% 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{mt13.tex} 
\versiondate{9.6.16} 
\copyrightdate{1995} 
 
\def\chaptername{Complements} 
\def\sectionname{Introduction} 
 
\newchapter{13} 
 
In this chapter I collect a number of results which do not lie in the 
direct line of the argument from 111A (the definition of 
`$\sigma$-algebra') to 123C (Lebesgue's Dominated Convergence 
Theorem), 
but which nevertheless demand inclusion in this volume, being both 
relatively elementary, essential for further developments and necessary 
to a proper comprehension of what has already been done. 
The longest section is \S134, dealing with a few of the elementary 
special properties of Lebesgue measure;  in particular, its 
translation-invariance, the existence of non-measurable sets and 
functions, and the Cantor set.   The other sections are comparatively 
lightweight.   \S131 discusses (measurable) subspaces and the 
interpretation of the formula $\int_Ef$, generalizing the idea of an 
integral $\int_a^bf$ of a function over an interval.   \S132 introduces 
the outer measure associated with a measure, a kind of inverse of 
\Caratheodory's construction of a measure from an outer measure. 
\S\S133 and 135 lay out suitable conventions for dealing with `infinity' 
and complex numbers (separately!  they don't mix well) as values either 
of integrands or of integrals;  at the same time I mention `upper' 
and `lower' integrals.   Finally, in \S136, I give some theorems on 
$\sigma$-algebras of sets, describing how they can (in some of the most 
important cases) be generated by relatively restricted operations. 
 
\discrpage 
 
\frfilename{mt131.tex}
\versiondate{18.3.05}
\copyrightdate{1995}

\def\chaptername{Complements}
\def\sectionname{Measurable subspaces}

\newsection{131}

Very commonly we wish to integrate a
function over a subset of a measure space;  for instance, to form an
integral $\int_a^bf(x)dx$, where $a<b$ in $\Bbb R$.   As often as not,
we wish to do this when the function is partly or wholly undefined
outside the subset, as in such expressions as $\int_0^1\ln x\,dx$.   The
natural framework in which to perform such operations is that of
`subspace measures'.   If $(X,\Sigma,\mu)$ is a measure space and
$H\in\Sigma$, there is a natural subspace measure $\mu_H$ on $H$,
which I describe in this section.
I begin with the definition of this subspace measure (131A-131C), with a
description of integration with respect to it (131E-131H);   this
gives a solid foundation for the concept of `integration over a
(measurable) subset' (131D).

\leader{131A}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$H\in\Sigma$.   Set $\Sigma_H=\{E:E\in\Sigma,\,E\subseteq H\}$ and let
$\mu_H$ be the restriction of $\mu$ to $\Sigma_H$.   Then
$(H,\Sigma_H,\mu_H)$ is a measure space.

\proof{ Of course $\Sigma_H$ is just $\{E\cap H:E\in\Sigma\}$, and I
have noted already (in 121A) that this is a $\sigma$-algebra of subsets
of $H$.   It is now obvious that $\mu_H$ satisfies (iii) of 112A, so
that $(H,\Sigma_H,\mu_H)$ is a measure space.
}%end of proof of 131A

\leader{131B}{Definition} If $(X,\Sigma,\mu)$ is any measure space and
$H\in\Sigma$, then $\mu_H$, as defined in 131A, is the {\bf
subspace measure} on $H$.

When $X=\BbbR^r$, where $r\ge 1$, and $\mu$ is Lebesgue measure on
$\BbbR^r$, I will call a subspace measure $\mu_H$ {\bf Lebesgue measure on}
$H$.

\cmmnt{It is worth noting the following elementary facts.}

\leader{131C}{Lemma} Let $(X,\Sigma,\mu)$ be a measure space,
$H\in\Sigma$, and $\mu_H$ the subspace measure on $H$, with domain
$\Sigma_H$. Then

(a) for any $A\subseteq H$, $A$ is $\mu_H$-negligible iff it is
$\mu$-negligible;

(b) if $G\in\Sigma_H$ then $(\mu_H)_G$, the subspace measure on $G$ when
$G$ is regarded as a measurable subset of $H$, is identical to $\mu_G$,
the subspace measure on $G$ when $G$ is regarded as a measurable subset
of $X$.

\leader{131D}{Integration over subsets:  Definition} Let
$(X,\Sigma,\mu)$ be a measure
space, $H\in\Sigma$ and $f$ a real-valued function defined on a
subset of $X$.   By $\int_H f$ (or $\int_H f(x)\mu(dx)$, etc.) I shall
mean $\int(f\restr H)d\mu_H$, if this exists\cmmnt{, following the definitions of
131A-131B and 122M, and taking
$\dom(f\restr H)=H\cap\dom f$, $(f\restr H)(x)=f(x)$ for
$x\in H\cap\dom f$}.


\leader{131E}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space,
$H\in\Sigma$, and $f$ a
real-valued function defined on a subset $\dom f$ of $H$.   Set
$\tilde f(x)=f(x)$ if $x\in\dom f$, $0$ if $x\in X\setminus H$.   Then $\int fd\mu_H=\int \tilde fd\mu$ if either is defined in $\Bbb R$.

\proof{{\bf (a)} If $f$ is $\mu_H$-simple, it is expressible as
$\sum_{i=0}^na_i\chi E_i$, where $E_0,\ldots,E_n\in\Sigma_H$,
$a_0,\ldots,a_n\in\Bbb R$ and $\mu_H E_i<\infty$ for each $i$.   Now
$\tilde f$ also is equal to $\sum_{i=0}^na_i\chi E_i$ if this is now
interpreted as a function from $X$ to $\Bbb R$.   So

\Centerline{$\sum_{i=0}^na_i\mu_HE_i
=\sum_{i=0}^na_i\mu E_i=\int\tilde fd\mu$.}

\medskip

{\bf (b)} If $f$ is a non-negative $\mu_H$-integrable function,
there is a non-decreasing sequence $\sequencen{f_n}$ of non-negative
$\mu_H$-simple functions converging to $f\,\,\mu_H$-almost everywhere;
now $\sequencen{\tilde f_n}$ is a non-decreasing sequence of
$\mu$-simple functions converging to $\tilde f\,\,\mu$-a.e.\ (131Ca), and

\Centerline{$ \sup_{n\in\Bbb N}\int \tilde f_nd\mu
=\sup_{n\in\Bbb N}\int f_nd\mu_H
=\int fd\mu_H<\infty$,}

\noindent so $\int\tilde fd\mu$ exists and is equal to $\int fd\mu_H$.

\medskip

{\bf (c)} If $f$ is $\mu_H$-integrable, it is expressible as $f_1-f_2$
where $f_1$ and $f_2$ are
non-negative $\mu_H$-integrable functions, so that
$\tilde f=\tilde f_1-\tilde f_2$ and

\Centerline{$ \int\tilde fd\mu=\int\tilde f_1d\mu-\int\tilde f_2d\mu=\int
f_1d\mu_H-\int f_2d\mu_H=\int fd\mu_H$.}

\medskip

{\bf (d)} Now suppose that $\tilde f$ is $\mu$-integrable.   In this
case there is a $\mu$-conegligible $E\in\Sigma$ such that
$E\subseteq\dom\tilde f$ and $\tilde f\restr E$ is
$\Sigma$-measurable (122P).   Of course $\mu(H\setminus E)=0$ so
$E\cap H$ is $\mu_H$-conegligible;  also, for any $a\in\Bbb R$,

\Centerline{$\{x:x\in E\cap H,\,f(x)\ge a\}
=H\cap\{x:x\in E,\,\tilde f(x)\ge a\}\in\Sigma_H$,}

\noindent so $f\restr E\cap H$ is $\Sigma_H$-measurable, and $f$ is
$\mu_H$-virtually measurable and defined $\mu_H$-a.e.   Next, for
$\epsilon>0$,

\Centerline{$\mu_H\{x:x\in E\cap H,\,|f(x)|\ge\epsilon\}
=\mu\{x:x\in E,\,|\tilde f(x)|\ge\epsilon\}<\infty$,}

\noindent while if $g$ is a $\mu_H$-simple function and
$g\le|f|\,\,\mu_H$-a.e.\ then $\tilde g\le|\tilde f|\,\,\mu$-a.e.\ and

\Centerline{$\int g\,d\mu_H=\int\tilde g\,d\mu
\le\int|\tilde  f|d\mu<\infty$.}

\noindent By the criteria of 122J and 122P, $f$ is
$\mu_H$-integrable, so that again we have
$\int fd\mu_H=\int\tilde fd\mu$.
}%end of proof of 131E


\leader{131F}{Corollary} Let $(X,\Sigma,\mu)$ be a measure space and
$f$ a real-valued function defined on a subset $\dom f$ of $X$.

(a) If $H\in\Sigma$ and $f$ is defined almost everywhere in $X$, then
$f\restr H$ is $\mu_H$-integrable iff $f\times\chi H$ is
$\mu$-integrable, and in this case $\int_Hf=\int f\times\chi H$.

(b) If $f$ is $\mu$-integrable, then $f\ge 0$ a.e.\ iff $\int_Hf\ge 0$
for every $H\in\Sigma$.

(c) If $f$ is $\mu$-integrable, then $f=0$ a.e.\ iff $\int_Hf=0$ for
every $H\in\Sigma$.

\proof{{\bf (a)} Because $\dom f$ is $\mu$-conegligible,
$(f\restr H)\ssptilde$, as defined in 131E, is equal to
$f\times\chi H\,\,\mu$-a.e., so that, by 131E,

\Centerline{$ \int_Hfd\mu
=\int(f\restr H)\ssptilde d\mu
=\int(f\times\chi H)d\mu$}

\noindent if any one of the integrals exists.

\medskip

{\bf (b)(i)} If $f\ge 0\,\,\mu$-a.e., then for any $H\in\Sigma$ we
must have $f\restr H\ge 0\,\,\mu_H$-a.e., so $\int_Hf=\int(f\restr
H)d\mu_H\ge 0$.

\medskip

\quad{\bf (ii)} If $\int_Hf\ge 0$ for every $H\in\Sigma$, let
$E\in\Sigma$ be a conegligible subset of $\dom f$ such that $f\restr E$
is measurable.
Set $F=\{x:x\in E,\,f(x)<0\}$.   Then $\int_Ff\ge 0$;  by 122Rc, it
follows that $f\restr F=0\,\,\mu_F$-a.e., which is possible only if
$\mu F=0$, in which case $f\ge 0\,\,\mu$-a.e.

\medskip

{\bf (c)} Apply (b) to $f$ and to $-f$ to see that $f\le 0\le f$ a.e.
}%end of proof of 131F

\leader{131G}{Corollary} Let $(X,\Sigma,\mu)$ be a measure space and
$H\in\Sigma$ a conegligible set.   If $f$ is any real-valued function
defined on a subset of $X$, $\int_Hf=\int f$ if either is defined.

\proof{ In the language of 131E, $f=(f\restr H)\ssptilde\,\,\mu$-almost
everywhere, so that

\Centerline{$\int f=\int(f\restr H)\ssptilde=\int_Hf$}

\noindent if any of the integrals is defined.
}%end of proof of 131G

\leader{131H}{Corollary} Let $(X,\Sigma,\mu)$ be a measure space and $f$,
$g$ two $\mu$-integrable real-valued functions.

(a) If $\int_Hf\ge\int_Hg$ for every $H\in\Sigma$ then $f\ge g$ a.e.

(b) If $\int_Hf=\int_Hg$ for every $H\in\Sigma$ then $f=g$ a.e.

\proof{ Apply 131Fb-131Fc to $f-g$.
}%end of proof of 131H

\exercises{
\leader{131X}{Basic exercises $\pmb{>}$(a)}
%\spheader 131Xa
Let $(X,\Sigma,\mu)$ be a measure space, and $f$ a
real-valued function which is integrable over $X$.   For $E\in\Sigma$
set $\nu E=\int_Ef$.   (i) Show that if $E$, $F$ are disjoint members of
$\Sigma$ then $\nu(E\cup F)=\nu E+\nu F$.   \Hint{131E.}
(ii) Show that if $\sequencen{E_n}$ is a disjoint sequence in $\Sigma$
then $\nu(\bigcup_{n\in\Bbb N}E_n)=\sum_{n=0}^{\infty}\nu E_n$.
\Hint{123C.} (iii) Show that if $f$ is non-negative then $(X,\Sigma,\nu)$
is a measure space.

\sqheader 131Xb Let $\mu$ be Lebesgue measure on $\Bbb R$.   (i) Show that whenever $a\le b$ in $\Bbb R$ and $f$ is a real-valued function with $\dom f\subseteq\Bbb R$, then

\Centerline{$fd\mu=\int_{\coint{a,b}}fd\mu
=\int_{\ocint{a,b}}fd\mu=\int_{[a,b]}fd\mu$}

\noindent if any of these is defined.   \Hint{apply 131E to four different
versions of $\tilde f$.}   Write $\int_a^bfd\mu$ for the common value.
(ii) Show that if $a\le b\le c$ in $\Bbb R$ then, for any real-valued
function $f$, $\int_a^cfd\mu=\int_a^bfd\mu+\int_b^cfd\mu$ if either side
is defined.   (iii) Show that if $f$ is integrable over $\Bbb R$, then
$(a,b)\mapsto\int_a^bfd\mu$ is continuous.   \Hint{{\it Either} consider
simple functions $f$ first {\it or} consider
$\lim_{n\to\infty}\int_{a_n}^bfd\mu$ for monotonic sequences
$\sequencen{a_n}$.}

\spheader 131Xc Let $g:\Bbb R\to\Bbb R$ be a non-decreasing
function and $\mu_g$ the associated Lebesgue-Stieltjes measure (114Xa).
(i) Show that if $a\le b\le c$ in $\Bbb R$ then, for any real-valued
function $f$,
$\int_{\coint{a,c}}fd\mu_g
=\int_{\coint{a,b}}fd\mu_g+\int_{\coint{b,c}}fd\mu_g$ if
either side is defined.   (ii) Show that if $f$ is $\mu_g$-integrable
over $\Bbb R$, then $(a,b)\mapsto\int_{\coint{a,b}}fd\mu_g$ is continuous
on $\{(a,b):a\le b,\,g$ is continuous at both $a$ and $b\}$.

\leader{131Y}{Further exercises (a)}
%\spheader 131Ya
Let $(X,\Sigma,\mu)$ be a measure
space and $E\in\Sigma$ a measurable set of finite measure.   Let
$\sequencen{f_n}$ be a sequence of measurable real-valued functions, with
measurable domains\footnote{I am grateful to P.Wallace Thompson for
pointing out that this clause, or something with similar effect, is
necessary.}, such that
$f=\lim_{n\to\infty}f_n$ is defined almost everywhere in $E$ (following
the conventions of 121Fa).   Show that for every $\epsilon>0$ there
is a measurable set $F\subseteq E$ such that
$\mu(E\setminus F)\le\epsilon$ and $\sequencen{f_n}$ converges uniformly
to $f$ on $F$.   (This is {\bf Egorov's theorem}.)

}%end of exercises

\endnotes{
\Notesheader{131} If you want a quick definition of $\int_Hf$ for
measurable $H$, the simplest seems to be that of 131E, which enables
you to avoid the concept of `subspace measure' entirely.   I think
however that we really do need to be able to speak of `Lebesgue
measure on $[0,1]$', for instance, meaning the subspace measure
$\mu_{[0,1]}$ where $\mu$ is Lebesgue measure on $\Bbb R$.

This section has a certain amount of detailed technical analysis.   The
point is that from 131A on we generally have at least two measures in
play, and the ordinary language of measure theory -- words like
`measurable' and `integrable'
-- becomes untrustworthy in such contexts, since it omits the crucial
declarations of which $\sigma$-algebras or measures are under
consideration.   Consequently I have to use less elegant and more
explicit terminology.   I hope however that once you have worked
carefully through such results as 131F you will feel that the pattern
formed is reasonably coherent.   The general rule is that for {\it
measurable} subspaces there are no serious surprises (131Cb, 131Fa).

I ought to remark that there is also a standard definition of subspace
measure on {\it non-measurable} subsets of a measure space.   I have
given the definition already in 113Yb;  for the theory of integration,
extending the results above, I will wait until \S214.   There are
significant extra difficulties and the extra generality is not often
needed in elementary applications.

Let me call your attention to 131Fb-131Fc and 131Xa-131Xc;   these are
first steps to understanding `indefinite integrals', the functionals
$E\mapsto\int_Ef:\Sigma\to\Bbb R$ where $f$ is an integrable function.   I will return to these in Chapters 22 and 23.
}%end of notes



\discrpage







\frfilename{mt132.tex}
\versiondate{6.4.05}
\copyrightdate{1995}

\def\chaptername{Complements}
\def\sectionname{Outer measures from measures}

\newsection{132}

The next topic I wish to mention is a simple construction with
applications everywhere in measure theory.   With any measure there is
associated, in a straightforward way, a standard outer measure
(132A-132B).   If we start with Lebesgue measure we just return to
Lebesgue outer measure (132C).   I take the opportunity to
introduce the idea of `measurable envelope' (132D-132E).

\leader{132A}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space.
Define $\mu^*:\Cal PX\to[0,\infty]$ by writing

\Centerline{$\mu^*A=\inf\{\mu E:E\in\Sigma,\,A\subseteq E\}$}

\noindent for every $A\subseteq X$.   Then

(a) for every $A\subseteq X$
there is an $E\in\Sigma$ such that $A\subseteq E$ and $\mu^*A=\mu E$;

(b) $\mu^*$ is an outer measure on $X$;

(c) $\mu^*E=\mu E$ for every $E\in\Sigma$;

(d) a set $A\subseteq X$ is $\mu$-negligible iff $\mu^*A=0$;

(e) $\mu^*(\bigcup_{n\in\Bbb N}A_n)=\lim_{n\to\infty}\mu^*A_n$ for
every non-decreasing sequence $\sequencen{A_n}$ of subsets of $X$;

(f) $\mu^*A=\mu^*(A\cap F)+\mu^*(A\setminus F)$ whenever
$A\subseteq X$ and $F\in\Sigma$.

\proof{{\bf (a)} For each $n\in\Bbb N$ we
can choose an $E_n\in\Sigma$ such that $A\subseteq E_n$ and
$\mu E_n\le\mu^*A+2^{-n}$;  now $E=\bigcap_{n\in\Bbb N}E_n\in\Sigma$,
$A\subseteq E$ and

\Centerline{$\mu^*A\le\mu E\le\inf_{n\in\Bbb N}\mu E_n\le\mu^*A$.}

\medskip

{\bf (b)}(i) $\mu^*\emptyset=\mu\emptyset=0$.   (ii) If
$A\subseteq B\subseteq X$ then $\{E:A\subseteq
E\in\Sigma\}\supseteq\{E:B\subseteq E\in\Sigma\}$ so
$\mu^*A\le\mu^*B$.
(iii) If $\sequencen{A_n}$ is any sequence in $\Cal PX$, then for
each $n\in\Bbb N$ there is an $E_n\in\Sigma$ such that
$A_n\subseteq E_n$ and $\mu E_n=\mu^*A_n$;  now
$\bigcup_{n\in\Bbb N}A_n\subseteq\bigcup_{n\in\Bbb N}E_n\in\Sigma$ so

\Centerline{$\mu^*(\bigcup_{n\in\Bbb N}A_n)
\le\mu(\bigcup_{n\in\Bbb N}E_n)
\le\sum_{n=0}^{\infty}\mu E_n=\sum_{n=0}^{\infty}\mu^*A_n$.}

\medskip

{\bf (c)} This is just because $\mu E\le\mu F$ whenever $E$,
$F\in\Sigma$ and $E\subseteq F$.

\medskip

{\bf (d)} By (a), $\mu^*A=0$ iff there is an $E\in\Sigma$ such
that $A\subseteq E$ and $\mu E=0$;  but this is the definition of
`negligible set'.

\medskip

{\bf (e)} Of
course $\sequencen{\mu^*A_n}$ is a non-decreasing sequence with limit
at most $\mu^*A$, writing $A=\bigcup_{n\in\Bbb N}A_n$, just because
$\mu^*B\le\mu^*C$ whenever $B\subseteq C\subseteq X$.    For each
$n\in\Bbb N$, let $E_n\in\Sigma$ be such that $A_n\subseteq E_n$ and
$\mu E_n=\mu^*A_n$.   Set $F_n=\bigcap_{m\ge n}E_m$ for each $n$;
then $\sequencen{F_n}$ is a non-decreasing sequence in $\Sigma$, and
$A_n\subseteq F_n\subseteq E_n$, so $\mu^*A_n=\mu F_n$ for each
$n\in\Bbb N$.   Set $F=\bigcup_{n\in\Bbb N}F_n$;  then $A\subseteq F$
so

\Centerline{$\mu^*A\le\mu F=\lim_{n\to\infty}\mu
F_n=\lim_{n\to\infty}\mu^*A_n$.}

\noindent Thus $\mu^*A=\lim_{n\to\infty}\mu^*A_n$, as claimed.

\medskip

{\bf (f)} Of course $\mu^*A\le\mu^*(A\cap F)+\mu^*(A\setminus F)$, by
(b).   On the other hand, there is an $E\in\Sigma$ such that
$A\subseteq E$ and $\mu E=\mu^*A$, by (a), and now
$A\cap F\subseteq E\cap F\in\Sigma$,
$A\setminus F\subseteq E\setminus F\in\Sigma$ so

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

}%end of proof of 132A

\leader{132B}{Definition} If $(X,\Sigma,\mu)$ is a measure space, I
will call $\mu^*$, as defined in 132A, {\bf the outer measure
defined from $\mu$}.

\cmmnt{\medskip

\noindent{\bf Remark} If we start with an outer measure $\theta$ on a
set $X$, construct a measure $\mu$ from $\theta$ by \Caratheodory's
method, and
then construct the outer measure $\mu^*$ from $\mu$, it is not
necessarily the case that $\mu^*=\theta$.   \prooflet{\Prf\ Take any
set $X$ with at least three members, and set $\theta A=0$ if
$A=\emptyset$, $1$ if $A=X$, $\bover12$ otherwise.
Then $\dom\mu=\{\emptyset,X\}$ and
$\mu^*A=1$ for every non-empty $A\subseteq X$.\ \Qed}

However, this problem does not arise with Lebesgue outer measure.   I
state the next proposition in terms of Lebesgue measure on $\BbbR^r$,
but if you skipped \S115 I hope that you will still be able to make
sense of this, and later results, in terms of Lebesgue measure on
$\Bbb R$, by setting $r=1$.
}%end of comment

\leader{132C}{Proposition} If $\theta$ is Lebesgue outer
measure on $\BbbR^r$
and $\mu$ is Lebesgue measure, then $\mu^*$, as defined in 132A, is
equal to $\theta$.

\proof{ Let $A\subseteq\BbbR^r$.

\medskip

{\bf (a)} If $E$ is measurable and
$A\subseteq E$, then $\theta A\le\theta E=\mu E$;  so
$\theta A\le\mu^*A$.

\medskip

{\bf (b)} If $\epsilon>0$, there is a sequence
$\sequencen{I_n}$ of half-open intervals, covering $A$, with
$\sum_{n=0}^{\infty}\mu I_n\le\theta A+\epsilon$ (using 114G/115G to
identify $\mu I_n$ with the volume $\lambda I_n$ used in the
definition of $\theta$), so

\Centerline{$\mu^*A\le\mu(\bigcup_{n\in\Bbb N}I_n)
\le\sum_{n=0}^{\infty}\mu I_n
\le\theta A+\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\mu^*A\le\theta A$.
}%end of proof of 132C

\cmmnt{\medskip

\noindent{\bf Remark}  Accordingly it will henceforth be unnecessary
to distinguish
$\theta$ from $\mu^*$ when speaking of `Lebesgue outer measure'.
(In the language of 132Xa below, Lebesgue outer measure is
`regular'.)   In
particular (using 132Aa), if $A\subseteq\BbbR^r$ there is a measurable
set $E\supseteq A$ such that $\mu E=\theta A$ (compare 134Fc).
}%end of comment

\leader{132D}{Measurable envelopes}
\cmmnt{The following is a useful concept in
this context.}   If $(X,\Sigma,\mu)$ is a measure space and
$A\subseteq X$, a {\bf measurable envelope} (or {\bf measurable
cover}) of $A$ is a set $E\in\Sigma$ such
that $A\subseteq E$ and $\mu(F\cap E)=\mu^*(F\cap A)$ for every
$F\in\Sigma$.   \cmmnt{In general, not every set in a measure space
has a measurable envelope (I suggest examples in 216Yc in Volume 2).
But we do have the following.}

\leader{132E}{Lemma} Let $(X,\Sigma,\mu)$ be a measure space.

(a) If $A\subseteq E\in\Sigma$, then $E$ is a measurable envelope of
$A$ iff $\mu F=0$ whenever $F\in\Sigma$ and $F\subseteq E\setminus A$.

(b) If $A\subseteq E\in\Sigma$ and $\mu E<\infty$ then $E$ is a
measurable envelope of $A$ iff $\mu E=\mu^*A$.

(c) If $E$ is a measurable envelope of $A$ and $H\in\Sigma$,
then $E\cap H$ is a measurable envelope of $A\cap H$.

(d) Let $\sequencen{A_n}$ be a sequence of subsets of $X$.   Suppose
that each $A_n$ has a measurable envelope $E_n$.   Then
$\bigcup_{n\in\Bbb N}E_n$ is
a measurable envelope of $\bigcup_{n\in\Bbb N}A_n$.

(e) If $A\subseteq X$ can be covered by a sequence of sets of finite
measure, then $A$ has a measurable envelope.

(f) In particular, if $\mu$ is Lebesgue measure on $\BbbR^r$, then
every subset of $\BbbR^r$ has a measurable envelope for $\mu$.

\proof{{\bf (a)} If $E$ is a measurable envelope of $A$, $F\in\Sigma$
and $F\subseteq E\setminus A$, then

\Centerline{$\mu F=\mu(F\cap E)=\mu^*(F\cap A)=0$.}

\noindent If $E$ is not a measurable envelope of $A$, there is an
$H\in\Sigma$ such that $\mu^*(A\cap H)<\mu(E\cap H)$.   Let
$G\in\Sigma$
be such that $A\cap H\subseteq G$ and $\mu G=\mu^*(A\cap H)$, and set
$F=E\cap H\setminus G$.   Since $\mu G<\mu(E\cap H)$, $\mu F>0$;   but
also $F\subseteq E$ and $F\cap A\subseteq H\cap A \setminus G$ is
empty.

\medskip

{\bf (b)} If $E$ is a measurable envelope of $A$ then we must have

\Centerline{$\mu^*A=\mu^*(A\cap E)=\mu(E\cap E)=\mu E$.}

\noindent If $\mu E=\mu^*A$, and $F\in\Sigma$ is a subset of
$E\setminus A$, then $A\subseteq E\setminus F$, so
$\mu(E\setminus F)=\mu E$;  because $\mu E$ is finite, it follows that
$\mu F=0$, so the condition of (a) is satisfied and $E$ is a
measurable envelope of $A$.

\medskip

{\bf (c)} If $F\in\Sigma$ and $F\subseteq E\cap H\setminus A$, then
$F\subseteq E\setminus A$, so $\mu F=0$, by (a);  as $F$ is arbitrary,
$E\cap H$ is a measurable envelope of $A\cap H$, by (a) again.

\medskip

{\bf (d)} Write $A$ for $\bigcup_{n\in\Bbb N}A_n$ and
$E$ for $\bigcup_{n\in\Bbb N}E_n$.
Then $A\subseteq E$.   If $F\in\Sigma$ and $F\subseteq E\setminus A$,
then, for every $n\in\Bbb N$, $F\cap E_n\subseteq E_n\setminus A_n$, so
$\mu(F\cap E_n)=0$, by (a).   Consequently
$F=\bigcup_{n\in\Bbb N}F\cap E_n$ is negligible;  as $F$ is arbitrary,
$E$ is a measurable envelope of $A$.

\medskip

{\bf (e)} Let $\sequencen{F_n}$ be a sequence of sets
of finite measure covering $A$.   For each $n\in\Bbb N$, let
$E_n\in\Sigma$ be such that $A\cap F_n\subseteq E_n$ and
$\mu E_n=\mu^*(A\cap F_n)$ (using 132Aa above);  by (b), $E_n$ is a
measurable envelope of $A\cap F_n$.   By (d), $\bigcup_{n\in\Bbb
N}E_n$ is a measurable envelope of $\bigcup_{n\in\Bbb N}A\cap F_n=A$.

\medskip

{\bf (f)} In the case of Lebesgue measure on $\BbbR^r$, of course, the
same sequence $\sequencen{B_n}$ will work for every $A$, if we take
$B_n$ to be the half-open interval $\coint{-\tbf{n},\tbf{n}}$ for each
$n\in\Bbb N$, writing $\tbf{n}=(n,\ldots,n)$ as in \S115.
}%end of proof of 132E

\leader{132F}{Full outer measure}\cmmnt{ This is a convenient
moment at which to introduce the following term.}  If $(X,\Sigma,\mu)$
is a measure space, a set $A\subseteq X$ is {\bf of full outer
measure}
or {\bf thick} if $X$ is a measurable envelope of $A$\cmmnt{;  that
is, if $\mu^*(F\cap A)=\mu F$ for every $F\in\Sigma$;  equivalently,
if $\mu F=0$ whenever $F\in\Sigma$ and $F\subseteq X\setminus A$.   If
$\mu X<\infty$, $A\subseteq X$ has full outer measure iff
$\mu^*A=\mu X$}.

\exercises{
\leader{132X}{Basic exercises $\pmb{>}$(a)}
%\spheader 132Xa
Let $X$ be a set and $\theta$ an outer measure on $X$;
let $\mu$ be the measure on $X$ defined by \Caratheodory's method from
$\theta$, and $\mu^*$ the outer measure defined from $\mu$ by the
construction of 132A.   (i) Show that $\mu^*A\ge\theta A$ for every
$A\subseteq X$.   (ii) $\theta$ is said to be a {\bf regular outer
measure} if $\theta=\mu^*$.
%Rao `Carathe\'odory regular'
Show that if there is any measure $\nu$
on $X$ such that $\theta=\nu^*$ then $\theta$ is regular.   (iii) Show
that if $\theta$ is regular and $\sequencen{A_n}$ is a non-decreasing
sequence of subsets of $X$, then
$\theta(\bigcup_{n\in\Bbb N}A_n)=\lim_{n\to\infty}\theta A_n$.
%132B

\spheader 132Xb Let $(X,\Sigma,\mu)$ be a measure space and $H$ any
member of $\Sigma$.   Let $\mu_H$ be the subspace measure on $H$
(131B) and $\mu^*$, $\mu_H^*$ the outer measures defined from $\mu$,
$\mu_H$ respectively.   Show that $\mu_H^*=\mu^*\restrp\Cal PH$.
%132B

\spheader 132Xc Give an example of a measure space $(X,\Sigma,\mu)$
such that the measure $\check\mu$ defined by \Caratheodory's method from
the outer measure $\mu^*$ is a proper extension of $\mu$.   \Hint{take
$\mu X=0$.}
%132B

\sqheader 132Xd Let $(X,\Sigma,\mu)$ be a measure space and $A$ a
subset of $X$.   Suppose that $\sequencen{E_n}$ is a sequence in $\Sigma$
such that $\sequencen{A\cap E_n}$ is disjoint.   Show that
$\mu^*(A\cap\bigcup_{n\in\Bbb N}E_n)=\sum_{n=0}^{\infty}\mu^*(A\cap E_n)$.
\Hint{replace $E_n$ by $E_n'=E_n\setminus\bigcup_{i<n}E_i$, and
use 132Ae-132Af.}
%132B used in 548G

\spheader 132Xe Let $(X,\Sigma,\mu)$ be a measure space and
$\sequencen{A_n}$ any sequence of subsets of $X$.   Show that the
outer measure of $\bigcup_{n\in\Bbb N}\bigcap_{i\ge n}A_i$ is at most
$\liminf_{n\to\infty}\mu^*A_n$.
%132B

\spheader 132Xf Let $(X,\Sigma,\mu)$ be a measure space and suppose that
$A\subseteq B\subseteq X$ are such that $\mu^*A=\mu^*B<\infty$.   Show
that $\mu^*(A\cap E)=\mu^*(B\cap E)$ for every $E\in\Sigma$.   \Hint{a
measurable envelope of $B$ is a measurable envelope of $A$.}
%132E

\sqheader 132Xg Let $\nu_g$ be a Lebesgue-Stieltjes measure on $\Bbb R$,
constructed as in 114Xa from a non-decreasing function
$g:\Bbb R\to\Bbb R$.   Show that
(i) the outer measure $\nu_g^*$ derived from $\nu_g$ (132A)
coincides with the outer measure $\theta_g$ of 114Xa;
(ii) if $A\subseteq\Bbb R$ is any set, then $A$ has a
measurable envelope for the measure $\nu_g$.
%132E

\sqheader 132Xh Let $A\subseteq\BbbR^r$ be a set which is not
measured by Lebesgue measure $\mu$.   Show that there is a bounded
measurable set $E$ such that $\mu^*(E\cap A)=\mu^*(E\setminus A)=\mu E>0$.
\Hint{take $E=E'\cap E''\cap B$, where $E'$ is a measurable envelope
for $A$, $E''$ is a measurable envelope for $\BbbR^r\setminus A$, and
$B$ is a suitable bounded set.}
%132E

\spheader 132Xi Let $\mu$ be Lebesgue measure on $\BbbR^r$ and
$\Sigma$ its domain, and $f$ a real-valued function, defined on a subset
of $\BbbR^r$, which is not $\Sigma$-measurable.
Show that there are  $q<q'$ in $\Bbb Q$ and a bounded measurable set
$E$ such that

\Centerline{$\mu^*\{x:x\in E\cap\dom f,\,f(x)\le q\}
=\mu^*\{x:x\in E\cap\dom f,\,f(x)\ge q'\}=\mu E>0$.}

\noindent \Hint{take $E_q$, $E'_q$ to be measurable envelopes for
$\{x:f(x)\le q\}$, $\{x:f(x)>q\}$ for each $q$.   Find $q$ such that
$\mu(E_q\cap E'_q)>0$ and $q'$ such that $\mu(E_q\cap E'_{q'})>0$.}
%132E

\spheader 132Xj Check that you can do exercise 113Yc.
%132+

\spheader 132Xk Let $(X,\Sigma,\mu)$ be a measure space and $\mu^*$ the
outer measure defined from $\mu$.
Show that $\mu^*(A\cup B)+\mu^*(A\cap B)\le\mu^*A+\mu^*B$
for all $A$, $B\subseteq X$.
%132B

\leader{132Y}{Further exercises (a)}
%\spheader 132Ya
Let $(X,\Sigma,\mu)$ be a measure space and $\sequencen{f_n}$
a sequence of real-valued functions defined almost everywhere in $X$.
Suppose that $\sequencen{\epsilon_n}$ is a sequence of
non-negative real numbers such that

\Centerline{$\sum_{n=0}^{\infty}\epsilon_n<\infty$,
\quad
$\sum_{n=0}^{\infty}\mu^*\{x:|f_{n+1}(x)-f_n(x)|\ge\epsilon_n\}<\infty$.}

\noindent Show that $\lim_{n\to\infty}f_n$ is defined (as a
real-valued function) almost everywhere.
%132B

\spheader 132Yb Let $(X,\Sigma,\mu)$ be a measure space, $Y$ a set and
$f:X\to Y$ a function.   Let $\nu$ be the image measure $\mu f^{-1}$
(112Xf).   Show that $\nu^*f[A]\ge\mu^*A$ for every $A\subseteq X$.
%132B

\spheader 132Yc Let $(X,\Sigma,\mu)$ be a measure space with
$\mu X<\infty$.   Let $\sequencen{A_n}$ be a sequence of subsets of
$X$ such that $\bigcup_{n\in\Bbb N}A_n$ has full outer measure in $X$.
Show that there is a
partition $\sequencen{E_n}$ of $X$ into measurable sets such that
$\mu E_n=\mu^*(A_n\cap E_n)$ for every $n\in\Bbb N$.
%132F

\spheader 132Yd Let $(X,\Sigma,\mu)$ be a measure space and $\Cal A$ a
family of subsets of $X$ such that
$\bigcap_{n\in\Bbb N}A_n$ has full outer measure for every sequence
$\sequencen{A_n}$
in $\Cal A$.   Show that there is a measure $\nu$ on $X$, extending
$\mu$, such that every member of $\Cal A$ is $\nu$-conegligible.
%132F

\spheader 132Ye Check that you can do exercises 113Yg-113Yh.
%132+

\spheader 132Yf\dvAnew{2010} Let $(X,\Sigma,\mu)$ be a measure space.
Show that
$\mu^*:\Cal PX\to[0,\infty]$ is {\bf alternating of all orders}, that is,

\Centerline{$\sum_{J\subseteq I,\#(J)\text{ is even}}
   \mu^*(A\cup\bigcup_{i\in J}A_i)
\le\sum_{J\subseteq I,\#(J)\text{ is odd}}\mu^*(A\cup\bigcup_{i\in J}A_i)$}

\noindent whenever $I$ is a
non-empty finite set, $\familyiI{A_i}$ is a family of
subsets of $X$ and $A$ is another subset of $X$.
%mt13bits

\spheader 132Yg\dvAnew{2015} Let $(X,\Sigma,\mu)$ be a measure space.   Suppose that $A\subseteq B\subseteq C\subseteq X$ and that
$\mu^*(B\setminus A)=\mu^*B$.   Show that $\mu^*(C\setminus A)=\mu^*C$.
%132E out of order query
% reduce to case \mu^*B<\infty, take measurable envelope E of B, split pb into
% part in E and part in X\setminus E
}%end of exercises

\endnotes{
\Notesheader{132} Almost the most fundamental fact in measure theory
is that in all important measure spaces there are non-measurable sets.
(For Lebesgue measure see 134B below.)   One can respond to this fact
in a variety of ways.   An approach which works quite well is just to
ignore it.   The point is that, for very deep reasons, the sets and
functions which arise in ordinary applications nearly always are
measurable, or can be made so by elementary manipulations;  the only
exceptions I know of in applied mathematics appear in generalized
control theory.   As a pure mathematician I am uncomfortable with such
an approach, and as a measure theorist I think it closes the door on
some of the most subtle ideas of the theory.   In this treatise,
therefore, non-measurable sets will always be present, if only
subliminally.   In this section I have described two of the basic
methods of dealing with them:  the move from a measure to an outer
measure, which at least assigns some sort of size to an arbitrary set,
and the idea of `measurable envelope', which (when defined) describes
the region in which the non-measurable set has to be taken into
account.   In both cases we seek to describe the non-measurable set from
the outside, so to speak.   There are no real difficulties, and the only
points to take note of are that (i) outside the boundary marked by
132Ee measurable envelopes need not exist (ii) \Caratheodory's construction
of a measure from an outer measure, and the construction here of an
outer measure from a measure, are closely related (132C, 132Xg, 113Yc,
132Xa(i)), but are not quite inverses of each other in general (132B,
132Xc).
}%end of notes

\discrpage

\frfilename{mt133.tex} 
\versiondate{29.3.10} 
\copyrightdate{1994} 
      
\def\chaptername{Complements} 
\def\sectionname{Wider concepts of integration} 
      
\newsection{133} 
      
There are various contexts in which it is useful to be able to assign a 
value to the integral of a function which is not quite covered by the 
basic definition in 122M.   In this section I offer suggestions 
concerning the assignment of the values $\pm\infty$ to integrals of 
real-valued functions (133A), the integration 
of complex-valued functions (133C-133H) and upper and lower integrals 
(133I-133L).   In \S135 below I will discuss a further elaboration of 
the ideas of Chapter 12. 
      
\leader{133A}{Infinite integrals} It is normal to restrict the phrase 
`$f$ is integrable' to functions $f$ to which a finite integral 
$\int f$ can 
be assigned\cmmnt{ (just as a series is called `summable' only when 
a finite 
sum can be assigned to it)}.   But for non-negative functions it is 
sometimes convenient to write `$\int f=\infty$' if, in some sense, the 
only way in which $f$ fails to be integrable is that the integral is too 
large;  that is, $f$ is defined almost everywhere, is $\mu$-virtually 
measurable, and either 
      
\Centerline{$\{x:x\in\dom f,\,f(x)\ge\epsilon\}$} 
      
\noindent includes a set of infinite measure for some $\epsilon>0$, or 
      
\Centerline{$\sup\{\int h:h\text{ is simple},\,h\leae f\}=\infty$.} 
      
\noindent\cmmnt{(Compare 122J.) }Under this rule,\cmmnt{ we shall still 
have}
      
\Centerline{$\int f_1+f_2=\int f_1+\int f_2$, 
\quad$\int cf=c\int f$} 
      
\noindent whenever $c\in\coint{0,\infty}$ and $f_1$, $f_2$, $f$ are 
non-negative functions for which $\int f_1$, $\int f_2$, $\int f$ are 
defined in $[0,\infty]$. 
      
We can therefore\cmmnt{ repeat the definition 122M and} say that 
      
\Centerline{$\int f_1-f_2=\int f_1-\int f_2$} 
      
\noindent whenever $f_1$, $f_2$ are real-valued functions such that 
$\int f_1$, $\int f_2$ are defined in $[0,\infty]$ and are not both 
infinite\cmmnt{;  the last condition being imposed to avoid the 
possibility of being asked to calculate $\infty-\infty$}. 
      
We still have the rules that 
      
\Centerline{$\int f+g=\int f+\int g$, 
\quad$\int(cf)=c\int f$, 
\quad$\int|f|\ge|\int f|$} 
      
\noindent at least when the right-hand-sides can be interpreted, 
allowing $0\cdot\infty=0$, but not allowing any interpretation of 
$\infty-\infty$;  and $\int f\le\int g$ whenever both integrals are 
defined and $f\leae g$.   \cmmnt{(But of course it is now possible 
to have $f\le g$ and $\int f=\int g=\pm\infty$ without $f$ and $g$  
being equal almost everywhere.)} 
      
Setting $f^+(x)=\max(f(x),0)$, $f^-(x)=\max(-f(x),0)$ for $x\in\dom f$, 
then 
      
\Centerline{$\int f=\infty\,\iff\,\int f^+=\infty$ and $f^-$ is 
integrable,} 
      
\Centerline{$\int f=-\infty\,\iff\,f^+$ is integrable and 
$\int f^-=\infty$.} 
 
\cmmnt{For further ideas in this direction, see \S135 below.} 
%J.Neveu is said to have called functions integrable in this sense
%quasi-integrable
          
\leader{133B}{Functions with exceptional values} It is also convenient 
to allow as `integrable' functions $f$ which take occasional values 
which are not real\cmmnt{ -- typically, where a formula for $f(x)$ 
allows the 
value `$\infty$' on some convention}.   For such a function I will 
write $\int f=\int\tilde f$ if $\int\tilde f$ is defined, where 
      
\Centerline{$\dom\tilde f=\{x:x\in\dom f,\,f(x)\in\Bbb R\}$, 
\quad$\tilde f(x)=f(x)$ for $x\in\dom\tilde f$.} 
      
\cmmnt{\noindent Since in this convention I still require $\tilde f$ 
to be 
defined almost everywhere in $X$, the set $\{x:x\in\dom 
f,\,f(x)\notin\Bbb R\}$ will have to be negligible.} 
      
\cmmnt{ 
\leader{133C}{Complex-valued functions} All the theory of 
measurable and 
integrable functions so far developed has been devoted to real-valued 
functions.   There are no substantial new ideas required to deal with 
complex-valued functions, but perhaps I should spell out some of the 
details, since 
there are many applications in which complex-valued functions are the 
most natural context in which to work. 
}%end of comment 
      
\leader{133D}{Definitions (a)} Let $X$ be a set and $\Sigma$ a 
$\sigma$-algebra of subsets of $X$.   If $D\subseteq X$ and 
$f:D\to\Bbb C$ is a function, then we 
say that $f$ is {\bf measurable} if its real and imaginary parts 
$\Real f$, $\Imag f$ are measurable\cmmnt{ in the sense of 121B-121C}. 
      
\medskip 
      
{\bf (b)} Let $(X,\Sigma,\mu)$ be a measure space.   If $f$ is a 
complex-valued function defined on a conegligible 
subset of $X$, we say that $f$ is {\bf integrable} if its real and 
imaginary parts are integrable, and then 
      
\Centerline{$\int f=\int\Real f+i\int\Imag f$.} 
      
\medskip 
      
{\bf (c)} Let $(X,\Sigma,\mu)$ be a measure space, $H\in\Sigma$ and $f$ 
a complex-valued function defined on a subset of $X$.   Then 
$\int_Hf$ is $\int(f\restr H)d\mu_H$ if this is defined in the sense of 
(b), taking the subspace measure $\mu_H$ to be that of 131A-131B. 
      
\ifnum\stylenumber=11\ifresultsonly\eject\fi\fi 
      
\leader{133E}{Lemma} (a) If $X$ is a set, $\Sigma$ is a $\sigma$-algebra 
of subsets of $X$, and $f$ and $g$ are measurable complex-valued 
functions with domains $\dom f$, $\dom g\subseteq X$, then 
      
\quad(i) $f+g:\dom f\cap\dom g\to\Bbb C$ is measurable; 
      
\quad(ii) $cf:\dom f\to\Bbb C$ is measurable, for every $c\in\Bbb C$; 
      
\quad(iii) $f\times g:\dom f\cap\dom g\to\Bbb C$ is measurable; 
      
\quad(iv) $f/g:\{x:x\in\dom f\cap\dom g,\,g(x)\ne 0\}\to\Bbb C$ is 
measurable; 
      
\quad(v) $|f|:\dom f\to\Bbb R$ is measurable. 
      
{(b)} If $\sequencen{f_n}$ is a sequence of measurable 
complex-valued functions defined on subsets of $X$, then 
$f=\lim_{n\to\infty}f_n$ is measurable, if we take $\dom f$ to be 
      
$$\eqalign{\{x:x\in\bigcup_{n\in\Bbb N}&\bigcap_{m\ge n}\dom f_m, 
  \,\lim_{n\to\infty}f_n(x)\text{ exists in }\Bbb C\}\cr 
&=\dom(\lim_{n\to\infty}\Real f_n) 
  \cap\dom(\lim_{n\to\infty}\Imag f_n).\cr}$$ 
      
\proof{{\bf (a)} All are immediate from 121E, if you write down 
the formulae for the real and imaginary parts of $f+g,\ldots,|f|$ in 
terms of the real and imaginary parts of $f$ and $g$. 
      
\medskip 
      
{\bf (b)} Use 121Fa. 
}%end of proof of 133E 
      
\leader{133F}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space. 
      
(a) If $f$ and $g$ are integrable 
complex-valued functions defined 
on conegligible subsets of $X$, then $f+g$ and $cf$ are integrable, 
$\int f+g=\int f+\int g$ and $\int cf=c\int f$, for 
every $c\in\Bbb C$.
      
{(b)} If $f$ is a complex-valued function defined on a conegligible 
subset of $X$, then $f$ is integrable iff $|f|$ is integrable and $f$ is 
$\mu$-virtually measurable\cmmnt{, that is, $\Real f$ and $\Imag f$ are
$\mu$-virtually measurable}. 
      
\proof{{\bf (a)} Use 122Oa-122Ob. 
      
\medskip 
      
{\bf (b)} The point is that $|\Real f|$, 
$|\Imag f|\le|f|\le|\Real f|+|\Imag f|$;  now we need only apply 122P an 
adequate number of times. 
}%end of proof of 133F 
      
\leader{133G}{Lebesgue's Dominated Convergence Theorem} Let 
$(X,\Sigma,\mu)$ 
be a measure space and $\langle f_n\rangle_{n\in\Bbb N}$ a sequence of 
integrable complex-valued functions on $X$ such that 
$f(x)=\lim_{n\to\infty}f_n(x)$ 
exists in $\Bbb C$ for almost every $x\in X$.   Suppose moreover that 
there is a real-valued 
integrable function $g$ on $X$ such that $|f_n|\leae g$ for each 
$n$.   Then $f$ is integrable and $\lim_{n\to\infty}\int f_n$ exists and is equal to $\int f$. 
      
\proof{Apply 123C to the sequences $\sequencen{\Real f_n}$ and 
$\sequencen{\Imag f_n}$. 
}%end of proof of 133G 
      
\leader{133H}{Corollary} Let $(X,\Sigma,\mu)$ be a measure space and 
$\ooint{a,b}$ a non-empty open interval in $\Bbb R$.   Let 
$f:X\times\ooint{a,b}\to\Bbb C$ be a function such that 
      
\inset{(i) the integral $F(t)=\int f(x,t)dx$ is defined for every 
$t\in\ooint{a,b}$;} 
      
\inset{(ii) the partial derivative $\pd{f}{t}$ of $f$ with respect to 
the second variable is defined everywhere in $X\times\ooint{a,b}$;} 
      
\inset{(iii) there is an integrable function $g:X\to\coint{0,\infty}$ 
such that $|\pd{f}{t}(x,t)|\le g(x)$ for every $x\in X$, $t\in\ooint{a,b}$.} 
      
\noindent Then the derivative $F'(t)$ and the integral 
$\biggerint\pd{f}{t}(x,t)dx$ exist for every $t\in\ooint{a,b}$, 
and are equal. 
      
\proof{Apply 123D to $\Real f$ and $\Imag f$.} 
      
      
\leader{133I}{Upper and lower integrals}\dvArevised{2007}\cmmnt{ I return  
now to real-valued functions.}   Let $(X,\Sigma,\mu)$ be a 
measure space and $f$ a real-valued function defined almost everywhere 
in $X$.   Its {\bf upper integral} is 
      
\Centerline{$\overlineint f 
=\inf\{\int g:\int g$ is defined in the sense of 133A and $f\leae g\}$,} 
      
\noindent allowing $\infty$ for $\inf\{\infty\}$ or $\inf\emptyset$ 
and $-\infty$ for 
$\inf\Bbb R$.   Similarly, the {\bf lower integral} of $f$ is 
      
\Centerline{$\underlineint f 
=\sup\{\int g:\int g$ is defined, $f\geae g\}$,} 
      
\noindent allowing $-\infty$ for $\sup\{-\infty\}$ or $\sup\emptyset$ 
and $\infty$ for $\sup\Bbb R$. 
      
\leader{133J}{Proposition}\dvArevised{2007} Let $(X,\Sigma,\mu)$ be a  
measure space. 
      
(a) Let $f$ be a real-valued function defined almost everywhere in $X$. 
      
\quad (i) If $\overline{\int}f$ is finite, then there is 
an integrable $g$ such that $f\leae g$ and $\int g=\overline{\int}f$. 
In this case, 
 
\Centerline{$\{x:x\in\dom f\cap\dom g$, $g(x)\le f(x)+g_0(x)\}$} 
 
\noindent has full outer measure for every measurable function 
$g_0:X\to\ooint{0,\infty}$. 
      
\quad (ii) If $\underline{\int}f$ is finite, 
then there is an integrable $h$ such that $h\leae f$ and 
$\int h=\underline{\int}f$.   In this case, 
 
\Centerline{$\{x:x\in\dom f\cap\dom h$, $f(x)\le h(x)+h_0(x)\}$} 
 
\noindent has full outer measure for every measurable function 
$h_0:X\to\ooint{0,\infty}$. 
      
{(b)} For any real-valued functions $f$, $g$ defined on conegligible 
subsets of $X$ and any $c\ge 0$, 
      
\vskip 2pt 
      
\quad(i) $\underline{\int}f\le\overline{\int}f$, 
      
\vskip2pt 
      
\quad (ii) $\overline{\int}f+g\le\overline{\int}f+\overline{\int}g$, 
      
\vskip2pt 
      
\quad (iii) $\overline{\int}cf=c\overline{\int}f$, 
      
\vskip2pt 
      
\quad (iv) $\underline{\int}(-f)=-\overline{\int}f$, 
      
\vskip2pt 
      
\quad (v) $\underline{\int}f+g\ge\underline{\int}f+\underline{\int}g$, 
      
\vskip2pt 
      
\quad (vi) $\underline{\int}cf=c\underline{\int}f$ 
      
\vskip2pt 
      
\noindent whenever the right-hand-sides do not involve adding $\infty$ 
to $-\infty$. 
      
(c) If $f\leae g$ then $\overline{\int}f\le\overline{\int}g$ and 
$\underline{\int}f\le\underline{\int}g$. 
      
(d) A real-valued function $f$ defined almost everywhere in $X$ is 
integrable iff 
      
\Centerline{$\overlineint f=\underlineint f=a\in\Bbb R$,} 
      
\noindent and in this case $\int f=a$. 
      
(e) $\mu^*A=\overline{\int}\chi A$ for every $A\subseteq X$. 
 
\proof{{\bf (a)(i)} For each 
$n\in\Bbb N$, choose a function $g_n$ such that $f\leae g_n$ 
and $\int g_n$ is defined and at most $2^{-n}+\overline{\int}f$;  as 
$\overline{\int}f\le\int g_n$, $\int g_n$ is finite,  
so $g_n$ is integrable.   Set  
$h_n=\inf_{i\le n}g_i$ for each $n$;  then $h_n$ is integrable (because 
$|h_n-g_0|\le\sum_{i=0}^n|g_i-g_0|$ on $\bigcap_{i\le n}\dom g_i$), and 
$f\leae h_n$, so 
      
\Centerline{$\overlineint f\le\int h_n 
\le\int g_n\le 2^{-n}+\overlineint f$.} 
      
\noindent By B.Levi's theorem (123A), applied to $\sequencen{-h_n}$, 
$g(x)=\inf_{n\in\Bbb N}h_n(x)\in\Bbb R$ for almost every $x$, and 
$\int g=\inf_{n\in\Bbb N}\int h_n=\overline{\int}f$;  also, of course, 
$f\leae g$. 
 
Now take a measurable function $g_0:X\to\ooint{0,\infty}$, 
and consider the set 
 
\Centerline{$A=\{x:x\in\dom f\cap\dom g$, $g(x)\le f(x)+g_0(x)\}$.} 
 
\noindent\Quer\ If $A$ does not have full outer measure, there is a 
non-negligible measurable set $F\subseteq X\setminus A$.   Since $g_0$ is 
strictly positive, $F=\bigcup_{n\in\Bbb N}F_n$ where  
$F_n=\{x:x\in F$, $g_0(x)\ge 2^{-n}\}$,  
and there is an $n\in\Bbb N$ such that $\mu F_n>0$.   Consider the 
function $g_1=g-2^{-n}\chi F$.   Then $f\leae g_1$.   Also  
$\int g_1=\int g-2^{-n}\mu F_n$ is 
strictly less than $\int g$, so $\overline{\int}f<\int g$.\ \Bang 
 
\medskip 
      
\quad{\bf (ii)} Argue similarly, or use (b-iv). 
      
\medskip 
      
{\bf (b)(i)} If either $\underline{\int}f=-\infty$ or 
$\overline{\int}f=\infty$ this is trivial.   Otherwise it follows at 
once from the fact that if $g\leae f\leae h$ 
then $\int g\le\int h$ if the integrals are defined (in the wide sense). 
      
\medskip 
      
\quad{\bf (ii)} If $a>\overline{\int}f+\overline{\int}g$, neither 
$\overline{\int}f$ nor $\overline{\int}g$ can be $\infty$, so there must be  
functions $f_1$, $g_1$ such that $f\leae f_1$, 
$g\leae g_1$ and $\int f_1+\int g_1\le a$.   Now $f+g\leae f_1+g_1$, so 
      
\Centerline{$ 
\overlineint f+g\le\int f_1+g_1\le a$.} 
      
\noindent As $a$ is arbitrary, we have the result. 
      
\medskip 
      
\quad{\bf (iii)($\pmb{\alpha}$)} If $c=0$ this is trivial.   {\bf 
($\pmb{\beta}$)} If $c>0$ and $a>c\overline{\int}f$, there must be an 
$f_1$ such that $f\leae f_1$ and $c\int f_1\le a$.   Now 
$cf\leae cf_1$ and $\int cf_1\le a$, so $\overline{\int}cf\le a$. 
As $a$ is arbitrary, $\overline{\int}cf\le c\overline{\int}f$. 
{\bf($\pmb{\gamma}$)}  Still supposing that $c>0$, we also have 
      
\Centerline{$c\overlineint f=c\overlineint c^{-1}cf 
\le cc^{-1}\overlineint cf=\overlineint cf$,} 
      
\noindent so we get equality. 
      
\medskip 
      
\quad{\bf (iv)} This is just because $\int(-f_1)=-\int f_1$ for any 
function $f_1$ for which either integral is defined. 
      
\medskip 
      
\quad{\bf (v)-(vi)} Use (iv) to turn $\underline{\int}$ into 
$\overline{\int}$, and apply (ii) or (iii). 
      
\medskip 
      
{\bf (c)} These are immediate from the definitions, because (for 
instance) if $g\leae h$ then $f\leae h$. 
      
\medskip 
      
{\bf (d)} If $f$ is integrable, then 
      
\Centerline{$\overlineint f=\int f=\underlineint f$} 
      
\noindent by 122Od.   If 
$\overline{\int}f=\underline{\int}f=a\in\Bbb R$, then, by (a), there are  
integrable $g$, $h$ such that $g\leae f\leae h$ 
and $\int g=\int h=a$, so that $g\eae h$, by 122Rc, $g\eae f\eae h$  
and $f$ is integrable, by 122Rb. 
 
\medskip 
 
{\bf (e)} If $E\supseteq A$ is measurable, then  
 
\Centerline{$\mu E=\int\chi E\ge\overlineint\chi A$;} 
 
\noindent as $E$ is arbitrary, $\mu^*A\ge\overline{\int}\chi A$.    
If $\int g$ is defined and $\chi A\leae g$, let $E\subseteq\dom g$ be a  
conegligible measurable set such that $g\restr E$ is measurable, and set  
$F=\{x:x\in E$, $g(x)\ge 1\}$.   Then $A\setminus F$ is negligible, so  
$\mu^*A\le\mu F\le\int g$;  as $g$ is arbitrary,  
$\mu^*A\le\overline{\int}\chi A$. 
}%end of proof of 133J 
      
\cmmnt{\medskip 
      
\noindent{\bf Remark} I hope that the formulae here remind you of 
$\limsup$, $\liminf$.} 
      
\leader{133K}{Convergence theorems for upper 
\dvrocolon{integrals}}\cmmnt{ We have the 
following versions of B.Levi's theorem and Fatou's Lemma. 
      
\medskip 
      
\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and 
$\sequencen{f_n}$ a sequence of real-valued functions defined almost 
everywhere in $X$. 
      
(a) If, for each $n$, $f_n\leae f_{n+1}$, and 
$-\infty<\sup_{n\in\Bbb N}\overline{\int}f_n<\infty$, then 
$f(x)=\sup_{n\in\Bbb N}f_n(x)$ is defined in $\Bbb R$ for almost every 
$x\in X$, and 
$\overline{\int}f=\sup_{n\in\Bbb N}\overline{\int}f_n$. 
      
(b) If, for each $n$, $f_n\ge 0$ a.e., and 
$\liminf_{n\to\infty}\overline{\int}f_n<\infty$, then 
$f(x)=\liminf_{n\to\infty}f_n(x)$ is defined in $\Bbb R$ for almost 
every $x\in X$, and 
$\overline{\int}f\le\liminf_{n\to\infty}\overline{\int}f_n$. 
      
\proof{{\bf (a)} Set $c=\sup_{n\in\Bbb N}\overline{\int}f_n$.    For 
each $n$, there is an integrable function $g_n$ such that $f_n\leae g_n$ 
and $\int g_n=\overline{\int}f_n$ (133J(a-i)).   
Set $g'_n=\min(g_n,g_{n+1})$; 
then $g'_n$ is integrable and $f_n\leae g'_n\leae g_n$, so 
      
\Centerline{$\overlineint f_n\le\int g'_n 
\le\int g_n=\overlineint f_n$} 
      
\noindent and $g'_n$ must be equal to $g_n$ a.e.   Consequently 
$g_n\leae g_{n+1}$, for each $n$, while 
$\sup_{n\in\Bbb N}\int g_n=c<\infty$. 
By B.Levi's theorem, $g=\sup_{n\in\Bbb N}g_n$ is defined, as a 
real-valued function, almost everywhere in $X$, and $\int g=c$.   Now of 
course $f(x)$ is defined, and not greater than $g(x)$, for any 
$x\in\dom g\cap\bigcap_{n\in\Bbb N}\dom f_n$ such that $f_n(x)\le g_n(x)$ 
for every $n$, that is, for almost every $x$;  so 
$\overline{\int}f\le\int g=c$.   On the other hand, $f_n\leae f$, so $\overline{\int}f_n\le\overline{\int}f$, 
for every $n\in\Bbb N$;  it follows that $\overline{\int}f$ must be at 
least $c$, and is therefore equal to $c$, as required. 
      
\medskip 
      
{\bf (b)} The argument follows that of 123B.   Set 
$c=\liminf_{n\to\infty}\overline{\int}f_n$.   For each $n$, set 
$g_n=\inf_{m\ge n}f_n$;  then  
$\overline{\int}g_n\le\inf_{m\ge n}\overline{\int}f_m\le c$.    
We have $g_n(x)\le g_{n+1}(x)$ for every 
$x\in\dom g_n$, that is, almost everywhere, for each $n$;  so, by (a), 
      
\Centerline{$\overlineint g 
=\sup_{n\in\Bbb N}\overlineint g_n\le c$,} 
      
\noindent where 
      
\Centerline{$g=\sup_{n\in\Bbb N}g_n\eae\liminf_{n\to\infty}f_n$,} 
      
\noindent and $\overline{\int}\liminf_{n\to\infty}f_n\le c$, as claimed. 
}%end of proof of 133K 
 
\leader{*133L}{}\dvAnew{2007}\cmmnt{ The following is at a less fundamental 
level than the results in 133J, but is still important. 
 
\medskip 
 
\noindent}{\bf Proposition}  
Let $(X,\Sigma,\mu)$ be a measure space and $f$ a 
real-valued function defined almost everywhere in $X$. 
Suppose that $h_1$, $h_2$ are non-negative virtually measurable 
functions defined almost everywhere in $X$.   Then  
 
\Centerline{$\overlineint f\times(h_1+h_2) 
=\overlineint f\times h_1+\overlineint f\times h_2$,}

\noindent where here, for once, we can interpret
$\infty+(-\infty)$ or $(-\infty)+\infty$ as $\infty$ if called for on
the right-hand side. 
 
\proof{{\bf (a)} If either $\overline{\int}f\times h_1=\infty$ or 
$\overline{\int}f\times h_2=\infty$ then
$\overline{\int}f\times(h_1+h_2)=\infty$.   \Prf\Quer\ Otherwise,
there is a $g$ such that $f\times(h_1+h_2)\leae g$ and $\int g<\infty$.   
In this case, 

\Centerline{$f\times h_1
\leae f^+\times h_1
\leae f^+\times(h_1+h_2)
=(f\times(h_1+h_2))^+
\leae g^+$}

\noindent so $\overline{\int}f\times h_1\le\int g^+<\infty$.   Similarly,
$\overline{\int}f\times h_2<\infty$;  contradicting our hypothesis.\
\Bang\QeD\   So in this case, under the local rule
$\infty+(-\infty)=(-\infty)+\infty=\infty$, we have the result.

\medskip

{\bf (b)} Now suppose that the upper integrals
$\overline{\int}f\times h_1$ and 
$\overline{\int}f\times h_2$ are both less than $\infty$, 
so that their sum can be interpreted by the usual rules.
By 133J(b-ii),  
$\overline{\int}f\times(h_1+h_2) 
\le\overline{\int}f\times h_1+\overline{\int}f\times h_2<\infty$.   
In the other direction, suppose that $g\geae f\times(h_1+h_2)$ 
and $\int g<\infty$.   For $i=1$, $2$ set 
 
$$\eqalign{g_i(x) 
&=\Bover{g(x)h_i(x)}{h_1(x)+h_2(x)} 
  \text{ if }x\in\dom g\cap\dom h_1\cap\dom h_2 
  \text{ and }h_1(x)+h_2(x)>0,\cr 
&=0\text{ for other }x\in X.\cr}$$ 
 
\noindent Then, for both $i$, $g_i$ is virtually measurable, 
$g_i^+\leae g^+$ and $g_i\geae f\times h_i$;  while $g\geae g_1+g_2$. 
\Prf\ The set 
 
\Centerline{$H=\{x:x\in\dom f\cap\dom g\cap\dom h_1\cap\dom h_2$, 
$g(x)\ge f(x)(h_1(x)+h_2(x))\}$} 
 
\noindent is conegligible, and for $x\in H$ 
 
$$\eqalign{g(x) 
&=g_1(x)+g_2(x)\text{ if }h_1(x)+h_2(x)>0,\cr 
&\ge 0=g_1(x)+g_2(x)\text{ if }h_1(x)+h_2(x)=0.  \text{ \Qed}\cr}$$ 
 
\noindent So 
 
\Centerline{$\overlineint f\times h_1+\overlineint f\times h_2 
\le\int g_1+\int g_2 
=\int g_1+g_2
\le\int g$} 
 
\noindent (because $\int g_1$ and $\int g_2$ are both at most 
$\int g^+<\infty$, so we can add them on the usual rules).   
As $g$ is arbitrary, 
$\overline{\int}f\times h_1+\overline{\int}f\times h_2 
\le\overline{\int}f\times(h_1+h_2)$ and we must have equality. 
}%end of proof of 133L 
 
\exercises{ 
\leader{133X}{Basic exercises $\pmb{>}$(a)} 
%\spheader 133Xa 
Let $(X,\Sigma,\mu)$ be a measure space, and $f:X\to\coint{0,\infty}$ a  
measurable function.   Show that 
      
$$\eqalign{\int fd\mu 
&=\sup_{n\in\Bbb N}2^{-n}\sum_{k=1}^{4^n}\mu\{x:f(x)\ge 2^{-n}k\}\cr 
&=\lim_{n\to\infty}2^{-n}\sum_{k=1}^{4^n}\mu\{x:f(x)\ge 2^{-n}k\}\cr}$$ 
      
\noindent in $[0,\infty]$. 
%133A 
      
\spheader 133Xb 
Let $(X,\Sigma,\mu)$ be a measure space and $f$ 
a complex-valued function defined on a subset of $X$.   (i) Show that if 
$E\in\Sigma$, then $f\restr E$ is $\mu_E$-integrable iff $\tilde f$ is 
$\mu$-integrable, writing $\mu_E$ for the subspace measure on $E$ and 
$\tilde f(x)=f(x)$ if $x\in E\cap\dom f$, $0$ if $x\in X\setminus E$; 
and in this case $\int_Efd\mu_E=\int \tilde fd\mu$.   (ii) Show that if 
$E\in\Sigma$ and $f$ is defined $\mu$-almost everywhere, then 
$f\restr E$ is 
$\mu_E$-integrable iff $f\times\chi E$ is $\mu$-integrable, and in this 
case $\int_Ef=\int f\times\chi E$.   (iii) Show that if $\int_Ef=0$ for 
every $E\in\Sigma$, then $f=0$ a.e. 
%133E 
      
\spheader 133Xc Suppose that $(X,\Sigma,\mu)$ is a measure space 
and that $G$ is an open subset of $\Bbb C$, that is, a set such that for 
every $w\in G$ there is a $\delta>0$ such that 
$\{z:|z-w|<\delta\}\subseteq G$.   Let $f:X\times G\to\Bbb C$ be a 
function, and suppose that the derivative $\pd{f}{z}$ of $f$ with 
respect to the 
second variable exists for all $x\in X$, $z\in G$.   Suppose 
moreover that (i) $F(z)=\int f(x,z)dx$ exists for every 
$z\in G$ (ii) there is an integrable function $g$ such that 
$|\pd{f}{z}(x,z)|\le g(x)$ for every $x\in X$, $z\in G$. 
Show that the derivative $F'$ of $F$ exists everywhere in $G$, and 
$F'(z)=\biggerint\pd{f}{z}(x,z)dx$ for every $z\in G$.   
({\it Hint\/}:  you will need to check that 
$|f(x,z)-f(x,w)|\le|z-w|g(x)$ whenever $x\in X$, $z\in G$ and $w$ is close to $z$.) 
%133F 
      
\sqheader 133Xd Let $f$ be a complex-valued function defined 
almost everywhere 
on $\coint{0,\infty}$, endowed as usual with Lebesgue measure.   Its 
{\bf Laplace transform} is the function $F$ defined by writing 
      
\Centerline{$F(s)=\int_0^{\infty}e^{-sx}f(x)dx$} 
      
\noindent for all those complex numbers $s$ for which the integral is 
defined in $\Bbb C$. 
      
\quad(i) Show that if $s\in\dom F$ and $\Real s'\ge\Real s$ then 
$s'\in\dom F$ (because $|e^{-s'x}e^{sx}|\le 1$ for all $x$). 
      
\quad(ii) Show that $F$ is analytic (that is, differentiable as a 
function of a complex variable) on the interior of its domain. 
\Hint{133Xc.} 
      
\quad(iii) Show that if $F$ is defined anywhere then $\lim_{\Real 
s\to\infty}F(s)=0$. 
      
\quad(iv) Show that if $f$, $g$ have Laplace transforms $F$, $G$ then 
the Laplace transform of $f+g$ is $F+G$, at least on $\dom F\cap \dom 
G$. 
%133Xc 133F 
      
\sqheader 133Xe Let $f$ be an integrable complex-valued function 
defined almost everywhere in $\Bbb R$, endowed as usual with Lebesgue 
measure.   Its {\bf Fourier transform} is the function $\varhatf$ 
defined by 
      
\Centerline{$\varhatf(s)=\Bover{1}{\sqrt{2\pi}} 
\int_{-\infty}^{\infty}e^{-isx}f(x)dx$} 
      
\noindent for all real $s$. 
      
\quad(i) Show that $\varhatf$ is continuous.   \Hint{use Lebesgue's 
Dominated Convergence Theorem on sequences of the form 
$f_n(x)=e^{-is_nx}f(x)$.} 
      
\quad(ii) Show that if $f$, $g$ have Fourier transforms $\varhatf$, 
$\varhat g$ then the Fourier transform of $f+g$ is $\varhatf+\varhat g$. 
      
\quad(iii) Show that if $\int xf(x)dx$ exists then $\varhatf$ is 
differentiable, with 
$\varhatf\vthsp'(s)=-\Bover{i}{\sqrt{2\pi}}\int xe^{-isx}f(x)dx$ for 
every $s$. 
%133Xc 133F 
      
\spheader 133Xf Let $(X,\Sigma,\mu)$ be a measure space and 
$\sequencen{f_n}$ a sequence of real-valued functions each defined 
almost everywhere in $X$.   Suppose that there is an integrable 
real-valued function $g$ such that $|f_n|\leae g$ for each $n$. 
Show that 
      
\Centerline{$\overlineint\liminf_{n\to\infty}f_n 
\le\liminf_{n\to\infty}\overlineint f_n$,\quad 
$\underlineint\limsup_{n\to\infty}f_n 
\ge\limsup_{n\to\infty}\underlineint f_n$.} 
%133K 
 
      
\leader{133Y}{Further exercises (a)} 
%\spheader 133Ya 
Use the ideas of 133C-133H to develop a theory of measurable and 
integrable functions taking values in $\BbbR^r$, where $r\ge 2$. 
%133H 
      
\spheader 133Yb Let $X$ be a set and $\Sigma$ a $\sigma$-algebra 
of subsets of $X$.   Let $Y$ be a subset of $X$ and $f:Y\to\Bbb C$ a 
$\Sigma_Y$-measurable function, where $\Sigma_Y=\{E\cap Y:E\in\Sigma\}$. 
Show that there is a $\Sigma$-measurable function $\tilde f:X\to\Bbb C$ 
extending $f$.  \Hint{121I.} 
%133F 
      
\spheader 133Yc Let $f$ be an integrable complex-valued function 
defined almost everywhere in $\BbbR^r$, endowed as usual with Lebesgue 
measure, where $r\ge 1$.   Its {\bf Fourier transform} is the function 
$\varhatf$ defined by 
      
\Centerline{$\varhatf(s)=\Bover{1}{{(\sqrt{2\pi})}^r} 
\int e^{-is\dotproduct x}f(x)dx$} 
      
\noindent for all $s\in\BbbR^r$, writing $s\dotproduct x$ for 
$\sigma_1\xi_1+\ldots+\sigma_r\xi_r$ if $s=(\sigma_1,\ldots,\sigma_r)$, 
$x=(\xi_1,\ldots,\xi_r)\in\BbbR^r$. 
      
\quad(i) Show that $\varhatf$ is continuous.   
      
\quad(ii) Show that if $f$, $g$ have Fourier transforms $\varhatf$, 
$\varhat g$ then the Fourier transform of $f+g$ is $\varhatf+\varhat g$. 
      
\quad (iii) Show that if $\int\|x\||f(x)|dx$ is finite
(taking $\|x\|=\sqrt{\xi_1^2+\ldots+\xi_r^2}$ if 
$x=(\xi_1,\dots,\xi_r)$), then $\varhatf$ is differentiable, with 
      
\Centerline{$\Pd{\varhatf}{\sigma_k}(s) 
=-\Bover{i}{{(\sqrt{2\pi})}^r}\int\xi_ke^{-is\dotproduct x}f(x)dx$} 
      
\noindent for every $s\in\BbbR^r$, $k\le r$. 
%133Xe 133F 
      
\spheader 133Yd Recall the definition of `quasi-simple' 
function from 122Yd.   Show that for any measure space $(X,\Sigma,\mu)$ 
and any real-valued function $f$ defined almost everywhere in $X$, 
      
\Centerline{$ 
\overlineint f=\inf\{\int g:g$ is quasi-simple, $f\leae g\}$,} 
      
\Centerline{$ 
\underlineint f=\sup\{\int g:g$ is quasi-simple, $f\geae g\}$,} 
      
\noindent allowing $\infty$ for $\inf\emptyset$ and $\sup\Bbb R$ and 
$-\infty$ for $\inf\Bbb R$ and $\sup\emptyset$. 
%133J 
      
\spheader 133Ye State and prove a similar result concerning the 
`pseudo-simple' functions of 122Ye. 
%133Yd 133J 
}%end of exercises 
      
\endnotes{ 
\Notesheader{133} I have spelt this section out in 
detail, even though there is nothing that can really be called a new 
idea in it, because it gives us an opportunity to review the previous 
work, and because the manipulations which are by now, I hope, becoming 
`obvious' to you are in fact justifiable only through difficult 
theorems, and I believe that it is at least some of the time right to 
look back to the exact points at which justifications were written out. 
      
You may have noticed similarities between results involving `upper 
integrals', as described here, and those of \S132 concerning `outer 
measure' (132Ae and 133Ka, for instance, or 132Xe and 133Kb).   These 
are not a coincidence;  an explanation of sorts can be found in 252Ym in 
Volume 2. 
}%end of notes 
      
\discrpage 
      
     
\frfilename{mt134.tex}
\versiondate{7.1.04}
\copyrightdate{1994}
     
%B(x,\delta) not B(x;\delta)
\def\chaptername{Complements}
\def\sectionname{More on Lebesgue measure}
     
\newsection{134}
     
The special properties of Lebesgue measure will take up a substantial
proportion of this treatise.   In this section I present a miscellany of
relatively easy basic results.    In
134A-134F, %134A 134B 134D 134E 134F
$r$ will be a fixed
integer greater than or equal to $1$, $\mu$ will be Lebesgue measure on
$\BbbR^r$ and $\mu^*$ will be Lebesgue outer measure\cmmnt{ (see
132C)};  when I say that a set or a function is `measurable', then it is
to be understood that (unless otherwise stated) this means `measurable
with respect to the $\sigma$-algebra of Lebesgue measurable sets', while
`negligible' means `negligible for Lebesgue measure'.   Most of the
results will be expressed in terms adapted to the
multi-dimensional case;  but if you are primarily interested in the
real line, you will miss none of the ideas if you read the
whole section as if $r=1$.
     
\leader{134A}{Proposition} Both Lebesgue outer measure and Lebesgue
measure are translation-invariant;  that is, setting
$A+x=\{a+x:a\in A\}$ for $A\subseteq\BbbR^r$, $x\in\BbbR^r$, we have
     
(a) $\mu^*(A+x)=\mu^* A$ for every $A\subseteq\BbbR^r$, $x\in\BbbR^r$;
     
(b) whenever $E\subseteq\BbbR^r$ is measurable and $x\in\BbbR^r$, then
$E+x$ is measurable, with $\mu(E+x)=\mu E$.
     
\proof{ The point is that if $I\subseteq\BbbR^r$ is a
half-open interval, as defined in 114Aa/115Ab, then so is $I+x$, and
$\lambda(I+x)=\lambda I$ for
every $x\in\BbbR^r$, where $\lambda$ is defined as in 114Ab/115Ac;
this is immediate from the definition, since
$\coint{a,b}+x=\coint{a+x,b+x}$.
     
\medskip
     
{\bf (a)} If $A\subseteq\BbbR^r$ and $x\in\BbbR^r$ and $\epsilon>0$,
we
can find a sequence $\langle I_j\rangle_{j\in\Bbb N}$ of half-open
intervals such that $A\subseteq\bigcup_{j\in\Bbb N}I_j$ and
$\sum_{j=0}^{\infty}\lambda I_j\le\mu^* A+\epsilon$.   Now
$A+x\subseteq\bigcup_{j\in\Bbb N}(I_j+x)$ so
     
     
\Centerline{$\mu^*(A+x)\le\sum_{j=0}^{\infty}\lambda
(I_j+x)=\sum_{j=0}^{\infty}\lambda I_j\le\mu^* A+\epsilon$.}
     
\noindent   As
$\epsilon$ is arbitrary, $\mu^* (A+x)\le\mu^* A$.   Similarly
     
\Centerline{$\mu^*A=\mu^*((A+x)+(-x))\le\mu^*(A+x)$,}
     
\noindent so $\mu^*(A+x)=\mu^* A$, as claimed.
     
\wheader{134A}{6}{2}{2}{12pt}
     
{\bf (b)} Now suppose that $E\subseteq\BbbR^r$ is measurable and
$x\in\BbbR^r$, and that $A\subseteq\BbbR^r$.   Then, using (a)
repeatedly,
     
$$\eqalign{\mu^*(A\cap(E+x))+\mu^*(A\setminus(E+x))
&=\mu^*(((A-x)\cap E)+x)+\mu^*(((A-x)\setminus E)+x)\cr
&=\mu^*((A-x)\cap E)+\mu^*((A-x)\setminus E)\cr
&=\mu^*(A-x)
=\mu^* A,\cr}$$
     
\noindent writing $A-x$ for $A+(-x)=\{a-x:a\in A\}$.   As $A$ is
arbitrary, $E+x$ is measurable.   Now
     
\Centerline{$\mu(E+x)=\mu^*(E+x)=\mu^* E=\mu
E$.}
}%end of proof of 134A
     
     
     
\leader{134B}{Theorem} Not every subset of $\BbbR^r$ is Lebesgue
measurable.
     
\proof{ Set $\tbf{0}=(0,\ldots,0)$, $\tbf{1}=(1,\ldots,1)\in\BbbR^r$.
On
     
\Centerline{$\coint{\tbf{0},\tbf{1}}
=\{(\xi_1,\ldots,\xi_r):\xi_i\in\coint{0,1}$ for every $i\le r\}$,}
     
\noindent consider the relation $\sim$,
defined by saying that $x\sim y$ iff $y-x\in\BbbQ^r$.   It is easy to
see
that this is an equivalence relation, so divides
$\coint{\tbf{0},\tbf{1}}$ into
equivalence classes.
Choose one point from each of these equivalence classes, and let $A$ be
the set of points obtained in this way.
Then $\mu^* A\le\mu^*\coint{\tbf{0},\tbf{1}}=1$.
     
Consider $A+\BbbQ^r=\{a+q:a\in A,\,q\in\BbbQ^r\}
=\bigcup_{q\in\BbbQ^r}A+q$.   This is equal to $\BbbR^r$.   \Prf\ If
$x\in\BbbR^r$,
there is
an $e\in\BbbZ^r$ such that $x-e\in\coint{\tbf{0},\tbf{1}}$;  there is
an $a\in A$
such that $a\sim x-e$, that is, $x-e-a\in\BbbQ^r$;  now
$x=a+(e+x-e-a)\in A+\BbbQ^r$.\ \Qed\   Next, $\BbbQ^r$ is countable
(111F(b-iv)), so we have
     
\Centerline{$\infty=\mu \BbbR^r\le\sum_{q\in\BbbQ^r}\mu^*(A+q)$,}
     
\noindent and there must be some $q\in\BbbQ^r$ such that
$\mu^*(A+q)>0$;
but as $\mu^*$ is translation-invariant (134A), $\mu^* A>0$.
     
Take $n\in\Bbb N$ such that $n>2^r/\mu^* A$, and distinct
$q_1,\ldots,q_n\in\coint{\tbf{0},\tbf{1}}\cap\BbbQ^r$.   If $a$, $b\in
A$ and $1\le
i<j\le n$, then $a+q_i\ne b+q_j$;  for if $a=b$ then $q_i\ne q_j$, while
if $a\ne b$ then $a\not\sim b$ so $b-a\ne q_i-q_j$.   Thus
$A+q_1,\ldots,A+q_n$ are disjoint.   On the other hand, all are subsets
of $\coint{\tbf{0},\tbf{2}}$.   So we have
     
\Centerline{$\sum_{i=1}^n\mu^*(A+q_i)
=n\mu^* A>2^r=\mu\coint{\tbf{0},\tbf{2}}\ge\mu^*(\bigcup_{1\le i\le
n}(A+q_i))$.}
     
\noindent It follows that not all the $A+q_i$ can be measurable;  as
Lebesgue measure is translation-invariant, we see that $A$ itself is not
measurable.   In any case we have found a non-measurable set.
}%end of proof of 134B
     
     
\cmmnt{
\leader{*134C}{Remark} 134B is known as `Vitali's construction'.

Observe that at the beginning of the proof I
asked you to {\it choose} one member of each of the equivalence classes
for $\sim$.   This is of course an appeal to the Axiom of Choice.   So
far I have made rather few appeals to the axiom of choice.   One was in
(a-iv) of the proof of 114D/115D;  an earlier one was in 112Db;
yet another in 121A.  See also 1A1F.   In all of
these, only `countable choice' was involved;  that is, I
needed to choose simultaneously one member of each of a named sequence
of sets.   Because there are surely uncountably many equivalence classes
for $\sim$, the form of choice needed for the example above is
essentially stronger than that needed for the positive results so far.
It is in fact the case that very large parts of measure theory can be
developed without appealing to the full strength of the axiom of
choice.
     
The significance of this is that it suggests the possibility that there
might be a consistent mathematical system in which enough of the axiom
of choice is valid to make measure theory possible, without having
enough to construct a non-Lebesgue-measurable set.   Such a system has
indeed been worked out by R.M.Solovay ({\smc Solovay 70}).   (In a
formal sense there is room for a residual doubt concerning its
consistency.   In my view this is of no importance.)   In Volume 5
I will return to the question of what Lebesgue measure looks like with a
weak axiom of choice, or none at all.   For
the moment, I have to say that 
nearly all measure theory continues to proceed in directions at least
consistent with the full axiom of choice, so that non-measurable sets
are constantly present, at least potentially;  and that will be my
normal position in this treatise.   But I mention the point at this
early stage because I believe that it could happen at any time that the
focus of interest might switch to systems in which the axiom of choice
is false;  and in this case measure theory without non-measurable sets
might become important to many pure mathematicians, and even to applied
mathematicians, who have no reason, other than the convenience of being
able to quote results from books like this one, for loyalty to the axiom
of choice.
     
I ought to remark that while we need a fairly strong form of the axiom
of choice to construct a non-Lebesgue-measurable set, a non-Borel set
can be constructed in much weaker set theories.   One possible
construction is outlined in \S423 in Volume 4.
     
Of course there is a non-Lebesgue-measurable subset of $\Bbb R$ iff
there is a non-Lebesgue-measurable function from $\Bbb R$ to $\Bbb R$;
for if every set is measurable, then the definition 121C makes it plain
that every real-valued function on any subset of $\Bbb R$ is measurable;
while if $A\subseteq\Bbb R$ is not measurable, then
$\chi A:\Bbb R\to\Bbb R$ is not measurable.
}%end of comment
     
\leader{*134D}{}\cmmnt{ In fact there are much stronger results
than 134B
concerning the existence of non-measurable sets (provided, of course,
that we allow ourselves to use the axiom of choice).   Here I give one
which can be reached by a slight refinement of the methods of 134B.
     
\medskip
     
\noindent}{\bf Proposition} There is a set $C\subseteq\BbbR^r$ such that
$F\cap C$ is not measurable for any measurable set $F$ of
non-zero measure;  so that both $C$ and its complement have full outer
measure in $\BbbR^r$.
     
\proof{{\bf (a)} Start from a set
$A\subseteq\coint{\tbf{0},\tbf{1}}\subseteq\BbbR^r$ such that
$\langle A+q\rangle_{q\in\BbbQ^r}$ is a partition of $\BbbR^r$, as
constructed in the proof of 134B.   As in 134B, the outer measure
$\mu^*A$ of $A$ must be greater than $0$.   The argument there shows in
fact that $\mu F=0$ for every measurable set $F\subseteq A$.   \Prf\
For every $n$ we can find distinct
$q_1,\ldots,q_n\in\coint{\tbf{0},\tbf{1}}\cap\BbbQ^r$, and now
     
\Centerline{$n\mu F=\mu(\bigcup_{1\le i\le
n}F+q_i)\le\mu\coint{\tbf{0},\tbf{2}}=2^r$,}
     
\noindent so that $\mu F\le 2^r/n$;  as $n$ is arbitrary, $\mu F=0$.
\Qed
     
\medskip
     
{\bf (b)} Now let $E\subseteq\coint{\tbf{0},\tbf{1}}$ be a measurable
envelope of $A$ (132Ef).   Then $E+q$ is a
measurable envelope of $A+q$ for any $q$.   \Prf\ I hope that this will
very soon be `an obvious consequence of the translation-invariance of
Lebesgue measure'.   In detail:  $A+q\subseteq E+q$, $E+q$ is measurable
and, for any measurable $F$,
     
$$\eqalign{\mu(F\cap(E+q))
&=\mu(((F-q)\cap E)+q)
=\mu((F-q)\cap E)\cr
&=\mu^*((F-q)\cap A)
=\mu^*(((F-q)\cap A)+q)
=\mu^*(F\cap(A+q)),\cr}$$
     
\noindent using 134A repeatedly.\ \Qed\  Also $E$ is a measurable
envelope of $A'=E\setminus A$.
\Prf\ Of course $E$ is a measurable set including $A'$.   If
$F\subseteq E\setminus A'$ is measurable then $F\subseteq A$, so
$\mu F=0$, by (a);  now 132Ea tells us that $E$ is a measurable envelope
of $A'$.\ \Qed\   It follows that $E+q$ is a
measurable envelope of $A'+q$ for every $q$.
     
\medskip
     
{\bf (c)} Let $\sequencen{q_n}$ be a sequence running over $\BbbQ^r$.
Then
     
\Centerline{$\bigcup_{n\in\Bbb N}E+q_n
\supseteq\bigcup_{n\in\Bbb N}A+q_n=\BbbR^r$.}
     
\noindent Write $E_n$ for $E+q_n\setminus\bigcup_{i<n}E+q_i$ for
$n\in\Bbb N$, so that $\sequencen{E_n}$ is disjoint and
$\bigcup_{n\in\Bbb N}E_n=\BbbR^r$.
     
Now set
     
\Centerline{$C=\bigcup_{n\in\Bbb N}E_n\cap(A+q_n)$.}
     
\noindent This is a set with the required properties.
     
\Prf\ {\bf (i)} Let $F\subseteq\BbbR^r$ be any non-negligible measurable
set.   Then there must be some $n\in\Bbb N$ such that
$\mu(F\cap E_n)>0$.   But this means that
     
\Centerline{$\mu^*(F\cap E_n\cap C)\ge\mu^*(F\cap E_n\cap(A+q_n))
=\mu(F\cap E_n\cap(E+q_n))=\mu(F\cap E_n)$,}
     
$$\eqalign{\mu^*(F\cap E_n\setminus C)
&\ge\mu^*(F\cap E_n\cap((E+q_n)\setminus(A+q_n)))\cr
&=\mu(F\cap E_n\cap(E+q_n))
=\mu(F\cap E_n).\cr}$$
     
\noindent Since
     
\Centerline{$\mu(F\cap E_n)\le\mu(E+q_n)=\mu E\le 1$,}
     
\noindent
$\mu^*(F\cap E_n\cap C)+\mu^*(F\cap E_n\setminus C)>\mu(F\cap E_n)$, and
$F\cap C$ cannot be measurable.
     
\medskip
     
\quad{\bf (ii)} In particular, no measurable subset of
$\BbbR^r\setminus C$ can have non-zero measure, and $C$ has full outer
measure;  similarly, $C$ has no measurable subset of non-zero measure,
and $\BbbR^r\setminus C$ has full outer measure.\ \Qed
}%end of proof of 134D
     
\cmmnt{\medskip
     
\noindent{\bf Remark} In fact it is the case that for any sequence
$\sequencen{D_n}$ of subsets of $\BbbR^r$ there is a set
$C\subseteq\BbbR^r$ such that
     
\Centerline{$\mu^*(E\cap D_n\cap C)=\mu^*(E\cap D_n\setminus C)
=\mu^*(E\cap D_n)$}
     
\noindent for every measurable set $E\subseteq\BbbR^r$ and every
$n\in\Bbb N$.   But for the
proof of this result we must wait for Volume 5.
}%end of comment
     
\cmmnt{
\leader{134E}{Borel sets and Lebesgue measure on $\BbbR^r$}
Recall from 111G
that the family $\Cal B$ of Borel sets in $\BbbR^r$ is the
$\sigma$-algebra generated by the family of open sets.   In 114G/115G I
showed that every Borel set in $\BbbR^r$ is Lebesgue measurable.   It
is time we returned to the topic and looked more closely at the very
intimate connexion between Borel and measurable sets.
     
Recall that a set $A\subseteq\BbbR^r$ is {\bf bounded} if
there is an $M$ such that $A\subseteq B(\tbf{0},M)=\{x:\|x\|\le M\}$;
equivalently, if $\sup_{x\in A}|\xi_j|<\infty$ for every $j\le r$
(writing $x=(\xi_1,\ldots,\xi_r)$, as in \S115).
}%end of comment
     
\leader{134F}{Proposition} (a) If $A\subseteq\BbbR^r$ is any set, then
     
\Centerline{$\mu^*A
=\inf\{\mu G:G\text{ is open},\,G\supseteq A\}
=\min\{\mu H:H\text{ is Borel},\,H\supseteq A\}$.}
     
(b)   If $E\subseteq\BbbR^r$ is measurable, then
     
\Centerline{$\mu E=\sup\{\mu F:F
\text{ is closed and bounded},\,F\subseteq E\}$,}
     
\noindent and there are Borel sets $H_1$, $H_2$ such that 
$H_1\subseteq E\subseteq H_2$ and
     
\Centerline{$\mu(H_2\setminus H_1)=\mu(H_2\setminus
E)=\mu(E\setminus H_1)=0$.}
     
(c) If $A\subseteq\BbbR^r$ is any set, then $A$ has a measurable
envelope which is a Borel set.
     
(d) If $f$ is a Lebesgue measurable real-valued function defined on a
subset of $\BbbR^r$, then there is a conegligible Borel set
$H\subseteq\BbbR^r$ such that $f\restr H$ is Borel measurable.
     
\proof{{\bf (a)(i)} First note that if $I\subseteq\BbbR^r$ is a
half-open interval, and $\epsilon>0$, then either $I=\emptyset$ is
already open, or $I$ is expressible as $\coint{a,b}$ where
$a=(\alpha_1,\ldots,\alpha_r)$, $b=(\beta_1,\ldots,\beta_r)$ and
$\alpha_i<\beta_i$ for every $i$.   In the latter case,
$G=\ooint{a-\epsilon(b-a),b}$ is an open set including $I$, and
     
\Centerline{$\mu G=\prod_{i=1}^r(1+\epsilon)(\beta_i-\alpha_i)
=(1+\epsilon)^r\mu I$,}
     
\noindent by the formula in 114G/115G.
     
\medskip
     
\quad{\bf (ii)} Now, given $\epsilon>0$, there is a sequence
$\sequencen{I_n}$ of half-open intervals, covering $A$, such that
$\sum_{n=0}^{\infty}\mu I_n\le\mu^*A+\epsilon$.   For each $n$, let
$G_n\supseteq I_n$ be an open set of measure at most
$(1+\epsilon)^r\mu I_n$.   Then $G=\bigcup_{n\in\Bbb N}G_n$ is open
(1A2Bd), and $A\subseteq G$;  also
     
\Centerline{$\mu G\le\sum_{n=0}^{\infty}\mu G_n
\le(1+\epsilon)^r\sum_{n=0}^{\infty}\mu I_n
\le(1+\epsilon)^r(\mu^*A+\epsilon)$.}
     
\noindent As $\epsilon$ is arbitrary,
$\mu^*A\ge\inf\{\mu G:G$ is open, $G\supseteq A\}$.
     
\medskip
     
\quad{\bf (iii)} Next, using (ii), we can choose for each $n\in\Bbb N$
an open set $G_n\supseteq A$ such that $\mu G_n\le\mu^*A+2^{-n}$.   Set
$H_0=\bigcap_{n\in\Bbb N}G_n$;  then $H_0$ is a Borel set,
$A\subseteq H_0$, and
     
\Centerline{$\mu H_0\le\inf_{n\in\Bbb N}\mu G_n\le\mu^*A$.}
     
\medskip
     
\quad{\bf (iv)} On the other hand, we surely have
$\mu^*A\le\mu^*H=\mu H$ for every Borel set $H\supseteq A$.   So we must
have
     
\Centerline{$\mu^*A\le\inf\{\mu G:G\text{ is open},\,G\supseteq A\}$,}
     
\noindent and
     
\Centerline{$\mu^*A=\mu H_0
=\min\{\mu H:H\text{ is Borel},\,H\supseteq A\}$.}
     
\medskip
     
     
{\bf (b)(i)} For each $n\in\Bbb N$, set $E_n=E\cap B(\tbf{0},n)$.
Let $G_n\supseteq E_n$ be an open set of measure at most
$\mu E_n+2^{-n}$;  then (because $\mu B(\tbf{0},n)<\infty$)
$\mu(G_n\setminus E_n)\le 2^{-n}$.
Now, for each $n$, set $G'_n=\bigcup_{m\ge n}G_m$;  then
$G'_n$ is open, $E=\bigcup_{m\ge n}E_m\subseteq G'_n$, and
     
\Centerline{$\mu(G'_n\setminus E)\le\sum_{m=n}^{\infty}\mu(G_m\setminus
E)\le\sum_{m=n}^{\infty}\mu(G_m\setminus E_m)
\le\sum_{m=n}^{\infty}2^{-m}=2^{-n+1}$.}
     
\noindent Setting $H_2=\bigcap_{n\in\Bbb N}G_n$, we see that $H_2$ is a
Borel set including $E$ and that $\mu(H_2\setminus E)=0$.
     
\medskip
     
\quad{\bf (ii)} Repeating the argument of (i) with $\BbbR^r\setminus E$
in place of $E$, we obtain
a Borel set $\tilde H_2\supseteq\BbbR^r\setminus E$ such that
$\mu(\tilde H_2\setminus(\BbbR^r\setminus E))=0$;  now
$H_1=\BbbR^r\setminus \tilde H_2$ is a Borel set included in $E$ and
     
\Centerline{$\mu(E\setminus H_1)
=\mu(\tilde H_2\setminus(\BbbR^r\setminus E))=0$.}
     
\noindent Of course we now also have
     
\Centerline{$\mu(H_2\setminus H_1)=\mu(H_2\setminus E)+\mu(E\setminus
H_1)=0$.}
     
\medskip
     
\quad{\bf (iii)} Again using the idea of (i), there is for each
$n\in\Bbb N$
an open set $\tilde G_n\supseteq B(\tbf{0},n)\setminus E$ such that
     
\Centerline{$\mu(\tilde G_n\cap E_n)\le\mu(\tilde
G_n\setminus(B(\tbf{0},n)\setminus E))\le 2^{-n}$.}
     
\noindent Set
     
\Centerline{$F_n=B(\tbf{0},n)\setminus\tilde G_n
=B(\tbf{0},n)\cap(\BbbR^r\setminus\tilde G_n)$;}
     
\noindent  then $F_n$ is
closed (1A2Fd) and bounded and $F_n\subseteq E_n\subseteq E$.
Also
     
\Centerline{$\mu E_n=\mu F_n+\mu(E_n\setminus F_n)
=\mu F_n+\mu(\tilde G_n\cap E_n)
\le\mu F_n+2^{-n}$.}
     
\noindent So
     
\Centerline{$\mu E=\lim_{n\to\infty}\mu E_n
\le\sup_{n\in\Bbb N}\mu F_n\le
\sup\{\mu F:F\text{ is closed and bounded},\,F\subseteq E\}$,}
     
\noindent and
     
\Centerline{$\mu E
=\sup\{\mu F:F\text{ is closed and bounded},\,F\subseteq E\}$.}
     
\medskip
     
{\bf (c)} Let $E$ be any measurable envelope of $A$ (132Ef), and
$H\supseteq E$ a Borel set such that $\mu(H\setminus E)=0$;  then
$\mu^*(F\cap A)=\mu(F\cap E)=\mu(F\cap H)$ for every measurable set $F$,
so $H$ is a measurable envelope of $A$.
     
\medskip
     
{\bf (d)} Set $D=\dom f$ and write $\Cal B$ for the $\sigma$-algebra of
Borel sets.   For each rational number $q$, let $E_q$ be a measurable
set such that $\{x:f(x)\le q\}=E_q\cap D$.   Let $H_q$, $H_q'\in\Cal B$
be such that $H_q\subseteq E_q\subseteq H'_q$ and
$\mu(H'_q\setminus H_q)=0$.   Let $H$ be the conegligible Borel set
$\BbbR^r\setminus\bigcup(H'_q\setminus H_q)$.   Then
     
\Centerline{$\{x:(f\restr H)(x)\le q\}=H\cap E_q\cap D=H_q\cap D\cap H$}
     
\noindent belongs to the subspace $\sigma$-algebra $\Cal B(D)$ for every
$q\in\Bbb Q$.   For irrational $a\in\Bbb R$, set
$H_a=\bigcap_{q\in\Bbb Q,q\ge a}H_q$;  then $H_a\in\Cal B$, and
     
\Centerline{$\{x:(f\restr H)(x)\le a\}=H_a\cap\dom(f\restr H)$.}
     
\noindent Thus $f\restr H$ is Borel measurable.
}%end of proof of 134F
     
\cmmnt{\medskip
     
\noindent{\bf Remark} The emphasis on closed {\it bounded} sets in part
(b) of this proposition is on account of their important topological
properties, in particular, the fact that they are `compact'.   This
is one of the most important facts about Lebesgue measure, as will
appear in Volume 4.   I will discuss `compactness' briefly in \S2A2 of
Volume 2.
}%end of comment
     
     
\vleader{48pt}{134G}{The Cantor set}\cmmnt{ One of the purposes of the
theory of Lebesgue measure and integration is to study rather more
irregular sets and functions than can be dealt with by more primitive
methods.   In the next few paragraphs
I discuss {\it measurable} sets and
functions which from the point of view of the present theory are
amenable without being trivial.   From now on, $\mu$ will be Lebesgue
measure on $\Bbb R$.
     
\medskip
     
}{\bf (a)} The `Cantor set'
$C\subseteq[0,1]$ is defined as the intersection  of a sequence
$\sequencen{C_n}$ of
sets, constructed as follows.   $C_0=[0,1]$.   Given that $C_n$ consists
of $2^n$ disjoint closed intervals each of length $3^{-n}$, take each of
these intervals and delete the middle third to produce two closed
intervals each of length $3^{-n-1}$;  take $C_{n+1}$ to be the union of
the $2^{n+1}$ closed intervals so formed, and continue.   Observe that
$\mu C_n=({2\over 3})^n$ for each $n$.

\medskip
     
\cmmnt{
\def\Caption{Approaching the Cantor set}
\picture{mt134g}{80pt}
% created with mt134g.for
% domain -0.1  1.1  scale  0.15
     
\noindent}The {\bf Cantor set} is $C=\bigcap_{n\in\Bbb N}C_n$.   Its
measure is
     
\Centerline{$\mu C=\lim_{n\to\infty}\mu C_n
=\lim_{n\to\infty}(\Bover23)^n=0$.}
     
\header{134Gb}{\bf (b)} Each $C_n$ can also be described as the set of
real numbers
expressible as $\sum_{j=1}^{\infty}3^{-j}\epsilon_j$ where every
$\epsilon_j$ is either $0$, $1$ or $2$, and $\epsilon_j\ne 1$ for $j\le
n$.   Consequently $C$ itself is the set of numbers expressible as
$\sum_{j=1}^{\infty}3^{-j}\epsilon_j$ where every $\epsilon_j$ is either
$0$ or $2$;  that is, the set of numbers between $0$ and $1$ expressible
in ternary form without 1's.   The expression in each case will be
unique, so we have a bijection $\phi:\{0,1\}^{\Bbb N}\to C$ defined by
writing
     
\Centerline{$\phi(z)=\Bover23\sum_{j=0}^{\infty}3^{-j}z(j)$}
     
\noindent for every $z\in\{0,1\}^{\Bbb N}$.
     
\leader{134H}{The Cantor function}\cmmnt{ Continuing from 134G, we
have the following construction.
     
     
\medskip
     
}{\bf (a)} For each $n\in\Bbb N$\cmmnt{ we} define a
function $f_n:[0,1]\to[0,1]$ by setting
     
\Centerline{$f_n(x)=(\Bover32)^n\mu(C_n\cap[0,x])$}
     
\noindent for each $x\in [0,1]$.   \cmmnt{Because $C_n$ is just a
finite union
of intervals,} $f_n$ is a polygonal function, with $f_n(0)=0$,
$f_n(1)=1$;  $f_n$ is constant on each of the $2^n-1$ open intervals
composing $[0,1]\setminus C_n$, and rises with slope $({3\over 2})^n$
on each of the $2^n$ closed intervals composing $C_n$.
     
\cmmnt{
\def\Caption{Approaching the Cantor function:
the functions $f_0$, $f_1$, $f_2$, $\pmb{f_3}$}
\picture{mt134ha1}{195pt}
% computed with mt134ha1.for
% domain -0.1  1.1  scale  0.15
%  dash/gap   0.3/0.3  0.267/0.233  0.233/0.167  0.2/0.1
}%end of comment
     
\cmmnt{If the $j$th interval of $C_n$, counting from the left, is
$[a_{nj},b_{nj}]$, then $f_n(a_{nj})=2^{-n}(j-1)$ and
$f_n(b_{nj})=2^{-n}j$.   Also, $a_{nj}=a_{n+1,2j-1}$ and
$b_{nj}=b_{n+1,2j}$;  hence, or otherwise, $f_{n+1}(a_{nj})=f_n(a_{nj})$
and $f_{n+1}(b_{nj})=f_{n}(b_{nj})$, and $f_{n+1}$ agrees with $f_n$ on
all the endpoints of the intervals of $C_n$, and therefore on
$[0,1]\setminus C_n$.
     
Within any particular interval $[a_{nj},b_{nj}]$ of $C_n$, the greatest
difference between $f_n(x)$ and $f_{n+1}(x)$ is at the new endpoints
within that interval, viz., $b_{n+1,2j-1}$ and $a_{n+1,2j}$;  and the
magnitude of the difference is ${1\over 6}2^{-n}$ (because, for
instance,
$f_n(b_{n+1,2j-1})={2\over 3}f_n(a_{nj})+{1\over 3}f_n(b_{nj})$, while
$f_{n+1}(b_{n+1,2j-1})={1\over 2}f_n(a_{nj})+{1\over 2}f_n(b_{nj})$).
Thus we have
$|f_{n+1}(x)-f_n(x)|\le{1\over 6}2^{-n}$ for every $n\in\Bbb N$,
$x\in[0,1]$.   Because $\sum_{n=0}^{\infty}{1\over 6}2^{-n}<\infty$,
}%end of comment
$\sequencen{f_n}$ is uniformly convergent to a function $f:[0,1]\to
[0,1]$, and $f$ will be continuous.
$f$ is the {\bf Cantor function} or {\bf Devil's Staircase}.
     
\def\Caption{The Cantor function}
\picture{mt134ha2}{187.43pt}
% computed with  mt134ha2.for
% domain  -0.1  1.1  scale  0.15
     
     
     
\header{134Hb}{\bf (b)} Because every $f_n$ is non-decreasing, so is
$f$.   \cmmnt{If
$x\in[0,1]\setminus C$, there is an $n$ such that $x\in[0,1]\setminus
C_n$;  let $I$ be the open interval of $[0,1]\setminus C_n$ containing
$x$;  then $f_{m+1}$ agrees on $I$ with $f_m$ for every $m\ge n$, so $f$
agrees on $I$ with $f_n$, and $f$ is constant on $I$.   Thus, in
particular, the derivative $f'(x)$ exists and is $0$ for every $x\in
[0,1]\setminus C$;  so} $f'$ is zero almost everywhere in $[0,1]$.
\cmmnt{Also, of course, $f(0)=0$ and $f(1)=1$, because $f_n(0)=0$,
$f_n(1)=1$
for every $n$.   It follows that} $f:[0,1]\to[0,1]$ is
surjective\cmmnt{ (by the Intermediate Value Theorem)}.
     
\header{134Hc}{\bf (c)} Let $\phi:\{0,1\}^{\Bbb N}\to C$ be the function
described in 134Gb.   Then
$f(\phi(z))=\bover12\sum_{j=0}^{\infty}2^{-j}z(j)$ for every
$z\in\{0,1\}^{\Bbb N}$.   \prooflet{\Prf\ Fix
$z=(\zeta_0,\zeta_1,\zeta_2,\ldots)$ in $\{0,1\}^{\Bbb N}$, and for each
$n$ take $I_n$ to be the component interval of $C_n$ containing
$\phi(z)$.   Then $I_{n+1}$ will be the left-hand third of $I_n$ if
$\zeta_n=0$ and the right-hand third if $\zeta_n=1$.   Taking $a_n$ to
be the left-hand endpoint of $I_n$, we see that
     
\Centerline{$a_{n+1}=a_n+\Bover233^{-n}\zeta_n$,
\quad$f_{n+1}(a_{n+1})=f_n(a_n)+\Bover122^{-n}\zeta_n$}
     
\noindent for each $n$.   Now
     
\Centerline{$\phi(z)=\lim_{n\to\infty}a_n$,
\quad$f(\phi(z))=\lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}f_n(a_n)
=\Bover12\sum_{j=0}^{\infty}2^{-j}\zeta_j$,}
     
\noindent as claimed.\ \Qed}%end of prooflet
\cmmnt{
     
In particular,} $f[C]=[0,1]$.   \prooflet{\Prf\ Any $x\in[0,1]$ is
expressible as $\sum_{j=0}^{\infty}2^{-j-1}z(j)=f(\phi(z))$ for some
$z\in\{0,1\}^{\Bbb N}$.\ \Qed}
     
\leader{134I}{The Cantor function modified}\cmmnt{ I continue the
argument of 134G-134H.
     
\medskip
     
}{\bf (a)} Consider the formula
     
\Centerline{$g(x)=\Bover12(x+f(x))$,}
     
\noindent where $f$ is the Cantor function, as defined in 134H;  this
defines a continuous function $g:[0,1]\to[0,1]$ which is strictly
increasing\cmmnt{ (because $f$ is non-decreasing)} and has $g(0)=0$,
$g(1)=1$;  \cmmnt{consequently, by the Intermediate Value Theorem,}
$g$ is bijective, and its inverse $g^{-1}:[0,1]\to[0,1]$ is
continuous.
\cmmnt{
     
Now} $g[C]$ is a closed set and $\mu g[C]=\bover12$.   \prooflet{\Prf\
Because $g$ is a permutation of the points of $[0,1]$, 
$[0,1]\setminus g[C]=g[\,[0,1]\setminus C]$.
For each of the open intervals $I_{nj}=\ooint{b_{nj},a_{n,j+1}}$ making
up $[0,1]\setminus C_n$, we see that
$g[I_{nj}]=\ooint{g(b_{nj}),g(a_{n,j+1})}$ has length just half the
length of $I_{nj}$.   Consequently $g[\,[0,1]\setminus C]=\bigcup_{n\ge
1,1\le j<2^n}g[I_{nj}]$ is open, and
     
$$\eqalign{\mu(g[\,[0,1]\setminus C_n])
&=\sum_{j=1}^{2^n-1}g(a_{n,j+1})-g(b_{nj})
=\Bover12\sum_{j=1}^{2^n-1}a_{n,j+1}-b_{nj}\cr
&=\Bover12\mu([0,1]\setminus C_n)
=\Bover12(1-(\Bover23)^n)\cr}$$
     
\noindent (134Ga).   Because $\sequencen{[0,1]\setminus C_n}$ is an
increasing sequence of sets with union $[0,1]\setminus C$,
     
\Centerline{$\mu g([\,[0,1]\setminus C])
=\lim_{n\to\infty}\mu g([\,[0,1]\setminus C_n])
=\Bover12$.}
     
\noindent So $g[C]=[0,1]\setminus g[\,[0,1]\setminus C]$ is closed and
$\mu g[C]=\bover12$.   \Qed}
     
\header{134Ib}{\bf (b)} By 134D there is a set $D\subseteq\Bbb R$ such
that
     
\Centerline{$\mu^*(g[C]\cap D)=\mu^*(g[C]\setminus D)
=\mu g[C]=\Bover12$;}
     
\noindent set $A=g[C]\cap D$.   \cmmnt{Of course} $A$ cannot be
measurable\cmmnt{, since $\mu^*A+\mu^*(g[C]\setminus A)>\mu g[C]$}.
However, $g^{-1}[A]\subseteq C$ must be measurable\cmmnt{, because
$\mu^*C=0$}.   This means that if we set
$h=\chi(g^{-1}[A]):[0,1]\to\Bbb R$, then $h$ is measurable;  but
$hg^{-1}\cmmnt{\mskip5mu =\chi A:[0,1]\to\Bbb R}$ is not.
     
Thus {\bf the composition of a measurable function with a continuous
function need not be measurable}.   \cmmnt{Contrast this with 121Eg.}
     
\leader{134J}{More examples}\cmmnt{ I think it is worth taking the
space to spell
out two more of the basic examples of Lebesgue measurable set in detail.
     
\medskip
     
}{\bf (a)}\cmmnt{ As already observed in 114G, every countable
subset of $\Bbb R$ is negligible.   In particular, $\Bbb Q$ is
negligible (111Eb).   We can say more.}   Let $\sequencen{q_n}$ be a
sequence running over $\Bbb Q$, and for each $n\in\Bbb N$ set
     
\Centerline{$I_n=\ooint{q_n-2^{-n},q_n+2^{-n}}$,}
     
\Centerline{$G_n=\bigcup_{k\ge n}I_k$.}
     
\noindent Then $G_n$ is an open set of measure at most
$\sum_{k=n}^{\infty}2\cdot 2^{-k}=4\cdot 2^{-n}$, and it contains all
but finitely
many points of $\Bbb Q$, so is dense\cmmnt{ (that is, meets every
non-trivial interval)}.   Set $F_n=\Bbb R\setminus G_n$;
then $F_n$ is closed,
$\mu(\Bbb R\setminus F_n)\le 4/2^n$, but $F_n$ does not\cmmnt{ contain
$q_k$ for any $k\ge n$, so $F_n$ cannot} include any non-trivial
interval.
\cmmnt{Observe that $\sequencen{G_n}$ is non-increasing so}
$\sequencen{F_n}$ is non-decreasing.
     
\header{134Jb}{\bf (b)}\cmmnt{ We can elaborate the above
construction, as follows.}   There is a measurable set
$E\subseteq\Bbb R$ such that $\mu(I\cap E)>0$ and $\mu(I\setminus E)>0$
for every
non-trivial interval $I\subseteq\Bbb R$.   \prooflet{\Prf\ First note
that if $k$, $n\in\Bbb N$, there is a $j\ge n$ such that
$q_j\in I_k$, so that $I_k\cap I_j\ne\emptyset$ and $\mu(I_k\setminus
F_n)>0$.   Now there must be an $l>n$ such that $\mu
G_l<\mu(I_k\setminus F_n)$, so that
     
\Centerline{$\mu(I_k\cap F_l\setminus F_n)
=\mu((I_k\setminus F_n)\setminus G_l)>0$.}
     
Choose $n_0<n_1<n_2<\ldots$ as follows.   Start with $n_0=0$.   Given
$n_{2k}$, where $k\in\Bbb N$, choose $n_{2k+1}$, $n_{2k+2}$ such that
     
\Centerline{$\mu(I_k\cap F_{n_{2k+1}}\setminus F_{n_{2k}})>0,\quad
\mu(I_k\cap F_{n_{2k+2}}\setminus F_{n_{2k+1}})>0$.}
     
\noindent Continue.
     
On completing the induction, set
     
\Centerline{$E=\bigcup_{k\in\Bbb N}F_{n_{2k+1}}\setminus F_{n_{2k}}$,
\quad$H=\bigcup_{k\in\Bbb N}F_{n_{2k+2}}\setminus F_{n_{2k+1}}$.}
     
\noindent   Because $\sequence{k}{F_k}$ is non-decreasing, 
$E\cap H=\emptyset$.   
If $k\in\Bbb N$, $E\cap I_k$ and $H\cap I_k$ both have
positive measure.   

Now suppose that $I\subseteq\Bbb R$ is an interval with more than one
point;  suppose that $a$, $b\in I$ and $a<b$.   Then 
there is an $m\in\Bbb N$
such that $4\cdot 2^{-m}\le b-a$;  now there is a $k\ge m$ such that
$q_k\in[a+2^{-m},b-2^{-m}]$, so that $I_k\subseteq I$ and
     
\Centerline{$\mu(I\cap E)\ge\mu(E\cap I_k)>0$,
\quad$\mu(I\setminus E)\ge\mu(H\cap I_k)>0$.  \Qed}}
     
\header{134Jc}{\bf (c)}
\cmmnt{This shows that} $E$ and its complement are measurable sets
which are\cmmnt{ not
merely both dense (like $\Bbb Q$ and $\Bbb R\setminus\Bbb Q$), but}
`essentially' dense in that they meet every non-empty open interval in a
set of positive measure, so that\cmmnt{ (for instance)} $E\setminus A$
is dense for every negligible set $A$.
     
\leader{*134K}{Riemann integration}\cmmnt{ I have tried, in writing
this book, to
assume as little prior knowledge as possible.   In particular, it is not
necessary to have studied Riemann integration.   Nevertheless, if you
have worked through the basic theory of the Riemann integral -- which
is, indeed, not only a splendid training in the techniques of
$\epsilon$-$\delta$ analysis, but also a continuing source of ideas for
the subject -- you will, I hope, wish to connect it
with the material we are looking at here;  both because you will not
want to feel that your labour has been wasted, and because you have
probably developed a number of intuitions which will continue to be
valuable, if suitably adapted to the new context.   I therefore give a
brief account of the relationship between the Riemann and Lebesgue
methods of integration on the real line.
     
\medskip
     
{\bf (a)} There are many ways of describing the Riemann
integral;  I choose
one of the popular ones.   If  $[a,b]$ is a non-trivial closed interval
in $\Bbb R$, then I say that a {\bf dissection} of $[a,b]$ is a finite
list $D=(a_0,a_1,\ldots,a_n)$, where $n\ge 1$, such that
$a=a_0<a_1<\ldots< a_n=b$.   If now $f$ is a real-valued function
defined (at least) on $[a,b]$ and bounded on $[a,b]$, the
{\bf upper sum} and {\bf lower sum} of $f$ on $[a,b]$ derived from $D$
are
     
\Centerline{$S_D(f)
=\sum_{i=1}^n(a_i-a_{i-1})\sup_{x\in\ooint{a_{i-1},a_i}}f(x)$,}
     
\Centerline{$s_D(f)
=\sum_{i=1}^n(a_i-a_{i-1})\inf_{x\in\ooint{a_{i-1},a_i}}f(x)$.}
     
\noindent You have to prove that if $D$ and $D'$ are two dissections of
$[a,b]$, then $s_D(f)\le S_{D'}(f)$.   Now define the {\bf upper Riemann
integral} and {\bf lower Riemann integral} of $f$ to be
     
\Centerline{$U_{[a,b]}(f)=\inf\{S_D(f):D$ is a dissection of $[a,b]\}$,}
     
\Centerline{$L_{[a,b]}(f)=\sup\{s_D(f):D$ is a dissection of $[a,b]\}$.}
     
\noindent Check that $L_{[a,b]}(f)$ is necessarily less than or equal to
$U_{[a,b]}(f)$.   Finally, declare $f$ to be {\bf Riemann integrable
over} $[a,b]$ if $U_{[a,b]}(f)=L_{[a,b]}(f)$, and in this case
take the common value to be the {\bf Riemann integral} $\Rint_a^bf$ of
$f$ over $[a,b]$.
     
\spheader 134Kb} %end of comment
If $f:[a,b]\to\Bbb R$ is Riemann integrable, it
is Lebesgue integrable, with the same integral.   \prooflet{\Prf\ For
any
dissection $D=(a_0,\ldots,a_n)$ of
$[a,b]$, define $g_D$, $h_D:[a,b]\to\Bbb R$ by saying
     
\Centerline{$g_D(x)=\inf\{f(y):y\in\ooint{a_{i-1},a_i}\}$ if
$a_{i-1}<x<a_i$, \quad $g_D(a_i)=f(a_i)$ for each $i$,}
     
\Centerline{$h_D(x)=\sup\{f(y):y\in\ooint{a_{i-1},a_i}\}$ if
$a_{i-1}<x<a_i$, \quad $h_D(a_i)=f(a_i)$ for each $i$.}
     
\noindent  Then $g_D$ and $h_D$ are constant on each interval
$\ooint{a_{i-1},a_i}$, so all sets $\{x:g_D(x)\le c\}$, $\{x:h_D(x)\le
c\}$ are finite unions of intervals, and $g_D$ and $h_D$ are measurable;
moreover,
     
\Centerline{$\int g_Dd\mu
=s_D(f)$,\quad $\int h_Dd\mu=S_D(f)$.}
     
\noindent Consequently
     
$$\eqalign{\Rint_a^bf=L_{[a,b]}(f)&=\sup_D\int g_Dd\mu
\le\underline{\int}fd\mu\cr
&\le\overline{\int}fd\mu
\le\inf_D\int h_Dd\mu=U_{[a,b]}(f)=\Rint_a^bf,\cr}$$
     
\noindent and $\overline{\int}fd\mu=\underline{\int}fd\nu=\Rint_a^bf$,
so that $\int fd\mu$ exists and is equal to $\Rint_a^bf$ (133Jd).\ \Qed
}%end of prooflet
     
     
\cmmnt{\header{134Kc}{\bf (c)} The discussion above is of the
`proper' Riemann integral,
of bounded functions on bounded intervals.   For unbounded functions and
unbounded intervals, one uses various forms of `improper' integral;
for instance, the improper Riemann integral
$\int_0^{\infty}\bover{\sin x}{x}dx$ is taken to be
$\lim_{a\to\infty}\int_0^a\bover{\sin x}{x}dx$,
while $\int_0^1\ln x\,dx$ is taken to be
$\lim_{a\downarrow 0}\int_a^1\ln x\,dx$.
Of these, the second exists as a Lebesgue integral, but the first does
not, because $\int_0^{\infty}|\bover{\sin x}{x}|dx=\infty$.   The power
of the Lebesgue integral to deal directly with `absolutely
integrable' unbounded functions on unbounded domains means that what one
might call `conditionally integrable' functions are pushed into the
background of the theory.   In Chapter 48 of Volume 4 I will discuss the
general theory of such functions, but for the time being I will deal
with them individually, on the rare occasions when they arise.
}%end of comment
     
\leader{*134L}{}\cmmnt{ There is in fact a beautiful
characterization of the Riemann integrable functions, as follows.
     
\medskip
     
\noindent}{\bf Proposition} If $a<b$ in $\Bbb R$, a bounded function
$f:[a,b]\to\Bbb R$ is Riemann integrable iff it is continuous almost
everywhere in $[a,b]$.
     
\proof{{\bf (a)}  Suppose that $f$ is Riemann integrable.   For
each $x\in[a,b]$, set
     
\Centerline{$g(x)
=\sup_{\delta>0}\inf_{y\in[a,b],|y-x|\le\delta}f(y)$,}
     
\Centerline{$h(x)
=\inf_{\delta>0}\sup_{y\in[a,b],|y-x|\le\delta}f(y)$,}
     
\noindent so that $f$ is continuous at $x$ iff $g(x)=h(x)$.    We have
$g\le f\le h$, so if $D$ is any dissection of $[a,b]$ then $S_D(g)\le
S_D(f)\le S_D(h)$ and $s_D(g)\le s_D(f)\le s_D(h)$.   But in fact
$S_D(f)=S_D(h)$ and $s_D(g)=s_D(f)$, because on any open interval
$\ooint{c,d}\subseteq[a,b]$ we must have
     
\Centerline{$\inf_{x\in\ooint{c,d}}g(x)=\inf_{x\in\ooint{c,d}}f(x)$,
\quad
$\sup_{x\in\ooint{c,d}}f(x)=\sup_{x\in\ooint{c,d}}h(x)$.}
     
\noindent It follows that
     
\Centerline{$L_{[a,b]}(f)=L_{[a,b]}(g)\le U_{[a,b]}(g)\le
U_{[a,b]}(f)$,}
     
\Centerline{$L_{[a,b]}(f)\le L_{[a,b]}(h)\le U_{[a,b]}(h)=
U_{[a,b]}(f)$.}
     
Because $f$ is Riemann integrable, both $g$ and
$h$ must be Riemann integrable, with integrals equal to
$\Rint_a^bf$.   By 134Kb, they are both Lebesgue integrable, with the
same integral.   But $g\le h$, so $g\eae h$, by 122Rd.   Now $f$ is
continuous at
any point where $g$ and $h$ agree, so $f$ is continuous a.e.
     
\medskip
     
{\bf (b)} Now suppose that $f$ is continuous a.e.   For each
$n\in\Bbb N$, let $D_n$ be the
dissection of $[a,b]$ into $2^n$ equal portions.   Set
     
\Centerline{$h_n(x)=\sup_{y\in\ooint{c,d}}f(y)$,
\quad$g_n(x)=\inf_{y\in\ooint{c,d}}f(y)$}
     
\noindent if $\ooint{c,d}$ is an open interval of $D_n$ containing $x$;
for definiteness, say $h_n(x)=g_n(x)=f(x)$ if $x$ is one of the points
of the list $D_n$.   Then $\sequencen{g_n}$, $\sequencen{h_n}$ are,
respectively, increasing and decreasing sequences of functions, each
function constant on each of a finite family of intervals covering
$[a,b]$;  and $s_{D_n}(f)=\int g_nd\mu$, $S_{D_n}(f)=\int h_nd\mu$.
Next,
     
\Centerline{$\lim_{n\to\infty}g_n(x)=\lim_{n\to\infty}h_n(x)=f(x)$}
     
\noindent at any point $x$ at which $f$ is continuous;  so
$f\eae\lim_{n\to\infty}g_n\eae\lim_{n\to\infty}h_n$.    By
Lebesgue's Dominated Convergence Theorem (123C),

\Centerline{$\lim_{n\to\infty}\int g_nd\mu
=\int fd\mu=\lim_{n\to\infty}\int h_nd\mu$;}

\noindent but this means that
     
\Centerline{$L_{[a,b]}(f)\ge\int fd\mu\ge U_{[a,b]}(f)$,}
     
\noindent so these are all equal and $f$ is Riemann integrable.
}%end of proof of 134L
     
     
\exercises{
\leader{134X}{Basic exercises $\pmb{>}$(a)}
%\spheader 134Xa
Show that if $f$ is an integrable real-valued
function on $\BbbR^r$, then $\int f(x+a)dx$ exists and is equal to
$\int f$ for every $a\in\BbbR^r$.   \Hint{start with simple functions
$f$.}
%134A
     
\spheader 134Xb More generally, show that if $E\subseteq\BbbR^r$ is
measurable and $f$ is a real-valued function which is integrable
over $E$ in the sense of 131D,
then $\int_{E-a}f(x+a)dx$ exists and is equal to $\int_Ef$ for every
$a\in\BbbR^r$.
%134Xa, 134A
     
\spheader 134Xc Show that if $C\subseteq\Bbb R$ is any
non-negligible set, it has a non-measurable subset.   \Hint{use the
method of 134B, taking the relation $\sim$ on a suitable bounded subset
of $C$ in place of $\coint{\tbf{0},\tbf{1}}$.}
%134B
     
\sqheader 134Xd Let $\nu_g$ be a Lebesgue-Stieltjes measure on
$\Bbb R$, constructed as in 114Xa from a non-decreasing function
$g:\Bbb R\to\Bbb R$, and $\Sigma_g$ its domain.   (See also 132Xg.)
Show that
     
\quad(i) if $A\subseteq\Bbb R$ is any set, then
     
$$\eqalign{\nu_g^*A&=\inf\{\nu_g G:G\text{ is open},\,G\supseteq A\}\cr
&=\min\{\nu_g H:H\text{ is Borel},\,H\supseteq A\};\cr}$$
     
\quad(ii) if $E\in\Sigma_g$, then
     
\Centerline{$\nu_g E=\sup\{\nu_g F:F
\text{ is closed and bounded},\,F\subseteq E\}$,}
     
\noindent and there are Borel sets $H_1$, $H_2$ such that
$H_1\subseteq E\subseteq H_2$ and
$\nu_g(H_2\setminus H_1)=\nu_g(H_2\setminus E)=\nu_g(E\setminus H_1)=0$;
     
\quad(iii) if $A\subseteq\Bbb R$ is any set, then $A$ has a
measurable envelope which is a Borel set;
     
\quad(iv) if $f$ is a $\Sigma_g$-measurable real-valued function defined
on a subset of $\Bbb R$, then there is a $\nu_g$-conegligible Borel set
$H\subseteq\Bbb R$ such that $f\restr H$ is Borel measurable.
%134F
     
\spheader 134Xe Let $E\subseteq\BbbR^r$ be a measurable set, and $\epsilon>0$.   (i) Show that there is an open set $G\supseteq E$ such that $\mu(G\setminus E)\le\epsilon$.   \Hint{apply 134Fa to each set
$E\cap B(\tbf{0},n)$.}   (ii) Show that there is a closed set
$F\subseteq E$ such that $\mu(E\setminus F)\le\epsilon$.
%134F
     
\spheader 134Xf Let $C\subseteq[0,1]$ be the Cantor set.   Show that
$\{x+y:x$, $y\in C\}=[0,2]$ and $\{x-y:x$, $y\in C\}=[-1,1]$.
%134G
     
\spheader 134Xg Let $f$, $g$ be functions from $\Bbb R$ to
itself.   Show that (i) if $f$ and $g$ are both Borel measurable, so is
their composition $fg$ (ii) if $f$ is Borel measurable and $g$ is
Lebesgue measurable, then $fg$ is Lebesgue measurable (iii) if $f$ is
Lebesgue measurable and $g$ is Borel measurable, then $fg$ need not be
Lebesgue measurable.
%134I
     
\spheader 134Xh Show that for any integer $r\ge 1$
there is a measurable set $E\subseteq\BbbR^r$ such that $E$
and $\BbbR^r\setminus E$ both meet every non-empty open interval in a
set of strictly positive measure.
%134J
     
\spheader 134Xi Give $[0,1]$ its subspace measure.   (i) Show that there
is a disjoint sequence $\sequencen{A_n}$ of subsets of $[0,1]$ all of
outer measure $1$.
(ii) Show that there is a function $f:[0,1]\to\ooint{0,1}$ such that
$\underline{\int}f=0$ and $\overline{\int}f=1$.
%134B
     
\spheader 134Xj Let $f$ be a measurable real function and $g$ a
real function such that $\dom g\setminus\dom f$ and 
$\{x:x\in\dom g\cap\dom f$, $g(x)\ne f(x)\}$ are both negligible.   
Show that $g$ is measurable.
%134Xg 134I 

\leader{134Y}{Further exercises (a)}
%\spheader 134Ya
Fix $c>0$.  For $A\subseteq\BbbR^r$ set
$cA=\{cx:x\in A\}$.   (i) Show that $\mu^*(cA)=c^r\mu^*A$ for every
$A\subseteq\BbbR^r$.   (ii) Show that $cE$ is measurable for every
measurable $E\subseteq\BbbR^r$.
%134A  
%this is in 115Xe;  can be dropped here query
     
\spheader 134Yb Let $\langle f_{mn}\rangle_{m,n\in\Bbb N}$,
$\sequence{m}{f_m}$, $f$ be real-valued measurable functions defined
almost everywhere in $\BbbR^r$ and such that
$f_m\eae\lim_{n\to\infty}f_{mn}$ for each $m$ and
$f\eae\lim_{m\to\infty}f_m$.   Show that there is a sequence
$\sequence{k}{n_k}$ such that $f\eae\lim_{k\to\infty}f_{k,n_k}$.
\Hint{take $n_k$ such that the measure of
$\{x:\|x\|\le k,\,|f_k(x)-f_{k,n_k}(x)|\ge 2^{-k}\}$ is at most $2^{-k}$
for each $k$.}
     
\spheader 134Yc Let $f$ be a measurable real-valued function
defined almost
everywhere in $\BbbR^r$.   Show that there is a sequence
$\sequencen{f_n}$ of continuous functions converging to $f$ almost
everywhere.   \Hint{Deal successively with the cases (i) 
$f=\chi I$ where $I$ is a half-open interval 
(ii) $f=\chi(\bigcup_{j\le n}I_j)$ where $I_0,\ldots,I_n$ are disjoint 
half-open intervals (iii)
$f=\chi E$ where $E$ is a measurable set of finite measure (iv) $f$ is a
simple function (v) general $f$, using 134Yb at steps (iii) and (v).}
%134Yb, 134F
     
\spheader 134Yd Let $f$ be a real-valued function defined on a
subset of $\BbbR^r$.   Show that the
following are equiveridical:   (i) $f$ is measurable (ii)
whenever $E\subseteq\BbbR^r$ is measurable and $\mu E>0$, there is a
measurable set $F\subseteq E$ such that $\mu F>0$ and $f\restr F$ is
continuous (iii) whenever $E\subseteq\BbbR^r$ is measurable and
$\gamma<\mu E$, there is a measurable $F\subseteq E$ such that
$\mu F\ge\gamma$ and $f\restr F$ is continuous.   \Hint{for
(i)$\Rightarrow$(iii), use 134Yc and 131Ya;  for (ii)$\Rightarrow$(i)
use 121D.
This is a version of {\bf Lusin's theorem}.}
%134F, 134Yc
     
\spheader 134Ye Let $\nu$ be a measure on $\Bbb R$ which is
translation-invariant in the sense of 134Ab, and such that $\nu[0,1]$ is
defined and equal to $1$.   Show that $\nu$ agrees with Lebesgue measure
on the Borel sets of $\Bbb R$.   \Hint{Show first that $[a,1]$
belongs to the domain of $\nu$ for every $a\in[0,1]$, and hence that
every half-open interval of length at most $1$ belongs to the domain of
$\nu$;  show that $\nu\coint{a,a+2^{-n}}=2^{-n}$ for every $a\in\Bbb R$,
$n\in\Bbb N$, and hence that $\nu\coint{a,b}=b-a$ whenever $a<b$.}
%134A, 134F
     
\spheader 134Yf Let $\nu$ be a measure on $\BbbR^r$ which is
translation-invariant in the sense of 134Ab, where $r>1$, and such that
$\nu[\tbf{0},\tbf{1}]$ is
defined and equal to $1$.   Show that $\nu$ agrees with Lebesgue measure
on the Borel sets of $\BbbR^r$.
%134Ye, 134F
     
\spheader 134Yg Show that if $f$ is any real-valued integrable
function on $\Bbb R$, and $\epsilon>0$, there is a continuous function
$g:\Bbb R\to\Bbb R$ such that $\{x:g(x)\ne 0\}$ is bounded and
$\int|f-g|\le\epsilon$.   \Hint{show that the set $\Phi$ of
functions $f$ with this property satisfies the conditions of 122Yb.}
%134F
     
\spheader 134Yh Repeat 134Yg for real-valued integrable
functions on $\BbbR^r$, where $r>1$.
%134Yg, 134F
     
\spheader 134Yi Repeat 134Fd, 134Xa, 134Xb, 134Yb, 134Yc, 134Yd, 134Yg
and 134Yh for complex-valued functions.
%134Yh, 134F
     
     
\spheader 134Yj Show that if $G\subseteq\BbbR^r$ is open and not empty,
it is expressible as a disjoint union of a sequence of half-open
intervals each of the form $\{x:2^{-m}n_i\le\xi_i<2^{-m}(n_i+1)$ for
every $i\le r\}$ where $m\in\Bbb N$, $n_1,\ldots,n_r\in\Bbb Z$.
     
\spheader 134Yk Show that a set $E\subseteq\BbbR^r$ is Lebesgue
negligible iff there is a sequence $\sequencen{C_n}$ of hypercubes in
$\BbbR^r$ such that $E\subseteq\bigcap_{n\in\Bbb N}\bigcup_{k\ge n}C_k$
and $\sum_{k=0}^{\infty}(\diam C_k)^r<\infty$, writing $\diam C_k$ for
the diameter of $C_k$.
%134Yj, 134F
     
\spheader 134Yl Show that there is a continuous function
$f:[0,1]\to[0,1]^2$ such that $\mu_1f^{-1}[E]=\mu_2E$ for every
measurable $E\subseteq[0,1]^2$, writing $\mu_1$, $\mu_2$ for Lebesgue
measure on $\Bbb R$, $\BbbR^2$ respectively.   \Hint{for each
$n\in\Bbb N$, express $[0,1]^2$ as the union of $4^n$ closed squares of
side $2^{-n}$;  call the set of these squares $\Cal D_n$.   Construct
continuous $f_n:[0,1]\to[0,1]^2$, families
$\langle I_D\rangle_{D\in\Cal D_n}$ inductively in such a way that each
$I_D$ is a closed interval of length $4^{-n}$ and 
$f_m[I_D]\subseteq D$ whenever $D\in\Cal D_n$ and $m\ge n$.
The induction will proceed more smoothly if you suppose that the path
$f_n$ enters each square in $\Cal D_n$ at a corner and leaves at an
adjacent corner.   Take $f=\lim_{n\to\infty}f_n$.  This is a special
kind of {\bf Peano} or {\bf space-filling} curve.}
%134H
     
\spheader 134Ym Show that if $r\le s$ there is a continuous
function $f:[0,1]^r\to[0,1]^s$ such that $\mu_rf^{-1}[E]=\mu_sE$ for
every measurable $E\subseteq[0,1]^s$, writing $\mu_r$, $\mu_s$ for
Lebesgue measure on $\BbbR^r$, $\BbbR^s$ respectively.
%134Yl, 134H
     
\spheader 134Yn Show that there is a continuous function
$f:\Bbb R\to\BbbR^2$ such that $\mu_1f^{-1}[E]=\mu_2E$ for every
measurable
$E\subseteq\BbbR^2$, writing $\mu_1$, $\mu_2$ for Lebesgue measure on
$\Bbb R$, $\BbbR^2$ respectively.
%134Yl, 134H
     
\spheader 134Yo Show that the function $f:[0,1]\to[0,1]^2$ of
134Yl may be chosen in such a way that $\mu_2f[E]=\mu_1E$ for every
Lebesgue measurable set $E\subseteq[0,1]$.   \Hint{using the
construction suggested in 134Yl, and setting
$H=f^{-1}[([0,1]\setminus\Bbb Q)^2]$, $f\restr H$ will be an isomorphism
between $(H,\mu_{1,H})$ and $(f[H],\mu_{2,f[H]})$, writing $\mu_{1,H}$
and $\mu_{2,f[H]}$ for the subspace measures.}
%134Yl, 134H
     
\spheader 134Yp Show that $\Bbb R$ can be expressed as the union
of a disjoint
sequence $\sequencen{E_n}$ of sets of finite measure such that
$\mu(I\cap E_n)>0$ for every non-empty open interval $I\subseteq\Bbb R$
and every $n\in\Bbb N$.
%134J
     
\spheader 134Yq Show that for any $r\ge 1$, $\BbbR^r$ can be
expressed as the union of a disjoint
sequence $\sequencen{E_n}$ of sets of finite measure such that
$\mu(G\cap E_n)>0$ for every non-empty open set $G\subseteq\BbbR^r$
and every $n\in\Bbb N$.
%134Yp, 134J
     
\spheader 134Yr Show that there is a disjoint sequence
$\sequencen{A_n}$ of subsets of $\Bbb R$ such that
$\mu^*(A_n\cap E)=\mu E$ for every measurable set $E$ and every
$n\in\Bbb N$.
%134Xi \query
({\it Remark\/}:  in fact there is a disjoint family
$\langle A_t\rangle_{t\in\Bbb R}$ with this property, but I think a new
idea is needed for this extension.   See 419I in Volume 4.)
%134J
     
\spheader 134Ys Repeat 134Yr for $\BbbR^r$,
where $r>1$.
%134Yr, 134J
     
\spheader 134Yt Describe a Borel measurable function $f:[0,1]\to[0,1]$
such that $f\restr A$ is discontinuous at every point of $A$ whenever
$A\subseteq[0,1]$ is a set of full outer measure.
%134J
     
\spheader 134Yu Let $\sequencen{E_n}$ be a sequence of non-negligible
measurable subsets of $\BbbR^r$.   Show that there is a measurable set
$E\subseteq\BbbR^r$ such that all the sets $E_n\cap E$, $E_n\setminus E$
are non-negligible.
%134J
}%end of exercises
     
\endnotes{
\Notesheader{134} Lebesgue measure enjoys an enormous variety of special
properties, corresponding to the richness of the real line, with its
algebraic and topological and order structures.   Here I have only been
able to hint at what is possible.
     
There are many methods of constructing non-measurable sets, all
significant;  the one I give in 134B is perhaps the most accessible, and
shows that translation-invariance is (subject to the axiom of choice) an
insuperable barrier to measuring every subset of $\Bbb R$.
     
In 134F I list some of the basic relationships between the measure and
the topology of Euclidean space.   Others are in 134Yc, 134Yd and 134Yg;
see also 134Xd.    A systematic analysis of these will take up a large
part of Volume 4.
     
The Cantor set and function (134G-134I) form one of the basic examples
in the theory.  Here I present them just as an interesting design and as
a counter-example to a natural conjecture.   But they will reappear in
three different chapters of Volume 2 as illustrations of three quite
different phenomena.
     
The relationship between the Lebesgue and Riemann integrals goes a good
deal deeper than I wish to explore just at present;  the fact that the
Lebesgue integral extends the Riemann integral (134Kb) is only a small
part of the story, and I should be sorry if you were left with the
impression that the Lebesgue integral therefore renders the Riemann
integral obsolete.   Without going into the details here, I hope that
134F and 134Yg make it plain that the Lebesgue integral is in some sense
the canonical extension of the Riemann integral.   (This, at least, I
shall return to in Chapter 43.)   Another way of
looking at this is 134Yf;  the Lebesgue integral is the basic
translation-invariant integral on $\BbbR^r$.
}%end of notes
     
\discrpage
     
     
\frfilename{mt135.tex}
\versiondate{14.9.04/14.7.07}
\copyrightdate{1994}

\def\chaptername{Complements}
\def\sectionname{The extended real line}

\newsection{135}

It is often convenient to allow
`$\infty$' into our formulae, and in the context of measure theory the
appropriate manipulations are sufficiently consistent for it to
be possible to develop a theory of the {\bf extended real line}, the set
$[-\infty,\infty]=\Bbb R\cup\{-\infty,\infty\}$, sometimes written
$\overline{\Bbb R}$.   I give a brief account without full proofs, as I
hope that by the time this material becomes necessary to the arguments I
use it will all appear thoroughly elementary.

\leader{135A}{The algebraic structure of $[-\infty,\infty]$ (a)} If we
write

\Centerline{$a+\infty=\infty+a=\infty$,
\quad$a+(-\infty)=(-\infty)+a=-\infty$}

\noindent for every $a\in\Bbb R$, and

\Centerline{$\infty+\infty=\infty$,\quad$(-\infty)+(-\infty)=-\infty$,}

\noindent but refuse to define $\infty+(-\infty)$ or $(-\infty)+\infty$,
we obtain a partially-defined binary operation on $[-\infty,\infty]$,
extending ordinary addition on $\Bbb R$.   This is {\it associative} in
the sense that

\inset{if $u$, $v$, $w\in[-\infty,\infty]$ and one of $u+(v+w)$,
$(u+v)+w$ is defined, so is the other, and they are then equal,}

\noindent and {\it commutative} in the sense that

\inset{if $u$, $v\in[-\infty,\infty]$ and one of $u+v$, $v+u$ is
defined, so is the other, and they are then equal.}

\noindent It has an {\it identity} $0$ such that $u+0=0+u=u$ for every
$u\in[-\infty,\infty]$;  but $\infty$ and $-\infty$ lack inverses.

\header{135Ab}{\bf (b)} If we define

\Centerline{$a\cdot\infty=\infty\cdot a=\infty$,
\quad$a\cdot(-\infty)=(-\infty)\cdot a=-\infty$}

\noindent for real $a>0$,

\Centerline{$a\cdot\infty=\infty\cdot a=-\infty$,
\quad$a\cdot(-\infty)=(-\infty)\cdot a=\infty$}

\noindent for real $a<0$,

\Centerline{$\infty\cdot\infty=(-\infty)\cdot(-\infty)=\infty$,
\quad$(-\infty)\cdot\infty=\infty\cdot(-\infty)=-\infty$,}

\Centerline{$0\cdot\infty=\infty\cdot 0=0\cdot(-\infty)
=(-\infty)\cdot0=0$}

\noindent then we obtain a binary operation on $[-\infty,\infty]$
extending ordinary multiplication on $\Bbb R$, which is associative and
commutative and has an identity 1;  $0$, $\infty$ and $-\infty$ lack
inverses.

\header{135Ac}{\bf (c)} We have a {\it distributive law}\cmmnt{, a little 
weaker than the associative and commutative laws of addition}:

\inset{if $u$, $v$, $w\in[-\infty,\infty]$ and both $u(v+w)$ and $uv+uw$
are defined, then they are equal.}

\cmmnt{\noindent (But note the problems which arise with such combinations as $\infty(1+(-2))$, $0\cdot\infty+0\cdot(-\infty)$.)}

\header{135Ad}{\bf (d)} While $\infty$ and $-\infty$ do not have
inverses in the semigroup $([-\infty,\infty],\cdot)$, there seems no harm 
in writing $a/\infty=a/(-\infty)=0$ for every $a\in\Bbb R$.  
\cmmnt{But of course
such an extension of the notion of division must be watched carefully in
such formulae as $u\cdot\bover{v}{u}$.}

\leader{135B}{The order structure of $[-\infty,\infty]$ (a)} If we write

\Centerline{$-\infty\le u\le \infty$ for every $u\in[-\infty,\infty]$,}

\noindent we obtain a relation on $[-\infty,\infty]$, extending the
usual ordering of $\Bbb R$, which is a {\it total} ordering, that is,

\inset{for any $u$, $v$, $w\in[-\infty,\infty]$, if $u\le v$ and $v\le
w$ then $u\le w$,}

\inset{$u\le u$ for every $u\in[-\infty,\infty]$,}

\inset{for any $u$, $v\in[-\infty,\infty]$, if $u\le v$ and $v\le u$
then $u=v$,}

\inset{for any $u$, $v\in[-\infty,\infty]$, either $u\le v$ or $v\le
u$.}

\noindent Moreover, every subset of $[-\infty,\infty]$ has a supremum
and an infimum, if we write $\sup\emptyset=-\infty$,
$\inf\emptyset=\infty$.

\header{135Bb}{\bf (b)} The ordering is `translation-invariant' in
the weak sense that

\inset{if $u$, $v$, $w\in[-\infty,\infty]$ and $v\le w$ and $u+v$, $u+w$
are both defined, then $u+v\le u+w$.}

\noindent It is preserved by non-negative multiplications in the sense
that

\inset{if $u$, $v$, $w\in[-\infty,\infty]$ and $0\le u$ and $v\le w$,
then $uv\le uw$,}

\noindent while it is reversed by non-positive multiplications in the
sense that

\inset{if $u$, $v$, $w\in[-\infty,\infty]$ and $u\le 0$ and $v\le w$,
then $uw\le uv$.}

\leader{135C}{The Borel structure of $[-\infty,\infty]$} We say that a
set $E\subseteq[-\infty,\infty]$ is a {\bf Borel set} in
$[-\infty,\infty]$ if $E\cap\Bbb R$ is a Borel subset of $\Bbb R$.
It is easy to check that the family of such sets is a
$\sigma$-algebra of subsets of  $[-\infty,\infty]$.  \cmmnt{See also 135Xb below.}

\ifnum\stylenumber=11\ifresultsonly\eject\fi\fi

\leader{135D}{Convergent sequences in $[-\infty,\infty]$} We can say
that a sequence $\sequencen{u_n}$ in $[-\infty,\infty]$  {\bf converges}
to $u\in[-\infty,\infty]$ if

\inset{whenever $v<u$ there is an $n_0\in\Bbb N$ such that $v\le u_n$
for every $n\ge n_0$, and whenever $u<v$ there is an $n_0\in\Bbb N$ such
that $u_n\le v$ for every $n\ge n_0$;}

\noindent alternatively,

\inset{{\it either} $u\in\Bbb R$ and for every $\delta>0$ there is an
$n_0\in\Bbb N$ such that $u_n\in[u-\delta,u+\delta]$ for every $n\ge
n_0$}

\inset{{\it or} $u=-\infty$ and for every $a\in\Bbb R$ there is an
$n_0\in\Bbb N$ such that $u_n\le a$ for every $n\ge n_0$}

\inset{{\it or} $u=\infty$ and for every $a\in\Bbb R$ there is an
$n_0\in\Bbb N$ such that $u_n\ge a$ for every $n\ge n_0$.}

\cmmnt{\noindent (Compare the notion of convergence in 112Ba.)}

\vleader{48pt}{135E}{Measurable functions} Let $X$ be any set and $\Sigma$ a
$\sigma$-algebra of subsets of $X$.

\header{135Ea}{\bf (a)} Let $D$ be a subset of $X$ and $\Sigma_D$ the
subspace $\sigma$-algebra (121A).   For any function
$f:D\to[-\infty,\infty]$, the following are equiveridical:

\quad (i) $\{x:f(x)<u\}\in\Sigma_D$ for every $u\in[-\infty,\infty]$;

\quad (ii) $\{x:f(x)\le u\}\in\Sigma_D$ for every
$u\in[-\infty,\infty]$;

\quad (iii) $\{x:f(x)>u\}\in\Sigma_D$ for every $u\in[-\infty,\infty]$;

\quad (iv) $\{x:f(x)\ge u\}\in\Sigma_D$ for every
$u\in[-\infty,\infty]$;

\quad (v) $\{x:f(x)\le q\}\in\Sigma_D$ for every $q\in\Bbb Q$.

\prooflet{\noindent\Prf\ The proof is almost identical to that of 121B.
The only modifications are:

-- in (i)$\Rightarrow$(ii), $\{x:f(x)\le\infty\}$ and
$\{x:f(x)\le -\infty\}$ are not necessarily equal to
$\bigcap_{n\in\Bbb N}\{x:f(x)<\infty+2^{-n}\}$,
$\bigcap_{n\in\Bbb N}\{x:f(x)<-\infty+2^{-n}\}$;  but the former is $D$,
so surely belongs to $\Sigma_D$, and the latter is
$\bigcap_{n\in\Bbb N}\{x:f(x)<-n\}$, so belongs to $\Sigma_D$.

-- In (iii)$\Rightarrow$(iv), similarly, we have to use the facts that

\Centerline{$\{x:f(x)\ge -\infty\}=D\in\Sigma_D$,
\quad$\{x:f(x)\ge\infty\}=\bigcap_{n\in\Bbb N}\{x:f(x)>n\}\in\Sigma_D$.}

-- Concerning the extra condition (v), of course we have
(ii)$\Rightarrow$(v), but also we have (v)$\Rightarrow$(i), because

\Centerline{$\{x:f(x)<u\}=\bigcup_{q\in\Bbb Q,q<u}\{x:f(x)\le q\}$}

\noindent for every $u\in[-\infty,\infty]$.
\Qed}


\header{135Eb}{\bf (b)} We may therefore say\cmmnt{, as in 121C,} that
a function taking values in $[-\infty,\infty]$ is {\bf measurable} if it
satisfies these equivalent conditions.

\header{135Ec}{\bf (c)} Note that if $f:D\to[-\infty,\infty]$ is
$\Sigma$-measurable, then

\Centerline{$E_{\infty}(f)=f^{-1}[\{\infty\}]=\{x:f(x)\ge\infty\}$,
\quad $E_{-\infty}(f)=f^{-1}[\{-\infty\}]=\{x:f(x)\le-\infty\}$}

\noindent must belong to $\Sigma_D$, while
$f_{\Bbb R}=f\restr D\setminus(E_{\infty}(f)\cup E_{-\infty}(f))$, the
`real-valued part of $f$', is measurable in the sense of 121C.

\header{135Ed}{\bf (d)} Conversely, if $E_{\infty}$ and $E_{-\infty}$
belong to $\Sigma_D$, and
$f_{\Bbb R}:D\setminus(E_{\infty}\cup E_{-\infty})\to\Bbb R$ is 
measurable, then $f:D\to[-\infty,\infty]$ will be
measurable, where $f(x)=\infty$ if $x\in E_{\infty}$, $f(x)=-\infty$ if
$x\in E_{-\infty}$ and $f(x)=f_{\Bbb R}(x)$ for other $x\in D$.

\header{135Ee}{\bf (e)} It follows that if $f$, $g$ are measurable
functions from subsets of $X$ to $[-\infty,\infty]$, then $f+g$,
$f\times g$ and $f/g$ are measurable.   \prooflet{\Prf\ This can be
proved either by adapting the arguments of 121Eb, 121Ed and 121Ee, or by
applying those results to $f_{\Bbb R}$ and $g_{\Bbb R}$ and considering
separately the sets on which one or both are infinite.   \Qed}

\header{135Ef}{\bf (f)} We can say that a function $h$ from a subset $D$ of $[-\infty,\infty]$ to $[-\infty,\infty]$ is {\bf Borel measurable} if it is
measurable\cmmnt{ (in the sense of (b) above)} with respect to the
Borel $\sigma$-algebra of $[-\infty,\infty]$\cmmnt{ (as defined in 135C)}.
Now if $X$ is a set, $\Sigma$ is a
$\sigma$-algebra of subsets of $X$, $f$ is a measurable function from a
subset of $X$ to $[-\infty,\infty]$ and $h$ is a Borel measurable
function from a subset of
$[-\infty,\infty]$ to $[-\infty,\infty]$, then $hf$ is measurable.
\prooflet{\Prf\ Apply 121Eg to $h^*f_{\Bbb R}$, where
$h^*=h\restr(\Bbb R\cap h^{-1}[\Bbb R])$, and then look separately at
the sets $\{x:f(x)=\pm\infty\}$, $\{x:hf(x)=\pm\infty\}$.   \Qed}

\header{135Eg}{\bf (g)} Let $X$ be a set and $\Sigma$ a
$\sigma$-algebra of subsets of $X$.   Let $\sequencen{f_n}$ be a
sequence of measurable functions from subsets of $X$ to
$[-\infty,\infty]$.   Then $\lim_{n\to\infty}f_n$,
$\sup_{n\in\Bbb N}f_n$ and $\inf_{n\in\Bbb N}f_n$ are measurable,
if\cmmnt{, following the principles set out in 121F,} we take their domains to be

\Centerline{$\{x:x\in\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}\dom f_m,\,
\lim_{n\to\infty}f_n(x)$ exists in $[-\infty,\infty]\}$,}

\Centerline{$\bigcap_{n\in\Bbb N}\dom f_n$.}

\prooflet{\noindent\Prf\ Follow the method of 121Fa-121Fc.   \Qed}

\leader{135F}{$[-\infty,\infty]$-valued integrable functions (a)} We are
surely not going to admit a function as `integrable' unless it is
finite almost everywhere, and for such functions the remarks in 133B are
already adequate.

\header{135Fb}{\bf (b)} However, it is possible to make a consistent
extension of the idea of an infinite integral, elaborating slightly the
ideas of 133A.   If $(X,\Sigma,\mu)$ is a measure space and $f$ is a
function, defined almost everywhere in $X$, taking values in
$[0,\infty]$, and virtually measurable (that is, such that $f\restr E$
is measurable\cmmnt{ in the sense of 135E} for some
conegligible set $E$), then we can safely write `$\int f=\infty$'
whenever $f$ is not integrable.   We shall find that for such functions
we have $\int f+g=\int f+\int g$ and $\int cf=c\int f$ for every
$c\in[0,\infty]$, using the definitions given above for addition and
multiplication on $[0,\infty]$.   Consequently, as in
122M-122O, %122M 122N 122O
we can say that for a general virtually measurable function $f$, defined
almost everywhere in $X$,
taking values in $[-\infty,\infty]$, $\int f=\int f_1-\int f_2$
whenever $f$ is expressible as a difference $f_1-f_2$ of non-negative
functions such that $\int f_1$ and $\int f_2$ are both defined and not
both infinite.   Now we have\cmmnt{, as always,} the basic formulae

\Centerline{$\int f+g=\int f +\int g$,\quad
$\int cf=c\int f$,\quad
$\int|f|\ge|\int f|$}

\noindent whenever the right-hand-sides are defined, and
$\int f\le\int g$ whenever $f\leae g$ and both integrals are defined.
\cmmnt{It is important to note that} $\int f$ can be finite, on this
definition, only when $f$ is finite almost everywhere.

\leader{135G}{}\cmmnt{ We now have versions of B.Levi's theorem and
Fatou's Lemma (compare 133K).

\wheader{135G}{6}{2}{2}{72pt}
\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$\sequencen{f_n}$ a sequence of $[-\infty,\infty]$-valued functions
defined almost everywhere in $X$ which have integrals defined in
$[-\infty,\infty]$.

(a) If $f_n\leae f_{n+1}$ for every $n$ and
$-\infty<\sup_{n\in\Bbb N}\int f_n$, then
$\int\sup_{n\in\Bbb N}f_n=\sup_{n\in\Bbb N}\int f_n$.

(b) If, for each $n$, $f_n\ge 0$ a.e., then
$\int\liminf_{n\to\infty}f_n\le\liminf_{n\to\infty}\int f_n$.

\proof{{\bf (a)} Note that $f=\sup_{n\in\Bbb N}f_n$ is defined
everywhere on $\bigcap_{n\in\Bbb N}\dom f_n$, which is almost
everywhere;  and that there is a conegligible set $E$ such that
$f_n\restr E$ is measurable for every $n$, so that $f\restr E$ is
measurable.   Now if $u=\sup_{n\in\Bbb N}\int f_n$ is finite, then all
but finitely many of the $f_n$ must be finite almost everywhere, and the
result is a consequence of B.Levi's theorem for
real-valued functions;  while if $u=\infty$ then surely
$\int\sup_{n\in\Bbb N}f_n$ is infinite.

\medskip

{\bf (b)} As in 123B or 133Kb, this now follows, applying (a) to
$g_n=\inf_{m\ge n}f_m$.
}%end of proof of 135G

\leader{135H}{Upper and lower integrals again (a)} To handle functions
taking values in $[-\infty,\infty]$ we need to adapt the definitions in
133I.   Let $(X,\Sigma,\mu)$ be a measure space and $f$ a
$[-\infty,\infty]$-valued function defined almost everywhere in $X$.    Its
{\bf upper integral} is

\Centerline{$\overlineint f
=\inf\{\int g:\int g$ is defined in the sense of 135F and $f\leae g\}$,}

\noindent allowing $\infty$ for $\inf\{\infty\}$ and $-\infty$ for
$\inf\ocint{-\infty,\infty}$ or $\inf[-\infty,\infty]$.
Similarly, the {\bf lower integral} of $f$ is

\Centerline{$\underlineint f
=\sup\{\int g:\int g$ is defined, $f\geae g\}$.}

\noindent With this modification, all the results of 133J
are valid for functions taking values in $[-\infty,\infty]$ rather than
in $\Bbb R$.

\header{135Hb}{\bf (b)}\cmmnt{ Corresponding to 133Ka, we have the
following.}   Let $(X,\Sigma,\mu)$ be a measure space, and
$\sequencen{f_n}$ a
sequence of $[-\infty,\infty]$-valued functions defined almost
everywhere in $X$.

\medskip

\quad{(i)} If $f_n\leae f_{n+1}$ for every $n$ and
$\sup_{n\in\Bbb N}\overline{\int}f_n>-\infty$, then
$\overline{\int}\sup_{n\in\Bbb N}f_n
=\sup_{n\in\Bbb N}\overline{\int}f_n$.

\medskip

\quad{(ii)} If, for each $n$, $f_n\ge 0$ a.e., then
$\overline{\int}\liminf_{n\to\infty}f_n
\le\liminf_{n\to\infty}\overline{\int}f_n$.

\leader{135I}{\bf Subspace \dvrocolon{measures}}\cmmnt{ We need to
re-examine the ideas of \S131 in the new context.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space,
and $H\in\Sigma$;  write $\Sigma_H$ for the subspace $\sigma$-algebra on
$H$ and $\mu_H$ for the subspace measure.   For any
$[-\infty,\infty]$-valued function $f$ defined on a subset of $H$, write
$\tilde f$ for the extension of $f$ defined by saying that
$\tilde f(x)=f(x)$ if $x\in\dom f$, $0$ if
$x\in X\setminus H$.

(a) Suppose that $f$ is a $[-\infty,\infty]$-valued function defined on a
subset of $H$.

\quad(i) $\dom f$ is $\mu_H$-conegligible iff $\dom\tilde f$ is
$\mu$-conegligible.

\quad(ii) $f$ is $\mu_H$-virtually measurable iff $\tilde f$ is
$\mu$-virtually measurable.

\quad(iii) $\int_Hfd\mu_H=\int_X\tilde fd\mu$ if either is
defined in $[-\infty,\infty]$.

(b) Suppose that $h$ is a $[-\infty,\infty]$-valued function defined almost
everywhere in $X$.   Then
$\int_H(h\restr H)d\mu_H=\int h\times\chi H\,d\mu$ if either is defined in
$[-\infty,\infty]$.

(c) If $h$ is a $[-\infty,\infty]$-valued function and $\int_Xh\,d\mu$ is
defined in $[-\infty,\infty]$, then $\int_H(h\restr H)d\mu_H$ is defined in
$[-\infty,\infty]$.

(d) Suppose that $h$ is a $[-\infty,\infty]$-valued function defined almost
everywhere in $X$.   Then

\Centerline{$\overlineint_H(h\restr H)d\mu_H
=\overlineint_Xh\times\chi Hd\mu$.}

\proof{{\bf (a)(i)} This is immediate from 131Ca, since
$H\setminus\dom f=X\setminus\dom\tilde f$.

\medskip

\quad{\bf (ii)}\grheada\ If $f$ is $\mu_H$-virtually measurable, there is a
$\mu_H$-conegligible $E\in\Sigma_H$ such that $f\restr E$ is
$\Sigma_H$-measurable.   There is an $F\in\Sigma$ such that $E=F\cap H$;
now $G=F\cup(X\setminus H)$ belongs to $\Sigma$ and $E=G\cap H$ and $G$ is
$\mu$-conegligible.   Also, for $q\in\Bbb Q$,

$$\eqalign{\{x:x\in G,\,\tilde f(x)\le q\}
&=\{x:x\in E,\,f(x)\le q\}\in\Sigma_H\subseteq\Sigma
   \text{ if }q<0,\cr
&=\{x:x\in E,\,f(x)\le q\}\cup(X\setminus H)\in\Sigma
   \text{ if }q\ge 0,\cr}$$

\noindent so $\tilde f\restr G$ is $\Sigma$-measurable and $\tilde f$ is
$\mu$-virtually measurable.

\medskip

\qquad\grheadb\ If $\tilde f$ is $\mu$-virtually measurable,
there is a $\mu$-conegligible $G\in\Sigma$ such that $\tilde f\restr G$ is
$\Sigma$-measurable.   Now $E=G\cap H$ belongs to $\Sigma_H$ and is
$\mu_H$-conegligible, and for $q\in\Bbb Q$

\Centerline{$\{x:x\in E,\,f(x)\le q\}
=H\cap\{x:x\in G,\,f(x)\le q\}\in\Sigma_H$.}

\noindent So $f\restr E$ is $\Sigma_H$-measurable and $f$ is
$\mu_H$-virtually measurable.

\medskip

\quad{\bf (iii)} Assume that at least one of the integrals is defined.
Then (ii) tells us that
there is a $\mu$-conegligible $E\in\Sigma$ such that $\tilde f\restr E$ is
$\Sigma$-measurable, in which case $f\restr H\cap E$ is
$\Sigma_H$-measurable.

\medskip

\qquad\grheada\ Suppose that $f$ is non-negative everywhere on its domain.
Then $\int_Hfd\mu_H$ and $\int_X\tilde fd\mu$ are both defined in
$[0,\infty]$.   If both are infinite, we can stop.   Otherwise,

\Centerline{$G=\{x:x\in E\cap H$, $f(x)<\infty\}
=\{x:x\in E$, $\tilde f(x)<\infty\}$}

\noindent must be conegligible.   Set $g=f\restr G\cap H$;  then
$\tilde g=\tilde f\restr G$, so $g=f\,\,\mu_H$-a.e.\ and
$\tilde g=\tilde f\,\,\mu$-a.e.   Accordingly
$\int_Hfd\mu_H=\int_Hg\,d\mu_H$ and
$\int_X\tilde fd\mu=\int_X\tilde g\,d\mu$.   Now we
are supposing that at least one of these is finite.   But in this case we
can apply 131E to see that $\int_Hg\,d\mu=\int_X\tilde g\,d\mu$, so
$\int_Hf\,d\mu=\int_X\tilde f\,d\mu$.

\medskip

\qquad\grheadb\ In general, express $f$ as $f^+-f^-$, where

\Centerline{$f^+(x)=\max(0,f(x))$,
\quad$f^-(x)=\max(0,-f(x))$}

\noindent for $x\in\dom f$.   Then $(f^+)\ssptilde=\tilde f^+$ and
$(f^-)\ssptilde=\tilde f^-$.   So

\Centerline{$\int_Hfd\mu_H
=\int_Hf^+d\mu_H-\int_Hf^{-}d\mu_H
=\int_X\tilde f^+d\mu-\int_X\tilde f^{-}d\mu
=\int_X\tilde fd\mu$}

\noindent if any of the four expressions is defined in $[-\infty,\infty]$.

\medskip

{\bf (b)} Set $f=h\restr H$;  then $(h\times\chi H)(x)=\tilde f(x)$ for
every $x\in\dom h$, so (a-iii) tells us that

\Centerline{$\int_Xh\times\chi H\,d\mu=\int_X\tilde fd\mu
=\int_H(h\restr H)d\mu_H$}

\noindent if any of the three is defined in $[-\infty,\infty]$.

\medskip

{\bf (c)} Setting $h^+(x)=\max(0,h(x))$ and $h^-(x)=\max(0,-h(x))$ for
$x\in\dom h$, both $\int_Xh^+d\mu$ and $\int_Xh^-d\mu$ are defined in
$[0,\infty]$, and at most one of them is infinite.   In particular, both
are $\mu$-virtually measurable and defined $\mu$-almost everywhere,
so the same is true of $h^+\times\chi H$ and $h^-\times\chi H$.    As
$\int_Xh^+\times\chi Hd\mu\le\int_Xh^+d\mu$ and
$\int_Xh^-\times\chi Hd\mu\le\int_Xh^-d\mu$, at most one of
$\int_Xh^+\times\chi Hd\mu$, $\int_Xh^-\times\chi Hd\mu$ is infinite, and

\Centerline{$\int_Xh\times\chi Hd\mu
=\int_Xh^+\times\chi Hd\mu-\int_Xh^-\times\chi Hd\mu$}

\noindent is defined in $[-\infty,\infty]$.   By (b) above,
$\int_H(h\restr H)d\mu_H$ is defined in $[-\infty,\infty]$.

\medskip

{\bf (d)(i)} Suppose that $\int_Xg\,d\mu$ is defined in $[-\infty,\infty]$
and that $h\times\chi H\le g\,\,\mu$-a.e.   Then

\Centerline{$\int_H(g\restr H)d\mu_H=\int_Xg\times\chi Hd\mu$}

\noindent is defined, by (c);  and as $g(x)\ge 0$ for $\mu$-almost every
$x\in X\setminus H$, $g\times\chi H\leae g$.   So

\Centerline{$\overlineint_H(h\restr H)d\mu_H
\le\int_H(g\restr H)d\mu_H=\int_Xg\times\chi Hd\mu\le\int_Xg\,d\mu$.}

\noindent As $g$ is arbitrary,
$\overline{\int}_H(h\restr H)d\mu_H
\le\overline{\int}_Xh\times\chi H\,d\mu$.

\medskip

\quad{\bf (ii)} Suppose that $\int_Hfd\mu_H$ is defined in
$[-\infty,\infty]$ and that $h\restr H\le f\,\,\mu_H$-a.e.   Then
$\int_X\tilde fd\mu$ is defined in $[-\infty,\infty]$ and
$h\times\chi H\le\tilde f\,\,\mu$-a.e., so

\Centerline{$\overlineint_Xh\times\chi H\,d\mu
\le\int_X\tilde fd\mu=\int_Hf\,d\mu_H$.}

\noindent As $f$ is arbitrary,
$\overline{\int}_Xh\times\chi H\,d\mu
\le\overline{\int}_H(h\restr H)d\mu_H$.
}%end of proof of 135I

\exercises{
\leader{135X}{Basic exercises (a)}  We say that a set
$G\subseteq[-\infty,\infty]$ is {\bf open} if (i) $G\cap \Bbb R$ is open
in the usual sense as a subset of $\Bbb R$ (ii) if $\infty\in G$, then
there is some $a\in\Bbb R$ such that $\ocint{a,\infty}\subseteq G$ (iii)
if $-\infty\in G$ then there is some $a\in\Bbb R$ such that
$\coint{-\infty,a}\subseteq G$.    Show that the family $\frak T$ of
open subsets of $[-\infty,\infty]$ has the properties corresponding to
(a)-(d) of 1A2B.

\header{135Xb}{\bf (b)} Show that the Borel sets of $[-\infty,\infty]$
as defined in 135C are precisely the members of the $\sigma$-algebra of
subsets of
$[-\infty,\infty]$ generated by the open sets as defined in 135Xa.

\sqheader 135Xc Define $\phi:[-\infty,\infty]\to[-1,1]$ by
setting

\Centerline{$\phi(-\infty)=-1$, \quad $\phi(x)=\tanh x
=\Bover{e^{2x}-1}{e^{2x}+1}$ if $-\infty<x<\infty$, \quad
$\phi(\infty)=1$.}

\noindent Show that (i) $\phi$ is an order-isomorphism between
$[-\infty,\infty]$ and $[-1,1]$ (ii) for any sequence $\sequencen{u_n}$ in $[-\infty,\infty]$, $\sequencen{u_n}\to u$ iff
$\sequencen{\phi(u_n)}\to\phi(u)$  (iii) for any set
$E\subseteq [-\infty,\infty]$, $E$ is Borel in $[-\infty,\infty]$ iff
$\phi[E]$ is a Borel subset of $\Bbb R$ (iv) a real-valued function
$h$ defined on a subset of $[-\infty,\infty]$ is Borel measurable iff
$h\phi^{-1}$ is Borel measurable.

\sqheader 135Xd Let $X$ be a set, $\Sigma$ a
$\sigma$-algebra of subsets of $X$ and $f$ a function from a subset of $X$
to $[-\infty,\infty]$.   Show that $f$ is measurable iff the composition
$\phi f$ is measurable, where $\phi$ is the function of 135Xc.   Use
this to reduce 135Ef and 135Eg to the corresponding results in
\S121.

\header{135Xe}{\bf (e)} Let $\phi:[-\infty,\infty]\to[-1,1]$ be the
function described in 135Xc.   Show that the functions

\Centerline{$(t,u)\mapsto\phi(\phi^{-1}(t)+\phi^{-1}(u)):
[-1,1]^2\setminus\{(-1,1),(1,-1)\}\to[-1,1]$,}

\Centerline{$(t,u)\mapsto\phi(\phi^{-1}(t)\phi^{-1}(u)):
[-1,1]^2\to[-1,1]$,}

\Centerline{$(t,u)\mapsto\phi(\phi^{-1}(t)/\phi^{-1}(u)):
\bigl([-1,1]\times([-1,1]\setminus\{0\})\bigr)
\setminus\{(\pm 1,\pm 1)\}\to[-1,1]$}

\noindent are Borel measurable.   Use this with 121K to prove 135Ee.

\header{135Xf}{\bf (f)} Following the conventions of 135Ab and 135Ad,
give  full descriptions of the cases in which $uu'/vv'=(u/v)(u'/v')$ and
in which $uw/vw=u/v$.

\spheader 135Xg Let $(X,\Sigma,\mu)$ be a measure space and suppose that 
$E\in\Sigma$ has non-zero finite measure.   Let $f$ be a virtually
measurable $[-\infty,\infty]$-valued function defined on a subset of $X$ and suppose that $f(x)$ is defined and greater than $\alpha$ for almost every $x\in E$.   Show that $\int_Ef>\alpha\mu E$.
%135F

\leader{135Y}{Further exercises (a)}
%\spheader 135Ya
Let $X$ be a set and $\Sigma$ a $\sigma$-algebra of subsets of $X$.   
Show that if $f:X\to[0,\infty]$ is $\Sigma$-measurable, there is a 
sequence $\sequencen{E_n}$ in $\Sigma$ such that 
$f=\sum_{n=0}^{\infty}\Bover1{n+1}\chi E_n$.

\spheader 135Yb Let $(X,\Sigma,\mu)$ be a measure space, and $f$, $g$ two
$[-\infty,\infty]$-valued functions, defined on subsets of $X$, such that
$\int f$ and $\int g$ are both defined in $[-\infty,\infty]$.   
(i) Show that
$\int f\vee g$ and $\int f\wedge g$ are defined in $[-\infty,\infty]$,
where $(f\vee g)(x)=\max(f(x),g(x))$, $(f\wedge g)(x)=\min(f(x),g(x))$ for
$x\in\dom f\cap\dom g$.   (ii) Show that 
$\int f\vee g+\int f\wedge g=\int f+\int g$ in the sense that if one of the
sums is defined in $[-\infty,\infty]$ so is the other, and they are then
equal.
%135F

\spheader 135Yc\dvAnew{2007} 
Let $(X,\Sigma,\mu)$ be a measure space, 
$f:X\to[-\infty,\infty]$ a function and $g:X\to[0,\infty]$, 
$h:X\to[0,\infty]$ measurable functions.   Show that 
$\overline{\int}f\times(g+h) 
=\overline{\int}f\times g+\overline{\int}f\times h$, where here
we interpret $\infty+(-\infty)$ as $\infty$, as in 133L. 
}%end of exercises

\endnotes{
\Notesheader{135} I have taken this exposition into a separate section
partly because of its length, and partly because I wish to emphasize
that these techniques are incidental to the principal ideas of this
volume.   Really all I am trying to do here is give a coherent account
of the language commonly used to deal with a variety of peripheral
cases.    As a general rule, `$\infty$' enters these arguments only as
a shorthand for certain types of triviality.   When we find ourselves
wishing to assign the values $\pm\infty$ to a function, either this
happens on a negligible set -- in which case it is often right, if
slightly less comforting, to think of the function as undefined on that
set -- or things have got completely out of hand, and the theory has
little useful to tell us.

Of course it is not difficult to incorporate the theory of the extended
real line directly into the arguments of Chapter 12, so that the results
of this section become the basic ones.   I have avoided this route
partly in an attempt to reduce the number of new ideas needed in the
technically very demanding material of Chapter 12 -- believing, as I do,
that independently of our treatment of $\pm\infty$ it is absolutely
necessary to be able to deal with partially-defined functions -- and
partly because I do not think that the real line should really be
regarded as a substructure of the extended real line.   I think that
they are different structures with different properties, and that the
original real line is overwhelmingly more important.
But it is fair to say that in terms of the ideas treated in this volume
they are so similar that when you are properly familiar with this work
you will be able to move freely from one to the other, so freely indeed
that you can safely leave the distinction to formal occasions, such as
when you are presenting the statement of a theorem.
}%end of notes

\discrpage

\frfilename{mt136.tex}
\versiondate{22.6.05}
\copyrightdate{2000}

\def\chaptername{Complements}
\def\sectionname{The Monotone Class Theorem}

\newsection{*136}

For the final section of this volume, I present two theorems on
$\sigma$-algebras, with some simple corollaries.   They are here because
I find no natural
home for them in Volume 2.   While they (especially 136B) are part of
the basic technique of measure theory, and have many and widespread
applications, they are not central to the particular
approach I have chosen, and can if you wish be left on one side until
they come to be needed.

\leader{136A}{Lemma} Let $X$ be a set, and $\Cal A$ a family of subsets
of $X$.   Then the following are equiveridical:

\inset{(i) $X\in\Cal A$, $B\setminus A\in\Cal A$  whenever $A$,
$B\in\Cal A$ and $A\subseteq B$, and $\bigcup_{n\in\Bbb N}A_n\in\Cal A$
whenever $\sequencen{A_n}$ is a non-decreasing sequence in $\Cal A$;

(ii) $\emptyset\in\Cal A$, $X\setminus A\in\Cal A$ for every
$A\in\Cal A$, and $\bigcup_{n\in\Bbb N}A_n\in\Cal A$ whenever
$\sequencen{A_n}$ is a disjoint sequence in $\Cal A$.}

\proof{{\bf (i)$\Rightarrow$(ii)} Suppose that (i) is true.   Then of
course $\emptyset=X\setminus X$ belongs to $\Cal A$ and
$X\setminus A\in\Cal A$ for every $A\in\Cal A$.
If $A$, $B\in\Cal A$ are disjoint,
then $A\subseteq X\setminus B\in\Cal A$, so
$(X\setminus B)\setminus A$ and its complement $A\cup B$ belong to
$\Cal A$.   So if $\sequencen{A_n}$ is a disjoint sequence in $\Cal A$,
$\bigcup_{i\le n}A_i\in\Cal A$ for every $n$, and
$\bigcup_{n\in\Bbb N}A_n$ is the union of a non-decreasing sequence in
$\Cal A$, so belongs to $\Cal A$.   Thus (ii) is true.

\medskip

{\bf (ii)$\Rightarrow$(i)} If (ii) is true, then of course
$X=X\setminus\emptyset$ belongs to $\Cal A$.   If $A$ and $B$ are
members of $\Cal A$ such that $A\subseteq B$, then $X\setminus B$
belongs to
$\Cal A$ and is disjoint from $A$, so $A\cup(X\setminus B)$ and its
complement $B\setminus A$ belong to $\Cal A$.   Thus the second clause
of (i) is satisfied.   As for the third, if $\sequencen{A_n}$ is a
non-decreasing sequence in $\Cal A$, then
$A_0,A_1\setminus A_0,A_2\setminus A_1,\ldots$ is a disjoint sequence in
$\Cal A$, so its union $\bigcup_{n\in\Bbb N}A_n$ belongs to $\Cal A$.
}%end of proof of 136A

\medskip

\noindent{\bf Definition} If $\Cal A\subseteq\Cal PX$ satisfies the
conditions of (i) and/or (ii) above, it is called a {\bf Dynkin class} of
subsets of $X$.

\leader{136B}{Monotone Class Theorem} Let $X$ be a set and $\Cal A$ a
Dynkin class of subsets of $X$.   Suppose that $\Cal I\subseteq\Cal A$
is such that $I\cap J\in\Cal I$ for all $I$, $J\in\Cal I$.   Then
$\Cal A$ includes the $\sigma$-algebra of subsets of $X$ generated by
$\Cal I$.

\proof{{\bf (a)} Let $\frak S$ be the family of Dynkin classes of
subsets of $X$ including $\Cal I$.   Then it is easy to check, using
either (i) or (ii) of 136A, that the intersection
$\Sigma=\bigcap\frak S$ also is a Dynkin class (compare 111Ga).
Because $\Cal A\in\frak S$, $\Sigma\subseteq\Cal A$.

\medskip

{\bf (b)} If $H\in\Sigma$, then

\Centerline{$\Sigma_H=\{E:E\in\Sigma,\,E\cap H\in\Sigma\}$}

\noindent is a Dynkin class.   \Prf\ ($\alpha$)
$X\cap H=H\in\Sigma$ so $X\in\Sigma_H$.   ($\beta$) If $A$,
$B\in\Sigma_H$ and $A\subseteq B$ then $A\cap H$, $B\cap H$ belong to
$\Sigma$ and $A\cap H\subseteq B\cap H$;  consequently

\Centerline{$(B\setminus A)\cap H=(B\cap H)\setminus(A\cap H)\in\Sigma$}

\noindent and $B\setminus A\in\Sigma_H$.   ($\gamma$) If
$\sequencen{A_n}$ is a non-decreasing sequence in $\Sigma_H$, then
$\sequencen{A_n\cap H}$ is a non-decreasing sequence in $\Sigma$, so

\Centerline{$(\bigcup_{n\in\Bbb N}A_n)\cap H
=\bigcup_{n\in\Bbb N}(A_n\cap H)\in\Sigma$}

\noindent and $\bigcup_{n\in\Bbb N}A_n\in\Sigma_H$.\ \Qed

It follows that if $I\cap H\in\Sigma$ for every $I\in\Cal I$, so that
$\Sigma_H\supseteq\Cal I$, then $\Sigma_H\in\frak S$ and must
be equal to $\Sigma$.

\medskip

{\bf (c)} We find next that $G\cap H\in\Sigma$ for all $G$,
$H\in\Sigma$.   \Prf\ Take  $I$, $J\in\Cal I$.   We know that
$I\cap J\in\Cal I$.   As $I$ is arbitrary, $\Sigma_J=\Sigma$ and
$H\in\Sigma_J$, that is, $H\cap J\in\Sigma$.   As $J$ is arbitrary,
$\Sigma_H=\Sigma$ and $G\in\Sigma_H$, that is, $G\cap H\in\Sigma$.\ \Qed

\medskip

{\bf (d)} Since $\Sigma$ is a Dynkin class,
$\emptyset=X\setminus X\in\Sigma$.   Also

\Centerline{$G\cup H
=X\setminus((X\setminus G)\cap(X\setminus H))\in\Sigma$}

\noindent for any $G$, $H\in\Sigma$ (using (c)).
So if $\sequencen{G_n}$ is any sequence in $\Sigma$,
$G'_n=\bigcup_{i\le n}G_i\in\Sigma$ for each $n$ (inducing on $n$).
But $\sequencen{G'_n}$
is now a non-decreasing sequence in $\Sigma$, so

\Centerline{$\bigcup_{n\in\Bbb N}G_n
=\bigcup_{n\in\Bbb N}G'_n\in\Sigma$.}

This means that $\Sigma$ satisfies all the conditions of 111A and is a
$\sigma$-algebra of subsets of $X$.   Since $\Cal I\subseteq\Sigma$,
$\Sigma$ must include the $\sigma$-algebra $\Sigma'$ of subsets of $X$
generated by $\Cal I$.   So $\Sigma'\subseteq\Sigma\subseteq\Cal A$, as
required.

(Actually, of course, $\Sigma=\Sigma'$, because $\Sigma'\in\frak S$.)
}%end of proof of 136B

\cmmnt{\medskip

\noindent{\bf Remark} I have seen this result called the
{\bf Sierpi\'nski Class Theorem} and the
{\bf $\pmb{\pi}$-$\pmb{\lambda}$ Theorem}.}

\leader{136C}{Corollary} Let $X$ be a set, and $\mu$, $\nu$ two measures
defined on $X$ with domains $\Sigma$, $\Tau$ respectively.   Suppose
that $\mu X=\nu X<\infty$, and that $\Cal I\subseteq\Sigma\cap\Tau$ is a
family of sets such that $\mu I=\nu I$ for every $I\in\Cal I$ and
$I\cap J\in\Cal I$ for all $I$, $J\in\Cal I$.   Then $\mu E=\nu E$ for
every $E$ in the $\sigma$-algebra of subsets of $X$ generated by
$\Cal I$.

\proof{ The point is that

\Centerline{$\Cal A=\{H:H\in\Sigma\cap\Tau,\,\mu H=\nu H\}$}

\noindent is a Dynkin class of subsets of $X$.   \Prf\ I work from (ii)
of 136A.   Of course $\emptyset\in\Cal A$.   If $A\in\Cal A$ then

\Centerline{$\mu(X\setminus A)=\mu X-\mu A=\nu X-\nu A
=\nu(X\setminus A)$}

\noindent (because $\mu X=\nu X<\infty$, so the subtraction is
safe), and $X\setminus A\in\Cal A$.   If $\sequencen{A_n}$ is a disjoint
sequence in $\Cal A$, then

\Centerline{$\mu A=\sum_{n=0}^{\infty}\mu A_n=\sum_{n=0}^{\infty}\nu A_n
=\nu A$,}

\noindent and $\bigcup_{n\in\Bbb N}A_n\in\Cal A$.\ \Qed

Since $\Cal I\subseteq\Cal A$, 136B tells us that the $\sigma$-algebra
$\Sigma'$ generated by $\Cal I$ is included in $\Cal A$, that is, $\mu$
and $\nu$ agree on $\Sigma'$.
}%end of proof of 136C

\vleader{72pt}{136D}{Corollary} Let $\mu$, $\nu$ be two measures on
$\BbbR^r$,
where $r\ge 1$, both defined, and agreeing, on all intervals of the form

\Centerline{$\ocint{-\infty,a}=\{x:x\le a\}=\{(\xi_1,\ldots,\xi_r):
\xi_i\le\alpha_i$ for every $i\le r\}$}

\noindent for $a=(\alpha_1,\ldots,\alpha_r)\in\BbbR^r$.   Suppose
further that $\mu\BbbR^r<\infty$.   Then $\mu$ and $\nu$ agree on all
the Borel subsets of $\BbbR^r$.

\proof{ In 136C, take $X=\BbbR^r$ and $\Cal I$ the set of intervals
$\ocint{-\infty,a}$.   Then $I\cap J\in\Cal I$ for all $I$,
$J\in\Cal I$, since
$\ocint{-\infty,a}\cap\ocint{-\infty,b}=\ocint{-\infty,a\wedge b}$,
writing
$a\wedge b=(\min(\alpha_1,\beta_1),\ldots,\min(\alpha_r,\beta_r))$ if
$a=(\alpha_1,\ldots,\alpha_r)$, $b=(\beta_1,\ldots,\beta_r)\in\BbbR^r$.
Also, setting $\tbf{n}=(n,\ldots,n)$ for $n\in\Bbb N$,

\Centerline{$\nu\BbbR^r=\lim_{n\to\infty}\nu\ocint{-\infty,\tbf{n}}
=\lim_{n\to\infty}\mu\ocint{-\infty,\tbf{n}}=\mu\BbbR^r$.}

\noindent So all the conditions of 136C are satisfied and $\mu$, $\nu$
agree on the $\sigma$-algebra $\Sigma$ generated by $\Cal I$.   But this
is just the $\sigma$-algebra of Borel sets, by 121J.
}%end of proof of 136D

\leader{136E}{Algebras of sets:  Definition} Let $X$ be a set.   A
family $\Cal E\subseteq\Cal PX$ is an {\bf algebra} or {\bf field} of
subsets of $X$ if

\inset{(i) $\emptyset\in\Cal E$;

(ii) for every $E\in\Cal E$, its complement $X\setminus E$
belongs to $\Cal E$;

(iii) for every $E$, $F\in\Cal E$, $E\cup F\in\Cal E$.}

\leader{136F}{Remarks}\cmmnt{ {\bf (a)} I could very well have
introduced this notion
in Chapter 11, along with `$\sigma$-algebras'.   I omitted it, apart
from some exercises, because
there seemed to be quite enough new definitions in \S111 already, and
because I had nothing substantial to say about algebras of sets.

\header{136Fb}} {\bf (b)} If $\Cal E$ is an algebra of subsets of $X$,
then

\Centerline{$E\cap F\cmmnt{\mskip5mu =X\setminus((X\setminus
E)\cup(X\setminus
F))}$,
\quad$E\setminus F\cmmnt{\mskip5mu =E\cap(X\setminus F)}$,}

\Centerline{$E_0\cup E_1\cup\ldots\cup E_n$,
\quad$E_0\cap E_1\cap\ldots\cap E_n$}

\noindent  belong to $\Cal E$ for all $E$, $F$,
$E_0,\ldots,E_n\in\Cal E$.   \cmmnt{(Induce on $n$ for the last.)}

\spheader 136Fc A $\sigma$-algebra of subsets of $X$ is\cmmnt{ (of
course)} an algebra of subsets of $X$.

\vleader{42pt}{136G}{Theorem} Let $X$ be a set and $\Cal E$ an algebra
of subsets of $X$.   Suppose that $\Cal A\subseteq\Cal PX$ is a family
of sets such that

\inset{($\alpha$) $\bigcup_{n\in\Bbb N}A_n\in\Cal A$ for every non-decreasing
sequence $\sequencen{A_n}$ in $\Cal A$,

($\beta$) $\bigcap_{n\in\Bbb N}A_n\in\Cal A$ for every non-increasing
sequence $\sequencen{A_n}$ in $\Cal A$,

($\gamma$) $\Cal E\subseteq\Cal A$.}

\noindent Then $\Cal A$ includes the $\sigma$-algebra of subsets of $X$
generated by $\Cal E$.

\proof{ I use the same ideas as in 136B.

\medskip

{\bf (a)} Let $\frak S$ be the family of all sets $\Cal S\subseteq\Cal
PX$ satisfying ($\alpha$)-($\gamma$).   Then its intersection
$\Sigma=\bigcap\frak S$ also satisfies the conditions.   Because $\Cal
A\in\frak S$, $\Sigma\subseteq\Cal A$.

\medskip

{\bf (b)} If $H\in\Sigma$, then

\Centerline{$\Sigma_H=\{E:E\in\Sigma,\,E\cap H\in\Sigma\}$}

\noindent satisfies conditions ($\alpha$)-($\beta$).   \Prf\ ($\alpha$)
If $\sequencen{A_n}$ is a non-decreasing sequence in $\Sigma_H$, then
$\sequencen{A_n\cap H}$ is a non-decreasing sequence in $\Sigma$, so

\Centerline{$(\bigcup_{n\in\Bbb N}A_n)\cap H
=\bigcup_{n\in\Bbb N}(A_n\cap
H)\in\Sigma$}

\noindent and $\bigcup_{n\in\Bbb N}A_n\in\Sigma_H$.  ($\beta$)
Similarly, if $\sequencen{A_n}$ is a non-increasing sequence in
$\Sigma_H$, then $\bigcap_{n\in\Bbb N}A_n\cap H\in\Sigma$ so
$\bigcap_{n\in\Bbb N}A_n\in\Sigma_H$.\ \Qed

It follows that if $E\cap H\in\Sigma$ for every $E\in\Cal E$, so that
$\Sigma_H$ also satisfies ($\gamma$), then $\Sigma_H\in\frak S$ and must
be equal to $\Sigma$.
\medskip

{\bf (c)} Consequently $G\cap H\in\Sigma$ for all $G$, $H\in\Sigma$.
\Prf\ Take  $E$, $F\in\Cal E$.   We know that $E\cap F\in\Cal E$.   As
$E$ is arbitrary, $\Sigma_F=\Sigma$ and $H\in\Sigma_F$, that is, $H\cap
F\in\Sigma$.   As $F$ is arbitrary, $\Sigma_H=\Sigma$ and
$G\in\Sigma_H$, that is, $G\cap H\in\Sigma$.\ \Qed

\medskip

{\bf (d)} Next, $\Sigma^*=\{X\setminus H:H\in\Sigma\}\in\frak S$.
\Prf\ ($\alpha$) If $\sequencen{A_n}$ is a non-decreasing sequence in
$\Sigma^*$, then $\sequencen{X\setminus A_n}$ is a non-increasing
sequence in $\Sigma$, so

\Centerline{$\bigcup_{n\in\Bbb N}A_n=X
\setminus\bigcap_{n\in\Bbb N}(X\setminus A_n)\in\Sigma^*$.}

\noindent ($\beta$) Similarly, if $\sequencen{A_n}$ is a non-increasing
sequence in $\Sigma^*$, then

\Centerline{$\bigcap_{n\in\Bbb N}A_n
=X\setminus\bigcup_{n\in\Bbb N}(X\setminus A_n)\in\Sigma^*$.}

\noindent ($\gamma$) If $E\in\Cal E$ then $X\setminus E\in\Cal E$ so
$X\setminus E\in\Sigma$ and $E\in\Sigma^*$.\ \QeD\  It follows that
$\Sigma\subseteq\Sigma^*$, that is, that $X\setminus H\in\Sigma$ for
every $H\in\Sigma$.

\medskip

{\bf (e)} Putting (c) and (d) together with the fact that $X\in\Sigma$
(because $X\in\Cal E$) and the union of a non-decreasing sequence in
$\Sigma$ belongs to $\Sigma$ (by condition ($\alpha$)), we see that the
same argument as in part (d) of the proof of 136B shows that $\Sigma$ is
a $\sigma$-algebra of subsets of $X$.   So, just as in 136B, we conclude
that the $\sigma$-algebra generated by $\Cal E$ is included in $\Sigma$
and therefore in $\Cal A$.
}%end of proof of 136G

\leader{*136H}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space such
that
$\mu X<\infty$, and $\Cal E$ a subalgebra of $\Sigma$;  let $\Sigma'$ be
the $\sigma$-algebra of subsets of $X$ generated by $\Cal E$.   If
$F\in\Sigma'$ and $\epsilon>0$, there is an $E\in\Cal E$ such that
$\mu(E\cap F)\le\epsilon$.

\proof{ Let $\Cal A$ be the family of sets $F\in\Sigma$ such that

\Centerline{for every $\epsilon>0$ there is an $E\in\Cal E$ such that
$\mu(F\symmdiff E)\le\epsilon$.}

\noindent Then $\Cal A$ is a Dynkin class.   \Prf\ I check the three
conditions of 136A(i).   ($\alpha$) $X\in\Cal A$ because
$X\in\Cal E$.   ($\beta$) If $F_1$, $F_2\in\Cal A$ and $\epsilon>0$,
there are $E_1$,
$E_2\in\Cal E$ such that $\mu(F_i\symmdiff E_i)\le\bover12\epsilon$ for
both $i$;  now $E_1\setminus E_2\in\Cal E$ and

\Centerline{$(F_1\setminus F_2)\symmdiff(E_1\setminus E_2)
\subseteq(F_1\symmdiff E_1)\cup(F_2\symmdiff E_2)$,}

\noindent so

\Centerline{$\mu((F_1\setminus F_2)\symmdiff(E_1\setminus E_2))
\le\mu(F_1\symmdiff E_1)+\mu(F_2\symmdiff E_2)\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $F_1\setminus F_2\in\Cal A$.
($\gamma$) If
$\sequencen{F_n}$ is a non-decreasing sequence in $\Cal A$, with union $F$,
and $\epsilon>0$, then

\Centerline{$\lim_{n\to\infty}\mu F_n=\mu F\le\mu X<\infty$,}

\noindent so there is an $n\in\Bbb N$ such that
$\mu(F\setminus F_n)\le\bover12\epsilon$.   Now there is an $E\in\Cal E$
such that $\mu(F_n\symmdiff E)\le\bover12\epsilon$;  as
$F\symmdiff E\subseteq(F\setminus F_n)\cup(F_n\symmdiff E)$,
$\mu(F\symmdiff E)\le\epsilon$.   As $\epsilon$ is arbitrary,
$F\in\Cal A$.\ \Qed

Since $\Cal E\subseteq\Cal A$ and $\Cal E$ is closed under $\cap$,
$\Cal A$ includes the $\sigma$-algebra $\Sigma'$ generated by $\Cal E$,
as claimed.
}%end of proof of 136H

\exercises{
\leader{136X}{Basic exercises $\pmb{>}$(a)}
%\spheader 136Xa
Let $X$ be a set and $\Cal A$ a family of subsets of $X$.   Show that
the following are equiveridical:

\quad(i) $X\in\Cal A$ and $B\setminus A\in\Cal A$ whenever $A$,
$B\in\Cal A$ and $A\subseteq B$;

\quad(ii) $\emptyset\in\Cal A$, $X\setminus A\in\Cal A$ for every
$A\in\Cal A$ and $A\cup B\in\Cal A$ whenever $A$,
$B\in\Cal A$ are disjoint.
%136A

\spheader 136Xb Suppose that $X$ is a set and $\Cal A\subseteq\Cal PX$.
Show that $\Cal A$ is a $\sigma$-algebra of subsets of $X$ iff it is a
Dynkin class and $A\cap B\in\Cal A$ whenever $A$, $B\in\Cal A$.
%136B

\spheader 136Xc\dvArevised{2010}
Let $X$ be a set, and $\Cal I$ a family of subsets of
$X$ such that $I\cap J\in\Cal I$ for all $I$, $J\in\Cal I$;  let $\Sigma$
be the $\sigma$-algebra of subsets of $X$ generated by $\Cal I$.   
Show that
$\mu E=\nu E$ whenever $E\in\Sigma$ is covered by a sequence in $\Cal I$.
({\it Hint\/}:  For $J\in\Cal I$, set
$\mu_JE=\mu(E\cap J)$,
$\nu_JE=\nu(E\cap J)$ for $E\in\Sigma$.    Use 136C to show that
$\mu_J=\nu_J$ for each $J$.)
%136C

\sqheader 136Xd Set $X=\{0,1,2,3\}$, $\Cal I=\{X,\{0,1\},\{0,2\}\}$.
Find two distinct measures $\mu$, $\nu$ on $X$, both defined on the
$\sigma$-algebra $\Cal PX$ and with $\mu I=\nu I<\infty$ for every
$I\in\Cal I$.
%136C

\spheader 136Xe Let $\Sigma$ be the family of subsets
of $\coint{0,1}$ expressible as finite unions of half-open intervals
$\coint{a,b}$.   Show that $\Sigma$ is an algebra of subsets of
$\coint{0,1}$.
%136E

\spheader 136Xf Let $X$ be a set, and $\Cal I$ a family of subsets of
$X$ such that $I\cap J\in\Cal I$ whenever $I$, $J\in\Cal I$.   Let
$\Sigma$ be the smallest family of sets such that $X\in\Sigma$,
$F\setminus E\in\Sigma$ whenever $E$, $F\in\Sigma$ and $E\subseteq F$,
and $\Cal I\subseteq\Sigma$.   Show that $\Sigma$ is an algebra of
subsets of $X$.
%136E

\spheader 136Xg Let $X$ be a set, and
$\Cal E$ an algebra of subsets of $X$.   A functional
$\nu:\Cal E\to\Bbb R$ is called ({\bf finitely}) {\bf additive} if
$\nu(E\cup F)=\nu E+\nu F$ whenever $E$,
$F\in\Cal E$ and $E\cap F=\emptyset$.   (i) Show that in this case
$\nu(E\cup F)+\nu(E\cap F)=\nu E+\nu F$ for all $E$, $F\in\Cal E$.
(ii) Show that if $\nu E\ge 0$ for every $E\in\Cal E$ then
$\nu(\bigcup_{i\le n}E_i)\le\sum_{i=0}^n\nu E_i$ for all
$E_0,\ldots,E_n\in\Cal E$.
%136F

\sqheader 136Xh Let $X$ be a set, and $\Cal A$ a family of subsets of
$X$ such that ($\alpha$) $\emptyset$, $X$ belong to $\Cal A$ ($\beta$)
$A\cap B\in\Cal A$ for all $A$, $B\in\Cal A$ ($\gamma$)
$A\cup B\in\Cal A$ whenever $A$, $B\in\Cal A$ and $A\cap B=\emptyset$.
Show that
$\{A:A\in\Cal A,\,X\setminus A\in\Cal A\}$ is an algebra of subsets of
$X$.
%136F

\sqheader 136Xi Let $X$ be a set, and $\Cal A$ a family of subsets of
$X$ such that ($\alpha$) $\emptyset$, $X$ belong to $\Cal A$ ($\beta$)
$\bigcap_{n\in\Bbb N}A_n\in\Cal A$ for every sequence $\sequencen{A_n}$
in $\Cal A$  ($\gamma$) $\bigcup_{n\in\Bbb N}A_n\in\Cal A$ for every
disjoint sequence $\sequencen{A_n}$ in $\Cal A$.   Show that
$\{A:A\in\Cal A,\,X\setminus A\in\Cal A\}$ is a $\sigma$-algebra of
subsets of $X$.
%136Xh, 136F

\sqheader 136Xj Let $\Cal A$ be a family of subsets of $\Bbb R$ such
that (i) $\bigcap_{n\in\Bbb N}A_n\in\Cal A$ for every sequence
$\sequencen{A_n}$ in $\Cal A$  (ii) $\bigcup_{n\in\Bbb N}A_n\in\Cal A$
for every disjoint sequence $\sequencen{A_n}$ in $\Cal A$ (iii) every
open interval $\ooint{a,b}$ belongs to $\Cal A$.   Show that every Borel
subset of $\Bbb R$ belongs to $\Cal A$.   ({\it Hint\/}:  show that
every half-open interval $\coint{a,b}$, $\ocint{a,b}$ belongs to
$\Cal A$, and
therefore all intervals $\ocint{-\infty,a}$, $\coint{a,\infty}$;  now
use 136Xi.)
%136Xi, 136F

\sqheader 136Xk Let $X$ be a set, $\Cal E$ an algebra of subsets of $X$,
and $\Cal A$ a family of subsets of $X$ such that ($\alpha$)
$\bigcap_{n\in\Bbb N}A_n\in\Cal A$ for every non-increasing sequence
$\sequencen{A_n}$ in $\Cal A$ ($\beta$)
$\bigcup_{n\in\Bbb N}A_n\in\Cal A$ for every disjoint sequence in
$\Cal A$ ($\gamma$) $\Cal E\subseteq\Cal A$.   Show that the
$\sigma$-algebra of sets generated by
$\Cal E$ is included in $\Cal A$.   \Hint{use the method of
136B to reduce to the case in which $A\cap B\in\Cal A$ for every $A$,
$B\in\Cal A$;  now use 136Xi.}
%136Xi, 136F

\leader{136Y}{Further exercises (a)} Let $X$ be a set and $\Cal E$ an
algebra of subsets of $X$.   Let $\nu:\Cal E\to\coint{0,\infty}$ be a
non-negative functional which is additive in the sense of 136Xg.
Define $\theta:\Cal PX\to\coint{0,\infty}$ by setting

\Centerline{$\theta A=\inf\{\sum_{n=0}^{\infty}\nu E_n:\sequencen{E_n}$
is a sequence in $\Cal E$ covering $A\}$}

\noindent for every $A\subseteq X$.   (i) Show that $\theta$ is an outer
measure on $X$ and that $\theta E\le\nu E$ for every $E\in\Cal E$.
(ii) Let $\mu$ be the measure on $X$ defined from $\theta$ by
\Caratheodory's method, and $\Sigma$ its domain.   Show that $\Cal
E\subseteq\Sigma$ and that $\mu E\le\nu E$ for every $E\in\Cal E$.
(iii) Show that the following are equiveridical:  ($\alpha$)
$\mu E=\nu E$
for every $E\in\Cal E$ ($\beta$) $\theta X=\nu X$ ($\gamma$) whenever
$\sequencen{E_n}$ is a non-increasing sequence in $\Cal E$ with empty
intersection, $\lim_{n\to\infty}\nu E_n=0$.

\spheader 136Yb Let $X$ be a set, $\Cal E$ an algebra of subsets of $X$,
and $\nu$ a non-negative additive functional on $\Cal E$.   Let $\Sigma$
be the $\sigma$-algebra of subsets of $X$ generated by $\Cal E$.   Show
that there is at most one measure $\mu$ on $X$ with domain $\Sigma$
extending $\nu$, and that there is such a measure iff
$\lim_{n\to\infty}\nu E_n=0$ for every non-increasing sequence
$\sequencen{E_n}$ in $\Cal E$ with empty intersection.

\spheader 136Yc Let $X$ be a set.   Let $\Cal G$ be a family of subsets
of $X$ such that (i)  $G\cap H\in\Cal G$ for all $G$, $H\in\Cal G$ (ii)
for every $G\in\Cal G$ there is a sequence $\sequencen{G_n}$ in $\Cal G$
such that $X\setminus G=\bigcap_{n\in\Bbb N}G_n$.   Let $\Cal A$ be a
family of subsets of $X$ such that ($\alpha$) $\emptyset$, $X\in\Cal A$
($\beta$) $\bigcap_{n\in\Bbb N}A_n\in\Cal A$
for every non-increasing sequence $\sequencen{A_n}$ in
$\Cal A$ ($\gamma$) $\bigcup_{n\in\Bbb N}A_n\in\Cal A$ for every
disjoint sequence in $\Cal A$ ($\delta$) $\Cal G\subseteq\Cal A$.   Show
that the $\sigma$-algebra of sets generated by $\Cal G$ is included in
$\Cal A$.
%136Xk, 136F
}%end of exercises

\endnotes{
\Notesheader{136} The most useful result here is 136B;  it will be
needed in Chapter 27, and helpful at
various other points in Volume 2, often through its corollaries 136C and
136Xc.   Of course 136C, like its corollary 136D and its special case
136Yb, can be used directly only on measures which do not take the value
$\infty$, since we have to know that $\mu(F\setminus E)=\mu F-\mu E$ for
measurable sets $E\subseteq F$;  that is why it comes into prominence
only when we specialize to probability measures (for which the whole
space has measure $1$).   So I include 136Xc to indicate a technique
that can take us a step farther.   I do not feel
that we are really ready for
general measures on the Borel sets of $\BbbR^r$, but I mention 136D to
show what kind of class $\Cal I$ can appear in 136B.

The two theorems here (136B, 136G) both address the question:
given a family of sets $\Cal I$, what operations must we perform in
order to build the $\sigma$-algebra $\Sigma$ generated by $\Cal I$?
For arbitrary $\Cal I$, of course, we expect to need complements and
unions of sequences.   The point of the theorems here is that if
$\Cal I$ has a certain amount of structure then we can reach $\Sigma$
with more limited operations;  thus if $\Cal I$ is an algebra of sets,
then {\it monotonic} unions and intersections are enough (136G).   Of
course there are innumerable variations on this theme.   I offer
136Xh-136Xj as
a typical result which will actually be used in Volume 4, and
136Xk and 136Yc as examples of possible modifications.   There is an
abstract version of 136B in 313G in Volume 3.

Having once started to consider the extension of an algebra of sets to a
$\sigma$-algebra, it is natural to ask for conditions under which a
functional on an algebra of sets can be extended to a measure.   The
condition of additivity (136Xg) is obviously necessary, and almost
equally obviously not sufficient.   I include 136Ya-136Yb as the most
important of many necessary and sufficient conditions for an additive
functional to be extendable to a measure.   We shall have to return to
this in Volume 4.
}%end of comment

\frnewpage

\frfilename{mt13conc.tex}
\versiondate{9.4.05}
\copyrightdate{2005}
     
\def\chaptername{Concordance}
\def\sectionname{Concordance}
     
\centerline{\bf Concordance for Chapter 13}
     
\bigskip
     
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
     
\leaveitout{1{}32E {\smc Fremlin 00} 132E let $\sigma_0'(L)$ be a 
measurable envelope of $f[\sigma(K)]$ included in $L$.   (Such an envelope exists because $\nu L$ is finite and
$f[\sigma(K)]\subseteq f[K]\subseteq L$.)
}
     
\leader{1{}32E}{Measurable envelopes} Parts (d) and (e) of 1{}32E %132E
in the 
2000 and 2001 editions, referred to in the 2001 edition of Volume 2 and 
the 2002 edition of Volume 3, are now parts (e) and (f). 


\leaveitout{1{}32F 132F {\smc K\"onig 04}, following 1.8:  
Definition of `thick' (=`full outer measure') quoted.
}

\leader{1{}32G}{Pull-back measures} Proposition 1{}32G, referred to in the
2006 edition of Volume 4, has been moved to 234F.

\leaveitout{1{}36G 136G $\Cal A$ includes the $\sigma$-algebra of subsets 
generated by $\Cal E$ {\smc Fremlin 03}
}
     
%\bigskip
     
%\noindent{\bf References}
     
%Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces', 
%Fundamenta Math.\ 165 (2000) 67-94.
   %n99821.tex

%K\"onig H.\ [04] `Projective limits via inner premeasures and the true 
%Wiener measure', Mediterranean J.\ Math.\ 1 (2004) 3-42.


\discrpage
     
     

\references{
\frfilename{mt1r.tex}
\versiondate{31.5.03}

% except for references in ...X., ...Y., ...Z. give locations in form
%  342L  rather than  342Lb

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
\gdef\chaptername{References}
\gdef\sectionname{References}
\gdef\headlinesectionname{References}

\centerline{\bf References for Volume 1}

\medskip

In addition to those (very few) works which I have mentioned in the
course of this volume, I list some of the books from which I myself
learnt measure theory, as a mark of grateful respect, and to
give you an opportunity to sample alternative approaches.

\medskip

Bartle R.G.\ [66] {\it The Elements of Integration.}  Wiley, 1966.

Cohn D.L.\ [80] {\it Measure Theory.}  Birkh\"auser, 1980.

Facenda J.A. \& Freniche F.J.\ [02]
{\it Integraci\'on de funciones de varias variables.}
Ediciones Pir\'amide, 2002.

Halmos P.R.\ [50] {\it Measure Theory.}   Van Nostrand, 1950.

Lebesgue H.\ [1901] `Sur une g\'en\'eralisation de l'int\'egrale
d\'efinie', C.R.Acad.Sci.\ (Paris) 132 (1901) 1025-1027.
Reprinted in {\smc Lebesgue 72}, vol.\ 1.

\leaveitout{Lebesgue H.\ [1902] {\it Int\'egrale, Longueur, Aire,}  th\`ese pr\'esent\'ee \`a la Facult\'e de Sciences de Paris,  Annali di Matematica Pura Appl.\ 7 (1902).   Reprinted in {\smc
Lebesgue 72}, vol.\ 1.}

Lebesgue H.\ [1904] {\it Le\c{c}ons sur l'int\'egration et la recherche
des fonctions primitives.} Gauthier-Villars, 1904.
Reprinted in {\smc Lebesgue 72}, vol.\ 2.
\cmmnt{[Vol.\ 1 {\it intro.}]}

Lebesgue H.\ [1918] `Remarques sur les th\'eories de la mesure et de l'int\'egration', Ann.\ Sci.\ Ecole Norm.\ Sup.\ 35 (1918) 191-250.
Reprinted in {\smc Lebesgue 72}, vol.\ 2.
\cmmnt{[Vol.\ 1 {\it intro.}]}

Lebesgue H.\ [72] {\it Oeuvres Scientifiques.}   L'Enseignement Math\'ematique, Institut de Math\'ematiques, Univ.\ de Gen\`eve, 1972.

Munroe M.E.\ [53] {\it Introduction to Measure and Integration.}
Addison-Wesley, 1953.

Royden H.L.\ [63] {\it Real Analysis.}  Macmillan, 1963.

Solovay R.M.\ [70] `A model of set theory in which every set of reals
is Lebesgue measurable', Annals of Math.\ 92 (1970) 1-56.   \cmmnt{[134C.]}

Tao T.\ [11] {\it An introduction to measure theory.}   Amer.\ Math.\ Soc.,
2011;   {\tt terrytao.files.wordpress.com/2011/01/measure-book1.pdf}.

Widom H.\ [69] {\it Lectures on Measure and Integration.}  Van Nostrand Reinhold, 1969.

Williamson J.H.\ [62] {\it Lebesgue Integration.}  Holt, Rinehart \& Winston, 1962.

\frnewpage

}%end of references

%---------------------------------------------------------------%

\end



