\subseteq\Sigma_{\text{um}}$ and
$\#(\Cal E)<\add\Cal N_{\omega}$.

\proof{{\bf (a)} Let $\Cal E\subseteq\CalUn$ be a set with cardinal less
than $\add\Cal N_{\omega}$, and $\mu$ a Borel probability measure on $X$
such that $\mu\{x\}=0$ for every $x\in X$.   Then
$\Cal E\subseteq\Cal N(\mu)$;  by 544K, $\bigcup\Cal E\in\Cal N(\mu)$;
as $\mu$ is arbitrary, $\bigcup\Cal E\in\CalUn$;  as $\Cal E$ is arbitrary,
$\add\CalUn\ge\add\Cal N_{\omega}$.

\medskip

{\bf (b)} Let $\mu$ be a totally finite Borel measure on $X$ and $\hat\mu$
its completion.   By 521Ad and 544K,

\Centerline{$\add\hat\mu
=\add\Cal N(\hat\mu)=\add\Cal N(\mu)\ge\add\Cal N_{\omega}
>\#(\Cal E)$.}

\noindent Since $\hat\mu$ measures every member of $\Cal E$, it
also measures $\bigcup\Cal E$ (521Aa);  as $\mu$ is arbitrary,
$\bigcup\Cal E\in\Sigma_{\text{um}}$.
}%end of proof of 544L

\leader{544M}{Theorem} Let $\kappa$ be an \am\ cardinal.
Then the following are equiveridical:

(i) for every family $\ofamily{\xi}{\kappa}{f_{\xi}}$ of regressive
functions defined on $\kappa\setminus\{0\}$ there is a family
$\ofamily{\xi}{\kappa}{\alpha_{\xi}}$ in $\kappa$ such that

\Centerline{$\{\kappa\setminus\zeta:\zeta<\kappa\}
  \cup\{f_{\xi}^{-1}[\{\alpha_{\xi}\}]:\xi<\kappa\}$}

\noindent has the finite intersection property;

(ii) for every family $\ofamily{\xi}{\kappa}{f_{\xi}}$ in
$\BbbN^{\kappa}$ there is a family
$\ofamily{\xi}{\kappa}{m_{\xi}}$ in $\Bbb N$ such that

\Centerline{$\{\kappa\setminus\zeta:\zeta<\kappa\}
  \cup\{f_{\xi}^{-1}[\{m_{\xi}\}]:\xi<\kappa\}$}

\noindent has the finite intersection property;

(iii) $\cov\Cal N_{\kappa}>\kappa$;

(iv) $\cov\Cal N(\mu)>\kappa$ whenever $(X,\mu)$ is a
locally compact semi-finite measure space and $\mu X>0$.

\wheader{544M}{0}{0}{0}{72pt}
\proof{  Let $\nu$ be a normal witnessing probability on $\kappa$.

\medskip

{\bf (i)$\Rightarrow$(ii)} Given a family $\ofamily{\xi}{\kappa}{f_{\xi}}$
as in (ii), apply (i) to $\ofamily{\xi}{\kappa}{f'_{\xi}}$ where
$f'_{\xi}(\eta)=0$ if $0<\eta<\omega$, $f_{\xi}(\eta)$ if
$\omega\le\eta<\kappa$.

\medskip

{\bf (ii)$\Rightarrow$(iii)} Let $\ofamily{\alpha}{\kappa}{A_{\alpha}}$
be a family in $\Cal N_{\kappa}$.
For each $\alpha<\kappa$ let $\langle F_{\alpha n}\rangle_{n\in\Bbb N}$
be a disjoint sequence of compact subsets of
$\{0,1\}^{\kappa}\setminus A_{\alpha}$
such that $\nu_{\kappa}(\bigcup_{n\in\Bbb N}F_{\alpha n})=1$.
By 543G there is a function
$h:\kappa\to\{0,1\}^{\kappa}$ which is \imp\ for $\nu$ and
$\nu_{\kappa}$.
Set $H_{\alpha}=h^{-1}(\bigcup_{n\in\Bbb N}F_{\alpha n})$;  then
$\nu H_{\alpha}=1$.
Let $H$ be the diagonal intersection of
$\ofamily{\alpha}{\kappa}{H_{\alpha}}$, so that $\nu H=1$.
Let $\langle\gamma_{\xi}\rangle_{\xi<\kappa}$
be the increasing enumeration of $H$.

For $\alpha$, $\xi<\kappa$ set

$$\eqalign{f_{\alpha}(\xi)
&=n\text{ if }h(\gamma_{\xi})\in F_{\alpha n},\cr
&=0\text{ if }\gamma_{\xi}\notin H_{\alpha}.\cr}$$

\noindent Then
there is a family $\ofamily{\alpha}{\kappa}{m_{\alpha}}$ in $\Bbb N$
such that
$\Cal E=\{\kappa\setminus\zeta:\zeta<\kappa\}
  \cup\{f_{\alpha}^{-1}[\{m_{\alpha}\}]:\alpha<\kappa\}$
has the finite intersection property.   Let $\Cal F\supseteq\Cal E$ be an
ultrafilter.
For any $\alpha<\kappa$ we have
$H\setminus H_{\alpha}\subseteq\alpha+1$, so that
$\{\xi:\gamma_{\xi}\notin H_{\alpha}\}$ is bounded above in $\kappa$ and
cannot belong to $\Cal F$.   Consequently
$\{\xi:h(\gamma_{\xi})\in F_{\alpha,m_{\alpha}}\}\in\Cal F$.   But this
implies at once that
$\langle F_{\alpha,m_{\alpha}}\rangle_{\alpha<\kappa}$ has the finite
intersection
property;  because all the  $F_{\alpha n}$ are compact, there is a
$y\in\bigcap_{\alpha<\kappa}F_{\alpha,m_{\alpha}}$, and now
$y\notin\bigcup_{\alpha<\kappa}A_{\alpha}$.

Because $\langle A_{\alpha}\rangle_{\alpha<\kappa}$ was arbitrary,
$\cov\Cal N_{\kappa}>\kappa$.

\medskip

{\bf(iii)$\Rightarrow$(iv)} As in 544B, this follows from 523F and 521Lb.

\medskip

{\bf (iv)$\Rightarrow$(i)} Let $(Z,\tilde\nu)$ be the Stone space of the
measure algebra $\frak A$ of $\nu$;
for $A\subseteq\kappa$ let $A^*$ be the
open-and-closed subset of $Z$ corresponding to
the image $A^{\ssbullet}$ of $A$ in $\frak{A}$.

Now let $\langle f_{\xi}\rangle_{\xi<\kappa}$ be a family of regressive
functions defined on $\kappa\setminus\{0\}$.
Because $\Cal N(\nu)$ is normal and $\omega_1$-saturated
and $f_{\xi}$ is regressive, there is for
each $\xi<\kappa$ a countable set $D_{\xi}\subseteq\kappa$ such that
$\nu f_{\xi}^{-1}[D_{\xi}]=1$ (541Ka).   For $\xi$, $\eta<\kappa$
set $A_{\xi\eta}=f_{\xi}^{-1}[\{\eta\}]$;  then
$\nu(\bigcup_{\eta\in D_{\xi}}A_{\xi\eta})=1$ so (because $D_{\xi}$
is countable) $\sup_{\eta\in D_{\xi}}A^{\ssbullet}_{\xi\eta}=1$ in
$\frak A$ and $\tilde\nu(\bigcup_{\eta\in D_{\xi}}A^*_{\xi\eta})=1$.
Set $E_{\xi}
=Z\setminus\bigcup_{\eta\in D_{\xi}}A^*_{\xi\eta}\in\Cal N(\tilde\nu)$.
By (iv), $Z\ne\bigcup_{\xi<\kappa}E_{\xi}$;
take $z\in Z\setminus\bigcup_{\xi<\kappa}E_{\xi}$.
Then for every $\xi<\kappa$ there must be an $\alpha_{\xi}\in D_{\xi}$
such that $z\in A^*_{\xi,\alpha_{\xi}}$.   But this implies that
$\{A^*_{\xi,\alpha_{\xi}}:\xi<\kappa\}$
is  a centered family of open subsets of $Z$.   It follows that
$\{A^{\ssbullet}_{\xi,\alpha_{\xi}}:\xi<\kappa\}$
is centered in $\frak{A}$.  Since $\nu\zeta=0$ for every $\zeta<\kappa$,
$\{A_{\xi,\alpha_{\xi}}:\xi<\kappa\}
\cup\{\kappa\setminus\zeta:\zeta<\kappa\}$ must have the finite
intersection property, as required.
}%end of proof of 544M

\leader{544N}{Cicho\'n's diagram and other cardinals (a)} Returning to the
concerns of Chapter 52,\cmmnt{ we see from the results above that}
any \am\
cardinal $\kappa$ is necessarily connected with the structures there.
\cmmnt{By 544B,} $\kappa\le\cov\Cal N_{\lambda}$ for every $\lambda$;
\cmmnt{by 544G, $\non\Cal N_{\omega}=\omega_1$, so}
all the cardinals on the bottom line of Cicho\'n's
diagram\cmmnt{ (522B)}, and\cmmnt{ therefore} the Martin numbers $\frak m$,
$\frak p$ etc.\cmmnt{\ (522T)}, must be $\omega_1$, while all the
cardinals on the top line must be at least $\kappa$.
\cmmnt{From 522Ub we see
also that} $\FN(\Cal P\Bbb N)$ must be at least $\kappa$.
\cmmnt{Concerning $\frak b$
and $\frak d$, the position is more complicated.}

\spheader 544Nb If $\kappa$ is an \am\ cardinal, then $\frak b<\kappa$.
\prooflet{\Prf\Quer\ Otherwise, we can choose inductively a family
$\ofamily{\xi}{\kappa}{f_{\xi}}$ in $\BbbN^{\Bbb N}$ such that
$\{n:f_{\xi}(n)\le f_{\eta}(n)\}$ is finite whenever $\eta<\xi<\kappa$.
Let $\nu$ be a witnessing probability measure on $\kappa$.   For $m$,
$i\in\Bbb N$ set $D_{mi}=\{\xi:\xi<\kappa$, $f_{\xi}(m)=i\}$.   Then

$$\eqalign{W
&=\{(\xi,\eta):\eta<\xi<\kappa\}
=\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}
  \{(\xi,\eta):f_{\eta}(m)<f_{\xi}(m)\}\cr
&=\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}
  \bigcup_{i<j}D_{mi}\times D_{mj}\cr}$$

\noindent belongs to $\Cal P\kappa\tensorhat\Cal P\kappa$.   But also

\Centerline{$\int\nu W[\{\xi\}]\nu(d\xi)=0<1
=\int\nu W^{-1}[\{\eta\}]\nu(d\eta)$,}

\noindent so this contradicts Fubini's theorem.\ \Bang\Qed
}%end of prooflet

\spheader 544Nc If $\kappa$ is an \am\ cardinal, then
$\cf\frak d\ne\kappa$.   \prooflet{\Prf\Quer\ Otherwise, let
$A\subseteq\BbbN^{\Bbb N}$ be a cofinal set with cardinal $\frak d$, and
express $A$ as $\bigcup_{\xi<\kappa}A_{\xi}$ where
$\ofamily{\xi}{\kappa}{A_{\xi}}$ is non-decreasing and
$\#(A_{\xi})<\frak d$
for every $\xi<\kappa$.   For each $\xi<\kappa$, we have an
$f_{\xi}\in\BbbN^{\Bbb N}$ such that $f_{\xi}\not\le g$ for any
$g\in A_{\xi}$.   Let $\nu$ be a witnessing probability on $\kappa$.   Then
for each $n\in\Bbb N$ we have an $h(n)\in\Bbb N$ such that
$\nu\{\xi:f_{\xi}(n)\ge h(n)\}\le 2^{-n-2}$.   This defines a function
$h\in\BbbN^{\Bbb N}$.   There must be a $g\in A$
such that $h\le g$;  let $\zeta<\kappa$ be such that $g\in A_{\zeta}$.
The set $\{\xi:f_{\xi}\le h\}$ has measure at least $\bover12$, so there is
some $\xi\ge\zeta$ such that

\Centerline{$f_{\xi}\le h\le g\in A_{\zeta}\subseteq A_{\xi}$,}

\noindent contrary to the choice of $f_{\xi}$.\ \Bang\Qed
}%end of prooflet

\spheader 544Nd\cmmnt{ As for the cardinals studied in \S523,
I have already noted
that $\cov\Cal N_{\lambda}\ge\kappa$ for any \am\ cardinal $\kappa$
and any $\lambda$, and we can say something about the possibility that
$\cov\Cal N_{\lambda}=\kappa$ (544M).
Recall that $\cff[\kappa]^{\le\omega}=\kappa$ (542Ia),
so that} $\cf\Cal N_{\kappa}=\max(\kappa,\cf\Cal N_{\omega})$ for any \qm\
cardinal $\kappa$.


\exercises{\leader{544X}{Basic exercises (a)}
%\spheader 544Xa
Let $\kappa$ be an \am\ cardinal, and $\nu$ a witnessing
probability on $\kappa$.   Show that there is a
set $C\subseteq\{0,1\}^{\kappa}\times\kappa$ such that
$\nu_{\kappa}C^{-1}[\{\xi\}]=0$ for every $\xi<\kappa$, but
$\nu_{\kappa}^*\{x:\nu C[\{x\}]=1\}=1$.
%543C

\spheader 544Xb Suppose that $\kappa$ is an \am\ cardinal.   Show that
$\BbbR^{\lambda}$ is measure-compact for every $\lambda<\kappa$.
\Hint{533J.}
%544B

\spheader 544Xc Let $\kappa$ be a \2vm\ cardinal, $\Cal I$ a normal maximal
ideal of $\Cal P\kappa$, $(X,\mu)$ a
quasi-Radon probability space of weight strictly less than $\kappa$,
and $f:[\kappa]^{<\omega}\to\Cal N(\mu)$ a function.   Show that there is a
$V\in\Cal I$ such that
$\bigcup\{f(I):I\in[\kappa\setminus V]^{<\omega}\}$ is $\mu$-negligible.
\Hint{541Xf.}
%544E  in fact  f  is constant on  \kappa\setminus V]^{<\omega}

\spheader 544Xd In 544F, show that if the magnitude of $\mu$ is less than
$\kappa$ and the augmented shrinking number $\shr^+(\mu)$ is at most
$\kappa$ then there is a $\nu$-conegligible $V\subseteq\kappa$ such that
$\mu_*(\bigcup_{I\in[V]^{<\omega}}f(I))=0$.
%544F

\spheader 544Xe Suppose that there is an \am\ cardinal.   Show that every
Radon measure on a first-countable compact Hausdorff space is uniformly
regular.   \Hint{533Hb.}
%544Ga 533Hb

\spheader 544Xf Suppose that $\kappa$ is an \am\ cardinal and that
$2^{\kappa}=\kappa^{(+n+1)}$.   Show that
$\non\Cal N_{2^{(\kappa^+)}}\le\kappa^+$.   \Hint{523I(a-iii).}
%544H

\spheader 544Xg Suppose that $\kappa$ is an \am\ cardinal and that
$(X,\rho)$ is a metric space.   Show that no subset of $X$ with
strong measure zero can have cardinal $\kappa$.
%544N

\spheader 544Xh Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space
such that every subset of $X^2$ is measured by the c.l.d.\
product measure $\mu\times\mu$.   Show that there is a countable subset of
$X$ with full outer measure.   \Hint{if singletons are negligible, consider
a well-ordering of $X$ as a subset of $X^2$.}
%?

\spheader 544Xi Let $\kappa$ be an \am\ cardinal, and $G$ a
group of permutations of $\kappa$  such that $\#(G)<\kappa$.
Show that there is a non-zero
strictly localizable atomless $G$-invariant measure with domain
$\Cal P\kappa$ and magnitude at most $\#(G)$.   \Hint{start with $G$
countable.}
%?
%   (Put a measure $\nu$ on a selector $S$ for the orbits
%of $G$, and set $\mu A=\sum_{g\in G}\nu(S\cap g^{-1}[A])$.)

\leader{544Y}{Further exercises (a)}
%\spheader 544Ya
Let $\kappa$ be a \rvm\ cardinal with witnessing
probability $\nu$.   Give $\kappa$ its discrete topology, so that $\nu$ is
a Borel measure and $\kappa^{\Bbb N}$ is metrizable.
Let $\lambda$ be the
Borel measure on $\kappa^{\Bbb N}$ constructed from $\nu$ by the method
of 434Ym.   (i) Show that if $\kappa$ is \am\ then
$\add\Cal N(\lambda)=\omega_1$.   (ii) Show that if $\kappa$ is \2vm\ then
$\add\Cal N(\lambda)=\kappa$.
%544K 54bits

\spheader 544Yb
Show that $\frak c$ does not have the property of 544M(ii).
%544M

\spheader 544Yc Show that a cardinal $\kappa$ is weakly compact iff it is
strongly inaccessible and has the property (i) of 544M.
%544M \query ref?  try Jech 78, Jech 03, Kanamori 03
}%end of exercises

\leader{544Z}{Problems (a)}
%\spheader 544Za
In 543C, can we replace `$w(X)<\add\nu$' with
`$\tau(\mu)<\add\nu$'?   More concretely, suppose that $(Z,\lambda)$ is the
Stone space of $(\frak B_{\omega},\bar\nu_{\omega})$, $\kappa$ is an \am\
cardinal and $\nu$ a
normal witnessing probability on $\kappa$\dvro{. }{, so that

\Centerline{$\tau(\lambda)=\omega<\add\nu\le\frak c=w(Z)$.}

\noindent}Let
$C\subseteq\kappa\times Z$ be such that $\lambda C[\{\xi\}]=0$ for every
$\xi<\kappa$.   \cmmnt{By 544C, we know that}
$\{z:\nu C^{-1}[\{z\}]>0\}$ has inner
measure zero.   But does it have to be negligible?
%544C

\spheader 544Zb Suppose that $\kappa$ is an \am\ cardinal.
Must there be a Sierpi\'nski set $A\subseteq\{0,1\}^{\omega}$
with cardinal $\kappa$?   (See 552E.)
%544G

\spheader 544Zc Suppose that $\kappa$ is an \am\ cardinal.   Can
$\non\Cal N_{\kappa}$ be greater than $\omega_1$?   What if
$\kappa=\frak c$?   (See 552H.)
%544H

\spheader 544Zd Can there be an \am\ cardinal less than $\frak d$?
(See the notes to \S555.)
%544N

\spheader 544Ze Can there be an \am\ cardinal less than or equal to
$\shr\Cal N_{\omega}$?   (See 555Yd.)
%544N

%to lift $\shr\Cal N_{\omega}$, start with a \2vm\ $\kappa$,
%force with ccc forcing to get $\frak m=\lambda$, now add $\kappa$ random
%reals;  the reals of the intermediate model give a set $A$ with cardinal
%$\lambda$ of full outer measure witnessing that
%$\shr\Cal N_{\omega}\ge\lambda$ (because every subset of $A$ in the final
%model is included in a countable union of subsets of $A$ in the
%intermediate model of the same cardinality).

\spheader 544Zf Suppose that there is an \am\ cardinal.   Does it follow
that $\cov\Cal N_{\omega}=\frak c$?   \cmmnt{(See 552Gc.)}

\endnotes{
\Notesheader{544} The vocabulary of this section (`locally compact
semi-finite measure space', `quasi-Radon probability space of weight less
than $\kappa$', `compact probability space with Maharam type less than
$\kappa$') makes significant demands on the reader, especially the reader
who really wants to know only what happens to Lebesgue measure.   But the
formulations I have chosen are not there
just on the off-chance that someone may
wish to apply the results in unexpected contexts.
I have tried to use the concepts established earlier in this treatise to
signal the nature of the arguments used at each stage.   Thus in 543C we
had an argument which depended on topological ideas, and could work only on
a space with a base which was
small compared with the \am\ cardinal in hand;  in 544C, the argument
depends on an \imp\ function from some power $\{0,1\}^{\lambda}$, so
requires a compact measure, but then finds a $\Delta$-nebula
with a countable root-cover $J$, so that 543C
can be applied to the usual measure on $\{0,1\}^J$, irrespective of the
size of $\lambda$.   Similar, but to my mind rather deeper, ideas
lead from 544E to 544F.   In both cases, there is a price to be paid for
moving to spaces $X$ of arbitrary complexity;  in one, an inequality
$\overline{\int}\int\le\int\overline{\int}$ becomes the weaker
$\underline{\int}\int\le\int\overline{\int}$;  in the other, a negligible
set turns into a set of inner measure zero (544Xd).

Another way to classify the results here is to ask which of them depend on
the Gitik-Shelah theorem.   The formulae in 544H betray such a dependence;
but it seems that the Gitik-Shelah theorem is also needed for the full
strength of 544B, 544G, 544J and 544M as written.
Historically this is significant,
because the idea behind 544G was worked out by K.Prikry and R.M.Solovay
before it was known for sure
that a witnessing measure on an \am\ cardinal could not
have countable Maharam type.   However 544B and 544J, for instance,
can be proved for Lebesgue measure without using the Gitik-Shelah theorem.

In the next chapter I will present a description of measure
theory in random real models.   Those already familiar with random real
forcing may recognise some of the theorems of this section (544G,
544N) as versions of characteristic results from this theory (552E, 552C).

544M is something different.
It was recognised in the 1960s that some of the ways in which \2vm\
cardinals are astonishing is that they are
`weakly $\Pi^1_1$-indescribable' (and, moreover, have many
weakly $\Pi^1_1$-indescribable cardinals below them;  see
{\smc Fremlin 93}, 4K).   I do
not give the `proper' definition of weak $\Pi^1_1$-indescribability, which
relies on concepts from model theory;  you may find it in {\smc Levy 71},
{\smc Baumgartner Taylor \& Wagon 77} or {\smc Fremlin 93};  for our
purposes here, the equivalent combinatorial definition in 544M(i) will
I think suffice.   For strongly inaccessible cardinals, it is the same
thing as weak compactness (544Yc).   Here I mention it only because it
turns out to be related to one of the standard questions I have been asking
in this volume (544M(iii)).   Of course the arguments above beg the
question, whether an \am\ cardinal can be weakly $\Pi^1_1$-indescribable,
especially in view of 544Yb;  see {\smc Fremlin 93}, 4R.

In 544K-544L I look at a question which seems to belong in Chapter 52,
or perhaps with the corresponding result in Hausdorff measures (534Bb).
But unless I am missing something,
the facts here depend on the Gitik-Shelah theorem via 544Ha.

This section has a longer list of problems than most.   In the last four
sections I have tried
to show something of the richness of the structures associated with any
\am\ cardinal;  I remain quite uncertain how much more we can hope to glean
from the combinatorial and measure-theoretic arguments available.
The problems of this chapter mostly have a special status.   They are
of course vacuous unless we suppose that there is an \am\ cardinal;  but
there is something else.   There is a well-understood process,
`Solovay's method', for building
models of set theory with \am\ cardinals from models with
\2vm\ cardinals (\S555).
In most cases, the problems have been solved for such models, and perhaps
they should be regarded as challenges to develop new forcing techniques.
}%end of notes

\discrpage




\frfilename{mt545.tex}
\versiondate{10.2.14}
\copyrightdate{2005}

\def\chaptername{Real-valued-measurable cardinals}
\def\sectionname{PMEA and NMA}

\newsection{545}

One of the reasons for supposing that it is consistent to assume that there
are measurable cardinals is that very much stronger axioms have been
studied at length without any contradiction appearing.   Here I mention two
such axioms which have obvious consequences in measure theory.

\leader{545A}{Theorem} The following are equiveridical:

(i) for every cardinal $\lambda$, there is a probability space
$(X,\Cal PX,\mu)$ with $\tau(\mu)\ge\lambda$ and
$\add\mu\ge\frak c$;

(ii) for every cardinal $\lambda$, there is an extension of the usual
measure $\nu_{\lambda}$ on $\{0,1\}^{\lambda}$ to a $\frak c$-additive
probability measure with domain $\Cal P(\{0,1\}^{\lambda})$;

(iii) for every semi-finite locally compact measure space
$(X,\Sigma,\mu)$\cmmnt{ (definition:  342Ad)},
there is an extension of $\mu$ to a
$\frak c$-additive measure with domain $\Cal PX$.

\proof{{\bf (i)$\Rightarrow$(ii)} Assume (i).   Let $\lambda$
be a cardinal;  of course (ii) is surely true for finite $\lambda$,
so we may take it that $\lambda\ge\omega$.   Let $(X,\Cal PX,\mu)$
be a probability space with Maharam type at least $\lambda^+$
and with
$\add\mu\ge\frak c$.   Taking $\frak A$ to be the measure algebra of $\mu$,
there is an $a\in\frak A$ such that the principal ideal $\frak A_a$ it
generates is homogeneous with Maharam type at least $\lambda$
(332S).   Let $E\in\Cal PX$ be
such that $E^{\ssbullet}=a$, so that the subspace measure $\mu_E$ is
\Mth\ with Maharam type at least $\lambda$.   Setting $\mu'A=\mu A/\mu E$
for $A\subseteq E$, $(E,\Cal PE,\mu')$ is a \Mth\
probability space with Maharam type at least $\lambda$, and
$\add\mu'\ge\add\mu\ge\frak c$.   By 343Ca, there is a function
$f:E\to\{0,1\}^{\lambda}$ which is \imp\ for $\mu'$ and $\nu_{\lambda}$.
Now the image measure $\nu=\mu'f^{-1}$ is a $\frak c$-additive
extension of $\nu_{\lambda}$ to $\Cal P(\{0,1\}^{\lambda})$.

\wheader{545A}{4}{2}{2}{60pt}

{\bf(ii)$\Rightarrow$(iii)} Assume (ii).

\medskip

\quad\grheada\ Suppose that $(X,\Sigma,\mu)$ is a compact probability
space.   Set $\lambda=\max(\omega,\tau(\mu))$.   Then there is
an \imp\ function $f:\{0,1\}^{\lambda}\to X$
(343Cd).  If $\nu$ is a $\frak c$-additive
extension of $\nu_{\lambda}$ to $\Cal P(\{0,1\}^{\lambda})$, then
$\nu f^{-1}$ is a $\frak c$-additive extension of $\mu$ to $\Cal PX$.

\medskip

\quad\grheadb\ Let $(X,\Sigma,\mu)$ be any semi-finite locally compact
measure space.   Set $\Sigma^{f+}=\{E:E\in\Sigma$, $0<\mu E<\infty\}$ and
let $\Cal E\subseteq\Sigma^{f+}$ be maximal subject to $E\cap F$ being
negligible for all distinct $E$, $F\in\Sigma$.   If
$H\in\Sigma$ and $\mu H<\infty$, then
$\Cal H=\{E:E\in\Cal E$, $\mu(E\cap H)>0\}$ is countable and
$E\setminus\bigcup\Cal H$ is negligible, so
$\mu H=\sum_{E\in\Cal E}\mu(E\cap H)$;  because $\mu$ is semi-finite,
$\mu H=\sum_{E\in\Cal E}\mu(E\cap H)$ for every $H\in\Sigma$.

For each $E\in\Cal E$, the subspace measure $\mu_E$ is compact;  applying
($\alpha$) to a normalization of $\mu_E$, we have an
extension $\mu'_E$ of $\mu_E$ to a  $\frak c$-additive measure with domain
$\Cal PE$.   Set $\mu'A=\sum_{E\in\Cal E}\mu'_E(A\cap E)$ for
$A\subseteq X$;  then $\mu':\Cal PX\to[0,\infty]$ is a $\frak c$-additive
measure extending $\mu$.

\medskip

{\bf(iii)$\Rightarrow$(ii)} and {\bf (ii)$\Rightarrow$(i)} are trivial.
}%end of proof of 545A

\leader{545B}{Definition} PMEA (the `{\bf product measure extension axiom}')
is the assertion that the statements (i)-(iii) of 545A are true.

\leader{545C}{Proposition} If PMEA is true, then $\frak c$ is
atomlessly-measurable.

\proof{ By 545A(ii) we have an extension of the usual measure on
$\{0,1\}^{\omega}$ to a $\frak c$-additive measure $\mu$ with domain
$\Cal P(\{0,1\}^{\omega})$.   Since $\mu$ is zero on singletons,
$\add\mu=\frak c$ exactly, so 543Ba and 543Bc tell
us that $\frak c$ is real-valued-measurable, therefore
atomlessly-measurable.
}

\leader{545D}{Definition} NMA (the `{\bf normal measure axiom}') is the
statement

\inset{\noindent For every set $I$ there is a $\frak c$-additive
probability measure $\nu$ on $S=[I]^{<\frak c}$, with domain $\Cal PS$,
such that

($\alpha$) $\nu\{s:i\in s\in S\}=1$ for every $i\in I$,

($\beta$) if $A\subseteq S$, $\nu A>0$ and $f:A\to I$ is such that
$f(s)\in s$ for every $s\in A$, then there is an $i\in I$ such that
$\nu\{s:s\in A$, $f(s)=i\}>0$.
}

\leader{545E}{Proposition} NMA implies PMEA.

\proof{ Assume NMA.   Let $\lambda$ be any cardinal.   Let $\kappa$ be a
regular
infinite cardinal greater than the cardinal power $\lambda^{\omega}$, and
$\nu$ a $\frak c$-additive
probability on $[\kappa]^{<\frak c}$ as in 545D.   For $\xi<\kappa$
define $f_{\xi}:[\kappa]^{<\frak c}\to\frak c$ by setting
$f_{\xi}(s)=\otp(s\cap\xi)$ for every $s\in[\kappa]^{<\frak c}$.   Then if
$\xi<\eta<\kappa$ we have $f_{\xi}(s)<f_{\eta}(s)$ whenever $\xi\in s$,
that is, for $\nu$-almost every $s$.

Let $g:\frak c\to\Cal P\Bbb N$ be any injection.   For $\xi<\kappa$ and
$n\in\Bbb N$ let $a_{\xi n}$ be the equivalence class
$\{s:n\in g(f_{\xi}(s))\}^{\ssbullet}$ in the measure algebra $\frak A$ of
$\nu$.   If $\xi<\eta<\kappa$ then $g(f_{\xi}(s))\ne g(f_{\eta}(s))$ for
$\nu$-almost every $s$, so
$\sup_{n\in\Bbb N}a_{\xi n}\Bsymmdiff a_{\eta n}=1$ in $\frak A$ and there
is an $n\in\Bbb N$ such that $a_{\xi n}\ne a_{\eta n}$.   Accordingly
$\#(\frak A)^{\omega}\ge\kappa>\lambda^{\omega}$ and
$\#(\frak A)>\lambda^{\omega}$.   As
$\#(\frak A)\le\max(4,\tau(\frak A)^{\omega})$ (4A1O/514De),
$\tau(\frak A)>\lambda$.   So $\nu$ witnesses that 545A(i) is true of
$\lambda$.
}%end of proof of 545E

\leader{545F}{Proposition} Suppose that NMA is true.   Let $\frak A$ be a
Boolean algebra such that whenever $s\in[\frak A]^{<\frak c}$ there is
a subalgebra $\frak B\subseteq\frak A$, including $s$, with a strictly
positive countably additive functional.   Then there is a strictly
positive countably additive functional on $\frak A$.

\cmmnt{\medskip

\noindent{\bf Remark} For the definition and elementary properties of
countably additive functionals on arbitrary Boolean algebras, see \S326.
}

\proof{ Of course we can suppose that $\frak A\ne\{0\}$.
Let $\nu$ be a $\frak c$-additive probability on
$S=[\frak A]^{<\frak c}$ as in 545D.   For each $s\in S$, let $\frak B_s$
be a subalgebra of $\frak A$ including $s$ with a strictly
positive countably additive functional $\mu_s$.   Normalizing
$\mu_s$ if necessary, we may suppose that $\mu_s1=1$.   Now, for
$a\in\frak A$, set $\mu(a)=\int\mu_s(a)\nu(ds)$;  because
$a\in s\subseteq\frak B_s=\dom\mu_s$ for $\nu$-almost every $s$, the
integral is well-defined.   Because every $\mu_s$ is additive, so is $\mu$;
because every $\mu_s$ is strictly positive, so is $\mu$.   If
$\sequencen{a_n}$ is a non-increasing sequence in $\frak A$ with infimum
$0$, then $\lim_{n\to\infty}\mu_s(a_n)=0$ whenever
$s\in[\frak A]^{<\frak c}$ contains every $a_n$, that is, for
$\nu$-almost every $s$;  so $\lim_{n\to\infty}\mu a_n=0$.   As
$\sequencen{a_n}$ is arbitrary, $\mu$ is countably additive
(326Ka).
}%end of proof of 545F

\leader{545G}{Corollary} Suppose that NMA is true.   Let $\frak A$ be a
Boolean algebra such that every $s\in[\frak A]^{<\frak c}$ is included in a
subalgebra of $\frak A$ which is, in itself, a measurable algebra.
Then $\frak A$ is a measurable algebra.

\proof{{\bf (a)} Because $\frak c$ is atomlessly-measurable,
it is surely greater
than $\omega_1$ (419G/438Cd/542C).   So a family in $\frak A\setminus\{0\}$
with cardinal $\omega_1$ lies within some measurable subalgebra of $\frak A$
and cannot be disjoint.   Thus $\frak A$ is ccc.

\medskip

{\bf (b)} If $A\subseteq\frak A$, set

\Centerline{$D_1=\{d:d\in\frak A$, $d\Bsubseteq a$ for some $a\in A\}$,}

\Centerline{$D_2=\{d:d\in\frak A$, $d\Bcap a=0$ for every $a\in A\}$.}

\noindent Then $D_1\cup D_2$ is order-dense in $\frak A$ so includes a
partition $D$ of unity in $\frak A$.   By (a), $D$ is countable, so lies
within a measurable subalgebra $\frak B$ of $\frak A$.   Now $D\cap D_1$
has a supremum $b$ in $\frak B$ which is disjoint from every member of
$D\cap D_2$.   But this means that $b$ is the supremum of $A$ in $\frak A$.
As $A$ is arbitrary, $\frak A$ is Dedekind complete.

\medskip

{\bf (c)} By 545F, $\frak A$ has a strictly positive countably additive
functional $\mu$;  but now $(\frak A,\mu)$ is a totally finite measure
algebra.
}%end of proof of 545G

%\query:  do we really need NMA here?  quite a lot works if we just
%say that every set with cardinal \le\omega_1  is in a measurable subalgebra

\exercises{\leader{545X}{Basic exercises (a)}
%\spheader 545Xa
Suppose that $I$ is a set, and that $\nu$
is a $\frak c$-additive probability measure with domain
$\Cal P([I]^{<\frak c})$ satisfying the conditions of 545D.   Suppose that
$A\subseteq[I]^{<\frak c}$ and $f:A\to I$ are such
that $f(s)\in s$ for every $s\in A$.   Show that there is a countable set
$D\subseteq I$ such that $f(s)\in D$ for $\nu$-almost every $s\in A$.
%545D

\leader{545Y}{Further exercises (a)}
%\spheader 545Ya
Suppose that $I$ is a set, and that $\nu$ is a
$\frak c$-additive probability measure with domain
$\Cal PS$, where $S=[I]^{<\frak c}$, satisfying the conditions of 545D.
Suppose that $f:[I]^{<\omega}\to S$ is any function.   Show that
$\nu\{s:s\in S$, $f(J)\subseteq s$ for every $J\in[s]^{<\omega}\}=1$.
%545D

\spheader 545Yb Suppose that NMA is true.   Show that
$\square_{\lambda}$ is false for every
$\lambda\ge\frak c$.   (Cf.\ 555Yf below.)
%545D 555Yf 55bits
}%end of exercises

\endnotes{
\Notesheader{545} I have given the sketchiest of accounts here.   The
main interest of PMEA and NMA has so far been
in their remarkable consequences in
general topology and (for NMA) its associated reflection principles;
see {\smc Fremlin 93} and the references there.   545G is such a reflection
principle.   Note that the measurable subalgebras declared to exist need
not be regularly embedded in the given algebra.
For K.Prikry's theorem that
it is consistent to assume NMA if it is consistent to suppose that there is
a supercompact cardinal, see 555N below.
}%end of notes

\discrpage



\frfilename{mt546.tex}
\versiondate{3.2.21}
\copyrightdate{2005}

\def\chaptername{Real-valued-measurable cardinals}
\def\sectionname{Power set $\sigma$-quotient algebras}

\def\kaqmc{\kappa_{\text{aqmc}}}
\def\pssqa{power set $\sigma$-quotient algebra}
\def\sphaT{\sphat\mskip5mu}

\newsection{546}

One way of interpreting the Gitik-Shelah theorem (543E) is to say that it
shows that `simple' atomless probability algebras cannot be of the form
$\Cal PX/\Cal N(\mu)$.
Similarly, the results of \S541-\S542 show that any ccc Boolean
algebra expressible as the quotient of a power set by a non-trivial
$\sigma$-ideal involves us
in dramatic complexities, though it is not clear when these must appear
in the quotient algebra itself.   In the next section I will present further
examples of algebras which cannot appear in this way.   To prepare
for these I collect some general facts about quotients of power set
algebras.

\leader{546A(a)}{Definition} A {\bf\pssqa} is a Boolean algebra which is
isomorphic to an algebra of the form $\Cal PX/\Cal I$ where $X$ is a set and
$\Cal I$ is a $\sigma$-ideal of subsets of $X$.

\spheader 546Ab
A {\bf normal \pssqa} is a Boolean algebra which is isomorphic
to an algebra of the form $\Cal P\kappa/\Cal I$ where $\kappa$ is a
regular uncountable cardinal and
$\Cal I$ is a normal ideal of $\Cal P\kappa$.

\spheader 546Ac\cmmnt{ I recall some notation which I will use in this
section.}   As in \S522, I will write $\non\Cal M$ for the uniformity of the
meager ideal of $\Bbb R$, $\non\Cal N$ for the uniformity of the
Lebesgue null ideal and $\cov\Cal N$ for the covering number of the
Lebesgue null ideal.   If $\kappa$ is a cardinal,
$\nu_{\kappa}$ will be the usual measure on $\{0,1\}^{\kappa}$.
$\Cal N_{\kappa}$ its null ideal and $\frak B_{\kappa}$ its measure
algebra;  $\frak G_{\kappa}$ will be the category
algebra of $\{0,1\}^{\kappa}$\cmmnt{ (4A3R-4A3S\formerly{4{}A3Q-4{}A3R})}.   
Note that\cmmnt{ we know that} the covering number and uniformity
of $\Cal N_{\omega}$ are $\cov\Cal N$ and $\non\Cal N$
respectively\cmmnt{ (522W(a-i))}, while
$\non\Cal M$ is the uniformity of the meager ideal $\Cal M_{\omega}$ of
$\{0,1\}^{\Bbb N}=\{0,1\}^{\omega}$.

\leader{546B}{Proposition} (a) Any \pssqa\ is Dedekind $\sigma$-complete.

(b) If $\frak A$ is a \pssqa, $\frak B$ is a Boolean algebra and
$\pi:\frak A\to\frak B$ is a surjective sequentially order-continuous
Boolean homomorphism, then $\frak B$ is a \pssqa.
In particular, any principal ideal of a \pssqa\ is a \pssqa.

(c) The simple product of any family of \pssqa{s} is a \pssqa.

\proof{{\bf (a)} 314C.

\medskip

{\bf (b)} Observe that by 313Qb a Boolean algebra $\frak A$ is a \pssqa\
iff there are
a set $X$ and a surjective sequentially order-continuous Boolean
homomorphism from $\Cal PX$ onto $\Cal A$.   It follows immediately
that an image of such
an algebra under a sequentially order-continuous homomorphism is again a
\pssqa.
And of course a principal ideal of $\frak A$ is an image of $\frak A$
under a
homomorphism $a\mapsto a\Bcap c$ which is actually order-continuous.

\medskip

{\bf (c)} If $\familyiI{\frak A_i}$ is a family of \pssqa{s}, then for
each $i\in I$
we have a set $X_i$ and a surjective sequentially order-continuous Boolean
homomorphism $\phi_i:\Cal PX_i\to\frak A_i$.
We can arrange that the $X_i$ are disjoint;  set $X=\bigcup_{i\in I}X_i$.
Now
$A\mapsto\familyiI{\phi_i(A\cap X_i)}:\Cal PX\to\prod_{i\in I}\frak A_i$
is a surjective sequentially order-continuous Boolean homomorphism, so
$\prod_{i\in I}\frak A_i$ is a \pssqa.
}%end of proof of 546B

\leader{546C}{Proposition} A non-zero
principal ideal of a normal \pssqa\ is a normal \pssqa.

\proof{ Let $\kappa$ be a regular uncountable cardinal, $\Cal I$ a
normal ideal on $\kappa$ and $a$ a non-zero element of
$\frak A=\Cal P\kappa/\Cal I$.   Let $A\subseteq\kappa$ be such that
$A^{\ssbullet}=a$, and set $\Cal J=\Cal I\cap\Cal PA$.
Then $\Cal J$ is a
proper ideal of subsets of $A$ and the principal ideal $\frak A_a$
is isomorphic to $\Cal PA/\Cal J$.   Because $\kappa$ is
regular and $\sup A=\kappa$, $\otp A=\kappa$ and we have an
order-isomorphism $h:\kappa\to A$.   If
$S\in\Cal PA\setminus\Cal J$ and $f:S\to A$ is regressive, then
$S\in\Cal P\kappa\setminus\Cal I$ so there is a $\beta<\kappa$ such that
$f^{-1}[\{\beta\}]\notin\Cal I$ (541H(iii));  now $\beta\in A$ and
$f^{-1}[\{\beta\}]\notin\Cal J$.   By 541H in the other direction,
$\Cal J^*=\{h^{-1}[J]:J\in\Cal J\}$ is a normal ideal on $\kappa$, and

\Centerline{$\frak A_a\cong\Cal PA/\Cal J\cong\Cal P\kappa/\Cal J^*$}

\noindent is a normal \pssqa.
}%end of proof of 546C

\leader{546D}{}\cmmnt{ You will observe that I do not claim that an
order-closed subalgebra of a \pssqa\ is a \pssqa.   See 546Xa.
However, our \pssqa{s} will often possess important
power set $\sigma$-quotient subalgebras, as in the following.

\medskip

\noindent}{\bf Lemma} Let $X$ be a set, $\Cal I$ a proper
$\sigma$-ideal of subsets of $X$ containing singletons, and $\frak A$ the
quotient algebra $\Cal PX/\Cal I$.   Write $\kappa$ for $\add\Cal I$.

(a) There are an $a\in\frak A\setminus\{0\}$, a $\sigma$-subalgebra
$\frak C$ of the principal ideal $\frak A_a$ and a $\kappa$-additive ideal
$\Cal J$ of $\Cal P\kappa$, containing singletons, such that
$\frak C\cong\Cal P\kappa/\Cal J$.

(b) If $\frak A$ is atomless and ccc then $\kappa\le\frak c$ is \qm,
$\frak C$ is atomless and we can arrange that $\Cal J$ should be a normal
ideal, so that $\frak C$ is a normal \pssqa.

\proof{{\bf (a)} Let $\ofamily{\xi}{\kappa}{Y_{\xi}}$ be a disjoint family
in $\Cal I$ with union $Y\notin\Cal I$.   Define $g:Y\to\kappa$ by setting
$g(x)=\xi$ when $x\in Y_{\xi}$.   Set

\Centerline{$\Cal J=\{B:B\subseteq\kappa$, $g^{-1}[B]\in\Cal I\}$.}

\noindent Then $\Cal J$ is a proper $\kappa$-additive ideal of subsets of
$\kappa$ containing singletons.
The map $B\mapsto g^{-1}[B]:\Cal P\kappa\to\Cal PY$ induces an injective
sequentially
order-continuous Boolean homomorphism from $\Cal P\kappa/\Cal J$ to
the principal ideal of $\Cal PX/\Cal I$ generated by
$a=Y^{\ssbullet}$, so we have a sequentially order-continuous embedding of
$\Cal P\kappa/\Cal J$ into $\frak A_a$;  of course $a\ne 0$.
By 314F(b-i), the image of
$\Cal P\kappa/\Cal J$ is a $\sigma$-subalgebra of $\frak A_a$.

\medskip

{\bf (b)} Of course $\kappa$ is uncountable (because $\Cal I$ is a
$\sigma$-ideal) and not $\infty$ (because $X=\bigcup\Cal I\notin\Cal I$),
so $\kappa$ is regular (513C(a-i)).   As $\frak A$ is ccc, $\Cal I$ is
$\omega_1$-saturated, by the definition in 541A, so $\kappa$ is \qm\ (542B);  
as $\frak A$ is atomless, $\kappa\le\frak c$ (541O).
So $\frak C\cong\Cal P\kappa/\Cal J$ will be atomless, by 541P.

Returning to the argument of (a), as $\frak A$ is ccc we have the option
of using 541J to give us a function $g:A\to\kappa$ such that
$\Cal J=\{B:B\subseteq\kappa$, $g^{-1}[B]\in\Cal I\}$
is a normal ideal, and then proceeding as before.
}%end of proof of 546D

\leader{546E}{Proposition} The
measure algebra of Lebesgue measure on $\Bbb R$ is not a \pssqa.

\proof{ This is really a very special case of 543E-543F.
Let $\frak A$ be the measure algebra of Lebesgue measure on
$\Bbb R$.   Then there is a functional $\bar\mu$ such that
$(\frak A,\bar\mu)$ is a probability algebra.
\Quer\ If $\frak A$ is a \pssqa,
let $X$, $\Cal I$ be such that $\Cal I$ is a
$\sigma$-ideal of $\Cal PX$ and there is an isomorphism
$\pi:\Cal PX\to\frak A$.   Set $\mu E=\bar\mu(\pi E^{\ssbullet})$ for
$E\subseteq X$, so that $(X,\Cal PX,\mu)$ is a probability space with null
ideal $\Cal I$ and measure algebra isomorphic to $\frak A$.
As $\frak A$ is atomless, so is $\mu$ (322Bg).   Write
$\kappa$ for $\add\mu$.   By 543B(f-ii), $\kappa$ is \am.
By 543F, the Maharam type $\tau(\mu)$ of $\mu$ is at least
$\min(\kappa^{(+\omega)},2^{\kappa})$;  but $\tau(\mu)$ is defined to
be the Maharam type of $\frak A$, which is $\omega$ (331P\Latereditions, %2016
331Xd).
As there is surely no \am\ cardinal less than $\omega$, we have a
contradiction.\ \Bang
}%end of proof of 546E

\leader{546F}{\bf Definition}
Let $\frak A$ be a Boolean algebra.
I will say that an {\bf e-h family} in $\frak A$ is a double sequence
$\langle e_{ij}\rangle_{i,j\in\Bbb N}$ in $\frak A$ such that

\Centerline{$\sequence{j}{e_{ij}}$ is disjoint for every $i\in\Bbb N$}

\noindent and

\Centerline{$\sup_{i\in\Bbb N}e_{i,f(i)}=1$}

\noindent for every $f\in\NN$.

\cmmnt{\medskip

\noindent{\bf Remark} I hope you will find this definition simple enough to
be manageable without any particular motivation.   It is here for the sake
of Lemma 546Ib, and will reappear in 547P.   See also 546Xc-546Xd.
}

\leader{546G}{Lemma} Let $\frak A$ be a Boolean algebra and
$\langle e_{ij}\rangle_{i,j\in\Bbb N}$ an e-h family in $\frak A$.
Then $\sup_{i\ge n}e_{i,f(i)}=1$ for every $f\in\NN$ and $n\in\Bbb N$.

\proof{\Quer\ Otherwise, take $f\in\NN$, $n\in\Bbb N$ and
$a\in\frak A\setminus\{0\}$ such that $a\Bcap e_{i,f(i)}=0$ for every
$i\ge n$.   Choose $\langle a_i\rangle_{i\le n}$,
$\ofamily{i}{n}{g(i)}$ such that $a_0=a$, $g(i)\in\Bbb N$ and
$a_{i+1}=a_i\Bsetminus e_{i,g(i)}$ is non-zero for every $i<n$.
Set $g(i)=f(i)$ for $i\ge n$;  then $g\in\NN$ and $a_n\Bcap e_{i,g(i)}=0$
for every $i\in\Bbb N$, which is supposed to be impossible.\ \Bang
}%end of proof of 546G

\leader{546H}{Free products and completed free products}
\cmmnt{We shall need the following ideas from Volume 3.}
For Boolean algebras $\frak A$, $\frak B$ I write
$\frak A\otimes\frak B$ for their free product\cmmnt{ (315N)} and
$\frak A\tensorhat\frak B$ for its Dedekind completion\cmmnt{ (314U);
for $a\in\frak A$ and $b\in\frak B$ I write $a\otimes b$ for the
intersection of their canonical images in $\frak A\otimes\frak B$
(315N again).   If $\frak A_0$, $\frak B_0$ are order-dense
subalgebras of $\frak A$ and $\frak B$ respectively, then
$\frak A_0\otimes\frak B_0$ is order-dense in $\frak A\otimes\frak B$
(315Kb) and therefore in the Dedekind complete Boolean algebra
$\frak A\tensorhat\frak B$, so $\frak A\tensorhat\frak B$
can be identified with $\frak A_0\tensorhat\frak B_0$ (314Ub).
Next, if $\frak A$, $\frak B$ and $\frak C$ are Boolean algebras, then
$(\frak A\otimes\frak B)\otimes\frak C
\cong\frak A\otimes(\frak B\otimes\frak C)$ (315L), so
$(\frak A\tensorhat\frak B)\tensorhat\frak C
\cong\frak A\tensorhat(\frak B\tensorhat\frak C)$}.   \cmmnt{Finally,
for any set $I$, the algebra $\Cal E_I$ of open-and-closed subsets of
$\{0,1\}^I$ is order-dense in the regular open algebra $\frak G_I$
of $\{0,1\}^I$ (314Uc\Latereditions), %2015
while if $I$ and $J$ are disjoint then
$\{0,1\}^{I\cup J}\cong\{0,1\}^I\times\{0,1\}^J$, so
$\Cal E_{I\cup J}\cong\Cal E_I\otimes\Cal E_J$ (315Ia) and
$\frak G_{I\cup J}\cong\frak G_I\tensorhat\frak G_J$;  consequently}
$\frak G_{\kappa}\cong\frak G_{\kappa}\tensorhat\frak G_{\omega}$ for any
infinite $\kappa$.

\leader{546I}{}\cmmnt{ In \S544 I discussed the consequences of supposing
that a non-trivial measurable algebra is a \pssqa, so that there is an
\am\ cardinal.   Other types of \pssqa\ also entail facts about the cardinals
$\non\Cal M$, $\non\Cal N$ and $\cov\Cal N$.

In the following, I will write $Z$ for $\{0,1\}^{\Bbb N}$,
$S_2$ for $\bigcup_{n\in\Bbb N}\{0,1\}^n$ 
and for $\sigma\in S_2$ I will set 
$I_{\sigma}=\{z:\sigma\subseteq z\in Z\}$.
%526H

\medskip

\noindent}{\bf Lemma} Let $X$ be a set, $\Cal I$ a proper
%$\omega_1$-saturated
$\sigma$-ideal of subsets of $X$, and $\frak A$ the
quotient algebra $\Cal PX/\Cal I$.   Write $\kappa$ for $\add\Cal I$.

(a)(i) If $\frak A$ has an atomless order-closed subalgebra which is a
measurable algebra, then

\qquad($\alpha$) $\kappa\le\non\Cal M$,

\qquad($\beta$) $\non\Cal N\le\#(X)$,

\qquad($\gamma$) $\cov\Cal N\ge\cov\Cal I\ge\kappa$.

\quad(ii) If $\frak G_{\omega}$ can be regularly embedded in $\frak A$,
then $\kappa\le\non\Cal N$.

(b) If $\frak A$ has a non-trivial
principal ideal with an e-h family, then $\non\Cal M\le\#(X)$.

(c) Suppose that $\kappa\le\non\Cal M$.   Then there is an
$a\in\frak A\setminus\{0\}$ such that for every family
$\langle a_{\xi n}\rangle_{\xi<\kappa,n\in\Bbb N}$ in $\frak A$
there is a family $\langle c_{m\sigma}\rangle_{m\in\Bbb N,\sigma\in S_2}$
in $\frak A$ such that

\Centerline{$\inf_{\sigma\in S_2}c_{m,\tau^{\smallfrown}\sigma}=c_{m\tau}$,
\quad$\sup_{\sigma\in S_2}c_{m,\tau^{\smallfrown}\sigma}=1$}

\noindent for every $\tau\in S_2$ and $m\in\Bbb N$, and

$$a\Bcap\inf_{m\in\Bbb N}\sup_{\sigma\in S_2}\bigl(c_{m\sigma}
  \Bcap\inf_{\Atop{i<\#(\sigma)}{\sigma(i)=1}}a_{\xi i}
  \Bsetminus\sup_{\Atop{i<\#(\sigma)}{\sigma(i)=0}}a_{\xi i}\bigr)
=0$$

\noindent for every $\xi<\kappa$.

(d) Suppose that $\kappa=\non\Cal N$.   Then $\frak A$ has a countably
generated order-closed subalgebra which is not a measurable algebra.

(e) Suppose that $\frak A$ is isomorphic to
$\frak C\tensorhat\frak G_{\omega}$ where $\frak C$ is a ccc
Boolean algebra.   Then $\cov\Cal N=\omega_1$.

\medskip

\proof{{\bf (a)(i)} Any non-trivial
atomless measurable algebra includes a closed subalgebra
which is isomorphic, as measure algebra, to
$\frak B_{\omega}$, so we have a regular embedding
$\pi:\frak B_{\omega}\to\frak A$.   Let $\sequence{i}{e_i}$ be the standard
generating family in $\frak B_{\omega}$ (525A);  for $i\in I$ choose
$E_i\subseteq X$ such that $\pi e_i=E_i^{\ssbullet}$ in $\frak A$.
Define $f:X\to Z$ by setting
$f(x)=\sequence{i}{\chi E_i(x)}$ for $x\in X$.   Then the set

\Centerline{$\{V:V\subseteq Z$ is Borel,
$f^{-1}[V]^{\ssbullet}=\pi V^{\ssbullet}\}$}

\noindent is a $\sigma$-algebra of sets containing every $E_i$, so is the
Borel $\sigma$-algebra of $Z$.

\medskip

\qquad\grheada\
Fix on a $\nu_{\omega}$-conegligible meager Borel subset $V_0$ of $Z$.
Let $D\subseteq Z$ be any set with cardinal less than
$\kappa$.   For $z\in D$ set $A_z=f^{-1}[z+V_0]$, where $+$ here is the
usual group operation of $Z$.   Because $\nu_{\omega}$ is
the Haar probability measure on the compact Hausdorff group
$(Z,+)$ (254Jd, 416U),
$\nu_{\omega}(z+V_0)=1$, $(z+V_0)^{\ssbullet}=1$ in
$\frak B_{\omega}$ and $A_z^{\ssbullet}=1$ in $\frak A$, that is,
$X\setminus A_z\in\Cal I$.   Because $\#(D)<\add\Cal I$,
$\bigcup_{z\in D}X\setminus A_z\in\Cal I$ and
$\bigcap_{z\in D}A_z\ne 0$.   Take $x\in\bigcap_{z\in D}A_z$.   If
$z\in D$ then $f(x)\in z+V_0$ so $z\in f(x)+V_0$;
thus $D\subseteq f(x)+V_0$.
But $z\mapsto f(x)+z: Z\to Z$ is a
homeomorphism so $f(x)+V_0$, like $V_0$, is meager, and $D$ is meager.
As $D$ is arbitrary, $\kappa\le\non\Cal M_{\omega}=\non\Cal M$.

\medskip

\qquad\grheadb\ On the other hand, if $W\subseteq Z$ is
a $\nu_{\omega}$-negligible Borel set, $f^{-1}[W]^{\ssbullet}=0$ and there
is an $x\in X$ such that $f(x)\notin W$.   So
$f[X]\notin\Cal N_{\omega}$ and $\non\Cal N=\non\Cal N_{\omega}\le\#(X)$.

\medskip

\qquad\grheadc\ If
$\lambda<\cov\Cal I$ and $\ofamily{\xi}{\lambda}{H_{\xi}}$ is a family of
negligible Borel sets in $Z$,
$f^{-1}[H_{\xi}]^{\ssbullet}
=\pi H_{\xi}^{\ssbullet}=0$ and $f^{-1}[H_{\xi}]\in\Cal I$
for every $\xi<\lambda$.   There is therefore an $x\in X$
such that $f(x)\notin H_{\xi}$ for every $\xi$.
As $\ofamily{\xi}{\lambda}{H_{\xi}}$ is arbitrary,

\Centerline{$\cov\Cal N=\cov\Cal N_{\omega}\ge\cov\Cal I
\ge\add\Cal I\ge\kappa$.}

\medskip

\quad{\bf (ii)} We can use essentially the same argument as in
(i-$\alpha$).   This time, we have a regular embedding
$\pi:\frak G_{\omega}\to\frak A$.   For $i\in\Bbb N$ set
$H_i=\{z:z\in Z$, $z(i)=1\}$, as before, but now take
$e_i$ to be the equivalence class $H_i^{\ssbullet}\in\frak G_{\omega}$;
as before, choose
$E_i\subseteq X$ such that $\pi e_i=E_i^{\ssbullet}$ in $\frak A$.
Again define $f:X\to Z$ by setting
$f(x)=\sequence{i}{\chi E_i(x)}$ for $x\in X$.   Once again,
$f^{-1}[V]^{\ssbullet}=\pi V^{\ssbullet}\}$ for every Borel set
$V\subseteq Z$.
Now take a comeager $\nu_{\omega}$-negligible Borel subset $V_0'$ of
$Z$ (e.g., the complement of the set $V_0$ chosen in
(i-$\alpha$)).

Once again, let $D\subseteq Z$ be any set with cardinal less than
$\kappa$, and for $z\in D$ set $A_z=f^{-1}[z+V_0']$.   As before,
$A_z^{\ssbullet}=1$ in $\frak A$ for every $z\in D$, so
there is an $x\in\bigcap_{z\in D}A_z$.   Now $D\subseteq f(x)+V_0$, which
is $\nu_{\omega}$-negligible.
As $D$ is arbitrary, $\non\Cal N=\non\Cal N_{\omega}\ge\kappa$.

\medskip

{\bf (b)} Let $\langle e_{ij}\rangle_{i,j\in\Bbb N}$ be an e-h
family in $\frak A_a$, where $a\in\frak A\setminus\{0\}$.   For each
$i\in\Bbb N$ let $\sequence{j}{E_{ij}}$ be a disjoint sequence of subsets
of $X$ such that $E_{ij}^{\ssbullet}=e_{ij}$ for every $j\in\Bbb N$.
For $x\in X$ set $z_x(i)=j$ if $j\in\Bbb N$ and $x\in E_{ij}$,
$0$ if $x\in X\setminus\bigcup_{j\in\Bbb N}E_{ij}$.   Then
$D=\{z_x:x\in X\}$ is a subset of $\NN$.   If
$g\in\NN$ and $E=\bigcap_{n\in\Bbb N}\bigcup_{i\ge n}E_{i,g(i)}$ then
$E^{\ssbullet}=\inf_{n\in\Bbb N}\sup_{i\ge n}e_{i,f(i)}=1$ (546G), so
$E\ne\emptyset$.   Take any $x\in E$;  then
$\{i:z_x(i)=g(i)\}=\{i:x\in E_{i,g(i)}\}$ is infinite.

Thus for every $g\in\NN$ there is an $f\in D$ such that $f\cap g$ is
infinite.   By 522Sc, $\non\Cal M\le\#(D)\le\#(X)$.

\medskip

{\bf (c)} Fix a disjoint family
$\ofamily{\xi}{\kappa}{Y_{\xi}}$ in $\Cal I$ such that
$Y=\bigcup_{\xi<\kappa}Y_{\xi}$ does not belong to $\Cal I$.
Set $a=Y^{\ssbullet}\in\frak A\setminus\{0\}$.
Given a family $\langle a_{\xi n}\rangle_{\xi<\kappa,n\in\Bbb N}$, choose
for each $\xi<\kappa$ and $n\in\Bbb N$ a set $A_{\xi n}\subseteq X$ such
that $A_{\xi n}^{\ssbullet}=a_{\xi n}$;  for $\xi<\kappa$, define
$f_{\xi}:X\to Z$ by setting
$f_{\xi}(x)=\sequencen{\chi A_{\xi n}(x)}$ for every $x\in X$.   

For $x\in X$, set

$$\eqalign{B_x
&=\{f_{\eta}(x):\eta<\xi\}\text{ if }\xi<\kappa\text{ and }x\in Y_{\xi},\cr
&=\emptyset\text{ otherwise}.\cr}$$

\noindent Because
$\kappa\le\non\Cal M=\non\Cal M_{\omega}$, $B_x$ is
meager in $Z$.   We can therefore find a sequence
$\sequence{m}{G_{mx}}$ of dense open subsets of
$Z$ such that $B_x\cap\bigcap_{m\in\Bbb N}G_{mx}=\emptyset$.

For $\sigma\in S_2$ and $m\in\Bbb N$, set

\Centerline{$C_{m\sigma}=\{x:I_{\sigma}\subseteq G_{mx}\}$,
\quad$c_{m\sigma}=C_{m\sigma}^{\ssbullet}$}

\noindent in $\frak A$.   If $\tau$, $\sigma\in S_2$ and $m\in\Bbb N$, then
$I_{\tau}\supseteq I_{\tau^{\smallfrown}\sigma}$,
$C_{m\tau}\subseteq C_{m,\tau^{\smallfrown}\sigma}$ and
$c_{m\tau}\Bsubseteq c_{m,\tau^{\smallfrown}\sigma}$.
If $\tau\in S_2$, $m\in\Bbb N$ and $x\in X$ then
there must be a $\sigma\in S_2$ such that
$G_{mx}\supseteq I_{\tau^{\smallfrown}\sigma}$, because
$G_{mx}$ is dense and open;  so
$\bigcup_{\sigma\in S_2}C_{m,\tau^{\smallfrown}\sigma}=X$ and
$\sup_{\sigma\in S_2}c_{m,\tau^{\smallfrown}\sigma}=1$, while
$\inf_{\sigma\in S_2}c_{m,\tau^{\smallfrown}\sigma}=c_{m\tau}$.

Now take any $\xi<\kappa$,
$x\in Y\setminus\bigcup_{\eta\le\xi}Y_{\eta}$ and $\sigma\in S_2$.   Then
$f_{\xi}(x)\in B_x$ and there is an $m\in\Bbb N$ such that
$f_{\xi}(x)\notin G_{mx}$.   But this means that if $x\in C_{m\sigma}$
then $\sigma\not\subseteq f_{\xi}(x)$, that is,
there is an $i<\#(\sigma)$ such that
$\sigma(i)\ne f_{\xi}(x)(i)=\chi A_{\xi i}(x)$, that is,

$$x
\notin C_{m\sigma}\cap\bigcap_{\Atop{i<\#(\sigma)}{\sigma(i)=1}}A_{\xi i}
   \setminus\bigcup_{\Atop{i<\#(\sigma)}{\sigma(i)=0}}A_{\xi i}.$$

\noindent Thus
$Y\setminus\bigcup_{\eta\le\xi}Y_{\eta}$ is disjoint from

$$\bigcap_{m\in\Bbb N}\bigcup_{\sigma\in S_2}
  \bigl(C_{m\sigma}\cap\bigcap_{\Atop{i<\#(\sigma)}{\sigma(i)=1}}A_{\xi i}
   \setminus\bigcup_{\Atop{i<\#(\sigma)}{\sigma(i)=0}}A_{\xi i}).$$

\noindent Because every $Y_{\eta}$ belongs to $\Cal I$ and
$\add\Cal I=\kappa$,
$(Y\setminus\bigcup_{\eta\le\xi}Y_{\eta})^{\ssbullet}=a$, so

$$a\Bcap\inf_{m\in\Bbb N}\sup_{\sigma\in S_2}
  \bigl(c_{m\sigma}\Bcap\inf_{\Atop{i<\#(\sigma)}{\sigma(i)=1}}a_{\xi i}
   \Bsetminus\sup_{\Atop{i<\#(\sigma)}{\sigma(i)=0}}a_{\xi i}\bigr)
=0,$$

\noindent as required.

\medskip

{\bf (d)} Take a non-negligible subset of $Z$ of
size $\kappa$, and enumerate it as $\ofamily{\xi}{\kappa}{z_{\xi}}$.   For
each $\xi<\kappa$, let $V_{\xi}$ be a $\nu_{\omega}$-negligible
G$_{\delta}$ set including $\{z_{\eta}:\eta\le\xi\}$;  express $V_{\xi}$
as $\bigcap_{n\in\Bbb N}\bigcup_{\sigma\in L_{n\xi}}I_{\sigma}$, where
$L_{n\xi}\subseteq S_2$ for each $n$;  we can do this in such a way that
$\sum_{\sigma\in L_{n\xi}}2^{-\#(\sigma)}\le 2^{-n}$ for every
$n\in\Bbb N$.

Again let $\ofamily{\xi}{\kappa}{Y_{\xi}}$ be a disjoint family in
$\Cal I$ such that
$Y=\bigcup_{\xi<\kappa}Y_{\xi}$ does not belong to $\Cal I$.
For $n\in\Bbb N$ and $\sigma\in S_2$,
set $A_{n\sigma}=\bigcup\{Y_{\xi}:\xi<\kappa$, $\sigma\in L_{n\xi}\}$.
Let $\frak B$ be the order-closed subalgebra of $\frak A$ generated by
$\{A_{n\sigma}^{\ssbullet}:n\in\Bbb N$, $\sigma\in S_2\}$.  Of course
$\frak B$ is countably generated.

\Quer\ Suppose, if possible, that $\frak B$ is a measurable algebra;  let
$\bar\mu$ be such that $(\frak B,\bar\mu)$ is a probability algebra.
Set $\Sigma=\{E:E\subseteq X$, $E^{\ssbullet}\in\frak B\}$ and
$\mu E=\bar\mu E^{\ssbullet}$ for $E\in\Sigma$.   Then
$(X,\Sigma,\mu)$ is a probability space.   In the product space
$X\times Z$ consider the set

\Centerline{$W
=\bigcap_{n\in\Bbb N}\bigcup_{\sigma\in S_2}A_{n\sigma}\times I_{\sigma}
\subseteq Y\times Z$.}

\noindent Then $W$ is measured by the product measure
$\mu\times\nu_{\omega}$.

If $\xi\le\zeta<\kappa$ and $n\in\Bbb N$, $z_{\xi}\in V_{\zeta}$ so there is a
$\sigma\in L_{n\zeta}$ such that $z_{\xi}\in I_{\sigma}$;  now
$\{z_{\xi}\}\times Y_{\zeta}\subseteq I_{\sigma}\times A_{n\sigma}$.
So for any $\xi<\kappa$,

\Centerline{$W^{-1}[\{z_{\xi}\}]
=\bigcap_{n\in\Bbb N}\bigcup_{\sigma\in S_2,z_{\xi}\in I_{\sigma}}
  A_{n\sigma}
\supseteq\bigcup_{\xi\le\zeta<\kappa}Y_{\zeta}
=Y\setminus\bigcup_{\eta<\xi}Y_{\eta}$}

\noindent has measure at least $\mu^*Y>0$, because
$Y^{\ssbullet}\ne 0$ and $(\bigcup_{\eta<\xi}Y_{\eta})^{\ssbullet}=0$ in
$\frak A$.   Since $\{z_{\xi}:\xi<\kappa\}\notin\Cal N_{\omega}$,
$W$ cannot be
$(\mu\times\nu_{\omega})$-negligible and there must be an
$x\in X$ such that $\mu W[\{x\}]>0$.   As $W[\{x\}]\ne\emptyset$, $x\in Y$;
let $\xi<\kappa$ be such that $x\in Y_{\xi}$.   Then

\Centerline{$W[\{x\}]
=\bigcap_{n\in\Bbb N}\bigcup_{\sigma\in S_2,x\in A_{n\sigma}}I_{\sigma}
=\bigcap_{n\in\Bbb N}\bigcup_{\sigma\in L_{n\xi}}I_{\sigma}
=V_{\xi}$;}

\noindent but $V_{\xi}$ is $\nu_{\omega}$-negligible.\ \Bang

So we have a countably generated order-closed subalgebra of $\frak A$ which
is not measurable, as required.

\medskip

{\bf (e)}\cmmnt{ (see {\smc Bartoszy\'nski \& Judah 95}, 3.3.12)}
As the regular open algebra $\frak G$ of $\NN$ is isomorphic to
$\frak G_{\omega}$ (use 515Oa), we have a surjective sequentially
order-continuous Boolean homomorphism
$\pi:\Cal PX\to\frak C\tensorhat\frak G$ with kernel $\Cal I$.
For each $n\in\Bbb N$ let $\sequence{i}{H_{ni}}$ enumerate the
open-and-closed subsets of $Z$ of measure at most $2^{-n}$;
let $\sequence{i}{E_{ni}}$ be a partition of $X$ such that
$1_{\frak C}\otimes\{\beta:\beta\in\NN$, $\beta(n)=i\}\in\frak C\otimes\frak G$
is equal to $\pi E_{ni}$ for each $i$.   For $x\in X$
set $H_x=\bigcap_{n\in\Bbb N}\bigcup_{i\ge n}H_{i,f_x(i)}$ where
$f_x(n)=i$ if $x\in E_{ni}$.   Then $H_x\in\Cal N_{\omega}$.
Write $S$ for $\bigcup_{m\in\Bbb N}\BbbN^m$, and for $\tau\in S$ write
$G_{\tau}$ for $\{\beta:\tau\subseteq\beta\in\NN\}\in\frak G$.
Let $D\subseteq Z$ be a set with cardinal $\omega_1$.

\Quer\ If $D+H_x\ne Z$ for every $x\in X$, take
$\alpha_x\in Z\setminus(D+H_x)$ for each $x$.
For each $d\in D$, $\alpha_x+d\notin H_x$ for every $x$, so
there is an $n_d\in\Bbb N$ such that
$A_d=\{x:\alpha_x+d\notin\bigcup_{i\ge n_d}H_{i,f_x(i)}\}$
does not belong to $\Cal I$;  let $a_d\in\frak C\setminus\{0\}$,
$\tau_d\in S$ be such that
$a_d\otimes G_{\tau_d}\Bsubseteq\pi A_d$.   Let
$n\in\Bbb N$, $m\in\Bbb N$, $\tau\in\BbbN^m$ be such that
$D'=\{d:d\in D$, $n_d=n$, $\tau_d\subseteq\tau\}$ is uncountable.
Extending $\tau$ if necessary, we can arrange that $m\ge n$.
Because $\frak C$ is ccc, there is an infinite $B\subseteq D'$ such that
$\inf_{d\in I}a_d\ne 0$ for every finite $I\subseteq B$ (this is trivial if
$\frak C$ is finite, and otherwise follows from 516Ld).

Let $k$ be so large that $\Bover{1+\ln k}{k}\le 2^{-m}$.   Let $r\ge m$ be
such that there is a set $F\subseteq\{d\restr r:d\in B\}$
with $\#(F)=k$;  let $I\in[B]^k$ be such that
$F=\{d\restr r:d\in I\}$.   Consider the set

\Centerline{$R=\{(u,v):u$, $v\in\{0,1\}^{r}$, $u+v\in F\}$}

\noindent where $+$ here is the group operation on
$\{0,1\}^{r}=\Bbb Z_2^{r}$.   Then
$\#(R[\{u\}])=\#(R^{-1}[\{v\}])=k$ for every $u$, $v\in\{0,1\}^{r}$.
By 5A1Q, there is a $K\subseteq\{0,1\}^{r}$ such that
$R[K]=\{0,1\}^{r}$ and $\#(K)\le\Bover{2^{r}(1+\ln k)}{k}\le 2^{r-m}$.
Now $\nu_{\omega}\{z:z\restr r\in K\}=2^{-r}\#(K)\le 2^{-m}$, so
there is a $j\in\Bbb N$ such that $H_{mj}=\{z:z\restr r\in K\}$.
But this means that $I+H_{mj}= Z$.
\Prf\ If $\alpha\in Z$, $\alpha\restr r\in R[K]$ and there
are $u\in K$, $v\in F$ such that $\alpha\restr r=u+v$.
Let $d\in I$ be such that $v=d\restr r$;  then

\Centerline{$(\alpha+d)\restr r=(u+v)+v=u\in K$}

\noindent so $\alpha+d\in H_{mj}$.\ \Qed

By the choice of $B$, $a=\inf_{d\in I}a_d$ is non-zero, so
$a\otimes\{\beta:\beta(m)=j\}\ne 0$ and
$A=\bigcap_{d\in I}A_d\cap E_{mj}$ does not belong to $\Cal I$.   We
therefore have an $x\in A$, and $f_x(m)=j$.
In this case, because $m\ge n=n_d$,
$\alpha_x\notin d+H_{m,f_x(m)}=d+H_{mj}$ for every $d\in I$.
But $I+H_{mj}=Z$.\ \Bang

So there is an $x\in X$ such that $D+H_x= Z$.
As $d+H_x\in\Cal N_{\omega}$ for every $d\in D$,
$\omega_1=\cov\Cal N_{\omega}=\cov\Cal N$.
}%end of proof of 546I

\leader{546J}{Theorem} Let $\frak A$ be a Boolean algebra such that
$\frak B_{\omega}$ can be regularly embedded in $\frak A$ and
$\frak A\cong\frak C\tensorhat\frak G_{\omega}$ for some ccc Boolean
algebra $\frak C$.   Then $\frak A$ is not a \pssqa.

\proof{ Note first that $\frak C\otimes\frak G_{\omega}$ is ccc
because $\frak C$ is ccc and $\frak G_{\omega}$ satisfies Knaster's
condition (516U), so $\frak A$ is ccc (514Ee).   Also $\frak A$ is atomless
(316Rb\formerly{3{}16Xi}).
\Quer\ If $\Cal I$ is a $\sigma$-ideal of subsets of $X$ and
$\Cal PX/\Cal I\cong\frak A$, then $\Cal I$ is $\omega_1$-saturated and
must contain all singleton
subsets of $X$, so $\add\Cal I$ is \qm\ (542B again) and greater
than $\omega_1$ (542C), and $\cov\Cal I>\omega_1$.
Because $\frak B_{\omega}$ can be regularly
embedded in $\frak A$, $\cov\Cal N>\omega_1$ (546I(a-i-$\gamma$)).
But this contradicts 546Ie.\ \Bang
}%end of proof of 546J

\leader{546K}{Corollary}\cmmnt{ (A.Kumar, private communication,
January 2016)} If $\lambda$, $\kappa$ are infinite cardinals then
$\cmmnt{\frak A=\mskip5mu}\frak B_{\lambda}\tensorhat\frak G_{\kappa}$ 
is not a \pssqa.

\proof{ $\frak B_{\omega}$ can be regularly embedded in $\frak B_{\lambda}$
which can be regularly embedded in $\frak A$ (315Kc),
so $\frak B_{\omega}$ can be
regularly embedded in $\frak A$ (313N).   On the other hand,
$\frak A$ is ccc (being the Dedekind completion of the free product of
Boolean algebras satisfying Knaster's condition) and
$\frak G_{\kappa}\cong\frak G_{\kappa}\tensorhat\frak G_{\omega}$.
Accordingly

\Centerline{$\frak A
=\frak B_{\lambda}\tensorhat\frak G_{\kappa}
\cong\frak B_{\lambda}
  \tensorhat(\frak G_{\kappa}\tensorhat\frak G_{\omega})
\cong(\frak B_{\lambda}\tensorhat\frak G_{\kappa})
  \tensorhat\frak G_{\omega}
=\frak A\tensorhat\frak G_{\omega}$}

\noindent and we can apply 546J.
}%end of proof of 546K


\exercises{\leader{546X}{Basic exercises (a)}
%\spheader 546Xa
Let $\frak A$ be any Dedekind complete ccc Boolean algebra.   Show that it
can be regularly embedded in a \pssqa.   \Hint{314M.}
%546B

\spheader 546Xb Show that the measure algebra of the usual measure on
$\{0,1\}^{\frak b}$ is not a \pssqa.   \Hint{544N.}
%546E 544N

\sqheader 546Xc
Show that $\frak G_{\omega}$ has an e-h family.
\Hint{for $i$, $j\in\Bbb N$ set $E_{ij}=\{x:x\in\{0,1\}^{\Bbb N}$,
$x(i+k)=0$ for $k<j$, $x(i+j)=1\}$.}
%546F

\spheader 546Xd
Let $\frak A$ be a \wsid\ Boolean algebra, not $\{0\}$.   Show that
there is no e-h family in $\frak A$.

\spheader 546Xe In 546I, show that if
$\kappa=\non\Cal M$ then $\frak A$ has a countably generated order-closed
subalgebra which is not isomorphic to $\frak G_{\omega}$.
%546Id

\leader{546Y}{Further exercises (a)}
%\spheader 546Ya
Let $\frak A$ be a Dedekind complete Boolean algebra, and
set $X=\NN$.   Let $\Bbb P$ be the forcing notion
$(\frak A\setminus\{0\},\Bsubseteqshort,1,\downarrow)$ (5A3M).
Show that $\frak A$ has an e-h family iff

\Centerline{$\VVdP$ there is an $\alpha\in\NN$ such that
$\alpha\cap\beta$ is infinite for every $\beta\in\check X$.}
%546F

\spheader 546Yb Suppose that $\frak A$ is a Boolean algebra, not $\{0\}$, 
and that $\kappa$ is a cardinal such that for every family
$\langle a_{\xi n}\rangle_{\xi<\kappa,n\in\Bbb N}$ in $\frak A$
there is a family $\langle c_{m\sigma}\rangle_{m\in\Bbb N,\sigma\in S_2}$
in $\frak A$ such that

\Centerline{$c_{m\tau}\Bsubseteq c_{m,\tau^{\smallfrown}\sigma}$ for every
$\sigma\in S_2$,
\quad$\sup_{\sigma\in S_2}c_{m,\tau^{\smallfrown}\sigma}=1$}

\noindent for every $\tau\in S_2$ and $m\in\Bbb N$, and

$$\inf_{m\in\Bbb N}\sup_{\sigma\in S_2}\bigl(c_{m\sigma}
  \Bcap\inf_{\Atop{i<\#(\sigma)}{\sigma(i)=1}}a_{\xi i}
  \Bsetminus\sup_{\Atop{i<\#(\sigma)}{\sigma(i)=0}}a_{\xi i}\bigr)
=0$$

\noindent for every $\xi<\kappa$.   Let $\Bbb P$ be the forcing notion
$(\frak A\setminus\{0\},\Bsubseteqshort,1,\downarrow)$.
Show that $\VVdP\,\,\non\Cal M>\check\kappa$.
%546I

\spheader 546Yc Let $\frak A$ be an atomless
ccc Dedekind complete Boolean algebra, not $\{0\}$.   Suppose
that every order-closed subalgebra of $\frak A$ with countable Maharam type
is purely atomic.   (See 539P.)  Show that $\frak A$ is not a \pssqa.
%mt54bits
}%end of exercises

% Kumar, message of 21.2.16, offers a model in which
% both $\frak B_{\omega}$ and $\frak G_{\omega}$ can be regularly
% embedded in a ccc \pssqa?

\endnotes{
\Notesheader{546} The substantial ideas of this section are all in 546I.
The results there were developed in the process of understanding the
forcing universes associated with \pssqa{s};  I have spelt out two of the
translations involved in 546Ya-546Yb.
I don't present them in this form in the main exposition because
none of the principal theorems of the present chapter rely on the Forcing
Theorem, and it is therefore not strictly necessary to know anything about
forcing to follow the proofs as written.   But it is hard to imagine that
anyone would have come to the formulae in 546Ic without having first
considered whether 546Yb could be true.   Let me emphasize that the methods
of 546I are very special to quotient algebras of power sets.

An obvious question arises immediately from the definition in 546Aa.
For a given \pssqa\ $\frak A$, how much scope for
variation is there in the pairs $(X,\Cal I)$ such that
$\frak A\cong\Cal PX/\Cal I$?   In view of 546I, it seems natural to begin
with the cardinals $\#(X)$ and $\add\Cal I$.   But I do not have even a
well-formed problem to pose here.   It may make a difference if we restrict
our attention to normal \pssqa{s}.   I introduce these because we shall
see in the next section that the category algebra $\frak G_{\omega_2}$ is not a
normal \pssqa\ (547F), but conceivably is a \pssqa.
}%end of notes

\discrpage
\frfilename{mt547.tex}
\versiondate{24.10.20}
\copyrightdate{2015}

\def\chaptername{Real-valued-measurable cardinals}
\def\sectionname{Cohen algebras and $\sigma$-measurable algebras}


\newsection{547}

I examine the conditions under which two classes of algebra can be
\pssqa{s}.   In the first, shorter, part of the section (547B-547G) I
look at Cohen algebras.   I then turn to `$\sigma$-measurable' algebras
(547H-547S).

\leader{547A}{Notation} If $I$ is a set, $\frak G_I$ will be
the category algebra of $\{0,1\}^I$.   
If $\Cal I$ is an ideal of subsets of $X$ and $\Cal J$ an
ideal of subsets of $Y$, then

\Centerline{$\Cal I\ltimes\Cal J
=\{W:W\subseteq X\times Y$, $\{x:W[\{x\}]\notin\Cal J\}\in\Cal I\}$,}

\Centerline{$\Cal I\rtimes\Cal J
=\{W:W\subseteq X\times Y$, $\{y:W^{-1}[\{y\}]\notin\Cal I\}\in\Cal J\}$
\ifresultsonly{.}\else{}\fi}

\noindent\cmmnt{(527B). }If $X$ is a topological space, $\Cal M(X)$ will be its
meager ideal, $\CalBa(X)$ its Borel $\sigma$-algebra
and $\widehat{\Cal B}(X)$ its Baire-property algebra.
If $\frak A$ is a Boolean algebra, $\frak C$ is a subalgebra of $\frak A$
and $a\in\frak A$, then $\upr(a,\frak C)=\min\{c:a\Bsubseteq c\in\frak C\}$
if this is defined\cmmnt{ (313S)}.

I will write $S_2$ for
$\bigcup_{n\in\Bbb N}\{0,1\}^n$, ordered by
$\subseteq$.   For $\sigma\in S_2$, set
$I_{\sigma}=\{z:\sigma\subseteq z\in\{0,1\}^{\Bbb N}\}$;  then
$\sigma$, $\tau\in S_2$ are (upwards) incompatible iff
neither extends the other iff $I_{\sigma}\cap I_{\tau}=\emptyset$,
while $\{I_{\sigma}:\sigma\in S_2\}$ is a base for the usual topology of
$\{0,1\}^{\Bbb N}$.

\leader{547B}{}\cmmnt{ In 527M I introduced `harmless' algebras.
Here we need to know a little about harmless \pssqa{s}.

\medskip

\noindent}{\bf Lemma} Suppose that $\kappa$ is a regular uncountable
cardinal and $\Cal I$ is a $\kappa$-additive
ideal of subsets of $\kappa$ such that $\Cal P\kappa/\Cal I$ is
harmless.   Then
$\cmmnt{\frak A=\mskip5mu}
\Cal P(\kappa\times\kappa)/\Cal I\ltimes\Cal I$ is harmless.

\proof{{\bf (a)} If $\kappa=\omega_1$ then $\Cal I$ is of the
form $\{I:A\cap I=\emptyset\}$ for some
countable $A\subseteq\kappa$.   \Prf\ Set
$A=\{\xi:\xi<\omega_1$, $\{\xi\}\notin\Cal I\}$.
Because $\Cal P\omega_1/\Cal I$ is harmless, it is ccc,
so $A$ is countable.   Set $\Cal J=\Cal I\cap\Cal P(\omega_1\setminus A)$,
so that $\Cal J$ is a
$\sigma$-ideal of subsets of $\omega_1\setminus A$ containing
singletons, and $\Cal P(\omega_1\setminus A)/\Cal J$ can be identified
with the principal ideal of $\Cal P\omega_1/\Cal I$ generated by
$(\omega_1\setminus A)^{\ssbullet}$, so is ccc.
Since $\omega_1$ is certainly
not weakly inaccessible, $\Cal J$ is not a proper ideal, by 541L, and
$\omega_1\setminus A\in\Cal I$.   It follows that
$\Cal I=\Cal P(\omega_1\setminus A)$, as stated.\ \Qed

In this case, $\frak A\cong\Cal P(A\times A)$ has a countable $\pi$-base
and is harmless, by 527Nd.
So let us suppose from now on that $\kappa>\omega_1$.

\medskip

{\bf (b)} By 527Bb, $\Cal I\ltimes\Cal I$ is $\kappa$-additive;  in
particular,
it is a $\sigma$-ideal.   Next, it is $\omega_1$-saturated.   \Prf\
Let $\ofamily{\alpha}{\omega_1}{V_{\alpha}}$ be a disjoint family in
$\Cal P(\kappa\times\kappa)$.   For each $\xi<\kappa$,
there is an $\alpha_{\xi}<\omega_1$ such that
$V_{\alpha}[\{\xi\}]\in\Cal I$ for every
$\alpha\ge\alpha_{\xi}$.   Because $\Cal I$ is $\omega_2$-additive and
$\omega_1$-saturated, there is an $\alpha^*<\omega_1$ such that
$\{\xi:\alpha_{\xi}>\alpha^*\}\in\Cal I$ (541E).   But now
$V_{\alpha}\in\Cal I\ltimes\Cal I$ for every $\alpha>\alpha^*$.\ \Qed

Consequently $\frak A$ is Dedekind complete (541B).

\medskip

{\bf (c)} Let $\frak C$ be an order-closed subalgebra of
$\frak A$ with countable Maharam type;  let $\sequencen{C_n}$ be a sequence of
subsets of $\kappa\times\kappa$ such that
$\frak C$ is the order-closed subalgebra of
$\frak A$ generated by $\{C_n^{\ssbullet}:n\in\Bbb N\}$.
For each $\xi<\kappa$, let $\frak B_{\xi}$ be the
order-closed subalgebra of $\Cal P\kappa/\Cal I$ generated by
$\{C_n[\{\xi\}]^{\ssbullet}:n\in\Bbb N\}$.   By 527Nb, $\frak B_{\xi}$ has
countable $\pi$-weight;  let $\Cal E_{\xi}$ be a countable subset
of $\Cal P\kappa$ such that $\{E^{\ssbullet}:E\in\Cal E_{\xi}\}$ is an order-dense
subset of $\frak B_{\xi}$;  let $\sequencen{E_{\xi n}}$ run over $\Cal E_{\xi}$.
We can of course suppose that $C_n[\{\xi\}]\in\Cal E_{\xi}$ and that
$E_{\xi,2n}=C_n[\{\xi\}]$ for each $n$.   For $n\in\Bbb N$ set
$E_n=\{(\xi,\eta):\xi<\kappa$, $\eta\in E_{\xi n}\}$;
we have $E_{2n}=C_n$ for each $n$.   Let $\frak B$ be the order-closed subalgebra of
$\Cal P\kappa/\Cal I$ generated by
$\{\{\xi:E_{\xi m}\cap E_{\xi n}\in\Cal I\}^{\ssbullet}:m$, $n\in\Bbb N\}$, and
$\Cal F$ a countable subset of $\Cal P\kappa$, containing $\kappa$ and
$\emptyset$,
such that $\{F^{\ssbullet}:F\in\Cal F\}$ is an order-dense set in $\frak B$.
Let $\Cal A$ be the countable set
$\{\emptyset\}\cup\{E_m\cap(F\times\kappa):m\in\Bbb N$, $F\in\Cal F\}$.

\medskip

{\bf (d)} If $\sequencen{A_n}$ is any sequence in $\Cal A$, there is a sequence
$\sequencen{\hat A_n}$ in $\Cal A$ such that
$\sup_{n\in\Bbb N}\hat A_n^{\ssbullet}$
is the complement of $\sup_{n\in\Bbb N}A_n^{\ssbullet}$ in $\frak A$.
\Prf\ Express
$A_n$ as $E_{m_n}\cap(F_n\times\kappa)$ where $m_n\in\Bbb N$ and
$F_n\in\Cal F$ for
each $n$.   Set $W=\bigcup_{n\in\Bbb N}A_n$.   If $\xi<\kappa$ then
$W[\{\xi\}]=\bigcup_{n\in\Bbb N,\xi\in F_n}E_{\xi m_n}$, so
$W[\{\xi\}]^{\ssbullet}\in\frak B_{\xi}$;  set
$J_{\xi}=\{j:W[\{\xi\}]\cap E_{\xi j}\in\Cal I\}$.   If
$G_j=\{\xi:\xi<\kappa$, $j\in J_{\xi}\}$ for each $j$ and
$\hat W=\bigcup_{j\in\Bbb N}E_j\cap(G_j\times\kappa)$,
$W[\{\xi\}]^{\ssbullet}$ and $\hat W[\{\xi\}]^{\ssbullet}$ are
complementary elements of
$\frak B_{\xi}$ for each $\xi$, so $W^{\ssbullet}$ and
$\hat W^{\ssbullet}$ are complementary elements of $\frak A$.

Now

\Centerline{$G_j=\{\xi:W[\{\xi\}]\cap E_{\xi j}\in\Cal I\}
=\bigcap_{n\in\Bbb N}\{\xi:\xi\notin F_n$ or
$E_{\xi m_n}\cap E_{\xi j}\in\Cal I\}$,}

\noindent so $G_j^{\ssbullet}\in\frak B$ and there is an
$\Cal F_j\subseteq\Cal F$
such that $G_j^{\ssbullet}=\sup_{F\in\Cal F_j}F^{\ssbullet}$.   Taking
$\sequencen{\hat A_n}$ to run over
$\{\emptyset\}\cup\{E_j\cap(F\times\kappa):j\in\Bbb N$, $F\in\Cal F_j\}$,
we get a sequence in $\Cal A$ such that
$\sup_{n\in\Bbb N}\hat A_n^{\ssbullet}=\hat W^{\ssbullet}$, as required.\
\Qed

\medskip

{\bf (e)} It follows that if we take $\frak D$ to be the set of those
$a\in\frak A$
expressible in the form $\sup_{n\in\Bbb N}A_n^{\ssbullet}$ for some
sequence in
$\Cal A$, the complement of an element of $\frak D$ belongs to $\frak D$;  as
$\frak D$ is certainly closed under countable suprema,
it is a $\sigma$-subalgebra of $\frak A$, therefore order-closed, because
$\frak A$ is ccc.   And $\{A_n^{\ssbullet}:n\in\Bbb N\}$ witnesses that
$\pi(\frak D)\le\omega$.

As $C_n=E_{2n}\cap(\kappa\times\kappa)\in\Cal A$ for each $n$,
$\frak C\subseteq\frak D$.   So
$\omega\ge\pi(\frak D)\ge\pi(\frak C)$, by 514Eb.

As $\frak C$ is an arbitrary countably generated order-closed subalgebra
of $\frak A$, $\frak A$ is harmless, by 527Nb in the other direction.
}%end of proof of 547B

\leader{547C}{}\cmmnt{ I wish to follow the lines of the argument in
543C-543E to
prove a similar theorem in which `measure' is replaced by `category'.
The lemma just
proved corresponds to the definition of $\tilde\nu$ in part (c) of the
proof of 543D.
The next result will play the role previously taken by 543C.

\medskip

\noindent}{\bf Proposition} Suppose that $\kappa$ is a regular uncountable
cardinal
and $\Cal I$ is a $\kappa$-additive ideal of subsets of $\kappa$ such that
$\Cal P\kappa/\Cal I$ is harmless.   Let $X$ be a
ccc topological space of $\pi$-weight less than $\kappa$.   Then
$\Cal M(X)\rtimes\Cal I\subseteq\Cal M(X)\ltimes\Cal I$.

\proof{{\bf (a)} Take $C\in\Cal M(X)\rtimes\Cal I$.
Set $A=\{\xi:\xi<\kappa$, $C^{-1}[\{\xi\}]\notin\Cal M(X)\}$ and
$B=\{x:x\in X$, $C[\{x\}]\notin\Cal I\}$, so that
$A\in\Cal I$ and I need to show that $B\in\Cal M(X)$.
For $\xi\in\kappa\setminus A$, let $\sequencen{F_{\xi n}}$ be a
sequence of nowhere dense sets in $X$ with union $C^{-1}[\{\xi\}]$;
for $\xi\in A$ set $F_{\xi n}=\emptyset$ for every $n$.   For each $n$,
set $C_n=\{(x,\xi):\xi<\kappa$, $x\in F_{\xi n}\}$ and
$B_n=\{x:C_n[\{x\}]\notin\Cal I\}$, so that $B=\bigcup_{n\in\Bbb N}B_n$ and it
will be enough to show that every $B_n$ is meager.   Fix $n\in\Bbb N$.

\medskip

{\bf (b)} Let $\ofamily{\alpha}{\pi(X)}{G_{\alpha}}$ enumerate a
$\pi$-base for the topology of $X$, and for $\alpha<\pi(X)$ let $D_{\alpha}$ be the
set of those $\xi<\kappa$ such that $G_{\alpha}\cap F_{\xi n}=\emptyset$.   Then
$W=\bigcup_{\alpha<\pi(X)}G_{\alpha}\times D_{\alpha}$ is disjoint from
$C_n$.   For each $\xi<\kappa$, set
$I_{\xi}=\{\alpha:\alpha<\pi(X)$, $\xi\in D_{\alpha}\}$;
then $\bigcup_{\alpha\in I_{\xi}}G_{\alpha}$ is dense in $X$.
Because $X$ is ccc,
there is a countable $J_{\xi}\subseteq I_{\xi}$ such that
$\bigcup_{\alpha\in J_{\xi}}G_{\alpha}$ is dense (5A4Bd).   Now $\Cal I$ is
$\omega_1$-saturated and $\pi(X)<\add\Cal I$, so there is a countable
$I\subseteq\pi(X)$
such that $A'=\{\xi:\xi<\kappa$, $J_{\xi}\not\subseteq I\}$ belongs to
$\Cal I$ (541D).

\medskip

{\bf (c)} Let $\frak B$ be the order-closed subalgebra of
$\Cal P\kappa/\Cal I$ generated by
$\{D_{\alpha}^{\ssbullet}:\alpha\in I\}$.
Because $\Cal P\kappa/\Cal I$ is harmless, $\pi(\frak B)\le\omega$
(527Nb again);  
let $\sequence{i}{F_i}$ be a sequence in $\Cal P\kappa$ such that
$\{F_i^{\ssbullet}:i\in\Bbb N\}$ is order-dense in $\frak B$.
Let $\Cal E$ be the countable subalgebra of $\Cal P\kappa$ generated by
$\{F_i:i\in\Bbb N\}\cup\{D_{\alpha}:\alpha\in I\}$, and set
$V=\bigcup(\Cal E\cap\Cal I)$, so that $V\in\Cal I$.
Give $Y=\kappa\setminus V$ the
second-countable topology generated by $\{E\setminus V:E\in\Cal E\}$.

If $H$ is a dense open set in $Y$, then $\kappa\setminus H\in\Cal I$.
\Prf\ Setting $\Cal E'=\{E:E\in\Cal E$, $E\setminus V\subseteq H\}$,
$H=(\bigcup\Cal E')\setminus V$, so
$H^{\ssbullet}=\sup_{E\in\Cal E'}E^{\ssbullet}$ in
$\Cal P\kappa/\Cal I$, and $H^{\ssbullet}\in\frak B$.   \Quer\ If
$\kappa\setminus H\notin\Cal I$, then $H^{\ssbullet}\ne 1$ and there is an
$i\in\Bbb N$ such that $F_i^{\ssbullet}$ is non-zero and disjoint from
$H^{\ssbullet}$.   In this case, $F_i\cap E\in\Cal I$ for every $E\in\Cal E'$, so
$F_i\setminus V$ is disjoint from $H$;  but $F_i\setminus V$ is a non-empty open
subset of $Y$.\ \Bang\Qed

Consequently $\Cal M(Y)\subseteq\Cal I$.

\medskip

{\bf (d)}
Set $W_0=\bigcup_{\alpha\in I}G_{\alpha}\times(D_{\alpha}\setminus V)$, 
so that $W_0$ is an open set in $X\times Y$.   Then $W_0$ is dense in 
$X\times Y$.
\Prf\Quer\ Otherwise, we have a non-empty open set $G\subseteq X$ and a 
non-empty open set $U\subseteq Y$ such that $I=I'\cup I''$, where
$I'=\{\alpha:\alpha\in I$, $G\cap G_{\alpha}=\emptyset\}$ and
$I''=\{\alpha:\alpha\in I$, $U\cap D_{\alpha}=\emptyset\}$.   As $U$ includes some
non-empty set of the form $E\setminus V$ where $E\in\Cal E$, $U\notin\Cal I$.
So there must be a $\xi\in U\setminus A'$.   In this case, $J_{\xi}\subseteq I$
while
$\bigcup_{\alpha\in J_{\xi}}G_{\alpha}$ is dense and meets $G$;  there is therefore
an $\alpha\in J_{\xi}\setminus I'$.   But now $\alpha\in I_{\xi}$ so
$\xi\in D_{\alpha}$, while also $\alpha\in I\setminus I'\subseteq I''$, 
so $U\cap D_{\alpha}=\emptyset$ and $\xi\notin D_{\alpha}$.\ \Bang\Qed

\medskip

{\bf (e)} $W'=(X\times Y)\setminus W$ is therefore meager in $X\times Y$
and belongs to $\Cal M(X)\ltimes\Cal M(Y)$, by 527Db.
If $x\in B_n$, then $C_n[\{x\}]\notin\Cal I$;  but
$C_n[\{x\}]\setminus V\subseteq W'[\{x\}]$,
so $W'[\{x\}]\notin\Cal I$ and $W'[\{x\}]\notin\Cal M(Y)$.
Accordingly

\Centerline{$B_n\subseteq\{x:W'[\{x\}]\notin\Cal M(Y)\}\in\Cal M(X)$,}

\noindent as required.
}%end of proof of 547C

\leader{547D}{Corollary} Suppose that $\kappa$ is a regular uncountable
cardinal
and $\Cal I$ is a proper $\kappa$-additive ideal of subsets of $\kappa$,
containing singletons, such that
$\Cal P\kappa/\Cal I$ is harmless.   Let $X$ be a
ccc topological space of $\pi$-weight less than $\kappa$.   

(a) Suppose that $\ofamily{\xi}{\kappa}{A_{\xi}}$ is a non-decreasing
family of subsets of $X$ with union $A$.   Then there is a
$\theta<\kappa$ such that $E\cap A_{\theta}$
is non-meager whenever $E\subseteq X$ is a set
with the Baire property and $E\cap A$ is not meager.

(b) If $\ofamily{\xi}{\kappa}{A_{\xi}}$ is a family in
$\Cal PX\setminus\Cal M(X)$ such that
$\#(\bigcup_{\xi<\kappa}A_{\xi})<\kappa$, then there are distinct $\xi$,
$\eta<\kappa$ such that $A_{\xi}\cap A_{\eta}\notin\Cal M(X)$.

(c) If we have a family $\ofamily{\xi}{\kappa}{h_{\xi}}$ of functions such that
$\dom h_{\xi}$ is a non-meager subset of $X$ for each $\xi$ and
$\#(\bigcup_{\xi<\kappa}h_{\xi})<\kappa$\cmmnt{ (identifying
each $h_{\xi}$ with its graph)}, then there are distinct
$\xi$, $\eta<\kappa$ such that
$\{x:h_{\xi}(x)$ and $h_{\eta}(x)$ are defined and equal$\}$ is non-meager.

\proof{{\bf (a)} Let $\frak G=\widehat{\Cal B}(X)/\Cal M(X)$
be the category algebra of $X$;  for $B\subseteq X$
set $\psi(B)=\inf\{E^{\ssbullet}:B\subseteq E\in\widehat{\Cal B}(X)\}$,
as in 514Ie.
Because $\frak G$ is ccc (514Ja), there is a sequence $\sequencen{\xi_n}$
in $\kappa$ such that
$\sup_{n\in\Bbb N}\psi(A_{\xi_n})=\sup_{\xi<\kappa}\psi(A_{\xi})$;  setting
$\theta=\sup_{n\in\Bbb N}\xi_n$, we see that $\theta<\kappa$ (because
$\cf\kappa>\omega$) and that $\psi(A_{\xi})\Bsubseteq\psi(A_{\theta})$ for every
$\xi<\kappa$.

\Quer\ Suppose, if possible, that there is a set $E\in\widehat{\Cal B}(X)$ such that
$E\cap A_{\theta}$ is meager but $E\cap A$ is non-meager.   Replacing
$E$ by $E\setminus A_{\theta}$ if
necessary, we may suppose that $E\cap A_{\theta}$ is empty.   If
$\xi<\kappa$, then

\Centerline{$\psi(E\cap A_{\xi})
\Bsubseteq E^{\ssbullet}\cap\psi(A_{\theta})
\Bsubseteq E^{\ssbullet}\cap(X\setminus E)^{\ssbullet}=0$,}

\noindent so $E\cap A_{\xi}$ is meager.

Define $f:E\cap A\to\kappa$ by setting $f(x)=\min\{\xi:x\in A_{\xi}\}$ for $x\in E$.
Consider the set

\Centerline{$C=\{(x,\xi):f(x)\le\xi<\kappa\}\subseteq(E\cap A)\times\kappa$.}

\noindent If $\xi<\kappa$, then

\Centerline{$C^{-1}[\{\xi\}]=\{x:x\in E$, $f(x)\le\xi\}
\subseteq E\cap A_{\xi}\in\Cal M(X)$;}

\noindent thus $C\in\Cal M(X)\rtimes\Cal I$.   By 547C,
$C\in\Cal M(X)\ltimes\Cal I$.   As $E\cap A$ is not meager, there is an $x\in E\cap A$
such that $C[\{x\}]\in\Cal I$.   But
$C[\{x\}]=\{\xi:f(x)\le\xi<\kappa\}\notin\Cal I$.\ \Bang

So $\theta$ has the required property.

\medskip

{\bf (b)} Write
$\Cal J=\Cal I\ltimes\Cal I\normalsubgroup\Cal P(\kappa\times\kappa)$.
By 547B, $\Cal P(\kappa\times\kappa)/\Cal J$ is harmless.   Set

\Centerline{$W=\{(x,\xi,\eta):\xi$, $\eta<\kappa$, $\xi\ne\eta$,
$x\in A_{\xi}\cap A_{\eta}\}$.}

\noindent\Quer\ If $A_{\xi}\cap A_{\eta}\in\Cal M(X)$ for all distinct $\xi$,
$\eta<\kappa$, then $W$, regarded as a subset of $X\times(\kappa\times\kappa)$,
belongs to $\Cal M(X)\rtimes\Cal J$;  by 547C, $W\in\Cal M(X)\ltimes\Cal J$.
For $x\in X$ set $C_x=\{\xi:\xi<\kappa$, $x\in A_{\xi}\}$.   Then
$W[\{x\}]=C_x^2\setminus\Delta$, where $\Delta=\{(\xi,\xi):\xi<\kappa\}$.   So
$W[\{x\}]\in\Cal J$ iff $C_x\in\Cal I$, and
$E=\{x:C_x\notin\Cal I\}$ is meager.   Next, $A=\bigcup_{\xi<\kappa}A_{\xi}$ is
supposed to have cardinal less than $\kappa$, so
$\bigcup_{x\in A\setminus E}C_x\in\Cal I$ and there is some
$\zeta\in\kappa\setminus\bigcup_{x\in A\setminus E}C_x$.   But in this case
$A_{\zeta}\subseteq E$ is meager.\ \BanG\  So we have the result.

\medskip

{\bf (c)(i)} For each $\xi<\kappa$,
set $A_{\xi}=\dom h_{\xi}$ and let $H_{\xi}$ be
the regular open set in $X$ such that $A_{\xi}\setminus H_{\xi}$
is meager and $G\cap H_{\xi}$ is empty whenever $G$ is open and
$G\cap A_{\xi}$ is meager (4A3Sa\formerly{4{}A3Ra}).   
Set $h'_{\xi}=h_{\xi}\restr H_{\xi}$ and
$Y=\bigcup_{\xi<\kappa}h'_{\xi}$;  let
$\pi_1:Y\to X$ be the first-coordinate projection.   Give $Y$ the topology
$\frak S=\{\pi_1^{-1}[G]:G\in\frak T\}$, where $\frak T$ is the topology
of $X$.

\medskip

\quad{\bf (ii)} If $\Cal U$ is any $\pi$-base for $\frak T$, then
$\Cal V=\{\pi_1^{-1}[U]:U\in\Cal U\}$ is a $\pi$-base for $\frak S$.
\Prf\ If $H\subseteq Y$ is open and not empty, take $G\in\frak T$ such that
$H=\pi_1^{-1}[G]$ and a $\xi<\kappa$ such that $H\cap h'_{\xi}\ne\emptyset$.
Then $G\cap H_{\xi}\cap A_{\xi}=G\cap\dom h'_{\xi}$ is non-empty;
by the choice of
$H_{\xi}$, $G\cap H_{\xi}\cap A_{\xi}$ is non-meager.   Set
$\Cal U'=\{U:U\in\Cal U$, $U\cap G\cap H_{\xi}\cap A_{\xi}=\emptyset\}$.   Then
$\bigcup\Cal U'$ cannot be dense and there is a non-empty $U\in\Cal U$
disjoint from
$\bigcup\Cal U'$.   But now $U\cap G\ne\emptyset$, so there is a non-empty
$U'\in\Cal U$ with $U'\subseteq U\cap G$;  in which case
$V=\pi_1^{-1}[U']$ belongs to
$\Cal V$, is included in $H$ and meets $h'_{\xi}$, so is not empty.
As $H$ is arbitrary, $\Cal V$ is a $\pi$-base for $\frak S$.\ \QeD\

\medskip

\quad{\bf (iii)}
It follows at once that $\pi(Y)\le\pi (X)<\kappa$.   We see also that if
$A\subseteq X$ is nowhere dense, then $\{G:G\in\frak T$, $G\cap A=\emptyset\}$ is a
$\pi$-base for $\frak T$,

\Centerline{$\{\pi_1^{-1}[G]:G\in\frak T$, $G\cap A=\emptyset\}
=\{H:H\in\frak S$, $H\cap\pi_1^{-1}[A]=\emptyset\}$}

\noindent is a $\pi$-base for $\frak S$ and $\pi_1^{-1}[A]$ is nowhere dense in $Y$.
Accordingly $\pi_1^{-1}[A]\in\Cal M(Y)$ for every $A\in\Cal M(X)$.

\medskip

\quad{\bf (iv)}
If $B\subseteq Y$ is nowhere dense in $Y$ then $\pi_1[B]$ is nowhere dense in $X$.
\Prf\ If $G\subseteq X$ is a non-empty open set, then either $\pi_1^{-1}[G]$ is empty
and $G\cap\pi_1[B]=\emptyset$, or $\pi_1^{-1}[G]$ is a non-empty open subset of $Y$.
In the latter case, $\pi_1^{-1}[G]\setminus\overline{B}$ must be of the form
$\pi_1^{-1}[G']$ for some open set $G'\subseteq X$, and $G'\cap G$ is a non-empty open
subset of $G$ disjoint from $\pi_1[B]$.\ \QeD\
It follows at once that $\pi_1[B]\in\Cal M(X)$ whenever $B\in\Cal M(Y)$.

\medskip

\quad{\bf (v)}
Since $\pi_1[h'_{\xi}]=A_{\xi}\cap H_{\xi}$ is non-meager in $X$, $h'_{\xi}$ is
non-meager in $Y$, for every $\xi$.   So (b) here tells us that there are distinct
$\xi$, $\eta<\kappa$ such that $h'_{\xi}\cap h'_{\eta}$ is non-meager in $Y$.
In this case, setting $A=\{x:h_{\xi}(x)$ and $h_{\xi}(y)$ are defined and equal$\}$,
$\pi_1^{-1}[A]$ includes $h'_{\xi}\cap h'_{\eta}$ so is non-meager, and $A$ is
non-meager, by (iii).
}%end of proof of 547D

\leader{547E}{}\cmmnt{ I separate an elementary fact from the argument of
the next theorem.

\medskip

\noindent}{\bf Lemma} Suppose that $X$, $I$ are sets, $\Cal I$ is a
$\sigma$-ideal of subsets of $X$ and $\phi$
is a sequentially order-continuous Boolean homomorphism from $\frak G_I$  
to $\Cal PX/\Cal I$.   Then there is a function $f:X\to\{0,1\}^I$ such that
$f^{-1}[E]^{\ssbullet}=\phi E^{\ssbullet}$ in $\frak A$
for every $E$ in the
Baire-property algebra $\widehat{\Cal B}$ of $\{0,1\}^I$.

\proof{ Write $\Cal M$ for the meager ideal of $\{0,1\}^I$.
For $i\in I$ set
$H_i=\{y:y\in\{0,1\}^I$, $y(i)=1\}\in\widehat{\Cal B}$ and choose
$F_i\subseteq X$ such that $F_i^{\ssbullet}=\phi H_i^{\ssbullet}$ in
$\frak G=\widehat{\Cal B}/\Cal M$.
Set $f(x)=\familyiI{\chi F_i(x)}$ for $x\in X$;  then
$f^{-1}[H_i]^{\ssbullet}=F_i^{\ssbullet}=\phi H_i^{\ssbullet}$ for every
$i$.   Since $H\mapsto f^{-1}[H]^{\ssbullet}$ and
$H\mapsto\phi H^{\ssbullet}$ are both sequentially order-continuous Boolean
homomorphisms, they must agree on the $\sigma$-algebra of subsets of
$\{0,1\}^I$ generated by $\{H_i:i\in I\}$, which is the Baire
$\sigma$-algebra $\CalBa$ of $\{0,1\}^I$ (4A3Of).
Next, as $\{0,1\}^I$ is completely regular and ccc, $\Cal M$ is generated
by $\CalBa\cap\Cal M$ (5A4E(d-ii)) and $f^{-1}[H]\in\Cal I$ for every
$H\in\Cal M$.   Now if $E\in\widehat{\Cal B}$ it is of the form
$G\symmdiff M$ where $G$ is a cozero set and $M$ is meager
(5A4E(d-iii)), so

\Centerline{$f^{-1}[E]^{\ssbullet}
=f^{-1}[G]^{\ssbullet}\Bsymmdiff f^{-1}[M]^{\ssbullet}
=\phi G^{\ssbullet}=\phi E^{\ssbullet}$.}
}%end of proof of 547E

\leader{547F}{The Gitik-Shelah theorem for
Cohen \dvrocolon{algebras}}\cmmnt{ I come
now to a companion result to the Gitik-Shelah Theorem in 543E.
I follow the proof I gave before as closely as I can.

\medskip

\noindent}{\bf Theorem}\cmmnt{ ({\smc Gitik \& Shelah 89},
{\smc Gitik \& Shelah 93})} Let $\kappa$ be a regular uncountable
cardinal and
$\Cal I$ a $\kappa$-additive ideal of subsets of $\kappa$ such that
$\cmmnt{\frak A=\mskip5mu}\Cal P\kappa/\Cal I$
is isomorphic to $\frak G_{\lambda}$ for some infinite
cardinal $\lambda$.   Then $\lambda\ge\min(\kappa^{(+\omega)},2^{\kappa})$.

\proof{{\bf (a)} Let $\phi:\frak G_{\lambda}\to\frak A$
be an isomorphism.   By 547E, we have a function
$f:\kappa\to X$ such that $f^{-1}[E]^{\ssbullet}=\phi E^{\ssbullet}$ in
$\frak A$ for every $E\in\widehat{\Cal B}(X)$.

\medskip

{\bf (b)}\Quer\ Suppose, if possible, that
$\lambda<\min(\kappa^{(+\omega)},2^{\kappa})$.

As $\frak A$ is atomless, ccc and not $\{0\}$, $\kappa$ is quasi-measurable
(542B) and at most $\frak c$ (541P).   So 543Ma tells us that if we set
$\zeta=\max(\lambda^+,\kappa^+)$, there are an infinite cardinal
$\delta<\kappa$, a stationary set $S\subseteq\zeta$, and a family
$\family{\alpha}{S}{g_{\alpha}}$ of functions from $\kappa$ to
$2^{\delta}$ such that $g_{\alpha}[\kappa]\subseteq\alpha$ for every
$\alpha\in S$ and $\#(g_{\alpha}\cap g_{\beta})<\kappa$ for distinct
$\alpha$, $\beta\in S$.   Moreover, we can arrange that

----- if $\lambda<\Tr(\kappa)$, then 
$g_{\alpha}[\kappa]\subseteq\kappa$ for every $\alpha\in S$;

----- if $\lambda\ge\Tr(\kappa)$, than
$g_{\alpha}\restr\gamma=g_{\beta}\restr\gamma$ whenever $\gamma<\kappa$
is a limit ordinal and $\alpha$, $\beta\in S$ are such that
$g_{\alpha}(\gamma)=g_{\beta}(\gamma)$.

\medskip

{\bf (c)} Fix an injective function $h:2^{\delta}\to\{0,1\}^{\delta}$.
For $\alpha\in S$ and $\iota<\delta$ set

\Centerline{$U_{\alpha\iota}
=\{\xi:\xi<\kappa$, $(hg_{\alpha}(\xi))(\iota)=1\}$,}

\noindent and let
$H_{\alpha\iota}\in\CalBa(X)$ be such that
$H_{\alpha\iota}^{\ssbullet}=\phi^{-1}(U^{\ssbullet}_{\alpha\iota})$ in
$\frak G_{\lambda}$;  then
$U_{\alpha\iota}\symmdiff f^{-1}[H_{\alpha\iota}]\in\Cal I$.   Define
$\tilde g_{\alpha}:X\to\{0,1\}^{\delta}$ by setting

$$\eqalign{(\tilde g_{\alpha}(x))(\iota)
&=1\text{ if }x\in H_{\alpha\iota},\cr
&=0\text{ otherwise}.\cr}$$

\noindent Then

$$\eqalign{\{\xi:
  \xi<\kappa,\,\tilde g_{\alpha}f(\xi)\ne hg_{\alpha}(\xi)\}
&=\bigcup_{\iota<\delta}\{\xi:
  (\tilde g_{\alpha}f(\xi))(\iota)\ne(hg_{\alpha}(\xi))(\iota)\}\cr
&=\bigcup_{\iota<\delta}
  U_{\alpha\iota}\symmdiff f^{-1}[H_{\alpha\iota}]
\in\Cal I\cr}$$

\noindent because $\delta<\kappa=\add\Cal I$.
Set $V_{\alpha}=\{\xi:\tilde g_{\alpha}f(\xi)=hg_{\alpha}(\xi)\}$, so
that $\kappa\setminus V_{\alpha}\in\Cal I$, for each $\alpha\in S$.

\medskip

{\bf (d)} Because every $H_{\alpha\iota}$ is determined by coordinates
in a countable
set, there is for each $\alpha\in S$ a set $I_{\alpha}\subseteq\lambda$
such that $\#(I_{\alpha})\le\delta$ and $H_{\alpha\iota}$ is determined
by coordinates in $I_{\alpha}$ for every $\iota<\delta$.
By 5A1K there is an $M\subseteq\lambda$ such that
$S_1=\{\alpha:\alpha\in S$, $I_{\alpha}\subseteq M\}$
is stationary in $\zeta$ and $\cf(\#(M))\le\delta$;  because
$\lambda<\kappa^{(+\omega)}$ and $\cf(\kappa)=\kappa>\delta$,
$\#(M)<\kappa$.   Set $\pi_M(z)=z\restr M$ for $z\in X$,
and $f_M=\pi_Mf:\kappa\to\{0,1\}^M$.

If $E\subseteq\{0,1\}^M$ has the Baire property,
then $\pi_M^{-1}[E]$ has the Baire
property in $X$ and $\pi_M^{-1}[E]$ is meager iff $E$ is
(5A4E(c-iii), applied to $\{0,1\}^M\times\{0,1\}^{\lambda\setminus M}$).
So $f_M^{-1}[E]\in\Cal I$ iff $E$ is meager.

\medskip

{\bf (e)} For each $\alpha\in S_1$, there is a $\theta_{\alpha}<\kappa$
such that $f_M[V_{\alpha}\cap\theta_{\alpha}]$ meets every non-empty open subset of
$\{0,1\}^M$ in a non-meager set.
\Prf\ Apply 547Da to $\ofamily{\xi}{\kappa}{f_M[V_{\alpha}\cap\xi]}$.
$\bigcup\Cal I=\kappa$ because $\frak A$ is atomless, and
$\kappa\notin\Cal I$ because $\frak A\ne\{0\}$;
while $\{0,1\}^M$ is certainly ccc, and has $\pi$-weight at most
$\max(\omega,\#(M))<\kappa$.   There is therefore
a $\theta_{\alpha}<\kappa$ such that $E\cap f_M[V_{\alpha}\cap\theta]$ is
non-meager whenever $E\subseteq\{0,1\}^M$ has the Baire property
and $E\cap f_M[V_{\alpha}]$ is non-meager.
If $G\subseteq\{0,1\}^M$ is a non-empty open set, then

\Centerline{$f_M^{-1}[G\setminus f_M[V_{\alpha}]]
\subseteq\kappa\setminus V_{\alpha}\in\Cal I$,}

\noindent so either
$G\setminus f_M[V_{\alpha}]$ is meager or it does not have the
Baire property;  in
either case, $G\cap f_M[V_{\alpha}]$ is non-meager so
$G\cap f_M[V_{\alpha}\cap\theta_{\alpha}]$ is non-meager.\ \Qed

Evidently we may take it that every $\theta_{\alpha}$ is a non-zero
limit ordinal.

\medskip

{\bf (f)} By 543Mb, there are a $\theta<\kappa$ and a 
$Y\in[2^{\delta}]^{<\kappa}$ such that
$S_2=\{\alpha:\alpha\in S_1$, $\theta_{\alpha}=\theta\}$,
$g_{\alpha}[\theta]\subseteq Y\}$ is stationary in $\zeta$.

\medskip

{\bf (g)} For each $\alpha\in S_2$, set

\Centerline{$Q_{\alpha}=f_M[V_{\alpha}\cap\theta]
=f_M[V_{\alpha}\cap\theta_{\alpha}]$,}

\noindent so that $Q_{\alpha}$ meets every non-empty open subset of
$\{0,1\}^M$ in a non-meager set.  Now every $H_{\alpha\iota}$ is
determined by coordinates in $I_{\alpha}\subseteq M$, so
we can express $\tilde g_{\alpha}$ as
$g^*_{\alpha}\pi_M$,
where $g^*_{\alpha}:\{0,1\}^M\to\{0,1\}^{\delta}$ is
Baire measurable in each coordinate.   If $y\in Q_{\alpha}$, take
$\xi\in V_{\alpha}\cap\theta$ such that $f_M(\xi)=y$;  then

\Centerline{$g_{\alpha}^*(y)=g_{\alpha}^*\pi_Mf(\xi)
=\tilde g_{\alpha}f(\xi)=hg_{\alpha}(\xi)\in h[Y]$.}

\noindent Thus
$g_{\alpha}^*\restr Q_{\alpha}\subseteq f_M[\theta]\times h[Y]$ for
every $\alpha\in S_2$, and we may apply 547Dc to $\{0,1\}^M$ and the
family
$\langle g^*_{\alpha}\restr Q_{\alpha}\rangle_{\alpha\in S^*}$,
where $S^*\subseteq S_2$ is a set with cardinal $\kappa$,
to see that there are distinct $\alpha$, $\beta\in S_2$ such that
$\{y:y\in Q_{\alpha}\cap Q_{\beta}$, $g^*_{\alpha}(y)=g^*_{\beta}(y)\}$ is non-meager.
Now, however, consider

\Centerline{$E=\{y:y\in\{0,1\}^M$, $g^*_{\alpha}(y)=g^*_{\beta}(y)\}$.}

\noindent Then $E=\bigcap_{\iota<\delta}E_{\iota}$, where

\Centerline{$E_{\iota}
=\{y:y\in\{0,1\}^M$, $g^*_{\alpha}(y)(\iota)=g^*_{\beta}(y)(\iota)\}$}

\noindent is a Baire subset of $\{0,1\}^M$ for each $\iota<\delta$.
Because $\delta<\kappa$ and $\Cal I$ is $\kappa$-additive and $\omega_1$-saturated,

$$\eqalign{f_M^{-1}[E]^{\ssbullet}
&=(\bigcap_{\iota<\delta}f_M^{-1}[E_{\iota}])^{\ssbullet}
=\inf_{\iota<\delta}f_M^{-1}[E_{\iota}]^{\ssbullet}\cr
&=\inf_{\iota\in K}f_M^{-1}[E_{\iota}]^{\ssbullet}
=f_M^{-1}[\bigcap_{\iota\in K}E_{\iota}]^{\ssbullet}\cr}$$

\noindent for some countable $K\subseteq\delta$.   In this case,
$E'=\bigcap_{\iota\in K}E_{\iota}$ is a Baire set including $E$, and
$f_M^{-1}[E'\setminus E]\in\Cal I$;  since $E'$ includes the non-meager set
$\{y:y\in Q_{\alpha}\cap Q_{\beta}$, $g^*_{\alpha}(y)=g^*_{\beta}(y)\}$, $E'$ is
non-meager and $f_M^{-1}[E']\notin\Cal I$, by (d) above;
accordingly $f_M^{-1}[E]\notin\Cal I$.

Consequently

$$\eqalign{\{\xi:g_{\alpha}(\xi)=g_{\beta}(\xi)\}^{\ssbullet}
&=\{\xi:hg_{\alpha}(\xi)=hg_{\beta}(\xi)\}^{\ssbullet}\cr
&=\{\xi:\xi\in V_{\alpha}\cap V_{\beta},
  \,\tilde g_{\alpha}f(\xi)=\tilde g_{\beta}f(\xi)\}^{\ssbullet}\cr
&=\{\xi:g^*_{\alpha}\pi_M f(\xi)=g^*_{\beta}\pi_M f(\xi)\}^{\ssbullet}
=f^{-1}_M[E]^{\ssbullet}
\ne 0\cr}$$

\noindent in $\frak A$.   But this is
absurd, because in (b) above we chose
$g_{\alpha}$, $g_{\beta}$ in such a way that
$\{\xi:g_{\alpha}(\xi)=g_{\beta}(\xi)\}$ would be bounded in $\kappa$.\
\BanG\

Thus we have the required contradiction, and the theorem is true.
}%end of proof of 547F

\leader{547G}{Corollary} (a) $\frak G_{\omega}$ is not a \pssqa.

(b) $\frak G_{\omega_1}$ is not a \pssqa.

\proof{ I shall be applying results from the last part of \S515, where I
spoke of regular open algebras;  so we shall need the fact that $\frak G_I$
is isomorphic to the regular open algebra of $\{0,1\}^I$
for every set $I$ (515Na).

\medskip

{\bf (a)} \Quer\ Suppose that $\frak G_{\omega}$ is a \pssqa.
By 546Db there are a regular uncountable
cardinal $\kappa$ and a normal ideal $\Cal J$ on $\kappa$ such that
$\Cal P\kappa/\Cal J$ is isomorphic to an atomless $\sigma$-subalgebra
$\frak D$ of a principal ideal $(\frak G_{\omega})_c$ of $\frak G_{\omega}$.
By 515Nc and 515Ob, $(\frak G_{\omega})_c$ and $\frak D$ are isomorphic to
$\frak G_{\omega}$.   But by 547F this is impossible.\ \Bang

\medskip

{\bf (b)} We can argue similarly, but using 515Q in place of 515Ob.
\Quer\ Suppose that $\frak G_{\omega_1}$ is a \pssqa.
Then there are a regular uncountable
cardinal $\kappa$ and a normal ideal $\Cal J$ on $\kappa$ such that
$\Cal P\kappa/\Cal J$ is isomorphic to an atomless $\sigma$-subalgebra
$\frak D$
of a principal ideal $(\frak G_{\omega_1})_c$ of $\frak G_{\omega_1}$.
By 515Nc, $(\frak G_{\omega_1})_c$ is isomorphic to
$\frak G_{\omega_1}$.   By 515Q, $\frak D$ is isomorphic to either
$\frak G_{\omega}$ or $\frak G_{\omega_1}$ or their simple product.
But this means that at least one of $\frak G_{\omega}$,
$\frak G_{\omega_1}$ is isomorphic to a principal ideal of the normal
\pssqa\ $\frak D$, and is therefore itself a normal \pssqa\ (546C);
and both are ruled out by 547F.\ \Bang
}%end of proof of 547G

\leader{547H}{Definitions} Let $\frak A$ be a Boolean algebra.

\spheader 547Ha I will say that $\frak A$ is
{\bf $\sigma$-measurable}, with {\bf witnessing sequence
$\sequencen{\frak B_n}$}, if $\frak A$ is Dedekind complete, every
$\frak B_n$ is an order-closed subalgebra of $\frak A$ which is, in itself,
a measurable algebra\cmmnt{ in the sense of \S391},
and $\bigcup_{n\in\Bbb N}\frak B_n$ is order-dense
in $\frak A$\cmmnt{ in the sense of 313J}.

\spheader 547Hb
If $\frak A$ is $\sigma$-measurable algebra, I will say that

\Centerline{$\tausm(\frak A)
=\min\{\sum_{n=0}^{\infty}\tau(\frak B_n):
\sequencen{\frak B_n}$ is a witnessing sequence for $\frak A\}$,}

\noindent where the sums here are cardinal sums\cmmnt{ (5A1F(b-i))}.

%547I
\leader{547I}{Examples (a)} Every measurable algebra is
$\sigma$-measurable.

\spheader 547Ib If $\frak A$ is Dedekind complete and
has countable $\pi$-weight, it is $\sigma$-measurable.   \prooflet{(We can
take every $\frak B_n$ to be of the form $\{0,b,1\Bsetminus b,1\}$.)}
In particular, $\frak G_{\omega}$ is $\sigma$-measurable\cmmnt{ (515Oa)}.

\spheader 547Ic If $\frak A$ is a measurable algebra and $\frak B$ is a
Boolean algebra with countable $\pi$-weight, then the Dedekind
completion $\frak C$ of the free product
$\frak A\otimes\frak B$\cmmnt{ (315I, 314U)} is $\sigma$-measurable, with
$\tausm(\frak C)\le\max(\omega,\tau(\frak A))$.
\prooflet{\Prf\ If $\frak B=\{0\}$ then $\frak C=\{0\}$ and the result is
trivial.   Otherwise, let $\sequencen{e_n}$ run over
an order-dense subset of
$\frak B\setminus\{0\}$;  set $c_n=1\otimes e_n\in\frak C$ for
each $n$.   For $n\in\Bbb N$ set

\Centerline{$\frak D_n=\{a\otimes e_n:a\in\frak A\}
=\{(a\otimes 1)\Bcap c_n:a\in\frak A\}$,}

\Centerline{$\frak B_n=\{b:b\in\frak C$, $b\Bcap c_n\in\frak D_n$,
$b\Bsetminus c_n\in\{0,1\Bsetminus c_n\}\}$.}

\noindent As $a\mapsto a\otimes e_n$ is injective (315K(e-ii)),
$\frak D_n$, regarded as a subalgebra of the principal ideal
$\frak C_{c_n}$ of $\frak C$, is a Boolean algebra isomorphic to $\frak A$,
so is in itself a measurable algebra.   Next,
$a\mapsto a\otimes 1$ is an order-continuous Boolean homomorphism
from $\frak A$ to $\frak A\otimes\frak B$ (315Kc) and therefore
from $\frak A$ to $\frak C$ (314Ta), while $c\mapsto c\Bcap c_n$ is an
order-continuous Boolean homomorphism from $\frak C$ to $\frak C_{c_n}$
(313Xi),
$a\mapsto a\otimes e_n$ is an order-continuous Boolean surjection from
$\frak A$ onto $\frak D_n$, and $\frak D_n$ is order-closed in
$\frak C_{c_n}$.   Next, $\frak B_n$ is isomorphic to the simple product
$\frak D_n\times\{0,1\Bsetminus c_n\}$, so again is a measurable algebra;
and as $\frak D_n\times\{0,1\Bsetminus c_n\}$ is order-closed in
$\frak C_{c_n}\times\frak C_{1\Bsetminus c_n}\cong\frak C$,
$\frak B_n$ is order-closed in $\frak C$.

Thus $\frak C$ is $\sigma$-measurable.   To estimate
$\tausm(\frak C)$ we have only to note that, for each $n$,
$\tau(\frak D_n)=\tau(\frak A)$, and if
$D\subseteq\frak D_n\,\,\tau$-generates $\frak D_n$ then
$D\cup\{c_n\}\,\,\tau$-generates $\frak B_n$.   So
$\tau(\frak B_n)\le\tau(\frak A)+1$ and
$\tausm(\frak C)$ is at most the cardinal product
$\omega\times(\tau(\frak A)+1)=\max(\omega,\tau(\frak A))$.\ \Qed
}

\leader{547J}{}\cmmnt{ Since we are looking at measurable subalgebras of
Boolean algebras, it will be helpful to have an elementary fact out in the
open.

\medskip

\noindent}{\bf Lemma} Let $\frak A$ be a Boolean algebra, and $\frak B$ a
regularly embedded subalgebra of $\frak A$ which is a measurable algebra.
Then $\frak B_a=\{b\Bcap a:b\in\frak B\}$ is a measurable algebra, with
$\tau(\frak B_a)\le\tau(\frak B)$, for every
$a\in\frak A$.

\proof{ Because $\frak B$ is regularly embedded,
the identity map from
$\frak B$ to $\frak A$ is order-continuous;  the map
$d\mapsto d\Bcap a:\frak A\to\frak A_a$ is an order-continuous Boolean
homomorphism;  so $b\mapsto a\Bcap b:\frak B\to\frak B_a$ is an
order-continuous Boolean homomorphism.   By 391Lc\Latereditions,
$\frak B_a$ is a
measurable algebra, isomorphic to a principal ideal of $\frak B$;  so
$\tau(\frak B_a)\le\tau(\frak B)$, by 331Hc or 514Ed.
}%end of proof of 547J

\leader{547K}{}\cmmnt{ I give some more or less straightforward properties
of $\sigma$-measurable algebras.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a $\sigma$-measurable Boolean
algebra with witnessing sequence $\sequencen{\frak B_n}$.

(a) $\frak A$ satisfies Knaster's condition, so is ccc.

(b) $a=\inf_{n\in\Bbb N}\upr(a,\frak B_n)$ for every $a\in\frak A$.

(c) $\tau(\frak A)\le\tausm(\frak A)$.

(d) If $a\in\frak A$ then the principal ideal $\frak A_a$ generated by $a$
is $\sigma$-measurable, with $\tausm(\frak A_a)\le\tausm(\frak A)$.

(e) If $\frak A$ is actually measurable, then
$\tausm(\frak A)=\tau(\frak A)$.

(f) $\pi(\frak A)\le\sum_{n=0}^{\infty}\pi(\frak B_n)$.

\proof{{\bf (a)} Let $\ofamily{\xi}{\omega_1}{a_{\xi}}$ be a family in
$\frak A\setminus\{0\}$.   For each $\xi<\omega_1$ choose a
non-zero $b_{\xi}\in\bigcup_{n\in\Bbb N}\frak B_n$ such that
$b_{\xi}\subseteq a_{\xi}$.   Then there is an $n\in\Bbb N$ such that
$D=\{\xi:\xi<\omega_1$, $b_{\xi}\in\frak B_n\}$ is uncountable.   Now
$\frak B_n$ satisfies Knaster's condition (525Tb), so there is an
uncountable $C\subseteq D$ such that $\family{\xi}{C}{b_{\xi}}$ is linked,
in which case $\family{\xi}{C}{a_{\xi}}$ will be linked.   As
$\ofamily{\xi}{\omega_1}{a_{\xi}}$ is arbitrary, $\frak A$ satisfies
Knaster's condition and is ccc (511Ef).

\medskip

{\bf (b)} \Quer\
Otherwise, $\inf_{n\in\Bbb N}\upr(a,\frak B_n)\Bsetminus a$ is
non-zero and includes a non-zero element $b$ of
$\bigcup_{n\in\Bbb N}\frak B_n$.   Let $k$ be such that $b\in\frak B_k$;
then $a\Bsubseteq 1\Bsetminus b\in\frak B_k$ so $\upr(a,\frak B_k)$ is
included in $1\Bsetminus b$ and cannot include $b$.\ \Bang

\medskip

{\bf (c)}
We can suppose that $\sequencen{\frak B_n}$ has been chosen such that
$\sum_{n=0}^{\infty}\tau(\frak B_n)=\tausm(\frak A)$.
For each $n\in\Bbb N$ let $B_n\subseteq\frak B_n$ be a set with cardinal
$\tau(\frak B_n)$ which $\tau$-generates $\frak B_n$.   Set
$B=\bigcup_{n\in\Bbb N}B_n$.   The order-closed subalgebra of $\frak A$
generated by $B$ includes every $\frak B_n$, so is order-dense in
$\frak A$, and therefore actually equal to $\frak A$.   Thus

\Centerline{$\tau(\frak A)\le\#(B)\le\sum_{n=0}^{\infty}\tau(\frak B_n)
=\tausm(\frak A)$.}

\medskip

{\bf (d)} Again suppose that $\sequencen{\frak B_n}$ was chosen such that
$\sum_{n=0}^{\infty}\tau(\frak B_n)=\tausm(\frak A)$.
For each $n\in\Bbb N$ set
$\frak B'_n=\{b\Bcap a:b\in\frak B_n\}$.   Then $\frak B'_n$ is an order-closed
subalgebra of $\frak A_a$ (314F(a-i)) and is a
measurable algebra with Maharam type at most $\tau(\frak B_n)$ (547J).
If $c\in\frak A_a$ is non-zero, there is a
$b\in\bigcup_{n\in\Bbb N}\frak B_n$ such that $0\ne b\Bsubseteq c$, and now
$b\in\bigcup_{n\in\Bbb N}\frak B'_n$.   So $\sequencen{\frak B'_n}$
witnesses that $\frak A_a$ is $\sigma$-measurable and that

\Centerline{$\tausm(\frak A_a)\le\sum_{n=0}^{\infty}\tau(\frak B'_n)
\le\sum_{n=0}^{\infty}\tau(\frak B_n)=\tausm(\frak A)$.}

\medskip

{\bf (e)} We already know from (c) that $\tau(\frak A)\le\tausm(\frak A)$.
But as $\frak A$ is itself measurable, $(\frak A,\{0,1\},\{0,1\},\ldots)$
is a witnessing sequence for $\frak A$ so
$\tausm(\frak A)\le\tau(\frak A)$.

\medskip

{\bf (f)} For each $n\in\Bbb N$ there is an order-dense subset $B_n$ of
$\frak B_n$ with $\#(B_n)=\pi(\frak B_n)$.   Now $\bigcup_{n\in\Bbb N}B_n$
is order-dense in $\frak A$, so

\Centerline{$\pi(\frak A)\le\#(\bigcup_{n\in\Bbb N}B_n)
\le\sum_{n=0}^{\infty}\#(B_n)=\sum_{n=0}^{\infty}\pi(\frak B_n)$.}
}%end of proof of 547K

\leader{547L}{}\cmmnt{ It will be useful to have an alternative definition
of $\sigma$-measurable algebra.

\medskip

\noindent}{\bf Lemma} Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\sequencen{\frak C_n}$ a sequence of $\sigma$-subalgebras of
$\frak A$.
For each $n\in\Bbb N$ let $\nu_n:\frak C_n\to\coint{0,\infty}$ be a
countably additive functional.   Suppose that for every $a\in\frak A$ there
are an $n\in\Bbb N$ and a $c\in\frak C_n$ such that $c\Bsubseteq a$ and
$\nu_nc>0$.   Then $\frak A$ is a $\sigma$-measurable algebra and
$\tausm(\frak A)
\le\max(\omega,\sup_{n\in\Bbb N}\tau(\frak C_n/\nu_n^{-1}[\{0\}]))$.

\proof{{\bf (a)} $\frak A$ is ccc.   \Prf\ Suppose that
$A\subseteq\frak A\setminus\{0\}$ is disjoint.   For each $a\in A$ choose
$n_A\in\Bbb N$ and $c_a\in\frak C_{n_a}$ such that $c_a\Bsubseteq a$ and
$\nu_{n_a}c_a>0$.   For $n\in\Bbb N$ set $A_n=\{a:a\in A$, $n_a=n\}$.
Then $\{c_a:a\in A_n\}$
is a disjoint family in $\frak C_n$, so
$\sum_{a\in A_n}\nu_nc_a\le\nu_n1$ is finite, and $A_n$ must be countable.
Accordingly $A=\bigcup_{n\in\Bbb N}A_n$ is countable.   As $A$ is
arbitrary, $\frak A$ is ccc.\ \Qed

Consequently $\frak A$ is Dedekind complete and every $\frak C_n$
is order-closed in $\frak A$ (316Fb).

\medskip

{\bf (b)} For each $n\in\Bbb N$, $\frak C_n$ is ccc and Dedekind complete,
and $\nu_n$ is completely additive (326P).   Set
$c_n=\sup\{c:c\in\frak C_n$, $\nu_nc=0\}$;  then $\nu_nc_n=0$.
Set $\frak B_n=\{a:a\in\frak A$, $a\Bsetminus c_n\in\frak C_n$,
$a\Bcap c_n\in\{0,c_n\}\}$.   Then $\frak B_n$ is an order-closed
subalgebra of $\frak A$ and the restriction of $\nu_n$ to
its principal ideal
$(\frak B_n)_{1\Bsetminus c_n}$ is a strictly positive countably additive
functional, so  $(\frak B_n)_{1\Bsetminus c_n}$ is a measurable algebra.
On the other hand, the complementary principal ideal
$(\frak B_n)_{c_n}=\{0,c_n\}$ is also a measurable algebra.   So
$\frak B_n$ is measurable.   Note also that
$(\frak B_n)_{1\Bsetminus c_n}\cong\frak C_n/\nu_n^{-1}[\{0\}]$,
so $\tau(\frak B_n)$ is at most the cardinal sum
$\tau(\frak C_n/\nu_n^{-1}[\{0\}])+1$.

\medskip

{\bf (c)} If $a\in\frak A$ there are an $n\in\Bbb N$ and a $c\in\frak C_n$
such that $c\Bsubseteq a$ and $\nu_nc>0$.   In this case,
$c\Bsetminus c_n\in\frak B_n$ and $0\ne c\Bsetminus c_n\Bsubseteq a$.
Thus $\bigcup_{n\in\Bbb N}\frak B_n$ is order-dense in $\frak A$, so
$\frak A$ is $\sigma$-measurable, with

\Centerline{$\tausm(\frak A)\le\sum_{n=0}^{\infty}\tau(\frak B_n)
\le\max(\omega,\sup_{n\in\Bbb N}\tau(\frak C_n/\nu_n^{-1}[\{0\}]))$,}

\noindent as claimed.
}%end of proof of 547L

%547M
\leader{547M}{}\cmmnt{ Turning to nowhere measurable algebras (391Bc),
the facts we need are the following.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Boolean algebra.

(a) If $\frak A$ is ccc and Dedekind complete and is
not a measurable algebra, then it has a
non-zero principal ideal which is nowhere measurable.

(b) If $\frak A$ is nowhere measurable and Dedekind complete
and $\frak B$ is an order-closed subalgebra of $\frak A$ which is a
measurable algebra, then for any $a\in\frak A$ there is a $d\Bsubseteq a$
such that $\upr(d,\frak B)=\upr(a\Bsetminus d,\frak B)
=\upr(a,\frak B)$.

(c) Suppose that $\frak A$ is nowhere measurable and Dedekind complete,
$\nu:\frak A\to\coint{0,\infty}$ is a submeasure\cmmnt{
(definition:  392A)}
and $\frak B_0,\ldots,\frak B_n$ are order-closed subalgebras of $\frak A$
which are all measurable algebras.   Then for any $a\in\frak A$ there is
a $d\Bsubseteq a$ such that
$\upr(a\Bsetminus d,\frak B_i)=\upr(a,\frak B_i)$ for every $i\le n$ and
$\nu d\ge 2^{-n-1}\nu a$.

(d) If $\frak A$ is nowhere measurable and Dedekind complete and
$\frak B_0,\ldots,\frak B_n$ are order-closed subalgebras of $\frak A$
which are all measurable algebras, then there are
disjoint $a_0,\ldots,a_n\in\frak A$ such that $\upr(a_i,\frak B_i)=1$
for every $i\le n$.

(e) If $\frak A$ is nowhere measurable and Dedekind complete,
$\frak B_0,\ldots,\frak B_n$ are order-closed subalgebras of $\frak A$
which are all measurable algebras and $a\in\frak A$, then there is
a $d\Bsubseteq a$ such that
$\upr(d,\frak B_i)=\upr(a\Bsetminus d,\frak B_i)=\upr(a,\frak B_i)$
for every $i\le n$.

\proof{{\bf (a)} Set $A=\{a:a\in\frak A$, $\frak A_a$ is a measurable
algebra$\}$.   Then $A$ is a $\sigma$-ideal of $\frak A$ so $a^*=\sup A$
belongs to $A$.   As $\frak A$ is not itself measurable, $a^*\ne 1$, and
$\frak A_{1\Bsetminus a^*}$ is a non-zero nowhere measurable principal
ideal of $\frak A$.

\medskip

{\bf (b)} Note first that if $c\in\frak A\setminus\{0\}$ then
$\frak B_c$ is a measurable algebra, by 547J, but $\frak A_c$ is not,
because we are supposing that $\frak A$ is nowhere measurable.   So
$\frak A_c\ne\frak B_c$.

Let $B$ be the set of those $b\in\frak B$ for which there is a
$d\Bsubseteq a$ such that

\Centerline{$\upr(d,\frak B)\Bcap\upr(a\Bsetminus d,\frak B)
\Bsupseteq b$.}

\noindent Let $B'\subseteq B$ be a maximal disjoint set.   \Quer\ If
$a\notBsubseteq\sup B'$, set $c=a\Bsetminus\sup B'$.   Then
there is a $d\Bsubseteq c$ such that
$d$ is not of the form $b\Bcap c$ for any $b\in\frak B$.   Consider
$b=\upr(d,\frak B)\Bcap\upr(a\Bsetminus d,\frak B)$.
By the maximality of $B'$, $b=0$.
But this means that
$\upr(d,\frak B)\Bcap a\Bsetminus d=0$ and
$d=c\Bcap\upr(d,\frak B)$.\ \Bang

Thus $a\Bsubseteq\sup B'$.   For each $b\in B'$ choose
$d_b\Bsubseteq a$ such that
$\upr(d_b,\frak B)\Bcap\upr(a\Bsetminus d_b,\frak B)$ includes
$b$.   Set
$d=\sup_{b\in B'}b\Bcap d_b$.   Then

$$\eqalign{\upr(d,\frak B)
&=\sup_{b\in B'}\upr(b\Bcap d_b,\frak B)
=\sup_{b\in B'}b\Bcap\upr(d_b,\frak B)
=\sup_{b\in B'}b
\Bsupseteq a\cr}$$

\noindent so $\upr(a,\frak B)\Bsubseteq\upr(d,\frak B)$;  as $d\Bsubseteq a$,
$\upr(a,\frak B)=\upr(d,\frak B)$.   Similarly

$$\eqalignno{\upr(a\Bsetminus d,\frak B)
&=\sup_{b\in B'}\upr(b\Bcap a\Bsetminus d,\frak B)
=\sup_{b\in B'}\upr(b\Bcap a\Bsetminus d_b,\frak B)
\Displaycause{because $B'$ is disjoint}
=\sup_{b\in B'}b\Bcap\upr(a\Bsetminus d_b,\frak B)
=\sup_{b\in B'}b
=\upr(a,\frak B).\cr}$$

\noindent So this $d$ serves.

\medskip

{\bf (c)} By (b), there is for each $i\le n$ an $a_i\Bsubseteq a$ such that
$\upr(a_i,\frak B_i)=\upr(a\Bsetminus a_i,\frak B_i)=\upr(a,\frak B_i)$.   
For $I\subseteq n+1$ set
$c_I=a\Bcap\inf_{i\in I}a_i\Bsetminus\sup_{i\le n,i\notin I}a_i$.
Then $a=\sup_{I\subseteq n+1}c_I$;  as $\nu$ is subadditive, there is
an $I\subseteq n+1$ such that $\nu c_I\ge 2^{-n-1}\nu a$.   If $i\le n$
then one of
$a_i$, $a\Bsetminus a_i$ is included in $a\Bsetminus c_I$, so
$\upr(a\Bsetminus c_I,\frak B_i)=\upr(a,\frak B_i)$.   Accordingly we can
take $d=c_I$.

\medskip

{\bf (d)} If $\frak A=\{0\}$, this is trivial;  suppose otherwise.
For each $i\le n$ let $\bar\mu_i$ be such that
$(\frak B_i,\bar\mu_i)$ is a probability algebra;  set
$\bar\mu_i^*a=\bar\mu_i(\upr(a,\frak B_i))$ for $a\in\frak A$.
Then $\bar\mu_i^*$ is a submeasure.   \Prf\ $\bar\mu_i^*0=0$ because
$\upr(0,\frak B_i)=0$.   If $a\Bsubseteq a'$ then
$\upr(a,\frak B_i)\Bsubseteq\upr(a',\frak B_i)$ so
$\bar\mu_i^*a\le\bar\mu_i^*a'$.   For any $a$, $a'\in\frak A$,
$\upr(a\Bcup a',\frak B_i)=\upr(a,\frak B_i)\Bcup\upr(a',\frak B_i)$, so
$\bar\mu_i^*(a\Bcup a')\le\bar\mu_i^*a+\bar\mu_i^*a'$.\ \QeD\

Furthermore, if $a\in\frak A$ and $b\in\frak B_i$,

$$\eqalign{\bar\mu_i^*a
&=\bar\mu_i(\upr(a,\frak B_i)\Bcap b)
  +\bar\mu_i(\upr(a,\frak B_i)\Bsetminus b)\cr&
=\bar\mu_i(\upr(a\Bcap b,\frak B_i))
  +\bar\mu_i(\upr(a\Bsetminus b,\frak B_i))
=\bar\mu_i^*(a\Bcap b)+\bar\mu^*(a\Bsetminus b).
\cr}$$

\noindent In addition, if
$\sequence{k}{a_k}$ is a non-decreasing sequence in $\frak A$ with supremum
$a$, $\sequence{k}{\upr(a_k,\frak B_i))}$ is a non-decreasing sequence
in $\frak B_i$ with supremum $\upr(a,\frak B_i)$, so
$\bar\mu_i^*a=\sup_{k\in\Bbb N}\bar\mu_i^*a_k$.

Choose $\sequence{m}{d_m}$,
$\langle a_{mi}\rangle_{m\in\Bbb N,i\le n}$ inductively, as follows.
The inductive hypothesis will be that $\upr(d_m,\frak B_i)=1$ for every
$i\le n$.   Start with $d_0=1$ and $a_{0i}=0$ for every $i\le n$.
For the inductive step to $m+1$, take $j\le n$ such that
$\bar\mu_j^*a_{mj}$ is minimal.
Set $c=\upr(a_{mj},\frak B_j)$ and
$\nu a=\bar\mu_j^*(a\Bsetminus c)$ for $a\in\frak A$.
Then $\nu:\frak A\to[0,1]$ is subadditive and $\nu d_m=1-\bar\mu_jc$.
By (d), there is a $d_{m+1}\Bsubseteq d_m$ such that
$\upr(d_{m+1},\frak B_i)=\upr(d_m\Bsetminus d_{m+1},\frak B_i)=1$ for every
$i\le n$ and $\nu(d_m\Bsetminus d_{m+1})\ge 2^{-n-1}\nu d_m$;  set

\Centerline{$a_{m+1,j}
=a_{mj}\Bcup((d_m\Bsetminus d_{m+1})\Bsetminus c)$,
\quad $a_{m+1,i}=a_{mi}$ for $i\ne j$.}

\noindent Note that

$$\eqalign{\bar\mu_j^*a_{m+1,j}
&=\bar\mu_j^*(a_{m+1,j}\Bcap c)+\bar\mu_j^*(a_{m+1,j}\Bsetminus c)
=\bar\mu_j^*a_{mj}+\bar\mu_j^*(d_m\Bsetminus d_{m+1})\Bsetminus c)\cr&
=\bar\mu_j^*a_{mj}+\nu(d_m\Bsetminus d_{m+1})
\ge\bar\mu_j^*a_{mj}+2^{-n-1}\nu d_m\cr&
=\bar\mu_j^*a_{mj}+2^{-n-1}(1-\bar\mu_jc)
\ge\bar\mu_j^*a_{mj}
  +2^{-n-1}(1-\Bover1{n+1}\sum_{i=0}^n\bar\mu_i^*a_{mi})
\cr}$$

\noindent because $\bar\mu_jc=\bar\mu_j^*a_{mj}$ was minimal.

Continue.   At the end of the induction, 
if we set $\gamma_m=\sum_{i=0}^n\bar\mu_i^*a_{mi}$ for
each $m$, we have

\Centerline{$\gamma_{m+1}\ge\gamma_m+2^{-n-1}(1-\Bover{\gamma_m}{n+1})$}

\noindent and

\Centerline{$n+1-\gamma_{m+1}
\le (n+1-\gamma_m)(1-\Bover{2^{-n-1}}{n+1})$}

\noindent for every $m$.   As surely $\gamma_m\le n+1$ for every $n$,
$\lim_{m\to\infty}\gamma_m=n+1$.

Set $a_i=\sup_{m\in\Bbb N}a_{mi}$ for each $i$.   Then
$\bar\mu_i^*a_i=\sup_{m\in\Bbb N}\bar\mu_i^*a_{mi}$ for each
$i$, so $\sum_{i=0}^n\bar\mu_i^*a_i\ge n+1$.
As $\bar\mu_i^*1=1$ for each $i$, we must have
$\bar\mu_i^*a_i=1$ for each $i$, that is,
$\upr(a_i,\frak B_i)=1$ for each $i$.

Finally, because $\sequence{m}{d_m}$ is non-increasing and for each $m$
$\langle a_{m+1,i}\Bsetminus a_{mi}\rangle_{i\le n}$ is a disjoint
family with
$\sup_{i\le n}a_{m+1,i}\Bsetminus a_{mi}\Bsubseteq d_m\Bsetminus d_{m+1}$,
we have $a_{mi}\Bcap a_{mj}=0$ whenever $m\in\Bbb N$ and $i\ne j$.
So $\langle a_i\rangle_{i\le n}$ is disjoint.
Thus we have found a suitable family.

\medskip

{\bf (e)} For $i\le n$ set
$\frak C_i=(\frak B_i)_a=\{a\Bcap b:b\in\frak B_i\}$, so that
$\frak C_i$ is an order-closed subalgebra of $\frak A_a$ and a 
measurable algebra,   We need to know that if $d\Bsubseteq a$ then

$$\eqalign{\upr(d,\frak C_i)
&=\inf\{c:c\in\frak C_i,\,d\Bsubseteq c\}
=\inf\{a\Bcap b:b\in\frak B_i,\,d\Bsubseteq b\}\cr&
=a\Bcap\inf\{b:b\in\frak B_i,\,d\Bsubseteq b\}
=a\Bcap\upr(d,\frak B_i).\cr}$$

Of course $\frak A_a$ is nowhere measurable.
By (d), we have disjoint $a_0,\ldots,a_n$ in $\frak A_a$ such that
$\upr(a_i,\frak C_i)=a$ for each $i$.   So

\Centerline{$a=a\Bcap\upr(a_i,\frak B_i)\Bsubseteq\upr(a_i,\frak B_i)
\Bsubseteq\upr(a,\frak B_i)$}

\noindent and $\upr(a,\frak B_i)=\upr(a_i,\frak B_i)$.

Next, for each $i$ we have a $d_i\Bsubseteq a$ such that
$\upr(d_i,\frak B_i)=\upr(a_i\Bsetminus d_i,\frak B_i)=\upr(a,\frak B_i)$,
by (b).   Set $d=\sup_{i\le n}d_i$.   Then, for any $i\le n$,
$d_i\Bsubseteq d$ and $a_i\Bsetminus d\Bsubseteq a\Bsetminus d$.
So

\Centerline{$\upr(a,\frak B_i)=\upr(a_i,\frak B_i)
=\upr(d_i,\frak B_i)
\Bsubseteq\upr(d,\frak B_i)\Bsubseteq\upr(a,\frak B_i)$,}

\Centerline{$\upr(a,\frak B_i)=\upr(a_i,\frak B_i)
=\upr(a_i\Bsetminus d_i,\frak B_i)
\Bsubseteq\upr(a\Bsetminus d,\frak B_i)\Bsubseteq\upr(a,\frak B_i)$}

\noindent and $\upr(d,\frak B_i)=\upr(a\Bsetminus d,\frak B_i)
=\upr(a,\frak B_i)$, as required.
}%end of proof of 547M

%547N
\leader{547N}{}\cmmnt{ Putting the concepts of $\sigma$-measurable algebra
and nowhere measurable algebra together, we have the following.

\medskip

\noindent}{\bf Proposition} If $\frak A$ is a non-trivial
nowhere measurable
$\sigma$-measurable algebra, then $\frak G_{\omega}$ can be regularly
embedded in $\frak A$.

\proof{{\bf (a)}
Let $\sequencen{\frak B_n}$ be a witnessing sequence of subalgebras
of $\frak A$.
Choose $\family{\sigma}{S_2}{a_{\sigma}}$ inductively
as follows.   Start with $a_{\emptyset}=1$.   Given $a_{\sigma}$ where
$\sigma\in\{0,1\}^n$, use 547Me to find $d\Bsubseteq a_{\sigma}$ such that
$\upr(d,\frak B_i)=\upr(a_{\sigma}\Bsetminus d,\frak B_i)
=\upr(a_{\sigma},\frak B_i)$ for every $i\le n$;  set
$a_{\sigma^{\smallfrown}\fraction{0}}=d$ and
$a_{\sigma^{\smallfrown}\fraction{1}}=a_{\sigma}\Bsetminus d$.

Observe that $\upr(a_{\sigma},\frak B_i)=\upr(a_{\tau},\frak B_i)$
whenever $\sigma\subseteq\tau$ in $S_2$ and $i\le\#(\sigma)$.

\medskip

{\bf (b)} Write $\Cal E$ for the algebra of open-and-closed subsets of
$\{0,1\}^{\Bbb N}$.
Because $a_{\emptyset}=1$ and we always have
$a_{\sigma^{\smallfrown}\fraction{0}}$,
$a_{\sigma^{\smallfrown}\fraction{1}}$ disjoint with supremum
$a_{\sigma}$, we have a Boolean homomorphism $\pi:\Cal E\to\frak A$
defined by setting $\pi I_{\sigma}=a_{\sigma}$ for $\sigma\in S_2$.
Because $a_{\sigma}\ne 0$ for every $\sigma$, $\pi$ is injective.
The point is that $\pi$ is
order-continuous.   \Prf\ Suppose that $\Cal G\subseteq\Cal E$ has
supremum
$1_{\Cal E}=\{0,1\}^{\Bbb N}$.   Then $\bigcup\Cal G$ is a dense open
subset of $\{0,1\}^{\Bbb N}$.
\Quer\ If $\sup_{G\in\Cal G}\pi G\ne 1_{\frak A}$, there are an
$n\in\Bbb N$
and a non-zero $b\in\frak B_n$ such that $b\Bcap\pi G=0$ for every
$G\in\Cal G$.   Now there is a $\sigma\in\{0,1\}^n$ such that
$b\Bcap a_{\sigma}\ne 0$.   There
must be a $G\in\Cal G$ such that $G\cap I_{\sigma}\ne\emptyset$, and
a $\tau$ extending $\sigma$ such that $I_{\tau}\subseteq G$.   In this
case, $a_{\tau}\Bsubseteq\pi G$ and $b\Bcap a_{\tau}=0$.   It follows
that $b$ is disjoint from
$\upr(a_{\tau},\frak B_n)=\upr(a_{\sigma},\frak B_n)$
and $b\Bcap a_{\sigma}=0$;  contrary to the choice of $\sigma$.\ \Bang

Thus $\sup\pi[\Cal G]=1_{\frak A}$ whenever $\sup\Cal G=1_{\Cal E}$.
By 313Lb, $\pi$ is order-continuous.\ \Qed

\medskip

{\bf (c)} Now $\Cal E$ is an order-dense subalgebra of the regular open
algebra $\RO(\{0,1\}^{\Bbb N})$ (314Uc\Latereditions).
Because $\frak A$ is Dedekind complete, we have an extension of
$\pi$ to an order-continuous Boolean homomorphism from
$\RO(\{0,1\}^{\Bbb N})$
to $\frak A$ (314Tb) which is still injective (313Xs\Latereditions).
Thus $\RO(\{0,1\}^{\Bbb N})$ is regularly
embedded in $\frak A$.   But
$\frak G_{\omega}\cong\RO(\{0,1\}^{\Bbb N})$ (515Oa), so
$\frak G_{\omega}$ is regularly embedded in $\frak A$, as claimed.
}%end of proof of 547N

%547O
\leader{547O}{}\cmmnt{ I come now to some less obvious properties of
$\sigma$-measurable algebras (547P, 547Q).
%547P prf b (defn of \perp), d (part c)

\medskip

\noindent}{\bf Lemma} Let $(\frak A,\bar\mu)$ be a probability algebra.
Suppose that $\family{\sigma}{S_2}{a_{\sigma}}$ is a family in $\frak A$
such that

\Centerline{$a_{\sigma}
=a_{\sigma^{\smallfrown}\fraction{0}}
   \Bcup a_{\sigma^{\smallfrown}\fraction{1}}
=\sup_{\tau\in S_2,\tau\supseteq\sigma}
  a_{\tau^{\smallfrown}\fraction{0}}\Bcap a_{\tau^{\smallfrown}\fraction{1}}$}

\noindent for every $\sigma\in S_2$.
For $A\subseteq S_2$ set $c_A=\sup_{\sigma\in A}a_{\sigma}$.

(a) For every $k\ge 1$ and $\epsilon>0$ there is an
$m\in\Bbb N$ such that

\Centerline{$\bar\mu(\sup_{I\in[\{0,1\}^m]^k}\inf_{\sigma\in I}a_{\sigma})
\ge\bar\mu a_{\emptyset}-\epsilon$.}

(b) For $A$, $B\subseteq S_2$ I will say that $A\perp B$ if $\sigma$ and
$\tau$ are incompatible whenever
$\sigma\in A$ and $\tau\in B$.   Now for any $\epsilon>0$ there are finite
$A$, $B\subseteq S_2$ such that $A\perp B$ and $\bar\mu c_A$,
$\bar\mu c_B$ are both at least $\bar\mu a_{\emptyset}-\epsilon$.

(c) For any $\epsilon>0$ there is a sequence $\sequencen{A_n}$ of 
finite subsets of $S_2$ such that 
$\bar\mu(\inf_{n\in\Bbb N}c_{A_n})\ge\bar\mu a_{\emptyset}-\epsilon$ and
$A_m\perp A_n$ whenever $m<n$ in $\Bbb N$.

\proof{{\bf (a)(i)} Note first that if $m\in\Bbb N$ and $\sigma\in\{0,1\}^m$, 
then $a_{\sigma}=\sup\{a_{\tau}:\sigma\subseteq\tau\in\{0,1\}^n\}$ for every
$n\ge m$.   (Induce on $n$.)   So if we set

\Centerline{$b_{\sigma n}=
\sup\{a_{\tau}\Bcap a_{\tau'}:\tau,\,\tau'\in\{0,1\}^n
     \text{ are different extensions of }\sigma\}$,}

\noindent then $b_{\sigma n}\Bsubseteq a_{\sigma}$ and

$$\eqalign{b_{\sigma n}&
=\sup\{a_{\tau}\Bcap a_{\tau'}:\tau,\,\tau'\in\{0,1\}^n
     \text{ are different extensions of }\sigma\}\cr&
=\sup\{a_{\upsilon}\Bcap a_{\upsilon'}:\upsilon,\,\upsilon'\in\{0,1\}^{n+1}
     \text{ and }\upsilon\restr n,\,\upsilon'\restr n
     \text{ are different extensions of }\sigma\}\cr&
\Bsubseteq\sup\{a_{\upsilon}\Bcap a_{\upsilon'}:
     \upsilon,\,\upsilon'\in\{0,1\}^{n+1}
     \text{ are different extensions of }\sigma\}\cr&
=b_{\sigma,n+1}\cr}$$

\noindent for every $n\in\Bbb N$.   As

$$a_{\sigma}
=\sup_{\tau\supseteq\sigma}
  a_{\tau^{\smallfrown}\fraction{0}}\Bcap a_{\tau^{\smallfrown}\fraction{1}}
=\sup_{n\in\Bbb N}\sup_{\Atop{\tau\supseteq\sigma}{\#(\tau)=n}}
  a_{\tau^{\smallfrown}\fraction{0}}\Bcap a_{\tau^{\smallfrown}\fraction{1}}
\Bsubseteq\sup_{n\in\Bbb N}b_{\sigma,n+1},$$

\noindent $\bar\mu a_{\sigma}=\sup_{n\in\Bbb N}\bar\mu b_{\sigma n}$.

\medskip

\quad{\bf (ii)} Induce on $k$.   For $k=1$ we can take $m=0$.
For the inductive step to $k+1\ge 2$, take $l$ such that
$d_0=\sup_{I\in[\{0,1\}^l]^k}\inf_{\sigma\in I}a_{\sigma})$ has measure at least
$\bar\mu a_{\emptyset}-\Bover12\epsilon$.
For each $\sigma\in\{0,1\}^l$ let $m_{\sigma}\ge l$ be such that

\Centerline{$\bar\mu(b_{\sigma m_{\sigma}})
\ge\bar\mu a_{\sigma}-2^{-l-2}\epsilon$.}

\noindent Set $m=\sup_{\sigma\in\{0,1\}^l}m_{\sigma}$.   Setting
$d_1=\sup_{\sigma\in\{0,1\}^l}b_{\sigma m}$,

$$\eqalign{\bar\mu(a_{\emptyset}\Bsetminus d_1)&
\le\sum_{\sigma\in\{0,1\}^l}
  \bar\mu(a_{\sigma}\Bsetminus b_{\sigma m})
\le\Bover12\epsilon.\cr}$$

Next, if we take any atom $e$ of the subalgebra of $\frak A$ generated by
$\{a_{\tau}:\tau\in\{0,1\}^m\}$ which is included in $d_0\Bcap d_1$,

\Centerline{$A=\{\sigma:\sigma\in\{0,1\}^l$, $e\Bsubseteq a_{\sigma}\}
=\{\sigma:\sigma\in\{0,1\}^l$, $e\Bcap a_{\sigma}\ne 0\}$}

\noindent has at least
$k$ members, and for each $\sigma\in A$ there are distinct
$\tau$, $\tau'\in\{0,1\}^m$ extending $\sigma$ such that
$e\Bcap a_{\tau}\Bcap a_{\tau'}\ne 0$ 
and $e\Bsubseteq a_{\tau}\Bcap a_{\tau'}$, so
$\{\tau:\tau\in\{0,1\}^m$, $e\Bsubseteq a_{\tau}\}$ has at least $2k$
members, and $e\Bsubseteq\inf_{\tau\in I}a_{\tau}$ for some
$I\in[\{0,1\}^m]^{2k}$.   As $e$ is arbitrary,

\Centerline{$\sup_{I\in[\{0,1\}^m]^{k+1}}\inf_{\tau\in I}a_{\tau}
\Bsupseteq d_0\Bcap d_1$}

\noindent has measure at least $\bar\mu a_{\emptyset}-\epsilon$, 
and the induction continues.

\medskip

{\bf (b)} If $a_{\emptyset}=0$ or $\epsilon>1$
the result is trivial, so suppose otherwise.   Set $\eta=\bover13\epsilon>0$.
Let $k\ge 1$ be so large that $2^{-k}\le\eta$.
By (a), there is an $m\in\Bbb N$ such that 
$\bar\mu d\ge\bar\mu a_{\emptyset}-\eta$, where

\Centerline{$d=\sup_{I\in[\{0,1\}^m]^k}\inf_{\sigma\in I}a_{\sigma}$.}

\noindent Of course $d\Bsubseteq a_{\emptyset}$ and
$\bar\mu(a_{\emptyset}\Bsetminus d)\le\eta$. 
Give $\{0,1\}^m$ its usual probability
measure $\nu$ (254J), so that singleton sets have measure $2^{-m}$.
Let $E$ be the set of atoms of the subalgebra $\frak B$ of $\frak A$ 
generated by $\{a_{\sigma}:\sigma\in\{0,1\}^m\}$;  note that
$d\in\frak B$ and $c_A\in\frak B$ for every $A\subseteq\{0,1\}^m$.
We see that if $e\in E$ and $e\Bsubseteq d$ then there is an
$I\in[\{0,1\}^m]^k$ such that $e\Bcap\inf_{\sigma\in I}a_{\sigma}\ne 0$, in
which case 

\Centerline{$\nu\{A:A\subseteq\{0,1\}^m$, $e\Bcap c_A=0\}
\le\nu\{A:A\cap I=\emptyset\}\le 2^{-k}\le\eta$}

\noindent so

$$\eqalign{\int\bar\mu(d\Bsetminus c_A)\nu(dA)&
=\sum_{e\in E,e\Bsubseteq d}\bar\mu e\cdot\nu\{A:e\Bcap c_A=0\}\cr&
\le\sum_{e\in E,e\Bsubseteq d}\eta\bar\mu e
=\eta\bar\mu d
\le\eta\cr}$$

\noindent and $\nu\{A:\bar\mu(d\Bsetminus c_A)>2\eta\}<\Bover12$.   There is
therefore an $A\subseteq\{0,1\}^m$ such that, setting $B=\{0,1\}^m\setminus A$,
both $\bar\mu(d\Bsetminus c_A)$ and $\bar\mu(d\Bsetminus c_B)$ are at most
$2\eta$.   But now $A\perp B$ and both
$\bar\mu c_A$ and $\bar\mu c_B$ are at least 
$\bar\mu d-2\eta\ge\bar\mu a_{\emptyset}-\epsilon$.

\medskip

{\bf (c)(i)} If $\epsilon>0$ and $A\subseteq S_2$ is finite
there are finite $B$, $B'\subseteq S_2$ such that every member of 
$B\cup B'$ extends a member of $A$, $B\perp B'$ and
$\min(\bar\mu c_B,\bar\mu c_{B'})\ge\bar\mu c_A-\epsilon$.
\Prf\ Let $A_0$ be the set of minimal members of $A$;  because $S_2$ is a tree,
every member of $A$ extends a member of $A_0$, so $c_{A_0}=c_A$, while no two
members of $A_0$ are compatible.   For each $\tau\in A_0$, we can apply (b) to
$\family{\sigma}{S_2}{a_{\tau\smallfrown\sigma}}$ to see that there are
finite $B_{\tau}$, $B'_{\tau}\subseteq S_2$ such that every member of
$B_{\tau}\cup B'_{\tau}$ extends $\tau$, $B_{\tau}\perp B'_{\tau}$ and

\Centerline{$\min(\bar\mu c_{B_{\tau}},\bar\mu c_{B'_{\tau}})
\ge\bar\mu a_{\tau}-\Bover{\epsilon}{1+\#(A)}$,}

\noindent that is,

\Centerline{$\max(\bar\mu(a_{\tau}\Bsetminus c_{B_{\tau}}),
\bar\mu(a_{\tau}\Bsetminus c_{B'_{\tau}}))\le\Bover{\epsilon}{1+\#(A)}$.}

\noindent Set $B=\bigcup_{\tau\in A_0}B_{\tau}$,
$B'=\bigcup_{\tau\in A_0}B'_{\tau}$;  these work.\
\Qed

\medskip

\quad{\bf (ii)} Now choose sequences $\sequencen{B_n}$, $\sequencen{A_n}$ 
of finite subsets of $S_2$ inductively such that 
$B_0=\{\emptyset\}$ and, for each $n\in\Bbb N$,

\Centerline{every member of $A_n\cup B_{n+1}$ extends a member of $B_n$,
\quad $A_n\perp B_{n+1}$,}

\Centerline{$\bar\mu c_{A_n}$ and $\bar\mu c_{B_{n+1}}$ are both at least
$\bar\mu c_{B_n}-2^{-n-2}\epsilon$.}

\noindent Then for each $n\ge 0$

$$\eqalign{\bar\mu(c_{B_{n+1}}\Bcap\inf_{i\le n}c_{A_i})
&\ge\bar\mu(c_{B_n}\Bcap\inf_{i<n}c_{A_i})
    -(\bar\mu c_{B_n}-\bar\mu c_{B_{n+1}})
    -(\bar\mu c_{B_n}-\bar\mu c_{A_n})\cr&
\ge\bar\mu(c_{B_n}\Bcap\inf_{i<n}c_{A_i})-2^{-n-1}\epsilon,
\cr}$$

\noindent so

\Centerline{$\bar\mu(\inf_{i\in\Bbb N}c_{A_i})
\ge\bar\mu a_{\emptyset}-\sum_{n=0}^{\infty}2^{-n-1}\epsilon
=\bar\mu a_{\emptyset}-\epsilon$.}

\noindent Also an easy induction shows that $B_n\perp A_i$ whenever $i<n$ in
$\Bbb N$ and therefore that $A_n\perp A_i$ whenever $i<n$.
}%end of proof of 547O

%547P
\leader{547P}{Proposition}\cmmnt{ (see {\smc Kumar \& Shelah 17}, 4.2\query)}
Let $\frak A$ be a $\sigma$-measurable Boolean algebra.
If $\frak C$ is an order-closed subalgebra of $\frak A$ of countable
Maharam type, there is a $c\in\frak C$ such that the principal ideal
$\frak C_c$ has an e-h family\cmmnt{ (definition:  546F)}
and its complement $\frak C_{1\Bsetminus c}$ is a measurable algebra.

\proof{{\bf (a)} The result is trivial if $\frak A=\{0\}$, so suppose otherwise.
Let $\sequencen{\frak B_n}$ be a witnessing sequence of
measurable subalgebras of $\frak A$.
For $n\in\Bbb N$ let $\bar\mu_n$ be such that $(\frak B_n,\bar\mu_n)$ is
a probability algebra.
Take a maximal disjoint set $D\subseteq\frak C^+$ such that
$\frak C_d$ is measurable for every $d\in D$.   Set $d^*=\sup D$.
By 547Ka, $D$ is countable;  consequently $\frak C_{d^*}$ is a measurable
algebra (391C, 391Xl).   Set $c=1\Bsetminus d^*$.

\medskip

{\bf (b)} If $c=0$ then of course $\frak C_c=\{0\}$ has an e-h family
in which every term is zero.   So suppose otherwise.
By the maximality of the set $D$ in (a) above, $\frak C_c$ has no
non-trivial measurable principal ideal;  in particular, it is atomless.
Since $\frak C_c$ also has countable Maharam type (514Ed),
there is a family $\family{\sigma}{S_2}{c_{\sigma}}$ in $\frak C$
such that $c_{\emptyset}=c$, $\frak C_c$ is generated by
$\{c_{\sigma}:\sigma\in S_2\}$ and

\Centerline{$c_{\sigma^{\smallfrown}\fraction 0}
   \Bcup c_{\sigma^{\smallfrown}\fraction 1}=c_{\sigma}$,
\quad$c_{\sigma^{\smallfrown}\fraction 0}
   \Bcap c_{\sigma^{\smallfrown}\fraction 1}=0$}

\noindent for every $\sigma\in S_2$.

For $A\subseteq S_2$ set $d_A=\sup_{\sigma\in A}c_{\sigma}$.
Then $d_A\Bcap d_B=0$ whenever $A$, $B\subseteq S_2$ and $A\perp B$ in the
sense of 547O.

\medskip

{\bf (c)} For $n\in\Bbb N$ and $\sigma\in S_2$ set 
$\tilde c^{(n)}_{\sigma}=\upr(c_{\sigma},\frak B_n)$.
Then $\tilde c^{(n)}_{\sigma}=\tilde c^{(n)}_{\sigma^{\smallfrown}\fraction 0}
\Bcup\tilde c^{(n)}_{\sigma^{\smallfrown}\fraction 1}$
for every $\sigma\in S_2$ (313Sb).   Also

\Centerline{$\tilde c^{(n)}_{\sigma}
\Bsubseteq\sup_{\tau\in S_2,\tau\supseteq\sigma}
  \tilde c^{(n)}_{\tau^{\smallfrown}\fraction{0}}
    \Bcap\tilde c^{(n)}_{\tau^{\smallfrown}\fraction{1}}$}

\noindent for every $\sigma\in S_2$.  \Prf\Quer\ Otherwise, since
$\sup_{\tau\supseteq\sigma}
  \tilde c^{(n)}_{\tau^{\smallfrown}\fraction{0}}
    \Bcap\tilde c^{(n)}_{\tau^{\smallfrown}\fraction{1}}\in\frak B_n$,

\Centerline{$e
=c_{\sigma}\Bsetminus\sup_{\tau\supseteq\sigma}
  \tilde c^{(n)}_{\tau^{\smallfrown}\fraction{0}}
     \Bcap\tilde c^{(n)}_{\tau^{\smallfrown}\fraction{1}}$}

\noindent is non-zero.   If $\tau\supseteq\sigma$ and
$e\Bcap\tilde c^{(n)}_{\tau}\subseteq c_{\tau}$, then 

$$\eqalignno{e\Bcap\tilde c^{(n)}_{\tau^{\smallfrown}\fraction{0}}&
=e\Bcap\tilde c^{(n)}_{\tau}\Bsetminus\tilde c^{(n)}_{\tau^{\smallfrown}\fraction{1}}
\Displaycause{because $e\Bcap\tilde c^{(n)}_{\tau^{\smallfrown}\fraction{0}}
\Bcap\tilde c^{(n)}_{\tau^{\smallfrown}\fraction{1}}=0$}
\Bsubseteq e\Bcap c_{\tau}\Bsetminus c_{\tau^{\smallfrown}\fraction{1}}
=e\Bcap c_{\tau^{\smallfrown}\fraction{0}}\cr}$$

\noindent and similarly $e\Bcap\tilde c^{(n)}_{\tau^{\smallfrown}\fraction{1}}
\Bsubseteq e\Bcap c_{\tau^{\smallfrown}\fraction{1}}$.   So
we see by induction on $\#(\tau)$ that
$e\Bcap\tilde c^{(n)}_{\tau}=e\Bcap c_{\tau}$ whenever $\tau\supseteq\sigma$.

Now $\frak D_0=\{e\Bcap b:b\in\frak B_n\}$ is an order-closed subalgebra
of $\frak A_e$ (314F(a-i) again)
containing $e\Bcap\tilde c^{(n)}_{\tau}$ whenever
$\sigma\subseteq\tau\in S_2$ ;  similarly,
$\frak D_1=\{e\Bcap c':c'\in\frak C\}$ is the order-closed subalgebra of
$\frak A_e$ generated by 

$$\eqalign{\{e\Bcap c_{\tau}:\tau\in S_2\}&
=\{e\Bcap c_{\sigma}\Bcap c_{\tau}:\tau\in S_2\}
\subseteq\{e\Bcap c_{\tau}:\sigma\subseteq\tau\in S_2\}\cup\{0\}\cr&
\subseteq\{e\Bcap\tilde c^{(n)}_{\tau}:\sigma\subseteq\tau\in S_2\}\cup\{0\}
\subseteq\frak D_0\cr}$$

\noindent (see 313Mc).   But this means that
$\frak D_1$ is included in $\frak D_0$ and is accordingly an order-closed
subalgebra of $\frak D_0$.   Since $\frak D_0$ is the image of the
measurable algebra $\frak B_n$ under an order-continuous homomorphism,
it is measurable (391Lc), so $\frak D_1$ also is measurable (391La).
Finally, $\frak D_1$ must be isomorphic to the principal ideal of
$\frak C$ generated by $\upr(e,\frak C)$.   So $\frak C_c$ has a
non-trivial measurable principal ideal, which we know is not the case.\
\Bang\Qed

\medskip

{\bf (d)} For $n$, $k\in\Bbb N$ we can choose a partition
$\sequence{j}{d_{nkj}}$ of unity in $\frak C_c$ such that
$b\Bcap d_{nkj}\ne 0$ whenever $b\in\frak B_n$, $b\Bsubseteq c$ and
$\bar\mu_nb>2^{-k}$.   \Prf\ Set

\Centerline{$a=1\Bsetminus\upr(1\Bsetminus c,\frak B_n)
=\sup\{b:b\in\frak B_n$, $b\Bsubseteq c\}$.}

\noindent For $\sigma\in S_2$, set 
$a_{\sigma}=a\Bcap\tilde c^{(n)}_{\sigma}=\upr(a\Bcap c_{\sigma},\frak B_n)$
(313Sc).   Using (c) we see that 

\Centerline{$a_{\sigma}
=a_{\sigma^{\smallfrown}\fraction{0}}
   \Bcup a_{\sigma^{\smallfrown}\fraction{1}}
=\sup_{\tau\in S_2,\tau\supseteq\sigma}
  a_{\tau^{\smallfrown}\fraction{0}}\Bcap a_{\tau^{\smallfrown}\fraction{1}}$}

\noindent for every $\sigma\in S_2$.   For $A\subseteq S_2$ set

\Centerline{$d_A=\sup_{\sigma\in A}c_{\sigma}$,
\quad$\tilde d_A=a\Bcap\upr(d_A,\frak B_n)
=\sup_{\sigma\in A}\tilde c^{(n)}_{\sigma}$}

\noindent (313Sb again).   By 547Oc, applied in $\frak B_n$, we have a sequence
$\sequence{j}{A_j}$ of subsets of $S_2$ such that
$A_j\perp A_{j'}$ whenever $j\ne j'$ and
$\bar\mu_n\tilde d_{A_j}\ge\bar\mu_na-2^{-k}$ for every $j\in\Bbb N$.   Set

\Centerline{$d_{nkj}=d_{A_j}$ for $j\ge 1$,
\quad$d_{nk0}=c\Bsetminus\sup_{j\ge 1}d_{A_j}\Bsupseteq d_{A_0}$.}

\noindent Then $\sequence{j}{d_{nkj}}$ is a partition of unity in
$\frak C_c$.   If $b\in\frak B_n$, $j\in\Bbb N$, $b\Bsubseteq c$ and
$\bar\mu_nb>2^{-k}$, then
$b\Bcap\upr(a\Bcap d_{A_j},\frak B_n)\ne 0$
so $b\Bcap d_{nkj}\ne 0$.\ \Qed

\medskip

{\bf (e)} $\frak C_c$ has an e-h family.   \Prf\
Take $\langle d_{nkj}\rangle_{n,k,j\in\Bbb N}$ from (d) above.
Suppose that $f:\Bbb N\times\Bbb N\to\Bbb N$ is a function
and that $e\in\frak C_c$ is
non-zero.   Take $n\in\Bbb N$, $b\in\frak B_n$ and $k\in\Bbb N$ such that
$0\ne b\Bsubseteq e$ and $2^{-k}<\bar\mu_n b$.
Then $0\ne b\Bcap d_{n,k,f(n,k)}\Bsubseteq e\Bcap d_{n,k,f(n,k)}$.
As $e$ is arbitrary,
$c=\sup\{d_{n,k,f(n,k)}:(n,k)\in\Bbb N\times\Bbb N\}$.
As $f$ is arbitrary,
$\langle\sequence{j}{d_{nkj}}\rangle_{n,k\in\Bbb N}$, suitably
re-enumerated, is an e-h family in $\frak C_c$.\ \Qed
}%end of proof of 547P

%547Q
\leader{547Q}{Lemma}\cmmnt{ (see {\smc Kumar \& Shelah 17}, 5.1\query)}
Let $\frak A$ be a $\sigma$-measurable algebra.
Set $\lambda=\max(\omega,\tausm(\frak A))$ and
$\kappa=\non\Cal M(\lambda^{\Bbb N})$, where in this product $\lambda$ is
given its discrete topology.   Then there is a family
$\langle a_{\xi n}\rangle_{\xi<\kappa,n\in\Bbb N}$ in $\frak A$ such that

\inset{\noindent
whenever $\langle d_{m\sigma}\rangle_{m\in\Bbb N,\sigma\in S_2}$
is a family in $\frak A$ such that

\Centerline{$d_{m\tau}\Bsubseteq d_{m,\tau^{\smallfrown}\sigma}$ for every
$\sigma\in S_2$,
\quad $\sup_{\sigma\in S_2}d_{m,\tau^{\smallfrown}\sigma}=1$}

\noindent for every
$\tau\in S_2$ and $m\in\Bbb N$, there is a $\xi<\kappa$ such that

$$\sup_{\sigma\in S_2}\bigl(d_{m\sigma}
  \Bcap\inf_{\Atop{i<\#(\sigma)}{\sigma(i)=1}}a_{\xi i}
  \Bsetminus\sup_{\Atop{i<\#(\sigma)}{\sigma(i)=0}}a_{\xi i}\bigr)
=1$$

\noindent for every $m\in\Bbb N$.}

\proof{{\bf (a)} Let $\sequencen{\frak B_n}$ be a witnessing sequence for
$\frak A$ with $\sum_{n=0}^{\infty}\tau(\frak B_n)=\lambda$.
Set $X=\{0,1\}^{\Bbb N}$.   For $i\in\Bbb N$, set
$E_i=\{x:x\in X$, $x(i)=1\}$.

For each $n\in\Bbb N$ such that $\frak B_n\ne\{0\}$
let $\mu_n$ be a Radon probability
measure on $Y_n=\{0,1\}^{\max(1,\tau(\frak B_n))}$ with measure algebra
isomorphic to $\frak B_n$ (524U);  when $\frak B_n=\{0\}$ take $\mu_n$ to
be the zero measure on $Y_n=\{0,1\}^{\emptyset}=\{\emptyset\}$.
In either case let $\bar\mu_n$ be the associated measure on
$\frak B_n$.   For $n\in\Bbb N$ and $E\in\dom\mu_n$,
write $\pi_nE$ for the corresponding
element of $\frak B_n$.   Let $\Cal E$ be the algebra of
open-and-closed subsets of $Y=\prod_{n\in\Bbb N}Y_n\cong\{0,1\}^{\lambda}$;
as $\lambda$ is infinite, $\#(\Cal E)=\lambda$.
Then we have a Boolean homomorphism
$\pi:\Cal E\to\frak A$ defined by setting $\pi(\{z:z(n)\in F\})=\pi_nF$
whenever $n\in\Bbb N$ and $F\subseteq Y_n$ is open-and-closed (cf.\
315I-315J).

Set $Z=C(Y;X)$ with its
compact-open topology (5A4J).   Giving $\Cal E$ its discrete topology,
$Z$ is homeomorphic to $\Cal E^{\Bbb N}\cong\lambda^{\Bbb N}$ (5A4Jb-5A4Jc), 
and there is a non-meager family $\ofamily{\xi}{\kappa}{g_{\xi}}$ in $Z$.
Write $\Cal J$ for the proper $\sigma$-ideal

\Centerline{$\{A:A\subseteq\kappa$, $\{g_{\xi}:\xi\in A\}$ is
meager in $Z\}$}

\noindent of subsets of $\kappa$.   Set

\Centerline{$a_{\xi i}=\pi(g_{\xi}^{-1}[E_i])$}

\noindent for $\xi<\kappa$ and $i\in\Bbb N$.

\medskip

{\bf (b)} Let $\langle d_{m\sigma}\rangle_{m\in\Bbb N,\sigma\in S_2}$
be a family in $\frak A$ such that

\Centerline{$d_{m\tau}\Bsubseteq d_{m,\tau^{\smallfrown}\sigma}$
for every $\sigma\in S_2$,
\quad$\sup_{\sigma\in S_2}d_{m,\tau^{\smallfrown}\sigma}=1$}

\noindent for every
$\tau\in S_2$ and $m\in\Bbb N$.   For $\xi<\kappa$ and $m\in\Bbb N$, set

$$e_{m\xi}
=\sup_{\sigma\in S_2}\bigl(d_{m\sigma}
  \Bcap\inf_{\Atop{i<\#(\sigma)}{\sigma(i)=1}}a_{\xi i}
  \Bsetminus\sup_{\Atop{i<\#(\sigma)}{\sigma(i)=0}}a_{\xi i}\bigr).$$

\medskip

{\bf (c)} We are approaching the heart of the argument.
\Quer\ Suppose, if possible, that

\Centerline{$D_0=\{\xi:\xi<\kappa$, $\inf_{m\in\Bbb N}e_{m\xi}\ne 1\}$}

\noindent does not belong to $\Cal J$.

\medskip

\quad{\bf (i)} There is an $m\in\Bbb N$ such that

\Centerline{$D_1=\{\xi:e_{m\xi}\ne 1\}$}

\noindent does not belong to $\Cal J$.   For $j\in\Bbb N$, set

\Centerline{$c_{j\sigma}
=\inf_{\tau\in\{0,1\}^j}d_{m,\tau^{\smallfrown}\sigma}$}

\noindent for $\sigma\in S_2$.
Then $\sup_{\sigma\in S_2}c_{j\sigma}=1$.   \Prf\ Enumerate
$\{0,1\}^j$ as $\ofamily{i}{k}{\tau_i}$.   Take
$a\in\frak A\setminus\{0\}$.   Choose $\ofamily{i}{k}{a_i}$ and
$\langle\sigma_i\rangle_{i\le k}$ such that

\Centerline{$a_0=a$, \quad$\sigma_0=\emptyset$,}

\inset{\noindent given that $i<k$, $a_i\in\frak A\setminus\{0\}$ and
$\sigma_i\in S_2$,
$\sigma_{i+1}\in S_2$ is to be an extension of $\sigma_i$ such that
$a_{i+1}=a_i\Bcap d_{m,\tau_i^{\smallfrown}\sigma_{i+1}}$ is non-zero;
such an extension exists because
$\sup_{\sigma\in S_2}
d_{m,\tau_i^{\smallfrown}\sigma_i^{\smallfrown}\sigma}=1$.}

\noindent At the end of the induction,

\Centerline{$0\ne a_{k}\Bsubseteq a_{i+1}
\Bsubseteq d_{m,\tau_i^{\smallfrown}\sigma_{i+1}}
\Bsubseteq d_{m,\tau_i^{\smallfrown}\sigma_{k}}$}

\noindent for every $i<k$, so $a_k\subseteq c_{j\sigma_k}$;  as
$a_k\subseteq a$, $a\Bcap c_{j\sigma_k}\ne 0$.
As $a$ is arbitrary,
$\sup_{\sigma\in S_2}c_{j\sigma}=1$.\ \Qed

\medskip

\quad{\bf (ii)} For $\xi\in D_1$ choose
$b_{\xi}\in\bigcup_{n\in\Bbb N}\frak B_n$ such that
$0\ne b_{\xi}\Bsubseteq 1\Bsetminus e_{m\xi}$;  let $n\in\Bbb N$ and
$\epsilon>0$ be such that
$D_2=\{\xi:\xi\in D_1$,
$b_{\xi}\in\frak B_n$, $\bar\mu_nb_{\xi}>\epsilon\}$
does not belong to $\Cal J$.   For $j\in\Bbb N$ and
$\sigma\in S_2$, write $\tilde c_{j\sigma}$ for
$\upr(c_{j\sigma},\frak B_n)$, and choose $H_{j\sigma}\in\dom\mu_n$
such that $\pi_nH_{j\sigma}=\tilde c_{j\sigma}$.
As $\sup_{\sigma\in S_2}\tilde c_{j\sigma}=1$ in $\frak B_n$,
$\bigcup_{\sigma\in S_2}H_{j\sigma}$ is $\mu_n$-conegligible.
Because $S_2$ is
countable, there is a set $K\subseteq Y_n$ such that

\Centerline{$K\subseteq\bigcup_{\sigma\in S_2}H_{j\sigma}$
for every $j\in\Bbb N$,}

\Centerline{$K\cap H_{j\sigma}$, $K\setminus H_{j\sigma}$ are compact for every
$j\in\Bbb N$ and $\sigma\in S_2$,}

\Centerline{$\mu_nK\ge 1-\epsilon$.}

\noindent As $K$ is compact and $H_{j\sigma}\cap K$ is relatively open
in $K$ for all $j$ and $\sigma$, we see that for every $j\in\Bbb N$
there is a finite set $J_j\subseteq S_2$ such that
$K\subseteq\bigcup_{\sigma\in J_j}H_{j\sigma}$.
Again because $K\cap H_{j\sigma}$ is relatively open in the compact set
$K$ for each $\sigma\in J_j$, we have a partition
$\family{\sigma}{J_j}{W_{j\sigma}}$ of $Y_n$ into open-and-closed sets such
that $K\cap W_{j\sigma}\subseteq H_{j\sigma}$
for every $\sigma\in J_j$.

\medskip

\quad{\bf (iii)} For $j\in\Bbb N$, set

$$\eqalign{G_j
&=\{g:g\in Z,\,g(z)(j+i)=\sigma(i)\text{ whenever }\sigma\in J_j,\cr&
\mskip100mu i<\#(\sigma),\,z\in Y\text{ and }z(n)\in W_{j\sigma}\}.
\cr}$$

\noindent Because every $W_{j\sigma}$ is compact, $G_j$ is open in $Z$.
Accordingly $G=\bigcup_{j\in\Bbb N}G_j$ is open.
Moreover, $G$ is dense in $Z$.
\Prf\ Suppose that $U\subseteq Z$ is a non-empty open set.   Take
$h\in U$.   Then there is a $j\in\Bbb N$ such that

\Centerline{$U_1=\{g:g^{-1}[E_i]=h^{-1}[E_i]$
for every $i<j\}$}

\noindent is an open neighbourhood of $h$ included in $U$ (5A4Jb-5A4Jc
again).   So if we define $g\in Z$ by saying that, for $z\in Y$ and
$i\in\Bbb N$,

$$\eqalign{g(z)(i)
&=\sigma(i-j)\text{ if }\sigma\in J_j,\,
    j\le i<j+\#(\sigma)\text{ and }z(n)\in W_{j\sigma},\cr
&=h(z)(i)\text{ otherwise},\cr}$$

\noindent we have a member of $G_j\cap U_1\subseteq G\cap U$.\ \Qed

\medskip

\quad{\bf (iv)} As $D_2\notin\Cal J$, $\{g_{\xi}:\xi\in D_2\}$ cannot be
nowhere dense and there is a $\xi\in D_2$ such that $g_{\xi}\in G$.
Let $j\in\Bbb N$ be such that $g_{\xi}\in G_j$.
For $\tau\in\{0,1\}^j$, $\sigma\in S_2$ set

$$a'_{\tau}=\inf_{\Atop{i<j}{\tau(i)=1}}a_{\xi i}
    \Bsetminus\sup_{\Atop{i<j}{\tau(i)=0}}a_{\xi i},
\quad b'_{\sigma}=\inf_{\Atop{i<\#(\sigma)}{\sigma(i)=1}}a_{\xi,j+i}
    \Bsetminus\sup_{\Atop{i<\#(\sigma)}{\sigma(i)=0}}a_{\xi,j+i}.$$

\noindent Of course $\sup_{\tau\in\{0,1\}^j}a'_{\tau}=1$.

Because $\bar\mu_nb_{\xi}+\bar\mu_n\pi_nK>1$,
$b_{\xi}\Bcap\pi_nK\ne 0$.   Now
$K\subseteq\bigcup_{\sigma\in J_j}W_{j\sigma}$ so there is a
$\sigma\in J_j$ such that
$b=b_{\xi}\Bcap\pi_n[K\cap W_{j\sigma}]\ne 0$.   Note that
$b\in\frak B_n$.   As $K\cap W_{j\sigma}\subseteq H_{j\sigma}$,

\Centerline{$b\Bsubseteq\pi_nH_{j\sigma}=\tilde c_{j\sigma}
=\upr(c_{j\sigma},\frak B_n)$}

\noindent and $b\Bcap c_{j\sigma}\ne 0$.

As $g_{\xi}\in G_j$, $g_{\xi}(z)(j+i)=\sigma(i)$ whenever
$z(n)\in W_{j\sigma}$ and $i<\#(\sigma)$, that is,

\inset{whenever $i<\#(\sigma)$ and $\sigma(i)=1$,
$g_{\xi}^{-1}[E_{j+i}]\supseteq\{z:z(n)\in W_{j\sigma}\}$
and $a_{\xi,j+i}\Bsupseteq\pi_nW_{j\sigma}$,

whenever $i<\#(\sigma)$ and $\sigma(i)=0$,
$g_{\xi}^{-1}[E_{j+i}]\cap\{z:z(n)\in W_{j\sigma}\}=\emptyset$
and $a_{\xi,j+i}\Bcap\pi_nW_{j\sigma}=0$,}

\noindent so $\pi_nW_{j\sigma}\Bsubseteq b'_{\sigma}$ and
$b\Bsubseteq b'_{\sigma}$.   Thus
$b_{\xi}\Bcap b'_{\sigma}\Bcap c_{j\sigma}\ne 0$.

We have

$$\eqalign{e_{m\xi}
&=\sup_{\upsilon\in S_2}\bigl(d_{m\upsilon}
  \Bcap\inf_{\Atop{i<\#(\upsilon)}{\upsilon(i)=1}}a_{\xi i}
  \Bsetminus\sup_{\Atop{i<\#(\upsilon)}{\upsilon(i)=0}}a_{\xi i}\bigr)\cr&
\Bsupseteq\sup_{\tau\in\{0,1\}^j}
  d_{m,\tau^{\smallfrown}\sigma}
  \Bcap\bigl(\inf_{\Atop{i<j}{\tau(i)=1}}a_{\xi i}
    \Bcap\inf_{\Atop{i<\#(\sigma)}{\sigma(i)=1}}a_{\xi,j+i}\bigr)
  \Bsetminus\bigl(\sup_{\Atop{i<j}{\tau(i)=0}}a_{\xi i}
    \Bcup\sup_{\Atop{i<\#(\sigma))}{\sigma(i)=0}}a_{\xi,j+i}\bigr)
    \cr&
=\sup_{\tau\in\{0,1\}^j}
  d_{m,\tau^{\smallfrown}\sigma}\Bcap a'_{\tau}\Bcap b'_{\sigma}
\Bsupseteq\sup_{\tau\in\{0,1\}^j}
  c_{j\sigma}\Bcap a'_{\tau}\Bcap b'_{\sigma}
=c_{j\sigma}\Bcap b'_{\sigma}.
\cr}$$

\noindent So $b_{\xi}\Bcap e_{m\xi}\ne 0$.   But we chose $b_{\xi}$ to be
disjoint from $e_{m\xi}$.\ \Bang

\medskip

\quad{\bf (v)} We conclude that $D_0\in\Cal J$.

\medskip

{\bf (d)} In particular, $D_0\ne\kappa$ and there is a $\xi<\kappa$ such
that $e_{m\xi}=1$ for every $\xi$, as required.
}%\end of proof of 547Q

\leader{547R}{Theorem} Suppose that $X$ is a set and $\Cal I$ is a proper
$\sigma$-ideal of subsets of $X$ such that the
quotient algebra $\frak A=\Cal PX/\Cal I$ is atomless and
$\sigma$-measurable.   Then $\tausm(\frak A)>\add\Cal I$.

\proof{{\bf (a)} Write $\kappa$ for $\add\Cal I$ and $\lambda$ for
$\max(\omega,\tausm(\frak A))$.   As $\kappa$ is surely infinite, it will be
enough to show that $\kappa<\lambda$.   For the time being
(down to the end of (g) below), suppose that $\frak A$ is nowhere
measurable and that there is a disjoint
family $\ofamily{\xi}{\kappa}{Y_{\xi}}$ in $\Cal I$ with union $X$.
Set $\Cal J=\{J:J\subseteq\kappa$,
$\bigcup_{\xi\in J}Y_{\xi}\in\Cal I\}$;  then $\Cal J$ is a proper
$\omega_1$-saturated $\sigma$-ideal of $\Cal P\kappa$ containing
singletons and $\add\Cal J=\kappa$.   Accordingly $\kappa$ is \qm.
Set $\frak C=\{(\bigcup_{\xi\in J}Y_{\xi})^{\ssbullet}:J\subseteq\kappa\}$;
then $\frak C$ is a $\sigma$-subalgebra of $\frak A$, so is order-closed in
$\frak A$, because $\frak A$ is ccc.   Also
$\frak C\cong\Cal P\kappa/\Cal J$.

\medskip

{\bf (b)} $\kappa\le\non\Cal N$.   \Prf\ By 547N, $\frak G_{\omega}$ can be
regularly embedded in $\frak A$, so 546I(a-ii) gives the result.\ \QeD\
It follows that $\frak C\cong\Cal P\kappa/\Cal J$ is atomless (541P).

\medskip

{\bf (c)} $\frak C$ has a countably generated order-closed subalgebra
$\frak D$ which
is not measurable.   \Prf\ Because $\frak A$ satisfies Knaster's condition,
so does $\frak C$ (516Sa).   By 516V, $\frak C$ has an atomless countably
generated order-closed subalgebra $\frak E$ say.   If $\frak E$ is not
measurable, we can stop.   Otherwise, $\non\Cal N\le\kappa$ (546I(a-i-$\beta$),
applied to $\kappa$ and $\Cal J$).   So $\kappa=\non\Cal N$ and
546Id tells us that $\frak C$ has a countably generated order-closed
subalgebra which is {\it not} measurable.\ \Qed

\medskip

{\bf (d)} $\non\Cal M\le\kappa$.   \Prf\ $\frak D$ is a countably generated
order-closed subalgebra of $\frak A$ which is not measurable.   By 547P,
there is a non-zero $d\in\frak D$ such that the corresponding principal
ideal of $\frak D$ has an e-h family, which is still an e-h family in
the principal ideal $\frak C_d$ of $\frak C$.
By 546Ib, applied to $\kappa$ and $\Cal J$, $\non\Cal M\le\kappa$.\ \Qed

\medskip

{\bf (e)} $\pi(\frak A)$ is uncountable.   \Prf\ $\frak A$ was set up as a
\pssqa, so cannot be isomorphic to $\frak G_{\omega}$, by 547G.   But
$\frak A$ is atomless, Dedekind complete and not $\{0\}$, so by 515Oa once more
it cannot have countable $\pi$-weight.\ \Qed

\medskip

{\bf (f)} $\kappa\le\non\Cal M$.   \Prf\
Let $\sequencen{\frak B_n}$ be a sequence witnessing that
$\frak A$ is $\sigma$-measurable.   Because
$\pi(\frak A)>\omega$, there is an $n\in\Bbb N$ such that
$\pi(\frak B_n)$ is uncountable (547Kf)
and $\frak B_n$ is not purely atomic;  let
$b\in\frak B_n$ be such that the corresponding principal ideal is atomless.
Now the principal ideal $\frak A_b$ of $\frak A$ is $\sigma$-measurable
(547Kd) and has an order-closed subalgebra $\frak B_n\cap\frak A_b$ which
is atomless and measurable.   At the same time, if $E\subseteq X$ is such
that $E^{\ssbullet}=b$, $\frak A_b\cong\Cal PE/\Cal I\cap\Cal PE$, so
546I(a-i-$\alpha$) tells us that 
$\non\Cal M\ge\add(\Cal I\cap\Cal PE)\ge\kappa$.\ \Qed

Thus $\kappa=\non\Cal M$.

\medskip

{\bf (g)} \Quer\ Suppose, if possible, that $\lambda\le\kappa$.
Because $\kappa$ is \qm, there is a family $\Cal A$ of countable sets
which is stationary over $\lambda$ and has cardinal at most $\lambda$ (542K).
Set $\theta=\non\Cal M(\lambda^{\Bbb N})$, where $\lambda$ here is given its
discrete topology.   By 5A4K,

\Centerline{$\theta\le\max(\lambda,\non\Cal M)=\kappa$.}

\noindent By 547Q, there is a family
$\langle a_{\xi n}\rangle_{\xi<\theta,n\in\Bbb N}$ such that

\inset{whenever $\langle d_{m\sigma}\rangle_{m\in\Bbb N,\sigma\in S_2}$
is a family in $\frak A$ such that

\Centerline{$d_{m\tau}\Bsubseteq d_{m,\tau^{\smallfrown}\sigma}$ for
every $\sigma\in S_2$,
\quad$\sup_{\sigma\in S_2}d_{m,\tau^{\smallfrown}\sigma}=1$}

\noindent for every
$\tau\in S_2$ and $m\in\Bbb N$, there is a $\xi<\theta$ such that

$$\sup_{\sigma\in S_2}\bigl(d_{m\sigma}
  \Bcap\inf_{\Atop{i<\#(\sigma)}{\sigma(i)=1}}a_{\xi i}
  \Bsetminus\sup_{\Atop{i<\#(\sigma)}{\sigma(i)=0}}a_{\xi i}\bigr)
=1$$

\noindent for every $m\in\Bbb N$.}

\noindent But this contradicts 546Ic, because $\theta\le\non\Cal M$.\ \Bang

Accordingly $\kappa<\lambda$, as required.

\medskip

{\bf (h)} Thus we have the result when $X$ is itself a union of $\kappa$
members of $\Cal I$ and $\frak A$ is nowhere measurable.   If $\frak A$ is
not nowhere measurable, that is, there is an $a\in\frak A\setminus\{0\}$
such that $\frak A_a$ is measurable, take $A\subseteq X$ such that
$A^{\ssbullet}=a$, and consider $\Cal I_A=\Cal I\cap\Cal PA$.
Then $\Cal PA/\Cal I_A\cong\frak A_a$ is an atomless measurable algebra.
By the Gitik-Shelah theorem (543E), $\tau(\frak A_a)>\add\Cal I_A$.
But now

\Centerline{$\kappa\le\add\Cal I_A<\tau(\frak A_a)=\tausm(\frak A_a)
\le\tausm(\frak A)\le\lambda$}

\noindent (547Kd, 547Kc), so $\kappa<\lambda$ in this case.

\medskip

{\bf (i)} So we need consider only the case in which $\frak A$ is nowhere
measurable, but perhaps $X$ is not itself a union of $\kappa$ members of
$\Cal I$.

Again let $\ofamily{\xi}{\kappa}{Y_{\xi}}$ be a family in
$\Cal I$ such that $Y=\bigcup_{\xi<\kappa}Y_{\xi}$ does not belong to
$\Cal I$, and set $b=Y^{\ssbullet}\in\frak A$.   Then
$\Cal I_Y=\Cal I\cap\Cal PY$ is a proper $\omega_1$-saturated
$\sigma$-ideal of subsets of $Y$ and $\add\Cal I_Y=\kappa$, while
$\frak A'=\Cal PY/\Cal I_Y$ is isomorphic to $\frak A_b$.
So $\frak A'$ is non-trivial, atomless, $\sigma$-measurable
and nowhere measurable with
$\tausm(\frak A')\le\tausm(\frak A)$ (547Kd again).
But now (a)-(g) tell us that $\add\Cal I_Y<\tausm(\frak A')$ so that
$\kappa<\lambda$ in this case also.   This completes the proof.
}%end of proof of 547R

\leader{547S}{Corollary} If a non-zero
$\sigma$-measurable algebra $\frak A$ is also an atomless \pssqa,
then there is a \qm\ cardinal less than $\tausm(\frak A)$.

\proof{ Let $X$ be a set and $\Cal I$ a $\sigma$-ideal of
subsets of $X$ such that $\frak A\cong\Cal PX/\Cal I$.   Then
$\Cal I$ must be a proper $\omega_1$-saturated ideal (547Ka), so
$\add\Cal I$ is \qm\ (542B);  but 547R tells us that
$\add\Cal I<\tausm(\frak A)$.
}%end of proof of 547S

\exercises{\leader{547X}{Basic exercises (a)}
%\spheader 547Xa
Show that a \pssqa\ with countable $\pi$-weight is purely
atomic.
%547K

\spheader 547Xb Show that the simple product of a countable family of
$\sigma$-measurable algebras is $\sigma$-measurable.
%547K

\spheader 547Xc Let $\frak A$ be a $\sigma$-measurable algebra.
Show that if a
cardinal $\kappa$ is a precaliber of every probability algebra it is a
precaliber of $\frak A$.
%547K

\spheader 547Xd Let $\frak A$ be a $\sigma$-measurable algebra.   Show that
$\tausm(\frak A)$ is at most $\pi(\frak A)$.
%547K

\spheader 547Xe(i) Show that every $\sigma$-measurable algebra is
chargeable (definition:  391Bb).
(ii) Show that a \wsid\ $\sigma$-measurable algebra is measurable.
%547K 391D

\spheader 547Xf Let $\frak A$ be a Dedekind complete Boolean algebra, and
suppose that there is a finite family $\Bbb B$ of order-closed
subalgebras of $\frak A$ such that every member of $\Bbb B$
is a measurable algebra and
$\bigcup\Bbb B$ is order-dense in $\frak A$.   Show that
$\frak A$ is a measurable algebra.   \Hint{show that $\frak A$ is \wsid.}
%547Xe 547K mt54bits

\spheader 547Xg Show that a $\sigma$-measurable Maharam algebra
is a measurable algebra.
%539O 547Xe 547K

\spheader 547Xh Suppose that $T$ is a Souslin tree (5A1Ed).  Show that
its regular open algebra $\RO^{\uparrow}(T)$ is not $\sigma$-measurable.
%547K 547Xe

\spheader 547Xi Let $\frak A$ be a $\sigma$-measurable algebra and $\frak B$ an
order-closed subalgebra of $\frak A$ which is $\sigma$-measurable.   Show that
$\tausm(\frak B)\le\tausm(\frak A)$.
\Hint{reduce to case in which $\frak A$ is infinite and
$\frak B$ is a homogeneous measure algebra with standard generating
family $\ofamily{\xi}{\kappa}{e_{\xi}}$;  take a witnessing sequence
$\sequencen{\frak B_n}$ for $\frak A$ and for $\xi<\kappa$ choose
$d_{\xi}\in\bigcup_{n\in\Bbb N}\frak B_n$ with
$0\ne d_{\xi}\Bsubseteq e_{\xi}$;  show that $\#(\{\xi:d_{\xi}\in\frak B_n\})$
is at most the topological weight of the metric space $\frak B_n$.}
%547K

\leader{547Y}{Further exercises (a)}
%\spheader 547Ya
Let $\frak A$ be a measurable algebra,
$\frak B$ a Boolean algebra, $\frak C$ the Dedekind completion of
$\frak A\otimes\frak B$, and $\frak D$ an order-closed subalgebra of $\frak C$.
Show that $\tau(\frak D)\le\max(\omega,\tau(\frak A),\pi(\frak B))$.
%547I

\spheader 547Yb Let $\frak A$ be a $\sigma$-measurable algebra.   Show that
there is a unique $a\in\frak A$ such that the principal ideal $\frak A_a$
has an
e-h family and every countably generated order-closed subalgebra of
$\frak A_{1\Bsetminus a}$ is a measurable algebra.   (See 539O.)
%547P

\spheader 547Yc Show that if $\kappa$ is an infinite cardinal less than
$\omega_{\omega}$, and we give $\kappa$ its discrete topology, then
$\non(\Cal M(\kappa^{\Bbb N}))=\max(\kappa,\non\Cal M)$.   (See 5A4K.)
%547R prf (g)

\spheader 547Yd Suppose that $X$ is a set, $\Cal I$ is a
proper $\sigma$-ideal of subsets of $X$ and $\Cal PX/\Cal I$ is \wsid.
(i) Show that $\add\Cal I\ne\frak b$.
(ii) Show that $\add\Cal I\ne\cf\frak d$.   (See 544N.)
%547 notes mt54bits
}%end of exercises

\leader{547Z}{Problems (a)}
%\spheader 547Za
Can $\frak G_{\omega_2}$ be a \pssqa?
%547G

\spheader 547Zb Can there be a \pssqa\ $\frak A$ such that
$c(\frak A)=\omega$ and $\pi(\frak A)=\omega_1$?
\cmmnt{Note that there seems to be no obstacle to an atomless
ccc \pssqa\ having Maharam type $\omega$ (555K).   
Similarly, it is generally supposed that it is possible for a \pssqa\ to have 
$\pi$-weight $\omega_1$ ({\smc Foreman 10}, Theorem 7.60).}
%547G

\spheader 547Zc
Let $\frak A$ be a non-purely-atomic Maharam algebra
with countable Maharam type.   Can $\frak A$ be a \pssqa?
What about algebras constructed as in 394B-394M?
%547G

\spheader 547Zd Find a $\sigma$-measurable algebra with an
order-closed subalgebra which is not $\sigma$-measurable.
%547K   what about  \frak B_{\kappa}\tensorhat\frak G_{\omega} ?

\spheader 547Ze Is there a $\sigma$-measurable algebra $\frak A$ such that
$\tau(\frak A)<\tausm(\frak A)$?
%547K

\endnotes{
\Notesheader{547}
The general message of the work here seems to be that `standard' algebras
cannot be \pssqa{s} (though see 555K).   It is true that the algebras we
are most interested in can be expressed as
order-closed subalgebras of \pssqa{s} (546Xa), but this is too easy to be
useful.   If there are no \qm\ cardinals, of course, this section becomes
redundant.   The point is that 547G, like 546E, is a theorem of ZFC;  and,
as I explained in the notes to \S541, I think it worth exploring
worlds in which there are \qm\ cardinals.

All the results as stated in 543F, 547G and 547R depend on moving from a
\pssqa\ to an order-closed subalgebra which is a normal \pssqa.
If we start from a measurable algebra, the subalgebra will again be
measurable, with Maharam type no greater than that of
the original algebra;  this is why 543E implies
543F.   For category algebras this doesn't work in the same way,
because an order-closed
subalgebra of $\frak G_{\omega_2}$, for instance,
can be very different in character.   (For examples see
{\smc Koppelberg \& Shelah 96} and {\smc Balcar Jech \& Zapletal 97}.)
So only 547G can be directly deduced from 547F.

I have every hope that there is a great deal more to be said about the
questions here, but I have to confess that I have no idea which way to go.
The problems in 547Z seem to me natural ones, but I don't know whether they
will turn out to be useful.   Looking at 546I and the argument of 547R,
it seems that it might be worth asking whether,
for our favourite ccc \pssqa{s} $\frak A$,
we can describe the types -- for instance, the combinatorial properties --
of \qm\ cardinals which can appear
as values of $\add\Cal I$ arising in an expression of $\frak A$ as
$\Cal PX/\Cal I$.   (See 547Yd.)
Note that as it seems that there can be a
proper class of \2vm\ cardinals,
and as every \2vm\ cardinal corresponds to a
representation of the algebra $\{0,1\}$ as a normal
\pssqa, we do not expect
any upper bounds on the complexity of the \qm\ cardinals arising in this way.
}%end of notes

\discrpage

\frfilename{mt548.tex}
\versiondate{30.6.21}
\copyrightdate{2016}

\def\chaptername{Real-valued-measurable cardinals}
\def\sectionname{Selectors and disjoint refinements}

\def\tausm{\tau_{\sigma\text{-m}}}
\def\pssqa{power set $\sigma$-quotient algebra}

\newsection{548}

We come now to a remarkable result (548C) which is a minor
extension of the principal theorem of {\smc Kumar \& Shelah 17}.
This leads directly to 548E, which is a corresponding elaboration of the
main theorem of {\smc Gitik \& Shelah 01}.
Both of these results apply to spaces whose Maharam types are not too
large, so give interesting facts
about Lebesgue measure not dependent on special axioms (548F).
A similar restriction on shrinking number leads to further results
of this kind (548G-548H) which are not necessarily applicable to Lebesgue
measure.   If we choose an easier target other methods are available
(548I-548K).

\cmmnt{Notation will follow that of \S\S546-547;
in particular, I will speak of
$\sigma$-measurable algebras and the associated cardinal function
$\tausm$ (547H).
As usual, I write $\Cal N(\mu)$ for the null ideal of a measure $\mu$.}

\leader{548A}{Lemma} Let $X$ be a set, $\Cal I$ a $\sigma$-ideal of subsets
of $X$, $\sequencen{(Y_n,\Tau_n,\nu_n)}$ a sequence of totally finite
measure spaces and $\sequencen{f_n}$ a sequence of functions, each $f_n$
being an injective
function from a subset of $X$ to $Y_n$.   Suppose that for every
$A\in\Cal PX\setminus\Cal I$ there are an $n\in\Bbb N$ and an $F\in\Tau_n$
such that $\nu_nF>0$ and $F\subseteq f_n[A]$.   Then
$\frak A=\Cal PX/\Cal I$ is $\sigma$-measurable and
$\tausm(\frak A)\le\max(\omega,\sup_{n\in\Bbb N}\tau(\nu_n))$.

\proof{{\bf (a)} Because $\Cal I$ is a $\sigma$-ideal of $\Cal PX$,
$\frak A$ is Dedekind $\sigma$-complete and the quotient map
$A\mapsto A^{\ssbullet}:\Cal PX\to\frak A$ is sequentially order-continuous
(313Qb).

\medskip

{\bf (b)} For each $n\in\Bbb N$ set
$\Cal E_n=\{F:F\in\Tau_n$, $f_n^{-1}[F]\in\Cal I\}$.   Then $\Cal E_n$ is
closed under countable unions so there is an $F_n\in\Cal E_n$ such that
$\nu_n(F\setminus F_n)=0$ for every $F\in\Cal E_n$ (215Ac).   Set
$\nu'_nF=\nu_n(F\setminus F_n)$ for $F\in\Tau_n$ and

\Centerline{$\frak C_n
=\{(f_n^{-1}[F])^{\ssbullet}:F\in\Tau_n\}
\subseteq\frak A$,}

\noindent so that $\nu'_n$ is a totally finite measure with domain
$\Tau_n$ and
$\frak C_n$ is a $\sigma$-subalgebra of $\frak A$.

The map $F\mapsto(f_n^{-1}[F])^{\ssbullet}:\Tau_n\to\frak C_n$ is a
surjective sequentially order-continuous Boolean homomorphism,
and $\nu'_n$ is zero on its kernel.   We therefore have a functional
$\hat\nu_n:\frak C_n\to\coint{0,\infty}$ defined by saying that
$\hat\nu_nc=\nu'_nF$ whenever $c\in\frak C_n$, $F\in\Tau_n$ and
$c=(f_n^{-1}[F])^{\ssbullet}$, and $\hat\nu_n$ is $\sigma$-additive because
$\nu'_n$ is.

\medskip

{\bf (c)} If $a\in\frak A\setminus\{0\}$,
let $A\in\Cal PX\setminus\Cal I$
be such that $a=A^{\ssbullet}$.   Then
$A_1=A\setminus\bigcup_{n\in\Bbb N}f_n^{-1}[F_n]$ does not belong to
$\Cal I$, so there are an $n\in\Bbb N$ and an
$F\in\Tau_n$ such that $\nu_nF>0$ and $F\subseteq f_n[A_1]$.
As $f_n$ is injective, $f_n^{-1}[F]\subseteq A_1$;
as $F\cap F_n=\emptyset$, $\nu'_nF>0$.
Consider $c=(f_n^{-1}[F])^{\ssbullet}$.   Then $\hat\nu_nc=\nu'_nF>0$ while

\Centerline{$c\Bsubseteq A_1^{\ssbullet}=A^{\ssbullet}=a$.}

\noindent Thus all the conditions of 547L are satisfied by
$\frak A$ and $\sequencen{(\frak C_n,\hat\nu_n})$, and
$\frak A$ is $\sigma$-measurable, with $\tausm(\frak A)\le\max(\omega,
\sup_{n\in\Bbb N}\tau(\frak C_n/\hat\nu_n^{-1}[\{0\}]))$.

\medskip

{\bf (d)} Next, $\tau(\frak C_n/\hat\nu_n^{-1}[\{0\}])\le\tau(\nu_n)$
for every $n\in\Bbb N$.
\Prf\  Write $\frak B_n$ for the measure
algebra of $\nu_n$, and let $B\subseteq\frak B_n$ be a $\tau$-generating
set with cardinal $\tau(\frak B_n)=\tau(\nu_n)$.   Now we have a set
$\Cal B\subseteq\Tau_n$, with cardinality $\tau(\nu_n)$, such that
$B=\{F^{\ssbullet}:F\in\Cal B\}$.   Let $\Tau'$ be the $\sigma$-subalgebra
of $\Tau_n$ generated by $\Cal B$;  then $\{F^{\ssbullet}:F\in\Tau'\}$ is a
$\sigma$-subalgebra of $\frak B_n$ including $B$.   Because $\frak B_n$ is
ccc, $\{F^{\ssbullet}:F\in\Tau'\}$ is order-closed and is the whole of
$\frak B_n$.   Thus for any $F\in\Tau_n$ there is an $F'\in\Tau'$ such that
$\nu_n(F\symmdiff F')=0$.

Now consider $\frak D=\{(f_n^{-1}[F])^{\ssbullet}:F\in\Tau'\}$.   This is a
$\sigma$-subalgebra of $\frak C_n$.   If $c\in\frak C_n$
there is an $F\in\Tau_n$ such that $c=(f_n^{-1}[F])^{\ssbullet}$ and
an $F'\in\Tau'$ such that $\nu_n(F\symmdiff F')=0$.   Of course
$\nu_n'(F\symmdiff F')=0$ so $\hat\nu_n(c\Bsymmdiff c')=0$ where
$c'=(f_n^{-1}[F])^{\ssbullet}\in\frak D$.   This means that the image of
$\frak D$ in $\frak C_n/\hat\nu_n^{-1}[\{0\}]$ is the whole of
$\frak C_n/\hat\nu_n^{-1}[\{0\}]$.

Because $\Tau'$ is the $\sigma$-subalgebra of itself generated by
$\Cal B$, $\frak D$ must be the $\sigma$-subalgebra of itself generated by
$D=(f_n^{-1}[F])^{\ssbullet}:F\in\Cal B\}$.   But now
$\frak C_n/\hat\nu_n^{-1}[\{0\}]$ is the $\sigma$-subalgebra of itself
generated by the image of $D$, and

\Centerline{$\tau(\frak C_n/\hat\nu_n^{-1}[\{0\}])
\le\#(D)\le\#(\Cal B)=\tau(\nu_n)$.  \Qed}

\medskip

{\bf (e)} It follows that

\Centerline{$\tausm(\frak A)
\le\max(\omega,\sup_{n\in\Bbb N}\tau(\frak C_n/\hat\nu_n^{-1}[\{0\}]))
\le\max(\omega,\sup_{n\in\Bbb N}\tau(\nu_n))$,}

\noindent as claimed.
}%end of proof of 548A

\leader{548B}{}\cmmnt{ The next argument will go more smoothly if we know the
following not-quite-trivial fact.

\medskip

\noindent}{\bf Lemma} Lat $(X,\Sigma,\mu)$ be a probability space and
$\sequence{i}{X_i}$ a partition of $X$.   Then $\mu$ can be extended to a
probability measure $\nu$ measuring every $X_i$ and with Maharam type
$\tau(\nu)\le\max(\omega,\tau(\mu))$.

\proof{ By 214P, there is a measure $\lambda$ on $X$, extending $\mu$, such that
$\lambda(E\cap\bigcup_{i\le n}X_i)$ is defined and equal to
$\mu^*(E\cap\bigcup_{i\le n}X_i)$ for every $E\in\Sigma$ and $n\in\Bbb N$.
In particular, $\bigcup_{i\le n}X_i$ is measured by $\lambda$ for every $n$, so
$X_n$ also is.   Let $\nu$ be the restriction of $\lambda$ to the
$\sigma$-algebra $\Tau$ of subsets of $X$ generated by
$\Sigma\cup\{X_i:i\in\Bbb N\}$;  then $\nu$ is a probability measure extending
$\mu$ and measuring every $X_i$.

Let $\frak A$, $\frak B$ be the measure algebras of $\mu$ and $\nu$, and
for $E\in\Sigma$, $F\in\Tau$ write $E^{\ssbullet}$, $F^{\frsmallcirc}$ for
their images in $\frak A$, $\frak B$ respectively.
There is a subset $B$ of the measure algebra $\frak A$ of $\mu$ such that
$\#(B)=\tau(\mu)$ and $B\,\,\tau$-generates $\frak A$.   Choose
$\Cal B\subseteq\Sigma$ such that $B=\{E^{\ssbullet}:E\in\Cal B\}$ and
$\#(\Cal B)=\#(B)$, and take $\Tau'$ to be the $\sigma$-algebra of subsets
of $X$ generated by $\Cal B\cup\{X_i:i\in\Bbb N\}$.   Then
$\Tau'\cap\Sigma$ is a $\sigma$-subalgebra of $\Sigma$ so its image
$\frak A'\subseteq\frak A$ is a $\sigma$-subalgebra of $\frak A$ including
$\frak B$.   Because $\frak A$ is ccc, $\frak A'$ is a closed subalgebra of
$\frak A$ and must be the whole of $\frak A$.   But this means that for every
$E\in\Sigma$ there is an $E'\in\Tau'\cap\Sigma$ such that
$0=\mu(E\symmdiff E')=\nu(E\symmdiff E')$.   Moving now to
$\frak B$, we see that $\frak B'=\{F^{\frsmallcirc}:F\in\Tau'\}$ is the
$\sigma$-subalgebra of $\frak B$ generated by
$\{F^{\frsmallcirc}:F\in\Cal B\}\cup\{X_i^{\frsmallcirc}:i\in\Bbb N\}$.
As $\frak B'$ contains $E^{\frsmallcirc}$ for every $E\in\Sigma$, it is in fact
the whole of $\frak B$, so

\Centerline{$\tau(\nu)=\tau(\frak B)
\le\#(\Cal B\cup\{X_i:i\in\Bbb N\})\le\max(\omega,\#(\Cal B))
=\max(\omega,\tau(\mu))$.}

}%end of proof of 548B

\leader{548C}{Theorem}\cmmnt{ (see {\smc Kumar \& Shelah 17})}
Suppose that $(X,\Sigma,\mu)$ is an atomless $\sigma$-finite measure space
and that there is no \qm\ cardinal less than the
Maharam type\cmmnt{ $\tau(\mu)$} of $\mu$.
Let $R\subseteq X\times X$ be an equivalence relation with
countable equivalence classes.
Then there is an $R$-free set\cmmnt{ (definition:  5A1Jd)} which
has full outer measure in $X$.

\proof{{\bf (a)} If $\mu X=0$ this is trivial, as we can take the empty set;
suppose otherwise.   As $\mu$ is atomless, $\tau(\mu)$ is infinite.
There is a probability measure on $X$ with the same measurable sets and the
same negligible sets as $\mu$ (215B);
this will have the same quotient algebra $\Sigma/\Cal N$ and
the same Maharam type, so we may take it that $\mu$ itself is a probability
measure.   By 5A1J(e-i) there is a partition $\sequence{i}{X_i}$ of $X$
into $R$-free sets.   By 548B there is a probability measure on $X$ extending
$\mu$, measuring every $X_i$ and with the same Maharam type;  as extending the
measure never increases the outer measure of a set, we can suppose that this
extension has been done, and that every $X_i$ is measured by $\mu$.
Completing $\mu$ does not change the sets of full outer measure
(212Eb) or the measure algebra (322Da),
so we may suppose that $\mu$ is complete.

For each $i\in\Bbb N$, set $f_i=R\cap(X\times X_i)$;  then
$f_i:R[X_i]\to X_i$ is a function which is injective on every $X_k$.

\medskip

{\bf (b)} Suppose that we are given $Y$, $A\subseteq X$ such that
$\mu^*A>0$.   Then there is a $B\subseteq A$ such that $\mu^*B>0$ and
$\mu^*(Y\setminus R[B])=\mu^*Y$.
\Prf\Quer\ Otherwise, take $\nu_i$ to be the subspace measure on
$Y\cap X_i$ for each $i\in\Bbb N$, and set
$\Cal I=\{B:B\subseteq A$, $\mu^*B=0\}$, the null ideal of the subspace
measure on $A$.   Then $\Cal I$ is a proper $\sigma$-ideal
of $\Cal PA$.   Because $\mu$ is atomless, $\Cal I$ contains every singleton
subset of $A$.

Take any $B\in\Cal PA\setminus\Cal I$.   We are supposing that
$\mu^*(Y\setminus R[B])<\mu^*Y$.  If $G$ is a measurable envelope of
$Y\setminus R[B]$, $\mu^*(Y\setminus G)>0$ and there is an $i\in\Bbb N$ such
that $\mu^*(Y\cap X_i\setminus G)>0$, in which case
$F=Y\cap X_i\setminus G$ belongs to the domain of $\nu_i$ and $\nu_iF>0$.
At the same time, $F\subseteq R[B]\cap X_i$, so if $y\in F$ there is an $x\in B$ such
that $(x,y)\in R$ and $f_i(x)=y$;  that is, $F\subseteq f_i[B]$.

Accordingly $A$, $\Cal I$, $\sequence{i}{(Y_i,\nu_i)}$ and
$\sequence{i}{f_i\restr A}$ satisfy the conditions of 548A and
$\frak A=\Cal PA/\Cal I$ is $\sigma$-measurable, with

\Centerline{$\tausm(\frak A)\le\max(\omega,\sup_{i\in\Bbb N}\tau(\nu_i))
\le\max(\omega,\tau(\mu))=\tau(\mu)$.}

\noindent (For the second inequality, note that as $\mu$ is a probability
measure $\nu_i$ is totally finite and $\tau(\nu_i)\le\tau(\mu)$ for
every $i\in\Bbb N$ (521Ff).)   Also $\Cal PA/\Cal I$ is
atomless, because if $C\in\Cal PA\setminus\Cal I$ the subspace measure on
$C$ is totally finite and atomless (214Q), so there is
a $C'\subseteq C$ such that neither $C'$ nor $C\setminus C'$ is
$\mu$-negligible.
So 547R tells us that $\tausm(\frak A)>\add\Cal I$.
But $\Cal PA/\Cal I$, being $\sigma$-measurable, is ccc,
so $\Cal I$ is $\omega_1$-saturated, and
$\add\Cal I$ is quasi-measurable and less than $\tau(\mu)$, contrary to
hypothesis.\ \Bang\Qed

\medskip

{\bf (c)} Now suppose that $Y$, $A\subseteq X$ and $A$ is $R$-free.
Then there is a $B\subseteq A$ such that
$\mu^*B\ge\bover12\mu^*A$ and $\mu^*(Y\setminus R[B])=\mu^*Y$.
\Prf\Quer\ Otherwise, let $\Cal B$ be the family of those sets
$B\subseteq A$ such that $\mu^*(Y\setminus R[B])=\mu^*Y$, and
choose $\sequencen{B_n}$,
$\sequencen{\gamma_n}$ inductively, as follows.   $B_0=\emptyset$.
Given that $B_n\in\Cal B$, set
$\gamma_n=\sup\{\mu^*B:B_n\subseteq B\in\Cal B\}$ and choose $B_{n+1}$ such
that $B_n\subseteq B_{n+1}\in\Cal B$ and $\mu^*B_{n+1}\ge\gamma_n-2^{-n}$.
Continue.

At the end of the induction, set $C=\bigcup_{n\in\Bbb N}B_n$.   As
$B_n\in\Cal B$, our counter-hypothesis declares that
$\mu^*B_n<\bover12\mu^*A$ for every $n$;  as
$\sequencen{B_n}$ is non-decreasing, $\mu^*C<\mu^*A$.
Let $E\in\Sigma$ be a
measurable envelope of $C$;  then $A\setminus E$ is non-negligible.
Set $Y'=Y\setminus R[C]$.   Applying (b) to $Y$ and $A\setminus E$, we
see that there is a
$B\subseteq A\setminus E$ such that $\mu^*B>0$ and
$\mu^*(Y'\setminus R[B])=\mu^*Y'$.   As $A$ is $R$-free and $B\cap C=\emptyset$,
$R[B]\cap R[C]=\emptyset$ (5A1J(e-ii)).

There must be an $n\in\Bbb N$ such that
$\mu^*B_{n+1}+2^{-n}<\mu^*B_n+\mu^*B$.   Now

$$\eqalign{\mu^*(B_n\cup B)
&=\mu^*((B_n\cup B)\cap E)+\mu^*((B_n\cup B)\setminus E)\cr&
=\mu^*B_n+\mu^*B
>\mu^*B_{n+1}+2^{-n}
\ge\gamma_n.\cr}$$

\noindent So $B_n\cup B\notin\Cal B$ and
$\mu^*(Y\setminus R[B_n\cup B])<\mu^*Y$.   Let $F\in\Sigma$ be the complement of
a measurable envelope of $Y\setminus R[B_n\cup B]$, so that $Y\cap F$ is
non-negligible and included in $R[B_n\cup B]=R[B_n]\cup R[B]$.
As $R[B_n]\subseteq R[C]$, $Y'\cap F\subseteq R[B]$.
Since $\mu^*(Y'\setminus R[B])=\mu^*Y'$,
$Y'\cap F$ must be negligible.   We are supposing that $\mu$ is complete, so
$F'=F\setminus Y'$ belongs to $\Sigma$ and $\mu(Y\cap F')>0$.
But

\Centerline{$Y\cap F'\subseteq(R[B_n]\cup R[B])\cap R[C]=R[B_n]$,}

\noindent so $\mu^*(Y\setminus R[B_n])<\mu^*Y$;
which is not so, because $B_n\in\Cal B$.\ \Bang\Qed

\medskip

{\bf (d)} Fix on a sequence $\sequencen{k_n}$ running over $\Bbb N$ with
cofinal repetitions, and choose a sequence $\sequencen{A_n}$
of subsets of $X$ inductively, as follows.
The inductive hypothesis will be that $\mu^*(X\setminus R[A_n])=1$
and that $A_n$ is $R$-free.   Start with $A_0=\emptyset$.   Given $A_n$,
choose a measurable envelope $G$ for $X_{k_n}\cap A_n$ and apply (c) to find a
$B\subseteq X_{k_n}\setminus(G\cup R[A_n])$ such that
$\mu^*B\ge\bover12\mu^*(X_{k_n}\setminus(G\cup R[A_n]))$ and

\Centerline{$\mu^*((X\setminus R[A_n])\setminus R[B])
=\mu^*(X\setminus R[A_n])=1$.}

\noindent Set $A_{n+1}=A_n\cup B$.   As $A_n$ and $B\subseteq X_{k_n}$ are
$R$-free and $B$ is disjoint from $R[A_n]$, $A_{n+1}$
is $R$-free, and the induction continues.

We also see that, whenever $k$, $n\in\Bbb N$ and $k_n=k$,
there is a measurable envelope $G$ of
$X_k\cap A_n$ such that $A_{n+1}\setminus A_n\subseteq X_k\setminus G$
and $\mu^*(A_{n+1}\setminus A_n)
\ge\bover12\mu^*(X_k\setminus(G\cup R[A_n]))$.   So

$$\eqalignno{\mu^*(A_{n+1}\cap X_k)
&=\mu^*(A_{n+1}\cap X_k\cap G)+\mu^*(A_{n+1}\cap X_k\setminus G)
  \cr&
=\mu^*(A_n\cap X_k) + \mu^*(A_{n+1}\setminus A_n)\cr&
\ge\mu^*(A_n\cap X_k)
   + \Bover12\mu^*((X_k\setminus R[A_n])\setminus G)\cr&
=\mu^*(A_n\cap X_k)
   + \Bover12\mu(X_k\setminus G)
\Displaycause{because $X\setminus R[A_n]$ has full outer measure and
$X_k\setminus G$ is measurable}
=\mu^*(A_n\cap X_k)
   + \Bover12(\mu X_k-\mu^*(X_k\cap A_n))
=\Bover12(\mu X_k+\mu^*(A_n\cap X_k)).
\cr}$$

\noindent But since $\{n:k_n=k\}$ is infinite, we see that
$\lim_{n\to\infty}\mu^*(A_n\cap X_k)=\mu X_k$;  and this is true for
every $k\in\Bbb N$.

\medskip

{\bf (d)} At the end of the induction, $A=\bigcup_{n\in\Bbb N}A_n$ is the
union of an upwards-directed family of $R$-free sets so is $R$-free.
Also, because $\sequence{k}{X_k}$ is a partition of $X$ into measurable sets,

\Centerline{$\mu^*A
=\sum_{k=0}^{\infty}\mu^*(A\cap X_k)
=\sum_{k=0}^{\infty}\lim_{n\to\infty}\mu^*(A_n\cap X_k)
=\sum_{k=0}^{\infty}\mu X_k
=1$.}

\noindent So we have found an $R$-free set of full outer measure.
}%end of proof of 548C

\leader{548D}{Lemma} Let $(X,\Sigma,\mu)$ be a measure space.
Then the following are equiveridical:

(i) whenever $\sequencen{A_n}$ is a sequence of subsets of $X$ there is a
disjoint sequence $\sequencen{A'_n}$ of sets such that $A'_n\subseteq A_n$ and
$\mu^*(A'_n)=\mu^*(A_n)$ for every $n\in\Bbb N$;

(ii) whenever $\sequencen{A_n}$ is a sequence of subsets of $X$ there is a
set $D\subseteq X$ such that
$\mu^*(A_n\cap D)=\mu^*(A_n\setminus D)=\mu^*A_n$ for every $n\in\Bbb N$.

\proof{{\bf (a)} If (i) is true and $\sequencen{A_n}$ is a sequence of subsets
of $X$, set $B_{2n}=B_{2n+1}=A_n$ for every $n\in\Bbb N$, and take a disjoint
sequence $\sequencen{B'_n}$ of sets such that $B'_n\subseteq B_n$
and $\mu^*B'_n=\mu^*B_n$ for every $n$.   Set $D=\bigcup_{n\in\Bbb N}B'_n$;
then $\mu^*(A_n\cap D)\ge\mu^*B'_{2n}=\mu^*A_n$ and
$\mu^*(A_n\setminus D)\ge\mu^*B'_{2n+1}=\mu^*A_n$, so
then $\mu^*(A_n\cap D)=\mu^*(A_n\setminus D)=\mu^*A_n$, for every $n\in\Bbb N$.

\medskip

{\bf (b)} If (ii) is true and $\sequencen{A_n}$ is a sequence of subsets
of $X$, choose $\sequencen{D_n}$ inductively so that

\Centerline{$\mu^*(A_m\cap D_n\setminus\bigcup_{i<n}D_i)
=\mu^*(A_m\setminus\bigcup_{i\le n}D_i)=\mu^*(A_m\setminus\bigcup_{i<n}D_i)$}

\noindent for every $m$, $n\in\Bbb N$.
Then in fact $\mu^*(A_m\cap D_n\setminus\bigcup_{i<n}D_i)=\mu^*A_m$ for all
$m$ and $n$, so if we set
$A'_n=A_n\cap D_n\setminus\bigcup_{i<n}D_i$ for each $n$, we see that
$\sequencen{A'_n}$ is disjoint and
$\mu^*A'_n=\mu^*A_n$ for every $n\in\Bbb N$.
}%end of proof of 548D

\leader{548E}{Theorem}\dvAformerly{5{}47Fa}\cmmnt{ (see
{\smc Gitik \& Shelah 01})} Let
$(X,\Sigma,\mu)$ be an atomless $\sigma$-finite measure space
such that there is no \qm\ cardinal less than the Maharam type of $\mu$.
Then for any sequence $\sequencen{A_n}$ of subsets of $X$ there is a
disjoint sequence $\sequencen{A'_n}$ such that $A'_n\subseteq A_n$
and $\mu^*A'_n=\mu^*A_n$ for every $n\in\Bbb N$.

\proof{ For each $n\in\Bbb N$ let $\nu_n$ be the subspace measure on $A_n$;
set $Y=\bigcup_{n\in\Bbb N}A_n\times\{n\}$ and
give $Y$ the $\sigma$-finite direct sum measure $\nu$, so that
$\nu E=\sum_{n=0}^{\infty}\nu_n(E^{-1}[\{n\}])$ whenever $E\subseteq Y$ is such
that $E^{-1}[\{n\}]$ is measured by $\nu_n$ for every $n$.   Then the Maharam
type of $\nu$ is at most $\sum_{n=0}^{\infty}\tau(\nu_n))
\le\max(\omega,\tau(\mu))$ (521G, 521Ff),
so there is no \qm\ cardinal less than
$\tau(\nu)$.   Also every $\nu_n$ is atomless (214Q again)
so $\nu$ is atomless (214Xh).

Let $R$ be the equivalence relation

\Centerline{$\{((x,m),(x,n)):m$, $n\in\Bbb N$, $x\in A_m\cap A_n\}
\subseteq Y\times Y$.}

\noindent By 548C, there is an $R$-free set $A\subseteq Y$ of full
outer measure in $Y$.   Set
$A'_n=\{x:(x,n)\in A\}$ for each $n$.   Then $\sequencen{A'_n}$ is disjoint
and

\Centerline{$\mu^*A'_n=\nu_n^*A'_n=\nu^*(A\cap(A_n\times\{n\}))
=\nu(A_n\times\{n\})=\nu_nA_n=\mu^*A_n$}

\noindent for every $n$, as required.
}%end of proof of 548E
%covers old 5{}46P, 5{}46Q;  renders old 5{}46J-5{}46Q redundant

\leader{548F}{Corollary} Let $\mu$ be Lebesgue measure on $\Bbb R$.

(a) Let $X$ be a subset of $\Bbb R$ and $\sim$ an
equivalence relation on $X$ with countable equivalence classes.   Then
there is a subset of $X$, with full outer measure for the subspace measure
on $X$, which meets each equivalence class in at most one point.

(b) Let $\sequencen{A_n}$ be any sequence of subsets of $\Bbb R$.   Then
there is a disjoint sequence $\sequencen{A'_n}$ such that $A'_n$ is a
subset of $A_n$, with the same outer measure as $A_n$, for every
$n$.

(c) Let $\sequencen{A_n}$ be any sequence of subsets of $\Bbb R$.   Then
there is a $D\subseteq\Bbb R$ such that
$\mu^*(A_n\cap D)=\mu^*(A_n\setminus D)=\mu^*A_n$ for every $n\in\Bbb N$.
%index  selector

\proof{ For (a)-(b), apply 548C and 548E,
noting that the measures here have countable
Maharam type and that there is surely no \qm\ cardinal less
than $\omega$.   For (c), use 548D.
}%end of proof of 548F

\leader{548G}{}\cmmnt{ The point of the formulation of 548C-548E in terms
of Maharam types is that we get non-trivial corollaries which are valid in
ZFC, just as we did in 547G.   With other hypotheses involving \qm\ cardinals
we can get further results, as in the following.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a totally finite
measure space in which singleton sets are negligible
and suppose that there is no \qm\ cardinal
less than or equal to the shrinking number
$\shr\Cal N(\mu)$.   Then for any $A\subseteq X$
there is a disjoint family
$\ofamily{\xi}{\omega_1}{A_{\xi}}$ of
subsets of $A$ such that $\mu^*A_{\xi}=\mu^*A$ for every $\xi<\omega_1$.

\cmmnt{\medskip

\noindent{\bf Remark} 548C-548E refer to atomless measure spaces rather
than those in which singletons are negligible.   Of course singletons are
negligible in any atomless totally finite measure space (215E).}

\proof{{\bf (a)} Let $\Cal E$ be the set of those
$F\in\Sigma$ such that $\mu F>0$ and there is
a disjoint family $\ofamily{\xi}{\omega_1}{B_{\xi}}$ of subsets of
$F\cap A$ such
that $\mu^*B_{\xi}=\mu F$ for every $\xi<\omega_1$.   Then whenever
$E\in\Sigma$ and $\mu^*(E\cap A)>0$ there is an
$F\in\Cal E$ included in $E$.
\Prf\ There is a $B\subseteq E\cap A$ such that $\#(B)\le\shr\Cal N(\mu)$
and $\mu^*B>0$.
Consider the ideal $\Cal N$ of negligible subsets of
$B$.   This is a $\sigma$-ideal containing every singleton set;  as
$\#(B)$
is not \qm, $\Cal N$ cannot be $\omega_1$-saturated and there is a
disjoint family $\ofamily{\xi}{\omega_1}{C_{\xi}}$ of non-negligible
subsets of $B$.
For each $\xi<\omega_1$ let $E_{\xi}\subseteq E$
be a measurable envelope of $C_{\xi}$
and $a_{\xi}=E_{\xi}^{\ssbullet}$ the corresponding element in the measure
algebra $\frak A$ of $\mu$.   For $\xi<\omega_1$ set
$b_{\xi}=\sup_{\eta\ge\xi}a_{\eta}$;  then
$\ofamily{\xi}{\omega_1}{b_{\xi}}$ is non-increasing.
As $\frak A$ is ccc, there is a
$\zeta<\omega_1$ such that $b_{\xi}=b_{\zeta}$ whenever
$\zeta\le\xi<\omega_1$.   Write $b$ for $b_{\zeta}$;  of course
$b\Bsupseteq a_{\zeta}$ is non-zero, and also $b\Bsubseteq E^{\ssbullet}$.

If $\xi<\omega_1$ there is an $\eta_{\xi}<\omega_1$ such that
$\sup_{\xi\le\eta<\eta_{\xi}}a_{\eta}=b_{\xi}\supseteq b$.   Let
$f:\omega_1\to\omega_1$ be a strictly increasing function such that
$f(\xi+1)=\eta_{f(\xi)}$ for every $\xi<\omega_1$.   Consider
$B_{\xi}=\bigcup_{f(\xi)\le\eta<f(\xi+1)}C_{\eta}$,
$F_{\xi}=\bigcup_{f(\xi)\le\eta<f(\xi+1)}E_{\eta}$
for $\xi<\omega_1$.
Then $\ofamily{\xi}{\omega_1}{B_{\xi}}$ is disjoint,
$F_{\xi}$ is a measurable envelope of $B_{\xi}$ for each $\xi$ (132Ed),
and $F_{\xi}^{\ssbullet}=\sup_{f(\xi)\le\eta<f(\xi+1)}a_{\eta}
\supseteq b$.   So if we take $F\in\Sigma$ such that $F\subseteq E$
and $F^{\ssbullet}=b$, $\ofamily{\xi}{\omega_1}{F\cap B_{\xi}}$ is a
disjoint family of subsets of $F\cap A$ witnessing that
$F\in\Cal E$.\ \Qed

\medskip

{\bf (b)} Let $\familyiI{F_i}$ be a maximal disjoint family in $\Cal E$;
set $G=X\setminus\bigcup_{i\in I}F_i$.   Then $I$ is countable and
$G\in\Sigma$.   By (a), $G\cap A$ must be negligible.
For each $i\in I$, choose a disjoint family
$\ofamily{\xi}{\omega_1}{B_{i\xi}}$ of subsets of $F_i\cap A$ such that
$\mu^*B_{i\xi}=\mu F_i$ for every $\xi$.   Set
$A_{\xi}=\bigcup_{i\in I}B_{i\xi}$ for each $\xi$.   Then
$\ofamily{\xi}{\omega_1}{A_{\xi}}$ is a disjoint family of subsets of
$A$.   Now, for each $\xi<\omega_1$,

$$\eqalignno{\mu^*A
&\le\sum_{i\in I}\mu F_i
=\sum_{i\in I}\mu^*B_{i\xi}
=\sum_{i\in I}\mu^*(A_{\xi}\cap F_i)
=\sum_{i\in I}\mu_{A_{\xi}}(A_{\xi}\cap F_i)
\Displaycause{where $\mu_{A_{\xi}}$ is the subspace measure on $A_{\xi}$}
\le\mu_{A_{\xi}}A_{\xi}
=\mu^*A_{\xi}
\le\mu^*A
\cr}$$

\noindent and we have $\mu^*A=\mu^*A_{\xi}$, as required.
}%end of proof of 548G

\cmmnt{\medskip

\noindent{\bf Remark} For cases in which one of the hypotheses

\Centerline{`there is no \qm\ cardinal less than $\tau(\mu)$',}

\Centerline{`there is no \qm\ cardinal less than or equal to
$\shr\Cal N(\mu)$'}

\noindent is satisfied and the other is not, see 548Ya and 555Yg.
}

\leader{548H}{Proposition} Let $(X,\Sigma,\mu)$ be a
totally finite measure space.   Suppose
that for every $A\subseteq X$ there is a partition
$\ofamily{\xi}{\omega_1}{A_{\xi}}$ of $A$ such that $\mu^*A_{\xi}=\mu^*A$
for every $\xi<\omega_1$.

(a) If $R$ is an equivalence relation on $X$ with countable equivalence classes,
there is an $R$-free set with full outer measure.

(b) For any sequence $\sequencen{A_n}$ of subsets of $X$ there is a
disjoint sequence $\sequencen{A'_n}$ such that $A'_n\subseteq A_n$
and $\mu^*A'_n=\mu^*A_n$ for every $n\in\Bbb N$.

(c) For any sequence $\sequencen{A_n}$ of subsets of $X$,
there is a $D\subseteq X$ such that
$\mu^*(A_n\cap D)=\mu^*(A_n\setminus D)=\mu^*A_n$ for every $n\in\Bbb N$.

\proof{{\bf (a)} By 5A1J(e-i), there is a partition $\sequencen{X_n}$ of
$X$ into $R$-free sets.   Now we can choose inductively
a sequence $\sequencen{A_n}$ of sets such
that $A_n\subseteq X_n\setminus\bigcup_{i<n}R[A_i]$
and $\mu^*A_n=\mu^*X_n$ for every $n$, while
$\mu^*(X_j\setminus\bigcup_{i<n}R[A_i])=\mu^*X_j$ whenever $j\ge n$.   \Prf\
Given $\ofamily{i}{n}{A_i}$ write $B_{jn}=X_j\setminus\bigcup_{i<n}R[A_i]$
for $j\in\Bbb N$ and choose a partition $\ofamily{\xi}{\omega_1}{C_{\xi}}$ of
$B_{nn}$ such that $\mu^*C_{\xi}=\mu^*B_{nn}$ for every $\xi<\omega_1$;
by the inductive hypothesis, $\mu^*C_{\xi}=\mu^*X_n$ for every $\xi$.
Now for any $j\in\Bbb N$, $\ofamily{\xi}{\omega_1}{R[C_{\xi}]}$ is disjoint
(5A1J(e-ii)), so $\mu^*(B_{jn}\setminus R[C_{\xi}])=\mu^*B_{jn}$ for all but
countably many $\xi$ (521Od).   There is therefore a $\xi<\omega_1$ such that
$\mu^*(B_{jn}\setminus R[C_{\xi}])=\mu^*B_{jn}$ for every $j\in\Bbb N$.
If $j>n$, then

\Centerline{$\mu^*(X_j\setminus(R[C_{\xi}]\cup\bigcup_{i<n}R[A_i]))
=\mu^*(B_{jn}\setminus R[C_{\xi}])
=\mu^*B_{jn}=\mu^*X_j$}

\noindent by the inductive hypothesis;  so if we set $A_n=C_{\xi}$ the
induction will continue.\ \Qed

At the end of the induction, we see that $A_n$ is $R$-free and
$A_n\cap\bigcup_{i<n}R[A_i]=\emptyset$ for every $n$, so
$A=\bigcup_{n\in\Bbb N}A_n$ is $R$-free.   And because
$X=\bigcup_{n\in\Bbb N}X_n$ and $\mu^*A_n=\mu^*X_n$ for every $n$,
$A$ has full outer measure.   \Prf\ If $F\in\Sigma$ and $\mu F>0$, there is an
$n\in\Bbb N$ such that $\mu^*(F\cap X_n)>0$;  moving to the subspace measure
$\mu_{X_n}$ on $X_n$, $\mu_{X_n}(F\cap X_n)>0$ so

$$\eqalignno{0&
<\mu_{X_n}^*(F\cap A_n)
=\mu^*(F\cap A_n)
\Displaycause{214Cd}
\le\mu^*(F\cap A). \text{ \Qed}\cr}$$
%214Xo

\medskip

{\bf (b)} As in the proof of 548E, we can apply (a) to the direct sum
measure $\nu$ on $Y=\bigcup_{n\in\Bbb N}A_n\times\{n\}$,
though this time we take the measure on $A_n$ to be
$\nu_n=2^{-n}\mu_{A_n}$ in order to ensure that
$\nu$ is totally finite.   Of course we have to
check that the hypothesis of this proposition is
satisfied by $(Y,\nu)$, but this is
easy.   Now we can use the same equivalence relation $R$ on $Y$ as in
548E, and if $A\subseteq Y$ is an $R$-free set of full outer measure
the last line will read

\Centerline{$\mu^*A'_n=2^n\nu_n^*A'_n=2^n\nu^*(A\cap(A_n\times\{n\})
=2^n\nu(A_n\times\{n\})=2^n\nu_nA_n=\mu^*A_n$}

\noindent for every $n$.

\medskip

{\bf (c)} This now follows from (b) by 548D.
}%end of proof of 548H

\leader{548I}{}\cmmnt{ I do not know how far we can hope to extend 548H(b-c)
to uncountable families in place of $\sequencen{A_n}$.   If in place of

\Centerline{$\mu^*(A_n\cap D)=\mu^*(A_n\setminus D)=\mu^*A_n$}

\noindent we ask rather for

\Centerline{$\min(\mu^*(A_n\cap D),\mu^*(A_n\setminus D))>0$}

\noindent we are led to rather different patterns, as follows.

\medskip

\noindent}{\bf Proposition} Let $(\frak A,\bar\mu)$ be a
probability algebra and $\kappa$ a cardinal.   Then the following are
equiveridical:

(i) $\kappa<\pi(\frak A_d)$ for every
$d\in\frak A^+=\frak A\setminus\{0\}$, writing
$\frak A_d$ for the principal ideal generated by $d$;

(ii) whenever $A\subseteq\frak A^+$ and $\#(A)\le\kappa$
there is a $b\in\frak A$ such that $a\Bcap b$ and
$a\Bsetminus b$ are both non-zero for every $a\in A$.

\proof{{\bf (a)} Suppose that (i) is false;  that there are a non-zero
$d\in\frak A$ and an order-dense set $A\subseteq\frak A_d^+$
such that $\#(A)\le\kappa$.   If $b\in\frak A$ and $b\Bcap d=0$ then
$a\Bcap b=0$ for every $a\in A$;  if $b\Bcap d\ne 0$ then there is an
$a\in A$ such that $a\Bsubseteq b\Bcap d$ and $a\Bsetminus b=0$.   So $A$
witnesses that (ii) is false.

For the rest of the proof, therefore, I suppose that (i) is true and seek
to prove (ii).

\medskip

{\bf (b)} We need an elementary calculation.   Let
$\ofamily{i}{n}{c_i}$ be a stochastically independent family in $\frak A$
such that $\bar\mu c_i=\gamma$ for every $i<n$, where $0<\gamma<1$.
Suppose that $a\in\frak A$ and that

\Centerline{$\sup_{i<n}\bar\mu(a\Bcap c_i)
\le\beta\gamma\bar\mu a$}

\noindent where $\beta<1$.
Then $n\le\Bover1{(1-\beta)^2\gamma\bar\mu a}$.

\Prf\ In $L^2(\frak A,\bar\mu)$ set

\Centerline{$e_i=\sqrt{\bover{\gamma}{1-\gamma}}\chi(1\Bsetminus c_i)
  -\sqrt{\bover{1-\gamma}{\gamma}}\chi c_i$}

\noindent for each $i<n$.   An easy calculation shows that
$\ofamily{i}{n}{e_i}$ is orthonormal.   Next, for each $i$,

$$\eqalign{(e_i|\chi a)
&=\sqrt{\Bover{\gamma}{1-\gamma}}\bar\mu(a\Bsetminus c_i)
  -\sqrt{\Bover{1-\gamma}{\gamma}}\bar\mu(a\Bcap c_i)\cr
&\ge\sqrt{\Bover{\gamma}{1-\gamma}}(\bar\mu a-\beta\gamma\bar\mu a)
  -\sqrt{\Bover{1-\gamma}{\gamma}}\beta\gamma\bar\mu a\cr
&=\sqrt{\Bover1{\gamma(1-\gamma)}}
  (\gamma(\bar\mu a-\beta\gamma\bar\mu a)-(1-\gamma)\beta\gamma\bar\mu a)\cr
&=\gamma\bar\mu a\sqrt{\Bover1{\gamma(1-\gamma)}}(1-\beta)
\ge(1-\beta)\bar\mu a\sqrt{\gamma}.\cr}$$

\noindent By 4A4Ji,

\Centerline{$\bar\mu a=\|\chi a\|_2^2
\ge\sum_{i<n}|(e_i|\chi a)|^2\ge n\gamma(1-\beta)^2(\bar\mu a)^2$}

\noindent and

\Centerline{$n\le\Bover1{(1-\beta)^2\gamma\bar\mu a}$}

\noindent as claimed.\ \Qed

\medskip

{\bf (c)} If $\frak A$ has an atom, then $\kappa=0$ and there is nothing to
prove.   So we may suppose henceforth that $\frak A$ is atomless.   Set
$\lambda=\min\{\tau(\frak A_d):d\in\frak A^+\}$.   Then the
measure algebra $(\frak B_{\lambda},\bar\nu_{\lambda})$ can be embedded in
$(\frak A,\bar\mu)$ (332P).   In particular, we can find, for each $n$,
a stochastically independent family $\ofamily{\xi}{\lambda}{c_{n\xi}}$ of
elements of $\frak A$ with $\bar\mu c_{n\xi}=\bover1{n!}$ for every $\xi$.

Set $q(n)=4n((2n)!+(2n+1)!)$ for each $n\in\Bbb N$, and let
$(\lambda^{\Bbb N},\subseteq^*,\Cal S^{(q)}_{\lambda})$ be the
corresponding version of the
$\lambda$-localization relation as described in 522L.   For
$a\in\frak A^+$ let $n_a\in\Bbb N$ be such that
$n_a\bar\mu a\ge 1$ and set

$$\eqalign{S_a
&=\{(n,\xi):n\ge n_a\text{ and
either }\bar\mu(a\Bcap c_{2n,\xi})\le\Bover{\bar\mu a}{2(2n)!}
\text{ or }\bar\mu(a\Bcap c_{2n+1,\xi})\le\Bover{\bar\mu a}{2(2n+1)!}\}\cr
&\subseteq\Bbb N\times\lambda.\cr}$$

\noindent Then $S_a\in\Cal S^{(q)}_{\lambda}$.   \Prf\ If $n\ge n_a$ then
(b), with $\beta=\bover12$, tells us that

\Centerline{$\#(\{\xi:\bar\mu(a\Bcap c_{2n,\xi})
\le\Bover{\bar\mu a}{2(2n)!}\})
\le\Bover{4(2n)!}{\bar\mu a}
\le 4n(2n)!$,}

\Centerline{$\#(\{\xi:\bar\mu(a\Bcap c_{2n+1,\xi})
\le\Bover{\bar\mu a}{2(2n+1)!}\})
\le\Bover{4(2n+1)!}{\bar\mu a}
\le 4n(2n+1)!$,}

\noindent so $\#(S_a[\{n\}])\le 4n((2n)!+(2n+1)!)=q(n)$.\ \Qed

\medskip

{\bf (d)} Now observe that

$$\eqalignno{\min\{\pi(\frak A_d):d\in\frak A^+\}
&=\pi(\frak B_{\lambda})\cr
\displaycause{see 524Mc}
&=\ci(\frak B_{\lambda}^+)
=\cov(\frak B_{\lambda}^+,\Bsupseteqshort,\frak B_{\lambda}^+)
=\cov(\frak B_{\lambda}^+,\Bsupseteqshort\strprime,
[\frak B_{\lambda}^+]^{\le\omega})
\Displaycause{512Gf}
=\cov(\lambda^{\Bbb N},\subseteq^*,\Cal S_{\lambda})
\Displaycause{where $(\lambda^{\Bbb N},\subseteq^*,\Cal S_{\lambda})$ is
the ordinary $\lambda$-localization relation, by 524H and 512Da}
=\cov(\lambda^{\Bbb N},\subseteq^*,\Cal S^{(q)}_{\lambda})\cr}$$

\noindent by 522L.

\medskip

{\bf (e)} Let $A\subseteq\frak A^+$ be a set with cardinal
less than  $\min\{\pi(\frak A_d):d\in\frak A^+\}$.   Then
$\#(A)<\cov(\lambda^{\Bbb N},\subseteq^*,\Cal S^{(q)}_{\lambda})$,
so there must be an
$f\in\lambda^{\Bbb N}$ such that $f\not\subseteq^*S_a$ for any $a\in A$.
Set

\Centerline{$b_{2n}=c_{2n,f(n)}$,
\quad$b_{2n+1}=c_{2n+1,f(n)}$,
\quad$b'_n=b_n\Bsetminus\sup_{i>n}b_i$ for $n\in\Bbb N$,}

\Centerline{$b=\sup_{n\in\Bbb N}b'_{2n}$.}

\noindent Then $a\Bcap b$ and $a\Bsetminus b$ are both non-zero
for every $a\in A$.   \Prf\ There is an
$n\ge n_a$ such that $(n,f(n))\notin S_a$, so that

\Centerline{$\bar\mu(a\Bcap b_{2n})
=\bar\mu(a\Bcap c_{2n,f(n)})
>\Bover{\bar\mu a}{2(2n)!}
\ge\Bover1{2n(2n)!}$,}

\Centerline{$\bar\mu(a\Bcap b_{2n+1})
=\bar\mu(a\Bcap c_{2n+1,f(n)})
>\Bover{\bar\mu a}{2(2n+1)!}
\ge\Bover1{2n(2n+1)!}$.}

\noindent But

$$\eqalignno{\bar\mu(b_{2n}\Bsetminus b'_{2n})
&\le\sum_{i=2n+1}^{\infty}\bar\mu b_i
=\sum_{i=2n+1}^{\infty}\Bover1{i!}\cr
&\le\Bover1{(2n)!}\sum_{j=1}^{\infty}\Bover1{(2n+1)^j}
=\Bover1{2n(2n)!}
<\bar\mu(a\Bcap b_{2n}),\cr
\bar\mu(b_{2n+1}\Bsetminus b'_{2n+1})
&\le\Bover1{(2n+1)(2n+1)!}<\bar\mu(a\Bcap b_{2n+1}),\cr}$$

\noindent so $a\Bcap b\Bsupseteq a\Bcap b'_{2n}$ and
$a\Bsetminus b\Bsupseteq a\Bcap b'_{2n+1}$ are both non-zero.\ \Qed

\medskip

{\bf (f)} As $A$ is arbitrary, (ii) is true.
}%end of proof of 548I

\leader{548J}{Proposition} Let $(X,\Sigma,\mu)$ be a strictly localizable
measure space with null ideal $\Cal N(\mu)$,
and $\kappa$ a cardinal such that

\doubleinset{(*)
whenever $\Cal E\in[\Sigma\setminus\Cal N(\mu)]^{\le\kappa}$ and
$F\in\Sigma\setminus\Cal N(\mu)$, there is a non-negligible measurable
$G\subseteq F$ such that $E\setminus G$ is non-negligible for every
$E\in\Cal E$.}

\noindent Then whenever $\ofamily{\xi}{\kappa}{A_{\xi}}$ is a family of
non-negligible subsets of $X$, there is a $G\in\Sigma$ such that
$A_{\xi}\cap G$ and $A_{\xi}\setminus G$ are non-negligible for every
$\xi<\kappa$.

\proof{{\bf (a)} Suppose to begin with that $\mu X=1$.   Let $\frak A$ be
the measure algebra of $\mu$.   Then (*) says just that
$\kappa<\pi(\frak A_d)$ for every $d\in\frak A^+$.   For each $\xi<\kappa$
let $E_{\xi}$ be a measurable envelope of $A_{\xi}$ and set
$a_{\xi}=E_{\xi}^{\ssbullet}$ in $\frak A$.   By 548I, there is a
$b\in\frak A$ such that $a_{\xi}\Bcap b$ and $a_{\xi}\Bsetminus b$ are
non-zero for every $\xi<\kappa$.   Let $G\in\Sigma$ be such that
$b=G^{\ssbullet}$;  then $E_{\xi}\cap G$ and $E_{\xi}\setminus G$ are
non-negligible for every $\xi$.   But this means that $A_{\xi}\cap G$ and
$A_{\xi}\setminus G$ are non-negligible for every $\xi$.

\medskip

{\bf (b)} If $\mu X=0$ the result is trivial.   For other totally
finite $\mu$, we get the result from (a)
if we replace $\mu$ by a suitable scalar
multiple.

\medskip

{\bf (c)} For the general case, let $\familyiI{X_i}$ be a decomposition of
$X$ and for $i\in I$ set
$J_i=\{\xi:\xi<\kappa$, $A_{\xi}\cap X_i$ is not negligible$\}$.   By (b),
applied to the subspace measure on $X_i$,
there is a measurable $G_i\subseteq X_i$ such that $A_{\xi}\cap G_i$ and
$A_{\xi}\cap X_i\setminus G_i$ are non-negligible for every $\xi\in J_i$.
Set $G=\bigcup_{i\in I}G_i$;  this works.
}%end of proof of 548J

\leader{548K}{Corollary} Let $(X,\Sigma,\mu)$ be an atomless
quasi-Radon measure space and
$\ofamily{\xi}{\omega_1}{A_{\xi}}$ a family of non-negligible subsets of
$X$.   Then there is a $D\subseteq X$ such that $A_{\xi}\cap D$ and
$A_{\xi}\setminus D$ are non-negligible for every $\xi<\omega_1$.

\proof{{\bf (a)} Suppose to begin with that $\mu$ is a \Mth\ probability
measure.   Let $\frak A$ be the measure algebra of $\mu$.   If
$\pi(\frak A)>\omega_1$ we can use 548J.   Otherwise, the $\pi$-weight
$\pi(\mu)$ of $\mu$ is $\omega_1$ (524Tb);
let $\ofamily{\xi}{\omega_1}{E_{\xi}}$ be a
coinitial family in $\Sigma\setminus\Cal N(\mu)$.   Then we can choose
$x_{\xi\eta}$ and $y_{\xi\eta}$, for $\xi$, $\eta<\omega_1$, so that

\inset{all the $x_{\xi\eta}$, $y_{\xi\eta}$ are different,

if $A_{\xi}\cap E_{\eta}\notin\Cal N(\mu)$ then $x_{\xi\eta}$ and
$y_{\xi\eta}$ belong to $A_{\xi}\cap E_{\eta}$.}

\noindent Set $D=\{x_{\xi\eta}:\xi$, $\eta<\omega_1\}$;  then
$\mu^*(A_{\xi}\cap D)=\mu^*A_{\xi}$ for every $\xi$.   \Prf\Quer\
Otherwise, let $E$ be a measurable envelope of $A_{\xi}$ and $F$ a
measurable envelope of $A_{\xi}\cap D$.   We have $\mu(E\setminus F)>0$, so
there is an $\eta<\omega_1$ such that $E_{\eta}\subseteq E\setminus F$, in
which case

\Centerline{$x_{\xi\eta}\in A_{\xi}\cap E_{\eta}\cap D\subseteq F$.\
\Bang\Qed}

\noindent Similarly, $\mu^*(A_{\xi}\setminus D)=\mu^*A_{\xi}$ for every
$\xi$, and we have a suitable set $D$.

\medskip

{\bf (b)} In general, $X$ has a decomposition into \Mth\ subspaces (as in
the proofs of 524J and 524P), so the full result follows as in
(b)-(c) of the proof of 548J.
}%end of proof of 548K

\exercises{\leader{548X}{Basic exercises (a)}
%\spheader 548Xa
Suppose that there is a \rvm\ cardinal.   Show that there are an atomless
probability space $X$ and an equivalence relation $\sim$ on $X$ with
countable equivalence classes such that no set of full outer measure in
$X$ can meet every equivalence class in a finite set.
%548C

\spheader 548Xb\cmmnt{ ({\smc Kumar 13})} Write $\mu$ for Lebesgue measure
on $\Bbb R$.   Show that for any $X\subseteq\Bbb R$ there is an
$A\subseteq X$ such that $\mu^*A=\mu^*X$ and $|x-y|\notin\Bbb Q$ for any
distinct $x$, $y\in A$.
%548C out of order query
%query:  what about  X\subseteq\BbbR^2 , \|x-y\|\notin\BbbQ ?
%   question by P Komjath

\spheader 548Xc (A.Kumar) Let $\mu$ be Lebesgue measure on $\Bbb R$.   Show
that if $\Cal A$ is any partition of $\Bbb R$ into countable sets, there
are disjoint subsets $\Cal A_0$, $\Cal A_1$ of $\Cal A$ such that
$\bigcup\Cal A_0$ and $\bigcup\Cal A_1$ both have full outer measure in
$\Bbb R$.
%548C problem GC

\spheader 548Xd Suppose that there is an \am\ cardinal.
Show that there are an atomless
probability space $X$ and a sequence $\sequencen{A_n}$ of subsets of $X$
such that whenever $A'_n\subseteq A_n$ is a set of full outer measure in
$A_n$ for each $n$, $\sequencen{A'_n}$ is not disjoint.
%548E

\spheader 548Xe
Let $(X,\Sigma,\mu)$ be a complete semi-finite
measure space with the measurable envelope property (definition:  213Xl),
and suppose that $\mu$ is nowhere all-measuring in the sense of
214Yd, that is, that whenever $A\subseteq X$ is not $\mu$-negligible there
is a subset of $A$ which is not measured by the subspace measure on $A$.
(i) Show that for any $A\subseteq X$ there is a $D\subseteq A$ such that
$\mu^*D=\mu^*(A\setminus D)=\mu^*A$.
(ii) Show that whenever
$\familyiI{A_i}$ is a finite family of subsets of $X$, there is a
disjoint family $\familyiI{A'_i}$ such that $A'_i\subseteq A_i$ and
$\mu^*A'_i=\mu^*A_i$ for every $i\in I$.
%548G

\spheader 548Xf Show that the following are equiveridical:
(i) there is no \qm\ cardinal;  (ii) if
$(X,\Sigma,\mu)$ is a probability space such that $\mu\{x\}=0$ for every
$x\in X$ then there is a disjoint
family $\ofamily{\xi}{\omega_1}{D_{\xi}}$ of subsets of $X$ such that
$\mu^*D_{\xi}=1$ for every $\xi<\omega_1$.
%548G  $\Cal I$ $\omega_1$-sat $\sigma$-ideal,
%\Sigma the \sigma-algebra generated by \Cal I

\spheader 548Xg Let $(X,\Sigma,\mu)$ be a measure space such that
$\non\Cal N(\mu)=\pi(\mu)=\kappa\ge\omega$.   Show that
($\alpha$) if $R$ is an equivalence relation on $X$ and all its equivalence
classes have size less than $\kappa$ then there is an $R$-free set of full
outer measure
($\beta$) whenever $\ofamily{\xi}{\kappa}{A_{\xi}}$ is a family of
subsets of $X$ there is a disjoint family $\ofamily{\xi}{\kappa}{A'_{\xi}}$
such that $A'_{\xi}\subseteq A_{\xi}$ and $\mu^*A'_{\xi}=\mu^*A_{\xi}$ for
every $\xi<\kappa$.
%548H 521D

\spheader 548Xh
Let $\ofamily{\xi}{\frak c}{E_{\xi}}$ be a family of Lebesgue measurable
subsets of $\Bbb R$.   Show that there is a disjoint family
$\ofamily{\xi}{\frak c}{A_{\xi}}$ of sets such that
$A_{\xi}\subseteq E_{\xi}$
and $\mu^*A_{\xi}=\mu E_{\xi}$ for every $\xi<\frak c$, where $\mu$ is
Lebesgue measure on $\Bbb R$.   \Hint{419I.}
%548H

\spheader 548Xi (M.R.Burke)
Let $\Cal N$ be the null ideal of Lebesgue measure $\mu$
on $\Bbb R$.   Show that if $2^{\non\Cal N}=\frak c$ then there is a family
$\ofamily{\xi}{\frak c}{A_{\xi}}$ of subsets of $\Bbb R$ such that for
every $D\subseteq\Bbb R$ there is some $\xi<\frak c$ such that
$\min(\mu^*(A_{\xi}\cap D),\mu^*(A_{\xi}\setminus D))<\mu^*A_{\xi}$.
%548H 548Xh

\spheader 548Xj Let $\frak A$ be a Boolean algebra.   (i) Show that the
following are equiveridical:  ($\alpha$)
$\pi(\frak A_a)>\omega$ for every $a\in\frak A^+$,
where $\frak A_a$ is the principal ideal generated
by $a$  ($\beta$) for every sequence $\sequencen{a_n}$ in
$\frak A^+$ there is a disjoint sequence $\sequencen{b_n}$ in
$\frak A^+$ such that $b_n\Bsubseteq a_n$ for every $n$.
(ii) Show that if $\frak A$ has the $\sigma$-interpolation property then
we can add ($\gamma$) whenever $A\subseteq\frak A^+$ is countable,
there is a $b\in\frak A$ such that $a\Bcap b$ and
$a\Bsetminus b$ are both non-zero for every $a\in A$.
%548I

\leader{548Y}{Further exercises (a)}
%\spheader 548Ya
Suppose that there is a \qm\ cardinal.   Show that there is
a probability space $(X,\mu)$ such that there is a \qm\ cardinal less than
$\shr\Cal N(\mu)$ but no \qm\ cardinal less than $\tau(\mu)$.
%548G 511Fc   \#(X)=\kappa^+   \Cal N=[X]^{\le\kappa}   \frak A=\{0,1\}
}%end of exercises

\leader{548Z}{Problems (a)}
Suppose that $\ofamily{\xi}{\omega_1}{A_{\xi}}$ is a
family of subsets of $[0,1]$.   Must there be a set $D\subseteq[0,1]$ such that
$\mu^*(A_{\xi}\cap D)=\mu^*(A_{\xi}\setminus D)=\mu^*A_{\xi}$ for every
$\xi<\omega_1$, where $\mu$ is Lebesgue measure on $[0,1]$?

\spheader 548Zb Suppose that there is no \qm\ cardinal.   Let
$(X,\Sigma,\mu)$ be an atomless probability space
and $\ofamily{\xi}{\omega_1}{A_{\xi}}$ a
family of subsets of $X$.   Must there be a disjoint family
$\ofamily{\xi}{\omega_1}{A'_{\xi}}$ such that $A'_{\xi}\subseteq A_{\xi}$
and $\mu^*A'_{\xi}=\mu^*A_{\xi}$ for every $\xi<\omega_1$?

\spheader 548Zc (P.Komjath) Suppose that $X\subseteq\BbbR^2$.   Must there
be a set $A\subseteq X$, of the same Lebesgue outer measure as $X$,
such that $\|x-y\|\notin\Bbb Q$ whenever $x$, $y\in A$ are distinct?
(See 548Xb.)

\endnotes{
\Notesheader{548} Of course 548F is much the most important case of
548C-548E, with facts about Lebesgue measure provable in ZFC,
whether or not there are \qm\ cardinals or special relationships
between the cardinals of \S522.   As far as I know there is no real
simplification available for this special case if we wish to avoid special
axioms.   In many models of set theory, of course,
there are other approaches, as in 548G and 548Xg;
and I note that it makes a difference that we start with
not-necessarily-measurable sets $A_n$ in 548Fb (548Xh).

The arguments here leave many obvious questions open.   The first group
concerns possible
extensions of 548Fc or 548E to uncountable families of sets, as in 548Z.
I remark that {\smc Shelah 03} describes a model in which there is a set
$A\in\Cal P\Bbb R\setminus\Cal N$
such that $\Cal PA\cap\Cal N$ is $\omega_1$-saturated in
$\Cal PA$, where $\Cal N$ is the null ideal of Lebesgue measure.
Elsewhere we can ask, in 548C and 548E, whether the hypotheses involving
\qm\ cardinals could be rewritten with \am\ cardinals.   Only in 548G is it
clear that non-\am\ \qm\ cardinals are relevant (548Xf).

The questions tackled in this section can be re-phrased as questions about
structures $(\Cal PX/\Cal I,\frak A)$ where $\Cal I$ is a $\sigma$-ideal of
subsets of $X$ and $\frak A$ is a $\sigma$-subalgebra of the \pssqa\
$\Cal PX/\Cal I$;   a requirement of the form
`$\mu^*(A\cap D)=\mu^*A$' becomes (in the context of a totally finite
measure $\mu$) `$\upr(a\Bcap d,\frak A)=\upr(a,\frak A)$', where
$\upr(a,\frak A)$ is the upper envelope of $a$ in $\frak A$\cmmnt{
(313S)}.

I include 548I-548K %548I 548J 548K
to show that if we are less ambitious then there are
quite different, and rather easier, arguments available.   The condition
(*) of 548J is exact if we are looking for a measurable splitting set $G$.
But I am not at all sure that 548K is in the right form.
}%end of notes

\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
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%W
\medskip

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\medskip%X\medskip%Y\medskip%Z

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\discrpage

}%end of references

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\end



