% 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{mt22.tex}
\versiondate{24.2.14}
\copyrightdate{1995}

\def\chaptername{The Fundamental Theorem of Calculus}
\def\sectionname{Introduction}

\newchapter{22}

In this chapter I address one of the most important properties of the
Lebesgue integral.   Given an integrable function $f:[a,b]\to\Bbb R$, we
can form its indefinite integral $F(x)=\int_a^xf(t)dt$ for $x\in[a,b]$.
Two questions immediately present themselves.   (i)  Can we expect to
have the derivative $F'$ of $F$ equal to $f$?   (ii)  Can we identify
which functions $F$ will appear as indefinite integrals?   Reasonably
satisfactory answers may be found for both of these questions:  $F'=f$
almost everywhere (222E) and indefinite integrals are the absolutely
continuous functions (225E).   In the course of dealing with them, we
need to develop a variety of techniques which lead to many striking
results both in the theory of Lebesgue measure and in other, apparently
unrelated, topics in real analysis.

The first step is `Vitali's theorem' (\S221), a remarkable
argument -- it is more a method than a theorem -- which uses the
geometric nature of the real line to extract disjoint subfamilies from
collections of intervals.   It is the foundation stone not only of the
results in \S222 but of all geometric measure theory, that is, measure
theory on spaces with a geometric structure.   I use it here to show
that monotonic functions are differentiable almost everywhere (222A).
Following this, Fatou's Lemma and Lebesgue's Dominated Convergence
Theorem are enough to show that the derivative of an indefinite integral
is almost everywhere equal to the integrand.   We find that some
innocent-looking manipulations of this fact take us surprisingly far;  I
present these in \S223.

I begin the second half of the chapter with a discussion of functions
`of bounded variation', that is, expressible as the difference of
bounded monotonic functions (\S224).   This is one of the least
measure-theoretic sections in the volume;  only in 224I and 224J are 
measure and integration even mentioned.   But this material is needed for 
Chapter 28 as well as for the next section, and is also
one of the basic topics of twentieth-century real analysis.   
\S225 deals with the
characterization of indefinite integrals as the `absolutely
continuous' functions.   In fact this is now quite easy;  it helps to
call on Vitali's theorem again, but everything else is a straightforward
application of methods previously used.   The second half of the section
introduces some new ideas in an attempt to give a deeper intuition into
the essential nature of absolutely continuous functions.   \S226 returns
to functions of bounded variation and their decomposition into
`saltus' and `absolutely continuous' and `singular' parts, the
first two being relatively manageable and the last looking something
like the Cantor function.



\discrpage

\frfilename{mt221.tex}
\versiondate{2.6.03}
\copyrightdate{1995}

\def\chaptername{The Fundamental Theorem of Calculus}
\def\sectionname{Vitali's theorem in $\Bbb R$}

\newsection{221}
\def\headlinesectionname{Vitali's theorem in $\eightBbb R$}

I give the first theorem of this chapter a section to itself.   It
occupies a position between measure theory and geometry (it is,
indeed, one of the fundamental results of `geometric measure
theory'), and its proof involves both the measure and the geometry of
the real line.

\leader{221A}{Vitali's theorem} Let $A$ be a bounded subset of $\Bbb R$
and $\Cal I$ a family of non-singleton closed
intervals in $\Bbb R$ such that every point of
$A$ belongs to arbitrarily short members of $\Cal I$.   Then there is a
countable set $\Cal I_0\subseteq\Cal I$ such that (i) $\Cal I_0$ is
disjoint, that is,
$I\cap I'=\emptyset$ for all distinct $I$, $I'\in\Cal I_0$ (ii)
$\mu(A\setminus\bigcup\Cal I_0)=0$, where $\mu$ is Lebesgue measure on
$\Bbb R$.

\proof{{\bf (a)} If there is a finite disjoint
set $\Cal I_0\subseteq\Cal I$
such that $A\subseteq\bigcup\Cal I_0$ (including the
possibility that $A=\Cal I_0=\emptyset$), we can stop.   So let us
suppose henceforth that there is no such $\Cal I_0$.

Let $\mu\sp*$ be Lebesgue outer
measure on $\Bbb R$.   Suppose that $|x|<M$ for every $x\in A$, and set

\Centerline{$\Cal I'=\{I:I\in\Cal I,\,I\subseteq[-M,M]\}$.}

\medskip

{\bf (b)} In this case, if $\Cal I_0$ is any finite disjoint subset of
$\Cal I'$, there is a $J\in\Cal I'$ which is disjoint from any member of
$\Cal I_0$.   \Prf\ Take $x\in A\setminus\bigcup\Cal I_0$.   Now there
is a $\delta>0$ such that
$[x-\delta,x+\delta]$ does not meet any member of $\Cal I_0$, and as
$|x|<M$ we can suppose that $[x-\delta,x+\delta]\subseteq[-M,M]$.   Let
$J$ be a member of $\Cal I$, containing $x$, and of length at most
$\delta$;  then $J\in\Cal I'$ and $J\cap\bigcup\Cal I_0=\emptyset$.\
\Qed

\medskip


{\bf (c)} We can now choose a sequence $\sequencen{\gamma_n}$ of real
numbers and a disjoint sequence $\sequencen{I_n}$ in $\Cal I'$
inductively, as follows.   Given $\langle I_j\rangle_{j<n}$ (if $n=0$,
this is the empty sequence, with no members), with $I_j\in\Cal I'$ for
each $j<n$, and $I_j\cap I_k=\emptyset$ for $j<k<n$, set

\Centerline{$\Cal J_n=\{I:I\in\Cal I',\,I\cap I_j=\emptyset$ for every
$j<n\}$.}

\noindent   We know from (b) that $\Cal J_n\ne\emptyset$.   Set

\Centerline{$\gamma_n=\sup\{\mu I:I\in\Cal J_n\}$;}

\noindent then $0<\gamma_n\le 2M$.   We may therefore choose a set
$I_n\in\Cal J_n$ such that $\mu I_n\ge{1\over 2}\gamma_n$, and this
continues the induction.

\medskip

{\bf (e)} Because the $I_n$ are disjoint Lebesgue measurable subsets of
$[-M,M]$, we have

\Centerline{$\sum_{n=0}^{\infty}\gamma_n
\le 2\sum_{n=0}^{\infty}\mu I_n\le 4M<\infty$,}

\noindent and $\lim_{n\to\infty}\gamma_n=0$.   Now define $I'_n$ to be
the closed
interval with the same midpoint as $I_n$ but five times the length, so
that it projects past each end of $I_n$ by at least $\gamma_n$.   I
claim that, for any $n$,

\Centerline{$A\subseteq\bigcup_{j<n}{I}_j\cup\bigcup_{j\ge n}I_j'$.}

\noindent\Prf\Quer\ Suppose, if possible, otherwise.   Take
any $x$ belonging to
$A\setminus(\bigcup_{j<n}{I}_j\cup\bigcup_{j\ge n}I'_j)$.   Let
$\delta>0$ be such that


\Centerline{$[x-\delta,x+\delta]
\subseteq[-M,M]\setminus\bigcup_{j<n}I_j$,}

\noindent and let $J\in\Cal I$ be such that

\Centerline{$x\in J\subseteq[x-\delta,x+\delta]$.}

\noindent   Then

\Centerline{$\mu J>0=\lim_{m\to\infty}\gamma_m$;}

\noindent  let $m$ be the least integer
greater than or equal to $n$ such that $\gamma_m<\mu J$.   In this case
$J$ cannot belong to $\Cal J_m$, so there must be some $k<m$ such that
$J\cap I_k\ne\emptyset$, because certainly $J\in\Cal I'$.   By the
choice of $\delta$, $k$ cannot be less than $n$, so $n\le k<m$, and
$\gamma_k\ge\mu J$.   In this case, the distance from $x$ to the nearest
endpoint of $I_k$ is at most $\mu J\le \gamma_k$.   But the ends of
$I'_k$ project beyond the ends of $I_k$ by at least $\gamma_k$, so $x\in
I'_k$;  which contradicts the choice of $x$. \Bang \Qed

\medskip

{\bf (f)} It follows that

\Centerline{$\mu\sp*(A\setminus\bigcup_{j<n}I_j)
\le\mu(\bigcup_{j\ge n}I'_j)\le\sum_{j=n}^{\infty}\mu I'_j
\le 5\sum_{j=n}^{\infty}\mu I_j$.}

\noindent As

\Centerline{$\sum_{j=0}^{\infty}\mu I_j\le 2M<\infty$,}

\noindent we must have

\Centerline{$\lim_{n\to\infty}\mu\sp*(A\setminus\bigcup_{j<n}I_j)=0,$}

\noindent and

\Centerline{$\mu(A\setminus\bigcup_{j\in\Bbb N}I_j)
=\mu\sp*(A\setminus\bigcup_{j\in\Bbb N}I_j)
\le\inf_{n\in\Bbb N}\mu^*(A\setminus\bigcup_{j<n}I_j)
=0$.}

Thus in this case we may set $\Cal I_0=\{I_n:n\in\Bbb N\}$ to obtain
a countable disjoint family in $\Cal I$ with
$\mu(A\setminus\bigcup\Cal I_0)=0$.
}%end of proof of 221A

\cmmnt{
\leader{221B}{Remarks (a)} I have expressed this theorem in the form 
`there is a countable set $\Cal I_0\subseteq\Cal I$ such that $\ldots$'
in an
attempt to find a concise way of expressing the three possibilities

\inset{(i) $A=\Cal I=\emptyset$, so that we must take
$\Cal I_0=\emptyset$;}

\inset{(ii) there are disjoint $I_0,\ldots,I_n\in\Cal I$ such that
$A\subseteq I_0\cup\ldots\cup I_n$, so that we can take
$\Cal I_0=\{I_0,\ldots,I_n\}$;}

\inset{(iii) there is a disjoint sequence $\sequencen{I_n}$ in $\Cal I$
such that $\mu(A\setminus\bigcup_{n\in\Bbb N}I_n)=0$, so that we can
take $\Cal I_0=\{I_n:n\in\Bbb N\}$.}

\noindent Of course many applications, like the proof of 221A itself,
will use forms of these three alternatives.

\header{221Bb}{\bf (b)} The actual theorem here, as stated, will be used
in the next section.   But quite as important as the statement of the
theorem is the principle of its proof.   The $I_n$ are chosen
`greedily', that
is, when we come to choose $I_n$ we look at the family $\Cal J_n$ of
possible intervals, given the choices $I_0,\ldots,I_{n-1}$ already made,
and choose an $I_n\in\Cal J_n$ which is `about' as big as it could be.
The supremum of the possibilities for $\mu I_n$ is $\gamma_n$;  but
since we do
not know that there is any $I\in\Cal J_n$ such that $\mu I=\gamma_n$, we
must settle for a little less.   I follow the standard formula in taking
$\mu I_n\ge\bover12\gamma_n$, but of course I could have taken
$\mu I_n\ge\bover{99}{100}\gamma_n$, or $\mu I_n\ge(1-2^{-n})\gamma_n$,
if that had helped later on.
The remarkable thing is that this works;  we can choose the $I_n$
without foresight and without considering their interrelationships (for
that matter, without examining the set $A$) beyond the minimal
requirement that $I_n\cap I_j=\emptyset$ for $j<n$, and even this
arbitrary and casual procedure yields a suitable sequence.

\header{221Bc}{\bf (c)} I have stated the theorem in terms of bounded
sets $A$ and closed intervals, which is adequate for our needs, but very
small changes in the proof suffice to deal with arbitrary
(non-singleton) intervals, and another refinement handles unbounded sets
$A$.   (See 221Ya.)
}%end of comment

\exercises{
\leader{221X}{Basic exercises (a)} Let $\alpha\in\ooint{0,1}$.
Suppose, in
part (c) of the proof of 221A, we take $\mu I_n\ge\alpha\gamma_n$ for
each $n\in\Bbb N$, rather than $\mu I_n\ge\bover12\gamma_n$.   What will
be the appropriate constant to take in place of $5$ in defining the sets
$I_j'$ of part (e)?

\leader{221Y}{Further exercises (a)}
Let $A$ be a subset of $\Bbb R$ and $\Cal I$ a family of non-singleton
intervals in $\Bbb R$ such that every point of $A$ belongs to
arbitrarily short members of $\Cal I$.   Show that there is a
countable disjoint set $\Cal I_0\subseteq\Cal I$ such that $A\setminus
\bigcup\Cal I_0$ is Lebesgue negligible.   \Hint{apply 221A to the
sets $A\cap\ooint{n,n+1}$, $\{\overline{I}:I\in\Cal
I,\,\overline{I}\subseteq\ooint{n,n+1}\}$, writing $\overline{I}$ for
the closed interval with the same endpoints as $I$.}

\header{221Yb}{\bf (b)} Let $\Cal J$ be any family of non-singleton
intervals in $\Bbb R$.   Show that $\bigcup\Cal J$ is Lebesgue
measurable.
\Hint{apply (a) to $A=\bigcup\Cal J$ and the family $\Cal I$ of
non-singleton subintervals of members of $\Cal J$.}

\header{221Yc}{\bf (c)} Let $(X,\rho)$ be a metric space, $A$ a subset
of $X$, and $\Cal I$ a family of closed balls of non-zero radius in $X$
such that every point of $A$ belongs to arbitrarily small members of
$\Cal I$.
(I say here that a set is a `closed ball of non-zero radius' if it is
expressible in the form $B(x,\delta)=\{y:\rho(y,x)\le\delta\}$ where
$x\in X$ and $\delta>0$.   Of course it is possible for such a ball to
be a singleton $\{x\}$.)   Show that either $A$ can be covered by a
finite disjoint family in $\Cal I$ or there is a disjoint sequence
$\sequencen{B(x_n,\delta_n)}$ in $\Cal I$ such that

\Centerline{$A\subseteq\bigcup_{m\le n}B(x_m,\delta_m)
  \cup\bigcup_{m>n}B(x_m,5\delta_m)$ for every $n\in\Bbb N$}

\noindent or there is a disjoint sequence $\sequencen{B(x_n,\delta_n)}$
in $\Cal I$ such that $\inf_{n\in\Bbb N}\delta_n>0$.

\spheader 221Yd Give an example of a family $\Cal I$ of open intervals
such that every point of $\Bbb R$ belongs to arbitrarily small members
of $\Cal I$, but if $\sequencen{I_n}$ is any disjoint sequence in
$\Cal I$, and for
each $n\in\Bbb N$ we write $I'_n$ for the closed interval with the same
centre as $I_n$ and ten times the length, then there is an $n$ such that
$\ooint{0,1}\not\subseteq\bigcup_{m<n}I_m\cup\bigcup_{m\ge n}I'_m$.
%mt22bits

\spheader 221Ye(i) Show that if $\Cal I$ is a {\it finite} family of
intervals in $\Bbb R$ there are $\Cal I_0$, $\Cal I_1\subseteq\Cal I$
such that $\bigcup(\Cal I_0\cup\Cal I_1)=\bigcup\Cal I$ and both
$\Cal I_0$ and $\Cal I_1$ are disjoint families.   \Hint{induce on
$\#(\Cal I)$.}   (ii) Suppose that $\Cal I$ is a family of non-singleton
intervals, of length at most $1$, covering a bounded set
$A\subseteq\Bbb R$, and that $\epsilon>0$.   Show that there is a
disjoint subfamily $\Cal I_0$ of $\Cal I$ such that
$\mu^*(A\setminus\bigcup\Cal I_0)\le\bover12\mu^*A+\epsilon$.
\Hint{replacing each member of $\Cal I$ by a slightly longer one with
rational endpoints, reduce to the case in which $\Cal I$ is countable
and thence to the case in which $\Cal I$ is finite;  now use (i).}
(iii) Use (ii) to prove Vitali's theorem.   (I learnt this argument from
J.Aldaz.)
%221A
}%end of exercises

\endnotes{
\Notesheader{221} I have headed this section `Vitali's theorem in
$\Bbb R$' because there is an $r$-dimensional version, which will appear
in Chapter 26 below.
There is an anomaly in the position of this theorem.   It is an
indispensable element of the proofs of some of the most important
theorems in measure theory;  on the other hand, the ideas involved in
its own proof
are not used elsewhere in the elementary theory.   I have therefore
myself sometimes omitted the proof when teaching this material, and
would not
reproach any student who left it to one side for the moment.   At some
stage, of course, any measure theorist must master the method, not just
for the sake of completeness, but in order to gain an intuition for
possible
variations.   I must emphasize that it is the {\it principle} of the
proof, rather than its details, which is important, because there are
innumerable
forms of `Vitali's theorem'.   (I offer some variations in the exercises
here and in \S261 below, and there are many others which are important
in more advanced work;  one will appear in \S472 in Volume 4.)   This
principle is, I suppose, that

\inset{(i) we choose the $I_n$ greedily, according to some more or less
natural criterion applicable to each $I_n$ as we come to choose it,
without attempting to look ahead;}

\inset{(ii) we prove that their sizes tend to zero, even though we
seemed to do nothing to ensure that they would (but note the shift from
$\Cal I$ to
$\Cal I'$ in part (a) of the proof of 221A, which is exactly what is
needed to make this step work);}

\inset{(iii) we check that for a suitable definition of $I_n'$,
enlarging $I_n$, we shall have
$A\subseteq\bigcup_{m<n}I_m\cup\bigcup_{m\ge n}I'_m$
for every $n$, while $\sum_{n=0}^{\infty}\mu I'_n<\infty$.}

\noindent In a way, we have to count ourselves lucky every time this
works.   The reason for studying as many variations as possible of a
technique of this kind is to learn to guess when we might be lucky.
}%end of comment

\discrpage

\frfilename{mt222.tex}
\versiondate{20.11.03/18.10.04}
\copyrightdate{2004}


%$(\DiniD^+f)(x)=(\DiniD^-f)(x)
%=(\Dinid^+f)(x)=(\Dinid^-f)(x)$

\def\chaptername{The Fundamental Theorem of Calculus}
\def\sectionname{Differentiating an indefinite integral}
     
\newsection{222}
     
I come now to the first of the two questions mentioned in the
introduction to this chapter:  if $f$ is an integrable function on
$[a,b]$, what is $\bover{d}{dx}\int_a^xf$?   It turns out that this
derivative exists and is equal to $f$ almost everywhere (222E).   The
argument is based on a striking property of monotonic functions:  they
are differentiable almost everywhere (222A), and we can bound the
integrals of their derivatives (222C).
     
\leader{222A}{Theorem} Let $I\subseteq\Bbb R$ be an interval and
$f:I\to\Bbb R$ a monotonic function.   Then $f$ is differentiable almost 
everywhere in $I$.
     
\cmmnt{\medskip
     
\noindent{\bf Remark} If I seem to be speaking of a measure on
$\Bbb R$ without naming it, as here, I mean Lebesgue measure.
}%end of comment
     
\proof{ As usual, write $\mu^*$ for Lebesgue outer measure
on $\Bbb R$, $\mu$ for Lebesgue measure.
     
\medskip
     
{\bf (a)} To begin with (down to the end of (c) below), let us suppose that $f$ is non-decreasing and $I$ is a bounded open interval
on which $f$ is bounded;  say $|f(x)|\le M$ for $x\in I$.
For any closed subinterval $J=[a,b]$ of $I$, write
$f^*(J)$ for the open interval $\ooint{f(a),f(b)}$.   For $x\in I$, write
     
\Centerline{$\DiniD f(x)=\limsup_{h\to 0}{1\over h}(f(x+h)-f(x))$,
\quad$\Dinid f(x)=\liminf_{h\to 0}{1\over h}(f(x+h)-f(x))$,}
     
\noindent allowing the value $\infty$ in both cases.   Then $f$ is
differentiable at $x$ iff $\DiniD f(x)=\Dinid f(x)\in\Bbb R$.
Because surely $\DiniD f(x)\ge \Dinid f(x)\ge 0$,  $f$ will be
differentiable at $x$ iff $\DiniD f(x)$ is finite and $\DiniD f(x)\le \Dinid f(x)$.
     
I therefore have to show that the sets
     
\Centerline{$\{x:x\in I,\,\DiniD f(x)=\infty\}$,
\quad$\{x:x\in I,\,\DiniD f(x)>\Dinid f(x)\}$}
     
\noindent are negligible.
     
\medskip
     
{\bf (b)} Let us take $A=\{x:x\in I,\,\DiniD f(x)=\infty\}$ first.   Fix an
integer $m\ge 1$ for the moment, and set
     
\Centerline{$A_m=\{x:x\in I,\,\DiniD f(x)>m\}\supseteq A$.}
     
\noindent Let $\Cal I$ be the family of non-trivial closed intervals
$[a,b]\subseteq I$ such that $f(b)-f(a)\ge m(b-a)$;   then
$\mu f^*(J)\ge m\mu J$ for every $J\in\Cal I$.   If $x\in A_m$,
then for any $\delta>0$ we have an $h$ with $0<|h|\le\delta$ and
${1\over h}(f(x+h)-f(x))>m$, so that
     
\Centerline{$[x,x+h]\in\Cal I$ if $h>0$,
\quad$[x+h,x]\in\Cal I$ if $h<0$;}
     
\noindent
thus every member of $A_m$ belongs to arbitrarily small intervals in $\Cal I$.   By Vitali's theorem (221A), there is a countable disjoint set $\Cal I_0\subseteq\Cal I$ such that
$\mu(A\setminus\bigcup\Cal I_0)=0$.
Now, because $f$ is non-decreasing,
$\langle f^*(J)\rangle_{J\in\Cal I_0}$ is disjoint, and all the $f^*(J)$ are included in $[-M,M]$, so $\sum_{J\in\Cal I_0}\mu f^*(J)\le 2M$
and $\sum_{J\in\Cal I_0}\mu J\le 2M/m$.   Because
$A_m\setminus\bigcup\Cal I_0$ is negligible,
     
     
\Centerline{$\mu^*A\le\mu^*A_m\le\Bover{2M}{m}$.}
     
\noindent As $m$ is arbitrary, $\mu^*A=0$ and $A$ is negligible.
     
\medskip
     
{\bf (c)} Now consider $B=\{x:x\in I,\,\DiniD f(x)>\Dinid f(x)\}$.   For $q$, $q'\in\Bbb Q$ with $0\le q<q'$, set
     
     
\Centerline{$B_{qq'}=\{x:x\in I,\,\Dinid f(x)< q,\,\DiniD f(x)>q'\}$.}
     
\noindent  Fix such $q$, $q'$ for the moment, and write
$\gamma=\mu^*B_{qq'}$.  Take any $\epsilon>0$, and let $G$ be an open
set including $B_{qq'}$ such that $\mu G\le\gamma+\epsilon$ (134Fa).
Let $\Cal J$ be the set of non-trivial closed intervals
$[a,b]\subseteq I\cap G$ such that $f(b)-f(a)\le q(b-a)$;  this time  $\mu f^*(J)\le q\mu J$ for $J\in\Cal J$.   Then every member of $B_{qq'}$ is included
in arbitrarily small members of $\Cal J$, so there is a countable
disjoint $\Cal J_0\subseteq\Cal J$ such that $B_{qq'}\setminus\bigcup\Cal
J_0$ is
negligible.   Let $L$ be the set of endpoints of members of $\Cal J_0$;
then $L$ is a countable union of doubleton sets, so is countable, therefore
negligible.  Set
     
\Centerline{$C=B_{qq'}\cap\bigcup\Cal J_0\setminus L$;}
     
\noindent  then
$\mu^*C=\gamma$.   Let $\Cal I$ be the set of non-trivial closed
intervals $J=[a,b]$ such that (i) $J$ is
included in one of the members of $\Cal J_0$ (ii)
$f(b)-f(a)\ge q'(b-a)$;  now $\mu f^*(J)\ge q'\mu J$ for every $J\in\Cal
I$.   Once again, because every member of $C$ is an interior point of
some member of $\Cal J_0$, every point of $C$ belongs to arbitrarily
small members of $\Cal I$;  so there is a countable disjoint $\Cal
I_0\subseteq\Cal I$ such that $\mu(C\setminus\bigcup\Cal I_0)=0$.
     
As in (b) above,
     
\Centerline{$\gamma q'\le q'\mu(\bigcup\Cal I_0)
=\sum_{I\in\Cal I_0}q'\mu I
\le\sum_{I\in\Cal I_0}\mu f^*(I)
=\mu(\bigcup_{I\in\Cal I_0}f^*(I))$.}
     
\noindent   On the other hand,
     
$$\eqalign{\mu(\bigcup_{J\in\Cal J_0}f^*(J))
&=\sum_{J\in\Cal J_0}\mu f^*(J)
\le q\sum_{J\in\Cal J_0}\mu J
=q\mu(\bigcup\Cal J_0)\cr
&\le q\mu(\bigcup\Cal J)
\le q\mu G
\le q(\gamma+\epsilon).\cr}$$
     
     
\noindent But $\bigcup_{I\in\Cal I_0}f^*(I)\subseteq\bigcup_{J\in\Cal
J_0}f^*(J)$, because every member of $\Cal I_0$ is included in a member
of $\Cal J_0$, so $\gamma q'\le q(\gamma+\epsilon)$ and
$\gamma\le \epsilon q/(q'-q)$.
As $\epsilon$ is arbitrary, $\gamma=0$.
     
Thus every $B_{qq'}$ is negligible.   Consequently $B=\bigcup_{q,q'
\in\Bbb Q,0\le q<q'}B_{qq'}$ is negligible.
     
\medskip
     
{\bf (d)} This deals with the case of a bounded open interval on which
$f$ is bounded and non-decreasing.
Still for non-decreasing $f$, but for an arbitrary interval $I$, observe
that $K=\{(q,q'):q,\,q'\in I\cap\Bbb Q,\,q<q'\}$ is countable and that
$I\setminus\bigcup_{(q,q')\in K}\ooint{q,q'}$ has at most two points
(the endpoints of $I$, if any), so is negligible.   If
we write $S$ for the set of points of $I$ at which $f$ is not
differentiable, then from (a)-(c) we see that $S\cap\ooint{q,q'}$ is
negligible for every
$(q,q')\in K$, so that $S\cap\bigcup_{(q,q')\in K}\ooint{q,q'}$ is
negligible and $S$ is negligible.
     
\medskip
     
{\bf (e)} Thus we are done if $f$ is non-decreasing.   For
non-increasing
$f$, apply the above to $-f$, which is differentiable at exactly the
same points as $f$.
}%end of proof of 222A
     
\leader{222B}{Remarks}\dvro{ If}{ {\bf (a)} I note that in the above argument I am 
using such formulae as $\sum_{J\in\Cal I_0}\mu f^*(J)$.
This is because Vitali's theorem leaves it open whether the families $\Cal I_0$ will 
be finite or infinite.   The sum must be
interpreted along the lines laid down in
112Bd in Volume 1;  generally, $\sum_{k\in K}a_k$, where $K$ is an
arbitrary set and every $a_k\ge 0$, is to be
$\sup_{L\subseteq K\text{ is finite}}\sum_{k\in L}a_k$, with the convention that $\sum_{k\in\emptyset}a_k=0$.
Now, in this context, if} $(X,\Sigma,\mu)$ is a measure space, $K$ is a
countable set, and $\family{k}{K}{E_k}$ is a family in $\Sigma$,
     
\Centerline{$\mu(\bigcup_{k\in K}E_k)\le\sum_{k\in K}\mu E_k$,}
     
\noindent with equality if $\family{k}{K}{E_k}$ is disjoint.
\prooflet{\Prf\ If $K=\emptyset$,
this is trivial.   Otherwise, let $n\mapsto k_n:\Bbb N\to K$ be a
surjection, and set
     
\Centerline{$K_n=\{k_i:i\le n\}$, \quad$G_n=\bigcup_{i\le n}E_{k_i}
=\bigcup_{k\in K_n}E_k$}
     
\noindent for each $n\in\Bbb N$.   Then $\sequencen{G_n}$
is a non-decreasing sequence with union $E=\bigcup_{k\in K}E_k$, so
     
\Centerline{$\mu E=\lim_{n\to\infty}\mu G_n=\sup_{n\in\Bbb N}\mu G_n$;}
     
\noindent and
     
\Centerline{$\mu G_n\le\sum_{k\in K_n}\mu E_k\le\sum_{k\in K}\mu E_k$}
     
\noindent for
every $n$, so $\mu E\le\sum_{k\in K}\mu E_k$.   If the $E_k$ are
disjoint, then $\mu G_n$ is precisely $\sum_{k\in K_n}\mu E_k$ for
each $n$;  but as $\sequencen{K_n}$ is a non-decreasing sequence of
sets with union $K$, every finite subset of $K$ is included in some
$K_n$, and
     
\Centerline{$\sum_{k\in K}\mu E_k=\sup_{n\in\Bbb N}\sum_{k\in K_n}
\mu E_k=\sup_{n\in\Bbb N}\mu G_n=\mu E$,}
     
\noindent as required.\ \Qed}
     
\cmmnt{
\spheader 222Bb Some readers will prefer to re-index sets
regularly, so that
all the sums they need to look at will be of the form $\sum_{i=0}^n$ or
$\sum_{i=0}^{\infty}$.   In effect, that is what I did in Volume 1, in
the proof of 114Da/115Da, when showing that Lebesgue outer measure is
indeed an outer measure.
The disadvantage of this procedure in the context of 222A is that we
must continually check that it doesn't matter whether we have a finite or 
infinite sum
at any particular moment.   I believe that it is worth taking the
trouble to learn the technique sketched here, because it very frequently
happens that we wish to consider unions of sets indexed by sets other 
than $\Bbb N$ and $\{0,\ldots,n\}$.
     
\spheader 222Bc Of course the argument above can be shortened if
you know a tiny bit
more about countable sets than I have explicitly stated so far.
But note that the value assigned to
$\sum_{k\in K}a_k$ must not depend on which enumeration
$\sequencen{k_n}$ we pick on.
}%end of comment
     
     
\leader{222C}{Lemma} Suppose that $a\le b$ in $\Bbb R$, and that
$F:[a,b]\to\Bbb R$ is a non-decreasing function.   Then $\int_a^bF'$
exists and is at most $F(b)-F(a)$.
     
\medskip
     
\def\tmpins{}
     
\noindent{\bf Remark}\cmmnt{ I discussed
integration over subsets at length in \S131 and \S214.   For measurable
subsets, which are sufficient for our needs in this
chapter, we have a simple description:  if $(X,\Sigma,\mu)$ is a measure
space, $E\in\Sigma$ and $f$ is a real-valued function, then
$\int_Ef=\int\tilde f$ if the latter integral exists, where
$\dom\tilde f=(E\cap\dom f)\cup(X\setminus E)$ and $\tilde f(x)=f(x)$ if 
$x\in E\cap\dom f$, $0$ if $x\in X\setminus E$ (apply 131Fa to
$\tilde f$).   It follows at once that if now $F\in\Sigma$ and $F\subseteq E$, $\int_Ff=\int_Ef\times\chi F$.
     
} I write $\int_a^xf$ to mean
$\int_{\coint{a,x}}f$\cmmnt{, which (because $\coint{a,x}$ is measurable) can be dealt with as described above}.
\cmmnt{Note that, as long as we are dealing with Lebesgue measure, so that $[a,x]\setminus\ooint{a,x}=\{a,x\}$ is negligible, there is no need to distinguish between $\int_{[a,x]}$, $\int_{\ooint{a,x}}$,
$\int_{\coint{a,x}}$, $\int_{\ocint{a,x}}$;  for other measures on
$\Bbb R$ we may
need to take more care.   I use half-open intervals to make it obvious
that $\int_a^xf+\int_x^yf=\int_a^yf$ if $a\le x\le y$, because
     
\Centerline{$f\times\chi\coint{a,y}
=f\times\chi\coint{a,x}\,+\,f\times\chi\coint{x,y}$.}
}%end of comment
     
\proof{{\bf (a)} The result is trivial if $a=b$;  let us suppose that
$a<b$.   By 222A, $F'$ is defined almost everywhere in $[a,b]$.
     
\medskip
     
{\bf (b)} For each $n\in\Bbb N$, define a simple function
$g_n:\coint{a,b}\to\Bbb R$ as follows.   For $0\le k<2^n$, set
$a_{nk}=a+2^{-n}k(b-a)$, $b_{nk}=a+2^{-n}(k+1)(b-a)$,
$I_{nk}=\coint{a_{nk},b_{nk}}$.   For each $x\in\coint{a,b}$, take that
$k<2^n$ such that $x\in I_{nk}$, and set
     
\Centerline{$g_n(x)=\Bover{2^n}{b-a}(F(b_{nk})-F(a_{nk}))$}
     
\noindent for $x\in I_{nk}$, so that $g_n$ gives the slope of the chord
of the graph of $F$ defined by the endpoints of $I_{nk}$.   Then
     
\Centerline{$\int_a^b g_n=\sum_{k=0}^{2^n-1}F(b_{nk})-F(a_{nk})
=F(b)-F(a)$.}
     
\medskip
     
{\bf (c)} On the other hand, if we set
     
\Centerline{$C=\{x:x\in\ooint{a,b},\,F'(x)$ exists$\}$,}
     
\noindent then $[a,b]\setminus C$ is negligible, by 222A, and
$F'(x)=\lim_{n\to\infty}g_n(x)$ for every $x\in C$.   \Prf\ Let
$\epsilon>0$.   Then there is a $\delta>0$ such that $x+h\in [a,b]$ and
$|F(x+h)-F(x))-hF'(x)|\le\epsilon|h|$ whenever $|h|\le\delta$.   Let
$n\in\Bbb N$ be such that $2^{-n}(b-a)\le\delta$.   Let
$k<2^n$ be such that $x\in I_{nk}$.   Then
     
\Centerline{$x-\delta\le a_{nk}\le x<b_{nk}\le x+\delta$,
\quad$g_n(x)=\Bover{2^n}{b-a}(F(b_{nk})-F(a_{nk}))$.}
     
\noindent Now we have
     
$$\eqalign{|g_n(x)-F'(x)|
&=|\Bover{2^n}{b-a}(F(b_{nk})-F(a_{nk}))-F'(x)|\cr
&=\Bover{2^n}{b-a}|F(b_{nk})-F(a_{nk})-(b_{nk}-a_{nk})F'(x)|\cr
&\le\Bover{2^n}{b-a}\bigl(|F(b_{nk})-F(x)-(b_{nk}-x)F'(x)|\cr
&\qquad\qquad\qquad\qquad   +|F(x)-F(a_{nk})-(x-a_{nk})F'(x)|\bigr)\cr
&\le \Bover{2^n}{b-a}(\epsilon|b_{nk}-x|+\epsilon|x-a_{nk}|)
=\epsilon.\cr}$$
     
\noindent And this is true whenever $2^{-n}\le\delta$,
that is, for all $n$ large enough.   As $\epsilon$ is arbitrary,
$F'(x)=\lim_{n\to\infty}g_n(x)$.\ \Qed
     
\medskip
     
{\bf (d)} Thus $g_n\to F'$ almost everywhere in $[a,b]$.   By Fatou's
Lemma,
     
\Centerline{$\int_a^bF'
=\int_a^b\liminf_{n\to\infty}g_n
\le\liminf_{n\to\infty}\int_a^bg_n
=\lim_{n\to\infty}\int_a^b g_n
=F(b)-F(a)$,}
     
\noindent as required.
}%end of proof of 222C
     
\cmmnt{
\medskip
     
\noindent{\bf Remark} There is a generalization of this result in 224I.
}
     
\leader{222D}{Lemma} Suppose that $a<b$ in $\Bbb R$, and that $f$, $g$
are real-valued functions, both integrable over $[a,b]$, such that
$\int_a^xf=\int_a^xg$ for
every $x\in[a,b]$.   Then $f=g$ almost everywhere in
$[a,b]$.
     
\proof{ The point is that
     
\Centerline{$\int_Ef=\int_a^bf\times\chi E
=\int_a^bg\times\chi E=\int_Eg$}
     
\noindent for any
measurable set $E\subseteq\coint{a,b}$.
     
\Prf\ {\bf (i)} If $E=\coint{c,d}$ where $a\le c\le d\le b$,
then
     
\Centerline{$\int_Ef=\int_a^df-\int_a^cf
=\int_a^dg-\int_a^cg=\int_Eg$.}
     
\medskip
     
\quad{\bf (ii)} If $E=\coint{a,b}\cap G$ for some open set
$G\subseteq\Bbb R$, then for each $n\in\Bbb N$ set
     
\Centerline{$K_n
=\{k:k\in\Bbb Z,\,|k|\le 4^n,\,\coint{2^{-n}k,2^{-n}(k+1)}\subseteq
G\}$,}
     
\Centerline{$H_n
=\bigcup_{k\in K_n}\coint{2^{-n}k,2^{-n}(k+1)}\cap\coint{a,b}$;}
     
\noindent then
$\sequencen{H_n}$ is a non-decreasing sequence of measurable sets with
union $E$, so $f\times\chi E=\lim_{n\to\infty}f\times\chi H_n$, and (by
Lebesgue's Dominated Convergence Theorem, because $|f\times\chi
H_n|\le|f|$ almost everywhere for every $n$, and $|f|$ is integrable)
     
\Centerline{$\int_Ef=\lim_{n\to\infty}\int_{H_n}f$.}
     
\noindent At the same time, each $H_n$ is a finite disjoint union of
half-open intervals in $\coint{a,b}$, so
     
\Centerline{$\int_{H_n}f
=\sum_{k\in K_n}\int_{\coint{2^{-n}k,2^{-n}(k+1)}\cap\coint{a,b}}f
=\sum_{k\in K_n}\int_{\coint{2^{-n}k,2^{-n}(k+1)}\cap\coint{a,b}}g
=\int_{H_n}g$,}
     
\noindent and
     
\Centerline{$\int_Eg=\lim_{n\to\infty}\int_{H_n}g
=\lim_{n\to\infty}\int_{H_n}f=\int_Ef$.}
     
\medskip
     
\quad{\bf (iii)} For general measurable $E\subseteq\coint{a,b}$, we can
choose for each $n\in\Bbb N$ an open set $G_n\supseteq E$ such that
$\mu G_n\le\mu E+2^{-n}$ (134Fa).   Set $G_n'=\bigcap_{m\le n}G_m$, $E_n=\coint{a,b}\cap G'_n$ for each $n$,
     
\Centerline{$F=\coint{a,b}\cap\bigcap_{n\in\Bbb N}G_n
=\bigcap_{n\in\Bbb N}\coint{a,b}\cap G_n'=\bigcap_{n\in\Bbb N}E_n$.}
     
\noindent Then $E\subseteq F$ and
     
\Centerline{$\mu F\le\inf_{n\in\Bbb N}\mu G_n=\mu E$,}
     
\noindent so $F\setminus E$ is negligible and
$f\times\chi(F\setminus E)$ is zero almost everywhere;  consequently $\int_{F\setminus E}f=0$
and $\int_Ff=\int_Ef$.  On the other hand,
     
\Centerline{$f\times\chi F=\lim_{n\to\infty}f\times\chi E_n$,}
     
\noindent so by Lebesgue's Dominated Convergence Theorem again
     
\Centerline{$\int_Ef=\int_Ff=\lim_{n\to\infty}\int_{E_n}f$.}
     
\noindent Similarly
     
\Centerline{$\int_Eg=\lim_{n\to\infty}\int_{E_n}g$.}
     
\noindent But by part (ii) we have $\int_{E_n}g=\int_{E_n}f$ for every
$n$, so $\int_Eg=\int_Ef$, as required.\ \Qed
     
By 131Hb, $f=g$ almost everywhere in $\coint{a,b}$, and therefore almost everywhere in $[a,b]$.
}%end of proof of 222D
     
\leader{222E}{Theorem} Suppose that $a\le b$ in $\Bbb R$ and that $f$ is
a real-valued function which is integrable over $[a,b]$.
Then $F(x)=\int_a^xf$ exists in $\Bbb R$ for every $x\in [a,b]$, and the derivative
$F'(x)$ exists and is equal to $f(x)$ for almost every $x\in[a,b]$.
     
\proof{{\bf (a)} For most of this proof (down to the end of
(c) below) I suppose that $f$ is non-negative.   In this case,
     
     
\Centerline{$F(y)=F(x)+\int_x^yf\ge F(x)$}
     
\noindent whenever $a\le x\le y\le b$;
thus $F$ is non-decreasing and therefore differentiable almost
everywhere in
$[a,b]$, by 222A.
     
By 222C we know also that $\int_a^xF'$ exists and is less than or equal
to $F(x)-F(a)=F(x)$ for every $x\in[a,b]$.
     
\medskip
     
{\bf (b)} Now suppose, in addition, that $f$ is bounded;  say $0\le
f(t)\le
M$ for every $t\in\dom f$.   Then $M-f$ is integrable over $[a,b]$;  let
$G$ be its indefinite integral, so that $G(x)=M(x-a)-F(x)$ for every
$x\in[a,b]$.   Applying (a) to $M-f$ and $G$, we have $\int_a^xG'\le
G(x)$ for every $x\in[a,b]$;  but of course $G'=M-F'$, so
$M(x-a)-\int_a^xF'\le M(x-a)-F(x)$, that is, $\int_a^xF'\ge F(x)$ for
every $x\in[a,b]$.   Thus $\int_a^xF'=\int_a^xf$ for every
$x\in[a,b]$.    Now 222D tells us that $F'=f$ almost
everywhere in $[a,b]$.
     
\medskip
     
{\bf (c)} Thus for bounded, non-negative $f$ we are done.   For
unbounded $f$, let $\sequencen{f_n}$ be a non-decreasing sequence of
non-negative simple functions converging to $f$ almost everywhere in
$[a,b]$, and let $\sequencen{F_n}$ be the corresponding indefinite
integrals.   Then for any $n$ and any $x$, $y$ with $a\le x\le y\le b$,
we have
     
\Centerline{$F(y)-F(x)=\int_x^yf\ge\int_x^yf_n=F_n(y)-F_n(x)$,}
     
\noindent so that
$F'(x)\ge F_n'(x)$ for any $x\in\ooint{a,b}$ where both are defined, and
$F'(x)\ge f_n(x)$ for almost every $x\in[a,b]$.   This is true for every
$n$, so $F'\ge f$ almost everywhere, and $F'-f\ge 0$ almost everywhere.
On the other hand, as noted in (a),
     
\Centerline{$\int_a^bF'\le F(b)-F(a)=\int_a^bf$,}
     
\noindent so $\int_a^bF'-f\le 0$.   It follows
that $F'\eae f$ (that is, that $F'=f$ almost everywhere in $[a,b]$)(122Rd).
     
\medskip
     
{\bf (d)} This completes the proof for non-negative $f$.   For general
$f$, we can express $f$ as $f_1-f_2$ where $f_1$, $f_2$ are
non-negative integrable functions;  now $F=F_1-F_2$ where $F_1$, $F_2$
are the corresponding indefinite integrals, so
$F'\eae F_1'-F_2'\eae f_1-f_2$, and $F'\eae f$.
}%end of proof of 222E
     
\leader{222F}{Corollary} Suppose that $f$ is any real-valued function which is integrable over $\Bbb R$, and set $F(x)=\int_{-\infty}^xf$ for every $x\in\Bbb R$.   Then $F'(x)$ exists and is equal to $f(x)$ for almost every $x\in\Bbb R$.
     
\proof{ For each $n\in\Bbb N$, set
     
\Centerline{$F_n(x)=\int_{-n}^xf$}
     
\noindent for $x\in[-n,n]$.   Then $F_n'(x)=f(x)$ for almost every
$x\in[-n,n]$.   But $F(x)=F(-n)+F_n(x)$ for every $x\in[-n,n]$, so
$F'(x)$ exists and is equal to $F'_n(x)$ for every $x\in\ooint{-n,n}$
for which $F'_n(x)$ is defined;  and $F'(x)=f(x)$ for almost every
$x\in[-n,n]$.   As $n$ is arbitrary, $F'\eae f$.
}%end of proof of 222F
     
\leader{222G}{Corollary} Suppose that $E\subseteq\Bbb R$ is a measurable
set and that $f$ is a real-valued function which is integrable over $E$.
Set $F(x)=\int_{E\cap\ooint{-\infty,x}}f$ for $x\in\Bbb R$.   Then
$F'(x)=f(x)$ for almost every $x\in E$, and $F'(x)=0$ for almost every
$x\in\Bbb R\setminus E$.
     
\proof{ Apply 222F to $\tilde f$, where $\tilde f(x)=f(x)$ for $x\in
E\cap\dom f$ and $\tilde f(x)=0$ for $x\in\Bbb R\setminus E$, so that
$F(x)=\int_{-\infty}^x\tilde f$ for every $x\in\Bbb R$.
}%end of proof of 222G
     
\leader{222H}{}\cmmnt{ The result that $\bover{d}{dx}\int_a^xf=f(x)$
for almost every $x$ is satisfying, but is no substitute for the more
elementary result that this equality is valid at any point at which $f$
is continuous.
     
\medskip
     
\noindent}{\bf Proposition} Suppose that $a\le b$ in $\Bbb R$ and that
$f$ is a real-valued function which is integrable over $[a,b]$.   Set
$F(x)=\int_a^xf$ for $x\in[a,b]$.   Then $F'(x)$ exists and is equal to
$f(x)$ at any point $x\in\dom(f)\cap\ooint{a,b}$ at which $f$ is
continuous.
     
\proof{ Set $c=f(x)$.  Let $\epsilon>0$.   Let $\delta>0$ be such that
$\delta\le\min(b-x,x-a)$ and $|f(t)-c|\le\epsilon$ whenever $t\in\dom f$
and $|t-x|\le\delta$.   If $x<y\le x+\delta$, then
     
\Centerline{$|\Bover{F(y)-F(x)}{y-x}-c|
=\Bover1{y-x}|\int_x^yf-c|
\le\Bover1{y-x}\int_x^y|f-c|
\le\epsilon$.}
     
\noindent Similarly, if $x-\delta\le y<x$,
     
\Centerline{$|\Bover{F(y)-F(x)}{y-x}-f(x)|
=\Bover1{x-y}|\int_y^xf-c|
\le\Bover1{x-y}\int_y^x|f-c|
\le\epsilon$.}
     
\noindent As $\epsilon$ is arbitrary,
     
\Centerline{$F'(x)=\lim_{y\to x}\Bover{F(y)-F(x)}{y-x}=c$,}
     
\noindent as required.
}%end of proof of 222H
     
     
\leader{222I}{Complex-valued functions} \cmmnt{So far in this section, I
have taken every $f$ to be real-valued.   The extension to
complex-valued $f$ is just a matter of applying the above results to the
real and imaginary parts of $f$.   Specifically, we have the following.

\medskip

}{\bf (a)} If $a\le b$ in $\Bbb R$ and $f$ is a
complex-valued function which is integrable over $[a,b]$, then
$F(x)=\int_a^xf$ is defined in $\Bbb C$ for every $x\in[a,b]$, and its
derivative $F'(x)$ exists and is equal to $f(x)$ for almost every
$x\in[a,b]$;  moreover, $F'(x)=f(x)$ whenever
$x\in\dom(f)\cap\ooint{a,b}$ and $f$ is continuous at $x$.
     
\spheader 222Ib If $f$ is a complex-valued function which is
integrable over $\Bbb R$, and $F(x)=\int_{-\infty}^xf$ for each 
$x\in\Bbb R$, then
$F'$ exists and is equal to $f$ almost everywhere in $\Bbb R$.
     
\spheader 222Ic If $E\subseteq\Bbb R$ is a measurable set and
$f$ is a complex-valued function which is integrable over $E$, and
$F(x)=\int_{E\cap\ooint{-\infty,x}}f$ for each $x\in\Bbb R$, then
$F'(x)=f(x)$ for almost every $x\in E$ and $F'(x)=0$ for almost every
$x\in\Bbb R\setminus E$.
     
\leader{*222J}{The Denjoy-Young-Saks \dvrocolon{theorem}}\cmmnt{ The 
next result 
will not be used, on present plans, anywhere in this treatise.   It
is however a classical part of the real analysis for which this volume is
supposed to be a foundation, and while the argument
requires a certain sophistication it is not really a large step from
Lebesgue's theorem 222A.   I must begin with some notation.
     
\medskip

\noindent}{\bf Definition} Let $f$ be any real function, and
$A\subseteq\Bbb R$ its domain.   Write  

\Centerline{$\tilde A^+
=\{x:x\in A$, $\ocint{x,x+\delta}\cap A\ne\emptyset$ 
for every $\delta>0\}$,}

\Centerline{$\tilde A^-
=\{x:x\in A$, $\coint{x-\delta,x}\cap A\ne\emptyset$ 
for every $\delta>0\}$.}
     
\noindent Set

\Centerline{$(\DiniD^+f)(x)
=\limsup_{y\in A,y\downarrow x}\Bover{f(y)-f(x)}{y-x}
=\inf_{\delta>0}\sup_{y\in A,x<y\le x+\delta}\Bover{f(y)-f(x)}{y-x}$,}

\Centerline{$(\Dinid^+f)(x)
=\liminf_{y\in A,y\downarrow x}\Bover{f(y)-f(x)}{y-x}
=\sup_{\delta>0}\inf_{y\in A,x<y\le x+\delta}\Bover{f(y)-f(x)}{y-x}$}

\noindent for $x\in\tilde A^+$, and

\Centerline{$(\DiniD^-f)(x)
=\limsup_{y\in A,y\uparrow x}\Bover{f(y)-f(x)}{y-x}
=\inf_{\delta>0}\sup_{y\in A,x-\delta\le y<x}\Bover{f(y)-f(x)}{y-x}$,}

\Centerline{$(\Dinid^-f)(x)
=\liminf_{y\in A,y\uparrow x}\Bover{f(y)-f(x)}{y-x}
=\sup_{\delta>0}\inf_{y\in A,x-\delta\le y<x}\Bover{f(y)-f(x)}{y-x}$}

\noindent for $x\in\tilde A^-$, all defined in $[-\infty,\infty]$.   
(These are the four {\bf Dini derivates} of $f$.\cmmnt{ You will also
see $D^+$, $d^+$, $D^-$, $d^-$ used in place of my $\DiniD^+$, 
$\Dinid^+$, $\DiniD^-$ and $\Dinid^-$.})

Note that we surely have $(\Dinid^+f)(x)\le(\DiniD^+f)(x)$
for every $x\in\tilde A^+$, while 
$(\Dinid^-f)(x)\le(\DiniD^-f)(x)$
for every $x\in\tilde A^-$.   The ordinary derivative
$f'(x)$ is defined and equal to $c\in\Bbb R$ iff 
($\alpha$) $x$ belongs to some open interval included in $A$ ($\beta$)
$(\DiniD^+f)(x)=(\Dinid^+f)(x)=(\DiniD^-f)(x)
=(\Dinid^-f)(x)=c$.

\leader{*222K}{Lemma} For $A\subseteq\Bbb R$, define
$\tilde A^+$ and $\tilde A^-$ as in 222J.   
Then $A\setminus\tilde A^+$ and
$A\setminus\tilde A^-$ are countable, therefore negligible.

\proof{ We have

\Centerline{$A\setminus\tilde A^+
=\bigcup_{q\in\Bbb Q}\{x:x\in A$, $x<q$, $A\cap\ocint{x,q}=\emptyset\}$.}

\noindent But for any $q\in\Bbb Q$ there can be at most one $x\in A$ such
that $x<q$ and $\ocint{x,q}$ does not meet $A$, 
so $A\setminus\tilde A^+$ is a
countable union of finite sets and is countable.   Similarly,

\Centerline{$A\setminus\tilde A^-
=\bigcup_{q\in\Bbb Q}\{x:x\in A$, $q<x$, $A\cap\coint{q,x}=\emptyset\}$}

\noindent is countable.
}%end of proof of 222K

\leader{*222L}{Theorem} Let $f$ be any real function, and $A$ its domain.  
Then for almost every $x\in A$

\qquad{\it either} all four Dini derivates of $f$ at $x$ are defined, 
finite and equal

\qquad{\it or} $(\DiniD^+f)(x)=(\Dinid^-f)(x)$ is finite,
$(\Dinid^+f)(x)=-\infty$ and $(\DiniD^-f)(x)=\infty$

\qquad{\it or} $(\Dinid^+f)(x)=(\DiniD^-f)(x)$ is finite,
$(\DiniD^+f)(x)=\infty$ and $(\Dinid^-f)(x)=-\infty$

\qquad{\it or} $(\DiniD^+f)(x)=(\DiniD^-f)(x)=\infty$ and
$(\Dinid^+f)(x)=(\Dinid^-f)(x)=-\infty$.

\proof{\footnote{I am indebted to P.Klinger Monteiro
for pointing out a blunder in the original version of this proof.}{\bf (a)(i)} 
Suppose that $n\in\Bbb N$ and $q\in\Bbb Q$ are such that 

\Centerline{$E_{qn}=\{x:x\in A$, $x<q$, 
  $f(y)\ge f(x)-n(y-x)\}$ for every $y\in A\cap[x,q]\}$}

\noindent is not
empty.   Set $\beta_{qn}=\sup E_{qn}\in\ocint{-\infty,q}$, 
$\alpha_{qn}=\inf E_{qn}\in[-\infty,\beta_{qn}]$ and 
$I_{qn}=\ooint{\alpha_{qn},\beta_{qn}}$;  now
for $x\in I_{qn}$ set
$f_{qn}(x)=\inf\{f(y)+ny:y\in A\cap[x,q]\}$.   Note that if 
$x\in E_{qn}\setminus\{\alpha_{qn},\beta_{qn}\}$ then $f_{qn}(x)=f(x)+nx$
is finite;  also $f_{qn}$ is non-decreasing, therefore finite everywhere in
$I_{qn}$, and of course $f_{qn}(x)\le f(x)+nx$ for
every $x\in A\cap I_{qn}$.   

Set $F_{qn}=E_{qn}\cap\dom f'_{qn}\subseteq I_{qn}$, and
$g_{qn}(x)=f_{qn}(x)-nx$ for $x\in I_{qn}$;  then
$g_{qn}$ is differentiable at every point of $F_{qn}$, while
$g_{qn}(x)\le f(x)$ for $x\in A\cap I_{qn}$ and
$g_{qn}(x)=f(x)$ for $x\in E_{qn}\cap I_{qn}$.

\medskip

\quad{\bf (ii)} Take $x\in\tilde F_{qn}^+\cap\tilde F_{qn}^-$.
Then $x\in I_{qn}\cap\tilde A^+\cap\tilde A^-$ so the Dini derivates
$(\Dinid^+f)(x)$ and $(\DiniD^-f)(x)$ are defined in $[-\infty,\infty]$,
while $g_{qn}(x)=f(x)$.

\medskip

\qquad\grheada\ $(\Dinid^+f)(x)=g'_{qn}(x)$.   \Prf\

$$\eqalignno{
(\Dinid^+f)(x)
&=\sup_{\delta>0}\inf_{y\in A\cap\ocint{x,x+\delta}}
    \Bover{f(y)-f(x)}{y-x}
\le\sup_{\delta>0}\inf_{y\in E_{qn}\cap\ocint{x,x+\delta}}
    \Bover{f(y)-f(x)}{y-x}\cr&
=\sup_{\delta>0}\inf_{y\in E_{qn}\cap\ocint{x,x+\delta}}
    \Bover{g_{qn}(y)-g_{qn}(x)}{y-x}
=g'_{qn}(x)
\Displaycause{because $x\in\tilde E_{qn}^+$}
=\sup_{0<\delta<\beta_{qn}-x}\inf_{y\in\ocint{x,x+\delta}}
  \Bover{g_{qn}(y)-g_{qn}(x)}{y-x}\cr&
\le\sup_{0<\delta<\beta_{qn}-x}\inf_{y\in A\cap\ocint{x,x+\delta}}
  \Bover{g_{qn}(y)-f(x)}{y-x}\cr&
\le\sup_{0<\delta<\beta_{qn}-x}\inf_{y\in A\cap\ocint{x,x+\delta}}
  \Bover{f(y)-f(x)}{y-x}
\Displaycause{because $g_{qn}(y)\le f(y)$ for $y\in A\cap I_{qn}$}
\le\sup_{\delta>0}\inf_{y\in A\cap\ocint{x,x+\delta}}
  \Bover{f(y)-f(x)}{y-x}
=(\Dinid^+f)(x).  \text{ \Qed}\cr}$$

\medskip
%\vfill\eject

\qquad\grheadb\ $(\DiniD^-f)(x)=g'_{qn}(x)$.   \Prf\

$$\eqalignno{(\DiniD^-f)(x)
&=\inf_{\delta>0}\sup_{y\in A\cap\coint{x-\delta,x}}
  \Bover{f(y)-f(x)}{y-x}
=\inf_{\delta>0}\sup_{y\in A\cap\coint{x-\delta,x}}
  \Bover{f(x)-f(y)}{x-y}\cr&
\le\inf_{0<\delta<x-\alpha_{qn}}\sup_{y\in A\cap\coint{x-\delta,x}}
  \Bover{f(x)-f(y)}{x-y}\cr&
\le\inf_{0<\delta<x-\alpha_{qn}}\sup_{y\in A\cap\coint{x-\delta,x}}
  \Bover{g_{qn}(x)-g_{qn}(y)}{x-y}
\Displaycause{because $g_{qn}(y)\le f(y)$ for $y\in A\cap I_{qn}$}
\le\inf_{0<\delta<x-\alpha_{qn}}\sup_{y\in\coint{x-\delta,x}}
  \Bover{g_{qn}(y)-g_{qn}(x)}{y-x}
=g_{qn}'(x)\cr&
=\inf_{\delta>0}\sup_{y\in E_{qn}\cap\coint{x-\delta,x}}
  \Bover{g_{qn}(y)-g_{qn}(x)}{y-x}\cr&
\Displaycause{because $x\in\tilde E_{qn}^-$}
=\inf_{\delta>0}\sup_{y\in E_{qn}\cap\coint{x-\delta,x}}
  \Bover{f(y)-f(x)}{y-x}\cr&
\le\inf_{\delta>0}\sup_{y\in A\cap\coint{x-\delta,x}}
  \Bover{f(y)-f(x)}{y-x}
=(\DiniD^-f)(x).  \text{ \Qed}\cr}$$

\medskip

\qquad\grheadc\  Putting these together, we see that if 
$x\in\tilde F_{qn}^+\cap\tilde F_{qn}^-$ then
$(\Dinid^+f)(x)=(\DiniD^-f)(x)=g'_{qn}(x)$ is finite.

\medskip

\quad{\bf (iii)} 
Conventionally setting $F_{qn}=\emptyset$ if $E_{qn}$ is empty, the last
sentence is true for all $q\in\Bbb Q$ and $n\in\Bbb N$.   Now we know that
$A\setminus\tilde A^+$ is negligible (222K), as are
$F_{qn}\setminus\tilde F^+_{qn}$,
$F_{qn}\setminus\tilde F^-_{qn}$ and $I_{qn}\setminus\dom f'_{qn}$ (222A), 
whenever $q\in\Bbb Q$ and $n\in\Bbb N$.
So $H
=(A\setminus\tilde A^+)\cup\bigcup_{q\in\Bbb Q,n\in\Bbb N}
   ((E_{qn}\setminus F_{qn})
      \cup(F_{qn}\setminus(\tilde F_{qn}^+\cap\tilde F_{qn}^-)))$
is negligible.   And if $x\in A\setminus H$ and $(\DiniD^+f(x))>-\infty$, then
$(\Dinid^+f)(x)=(\DiniD^-f)(x)\in\Bbb R$.   \Prf\ As 
$A\setminus\tilde A^+\subseteq H$, $x\in\tilde A^+$.   Let $n\in\Bbb N$ be such that
$(\DiniD^+f)(x)>-n$.   Then there is a $\delta>0$ such that
$\Bover{f(y)-f(x)}{y-x}>-n$ whenever $y\in A$ and $x<y\le x+\delta$.
Let $q\in\Bbb Q$ be such that $x<q\le x+\delta$;  then
$f(y)\ge f(x)-n(y-x)$ whenever $y\in A\cap[x,q]$, and $x\in E_{qn}\setminus H$.
So $x\in F_{qn}\setminus H\subseteq\tilde F_{qn}^+\cap\tilde F_{qn}^-$.
So (ii) above tells us that 
$(\Dinid^+f)(x)=(\DiniD^-f)(x)=g'_{qn}(x)$ is finite.\ \Qed

\medskip

{\bf (b)} Thus for almost every $x\in A$,

\Centerline{either $(\DiniD^+f)(x)=-\infty$ or
$(\DiniD^+f)(x)=(\Dinid^-f)(x)\in\Bbb R$.}

\noindent Applying (a) to $x\mapsto f(-x):-A\to\Bbb R$,
$x\mapsto-f(x):A\to\Bbb R$ and $x\mapsto-f(-x):-A\to\Bbb R$, we see
that, for almost every $x\in A$,

\Centerline{either $(\DiniD^-f)(x)=\infty$ or
$(\DiniD^-f)(x)=(\Dinid^+f)(x)\in\Bbb R$,}

\Centerline{either $(\DiniD^+f)(x)=\infty$ or
$(\DiniD^+f)(x)=(\Dinid^-f)(x)\in\Bbb R$,}

\Centerline{either $(\Dinid^-f)(x)=-\infty$ or
$(\Dinid^-f)(x)=(\DiniD^+f)(x)\in\Bbb R$.}

\noindent For almost every $x\in A$, therefore,

\Centerline{$(\Dinid^+f)(x)>-\infty
\Longrightarrow (\DiniD^-f)(x)<\infty
\Longrightarrow (\Dinid^+f)(x)<\infty$}

\noindent and in this case $(\Dinid^+f)(x)=(\DiniD^-f)(x)$ is finite;
similarly,

\Centerline{$(\Dinid^-f)(x)>-\infty
\iff(\DiniD^+f)(x)<\infty\iff(\Dinid^-f)(x)=(\DiniD^+f)(x)\in\Bbb R$.}

\noindent Thus we have

\Centerline{either $(\Dinid^+f)(x)=(\DiniD^-f)(x)$ is finite
or $(\Dinid^+f)(x)=-\infty$ and
$(\DiniD^-f)(x)=\infty$,}

\Centerline{either $(\Dinid^-f)(x)=(\DiniD^+f)(x)$ is finite
or $(\Dinid^-f)(x)=-\infty$ and
$(\DiniD^+f)(x)=\infty$.}

\noindent These two dichotomies lead to four possibilities;  and since 

\Centerline{$(\Dinid^+f)(x)=(\DiniD^-f)(x)$ is finite,
\quad$(\Dinid^-f)(x)=(\DiniD^+f)(x)$ is finite}

\noindent together imply that

\Centerline{$(\Dinid^-f)(x)\le(\DiniD^-f)(x)=(\Dinid^+f)(x)\le(\DiniD^+f)(x)
=(\Dinid^-f)(x)$,}

\noindent this can happen only when all four derivates are equal and
finite, so we have the four cases listed in the statement of the theorem.
}%end of proof of 222L

\exercises{
\leader{222X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 222Xa
Let $F:[0,1]\to[0,1]$ be the Cantor function (134H).
Show that $\int_0^1F'=0<F(1)-F(0)$.
%222C
     
\sqheader 222Xb Suppose that $a<b$ in $\Bbb R$ and that $h$ is a
real-valued function such that $\int_x^yh$ exists and is non-negative
whenever $a\le x\le y\le b$.   Show that $h\ge 0$ almost everywhere in $[a,b]$.
%222D
     
\sqheader 222Xc Suppose that $a<b$ in $\Bbb R$ and that $f$, $g$ are
integrable complex-valued functions on $[a,b]$ such that
$\int_a^xf=\int_a^xg$ for every $x\in [a,b]$.   Show that $f=g$ almost everywhere in $[a,b]$.
%222I
     
\sqheader 222Xd Suppose that $a<b$ in
$\Bbb R$ and that $f$ is a
real-valued function which is integrable over $[a,b]$.   Show that the
indefinite integral $x\mapsto\int_a^xf$ is continuous.
%222+
     
\leader{222Y}{Further exercises (a)} 
%\spheader 222Ya
Let $\sequencen{F_n}$ be a sequence
of non-negative, non-decreasing functions on $[0,1]$ such that
$F(x)=\sum_{n=0}^{\infty}F_n(x)$ is finite for every $x\in[0,1]$.   Show
that $\sum_{n=0}^{\infty}F'_n(x)=F'(x)$ for almost every $x\in[0,1]$.
({\it Hint\/}:  take $\sequence{k}{n_k}$ such that
$\sum_{k=0}^{\infty}F(1)-G_k(1)<\infty$, where
$G_k=\sum_{j=0}^{n_k}F_j$, and set
$H(x)=\sum_{k=0}^{\infty}F(x)-G_k(x)$.   Observe that
$\sum_{k=0}^{\infty}F'(x)-G'_k(x)\le H'(x)$ whenever all the derivatives
are defined, so that $F'=\lim_{k\to\infty}G'_k$ almost everywhere.)
%222A
     
\spheader 222Yb Let $F:[0,1]\to\Bbb R$ be a continuous non-decreasing
function.   (i)   Show that if $c\in\Bbb R$ then
$C=\{(x,y):x,\,y\in[0,1],\,F(y)-F(x)=c\}$ is connected.   ({\it Hint\/}:
A set $A\subseteq\BbbR^r$ is {\bf connected} if there is no continuous
surjection $h:A\to\{0,1\}$.   Show that if $h:C\to\{0,1\}$ is continuous
then it is of the form $(x,y)\mapsto h_1(x)$ for some continuous
function $h_1$.)  (ii) Now suppose that $F(0)=0$, $F(1)=1$ and that
$G:[0,1]\to[0,1]$ is a second continuous non-decreasing function with
$G(0)=0$, $G(1)=1$.   Show that for any $n\ge 1$ there are $x$,
$y\in[0,1]$ such that $F(y)-F(x)=G(y)-G(x)=\bover1n$.
%222+
     
\spheader 222Yc Let $f$, $g$ be non-negative integrable functions on
$\Bbb R$, and $n\ge 1$.   Show that there are $u<v$ in
$[-\infty,\infty]$ such that $\int_u^vf=\bover1n\int f$ and
$\int_u^vg=\bover1n\int g$.
%222Yb 222+
     
\spheader 222Yd Let $f:\Bbb R\to\Bbb R$ be measurable.   Show that 
$H=\dom f'$ is a measurable set and that $f'$ is a measurable function.
%222+

\spheader 222Ye Suppose that $A\subseteq B\subseteq\Bbb R$, 
$f:B\to\Bbb R$ is a function and $f_A=f\restr A$.   Show that
(i) if $x\in\tilde A^+$, as defined in 222J, then
$(\Dinid^+f)(x)\le(\Dinid^+f_A)(x)\le(\DiniD^+f_A)(x)\le(\DiniD^+f)(x)$
(ii) if $x\in\tilde A^-$, then
$(\Dinid^-f)(x)\le(\Dinid^-f_A)(x)\le(\DiniD^-f_A)(x)\le(\DiniD^-f)(x)$.
%222J

\spheader 222Yf Construct a Borel measurable
function $f:[0,1]\to\{-1,0,1\}$ such that
each of the four possibilities described in Theorem 222L occurs on a set 
of measure $\bover14$.
%222L


}%end of exercises
     
\endnotes{
\Notesheader{222} I have relegated to an exercise (222Xd) the
fundamental fact that an indefinite integral $x\mapsto\int_a^xf$ is
always continuous;  this is not strictly speaking needed in this
section, and a much stronger result is given in 225E.   There is also
much more to be said about monotonic functions, to which I will return
in \S224.   What we need here is the fact that they are differentiable
almost everywhere (222A), which I prove by applying Vitali's theorem
three times, once in part (b) of the proof and twice in part (c).
Following this, the arguments of 222C-222E form a fine series of
exercises in the central ideas of Volume 1, using the concept of
integration over a (measurable) subset, Fatou's Lemma (part (d) of the
proof of 222C), Lebesgue's Dominated Convergence Theorem (parts (ii)
and (iii) of the proof of 222D) and the approximation of Lebesgue
measurable sets by open sets (part (iii) of the proof of 222D).   Of
course knowing that $\bover{d}{dx}\int_a^xf=f(x)$ almost everywhere is not at all
the same thing as knowing that this holds for any particular $x$, and
when we come to differentiate any particular indefinite integral we
generally turn to 222H first;  the point of 222E is that it applies to
wildly discontinuous functions, for which more primitive methods give no
information at all.

}%end of comment
     
\discrpage
     
\frfilename{mt223.tex}
\versiondate{9.9.04}
\copyrightdate{1995}
     
\def\chaptername{The Fundamental Theorem of Calculus}
\def\sectionname{Lebesgue's density theorems}
\def\undpsi{\underline{\psi}}
     
\newsection{223}
     
I now turn to a group of results which may be thought of as corollaries
of Theorem 222E, but which also have a vigorous life of their own,
including the possibility of significant generalizations which will be
treated in Chapter 26.   The idea is that any measurable function $f$ on
$\Bbb R$ is almost everywhere `continuous' in a variety of very weak
senses;  for almost every $x$, the value $f(x)$ is determined by the
behaviour of $f$ near $x$, in the sense that $f(y)\bumpeq f(x)$ for
`most' $y$ near $x$.    I should perhaps say that while I recommend this
work as a preparation for Chapter 26, and I also rely on it in Chapter
28, I shall not refer to it again in the present chapter, so that
readers in a hurry to characterize indefinite integrals may proceed
directly to \S224.
     
\leader{223A}{Lebesgue's Density Theorem:  integral form} Let  $I$ be an
interval in $\Bbb R$, and let $f$ be a real-valued function which is
integrable over $I$.   Then
     
$$\eqalign{f(x)
&=\lim_{h\downarrow 0}\Bover1{h}\int_{x}^{x+h}f
=\lim_{h\downarrow 0}\Bover1h\int_{x-h}^xf
=\lim_{h\downarrow 0}\Bover1{2h}\int_{x-h}^{x+h}f\cr}$$
     
\noindent for almost every $x\in I$.
     
     
\proof{ Setting $F(x)=\int_{I\cap\ooint{-\infty,x}}f$, we
know from 222G that
     
$$\eqalign{f(x)=F'(x)
&=\lim_{h\downarrow 0}\Bover1{h}(F(x+h)-F(x))
   =\lim_{h\downarrow 0}\Bover1{h}\int_{x}^{x+h}f\cr
&=\lim_{h\downarrow 0}\Bover1h(F(x)-F(x-h))
   =\lim_{h\downarrow 0}\Bover1h\int_{x-h}^xf\cr
&=\lim_{h\downarrow 0}\Bover1{2h}(F(x+h)-F(x-h))
   =\lim_{h\downarrow 0}\Bover1{2h}\int_{x-h}^{x+h}f\cr}$$
     
\noindent for almost every $x\in I$.
}%end of proof of 223A
     
\leader{223B}{Corollary} Let $E\subseteq\Bbb R$ be a measurable set.
Then
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}\mu(E\cap[x-h,x+h])=1$ for
almost every $x\in E$,}
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}\mu(E\cap[x-h,x+h])=0$ for
almost every $x\in \Bbb R\setminus E$.}
     
     
\proof{ Take $n\in \Bbb N$.   Applying 223A to
$f=\chi(E\cap[-n,n])$, we see that
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}\int_{x-h}^{x+h}f
=\lim_{h\downarrow 0}\Bover1{2h}\mu(E\cap[x-h,x+h])$}
     
\noindent whenever $x\in\ooint{-n,n}$ and either limit exists, so that
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}\mu(E\cap[x-h,x+h])=1$ for
almost every $x\in E\cap[-n,n]$,}
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}\mu(E\cap[x-h,x+h])=0$ for
almost every $x\in [-n,n]\setminus E$.}
     
\noindent As $n$ is arbitrary, we have the result.
}%end of proof of 223B
     
\cmmnt{\medskip
     
\noindent{\bf Remark} For a measurable set $E\subseteq\Bbb R$, a point
$x$ such that $\lim_{h\downarrow 0}\Bover1{2h}\mu(E\cap[x-h,x+h])=1$ is
sometimes called a {\bf density point} of $E$.
}%end of comment
     
\leader{223C}{Corollary} Let $f$ be a measurable real-valued function
defined almost everywhere in $\Bbb R$.   Then for almost every
$x\in\Bbb R$,
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}\mu\{y:y\in\dom f,\,|y-x|\le
h,\,|f(y)-f(x)|\le\epsilon\}=1$,}
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}\mu\{y:y\in\dom f,\,|y-x|\le
h,\,|f(y)-f(x)|\ge\epsilon\}=0$}
     
\noindent for every $\epsilon>0$.
     
\proof{ For $q$, $q'\in\Bbb Q$, set
     
\Centerline{$D_{qq'}=\{x:x\in\dom f,\,q\le f(x)<q'\}$,}
     
\noindent so that $D_{qq'}$ is measurable,
     
\Centerline{$C_{qq'}=\{x:x\in D_{qq'},\,\lim_{h\downarrow 0}
\Bover1{2h}\mu(D_{qq'}\cap [x-h,x+h])=1\}$,}
     
     
\noindent so that $D_{qq'}\setminus C_{qq'}$ is negligible, by 223B;
now
set
     
\Centerline{$C=\dom f\setminus\bigcup_{q,q'\in\Bbb Q}(D_{qq'}\setminus
C_{qq'})$,}
     
\noindent so that $C$ is conegligible.   If $x\in C$ and
$\epsilon>0$, then there are $q$, $q'\in \Bbb Q$ such that
$f(x)-\epsilon\le q\le f(x)<q'\le f(x)+\epsilon$, so that $x$ belongs to
$D_{qq'}$ and therefore to $C_{qq'}$, and now
     
$$\eqalign{\liminf_{h\downarrow 0}\Bover1{2h}
&\mu\{y:y\in \dom f\cap [x-h,x+h],\,|f(y)-f(x)|\le\epsilon\}\cr
&\ge\liminf_{h\downarrow 0}
\Bover1{2h}\mu(D_{qq'}\cap [x-h,x+h])
=1,\cr}$$
     
\noindent so
     
\Centerline{$\lim_{h\downarrow 0}
\Bover1{2h}\mu\{y:y\in \dom f\cap [x-h,x+h],\,|f(y)-f(x)|\le\epsilon\}
=1$.}
     
It follows at once that
     
\Centerline{$\lim_{h\downarrow 0}
\Bover1{2h}\mu\{y:y\in \dom f\cap [x-h,x+h],\,|f(y)-f(x)|>\epsilon\}
=0$}
     
\noindent for almost every $x$;  but since $\epsilon$ is arbitrary, this
is also true of $\bover12\epsilon$, so in fact
     
\Centerline{$\lim_{h\downarrow 0}
\Bover1{2h}\mu\{y:y\in \dom f\cap [x-h,x+h],\,|f(y)-f(x)|\ge\epsilon\}
=0$}
     
\noindent for almost every $x$.
}%end of proof of 223C
     
\leader{223D}{Theorem} Let $I$ be an interval in $\Bbb R$, and let $f$
be a real-valued function which is integrable over $I$.   Then
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}
\int_{x-h}^{x+h}|f(y)-f(x)|dy=0$}
     
\noindent for almost every $x\in I$.
     
\proof{{\bf (a)} Suppose first that $I$ is a bounded open
interval $\ooint{a,b}$.   For each $q\in\Bbb Q$, set $g_q(x)=|f(x)-q|$
for $x\in I\cap\dom f$;  then $g$ is integrable over $I$, and
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}
\int_{x-h}^{x+h}g_q=g_q(x)$}
     
\noindent for almost every $x\in I$, by 223A.   Setting
     
\Centerline{$E_q=\{x:x\in I\cap\dom f,\,\lim_{h\downarrow 0}
\Bover1{2h}\int_{x-h}^{x+h}g_q=g_q(x)\}$,}
     
\noindent we have $I\setminus E_q$ negligible, so $I\setminus E$ is
negligible, where $E=\bigcap_{q\in\Bbb Q}E_q$.   Now
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}
\int_{x-h}^{x+h}|f(y)-f(x)|dy=0$}
     
\noindent for every $x\in E$.   \Prf\ Take $x\in E$ and $\epsilon>0$.
Then there is a $q\in\Bbb Q$ such that $|f(x)-q|\le\epsilon$, so that
     
\Centerline{$|f(y)-f(x)|\le|f(y)-q|+\epsilon=g_q(y)+\epsilon$}
     
\noindent for every $y\in I\cap\dom f$, and

$$\eqalign{\limsup_{h\downarrow 0}\Bover1{2h}
\int_{x-h}^{x+h}|f(y)-f(x)|dy
&\le\limsup_{h\downarrow 0}\Bover1{2h}
\int_{x-h}^{x+h}g_q(y)+\epsilon\,dy\cr
&=\epsilon+g_q(x)
\le 2\epsilon.\cr}$$
     
\noindent As $\epsilon$ is arbitrary,
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}
\int_{x-h}^{x+h}|f(y)-f(x)|dy=0$,}
     
\noindent as required.\ \Qed
     
\medskip
     
{\bf (b)}  If $I$ is an unbounded open interval, apply (a) to the
intervals $I_n=I\cap\ooint{-n,n}$ to see that the limit is zero almost
everywhere in every $I_n$, and therefore on $I$.   If $I$ is an
arbitrary interval, note that it differs by at most two points from an
open interval, and that since we are looking only for something to
happen almost everywhere we can ignore these points.
}%end of proof of 223D
     
\medskip
     
\noindent{\bf Remark} The set
     
\Centerline{$\{x:x\in\dom f,\,\lim_{h\downarrow 0}\Bover1{2h}
\int_{x-h}^{x+h}|f(y)-f(x)|dy=0\}$}
     
\noindent is sometimes called the {\bf Lebesgue set} of $f$.
     
\leader{223E}{Complex-valued functions}\cmmnt{ I have expressed
the results above in terms of real-valued functions, this being the most
natural vehicle for the ideas.   However there are applications of great
importance in which the functions involved are complex-valued, so I
spell out the relevant statements here.   In all cases the proof is
elementary, being nothing more than applying the corresponding result
(223A, 223C or 223D) to the real and imaginary parts of the function
$f$.}
     
\header{223Ea}{\bf (a)} Let  $I$ be an interval in $\Bbb R$, and let $f$
be a complex-valued function which is integrable over $I$.   Then
     
$$\eqalign{f(x)
=\lim_{h\downarrow 0}\Bover1{h}\int_{x}^{x+h}f
=\lim_{h\downarrow 0}\Bover1h\int_{x-h}^xf
=\lim_{h\downarrow 0}\Bover1{2h}\int_{x-h}^{x+h}f\cr}$$
     
\noindent for almost every $x\in I$.
     
\header{223Eb}{\bf (b)} Let $f$ be a measurable complex-valued function
defined almost everywhere in $\Bbb R$.   Then for almost every
$x\in\Bbb R$,
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}\mu\{y:y\in\dom f,\,|y-x|\le
h,\,|f(y)-f(x)|\ge\epsilon\}=0$}
     
\noindent for every $\epsilon>0$.
     
\header{223Ec}{\bf (c)} Let $I$ be an interval in $\Bbb R$, and let $f$
be a complex-valued function which is integrable over $I$.   Then
     
\Centerline{$\lim_{h\downarrow 0}\Bover1{2h}
\int_{x-h}^{x+h}|f(y)-f(x)|dy=0$}
     
\noindent for almost every $x\in I$.
     
\exercises{
\leader{223X}{Basic exercises $\pmb{>}$(a)} Let $E\subseteq[0,1]$ be a 
measurable set for which there is an $\alpha>0$ such that
$\mu(E\cap[a,b])\ge\alpha(b-a)$ whenever $0\le a\le b\le 1$.
Show that $\mu E=1$.
%223B     
     
\spheader 223Xb Let $A\subseteq\Bbb R$ be any set.   Show that
$\lim_{h\downarrow 0}\Bover1{2h}\mu^*(A\cap[x-h,x+h])=1$ for
almost every $x\in A$.  \Hint{apply 223B to a measurable envelope
$E$ of $A$.}
%223B
     
\spheader 223Xc Let $E$, $F\subseteq\Bbb R$ be measurable sets, and
$x\in\Bbb R$ a point which is a density point of both.   Show that $x$
is a density point of $E\cap F$.
%223B
     
\spheader 223Xd Let $E\subseteq\Bbb R$ be a non-negligible measurable
set.   Show that for any $n\in\Bbb N$ there is a $\delta>0$ such that
$\bigcap_{i\le n}E+x_i$ is non-empty whenever $x_0,\ldots,x_n\in\Bbb R$
are such that $|x_i-x_j|\le\delta$ for all $i$, $j\le n$.   \Hint{find a
non-trivial interval $I$ such that $\mu(E\cap I)>\bover{n}{n+1}\mu I$.}
%223B
     
\spheader 223Xe Let $f$ be any real-valued function defined almost
everywhere in $\Bbb R$.   Show that $\lim_{h\downarrow
0}\Bover1{2h}\mu^*\{y:y\in\dom f,\,|y-x|\le
h,\,|f(y)-f(x)|\le\epsilon\}=1$ for almost every $x\in\Bbb R$.
\Hint{use the argument of 223C, but with 223Xb in place of 223B.}
%223C
     
\sqheader 223Xf Let $I$ be an interval in $\Bbb R$, and let $f$
be a real-valued function which is integrable over $I$.   Show that
$\lim_{h\downarrow 0}\bover1{h}
\int_{x}^{x+h}|f(y)-f(x)|dy=0$ for almost every $x\in I$.
%223D
     
\spheader 223Xg Let $E$, $F\subseteq\Bbb R$ be
measurable sets, and suppose that $F$ is bounded and of non-zero
measure.   Let $x\in\Bbb R$ be such that
$\lim_{h\downarrow 0}\Bover1{2h}\mu(E\cap[x-h,x+h])=1$.   Show that
$\lim_{h\downarrow 0}\Bover{\mu(E\cap(x+hF))}{h\,\mu F}=1$.
({\it Hint:\/} it helps to know that $\mu(hF)=h\mu F$
(134Ya, 263A).   Show that if $F\subseteq[-M,M]$, then
     
\Centerline{$\Bover1{2hM}\mu(E\cap[x-hM,x+hM])
\le 1-\Bover{\mu F}{2M}
  \bigl(1-\Bover{\mu(E\cap(x+hF))}{h\,\mu F}\bigr)$.)}
     
\noindent (Compare 223Ya.)
%223B
     
\spheader 223Xh Let $f$ be a real-valued function which is integrable
over $\Bbb R$, and $E$ be the Lebesgue set of $f$.   Show that
$\lim_{h\downarrow 0}\bover1{2h}\int_{x-h}^{x+h}|f(t)-c|dt=|f(x)-c|$ 
for every $x\in E$ and $c\in\Bbb R$.
%223D
     
\spheader 223Xi Let $f$ be an integrable real-valued function defined
almost everywhere in $\Bbb R$.   Let $x\in\dom f$ be such that
$\lim_{n\to\infty}\Bover{n}2\int^{x+1/n}_{x-1/n}|f(y)-f(x)|=0$.   Show
that $x$ belongs to the Lebesgue set of $f$.
%223D
     
\spheader 223Xj Let $f$ be an integrable real-valued function defined
almost everywhere in $\Bbb R$, and $x$ any point of the Lebesgue set of
$f$.   Show
that for every $\epsilon>0$ there is a $\delta>0$ such that whenever $I$
is a non-trivial interval and $x\in I\subseteq[x-\delta,x+\delta]$, then
$|f(x)-\Bover1{\mu I}\int_If|\le\epsilon$.
%223D
     
\leader{223Y}{Further exercises (a)} Let $E$, $F\subseteq\Bbb R$ be
measurable sets, and suppose that $0<\mu F<\infty$.   Let $x\in\Bbb R$
be such that $\lim_{h\downarrow 0}\Bover1{2h}\mu(E\cap[x-h,x+h])=1$.
Show that
     
\Centerline{$\lim_{h\downarrow 0}\Bover{\mu(E\cap(x+hF))}{h\,\mu F}=1$.}
     
\noindent \Hint{apply 223Xg to sets of the form $F\cap[-M,M]$.}
%223Xg 223B
     
\header{223Yb}{\bf (b)} Let $\frak T$ be the family of measurable sets
$G\subseteq\Bbb R$ such that every point of $G$ is a density point of
$G$.   {(i)} Show that $\frak T$ is a topology on $\Bbb R$.   ({\it
Hint\/}: take $\Cal G\subseteq\frak T$.   By 215B(iv) there is a
countable
$\Cal G_0\subseteq\Cal G$ such that $\mu(G\setminus\bigcup\Cal G_0)=0$
for every $G\in\Cal G$.   Show that
     
\Centerline{$\bigcup\Cal G\subseteq\{x:\limsup_{h\downarrow 0}
  \Bover1{2h}\mu(\bigcup\Cal G_0\cap[x-h,x+h])>0\}$,}
     
\noindent so that $\mu(\bigcup\Cal G\setminus\bigcup\Cal G_0)=0$.)   {(ii)} Show that a function
$f:\Bbb R\to\Bbb R$ is measurable iff it is $\frak T$-continuous at
almost every $x\in\Bbb R$.   ($\frak T$ is the {\bf density topology} on
$\Bbb R$.   See 414P in Volume 4.)
%223B
     
\spheader 223Yc Show that if $f:[a,b]\to\Bbb R$ is bounded and
continuous for the density topology on $\Bbb R$, then
$f(x)=\bover{d}{dx}\int_a^xf$ for every $x\in\ooint{a,b}$.
%223Yb 223B
     
\spheader 223Yd Show that a function $f:\Bbb R\to\Bbb R$ is continuous
for the density topology at $x\in\Bbb R$ iff 
$\lim_{h\downarrow 0}\bover1{2h}
  \mu^*\{y:y\in[x-h,x+h],\,|f(y)-f(x)|\ge\epsilon\}=0$ for
every $\epsilon>0$.
%223Yb 223B
     
\spheader 223Ye A set $A\subseteq\Bbb R$ is {\bf porous} at a point
$x\in\Bbb R$ if $\limsup_{y\to x}\Bover{\rho(y,A)}{|y-x|}>0$, where
$\rho(y,A)=\inf_{a\in A}|y-a|$.   (Take $\rho(y,\emptyset)=\infty$.)
Show that if $A$ is porous at every $x\in A$ then $A$ is negligible.
%223B
     
\spheader 223Yf For a measurable set $E\subseteq\Bbb R$ write
$\intstar E$ for the set of its density points.   Show that if $E$, 
$F\subseteq\Bbb R$ are measurable then (i)
$\intstar(E\cap F)=\intstar E\cap\intstar F$ 
(ii) $\intstar E\subseteq\intstar F$
iff $\mu(E\setminus F)=0$ (iii) $\mu(E\symmdiff\intstar E)=0$ (iv)
$\intstar(\intstar E)=\intstar E$ (v) for
every compact set $K\subseteq\intstar E$ there is a compact set
$L\subseteq K\cup E$ such that $K\subseteq\intstar L$.
%223B
     
\spheader 223Yg Let $f$ be an integrable real-valued function defined
almost everywhere in $\Bbb R$, and $x$ any point of the Lebesgue set of
$f$.   Show that for every $\epsilon>0$ there is a $\delta>0$ such that
$|f(x)\int g-\int f\times g|\le\epsilon\int g$ whenever
$g:\Bbb R\to\coint{0,\infty}$ is such that $g$ is non-decreasing on
$\ocint{-\infty,x}$, non-increasing on $\coint{x,\infty}$ and zero
outside $[x-\delta,x+\delta]$.   \Hint{express $g$ as a limit almost
everywhere of functions of the form
$\Bover{g(x)}{n+1}\sum_{i=0}^n\chi\ooint{a_i,b_i}$, where
$x-\delta\le a_0\le\ldots\le a_n\le x\le b_n\le\ldots\le b_0
\le x+\delta$.}
%223D

\spheader 223Yh For each integrable real-valued function $f$ defined almost
everywhere in $\Bbb R$, let $E_f$ be the Lebesgue set of $f$.   Show that
$E_f\cap E_g\subseteq E_{f+g}$, $E_f\subseteq E_{|f|}$ for all integrable
$f$, $g$.
%223D

\spheader 223Yi\dvAnew{2014} 
Let $E\subseteq\Bbb R$ be a non-negligible measurable set. 
Show that $0$ belongs to the interior of $E-E=\{x-y:x$, $y\in E\}$.
%223B
}%end of exercises
     
\endnotes{
\Notesheader{223} The results of this section can be
thought of as saying that a measurable function is in some sense 
`almost continuous';  indeed, 223Yb is an attempt to make this notion
precise.   For an integrable function we have stronger results, of which
the furthest-reaching seems to be 223D/223Ec.
     
There are $r$-dimensional versions of all these theorems, using balls
centered on $x$ in place of intervals $[x-h,x+h]$;  I give these in
261C-261E.   %261C 261D 261E
A new idea is needed for the $r$-dimensional version of
Lebesgue's density theorem (261C), but the rest of the
generalization is straightforward.   A less natural, and less important,
extension, also in \S261, involves functions defined on
non-measurable sets (compare 223Xb-223Xe).
}%end of comment
     
\discrpage

\frfilename{mt224.tex}
\versiondate{29.9.04}
\copyrightdate{1997}

\def\chaptername{The Fundamental Theorem of Calculus}
\def\sectionname{Functions of bounded variation}

\newsection{224}

I turn now to the second of the two problems to which this chapter is
devoted:  the identification of those real functions which are
indefinite integrals.   I take the opportunity to offer a brief
introduction to the theory of functions of bounded variation, which are
interesting in themselves and will be important in Chapter 28.   I give
the
basic characterization of these functions as differences of monotonic
functions (224D), with a representative sample of their elementary
properties.

\leader{224A}{Definition} Let $f$ be a
real-valued function  and  $D$ a subset of $\Bbb R$.   I define
$\Var_D(f)$, the {\bf (total) variation of $f$ on $D$}, as follows.   If
$D\cap\dom f=\emptyset$, $\Var_D(f)=0$.   Otherwise, $\Var_D(f)$ is

\Centerline{$\sup\{\sum_{i=1}^n|f(a_i)-f(a_{i-1})|:
  a_0,a_1,\ldots,a_n\in D\cap\dom f,\,a_0\le a_1\le\ldots\le a_n\}$,}

\noindent allowing $\Var_{D}(f)=\infty$.
If $\Var_{D}(f)$ is finite, we say that $f$ is {\bf of bounded
variation} on $D$.   \cmmnt{If the context seems clear,} I may write
$\Var f$ for $\Var_{\dom f}(f)$, and say that $f$ is simply `{\bf of
bounded variation}' if this is finite.

\leader{224B}{Remarks}\cmmnt{ {\bf (a)} In the present chapter, we
shall virtually exclusively be concerned with the case in which $D$ is a
bounded closed interval included in $\dom f$.   The general formulation
will be useful for some technical questions arising in Chapter 28;  but
if it makes
you more comfortable, you will lose nothing by supposing for the moment
that $D$ is an interval.

\spheader 224Bb Clearly}

\Centerline{$\Var_D(f)=\Var_{D\cap\dom f}(f)=\Var(f\restr D)$}

\noindent for all $D$, $f$.

\leader{224C}{Proposition} (a) If $f$, $g$ are two real-valued functions
and $D\subseteq\Bbb R$, then

\Centerline{$\Var_{D}(f+g)\le\Var_{D}(f)+\Var_{D}(g)$.}

(b) If $f$ is a real-valued function, $D\subseteq\Bbb R$ and
$c\in\Bbb R$ then $\Var_{D}(cf)=|c|\Var_{D}(f)$.

(c) If $f$ is a real-valued function, $D\subseteq\Bbb R$ and
$x\in \Bbb R$ then

\Centerline{$\Var_{D}(f)
\ge\Var_{D\cap\ocint{-\infty,x}}(f)+\Var_{D\cap\coint{x,\infty}}(f)$,}

\noindent with equality if $x\in D\cap\dom f$.

(d) If $f$ is a real-valued function and $D\subseteq D'\subseteq\Bbb R$
then $\Var_D(f)\le\Var_{D'}(f)$.

(e) If $f$ is a real-valued function and $D\subseteq\Bbb R$, then
$|f(x)-f(y)|\le\Var_D(f)$ for all $x$, $y\in D\cap\dom f$;  so if $f$ is
of bounded variation on $D$ then $f$ is bounded on $D\cap\dom f$ and (if
$D\cap\dom f\ne\emptyset$)

\Centerline{$\sup_{y\in D\cap\dom f}|f(y)|\le |f(x)|+\Var_D(f)$}

\noindent for every $x\in D\cap\dom f$.

(f) If $f$ is a monotonic real-valued function and $D\subseteq\Bbb R$
meets $\dom f$, then
\discrcenter{468pt}
{$\Var_D(f)=\sup_{x\in D\cap\dom f}f(x)-\inf_{x\in D\cap\dom f}f(x)$.}

\proof{{\bf (a)}  If $D\cap\dom(f+g)=\emptyset$ this is trivial,
because $\Var_D(f)$ and $\Var_D(g)$  are surely non-negative.
Otherwise, if
$a_0\le\ldots\le a_n$ in $D\cap\dom(f+g)$, then

$$\eqalign{\sum_{i=1}^n|(f+g)(a_i)-(f+g)(a_{i-1})|
&\le \sum_{i=1}^n|f(a_i)-f(a_{i-1})|
+\sum_{i=1}^n|g(a_i)-g(a_{i-1})|\cr
&\le\Var_{D}(f)+\Var_{D}(g);\cr}$$

\noindent as $a_0,\ldots,a_n$ are arbitrary,
$\Var_{D}(f+g)\le\Var_{D}(f)+\Var_{D}(g)$.

\medskip

{\bf (b)}

\Centerline{$\sum_{i=1}^n|(cf)(a_i)-(cf)(a_{i-1})|
=|c|\sum_{i=1}^n|f(a_i)-f(a_{i-1})|$}

\noindent whenever
$a_0\le\ldots\le a_n$ in $D\cap\dom f$.

\medskip

{\bf (c)(i)} If either $D\cap\ocint{-\infty,x}\cap\dom f$ or
$D\cap\coint{x,\infty}\cap\dom f$ is empty, this is trivial.   If
$a_0\le\ldots\le a_m$ in $D\cap\ocint{-\infty,x}\cap\dom f$,
$b_0\le\ldots\le b_n$ in $D\cap\coint{x,\infty}\cap\dom f$, then

$$\eqalign{\sum_{i=1}^m|f(a_i)-f(a_{i-1})|
+\sum_{j=1}^n|f(b_i)-f(b_{i-1})|
&\le\sum_{i=1}^{m+n+1}|f(a_i)-f(a_{i-1})|\cr
&\le\Var_D(f),\cr}$$

\noindent if we write $a_i=b_{i-m-1}$ for $m+1\le i\le m+n+1$.
So

\Centerline{$\Var_{D\cap\ocint{-\infty,x}}(f)
+\Var_{D\cap\coint{x,\infty}}(f)
\le\Var_D(f)$.}

\medskip

\quad{\bf (ii)} Now suppose that $x\in D\cap\dom f$.
If $a_0\le\ldots\le a_n$ in $D\cap\dom f$, and $a_0\le x\le a_n$, let
$k$ be such that $x\in [a_{k-1},a_k]$;
then

$$\eqalign{\sum_{i=1}^n|f(a_i)-f(a_{i-1})|
&\le\sum_{i=1}^{k-1}|f(a_i)-f(a_{i-1})|
+|f(x)-f(a_{k-1})|\cr
&\mskip150mu +|f(a_k)-f(x)|+\sum_{i=k+1}^n|f(a_i)-f(a_{i-1})|\cr
&\le\Var_{D\cap\ocint{-\infty,x}}(f)
+\Var_{D\cap\coint{x,\infty}}(f)
\cr}$$

\noindent (counting empty sums $\sum_{i=1}^0$,
$\sum_{i=n+1}^n$ as $0$).   If $x\le a_0$ then
$\sum_{i=1}^n|f(a_i)-f(a_{i-1})|\le\Var_{D\cap\coint{x,\infty}}(f)$;
if $x\ge a_n$ then
$\sum_{i=1}^n|f(a_i)-f(a_{i-1})|\le\Var_{D\cap\ocint{-\infty,x}}(f)$.
Thus

\Centerline{$\sum_{i=1}^n|f(a_i)-f(a_{i-1})|
\le\Var_{D\cap\ocint{-\infty,x}}(f)
+\Var_{D\cap\coint{x,\infty}}(f)
$}

\noindent in all cases;  as $a_0,\ldots,a_n$ are arbitrary,

\Centerline{$\Var_D(f)
\le\Var_{D\cap\ocint{-\infty,x}}(f)
+\Var_{D\cap\coint{x,\infty}}(f)$.}

\noindent So the two sides are equal.

\medskip

{\bf (d)} is trivial.

\medskip

{\bf (e)} If $x$, $y\in D\cap\dom f$ and $x\le y$ then

\Centerline{$|f(x)-f(y)|=|f(y)-f(x)|\le \Var_D(f)$}

\noindent by the definition of $\Var_D$;  and the same is true if
$y\le x$.   So of course $|f(y)|\le|f(x)|+\Var_D(f)$.

\medskip

{\bf (f)} If $f$ is non-decreasing, then

$$\eqalign{\Var_D(f)
&=\sup\{\sum_{i=1}^n|f(a_i)-f(a_{i-1})|:a_0,a_1,\ldots,a_n\in
D\cap\dom f,\,a_0\le
a_1\le\ldots\le a_n\}\cr
&=\sup\{\sum_{i=1}^nf(a_i)-f(a_{i-1}):a_0,a_1,\ldots,a_n\in
D\cap\dom f,\,a_0\le
a_1\le\ldots\le a_n\}\cr
&=\sup\{f(b)-f(a):a,\,b\in D\cap\dom f,\,a\le b\}\cr
&=\sup_{b\in D\cap\dom f}f(b)-\inf_{a\in D\cap\dom f}f(a).\cr}$$

\noindent If $f$ is non-increasing then

$$\eqalign{\Var_D(f)
&=\sup\{\sum_{i=1}^n|f(a_i)-f(a_{i-1})|:a_0,a_1,\ldots,a_n\in
D\cap\dom f,\,a_0\le
a_1\le\ldots\le a_n\}\cr
&=\sup\{\sum_{i=1}^nf(a_{i-1})-f(a_i):a_0,a_1,\ldots,a_n\in
D\cap\dom f,\,a_0\le
a_1\le\ldots\le a_n\}\cr
&=\sup\{f(a)-f(b):a,\,b\in D\cap\dom f,\,a\le b\}\cr
&=\sup_{a\in D\cap\dom f}f(a)-\inf_{b\in D\cap\dom f}f(b).\cr}$$
}%end of proof of 224C

\leader{224D}{Theorem} For any real-valued function $f$ and any set
$D\subseteq \Bbb R$, the following are equiveridical:

\quad (i) there are two bounded
non-decreasing functions $f_1$, $f_2:\Bbb R\to\Bbb R$ such that
$f=f_1-f_2$ on $D\cap\dom f$;

\quad (ii) $f$ is of bounded variation on $D$;

\quad (iii) there are bounded
non-decreasing functions $f_1$, $f_2:\Bbb R\to\Bbb R$ such that
$f=f_1-f_2$ on $D\cap\dom f$ and $\Var_D(f)=\Var f_1+\Var f_2$.

\proof{{\bf (i)$\Rightarrow$(ii)} If $f:\Bbb R\to\Bbb R$ is
bounded and non-decreasing, then
$\Var f=\sup_{x\in\Bbb R}f(x)-\inf_{x\in\Bbb R}f(x)$ is
finite.   So if $f$ agrees on $D\cap\dom f$ with $f_1-f_2$
where $f_1$ and $f_2$ are bounded and non-decreasing, then

$$\eqalign{\Var_D(f)
&=\Var_{D\cap\dom f}(f)
\le\Var_{D\cap\dom f}(f_1)+\Var_{D\cap\dom f}(f_2)\cr
&\le\Var f_1+\Var f_2
<\infty,}$$

\noindent using (a), (b) and (d) of 224C.

\medskip

{\bf(ii)$\Rightarrow$(iii)} Suppose that $f$ is of bounded variation on
$D$.    Set $D'=D\cap\dom f$.   If $D'=\emptyset$ we can take both $f_j$
to be the zero function, so henceforth suppose that $D'\ne\emptyset$.
Write

\Centerline{$g(x)=\Var_{D\cap\ocint{-\infty,x}}(f)$}

\noindent for $x\in D'$.   Then $g_1=g+f$ and $g_2=g-f$ are both
non-decreasing.   \Prf\ If $a$, $b\in D'$ and $a\le b$, then

\Centerline{$g(b)=g(a)+\Var_{D\cap[a,b]}(f)\ge g(a)+|f(b)-f(a)|$.}

\noindent So

\Centerline{$g_1(b)-g_1(a)=g(b)-g(a)+f(b)-f(a)$,
\quad$g_2(b)-g_2(a)=g(b)-g(a)-f(b)+f(a)$}

\noindent are both non-negative.\ \Qed

Now there are non-decreasing functions $h_1$, $h_2:\Bbb R\to\Bbb R$,
extending $g_1$, $g_2$ respectively, such that $\Var h_j=\Var g_j$ for
both $j$.   \Prf\  $f$
is bounded on $D$, by 224Ce, and $g$ is bounded just because
$\Var_D(f)<\infty$, so that $g_j$ is bounded.   Set
$c_j=\inf_{x\in D'}g_j(x)$ and

\Centerline{$h_j(x)=\sup(\{c_j\}\cup\{g_j(y):y\in D',\,y\le x\})$}

\noindent for every $x\in\Bbb R$;  this works.\ \QeD\   Observe that for
$x\in D'$,

\Centerline{$h_1(x)+h_2(x)=g_1(x)+g_2(x)=g(x)+f(x)+g(x)-f(x)=2g(x)$,}

\Centerline{$h_1(x)-h_2(x)=2f(x)$.}

Now, because $g_1$ and $g_2$ are non-decreasing,

\Centerline{$\sup_{x\in D'}g_1(x)+\sup_{x\in D'}g_2(x)=\sup_{x\in
D'}g_1(x)+g_2(x)=2\sup_{x\in D'}g(x)$,}

\Centerline{$\inf_{x\in D'}g_1(x)+\inf_{x\in D'}g_2(x)=\inf_{x\in
D'}g_1(x)+g_2(x)=2\inf_{x\in D'}g(x)\ge 0$.}

\noindent But this means that

\Centerline{$\Var h_1+\Var h_2
=\Var g_1+\Var g_2
=2\Var g
\le 2\Var_D(f)$,}

\noindent using 224Cf three times.   So if we set
$f_j(x)=\bover12h_j(x)$
for $j\in\{1,2\}$ and $x\in\Bbb R$, we shall have non-decreasing functions
such that

\Centerline{$f_1(x)-f_2(x)=f(x)$ for $x\in D'$,
\quad$\Var f_1+\Var f_2=\Bover12\Var h_1+\Bover12\Var h_2\le\Var_D(f)$.}

\noindent Since we surely also have

\Centerline{$\Var_D(f)\le\Var_D(f_1)+\Var_D(f_2)\le\Var f_1+\Var f_2$,}

\noindent we see that $\Var_D(f)=\Var f_1+\Var f_2$, and (iii) is true.

\medskip

{\bf (iii)$\Rightarrow$(i)} is trivial.
}%end of proof of 224D

\leader{224E}{Corollary} Let $f$ be a real-valued function and $D$ any
subset of $\Bbb R$.   If $f$ is of bounded variation on $D$, then

\Centerline{$\lim_{x\downarrow a}\Var_{D\cap\ocint{a,x}}(f)
=\lim_{x\uparrow a}\Var_{D\cap\coint{x,a}}(f)=0$}
\noindent for every $a\in\Bbb R$, and

\Centerline{$\lim_{a\to-\infty}\Var_{D\cap\ocint{-\infty,a}}(f)
=\lim_{a\to\infty}\Var_{D\cap\coint{a,\infty}}(f)=0$.}

\proof{{\bf (a)} Consider first the case in which
$D=\dom f=\Bbb R$ and $f$ is a bounded non-decreasing function.   Then

\Centerline{$\Var_{D\cap\ocint{a,x}}(f)
=\sup_{y\in\ocint{a,x}}f(x)-f(y)
=f(x)-\inf_{y>a}f(y)
=f(x)-\lim_{y\downarrow a}f(y)$,}

\noindent so of course

\Centerline{$\lim_{x\downarrow a}\Var_{D\cap\ocint{a,x}}(f)
=\lim_{x\downarrow a}f(x)-\lim_{y\downarrow a}f(y)=0$.}

\noindent In the same way

\Centerline{$\lim_{x\uparrow a}\Var_{D\cap\coint{x,a}}(f)
=\lim_{y\uparrow a}f(y)-\lim_{x\uparrow a}f(x)=0$,}

\Centerline{$\lim_{a\to-\infty}\Var_{D\cap\ocint{-\infty,a}}(f)
=\lim_{a\to-\infty}f(a)-\lim_{y\to-\infty}f(y)=0$,}

\Centerline{$\lim_{a\to\infty}\Var_{D\cap\coint{a,\infty}}(f)
=\lim_{y\to\infty}f(y)-\lim_{a\to\infty}f(a)=0$.}

\medskip

{\bf (b)} For the general case, define $f_1$, $f_2$ from $f$ and $D$ as
in 224D.   Then for every interval $I$ we have

\Centerline{$\Var_{D\cap I}(f)\le\Var_I(f_1)+\Var_I(f_2)$,}

\noindent so the results for $f$ follow from those for $f_1$ and $f_2$
as established in part (a) of the proof.
}%end of proof of 224E

\leader{224F}{Corollary} Let $f$ be a real-valued function of bounded
variation on $[a,b]$, where $a<b$.   If $\dom f$ meets every interval
$\ocint{a,a+\delta}$ with $\delta>0$,   then

\Centerline{$\lim_{t\in\dom f,t\downarrow a}f(t)$}

\noindent is defined in $\Bbb R$.   If $\dom f$ meets
$\coint{b-\delta,b}$ for every $\delta>0$, then

\Centerline{$\lim_{t\in\dom f,t\uparrow b}f(t)$}

\noindent is defined in $\Bbb R$.

\wheader{224F}{0}{0}{0}{36pt}

\proof{ Let $f_1$, $f_2:\Bbb R\to\Bbb R$ be non-decreasing
functions such that $f=f_1-f_2$ on $[a,b]\cap\dom f$.   Then

\Centerline{$\lim_{t\in\dom f,t\downarrow a}f(t)
=\lim_{t\downarrow a}f_1(t)-\lim_{t\downarrow a}f_2(t)
=\inf_{t>a}f_1(t)-\inf_{t>a}f_2(t)$,}

\Centerline{$\lim_{t\in\dom f,t\uparrow b}f(t)
=\lim_{t\uparrow b}f_1(t)-\lim_{t\uparrow b}f_2(t)
=\sup_{t<b}f_1(t)-\sup_{t<b}f_2(t)$.}
}%end of proof of 224F

\leader{224G}{Corollary} Let $f$, $g$ be real functions and $D$ a subset
of $\Bbb R$.   If $f$ and $g$ are of bounded variation on $D$, so is
$f\times g$.

\proof{{\bf (a)} The point is that there are {\it non-negative}
bounded non-decreasing functions $f_1$, $f_2:\Bbb R\to\Bbb R$ such that
$f=f_1-f_2$ on $D\cap\dom f$.   \Prf\ We know that there are bounded
non-decreasing $h_1$, $h_2$ such that $f=h_1-h_2$ on $D\cap\dom f$.
Set $\gamma_i=\inf_{x\in\Bbb R}h_i(x)$ for $i=1$, $2$,

\Centerline{$\beta_1=\max(\gamma_1-\gamma_2,0)$,
\quad$\beta_2=\max(\gamma_2-\gamma_1,0)$,}

\Centerline{$f_1=h_1-\gamma_1+\beta_1$,
\quad$f_2=h_1-\gamma_2+\beta_2$;}

\noindent this works.\ \Qed

\medskip

{\bf (b)} Now taking similar functions $g_1$, $g_2$ such that
$g=g_1-g_2$ on $D\cap\dom g$, we have

\Centerline{$f\times g=f_1\times g_1-f_2\times g_1-f_1\times g_2
+f_2\times g_2$}

\noindent everywhere in $D\cap\dom(f\times g)=D\cap\dom f\cap\dom g$;
but all the $f_i\times g_j$ are bounded non-decreasing functions, so of
bounded variation, and $f\times g$ must be of bounded variation on $D$.
}%end of proof of 224G

\leader{224H}{Proposition} Let $f:D\to\Bbb R$ be a function of bounded
variation, where $D\subseteq\Bbb R$.   Then $f$ is continuous at all
except countably many points of $D$.

\proof{ For $n\ge 1$ set

$$\eqalign{A_n
&=\{x:x\in D,\text{ for every }\delta>0\text{ there is a }y\in
D\cap[x-\delta,x+\delta]\cr
&\qquad\qquad\qquad\qquad\text{ such that }|f(y)-f(x)|
  \ge\Bover1n\}.\cr}$$

\noindent Then  $\#(A_n)\le n\Var f$.   \Prf\Quer\ Otherwise, we can
find distinct $x_0,\ldots,x_k\in A_n$ with $k+1>n\Var f$.   Order these
so that $x_0<x_1<\ldots<x_k$.   Set
$\delta=\bover12\min_{1\le i\le k}x_i-x_{i-1}>0$.   For each $i$, there
is a $y_i\in D\cap[x_i-\delta,x_i+\delta]$ such that
$|f(y_i)-f(x_i)|\ge\bover1n$.   Take $x'_i$, $y'_i$ to be $x_i$, $y_i$
in order, so that $x'_i<y'_i$.   Now

\Centerline{$x'_0\le y'_0\le x'_1\le y'_1\le\ldots\le x'_k\le y'_k$,}
\noindent and

\Centerline{$\Var f
\ge\sum_{i=0}^k|f(y'_i)-f(x'_i)|
=\sum_{i=0}^k|f(y_i)-f(x_i)|
\ge \Bover1n(k+1)
>\Var f$,}

\noindent which is impossible.\ \Bang\Qed

It follows that $A=\bigcup_{n\in\Bbb N}A_n$ is countable, being a
countable
union of finite sets.   But $A$ is exactly the set of points of $D$ at
which $f$ is not continuous.
}%end of proof of 224H

\leader{224I}{Theorem} Let $I\subseteq\Bbb R$ be an interval, and
$f:I\to\Bbb R$ a function of bounded variation.   Then $f$ is
differentiable almost everywhere in $I$, and $f'$ is integrable over
$I$, with

\Centerline{$\int_I|f'|\le\Var_I(f)$.}

\proof{{\bf (a)} Let $f_1$ and $f_2$ be non-decreasing functions such
that $f=f_1-f_2$ everywhere in $I$ (224D).   Then $f_1$ and $f_2$ are
differentiable almost everywhere (222A).   At any point of $I$ except
possibly its endpoints, if any, $f$ will be differentiable if $f_1$ and
$f_2$ are, so $f'(x)$ is defined for almost every $x\in I$.

\medskip

{\bf (b)} Set $F(x)=\Var_{I\cap\ocint{-\infty,x}}f$ for $x\in\Bbb R$.
If $x$, $y\in I$ and $x\le y$, then

\Centerline{$F(y)-F(x)=\Var_{[x,y]}f\ge|f(y)-f(x)|$,}

\noindent by 224Cc;  so $F'(x)\ge|f'(x)|$ whenever $x$ is an interior
point of $I$ and both derivatives exist, which is almost everywhere.
So $\int_I|f'|\le\int_IF'$.   But if $a$, $b\in I$ and $a\le b$,

\Centerline{$\int_a^bF'\le F(b)-F(a)\le F(b)\le\Var f$.}

\noindent Now $I$ is expressible as $\bigcup_{n\in\Bbb N}[a_n,b_n]$
where $a_{n+1}\le a_n\le b_n\le b_{n+1}$ for every $n$.   So

$$\eqalignno{\int_I|f'|
&\le\int_IF'
=\int F'\times\chi I\cr
&=\int\sup_{n\in\Bbb N}F'\times\chi[a_n,b_n]
=\sup_{n\in\Bbb N}\int F'\times\chi[a_n,b_n]\cr
\noalign{\noindent (by B.Levi's theorem)}
&=\sup_{n\in\Bbb N}\int_{a_n}^{b_n}F'
\le\Var_I(f).\cr}$$
}%end of proof of 224I

\leader{224J}{}\cmmnt{ The next result is not needed in this chapter,
but is one of the most useful properties of functions of bounded
variation, and will be used repeatedly in Chapter 28.

\medskip

\noindent}{\bf Proposition} Let $f$, $g$ be real-valued
functions defined on subsets of $\Bbb R$, and suppose that $g$ is
integrable over an interval $[a,b]$, where $a<b$, and $f$ is of bounded
variation on $\ooint{a,b}$ and defined almost everywhere in
$\ooint{a,b}$.   Then $f\times g$ is integrable over $[a,b]$, and

$$\bigl|\int_a^bf\times g\bigr|
\le\bigl(\lim_{x\in\dom f,x\uparrow b}|f(x)|+\Var_{\ooint{a,b}}(f)\bigr)
  \sup_{c\in[a,b]}\bigl|\int_a^cg\bigr|.$$

\proof{{\bf (a)} By 224F, $l=\lim_{x\in\dom f,x\uparrow b}f(x)$ is
defined.   Write $M=|l|+\Var_{\ooint{a,b}}(f)$.   Note that if $y$ is
any point of $\dom f\cap\ooint{a,b}$,

\Centerline{$|f(y)|\le|f(x)|+|f(x)-f(y)|
\le|f(x)|+\Var_{\ooint{a,b}}(f)\to M$}

\noindent as $x\uparrow b$ in $\dom f$, so
$|f(y)|\le M$.   Moreover, $f$ is measurable
on $\ooint{a,b}$, because there are bounded monotonic functions
$f_1$, $f_2:\Bbb R\to\Bbb R$ such that $f=f_1-f_2$
everywhere in $\ooint{a,b}\cap\dom f$.   So $f\times g$ is measurable
and dominated by $M|g|$, and is integrable over $\ooint{a,b}$ or
$[a,b]$.

\medskip

{\bf (b)} For $n\in\Bbb N$, $k\le 2^n$ set
$a_{nk}=a+2^{-n}k(b-a)$, and for $1\le k\le 2^n$ choose
$x_{nk}\in\dom f\cap\ocint{a_{n,k-1},a_{nk}}$.    Define
$f_n:\ocint{a,b}\to\Bbb R$ by
setting $f_n(x)=f(x_{nk})$ if $1\le k\le 2^n$ and
$x\in\ocint{a_{n,k-1},a_{nk}}$.   Then $f(x)=\lim_{n\to\infty}f_n(x)$
whenever $x\in\ooint{a,b}\cap\dom f$ and $f$ is continuous at $x$, which
must be almost everywhere (224H).   Note next that all the $f_n$ are
measurable, and that they are uniformly bounded, in
modulus, by $M$.    So $\{f_n\times g:n\in\Bbb N\}$ is dominated by the
integrable function $M|g|$, and Lebesgue's
Dominated Convergence Theorem tells us that

\Centerline{$\int_a^bf\times g=\lim_{n\to\infty}\int_a^bf_n\times g$.}

\medskip

{\bf (c)} Fix $n\in\Bbb N$ for the moment.    Set
$K=\sup_{c\in[a,b]}|\int_a^cg|$.   (Note that $K$ is finite because
$c\mapsto\int_a^cg$ is continuous.)  Then

$$\eqalignno{\bigl|\int_a^bf_n\times g\bigr|
&=\bigl|\sum_{k=1}^{2^n}\int_{a_{n,k-1}}^{a_{nk}}f_n\times g\bigr|\cr
&=\bigl|\sum_{k=1}^{2^n}
    f(x_{nk})(\int_a^{a_{nk}}g-\int_a^{a_{n,k-1}}g)\bigr|\cr
&=\bigl|\sum_{k=1}^{2^n-1}
   (f(x_{nk})-f(x_{n,k+1}))\int_a^{a_{nk}}g
   +f(x_{n,2^n})\int_a^bg\bigr|\cr
&\le\bigl|f(x_{n,2^n})\bigr|\bigl|\int_a^bg\bigr|+\sum_{k=1}^{2^n-1}
   \bigl|f(x_{n,k+1})-f(x_{nk})\bigr|\bigl|\int_a^{a_{nk}}g\bigr|\cr
&\le(|f(x_{n,2^n})|+\Var_{\ooint{a,b}}(f))K
\to MK\cr}$$

\noindent as $n\to\infty$.

\medskip

{\bf (d)} Now

\Centerline{$|\int_a^bf\times g|
=\lim_{n\to\infty}|\int_a^bf_n\times g|\le MK$,}

\noindent as required.
}%end of proof of 224J

\leader{224K}{Complex-valued functions}\cmmnt{ So far I have taken all
functions to be real-valued.   This is adequate for the needs of the
present chapter, but in Chapter 28 we shall need to look at
complex-valued functions of bounded variation, and I should perhaps
spell out the (elementary) adaptations involved in the extension to the
complex case.
}

\spheader 224Ka Let $D$ be a subset of $\Bbb R$ and $f$ a
complex-valued function.   The
{\bf variation} of $f$ on $D$, $\Var_D(f)$, is zero if
$D\cap\dom f=\emptyset$, and otherwise is

\Centerline{$\sup\{\sum_{j=1}^n|f(a_j)-f(a_{j-1})|:
 a_0\le a_1\le\ldots\le a_n$ in $D\cap\dom f\}$,}

\noindent allowing $\infty$.
If $\Var_{D}(f)$ is finite, we say that $f$ is {\bf of bounded
variation} on $D$.

\spheader 224Kb \dvro{A}{Just as in the real case, a} complex-valued
function of bounded variation must be bounded, and
\Centerline{$\Var_{D}(f+g)\le\Var_{D}(f)+\Var_{D}(g)$,}

\Centerline{$\Var_{D}(cf)=|c|\Var_{D}(f)$,}

\Centerline{$\Var_{D}(f)\ge\Var_{D\cap\ocint{-\infty,x}}(f)
+\Var_{D\cap\coint{x,\infty}}(f)$}

\noindent for every
$x\in\Bbb R$, with equality if $x\in D\cap\dom f$,

\Centerline{$\Var_D(f)\le\Var_{D'}(f)$ whenever $D\subseteq
D'$\dvro{.}{;}}

\cmmnt{\noindent the arguments of 224C go through unchanged.}

\spheader 224Kc A complex-valued function is of bounded
variation iff its real and imaginary parts are both of bounded
variation\dvro{. }{ (because

\Centerline{$\max(\Var_{D}(\Real f),\Var_{D}(\Imag f))
\le\Var_{D}(f)\le\Var_{D}(\Real f)+\Var_{D}(\Imag f)$.)}

\noindent}So a complex-valued function $f$ is of bounded variation on
$D$ iff there are bounded non-decreasing functions
$f_1,\ldots,f_4:\Bbb R\to\Bbb R$ such that
$f=f_1-f_2+if_3-if_4$ on $D$\cmmnt{ (224D)}.

\spheader 224Kd Let $f$ be a complex-valued function and $D$ any
subset of $\Bbb R$.   If $f$ is of bounded variation on $D$, then

\Centerline{$\lim_{x\downarrow a}\Var_{D\cap\ocint{a,x}}(f)
=\lim_{x\uparrow a}\Var_{D\cap\coint{x,a}}(f)=0$}

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

\Centerline{$\lim_{a\to-\infty}\Var_{D\cap\ocint{-\infty,a}}(f)
=\lim_{a\to\infty}\Var_{D\cap\coint{a,\infty}}(f)=0$.}

\prooflet{\noindent (Apply 224E to the real and imaginary parts of
$f$.)}

\spheader 224Ke Let $f$ be a complex-valued function of bounded
variation on $[a,b]$, where $a<b$.   If $\dom f$ meets every interval
$\ocint{a,a+\delta}$ with $\delta>0$,   then
 $\lim_{t\in\dom f,t\downarrow a}f(t)$
is defined in $\Bbb C$.   If $\dom f$ meets
$\coint{b-\delta,b}$ for every $\delta>0$, then
$\lim_{t\in\dom f,t\uparrow b}f(t)$
is defined in $\Bbb C$.   \prooflet{(Apply 224F to the real
and imaginary parts of $f$.)}

\spheader 224Kf Let $f$, $g$ be complex functions and $D$ a subset
of $\Bbb R$.   If $f$ and $g$ are of bounded variation on $D$, so is
$f\times g$.   \prooflet{(For $f\times g$ is expressible as a linear
combination of the four products
$\Real f\times\Real g,\ldots,\Imag f\times\Imag g$,
to each of which we can apply 224G.)}

\spheader 224Kg Let $I\subseteq\Bbb R$ be an interval, and
$f:I\to\Bbb C$ a function of bounded variation.   Then $f$ is
differentiable almost everywhere in $I$, and $\int_I|f'|\le\Var_I(f)$.
\prooflet{(As 224I.)}

\spheader 224Kh Let $f$ and $g$ be complex-valued
functions defined on subsets of $\Bbb R$, and suppose that $g$ is
integrable over an interval $[a,b]$, where $a<b$, and $f$ is of bounded
variation on $\ooint{a,b}$ and defined almost everywhere in
$\ooint{a,b}$.   Then $f\times g$ is integrable over $[a,b]$, and

$$\bigl|\int_a^bf\times g\bigr|
\le\bigl(\lim_{x\in\dom f,x\uparrow b}|f(x)|+\Var_{\ooint{a,b}}(f)\bigr)
  \sup_{c\in[a,b]}\bigl|\int_a^cg\bigr|.$$

\prooflet{\noindent (The argument of 224J applies virtually unchanged.)}

\exercises{
\leader{224X}{Basic exercises $\pmb{>}$(a)}
%\spheader 224Xa
Set $f(x)=x^2\sin\Bover1{x^2}$ for $x\ne 0$, $f(0)=0$.   Show that
$f:\Bbb R\to\Bbb R$ is differentiable everywhere and uniformly
continuous, but is not of bounded variation on any non-trivial interval
containing $0$.
%224A

\spheader 224Xb Give an example of a non-negative function
$g:[0,1]\to[0,1]$, of bounded variation, such that $\sqrt{g}$ is not of
bounded variation.
%224A

\spheader 224Xc Show that if $f$ is any real-valued function
defined on a subset of $\Bbb R$, there is a function
$\tilde f:\Bbb R\to\Bbb R$, extending $f$, such that
$\Var\tilde f=\Var f$.   Under what circumstances is $\tilde f$ unique?
%224D

\spheader 224Xd Let $f:D\to\Bbb R$ be a function of bounded variation,
where $D\subseteq\Bbb R$ is a non-empty set.   Show that if $\inf_{x\in
D}|f(x)|>0$ then $1/f$ is of bounded variation.
%224C

\spheader 224Xe Let $f:[a,b]\to\Bbb R$ be a continuous function, where
$a\le b$ in $\Bbb R$.   Show that if $c<\Var f$ then there is a
$\delta>0$
such that $\sum_{i=1}^n|f(a_i)-f(a_{i-1})|\ge c$ whenever
$a=a_0\le a_1\le\ldots\le a_n=b$ and
$\max_{1\le i\le n}a_i-a_{i-1}\le\delta$.
%224C

\spheader 224Xf Let $\sequencen{f_n}$ be a sequence of real functions,
and set $f(x)=\lim_{n\to\infty}f_n(x)$ whenever the limit is defined.
Show that $\Var f\le\liminf_{n\to\infty}\Var f_n$.
%224A

\spheader 224Xg Let $f$ be a real-valued function which is integrable
over an interval $[a,b]\subseteq\Bbb R$.   Set $F(x)=\int_a^xf$ for
$x\in[a,b]$.
Show that $\Var F=\int_a^b|f|$.   ({\it Hint\/}: start by
checking that $\Var F\le\int|f|$;  for the reverse inequality, consider
the case $f\ge 0$ first.)
%224D

\spheader 224Xh Show that if $f$ is a real-valued function defined on a
non-empty set $D\subseteq\Bbb R$, then

\Centerline{$\Var f=\sup\{|\sum_{i=1}^n(-1)^i(f(a_i)-f(a_{i-1}))|:
  a_0\le a_1\le\ldots\le a_n$ in $D\}$.}
%224A

\spheader 224Xi Let $f$ be a real-valued function which is integrable
over a bounded interval $[a,b]\subseteq\Bbb R$.   Show that

\Centerline{$\int_a^b|f|
=$$\sup\{|\sum_{i=1}^n(-1)^i\int_{a_{i-1}}^{a_i}f|:
  a=a_0\le a_1\le a_2\le\ldots\le a_n=b\}$.}

\noindent({\it Hint\/}: put 224Xg and 224Xh together.)
%224Xg, 224Xh, 224D

\spheader 224Xj Let $f$ and $g$ be real-valued
functions defined on subsets of $\Bbb R$, and suppose that $g$ is
integrable over an interval $[a,b]$, where $a<b$, and $f$ is of bounded
variation on $\ooint{a,b}$ and defined almost everywhere in
$\ooint{a,b}$.   Show that

\Centerline{$|\int_a^bf\times g|
\le(\lim_{x\in\dom f,x\downarrow a}|f(x)|
  +\Var_{\ooint{a,b}}(f))\sup_{c\in[a,b]}|\int_c^bg|$.}
%224J

\spheader 224Xk\dvAnew{2012} Suppose that $D\subseteq\Bbb R$ and
$f:D\to\Bbb R$ is a function.   Show that $f$ is
expressible as a difference of non-decreasing functions iff
$\Var_{D\cap[a,b]}(f)$ is finite whenever $a\le b$ in $D$.
%224D out of order query

\spheader 224Xl\dvAnew{2012}
Suppose that $D\subseteq\Bbb R$ and that $f:D\to\Bbb R$ is a continuous
function of bounded variation.   Show that $f$ is expressible as the
difference of two continuous non-decreasing functions.
%224E out of order query

\spheader 224Xm\dvAnew{2013}
Suppose that $D\subseteq\Bbb R$ and that $f:D\to\Bbb R$ is a
function of bounded variation which is continuous on the right, that is,
whenever $x\in D$ and $\epsilon>0$ there is a $\delta>0$ such that
$|f(t)-f(x)|\le\epsilon$ for every $t\in D\cap[x,x+\delta]$.
Show that $f$ is expressible as the
difference of two non-decreasing functions which are continuous on the
right.
%224E out of order query

\leader{224Y}{Further exercises (a)}
%\spheader 224Ya
Show that if $f$ is any complex-valued function
defined on a subset of $\Bbb R$, there is a function
$\tilde f:\Bbb R\to\Bbb C$, extending $f$, such that
$\Var\tilde f=\Var f$.   Under what circumstances is $\tilde f$ unique?

\spheader 224Yb Let $D$ be any non-empty subset of $\Bbb R$, and
let $\eusm V$ be the space of functions $f:D\to\Bbb R$ of bounded
variation.   For $f\in\eusm V$ set

\Centerline{$\|f\|=\sup\{|f(t_0)|+\sum_{i=1}^n|f(t_i)-f(t_{i-1})|:
t_0\le\ldots\le t_n\in D\}$.}

\noindent Show that (i) $\|\,\|$ is a norm on $\eusm V$ (ii) $\eusm V$
is complete under $\|\,\|$ (iii) $\|f\times g\|\le\|f\|\|g\|$ for all
$f$, $g\in\eusm V$, so that $\eusm V$ is a Banach algebra.

\spheader 224Yc Let $f:\Bbb R\to\Bbb R$ be a function of bounded
variation.
Show that there is a sequence $\sequencen{f_n}$ of differentiable
functions
such that $\lim_{n\to\infty}f_n(x)=f(x)$ for every $x\in\Bbb R$,
$\lim_{n\to\infty}\int|f_n-f|=0$, and $\Var(f_n)\le\Var(f)$ for every
$n\in\Bbb N$.   \Hint{start with non-decreasing $f$.}

\spheader 224Yd For any partially ordered set $X$ and any
function $f:X\to\Bbb R$, say that $\Var_X(f)=0$ if $X=\emptyset$ and
otherwise

\Centerline{$\Var_X(f)
=\sup\{\sum_{i=1}^n|f(a_i)-f(a_{i-1})|:a_0,a_1,\ldots,a_n\in X,\,a_0\le
a_1\le\ldots\le a_n\}$.}

\noindent State and prove results in this framework generalizing 224D
and 224Yb.   ({\it Hints\/}:  $f$ will be `non-decreasing' if
$f(x)\le f(y)$ whenever $x\le y$;  interpret $\ocint{-\infty,x}$ as
$\{y:y\le x\}$.)

\spheader 224Ye Let $(X,\rho)$ be a metric space and $f:[a,b]\to X$ a
function, where $a\le b$ in $\Bbb R$.   Set
\ifnum\stylenumber=12

\Centerline{$\Var_{[a,b]}(f)
=\sup\{\sum_{i=1}^n\rho(f(a_i),f(a_{i-1})):
a\le a_0\le\ldots\le a_n\le b\}$.}

\noindent\else
$\Var_{[a,b]}(f)
=\sup\{\sum_{i=1}^n\rho(f(a_i),f(a_{i-1})):
a\le a_0\le\ldots\le a_n\le b\}$. \fi
(i) Show that
$\Var_{[a,b]}(f)=\Var_{[a,c]}(f)+\Var_{[c,b]}(f)$ for every $c\in[a,b]$.
(ii) Show that if $\Var_{[a,b]}(f)$ is finite
then $f$ is continuous at all but countably many points of $[a,b]$.
(iii) Show that if $X$ is complete and $\Var_{[a,b]}(f)<\infty$ then
$\lim_{t\uparrow x}f(t)$ is defined for every $x\in\ocint{a,b}$.
(iv) Show that if $X$ is complete then
$\Var_{[a,b]}(f)$ is finite iff $f$ is expressible as a
composition $gh$, where $h:[a,b]\to\Bbb R$ is non-decreasing and
$g:\Bbb R\to X$ is $1$-Lipschitz, that is,
$\rho(g(c),g(d))\le|c-d|$ for all $c$, $d\in\Bbb R$.

\spheader 224Yf Let $U$ be a normed space and $a\le b$ in $\Bbb R$.
For functions $f:[a,b]\to U$ define $\Var_{[a,b]}(f)$ as in 224Ye, using
the standard metric $\rho(x,y)=\|x-y\|$ for $x$, $y\in U$.    (i) Show
that $\Var_{[a,b]}(f+g)\le\Var_{[a,b]}(f)+\Var_{[a,b]}(g)$,
$\Var_{[a,b]}(cf)=|c|\Var_{[a,b]}(f)$ for all $f$, $g:[a,b]\to U$ and
all $c\in\Bbb R$.   (ii) Show that if $V$ is another normed space and
$T:U\to V$ is a bounded linear operator then
$\Var_{[a,b]}(Tf)\le\|T\|\Var_{[a,b]}(f)$ for every $f:[a,b]\to U$.

\spheader 224Yg Let $f:[0,1]\to\Bbb R$ be a continuous function.   For
$y\in\Bbb R$ set $h(y)=\#(f^{-1}[\{y\}])$ if this is finite, $\infty$
otherwise.   Show that (if we allow $\infty$ as a value of the integral)
$\Var_{[0,1]}(f)=\int h$.   ({\it Hint\/}:  for $n\in\Bbb N$, $i<2^n$
set $c_{ni}=\sup\{f(x)-f(y):x,y\in[2^{-n}i,2^{-n}(i+1)]\}$,
$h_{ni}(y)=1$ if $y\in f[\,\coint{2^{-n}i,2^{-n}(i+1)}\,]$, $0$
otherwise.   Show that $c_{ni}=\int h_{ni}$,
$\lim_{n\to\infty}\sum_{i=0}^{2^n-1}c_{ni}=\Var f$,
$\lim_{n\to\infty}\sum_{i=0}^{2^n-1}h_{ni}=h$.)
(See also 226Yc.)

\spheader 224Yh Let $\nu$ be any Lebesgue-Stieltjes measure on $\Bbb R$,
$I\subseteq \Bbb R$ an interval (which may be either open or closed,
bounded or unbounded), and $D\subseteq I$ a non-empty
set.  Let $\eusm V$ be the space of functions of bounded variation from
$D$ to $\Bbb R$, and $\|\,\|$ the norm of 224Yb on $\eusm V$.   Let
$g:D\to\Bbb R$ be a function such that $\int_{[a,b]\cap D}g\,d\nu$
exists whenever $a\le b$ in $I$, and
$K=\sup_{a,b\in I,a\le  b}|\int_{[a,b]\cap D}g\,d\nu|$.
Show that $|\int_Df\times g\,d\nu|\le K\|f\|$ for every $f\in\eusm V$.
%224K

\spheader 224Yi Explain how to apply 224Yh with $D=\Bbb N$ to
obtain Abel's theorem that the product of a monotonic sequence
converging to $0$ with a series which has bounded partial sums is
summable.
%224Yh 224K

\spheader 224Yj Suppose that $I\subseteq\Bbb R$ is an interval, and
that $\sequencen{A_n}$ is a sequence of sets covering $I$.
Let $f:I\to\Bbb R$ be continuous.   Show that
$\Var f\le\sum_{n=0}^{\infty}\Var_{A_n}f$.
\Hint{reduce to the case of closed sets $A_n$;
use Baire's theorem (4A2Ma).}

\spheader 224Yk\dvAnew{2023} 
Let $f:D\to\Bbb R$ be a function, where $D\subseteq\Bbb R$.  
Show that the following are equiveridical:  ($\alpha$) $\lim_{n\to\infty}f(t_n)$
is defined for every montonic sequence $\sequencen{t_n}$ in $D$;  ($\beta$)
for every $\epsilon>0$ there is a function $g:D\to\Bbb R$ of bounded variation
such that $|f(t)-g(t)|\le\epsilon$ for every $t\in D$.
}%end of exercises

\endnotes{
\Notesheader{224} I have taken the ideas above rather farther than we
need immediately;  for the present chapter, it is enough to consider the
case in which $D=\dom f=[a,b]$ for some interval $[a,b]\subseteq\Bbb R$.
The extension to functions
with irregular domains will be useful in Chapter 28, and the extension
to irregular sets $D$, while not important to us here, is of some
interest -- for instance, taking $D=\Bbb N$, we obtain the notion of
`sequence of bounded variation', which is surely relevant to problems of
convergence and summability.

The central result of the section is of course the fact that a function
of bounded variation can be expressed as the difference of monotonic
functions (224D);  indeed, one of the objects of the concept is to
characterize the
linear span of the monotonic functions.   Nearly everything else here
can be derived as easy consequences of this, as in 224E-224G.   In 224I
and 224Xg we go a little deeper, and indeed some measure theory appears;
this is where the ideas here begin to connect with the real business of
this  chapter, to be continued in the next section.
Another result which is
easy enough in itself, but contains the germs of important ideas, is
224Yg.

In 224Yb I mention a natural development in functional analysis, and in
224Yd-224Yf %224Yd 224Ye 224Yf
I suggest further wide-ranging generalizations.
}%end of notes

\discrpage
\frfilename{mt225.tex}
\versiondate{16.8.15}
\copyrightdate{1996}

\def\chaptername{The Fundamental Theorem of Calculus}
\def\sectionname{Absolutely continuous functions}

\newsection{225}

We are now ready for a full characterization of the functions that can
appear as indefinite integrals (225E, 225Xf).   The essential idea
is that of `absolute continuity' (225B).   In the second half of the
section (225G-225N) I describe some of the relationships between
this concept and those we have already seen.

\leader{225A}{Absolute continuity of the
indefinite \dvrocolon{integral}}\cmmnt{ I
begin with an easy fundamental result from general measure theory.

\medskip

\noindent}{\bf Theorem} Let
$(X,\Sigma,\mu)$ be any measure space and $f$ an integrable real-valued
function defined on a conegligible subset of $X$.
Then for any $\epsilon>0$ there are a measurable set
$E$ of finite measure and a real number
$\delta>0$ such that $\int_F|f|\le\epsilon$ whenever $F\in\Sigma$ and
$\mu(F\cap E)\le\delta$.

\proof{ There is a non-decreasing sequence $\sequencen{g_n}$ of
non-negative simple functions such that $|f|\eae\lim_{n\to\infty}g_n$
and $\int|f|=\lim_{n\to\infty}\int g_n$.   Take $n\in\Bbb N$ such that
$\int g_n\ge\int|f|-\bover12\epsilon$.
Let $M>0$, $E\in\Sigma$ be such that $\mu E<\infty$ and
$g_n\le M\chi E$;  set $\delta=\epsilon/2M$.
If $F\in\Sigma$ and $\mu(F\cap E)\le\delta$, then

\Centerline{$\int_Fg_n=\int g_n\times\chi F
\le M\mu(F\cap E)\le\Bover12\epsilon$;}

\noindent  consequently

\Centerline{$\int_F|f|
=\int_Fg_n+\int_F|f|-g_n\le\Bover12\epsilon+\int|f|-g_n\le\epsilon$.}
}%end of proof of 225A

\leader{225B}{Absolutely continuous functions on $\Bbb R$:  Definition}
If $[a,b]$ is a non-empty
closed interval in $\Bbb R$ and $f:[a,b]\to\Bbb R$ is a function, we say
that $f$ is {\bf absolutely continuous} if for every $\epsilon>0$ there
is a $\delta>0$ such that $\sum_{i=1}^n|f(b_i)-f(a_i)|\le\epsilon$
whenever $a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta$.

\cmmnt{\medskip

\noindent{\bf Remark} The phrase `absolutely continuous' is used in
various senses in measure theory, closely related (if you look at them
in the right way) but not identical;  you will need to keep the context
of each definition in clear focus.
}%end of comment

\leader{225C}{Proposition} Let $[a,b]$ be a non-empty closed interval in
$\Bbb R$.

(a) If $f:[a,b]\to\Bbb R$ is absolutely continuous, it is uniformly
continuous.

(b) If $f:[a,b]\to\Bbb R$ is absolutely continuous it is of bounded
variation on $[a,b]$, so is differentiable almost everywhere in $[a,b]$,
and its derivative is integrable over $[a,b]$.

(c) If $f$, $g:[a,b]\to\Bbb R$ are absolutely continuous, so are
$f+g$ and $cf$, for every $c\in\Bbb R$.

(d) If $f$, $g:[a,b]\to\Bbb R$ are absolutely continuous so is $f\times
g$.

(e) If $g:[a,b]\to [c,d]$ and $f:[c,d]\to\Bbb R$ are absolutely
continuous, and $g$ is non-decreasing, then the composition
$fg:[a,b]\to\Bbb R$ is absolutely continuous.

\proof{{\bf (a)} Let $\epsilon>0$.   Then there is a $\delta>0$ such
that $\sum_{i=1}^n|f(b_i)-f(a_i)|\le\epsilon$
whenever $a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta$;  but of course now
$|f(y)-f(x)|\le\epsilon$ whenever $x$, $y\in[a,b]$ and
$|x-y|\le\delta$.   As $\epsilon$ is arbitrary, $f$ is uniformly
continuous.



\medskip

{\bf (b)} Let $\delta>0$ be such that
$\sum_{i=1}^n|f(b_i)-f(a_i)|\le 1$
whenever $a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta$.   If
$a\le c=c_0\le c_1\le\ldots\le c_n\le d\le\min(b,c+\delta)$, then
$\sum_{i=1}^n|f(c_i)-f(c_{i-1})|\le 1$, so $\Var_{[c,d]}(f)\le 1$;
accordingly (inducing on $k$, using
224Cc for the inductive step) $\Var_{[a,\min(a+k\delta,b)]}(f)\le k$
for every $k$, and

\Centerline{$\Var_{[a,b]}(f)\le\lceil(b-a)/\delta\rceil<\infty$.}

It follows that $f'$ is integrable, by 224I.

\medskip

{\bf (c)(i)} Let $\epsilon>0$.   Then there are $\delta_1$,
$\delta_2>0$ such that

\Centerline{$\sum_{i=1}^n|f(b_i)-f(a_i)|\le\Bover12\epsilon$}

\noindent whenever
$a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta_1$,

\Centerline{$\sum_{i=1}^n|g(b_i)-g(a_i)|\le\Bover12\epsilon$}

\noindent whenever $a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le
b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta_2$.   Set
$\delta=\min(\delta_1,\delta_2)>0$,
and suppose that $a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le
b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta$.   Then


\Centerline{$\sum_{i=1}^n|(f+g)(b_i)-(f+g)(a_i)|\le
\sum_{i=1}^n|f(b_i)-f(a_i)|+\sum_{i=1}^n|g(b_i)-g(a_i)|\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $f+g$ is absolutely
continuous.

\medskip

\quad{\bf (ii)} Let $\epsilon>0$.   Then there is a
$\delta>0$ such that

\Centerline{$\sum_{i=1}^n|f(b_i)-f(a_i)|\le\Bover{\epsilon}{1+|c|}$}

\noindent whenever
$a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta$.   Now

\Centerline{$\sum_{i=1}^n|(cf)(b_i)-(cf)(a_i)|\le\epsilon$}

\noindent whenever
$a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta$.   As $\epsilon$ is arbitrary, $cf$ is
absolutely continuous.

\medskip

{\bf (d)} By either (a) or (b), $f$ and $g$ are bounded;  set
$M=\sup_{x\in[a,b]}|f(x)|$, $M'=\sup_{x\in[a,b]}|g(x)|$.   Let
$\epsilon>0$.   Then there are $\delta_1$, $\delta_2>0$ such that

\inset{$\sum_{i=1}^n|f(b_i)-f(a_i)|\le\epsilon$
whenever $a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta_1$,}

\inset{$\sum_{i=1}^n|g(b_i)-g(a_i)|\le\epsilon$
whenever $a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta_2$.}

\noindent Set $\delta=\min(\delta_1,\delta_2)>0$ and suppose that $a\le
a_1\le b_1\le\ldots\le b_n\le b$ and $\sum_{i=1}^nb_i-a_i\le\delta$.
Then

$$\eqalign{\sum_{i=1}^n|f(b_i)g(b_i)-f(a_i)g(a_i)|
&=\sum_{i=1}^n|(f(b_i)-f(a_i))g(b_i)+f(a_i)(g(b_i)-g(a_i))|\cr
&\le\sum_{i=1}^n|f(b_i)-f(a_i)||g(b_i)|+|f(a_i)||g(b_i)-g(a_i)|\cr
&\le\sum_{i=1}^n|f(b_i)-f(a_i)|M'+M|g(b_i)-g(a_i)|\cr
&\le\epsilon M'+M\epsilon
=\epsilon(M+M').\cr}$$

\noindent As $\epsilon$ is arbitrary, $f\times g$ is absolutely
continuous.

\medskip

{\bf (e)} Let $\epsilon>0$.   Then there is a $\delta>0$ such that
$\sum_{i=1}^n|f(d_i)-f(c_i)|\le\epsilon$ whenever $c\le c_1\le
d_1\le\ldots\le c_n\le d_n\le d$ and $\sum_{i=1}^nd_i-c_i\le\delta$;
and there is an $\eta>0$ such that
$\sum_{i=1}^n|g(b_i)-g(a_i)|\le\delta$ whenever $a\le a_1\le
b_1\le\ldots\le a_n\le b_n\le b$ and $\sum_{i=1}^nb_i-a_i\le\eta$.   Now
suppose that $a\le a_1\le b_1\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\eta$.   Because $g$ is non-decreasing, we have
$c\le g(a_1)\le\ldots\le g(b_n)\le d$ and
$\sum_{i=1}^ng(b_i)-g(a_i)\le\delta$, so $\sum_{i=1}^n|f(g(b_i))
-f(g(a_i))|\le\epsilon$;  as $\epsilon$ is arbitrary, $fg$ is absolutely
continuous.
}%end of proof of 225C

\leader{225D}{Lemma} Let $[a,b]$ be a non-empty closed interval in
$\Bbb R$ and $f:[a,b]\to\Bbb R$ an absolutely continuous function which
has zero derivative almost everywhere in $[a,b]$.
Then $f$ is constant on $[a,b]$.

\proof{ Let $x\in[a,b]$, $\epsilon>0$.   Let $\delta>0$ be
such that $\sum_{i=1}^n|f(b_i)-f(a_i)|\le\epsilon$
whenever $a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta$.   Set $A=\{t:a<t<x,\,f'(t)$ exists
$=0\}$;  then $\mu A=x-a$, writing $\mu$ for Lebesgue measure.
Let $\Cal I$ be the set of non-empty non-singleton closed intervals
$[c,d]\subseteq[a,x]$ such that $|f(d)-f(c)|\le\epsilon (d-c)$;  then
every member of $A$ belongs to arbitrarily short members of $\Cal I$.
By Vitali's theorem (221A), there is a countable disjoint family
$\Cal I_0\subseteq\Cal I$ such that $\mu(A\setminus\bigcup\Cal I_0)=0$,
that is,

\Centerline{$x-a=\mu(\bigcup\Cal I_0)=\sum_{I\in\Cal I_0}\mu I$.}

\noindent Now there is a finite $\Cal I_1\subseteq\Cal I_0$ such that

\Centerline{$\mu(\bigcup\Cal I_1)
=\sum_{I\in\Cal I_1}\mu I\ge x-a-\delta$.}

\noindent If $\Cal I_1=\emptyset$, then $x\le a+\delta$ and
$|f(x)-f(a)|\le\epsilon$.   Otherwise, express $\Cal I_1$ as
$\{[c_0,d_0],\ldots,[c_n,d_n]\}$, where
$a\le c_0<d_0<c_1<d_1<\ldots<c_n<d_n\le x$.   Then

\Centerline{$(c_0-a)+\sum_{i=1}^n(c_i-d_{i-1})
+(x-d_n)=\mu([a,x]\setminus\bigcup\Cal I_1)\le\delta$,}

\noindent so

\Centerline{$|f(c_0)-f(a)|+\sum_{i=1}^n|f(c_i)-f(d_{i-1})|+|f(x)-f(d_n)|
\le\epsilon$.}

\noindent On the other hand, $|f(d_i)-f(c_i)|\le\epsilon(d_i-c_i)$ for
each $i$, so

\Centerline{$\sum_{i=0}^n|f(d_i)-f(c_i)|\le\epsilon\sum_{i=0}^nd_i-c_i
\le\epsilon(x-a)$.}

\noindent Putting these together,

$$\eqalign{|f(x)-f(a)|
&\le|f(c_0)-f(a)|+|f(d_0)-f(c_0)|+|f(c_1)-f(d_0)|+\ldots\cr
&\hskip10em+|f(d_n)-f(c_n)|+|f(x)-f(d_n)|\cr
&=|f(c_0)-f(a)|+\sum_{i=1}^n|f(c_i)-f(d_{i-1})|\cr
&\hskip10em+|f(x)-f(d_n)|+\sum_{i=0}^n|f(d_i)-f(c_i)|\cr
&\le\epsilon+\epsilon(x-a)
=\epsilon(1+x-a).\cr}$$

\noindent As $\epsilon$ is arbitrary, $f(x)=f(a)$.   As $x$ is
arbitrary, $f$ is constant.
}%end of proof of 225D

\leader{225E}{Theorem}   Let $[a,b]$ be a non-empty closed interval in
$\Bbb R$ and $F:[a,b]\to\Bbb R$ a function.   Then the following are
equiveridical:

(i) there is an integrable real-valued function $f$ such that
$F(x)=F(a)+\int_a^xf$ for every $x\in[a,b]$;

(ii) $\int_a^xF'$ exists and is
equal to $F(x)-F(a)$ for every $x\in[a,b]$;

(iii) $F$ is absolutely continuous.

\cmmnt{\medskip

\noindent{\bf Remark} Here, and for the rest of the section (except in
225Oa), integrals will
be taken with respect to Lebesgue measure on $\Bbb R$.
}

\proof{{\bf (i)$\Rightarrow$(iii)}  Assume (i).
Let $\epsilon>0$.   By 225A, there is a
$\delta>0$ such that $\int_H|f|\le\epsilon$ whenever $H\subseteq[a,b]$
and $\mu H\le\delta$, writing $\mu$ for Lebesgue measure as usual.   Now
suppose that $a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le b_n\le b$
and $\sum_{i=1}^nb_i-a_i\le\delta$.   Consider $H=\bigcup_{1\le i\le
n}\coint{a_i,b_i}$.   Then $\mu H\le\delta$ and

\Centerline{$\sum_{i=1}^n|F(b_i)-F(a_i)|
=\sum_{i=1}^n|\int_{\coint{a_i,b_i}}f|
\le\sum_{i=1}^n\int_{\coint{a_i,b_i}}|f|
=\int_F|f|\le\epsilon$.}

\noindent   As
$\epsilon$ is arbitrary, $F$ is absolutely continuous.

\medskip

{\bf (iii)$\Rightarrow$(ii)} If $F$ is absolutely continuous, then it is
of bounded variation (225Cb),  so $\int_a^bF'$ exists (224I).   Set
$G(x)=\int_a^xF'$ for $x\in [a,b]$;
then $G'\eae F'$ (222E) and $G$ is absolutely continuous (by
(i)$\Rightarrow$(iii) just proved).   Accordingly $G-F$ is absolutely
continuous (225Cc) and is differentiable, with zero derivative, almost
everywhere.   It follows that $G-F$ must be constant (225D).   But as
$G(a)=0$, $G=F-F(a)$;  just as required by (ii).

\medskip

{\bf (ii)$\Rightarrow$(i)} is trivial.
}%end of proof of 225E

\leader{225F}{Integration by \dvrocolon{parts}}\cmmnt{ As an
application of this
result, I give a justification of a familiar formula.

\medskip

\noindent}{\bf Theorem} Let $f$ be a real-valued function which is
integrable over
an interval $[a,b]\subseteq\Bbb R$, and $g:[a,b]\to\Bbb R$ an absolutely
continuous function.   Suppose that $F$ is an indefinite integral of
$f$, so that $F(x)-F(a)=\int_a^xf$ for $x\in[a,b]$.   Then

\Centerline{$\int_a^bf\times g
=F(b)g(b) - F(a)g(a) -\int_a^bF\times g'$.}

\proof{ Set $h=F\times g$.   Because $F$ is absolutely continuous
(225E), so is $h$ (225Cd).   Consequently
$h(b)-h(a)=\int_a^bh'$, by (iii)$\Rightarrow$(ii) of 225E.   But
$h'=F'\times g+F\times g'$ wherever $F'$ and $g'$ are defined, which is
almost everywhere, and $F'\eae f$, by 222E again;  so
$h'\eae f\times g+F\times g'$.   Finally, $g$ and $F$ are continuous, therefore measurable,
and bounded, while $f$ and $g'$ are integrable (using 225E yet again),
so $f\times g$ and $F\times g'$ are integrable, and

\Centerline{$F(b)g(b)-F(a)g(a)
=h(b)-h(a)
=\int_a^bh'
=\int_a^bf\times g+\int_a^bF\times g'$,}

\noindent as required.
}%end of proof of 225F

\leader{225G}{}\cmmnt{ I come now to a group of results at a rather
deeper level than most of the work of this chapter, being closer to the
ideas of Chapter 26.

\medskip

\noindent}{\bf Proposition} Let $[a,b]$ be a non-empty closed interval
in $\Bbb R$ and $f:[a,b]\to\Bbb R$ an absolutely continuous function.

(a) $f[A]$ is negligible for every negligible set $A\subseteq\Bbb R$.

(b)\dvAnew{2013}
$f[E]$ is measurable for every measurable set $E\subseteq\Bbb R$.

\proof{{\bf (a)} Let $\epsilon>0$.   Then there is a $\delta>0$ such that
$\sum_{i=1}^n|f(b_i)-f(a_i)|\le\epsilon$ whenever
$a\le a_1\le b_1\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta$.   Now
there is a sequence $\sequence{k}{I_k}$ of closed intervals, covering
$A$, with $\sum_{k=0}^{\infty}\mu I_k\le\delta$.   For each
$m\in\Bbb N$, let $F_m$ be $[a,b]\cap\bigcup_{k\le m}I_k$.   Then
$\mu f[F_m]\le\epsilon$.   \Prf\  $F_m$ must be expressible as
$\bigcup_{i\le n}[c_i,d_i]$ where $n\le m$ and
$a\le c_0\le d_0\le\ldots\le c_n\le d_n\le b$.
For each $i\le n$ choose $x_i$, $y_i$ such that
$c_i\le x_i$, $y_i\le d_i$ and

\Centerline{$f(x_i)=\min_{x\in[c_i,d_i]}f(x)$,\quad
$f(y_i)=\max_{x\in[c_i,d_i]}f(x)$;}

\noindent such exist because $f$ is continuous (225Ca), so is bounded and
attains its bounds on $[c_i,d_i]$.   Set $a_i=\min(x_i,y_i)$,
$b_i=\max(x_i,y_i)$, so that $c_i\le a_i\le b_i\le d_i$.   Then

\Centerline{$\sum_{i=0}^nb_i-a_i\le\sum_{i=0}^nd_i-c_i=\mu F_m
\le\mu(\bigcup_{k\in\Bbb N}I_k)\le\delta$,}

\noindent so

$$\eqalign{\mu f[F_m]
&=\mu(\bigcup_{i\le m}f[\,[c_i,d_i]\,])
\le\sum_{i=0}^n\mu(f[\,[c_i,d_i]\,])\cr
&=\sum_{i=0}^n\mu[f(x_i),f(y_i)]
=\sum_{i=0}^n|f(b_i)-f(a_i)|
\le\epsilon.  \text{ \Qed}\cr}$$

\noindent But $\sequence{m}{f[F_m]}$ is a non-decreasing sequence
covering $f[A]$, so

\Centerline{$\mu^*f[A]\le\mu(\bigcup_{m\in\Bbb N}f[F_m])
=\sup_{m\in\Bbb N}\mu f[F_m]\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $f[A]$ is negligible, as claimed.

\medskip

{\bf (b)} By 134Fb, there is a sequence $\sequencen{F_n}$ of closed subsets
of $E\cap[a,b]$ such that $\lim_{n\to\infty}\mu F_n=\mu(E\cap[a,b])$.
For each $n$, $F_n$ is closed and bounded, therefore compact (2A2F);  as
$f$ is continuous, $f[F_n]$ is compact (2A2Eb), therefore closed
(2A2F, in the other direction) and measurable (114G).   Next,
setting $A=E\cap[a,b]\setminus\bigcup_{n\in\Bbb N}F_n$,
$A$ is negligible, so $f[A]$ is negligible, by (a) here,
therefore measurable.   Consequently

\Centerline{$f[E]=f[E\cap[a,b]]=f[\bigcup_{n\in\Bbb N}F_n\cup A]
=\bigcup_{n\in\Bbb N}f[F_n]\cup f[A]$}

\noindent is measurable, as claimed.
}%end of proof of 225G

\leader{225H}{Semi-continuous functions}\cmmnt{ In preparation for the
last main result of this section, I give a general result concerning
measurable real-valued functions on subsets of $\Bbb R$.   It will be
convenient here, for once, to consider functions taking values in
$[-\infty,\infty]$.}   If $D\subseteq\BbbR^r$, a function
$g:D\to[-\infty,\infty]$ is {\bf lower
semi-continuous} if $\{x:g(x)>u\}$ is an open subset of
$D$\cmmnt{ (for the subspace topology, see 2A3C)} for
every $u\in[-\infty,\infty]$.   Any lower semi-continuous function is
Borel measurable, therefore Lebesgue
measurable\cmmnt{ (121B-121D)}. %121B 121C 121D
\cmmnt{Now we have the following result.}

\leader{225I}{Proposition}  Suppose that $r\ge 1$ and that $f$ is a
real-valued function, defined on
a subset $D$ of $\BbbR^r$, which is integrable over $D$.      Then for
any $\epsilon>0$ there is a lower semi-continuous function
$g:\BbbR^r\to[-\infty,\infty]$ such that $g(x)\ge f(x)$ for every
$x\in D$ and $\int_Dg$ is defined and not greater than
$\epsilon+\int_Df$.

\cmmnt{\medskip

\noindent{\bf Remarks} This is a result of great general importance, so
I give it in a fairly general form;  but for the present chapter all we
need is the case $r=1$, $D=[a,b]$ where $a\le b$.
}%end of comment

\proof{{\bf (a)} We can enumerate $\Bbb Q$ as $\sequencen{q_n}$.    By
225A, there is a $\delta>0$ such that $\int_F|f|\le\bover12{\epsilon}$
whenever $\mu_DF\le\delta$, where $\mu_D$ is the subspace measure on
$D$, so that $\mu_DF=\mu^*F$, the outer Lebesgue measure of $F$, for
every $F\in\Sigma_D$, the domain of $\mu_D$ (214A-214B).     For each
$n\in\Bbb N$, set

\Centerline{$\delta_n=2^{-n-1}\min(\Bover{\epsilon}{1+2|q_n|},\delta)$,}

\noindent so that
$\sum_{n=0}^{\infty}\delta_n|q_n|\le\bover12{\epsilon}$ and
$\sum_{n=0}^{\infty}\delta_n\le\delta$.   For each $n\in\Bbb N$, let
$E_n\subseteq\BbbR^r$ be a Lebesgue measurable set such that
$\{x:f(x)\ge q_n\}=D\cap E_n$, and choose an open set
$G_n\supseteq E_n\cap B(\tbf{0},n)$ such that
$\mu G_n\le\mu(E_n\cap B(\tbf{0},n))+\delta_n$ (134Fa), writing
$B(\tbf{0},n)$ for the ball
$\{x:\|x\|\le n\}$.   For $x\in\BbbR^r$, set

\Centerline{$g(x)=\sup\{q_n:x\in G_n\}$,}

\noindent allowing $-\infty$ as $\sup\emptyset$ and $\infty$ as the
supremum of a set with no upper bound in $\Bbb R$.

\medskip

{\bf (b)} Now check the properties of $g$.

\medskip

\quad{\bf (i)} $g$ is lower semi-continuous.   \Prf\ If
$u\in[-\infty,\infty]$, then

\Centerline{$\{x:g(x)>u\}=\bigcup\{G_n:q_n>u\}$}

\noindent is a union of open sets, therefore open.\ \Qed

\medskip

\quad{\bf (ii)} $g(x)\ge f(x)$ for every $x\in D$.   \Prf\ If
$x\in D$ and $\eta>0$, there is an $n\in\Bbb N$ such that
$\|x\|\le n$ and $f(x)-\eta\le q_n\le f(x)$;   now
$x\in E_n\subseteq G_n$
so $g(x)\ge q_n\ge f(x)-\eta$.   As $\eta$ is arbitrary,
$g(x)\ge f(x)$.\ \Qed

\medskip

\quad{\bf (iii)}  Consider the functions
$h_1$, $h_2:D\to\ocint{-\infty,\infty}$ defined by setting

$$\eqalign{h_1(x)
&=|f(x)|\text{ if }x\in D\cap\bigcup_{n\in\Bbb N}(G_n\setminus E_n),\cr
&=0\text{ for other }x\in D,\cr
h_2(x)&=\sum_{n=0}^{\infty}|q_n|\chi(G_n\setminus E_n)(x)
\text{ for every }x\in D.\cr}$$

\noindent Setting $F=\bigcup_{n\in\Bbb N}G_n\setminus E_n$,

\Centerline{$\mu F\le\sum_{n=0}^{\infty}\mu(G_n\setminus E_n)
\le\delta$,}

\noindent so

\Centerline{$\int_Dh_1
=\int_{D\cap F}|f|\le\Bover12\epsilon$}

\noindent by the choice of $\delta$.   As for $h_2$, we have
(by B.Levi's theorem)

\Centerline{$\int_Dh_2
=\sum_{n=0}^{\infty}|q_n|\mu_D(D\cap G_n\setminus F_n)
\le\sum_{n=0}^{\infty}|q_n|\mu(G_n\setminus F_n)
\le\Bover12\epsilon$}

\noindent -- because this is finite, $h_2(x)<\infty$ for almost every
$x\in D$.   Thus $\int_Dh_1+h_2\le\epsilon$.

\medskip

\quad{\bf (iv)} The point is that $g\le f+h_1+h_2$ everywhere in
$D$.   \Prf\ Take any $x\in D$.   If $n\in\Bbb N$
and $x\in G_n$, then either $x\in E_n$, in which case

\Centerline{$f(x)+h_1(x)+h_2(x)\ge f(x)\ge q_n$,}

\noindent or $x\in G_n\setminus E_n$, in which case

\Centerline{$f(x)+h_1(x)+h_2(x)\ge f(x)+|f(x)|+|q_n|\ge q_n$.}

\noindent Thus

\Centerline{$f(x)+h_1(x)+h_2(x)\ge\sup\{q_n:x\in G_n\}\ge g(x)$.   \Qed}

\noindent So $g\le f+h_1+h_2$ everywhere in $D$.

\medskip

\quad{\bf (v)} Putting (iii) and (iv) together,

\Centerline{$\int_Dg\le\int_Df+h_1+h_2\le\epsilon+\int_Df$,}

\noindent as required.
}%end of proof of 225I


\leader{225J}{}\cmmnt{ We need some results on Borel measurable sets
and functions which are of independent interest.

\medskip

\noindent}{\bf Theorem} Let $D$ be a subset of $\Bbb R$ and
$f:D\to\Bbb R$ any function.   Then

\Centerline{$E=\{x:x\in D,\,f$ is continuous at $x\}$}

\noindent is relatively Borel measurable in $D$, and

\Centerline{$F=\{x:x\in D,\,f$ is differentiable at $x\}$}

\noindent is Borel measurable in $\Bbb R$;  moreover, $f':F\to\Bbb R$ is
Borel measurable.

\wheader{225J}{0}{0}{0}{36pt}

\proof{{\bf (a)} For $k\in\Bbb N$ set

\Centerline{$\Cal G_k=\{\ooint{a,b}:a,\,b\in\Bbb R,\,
|f(x)-f(y)|\le 2^{-k}$ for all $x$, $y\in D\cap\ooint{a,b}\}$.}

\noindent Then $G_k=\bigcup\Cal G_k$ is an open set, so
$E_0=\bigcap_{k\in\Bbb N}G_k$ is a Borel set.   But $E=D\cap E_0$, so
$E$ is a relatively Borel subset of $D$.

\medskip

{\bf (b)(i)} I should perhaps say at once that when interpreting the
formula $f'(x)=\lim_{h\to 0}(f(x+h)-f(x))/h$, I insist on the
restrictive definition

\Centerline{$a=\lim_{h\to 0}\bover{f(x+h)-f(x)}{h}$}

\noindent if

\qquad\quad for every $\epsilon>0$ there is a $\delta>0$ such that
$\Bover{f(x+h)-f(x)}{h}$ is defined and

\hfill$|\Bover{f(x+h)-f(x)}{h}-a|\le\epsilon$ whenever
$0<|h|\le\delta$.\qquad\qquad\quad

\noindent So $f'(x)$ can be defined only if there is some $\delta>0$
such that the whole interval $[x-\delta,x+\delta]$ lies within the
domain $D$ of $f$.

\medskip

\quad{\bf (ii)} For $p$, $q$, $q'\in\Bbb Q$ and $k\in\Bbb N$ set

$$\eqalign{H(k,p,q,q')
&=\{x:x\in E\cap\ooint{q,q'},\,|f(y)-f(x)-p(y-x)|\le 2^{-k}|y-x|
\text{ for every }y\in\ooint{q,q'}\}\cr&
\mskip300mu\text{ if }\ooint{q,q'}\subseteq D\cr
&=\emptyset\text{ otherwise}.\cr}$$
%display works in smallprint version

\noindent Then
$H(k,p,q,q')=E\cap\ooint{q,q'}\cap\overline{H(k,p,q,q')}$.   \Prf\ If
$x\in E\cap\ooint{q,q'}\cap\overline{H(k,p,q,q')}$ there is a sequence
$\sequencen{x_n}$ in $H(k,p,q,q')$ converging to $x$.
Because $f$ is continuous at $x$,

$$\eqalign{|f(y)-f(x)-p(y-x)|
&=\lim_{n\to\infty}|f(y)-f(x_n)-p(y-x_n)|\cr&
\le 2^{-k}\lim_{n\to\infty}2^{-k}|y-x_n|
=2^{-k}|y-x|\cr}$$

\noindent for every $y\in\ooint{q,q'}$, so that
$x\in H(k,p,q,q')$.\ \QeD\   Consequently $H(k,p,q,q')$ is a Borel set.
\Prf\ There is a Borel set $E_0$ such that $E=E_0\cap D$, by (a), so that
if $\ooint{q,q'}\subseteq D$ then

\Centerline{$H(k,p,q,q')=E\cap\ooint{q,q'}\cap\overline{H(k,p,q,q')}
=E_0\cap\ooint{q,q'}\cap\overline{H(k,p,q,q')}$}

\noindent is Borel.   Otherwise, of course, $H(k,p,q,q')$ is Borel because
it is empty.\ \Qed

\medskip

\quad{\bf (iii)} Now

\Centerline{$F
=\bigcap_{k\in\Bbb N}\bigcup_{p,q,q'\in\Bbb Q}H(k,p,q,q')$.}

\noindent \Prf\ ($\alpha$) Suppose $x\in F$, that is, $f'(x)$ is
defined;  say $f'(x)=a$.   Take any $k\in\Bbb N$.   Then there are
$p\in\Bbb Q$, $\delta>0$ such that
$|p-a|\le 2^{-k-1}$ and $[x-\delta,x+\delta]\subseteq D$ and
$|\bover{f(x+h)-f(x)}{h}-a|\le 2^{-k-1}$ whenever $0<|h|\le\delta$;
now take
$q\in\Bbb Q\cap\coint{x-\delta,x}$, $q'\in\Bbb Q\cap\ocint{x,x+\delta}$
and see that $x\in H(k,p,q,q')$.   As $x$ is arbitrary,
$F\subseteq\bigcap_{k\in\Bbb N}\bigcup_{p,q,q'\in\Bbb Q}H(k,p,q,q')$.
($\beta$) If
$x\in\bigcap_{k\in\Bbb N}\bigcup_{p,q,q'\in\Bbb Q}H(k,p,q,q')$, then for
each $k\in\Bbb N$ choose $p_k$, $q_k$,
$q'_k\in\Bbb Q$ such that $x\in H(k,p_k,q_k,q'_k)$.   If $h\ne 0$,
$x+h\in\ooint{q_k,q'_k}$ then $|\bover{f(x+h)-f(x)}{h}-p_k|\le 2^{-k}$.
But this means, first, that $|p_k-p_l|\le 2^{-k}+2^{-l}$ for every $k$,
$l$ (since surely there is some $h\ne 0$ such that
$x+h\in\ooint{q_k,q'_k}\cap\ooint{q_l,q'_l}$), so that
$\sequence{k}{p_k}$ is a Cauchy sequence, with limit $a$ say;  and,
second, that $|\bover{f(x+h)-f(x)}{h}-a|\le 2^{-k}+|a-p_k|$ whenever
$h\ne 0$ and $x+h\in\ooint{q_k,q'_k}$, so that $f'(x)$ is defined and
equal to $a$.\ \Qed

\medskip

\quad{\bf (iv)} Because $\Bbb Q$ is countable, all the unions
$\bigcup_{p,q,q'\in\Bbb Q}H(k,p,q,q')$ are Borel sets, so $F$ also is.

\medskip

\quad{\bf (v)} Now enumerate $\BbbQ^3$ as
$\sequence{i}{(p_i,q_i,q'_i)}$, and set
$H'_{ki}=H(k,p_i,q_i,q'_i)\setminus\bigcup_{j<i}H(k,p_j,q_j,q'_j)$ for
each $k$, $i\in\Bbb N$.   Every $H'_{ki}$ is Borel measurable,
$\sequence{i}{H'_{ki}}$ is disjoint, and

\Centerline{$\bigcup_{i\in\Bbb N}H'_{ki}
=\bigcup_{i\in\Bbb N}H(k,p_i,q_i,q'_i)\supseteq F$}

\noindent for each $k$.   Note that
$|f'(x)-p|\le 2^{-k}$ whenever $x\in F\cap H(k,p,q,q')$, so if we set
$f_k(x)=p_i$ for every $x\in H'_{ki}$ we shall have a Borel
measurable function
$f_k$ such that $|f(x)-f_k(x)|\le 2^{-k}$ for every $x\in F$.
Accordingly $f'=\lim_{k\to\infty}f_k\restr F$ is Borel measurable.
}%end of proof of 225J

\leader{225K}{Proposition} Let $[a,b]$ be a non-empty closed interval in
$\Bbb R$, and $f:[a,b]\to\Bbb R$ a function.   Set
$F=\{x:x\in\ooint{a,b},\,f'(x)$ is defined$\}$.   Then $f$ is absolutely
continuous iff (i) $f$ is continuous (ii) $f'$ is integrable over $F$
(iii) $f[\,[a,b]\setminus F]$ is negligible.

\proof{{\bf (a)} Suppose first that $f$ is absolutely continuous.   Then
$f$ is surely continuous (225Ca) and $f'$ is integrable over $[a,b]$,
therefore over $F$ (225E);  also $[a,b]\setminus F$ is negligible, so
$f[\,[a,b]\setminus F]$ is negligible, by 225G.

\medskip

{\bf (b)} So now suppose that $f$ satisfies the conditions.   Set
$f^*(x)=|f'(x)|$ for $x\in F$, $0$ for
$x\in[a,b]\setminus F$.   Then $f(b)\le f(a)+\int_a^bf^*$.

\medskip

\Prf\ {\bf (i)} Because $F$ is a Borel set and $f'$ is a Borel
measurable function (225J), $f^*$ is measurable.   Let $\epsilon>0$.
Let $G$ be an open subset of
$\Bbb R$ such that $f[\,[a,b]\setminus F]\subseteq G$ and
$\mu G\le\epsilon$ (134Fa again).   Let $g:\Bbb R\to[0,\infty]$ be a lower
semi-continuous function such that $f^*(x)\le g(x)$ for every
$x\in[a,b]$ and $\int_a^bg\le\int_a^bf^*+\epsilon$ (225I).   Consider

\Centerline{$A=\{x:a\le x\le b,\,\mu([f(a),f(x)]\setminus G)
  \le 2\epsilon(x-a)+\int_a^xg\}$,}

\noindent interpreting $[f(a),f(x)]$ as $\emptyset$ if $f(x)<f(a)$.
Then $a\in A\subseteq[a,b]$, so $c=\sup A$ is defined and belongs to
$[a,b]$.

Because $f$ is continuous, the function
$x\mapsto\mu([f(a),f(x)]\setminus G)$ is continuous;  also
$x\mapsto 2\epsilon(x-a)+\int_a^xg$ is certainly continuous, so
$c\in A$.

\medskip

\quad{\bf (ii)} \Quer\ If $c\in F$, so that $f^*(c)=|f'(c)|$, then there
is a $\delta>0$ such that

\Centerline{$a\le c-\delta\le c+\delta\le b$,}

\Centerline{$g(x)\ge g(c)-\epsilon\ge|f'(c)|-\epsilon$ whenever
$|x-c|\le\delta$,}

\Centerline{$|\Bover{f(x)-f(c)}{x-c}-f'(c)|\le\epsilon$ whenever
$|x-c|\le\delta$.}

\noindent Consider $x=c+\delta$.   Then $c<x\le b$ and

$$\eqalignno{\mu([f(a),f(x)]\setminus G)
&\le\mu([f(a),f(c)]\setminus G)+|f(x)-f(c)|\cr
&\le2\epsilon(c-a)+\int_a^cg+\epsilon(x-c)+|f'(c)|(x-c)\cr
&\le2\epsilon(c-a)+\int_a^cg
         +\epsilon(x-c)+\int_c^x(g+\epsilon)\cr
\noalign{\noindent (because $g(t)\ge |f'(c)|-\epsilon$ whenever
$c\le t\le x$)}
&=2\epsilon(x-a)+\int_a^xg,\cr}$$

\noindent so $x\in A$;  but $c$ is supposed to be an upper bound of
$A$.\ \Bang

Thus $c\in[a,b]\setminus F$.

\medskip

\quad{\bf (iii)} \Quer\ Now suppose, if possible, that
$c<b$.   We know that $f(c)\in G$, so there is an $\eta>0$ such that
$[f(c)-\eta,f(c)+\eta]\subseteq G$;  now there is a $\delta>0$
such that $|f(x)-f(c)|\le\eta$ whenever $x\in[a,b]$ and
$|x-c|\le\delta$.   Set $x=\min(c+\delta,b)$;  then $c<x\le b$ and
$[f(c),f(x)]\subseteq G$, so

\Centerline{$\mu([f(a),f(x)]\setminus G)
=\mu([f(a),f(c)]\setminus G)
\le 2\epsilon(c-a)+\int_a^cg
\le 2\epsilon(x-a)+\int_a^xg$}

\noindent and once again $x\in A$, even though $x>\sup A$.\ \Bang

\medskip

\quad{\bf (iv)} We conclude that $c=b$, so that $b\in A$.   But this
means that

$$\eqalign{f(b)-f(a)
&\le\mu([f(a),f(b)])
\le\mu([f(a),f(b)]\setminus G)+\mu G\cr
&\le 2\epsilon(b-a)+\int_a^bg+\epsilon
\le 2\epsilon(b-a)+\int_a^bf^*+\epsilon+\epsilon\cr
&=2\epsilon(1+b-a)+\int_a^bf^*.\cr}$$

\noindent As $\epsilon$ is arbitrary, $f(b)-f(a)\le\int_a^bf^*$, as
claimed.\ \Qed

\medskip

{\bf (c)} Similarly, or applying (b) to $-f$,
$f(a)-f(b)\le\int_a^bf^*$, so that $|f(b)-f(a)|\le\int_a^bf^*$.

Of course the argument applies equally to any subinterval of $[a,b]$, so
$|f(d)-f(c)|\le\int_c^df^*$ whenever
$a\le c\le d\le b$.   Now let $\epsilon>0$.   By 225A once more, there
is a $\delta>0$ such that $\int_Ef^*\le\epsilon$ whenever
$E\subseteq[a,b]$ and $\mu E\le\delta$.   Suppose that
$a\le a_1\le b_1\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta$.   Then

\Centerline{$\sum_{i=1}^n|f(b_i)-f(a_i)|
\le\sum_{i=1}^n\int_{a_i}^{b_i}f^*
=\int_{\bigcupop_{i\le n}[a_i,b_i]}f^*
\le\epsilon$.}

\noindent So $f$ is absolutely continuous, as claimed.
}%end of proof of 225K

\leader{225L}{Corollary} Let $[a,b]$ be a non-empty closed interval in
$\Bbb R$.   Let $f:[a,b]\to\Bbb R$ be a continuous function which is
differentiable on the open interval $\ooint{a,b}$.   If its derivative
$f'$ is integrable over $[a,b]$, then $f$ is absolutely continuous, and
$f(b)-f(a)=\int_a^bf'$.

\proof{ $f[\,[a,b]\setminus F]=\{f(a),f(b)\}$ is surely negligible, so
$f$ is absolutely continuous, by 225K;  consequently
$f(b)-f(a)=\int_a^bf'$, by 225E.
}%end of proof of 225L

\leader{225M}{Corollary} Let $[a,b]$ be a non-empty closed interval in
$\Bbb R$, and $f:[a,b]\to\Bbb R$ a continuous function.   Then $f$ is
absolutely continuous iff it is continuous and of bounded variation and
$f[A]$ is negligible for every negligible $A\subseteq[a,b]$.

\proof{{\bf (a)} Suppose that $f$ is absolutely continuous.   By
225C(a-b) it is continuous and of bounded variation, and by 225G we have
$f[A]$ negligible for every negligible $A\subseteq [a,b]$.

\medskip

{\bf (b)} So now suppose that $f$ satisfies the conditions.   Set
$F=\{x:x\in\ooint{a,b},\,f'(x)$ is defined$\}$.   By 224I once more,
$[a,b]\setminus F$ is negligible, so $f[\,[a,b]\setminus F]$ is
negligible.   Moreover, also by 224I, $f'$ is integrable over $[a,b]$
or $F$.   So the conditions of 225K are satisfied and $f$ is absolutely
continuous.
}%end of proof of 225M

\leader{225N}{The Cantor function}\cmmnt{ I should mention the
standard example of a continuous function of bounded variation which is
not absolutely continuous.}   Let $C\subseteq[0,1]$ be the Cantor
set\cmmnt{ (134G)}.
Recall that the `Cantor function' is a non-decreasing continuous
function $f:[0,1]\to[0,1]$ such that $f'(x)$ is defined and equal to
zero for every $x\in[0,1]\setminus C$,
but $f(0)=0<1=f(1)$\cmmnt{ (134H)}.   \cmmnt{Of
course} $f$ is of bounded variation and not absolutely continuous.   $C$
is negligible and $f[C]=[0,1]$ is not.   If $x\in C$, then for every
$n\in\Bbb N$ there is an interval of length $3^{-n}$, containing $x$, on
which $f$ increases by $2^{-n}$;  so $f$ cannot be differentiable at
$x$, and the set $F=\dom f'$ of 225K is precisely $[0,1]\setminus C$, so
that $f[\,[0,1]\setminus F]=[0,1]$.

\vleader{48pt}{225O}{Complex-valued functions}\cmmnt{ As usual, I
spell out
the results above in the forms applicable to complex-valued functions.

\medskip

} {\bf (a)} Let
$(X,\Sigma,\mu)$ be any measure space and $f$ an integrable
complex-valued
function defined on a conegligible subset of $X$.
Then for any $\epsilon>0$ there are a measurable set
$E$ of finite measure and a real number
$\delta>0$ such that $\int_F|f|\le\epsilon$ whenever $F\in\Sigma$ and
$\mu(F\cap E)\le\delta$.  \prooflet{(Apply 225A to $|f|$.)}

\spheader 225Ob If $[a,b]$ is a non-empty
closed interval in $\Bbb R$ and $f:[a,b]\to\Bbb C$ is a function, we say
that $f$ is {\bf absolutely continuous} if for every $\epsilon>0$ there
is a $\delta>0$ such that $\sum_{i=1}^n|f(b_i)-f(a_i)|\le\epsilon$
whenever $a\le a_1\le b_1\le a_2\le b_2\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\delta$.   Observe that $f$ is absolutely
continuous iff its real and imaginary parts are both absolutely
continuous.

\spheader 225Oc Let $[a,b]$ be a non-empty closed interval in
$\Bbb R$.

\quad(i) If $f:[a,b]\to\Bbb C$ is absolutely continuous it is of bounded
variation on $[a,b]$, so is differentiable almost everywhere in $[a,b]$,
and its derivative is integrable over $[a,b]$.

\quad(ii) If $f$, $g:[a,b]\to\Bbb C$ are absolutely continuous, so are
$f+g$ and $\zeta f$, for any $\zeta\in\Bbb C$, and $f\times g$.

\quad(iii) If $g:[a,b]\to[c,d]$ is monotonic and absolutely continuous,
and
$f:[c,d]\to\Bbb C$ is absolutely continuous, then $fg:[a,b]\to\Bbb C$ is
absolutely continuous.

\spheader 225Od Let $[a,b]$ be a non-empty closed interval in
$\Bbb R$ and $F:[a,b]\to\Bbb C$ a function.   Then the following are
equiveridical:

\quad(i) there is an integrable complex-valued function $f$ such that
$F(x)=F(a)+\int_a^xf$ for every $x\in[a,b]$;

\quad(ii) $\int_a^xF'$ exists and is
equal to $F(x)-F(a)$ for every $x\in[a,b]$;

\quad(iii) $F$ is absolutely continuous.

\cmmnt{\noindent (Apply 225E to the real and imaginary parts of $F$.)}

\spheader 225Oe Let $f$ be an integrable complex-valued function
on an interval $[a,b]\subseteq\Bbb R$, and $g:[a,b]\to\Bbb C$ an
absolutely continuous function.   Set $F(x)=\int_a^xf$ for $x\in[a,b]$.
Then

\Centerline{$\int_a^bf\times g
=F(b)g(b) - F(a)g(a) -\int_a^bF\times g'$.}

\cmmnt{\noindent (Apply 225F to the real and imaginary parts of $f$
and $g$.)}

\spheader 225Of Let $f$ be a continuous complex-valued function
on a closed interval $[a,b]\subseteq\Bbb R$, and suppose that $f$ is
differentiable at every point of the open interval $\ooint{a,b}$, with
$f'$ integrable over $[a,b]$.   Then $f$ is absolutely continuous.
\cmmnt{(Apply 225L to the real and imaginary parts of $f$.)}

\cmmnt{\spheader 225Og For a result corresponding to 225M, see 264Yp.}

\exercises{
\leader{225X}{Basic exercises (a)} 
%\spheader 225Xa
Show directly from the definition in
225B (without appealing to
225E) that any absolutely continuous real-valued function on a closed
interval $[a,b]$ is expressible as the difference of non-decreasing
absolutely continuous functions.
%225B

\spheader 225Xb Show directly from the definition in 225B and
the Mean Value Theorem (without appealing to 225K) that if a function
$f$ is continuous on a closed interval $[a,b]$, differentiable on the
open interval $\ooint{a,b}$, and has bounded derivative in
$\ooint{a,b}$, then $f$ is absolutely continuous, so that
$f(x)=f(a)+\int_a^xf'$ for every $x\in[a,b]$.
%225E

\spheader 225Xc Show that if $f:[a,b]\to\Bbb R$ is absolutely
continuous, then $\Var f=\int_a^b|f'|$.   \Hint{put 224I and
225E together.}
%225E

\spheader 225Xd Let $g:\Bbb R\to\Bbb R$ be a
non-decreasing function which is absolutely continuous on every bounded
interval; let $\mu_g$ be the associated Lebesgue-Stieltjes measure
(114Xa), and $\Sigma_g$ its domain.   Show that $\int_Eg'=\mu_gE$  for
any $E\in\Sigma_g$, if we allow $\infty$ as a value of the
integral.   \Hint{start with intervals $E$.}
%225E

\spheader 225Xe Let $g:[a,b]\to\Bbb R$ be a non-decreasing
absolutely continuous function, and $f:[g(a),g(b)]\to\Bbb R$ a
continuous function.   Show that $\int_{g(a)}^{g(b)}f(t)dt=
\int_a^bf(g(t))g'(t)dt$.   \Hint{set $F(x)=\int_{g(a)}^xf$,
$G=Fg$ and consider $\int_a^bG'(t)dt$.   See also 263J.}
%225E

\spheader 225Xf Suppose that $I\subseteq\Bbb R$ is any
non-trivial interval (bounded or unbounded, open, closed or half-open,
but not empty or a singleton), and $f:I\to\Bbb R$ a function.   Show
that
$f$ is absolutely continuous on every closed bounded subinterval of $I$
iff there is a function $g$ such that $\int_a^bg=f(b)-f(a)$ whenever
$a\le b$ in $I$, and in this case $g$ is integrable iff $f$ is of
bounded variation on $I$.
%225E

\spheader 225Xg Show that
$\biggerint_0^1\Bover{\ln x}{x-1}dx=\sum_{n=1}^{\infty}\Bover1{n^2}$.
\Hint{use 225F to find $\int_0^1x^n\ln x\,dx$, and recall that
$\bover1{1-x}=\sum_{n=0}^{\infty}x^n$ for $0\le x<1$.}
%225F

\spheader 225Xh(i) Show that $\int_0^1t^adt$ is finite for
every $a>-1$.   (ii) Show that $\int_1^{\infty}t^ae^{-t}dt$ is finite
for every $a\in\Bbb R$.   \Hint{show that there is an $M$ such
that $t^a\le Me^{t/2}$ for $t\ge 1$.}   (iii) Show that
$\Gamma(a)=\int_0^{\infty}t^{a-1}e^{-t}dt$ is defined for every $a>0$.
(iv) Show that $\Gamma(a+1)=a\Gamma(a)$ for every $a>0$.   (v) Show that
$\Gamma(n+1)=n!$ for every $n\in\Bbb N$.

($\Gamma$ is of course the {\bf gamma function}.)
%225F

\spheader 225Xi Show that if $b>0$ then
$\int_0^{\infty}u^{b-1}e^{-u^2/2}du=2^{(b-2)/2}\Gamma(\bover{b}2)$.
\Hint{consider $f(t)=t^{(b-2)/2}e^{-t}$, $g(u)=u^2/2$ in 225Xe.}
%225Xe, 225Xh, 225F

\spheader 225Xj Suppose that $f$, $g$ are
lower semi-continuous functions, defined on subsets of $\BbbR^r$,
and taking values in $\ocint{-\infty,\infty}$.   (i) Show that $f+g$,
$f\wedge g$ and $f\vee g$ are lower semi-continuous, and that
$\alpha f$ is lower
semi-continuous for every $\alpha\ge 0$.   (ii) Show that if $f$ and $g$
are non-negative, then $f\times g$ is lower semi-continuous.
(iii) Show that if $f$ is non-negative and $g$ is continuous, then
$f\times g$ is lower semi-continuous.   (iv) Show that if $f$ is
non-decreasing then the composition $fg$ is lower semi-continuous.
%225H

\spheader 225Xk Let $A$ be a non-empty family of lower
semi-continuous functions defined on subsets of $\BbbR^r$ and taking
values in $[-\infty,\infty]$.   Set $g(x)=\sup\{f(x):f\in A,\,x\in\dom
f\}$ for $x\in
D=\bigcup_{f\in A}\dom f$.   Show that $g$ is lower
semi-continuous.
%225H

\spheader 225Xl Let $f:[a,b]\to\Bbb R$ be an absolutely continuous
function, where $a\le b$.   (i) Show that $|f|:[a,b]\to\Bbb R$ is
absolutely continuous.   (ii)  Show that $gf$ is absolutely continuous
whenever $g:f[\,[a,b]\,]\to\Bbb R$ is absolutely continuous and
$g'$ is bounded.   (iii) Show that if $g:[a,b]\to\Bbb R$ is absolutely
continuous and $\inf_{x\in[a,b]}|g(x)|>0$ then
$f/g$ is absolutely continuous.
%225K

\spheader 225Xm Suppose that $f:[a,b]\to\Bbb R$ is continuous, and
differentiable at all but countably many
points of $[a,b]$.   Show that $f$ is absolutely continuous iff it is of
bounded variation.
%225K

\spheader 225Xn Let $f:\coint{0,\infty}\to\Bbb C$ be a function
which is absolutely continuous on $[0,a]$ for every
$a\in\coint{0,\infty}$ and has Laplace transform
$F(s)=\int_0^{\infty}e^{-sx}f(x)dx$ defined on $\{s:\Real s>S\}$.
Suppose also that $\lim_{x\to\infty}e^{-Sx}f(x)=0$.   Show
that $f'$ has Laplace transform $sF(s)-f(0)$ defined whenever $\Real
s>S$.   ({\it Hint\/}:  show that

\Centerline{$f(x)e^{-sx}-f(0)
=\int_0^x\Bover{d}{dt}(f(t)e^{-st})dt$}

\noindent for every $x\ge 0$.)
%225O 225F

\leader{225Y}{Further exercises (a)}
%\spheader 225Ya
Show that the composition of two absolutely
continuous functions need not be absolutely continuous.   \Hint{224Xb.}
%225B

\spheader 225Yb Let $f:[a,b]\to\Bbb R$ be a continuous function, where
$a<b$.   Set $G=\{x:x\in\ooint{a,b},\,\exists\,y\in\ocint{x,b}$ such
that $f(x)<f(y)\}$.   Show that $G$ is open and is expressible as a
disjoint union of intervals $\ooint{c,d}$ where $f(c)\le f(d)$.   Use
this to prove 225D without calling on Vitali's theorem.
%225D 2A2I

\spheader 225Yc Let $f:[a,b]\to\Bbb R$ be a function of bounded
variation and $\gamma>0$.   Show that there is an absolutely continuous
function $g:[a,b]\to\Bbb R$ such that $|g'(x)|\le\gamma$ wherever the
derivative is defined and $\{x:x\in[a,b],\,f(x)\ne g(x)\}$ has measure
at most $\bover1{\gamma}\Var f$.   \Hint{reduce to the case of
non-decreasing $f$.   Apply 225Yb to the function
$x\mapsto f(x)-\gamma x$ and show that $\gamma\mu G\le\Var f$.
Set $g(x)=f(x)$ for $x\in\ooint{a,b}\setminus G$.}
%225Yb 225D

\spheader 225Yd\dvAnew{2013}
Let $f:\Bbb R\to\Bbb R$ be a function which is absolutely
continuous on every bounded interval.   Show that
$\Var f\le\bover12\Var f'+\int|f|$.
%225E mt22bits

\spheader 225Ye Let $f$ be a non-negative measurable real-valued
function defined on a subset $D$ of $\BbbR^r$, where $r\ge 1$.   Show
that for any $\epsilon>0$ there is a lower semi-continuous
function $g:\BbbR^r\to[-\infty,\infty]$ such that $g(x)\ge f(x)$ for
every $x\in D$ and $\int_Dg-f\le\epsilon$.
%225I

\spheader 225Yf Let $f$ be a measurable real-valued function
defined on a subset $D$ of $\BbbR^r$, where $r\ge 1$.   Show that for
any $\epsilon>0$ there is a lower semi-continuous function
$g:\BbbR^r\to[-\infty,\infty]$ such that $g(x)\ge f(x)$ for every
$x\in D$ and
$\mu^*\{x:x\in D,\,g(x)>f(x)\}\le\epsilon$.   \Hint{134Yd, 134Fb.}
%225I

\spheader 225Yg(i) Show that if $f$ is a Lebesgue measurable real function
then all its Dini derivates are Lebesgue measurable.
(ii) Show that if $f$ is a Borel measurable real function
then all its Dini derivates are Borel measurable.
%225J mt22bits
}%end of exercises

\endnotes{
\Notesheader{225} There is a good deal more to say about absolutely
continuous functions;  I will return to the topic in the next section
and in Chapter 26.   I shall rarely make direct use of the results
from 225H on in their full strengths,
but it seems to me that this kind of investigation is
necessary for any clear picture of the relationships between such
concepts as absolute continuity and bounded variation.    Of course, in
order to apply these results, we do need a store of simple kinds
of absolutely continuous function, differentiable functions with
bounded derivative forming the most important class (225Xb).   A larger
family of the same kind is the class of `Lipschitz' functions
(262Bc).

The definition of `absolutely continuous function' is ordinarily set
out for closed bounded intervals, as in 225B.   The point is that for
other intervals the simplest generalizations of this formulation do not
seem quite appropriate.   In 225Xf I try to suggest the kind of demands
one might make on functions defined on other types of interval.

I should remark that the real prize is still not quite within our grasp.
I have been able to give a reasonably satisfactory formulation of simple
integration by parts (225F), at least for bounded intervals -- a further
limiting process is necessary to deal with unbounded intervals.   But a
companion method from advanced calculus, integration by substitution,
remains elusive.   The best I think we can do at this point is 225Xe,
which insists on a continuous integrand $f$.   It is the case that the
result is valid for general integrable $f$, but there are some further
subtleties to be mastered on the way;  the necessary ideas are given in
the much more general results 235A and 263D below, and applied to the
one-dimensional case in 263J.

On the way to the characterization of absolutely continuous functions in
225K, I find myself calling on one of the fundamental relationships
between Lebesgue measure and the topology of $\BbbR^r$ (225I).   The
technique here can be adapted to give many variations of the result;
see 225Ye-225Yf.   If you have not seen semi-continuous functions
before, 225Xj-225Xk give a partial idea of their properties.    In 225J
I give a fragment of `descriptive set theory', the study of the kinds of
set which can arise from the formulae of analysis.   These ideas too
will re-surface elsewhere (compare 225Yg and also the proof of 262M
below) and will be of great importance in Volumes 4 and 5.
}%end of notes

\discrpage

\frfilename{mt226.tex}
\versiondate{16.11.13}
\copyrightdate{2000}

\def\chaptername{The Fundamental Theorem of Calculus}
\def\sectionname{The Lebesgue decomposition of a function of bounded
variation}

\newsection{226}

I end this chapter with some notes on a method of analysing a general
function of bounded variation which may help to give a picture of what
such functions can be, though (apart from 226A)
it is hardly needed in this volume.

\leader{226A}{Sums over arbitrary index sets} \cmmnt{To get a full
picture of this
fragment of real analysis, a bit of preparation will be
helpful.   This concerns the notion of a sum over an arbitrary
index set, which I have rather been skirting around so far.

\header{226Aa}}{\bf (a)} If $I$ is any set and
$\langle a_i\rangle_{i\in I}$ any family in $[0,\infty]$, we set

\Centerline{$\sum_{i\in I}a_i
=\sup\{\sum_{i\in K}a_i:K\text{ is a finite subset of }I\}$,}

\noindent with the convention that $\sum_{i\in\emptyset}a_i=0$.
\cmmnt{(See 112Bd, 222Ba.)}   For general $a_i\in[-\infty,\infty]$, we
can set

\Centerline{$\sum_{i\in I}a_i=\sum_{i\in I}a_i^+-\sum_{i\in I}a_i^-$}

\noindent if this is defined in $[-\infty,\infty]$\cmmnt{, that is, at
least one of $\sum_{i\in I}a_i^+$, $\sum_{i\in I}a_i^-$ is finite},
where $a^+=\max(a,0)$ and $a^-=\max(-a,0)$ for each $a$.   If
$\sum_{i\in I}a_i$ is defined and finite, we say
that $\familyiI{a_i}$ is {\bf summable}.

\header{226Ab}{\bf (b)}\cmmnt{ Since this is a book on measure theory,
I will immediately describe the relationship between this kind of
summability
and an appropriate notion of integration.}   For any set $I$, we have
the corresponding `counting measure' $\mu$ on $I$\cmmnt{ (112Bd)}.
Every\cmmnt{ subset
of $I$ is measurable, so every} family $\langle a_i\rangle_{i\in I}$ of
real numbers is a measurable real-valued function on $I$.
A\cmmnt{ subset of $I$ has finite measure iff it is finite;
so a} real-valued
function $f$ on $I$ is `simple' if $K=\{i:f(i)\ne 0\}$ is finite.
\cmmnt{In this case,

\Centerline{$\int fd\mu=\sum_{i\in K}f(i)=\sum_{i\in I}f(i)$}

\noindent as defined in part (a).  The measure $\mu$ is semi-finite
(211Nc) so a non-negative function $f$ is integrable iff
$\int f=\sup_{\mu K<\infty}\int_Kf$ is finite (213B);   but of course
this supremum is precisely

\Centerline{$\sup\{\sum_{i\in K}f(i):K\subseteq I\text{ is finite}\}
=\sum_{i\in I}f(i)$.}


\noindent}%end of comment
Now a general function $f:I\to\Bbb R$ is integrable\cmmnt{ iff it is
measurable and $\int|f|d\mu<\infty$, that is,} iff
$\sum_{i\in I}|f(i)|<\infty$, and in this case

\dvro{\Centerline{$\int fd\mu=\sum_{i\in I}f(i)$,}}
{\Centerline{$\int fd\mu
=\int f^+d\mu-\int f^-d\mu
=\sum_{i\in I}f(i)^+-\sum_{i\in I}f(i)^-
=\sum_{i\in I}f(i)$,}}

\noindent\cmmnt{writing $f^{\pm}(i)=f(i)^{\pm}$ for each $i$.  }Thus
we have

\Centerline{$\sum_{i\in I}a_i=\int_Ia_i\mu(di)$,}

\noindent and the standard rules under which we allow $\infty$ as the
value of an integral\cmmnt{ (133A, 135F)} match\cmmnt{ well with} the
interpretations in (a) above.

\header{226Ac}{\bf (c)}\cmmnt{ Accordingly, and unsurprisingly, the
operation
of summation is a linear operation on the linear space of summable
families of real numbers.

}  I observe here that this notion of summability is `absolute';  a
family $\langle a_i\rangle_{i\in I}$ is summable iff it is absolutely
summable.   \cmmnt{This is necessary because it must also be
`unconditional';  we have no structure on an arbitrary set $I$ to guide
us to take the sum in any particular order.   See 226Xa.   In
particular, I distinguish between `$\sum_{n\in\Bbb N}a_n$', which in
this book will always be interpreted as in 226A above, and
`$\sum_{n=0}^{\infty}a_n$' which (if it makes a difference) should be
interpreted as $\lim_{m\to\infty}\sum_{n=0}^ma_n$.   So, for instance,
$\sum_{n=0}^{\infty}\Bover{(-1)^n}{n+1}=\ln 2$, while
$\sum_{n\in\Bbb N}\Bover{(-1)^n}{n+1}$ is undefined.   Of course
$\sum_{n=0}^{\infty}a_n=\sum_{n\in\Bbb N}a_n$ whenever the latter is
defined in $[-\infty,\infty]$.}

\header{226Ad}{\bf (d)}\cmmnt{ There is another, and very important,
approach
to the sum described here.}   If $\langle a_i\rangle_{i\in I}$ is an
(absolutely) summable family of real numbers, then for every
$\epsilon>0$ there is a finite $K\subseteq I$ such that
$\sum_{i\in I\setminus K}|a_i|\le\epsilon$.   \prooflet{\Prf\ This is
nothing but a
special case of 225A;  there is a set $K$ with $\mu K<\infty$ such that
$\int_{I\setminus K}|a_i|\mu(di)\le\epsilon$, but

\Centerline{$\int_{I\setminus K}|a_i|\mu(di)
=\sum_{i\in I\setminus K}|a_i|$.   \Qed}

\noindent}\cmmnt{(Of course there are `direct' proofs of this
result from
the definition in (a), not mentioning measures or integrals.   But
I think you will see that they rely on the same idea as that in the
proof of 225A.)  }Consequently, for any family
$\langle a_i\rangle_{i\in I}$ of real numbers and any $s\in\Bbb R$, the
following are equiveridical:

\inset{(i) $\sum_{i\in I}a_i=s$;}

\inset{(ii) for every $\epsilon>0$ there is a finite $K\subseteq I$ such
that $|s-\sum_{i\in J}a_i|\le\epsilon$ whenever $J$ is finite and
$K\subseteq J\subseteq I$.}

\noindent\prooflet{\Prf\ {\bf (i)$\Rightarrow$(ii)} Take $K$ such that
$\sum_{i\in
I\setminus K}|a_i|\le\epsilon$.   If $K\subseteq J\subseteq I$, then

\Centerline{$|s-\sum_{i\in J}a_i|=|\sum_{i\in I\setminus J}a_i|
\le\sum_{i\in I\setminus K}|a_i|\le\epsilon$.}

{\bf (ii)$\Rightarrow$(i)} Let $\epsilon>0$, and let $K\subseteq I$ be
as described in (ii).   If $J\subseteq I\setminus K$ is any finite set,
then set $J_1=\{i:i\in J,\,a_i\ge 0\}$, $J_2=J\setminus J_1$.   We have

$$\eqalign{\sum_{i\in J}|a_i|
&=|\sum_{i\in J_1\cup K}a_i-\sum_{i\in J_2\cup K}a_i|\cr
&\le|s-\sum_{i\in J_1\cup K}a_i|+|s-\sum_{i\in J_2\cup K}a_i|
\le 2\epsilon.\cr}$$

\noindent As $J$ is arbitrary, $\sum_{i\in I\setminus K}|a_i|\le
2\epsilon$ and

\Centerline{$\sum_{i\in I}|a_i|\le\sum_{i\in K}|a_i|+2\epsilon<\infty$.}

\noindent Accordingly $\sum_{i\in I}a_i$ is well-defined in $\Bbb R$.
Also

\Centerline{$|s-\sum_{i\in I}a_i|
\le|s-\sum_{i\in K}a_i|+|\sum_{i\in I\setminus K}a_i|
\le\epsilon+\sum_{i\in I\setminus K}|a_i|
\le 3\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\sum_{i\in I}a_i=s$, as required.
\Qed}

\cmmnt{In this way, we express $\sum_{i\in I}a_i$ directly as a limit;
we could write it as

\Centerline{$\sum_{i\in I}a_i=\lim_{K\uparrow I}\sum_{i\in K}a_i$,}

\noindent on the understanding that we look at finite sets $K$ in the
right-hand formula.
}%end of comment

\header{226Ae}{\bf (e)}\cmmnt{ Yet another approach is through the
following
fact.}   If $\sum_{i\in I}|a_i|<\infty$, then\cmmnt{ for any
$\delta>0$ the
set $\{i:|a_i|\ge\delta\}$ is finite, indeed can have at most
${1\over{\delta}}\sum_{i\in I}|a_i|$ members;  consequently}

\Centerline{$J=\{i:a_i\ne 0\}
=\bigcup_{n\in\Bbb N}\{i:|a_i|\ge 2^{-n}\}$}

\noindent is countable\cmmnt{ (1A1F)}.   If $J$ is finite,
then\cmmnt{ of course} $\sum_{i\in I}a_i=\sum_{i\in J}a_i$ reduces to a finite sum.   Otherwise, we can
enumerate $J$ as $\sequencen{j_n}$, and we shall have

\Centerline{$\sum_{i\in I}a_i=\sum_{i\in
J}a_i=\lim_{n\to\infty}\sum_{k=0}^na_{j_k}=\sum_{n=0}^{\infty}a_{j_n}$
\dvro{.}{}}

\noindent\cmmnt{(using (d) to reduce the sum $\sum_{i\in J}a_i$ to a
limit of
finite sums).  }Conversely, if $\langle a_i\rangle_{i\in I}$ is such
that there is a countably infinite $J\subseteq\{i:a_i\ne 0\}$ enumerated
as $\sequencen{j_n}$, and if $\sum_{n=0}^{\infty}|a_{j_n}|<\infty$, then
$\sum_{i\in I}a_i$ will be $\sum_{n=0}^{\infty}a_{j_n}$.

\spheader 226Af
%new 2008
\cmmnt{It will be useful later to have a fragment of general theory.}
Let $I$ and
$J$ be sets and $\langle a_{ij}\rangle_{i\in I,j\in J}$ a family in
$[0,\infty]$.   Then

\Centerline{$\sum_{(i,j)\in I\times J}a_{ij}
=\sum_{i\in I}(\sum_{j\in J}a_{ij})
=\sum_{j\in J}(\sum_{i\in I}a_{ij})$.}

\prooflet{\noindent\Prf\ (i) If
$\sum_{(i,j)\in I\times J}a_{ij}>u$, then there is a finite set
$M\subseteq I\times J$ such that
$\sum_{(i,j)\in M}a_{ij}>u$.   Now $K=\{i:(i,j)\in M\}$ and
$L=\{j:(i,j)\in M\}$ are finite, so

$$\eqalignno{\sum_{i\in I}\sum_{j\in J}a_{ij}
&\ge\sum_{i\in K}\sum_{j\in J}a_{ij}
\ge\sum_{i\in K}\sum_{j\in L}a_{ij}\cr
\displaycause{because $\sum_{j\in J}a_{ij}\ge\sum_{j\in L}a_{ij}$ for every
$i$}
&=\sum_{(i,j)\in K\times L}a_{ij}
\ge\sum_{(i,j)\in M}a_{ij}
>u.\cr}$$

\noindent As $u$ is arbitrary,
$\sum_{i\in I}\sum_{j\in J}a_{ij}\ge\sum_{(i,j)\in I\times J}a_{ij}$.
(ii) If $\sum_{i\in I}\sum_{j\in J}a_{ij}>u\ge 0$, there is a
non-empty finite
set $K\subseteq I$ such that $\sum_{i\in K}\sum_{j\in J}a_{ij}>u$.
Let $\epsilon\in\ooint{0,1}$ be such that
$\sum_{i\in K}\sum_{j\in J}a_{ij}>u+\epsilon$, and set
$\delta=\Bover{\epsilon}{\#(K)}$.   For each $i\in K$ set
$\gamma_i=\min(u+1,\sum_{j\in J}a_{ij})-\delta$;
then

\Centerline{$\epsilon+\sum_{i\in K}\gamma_i
=\sum_{i\in K}\min(u+1,\sum_{j\in J}a_{ij})
\ge\min(u+1,\sum_{i\in K}\sum_{j\in J}a_{ij})
>u+\epsilon$,}

\noindent so $\sum_{i\in K}\gamma_i>u$.   For each $i\in K$,
$\gamma_i<\sum_{j\in J}a_{ij}$, so there is a finite
$L_i\subseteq J$ such that $\sum_{j\in L_i}a_{ij}\ge\gamma_i$.
Set $M=\{(i,j):i\in K$, $j\in L_i\}$, so that $M$ is a finite subset of
$I\times J$;  then

\Centerline{$\sum_{(i,j)\in I\times J}a_{ij}
\ge\sum_{(i,j)\in M}a_{ij}
=\sum_{i\in K}\sum_{j\in L_i}a_{ij}
\ge\sum_{i\in K}\gamma_i
>u$.}

\noindent As $u$ is arbitrary,
$\sum_{(i,j)\in I\times J}a_{ij}\ge\sum_{i\in I}\sum_{j\in J}a_{ij}$ and
these two sums are equal.   (iii) Similarly,
$\sum_{(i,j)\in I\times J}a_{ij}=\sum_{j\in J}\sum_{i\in I}a_{ij}$.\ \Qed}

\leader{226B}{Saltus functions}\cmmnt{ Now we are ready for a special
type of
function of bounded variation on $\Bbb R$.}   Suppose that $a<b$ in
$\Bbb R$.

\header{226Ba}{\bf (a)} A (real) {\bf saltus function} on $[a,b]$ is
a function $F:[a,b]\to\Bbb R$ expressible in the form

\Centerline{$F(x)=\sum_{t\in\coint{a,x}}u_t+\sum_{t\in[a,x]}v_t$}

\noindent for $x\in [a,b]$, where $\langle
u_t\rangle_{t\in\coint{a,b}}$,
$\langle v_t\rangle_{t\in[a,b]}$ are real-valued families such that
$\sum_{t\in\coint{a,b}}|u_t|$ and
$\sum_{t\in[a,b]}|v_t|$ are finite.

\header{226Bb}{\bf (b)} For any function $F:[a,b]\to\Bbb R$ we can write

\Centerline{$F(x^+)=\lim_{y\downarrow x}F(y)$ if $x\in\coint{a,b}$ and
the limit exists,}

\Centerline{$F(x^-)=\lim_{y\uparrow x}F(y)$ if $x\in\ocint{a,b}$ and the
limit exists.}

\noindent\cmmnt{(I hope that this will not lead to confusion with the
alternative interpretation of $x^+$ as $\max(x,0)$.)  }Observe that if
$F$ is a saltus function, as defined in (b),
with associated families $\langle u_t\rangle_{t\in\coint{a,b}}$ and
$\langle v_t\rangle_{t\in[a,b]}$, then $v_a=F(a)$, $v_x=F(x)-F(x^-)$ for
$x\in\ocint{a,b}$ and $u_x=F(x^+)-F(x)$ for $x\in\coint{a,b}$.
\prooflet{\Prf\ Let
$\epsilon>0$.   As remarked in 226Ad, there is a finite
$K\subseteq[a,b]$ such that

\Centerline{$\sum_{t\in\coint{a,b}\setminus
K}|u_t|+\sum_{t\in[a,b]\setminus K}|v_t|\le\epsilon$.}

\noindent Given $x\in[a,b]$, let $\delta>0$ be such that
$[x-\delta,x+\delta]$ contains no point of $K$ except perhaps $x$.
In this case, if $\max(a,x-\delta)\le y<x$, we must have

$$\eqalign{|F(y)-(F(x)-v_x)|
&=|\sum_{t\in\coint{y,x}}u_t+\sum_{t\in\ooint{y,x}}v_t|\cr
&\le\sum_{t\in\coint{a,b}\setminus K}|u_t|
  +\sum_{t\in[a,b]\setminus K}|v_t|
\le\epsilon,\cr}$$

\noindent while if $x<y\le\min(b,x+\delta)$ we shall have

$$\eqalign{|F(y)-(F(x)+u_x)|
&=|\sum_{t\in\ooint{x,y}}u_t+\sum_{t\in\ocint{x,y}}v_t|\cr
&\le\sum_{t\in\coint{a,b}\setminus K}|u_t|
  +\sum_{t\in[a,b]\setminus K}|v_t|
\le\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, we get $F(x^-)=F(x)-v_x$ (if
$x>a$) and $F(x^+)=F(x)+u_x$ (if $x<b$).   \Qed}

\cmmnt{It follows that }$F$ is continuous at $x\in\ooint{a,b}$
iff $u_x=v_x=0$,
while $F$ is continuous at $a$ iff $u_a=0$ and $F$ is continuous at $b$
iff $v_b=0$.   In particular, $\{x:x\in[a,b],\,F\text{ is not continuous
at }x\}$ is countable\cmmnt{ (see 226Ae)}.

\header{226Bc}{\bf (c)} If $F$ is a saltus function defined on $[a,b]$,
with associated families $\langle u_t\rangle_{t\in\coint{a,b}}$ and
$\langle v_t\rangle_{t\in[a,b]}$, then $F$ is of bounded variation on
$[a,b]$, and

\Centerline{$\Var_{[a,b]}(F)
\le\sum_{t\in\coint{a,b}}|u_t|+\sum_{t\in\ocint{a,b}}|v_t|$.}

\noindent\prooflet{\Prf\ If $a\le x<y\le b$, then

\Centerline{$F(y)-F(x)=u_x+\sum_{t\in\ooint{x,y}}(u_t+v_t)+v_y$,}

\noindent so

\Centerline{$|F(y)-F(x)|
\le\sum_{t\in\coint{x,y}}|u_t|+\sum_{t\in\ocint{x,y}}|v_t|$.}

\noindent If $a\le a_0\le a_1\le\ldots\le a_n\le b$, then

$$\eqalign{\sum_{i=1}^n|F(a_i)-F(a_{i-1})|
&\le\sum_{i=1}^n\bigl(\sum_{t\in\coint{a_{i-1},a_i}}|u_t|
   +\sum_{t\in\ocint{a_{i-1},a_i}}|v_t|\bigr)\cr
&\le\sum_{t\in\coint{a,b}}|u_t|+\sum_{t\in\ocint{a,b}}|v_t|.\cr}$$

\noindent Consequently

\Centerline{$\Var_{[a,b]}(F)
\le\sum_{t\in\coint{a,b}}|u_t|+\sum_{t\in\ocint{a,b}}|v_t|<\infty$. \Qed}}

\header{226Bd}{\bf (d)} The inequality in (c) is actually an equality.
\prooflet{To see this, note first that if $a\le x<y\le b$, then
$\Var_{[x,y]}(F)\ge|u_x|+|v_y|$.   \Prf\ I noted in (b) that
$u_x=\lim_{t\downarrow x}F(t)-F(x)$ and
$v_y=F(y)-\lim_{t\uparrow y}F(t)$.
So, given $\epsilon>0$, we can find $t_1$, $t_2$ such that
$x<t_1\le t_2<y$ and

\Centerline{$|F(t_1)-F(x)|\ge|u_x|-\epsilon$,
\quad$|F(y)-F(t_2)|\ge|v_y|-\epsilon$.}

\noindent Now

\Centerline{$\Var_{[x,y]}(F)
\ge|F(t_1)-F(x)|+|F(t_2)-F(t_1)|+|F(y)-F(t_2)|
\ge|u_x|+|v_y|-2\epsilon$.}

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

Now, given $a\le t_0<t_1<\ldots<t_n\le b$, we must have

$$\eqalignno{\Var_{[a,b]}(F)
&\ge\sum_{i=1}^n\Var_{[t_{i-1},t_i]}(F)\cr
\noalign{\noindent(using 224Cc)}
&\ge\sum_{i=1}^n|u_{t_{i-1}}|+|v_{t_i}|.\cr}$$

\noindent As $t_0,\ldots,t_n$ are arbitrary,

\Centerline{$\Var_{[a,b]}(F)
\ge\sum_{t\in\coint{a,b}}|u_t|+\sum_{t\in\ocint{a,b}}|v_t|$,}

\noindent as required.}

\header{226Be}{\bf (e)} Because a saltus function is of bounded
variation\cmmnt{ ((c) above)}, it is differentiable almost
everywhere\cmmnt{ (224I)}.
In fact its derivative is zero almost everywhere.   \prooflet{\Prf\ Let
$F:[a,b]\to\Bbb R$ be a saltus function, with associated families
$\langle u_t\rangle_{t\in\coint{a,b}}$, $\langle
v_t\rangle_{t\in[a,b]}$.
Let $\epsilon>0$.   Let $K\subseteq[a,b]$ be a finite set such that

\Centerline{$\sum_{t\in\coint{a,b}\setminus
K}|u_t|+\sum_{t\in[a,b]\setminus K}|v_t|\le\epsilon$.}

\noindent Set

$$\eqalign{u'_t
&=u_t\text{ if }t\in\coint{a,b}\cap K,\cr
&=0\text{ if }t\in\coint{a,b}\setminus K,\cr
v'_t&=v_t\text{ if }t\in K,\cr
&=0\text{ if }t\in[a,b]\setminus K,\cr
u''_t&=u_t-u'_t\text{ for }t\in\coint{a,b},\cr
v''_t&=v_t-v'_t\text{ for }t\in[a,b].\cr}$$

\noindent Let $G$, $H$ be the saltus functions corresponding to
$\langle u'_t\rangle_{t\in\coint{a,b}}$,
$\langle v'_t\rangle_{t\in[a,b]}$ and
$\langle u''_t\rangle_{t\in\coint{a,b}}$
$\langle v''_t\rangle_{t\in[a,b]}$, so that $F=G+H$.   Then $G'(t)=0$
for every $t\in\ooint{a,b}\setminus K$, since $\ooint{a,b}\setminus K$
comprises a finite number of open intervals on each of which $G$ is
constant.   So $G'=0$ a.e.\ and $F'\eae H'$.   On the other hand,

\Centerline{$\int_a^b|H'|\le\Var_{[a,b]}(H)
=\sum_{t\in\coint{a,b}\setminus K}|u_t|
  +\sum_{t\in\ocint{a,b}\setminus K}|v_t|
\le\epsilon$,}

\noindent using 224I and (d) above.   So

\Centerline{$\int_a^b|F'|=\int_a^b|H'|\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\int_a^b|F'|=0$ and $F'=0$ a.e.,
as claimed.   \Qed}

\leader{226C}{The Lebesgue decomposition of a function of bounded
variation}  Take $a$, $b\in\Bbb R$ with $a<b$.

\header{226Ca}{\bf (a)} If $F:[a,b]\to\Bbb R$ is non-decreasing, set
$v_a=0$, $v_t=F(t)-F(t^-)$ for $t\in\ocint{a,b}$, $u_t=F(t^+)-F(t)$ for
$t\in\coint{a,b}$\cmmnt{, defining $F(t^+)$, $F(t^-)$ as in 226Bb}.
Then all the $v_t$, $u_t$ are
non-negative, and\cmmnt{ if $a<t_0<t_1<\ldots<t_n< b$, then

\Centerline{$\sum_{i=0}^n(u_{t_i}+v_{t_i})
=\sum_{i=0}^n(F(t_i^+)-F(t_i^-))\le F(b)-F(a)$.}

\noindent   Accordingly}
$\sum_{t\in\coint{a,b}}u_t$ and $\sum_{t\in[a,b]}v_t$ are both finite.
Let $F_p$ be the corresponding saltus function\cmmnt{, as defined in
226Ba, so that

\Centerline{$F_p(x)
=F(a^+)-F(a)+\sum_{t\in\ooint{a,x}}(F(t^+)-F(t^-))+F(x)
-F(x^-)$}

\noindent if $a<x\le b$}.   \cmmnt{If $a\le x<y\le b$ then

$$\eqalign{F_p(y)-F_p(x)
&=F(x^+)-F(x)+\sum_{t\in\ooint{x,y}}(F(t^+)-F(t^-))+F(y)-F(y^-)\cr
&\le F(y)-F(x)\cr}$$

\noindent because if $x=t_0<t_1<\ldots<t_n<t_{n+1}=y$ then

$$\eqalign{F(x^+)-F(x)
&+\sum_{i=1}^{n}(F(t_i^+)-F(t_i^-))+F(y)-F(y^-)\cr
&=F(y)-F(x)-\sum_{i=1}^{n+1}(F(t_i^-)-F(t_{i-1}^+))
\le F(y)-F(x).\cr}$$

\noindent Accordingly both }$F_p$ and $F_c=F-F_p$ are
non-decreasing.   \cmmnt{Also, because

\Centerline{$F_p(a)=0=v_a$,}

\Centerline{$F_p(t)-F_p(t^-)=v_t=F(t)-F(t^-)$ for
$t\in\ocint{a,b}$,}

\Centerline{$F_p(t^+)-F_p(t)=u_t=F(t^+)-F(t)$ for
$t\in\coint{a,b}$,}

\noindent we shall have

\Centerline{$F_c(a)=F(a)$,}

\Centerline{$F_c(t)=F_c(t^-)$ for
$t\in\ocint{a,b}$,}

\Centerline{$F_c(t)=F_c(t^+)$ for $t\in\coint{a,b}$,}

\noindent and }$F_c$ is continuous.

Clearly this expression of $F=F_p+F_c$ as the sum of a saltus function
and a continuous function is unique, except that we can freely add a
constant to one if we subtract it from the other.

\header{226Cb}{\bf (b)}\cmmnt{ Still taking $F:[a,b]\to\Bbb R$ to be
non-decreasing, we know that $F'$ is integrable (222C);
moreover,
$F'\eae F'_c$, by 226Be.}   Set $F_{ac}(x)=F(a)+\int_a^xF'$
for each $x\in[a,b]$.   \cmmnt{We have

\Centerline{$F_{ac}(y)-F_{ac}(x)=\int_x^yF_c'\le F_c(y)-F_c(x)$}

\noindent for $a\le x\le y\le b$ (222C again), so} $F_{cs}=F_c-F_{ac}$
is still non-decreasing;  \cmmnt{$F_{ac}$ is continuous (225A),
so }$F_{cs}$ is
continuous;  \cmmnt{$F_{ac}'\eae F'\eae F_c'$ (222E),
so }$F_{cs}'=0$ a.e.

Again, the expression of $F_c=F_{ac}+F_{cs}$ as the sum of an absolutely
continuous function and a function with zero derivative almost
everywhere is unique, except for the possibility of moving a constant
from one to the other\cmmnt{, because two absolutely continuous
functions whose
derivatives are equal almost everywhere must differ by a constant
(225D)}.

\header{226Cc}{\bf (c)} Putting\cmmnt{ all} these together:  if
$F:[a,b]\to\Bbb R$ is any
non-decreasing function, it is expressible as $F_p+F_{ac}+F_{cs}$, where
$F_p$ is a saltus function, $F_{ac}$ is absolutely continuous, and
$F_{cs}$ is continuous and differentiable, with zero derivative, almost
everywhere;  all three components are non-decreasing;  and the
expression is unique if we say that $F_{ac}(a)=F(a)$ and
$F_p(a)=F_{cs}(a)=0$.

The Cantor function $f:[0,1]\to[0,1]$\cmmnt{ (134H)} is continuous and
$f'=0$ a.e.\cmmnt{ (134Hb)}, so $f_p=f_{ac}=0$ and $f=f_{cs}$.
Setting $g(x)=\bover12(x+f(x))$ for $x\in[0,1]$\cmmnt{, as in 134I},
we get $g_p(x)=0$, $g_{ac}(x)={x\over 2}$ and
$g_{cs}(x)={1\over 2}f(x)$.

\header{226Cd}{\bf (d)} Now suppose that $F:[a,b]\to\Bbb R$ is of
bounded variation.
Then it is expressible as a difference $G-H$ of
non-decreasing functions\cmmnt{ (224D)}.   So writing $F_p=G_p-H_p$,
etc., we can express $F$ as a sum $F_p+F_{cs}+F_{ac}$, where $F_p$ is a
saltus function, $F_{ac}$ is absolutely continuous, $F_{cs}$ is
continuous, $F_{cs}'=0$ a.e., $F_{ac}(a)=F(a)$ and $F_{cs}(a)=F_{p}(a)=0$.
Under these conditions the expression is unique\dvro{.}{, because (for
instance) $F_p(t^+)-F_p(t)=F(t^+)-F(t)$ for $t\in\coint{a,b}$, while
$F_{ac}'\eae(F-F_p)'\eae F'$.}

This is a {\bf Lebesgue decomposition} of the function $F$.
\cmmnt{(I have to say `a' Lebesgue decomposition because of course
the assignments $F_{ac}(a)=F(a)$, $F_{p}(a)=F_{cs}(a)=0$ are arbitrary.)}   I will call $F_p$ the {\bf saltus part} of $F$.

\leader{226D}{Complex functions} \cmmnt{The modifications needed to
deal with complex functions are elementary.

\medskip

}{\bf (a)} If $I$ is any set and
$\langle a_j\rangle_{j\in I}$ is a family of complex numbers, then the
following are equiveridical:

\inset{(i) $\sum_{j\in I}|a_j|<\infty$;}

\inset{(ii) there is an $s\in\Bbb C$ such that for every $\epsilon>0$
there is a finite $K\subseteq I$ such that
$|s-\sum_{j\in J}a_j|\le\epsilon$ whenever $J$ is finite and
$K\subseteq J\subseteq I$.}

\noindent In this case

\Centerline{$s=\sum_{j\in I}\Real(a_j)+i\sum_{j\in
I}\Imag(a_j)=\int_Ia_j\mu(dj)$,}

\noindent where $\mu$ is counting measure on $I$, and we write
$s=\sum_{j\in I}a_j$.

\header{226Db}{\bf (b)} If $a<b$ in $\Bbb R$, a complex {\bf saltus
function} on $[a,b]$ is a function $F:[a,b]\to\Bbb C$ expressible in the
form

\Centerline{$F(x)=\sum_{t\in\coint{a,x}}u_t+\sum_{t\in[a,x]}v_t$}

\noindent for $x\in [a,b]$, where $\langle
u_t\rangle_{t\in\coint{a,b}}$,
$\langle v_t\rangle_{t\in[a,b]}$ are complex-valued families such that
$\sum_{t\in\coint{a,b}}|u_t|$ and
$\sum_{t\in[a,b]}|v_t|$ are finite\cmmnt{;  that is, if the real and
imaginary parts of $F$ are saltus functions}.   In this case $F$ is
continuous except at countably many points and differentiable, with zero
derivative, almost everywhere in $[a,b]$, and

\Centerline{$u_x=\lim_{t\downarrow x}F(t)-F(x)$ for every
$x\in\coint{a,b}$,}

\Centerline{$v_x=\lim_{t\uparrow x}F(x)-F(t)$ for every
$x\in\ocint{a,b}$\dvro{.}{}}

\noindent\cmmnt{(apply the results of 226B to the real and imaginary
parts of $F$).   }$F$ is of bounded variation, and its variation is

\Centerline{$\Var_{[a,b]}(F)
=\sum_{t\in\coint{a,b}}|u_t|+\sum_{t\in\ocint{a,b}}|v_t|$\dvro{.}{}}

\cmmnt{\noindent (repeat the arguments of 226Bc-d).}

\header{226Dc}{\bf (c)} If $F:[a,b]\to\Bbb C$ is a function of bounded
variation, where $a<b$ in $\Bbb R$, it is uniquely expressible as
$F=F_p+F_{cs}+F_{ac}$, where $F_p$ is a saltus function, $F_{ac}$ is
absolutely continuous, $F_{cs}$ is continuous and has zero derivative
almost everywhere, and $F_{ac}(a)=F(a)$, $F_p(a)=F_{cs}(a)=0$.
\cmmnt{(Apply
226C to the real and imaginary parts of $F$.)}

\leader{226E}{}\cmmnt{ As an elementary exercise in the language of
226A, I interpolate a version of a theorem of B.Levi
%Levi 1906
which is sometimes useful.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space, $I$
a {\it countable} set, and $\familyiI{f_i}$ a family of $\mu$-integrable
real- or complex-valued functions such that $\sum_{i\in I}\int|f_i|d\mu$
is finite.   Then
$f(x)=\sum_{i\in I}f_i(x)$ is defined almost everywhere and
$\int fd\mu=\sum_{i\in I}\int f_id\mu$.

\proof{ If $I$ is finite this is elementary.   Otherwise, since there
must be a bijection between $I$ and $\Bbb N$, we may take it that
$I=\Bbb N$.   Setting $g_n=\sum_{i=0}^n|f_i|$ for each $n$, we have a
non-decreasing sequence $\sequencen{g_n}$ of integrable functions such
that $\int g_n\le\sum_{i\in\Bbb N}\int|f_i|$ for every $n$, so that
$g=\sup_{n\in\Bbb N}g_n$ is integrable, by B.Levi's theorem as stated in
123A.   In particular, $g$ is finite almost everywhere.   Now if
$x\in X$ is such that $g(x)$ is defined and finite,
$\sum_{i\in J}|f_i(x)|\le g(x)$ for every finite $J\subseteq\Bbb N$, so
$\sum_{i\in\Bbb N}|f_i(x)|$ and $\sum_{i\in\Bbb N}f_i(x)$ are defined.
In this case, of course,
$\sum_{i\in\Bbb N}f_i(x)=\lim_{n\to\infty}\sum_{i=0}^nf_i(x)$.   But
$|\sum_{i=0}^nf_i|\leae g$ for each $n$, so Lebesgue's Dominated
Convergence Theorem tells us that

\Centerline{$\int\sum_{i\in\Bbb N}f_i
=\lim_{n\to\infty}\int\sum_{i=0}^nf_i
=\lim_{n\to\infty}\sum_{i=0}^n\int f_i
=\sum_{i\in\Bbb N}\int f_i$.}
}%end of proof of 226E

\exercises{
\leader{226X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 226Xa
Suppose that $I$ and $J$ are sets and that
$\familyiI{a_i}$ is a summable family of real numbers.   (i) Show that
if $f:J\to I$ is injective then $\family{j}{J}{a_{f(j)}}$ is summable.
(ii) Show that if $g:I\to J$ is any function, then
$\sum_{j\in J}\sum_{i\in g^{-1}[\{j\}]}a_i$ is defined and equal to
$\sum_{i\in I}a_i$.
%226A

\sqheader 226Xb A {\bf step-function} on an
interval $[a,b]$ is a function $F$
such that, for suitable $t_0,\ldots,t_n$ with $a=t_0\le\ldots\le t_n=b$,
$F$ is constant on each interval $\ooint{t_{i-1},t_i}$.   Show that
$F:[a,b]\to\Bbb R$ is a saltus function iff for every $\epsilon>0$ there
is a step-function $G:[a,b]\to\Bbb R$ such that
$\Var_{[a,b]}(F-G)\le\epsilon$.
%226B

\spheader 226Xc Let $F$, $G$ be real-valued functions of bounded
variation defined on an interval $[a,b]\subseteq\Bbb R$.   Show that, in
the language of 226C,

\Centerline{$(F+G)_p=F_p+G_p$,\quad $(F+G)_c=F_c+G_c$,}

\Centerline{$(F+G)_{cs}=F_{cs}+G_{cs}$,
\quad$(F+G)_{ac}=F_{ac}+G_{ac}$.}
%226C

\sqheader 226Xd Let $F$ be a real-valued function of bounded
variation on an interval $[a,b]\subseteq\Bbb R$.   Show that, in the
language of 226C,

\Centerline{$\Var_{[a,b]}(F)
=\Var_{[a,b]}(F_p)+\Var_{[a,b]}(F_c)
=\Var_{[a,b]}(F_p)+\Var_{[a,b]}(F_{cs})+\Var_{[a,b]}(F_{ac})$.}
%226C

\spheader 226Xe Let $F$ be a real-valued function of bounded
variation on an interval $[a,b]\subseteq\Bbb R$.   Show that $F$ is
absolutely continuous iff $\Var_{[a,b]}(F)=\int_a^b|F'|$.
%226Xd

\spheader 226Xf Consider the function $g$ of 134I/226Cc.   Show
that $g^{-1}:[0,1]\to[0,1]$ is differentiable almost everywhere in
$[0,1]$, and find $\mu\{x:(g^{-1})'(x)\le a\}$ for each $a\in\Bbb R$.
%226C

\sqheader 226Xg(i) Show that a continuous bijection $f:[0,1]\to[0,1]$ is either
strictly increasing or strictly decreasing, and that its inverse is
continuous.   (ii) Show that if
$f:[0,1]\to[0,1]$ is a continuous bijection, then $f'=0$ a.e.\ in $[0,1]$
iff there is a
conegligible set $E\subseteq[0,1]$ such that $f[E]$ is negligible, and that in
this case $f^{-1}$ has the same property.
(iii) Construct a function satisfying the conditions of (ii).
\Hint{try $f=\sum_{n=0}^{\infty}2^{-n-1}f_n$ where each $f_n$ is a variation on
the Cantor function.}
(iv) Repeat (iii) with $f=f^{-1}$.

\leader{226Y}{Further exercises (a)}
%\spheader 226Ya
Show that a set $I$ is countable iff there is a summable
family $\familyiI{a_i}$ of non-zero real numbers.
%226A

\spheader 226Yb
Explain modifications which might be appropriate in the description
of the Lebesgue decomposition of a function of bounded variation
if we wish to consider functions on open or
half-open intervals, including unbounded intervals.
%226C

\spheader 226Yc Suppose that $F:[a,b]\to\Bbb R$ is a function of bounded
variation, and set $h(y)=\#(F^{-1}[\{y\}])$ for $y\in\Bbb R$.   Show
that $\int h=\Var_{[a,b]}(F_c)$, where $F_c$ is the `continuous part'
of $F$ as defined in 226Ca/226Cd.
%226C

\spheader 226Yd Suppose that $a<b$ in $\Bbb R$, and that
$F:[a,b]\to\Bbb R$ is a function of bounded variation;  let $F_p$ be its saltus part.   Show that
$|F(b)-F(a)|\le\mu F[\,[a,b]\,]+\Var_{[a,b]}F_p$, where $\mu$ is Lebesgue measure on $\Bbb R$.
%226C
}%end of exercises


\endnotes{
\Notesheader{226} In 232I and 232Yh below I will revisit these ideas,
linking them to a decomposition of the Lebesgue-Stieltjes measure
corresponding to a non-decreasing real function, and thence to more
general measures.   All this work is peripheral to the main concerns of
this volume, but I think it is illuminating, and certainly it is part of
the basic knowledge assumed of anyone working in real analysis.
}%end of notes

\frnewpage

\ifresultsonly\else
\frfilename{mt22conc.tex}
\versiondate{17.8.15}
\copyrightdate{2015}

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}

\def\chaptername{The Fundamental Theorem of Calculus}
\def\sectionname{Concordance}

\newsection{}

I list here the section and paragraph numbers which have (to my 
knowledge) appeared in print in references to this chapter, and which 
have since been changed.

\medskip

\leader{2{}25Xi}{} This exercise, referred to in 
%2012 372Yh 2002 2004 372Yf
the 2002, 2004 and 2012 printings of Volume 3, is now 225Xg.

\leader{2{}25Xj}{} This exercise, referred to in 
%2013 455Xj 2006 2003 455Xe
the 2003, 2006 and 2013 printings of Volume 4, is now 225Xh.

\leaveitout{\leader{2{}2}{}

\leader{2{}2}{}
}

\fi
\references{
\frfilename{mt2r.tex}
\versiondate{17.12.12}
\copyrightdate{2000}

% except for references in ...X., ...Y., ...Z. give locations in form
%  342L  rather than  342Lb
%  \S511 {\it notes\/}, \S512 {\it intro.}]

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

\centerline{\bf References for Volume 2}

\medskip

%A

Alexits G.\ [78] (ed.) {\it Fourier Analysis and Approximation Theory.}
North-Holland, 1978 (Colloq.\ Math.\ Soc.\ Janos Bolyai 19).

Antonov N.Yu.\ [96] `Convergence of Fourier series', East J.\ Approx.\ 7 (1996) 187-196.
\cmmnt{[\S286 {\it notes\/}.]}

Arias de Reyna J.\ [02] {\it Pointwise Convergence of Fourier Series.} Springer, 2002 (Lecture Notes in Mathematics 1785).
\cmmnt{[\S286 {\it notes\/}.]}

\medskip%B

Baker R.\ [04] `{}``Lebesgue measure'' on $\Bbb R^{\infty}$, II',
Proc.\ Amer.\ Math.\ Soc.\ 132 (2004) 2577-2591.
\cmmnt{[254Yb.]}

Bergelson V., March P.\ \& Rosenblatt J.\ [96] (eds.) {\it Convergence
in Ergodic Theory and Probability.}  de Gruyter, 1996.

\indexheader{Bogachev}
Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.

du Bois-Reymond P.\ [1876] `Untersuchungen \"uber die Convergenz und
Divergenz der Fouriersche Darstellungformeln', Abh.\ Akad.\ M\"unchen 12
(1876) 1-103.   \cmmnt{[\S282 {\it notes}.]}

Bourbaki N.\ [66] {\it General Topology.}  Hermann/Addison-Wesley, 1968.
\cmmnt{[2A5F.]}

Bourbaki N.\ [87] {\it Topological Vector Spaces.}  Springer, 1987.
\cmmnt{[2A5E.]}

\medskip%C

Carleson L.\ [66] `On convergence and growth of partial sums of
Fourier series', Acta Math.\ 116 (1966) 135-157.  \cmmnt{[\S282 {\it
notes}, \S286 {\it intro.}, \S286 {\it notes}.]}

Clarkson J.A.\ [1936] `Uniformly convex spaces', Trans.\ Amer.\ Math.\ Soc.\
40 (1936) 396-414.
\cmmnt{[244O.]}

\medskip%D

Defant A.\ \& Floret K.\ [93] {\it Tensor Norms and Operator Ideals},
North-Holland, 1993.   \cmmnt{[\S253 {\it notes}.]}

Doob J.L.\ [53] {\it Stochastic Processes.}  Wiley, 1953.
%used for 275Ye

Dudley R.M.\ [89] {\it Real Analysis and Probability.} Wadsworth \&
Brooks/Cole, 1989.   \cmmnt{[\S282 {\it notes}.]}

Dunford N.\ \& Schwartz J.T.\ [57] {\it Linear Operators I.}  Wiley,
1957 (reprinted 1988).  \cmmnt{[\S244 {\it notes}, 2A5J.]}

\medskip%E

Enderton H.B.\ [77] {\it Elements of Set Theory.}   Academic, 1977.
\cmmnt{[\S2A1.]}

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).   \cmmnt{[2A5F.]}

Etemadi N.\ [96] `On convergence of partial sums of independent random
variables', pp.\ 137-144 in {\smc Bergelson March \& Rosenblatt 96}.
\cmmnt{[272V.]}

Evans L.C.\ \& Gariepy R.F.\ [92] {\it Measure Theory and Fine
Properties of Functions.}   CRC Press, 1992.
\cmmnt{[263Ec, \S265 {\it notes}.]}
%Kirszbraun's theorem mentioned but not proved

\medskip%F

Federer H.\ [69] {\it Geometric Measure Theory.}  Springer, 1969
(reprinted 1996).
\cmmnt{[262C, 263Ec, \S264 {\it notes}, \S265 {\it notes},
\S266 {\it notes}.]}

Feller W.\ [66] {\it An Introduction to Probability Theory and its
Applications,} vol.\ II.   Wiley, 1966.  \cmmnt{[Chap.\ 27 {\it
intro.}, 274H, 275Xc, 285N.]}

Fremlin D.H.\ [74] {\it Topological Riesz Spaces and
Measure Theory.}   Cambridge U.P., 1974.
\cmmnt{[\S232 {\it notes}, 241F, \S244 {\it notes}, \S245 {\it notes},
\S247 {\it notes}.]}

Fremlin D.H.\ [93] `Real-valued-measurable cardinals', pp.\ 151-304 in
{\smc Judah 93}.  \cmmnt{[232H.]}

\medskip%G\medskip%H

Haimo D.T.\ [67] (ed.) {\it Orthogonal Expansions and their Continuous
Analogues.}   Southern Illinois University Press, 1967.

Hall P.\ [82] {\it Rates of Convergence in the Central Limit Theorem.}
Pitman, 1982.  \cmmnt{[274H.]}

Halmos P.R.\ [50] {\it Measure Theory.}  Van Nostrand, 1950.
\cmmnt{[\S251 {\it notes}, \S252 {\it notes}, 255Yn.]}

Halmos P.R.\ [60] {\it Naive Set Theory.} Van Nostrand, 1960.
\cmmnt{[\S2A1.]}

Hanner O.\ [56] `On the uniform convexity of $L^p$ and $l^p$',
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\cmmnt{[\S2A1.]}

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\medskip%I\medskip%J

Jorsb{\o}e O.G.\ \& Mejlbro L.\ [82] {\it The Carleson-Hunt Theorem on
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\medskip%K

\indexheader{Kelley}
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\medskip%L

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%expanded from 
% Publ.\ Mat.\ 48 (2004) 251-307
\cmmnt{[\S286 {\it notes}.]}

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\cmmnt{[241F.]}


\medskip%M

Mozzochi C.J.\ [71] {\it On the Pointwise Convergence of Fourier
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\medskip%N

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%anon04.PDF

\medskip%O\medskip%P\medskip%Q\medskip%R

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\medskip%S

%Saks S.\ [1924] 'Sur les nombres d\'eriv\'es des fonctions', 
%  Fundamenta Math.\ 5 (1924) 98-104.
% D-Y-S theorem
% books/saks.24.pdf   http://matwbn.icm.edu.pl/ksiazki/fm/fm5/fm5112.pdf

Saks S.\ [1924] 'Sur les nombres d\'eriv\'es des fonctions', Fund

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\medskip%T\medskip%U\medskip%V\medskip%W\medskip%Y\medskip%Z

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Zygmund A. [59] {\it Trigonometric Series.}   Cambridge U.P, 1959.
\cmmnt{[\S244 {\it notes}, \S282 {\it notes}, 284Xk.]}

\frnewpage

}%end of references

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