% 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{mt2.tex}
\versiondate{24.2.16}

\noindent{\bf Introduction to Volume 2}

\medskip

For this second volume I have chosen seven topics through which to
explore the insights and challenges offered by measure theory.   Some,
like the Radon-Nikod\'ym theorem (Chapter 23) are necessary for any
understanding of the structure of the subject;   others, like Fourier
analysis (Chapter 28) and the discussion of function spaces (Chapter 24)
demonstrate the power of measure theory to attack problems in general
real and functional analysis.   But all have applications outside
measure theory, and all have influenced its development.   These are the
parts of measure theory which any analyst may find himself using.

Every topic is one which ideally one would wish undergraduates to have
seen, but the length of this volume makes it plain that no ordinary
undergraduate
course could include very much of it.   It is directed rather at graduate
level, where I hope it will be found adequate to support all but the most
ambitious courses in measure theory, though it is perhaps a bit too solid
to be suitable for direct
use as a course text, except with careful selection of the parts to be
covered.   If you are using it to teach yourself measure theory, I
strongly recommend an eclectic approach, looking for particular subjects
and theorems that seem startling or useful, and working backwards from
them.   My other objective, of course, is to provide an account of the
central ideas at this level in measure theory, rather fuller than can
easily be found in one volume elsewhere.   I cannot claim that it is
`definitive', but I do think I cover a good deal of ground in ways that
provide a firm foundation for further study.   As in Volume 1, I usually
do not shrink from giving `best' results, like Lindeberg's condition
for the Central Limit Theorem (\S274), or the theory of products of
arbitrary measure spaces (\S251).
If I were teaching this material to students in a PhD programme
I would rather accept a limitation in the breadth of the course than
leave them unaware of what could be done in the areas discussed.

The topics interact in complex ways -- one of the purposes of this book
is to exhibit their relationships.   There is no canonical linear
ordering in which they should be taken.   Nor do I think organization
charts are very helpful, not least because it may be only two or three
paragraphs in a section which are needed for a given chapter later on.
I do at least try to lay the material of each section out in an order
which makes initial segments useful by themselves.   But the order in
which to take the chapters is to a considerable extent for you to
choose, perhaps after a glance at their individual introductions.   I
have done my best to pitch the exposition at much the same level
throughout the volume, sometimes allowing gradients to steepen in the
course of a chapter or a section, but always trying to return to
something which anyone who has mastered Volume 1 ought to be able to
cope with.   (Though perhaps the main theorems of Chapter 26 are harder
work than the principal results elsewhere, and \S286 is only for the most
determined.)

I said there were seven topics, and you will see eight chapters ahead of
you.   This is because Chapter 21 is rather different from the rest.
It is the purest of pure measure theory, and is here only because there
are places later in the volume where (in my view) the theorems make
sense only in the light of some abstract concepts which are not
particularly difficult, but are also not obvious.   However it is fair
to say that the most important ideas of this volume do not really depend
on the work of Chapter 21.

As always, it is a puzzle to know how much prior knowledge to assume in
this volume.   I do of course call on the results of Volume 1 of this
treatise whenever they seem to be relevant.   I do not doubt, however,
that there will be readers who have learnt the elementary theory from
other sources.   Provided you can, from first principles, construct
Lebesgue measure and prove the basic convergence theorems for integrals
on arbitrary measure spaces, you ought to be able to embark on the
present volume.   Perhaps it would be helpful to have in hand the
results-only version of Volume 1, since that includes the most important
definitions as well as statements of the theorems.

There is also the question of how much material from outside measure
theory is needed.   Chapter 21 calls for some non-trivial set theory
(given in \S2A1), but the more advanced ideas are needed only for the
counter-examples in \S216, and can be passed over to begin with.   The
problems become acute in Chapter 24.   Here we need a variety of results
from functional analysis, some of them depending on non-trivial ideas in
general topology.   For a full understanding of this material there is
no substitute for a course in normed spaces up to and including a study
of weak compactness.   But I do not like to insist on such a
preparation, because it is likely to be simultaneously too much and too
little.   Too much, because I hardly mention linear operators at this
stage;  too little, because I do ask for some of the theory of
non-locally-convex spaces, which is often omitted in first courses on
functional analysis.   At the risk, therefore, of wasting paper, I have
written out condensed accounts of the essential facts
(\S\S2A3-2A5).  %\S2A3 \S2A4 \S2A5

\bigskip

\noindent{\bf Note on second printing, April 2003}

\medskip

For the second printing of this volume, I have made two substantial
corrections to inadequate proofs and a large number of minor amendments;
I am most grateful to T.D.Austin for his careful reading of the first
printing.   In addition, I have added a dozen exercises and a handful of
straightforward results which turn out to be relevant to the work of
later volumes and fit naturally here.

The regular process of revision of this work has led me to make a couple
of notational innovations not described explicitly in the early printings
of Volume 1.   I trust that most readers will find these immediately
comprehensible.   If, however, you find that there is a puzzling
cross-reference which you are unable to match with anything in the
version of Volume 1 which you are using, it may be worth while checking
the errata pages in {\tt
http://www1.essex.ac.uk/maths/people/fremlin/mterr.htm}.

\bigskip

\noindent{\bf Note on hardback edition, January 2010}

\medskip

For the new (`Lulu') edition of this volume, I have eliminated a number of
further errors;  no doubt many remain.   There are many new exercises,
several new theorems
%214P, 222L Dini derivates, half of \S234, 244O, 251L, \S266, 272Q, 272W
and some corresponding rearrangements of material.
%\S234
The new results are mostly additions with little effect on the structure of
the work, but there is a short new section (\S266) on the Brunn-Minkowski
inequality.

\bigskip

\noindent{\bf Note on second printing of hardback edition, April 2016}

\medskip

There is the usual crop of small mistakes to be corrected, and assorted minor
amendments and additions.   But my principal reason for issuing a new printed
version is a major fault in the proof of Carleson's theorem, where an imprudent
move to simplify the argument of {\smc Lacey \& Thiele 00} was based on an
undergraduate error\footnote{I am most
grateful to A.Derighetti for bringing this to my attention.}.
While the blunder is conspicuous enough, a resolution
seems to require an adjustment in a definition, and is not a fair demand on
a graduate seminar, the intended readership for this material.
Furthermore, the proof was supposed to be a distinguishing feature of not
only this volume, but of the treatise as a whole.   So, with apologies to any
who retired hurt from an encounter with the original version, I present a
revision which I hope is essentially sound.

\frnewpage

\frfilename{mt21.tex} 
\versiondate{17.1.15} 
\copyrightdate{1994} 
 
\def\chaptername{Taxonomy of measure spaces} 
 
\gdef\topparagraph{} 
   \gdef\bottomparagraph{Chap.\ 21 {\it intro.}} 
   \gdef\newparagraph{Chap.\ 21 {\it intro.}} 
   \gdef\sectionname{Introduction} 
   \gdef\headlinesectionname{Introduction} 
   \centerline{\bf *Chapter 21} 
   \medskip 
   \centerline{\bf \chaptername} 
   \medskip 
 
I begin this volume with a `starred chapter'.   The point is that I do 
not really recommend this chapter for beginners.   It deals with a 
variety of technical questions which are of great importance for the 
later development of the subject, but are likely to be both abstract and 
obscure for anyone who has not encountered the problems these techniques 
are designed to solve.   On the other hand, if (as is customary) this 
work is omitted, and the ideas are introduced only when urgently needed, 
the student is likely to finish with very vague ideas on which theorems 
can be expected to apply to which types of measure space, and with no 
vocabulary in which to express those ideas.   I therefore 
take a few pages to introduce the terminology and concepts which can be 
used to distinguish `good' measure spaces from others, with a few of 
the basic relationships.   The only paragraphs which are immediately 
relevant to the theory set out in Volume 1 are those on `complete',  
`$\sigma$-finite' and 
`semi-finite' measure spaces (211A, 211D, 211F, 211Lc, \S212,  
213A-213B, 215B), and on Lebesgue measure (211M).   For the rest, I think  
that a newcomer 
to the subject can very well pass over this chapter for the time being, and return to it 
for particular items when the text of later chapters refers to it. 
On the other hand, it can also be used as an introduction to the flavour  
of the `purest' kind of measure theory, the study of measure spaces for  
their own sake, with a systematic discussion of a few of the elementary  
constructions. 
 
\discrpage 
 
\frfilename{mt211.tex}
\versiondate{20.11.03}

\def\chaptername{Taxonomy of measure spaces}
\def\sectionname{Definitions}

\newsection{211}

I start with a list of definitions, corresponding to the concepts which
I have found to be of value in distinguishing different types of measure
space.   Necessarily, the significance of many of these ideas is likely
to be obscure until you have encountered some of the obstacles which
arise later on.   Nevertheless, you will I hope be
able to deal with these definitions on a formal, abstract basis, and to
follow the elementary arguments involved in establishing the
relationships between them (211L).

In 216C-216E %216C 216D 216E
below I will give three substantial examples to demonstrate the rich variety of objects which the definition of `measure space'
encompasses.   In the present section, therefore, I content myself with
very brief descriptions of sufficient cases to show at least that
each of the definitions here discriminates between different spaces
(211M-211R).

\leader{211A}{Definition} Let $(X,\Sigma,\mu)$ be a measure space.
Then $\mu$, or $(X,\Sigma,\mu)$, is ({\bf \Caratheodory}) {\bf complete} if whenever
$A\subseteq E\in\Sigma$ and $\mu E=0$ then $A\in\Sigma$\cmmnt{;  that is, if every negligible subset of $X$ is measurable}.

\leader{211B}{Definition} Let $(X,\Sigma,\mu)$ be a measure space.
Then $(X,\Sigma,\mu)$ is a {\bf probability space} if $\mu X=1$.
In this case $\mu$ is called a {\bf probability} or {\bf probability
measure}.

\leader{211C}{Definition} Let $(X,\Sigma,\mu)$ be a measure space.
Then $\mu$, or $(X,\Sigma,\mu)$, is {\bf totally finite} if $\mu
X<\infty$.

\leader{211D}{Definition} Let $(X,\Sigma,\mu)$ be a measure space.
Then $\mu$, or $(X,\Sigma,\mu)$, is {\bf $\sigma$-finite} if there is a
sequence $\sequencen{E_n}$ of measurable sets of finite measure such
that $X=\bigcup_{n\in\Bbb N}E_n$.

\cmmnt{\medskip

\noindent{\bf Remark} Note that in this case we can set

\Centerline{$F_n=E_n\setminus\bigcup_{i<n}E_i$,\quad $G_n=\bigcup_{i\le
n}E_i$}

\noindent for each $n$, to obtain a partition $\sequencen{F_n}$ of
$X$ (that is, a disjoint cover of $X$) into measurable sets of finite measure, and a non-decreasing sequence
$\sequencen{G_n}$ of sets of finite measure covering $X$.
}%end of comment

\leader{211E}{Definition} Let $(X,\Sigma,\mu)$ be a measure space.
Then $\mu$, or $(X,\Sigma,\mu)$, is {\bf strictly localizable} or {\bf
decomposable} if there is a partition
$\langle X_i\rangle_{i\in I}$ of $X$ into measurable sets of finite
measure such that

\Centerline{$\Sigma
=\{E:E\subseteq X,\,E\cap X_i\in\Sigma\Forall i\in I\}$,}

\Centerline{$\mu E=\sum_{i\in I}\mu(E\cap X_i)$ for every $E\in\Sigma$.}

\noindent I will call such a family $\langle X_i\rangle_{i\in I}$ a {\bf
decomposition} of $X$.

\cmmnt{\medskip

\noindent{\bf Remark} In this context, we can interpret the sum
$\sum_{i\in I}\mu(E\cap X_i)$ simply as

\Centerline{$\sup\{\sum_{i\in J}\mu(E\cap X_i):J$ is a finite subset of
$I\}$,}

\noindent taking $\sum_{i\in\emptyset}\mu(E\cap X_i)=0$, because we are
concerned only with sums of non-negative terms (cf.\ 112Bd).
}%end of comment

\leader{211F}{Definition} Let $(X,\Sigma,\mu)$ be a measure space.
Then $\mu$, or $(X,\Sigma,\mu)$, is {\bf semi-finite} if whenever
$E\in\Sigma$ and $\mu E=\infty$ there is an $F\subseteq E$ such that
$F\in\Sigma$ and $0<\mu F<\infty$.

\leader{211G}{Definition} Let $(X,\Sigma,\mu)$ be a measure space.
Then $\mu$, or $(X,\Sigma,\mu)$, is {\bf localizable} if it
is semi-finite and, for every $\Cal E\subseteq\Sigma$, there is an
$H\in\Sigma$ such that (i) $E\setminus H$ is negligible for every
$E\in\Cal E$ (ii) if $G\in\Sigma$ and $E\setminus G$ is negligible for
every $E\in\Cal E$, then $H\setminus G$ is negligible.
It will be
convenient to call such a set $H$  {\bf an essential supremum} of
$\Cal E$ in $\Sigma$.

\cmmnt{\medskip

\noindent{\bf Remark} The definition  here is clumsy,
because really the concept applies to measure {\it algebras}
rather than to measure {\it spaces} (see 211Ya-211Yb).   However, the
present definition  can be made to work (see 213N, 241G, 243G below) and
enables us to proceed
without a formal introduction to the concept of `measure algebra'
before the time comes to do the job properly in Volume 3.
}

\leader{211H}{Definition} Let $(X,\Sigma,\mu)$ be a measure space.
Then $\mu$, or $(X,\Sigma,\mu)$, is {\bf locally determined} if it is
semi-finite and

\Centerline{$\Sigma=\{E:E\subseteq X,\,E\cap F\in\Sigma$ whenever
$F\in\Sigma$ and $\mu F<\infty\}$\dvro{.}{;}}

\cmmnt{\noindent that is to say, for any $E\in\Cal PX\setminus\Sigma$
there is
an $F\in\Sigma$ such that $\mu F<\infty$ and $E\cap F\notin\Sigma$.}

\leader{211I}{Definition} Let $(X,\Sigma,\mu)$ be a measure space.    A
set $E\in\Sigma$ is an {\bf atom} for $\mu$ if $\mu E>0$ and
whenever $F\in\Sigma$, $F\subseteq E$ one of $F$, $E\setminus F$ is
negligible.

\leader{211J}{Definition} Let $(X,\Sigma,\mu)$ be a measure space.
Then $\mu$, or $(X,\Sigma,\mu)$, is {\bf atomless} or {\bf diffused} if
there are no atoms for $\mu$.
\cmmnt{(Note that this is {\it not} the same thing as saying that all
finite sets are negligible;  see 211R below.   Some authors use the word
{\bf continuous} in this context.)}

\leader{211K}{Definition} Let $(X,\Sigma,\mu)$ be a measure space.
Then $\mu$, or $(X,\Sigma,\mu)$, is {\bf purely atomic} if whenever
$E\in\Sigma$ and $E$ is not negligible there is an atom for $\mu$
included in $E$.

\medskip

\noindent{\bf Remark} \cmmnt{Recall that a measure $\mu$ on a set $X$ is
{\bf point-supported} if $\mu$ measures every subset of $X$
and $\mu E=\sum_{x\in E}\mu\{x\}$ for every $E\subseteq X$ (112Bd).}
Every point-supported measure is purely atomic\prooflet{, because
$\{x\}$ must be an atom whenever
$\mu\{x\}>0$}, but not every purely atomic measure is
point-supported\cmmnt{ (211R)}.

% {\smc Ramachandran 02} uses `discrete' for `purely atomic'

\leader{211L}{}\cmmnt{ The relationships between the concepts above are in
a sense very straightforward;  all the direct implications in which one
property implies another are given in the next theorem.

\medskip

\noindent}{\bf Theorem} (a) A probability space is
totally finite.

(b) A totally finite measure space is $\sigma$-finite.

(c) A $\sigma$-finite measure space is strictly localizable.

(d) A strictly localizable measure space is localizable and locally
determined.

(e) A localizable measure space is semi-finite.

(f) A locally determined measure space is semi-finite.

\proof{ (a), (b), (e) and (f) are trivial.

\medskip

{\bf (c)} Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space;  let
$\sequencen{F_n}$ be a disjoint sequence of measurable sets of finite
measure covering $X$ (see the remark in 211D).   If $E\in\Sigma$, then
of course $E\cap F_n\in\Sigma$ for every $n\in \Bbb N$, and

\Centerline{$\mu E=\sum_{n=0}^{\infty}\mu(E\cap
F_n)=\sum_{n\in \Bbb N}\mu(E\cap F_n)$.}

\noindent   If $E\subseteq X$ and $E\cap
F_n\in\Sigma$ for every $n\in \Bbb N$, then

\Centerline{$E=\bigcup_{n\in \Bbb N}E\cap F_n\in \Sigma$.}

\noindent So $\sequencen{F_n}$ is a decomposition of $X$
and $(X,\Sigma,\mu)$ is strictly localizable.

\medskip

{\bf (d)} Let $(X,\Sigma,\mu)$ be a strictly localizable measure space;
let $\langle X_i\rangle_{i\in I}$ be a decomposition of $X$.

\medskip

\quad{\bf (i)} Let $\Cal E$ be a family of measurable subsets of $X$.
Let $\Cal F$ be the family of measurable sets $F\subseteq X$ such that
$\mu(F\cap E)=0$ for every $E\in\Cal E$.   Note that
$\emptyset\in\Cal F$ and, if
$\sequencen{F_n}$ is any sequence in $\Cal F$, then
$\bigcup_{n\in\Bbb N}F_n\in\Cal F$.   For each $i\in I$, set $\gamma_i=\sup\{\mu(F\cap X_i):F\in\Cal F\}$ and choose a sequence $\sequencen{F_{in}}$ in $\Cal F$ such that $\lim_{n\to\infty}\mu(F_{in}\cap X_i)=\gamma_i$;  set

\Centerline{$F_i=\bigcup_{n\in\Bbb N}F_{in}\in\Cal F$.}

\noindent   Set

\Centerline{$F=\bigcup_{i\in I}F_i\cap X_i\subseteq X$}

\noindent and $H=X\setminus F$.

We see that $F\cap X_i=F_i\cap X_i$ for each $i\in I$ (because
$\langle X_i\rangle_{i\in I}$ is disjoint), so $F\in\Sigma$ and
$H\in \Sigma$.   For any $E\in\Cal E$,

\Centerline{$\mu(E\setminus H)=\mu(E\cap F)
=\sum_{i\in I}\mu(E\cap F\cap X_i)
=\sum_{i\in I}\mu(E\cap F_i\cap X_i)=0$}

\noindent because every $F_i$ belongs to $\Cal F$.   Thus $F\in\Cal F$.
If $G\in\Sigma$ and
$\mu(E\setminus G)=0$ for every $E\in\Cal E$, then $X\setminus G$ and
$F'=F\cup(X\setminus G)$ belong to $\Cal F$.   So
$\mu(F'\cap X_i)\le\gamma_i$ for each $i\in I$.   But also
$\mu(F\cap X_i)\ge\sup_{n\in\Bbb N}\mu(F_{in}\cap X_i)=\gamma_i$, so $\mu(F\cap X_i)=\mu(F'\cap X_i)$ for each $i$.   {\it Because $\mu X_i$ is finite,}
it follows that $\mu((F'\setminus F)\cap X_i)=0$, for each $i$.
Summing over $i$, $\mu(F'\setminus F)=0$, that is,
$\mu(H\setminus G)=0$.

Thus $H$ is an essential supremum for $\Cal E$ in
$\Sigma$.   As $\Cal E$ is arbitrary, $(X,\Sigma,\mu)$ is localizable.

\medskip

\quad{\bf (ii)}
If $E\in\Sigma$ and $\mu E=\infty$, then there is some $i\in I$ such
that

\Centerline{$0<\mu(E\cap X_i)\le\mu X_i<\infty$;}

\noindent  so $(X,\Sigma,\mu)$ is
semi-finite.    If $E\subseteq X$ and $E\cap F\in\Sigma$ whenever $\mu
F<\infty$, then, in particular, $E\cap X_i\in\Sigma$ for every $i\in I$,
so $E\in\Sigma$;  thus $(X,\Sigma,\mu)$ is locally determined.
}%end of proof of 211L

\leader{211M}{Example:  Lebesgue measure}\cmmnt{ Let us consider Lebesgue measure in the light of the concepts above.}   Write $\mu$ for Lebesgue measure on $\BbbR^r$\cmmnt{ and $\Sigma$ for its domain}.

\spheader 211Ma $\mu$ is complete\prooflet{, because it is
constructed by \Caratheodory's method;  if $A\subseteq E$ and $\mu
E=0$, then $\mu^*A=\mu^*E=0$ (writing $\mu^*$ for Lebesgue outer
measure), so, for any $B\subseteq\Bbb R$,

\Centerline{$\mu^*(B\cap A)+\mu^*(B\setminus A)\le 0+\mu^*B=\mu^*B$,}

\noindent and $A$ must be measurable}.

\spheader 211Mb $\mu$ is $\sigma$-finite\prooflet{, because
$\BbbR^r=\bigcup_{n\in\Bbb N}[-\tbf{n},\tbf{n}]$, writing $\tbf{n}$ for the vector
$(n,\ldots,n)$, and $\mu[-\tbf{n},\tbf{n}]=(2n)^r<\infty$ for every $n$.
Of course $\mu$ is neither totally finite nor a probability measure}.

\spheader 211Mc\cmmnt{ Because} $\mu$
is\cmmnt{ $\sigma$-finite, it is} strictly
localizable\cmmnt{ (211Lc)}, localizable\cmmnt{ (211Ld)}, locally
determined\cmmnt{ (211Ld)} and semi-finite\cmmnt{ (211Le or 211Lf)}.

\spheader 211Md $\mu$ is atomless.   \prooflet{\Prf\ Suppose that
$E\in\Sigma$.  Consider the function

\Centerline{$a\mapsto f(a)
=\mu(E\cap[-\tbf{a},\tbf{a}]):\coint{0,\infty}\to\Bbb R$}

\noindent We have

\Centerline{$f(a)
\le f(b)\le f(a)+\mu[-\tbf{b},\tbf{b}]-\mu[-\tbf{a},\tbf{a}]
=f(a)+(2b)^r-(2a)^r$}

\noindent whenever $a\le b$ in $\coint{0,\infty}$, so $f$ is
continuous.   Now $f(0)=0$ and $\lim_{n\to\infty}f(n)=\mu E>0$.   By the
Intermediate Value Theorem there is an $a\in\coint{0,\infty}$ such that
$0<f(a)<\mu E$.   So we have

\Centerline{$0<\mu(E\cap[-\tbf{a},\tbf{a}])<\mu E$.}

\noindent As $E$ is arbitrary, $\mu$ is atomless.\ \Qed}

\header{211Me}{\bf (e)} It is now a trivial observation that $\mu$
cannot be purely atomic\cmmnt{, because $\BbbR^r$ itself is a set of
positive measure not including any atom}.

\leader{211N}{Counting measure} Take $X$ to be any uncountable set\cmmnt{ (e.g.,
$X=\Bbb R$)}, and $\mu$ to be counting measure on $X$\cmmnt{ (112Bd)}.

\header{211Na}{\bf (a)} $\mu$ is
complete\ifwithproofs{, because if $A\subseteq E$ and
$\mu E=0$ then

\Centerline{$A=E=\emptyset\in\Sigma$.}
}\else{.}\fi

\header{211Nb}{\bf (b)} $\mu$ is not $\sigma$-finite\cmmnt{, because if $\sequencen{E_n}$ is any sequence of sets of finite measure then every $E_n$ is finite, therefore countable, and
$\bigcup_{n\in\Bbb N}E_n$ is countable (1A1F), so cannot be $X$}.   \cmmnt{{\it A fortiori,}} $\mu$ is not a
probability measure nor totally finite.

\spheader 211Nc $\mu$ is strictly localizable.   \prooflet{\Prf\ Set $X_x=\{x\}$ for every $x\in X$.  Then $\langle X_x\rangle_{x\in X}$ is a partition of $X$, and for any $E\subseteq X$

\Centerline{$\mu(E\cap X_x)=1$ if $x\in E$,\quad$0$ otherwise.}

\noindent By the definition of $\mu$,

\Centerline{$\mu E=\sum_{x\in X}\mu(E\cap X_x)$}

\noindent for every $E\subseteq X$, and $\langle X_x\rangle_{x\in X}$ is
a decomposition of $X$.\ \Qed

Consequently} $\mu$ is localizable, locally determined and semi-finite.

\header{211Nd}{\bf (d)} $\mu$ is purely atomic.   \prooflet{\Prf\ $\{x\}$
is an atom for every $x\in X$, and if $\mu E>0$ then surely $E$ includes
$\{x\}$ for some $x$.\ \QeD}   \cmmnt{Obviously,} $\mu$ is not atomless.

\leader{211O}{A non-semi-finite space} Set $X=\{0\}$,
$\Sigma=\{\emptyset,X\}$, $\mu\emptyset=0$ and $\mu X=\infty$.   Then
$\mu$ is not semi-finite,\cmmnt{ as $\mu X=\infty$ but $X$ has no subset of non-zero finite measure.   It follows that $\mu$ cannot be} localizable,
locally determined, $\sigma$-finite, totally finite nor a probability
measure.   \cmmnt{Because $\Sigma=\Cal PX$,} $\mu$ is complete.
\cmmnt{$X$ is an atom
for $\mu$, so} $\mu$ is purely atomic (indeed, it is point-supported).

\leader{211P}{A non-complete space} Write $\Cal B$ for
the $\sigma$-algebra of Borel subsets of $\Bbb R$\cmmnt{ (111G)}, and
$\mu$ for the
restriction of Lebesgue measure to $\Cal B$\cmmnt{ (recall that by 114G
every Borel subset of $\Bbb R$ is Lebesgue measurable)}.
Then $(\Bbb R,\Cal B,\mu)$ is atomless, $\sigma$-finite and not complete.

\proof{{\bf (a)} To see that $\mu$ is not complete, recall that there is a
continuous, strictly increasing permutation
$g:[0,1]\to[0,1]$ such that $\mu g[C]>0$, where $C$ is the Cantor set,
so that there is a
set $A\subseteq g[C]$ which is not Lebesgue measurable (134Ib).   Now
$g^{-1}[A]\subseteq C$ cannot be a Borel set, since
$\chi A=\chi(g^{-1}[A])\smallcirc g^{-1}$ is not Lebesgue
measurable, therefore not Borel measurable, and the composition of
two Borel measurable functions is Borel measurable (121Eg);
so $g^{-1}[A]$ is a non-measurable subset of the negligible set $C$.

\medskip

{\bf (b)} The rest of the arguments of 211M apply to $\mu$ just as
well as to true Lebesgue measure, so $\mu$ is $\sigma$-finite and
atomless.
}%end of proof of 211P

\cmmnt{\medskip

\noindent{\bf *Remark} The argument offered in (a) could give rise to
a seriously false impression.   The set $A$ referred to there can be
constructed only with the use of a strong form of the axiom of choice.
No such device is necessary for the result here.
There are many methods of
constructing non-Borel subsets of the Cantor set, all illuminating in
different ways.   In the absence of any form of the axiom of choice, there
are difficulties with the concept of `Borel set', and others with the
concept of `Lebesgue measure', which I will come to in
Chapter 56;  but countable choice is quite sufficient for the existence of
a non-Borel subset of $\Bbb R$.   For details of a possible approach
see 423M in Volume 4.
}

\leader{211Q}{Some probability spaces} Two\cmmnt{ obvious} constructions
of probability spaces\cmmnt{, restricting myself to the methods described in Volume 1,} are

(a) the subspace measure induced by Lebesgue measure on
$[0,1]$\cmmnt{ (131B)};

(b) the point-supported measure induced on a set $X$ by a function $h:X\to[0,1]$ such
that $\sum_{x\in X}h(x)=1$, writing $\mu E=\sum_{x\in E}h(x)$ for every
$E\subseteq X$;  for instance, if $X$ is a singleton
$\{x\}$ and $h(x)=1$, or if $X=\Bbb N$ and $h(n)=2^{-n-1}$.

Of these two, (a) gives an atomless probability measure and (b) gives a
purely atomic probability measure.

\leader{211R}{Countable-cocountable measure}\cmmnt{ The following is one of the basic constructions to keep in mind when considering abstract
measure spaces.

\medskip

}{\bf (a)} Let $X$ be any set.
Let $\Sigma$ be the family of those sets $E\subseteq X$
such that either $E$ or $X\setminus E$ is
countable.   Then $\Sigma$ is a $\sigma$-algebra of subsets of $X$.
\prooflet{\Prf\  {(i)} $\emptyset$ is countable, so belongs to $\Sigma$.
{(ii)} The condition for $E$ to belong to $\Sigma$ is symmetric
between $E$ and $X\setminus E$, so $X\setminus E\in\Sigma$ for every
$E\in\Sigma$.   {(iii)} Let $\sequencen{E_n}$ be any sequence in
$\Sigma$,
and set $E=\bigcup_{n\in\Bbb N}E_n$.   If every $E_n$ is countable, then
$E$ is countable, so belongs to $\Sigma$.   Otherwise, there is
some $n$ such that $X\setminus E_n$ is countable, in which case
$X\setminus E\subseteq X\setminus E_n$ is countable, so again
$E\in\Sigma$.\ \QeD}   $\Sigma$ is called the
{\bf countable-cocountable $\sigma$-algebra} of $X$.

\header{211Rb}{\bf (b)} Now consider the function $\mu:\Sigma\to\{0,1\}$
defined by writing $\mu E=0$ if $E$ is countable, $\mu E=1$ if
$E\in\Sigma$ and $E$ is not countable.   Then $\mu$ is a measure.
\prooflet{\Prf\ {(i)} $\emptyset$ is countable so $\mu\emptyset=0$.   {(ii)}
Let $\sequencen{E_n}$ be a disjoint sequence in $\Sigma$, and $E$
its union.   {(${\alpha}$)} If every $E_m$ is countable, then so
is $E$, so

\Centerline{$\mu E=0=\sum_{n=0}^{\infty}\mu E_n$.}

\noindent{(${\beta}$)} If some $E_m$ is uncountable, then
$E\supseteq E_m$ also is uncountable, and $\mu E=\mu E_m=1$.   But in
this case, because $E_m\in\Sigma$, $X\setminus E_m$ is countable, so
$E_n$, being a subset of $X\setminus E_m$, is countable for every $n\ne
m$;  thus $\mu E_n=0$ for every $n\ne m$, and

\Centerline{$\mu E=1=\sum_{n=0}^{\infty}\mu E_n$.}

\noindent As $\sequencen{E_n}$ is arbitrary, $\mu$ is a measure.\ \Qed}
This is the {\bf countable-cocountable measure} on $X$.

\header{211Rc}{\bf (c)} If $X$ is any uncountable set and
$\mu$ is the countable-cocountable measure on $X$, then $\mu$ is a
complete, purely atomic probability measure, but is not point-supported.   \prooflet{\Prf\ {(i)} If
$A\subseteq E$ and $\mu E=0$, then $E$ is countable, so $A$ also is
countable and belongs to $\Sigma$.   Thus $\mu$ is complete.   {(ii)}
Because $X$ is uncountable, $\mu X=1$ and $\mu$ is a probability
measure.   {(iii)} If $\mu E>0$, then $\mu F=\mu E=1$ whenever $F$
is a non-negligible measurable subset of $E$, so $E$ is itself an atom;
thus $\mu$ is purely atomic.   (iv) $\mu X=1>0=\sum_{x\in X}\mu\{x\}$, so $\mu$ is not point-supported.\ \Qed}

\exercises{\leader{211X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 211Xa
Let $(X,\Sigma,\mu)$ be a semi-finite measure space.   Show
that $\mu$ is $\sigma$-finite iff there is a totally finite measure
$\nu$ on $X$ with the same measurable sets and the same negligible sets
as $\mu$.
%211D

\sqheader 211Xb Let $g:\Bbb R\to\Bbb R$ be a non-decreasing
function and $\mu_g$ the associated Lebesgue-Stieltjes measure (114Xa).
Show that $\mu_g$ is complete and $\sigma$-finite.   Show that

\quad (i) $\mu_g$ is totally finite iff $g$ is bounded;

\quad (ii) $\mu_g$ is a probability measure iff $\lim_{x\to\infty}g(x)
-\lim_{x\to-\infty}g(x)=1$;

\quad (iii) $\mu_g$ is atomless iff $g$ is continuous;

\quad (iv) if $E$ is any atom for $\mu_g$, there is a point $x\in E$
such that $\mu_gE=\mu_g\{x\}$;

\quad (v) $\mu_g$ is purely atomic iff it is point-supported.
%211M

\sqheader 211Xc Let $\mu$ be counting measure on a set $X$.
Show that $\mu$ is always complete, strictly localizable and purely atomic, and
that it is $\sigma$-finite iff $X$ is countable, totally finite iff $X$
is finite, a probability measure iff $X$ is a singleton, and atomless
iff $X$ is empty.
%211N

\spheader 211Xd Show that a point-supported measure is always complete,
and is strictly localizable iff it is semi-finite.
%211O

\spheader 211Xe Let $X$ be a set.   Show that for any $\sigma$-ideal
$\Cal I$ of subsets of $X$ (definition:  112Db), the set

\Centerline{$\Sigma=\Cal I\cup\{X\setminus A:A\in\Cal I\}$}

\noindent is a $\sigma$-algebra of subsets of $X$, and that there is a
measure $\mu:\Sigma\to\{0,1\}$ given by setting

\Centerline{$\mu E=0$ if $E\in\Cal I$,
\quad $\mu E=1$ if $E\in\Sigma\setminus\Cal I$.}

\noindent Show that $\Cal I$ is precisely the null ideal of $\mu$, that $\mu$ is complete, totally finite and purely atomic, and is a
probability measure iff $X\notin\Cal I$.
%211R

\spheader 211Xg Let $(X,\Sigma,\mu)$ be a measure space such that
$\mu X>0$.    Show that the set of conegligible subsets of $X$ is a filter
on $X$.

\leader{211Y}{Further exercises (a)}
%\spheader 211Ya
Let $(X,\Sigma,\mu)$ be a measure space, and for
$E$, $F\in\Sigma$ write $E\sim F$ if $\mu(E\symmdiff F)=0$.   Show that
$\sim$ is an equivalence relation on $\Sigma$.   Let $\frak A$ be the
set of equivalence classes in $\Sigma$ for $\sim$;  for $E\in\Sigma$,
write $E^{\ssbullet}\in\frak A$ for its equivalence class.   Show that
there is a partial ordering $\Bsubseteq$ on $\frak A$ defined by saying
that, for $E$, $F\in\Sigma$,

\Centerline{$E^{\ssbullet}\Bsubseteq F^{\ssbullet}
\,\iff\,\mu(E\setminus F)=0$.}

\noindent Show that $\mu$ is localizable iff for every
$A\subseteq\frak A$ there is an $h\in\frak A$ such that (i)
$a\Bsubseteq h$ for every
$a\in A$ (ii) whenever $g\in\frak A$ is such that $a\Bsubseteq g$ for
every $a\in\frak A$, then $h\Bsubseteq g$.
%211G

\spheader 211Yb Let $(X,\Sigma,\mu)$ be a measure space, and
construct $\frak A$ as in 211Ya.   Show that there are operations
$\Bcup$, $\Bcap$, $\Bsetminus$ on $\frak A$ defined by saying that

\Centerline{$E^{\ssbullet}\Bcap F^{\ssbullet}=(E\cap F)^{\ssbullet}$,
\quad$E^{\ssbullet}\Bcup F^{\ssbullet}=(E\cup F)^{\ssbullet}$,
\quad$E^{\ssbullet}\Bsetminus F^{\ssbullet}=(E\setminus F)^{\ssbullet}$}

\noindent for all $E$, $F\in\Sigma$.
Show that if $A\subseteq\frak A$ is any {\it countable} set, then there
is an $h\in\frak A$ such that (i) $a\Bsubseteq h$ for every
$a\in A$ (ii) whenever $g\in\frak A$ is such that $a\Bsubseteq g$ for
every $a\in\frak A$, then $h\subseteq g$.   Show that there is a
functional $\bar\mu:\frak A\to[0,\infty]$ defined by saying that
$\bar\mu(E^{\ssbullet})=\mu E$ for every $E\in\Sigma$.
($(\frak A,\bar\mu)$ is called the {\bf measure algebra} of
$(X,\Sigma,\mu)$.)
%211Ya 211G

\spheader 211Yc Let $(X,\Sigma,\mu)$ be a semi-finite measure
space.   Show
that it is atomless iff whenever $\epsilon>0$, $E\in\Sigma$ and
$\mu E<\infty$, then there is a finite partition of $E$ into measurable sets of measure at most $\epsilon$.
%211J

\spheader 211Yd Let $(X,\Sigma,\mu)$ be a strictly localizable
measure space.   Show that the following are equiveridical:
(i) $(X,\Sigma,\mu)$ is atomless;  (ii) for every $\epsilon>0$
there is a decomposition of $X$ consisting of sets of measure at most
$\epsilon$;  (iii)\dvAnew{2018} 
there is a measurable function $f:X\to\Bbb R$ such
that $\mu f^{-1}[\{t\}]=0$ for every $t\in\Bbb R$.
%211Yc 211J

\spheader 211Ye Let $\Sigma$ be the countable-cocountable
$\sigma$-algebra of $\Bbb R$.   Show that
$\coint{0,\infty}\notin\Sigma$.   Let $\mu$ be the restriction of
counting measure to $\Sigma$.   Show that $(\Bbb R,\Sigma,\mu)$ is
complete, semi-finite and purely atomic, but not localizable nor locally
determined.
%211R
}%end of exercises

\endnotes{
\Notesheader{211} The list of definitions in 211A-211K
probably strikes you as quite long enough, even though I have omitted
many occasionally useful ideas.   The concepts here vary
widely in importance, and the importance of each varies widely with
context.   My own view is that it is absolutely necessary, when studying
any measure space, to know its classification under the eleven
discriminating features listed here, and to be able to describe any
atoms which are present.   Fortunately, for most `ordinary' measure
spaces, the classification is fairly quick, because if (for instance)
the space is $\sigma$-finite, and you know the measure of the whole
space, the only remaining questions concern completeness and atoms.
The distinctions between spaces which are, or are not,
strictly localizable, semi-finite, localizable and locally determined
are relevant only for spaces which are not $\sigma$-finite, and do not
arise in elementary applications.

I think it is also fair to say that the notions of `complete' and
`locally determined' measure space are technical;  I mean, that they do
not correspond to significant features of the essential structure of a
space, though there are some interesting problems concerning incomplete
measures.   One manifestation of this is the existence of canonical
constructions for rendering spaces complete or complete and locally
determined (212C, 213D-213E).   In addition,  measure spaces which
are not semi-finite do not really belong to measure theory, but rather
to the more general study of $\sigma$-algebras and $\sigma$-ideals.
The most important classifications, in terms of the behaviour of a
measure space, seem to me to be `$\sigma$-finite', `localizable'
and `strictly localizable';  these are the critical features which
cannot be forced by elementary constructions.

If you know anything about Borel subsets of the real line, the argument of part (a) of the proof of 211P must look very clumsy.   But `better' proofs
rely on ideas which we shall not need until Volume 4, and the proof here is based on a construction which we have to understand for other reasons.
}%end of notes

\discrpage


\frfilename{mt212.tex}
\versiondate{10.9.04}

\def\chaptername{Taxonomy of measure spaces}
\def\sectionname{Complete spaces}

\newsection{212}

In the next two sections of this chapter I give brief accounts of the
theory of measure spaces possessing certain of the properties described
in \S211.   I begin with `completeness'.   I give the elementary
facts about complete measure spaces in 212A-212B;  then I turn to the notion of
`completion' of a measure (212C) and its relationships with the other
concepts of measure theory introduced so far (212D-212G).

\leader{212A}{Proposition} Any measure space constructed
by \Caratheodory's method is complete.

\proof{ Recall that `\Caratheodory's method' starts from an arbitrary
outer measure $\theta:\Cal PX\to[0,\infty]$ and sets

\Centerline{$\Sigma=\{E:E\subseteq X,\,\theta A
=\theta(A\cap E)+\theta(A\setminus E)$ for every $A\subseteq X\}$,
\quad$\mu=\theta\restr\Sigma$}

\noindent (113C).    In this case, if $B\subseteq E\in\Sigma$ and
$\mu E=0$, then $\theta B=\theta E=0$ (113A(ii)), so for any
$A\subseteq X$ we have

\Centerline{$\theta(A\cap B)+\theta(A\setminus B)
=\theta(A\setminus B)\le\theta A
\le\theta(A\cap B)+\theta(A\setminus B)$,}

\noindent and $B\in\Sigma$.
}%end of proof of 212A

\leader{212B}{Proposition} (a) If $(X,\Sigma,\mu)$ is a complete measure
space, then any conegligible subset of $X$ is measurable.

(b) Let $(X,\Sigma,\mu)$ be a complete measure space, and $f$ a
$[-\infty,\infty]$-valued function defined on a subset of $X$.   If $f$ is
virtually measurable\cmmnt{ (that is, there is a
conegligible set $E\subseteq X$ such that $f\restr E$ is measurable)},
then $f$ is measurable.

(c) Let $(X,\Sigma,\mu)$ be a complete measure space, and $f$ a
real-valued function defined on a conegligible subset of $X$.   Then the
following are equiveridical, that is, if one is true so are the others:

\quad (i) $f$ is integrable;

\quad(ii) $f$ is measurable and $|f|$ is integrable;

\quad(iii) $f$ is measurable and there is an integrable function $g$
such that $|f|\leae g$.

\proof{{\bf (a)} If $E$ is conegligible, then $X\setminus E$ is
negligible, therefore measurable, and $E$ is measurable.

\medskip

{\bf (b)} Let $a\in\Bbb R$.   Then there is an $H\in\Sigma$ such that


\Centerline{$\{x:(f\restr E)(x)\le a\}=H\cap\dom(f\restr E)
=H\cap E\cap\dom f$.}

\noindent Now $F=\{x:x\in\dom f\setminus E,\,f(x)\le a\}$ is a subset of
the negligible set $X\setminus E$, so is measurable, and

\Centerline{$\{x:f(x)\le a\}=(F\cup H)\cap\dom f\in\Sigma_{\dom f}$,}

\noindent writing $\Sigma_D=\{D\cap E:E\in\Sigma\}$, as in 121A.   As
$a$ is arbitrary, $f$ is measurable (135E).

\medskip

{\bf (c)(i)$\Rightarrow$(ii)} If $f$ is integrable, then by 122P $f$ is
virtually measurable and by 122Re $|f|$ is integrable.   By (b) here,
$f$ is measurable, so (ii) is true.

\medskip

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

\medskip

\quad{\bf (iii)$\Rightarrow$(i)} If $f$ is measurable and $g$ is integrable
and $|f|\leae g$, then $f$ is virtually measurable, $|g|$ is integrable and
$|f|\leae|g|$, so 122P tells us that $f$ is integrable.
}%end of proof of 212B

\leader{212C}{The completion of a measure} Let $(X,\Sigma,\mu)$ be any
measure space.

\header{212Ca}{\bf (a)} Let $\hat\Sigma$ be the family of those sets
$E\subseteq X$ such that there are $E'$, $E''\in\Sigma$ with
$E'\subseteq E\subseteq E''$ and $\mu(E''\setminus E')=0$.   Then
$\hat\Sigma$ is a $\sigma$-algebra of subsets of $X$.   \prooflet{\Prf\
{(i)} Of course $\emptyset$ belongs to
$\hat\Sigma$, because we can take $E'=E''=\emptyset$.  {(ii)} If
$E\in\hat\Sigma$, take $E'$, $E''\in\Sigma$ such that
$E'\subseteq E\subseteq E''$ and $\mu(E''\setminus E')=0$.
Then

\Centerline{$X\setminus E''\subseteq X\setminus E
\subseteq X\setminus E'$,
\quad $\mu((X\setminus E')\setminus(X\setminus E''))
=\mu(E''\setminus E')=0$,}

\noindent so $X\setminus E\in\hat\Sigma$. {(iii)} If
$\sequencen{E_n}$ is a sequence in
$\hat\Sigma$, then for each $n$ choose $E'_n$, $E_n''\in\Sigma$ such
that $E'_n\subseteq E_n\subseteq E_n''$ and
$\mu(E_n''\setminus E'_n)=0$.   Set

\Centerline{$E=\bigcup_{n\in\Bbb N}E_n$,
\quad$E'=\bigcup_{n\in\Bbb N}E'_n$,
\quad$E''=\bigcup_{n\in\Bbb N}E_n''$;}

\noindent  then $E'\subseteq E\subseteq E''$ and
$E''\setminus E'\subseteq\bigcup_{n\in\Bbb N}(E_n''\setminus E'_n)$ is
negligible, so $E\in\hat\Sigma$.  \Qed}

\spheader 212Cb For $E\in\hat\Sigma$, set

\Centerline{$\hat\mu E=\mu^*E
=\min\{\mu F:E\subseteq F\in\Sigma\}$\dvro{.}{}}

\cmmnt{
\noindent (132A).   It is worth remarking at once that if $E\in\hat\Sigma$,
$E'$, $E''\in\Sigma$, $E'\subseteq E\subseteq E''$ and
$\mu(E''\setminus  E')=0$, then $\mu E'=\hat\mu E=\mu E''$;  this is
because

\Centerline{$\mu E'=\mu^*E'\le\mu^*E\le\mu^*E''=\mu E''
=\mu E'+\mu(E''\setminus E)=\mu E'$}

\noindent (recalling from 132A, or noting now, that $\mu^*A\le\mu^*B$
whenever $A\subseteq B\subseteq X$, and that $\mu^*$ agrees with $\mu$
on $\Sigma$).
}%end of comment

\header{212Cc}{\bf (c)}\cmmnt{ We now find that}
$(X,\hat\Sigma,\hat\mu)$ is a
measure space.   \prooflet{\Prf\ (i) Of course $\hat\mu$, like $\mu$,
takes values in $[0,\infty]$.   (ii)  $\hat\mu\emptyset=\mu\emptyset=0$.
(iii)  Let $\sequencen{E_n}$ be a disjoint sequence in
$\hat\Sigma$, with union $E$.  For each $n\in\Bbb N$ choose $E_n'$,
$E_n''\in\Sigma$ such that $E_n'\subseteq E_n\subseteq E_n''$ and
$\mu(E_n''\setminus E_n')=0$.    Set $E'=\bigcup_{n\in\Bbb N}E_n'$,
$E''=\bigcup_{n\in\Bbb N}E_n''$.   Then (as in (a-iii) above)
$E'\subseteq E\subseteq E''$ and $\mu(E''\setminus E')=0$, so

\Centerline{$\hat\mu E=\mu E'=\sum_{n=0}^{\infty}\mu E_n'
=\sum_{n=0}^{\infty}\hat\mu E_n$}

\noindent because $\sequencen{E_n'}$, like $\sequencen{E_n}$, is
disjoint. \Qed}

\header{212Cd}{\bf (d)}\cmmnt{ The measure space}
$(X,\hat\Sigma,\hat\mu)$
is called the {\bf completion} of the measure space $(X,\Sigma,\mu)$;
\cmmnt{equally,} I will call $\hat\mu$ the {\bf completion} of $\mu$,
and occasionally\cmmnt{ (if it seems plain which null ideal is
under consideration)} I will call $\hat\Sigma$ the completion of
$\Sigma$.   Members of $\hat\Sigma$ are sometimes called
{\bf $\mu$-measurable}.

\leader{212D}{}\cmmnt{There is something I had better check at once.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be any measure space.
Then $(X,\hat\Sigma,\hat\mu)$, as defined in 212C, is a complete measure
space and $\hat\mu$ is an extension of $\mu$;  and
$(X,\hat\Sigma,\hat\mu)=(X,\Sigma,\mu)$ iff $(X,\Sigma,\mu)$ is
complete.

\proof{{\bf (a)} Suppose that $A\subseteq E\in\hat\Sigma$ and
$\hat\mu E=0$.   Then (by 212Cb) there is an $E''\in\Sigma$ such that
$E\subseteq E''$ and $\mu E''=0$.   Accordingly we have

\Centerline{$\emptyset\subseteq A\subseteq E''$,\quad
$\mu(E''\setminus\emptyset)=0$,}

\noindent so $A\in\hat\Sigma$.   As $A$ is arbitrary, $\hat\mu$ is
complete.

\medskip

{\bf (b)} If $E\in\Sigma$, then of course $E\in\hat\Sigma$, because
$E\subseteq E\subseteq E$ and $\mu(E\setminus E)=0$;  and $\hat\mu
E=\mu^*E=\mu E$.   Thus $\Sigma\subseteq\hat\Sigma$ and $\hat\mu$
extends $\mu$.

\medskip

{\bf (c)} If $\mu=\hat\mu$ then of course
$\mu$ must be complete.   If $\mu$ is complete,
and $E\in\hat\Sigma$, then there are $E'$, $E''\in\Sigma$ such that
$E'\subseteq E\subseteq E''$ and $\mu(E''\setminus E')=0$.   But now
$E\setminus E'\subseteq E''\setminus E'$, so (because $(X,\Sigma,\mu)$
is complete) $E\setminus E'\in\Sigma$ and $E=E'\cup(E\setminus
E')\in\Sigma$.   As $E$ is arbitrary, $\hat\Sigma\subseteq\Sigma$ and
$\hat\Sigma=\Sigma$ and $\mu=\hat\mu$.
}%end of proof of 212D

\vleader{72pt}{212E}{}\cmmnt{The importance of this construction is such that
it is worth spelling out some further elementary properties.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$(X,\hat\Sigma,\hat\mu)$ its completion.

(a) The outer measures $\hat\mu^*$, $\mu^*$ defined from $\hat\mu$ and
$\mu$ coincide.

(b) $\mu$, $\hat\mu$ give rise to the same negligible and conegligible
sets and the same sets of full outer measure.

(c) $\hat\mu$ is the only measure with domain $\hat\Sigma$ which agrees
with $\mu$ on $\Sigma$.

(d) A subset of $X$ belongs to $\hat\Sigma$ iff it is expressible as
$F\symmdiff A$ where $F\in\Sigma$ and $A$ is $\mu$-negligible.

\proof{{\bf (a)} Take any $A\subseteq X$.   (i) If
$A\subseteq F\in\Sigma$, then $F\in\hat\Sigma$ and $\mu F=\hat\mu F$, so

\Centerline{$\hat\mu^*A\le\hat\mu F=\mu F$;}

\noindent as $F$ is arbitrary, $\hat\mu^*A\le\mu^*A$.   (ii) If
$A\subseteq E\in\hat\Sigma$, there is an $E''\in\Sigma$ such that
$E\subseteq E''$ and $\mu E''=\hat\mu E$, so

\Centerline{$\mu^*A\le\mu E''=\hat\mu E$;}

\noindent as $E$ is arbitrary, $\mu^*A\le\hat\mu^*A$.

\medskip
{\bf (b)} Now, for $A\subseteq X$,

\Centerline{$A$ is $\mu$-negligible$\,\iff\mu^*A=0
\iff\hat\mu^*A=0\iff A$ is $\hat\mu$-negligible,}

$$\eqalign{A\text{ is }\mu&\text{-conegligible}
\,\iff\,\mu^*(X\setminus A)=0\cr
&\iff\,\hat\mu^*(X\setminus A)=0
\iff A\text{ is }\hat\mu\text{-conegligible}.\cr}$$

\noindent If $A$ has full outer measure for $\mu$, $F\in\hat\Sigma$ and $F\cap A=\emptyset$, then there is an $F'\in\Sigma$ such that $F'\subseteq F$ and $\mu F'=\hat\mu F$;  as $F'\cap A=\emptyset$,
$F'$ is $\mu$-negligible and $F$ is $\hat\mu$-negligible;  as $F$ is arbitrary, $A$ has full outer measure for $\hat\mu$.   In the other direction, of course, if $A$ has full outer measure for
$\hat\mu$ then

\Centerline{$\mu^*(F\cap A)=\hat\mu^*(F\cap A)=\hat\mu F=\mu F$}

\noindent for every $F\in\Sigma$, so $A$ has full outer measure for $\mu$.

\medskip

{\bf (c)} If $\tilde\mu$ is any measure with domain $\hat\Sigma$
extending $\mu$, we must have

\Centerline{$\mu E'\le\tilde\mu E\le\mu E''$,
\quad $\mu E'=\hat\mu E=\mu E''$,}

\noindent so $\tilde\mu E=\hat\mu E$, whenever $E'$, $E''\in\Sigma$,
$E'\subseteq E\subseteq E''$ and $\mu(E''\setminus E')=0$.

\medskip

{\bf (d)(i)} If $E\in\hat\Sigma$, take $E'$, $E''\in\Sigma$ such that
$E'\subseteq E\subseteq E''$ and $\mu(E''\setminus E')=0$.   Then
$E\setminus E'\subseteq E''\setminus E'$, so $E\setminus E'$ is
$\mu$-negligible, and $E=E'\symmdiff(E\setminus E')$ is the symmetric
difference of a member of $\Sigma$ and a negligible set.

\medskip

\quad{\bf (ii)} If $E=F\symmdiff A$, where $F\in\Sigma$ and
$A$ is $\mu$-negligible, take $G\in\Sigma$ such that $\mu G=0$ and
$A\subseteq G$;  then $F\setminus G\subseteq E\subseteq F\cup G$ and
$\mu((F\cup G)\setminus(F\setminus G))=\mu G=0$, so $E\in\hat\Sigma$.
}%end of proof of 212E

\vleader{60pt}{212F}{}\cmmnt{Now let us consider integration with respect to
the completion of a measure.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space and
$(X,\hat\Sigma,\hat\mu)$ its completion.

(a) A $[-\infty,\infty]$-valued function $f$ defined on a subset of $X$
is $\hat\Sigma$-measurable iff it is $\mu$-virtually measurable.

(b) Let $f$ be a $[-\infty,\infty]$-valued function defined on a subset
of $X$.   Then
$\int fd\mu=\int fd\hat\mu$ if either is defined in $[-\infty,\infty]$;
in particular, $f$ is $\mu$-integrable iff it is $\hat\mu$-integrable.

\proof{{\bf (a)(i)}  Suppose that $f$ is a $[-\infty,\infty]$-valued
$\hat\Sigma$-measurable function.    For $q\in\Bbb Q$ let
$E_q\in\hat\Sigma$ be such that $\{x:f(x)\le q\}=\dom f\cap E_q$, and
choose $E_q'$, $E_q''\in\Sigma$ such that
$E_q'\subseteq E_q\subseteq E_q''$ and $\mu(E_q''\setminus E_q')=0$.
Set $H=X\setminus\bigcup_{q\in\Bbb Q}(E_q''\setminus E_q')$;  then $H$
is $\mu$-conegligible.   For $a\in\Bbb R$ set

\Centerline{$G_a=\bigcup_{q\in\Bbb Q,q<a}E_q'\in\Sigma$;}

\noindent  then

\Centerline{$\{x:x\in\dom(f\restr H),\,(f\restr H)(x)<a\}
=G_a\cap\dom(f\restr H)$.}

\noindent This shows that $f\restr H$ is
$\Sigma$-measurable, so that $f$ is $\mu$-virtually measurable.

\medskip

\quad{\bf (ii)} If $f$ is $\mu$-virtually measurable, then there is a
$\mu$-conegligible set $H\subseteq X$ such that $f\restr H$ is
$\Sigma$-measurable.   Since $\Sigma\subseteq\hat\Sigma$, $f\restr H$ is
also $\hat\Sigma$-measurable.   And $H$ is $\hat\mu$-conegligible, by
212Eb.   But this means that $f$ is $\hat\mu$-virtually measurable,
therefore $\hat\Sigma$-measurable, by 212Bb.

\medskip

{\bf (b)(i)}  Let $f:D\to[-\infty,\infty]$ be a function, where
$D\subseteq X$.
If either of $\int fd\mu$, $\int fd\hat\mu$ is defined in
$[-\infty,\infty]$, then $f$ is virtually measurable, and defined almost
everywhere, for one of the appropriate measures, and therefore for both
(putting (a) above together with 212Bb).

\medskip

\quad{\bf (ii)} Now suppose that $f$ is non-negative and
integrable either with respect to $\mu$ or with respect to $\hat\mu$.
Let $E\in\Sigma$ be a conegligible set included in $\dom f$ such
that $f\restr E$ is $\Sigma$-measurable.    For $n\in\Bbb N$, $k\ge 1$ set

\Centerline{$E_{nk}=\{x:x\in E$, $f(x)\ge 2^{-n}k\}$;}

\noindent then each $E_{nk}$ belongs to $\Sigma$ and is of finite
measure for both $\mu$ and $\hat\mu$.   (If $f$ is $\mu$-integrable,

\Centerline{$\hat\mu E_{nk}=\mu E_{nk}\le 2^n\int fd\mu$;}

\noindent if $f$ is $\hat\mu$-integrable,

\Centerline{$\mu E_{nk}=\hat\mu E_{nk}\le 2^n\int fd\hat\mu$.)}

\noindent So

\Centerline{$f_n=\sum_{k=1}^{4^n}2^{-n}\chi E_{nk}$}

\noindent is both $\mu$-simple and $\hat\mu$-simple, and
$\int f_nd\mu=\int f_nd\hat\mu$.   Observe that, for $x\in E$,

$$\eqalign{f_n(x)
&=2^{-n}k\text{ if }k<4^n\text{ and }2^{-n}k\le f(x)<2^{-n}(k+1),\cr
&=2^n\text{ if }f(x)\ge 2^n.\cr}$$

\noindent Thus $\sequencen{f_n}$ is a non-decreasing
sequence of functions converging to $f$ at every point of $E$, that is,
both $\mu$-almost everywhere and $\hat\mu$-almost everywhere.   So we
have, for any $c\in\Bbb R$,

$$\eqalign{\int fd\mu=c
&\iff\lim_{n\to\infty}\int f_nd\mu=c\cr
&\iff\lim_{n\to\infty}\int f_nd\hat\mu=c
\iff\int fd\hat\mu=c.\cr}$$

\medskip

\quad{\bf (iii)} As for infinite integrals, recall that for a non-negative
function I write `$\int f=\infty$' just when $f$ is defined almost
everywhere, is virtually measurable, and is not integrable.   So (i) and
(ii) together show that $\int fd\mu=\int fd\hat\mu$ whenever $f$ is
non-negative and either integral is defined in $[0,\infty]$.

\medskip

\quad{\bf (iv)} Since both $\mu$, $\hat\mu$ agree that $\int f$ is to be
interpreted as $\int f^+-\int f^-$ just when this can be defined in
$[-\infty,\infty]$, writing $f^+(x)=\max(f(x),0)$,
$f^-(x)=\max(-f(x),0)$
for $x\in\dom f$, the result for general real-valued $f$ follows at
once.
}%end of proof of 212F

\leader{212G}{}\cmmnt{I turn now to the question of the effect of the
construction on the properties listed in
211B-211K. %211B 211C 211D 211E 211F 211G 211H 211I 211J 211K
\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$(X,\hat\Sigma,\hat\mu)$ its completion.

(a) $(X,\hat\Sigma,\hat\mu)$ is a probability space, or totally finite,
or $\sigma$-finite, or semi-finite, or localizable,  iff
$(X,\Sigma,\mu)$ is.

(b) $(X,\hat\Sigma,\hat\mu)$ is strictly localizable if $(X,\Sigma,\mu)$
is, and any decomposition of $X$ for $\mu$ is a decomposition for
$\hat\mu$.

(c) A set $H\in\hat\Sigma$ is an atom for $\hat\mu$ iff there is an
$E\in\Sigma$ such that $E$ is an atom for $\mu$ and $\hat\mu(H\symmdiff
E)=0$.

(d) $(X,\hat\Sigma,\hat\mu)$ is atomless or purely atomic iff
$(X,\Sigma,\mu)$ is.

\proof{{\bf (a)(i)} Because $\hat\mu X=\mu X$,
$(X,\hat\Sigma,\hat\mu)$ is a
probability space, or totally finite, iff $(X,\Sigma,\mu)$ is.

\medskip

\quad{\bf (ii)}\grheada\ If $(X,\Sigma,\mu)$ is $\sigma$-finite,
there is a sequence $\sequencen{E_n}$, covering $X$, with $\mu
E_n<\infty$
for each $n$.   Now $\hat\mu E_n<\infty$ for each $n$, so
$(X,\hat\Sigma,\hat\mu)$ is $\sigma$-finite.

\medskip

\qquad\grheadb\ If $(X,\hat\Sigma,\hat\mu)$ is $\sigma$-finite,
there is a sequence $\sequencen{E_n}$, covering $X$, with
$\hat\mu E_n<\infty$ for each $n$.   Now we can find, for each $n$, an
$E_n''\in\Sigma$ such that $\mu E_n''<\infty$ and $E_n\subseteq E_n''$;
so that $\sequencen{E_n''}$ witnesses that $(X,\Sigma,\mu)$ is
$\sigma$-finite.

\medskip

\quad{\bf (iii)}\grheada\ If $(X,\Sigma,\mu)$ is semi-finite and
$\hat\mu E=\infty$, then there is an $E'\in\Sigma$ such
that $E'\subseteq E$ and $\mu E'=\infty$.  Next, there is an
$F\in\Sigma$
such that $F\subseteq E'$ and $0<\mu F<\infty$.   Of course we now have
$F\in\hat\Sigma$, $F\subseteq E$ and $0<\hat\mu F<\infty$.   As $E$ is
arbitrary, $(X,\hat\Sigma,\mu)$ is semi-finite.

\medskip

\qquad\grheadb\ If $(X,\hat\Sigma,\hat\mu)$ is semi-finite
and $\mu E=\infty$, then $\hat\mu E=\infty$, so there is an $F\subseteq
E$
such that $0<\hat\mu F<\infty$.   Next, there is an $F'\in\Sigma$ such
that $F'\subseteq F$ and $\mu F'=\hat\mu F$.   Of course we now have
$F'\subseteq E$ and $0<\mu F'<\infty$.   As $E$ is arbitrary,
$(X,\Sigma,\mu)$ is semi-finite.

\medskip

\quad{\bf (iv)}\grheada\ If $(X,\Sigma,\mu)$ is localizable and
$\Cal E\subseteq \hat\Sigma$, then set

\Centerline{$\Cal F=\{F:F\in\Sigma,\,\exists\enskip
E\in\Cal E,\, F\subseteq E\}$.}

\noindent Let $H$ be an essential supremum of
$\Cal F$ in $\Sigma$, as in 211G.

If $E\in\Cal E$, there is an $E'\in\Sigma$ such that $E'\subseteq E$ and
$E\setminus E'$ is negligible;  now $E'\in\Cal F$, so

\Centerline{$\hat\mu(E\setminus H)
\le\hat\mu(E\setminus E')+\mu(E'\setminus H)=0$.}

\noindent If $G\in\hat\Sigma$ and
$\hat\mu(E\setminus G)=0$ for every $E\in\Cal E$, let $G''\in\Sigma$ be
such that $G\subseteq G''$ and $\hat\mu(G''\setminus G)=0$;  then, for
any $F\in\Cal F$, there is an
$E\in\Cal E$ including $F$, so that

\Centerline{$\mu(F\setminus G'')\le\hat\mu(E\setminus G)=0.$}

\noindent As $F$ is arbitrary, $\mu(H\setminus G'')=0$ and
$\hat\mu(H\setminus G)=0$.   This shows that $H$ is an essential
supremum of $\Cal E$ in $\hat\Sigma$.   As $\Cal E$ is arbitrary,
$(X,\hat\Sigma,\hat\mu)$ is localizable.

\medskip

\qquad\grheadb\ Suppose that $(X,\hat\Sigma,\hat\mu)$ is
localizable and that $\Cal E\subseteq\Sigma$.   Working in
$(X,\hat\Sigma,\hat\mu)$, let $H$ be an essential supremum for $\Cal E$
in $\hat\Sigma$.   Let $H'\in\Sigma$ be such that $H'\subseteq H$ and
$\hat\mu(H\setminus H')=0$.   Then

\Centerline{$\mu(E\setminus H')
\le\hat\mu(E\setminus H)+\hat\mu(H\setminus H')=0$}

\noindent for every $E\in\Cal E$;  while if $G\in\Sigma$ and
$\mu(E\setminus G)=0$ for every $E\in\Cal E$, we must have

\Centerline{$\mu(H'\setminus G)\le\hat\mu(H\setminus G)=0$.}

\noindent   Thus $H'$ is an essential supremum of $\Cal E$ in $\Sigma$.
As $\Cal E$ is arbitrary, $(X,\Sigma,\mu)$ is localizable.

\medskip

{\bf (b)} Let $\langle X_i\rangle_{i\in I}$ be a decomposition of $X$
for $\mu$, as in 211E.    Of course it is a partition of $X$ into
sets of finite $\hat\mu$-measure.   If $H\subseteq X$ and
$H\cap X_i\in\hat\Sigma$ for every $i$, choose for each $i\in I$ sets $E_i'$, $E_i''\in\Sigma$ such that

\Centerline{$E_i'\subseteq H\cap X_i\subseteq E_i''$,\quad
$\mu(E_i''\setminus E_i')=0$.}

\noindent   Set $E'=\bigcup_{i\in I}E_i'$,
$E''=\bigcup_{i\in I}(E_i''\cap X_i)$.   Then $E'\cap X_i=E_i'$,
$E''\cap X_i=E_i''\cap X_i$ for
each $i$, so $E'$ and $E''$ belong to $\Sigma$.   Also

\Centerline{$\mu(E''\setminus E')
=\sum_{i\in I}\mu(E_i''\cap X_i\setminus E_i')=0$.}

\noindent As $E'\subseteq H\subseteq E''$, $H\in\hat\Sigma$ and

\Centerline{$\hat\mu H=\mu E'
=\sum_{i\in I}\mu E'_i=\sum_{i\in I}\hat\mu(H\cap X_i)$.}

\noindent As $H$ is arbitrary, $\langle X_i\rangle_{i\in I}$
is a decomposition of $X$ for $\hat\mu$.

Accordingly, $(X,\hat\Sigma,\hat\mu)$ is strictly localizable if such a
decomposition exists, which is so if $(X,\Sigma,\mu)$ is strictly
localizable.

\medskip

{\bf (c)-(d)(i)} Suppose that $E\in\hat\Sigma$ is an atom for $\hat\mu$.
Let $E'\in\Sigma$ be such that $E'\subseteq E$ and
$\hat\mu(E\setminus E')=0$.   Then $\mu E'=\hat\mu E>0$.   If $F\in\Sigma$ and $F\subseteq E'$, then $F\subseteq E$, so either
$\mu F=\hat\mu F=0$ or $\mu(E'\setminus F)=\hat\mu(E\setminus F)=0$.   As
$F$ is arbitrary, $E'$ is an atom for $\mu$, and
$\hat\mu(E\symmdiff E')=\hat\mu(E\setminus E')=0$.

\medskip

\quad{\bf (ii)} Suppose that $E\in\Sigma$ is an atom for $\mu$, and that
$H\in\hat\Sigma$ is such that $\hat\mu(H\symmdiff E)=0$.   Then
$\hat\mu H=\mu E>0$.   If $F\in\hat\Sigma$ and $F\subseteq H$, let
$F'\subseteq F$ be such that $F'\in\Sigma$ and
$\hat\mu(F\setminus F')=0$.   Then
$E\cap F'\subseteq E$ and $\hat\mu(F\symmdiff(E\cap F'))=0$, so either
$\hat\mu F=\mu(E\cap F')=0$ or
$\hat\mu(H\setminus F)=\mu(E\setminus F')=0$.   As $F$ is arbitrary, $H$
is an atom for $\hat\mu$.

\medskip

\quad{\bf (iii)} It follows at once that $(X,\hat\Sigma,\hat\mu)$ is
atomless iff $(X,\Sigma,\mu)$ is.

\medskip

\quad{\bf (iv)}\grheada\ On the other hand, if $(X,\Sigma,\mu)$
is purely atomic and $\hat\mu H>0$, there is an $E\in\Sigma$ such that
$E\subseteq H$ and $\mu E>0$, and an atom $F$ for $\mu$ such that
$F\subseteq E$;  but $F$ is also an atom for $\hat\mu$.   As $H$ is
arbitrary, $(X,\hat\Sigma,\hat\mu)$ is purely atomic.

\medskip

\qquad\grheadb\ And if $(X,\hat\Sigma,\hat\mu)$ is purely
atomic and $\mu E>0$, then there is an $H\subseteq E$ which is an atom
for
$\hat\mu$;  now let $F\in\Sigma$ be such that $F\subseteq H$ and
$\hat\mu(H\setminus F)=0$, so that $F$ is an atom for $\mu$ and
$F\subseteq E$.   As $E$ is arbitary, $(X,\Sigma,\mu)$ is purely atomic.
}%end of proof of 212G

\exercises{
\leader{212X}{Basic exercises $\pmb{>}$(a)} 
%\spheader 212Xa
Let $(X,\Sigma,\mu)$ be a
complete measure space.   Suppose that $A\subseteq E\in\Sigma$ and that
$\mu^*A+\mu^*(E\setminus A)=\mu E<\infty$.   Show that $A\in\Sigma$.

\sqheader 212Xb Let $\mu$ and $\nu$ be two measures on a set $X$, with
completions $\hat\mu$ and $\hat\nu$.   Show that the following are
equiveridical:  (i) the outer measures $\mu^*$, $\nu^*$ defined from $\mu$
and $\nu$ coincide;  (ii) $\hat\mu E=\hat\nu E$ whenever either is
defined and finite;   (iii) $\int fd\mu=\int fd\nu$ whenever $f$ is a
real-valued function such that either integral is defined and finite.
\Hint{for (i)$\Rightarrow$(ii), if $\hat\mu E<\infty$, take a measurable
envelope $F$ of $E$ for $\nu$ and calculate
$\nu^*E+\nu^*(F\setminus E)$.}

\spheader 212Xc Let $\mu$ be the restriction of Lebesgue measure
to the Borel $\sigma$-algebra of $\Bbb R$, as in 211P.   Show that its
completion is Lebesgue measure itself.   \Hint{134F.}

\spheader 212Xd Repeat 212Xc for (i) Lebesgue measure on
$\BbbR^r$ (ii) Lebesgue-Stieltjes measures on $\Bbb R$ (114Xa).

\spheader 212Xe
Let $X$ be a set and $\Sigma$ a $\sigma$-algebra
of subsets of $X$.   Let $\Cal I$ be a $\sigma$-ideal of subsets of $X$
(112Db).
(i) Show that $\Sigma_1=\{E\symmdiff A:E\in\Sigma,\,A\in\Cal I\}$ is a
$\sigma$-algebra of subsets of $X$.
(ii) Let $\Sigma_2$ be the family of sets $E\subseteq X$ such
that there are $E'$, $E''\in\Sigma$ with $E'\subseteq E\subseteq E''$
and $E''\setminus E'\in\Cal I$.   Show that $\Sigma_2$ is a
$\sigma$-algebra of subsets of $X$ and that $\Sigma_2\subseteq\Sigma_1$.
(iii) Show that $\Sigma_2=\Sigma_1$ iff every member of
$\Cal I$ is included in a member of $\Sigma\cap\Cal I$.

\spheader 212Xf
Let $(X,\Sigma,\mu)$ be a measure space, $Y$ any set and
$\phi:X\to Y$ a function.   Set $\theta B=\mu^*\phi^{-1}[B]$ for every
$B\subseteq Y$.   (i) Show that $\theta$ is an outer measure on $Y$.
(ii) Let $\nu$ be the measure defined from $\theta$ by \Caratheodory's
method, and $\Tau$ its domain.   Show that if $C\subseteq Y$ and
$\phi^{-1}[C]\in\Sigma$ then $C\in\Tau$.   (iii) Suppose that
$(X,\Sigma,\mu)$ is complete
and totally finite.   Show that $\nu$ is the image measure
$\mu\phi^{-1}$.

\spheader 212Xg
Let $g$, $h$ be two non-decreasing functions from
$\Bbb R$ to itself, and $\mu_g$, $\mu_h$ the associated
Lebesgue-Stieltjes
measures.   Show that a real-valued function $f$ defined on a subset of
$\Bbb R$ is $\mu_{g+h}$-integrable iff it is both
$\mu_g$-integrable and $\mu_h$-integrable, and that then
$\int fd\mu_{g+h}=\int fd\mu_g+\int fd\mu_h$.   \Hint{114Yb}.

\spheader 212Xh Let $(X,\Sigma,\mu)$ be a measure 
space, and $\Cal I$ a
$\sigma$-ideal of subsets of $X$;  set
$\Sigma_1=\{E\symmdiff A:E\in\Sigma,\,A\in\Cal I\}$, as in 212Xe.
Show that if every member of $\Sigma\cap\Cal I$
is $\mu$-negligible, then there is a unique extension of $\mu$ to a
measure $\mu_1$ with domain $\Sigma_1$ such that $\mu_1A=0$ for every
$A\in\Cal I$.
%212C

\spheader 212Xi\dvAnew{2009}
Let $(X,\Sigma,\mu)$ be a complete measure space such that
$\mu X>0$, $Y$ a set, $f:X\to Y$ a function and $\mu f^{-1}$ the image
measure on $Y$.   Show that if $\Cal F$ is the filter of
$\mu$-conegligible subsets of $X$, then the image filter $f[[\Cal F]]$
(2A1Ib) is the filter of $\mu f^{-1}$-conegligible subsets of $Y$.

\spheader 212Xj\dvAnew{2010}
Let $(X,\Sigma,\mu)$ be a complete measure space and $f:X\to\Bbb R$ a
function such that $\overline{\int}fd\mu<\infty$.   Show that there is a
measurable function $g:X\to\Bbb R$ such that $f(x)\le g(x)$ for every 
$x\in X$ and $\int g\,d\mu=\overline{\int}fd\mu$.

\leader{212Y}{Further exercises (a)}
%\spheader 212Ya
Let $X$ be a set and $\phi$ an inner measure on $X$,
that is, a functional from $\Cal PX$ to $[0,\infty]$ such that

\inset{$\phi\emptyset=0$,}
\inset{$\phi(A\cup B)\ge\phi A+\phi B$ if $A\cap B=\emptyset$,}

\inset{$\phi(\bigcap_{n\in\Bbb N}A_n)=\lim_{n\to\infty}\phi A_n$
whenever $\sequencen{A_n}$ is a non-increasing sequence of subsets of
$X$ and $\phi A_0<\infty$,}

\inset{if $\phi A=\infty$, $a\in\Bbb R$ there is a $B\subseteq A$ such
that $a\le\phi B<\infty$.}

\noindent Let $\mu$ be the measure defined from $\phi$, that is,
$\mu=\phi\restr\Sigma$, where

\Centerline{$\Sigma=\{E:\phi(A)
  =\phi(A\cap E)+\phi(A\setminus E)\Forall A\subseteq X\}$}

\noindent (113Yg).   Show that $\mu$ must be complete.

\spheader 212Yb\dvAnew{2017}
Let $(X,\Sigma,\mu)$ be a strictly localizable measure
space.   Suppose that for every $n\in\Bbb N$ there is a disjoint family
$\ofamily{i}{n}{D_i}$ of subsets of full outer measure.    Show that
there is a disjoint sequence $\sequencen{D_n}$ of sets of full outer measure.
%+ n17910
}%end of exercises

\endnotes{
\Notesheader{212} The process of completion is so natural, and so
universally applicable,
and so convenient, that over large parts of measure theory it is
reasonable to use only complete measure spaces.   Indeed many authors so
phrase their definitions that, explicitly or implicitly, only complete
measure spaces are considered.   In this treatise I avoid taking quite
such a large step, even though it would simplify the statements of many
of the theorems in this volume (for instance).   I did take the trouble, in Volume 1, to give a definition of `integrable function' which, in effect, looks at integrability with respect to the
completion of a measure (212Fb).   There are non-complete
measure spaces which are worthy of study (for example, the restriction
of Lebesgue measure to the Borel $\sigma$-algebra of $\Bbb R$ -- see
211P), and some
interesting questions to be dealt with in Volumes 3 and 5 apply to
them.   At the cost of rather a lot of verbal manoeuvres, therefore, I
prefer to write theorems out in a form in which they can be applied to
arbitrary measure spaces, without assuming completeness.   But it would
be reasonable, and indeed would sharpen your technique, if you regularly
sought the alternative formulations which become natural if you are
interested only in complete spaces.
}

\discrpage


\frfilename{mt213.tex}
\versiondate{13.9.13}

\copyrightdate{2000}
\def\chaptername{Taxonomy of measure spaces}
\def\sectionname{Semi-finite, locally determined and localizable spaces}

\newsection{213}

In this section I collect a variety of useful facts concerning these
types of measure space.   I start with the characteristic properties of
semi-finite spaces (213A-213B), and continue with complete locally
determined spaces (213C) and the concept of
`c.l.d.\ version' (213D-213H),
%213D 213E 213F 213G 213H
the most powerful of the universally
available methods of modifying a measure space into a better-behaved
one.   I briefly discuss `locally determined negligible sets'
(213I-213L),
%213I 213J 213K 213L
and measurable envelopes (213L-213M),
and end with results on localizable spaces (213N) and strictly
localizable spaces (213O).

\leader{213A}{Lemma} Let $(X,\Sigma,\mu)$ be a semi-finite measure
space.   Then

\Centerline{$\mu E
=\sup\{\mu F:F\in\Sigma,\,F\subseteq E,\,\mu F<\infty\}$}

\noindent for every $E\in\Sigma$.

\proof{ Set
$c=\sup\{\mu F:F\in\Sigma,\,F\subseteq E,\,\mu F<\infty\}$.   Then
surely $c\le\mu E$, so if
$c=\infty$ we can stop.   If $c<\infty$, let $\sequencen{F_n}$
be a sequence of measurable subsets of $E$, of finite measure, such that
$\lim_{n\to\infty}\mu F_n=c$;  set $F=\bigcup_{n\in\Bbb N}F_n$.
For each $n\in\Bbb N$, $\bigcup_{k\le n}F_k$ is a measurable set of
finite measure included in $E$, so $\mu(\bigcup_{k\le n}F_k)\le c$,
and

\Centerline{$\mu F=\lim_{n\to\infty}\mu(\bigcup_{k\le n}F_k)\le c$.}

\noindent Also


\Centerline{$\mu F\ge\sup_{n\in\Bbb N}\mu F_n\ge c$,}

\noindent so $\mu F=c$.

If $F'$ is a measurable subset of $E\setminus F$ and $\mu F'<\infty$,
then $F\cup F'$ has finite measure and is included in $E$, so has
measure at most $c=\mu F$;  it follows that $\mu F'=0$.   But this
means that $\mu(E\setminus F)$ cannot be infinite, since then, because
$(X,\Sigma,\mu)$ is semi-finite, it would have to include a measurable
set of non-zero finite measure.   So $E\setminus F$ has finite measure,
and is therefore in fact negligible;  and $\mu E=c$, as claimed.
}%end of proof of 213A

\vleader{72pt}{213B}{Proposition} Let $(X,\Sigma,\mu)$ be a semi-finite measure
space.   Let $f$ be a $\mu$-virtually measurable $[0,\infty]$-valued
function defined almost everywhere in $X$.   Then

$$\eqalign{\int f
&=\sup\{\int g:g\text{ is a simple function},\,g\leae f\}\cr
&=\sup_{F\in\Sigma,\mu F<\infty}\int_Ff}$$

\noindent in $[0,\infty]$.

\proof{{\bf (a)} For any measure space $(X,\Sigma,\mu)$, a
$[0,\infty]$-valued function defined on a subset of $X$ is
integrable iff there is a conegligible set $E$ such that

\inset{($\alpha$) $E\subseteq\dom f$ and $f\restr E$ is measurable,}

\inset{($\beta$)
$\sup\{\int g:g$ is a simple function, $g\leae f\}$ is finite,}

\inset{($\gamma$) for every $\epsilon>0$,
$\{x:x\in E,\,f(x)\ge\epsilon\}$ has finite measure,}

\inset{($\delta$) $f$ is finite almost everywhere}

\noindent (see 122Ja, 133B).   But if $\mu$ is semi-finite, ($\gamma$)
and ($\delta$) are consequences of the rest.   \Prf\ Let $\epsilon>0$.
Set

\Centerline{$E_{\epsilon}=\{x:x\in E,\,f(x)\ge\epsilon\}$,}

\Centerline{$c
=\sup\{\int g:g$ is a simple function, $g\leae f\}$;}

\noindent we are supposing that $c$ is finite.   If
$F\subseteq E_{\epsilon}$ is measurable and $\mu F<\infty$, then
$\epsilon\chi F$ is a
simple function and $\epsilon\chi F\leae f$, so

\Centerline{$\epsilon\mu F=\int\epsilon\chi F\le c$,
\quad$\mu F\le\Bover{c}{\epsilon}$.}

\noindent As $F$ is arbitrary, 213A tells us that
$\mu E_{\epsilon}\le\bover{c}{\epsilon}$ is finite.   As $\epsilon$ is
arbitrary, ($\gamma$) is satisfied.

As for ($\delta$), if $F=\{x:x\in E,\,f(x)=\infty\}$ then $\mu F$ is
finite (by ($\gamma$)) and $n\chi F\leae f$, so $n\mu F\le c$, for
every $n\in\Bbb N$, so $\mu F=0$.\ \Qed

\medskip

{\bf (b)} Now suppose that $f:D\to[0,\infty]$ is a $\mu$-virtually
measurable function, where $D\subseteq X$ is conegligible, so that
$\int f$ is defined in $[0,\infty]$ (135F).   Then (a) tells us that

\leaveitout{$$\eqalignno{\int f
&=\sup_{g\text{ is simple},g\le f\text{ a.e.}}\int g\cr
\displaycause{if either is finite, and therefore also if either is
infinite}
&=\sup_{g\text{ is simple},g\le f\text{ a.e.},\mu F<\infty}\int_Fg
\le\sup_{\mu F<\infty}\int_Ff
\le\int f,\cr}$$}

$$\eqalignno{\int f
&=\sup_{\Atop{g\text{ is simple}}{g\le f\text{ a.e.}}}\int g\cr
\displaycause{if either is finite, and therefore also if either is
infinite}
&=\sup_{\Atop{g\text{ is simple}}{\Atop{g\le f\text{ a.e.}}
{\mu F<\infty}}}\int_Fg
\le\sup_{\mu F<\infty}\int_Ff
\le\int f,\cr}$$

\noindent so we have the equalities we seek.
}%end of proof of 213B

\leader{*213C}{Proposition} Let $(X,\Sigma,\mu)$ be a complete
locally determined measure space, and $\mu^*$ the outer measure derived
from $\mu$\cmmnt{ (132A-132B)}.   Then the measure defined from
$\mu^*$ by \Caratheodory's method is $\mu$ itself.

\proof{ Write $\check\mu$ for the measure defined by \Caratheodory's
method from $\mu^*$, and $\check\Sigma$ for its domain.

\medskip

{\bf (a)} If $E\in\Sigma$ and $A\subseteq X$ then
$\mu^*(A\cap E)+\mu^*(A\setminus E)=\mu^*A$ (132Af), so
$E\in\check\Sigma$.   Now
$\check\mu E=\mu^*E=\mu E$ (132Ac).   Thus $\Sigma\subseteq\check\Sigma$
and $\mu=\check\mu\restr\Sigma$.

\medskip

{\bf (b)} Now suppose that $H\in\check\Sigma$.   Let $E\in\Sigma$ be
such that $\mu E<\infty$.   Then $H\cap E\in\Sigma$.   \Prf\ Let $E_1$,
$E_2\in\Sigma$ be measurable envelopes of $E\cap H$, $E\setminus H$
respectively, both included in $E$ (132Ee).   Because $H\in\check\Sigma$,

\Centerline{$\mu E_1+\mu E_2=\mu^*(E\cap H)+\mu^*(E\setminus H)
=\mu^*E=\mu E$.}

\noindent As $E_1\cup E_2=E$,

\Centerline{$\mu(E_1\cap E_2)=\mu E_1+\mu E_2-\mu E=0$.}

\noindent Now $E_1\setminus(E\cap H)\subseteq E_1\cap E_2$;  because
$\mu$ is complete, $E_1\setminus(E\cap H)$ and $E\cap H$ belong to
$\Sigma$.\ \Qed

As $E$ is arbitrary, and $\mu$ is locally determined, $H\in\Sigma$.   As
$H$ is arbitrary, $\check\Sigma=\Sigma$ and $\check\mu=\mu$.
}%end of proof of 213C

\leader{213D}{C.l.d.\ versions:  Proposition} Let $(X,\Sigma,\mu)$ be a
measure space.    Write $(X,\hat\Sigma,\hat\mu)$ for its
completion\cmmnt{ (212C)} and $\Sigma^f$ for
$\{E:E\in\Sigma,\,\mu E<\infty\}$.   Set

\Centerline{$\tilde\Sigma=\{H:H\subseteq X,\,H\cap E\in\hat\Sigma$ for
every $E\in\Sigma^f\}$,}

\noindent and for $H\in\tilde\Sigma$ set

\Centerline{$\tilde\mu H=\sup\{\hat\mu(H\cap E):E\in\Sigma^f\}$.}

\noindent Then
$(X,\tilde\Sigma,\tilde\mu)$ is a complete locally determined
measure space.

\proof{{\bf (a)} I check first that $\tilde\Sigma$ is a
$\sigma$-algebra.   \Prf\ {(i)} $\emptyset\cap E=\emptyset\in\hat\Sigma$
for every $E\in\Sigma^f$, so $\emptyset\in\tilde\Sigma$.   {(ii)} If
$H\in\tilde\Sigma$ then

\Centerline{$(X\setminus H)\cap E
=E\setminus(E\cap H)\in\hat\Sigma$}

\noindent for every $E\in\Sigma^f$, so $X\setminus H\in\tilde\Sigma$.
{(iii)} If $\sequencen{H_n}$ is a sequence in $\tilde\Sigma$ with union
$H$, then

\Centerline{$H\cap E=\bigcup_{n\in\Bbb N}H_n\cap E\in\hat\Sigma$}

\noindent for every
$E\in\Sigma^f$, so $H\in\tilde\Sigma$.\ \Qed

\medskip

{\bf (b)} It is obvious that $\tilde\mu \emptyset=0$.   If
$\sequencen{H_n}$ is a disjoint sequence in $\tilde\Sigma$ with union
$H$, then

$$\eqalign{\tilde\mu H
&=\sup\{\hat\mu(H\cap E):E\in\Sigma^f\}\cr
&=\sup\{\sum_{n=0}^{\infty}\hat\mu(H_n\cap E):E\in\Sigma^f\}
\le\sum_{n=0}^{\infty}\tilde\mu H_n.\cr}$$

\noindent On the other hand, given $a<\sum_{n=0}^{\infty}\tilde\mu H_n$,
there is an $m\in\Bbb N$ such that $a<\sum_{n=0}^m\tilde\mu H_n$;  now
we can find $E_0,\ldots,E_m\in\Sigma^f$ such that
$a\le\sum_{n=0}^m\hat\mu(H_n\cap E_n)$.   Set $E=\bigcup_{n\le
m}E_n\in\Sigma^f$;  then

\Centerline{$\tilde\mu H\ge\hat\mu(H\cap E)
=\sum_{n=0}^{\infty}\hat\mu(H_n\cap E)
\ge\sum_{n=0}^m\hat\mu(H_n\cap E_n)
\ge a$.}

\noindent As $a$ is arbitrary, $\tilde\mu
H\ge\sum_{n=0}^{\infty}\tilde\mu H_n$ and $\tilde\mu
H=\sum_{n=0}^{\infty}\tilde\mu H_n$.

\medskip

{\bf (c)} Thus $(X,\tilde\Sigma,\tilde\mu)$ is a measure space.   To
see that it is semi-finite, note first that
$\hat\Sigma\subseteq\tilde\Sigma$ (because if $H\in\hat\Sigma$ then
surely $H\cap E\in\hat\Sigma$ for every $E\in\Sigma^f$), and that
$\tilde\mu H=\hat\mu H$ whenever $\hat\mu H<\infty$ (because then, by
the definition in 212Ca, there is an $E\in\Sigma^f$ such that
$H\subseteq E$, so that
$\tilde\mu H=\hat\mu(H\cap E)=\hat\mu H$).   Now suppose that
$H\in\tilde\Sigma$ and that $\tilde\mu H=\infty$.   There is surely an
$E\in\Sigma^f$ such that $\hat\mu(H\cap E)>0$;  but now $0<\hat\mu(H\cap
E)<\infty$, so $0<\tilde\mu(H\cap E)<\infty$.

\medskip

{\bf (d)} Thus $(X,\tilde\Sigma,\tilde\mu)$ is a semi-finite measure space.
To see that it is locally determined, let $H\subseteq X$ be such that
$H\cap G\in\tilde\Sigma$ whenever $G\in\tilde\Sigma$ and
$\tilde\mu G<\infty$.   Then, in particular, we must have
$H\cap E\in\tilde\Sigma$
for every $E\in\Sigma^f$.   But this means in fact that
$H\cap E\in\hat\Sigma$ for every $E\in\Sigma^f$, so that
$H\in\tilde\Sigma$.
As $H$ is arbitrary, $(X,\Sigma,\mu)$ is locally determined.

\medskip

{\bf (e)} To see that $(X,\tilde\Sigma,\tilde\mu)$ is complete,
suppose that $A\subseteq H\in\tilde\Sigma$ and that $\tilde\mu H=0$.
Then for every $E\in\Sigma^f$ we must have $\hat\mu(H\cap E)=0$.
Because $(X,\hat\Sigma,\hat\mu)$ is complete, and
$A\cap E\subseteq H\cap E$, $A\cap E\in\hat\Sigma$.   As $E$ is
arbitrary, $A\in\tilde\Sigma$.
}%end of proof of 213D

\leader{213E}{Definition} For any measure space $(X,\Sigma,\mu)$, I will
call $(X,\tilde\Sigma,\tilde\mu)$, as constructed in 213D, the
{\bf c.l.d.\ version} (`complete locally determined version') of $(X,\Sigma,\mu)$;  and $\tilde\mu$ will be the
{\bf c.l.d.\ version} of $\mu$.

\leader{213F}{}\cmmnt{ Following the same pattern as in
212E-212G, %212E 212F 212G
I start with some elementary remarks to facilitate manipulation of this
construction.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be any measure space
and $(X,\tilde\Sigma,\tilde\mu)$ its c.l.d.\ version.

(a) $\Sigma\subseteq\tilde\Sigma$ and $\tilde\mu E=\mu E$ whenever
$E\in\Sigma$ and $\mu E<\infty$ -- in fact, if $(X,\hat\Sigma,\hat\mu)$
is the completion of $(X,\Sigma,\mu)$, $\hat\Sigma\subseteq\tilde\Sigma$
and $\tilde\mu E=\hat\mu E$ whenever $\hat\mu E<\infty$.

(b) Writing $\tilde\mu^*$ and $\mu^*$ for the outer measures defined
from $\tilde\mu$ and $\mu$ respectively, $\tilde\mu^*A\le\mu^*A$ for
every $A\subseteq X$, with equality if
$\mu^*A$ is finite.   In particular, $\mu$-negligible sets are
$\tilde\mu$-negligible;   consequently, $\mu$-conegligible sets are
$\tilde\mu$-conegligible.

(c) If $H\in\tilde\Sigma$,

\quad (i) $\tilde\mu H=\sup\{\mu F:E\in\Sigma$, $\mu F<\infty$,
$F\subseteq H\}$;

\quad (ii) there is an $E\in\Sigma$ such that
$E\subseteq H$ and $\mu E=\tilde\mu H$, so that
if $\tilde\mu H<\infty$ then $\tilde\mu(H\setminus E)=0$.

\proof{{\bf (a)} This is already covered by remarks in the proof of
213D.

\medskip

{\bf (b)} If $\mu^*A=\infty$ then surely $\tilde\mu^*A\le\mu^*A$.   If
$\mu^*A<\infty$, take $E\in\Sigma$ such that $A\subseteq E$ and
$\mu E=\mu^*A$ (132Aa).   Then

\Centerline{$\tilde\mu^*A\le\tilde\mu E=\mu E=\mu^*A$.}

\noindent On the other hand, if $A\subseteq H\in\tilde\Sigma$, then

\Centerline{$\tilde\mu H\ge\hat\mu(H\cap E)\ge\hat\mu^*A=\mu^*A$,}

\noindent using 212Ea.   So $\mu^*A\le\tilde\mu^*A$ and
$\mu^*A=\tilde\mu^*A$.

\medskip

{\bf (c)} Write $\Sigma^f$ for $\{E:E\in\Sigma,\,\mu E<\infty\}$;  then,
by the definition in 213D,
$\tilde\mu H=\sup\{\hat\mu(H\cap E):E\in\Sigma^f\}$.   Let
$\sequencen{E_n}$ be a sequence in $\Sigma^f$
such that $\tilde\mu H=\sup_{n\in\Bbb N}\hat\mu(H\cap E_n)$.   For each
$n\in\Bbb N$ there is an $F_n\in\Sigma$ such that
$F_n\subseteq H\cap E_n$ and
$\mu F_n=\hat\mu(H\cap E_n)$ (212C).   Set $E=\bigcup_{n\in\Bbb N}F_n$.
Then $E\in\Sigma$, $E\subseteq H$ and

\Centerline{$\tilde\mu H
=\sup_{n\in\Bbb N}\mu F_n
\le\lim_{n\to\infty}\mu(\bigcup_{i\le n}F_i)
=\mu E
=\lim_{n\to\infty}\tilde\mu(\bigcup_{i\le n}F_i)
\le\tilde\mu H,$}

\noindent so $\mu E=\tilde\mu H$, and if $\tilde\mu H<\infty$ then
$\tilde\mu(H\setminus E)=0$.    At the same time,

$$\eqalignno{\tilde\mu H
&=\sup_{n\in\Bbb N}\mu F_n
\le\sup_{F\in\Sigma^f,F\subseteq H}\mu F
=\sup_{F\in\Sigma^f,F\subseteq H}\tilde\mu F\cr
\displaycause{by (a) again}
&\le\tilde\mu H,\cr}$$

\noindent so we have equality here too.
}%end of proof of 213F

\leader{213G}{}\cmmnt{ The next step is to look at functions which are
measurable or integrable with respect to $\tilde\mu$.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$(X,\tilde\Sigma,\tilde\mu)$ its c.l.d.\ version.

(a) If a real-valued function $f$ defined on a subset of $X$ is
$\mu$-virtually measurable, it is $\tilde\Sigma$-measurable.
%should we have $[-\infty,\infty]$-valued here?

(b) If a real-valued function is $\mu$-integrable, it is
$\tilde\mu$-integrable with the same integral.

(c) If $f$ is a $\tilde\mu$-integrable real-valued function, there is a
$\mu$-integrable real-valued function which is equal to
$f\,\,\tilde\mu$-almost everywhere.

\proof{ Write $\Sigma^f$ for $\{E:E\in\Sigma,\,\mu E<\infty\}$.   By
213Fa, $\tilde\mu$ and $\mu$ agree on $\Sigma^f$.

\medskip

{\bf (a)} By 212Fa, $f$ is $\hat\Sigma$-measurable, where
$\hat\Sigma$ is the domain of the completion of $\mu$;  but since
$\hat\Sigma\subseteq\tilde\Sigma$, $f$ is $\tilde\Sigma$-measurable.

\medskip

{\bf (b)(i)} If $f$ is a $\mu$-simple function it is
$\tilde\mu$-simple,
and $\int fd\mu=\int fd\tilde\mu$, because $\tilde\mu E=\mu E$ for every
$E\in\Sigma^f$.

\medskip

\quad{\bf (ii)} If $f$ is a non-negative $\mu$-integrable function,
there is a non-decreasing sequence $\sequencen{f_n}$ of $\mu$-simple
functions converging to $f\,\,\mu$-almost everywhere;  now (by 213Fb)
$\mu$-negligible sets are $\tilde\mu$-negligible, so $\sequencen{f_n}$
converges to $f\,\,\tilde\mu$-a.e.\ and (by B.Levi's theorem, 123A) $f$
is $\tilde\mu$-integrable, with

\Centerline{$\int fd\tilde\mu=\lim_{n\to\infty}\int f_nd\tilde\mu
=\lim_{n\to\infty}\int f_nd\mu=\int fd\mu$.}

\medskip

\quad{\bf (iii)} In general, if $\int fd\mu$ is defined in
$\Bbb R$, we have

\Centerline{$\int fd\tilde\mu
=\int f^+d\tilde\mu-\int f^-d\tilde\mu
=\int f^+d\mu-\int f^-d\mu=\int fd\mu$,}

\noindent writing $f^+$ for $f\vee 0$ and $f^-$ for $(-f)\vee 0$.

\medskip

{\bf (c)(i)} Let $f$ be a $\tilde\mu$-simple function.   Express it as
$\sum_{i=0}^na_i\chi H_i$ where $\tilde\mu H_i<\infty$ for each $i$.
Choose $E_0,\ldots,E_n\in\Sigma$ such that $E_i\subseteq H_i$ and
$\tilde\mu(H_i\setminus E_i)=0$ for each $i$ (using 213Fc above).   Then
$g=\sum_{i=0}^na_i\chi E_i$ is $\mu$-simple,
$g=f\,\,\tilde\mu$-a.e., and $\int g\,d\mu=\int fd\tilde\mu$.

\medskip

\quad{\bf (ii)} Let $f$ be a non-negative $\tilde\mu$-integrable
function.   Let $\sequencen{f_n}$ be a non-decreasing sequence of
$\tilde\mu$-simple functions converging $\tilde\mu$-almost everywhere to
$f$.   For each $n$, choose a $\mu$-simple function $g_n$ equal
$\tilde\mu$-almost everywhere to $f_n$.   Then $\{x:g_{n+1}(x)<g_n(x)\}$
belongs to $\Sigma^f$ and is $\tilde\mu$-negligible, therefore
$\mu$-negligible.   So $\sequencen{g_n}$ is non-decreasing $\mu$-almost
everywhere.   Because

\Centerline{$\lim_{n\to\infty}\int g_nd\mu
=\lim_{n\to\infty}\int f_nd\tilde\mu=\int fd\tilde\mu$,}

\noindent B.Levi's theorem tells us that $\sequencen{g_n}$ converges
$\mu$-almost everywhere to a $\mu$-integrable function $g$;  because
$\mu$-negligible sets are $\tilde\mu$-negligible,

$$\eqalign{(X\setminus\dom f)
&\cup(X\setminus\dom g)\cr
&\cup\bigcup_{n\in\Bbb N}\{x:f_n(x)\ne g_n(x)\}\cr
&\cup\{x:x\in\dom
f,\,f(x)\ne\sup_{n\in\Bbb N}f_n(x)\}\cr
&\cup\{x:x\in\dom
g,\,g(x)\ne\sup_{n\in\Bbb N}g_n(x)\}\cr}$$

\noindent is $\tilde\mu$-negligible, and $f=g\,\,\tilde\mu$-a.e.

\medskip

\quad{\bf (iii)} If $f$ is $\tilde\mu$-integrable, express it as
$f_1-f_2$ where $f_1$ and $f_2$ are $\tilde\mu$-integrable and
non-negative;  then there
are $\mu$-integrable functions $g_1$, $g_2$ such that $f_1=g_1$,
$f_2=g_2\,\,\tilde\mu$-a.e., so that $g=g_1-g_2$ is $\mu$-integrable and
equal to $f\,\,\tilde\mu$-a.e.
}%end of proof of 213G

\leader{213H}{}\cmmnt{ Thirdly, I turn to the effect of the
construction here on the other properties being considered in this
chapter.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space,
$(X,\hat\Sigma,\hat\mu)$ its completion and $(X,\tilde\Sigma,\tilde\mu)$
its c.l.d.\ version.

(a) If $(X,\Sigma,\mu)$ is a probability space, or totally finite, or
$\sigma$-finite, or strictly localizable, so is
$(X,\tilde\Sigma,\tilde\mu)$, and in all these cases
$\tilde\mu=\hat\mu$;

(b) if $(X,\Sigma,\mu)$ is localizable, so is
$(X,\tilde\Sigma,\tilde\mu)$, and for every $H\in\tilde\Sigma$ there is
an $E\in\Sigma$ such that $\tilde\mu(E\symmdiff H)=0$;

(c) $(X,\Sigma,\mu)$ is semi-finite iff $\tilde\mu F=\mu F$ for every
$F\in\Sigma$, and in this case $\int fd\tilde\mu=\int fd\mu$ whenever
the latter is defined in $[-\infty,\infty]$;

(d) a set $H\in\tilde\Sigma$ is an atom for $\tilde\mu$ iff there is an
atom $E$ for $\mu$ such that $\mu E<\infty$ and
$\tilde\mu(H\symmdiff E)=0$;

(e) if $(X,\Sigma,\mu)$ is atomless or purely atomic, so is
$(X,\tilde\Sigma,\tilde\mu)$;

(f) $(X,\Sigma,\mu)$ is complete and locally determined iff
$\tilde\mu=\mu$.


\proof{{\bf (a)(i)} I start by showing that if $(X,\Sigma,\mu)$ is
strictly localizable, then
$\tilde\mu=\hat\mu$.   \Prf\ Let $\langle X_i\rangle_{i\in I}$ be a
decomposition of $X$ for $\mu$;  then it is also a decomposition for
$\hat\mu$ (212Gb).   If $H\in\tilde\Sigma$, we shall have
$H\cap X_i\in\hat\Sigma$ for every $i$, and therefore $H\in\hat\Sigma$;
moreover,

$$\eqalign{\hat\mu H
&=\sum_{i\in I}\hat\mu(H\cap X_i)
=\sup\{\sum_{i\in J}\hat\mu(H\cap X_i):J\subseteq I
\text{ is finite}\}\cr
&\le\sup\{\hat\mu(H\cap E):E\in\Sigma,\,\mu E<\infty\}
=\tilde\mu H
\le\hat\mu H.\cr}$$

\noindent So $\hat\mu H=\tilde\mu H$ for every $H\in\tilde\Sigma$ and
$\hat\mu=\tilde\mu$.\ \Qed

\medskip

\quad{\bf (ii)} Consequently, if $(X,\Sigma,\mu)$ is a probability
space, or totally
finite, or $\sigma$-finite, or strictly localizable, so is
$(X,\tilde\Sigma,\tilde\mu)$, using 212Ga-212Gb to see that
$(X,\hat\Sigma,\hat\mu)$ has the property involved.

\medskip

{\bf (b)} If $(X,\Sigma,\mu)$ is localizable, let $\Cal H$ be any subset
of $\tilde\Sigma$.   Set

\Centerline{$\Cal E=\{E:E\in\Sigma^f,\,\exists\enskip
H\in\Cal H,\,E\subseteq H\}$}

\noindent where $\Sigma^f=\{E:\mu E<\infty\}$ as usual.
Working in $(X,\Sigma,\mu)$, let $F\in\Sigma$ be an essential
supremum for $\Cal E$.

\medskip

\quad{\bf (i)} \Quer\ Suppose, if possible, that there is an
$H\in\Cal H$ such that $\tilde\mu(H\setminus F)>0$.   By 213F(c-i),
there is an
$E\in\Sigma^f$ such that $E\subseteq H\setminus F$ and $\mu E>0$.
This $E$ belongs to $\Cal E$
and $\mu(E\setminus F)=\mu E>0$;  which is impossible if $F$ is an
essential supremum of $\Cal E$.\ \Bang

\medskip

\quad{\bf (ii)} Thus $\tilde\mu(H\setminus F)=0$ for every $H\in\Cal H$.
Now take any $G\in\tilde\Sigma$ such that $\tilde\mu(H\setminus G)=0$
for every $H\in\Cal H$.   \Quer\ If $\tilde\mu(F\setminus G)>0$, there
is an $E_0\in\Sigma^f$ such that $E_0\subseteq F\setminus G$ and
$\mu E_0>0$.   If $E\in\Cal E$, there is an
$H\in\Cal H$ such that $E\subseteq H$, so that
$E\setminus(F\setminus E_0)\subseteq H\setminus(F\cap G)$, while
$\mu(E\setminus(F\setminus E_0))<\infty$;  so

\Centerline{$\mu(E\setminus(F\setminus E_0))
\le\tilde\mu(H\setminus(F\cap G))
\le\tilde\mu(H\setminus F)+\tilde\mu(H\setminus G)
=0$.}

\noindent Because $F$ is an essential supremum for $\Cal E$ in $\Sigma$,

\Centerline{$0
=\mu(F\setminus(F\setminus E_0))
=\mu E_0$.  \Bang}

\noindent This shows that $F$ is an essential supremum for $\Cal H$ in
$\tilde\Sigma$.   As $\Cal H$ is arbitrary, $(X,\tilde\Sigma,\tilde\mu)$
is localizable.

\medskip

\quad{\bf (iii)} The argument of (i)-(ii) shows in fact that if $\Cal
H\subseteq\tilde\Sigma$ then $\Cal H$ has an essential supremum $F$ in
$\tilde\Sigma$ such that $F$ actually belongs to $\Sigma$.   Taking
$\Cal H=\{H\}$, we see that if $H\in\tilde\Sigma$ there is an
$F\in\Sigma$ such that $\mu(H\symmdiff F)=0$.

\medskip

{\bf (c)} We already know that $\tilde\mu E\le\mu E$ for every
$E\in\Sigma$, with equality if $\mu E<\infty$, by 213Fa.

\medskip

\quad{\bf (i)} If $(X,\Sigma,\mu)$ is semi-finite, then 213A
and 213F(c-i) tell us that for any $F\in\Sigma$ we have

\Centerline{$\mu F
=\sup\{\mu E:E\in\Sigma,\,E\subseteq F,\,\mu E<\infty\}
=\tilde\mu F$.}

\medskip

\quad{\bf (ii)} Suppose that $\tilde\mu F=\mu F$ for every $F\in\Sigma$.
If $\mu F=\infty$, then $\tilde\mu F=\infty$ so (by 213F(c-i) again)
there must be an $E\in\Sigma^f$ such that $E\subseteq F$ and
$\mu E>0$;  as $F$ is arbitrary, $(X,\Sigma,\mu)$ is
semi-finite.

\medskip

\quad{\bf (iii)} If $f$ is non-negative and $\int fd\mu=\infty$, then
$f$ is $\mu$-virtually measurable, therefore $\tilde\Sigma$-measurable
(213Ga), and defined $\mu$-almost everywhere, therefore
$\tilde\mu$-almost everywhere.   Now

$$\eqalign{\int fd\tilde\mu
&=\sup\{\int g\,d\tilde\mu:g\text{ is }\tilde\mu\text{-simple},\,
   0\le g\le f\,\,\tilde\mu\text{-a.e.}\}\cr
&\ge\sup\{\int g\,d\mu:g\text{ is }\mu\text{-simple},\,
   0\le g\le f\,\,\mu\text{-a.e.}\}=\infty\cr}$$

\noindent by 213B.   With 213Gb, this shows that
$\int fd\tilde\mu=\int fd\mu$ whenever $f$ is non-negative and
$\int fd\mu$ is defined in $[0,\infty]$.   Applying this to the positive
and negative parts of $f$, we see that $\int fd\tilde\mu=\int fd\mu$
whenever the latter is defined in $[-\infty,\infty]$.

\medskip

{\bf (d)(i)} If $H\in\tilde\Sigma$ is an atom for $\tilde\mu$, then
there is an $E\in\Sigma^f$ such that $E\subseteq H$ and $0<\mu E<\infty$.
In this case, $\tilde\mu E>0$ so $\tilde\mu(H\setminus E)$ must be zero.
If $F\in\Sigma$ and $F\subseteq E$, then either
$\mu F=\tilde\mu F=0$ or $\mu(E\setminus F)=\tilde\mu(H\setminus F)=0$.
Thus $E\in\Sigma$ is an atom for $\mu$ with
$\tilde\mu(H\symmdiff E)=0$ and $\mu E<\infty$.

\medskip

\quad{\bf (ii)} If $H\in\tilde\Sigma$ and there is an atom $E$ for $\mu$
such that $\mu E<\infty$ and $\tilde\mu(H\symmdiff E)=0$, let
$G\in\tilde\Sigma$ be a subset of $H$ with $\tilde\mu G>0$.
We have $\tilde\mu(E\cap G)=\tilde\mu(H\cap G)>0$, so there is an
$F\in\Sigma$ such that $F\subseteq E\cap G$ and $\mu F>0$.   Now
$\mu(E\setminus F)$ must be zero, so

\Centerline{$\tilde\mu(H\setminus G)\le\tilde\mu(H\setminus F)
=\tilde\mu(E\setminus F)=\mu(E\setminus F)=0$.}

\noindent As $G$ is arbitrary, $H$ is an atom for $\tilde\mu$.

\medskip

{\bf (e)} If $(X,\Sigma,\mu)$ is atomless, then
$(X,\tilde\Sigma,\tilde\mu)$ must be atomless, by (d).

If $(X,\Sigma,\mu)$ is purely atomic, $H\in\tilde\Sigma$ and
$\tilde\mu H>0$, then there is an $E\in\Sigma^f$ such that
$E\subseteq H$ and $\mu E>0$.
There is an atom $F$ for $\mu$ such that $F\subseteq E$;   now
$\mu F<\infty$ so $F$ is an atom for $\tilde\mu$, by (d).   Also
$F\subseteq H$.   As $H$ is arbitrary, $(X,\tilde\Sigma,\tilde\mu)$ is
purely atomic.

\medskip

{\bf (f)} If $\mu=\tilde\mu$, then of course $(X,\Sigma,\mu)$ must be
complete and locally determined, because $(X,\tilde\Sigma,\tilde\mu)$
is.   If $(X,\Sigma,\mu)$ is complete and locally determined, then
$\hat\mu=\mu$ so (using the definition in 213D)
$\tilde\Sigma\subseteq\Sigma$ and $\tilde\mu=\mu$, by (c) above.
}%end of proof of 213H

\leader{213I}{Locally determined negligible \dvrocolon{sets}}\cmmnt{
The following simple idea is occasionally useful.

\medskip

\noindent}{\bf Definition} A measure space $(X,\Sigma,\mu)$ has
{\bf locally determined negligible sets} if for every non-negligible
$A\subseteq X$ there is an $E\in\Sigma$ such that $\mu E<\infty$ and
$A\cap E$ is not negligible.

\leader{213J}{Proposition} If a measure space $(X,\Sigma,\mu)$ is
{\it either} strictly localizable {\it or} complete and locally
determined, it has locally determined negligible sets.

\proof{ Let $A\subseteq X$ be a set such that $A\cap E$ is negligible
whenever $\mu E<\infty$;  I need to show that $A$ is negligible.

\medskip

{\bf (i)} If $\mu$ is strictly localizable, let $\familyiI{X_i}$ be a
decomposition of $X$.   For each $i\in I$, $A\cap X_i$ is negligible, so
there we can choose
a negligible $E_i\in\Sigma$ such that $A\cap X_i\subseteq E_i$.
Set $E=\bigcup_{i\in I}E_i\cap X_i$.   Then
$\mu E=\sum_{i\in I}\mu(E_i\cap X_i)=0$ and $A\subseteq E$, so $A$ is
negligible.

\medskip

{\bf (ii)} If $\mu$ is complete and locally determined, take any
measurable set $E$ of finite measure.   Then $A\cap E$ is negligible,
therefore measurable;  as $E$ is arbitrary, $A$ is measurable;  as $\mu$
is semi-finite, $A$ is negligible.
}%end of proof of 213J

\leader{*213K}{Lemma} If a measure space $(X,\Sigma,\mu)$ has locally
determined negligible sets, and $\Cal E\subseteq\Sigma$ has an essential
supremum $H\in\Sigma$\cmmnt{ in the sense of 211G}, then
$H\setminus\bigcup\Cal E$ is negligible.

\proof{ Set $A=H\setminus\bigcup\Cal E$.   Take any $F\in\Sigma$ such
that $\mu F<\infty$.   Then
$F\cap A$ has a measurable envelope $V$ say (132Ee again).
If $E\in\Cal E$, then

\Centerline{$\mu(E\setminus(X\setminus V))
=\mu(E\cap V)=\mu^*(E\cap F\cap A)=0$,}

\noindent so $H\cap V=H\setminus(X\setminus V)$ is negligible and
$F\cap A$ is negligible.   As $F$ is arbitrary and $\mu$ has locally
determined negligible sets, $A$ is negligible, as claimed.
}%end of proof of 213K

\leader{213L}{Proposition} Let $(X,\Sigma,\mu)$ be a localizable measure
space with locally determined negligible sets.
Then every subset $A$ of $X$ has a measurable envelope.

\proof{ Set

\Centerline{$\Cal E=\{E:E\in\Sigma,\,\mu^*(A\cap E)=\mu E<\infty\}$.}

\noindent Let $G$ be an essential supremum for $\Cal E$ in $\Sigma$.

\medskip

\quad{\bf (i)} $A\setminus G$ is negligible.   \Prf\ Let $F$ be any set
of finite measure for $\mu$.   Let $E$ be a measurable envelope of
$A\cap F$.   Then $E\in\Cal E$ so $E\setminus G$ is negligible.   But
$F\cap A\setminus G\subseteq E\setminus G$, so $F\cap A\setminus G$ is
negligible.   Because $\mu$ has locally determined negligible sets,
this is enough to show that $A\setminus G$ is negligible.\ \Qed

\medskip

\quad{\bf (ii)} Let $E_0$ be a negligible measurable set including
$A\setminus G$, and set $\tilde G=E_0\cup G$, so that
$\tilde G\in\Sigma$,
$A\subseteq\tilde G$ and $\mu(\tilde G\setminus G)=0$.   \Quer\
Suppose, if
possible, that there is an $F\in\Sigma$ such that
$\mu^*(A\cap F)<\mu(\tilde G\cap F)$.   Let $F_1\subseteq F$ be a
measurable envelope of $A\cap F$.   Set $H=X\setminus(F\setminus F_1)$;
then $A\subseteq H$.   If $E\in\Cal E$ then

\Centerline{$\mu E=\mu^*(A\cap E)\le\mu(H\cap E)$,}

\noindent so $E\setminus H$ is negligible;  as $E$ is arbitrary,
$G\setminus H$ is negligible and $\tilde G\setminus H$ is negligible.
But $\tilde G\cap F\setminus F_1\subseteq\tilde G\setminus H$ and

\Centerline{$\mu(\tilde G\cap F\setminus F_1)
=\mu(\tilde G\cap F)-\mu^*(A\cap F)>0$. \Bang}

This shows that $\tilde G$ is a measurable envelope of $A$, as required.
}%end of proof of 213L

\leader{213M}{Corollary} (a) If $(X,\Sigma,\mu)$ is $\sigma$-finite, then
every subset of $X$ has a measurable envelope for $\mu$.

(b) If $(X,\Sigma,\mu)$ is localizable, then every subset of $X$ has a
measurable envelope for the c.l.d.\ version of $\mu$.

\proof{{\bf (a)} Use 132Ee, or 213L, 211Lc and 213J.

\medskip

{\bf (b)} Use 213L and the fact that the c.l.d.\ version of $\mu$ is
localizable as well as being complete and locally determined (213Hb).
}%end of proof of 213M

\leader{213N}{}\cmmnt{ When we come to use the concept of
`localizability', it will frequently be through the following
property, which in fact characterizes localizable spaces (213Xm).

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ be a localizable measure
space.   Suppose that $\Phi$ is a family of measurable real-valued
functions, all defined on measurable subsets of $X$, such that whenever
$f$, $g\in\Phi$ then $f=g$ almost everywhere in $\dom f\cap\dom g$.
Then there is a measurable function $h:X\to\Bbb R$ such that every
$f\in\Phi$ agrees with $h$ almost everywhere in $\dom f$.

\proof{ For $q\in\Bbb Q$, $f\in\Phi$ set

\Centerline{$E_{fq}=\{x:x\in\dom f,\,f(x)\ge q\}\in\Sigma$.}

\noindent For each $q\in\Bbb Q$, let $E_q$ be an essential supremum of
$\{E_{fq}:f\in\Phi\}$ in $\Sigma$.   Set

\Centerline{$h^*(x)=\sup\{q:q\in\Bbb Q,\,x\in E_q\}\in[-\infty,\infty]$}

\noindent for $x\in X$, taking $\sup\emptyset=-\infty$ if necessary.

If $f$, $g\in\Phi$ and $q\in\Bbb Q$, then

$$\eqalign{E_{fq}\setminus(X\setminus(\dom g\setminus E_{gq}))
&=E_{fq}\cap\dom g\setminus E_{gq}\cr
&\subseteq\{x:x\in\dom f\cap\dom g,\,f(x)\ne g(x)\}\cr}$$

\noindent is negligible;  as $f$ is arbitrary,

\Centerline{$E_q\cap\dom g\setminus E_{gq}=E_q\setminus(X\setminus(\dom
g\setminus E_{gq}))$}

\noindent is negligible.   Also $E_{gq}\setminus E_q$ is negligible, so
$E_{gq}\symmdiff(E_q\cap\dom g)$ is negligible.   Set
$H_g=\bigcup_{q\in\Bbb Q}E_{gq}\symmdiff(E_q\cap\dom g)$;  then $H_g$
is negligible.   But if $x\in\dom g\setminus H_g$, then, for every
$q\in\Bbb Q$, $x\in E_q\iff x\in E_{gq}$;  it follows that for such $x$,
$h^*(x)=g(x)$.   Thus $h^*=g$ almost everywhere in $\dom g$;  and this
is true for every $g\in\Phi$.

The function $h^*$ is not necessarily real-valued.   But it is
measurable, because
\Centerline{$\{x:h^*(x)>a\}=\bigcup\{E_q:q\in\Bbb Q,\,q>a\}\in\Sigma$}

\noindent for every real $a$.   So if we modify it by setting

$$\eqalign{h(x)&=h^*(x)\text{ if }h(x)\in\Bbb R,\cr
&=0\text{ if }h^*(x)\in\{-\infty,\infty\},\cr}$$

\noindent we shall get a measurable real-valued function $h:X\to\Bbb R$;
and for any $g\in\Phi$, $h(x)$ will be equal to $g(x)$ at least whenever
$h^*(x)=g(x)$, which is true for almost every $x\in\dom g$.   Thus $h$
is a suitable function.
}%end of proof of 213N

\leader{213O}{}\cmmnt{ There is an interesting and useful criterion
for a space to be strictly localizable which I introduce at this point,
though it will be used rarely in this volume.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a complete locally
determined space.

(a) Suppose that there is a disjoint family $\Cal E\subseteq\Sigma$
such that ($\alpha$) $\mu E<\infty$ for every $E\in\Cal E$
($\beta$) whenever $F\in\Sigma$ and $\mu F>0$ then there is an
$E\in\Cal E$
such that $\mu(E\cap F)>0$.   Then $(X,\Sigma,\mu)$ is strictly
localizable, $\bigcup\Cal E$ is conegligible, and
$\Cal E\cup\{X\setminus\bigcup\Cal E\}$ is a
decomposition of $X$.

(b) Suppose that $\langle X_i\rangle_{i\in I}$ is a partition of $X$
into
measurable sets of finite measure such that whenever $E\in\Sigma$ and
$\mu E>0$ there is an $i\in I$ such that $\mu(E\cap X_i)>0$.   Then
$(X,\Sigma,\mu)$ is strictly localizable, and
$\langle X_i\rangle_{i\in I}$ is a decomposition of $X$.

\proof{{\bf (a)(i)}
The first thing to note is that if $F\in\Sigma$ and
$\mu F<\infty$, there is a countable
$\Cal E'\subseteq\Cal E$ such that
$\mu(F\setminus\bigcup\Cal E')=0$.   \Prf\ Set

\Centerline{$\Cal E'_n=\{E:E\in\Cal E,\,\mu(F\cap E)\ge 2^{-n}\}$ for
each $n\in\Bbb N$,}

\Centerline{$\Cal E'=\bigcup_{n\in\Bbb N}\Cal E'_n=\{E:E\in\Cal
E,\,\mu(F\cap E)>0\}$.}

\noindent Because $\Cal E$ is disjoint, we must have

\Centerline{$\#(\Cal E'_n)\le 2^n\mu F$}

\noindent for every $n\in\Bbb N$, so that every $\Cal E'_n$ is finite
and $\Cal E'$, being the union of a sequence of countable sets, is
countable.   Set $E'=\bigcup\Cal E'$ and $F'=F\setminus E'$, so that
both $E'$ and $F'$ belong to $\Sigma$.   If $E\in\Cal E'$, then
$E\subseteq E'$ so $\mu(E\cap F')=\mu\emptyset=0$;  if $E\in\Cal
E\setminus\Cal E'$, then $\mu(E\cap F')=\mu(E\cap F)=0$.   Thus
$\mu(E\cap F')=0$ for every $E\in\Cal E$.   By the hypothesis ($\beta$)
on
$\Cal E$, $\mu F'=0$, so $\mu(F\setminus\bigcup\Cal E')=0$, as
required.\ \Qed

\medskip

\quad{\bf (ii)} Now suppose that $H\subseteq X$ is such that $H\cap
E\in\Sigma$ for every $E\in\Cal E$.   In this case $H\in\Sigma$.   \Prf\
Let $F\in\Sigma$ be such that $\mu F<\infty$.   Let $\Cal
E'\subseteq\Cal E$ be a countable set such that $\mu(F\setminus E')=0$,
where $E'=\bigcup\Cal E'$.  Then $H\cap(F\setminus E')\in\Sigma$
because $(X,\Sigma,\mu)$ is complete.   But also $H\cap
E'=\bigcup_{E\in\Cal E'}H\cap E\in\Sigma$.   So

\Centerline{$H\cap F=(H\cap(F\setminus E'))\cup(F\cap(H\cap
E'))\in\Sigma$.}

\noindent As $F$ is arbitrary and $(X,\Sigma,\mu)$ is locally
determined, $H\in\Sigma$.\ \Qed

\medskip

\quad{\bf (iii)} We find also that $\mu H=\sum_{E\in\Cal E}\mu(H\cap E)$
for every $H\in\Sigma$.   \Prf\ {($\alpha$)} Because $\Cal E$ is
disjoint, we
must have $\sum_{E\in\Cal E'}\mu(H\cap E)\le\mu H$ for every finite
$\Cal E'\subseteq\Cal E$, so

\Centerline{$\sum_{E\in\Cal E}\mu(H\cap E)
=\sup\{\sum_{E\in\Cal E'}
  \mu(H\cap E):\Cal E'\subseteq\Cal E$ is finite$\}\le\mu H$.}

\noindent ($\beta$) For the reverse inequality, consider first the case
$\mu H<\infty$.   By (i), there is a countable $\Cal E'\subseteq\Cal E$
such that $\mu(H\setminus\bigcup\Cal E')=0$, so that

\Centerline{$\mu H=\mu(H\cap\bigcup\Cal E')
=\sum_{E\in\Cal E'}\mu(H\cap E)\le\sum_{E\in\Cal E}\mu(H\cap E)$.}

\noindent{($\gamma$)} In general, because $(X,\Sigma,\mu)$ is
semi-finite,

$$\eqalign{\mu H
&=\sup\{\mu F:F\subseteq H,\,\mu F<\infty\}\cr
&\le\sup\{\sum_{E\in\Cal E}\mu(F\cap E):F\subseteq H,\,\mu F<\infty\}
\le\sum_{E\in\Cal E}\mu(H\cap E).\cr}$$

\noindent So in all cases we have
$\mu H\le\sum_{E\in\Cal E}\mu(H\cap E)$, and the two are equal.\ \Qed

\medskip

\quad{\bf (iv)} In particular, setting $E_0=X\setminus\bigcup\Cal E$,
$E_0\in\Sigma$ and $\mu E_0=0$;  that is, $\bigcup\Cal E$ is
conegligible.
Consider $\Cal E^*=\Cal E\cup\{E_0\}$.
This is a partition of $X$ into sets of finite measure (now using the
hypothesis ($\alpha$) on $\Cal E$).   If $H\subseteq X$ is such that
$H\cap E\in\Sigma$ for every $E\in\Cal E^*$, then $H\in\Sigma$ and

\Centerline{$\mu H=\sum_{E\in\Cal E}\mu(H\cap E)
=\sum_{E\in\Cal E^*}\mu(H\cap E)$.}

\noindent Thus $\Cal E^*$ (or, if you prefer, the indexed family
$\langle E\rangle_{E\in\Cal E^*}$) is a decomposition witnessing that
$(X,\Sigma,\mu)$ is strictly localizable.

\medskip

{\bf (b)} Apply (a) with $\Cal E=\{X_i:i\in I\}$, noting that $E_0$ in
(iv) is empty, so can be dropped.
}%end of proof of 213O

\exercises{
\leader{213X}{Basic exercises (a)}
%\spheader 213Xa
Let $(X,\Sigma,\mu)$ be any measure space, $\mu^*$ the outer measure
defined from $\mu$, and $\check\mu$ the measure defined by
\Caratheodory's method from $\mu^*$;  write $\check\Sigma$ for the
domain of $\check\mu$.   Show that
(i) $\check\mu$ extends the completion $\hat\mu$ of $\mu$;
(ii) if $H\subseteq X$ is such that
$H\cap F\in\check\Sigma$ whenever $F\in\Sigma$ and $\mu F<\infty$, then
$H\in\check\Sigma$;
(iii) $(\check\mu)^*=\mu^*$, so that the integrable functions
for $\check\mu$ and $\mu$ are the same (212Xb);  (iv) if $\mu$ is
strictly localizable then $\check\mu=\hat\mu$.
%213C 213Yb

\sqheader 213Xb
Let $\mu$ be counting measure restricted to the
countable-cocountable $\sigma$-algebra of a set $X$ (211R, 211Ye).
(i) Show that the c.l.d.\ version $\tilde\mu$ of $\mu$ is just counting
measure on $X$.   (ii) Show that $\check\mu$, as defined in 213Xa, is
equal to $\tilde\mu$, and in particular strictly extends the completion
of $\mu$ if $X$ is uncountable.
%213E

\spheader 213Xc Let $(X,\Sigma,\mu)$ be any measure space.   For
$E\in\Sigma$ set

\Centerline{$\mu_{\text{sf}}E
=\sup\{\mu(E\cap F):F\in\Sigma,\,\mu F<\infty\}$.}

\quad{(i)} Show that $(X,\Sigma,\mu_{\text{sf}})$ is a semi-finite measure
space, and is equal to $(X,\Sigma,\mu)$ iff $(X,\Sigma,\mu)$ is
semi-finite.

\quad{(ii)} Show that a $\mu$-integrable real-valued function $f$ is
$\mu_{\text{sf}}$-integrable, with the same integral.

\quad(iii) Show that if $E\in\Sigma$ then $E$
can be expressed as $E_1\cup E_2$ where $E_1$, $E_2\in\Sigma$,
$\mu E_1=\mu_{\text{sf}}E_1$ and $\mu_{\text{sf}}E_2=0$.

\quad{(iv)} Show that if $f$ is a $\mu_{\text{sf}}$-integrable real-valued
function on $X$, it is equal $\mu_{\text{sf}}$-almost everywhere to a
$\mu$-integrable function.

\quad{(v)} Show that if $(X,\Sigma,\mu_{\text{sf}})$ is complete, so is
$(X,\Sigma,\mu)$.

\quad{(vi)} Show that $\mu$ and $\mu_{\text{sf}}$ have identical c.l.d.\
versions.
%213F

\spheader 213Xd Let $(X,\Sigma,\mu)$ be any measure space.   Define
$\check\mu$ as in 213Xa.   Show that $(\check\mu)_{\text{sf}}$, as constructed
in 213Xc, is precisely the c.l.d.\ version $\tilde\mu$ of $\mu$, so that
$\check\mu=\tilde\mu$ iff $\check\mu$ is semi-finite.
%213F, 213Xa, 213Xc

\spheader 213Xe Let $(X,\Sigma,\mu)$ be a measure space.   For
$A\subseteq X$ set
$\mu_* A=\sup\{\mu E:E\in\Sigma,\,\mu E<\infty,\,E\subseteq A\}$, as in
113Yh.   (i) Show that the measure constructed from $\mu_*$
by the method of 113Yg/212Ya is just the c.l.d.\ version $\tilde\mu$ of
$\mu$.
(ii) Show that $\tilde\mu_*=\mu_*$.   (iii) Show that if $\nu$ is
another measure
on $X$, with domain $\Tau$, then $\tilde\mu=\tilde\nu$ iff
$\mu_*=\nu_*$.
%213F

\spheader 213Xf Let $X$ be a set and $\theta$ an outer measure on $X$.
Show that $\theta_{\text{sf}}$, defined by writing

\Centerline{$\theta_{\text{sf}}A=\sup\{\theta B:B\subseteq A,\,\theta
B<\infty\}$}

\noindent is also an outer measure on $X$.   Show that the measures
defined by \Caratheodory's method from $\theta$, $\theta_{\text{sf}}$ have the
same domains.
%213F

\spheader 213Xg Let $(X,\Sigma,\mu)$ be any measure space.   Set

\Centerline{$\mu^*_{\text{sf}}A
=\sup\{\mu^*(A\cap E):E\in\Sigma,\,\mu E<\infty\}$}

\noindent for every $A\subseteq X$.

\quad{(i)} Show that

\Centerline{$\mu^*_{\text{sf}}A=\sup\{\mu^*B:B\subseteq A,\,\mu^*B<\infty\}$}

\noindent for every $A$.

\quad{(ii)} Show that $\mu^*_{\text{sf}}$ is an outer measure.

\quad(iii) Show that if $A\subseteq X$ and $\mu^*_{\text{sf}}A<\infty$, there
is an $E\in\Sigma$ such that $\mu^*_{\text{sf}}A=\mu^*(A\cap E)=\mu E$,
$\mu^*_{\text{sf}}(A\setminus E)=0$.   \Hint{take a non-decreasing sequence
$\sequencen{E_n}$ of measurable sets of finite measure such that
$\mu^*_{\text{sf}}A=\lim_{n\to\infty}\mu^*(A\cap E_n)$, and let
$E\subseteq\bigcup_{n\in\Bbb N}E_n$ be a measurable envelope of
$A\cap\bigcup_{n\in\Bbb N}E_n$.}

\quad{(iv)} Show that the measure defined from $\mu^*_{\text{sf}}$ by
\Caratheodory's method is precisely the c.l.d.\ version $\tilde\mu$ of
$\mu$.

\quad{(v)} Show that $\mu^*_{\text{sf}}=\tilde\mu^*$, so that if $\mu$ is
complete and locally determined then $\mu^*_{\text{sf}}=\mu^*$.
%213F, 213Xc, 213Xf

\sqheader 213Xh Let $(X,\Sigma,\mu)$ be a measure space
with locally determined measurable sets.   Show that it is semi-finite.
%213I

\sqheader 213Xi Let $(X,\Sigma,\mu)$ be a measure space, $\hat\mu$ the
completion of $\mu$,
$\tilde\mu$ the c.l.d.\ version of $\mu$
and $\check\mu$ the measure defined by \Caratheodory's method
from $\mu^*$.   Show that the following are equiveridical:
(i) $\mu$ has
locally determined negligible sets;
(ii) $\mu$ and $\tilde\mu$ have the
same negligible sets;
(iii) $\check\mu=\tilde\mu$;
(iv) $\hat\mu$ and $\tilde\mu$ have the same sets of finite measure;
(v) $\mu$ and $\tilde\mu$ have the same integrable functions;
(vi) $\tilde\mu^*=\mu^*$;
(vii) the outer measure $\mu^*_{\text{sf}}$ of 213Xg is equal to $\mu^*$.
%213J 213Xi

\sqheader 213Xj Let $(X,\Sigma,\mu)$ be a strictly localizable measure
space with a decomposition $\langle X_i\rangle_{i\in I}$.   Show that
$\mu^*A=\sum_{i\in I}\mu^*(A\cap X_i)$ for every $A\subseteq X$.
%213L

\sqheader 213Xk Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space, and let $A\subseteq X$ be such that
$\min(\mu^*(E\cap A),\mu^*(E\setminus A))<\mu E$ whenever $E\in\Sigma$
and $0<\mu E<\infty$.
Show that $A\in\Sigma$.   \Hint{given $\mu F<\infty$, consider
the intersection $E$ of measurable envelopes of $F\cap A$, $F\setminus
A$ to see that $\mu^*(F\cap A)+\mu^*(F\setminus A)=\mu F$.}
%213M

\spheader 213Xl Let us say that a measure space $(X,\Sigma,\mu)$ has the
{\bf measurable envelope property} if every subset of $X$ has a
measurable envelope.   (i) Show that a semi-finite space with the
measurable envelope property has locally determined negligible sets.
(ii) Show that a complete semi-finite space with
the measurable envelope property is locally determined.
%213M

\spheader 213Xm Let $(X,\Sigma,\mu)$ be a semi-finite measure space, and
suppose that it satisfies the conclusion of Theorem 213N.   Show that it
is localizable.   \Hint{given $\Cal E\subseteq\Sigma$, set
$\Cal F=\{F:F\in\Sigma,\,E\cap F$ is negligible for every
$E\in\Cal E\}$.
Let $\Phi$ be the set of functions $f$ from subsets of $X$ to $\{0,1\}$
such that $f^{-1}[\{1\}]\in \Cal E$ and $f^{-1}[\{0\}]\in\Cal F$.}
%213N

\spheader 213Xn Let $(X,\Sigma,\mu)$ be a measure space.   Show that its
c.l.d.\ version is strictly localizable iff there is a disjoint family
$\Cal E\subseteq\Sigma$ such that $\mu E<\infty$ for every $E\in\Cal E$
and whenever $F\in\Sigma$ and $0<\mu F<\infty$ there is an $E\in\Cal E$
such that $\mu(E\cap F)>0$.
%213O

\spheader 213Xo Show that the c.l.d.\ version of any point-supported
measure is point-supported.

\vleader{48pt}{213Y}{Further exercises (a)}
%\spheader 213Ya
Let $(X,\Sigma,\mu)$ be a measure space.   Show that $\mu$
is semi-finite iff there is a family $\Cal E\subseteq\Sigma$ such that
$\mu E<\infty$ for every $E\in\Cal E$ and
$\mu F=\sum_{E\in\Cal E}\mu(F\cap E)$ for every $F\in\Sigma$.
\Hint{take $\Cal E$ maximal subject to the intersection of any two elements
being negligible.}
%213A  would be useful in 552M, (i) => (iii)

\spheader 213Yb
Set $X=\Bbb N$, and for $A\subseteq X$ set

\Centerline{$\theta A=\sqrt{\#(A)}$ if $A$ is finite, $\infty$ if $A$ is
infinite.}

\noindent Show that $\theta$ is an outer measure on $X$, that
$\theta A=\sup\{\theta B:B\subseteq A,\,\theta B<\infty\}$ for every
$A\subseteq X$, but that the measure $\mu$ defined from $\theta$ by
\Caratheodory's
method is not semi-finite.   Show that if $\check\mu$ is the measure
defined by \Caratheodory's method from $\mu^*$ (213Xa), then
$\check\mu\ne\mu$.
%213C

\spheader 213Yc Set $X=[0,1]\times\{0,1\}$, and let
$\Sigma$ be the family of those subsets $E$ of $X$ such that

\Centerline{$\{x:x\in[0,1],\,E[\{x\}]\ne\emptyset,\,E[\{x\}]\ne\{0,1\}\}
$}

\noindent is countable, writing $E[\{x\}]=\{y:(x,y)\in E\}$ for each
$x\in[0,1]$.   Show that $\Sigma$ is a $\sigma$-algebra of subsets of
$X$.   For $E\in\Sigma$, set $\mu E=\#(\{x:(x,1)\in E\})$ if this is
finite, $\infty$ otherwise.   Show that $\mu$ is a complete semi-finite
measure.   Show that the measure $\check\mu$ defined from $\mu^*$ by
\Caratheodory's method (213Xa) is not semi-finite.   Show that the
domain of the c.l.d.\ version of $\mu$ is the whole of $\Cal PX$.
%213Xa, 213F

\spheader 213Yd Set $X=\Bbb N$, and for $A\subseteq X$ set

\Centerline{$\phi A=\#(A)^2$ if $A$ is finite, $\infty$ if $A$ is
infinite.}

\noindent Show that $\phi$ satisfies the conditions of 212Ya,
but that the measure defined from $\phi$ by the method there is not
semi-finite.
%213Xe, 213F

\spheader 213Ye Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space.   Suppose that $D\subseteq X$ and that
$f:D\to\Bbb R$ is a function.   Show that the following are equiveridical:
(i) $f$ is measurable;  (ii)

\Centerline{$\mu^*\{x:x\in D\cap E,\,f(x)\le
a\}+\mu^*\{x:x\in D\cap E,\,f(x)\ge b\}\le\mu E$}

\noindent whenever $a<b$ in $\Bbb R$, $E\in\Sigma$ and $\mu E<\infty$ (iii)

\Centerline{$\min(\mu^*\{x:x\in D\cap E,\,f(x)\le
a\},\mu^*\{x:x\in D\cap E,\,f(x)\ge b\})<\mu E$}

\noindent  whenever $a<b$ in $\Bbb R$ and $0<\mu E<\infty$.  ({\it
Hint\/}:
for (iii)$\Rightarrow$(i), show that if $E\subseteq X$ then

\Centerline{$\mu^*\{x:x\in D\cap E,\,f(x)>a\}
=\sup_{b>a}\mu^*\{x:x\in D\cap E,\,f(x)\ge b\}$,}

\noindent and use 213Xk above.)
%213Xk, 213M

\spheader 213Yf Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and suppose that $\Cal E\subseteq\Sigma$ is such that
$\mu E<\infty$ for every $E\in\Cal E$ and whenever $F\in\Sigma$ and
$\mu F<\infty$ there is a countable $\Cal E_0\subseteq\Cal E$ such that
$F\setminus\bigcup\Cal E_0$, $F\cap\bigcup(\Cal E\setminus\Cal E_0)$ are
negligible.   Show that $(X,\Sigma,\mu)$ is strictly localizable.
%213O
}%end of exercises

\endnotes{
\Notesheader{213} I think it is fair to say that if the definition of
`measure space' were re-written to exclude all spaces which are not
semi-finite, nothing significant would be lost from the theory.   There
are solid reasons for not taking such a drastic step, starting with the
fact that it would confuse everyone (if you say to an unprepared
audience `let $(X,\Sigma,\mu)$ be a measure space', there is a danger
that some will imagine that you mean `$\sigma$-finite measure space',
but very few will suppose that you mean `semi-finite measure space').
But the whole point of measure theory is that we distinguish between
sets by their measures, and if every subset of $E$ is either
non-measurable, or negligible, or of infinite measure, the
classification is
too crude to support most of the usual ideas, starting, of course, with
ordinary integration.

Let us say that a measurable set $E$ is {\bf purely infinite} if $E$
itself and all its non-negligible measurable subsets have infinite
measure.   On the definition of the integral which I chose in Volume 1,
every simple function, and therefore every integrable function, must be
zero almost everywhere in $E$.   This means that the whole theory of
integration will ignore $E$ entirely.   Looking at the definition of
 `c.l.d.\ version' (213D-213E), you will see that the c.l.d.\ version
of the measure will render $E$ negligible, as does the `semi-finite
version' described in 213Xc.   These amendments do not, however, affect
sets of finite measure, and consequently leave integrable functions
integrable, with the same integrals.

The strongest reason we have yet seen for admitting non-semi-finite
spaces into consideration is that \Caratheodory's method does not
always produce semi-finite spaces.   (I give examples in 213Yb-213Yc;
more important ones are the Hausdorff measures of
\S\S264-265 below.)   In practice the right thing to do
is often to take the c.l.d.\ version of the measure produced by
\Caratheodory's construction.

It is a reasonable general philosophy, in measure theory, to say that we
wish to measure as many sets, and integrate as many functions, as we can
manage in a canonical way -- I mean, without making blatantly arbitrary
choices about the values we assign to our measure or integral.   The
revision of a measure $\mu$ to its c.l.d.\ version $\tilde\mu$ is about
as far as we can go with an arbitrary measure space in which
we have no other structure to guide our choices.

You will observe that $\tilde\mu$ is not as close to $\mu$ as the
completion $\hat\mu$ of $\mu$
is;  naturally so, because if $E\in\Sigma$ is purely infinite for
$\mu$ then we have to choose between
setting $\tilde\mu E=0\ne\mu E$ and finding some way of fitting many
sets of finite measure into $E$;  which if $E$ is a singleton will be
actually impossible, and in any case would be an arbitrary process.
However the integrable functions for $\tilde\mu$, while not always the
same as those for $\mu$ (since $\tilde\mu$ turns purely infinite sets
into negligible ones, so that their indicator functions become
integrable), are `nearly' the same, in the sense that any
$\tilde\mu$-integrable function can be changed into a $\mu$-integrable
function by adjusting it on a $\tilde\mu$-negligible set.   This
corresponds, of course, to the fact that any set of finite measure for
$\tilde\mu$ is the symmetric difference of a set of finite measure for
$\mu$ and a $\tilde\mu$-negligible set.   For sets of infinite measure
this can fail, unless $\mu$ is localizable (213Hb, 213Xb).

If $(X,\Sigma,\mu)$ is semi-finite, or localizable, or strictly
localizable, then of course it is correspondingly closer to
$(X,\tilde\Sigma,\tilde\mu)$, as detailed in 213Ha-c.

It is worth noting that while the measure $\check\mu$ obtained by
\Caratheodory's method directly from the outer measure $\mu^*$ defined
from $\mu$ may fail to be semi-finite, even when $\mu$ is (213Yc), a
simple modification of $\mu^*$ (213Xg) yields the c.l.d.\ version
$\tilde\mu$ of $\mu$, which can also be obtained from an appropriate
inner measure (213Xe).   The measure $\check\mu$ is of course related
in other ways to $\tilde\mu$;  see 213Xd.
}%end of notes

\discrpage


\frfilename{mt214.tex}
\versiondate{22.5.09}
\copyrightdate{2009}

\def\chaptername{Taxonomy of measure spaces}
\def\sectionname{Subspaces}

\newsection{214}

In \S131 I described a construction for subspace measures on measurable
subsets.   It is now time to give the generalization to subspace
measures
on arbitrary subsets of a measure space.   The relationship between this
construction and the properties listed in \S211 is not quite as
straightforward as one might imagine, and in this section I try to give
a full account of what can be expected of subspaces in general.   I
think that for the present volume only (i) general subspaces of
$\sigma$-finite spaces and
(ii) measurable subspaces of general measure spaces will be
needed in any essential way, and these do not give any difficulty;  but
in later volumes we shall need the full theory.

I begin with a general construction for `subspace measures'
(214A-214C), %214A 214B 214C
with an account of integration with respect to a subspace
measure (214E-214G); %214E 214F 214G
these (with 131E-131H) %131E 131F 131G 131H
give a solid foundation for
the concept of `integration over a subset' (214D).   I present
this work in its full natural generality, which will eventually be
essential, but even for Lebesgue measure alone it is important to be
aware of the ideas here.    I continue with answers to some obvious
questions concerning subspace measures and the properties of measure
spaces so far considered, both for general subspaces (214I) and for
measurable subspaces (214K), and I mention a basic construction for
assembling measure spaces side-by-side, the `direct sums' of 214L-214M.
At the end of the section I discuss a measure extension problem
(214O-214P).

\leader{214A}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$Y$ any subset of $X$.   Let $\mu^*$ be the outer measure defined from
$\mu$\cmmnt{ (132A-132B)}, and set $\Sigma_Y=\{E\cap Y:E\in\Sigma\}$;  let
$\mu_Y$ be the restriction of $\mu^*$ to $\Sigma_Y$.   Then
$(Y,\Sigma_Y,\mu_Y)$ is a measure space.

\proof{{\bf (a)} I have noted in 121A that $\Sigma_Y$ is a
$\sigma$-algebra of subsets of $Y$.

\medskip

{\bf (b)} Of course $\mu_YF\in[0,\infty]$ for
every $F\in\Sigma_Y$.

\medskip

{\bf (c)}  $\mu_Y\emptyset=\mu^*\emptyset=0$.


\medskip

{\bf (d)} If $\sequencen{F_n}$ is a disjoint sequence in $\Sigma_Y$ with
union $F$,  then choose $E_n$, $E'_n$, $E\in\Sigma$ such that
$F_n=Y\cap E_n$, $F_n\subseteq E'_n$ and $\mu_YF_n=\mu E'_n$ for each
$n$, $F\subseteq E$ and $\mu_YF=\mu E$ (using 132Aa repeatedly).   Set
$G_n=E_n\cap E'_n\cap E\setminus\bigcup_{m<n}E_m$ for each $n\in\Bbb N$;
then $\sequencen{G_n}$ is disjoint and $F_n\subseteq G_n\subseteq E'_n$
for each $n$, so $\mu_YF_n=\mu G_n$.   Also
$F\subseteq\bigcup_{n\in\Bbb N}G_n\subseteq E$, so

\Centerline{$\mu_YF=\mu(\bigcup_{n\in\Bbb N}G_n)
=\sum_{n=0}^{\infty}\mu G_n=\sum_{n=0}^{\infty}\mu_YF_n$.}

\noindent As $\sequencen{F_n}$ is
arbitrary, $\mu_Y$ is a measure.
}%end of proof of 214A

\leader{214B}{Definition} If $(X,\Sigma,\mu)$ is any measure space and
$Y$ is any subset of $X$, then $\mu_Y$, defined as in 214A, is the {\bf
subspace measure} on $Y$.

\cmmnt{It is worth noting the following.}

\leader{214C}{Lemma} Let $(X,\Sigma,\mu)$ be a measure space, $Y$ a
subset of $X$, $\mu_Y$ the subspace measure on $Y$ and $\Sigma_Y$ its
domain.  Then

(a) for any $F\in\Sigma_Y$, there is an $E\in\Sigma$ such that
$F=E\cap Y$ and $\mu E=\mu_YF$;

(b) for any $A\subseteq Y$, $A$ is $\mu_Y$-negligible iff it is
$\mu$-negligible;

(c)(i) if $A\subseteq X$ is $\mu$-conegligible, then $A\cap Y$
is $\mu_Y$-conegligible;

\quad (ii) if $A\subseteq Y$ is
$\mu_Y$-conegligible, then $A\cup(X\setminus Y)$ is $\mu$-conegligible;

(d) $(\mu_Y)^*$\cmmnt{, the outer measure on $Y$ defined from
$\mu_Y$,} agrees with $\mu^*$ on $\Cal PY$;

(e) if $Z\subseteq Y\subseteq X$, then
$\Sigma_Z=(\Sigma_Y)_Z$, the subspace $\sigma$-algebra of
subsets of $Z$ regarded as a subspace of $(Y,\Sigma_Y)$, and
$\mu_Z=(\mu_Y)_Z$ is the subspace measure on $Z$
regarded as a subspace of $(Y,\mu_Y)$;

(f) if $Y\in\Sigma$, then $\mu_Y$, as defined here, is exactly the
subspace measure on $Y$ defined in 131A-131B\cmmnt{;  that is,
$\Sigma_Y=\Sigma\cap\Cal PY$ and $\mu_Y=\mu\restr\Sigma_Y$}.

\proof{{\bf (a)} By the definition of $\Sigma_Y$, there is an
$E_0\in\Sigma$ such that $F=E_0\cap Y$.   By 132Aa, there is an
$E_1\in\Sigma$ such that $F\subseteq E_1$ and $\mu^*F=\mu E_1$.   Set
$E=E_0\cap E_1$;  this serves.

\medskip

{\bf (b)}(i) If $A$ is $\mu_Y$-negligible, there
is a set $F\in\Sigma_Y$ such that $A\subseteq F$ and $\mu_YF=0$;   now
$\mu^*A\le\mu^*F=0$ so $A$ is $\mu$-negligible, by 132Ad.   (ii)
If $A$ is
$\mu$-negligible, there is an $E\in\Sigma$ such that $A\subseteq E$ and
$\mu E=0$;  now $A\subseteq E\cap Y\in\Sigma_Y$ and $\mu_Y(E\cap Y)=0$,
so $A$ is $\mu_Y$-negligible.

\medskip

{\bf (c)} If $A\subseteq X$ is $\mu$-conegligible, then $A\cap Y$
is $\mu_Y$-conegligible, because $Y\setminus A=Y\cap(X\setminus A)$ is
$\mu$-negligible, therefore $\mu_Y$-negligible.   If $A\subseteq Y$ is
$\mu_Y$-conegligible, then $A\cup(X\setminus Y)$ is $\mu$-conegligible
because $X\setminus(A\cup(X\setminus Y))=Y\setminus A$ is
$\mu_Y$-negligible, therefore $\mu$-negligible.

\medskip

{\bf (d)} Let $A\subseteq Y$.   (i) If
$A\subseteq E\in\Sigma$, then $A\subseteq E\cap Y\in\Sigma_Y$, so
$\mu_Y^*A\le\mu_Y(E\cap Y)\le\mu E$;  as $E$ is arbitrary,
$\mu_Y^*A\le\mu^*A$.    (ii) If $A\subseteq F\in\Sigma_Y$, there is
an $E\in\Sigma$ such that $F\subseteq E$ and $\mu_YF=\mu^*F=\mu E$;
now $A\subseteq E$ so $\mu^*A\le\mu E=\mu_YF$.   As $F$ is arbitrary,
$\mu^*A\le\mu_Y^*A$.

\medskip

{\bf (e)} That $\Sigma_Z=(\Sigma_Y)_Z$ is immediate from the definition
of $\Sigma_Y$, etc.;   now

\Centerline{$(\mu_Y)_Z=
\mu_Y^*\restr\Sigma_Z=\mu^*\restr\Sigma_Z=\mu_Z$}

\noindent by (d).

\medskip

{\bf (f)}  This is elementary, because $E\cap Y\in\Sigma$ and
$\mu^*(E\cap Y)=\mu(E\cap Y)$ for every $E\in\Sigma$.
}%end of proof of 214C

\leader{214D}{Integration over subsets:  Definition} Let
$(X,\Sigma,\mu)$ be a measure
space, $Y$ a subset of $X$ and $f$ a $[-\infty,\infty]$-valued function
defined on a
subset of $X$.   By $\int_Y f$\cmmnt{ (or $\int_Y f(x)\mu(dx)$, etc.)}
I mean $\int(f\restrp Y)d\mu_Y$, if this exists in
$[-\infty,\infty]$\cmmnt{, following the definitions of
214A-214B, 133A and 135F, and taking $\dom(f\restrp Y)=Y\cap\dom f$,
$(f\restrp Y)(x)=f(x)$ for $x\in Y\cap\dom f$}.   \cmmnt{(Compare
131D.)}

\leader{214E}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space,
$Y\subseteq X$, and $f$ a
$[-\infty,\infty]$-valued function defined on a subset $\dom f$ of $X$.

(a) If $f$ is $\mu$-integrable then $f\restrp Y$ is $\mu_Y$-integrable, and
$\int_Y f\le\int f$ if $f$ is non-negative.

(b) If $\dom f\subseteq Y$ and $f$ is $\mu_Y$-integrable, then there is
a $\mu$-integrable function $\tilde f$ on $X$, extending $f$, such that
$\int_F\tilde f=\int_{F\cap Y}f$ for every $F\in\Sigma$.

\proof{{\bf (a)(i)} If $f$ is $\mu$-simple, it is expressible as
$\sum_{i=0}^na_i\chi E_i$, where $E_0,\ldots,E_n\in\Sigma$,
$a_0,\ldots,a_n\in\Bbb R$ and $\mu E_i<\infty$ for each $i$.   Now
$f\restrp Y=\sum_{i=0}^na_i\chi_Y(E_i\cap Y)$, where $\chi_Y(E_i\cap
Y)=(\chi E_i)\restrp Y$ is the indicator function of $E_i\cap Y$
regarded as a subset of $Y$;  and each $E_i\cap Y$ belongs to
$\Sigma_Y$, with $\mu_Y(E_i\cap Y)\le\mu E_i<\infty$, so $f\restr
Y:Y\to\Bbb R$ is $\mu_Y$-simple.

If $f:X\to \Bbb R$ is a non-negative simple function, it is expressible
as $\sum_{i=0}^na_i\chi E_i$ where $E_0,\ldots,E_n$ are disjoint sets of
finite measure (122Cb).   Now
$f\restrp Y=\sum_{i=0}^na_i\chi_Y(E_i\cap Y)$ and

\Centerline{$\int (f\restrp Y)d\mu_Y
=\sum_{i=0}^na_i\mu_Y(E_i\cap Y)\le\sum_{i=0}^na_i\mu E_i=\int fd\mu$}

\noindent because $a_i\ge 0$ whenever $E_i\ne\emptyset$, so that
$a_i\mu_Y(E_i\cap Y)\le a_i\mu E_i$ for every $i$.

\medskip

\quad{\bf (ii)} If $f$ is a non-negative $\mu$-integrable function,
there is a non-decreasing sequence $\sequencen{f_n}$ of non-negative
$\mu$-simple functions converging to $f\,\,\mu$-almost everywhere;  now
$\sequencen{f_n\restrp Y}$ is a non-decreasing sequence of $\mu_Y$-simple
functions increasing to $f\restrp Y\,\,\mu_Y$-a.e.\ (by 214Cb), and

\Centerline{$\sup_{n\in\Bbb N}\int(f_n\restrp Y)d\mu_Y
\le\sup_{n\in\Bbb N}\int f_nd\mu
=\int fd\mu<\infty$,}

\noindent so $\int(f\restrp Y)d\mu_Y$ exists and is at most $\int fd\mu$.

\medskip

\quad{\bf (iii)} Finally, if $f$ is any $\mu$-integrable real-valued
function, it is expressible as $f_1-f_2$ where $f_1$ and $f_2$ are
non-negative $\mu$-integrable functions, so that
$f\restrp Y=(f_1\restrp Y)-(f_2\restrp Y)$ is $\mu_Y$-integrable.

\medskip

{\bf (b)} Let us say that if $f$ is a $\mu_Y$-integrable function, then
an  `enveloping extension' of $f$ is a $\mu$-integrable function
$\tilde f$, extending $f$, real-valued on $X\setminus Y$, such that
$\int_F\tilde f=\int_{F\cap Y}f$ for every $F\in\Sigma$.

\medskip

\quad{\bf (i)} If $f$ is of the form $\chi H$, where $H\in\Sigma_Y$ and
$\mu_YH<\infty$, let $E_0\in\Sigma$ be such that $H=Y\cap E_0$ and
$E_1\in\Sigma$ a measurable envelope for $H$ (132Ee);   then
$E=E_0\cap E_1$ is a measurable envelope for $H$ and $H=E\cap Y$.   Set
$\tilde f=\chi E$, regarded as a function from $X$ to $\{0,1\}$.   Then
$\tilde f\restrp Y=f$, and for any $F\in\Sigma$ we have

\Centerline{$\int_F\tilde f=\mu_F(E\cap F)=\mu(E\cap F)
=\mu^*(H\cap F)=\mu_{Y\cap F}(H\cap F)=\int_{Y\cap F}f$.}

\noindent So $\tilde f$ is an enveloping extension of $f$.

\medskip

\quad{\bf (ii)} If $f$, $g$ are $\mu_Y$-integrable functions with
enveloping extensions $\tilde f$, $\tilde g$, and $a$, $b\in\Bbb R$,
then $a\tilde f+b\tilde g$ extends $af+bg$ and

$$\eqalign{\int_Fa\tilde f+b\tilde g
&=a\int_F\tilde f+b\int_F\tilde g\cr
&=a\int_{F\cap Y}f+b\int_{F\cap Y}g
=\int_{F\cap Y}af+bg\cr}$$

\noindent for every $F\in\Sigma$, so $a\tilde f+b\tilde g$ is an
enveloping extension of $af+bg$.

\medskip

\quad{\bf (iii)} Putting (i) and (ii) together, we see that every
$\mu_Y$-simple function $f$ has an enveloping extension.

\medskip

\quad{\bf (iv)} Now suppose that $\sequencen{f_n}$ is a non-decreasing
sequence of non-negative $\mu_Y$-simple functions converging
$\mu_Y$-almost everywhere to a $\mu_Y$-integrable function $f$.   For
each $n\in\Bbb N$ let $\tilde f_n$ be an enveloping extension of $f_n$.
Then $\tilde f_n\leae\tilde f_{n+1}$.   \Prf\ If $F\in\Sigma$ then

\Centerline{$\int_F\tilde f_n=\int_{F\cap Y}f_n
\le\int_{F\cap Y}f_{n+1}=\int_F\tilde f_{n+1}$.}

\noindent So $\tilde f_n\leae\tilde f_{n+1}$, by 131Ha.\ \QeD\   Also

\Centerline{$\lim_{n\to\infty}\int_F\tilde f_n
=\lim_{n\to\infty}\int_{F\cap Y}f_n
=\int_{F\cap Y}f$}

\noindent for every $F\in\Sigma$.   Taking $F=X$ to begin with, B.Levi's
theorem tells us that $h=\lim_{n\to\infty}\tilde f_n$ is defined (as a
real-valued function) $\mu$-almost everywhere;  now letting $F$ vary, we
have $\int_Fh=\int_{F\cap Y}f$ for every $F\in\Sigma$, because
$h\restr F=\lim_{n\to\infty}\tilde f_n\restr F\,\,\mu_F$-a.e.   (I seem
to be using 214Cb here.)   Now $h\restrp Y=f\,\,\mu_Y$-a.e., by 214Cb
again.   If we define $\tilde f$ by setting

\Centerline{$\tilde f(x)=f(x)$ for $x\in\dom f$, $h(x)$ for
$x\in\dom h\setminus\dom f$, $0$ for other $x\in X$,}

\noindent then $\tilde f$ is defined everywhere in $X$ and is equal to
$h\,\,\mu$-almost everywhere;  so that if $F\in\Sigma$,
$\tilde f\restr F$ will be equal to $h\restr F\,\,\mu_F$-almost
everywhere, and

\Centerline{$\int_F\tilde f=\int_Fh=\int_{F\cap Y}f$.}

\noindent As $F$ is arbitrary, $\tilde f$ is an enveloping
extension of $f$.

\medskip

\quad{\bf (v)} Thus every non-negative $\mu_Y$-integrable function has
an enveloping extension.   Using (ii) again, every $\mu_Y$-integrable
function has an enveloping extension, as claimed.
}%end of proof of 214E

\leader{214F}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space, $Y$
a subset of $X$, and $f$ a $[-\infty,\infty]$-valued function such that
$\int_Xf$ is defined in $[-\infty,\infty]$.   If {\it either} $Y$ has
full outer measure in $X$ {\it or} $f$ is zero almost everywhere in
$X\setminus Y$, then $\int_Yf$ is defined and equal to $\int_Xf$.

\proof{{\bf (a)} Suppose first that $f$ is non-negative,
$\Sigma$-measurable and defined everywhere in $X$.   In this case
$f\restrp Y$ is $\Sigma_Y$-measurable.   Set
$F_{nk}=\{x:x\in X,\,f(x)\ge 2^{-n}k\}$ for $k$, $n\in\Bbb N$,
$f_n=\sum_{k=1}^{4^n}2^{-n}\chi F_{nk}$ for $n\in\Bbb N$, so that
$\sequencen{f_n}$ is a non-decreasing sequence of real-valued measurable
functions converging everywhere to $f$, and
$\int_Xf=\lim_{n\to\infty}\int_Xf_n$.   For each $n\in\Bbb N$ and $k\ge 1$,

\Centerline{$\mu_Y(F_{nk}\cap Y)=\mu^*(F_{nk}\cap Y)=\mu F_{nk}$}

\noindent either because $F_{nk}\setminus Y$ is negligible or because
$X$ is a measurable envelope of $Y$.   So

$$\eqalignno{\int_Yf
&=\lim_{n\to\infty}\int_Yf_n
=\lim_{n\to\infty}\sum_{k=1}^{4^n}2^{-n}\mu_Y(F_{nk}\cap Y)\cr
&=\lim_{n\to\infty}\sum_{k=1}^{4^n}2^{-n}\mu F_{nk}
=\lim_{n\to\infty}\int_Xf_n
=\int_Xf.\cr}$$

\medskip

{\bf (b)} Now suppose that $f$ is non-negative, defined almost
everywhere in $X$ and $\mu$-virtually
measurable.   In this case there is a conegligible measurable set
$E\subseteq\dom f$ such that $f\restr E$ is measurable.   Set
$\tilde f(x)=f(x)$ for $x\in E$, $0$ for $x\in X\setminus E$;  then
$\tilde f$
satisfies the conditions of (a) and $f=\tilde f\,\,\mu$-a.e.
Accordingly $f\restrp Y=\tilde f\restrp Y\,\,\mu_Y$-a.e.\ (214Cc), and

\Centerline{$\int_Yf=\int_Y\tilde f=\int_X\tilde f=\int_Xf$.}

\medskip

{\bf (c)} Finally, for the general case, we can apply (b) to the
positive and negative parts $f^+$, $f^-$ of $f$ to get

\Centerline{$\int_Yf=\int_Yf^+-\int_Yf^-=\int_Xf^+-\int_Xf^-=\int_Xf$.}
}%end of proof of 214F

\leader{214G}{Corollary} Let $(X,\Sigma,\mu)$ be a measure space, $Y$ a
subset of $X$, and $E\in\Sigma$ a measurable envelope of $Y$.   If $f$
is a $[-\infty,\infty]$-valued function such that $\int_Ef$ is defined
in $[-\infty,\infty]$, then $\int_Yf$ is defined and equal to $\int_Ef$.

\proof{ By 214Ce, we can identify the subspace measure $\mu_Y$ with the
subspace measure $(\mu_E)_Y$ induced by the subspace measure on $E$.
Now, regarded as a subspace of $E$, $Y$ has full outer measure, so
214F gives the result.
}%end of proof of 214G

\leader{214H}{Subspaces and \Caratheodory's
\dvrocolon{method}}\cmmnt{ The following easy technical results will
occasionally be useful.

\medskip

\noindent}{\bf Lemma} Let $X$ be a set, $Y\subseteq X$ a subset, and
$\theta$ an outer measure on $X$.

(a) $\theta_Y=\theta\restrp\Cal PY$ is an outer measure on $Y$.

(b) Let $\mu$, $\nu$ be the measures on $X$, $Y$ defined by
\Caratheodory's method from the outer measures $\theta$, $\theta_Y$, and
$\Sigma$, $\Tau$ their domains;  let $\mu_Y$ be the subspace measure on
$Y$ induced by $\mu$, and $\Sigma_Y$ its domain.   Then

\quad (i) $\Sigma_Y\subseteq\Tau$ and $\nu F\le\mu_YF$ for every
$F\in\Sigma_Y$;

\quad(ii) if $Y\in\Sigma$ then $\nu=\mu_Y$;

\quad(iii) if $\theta=\mu^*$\cmmnt{ (that is, $\theta$ is
`regular')} then $\nu$ extends $\mu_Y$;

\quad(iv) if $\theta=\mu^*$ and $\theta Y<\infty$ then $\nu=\mu_Y$.

\proof{{\bf (a)} You have only to read the definition of `outer
measure' (113A).

\medskip

{\bf (b)(i)} Suppose that $F\in\Sigma_Y$.   Then it is of the form
$E\cap Y$ where $E\in\Sigma$.   If $A\subseteq Y$, then

\Centerline{$\theta_Y(A\cap F)+\theta_Y(A\setminus F)
=\theta(A\cap F)+\theta(A\setminus F)
=\theta(A\cap E)+\theta(A\setminus E)
=\theta A=\theta_YA$,}

\noindent so $F\in\Tau$.   Now

\Centerline{$\nu F=\theta_YF=\theta F\le\mu^*F=\mu_YF$.}

\medskip

\quad{\bf (ii)} Suppose that $F\in\Tau$.   If $A\subseteq X$, then

$$\eqalign{\theta A
&=\theta(A\cap Y)+\theta(A\setminus Y)
=\theta_Y(A\cap Y)+\theta(A\setminus Y)\cr
&=\theta_Y(A\cap Y\cap F)+\theta_Y(A\cap Y\setminus F)
  +\theta(A\setminus Y)\cr
&=\theta(A\cap F)+\theta(A\cap Y\setminus F)+\theta(A\setminus Y)\cr
&=\theta(A\cap F)+\theta((A\setminus F)\cap Y)
  +\theta((A\setminus F)\setminus Y)
=\theta(A\cap F)+\theta(A\setminus F);\cr}$$

\noindent as $A$ is arbitrary, $F\in\Sigma$ and therefore
$F\in\Sigma_Y$.
Also

\Centerline{$\mu_YF=\mu F=\theta F=\theta_YF=\nu F$.}
\noindent Putting this together with (i), we see that $\mu_Y$ and $\nu$
are identical.

\medskip

\quad{\bf (iii)} Let $F\in\Sigma_Y$.   Then $F\in\Tau$, by (i).
Now $\nu F=\theta F=\mu^*F=\mu_YF$.   As $F$ is arbitrary,
$\nu$ extends $\mu_Y$.

\medskip

\quad{\bf (iv)} Now suppose that $F\in\Tau$.   Because
$\mu^*Y=\theta Y<\infty$, we have measurable envelopes $E_1$, $E_2$ of
$F$ and $Y\setminus F$ for $\mu$ (132Ee).   Then

$$\eqalign{\theta Y
&=\theta_YY
=\theta_YF+\theta_Y(Y\setminus F)
=\theta F+\theta(Y\setminus F)\cr
&=\mu^*F+\mu^*(Y\setminus F)
=\mu E_1+\mu E_2\ge\mu(E_1\cup E_2)
=\theta(E_1\cup E_2)\ge\theta Y,\cr}$$

\noindent so $\mu E_1+\mu E_2=\mu(E_1\cup E_2)$ and

\Centerline{$\mu(E_1\cap E_2)=\mu E_1+\mu E_2-\mu(E_1\cup E_2)=0$.}

\noindent   As $\mu$ is complete (212A) and
$E_1\cap Y\setminus F\subseteq E_1\cap E_2$ is $\mu$-negligible,
therefore belongs to
$\Sigma$, $F=Y\cap(E_1\setminus(E_1\cap Y\setminus F))$ belongs to
$\Sigma_Y$.   Thus $\Tau\subseteq\Sigma_Y$;  putting this together
with (iii), we see that $\nu=\mu_Y$.
}%end of proof of 214H

\leader{214I}{}\cmmnt{ I now turn to the relationships between
subspace measures and the classification of measure spaces developed in
this chapter.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ be a measure
space and $Y$ a subset of $X$.   Let $\mu_Y$ be the subspace measure on
$Y$ and $\Sigma_Y$ its domain.

(a) If $(X,\Sigma,\mu)$ is complete, or totally finite, or
$\sigma$-finite, or strictly localizable, so is $(Y,\Sigma_Y,\mu_Y)$.
If $\langle X_i\rangle_{i\in I}$ is a decomposition of $X$ for $\mu$,
then $\langle X_i\cap Y\rangle_{i\in I}$ is a decomposition of $Y$ for
$\mu_Y$.

(b)
Writing $\hat\mu$ for the completion of $\mu$, the subspace measure
$\hat\mu_Y=(\hat\mu)_Y$ is the completion of $\mu_Y$.

(c) If $(X,\Sigma,\mu)$ has locally determined negligible sets,
then $\mu_Y$ is semi-finite.

(d) If $(X,\Sigma,\mu)$ is complete and locally determined, then
$(Y,\Sigma_Y,\mu_Y)$ is complete and semi-finite.

(e) If $(X,\Sigma,\mu)$ is complete, locally determined and localizable
then so is $(Y,\Sigma_Y,\mu_Y)$.

\proof{{\bf (a)(i)} Suppose that $(X,\Sigma,\mu)$ is complete.
If $A\subseteq U\in\Sigma_Y$ and $\mu_YU=0$, there is an $E\in\Sigma$
such that $U=E\cap Y$ and $\mu E=\mu_YU=0$;  now $A\subseteq E$ so
$A\in\Sigma$ and $A=A\cap Y\in\Sigma_Y$.

\medskip

\quad{\bf (ii)}
$\mu_YY=\mu^*Y\le\mu X$, so $\mu_Y$ is totally finite if $\mu$ is.

\medskip

\quad{\bf (iii)} If $\sequencen{X_n}$ is a sequence of sets of finite
measure for $\mu$ which covers $X$, then $\sequencen{X_n\cap Y}$ is a
sequence of sets of finite measure for $\mu_Y$ which covers $Y$.   So
$(Y,\Sigma_Y,\mu_Y)$ is $\sigma$-finite if $(X,\Sigma,\mu)$ is.

\medskip

\quad{\bf (iv)} Suppose that $\langle X_i\rangle_{i\in I}$ is a
decomposition of $X$ for $\mu$.   Then $\langle X_i\cap
Y\rangle_{i\in I}$ is a decomposition of $Y$ for $\mu_Y$.   \Prf\
Because $\mu_Y(X_i\cap Y)\le\mu X_i<\infty$ for each $i$,
$\langle X_i\cap Y\rangle_{i\in I}$ is a partition of $Y$ into sets of finite measure.   Suppose that $U\subseteq Y$ is such that
$U_i=U\cap X_i\cap Y\in\Sigma_Y$ for every $i$.   For each $i\in I$, choose $E_i\in\Sigma$
such that $U_i=E_i\cap Y$ and $\mu E_i=\mu_YU_i$;  we may of course
suppose that $E_i\subseteq X_i$.   Set $E=\bigcup_{i\in I}E_i$.   Then
$E\cap X_i=E_i\in\Sigma$ for every $i$, so $E\in\Sigma$ and
$\mu E=\sum_{i\in I}\mu E_i$.   Now $U=E\cap Y$ so $U\in\Sigma_Y$ and

\Centerline{$\mu_YU\le\mu E=\sum_{i\in I}\mu E_i
=\sum_{i\in I}\mu_YU_i$.}

\noindent On the other hand, $\mu_YU$ is surely greater than or equal to
$\sum_{i\in I}\mu_YU_i
=\sup_{J\subseteq I\text{ is finite}}\sum_{i\in J}\mu_YU_i$, so they are
equal.   As $U$ is arbitrary,
$\langle X_i\cap Y\rangle_{i\in I}$ is a decomposition of $Y$ for
$\mu_Y$.\ \Qed

Consequently $(Y,\Sigma_Y,\mu_Y)$ is strictly localizable if
$(X,\Sigma,\mu)$ is.

\medskip

{\bf (b)} The domain of the completion $(\mu_Y)\sphat\mskip5mu$ is

$$\eqalignno{\hat\Sigma_Y
&=\{F\symmdiff A:F\in\Sigma_Y,\,A\subseteq Y
   \text{ is }\mu_Y\text{-negligible}\}\cr
&=\{(E\cap Y)\symmdiff(A\cap Y):E\in\Sigma,\,
   A\subseteq X\text{ is }\mu\text{-negligible}\}\cr
\displaycause{214Cb}
&=\{(E\symmdiff A)\cap Y:E\in\Sigma,\,
   A\text{ is }\mu\text{-negligible}\}
=\dom\hat\mu_Y.\cr}$$

\noindent If $H\in\hat\Sigma_Y$ then

\Centerline{$(\mu_Y)\sphat\mskip3mu(H)
=\mu_Y^*H=\mu^*H=(\hat\mu)^*H=\hat\mu_YH$,}

\noindent using 214Cd for the second step, and 212Ea for the third.

\medskip

{\bf (c)} Take $U\in\Sigma_Y$ such
that $\mu_YU>0$.   Then there is an $E\in\Sigma$ such that
$\mu E<\infty$
and $\mu^*(E\cap U)>0$.   \Prf\Quer\ Otherwise, $E\cap U$ is
$\mu$-negligible whenever $\mu E<\infty$;  because $\mu$ has locally determined negligible sets, $U$ is $\mu$-negligible and $\mu_YU=\mu^*U=0$.\ \Bang\QeD   Now
$E\cap U\in\Sigma_Y$ and

\Centerline{$0<\mu^*(E\cap U)=\mu_Y(E\cap U)\le\mu E<\infty$.}

\medskip

{\bf (d)} By (a), $\mu_Y$ is complete;  by 213J and (c) here, it is
semi-finite.

\medskip

{\bf (e)} By (d), $\mu_Y$ is complete and semi-finite.
To see that it is locally
determined, take any $U\subseteq Y$ such that $U\cap V\in\Sigma_Y$
whenever $V\in\Sigma_Y$ and $\mu_YV<\infty$.   By 213J and 213L,
there is a
measurable envelope $E$ of $U$ for $\mu$;  of course
$E\cap Y\in\Sigma_Y$.

I claim that $\mu(E\cap Y\setminus U)=0$.   \Prf\ Take any $F\in\Sigma$
with $\mu F<\infty$.   Then $F\cap U\in\Sigma_Y$, so
\Centerline{$\mu_Y(F\cap E\cap Y)\le\mu(F\cap E)=\mu^*(F\cap
U)=\mu_Y(F\cap U)\le\mu_Y(F\cap E\cap Y)$;}

\noindent thus $\mu_Y(F\cap E\cap Y)=\mu_Y(F\cap U)$ and

\Centerline{$\mu^*(F\cap E\cap Y\setminus U)
=\mu_Y(F\cap E\cap Y\setminus U)=0$.}

\noindent Because $\mu$ is complete, $\mu(F\cap E\cap Y\setminus U)=0$;
because $\mu$ is locally determined and $F$ is arbitrary,
$\mu(E\cap Y\setminus U)=0$.\ \QeD\  But this means that
$E\cap Y\setminus U\in\Sigma_Y$ and $U\in\Sigma_Y$.   As $U$ is
arbitrary, $\mu_Y$ is locally determined.

To see that $\mu_Y$ is localizable,
let $\Cal U$ be any family in $\Sigma_Y$.   Set

\Centerline{$\Cal E=\{E:E\in\Sigma,\,\mu E<\infty,\,
  \mu E=\mu^*(E\cap U)$ for some $U\in\Cal U\}$,}

\noindent and let $G\in\Sigma$ be an essential supremum for $\Cal E$ in
$\Sigma$.   I claim that $G\cap Y$ is an essential supremum for $\Cal U$
in $\Sigma_Y$.   \Prf\ (i) \Quer\ If $U\in\Cal U$ and
$U\setminus(G\cap Y)$ is not negligible, then (because $\mu_Y$ is
semi-finite) there is a
$V\in\Sigma_Y$ such that $V\subseteq U\setminus G$ and
$0<\mu_YV<\infty$.   Now there is an $E\in\Sigma$ such that
$V\subseteq E$ and $\mu E=\mu^*V$.
We have $\mu^*(E\cap U)\ge\mu^*V=\mu E$, so $E\in\Cal E$ and
$E\setminus G$
must be negligible;  but $V\subseteq E\setminus G$ is not negligible.\
\BanG\   Thus $U\setminus(G\cap Y)$ is negligible for every
$U\in\Cal U$.
(ii)  If $W\in\Sigma_Y$ is such that $U\setminus W$ is negligible for
every $U\in\Cal U$, express $W$ as $H\cap Y$ where $H\in\Sigma$.   If
$E\in\Cal E$, there is a $U\in\Cal U$ such that $\mu E=\mu^*(E\cap U)$;
now $\mu^*(E\cap U\setminus W)=0$, so
$\mu E=\mu^*(E\cap U\cap W)\le\mu(E\cap H)$ and $E\setminus H$ is
negligible.   As $E$ is
arbitrary, $H$ is an essential upper bound for $\Cal E$ and
$G\setminus H$ is negligible;  but this means that $G\cap Y\setminus W$
is negligible.   As $W$ is arbitrary, $G\cap Y$ is an essential supremum
for $\Cal U$.\ \Qed

As $\Cal U$ is arbitrary, $\mu_Y$ is localizable.
}%end of proof of 214I

\leader{214J}{Upper and
lower \dvrocolon{integrals}}\cmmnt{ The
following elementary facts are sometimes useful.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space,
$A$ a subset of $X$ and $f$ a real-valued function defined almost
everywhere in $X$.   Then

(a)  if {\it either} $f$ is non-negative {\it or}
$A$ has full outer measure in $X$,
$\overline{\intop}(f\restr A)d\mu_A\le\overline{\intop}fd\mu$;

(b) if $A$ has full outer measure in $X$,
$\underline{\int}fd\mu\le\underline{\int}(f\restr A)d\mu_A$.

\proof{{\bf (a)(i)} Suppose that $f$ is non-negative.
If $\overline{\intop}fd\mu=\infty$, the result is trivial.
Otherwise, there is a $\mu$-integrable function $g$ such that
$f\le g\,\,\mu$-a.e.\ and $\overline{\intop}fd\mu=\int g\,d\mu$, by
133J(a-i).   Now $f\restr A\le g\restr A\,\,\mu_A$-a.e., by 214Cb, and
$\int(g\restr A)\,d\mu_A$
is defined and less than or equal to $\int g\,d\mu$, by 214Ea;  so

\Centerline{$\overlineint(f\restr A)d\mu_A\le\int(g\restr A)d\mu_A
\le\int g\,d\mu=\overlineint fd\mu$.}

\medskip

\quad{\bf (ii)} Now suppose that $A$ has full outer measure in $X$.
If $g$ is such that $f\le g\,\,\mu$-a.e.\ and $\int g\,d\mu$ is defined in
$[-\infty,\infty]$, then $f\restr A\le g\restr A\,\,\mu_A$-a.e.\ and
$\int(g\restr A)d\mu_A=\int g\,d\mu$, by 214F.   So
$\overline{\int}(f\restr A)d\mu_A\le\int g\,d\mu$.   As $g$ is arbitrary,
$\overline{\int}(f\restr A)d\mu_A\le\overline{\int}f\,d\mu$.

\medskip

{\bf (b)} Apply (a) to $-f$, and use 133J(b-iv).
}%end of proof of 214J

\leader{214K}{Measurable subspaces:  Proposition}
Let $(X,\Sigma,\mu)$ be a measure space.

(a) Let $E\in\Sigma$ and let $\mu_E$ be the subspace measure, with
$\Sigma_E$ its domain.   If $(X,\Sigma,\mu)$ is complete, or totally
finite, or $\sigma$-finite, or strictly localizable, or semi-finite, or
localizable, or locally determined, or atomless, or purely atomic, so is
$(E,\Sigma_E,\mu_E)$.

(b) Suppose that $\langle X_i\rangle_{i\in I}$ is a partition of
$X$ into measurable sets\cmmnt{ (not necessarily of finite measure)} such that

\Centerline{$\Sigma=\{E:E\subseteq X,\,
E\cap  X_i\in\Sigma$ for every $i\in I\}$,}

\Centerline{$\mu E=\sum_{i\in I}\mu(E\cap X_i)$ for every $E\in\Sigma$.}
\noindent Then $(X,\Sigma,\mu)$ is complete, or strictly localizable, or
semi-finite, or localizable, or locally determined, or atomless, or
purely atomic, iff $(X_i,\Sigma_{X_i},\mu_{X_i})$ has that property for
every
$i\in I$.

\proof{ I really think that if you have read attentively up
to this point, you ought to find this easy.   If you are in any doubt,
this makes a very suitable set of sixteen exercises to do.
}%end of proof of 214K

\leader{214L}{Direct sums} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be any indexed family of
measure spaces.   Set
$X=\bigcup_{i\in I}(X_i\times\{i\})$;  for $E\subseteq X$, $i\in I$ set
$E_i=\{x:(x,i)\in E\}$.   Write

\Centerline{$\Sigma
=\{E:E\subseteq X,\,E_i\in\Sigma_i$ for every $i \in I\}$,}

\Centerline{$\mu E=\sum_{i\in I}\mu_iE_i$ for every $E\in\Sigma$.}

\noindent Then it is easy to check that $(X,\Sigma,\mu)$ is a measure
space;  I will call it the {\bf direct sum} of the family
$\familyiI{(X_i,\Sigma_i,\mu_i)}$.   Note that if $(X,\Sigma,\mu)$ is
any strictly localizable measure space, with decomposition
$\familyiI{X_i}$,
then we have a natural isomorphism between $(X,\Sigma,\mu)$ and the
direct sum
$(X',\Sigma',\mu')=\bigoplus_{i\in I}(X_i,\Sigma_{X_i},\mu_{X_i})$ of
the subspace measures, if we match $(x,i)\in X'$ with $x\in X$ for every
$i\in I$ and $x\in X_i$.

\cmmnt{For some of the elementary properties (to put
it plainly, I know of no properties which are not elementary) of direct
sums, see 214M and 214Xh-214Xk.}
% 214Xh, 214Xi, 214Xj, 214Xk

\leader{214M}{Proposition}
Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a
family of measure spaces, with direct sum $(X,\Sigma,\mu)$.   Let $f$ be
a real-valued function defined on a subset of $X$.   For each $i\in I$,
set $f_i(x)=f(x,i)$ whenever $(x,i)\in\dom f$.

(a) $f$ is measurable iff $f_i$ is measurable for every $i\in I$.

(b) If $f$ is non-negative, then
$\int fd\mu=\sum_{i\in I}\int f_id\mu_i$ if either is defined in
$[0,\infty]$.

\proof{{\bf (a)} For $a\in\Bbb R$, set
$F_a=\{(x,i):(x,i)\in\dom f,\,f(x,i)\ge a\}$.   (i) If $f$ is
measurable, $i\in I$ and $a\in\Bbb R$, then there is an $E\in\Sigma$
such that $F_a=E\cap\dom f$;  now

\Centerline{$\{x:f_i(x)\ge a\}
=\dom f_i\cap\{x:(x,i)\in E\}$}

\noindent belongs to the subspace $\sigma$-algebra on $\dom f_i$ induced
by $\Sigma_i$.   As $a$ is arbitrary, $f_i$ is measurable.   (ii) If
every $f_i$ is measurable and $a\in\Bbb R$, then for each $i\in I$ there
is an $E_i\in\Sigma_i$ such that $\{x:(x,i)\in F_a\}=E_i\cap\dom f$;
setting $E=\{(x,i):i\in I,\,x\in E_i\}$, $F_a=\dom f\cap E$ belongs to
the subspace $\sigma$-algebra on $\dom f$.   As $a$ is arbitrary, $f$ is
measurable.

\medskip

{\bf (b)(i)} Suppose first that $f$ is measurable and defined
everywhere.   Set $F_{nk}=\{(x,i):(x,i)\in X,\,f(x,i)\ge 2^{-n}k\}$ for
$k$, $n\in\Bbb N$, $g_n=\sum_{k=1}^{4^n}2^{-n}\chi F_{nk}$ for
$n\in\Bbb N$, $F_{nki}=\{x:(x,i)\in F_{nk}\}$ for $k$, $n\in\Bbb N$ and
$i\in I$, $g_{ni}(x)=g_n(x,i)$ for $i\in I$, $x\in X_i$.   Then

$$\eqalign{\int fd\mu
&=\lim_{n\to\infty}\int g_nd\mu
=\sup_{n\in\Bbb N}\sum_{k=1}^{4^n}2^{-n}\mu F_{nk}\cr
&=\sup_{n\in\Bbb N}\sum_{k=1}^{4^n}\sum_{i\in I}2^{-n}\mu F_{nki}
=\sum_{i\in I}\sup_{n\in\Bbb N}\sum_{k=1}^{4^n}2^{-n}\mu F_{nki}\cr
&=\sum_{i\in I}\sup_{n\in\Bbb N}\int g_{ni}d\mu_i
=\sum_{i\in I}\int f_id\mu_i.\cr}$$

\medskip

\quad{\bf (ii)} Generally, if $\int fd\mu$ is defined, there are a
measurable $g:X\to\coint{0,\infty}$ and a conegligible measurable set
$E\subseteq\dom f$ such that $g=f$ on $E$.   Now
$E_i=\{x:(x,i)\in X_i\}$ belongs to $\Sigma_i$ for each $i$, and
$\sum_{i\in I}\mu_i(X_i\setminus E_i)=\mu(X\setminus E)=0$, so $E_i$ is
$\mu_i$-conegligible for every $i$.   Setting $g_i(x)=g(x,i)$ for
$x\in X_i$, (i) tells us that

\Centerline{$\sum_{i\in I}\int f_id\mu_i
=\sum_{i\in I}\int g_id\mu_i=\int g\,d\mu=\int fd\mu$.}

\medskip

\quad{\bf (iii)} On the other hand, if $\int f_id\mu_i$ is defined for
every $i\in I$, then for each $i\in I$ we can find a measurable function
$g_i:X_i\to\coint{0,\infty}$ and a $\mu_i$-conegligible measurable set
$E_i\subseteq\dom f_i$ such that $g_i=f_i$ on $E_i$.   Setting
$g(x,i)=g_i(x)$ for $i\in I$, $x\in X_i$, (a) tells us that $g$ is
measurable, while $g=f$ on $\{(x,i):i\in I,\,x\in E_i\}$, which is
conegligible (by the calculation in (ii) just above);  so

\Centerline{$\int fd\mu=\int g\,d\mu
=\sum_{i\in I}\int g_id\mu_i=\sum_{i\in I}\int f_id\mu_i$,}

\noindent again using (i) for the middle step.
}%end of proof of 214M

\leader{214N}{Corollary}
Let $(X,\Sigma,\mu)$ be a measure space with a
decomposition $\familyiI{X_i}$.   If $f$ is a real-valued function
defined on a subset of $X$, then

(a) $f$ is measurable iff $f\restr X_i$ is measurable for every
$i\in I$,

(b) if $f\ge 0$, then $\int f=\sum_{i\in I}\int_{X_i}f$ if either is
defined in $[0,\infty]$.

\proof{ Apply 214M to the direct sum of
$\familyiI{(X_i,\Sigma_{X_i},\mu_{X_i})}$, identified with
$(X,\Sigma,\mu)$ as in 214L.
}%end of proof of 214N

\leader{*214O}{}\cmmnt{ I make space here for a general theorem
which puts rather heavy demands on the reader.   So I ought to say that I
advise skipping it on first reading.   It
will not be quoted in this volume, in
the full form here I do not expect to use it anywhere in this treatise,
only the special case of 214Xm is at all often applied, and the proof
depends on a concept (`ideal of sets') and a technique
(`transfinite induction', part (d) of the proof of 214P) which are
used nowhere else in this volume.   However,
`extension of measures' is one of the central themes of Volume 4, and this
result may help to make sense of some of the patterns which will
appear there.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a measure space, and $\Cal I$
an ideal of subsets of $X$, that is, a family of subsets of $X$ such that
$\emptyset\in\Cal I$, $I\cup J\in\Cal I$ for all $I$, $J\in\Cal I$, and
$I\in\Cal I$ whenever $I\subseteq J\in\Cal I$.   Then there is a measure
$\lambda$ on $X$ such that $\Sigma\cup\Cal I\subseteq\dom\lambda$,
$\mu E=\lambda E+\sup_{I\in\Cal I}\mu^*(E\cap I)$ for every $E\in\Sigma$,
and $\lambda I=0$ for every $I\in\Cal I$.

\proof{{\bf (a)} Let $\Lambda$ be the set of those $F\subseteq X$ such that
there are $E\in\Sigma$ and a countable $\Cal J\subseteq\Cal I$ such that
$E\symmdiff F\subseteq\bigcup\Cal J$.   Then $\Lambda$ is a $\sigma$-algebra
of subsets of $X$ including $\Sigma\cup\Cal I$.   \Prf\
$\Sigma\subseteq\Lambda$ because
$E\symmdiff E\subseteq\bigcup\emptyset$ for every $E\in\Sigma$.   
$\Cal I\subseteq\Lambda$ because $\emptyset\symmdiff I\subseteq\bigcup\{I\}$ for
every $I\in\Cal I$.   In particular, $\emptyset\in\Lambda$.   If $F\in\Lambda$,
let $E\in\Sigma$ and $\Cal J\subseteq\Cal I$ be such that $\Cal J$ is
countable and $F\symmdiff E\subseteq\bigcup\Cal J$;  then
$(X\setminus F)\symmdiff(X\setminus E)\subseteq\bigcup\Cal J$ so
$X\setminus F\in\Lambda$.   If $\sequencen{F_n}$ is a sequence in $\Lambda$ with
union $F$, then for each $n\in\Bbb N$ choose $E_n\in\Sigma$,
$\Cal J_n\subseteq\Cal I$ such that $\Cal J_n$ is countable and
$E_n\symmdiff F_n\subseteq\bigcup\Cal J_n$;  then
$E=\bigcup_{n\in\Bbb N}E_n$ belongs to $\Sigma$,
$\Cal J=\bigcup_{n\in\Bbb N}\Cal J_n$ is a countable subset of $\Cal I$ and
$E\symmdiff F\subseteq\bigcup\Cal J$, so $F\in\Lambda$.   Thus $\Lambda$ is a
$\sigma$-algebra.\ \Qed

\medskip

{\bf (b)} For $F\in\Lambda$ set

\Centerline{$\lambda F=\sup\{\mu E:E\in\Sigma$,
$E\subseteq F$, $\mu^*(E\cap I)=0$ for every $I\in\Cal I\}$.}

\noindent Then $\lambda$ is a measure.   \Prf\ The only subset of
$\emptyset$ is $\emptyset$, so $\lambda\emptyset=0$.   Let
$\sequencen{F_n}$ be a disjoint sequence in $\Lambda$ with union $F$ and
set $u=\sum_{n=0}^{\infty}\lambda F_n$.
(i) For each $n\in\Bbb N$ take $E_n\in\Sigma$ and a countable
$\Cal J_n\subseteq\Cal I$ such that $E_n\symmdiff F_n\subseteq\bigcup\Cal J_n$.
Set $\Cal J=\bigcup_{n\in\Bbb N}\Cal J_n$.   If $E\in\Sigma$, $E\subseteq F$ and
$\mu^*(E\cap I)=0$ for every $I\in\Cal I$, then $\mu^*(E\cap\bigcup\Cal J)=0$;
let $H\in\Sigma$ be a $\mu$-negligible set including $E\cap\bigcup\Cal J$.
If $n\in\Bbb N$, then $(E\cap E_n)\symmdiff(E\cap F_n)\subseteq H$ so
$E\cap F_n\setminus H=E\cap E_n\setminus H$ belongs to $\Sigma$, while
$\mu^*((E\cap F_n\setminus H)\cap I)=0$ for every $I\in\Cal I$.   Now

\Centerline{$\mu E
=\mu(E\cap F\setminus H)
=\sum_{n=0}^{\infty}\mu(E\cap F_n\setminus H)
\le\sum_{n=0}^{\infty}\lambda F_n
=u$.}

\noindent As $E$ is arbitrary, $\lambda F\le u$.
(ii) Take any $\gamma<u$.   For $n\in\Bbb N$, set
$\gamma_n=\lambda F_n-2^{-n-1}\min(1,u-\gamma)$ if $\lambda F_n$ is finite,
$\gamma$ otherwise.   For each $n$, we
can find an $E_n\in\Sigma$ such that $E_n\subseteq F_n$,
$\mu^*(E_n\cap I)=0$
for every $I\in\Cal I$, and $\mu E_n\ge\gamma_n$.   Set
$E=\bigcup_{n\in\Bbb N}E_n$;  then $E\subseteq F$ and
$E\cap I=\bigcup_{n\in\Bbb N}E_n\cap I$ is $\mu$-negligible for every
$I\in\Cal I$, so $\lambda F\ge\mu E=\sum_{n=0}^{\infty}\mu E_n\ge\gamma$.
As $\gamma$ is arbitrary, $\lambda F\ge u$.   (iii) As
$\sequencen{F_n}$ is arbitrary, $\lambda$ is a measure.\ \Qed

\medskip

{\bf (c)} Now take any $E\in\Sigma$ and set
$u=\sup_{I\in\Cal I}\mu^*(E\cap I)$.   If $u=\infty$ then
we certainly have
$\mu E=\infty=\lambda E+u$.
Otherwise, let $\sequencen{I_n}$ be a sequence in $\Cal I$ such that
$\lim_{n\to\infty}\mu^*(E\cap I_n)=u$;  replacing $I_n$ by
$\bigcup_{m\le n}I_m$ for each $n$ if necessary, we may suppose that
$\sequencen{I_n}$ is non-decreasing.   Set
$A=E\cap\bigcup_{n\in\Bbb N}I_n$;   because $E\cap I_n$ has finite outer
measure for each $n$, $A$ can be covered by a sequence of sets of finite
measure, and has a measurable envelope $H$ for $\mu$ included in
$E$ (132Ee).   Observe that

\Centerline{$\mu H=\mu^*A=\sup_{n\in\Bbb N}\mu^*(E\cap I_n)=u$}

\noindent by 132Ae.

Set $G=E\setminus H$.   Then $\mu^*(G\cap I)=0$ for every $I\in\Cal I$.
\Prf\ For any $n\in\Bbb N$ there is an $F\in\Sigma$
such that $F\supseteq E\cap(I_n\cup I)$ and $\mu F\le u$;  in which case

\Centerline{$\mu^*(G\cap I)+\mu^*(E\cap I_n)
\le\mu(F\setminus H)+\mu(F\cap H)\le u$.}

\noindent As $n$ is arbitrary, $\mu^*(G\cap I)=0$.\ \QeD\   Accordingly

\Centerline{$u+\lambda E\ge\mu H+\mu G=\mu E$.}

\noindent On the other hand, if $F\in\Sigma$ is such that $F\subseteq E$
and $\mu^*(F\cap I)=0$ for every $I\in\Cal I$, then

\Centerline{$\mu^*(E\cap I_n)
\le\mu(E\setminus F)+\mu^*(F\cap I_n)=\mu(E\setminus F)$}

\noindent for every $n$, so

\Centerline{$u+\mu F\le\mu(E\setminus F)+\mu F=\mu E$;}

\noindent as $F$ is arbitrary, $u+\lambda E\le\mu E$.

\medskip

{\bf (d)} If $J\in\Cal I$, $F\in\Sigma$, $F\subseteq J$ and
$\mu^*(F\cap I)=0$ for every $I\in\Cal I$, then $F\cap J=F$ is
$\mu$-negligible;  as $F$ is arbitrary, $\lambda J=0$.
Thus $\lambda$ has all the required properties.
}%end of proof of 214O

\leader{*214P}{Theorem}
Let $(X,\Sigma,\mu)$ be a measure space, and
$\Cal A$ a family of subsets of $X$ which is well-ordered by the relation
$\subseteq$.   Then there is an extension of $\mu$ to a measure $\lambda$
on $X$ such that $\lambda(E\cap A)$ is defined and equal to
$\mu^*(E\cap A)$ whenever $E\in\Sigma$ and $A\in\Cal A$.

\proof{{\bf (a)} Adding $\emptyset$ and $X$ to $\Cal A$ if necessary, we
may suppose that
$\Cal A$ has $\emptyset$ as its least member and $X$ as its greatest
member.   By 2A1Dg, $\Cal A$ is isomorphic, as
ordered set, to some ordinal;  since $\Cal A$ has a greatest member, this
ordinal is a successor, expressible as $\zeta+1$;  let
$\xi\mapsto A_{\xi}:\zeta+1\to\Cal A$ be the order-isomorphism, so that
$\langle A_{\xi}\rangle_{\xi\le\zeta}$ is a non-decreasing family of
subsets of $X$, $A_0=\emptyset$ and $A_{\zeta}=X$.

\medskip

{\bf (b)} For each ordinal $\xi\le\zeta$, write $\mu_{\xi}$ for
the subspace measure on $A_{\xi}$, $\Sigma_{\xi}$ for its domain
and $\Cal I_{\xi}$ for $\bigcup_{\eta<\xi}\Cal PA_{\eta}$.
Because $A_{\eta}\cup A_{\eta'}=A_{\max(\eta,\eta')}$ for $\eta$,
$\eta'<\xi$, $\Cal I_{\xi}$ is an ideal of subsets of $A_{\xi}$.
By 214O, we have a measure $\lambda_{\xi}$ on $A_{\xi}$, with domain
$\Lambda_{\xi}\supseteq\Sigma_{\xi}\cup\Cal I_{\xi}$, such that
$\mu_{\xi}E=\lambda_{\xi}E+\sup_{I\in\Cal I_{\xi}}\mu_{\xi}^*(E\cap I)$
for every $E\in\Sigma_{\xi}$ and $\lambda_{\xi}I=0$ for every
$I\in\Cal I_{\xi}$.   Because every member of $\Cal I_{\xi}$ is
included in $A_{\eta}$ for some $\eta<\xi$, we have

\Centerline{$\mu^*(E\cap A_{\xi})
=\lambda_{\xi}(E\cap A_{\xi})+\sup_{\eta<\xi}\mu_{\xi}^*(E\cap A_{\eta})
=\lambda_{\xi}(E\cap A_{\xi})+\sup_{\eta<\xi}\mu^*(E\cap A_{\eta})$}

\noindent (214Cd) for every $E\in\Sigma$.   Also, of course,
$\lambda_{\xi}A_{\eta}=0$ for every $\eta<\xi$.

\medskip

{\bf (c)} Now set

\Centerline{$\Lambda=\{F:F\subseteq X$, $F\cap A_{\xi}\in\Lambda_{\xi}$ for
every $\xi\le\zeta\}$,}

\Centerline{$\lambda F=\sum_{\xi\le\zeta}\lambda_{\xi}(F\cap A_{\xi})$}

\noindent for every $F\in\Lambda$.   Because $\Lambda_{\xi}$ is a
$\sigma$-algebra of subsets of $A_{\xi}$ for each $\xi$, $\Lambda$ is a
$\sigma$-algebra of subsets of $X$;  because every $\lambda_{\xi}$ is a
measure, so is $\lambda$.   If $E\in\Sigma$, then

\Centerline{$E\cap A_{\xi}\in\Sigma_{\xi}\subseteq\Lambda_{\xi}$}

\noindent for each $\xi$, so $E\in\Lambda$.   If $\eta\le\zeta$, then for
each $\xi\le\zeta$ either $\eta<\xi$ and

\Centerline{$A_{\eta}\cap A_{\xi}=A_{\eta}
\in\Cal I_{\xi}\subseteq\Lambda_{\xi}$}

\noindent or $\eta\ge\xi$ and $A_{\eta}\cap A_{\xi}=A_{\xi}$ belongs to
$\Lambda_{\xi}$.   So $A_{\eta}\in\Lambda$ for every $\eta\le\zeta$.

\medskip

{\bf (d)} Finally, $\lambda(E\cap A_{\xi})=\mu^*(E\cap A_{\xi})$ whenever
$E\in\Sigma$ and $\xi\le\zeta$.   \Prf\Quer\ Otherwise, because the ordinal
$\zeta+1$ is well-ordered, there is a least
$\xi$ such that $\lambda(E\cap A_{\xi})\ne\mu^*(E\cap A_{\xi})$.
As $A_0=\emptyset$ we surely have $\lambda(E\cap A_0)=\mu^*(E\cap A_0)$
and $\xi>0$.
Note that if $\eta>\xi$, then $\lambda_{\eta}(E\cap A_{\xi})=0$;  so

\Centerline{$\lambda(E\cap A_{\xi})
=\sum_{\eta\le\xi}\lambda_{\eta}(E\cap A_{\xi}\cap A_{\eta})
=\sum_{\eta\le\xi}\lambda_{\eta}(E\cap A_{\eta})$.}

Now

$$\eqalignno{\mu^*(E\cap A_{\xi})
&=\lambda_{\xi}(E\cap A_{\xi})+\sup_{\xi'<\xi}\mu^*(E\cap A_{\xi'})\cr
\displaycause{(b) above}
&=\lambda_{\xi}(E\cap A_{\xi})
  +\sup_{\xi'<\xi}\sum_{\eta\le\xi'}\lambda_{\eta}(E\cap A_{\eta})\cr
\displaycause{because $\xi$ was the first problematic ordinal}
&=\lambda_{\xi}(E\cap A_{\xi})
  +\sup_{\xi'<\xi}\sup_{K\subseteq\xi'+1\text{ is finite}}
  \sum_{\eta\in K}\lambda_{\eta}(E\cap A_{\eta})\cr
\displaycause{see the definition of `sum' in 112Bd, or 226A below}
&=\lambda_{\xi}(E\cap A_{\xi})
  +\sup_{K\subseteq\xi\text{ is finite}}
  \sum_{\eta\in K}\lambda_{\eta}(E\cap A_{\eta})\cr
&=\sup_{K\subseteq\xi+1\text{ is finite}}
  \sum_{\eta\in K}\lambda_{\eta}(E\cap A_{\eta})
=\sum_{\eta\le\xi}\lambda_{\eta}(E\cap A_{\eta})
\ne\mu^*(E\cap A_{\xi})\cr}$$

\noindent by the choice of $\xi$;  but this is absurd.\ \Bang\Qed

In particular,

\Centerline{$\lambda E=\lambda(E\cap A_{\zeta})=\mu^*(E\cap A_{\zeta})
=\mu E$}

\noindent for every $E\in\Sigma$.
This completes the proof of the theorem.
}%end of proof of 214P

\leader{*214Q}{Proposition}\dvAformerly{2{}14Xd}
Suppose that $(X,\Sigma,\mu)$ is an atomless
measure space and $Y$ a subset of $X$ such that
the subspace measure $\mu_Y$ is semi-finite.   Then $\mu_Y$ is atomless.
%for 548C 548E
%\query what about general $Y$?

\proof{ Let $F\subseteq Y$ be such that $\mu_YF$ is defined and not $0$.
Because $\mu_Y$ is semi-finite, there is an $F'\subseteq F$ such that
$\mu_YF'$ is defined, finite and not zero.   In this case,
$\mu^*F'=\mu_YF'$
is finite, so $F'$ has a measurable envelope $E$ say with respect to $\mu$.
Because $\mu$ is atomless, there is an $E_1\in\Sigma$ such that
$E_1\subseteq E$ and neither $E_1$ nor
$E\setminus E_1$ is
$\mu$-negligible.   Now $F\cap E_1$ is measured by $\mu_Y$ and

\Centerline{$\mu_Y(F\cap E_1)\ge\mu^*(F'\cap E_1)=\mu(E\cap E_1)>0$,}

\Centerline{$\mu_Y(F\setminus E_1)\ge\mu^*(F'\setminus E_1)
=\mu(E\setminus E_1)>0$.}

\noindent As $F$ is arbitrary, $\mu_Y$ is atomless.
}%end of proof of 214Q

\exercises{
\leader{214X}{Basic exercises (a)}
%\spheader 214Xa
Let $(X,\Sigma,\mu)$ be a localizable measure space.   Show
that there is an $E\in\Sigma$ such that the subspace measure $\mu_E$
is purely atomic and $\mu_{X\setminus E}$ is atomless.
%214/

\spheader 214Xb Let $X$ be a set, $\theta$ a regular outer measure on
$X$, and $Y$ a subset of $X$.   Let $\mu$ be the measure on $X$ defined
by \Caratheodory's method from $\theta$, $\mu_Y$ the subspace measure on
$Y$, and $\nu$ the measure on $Y$ defined by \Caratheodory's method from
$\theta\restrp\Cal PY$.   Show that if $\mu_Y$ is locally determined
(in particular, if $\mu$ is locally
determined and localizable) then $\nu=\mu_Y$.
%214I

\spheader 214Xc Let $(X,\Sigma,\mu)$ be a localizable measure space, and
$Y$ a subset of $X$ such that the subspace measure $\mu_Y$ is
semi-finite.   Show that $\mu_Y$ is localizable.
%214I

\sqheader 214Xd Let $(X,\Sigma,\mu)$ be a measure space, and $Y$ a
subset of $X$ such that the subspace measure $\mu_Y$ is semi-finite.
(i) Show that a set $F\subseteq Y$ is an atom for $\mu_Y$ iff it is of the
form $E\cap Y$ where $E$ an atom for $\mu$.
(ii) Show that if $\mu$ is purely atomic, so is $\mu_Y$.
%214I  used in 5x47A

\spheader 214Xe Let $(X,\Sigma,\mu)$ be a localizable measure
space, and $Y$ any subset of $X$.   Show that the c.l.d.\ version of the
subspace measure on $Y$ is localizable.
%214I

\spheader 214Xf Let $(X,\Sigma,\mu)$ be a measure space with locally
determined negligible sets, and $Y$ a
subset of $X$, with its subspace measure $\mu_Y$.
Show that $\mu_Y$ has locally determined negligible sets.
%214I

\sqheader 214Xg Let $(X,\Sigma,\mu)$ be a measure space.   Show that
$(X,\Sigma,\mu)$ has locally determined
negligible sets iff the subspace measure $\mu_Y$ is semi-finite for
every $Y\subseteq X$.
%214I, 214Xf

\sqheader 214Xh Let $\langle (X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a
family
of measure spaces, with direct sum $(X,\Sigma,\mu)$ (214L).   Set
$X'_i=X_i\times\{i\}\subseteq X$ for each $i\in I$.   Show that $X'_i$,
with the subspace measure, is isomorphic to $(X_i,\Sigma_i,\mu_i)$.
Under what circumstances is $\langle X'_i\rangle_{i\in I}$ a
decomposition of $X$?   Show that $\mu$ is complete, or strictly
localizable, or localizable, or locally determined, or semi-finite, or
atomless, %used in 548E
or purely atomic iff every $\mu_i$ is.
Show that a measure space is strictly localizable iff it is isomorphic
to a direct sum of totally finite spaces.
%214L

\sqheader 214Xi Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a
family of measure spaces, and $(X,\Sigma,\mu)$ their direct sum.   Show
that the completion of $(X,\Sigma,\mu)$ can be identified with the
direct sum of the completions of the $(X_i,\Sigma_i,\mu_i)$, and that
the c.l.d.\ version of $(X,\Sigma,\mu)$ can be identified with the
direct sum of the c.l.d.\ versions of the $(X_i,\Sigma_i,\mu_i)$.
%214L

\spheader 214Xj Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of measure
spaces.   Show that their direct sum has
locally determined negligible sets iff every $\mu_i$ has.
%214L, 214Xf

\spheader 214Xk
Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of measure
spaces, and $(X,\Sigma,\mu)$ their direct sum.   Show that
$(X,\Sigma,\mu)$ has the measurable envelope property (213Xl) iff every
$(X_i,\Sigma_i,\mu_i)$ has.
%214L

\spheader 214Xl
Let $(X,\Sigma,\mu)$ be a measure space, $Y$ a subset of
$X$, and $f:X\to[0,\infty]$
a function such that $\int_Yf$ is defined in $[0,\infty]$.
Show that $\int_Yf=\overline{\int}f\times\chi Y\,d\mu$.
%214E

\sqheader 214Xm Write out a direct proof of 214P in the special case in which
$\Cal A=\{A\}$.   ({\it Hint\/}:  for $E$, $F\in\Sigma$,

\Centerline{$\lambda((E\cap A)\cup(F\setminus A))
=\mu^*(E\cap A)+\sup\{\mu G:G\in\Sigma$, $G\subseteq F\setminus A\}$.)}
%214P

\sqheader 214Xn
Let $(X,\Sigma,\mu)$ be a measure space and $\Cal A$ a finite family of
subsets of $X$.   Show that there is a measure on $X$,
extending $\mu$, which measures every member of $\Cal A$.
%214Xm 214P

\sqheader 214Xo\dvAnew{2021}
Let $(X,\Sigma,\mu)$ be a measure space and $\sequencen{X_n}$ a sequence of
subsets of $X$ such that $\bigcup_{n\in\Bbb N}X_n$ has full outer measure on
$X$.   Suppose that for each $n\in\Bbb N$ we have a set $A_n\subseteq X_n$ of
full outer measure for the subspace measure on $X_n$,   Show that
$\bigcup_{n\in\Bbb N}A_n$ has full outer measure in $X$.
%214B out of order query
%for 548H

\leader{214Y}{Further exercises (a)}
%\spheader 214Ya
Let $(X,\Sigma,\mu)$ be a measure space and $A$ a subset of $X$
such that the subspace measure on $A$ is semi-finite.   Set
$\alpha=\sup\{\mu E:E\in\Sigma$, $E\subseteq A\}$.
Show that if $\alpha\le\gamma\le\mu^*A$ then there is a measure
$\lambda$ on $X$, extending $\mu$, such that $\lambda A=\gamma$.
%214P 214Xm

\spheader 214Yb
Let $(X,\Sigma,\mu)$ be a measure space and
$\family{n}{\Bbb Z}{A_n}$ a double-ended sequence of subsets of $X$ such
that $A_m\subseteq A_n$ whenever $m\le n$ in $\Bbb Z$.   Show that there
is a measure on $X$, extending $\mu$, which measures every $A_n$.
\Hint{use 214P twice.}
%214P

\spheader 214Yc
Let $X$ be a set and $\Cal A$ a family of subsets of $X$.
Show that the following are equiveridical:  (i) for every measure $\mu$ on
$X$ there is a measure on $X$ extending $\mu$ and measuring every member of
$\Cal A$;  (ii) for every totally finite measure $\mu$ on
$X$ there is a measure on $X$ extending $\mu$ and measuring every member of
$\Cal A$.   \Hint{213Xa.}

\spheader 214Yd\dvAnew{2015}
For this exercise only, I will say that a measure $\mu$ on a set $X$ is
{\bf nowhere all-measuring} if whenever $A\subseteq X$ is not
$\mu$-negligible there is a subset of $A$ which is not measured by the
subspace measure on $A$.   Show that if $X$ is a set and
$\mu_0,\ldots,\mu_n$ are nowhere all-measuring complete totally finite
measures on $X$, then there are disjoint
$A_0,\ldots,A_n\subseteq X$ such that $\mu_i^*A_i=\mu_iX$ for every
$i\le n$.   \Hint{start with the case $n=1$, $\mu_0=\mu_1$.}
%mt21bits

\spheader 214Ye\dvAnew{2016}
Let $(X,\Sigma,\mu)$ be a measure space and $\Cal A$ a disjoint family of
subsets of $X$.   Show that there is a measure on $X$,
extending $\mu$, which measures every member of $\Cal A$.
%214P

}%end of exercises

\endnotes{
\Notesheader{214} I take the first part of the section, down to 214H,
slowly and carefully, because while none of the arguments are deep
(214Eb is the longest) the patterns formed by the results are not always
easy to predict.   There is a counter-example to a tempting extension of
214H/214Xb in 216Xb.

The message of the second part of the section (214I-214L, 214Q) is that
subspaces inherit
many, but not all, of the properties of a measure space;  and in
particular there can be a difficulty with semi-finiteness, unless we have
locally determined negligible sets (214Xg).   (I give an example in
216Xa.)   Of course
213Hb shows that if we start with a localizable space, we can convert
it into a complete locally determined localizable space without doing
great violence to the structure of the space, so the difficulty is
ordinarily superable.

By far the most important case of 214P is when $\Cal A=\{A\}$ is a
singleton, so that the argument simplifies dramatically (214Xm).   In
\S439 of Volume 4 I will return to the problem of extending a measure to
a given larger $\sigma$-algebra in the absence of any helpful
auxiliary structure.   That section will mostly offer counter-examples,
in particular showing that there is no general theorem extending
214Xn from finite families to countable families, and that the special
conditions in 214P and 214Yb
are there for good reasons.   But in Chapter 54 and \S552 of Volume 5 I
will discuss mathematical systems in which much stronger extension
theorems are true, at least if we start from Lebesgue measure.
}%end of notes

\discrpage


\frfilename{mt215.tex} 
\versiondate{13.11.13} 
\copyrightdate{2000} 
      
\def\chaptername{Taxonomy of measure spaces} 
\def\sectionname{$\sigma$-finite spaces and the principle of exhaustion} 
      
\newsection{215} 
      
I interpolate a short section to deal with some useful facts which might 
get lost if buried in one of the longer sections of this chapter.   The 
great majority of the applications of measure theory involve 
$\sigma$-finite spaces, to the point that many authors skim over any 
others.   I 
myself prefer to signal the importance of such concepts by explicitly 
stating just which theorems apply only to the restricted class of 
spaces.   But undoubtedly some facts about $\sigma$-finite spaces need 
to be grasped early on.   In 215B I give a list of properties 
characterizing $\sigma$-finite spaces.   Some of these make better sense 
in the light of the principle of exhaustion (215A).   I take the 
opportunity to include a fundamental fact about atomless measure spaces 
(215D). 
      
\leader{215A}{The principle of \dvrocolon{exhaustion}}\cmmnt{ The 
following is an example of the use of one of the most important methods 
in measure theory. 
      
\medskip 
      
\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be any measure space and 
$\Cal E\subseteq\Sigma$ a non-empty set such that 
$\sup_{n\in\Bbb N}\mu F_n$ is finite for every non-decreasing sequence 
$\sequencen{F_n}$ in $\Cal E$. 
      
(a) There is a 
non-decreasing sequence $\sequencen{F_n}$ in $\Cal E$ such that, for 
every $E\in\Sigma$, {\it either} there is an $n\in\Bbb N$ such that 
$E\cup F_n$ is not included in any member of $\Cal E$ {\it or}, setting 
$F=\bigcup_{n\in\Bbb N}F_n$, 
      
\Centerline{$\lim_{n\to\infty}\mu(E\setminus F_n) 
=\mu(E\setminus F)=0$.} 
      
\noindent In particular, if $E\in\Cal E$ and 
$E\supseteq F$, then 
$E\setminus F$ is negligible. 
      
(b) If $\Cal E$ is upwards-directed, then there is a non-decreasing 
sequence $\sequencen{F_n}$ in $\Cal E$ such that, setting 
$F=\bigcup_{n\in\Bbb N}F_n$, $\mu F=\sup_{E\in\Cal E}\mu E$ and 
$E\setminus F$ is negligible for every $E\in\Cal E$, so that $F$ is an 
essential supremum of $\Cal E$ in $\Sigma$\cmmnt{ in the sense of 
211G}. 
      
(c) If the union of any non-decreasing sequence in $\Cal E$ belongs to 
$\Cal E$, then there is an $F\in\Cal E$ such that $E\setminus F$ is 
negligible whenever $E\in\Cal E$ and $F\subseteq E$. 
      
\proof{{\bf (a)} Choose $\sequencen{F_n}$, $\sequencen{\Cal E_n}$ and 
$\sequencen{u_n}$ inductively, as follows.   Take $F_0$ to be any member 
of $\Cal E$.   Given $F_n\in\Cal E$, set 
$\Cal E_n=\{E:F_n\subseteq E\in\Cal E\}$ and 
$u_n=\sup\{\mu E:E\in\Cal E_n\}$ in $[0,\infty]$, and choose 
$F_{n+1}\in\Cal E_n$ such that $\mu F_{n+1}\ge\min(n,u_n-2^{-n})$; 
continue. 
      
Observe that this construction yields a non-decreasing sequence 
$\sequencen{F_n}$ in $\Cal E$.   Since $\Cal E_{n+1}\subseteq\Cal E_n$ 
for every $n$, $\sequencen{u_n}$ is non-increasing, and has a limit $u$ 
in $[0,\infty]$.   Since $\min(n,u-2^{-n})\le\mu F_{n+1}\le u_n$ for 
every $n$, $\lim_{n\to\infty}\mu F_n=u$.   Our hypothesis on $\Cal E$ 
now tells us that $u$ is finite. 
      
If $E\in\Sigma$ is such that for every $n\in\Bbb N$ there is an 
$E_n\in\Cal E$ such that 
$E\cup F_n\subseteq E_n$, then $E_n\in\Cal E_n$, so 
      
\Centerline{$\mu F_n\le\mu(E\cup F_n)\le\mu E_n\le u_n$} 
      
\noindent for every $n$, and $\lim_{n\to\infty}\mu(E\cup F_n)=u$.   But 
this means that 
      
\Centerline{$\mu(E\setminus F) 
\le\lim_{n\to\infty}\mu(E\setminus F_n) 
=\lim_{n\to\infty}\mu(E\cup F_n)-\mu F_n 
=0$,} 
      
\noindent as stated.   In particular, this is so if $E\in\Cal E$ and 
$E\supseteq F$. 
      
\medskip 
      
{\bf (b)} Take $\sequencen{F_n}$ from (a).   If $E\in\Cal E$, then 
(because $\Cal E$ is upwards-directed) $E\cup F_n$ is included in some 
member of $\Cal E$ for every $n\in\Bbb N$;  so we must have the second 
alternative of (a), and $E\setminus F$ is negligible.    It follows that 
      
\Centerline{$\sup_{E\in\Cal E}\mu E 
\le\mu F 
=\lim_{n\to\infty}\mu F_n 
\le\sup_{E\in\Cal E}\mu E$,} 
      
\noindent so $\mu F=\sup_{E\in\Cal E}\mu E$. 
      
If $G$ is any measurable set such that $E\setminus G$ is negligible for 
every $E\in\Cal E$, then $F_n\setminus G$ is negligible for every $n$, 
so that $F\setminus G$ is negligible;  thus $F$ is an essential supremum 
for $\Cal E$. 
      
\medskip 
      
{\bf (c)} Again take $\sequencen{F_n}$ from (a), and set 
$F=\bigcup_{n\in\Bbb N}E_n$.   Our hypothesis now is that $F\in\Cal E$, 
so has both the properties declared. 
}%end of proof of 215A 
      
\leader{215B}{}\cmmnt{ $\sigma$-finite spaces are so important that I 
think it is worth spelling out the following facts. 
      
\medskip 
      
\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a semi-finite 
measure space.   Write $\Cal N$ for the family of $\mu$-negligible sets 
and $\Sigma^f$ for the family of measurable sets of finite measure. 
Then the following are equiveridical: 
      
(i) $(X,\Sigma,\mu)$ is $\sigma$-finite; 
      
(ii) every disjoint family in $\Sigma^f\setminus\Cal N$ is countable; 
      
(iii) every disjoint family in $\Sigma\setminus\Cal N$ is countable; 
      
(iv) for every $\Cal E\subseteq\Sigma$ there is a countable set 
$\Cal E_0\subseteq\Cal E$ such that $E\setminus\bigcup\Cal E_0$ is 
negligible for every $E\in\Cal E$; 
      
(v) for every non-empty upwards-directed $\Cal E\subseteq\Sigma$ there 
is a non-decreasing sequence $\sequencen{F_n}$ in $\Cal E$ such that 
$E\setminus\bigcup_{n\in\Bbb N}F_n$ is negligible for every 
$E\in\Cal E$; 
      
(vi) for every non-empty $\Cal E\subseteq\Sigma$, there is a 
non-decreasing sequence $\sequencen{F_n}$ in $\Cal E$ such that 
$E\setminus\bigcup_{n\in\Bbb N}F_n$ is negligible whenever $E\in\Cal E$ 
and $E\supseteq F_n$ for every $n\in\Bbb N$; 
      
(vii) {\it either} $\mu X=0$ {\it or} there is a probability measure 
$\nu$ on $X$ with the same domain and the same negligible sets as $\mu$; 
      
(viii) there is a measurable integrable function $f:X\to\ocint{0,1}$; 
      
(ix) either $\mu X=0$ or there is a measurable function 
$f:X\to\ooint{0,\infty}$ such that $\int fd\mu=1$. 
      
\proof{{\bf (i)$\Rightarrow$(vii) and (viii)} If $\mu X=0$, (vii) is 
trivial and we can take $f=\chi X$ in (viii).   Otherwise, let 
$\sequencen{E_n}$ be a disjoint sequence in 
$\Sigma^f$ covering $X$.   Then it is easy to see that there is a 
sequence $\sequencen{\alpha_n}$ of strictly positive real numbers such 
that $\sum_{n=0}^{\infty}\alpha_n\mu E_n=1$.   Set 
$\nu E=\sum_{n=0}^{\infty}\alpha_n\mu(E\cap E_n)$ for $E\in\Sigma$; 
then $\nu$ is a probability measure with domain $\Sigma$ and the same 
negligible sets as $\mu$.   Also 
$f=\sum_{n=0}^{\infty}\min(1,\alpha_n)\chi E_n$ is a strictly positive 
measurable integrable function. 
      
\medskip 
      
{\bf (vii)$\Rightarrow$(vi) and (v)} Assume (vii), and let $\Cal E$ be a 
non-empty family of measurable sets.   If $\mu X=0$ then (vi) and (v) 
are certainly true.   Otherwise, let $\nu$ be a probability measure with 
domain $\Sigma$ and the same negligible sets as $\mu$.   Since 
$\sup_{E\in\Cal E}\nu E\le 1$ is finite, we can apply 215Aa to find a 
non-decreasing sequence $\sequencen{F_n}$ in $\Cal E$ such that 
$E\setminus\bigcup_{n\in\Bbb N}F_n$ is negligible whenever $E\in\Cal E$ 
includes $\bigcup_{n\in\Bbb N}F_n$;  and if $\Cal E$ is 
upwards-directed, $E\setminus\bigcup_{n\in\Bbb N}F_n$ will be negligible 
for every $E\in\Cal E$, as in 215Ab. 
      
\medskip 
      
{\bf (vi)$\Rightarrow$(iv)} Assume (vi), and let $\Cal E$ be any subset 
of $\Sigma$.   Set 
      
\Centerline{$\Cal H=\{\bigcup\Cal E_0:\Cal E_0\subseteq\Cal E$ is 
countable$\}$.} 
      
\noindent By (vi), there is a sequence $\sequencen{H_n}$ in $\Cal H$ 
such that $H\setminus\bigcup_{n\in\Bbb N}H_n$ is negligible whenever 
$H\in\Cal H$ and $H\supseteq H_n$ for every $n\in\Bbb N$.   Now we can 
express each $H_n$ as $\bigcup\Cal E'_n$, where 
$\Cal E'_n\subseteq\Cal E$ is countable;  setting 
$\Cal E_0=\bigcup_{n\in\Bbb N}\Cal E'_n$, $\Cal E_0$ is countable. 
If $E\in\Cal E$, then 
$E\cup\bigcup_{n\in\Bbb N}H_n=\bigcup(\{E\}\cup\Cal E_0)$ belongs to 
$\Cal H$ and includes every $H_n$, so that 
$E\setminus\bigcup\Cal E_0=E\setminus\bigcup_{n\in\Bbb N}H_n$ is 
negligible.   So $\Cal E_0$ has the property we need, and (iv) is true. 
      
\medskip 
      
{\bf (iv)$\Rightarrow$(iii)} Assume (iv).   If $\Cal E$ is a disjoint 
family in $\Sigma\setminus\Cal N$, take a countable 
$\Cal E_0\subseteq\Cal E$ such that $E\setminus\bigcup\Cal E_0$ is 
negligible for every $E\in\Cal E$.   Then $E=E\setminus\bigcup\Cal E_0$ 
is negligible for every $E\in\Cal E\setminus\Cal E_0$;  but this just 
means that $\Cal E\setminus\Cal E_0$ is empty, so that 
$\Cal E=\Cal E_0$ is countable. 
      
\medskip 
      
{\bf (iii)$\Rightarrow$(ii)} is trivial. 
      
\medskip 
      
{\bf (ii)$\Rightarrow$(i)} Assume (ii).   Let $\frak P$ be the set of 
all disjoint subsets of $\Sigma^f\setminus\Cal N$, ordered by 
$\subseteq$.   Then $\frak P$ is a partially ordered set, not empty (as 
$\emptyset\in \frak P$), and if $\frak Q\subseteq \frak P$ is non-empty 
and totally ordered then it has an upper bound in $\frak P$.   \Prf\ Set 
$\Cal E=\bigcup \frak Q$, the union of all the disjoint families 
belonging to $\frak Q$.   If $E\in\Cal E$ then $E\in\Cal C$ for some 
$\Cal C\in \frak Q$, so $E\in\Sigma^f\setminus\Cal N$.  If $E$, 
$F\in\Cal E$ and $E\ne F$, then there are $\Cal C$, $\Cal D\in \frak Q$ 
such that $E\in\Cal C$, $F\in\Cal D$;  now $\frak Q$ is totally ordered, 
so one of $\Cal C$, $\Cal D$ is larger than the other, and in either 
case $\Cal C\cup\Cal D$ is a member of $\frak Q$ containing both $E$ and 
$F$.   But since any member of $\frak Q$ is a disjoint collection of 
sets, $E\cap F=\emptyset$.   As $E$ and $F$ are arbitrary, $\Cal E$ is 
a disjoint family of sets and belongs to $\frak P$.   And of course 
$\Cal C\subseteq\Cal E$ for every $\Cal C\in \frak Q$, so $\Cal E$ is an 
upper bound for $\frak Q$ in $\frak P$.\ \Qed 
      
By Zorn's Lemma (2A1M), $\frak P$ has a maximal element $\Cal E$ say. 
By (ii), $\Cal E$ must be countable, so $\bigcup\Cal E\in\Sigma$.   Now 
$H=X\setminus\bigcup\Cal E$ is negligible.   \Prf\Quer\ Suppose, if 
possible, otherwise.   Because $(X,\Sigma,\mu)$ is semi-finite, there is 
a set $G$ of finite measure such that $G\subseteq H$ and $\mu G>0$, that 
is, $G\in\Sigma^f\setminus\Cal N$ and $G\cap E=\emptyset$ for every 
$E\in\Cal E$.  But this means that $\{G\}\cup\Cal E$ is a member of 
$\frak P$ strictly larger than $\Cal E$, which is supposed to be 
impossible.\ \Bang\Qed 
      
Let $\sequencen{X_n}$ be a sequence running over $\Cal E\cup\{H\}$. 
Then $\sequencen{X_n}$ is a cover of $X$ by a sequence of measurable 
sets of finite measure, so $(X,\Sigma,\mu)$ is $\sigma$-finite. 
      
\medskip 
      
{\bf (v)$\Rightarrow$(i)} If (v) is true, then we have a sequence 
$\sequencen{E_n}$ in $\Sigma^f$ such that 
$E\setminus\bigcup_{n\in\Bbb N}E_n$ is negligible for every 
$E\in\Sigma^f$.   Because $\mu$ is semi-finite, 
$X\setminus\bigcup_{n\in\Bbb N}E_n$ must be negligible, so $X$ is 
covered by a countable family of sets of finite measure and $\mu$ is 
$\sigma$-finite. 
      
\medskip 
      
{\bf (viii)$\Rightarrow$(ix)} If $\mu X=0$ this is trivial.   Otherwise, 
if $f$ is a strictly positive measurable integrable function, then 
$c=\int f>0$ (122Rc), so $\Bover1cf$ is a strictly positive measurable 
function with integral $1$. 
      
{\bf (ix)$\Rightarrow$(i)} If $f:X\to\ooint{0,\infty}$ is measurable and 
integrable, $\sequencen{\{x:f(x)\ge 2^{-n}\}}$ is a sequence of sets of 
finite measure covering $X$. 
}%end of proof of 215B 
      
\leader{215C}{Corollary} Let $(X,\Sigma,\mu)$ be a $\sigma$-finite 
measure space, and suppose that $\Cal E\subseteq\Sigma$ is any non-empty 
set. 
      
(a) There is a non-decreasing sequence $\sequencen{F_n}$ in $\Cal E$ 
such that, for every $E\in\Sigma$, {\it either} there is an $n\in\Bbb N$ 
such that $E\cup F_n$ is not included in any member of $\Cal E$ {\it or} 
$E\setminus\bigcup_{n\in\Bbb N}F_n$ is negligible. 
      
(b) If $\Cal E$ is upwards-directed, then there is a non-decreasing 
sequence $\sequencen{F_n}$ in $\Cal E$ such that 
$\bigcup_{n\in\Bbb N}F_n$ is an essential supremum of $\Cal E$ in 
$\Sigma$. 
      
(c) If the union of any non-decreasing sequence in $\Cal E$ belongs to 
$\Cal E$, then there is an $F\in\Cal E$ such that $E\setminus F$ is 
negligible whenever $E\in\Cal E$ and $F\subseteq E$. 
      
\proof{ By 215B, there is a totally finite measure $\nu$ on $X$ with the 
same measurable sets and the same negligible sets as $\mu$.   Since 
$\sup_{E\in\Cal E}\nu E$ is finite, we can apply 215A to $\nu$ to obtain 
the results. 
}%end of proof of 215C 
      
\leader{215D}{}\cmmnt{ As a further example of the use of the 
principle of exhaustion, I give a fundamental fact about atomless 
measure spaces. 
      
\medskip 
      
\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be an atomless measure 
space.   If $E\in\Sigma$ and $0\le\alpha\le\mu E<\infty$, there is an 
$F\in\Sigma$ such that $F\subseteq E$ and $\mu F=\alpha$. 
      
\proof{{\bf (a)} We need to know that if $G\in\Sigma$ is non-negligible 
and $n\in\Bbb N$, then there is an $H\subseteq G$ such that 
$0<\mu H\le 2^{-n}\mu G$.   \Prf\ Induce on $n$.   For $n=0$ this is 
trivial.   For the inductive step to $n+1$, use the inductive hypothesis 
to find $H\subseteq G$ such that $0<\mu H\le 2^{-n}\mu G$.   Because 
$\mu$ is atomless, there is an $H'\subseteq H$ such that $\mu H'$, 
$\mu(H\setminus H')$ are both defined and non-zero.   Now at least one 
of them has measure less than or equal to $\bover12\mu H$, so gives us a 
subset of $G$ of non-zero measure less than or equal to 
$2^{-n-1}\mu G$.\ \Qed 
      
It follows that if $G\in\Sigma$ has non-zero finite measure and 
$\epsilon>0$, there is a measurable set $H\subseteq G$ such that 
$0<\mu H\le\epsilon$. 
      
\medskip 
      
{\bf (b)} Let $\Cal H$ be the family of all those $H\in\Sigma$ such that 
$H\subseteq E$ and $\mu H\le\alpha$.   If $\sequencen{H_n}$ is any 
non-decreasing sequence in $\Cal H$, then 
$\mu(\bigcup_{n\in\Bbb N}H_n)=\lim_{n\to\infty}\mu H_n\le\alpha$, so 
$\bigcup_{n\in\Bbb N}H_n\in\Cal H$.   So 215Ac tells us that there is an 
$F\in\Cal H$ such that $H\setminus F$ is negligible whenever 
$H\in\Cal H$ and $F\subseteq H$.   \Quer\ Suppose, if possible, that 
$\mu F<\alpha$.   By (a), there is an 
$H\subseteq E\setminus F$ such that $0<\mu H\le\alpha-\mu F$.   But in 
this case $H\cup F\in\Cal H$ and $\mu((H\cup F)\setminus F)>0$, which is 
impossible.\ \Bang 
      
So we have found an appropriate set $F$. 
}%end of proof of 215D 
      
\leader{*215E}{}\cmmnt{ One of the basic properties of Lebesgue measure is
that singleton subsets (and therefore countable subsets) are negligible.
This is of course associated with the fact that Lebesgue measure is
atomless (211Md).   It is easy to construct measures for which singleton
sets are negligible but there are atoms (e.g., the countable-cocountable
measures of 211R).   It takes a little more ingenuity to construct
atomless measures in which not all singleton sets are negligible
(216Ye).   The following result gives conditions under which this
can't happen.

\medskip

\noindent}{\bf Proposition}\dvAnew{2015} 
Let $(X,\Sigma,\mu)$ be an atomless measure space and $x\in X$.   

(a) If $\mu^*\{x\}$ is finite then $\{x\}$ is negligible.

(b) If $\mu$ has locally determined negligible sets then $\{x\}$ is
negligible.

(c) If $\mu$ is localizable then $\{x\}$ is negligible.

\proof{{\bf (a)} \Quer\ Otherwise, $0<\mu^*\{x\}<\infty$.   Let $E$ be a
set containing $x$ such that $\mu E<2\mu^*\{x\}$.   By 215D, there is an
$F\subseteq E$ such that 

\Centerline{$\mu F=\bover12\mu E=\mu(E\setminus F)<\mu^*\{x\}$.}

\noindent So neither $F$ nor $E\setminus F$ can contain $x$;  but
$x\in E$.\ \Bang

\medskip

{\bf (b)} For any set $E$ of finite measure, $E\cap\{x\}$ is negligible; 
for if it is not empty then $x\in E$ and we can apply (a).   As
$\mu$ has locally determined negligible sets, $\{x\}$ is negligible.

\medskip

{\bf (c)} Let $\Cal E\subseteq\Sigma^f$ be a maximal family such that
$\mu E<\infty$ for every $E\in\Cal E$ and $\mu(E\cap F)=\emptyset$ whenever
$E$, $F\in\Cal E$ are distinct.   For each $E\in\Cal E$ and $n\in\Bbb N$,
use 215D $n$ times to find a partition
$E_{n0}$, $E_{n1},\ldots,E_{nn}$ of $E$
such that $\mu E_{ni}=\bover1{n+1}\mu E$ for every $i\le n$.   Next, for
$i\le n\in\Bbb N$, let $H_{ni}$ be an essential supremum of 
$\{E_{ni}:E\in\Cal E\}$.   For each $n\in\Bbb N$ and $E\in\Cal E$.

\Centerline{$\mu(E\setminus\bigcup_{i\le n}H_{ni})
\le\sum_{i=0}^n\mu(E_{ni}\setminus H_{ni})=0$.}

\noindent So if $F\in\Sigma$, $\mu F<\infty$
and $F\cap\bigcup_{i\le n}H_{ni}=\emptyset$, we shall have $\mu(E\cap F)=0$
for every $E\in\Cal E$;  as $\Cal E$ was maximal, $F$ must be negligible.
We are supposing that $\mu$ is semi-finite, so 
$X\setminus\bigcup_{i\le n}H_{ni}$ is negligible.

It follows that if there is any $n$ such that 
$x\notin\bigcup_{i\le n}H_{ni}$ then $\{x\}$ is negligible and we can stop.
Consider $H=\bigcap\{H_{ni}:i\le n\in\Bbb N$, $x\in H_{ni}\}$.
\Quer\ If $\mu H>0$, there is an $F\subseteq H$ such that $0<\mu F<\infty$.
By the maximality of $\Cal E$, there is an $E^*\in\Cal E$ such that
$\mu(F\cap E^*)>0$.   Let $n$ be such that 
$\mu(F\cap E^*)>\bover1{n+1}\mu E^*$.
There is an $i\le n$ such that $x\in H_{ni}$, so that
$F\subseteq H_{ni}$, while $\mu(F\cap E^*_{ni})\le\bover1{n+1}\mu E^*$,
so $F'=F\cap E^*\setminus E^*_{ni}$ has non-zero measure.   Consider
$G=H_{ni}\setminus F'$.   If $E\in\Cal E$, then either
$E=E^*$ and 

\Centerline{$E_{ni}\setminus G=E^*_{ni}\setminus G
=E^*_{ni}\setminus H_{ni}$}

\noindent is negligible, or $E\ne E^*$ and 

\Centerline{$E_{ni}\setminus G
\subseteq(E_{ni}\setminus H_{ni})\cup(E\cap E^*)$}

\noindent is negligible.   Because $H_{ni}$ is an essential supremum of
$\{E_{ni}:E\in\Cal E\}$, $H_{ni}\setminus G$ must be negligible and
$F'$ is negligible;  which is impossible.\ \Qed

Thus $H$ is a negligible set containing $x$ and $\{x\}$ is negligible in
this case also.
}%end of proof of 215E

\exercises{\leader{215X}{Basic exercises (a)} 
%\spheader 215Xa 
Let $(X,\Sigma,\mu)$ be a measure space and $\Phi$ a 
non-empty set of $\mu$-integrable real-valued functions from $X$ to 
$\Bbb R$.   Suppose that $\sup_{n\in\Bbb N}\int f_n$ is finite for every 
sequence $\sequencen{f_n}$ in $\Phi$ such that $f_n\leae f_{n+1}$ 
for every $n$.   Show that there is a sequence $\sequencen{f_n}$ in 
$\Phi$ such that $f_n\leae f_{n+1}$ for every $n$ and, for every 
integrable real-valued function $f$ on $X$, {\it either} 
$f\leae\sup_{n\in\Bbb N}f_n$ {\it or} there is an 
$n\in\Bbb N$ such that no member of $\Phi$ is greater than or equal to 
$\max(f,f_n)$ almost everywhere. 
%215A 
      
\sqheader 215Xb Let $(X,\Sigma,\mu)$ be a measure space.   (i) Suppose 
that $\Cal E$ is a non-empty upwards-directed subset of $\Sigma$ such 
that $c=\sup_{E\in\Cal E}\mu E$ is finite.   Show that 
$E\setminus\bigcup_{n\in\Bbb N}F_n$ is negligible whenever $E\in\Cal E$ 
and $\sequencen{F_n}$ is a sequence in $\Cal E$ such that 
$\lim_{n\to\infty}\mu F_n=c$.   (ii) Let $\Phi$ be a non-empty set of 
integrable functions on $X$ which is upwards-directed in the sense that 
for all $f$, $g\in\Phi$ there is an $h\in\Phi$ such that 
$\max(f,g)\leae h$, and suppose that $c=\sup_{f\in\Phi}\int f$ is 
finite.   Show that $f\leae\sup_{n\in\Bbb N}f_n$ whenever $f\in\Phi$ 
and $\sequencen{f_n}$ is a sequence in $\Phi$ such that 
$\lim_{n\to\infty}\int f_n=c$. 
%215A 
      
\spheader 215Xc Use 215A to shorten the proof of 211Ld. 
      
\spheader 215Xd Give an example of a (non-semi-finite) measure space 
$(X,\Sigma,\mu)$ satisfying conditions (ii)-(iv) of 215B, but not (i). 
%215B 
      
\sqheader 215Xe Let $(X,\Sigma,\mu)$ be an atomless $\sigma$-finite 
measure space.   Show that for any $\epsilon>0$ there is a disjoint 
sequence $\sequencen{E_n}$ of measurable sets with measure at most 
$\epsilon$ such that $X=\bigcup_{n\in\Bbb N}E_n$. 
%215D 
      
\spheader 215Xf Let $(X,\Sigma,\mu)$ be an atomless strictly localizable 
measure space.   Show that for any $\epsilon>0$ there is a decomposition 
$\familyiI{X_i}$ of $X$ such that $\mu X_i\le\epsilon$ for every 
$i\in I$. 
      
\leader{215Y}{Further exercises (a)} 
%\spheader 215Ya 
Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space 
and $\langle f_{mn}\rangle_{m,n\in\Bbb N}$, $\sequence{m}{f_m}$, $f$ 
measurable real-valued functions defined almost everywhere in $X$ and 
such that $\sequencen{f_{mn}}\to f_m$ a.e.\ for 
each $m$ and $\sequence{m}{f_m}\to f$ a.e.   Show that there is a 
strictly increasing sequence $\sequence{m}{n_m}$ in $\Bbb N$ such that 
$\sequence{m}{f_{m,n_m}}\to f$ a.e.   (Compare 134Yb.) 
%215B 
      
\spheader 215Yb Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space. 
Let $\sequencen{f_n}$ be a sequence of measurable real-valued functions 
such that $f=\lim_{n\to\infty}f_n$ is defined almost everywhere in $X$. 
Show that there is a non-decreasing sequence $\sequence{k}{X_k}$ of 
measurable subsets of $X$ such that $\bigcup_{k\in\Bbb N}X_k$ is 
conegligible in $X$ and $\sequencen{f_n}\to f$ uniformly on every $X_k$, 
in the sense that for any $\epsilon>0$ there is an $m\in\Bbb N$ such 
that $|f_j(x)-f(x)|$ is defined and less than or equal to $\epsilon$ 
whenever $j\ge m$ and $x\in X_k$. 
      
(This is a version of Egorov's theorem.) 
%215B 
      
\spheader 215Yc Let $(X,\Sigma,\mu)$ be a totally finite measure space 
and $\sequencen{f_n}$, $f$ measurable real-valued functions defined 
almost everywhere in $X$.   Show that $\sequencen{f_n}\to f$ a.e.\ iff 
there is a sequence $\sequencen{\epsilon_n}$ of strictly positive real 
numbers, converging to $0$, such that 
      
\Centerline{$\lim_{n\to\infty}\mu^*(\bigcup_{k\ge n}\{x:x\in\dom 
f_k\cap\dom f,\,|f_k(x)-f(x)|\ge\epsilon_n\})=0$.} 
%215B 
      
\spheader 215Yd Find a direct proof of (v)$\Rightarrow$(vi) in 215B. 
\Hint{given $\Cal E\subseteq\Sigma$, use Zorn's Lemma to find a maximal 
totally ordered $\Cal E'\subseteq\Cal E$ such that 
$E\symmdiff F\notin\Cal N$ for any distinct $E$, $F\in\Cal E'$, and 
apply (v) to $\Cal E'$.} 

}%end of exercises 
      
\endnotes{ 
\Notesheader{215} 
The common ground of 215A, 215B(vi), 215C and 215Xa is actually one of 
the most fundamental ideas in measure theory.   It appears in such 
various forms that it is often easier to prove an application from first 
principles than to explain how it can be reduced to the versions here. 
But I will try henceforth to signal such applications as they arise, 
calling the 
method (the proof of 215Aa or 215Xa) the `principle of exhaustion'. 
One point which is perhaps worth noting here is the inductive 
construction of the sequence $\sequencen{F_n}$ in the proof of 215Aa. 
Each $F_{n+1}$ is chosen {\it after} the preceding one.   It is this 
which makes it possible, in the proof of 215B(vii)$\Rightarrow$(vi), to 
extract a suitable sequence $\sequencen{F_n}$ directly.   In many 
applications (starting with what is surely the most important one in the 
elementary theory, the Radon-Nikod\'ym theorem of \S232, or with part 
(i) of the proof of 211Ld), this 
refinement is not needed;  we are dealing with an upwards-directed set, 
as in 215B(v), and can choose the whole sequence $\sequencen{F_n}$ at 
once, no term interacting with any other, as in 215Xb.   The axiom of 
`dependent choice', which asserts that we can construct sequences 
term-by-term, is known to be stronger than the axiom of `countable 
choice', which asserts only that we can choose countably many objects 
simultaneously. 
      
In 215B I try to indicate the most characteristic properties of 
$\sigma$-finiteness;  in particular, the properties which distinguish 
$\sigma$-finite measures from other strictly localizable measures. 
This result is in a way more abstract than the manipulations in the 
rest of the section.   Note that it makes an essential use of the axiom 
of choice in the form of Zorn's Lemma.   I spent a paragraph in 134C 
commenting on the distinction between `countable choice', which is 
needed for anything which looks like the standard theory of Lebesgue 
measure, and the full axiom of choice, which is relatively little used 
in the elementary theory.   The implication (ii)$\Rightarrow$(i) of 215B 
is one of the points where we do need something beyond countable choice. 
(I should perhaps remark that the whole theory of 
non-$\sigma$-finite measure spaces looks very odd without the general 
axiom of choice.) 
Note also that in 215B the proofs of (i)$\Rightarrow$(vii) and 
(vii)$\Rightarrow$(vi) are the only points where anything so vulgar as a 
number appears.   The conditions (iii), (iv), (v) and (vi) are linked in 
ways that have nothing to do with measure theory, and involve only with 
the structure $(X,\Sigma,\Cal N)$.   (See 215Yd here, and 316D-316E in 
Volume 3.)   There are similar conditions relating to measurable 
functions 
rather than measurable sets;  for a fairly abstract example, see 241Ye. 
      
In 215Ya-215Yc are three more standard theorems on 
almost-everywhere-convergent sequences which depend on 
$\sigma$- or total finiteness. 
}%end of notes 
      
\discrpage 
      
\frfilename{mt216.tex}
\versiondate{25.9.04}
\copyrightdate{1994}

\def\chaptername{Taxonomy of measure spaces}
\def\sectionname{Examples}

\newsection{216}

It is common practice -- and, in my view, good
practice -- in books on pure mathematics, to provide discriminating
examples;  I mean that whenever we are given a list of new concepts, we
expect to be provided with examples to show that we have a fair picture
of the relationships between them, and in particular that we are not
being kept ignorant of some startling implication.   Concerning the
concepts listed in
211A-211K, %211A 211B 211C 211D 211E 211F 211G 211H 211I 211J 211K
we have ten different properties which some, but
not all, measure spaces possess, giving a conceivable total of $2^{10}$
different types of measure space, classified according to which of these
ten properties they have.   The list of basic relationships in 211L
reduces these 1024 possibilities to 72.   Observing that a space can be
simultaneously atomless and purely atomic only when the measure of the
whole space is $0$, we find ourselves with 56 possibilities, being two
trivial cases with $\mu X=0$ (because such a measure may or may not be
complete) together with $9\times 2\times 3$ cases, corresponding to the
nine classes

\quad probability spaces,

\quad spaces which are totally finite, but not probability spaces,

\quad spaces which are $\sigma$-finite, but not totally finite,

\quad spaces which are strictly localizable, but not $\sigma$-finite,

\quad spaces which are localizable and locally determined, but not
strictly localizable,

\quad spaces which are localizable, but not locally determined,

\quad spaces which are locally determined, but not localizable,

\quad spaces which are semi-finite, but neither locally determined nor
localizable,

\quad spaces which are not semi-finite;

\noindent the two classes

\quad spaces which are complete,

\quad spaces which are not complete;

\noindent and the three classes

\quad spaces which are atomless, not of measure $0$,

\quad spaces which are purely atomic, not of measure $0$,

\quad spaces which are neither atomless nor purely atomic.

I do not propose to give a complete set of fifty-six examples,
particularly as rather fewer than fifty-six different ideas are
required.   However, I do think that for a proper understanding of
abstract measure spaces it is necessary to have seen realizations of
some of the critical combinations of properties.   I therefore take a
few paragraphs to describe three special examples to add to those of
211M-211R. %211M 211N 211O 211P 211Q 211R

\leader{216A}{Lebesgue measure}\cmmnt{ Before turning to the new ideas, let me mention Lebesgue measure again.
As remarked in 211M, 211P and 211Qa,

}(a) Lebesgue measure $\mu$ on $\Bbb R$ is complete, atomless and
$\sigma$-finite, therefore
strictly localizable, localizable and locally determined.

(b) The subspace measure $\mu_{[0,1]}$ on $[0,1]$ is a complete,
atomless probability measure.

(c) The restriction $\mu\restr\Cal B$ of $\mu$ to the Borel
$\sigma$-algebra $\Cal B$ of $\Bbb R$ is atomless,
$\sigma$-finite and not complete.

\cmmnt{
\leader{216B}{}I now embark on the description of three
`counter-examples';  meaning
spaces built specifically for the purpose of showing that there are no
unexpected implications among the ten properties under consideration
here.   Even by the standards of this chapter these must be regarded as
dispensable by the student who wants to get on with the real business of
understanding the big theorems of the subject.   Neither the existence
of these examples, nor the techniques needed in constructing them, are
vital for anything else we shall look at before Volume 5.   But if you
are going to take abstract measure theory seriously at all, sooner or
later you will need to form some kind of mental picture of the nature of
the spaces possessing the different properties here, and a minimal
requirement of such a picture is that it should include the
discriminations witnessed by these examples.
}%end of comment

\leader{*216C}{A complete, localizable, non-locally-determined
space}\cmmnt{ The
first example hardly needs an idea beyond what we already have, but it
does call for more manipulations than it seems fair to set as an
exercise, and may therefore be useful as a demonstration of technique.

\medskip

}{\bf (a)} Let $I$ be any uncountable set, and set $X=\{0,1\}\times I$.
For $E\subseteq X$, $y\in\{0,1\}$ set
$E[\{y\}]=\{i:(y,i)\in E\}\subseteq I$.   Set

\Centerline{$\Sigma=\{E:E\subseteq X,\,E[\{0\}]\symmdiff E[\{1\}]$ is
countable$\}$.}

\noindent Then $\Sigma$ is a $\sigma$-algebra of subsets of $X$.
\prooflet{\Prf\ (i)
$\emptyset[\{0\}]\symmdiff\emptyset[\{1\}]=\emptyset$ is countable, so
$\emptyset\in\Sigma$.   (ii) If $E\in\Sigma$ then

\Centerline{$(X\setminus E)[\{0\}]
\symmdiff(X\setminus E)[\{1\}]=E[\{0\}]\symmdiff E[\{1\}]$}

\noindent is countable.   (iii) If $\sequencen{E_n}$ is a
sequence in
$\Sigma$ and $E=\bigcup_{n\in\Bbb N}E_n$, then

\Centerline{$E[\{0\}]\symmdiff E[\{1\}]
\subseteq\bigcup_{n\in\Bbb N}E_n[\{0\}]\symmdiff E_n[\{1\}]$}

\noindent is countable.\ \Qed}

For $E\in\Sigma$, set $\mu E=\#(E[\{0\}])$ if this is finite, $\infty$
otherwise;  then $(X,\Sigma,\mu)$ is a measure space.

\medskip

{\bf (b)} $(X,\Sigma,\mu)$ is complete.   \prooflet{\Prf\ If $A\subseteq
E\in\Sigma$ and $\mu E=0$, then $(0,i)\notin E$ for every $i$.   So

\Centerline{$A[\{0\}]\symmdiff A[\{1\}]=A[\{1\}]\subseteq E[\{1\}]
=E[\{1\}]\symmdiff E[\{0\}]$}

\noindent must be countable, and $A\in\Sigma$.\ \Qed}

\medskip

{\bf (c)} $(X,\Sigma,\mu)$ is semi-finite.   \prooflet{\Prf\ If $E\in\Sigma$
and $\mu E>0$, there is an $i\in I$ such that $(0,i)\in E$;  now
$F=\{(0,i)\}\subseteq E$ and $\mu F=1$.\ \Qed}

\medskip

{\bf (d)} $(X,\Sigma,\mu)$ is localizable.   \prooflet{\Prf\ Let $\Cal E$ be
any subset of $\Sigma$.   Set

\Centerline{$J=\bigcup_{E\in\Cal E}E[\{0\}]$,
\quad $G=\{0,1\}\times J$.}

\noindent Then $G\in\Sigma$.   If $H\in\Sigma$, then

$$\eqalign{\mu(E\setminus H)&=0 \text{ for every }E\in\Cal E\cr
&\iff E[\{0\}]\subseteq H[\{0\}]\text{ for every }E\in\Cal E\cr
&\iff (0,i)\in H\text{ for every }i\in J\cr
&\iff \mu(G\setminus H)=0.\cr}$$

\noindent Thus $G$ is an essential supremum for $\Cal E$ in $\Sigma$;  as
$\Cal E$ is arbitrary, $\mu$ is localizable.\ \Qed}

\medskip

{\bf (e)} $(X,\Sigma,\mu)$ is not locally determined.   \prooflet{\Prf\
Consider $H=\{0\}\times I$.   Then $H\notin\Sigma$ because
$H[\{0\}]\symmdiff H[\{1\}]=I$ is uncountable.   But let $E\in\Sigma$ be any
set such that $\mu E<\infty$.   Then

\Centerline{$(E\cap H)[\{0\}]\symmdiff (E\cap H)[\{1\}]
=(E\cap H)[\{0\}]\subseteq E[\{0\}]$}

\noindent is finite, so $E\cap H\in\Sigma$.   As $E$ is arbitrary, $H$
witnesses that $\mu$ is not locally determined.\ \Qed}
\medskip

{\bf (f)} $(X,\Sigma,\mu)$ is
purely atomic.   \prooflet{\Prf\ Let $E\in\Sigma$ be any set of non-zero
measure.   Let $i\in I$ be such that $(0,i)\in E$.   Then $(0,i)\in E$ and
$F=\{(0,i)\}$ is a set of measure $1$, included in $E$;  because $F$
is a singleton set, it must be an atom for $\mu$;  as $E$ is arbitrary,
$\mu$ is purely atomic.\ \Qed}

\cmmnt{\medskip

{\bf (g)} Thus the construction here yields a complete, localizable, purely
atomic, non-locally-determined space.
}%end of comment

\leader{*216D}{A complete, locally determined space which is not
localizable}\cmmnt{ The next construction requires a little set theory.}
We need two sets $I$, $J$ such that $I$ is uncountable\cmmnt{ (more
strictly, $I$
cannot be expressed as the union of countably many countable sets)},
$I\subseteq J$ and $J$ cannot be expressed as $\bigcup_{i\in I}K_i$
where every $K_i$ is countable.   \cmmnt{The most natural way of doing
this, subject to the axiom of choice, is to take $I=\omega_1$, the first
uncountable ordinal, and $J$ to be $\omega_2$, the first ordinal from
which there is no injection into $\omega_1$ (see 2A1Fc);  but in case
you prefer other formulations (e.g., $I=\{\{x\}:x\in\Bbb R\}$ and
$J=\Cal P\Bbb R$), I will write the following argument in
terms of $I$ and $J$, and you can pick your own pair.}%end of comment

\medskip

{\bf (a)} Let $\Tau$ be the countable-cocountable $\sigma$-algebra
of $J$ and $\nu$ the countable-cocountable measure on
$J$\cmmnt{ (211R)}.   Set $X=J\times J$ and for $E\subseteq X$ set

\Centerline{$E[\{\xi\}]=\{\eta:(\xi,\eta)\in E\}$,\quad
$E^{-1}[\{\xi\}]=\{\eta:(\eta,\xi)\in E\}$}

\noindent for every $\xi\in J$.   Set

\Centerline{$\Sigma=\{E:E[\{\xi\}]$ and $E^{-1}[\{\xi\}]$ belong to
$\Tau$ for every $\xi\in J\}$,}

\Centerline{$\mu E=\sum_{\xi\in J}\nu E[\{\xi\}]
+\sum_{\xi\in J}\nu E^{-1}[\{\xi\}]$}

\noindent for every $E\in\Sigma$.   \cmmnt{It is easy to check that}
$\Sigma$ is a $\sigma$-algebra and\cmmnt{ that} $\mu$ is a measure.

\medskip

{\bf (b)} $(X,\Sigma,\mu)$ is complete.   \prooflet{\Prf\ If $A\subseteq
E\in\Sigma$ and $\mu E=0$, then all the sets
$E[\{\xi\}]$ and $E^{-1}[\{\xi\}]$ are countable, so the same is true of
all the sets $A[\{\xi\}]$ and $A^{-1}[\{\xi\}]$, and $A\in\Sigma$.\ \Qed}

\medskip

{\bf (d)} $(X,\Sigma,\mu)$ is semi-finite.   \prooflet{\Prf\ For each
$\zeta\in J$, set

\Centerline{$G_{\zeta}=\{\zeta\}\times J$,\quad
$\tilde G_{\zeta}=J\times\{\zeta\}$.}

\noindent Then all the sections $G_{\zeta}[\{\xi\}]$,
$G_{\zeta}^{-1}[\{\xi\}]$, $\tilde G_{\zeta}[\{\xi\}]$ and
$\tilde  G_{\zeta}^{-1}[\{\xi\}]$ are either $J$ or $\emptyset$ or
$\{\zeta\}$, so belong to $\Tau$, and all the $G_{\zeta}$,
$\tilde G_{\zeta}$ belong to $\Sigma$, with $\mu$-measure $1$.

Suppose that $E\in\Sigma$ is a set of strictly positive measure.   Then
there must be some $\xi\in J$ such that

\Centerline{$0<\nu E[\{\xi\}]+\nu E^{-1}[\{\xi\}]
=\mu(E\cap G_{\xi})+\mu(E\cap \tilde G_{\xi})<\infty$,}

\noindent and one of the sets $E\cap G_{\xi}$, $E\cap\tilde G_{\xi}$ is
a set of non-zero finite measure included in $E$.\ \Qed}

\medskip

{\bf (e)} $(X,\Sigma,\mu)$ is locally determined.   \prooflet{\Prf\ Suppose
that $H\subseteq X$ is such that $H\cap E\in\Sigma$ whenever
$E\in\Sigma$ and $\mu E<\infty$.   Then, in particular, $H\cap
G_{\zeta}$ and $H\cap \tilde G_{\zeta}$ belong to $\Sigma$, so

\Centerline{$H[\{\zeta\}]
=(H\cap \tilde G_{\zeta})[\{\zeta\}]\in\Tau$,}

\Centerline{$H^{-1}[\{\zeta\}]
=(H\cap G_{\zeta})^{-1}[\{\zeta\}]\in\Tau$,}

\noindent for every $\zeta\in J$.   This shows that $H\in\Sigma$.   As
$H$ is arbitrary, $\mu$ is locally determined.\ \Qed}

\medskip

{\bf (f)} $(X,\Sigma,\mu)$ is not localizable.
\prooflet{\Prf\ Set $\Cal E=\{G_{\zeta}:\zeta\in J\}$.
\Quer\ Suppose, if possible, that
$G\in\Sigma$ is an essential supremum for $\Cal E$.   Then

\Centerline{$\nu(J\setminus G[\{\xi\}])
=\mu(G_{\xi}\setminus G)=0$}

\noindent and $J\setminus G[\{\xi\}]$ is countable, for every $\xi\in
J$.   Consequently $J\ne\bigcup_{\xi\in I}(J\setminus G[\{\xi\}])$, and
there is an $\eta$ belonging to $J\setminus\bigcup_{\xi\in I}(J\setminus
G[\{\xi\}])=\bigcap_{\xi\in I}G[\{\xi\}]$.
This means just that $(\xi,\eta)\in G$ for every $\xi\in I$, that is,
that $I\subseteq G^{-1}[\{\eta\}]$.   Accordingly
$G^{-1}[\{\eta\}]$ is uncountable, so that
$\nu G^{-1}[\{\eta\}]=\mu(G\cap\tilde G_{\eta})=1$.   But observe that
$\mu(G_{\xi}\cap \tilde G_{\eta})=\mu\{(\xi,\eta)\}=0$ for every $\xi\in
J$.   This means that, setting $H=X\setminus \tilde G_{\eta}$,
$E\setminus H$ is negligible, for every $E\in\Cal E$;  so that we must
have $0=\mu(G\setminus H)=\mu(G\cap\tilde G_{\eta})=1$, which is
absurd.\ \Bang

Thus $\Cal E$ has no essential supremum in $\Sigma$, and $\mu$ cannot be
localizable.\ \Qed}
\medskip

{\bf (g)} $(X,\Sigma,\mu)$ is purely atomic.   \prooflet{\Prf\ If
$E\in\Sigma$ has
non-zero measure, there must be some $\xi\in J$ such that one of
$E[\{\xi\}]$, $E^{-1}[\{\xi\}]$ is not countable;  that is, such that
one of $E\cap G_{\xi}$, $E\cap \tilde G_{\xi}$ is not negligible.   But
if now $H\in\Sigma$ and $H\subseteq E\cap G_{\xi}$,
either $H[\{\xi\}]$ is countable, and
$\mu H=0$, or $J\setminus H[\{\xi\}]$ is countable, and
$\mu(G_{\xi}\setminus H)=0$;  similarly, if
$H\subseteq E\cap\tilde G_{\xi}$,
one of $\mu H$, $\mu (\tilde G_{\xi}\setminus H)$ must be $0$,
according to whether $H^{-1}[\{\xi\}]$ is countable or not.   Thus
$E\cap G_{\xi}$ and $E\cap\tilde G_{\xi}$, if not negligible, must be
atoms, and $E$ must include an atom.   As $E$ is arbitrary, $\mu$ is purely
atomic.\ \Qed}

\cmmnt{\medskip

{\bf (h)} Thus $(X,\Sigma,\mu)$ is complete, locally determined and
purely atomic, but is not localizable.
}%end of comment

\ifdim\pagewidth>467pt\fontdimen3\tenbf=3pt\fontdimen4\tenbf=2pt\fi
\leader{*216E}{}{\bf A complete, locally determined, localizable space
which is not strictly localizable}\cmmnt{ For the last, and most
interesting, construction, we need a non-trivial result in infinitary
combinatorics, which I have written out in 2A1P:  if $I$ is any set, and
$\langle f_{\alpha}\rangle_{\alpha\in A}$ is a family in $\{0,1\}^I$, the set of
functions from $I$ to $\{0,1\}$, with $\#(A)$ strictly greater than
$\frak c$, the cardinal of the continuum, and if
$\langle K_{\alpha}\rangle_{\alpha\in A}$ is any family of countable subsets of
$I$, then there must be distinct $\alpha$, $\beta\in A$ such that
$f_{\alpha}$ and $f_{\beta}$ agree on $K_{\alpha}\cap K_{\beta}$.
\fontdimen3\tenbf=1.92pt
\fontdimen4\tenbf=1.28pt

Armed with this fact, I proceed as follows.

\medskip

}{\bf (a)} Let $C$ be any set with cardinal greater than $\frak c$.   Set
$I=\Cal PC$ and $X=\{0,1\}^I$.   For
$\gamma\in C$, define $x_{\gamma}\in X$ by saying that
$x_{\gamma}(\Gamma)=1$ if $\gamma\in\Gamma\subseteq C$ and
$x_{\gamma}(\Gamma)=0$ if $\gamma\notin\Gamma\subseteq C$.   Let $\Cal
K$ be the family of countable subsets of $I$, and for $K\in\Cal K$,
$\gamma\in C$ set

\Centerline{$F_{\gamma K}=\{x:x\in X,\,x\restr K=x_{\gamma}\restr
K\}\subseteq X$.}

\noindent Let

$$\eqalign{\Sigma_{\gamma}=\{E:E\subseteq X,
&\text{ either there is a }K\in\Cal K\text{ such that }F_{\gamma
K}\subseteq E\cr
&\text{ or there is a }K\in\Cal K\text{ such that }F_{\gamma K}\subseteq
X\setminus E\}.\cr}$$

\noindent Then $\Sigma_{\gamma}$ is a $\sigma$-algebra of subsets of
$X$.   \prooflet{\Prf\ (i) $F_{\gamma\emptyset}\subseteq
X\setminus\emptyset$ so
$\emptyset\in\Sigma_{\gamma}$.   (ii) The definition of
$\Sigma_{\gamma}$ is symmetric between $E$ and $X\setminus E$, so
$X\setminus E\in\Sigma_{\gamma}$ whenever $E\in\Sigma_{\gamma}$.   (iii)
Let $\sequencen{E_n}$ be a sequence in $\Sigma_{\gamma}$, with union
$E$.   ($\alpha$) If there are $n\in\Bbb N$, $K\in\Cal K$ such that
$F_{\gamma K}\subseteq E_n$, then $F_{\gamma K}\subseteq E$, so
$E\in\Sigma_{\gamma}$.   ($\beta$) Otherwise, there is for each
$n\in\Bbb N$ a $K_n\in\Cal K$ such that $F_{\gamma,K_n}\subseteq
X\setminus E_n$.   Set $K=\bigcup_{n\in\Bbb N}K_n\in\Cal K$.   Then

$$\eqalign{F_{\gamma K}
&=\{x:x\restr K=x_{\gamma}\restr K\}
=\{x:x\restr K_n=x_{\gamma}\restr K_n\text{ for every }n\in\Bbb N\}\cr
&=\bigcap_{n\in\Bbb N}F_{\gamma,K_n}
\subseteq\bigcap_{n\in\Bbb N}X\setminus E_n
=X\setminus E,\cr}$$

\noindent so again $E\in\Sigma_{\gamma}$.
As $\sequencen{E_n}$ is arbitrary, $\Sigma_{\gamma}$ is a
$\sigma$-algebra.\ \Qed}

\medskip

{\bf (b)} Set

\Centerline{$\Sigma=\bigcap_{\gamma\in C}\Sigma_{\gamma}$;}

\noindent then $\Sigma$\cmmnt{, being an intersection of
$\sigma$-algebras,} is
a $\sigma$-algebra of subsets of $X$\cmmnt{ (see 111Ga)}.
Define $\mu:\Sigma\to[0,\infty]$ by setting

$$\eqalign{\mu E&=\#(\{\gamma:x_{\gamma}\in E\})
\text{ if this is finite},\cr
&=\infty\text{ otherwise};\cr}$$

\noindent then $\mu$ is a measure.

\medskip

{\bf (c)} It will be convenient later to know something about the sets

\Centerline{$G_D=\{x:x\in X,\,x(D)=1\}$}

\noindent for $D\subseteq C$.   In particular, every $G_D$ belongs to
$\Sigma$.   \prooflet{\Prf\  If $\gamma\in D$, then $x_{\gamma}(D)=1$ so
$G_D=F_{\gamma,\{D\}}\in\Sigma_{\gamma}$.   If $\gamma\in C\setminus D$,
then $x_{\gamma}(D)=0$ so $G_D=X\setminus
F_{\gamma,\{D\}}\in\Sigma_{\gamma}$.\ \QeD}
Also\cmmnt{, of course,} $\{\gamma:x_{\gamma}\in G_D\}=D$.

\medskip

{\bf (d)} $(X,\Sigma,\mu)$ is complete.   \prooflet{\Prf\ Suppose that
$A\subseteq
E\subseteq\Sigma$ and that $\mu E=0$.   For every $\gamma\in C$,
$E\in\Sigma_{\gamma}$ and $x_{\gamma}\notin E$, so $F_{\gamma
K}\not\subseteq E$ for any $K\in\Cal K$ and there is a $K\in\Cal K$ such
that
\Centerline{$F_{\gamma K}\subseteq X\setminus E\subseteq X\setminus A$.}

\noindent Thus $A\in\Sigma_{\gamma}$;  as $\gamma$ is arbitrary,
$A\in\Sigma$.   As $A$ is arbitrary, $\mu$ is complete.\ \Qed}

\medskip

{\bf (e)} $(X,\Sigma,\mu)$ is semi-finite.   \prooflet{\Prf\ Let
$E\in\Sigma$ be a
set of positive measure.   Then there must be some $\gamma\in C$ such
that $x_{\gamma}\in E$.   Consider $E'=E\cap G_{\{\gamma\}}$.   As
$x_{\gamma}\in E'$, $\mu E'\ge 1>0$.   On the other hand, $\mu
G_{\{\gamma\}}=\#(\{\delta:\delta\in\{\gamma\}\})=1$, so $\mu E'=1$.
As $E$ is arbitrary, $\mu$ is
semi-finite.\ \Qed}

\medskip

{\bf (f)} $(X,\Sigma,\mu)$ is localizable.   \prooflet{\Prf\ Let $\Cal E$ be
any subset of $\Sigma$.   Set $D=\{\delta:\delta\in
C,\,x_{\delta}\in\bigcup\Cal E\}$.   Consider $G_D$.
For $H\in\Sigma$,

$$\eqalign{\mu(E\setminus H)&=0\text{ for every }E\in\Cal E\cr
&\iff x_{\gamma}\notin E\setminus H\text{ for every }E\in\Cal E,
\gamma\in C\cr
&\iff x_{\gamma}\in H\text{ for every }\gamma\in D\cr
&\iff x_{\gamma}\notin G_D\setminus H\text{ for every }\gamma\in C\cr
&\iff \mu(G_D\setminus H)=0.\cr}$$

\noindent Thus $G_D$ is an essential supremum for $\Cal E$ in
$\Sigma$.   As $\Cal E$ is arbitrary, $\mu$ is localizable.\ \Qed}

\medskip

{\bf (g)} $(X,\Sigma,\mu)$ is not strictly localizable.
\prooflet{\Prf\Quer\
Suppose, if possible, that $\langle X_j\rangle_{j\in J}$ is a
decomposition of $(X,\Sigma,\mu)$.   Set $J'=\{j:j\in J,\,\mu X_j>0\}$.
For each $j\in J'$, the set $C_j=\{\gamma:x_{\gamma}\in X_j\}$ must be
finite and non-empty.   Moreover, for each $\gamma\in C$, there must be
some $j\in J$ such that $\mu(G_{\{\gamma\}}\cap X_j)>0$, and in this
case $j\in J'$ and $\gamma\in C_j$.   Thus $C=\bigcup_{j\in J'}C_j$.
Because $\#(C)>\frak c$, $\#(J')>\frak c$ (2A1Ld).

For each $j\in J'$, choose $\gamma_j\in C_j$.   Then

\Centerline{$x_{\gamma_j}\in X_j\in\Sigma\subseteq\Sigma_{\gamma_j}$,}

\noindent so there must be a $K_j\in\Cal K$ such that
$F_{\gamma_j,K_j}\subseteq X_j$.

At this point I finally turn to the result cited at the start of this
example.   Because $\#(J')>\frak c$, there must be distinct $j$, $k\in
J'$ such that $x_{\gamma_j}$ and $x_{\gamma_k}$ agree on $K_j\cap K_k$.
We may therefore define $x\in X$ by saying that

$$\eqalign{x(\delta)&=x_{\gamma_j}(\delta)\text{ if }\delta\in K_j,\cr
&=x_{\gamma_k}(\delta)\text{ if }\delta \in K_k,\cr
&=0\text{ if }\delta\in C\setminus(K_j\cup K_j).\cr}$$

\noindent Now

\Centerline{$x\in F_{\gamma_j,K_j}\cap F_{\gamma_k,K_k}
\subseteq X_j\cap X_k$,}

\noindent and $X_j\cap X_k\ne\emptyset$;  contradicting the assumption
that the $X_j$ formed a decomposition of $X$.\ \Bang\Qed}

\medskip

{\bf (h)} $(X,\Sigma,\mu)$ is purely atomic.   \prooflet{\Prf\ If
$E\in\Sigma$ and
$\mu E>0$, then (as remarked in (e) above) there is a $\gamma\in C$ such
that $\mu(E\cap G_{\{\gamma\}})=1$;  now $E\cap G_{\{\gamma\}}$ must be
an atom.\ \Qed}

\cmmnt{\medskip

{\bf (i)} Accordingly $(X,\Sigma,\mu)$ is a complete, locally
determined, localizable, purely atomic measure space which is not
strictly localizable.
}%end of comment

\exercises{
\leader{216X}{Basic exercises (a)}
%\spheader 216Xa
In the construction of 216C, show that the
subspace measure on $\{1\}\times I$ is not semi-finite.
%216C

\spheader 216Xb Suppose, in 216D, that $I=\omega_1$.   (i) Show that the set $\{(\xi,\eta):\xi\le\eta<\omega_1\}$ is
measured by the measure
constructed by \Caratheodory's method from
$\mu^*\restrp\Cal P(I\times I)$,
but not by the subspace measure on $I\times I$.   (ii) Hence, or otherwise,
show that the subspace measure on $I\times I$ is not locally determined.
%216D

\spheader 216Xc In 216Ya, 252Yq and 252Ys below, I indicate how to
construct atomless versions of 216C, 216D and 216E, that is, atomless
complete measure
spaces of which the first is localizable but not locally determined, the
second is locally determined spaces but not localizable,
and the third is locally determined and localizable
but not strictly localizable.   Show how direct sums of these,
together with
counting measure and the examples described in this chapter, can be
assembled to provide all 56 examples called for by the discussion in
the introduction to this section.
%216E

\leader{216Y}{Further exercises (a)}
%\spheader 216Ya
Let $\lambda$ be Lebesgue measure on $[0,1]$, and $\Lambda$ its domain.
Set $Y=[0,1]\times\{0,1\}$ and write

\Centerline{$\Tau=\{F:F\subseteq Y,\,F^{-1}[\{0\}]\in\Lambda\}$,}

\Centerline{$\nu F=\lambda F^{-1}[\{0\}]$ for every $F\in\Tau$.}

\noindent Set

\Centerline{$\Tau_0=\{F:F\in\Tau,\,F^{-1}[\{0\}]\symmdiff
F^{-1}[\{1\}]$ is $\lambda$-negligible$\}$.}

\noindent Let $I$ be an uncountable set.   Set $X=Y\times I$,

\Centerline{$\Sigma=\{E:E\subseteq X,\,E^{-1}[\{i\}]\in\Tau$ for every $i\in
I$, $\{i:E^{-1}[\{i\}]\notin \Tau_0\}$ is countable$\}$,}

\Centerline{$\mu E=\sum_{i\in I}\nu E^{-1}[\{i\}]$ for $E\in\Sigma$.}

\noindent (i) Show that $(Y,\Tau,\nu)$ and $(Y,\Tau_0,\nu\restrp\Tau_0)$ are
complete probability spaces, and that for every $F\in\Tau$ there is an
$F'\in\Tau_0$ such
that $\nu(F\symmdiff F')=0$.   (ii) Show that $(X,\Sigma,\mu)$ is an
atomless complete localizable measure space which is not locally determined.
%216C

\spheader 216Yb Define a measure $\mu$ on $X=\omega_2\times\omega_2$
as follows.   Take $\Sigma$ to be the $\sigma$-algebra of subsets of $X$
generated by

\Centerline{$\{A\times\omega_2:A\subseteq\omega_2\}
\cup\{\omega_2\times\alpha:\alpha<\omega_2\}$.}

\noindent For $E\in\Sigma$ set

\Centerline{$W(E)=\{\xi:\xi<\omega_2$, $\sup E[\{\xi\}]=\omega_2\}$,}

\noindent and set $\mu E=\#(W(E))$ if this is finite, $\infty$ otherwise.
Show that $\mu$ is a measure on $X$, is localizable and locally determined,
but does not have locally determined negligible sets.   Find a subspace $Y$
of $X$ such that the subspace measure on $Y$ is not semi-finite.
%216D

\spheader 216Yc Show that in the space described in 216E every set has
a measurable envelope, but that this is not true in the spaces of 216C and 216D.
%216E

\spheader 216Yd Set $X=\omega_1\times\omega_2$.   For $E\subseteq X$ set

\Centerline{$A(E)=\{\zeta:$ for some $\xi$, just one of
$(\xi,\zeta)$, $(\xi,\zeta+1)$ belongs to $E\}$,}

\ifdim\pagewidth>400pt
\Centerline{$B(E)=\{\zeta:$ there are $\xi$, $\zeta'$ such that
$\zeta<\zeta'<\omega_2$ and just one of
$(\xi,\zeta)$, $(\xi,\zeta')$ belongs to $E\}$,}
\else
$$\eqalign{B(E)
&=\{\zeta:\text{ there are }\xi,\,\zeta'\text{ such that }
\zeta<\zeta'<\omega_2\cr&
\mskip100mu\text{ and just one of }
(\xi,\zeta),\,(\xi,\zeta')\text{ belongs to }E\},
\cr}$$
\fi

\Centerline{$W(E)=\{\xi:\#(E[\{\xi\}])=\omega_2\}$.}

\noindent Let $\Sigma$ be the set of subsets $E$ of $X$ such that $A(E)$ is
countable and $\#(B(E))\le\omega_1$.   For $E\in\Sigma$, set
$\mu E=\#(W(E))$ if this is finite, $\infty$ otherwise.   (i) Show that
$(X,\Sigma,\mu)$ is a measure space.   (ii) Show that if $\hat\mu$ is the
completion of $\mu$, then its domain is the set of subsets $E$ of $X$ such
that $A(E)$ is countable, and $\hat\mu$ is strictly localizable.   (iii)
Show that $\mu$ is not strictly localizable.
%216Yb 216D

\spheader 216Ye\dvAnew{2015} Show that there is a
complete atomless semi-finite measure space
with a singleton subset which is not negligible.   
\Hint{set $X=(\omega_1\times[0,1])\cup\{\omega_1\}$ and let
$\Sigma$ be the $\sigma$-algebra of subsets of $X$ generated by
$\{\{\xi\}\times E:\xi<\omega_1$, 
$E\subseteq[0,1]$ is Lebesgue measurable$\}$}.
%215E

}%end of exercises

\endnotes{
\Notesheader{216} The examples 216C-216E %216C 216D 216E
are designed to
form, with Lebesgue measure, a basis for constructing a complete set of
examples for the concepts listed in
211A-211K.   %211A 211B 211C 211D 211E 211F 211G 211H 211I 211J 211K
One does not really
expect to encounter these phenomena in applications, but a clear
understanding of the possibilities demonstrated by these examples is
part of a proper appreciation of their rarity.   Of course, if we add
further properties to our list -- for instance, the property of having
locally determined negligible sets (213I), or the property that every
subset should have a measurable envelope
(213Xl) -- then there are further positive results to complement 211L, and more examples to hunt for, like 216Yb.
But it is
time, perhaps past time, that we returned to the classical theorems
which apply to the measure spaces at the centre of the subject.
}%end of comment

\discrpage

\ifresultsonly\else
\frfilename{mt21conc.tex}
\versiondate{10.4.00/29.9.04}
\copyrightdate{2000}

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}

\def\chaptername{Taxonomy of measure spaces}
\def\sectionname{Concordance}

\newsection{}

I list here the section and paragraph numbers which have (to my 
knowledge) appeared in print in references to this chapter, and which 
have since been changed.

\medskip

\leader{2{}11Ya}{Countable-cocountable algebra of $\Bbb R$} 
This exercise, referred to in the 2002 edition of Volume 3, has been
moved to 211Ye.

\leaveitout{2{}13O (sufficient condition for strict localizability)
referred to in {\smc Bogachev 07}, is now in 213O.}

\leaveitout{2{}15A-2{}15{}B {\smc Fremlin 00}.  215A-215B
By the principle of exhaustion there is a countable set 
$\Cal F_0\subseteq\Cal F$ such that $F\setminus F_1$ is negligible for every $F\in\Cal F$, where $F_1=\bigcup\Cal F_0$.}

\leaveitout{{\smc Fremlin 00} \S214 subspace measures    
}

\leader{2{}13Y}{Inner measures} Exercise 2{}13Yc, referred to in the
2003 and 2006 editions of Volume 4, is now 213Yd.   %413C

\leader{2{}14J}{Subspace measures on measurable subspaces, direct sums} 
2{}14J-2{}14M, referred to in the 2002 and 2004
editions of Volume 3, the 2003 and 2006 editions of Volume 4, and the 2008
edition of Volume 5, have been moved to
214K-214N.  %214K 214L 214M 214N

\leader{2{}14N}{Upper and lower integrals} This result, referred to in the
2008 edition of Volume 5, has been moved to 214J.
%5{}37

\leader{2{}15Yc}{Measurable envelopes} This exercise, 
referred to in the 2000 edition of Volume 1, has been moved to 216Yc.
%132D

\leaveitout{
\leader{2{}16D}{A complete, locally determined space which is not localizable}
This example has been referred to in {\smc Okada \& Ricker 19}.
}

\leaveitout{\leader{2{}16}{Examples} referred to in {\smc Bogachev 07} are
now in \S216.}

\leaveitout{\bigskip

\noindent{\bf Reference}

Buskes G., de Jeu M., Dodds P., Schep A., Sukochev F., van Neerven J.
\& Wickstead A.\ [19] (eds.) 
{\it Positivity and Noncommutative Analysis}, Birkh\"auser 2019.

Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces', 
to appear in Fundamenta Math.

Okada S.\ \& Ricker W.J.\ [19] `Classes of Localizable Measure Spaces',
pp.\ 425-469 in 
{\smc Buskes de Jeu Dodds Schep Sukochev van Neerven Wickstead 19}.

}%end of leaveitout

\frnewpage

\fi
\references{
\frfilename{mt2r.tex}
\versiondate{17.12.12}
\copyrightdate{2000}

% 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 2}

\medskip

%A

Alexits G.\ [78] (ed.) {\it Fourier Analysis and Approximation Theory.}
North-Holland, 1978 (Colloq.\ Math.\ Soc.\ Janos Bolyai 19).

Antonov N.Yu.\ [96] `Convergence of Fourier series', East J.\ Approx.\ 7 (1996) 187-196.
\cmmnt{[\S286 {\it notes\/}.]}

Arias de Reyna J.\ [02] {\it Pointwise Convergence of Fourier Series.} Springer, 2002 (Lecture Notes in Mathematics 1785).
\cmmnt{[\S286 {\it notes\/}.]}

\medskip%B

Baker R.\ [04] `{}``Lebesgue measure'' on $\Bbb R^{\infty}$, II',
Proc.\ Amer.\ Math.\ Soc.\ 132 (2004) 2577-2591.
\cmmnt{[254Yb.]}

Bergelson V., March P.\ \& Rosenblatt J.\ [96] (eds.) {\it Convergence
in Ergodic Theory and Probability.}  de Gruyter, 1996.

\indexheader{Bogachev}
Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.

du Bois-Reymond P.\ [1876] `Untersuchungen \"uber die Convergenz und
Divergenz der Fouriersche Darstellungformeln', Abh.\ Akad.\ M\"unchen 12
(1876) 1-103.   \cmmnt{[\S282 {\it notes}.]}

Bourbaki N.\ [66] {\it General Topology.}  Hermann/Addison-Wesley, 1968.
\cmmnt{[2A5F.]}

Bourbaki N.\ [87] {\it Topological Vector Spaces.}  Springer, 1987.
\cmmnt{[2A5E.]}

\medskip%C

Carleson L.\ [66] `On convergence and growth of partial sums of
Fourier series', Acta Math.\ 116 (1966) 135-157.  \cmmnt{[\S282 {\it
notes}, \S286 {\it intro.}, \S286 {\it notes}.]}

Clarkson J.A.\ [1936] `Uniformly convex spaces', Trans.\ Amer.\ Math.\ Soc.\
40 (1936) 396-414.
\cmmnt{[244O.]}

\medskip%D

Defant A.\ \& Floret K.\ [93] {\it Tensor Norms and Operator Ideals},
North-Holland, 1993.   \cmmnt{[\S253 {\it notes}.]}

Doob J.L.\ [53] {\it Stochastic Processes.}  Wiley, 1953.
%used for 275Ye

Dudley R.M.\ [89] {\it Real Analysis and Probability.} Wadsworth \&
Brooks/Cole, 1989.   \cmmnt{[\S282 {\it notes}.]}

Dunford N.\ \& Schwartz J.T.\ [57] {\it Linear Operators I.}  Wiley,
1957 (reprinted 1988).  \cmmnt{[\S244 {\it notes}, 2A5J.]}

\medskip%E

Enderton H.B.\ [77] {\it Elements of Set Theory.}   Academic, 1977.
\cmmnt{[\S2A1.]}

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).   \cmmnt{[2A5F.]}

Etemadi N.\ [96] `On convergence of partial sums of independent random
variables', pp.\ 137-144 in {\smc Bergelson March \& Rosenblatt 96}.
\cmmnt{[272V.]}

Evans L.C.\ \& Gariepy R.F.\ [92] {\it Measure Theory and Fine
Properties of Functions.}   CRC Press, 1992.
\cmmnt{[263Ec, \S265 {\it notes}.]}
%Kirszbraun's theorem mentioned but not proved

\medskip%F

Federer H.\ [69] {\it Geometric Measure Theory.}  Springer, 1969
(reprinted 1996).
\cmmnt{[262C, 263Ec, \S264 {\it notes}, \S265 {\it notes},
\S266 {\it notes}.]}

Feller W.\ [66] {\it An Introduction to Probability Theory and its
Applications,} vol.\ II.   Wiley, 1966.  \cmmnt{[Chap.\ 27 {\it
intro.}, 274H, 275Xc, 285N.]}

Fremlin D.H.\ [74] {\it Topological Riesz Spaces and
Measure Theory.}   Cambridge U.P., 1974.
\cmmnt{[\S232 {\it notes}, 241F, \S244 {\it notes}, \S245 {\it notes},
\S247 {\it notes}.]}

Fremlin D.H.\ [93] `Real-valued-measurable cardinals', pp.\ 151-304 in
{\smc Judah 93}.  \cmmnt{[232H.]}

\medskip%G\medskip%H

Haimo D.T.\ [67] (ed.) {\it Orthogonal Expansions and their Continuous
Analogues.}   Southern Illinois University Press, 1967.

Hall P.\ [82] {\it Rates of Convergence in the Central Limit Theorem.}
Pitman, 1982.  \cmmnt{[274H.]}

Halmos P.R.\ [50] {\it Measure Theory.}  Van Nostrand, 1950.
\cmmnt{[\S251 {\it notes}, \S252 {\it notes}, 255Yn.]}

Halmos P.R.\ [60] {\it Naive Set Theory.} Van Nostrand, 1960.
\cmmnt{[\S2A1.]}

Hanner O.\ [56] `On the uniform convexity of $L^p$ and $l^p$',
Arkiv f\"or Matematik 3 (1956) 239-244.
\cmmnt{[244O.]}

Henle J.M.\ [86] {\it An Outline of Set Theory.}  Springer, 1986.
\cmmnt{[\S2A1.]}

Hoeffding W.\ [63]
`Probability inequalities for sums of bounded random variables',
J.\ Amer.\ Statistical Association 58 (1963) 13-30.
\cmmnt{[272W.]}

Hunt R.A.\ [67] `On the convergence of Fourier series', pp.\ 235-255
in {\smc Haimo 67}.  \cmmnt{[\S286 {\it notes}.]}

\medskip%I\medskip%J

Jorsb{\o}e O.G.\ \& Mejlbro L.\ [82] {\it The Carleson-Hunt Theorem on
Fourier Series.}  Springer, 1982 (Lecture Notes in Mathematics 911).
\cmmnt{[\S286 {\it notes}.]}

Judah H.\ [93] (ed.) {Proceedings of the Bar-Ilan Conference on Set
Theory and the Reals, 1991.} Amer.\ Math.\ Soc.\ (Israel Mathematical
Conference Proceedings 6), 1993.

\medskip%K

\indexheader{Kelley}
Kelley J.L.\ [55] {\it General Topology.} Van Nostrand, 1955.
\cmmnt{[2A5F.]}

Kelley J.L.\ \& Namioka I.\ [76] {\it Linear Topological Spaces.}
Springer, 1976.   \cmmnt{[2A5C.]}

\indexheader{Kirszbraun}
Kirszbraun M.D.\ [1934] `\"Uber die zusammenziehenden und Lipschitzian
Transformationen', Fund.\ Math.\ 22 (1934) 77-108.  \cmmnt{[262C.]}

Kolmogorov A.N.\ [1926] `Une s\'erie de Fourier-Lebesgue divergente
partout', C.\ R.\ Acad.\ Sci.\ Paris 183 (1926) 1327-1328.
\cmmnt{[\S282 {\it notes}.]}

Koml\'os J.\ [67] `A generalization of a problem of Steinhaus', Acta
Math.\ Acad.\ Sci.\ Hung.\ 18 (1967) 217-229.  \cmmnt{[276H.]}

K\"orner T.W.\ [88] {\it Fourier Analysis.}  Cambridge U.P., 1988.
\cmmnt{[\S282 {\it notes}.]}

K\"othe G.\ [69] {\it Topological Vector Spaces I.}   Springer, 1969.
\cmmnt{[2A5C, 2A5E, 2A5J.]}

Krivine J.-L.\ [71] {\it Introduction to Axiomatic Set Theory.}
D.~Reidel, 1971.  \cmmnt{[\S2A1.]}

\medskip%L

Lacey M.T.\ [05] {\it Carleson's Theorem: Proof, Complements, Variations.}
{\tt http://arxiv.org/pdf/math{\bsp}0307008v4{\penalty-100}.pdf}.
%expanded from 
% Publ.\ Mat.\ 48 (2004) 251-307
\cmmnt{[\S286 {\it notes}.]}

Lacey M.T.\ \& Thiele C.M.\ [00] `A proof of boundedness of the Carleson
operator', Math.\ Research Letters 7 (2000) 1-10.
\cmmnt{[\S286 {\it intro.}, 286H.]}

Lebesgue H.\ [72] {\it Oeuvres Scientifiques.}   L'Enseignement
Math\'ematique, Institut de Math\'ematiques, Univ.\ de Gen\`eve, 1972.
\cmmnt{[Chap.\ 27 {\it intro.}]}

Liapounoff A.\ [1901] `Nouvelle forme du th\'eor\`eme sur la limite
de probabilit\'e', M\'em.\ Acad.\ Imp.\ Sci.\ St-P\'etersbourg 12(5)
(1901) 1-24. \cmmnt{[274Xh.]}

Lighthill M.J.\ [59] {\it Introduction to Fourier Analysis and
Generalised Functions.}  Cambridge U.P., 1959.
\cmmnt{[\S284 {\it notes}.]}

Lindeberg J.W.\ [1922] `Eine neue Herleitung des Exponentialgesetzes in
der Wahrscheinlichkeitsrechnung', Math.\ Zeitschrift 15 (1922) 211-225.
\cmmnt{[274H, \S274 {\it notes}.]}

Lipschutz S.\ [64] {\it Set Theory and Related Topics.}  McGraw-Hill,
1964 (Schaum's Outline Series).  \cmmnt{[\S2A1.]}

Lo\`eve M.\ [77] {\it Probability Theory I.} Springer, 1977.
\cmmnt{[Chap.\ 27 {\it intro.}, 274H.]}

Luxemburg W.A.J.\ \& Zaanen A.C.\ [71] {\it Riesz Spaces
I.} North-Holland, 1971.
\cmmnt{[241F.]}


\medskip%M

Mozzochi C.J.\ [71] {\it On the Pointwise Convergence of Fourier
Series.}   Springer, 1971 (Lecture Notes in Mathematics 199).
\cmmnt{[\S286 {\it notes}.]}

\medskip%N

Naor A.\ [04] `Proof of the uniform convexity lemma',
{\tt http://www.cims.nyu.edu/\penalty-100$^{\sim}$naor/homepage files/\ifnum\stylenumber=12\break\fi{\vthsp}inequality\discretionary{}{}{}.pdf},
26.2.04.
\cmmnt{[244O.]}
%anon04.PDF

\medskip%O\medskip%P\medskip%Q\medskip%R

R\'enyi A.\ [70] {\it Probability Theory.}  North-Holland, 1970.
\cmmnt{[274H.]}

Roitman J.\ [90] {\it  An Introduction to Set Theory.}  Wiley, 1990.
\cmmnt{[\S2A1.]}

Roselli P.\ \& Willem M.\ [02] `A convexity inequality',
Amer.\ Math.\ Monthly 109 (2002) 64-70.
\cmmnt{[244Ym.]}

\medskip%S

%Saks S.\ [1924] 'Sur les nombres d\'eriv\'es des fonctions', 
%  Fundamenta Math.\ 5 (1924) 98-104.
% D-Y-S theorem
% books/saks.24.pdf   http://matwbn.icm.edu.pl/ksiazki/fm/fm5/fm5112.pdf

Saks S.\ [1924] 'Sur les nombres d\'eriv\'es des fonctions', Fund

Schipp F.\ [78] `On Carleson's method', pp.\ 679-695 in
{\smc Alexits 78}.   \cmmnt{[\S286 {\it notes}.]}

Semadeni Z.\ [71] {\it Banach spaces of continuous functions I.}
Polish Scientific Publishers, 1971.  \cmmnt{[\S253 {\it notes}.]}

Shiryayev A.N.\ [84] {\it Probability.} Springer, 1984. \cmmnt{[285N.]}

Steele J.M.\ [86] `An Efron-Stein inequality of nonsymmetric statistics',
Annals of Statistics 14 (1986) 753-758.
\cmmnt{[274Ya.]}

\medskip%T\medskip%U\medskip%V\medskip%W\medskip%Y\medskip%Z

Zaanen A.C.\ [83] {\it Riesz Spaces II.} North-Holland, 1983.
\cmmnt{[241F.]}

Zygmund A. [59] {\it Trigonometric Series.}   Cambridge U.P, 1959.
\cmmnt{[\S244 {\it notes}, \S282 {\it notes}, 284Xk.]}

\frnewpage

}%end of references

%---------------------------------------------------------------%

\end



