% 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{mt26.tex}
\versiondate{5.9.03/24.7.04}
\copyrightdate{2004}

\def\diam{\mathop{\text{diam}}}
\def\dist{\mathop{\text{dist}}}

\def\chaptername{Change of Variable in the Integral}
\newchapter{26}
\def\chaptername{Change of variable in the integral}

I suppose most courses on basic calculus still devote a
substantial amount of time to practice in the techniques of integrating
standard functions.   Surely the most powerful single technique is that
of substitution:  replacing $\int g(y)dy$ by $\int g(\phi(x))\phi'(x)dx$
for an appropriate function $\phi$.   At this level one usually
concentrates on the skills of guessing at appropriate $\phi$ and getting
the formulae right.   I will not address such questions here, except for
rare special cases;  in this book I am concerned rather with validating
the process.   For functions of one variable, it can usually be
justified by an appeal to the Fundamental Theorem of Calculus, 
and for any particular case I would normally
go first to \S225 in the hope that the results there would
cover it.   But for functions of two or
more variables some much deeper ideas are necessary.

I have already treated the general problem of
integration-by-substitution in abstract measure spaces in \S235.
There I described conditions under which $\int g(y)dy=\int
g(\phi(x))J(x)dx$ for an appropriate function $J$.   The context there
gave very little scope for suggestions as to how to compute $J$;  at
best, it could be presented as a Radon-Nikod\'ym derivative (235M).   In
this chapter I give a form of the fundamental theorem for the case of
Lebesgue measure, in which $\phi$ is a more or less differentiable
function between Euclidean spaces, and $J$ is a `Jacobian', the
modulus of the determinant of the derivative of $\phi$ (263D).   This
necessarily depends on a serious investigation of the relationship
between Lebesgue measure and geometry.
The first step is to establish a form of Vitali's theorem for
$r$-dimensional space, together with $r$-dimensional density theorems;
I do this in \S261, following closely the scheme of \S\S221 and 223
above.   We need to know quite a lot about differentiable functions
between Euclidean spaces, and it turns out that the theory is
intertwined with that of `Lipschitz' functions;  I treat these in
\S262.

In the next two sections of the chapter, I turn to a separate problem
for which some of the same techniques turn out to be appropriate:  the
description of surface measure on (smooth) surfaces in Euclidean space,
like the surface of a cone or sphere.   I suppose there is no difficulty
in forming a robust intuition as to what is meant by the `area' of
such a surface and of suitably simple regions within it, and there is a
very strong presumption that there ought to be an expression for this
intuition in terms of measure theory as presented in this book;  but the
details are not I think straightforward.   The first point to note is
that for any calculation of the area of a region $G$ in a surface $S$,
one would always turn at once to a parametrization of the region, that
is, a bijection $\phi:D\to G$ from some subset $D$ of Euclidean space.
But obviously one needs to be sure that the result of the calculation is
independent of the parametrization chosen, and while it would be
possible to base the theory on results showing such independence
directly, that does not seem to me to be a true reflection of the
underlying intuition, which is that the area of simple surfaces, at
least, is something intrinsic to their geometry.   I therefore see no
acceptable alternative to a theory of `$r$-dimensional measure' which
can be described in purely geometric terms.   This is the burden of
\S264, in which I give the definition and most fundamental properties
of Hausdorff $r$-dimensional measure in Euclidean spaces.   With this
established, we find that the techniques of \S\S261-263 are sufficient
to relate it to calculations through parametrizations, which is what I
do in \S265.

The chapter ends with a brief account of the Brunn-Minkowski inequality 
(266C), which is an essential tool for the geometric measure theory of 
convex sets.

\discrpage

\frfilename{mt261.tex} 
\versiondate{11.12.12} 
\copyrightdate{2000} 
      
\def\dist{\mathop{\text{dist}}} 
      
\def\chaptername{Change of variable in the integral} 
\def\sectionname{Vitali's theorem in $\BbbR^r$} 
      
\newsection{261} 
      
\def\headlinesectionname{Vitali's theorem in $\eightBbb R^r$} 
      
The main aim of this section is to give $r$-dimensional versions of 
Vitali's theorem and Lebesgue's Density Theorem, following 
ideas already presented in \S\S221 and 223.   I end with a proof that
Lebesgue outer measure can be defined in terms of coverings by balls
instead of by intervals (261F).
      
\cmmnt{ 
\leader{261A}{Notation} For most of this chapter, we shall be dealing 
with the geometry and 
measure of Euclidean space;  it will save space to fix some notation. 
      
Throughout this section and the two following, $r\ge 1$ will be an 
integer.  I will use Roman letters for members of $\BbbR^r$ and 
Greek letters for their coordinates, so that 
$a=(\alpha_1,\ldots,\alpha_r)$, etc.;  if you see any Greek letter with 
a subscript you should look first for a nearby vector of which it might 
be a coordinate.   The measure under consideration will nearly always be 
Lebesgue measure on $\Bbb R^r$;  so unless otherwise indicated  $\mu$ 
should be interpreted as 
Lebesgue measure, and $\mu^*$ as Lebesgue outer measure.   Similarly, 
$\int\ldots dx$ will always be integration with respect to Lebesgue 
measure (in a dimension determined by the context). 
      
For $x=(\xi_1,\ldots,\xi_r)\in\BbbR^r$, write 
$\|x\|=\sqrt{\xi_1^2+\ldots+\xi_r^2}$.   Recall that 
$\|x+y\|\le\|x\|+\|y\|$ (1A2C) and that $\|\alpha x\|=|\alpha|\|x\|$ 
for any vectors $x$, $y$ and scalar $\alpha$. 
      
I will use the same notation as in \S115 for `intervals', so 
that, in particular, 
      
\Centerline{$\coint{a,b} 
=\{x:\alpha_i\le\xi_i<\beta_i\Forall i\le r\}$,} 
      
\Centerline{$\ooint{a,b} 
=\{x:\alpha_i<\xi_i<\beta_i\Forall i\le r\}$,} 
      
\Centerline{$[a,b] 
=\{x:\alpha_i\le \xi_i\le \beta_i\Forall i\le r\}$} 
      
\noindent whenever $a$, $b\in\BbbR^r$. 
      
$\tbf{0}=(0,\ldots,0)$ will be the zero vector in $\BbbR^r$, and 
$\tbf{1}$ will be $(1,\ldots,1)$.   If 
$x\in\BbbR^r$ and $\delta>0$, $B(x,\delta)$ will be the closed ball 
with centre $x$ and radius $\delta$, that is, 
$\{y:y\in\BbbR^r,\,\|y-x\|\le\delta\}$. 
Note that $B(x,\delta)=x+B(\tbf{0},\delta)$;  so that by the 
translation-invariance of Lebesgue measure we have 
      
\Centerline{$\mu B(x,\delta)=\mu B(\tbf{0},\delta) 
=\beta_r\delta^r$,} 
      
\noindent where 
      
$$\eqalign{\beta_r 
&=\Bover{1}{k!}\pi^k\text{ if }r=2k\text{ is even},\cr 
&=\Bover{2^{2k+1}k!}{(2k+1)!}\pi^k\text{ if }r=2k+1\text{ is odd}\cr}$$ 
      
\noindent (252Q). 
}%end of comment 
      
\leader{261B}{Vitali's theorem in $\BbbR^r$} Let $A\subseteq\BbbR^r$ 
be any set, and $\Cal I$ a family of closed non-trivial\cmmnt{ (that 
is, non-singleton, or, equivalently, non-negligible)} 
balls in $\BbbR^r$ such that every point of $A$ is contained in 
arbitrarily small members of $\Cal I$.   Then there is a countable 
disjoint set $\Cal I_0\subseteq \Cal I$ such that 
$\mu(A\setminus\bigcup\Cal I_0)=0$. 
      
\proof{{\bf (a)} To begin with (down to the end of (f) below), suppose 
that $\|x\|<M$ for every $x\in A$, and 
set 
      
\Centerline{$\Cal 
I'=\{I:I\in\Cal I,\,I\subseteq B(\tbf{0},M)\}$.} 
      
\noindent   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 $\Cal I_0$. 
      
\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 $x\in A\setminus\bigcup\Cal I_0$.   Because 
every member of $\Cal I_0$ is closed, there is a $\delta>0$ such that 
$B(x,\delta)$ does not meet any member of $\Cal I_0$, and as 
$\|x\|<M$ we can suppose that $B(x,\delta)\subseteq B(\tbf{0},M)$.   Let 
$J$ be a member of $\Cal I$, containing $x$, and of diameter at most 
$\delta$;  then $J\in\Cal I'$ and $J\cap\bigcup\Cal I_0=\emptyset$.\ 
\Qed 
      
\medskip 
      
{\bf (c)} We can therefore 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 $\Cal 
J_n=\{I:I\in\Cal I',\,I\cap I_j=\emptyset$ for every $j<n\}$.   We know 
from (b) that $\Cal J_n\ne\emptyset$.   Set 
      
\Centerline{$\gamma_n=\sup\{\diam I:I\in\Cal J_n\}$;} 
      
\noindent then $\gamma_n\le 2M$, because every member of $\Cal J_n$ is 
included in $B(\tbf{0},M)$.   We can therefore find a set 
$I_n\in\Cal J_n$ such that $\diam I_n\ge{1\over 2}\gamma_n$, and this 
continues the induction. 
      
\medskip 
      
{\bf (e)} Because the $I_n$ are disjoint measurable subsets of 
the bounded set $B(\tbf{0},M)$, we have 
      
\Centerline{$\sum_{n=0}^{\infty}\mu I_n 
\le\mu B(\tbf{0},M)<\infty$,} 
      
\noindent and $\lim_{n\to\infty}\mu I_n=0$.   Also 
$\mu I_n\ge\beta_r(\bover14\gamma_n)^r$ for each $n$, so 
$\lim_{n\to\infty}\gamma_n=0$. 
      
Now define $I'_n$ to be the closed ball with the same centre as $I_n$ 
but five times the diameter, so 
that it contains every point within a distance $\gamma_n$ of $I_n$.   I 
claim that, for any $n$, 
$A\subseteq\bigcup_{j<n}{I}_j\cup\bigcup_{j\ge n}I_j'$. 
\Prf\Quer\ Suppose, if possible, otherwise.   Take any 
$x\in A\setminus(\bigcup_{j<n}{I}_j\cup\bigcup_{j\ge n}I'_j)$.   Let 
$\delta>0$ be such that 
      
      
\Centerline{$B(x,\delta) 
  \subseteq B(\tbf{0},M)\setminus\bigcup_{j<n}I_j$,} 
      
\noindent and let 
$J\in\Cal I$ be such that $x\in J\subseteq B(x,\delta)$.   Then 
      
\Centerline{$\lim_{m\to\infty}\gamma_m=0<\diam J$} 
      
\noindent (this is where we use the hypothesis that all the balls in 
$\Cal I$ are non-trivial);  let $m$ be the least integer 
greater than or equal to $n$ such that $\gamma_m<\diam J$.   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 
$\gamma_k\ge\diam J$.   So the distance from $x$ to the 
nearest point of $I_k$ is at most $\diam J\le \gamma_k$.   But this 
means 
that $x\in 
I'_k$;  which contradicts the choice of $x$.\ \Bang\Qed 
      
\medskip 
      
{\bf (f)} It follows that 
      
\Centerline{$\mu\sp*(A\setminus\bigcup_{j<n}I_j)\le\mu(\bigcup_{j\ge 
n}I'_j)\le\sum_{j=n}^{\infty}\mu I'_j\le 5^r\sum_{j=n}^{\infty}\mu 
I_j$.} 
      
\noindent As 
      
\Centerline{$\sum_{j=0}^{\infty}\mu I_j\le \mu B(\tbf{0},M)<\infty$,} 
      
\noindent $\lim_{n\to\infty}\mu\sp*(A\setminus\bigcup_{j<n}I_j)=0$ and 
      
\Centerline{$\mu(A\setminus\bigcup_{j\in\Bbb N}I_j)= 
\mu\sp*(A\setminus\bigcup_{j\in\Bbb N}I_j)=0$.} 
      
      
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 $\mu(A\setminus\bigcup 
\Cal I_0)=0$. 
      
      
\medskip 
      
{\bf (g)}  This completes the proof if $A$ is bounded.   In general, set 
      
\Centerline{$U_n=\{x:x\in \BbbR^r,\,n<\|x\|<n+1\}$, 
\quad $A_n=A\cap U_n$, 
\quad$\Cal J_n=\{I:I\in\Cal I,\,I\subseteq U_n\}$,} 
      
\noindent for each $n\in\Bbb N$.   Then for each $n$ we see that every 
point of $A_n$ belongs to arbitrarily small members of $\Cal J_n$, so 
there is a countable disjoint $\Cal J'_n\subseteq\Cal J_n$ such that 
$A_n\setminus\bigcup\Cal J'_n$ is negligible.   Now (because the $U_n$ 
are disjoint) $\Cal I_0=\bigcup_{n\in\Bbb N}\Cal J'_n$ is disjoint, and 
of course $\Cal I_0$ is a countable subset of $\Cal I$;  moreover, 
      
\Centerline{$A\setminus\bigcup\Cal I_0 
\subseteq(\BbbR^r\setminus\bigcup_{n\in\Bbb N}U_n) 
\cup\bigcup_{n\in\Bbb N}(A_n\setminus\bigcup\Cal J'_n)$} 
      
\noindent is negligible.   (To see that $\Bbb 
R^r\setminus\bigcup_{n\in\Bbb N}U_n=\{x:\|x\|\in\Bbb N\}$ is negligible, 
note that for any $n\in\Bbb N$ the set 
      
\Centerline{$\{x:\|x\|=n\}\subseteq B(\tbf{0},n)\setminus 
B(\tbf{0},\delta n)$} 
      
\noindent has measure at most $\beta_rn^r-\beta_r(\delta n)^r$ for every 
$\delta\in\coint{0,1}$, so must be negligible.) 
}%end of proof of 261B 
      
      
\leader{261C}{}\cmmnt{ Just as in \S223, we can use the 
$r$-dimensional Vitali theorem to prove theorems on the approximation of 
functions by their local mean values. 
      
\medskip 
      
\noindent}{\bf Density Theorem in $\BbbR^r$:  integral form} Let $D$ be 
a subset of $\BbbR^r$, and $f$ a real-valued function which is 
integrable over $D$.   Then 
      
\Centerline{$f(x)=\lim_{\delta\downarrow 0}\Bover{1}{\mu B(x,\delta)} 
  \int_{D\cap B(x,\delta)}fd\mu$} 
      
\noindent for almost every $x\in D$. 
      
\proof{{\bf (a)} To begin with (down to the end of (b)), let us 
suppose that $D=\dom f=\BbbR^r$. 
      
Take $n\in\Bbb N$ and $q$, $q'\in\Bbb Q$ with $q<q'$, and set 
      
\Centerline{$A=A_{nqq'} 
=\{x:\|x\|\le n,\,f(x)\le q,\, 
  \limsup_{\delta\downarrow 0}\Bover{1}{\mu B(x,\delta)} 
  \int_{B(x,\delta)}fd\mu>q'\}$.} 
      
\noindent\Quer\ Suppose, if possible, that $\mu^*A>0$.      Let 
$\epsilon>0$ be such that $\epsilon(1+|q|)<(q'-q)\mu^* A$, and let 
$\eta\in\ocint{0,\epsilon}$ be such that $\int_{E}|f|\le\epsilon$ 
whenever $\mu E\le\eta$ (225A).   Let $G\supseteq A$ be an open set of 
measure at most 
$\mu^*A+\eta$ (134Fa).   Let $\Cal I$ be the set of 
non-trivial closed 
balls $B\subseteq G$ such that $\bover{1}{\mu B}\int_Bfd\mu\ge q'$. 
Then every point of $A$ is contained in (indeed, is the centre 
of) arbitrarily small members of $\Cal I$.   So there is a countable 
disjoint set $\Cal I_0\subseteq\Cal I$ such that 
$\mu(A\setminus\bigcup\Cal I_0)=0$, by 261B;  set $H=\bigcup\Cal I_0$. 
      
Because $\int_{I}fd\mu\ge q'\mu I$ for each $I\in\Cal I_0$, we 
have 
      
\Centerline{$\int_{H}fd\mu=\sum_{I\in\Cal I_0}\int_Ifd\mu 
\ge q'\sum_{I\in\Cal I_0}\mu I=q'\mu H\ge q'\mu^* A$.} 
      
\noindent Set 
      
\Centerline{$E=\{x:x\in G,\,f(x)\le q\}$.} 
      
\noindent Then $E$ is measurable, and $A\subseteq E\subseteq G$; 
so 
      
\Centerline{$\mu^* A\le\mu E\le\mu G\le\mu^* A+\eta 
\le\mu^* A+\epsilon$.} 
      
\noindent  Also 
      
\Centerline{$\mu(H\setminus E)\le\mu G-\mu E\le\eta$,} 
      
\noindent so by 
the choice of $\eta$, $\int_{H\setminus E}f\le\epsilon$ and 
      
$$\eqalignno{\int_{H}f 
&\le\epsilon+\int_{H\cap E}f 
\le\epsilon+q\mu(H\cap E)\cr 
&\le\epsilon+q\mu^*A+|q|(\mu(H\cap E)-\mu^* A) 
\le q\mu^* A+\epsilon(1+|q|)\cr 
\displaycause{because $\mu^*A=\mu^*(A\cap H)\le\mu(H\cap E)\le\mu E$} 
&<q'\mu^* A 
\le\int_{H}f,\cr}$$ 
      
\noindent which is impossible.\ \Bang 
      
Thus $A_{nqq'}$ is negligible.   This is true for all $q<q'$ and all 
$n$, so 
      
\Centerline{$A^*=\bigcup_{q,q'\in\Bbb Q,q<q'} 
  \bigcup_{n\in\Bbb N}A_{nqq'}$} 
      
\noindent is negligible.   But 
      
\Centerline{$f(x) 
\ge\limsup_{\delta\downarrow 0}\Bover{1}{\mu B(x,\delta)} 
  \int_{B(x,\delta)}f$} 
      
\noindent for every $x\in\BbbR^r\setminus A^*$, that is, for almost 
all $x\in \BbbR^r$. 
      
\medskip 
      
{\bf (b)} Similarly, or applying this result to $-f$. 
      
\Centerline{$f(x) 
\le\liminf_{\delta\downarrow 0}\Bover{1}{\mu B(x,\delta)} 
  \int_{B(x,\delta)}f$} 
      
\noindent for almost every $x$, so 
      
\Centerline{$f(x) 
=\lim_{\delta\downarrow 0}\Bover{1}{\mu B(x,\delta)} 
  \int_{B(x,\delta)}f$} 
      
\noindent for almost every $x$. 
      
\medskip 
      
{\bf (c)} For the (superficially) more general case enunciated in the 
theorem, let $\tilde f$ be a $\mu$-integrable function extending 
$f\restr D$, defined everywhere on $\BbbR^r$, and such that 
$\int_F\tilde f=\int_{D\cap F}f$ for every measurable 
$F\subseteq\BbbR^r$ (applying 214Eb to $f\restr D$).   Then 
      
\Centerline{$f(x)=\tilde f(x) 
=\lim_{\delta\downarrow 0}\Bover{1}{\mu B(x,\delta)} 
  \int_{B(x,\delta)}\tilde f 
=\lim_{\delta\downarrow 0}\Bover{1}{\mu B(x,\delta)} 
  \int_{D\cap B(x,\delta)}f$} 
      
\noindent for almost every $x\in D$. 
}%end of proof of 261C 
      
\leader{261D}{Corollary} (a) If $D\subseteq\BbbR^r$ is any set, then 
      
\Centerline{$\lim_{\delta\downarrow 0}\Bover{\mu^*(D\cap B(x,\delta))} 
{\mu B(x,\delta)}=1$} 
      
\noindent for almost every $x\in D$. 
      
(b) If $E\subseteq\BbbR^r$ is a measurable set, then 
      
\Centerline{$\lim_{\delta\downarrow 0}\Bover{\mu(E\cap B(x,\delta))} 
{\mu B(x,\delta)}=\chi E(x)$} 
      
\noindent for almost every $x\in \BbbR^r$. 
      
      
(c) If $D\subseteq\BbbR^r$ and $f:D\to\Bbb R$ is any function, then for 
almost every $x\in D$, 
      
\Centerline{$\lim_{\delta\downarrow 0} 
\Bover{\mu^*(\{y:y\in D,\,|f(y)-f(x)|\le\epsilon\}\cap 
B(x,\delta))} 
{\mu B(x,\delta)}=1$} 
      
\noindent for every $\epsilon>0$. 
      
(d) If $D\subseteq\BbbR^r$ and $f:D\to\Bbb R$ is measurable, then for 
almost every $x\in D$, 
      
\Centerline{$\lim_{\delta\downarrow 0} 
\Bover{\mu^*(\{y:y\in D,\,|f(y)-f(x)|\ge\epsilon\}\cap 
B(x,\delta))} 
{\mu B(x,\delta)}=0$} 
      
\noindent for every $\epsilon>0$. 
      
\proof{{\bf (a)} Apply 261C with $f=\chi B(\tbf{0},n)$ to see that, for 
any $n\in\Bbb N$, 
      
\Centerline{$\lim_{\delta\downarrow 0}\Bover{\mu^*(D\cap B(x,\delta))} 
{\mu B(x,\delta)}=1$} 
      
\noindent for almost every $x\in D$ with $\|x\|<n$. 
      
\medskip 
      
{\bf (b)} Apply (a) to $E$ to see that 
      
\Centerline{$\liminf_{\delta\downarrow 0}\Bover{\mu(E\cap 
B(x,\delta))}{\mu B(x,\delta)}\ge\chi E(x)$} 
      
\noindent for almost every $x\in\BbbR^r$, and to $E'=\BbbR^r\setminus 
E$ to see that 
      
\Centerline{$\limsup_{\delta\downarrow 0}\Bover{\mu(E\cap 
B(x,\delta))}{\mu B(x,\delta)} 
=1-\liminf_{\delta\downarrow 0} 
  \Bover{\mu(E'\cap B(x,\delta))}{\mu B(x,\delta)} 
\le 1-\chi E'(x)=\chi E(x)$} 
      
\noindent for almost every $x$. 
      
\medskip 
      
{\bf (c)} For $q$, $q'\in\Bbb Q$, set 
      
\Centerline{$D_{qq'}=\{x:x\in D,\,q\le f(x)\le q'\}$,} 
      
\Centerline{$C_{qq'}=\{x:x\in D_{qq'},\,\lim_{\delta\downarrow 0} 
\Bover{\mu^*(D_{qq'}\cap B(x,\delta))}{\mu B(x,\delta)}=1\}$;} 
      
\noindent now set 
      
\Centerline{$C 
=D\setminus\bigcup_{q,q'\in\Bbb Q}(D_{qq'}\setminus C_{qq'})$,} 
      
\noindent so that $D\setminus C$ is negligible.   If $x\in C$ and 
$\epsilon>0$, then there are $q$, $q'\in\Bbb Q$ such that 
$f(x)-\epsilon\le q\le f(x)\le q'\le f(x)+\epsilon$, and now 
$x\in C_{qq'}$;  accordingly
      
\Centerline{$\liminf_{\delta\downarrow 0} 
\Bover{\mu^*\{y:y\in D\cap B(x,\delta),\,|f(y)-f(x)|\le\epsilon\}} 
  {\mu B(x,\delta)} 
\ge\liminf_{\delta\downarrow 0} 
  \Bover{\mu^*(D_{qq'}\cap B(x,\delta))}{\mu B(x,\delta)}=1$,} 
      
\noindent so 
      
\Centerline{$\lim_{\delta\downarrow 0} 
\Bover{\mu^*\{y:y\in D\cap B(x,\delta),\,|f(y)-f(x)|\le\epsilon\}} 
{\mu B(x,\delta)} 
=1$.} 
      
\medskip 
      
{\bf (d)} Define $C$ as in (c).   We know from (a) that $\mu(D\setminus 
C')=0$, where 
      
\Centerline{$C'=\{x:x\in D,\lim_{\delta\downarrow 0}\Bover{\mu^*(D\cap 
B(x,\delta))} 
{\mu B(x,\delta)}=1\}$. 
} 
      
\noindent If $x\in C\cap C'$ and $\epsilon>0$, we know from (c) that 
      
 \Centerline{$\lim_{\delta\downarrow 0} 
\Bover{\mu^*\{y:y\in D\cap B(x,\delta),\,|f(y)-f(x)|\le\epsilon/2\}} 
{\mu B(x,\delta)} 
=1$.} 
      
\noindent But because $f$ is measurable, we have 
      
$$\eqalign{\mu^*\{y:y\in &D\cap B(x,\delta),\, 
|f(y)-f(x)|\ge\epsilon\}\cr 
&+\mu^*\{y:y\in D\cap B(x,\delta),\,|f(y)-f(x)|\le\Bover12\epsilon\} 
\le\mu^*(D\cap B(x,\delta))\cr}$$ 
      
\noindent for every $\delta>0$.   Accordingly 
      
$$\eqalign{\limsup_{\delta\downarrow 0} 
&\Bover{\mu^*\{y:y\in D\cap B(x,\delta),\,|f(y)-f(x)|\ge\epsilon\}} 
{\mu B(x,\delta)}\cr 
&\le\lim_{\delta\downarrow 0} 
  \Bover{\mu^*(D\cap B(x,\delta))}{\mu B(x,\delta)} 
-\lim_{\delta\downarrow 0} 
  \Bover{\mu^*\{y:y\in D\cap B(x,\delta),\,|f(y)-f(x)|\le\epsilon/2\}} 
  {\mu B(x,\delta)} 
=0,\cr}$$ 
      
\noindent and 
      
\Centerline{$\lim_{\delta\downarrow 0} 
\Bover{\mu^*\{y:y\in D\cap B(x,\delta),\,|f(y)-f(x)|\ge\epsilon\}} 
{\mu B(x,\delta)}=0$} 
      
\noindent for every $x\in C\cap C'$, that is, for almost every $x\in D$. 
}%end of proof of 261D 
      
\leader{261E}{Theorem} Let $f$ be a locally integrable function defined on 
a conegligible subset of $\BbbR^r$.   Then 
      
\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\mu B(x,\delta)} 
\int_{B(x,\delta)}|f(y)-f(x)|dy=0$} 
      
\noindent for almost every $x\in\BbbR^r$. 
      
\proof{ (Compare 223D.) 
      
\medskip 
      
{\bf (a)} Fix $n\in\Bbb N$ for the moment, and set $G=\{x:\|x\|<n\}$. 
For each $q\in\Bbb Q$, set $g_q(x)=|f(x)-q|$ 
for $x\in G\cap\dom f$;  then $g_q$ is integrable over $G$, and 
      
\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\mu B(x,\delta)} 
\int_{G\cap B(x,\delta)}g_q=g_q(x)$} 
      
\noindent for almost every $x\in G$, by 261C.   Setting 
      
\Centerline{$E_q=\{x:x\in G\cap\dom f,\,\lim_{\delta\downarrow 0} 
\Bover1{\mu B(x,\delta)}\int_{G\cap B(x,\delta)}g_q=g_q(x)\}$,} 
      
\noindent we have $G\setminus E_q$ negligible for every $q$, so 
$G\setminus E$ is 
negligible, where $E=\bigcap_{q\in\Bbb Q}E_q$.   Now 
      
\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\mu B(x,\delta)} 
\int_{G\cap B(x,\delta)}|f(y)-f(x)|dy=0$} 
      
\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 $y\in G\cap\dom f$, and 
      
$$\eqalign{\limsup_{\delta\downarrow 0}\Bover1{\mu B(x,\delta)} 
\int_{G\cap B(x,\delta)}|f(y)-f(x)|dy 
&\le\limsup_{\delta\downarrow 0}\Bover1{\mu B(x,\delta)} 
\int_{G\cap B(x,\delta)}g_q(y)+\epsilon\,dy\cr 
&=\epsilon+g_q(x) 
\le 2\epsilon.\cr}$$ 
      
\noindent As $\epsilon$ is arbitrary, 
      
\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\mu B(x,\delta)} 
\int_{G\cap B(x,\delta)}|f(y)-f(x)|dy=0$,} 
      
\noindent as required.\ \Qed 
      
\medskip 
      
{\bf (b)} Because $G$ is open, 
 
\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\mu B(x,\delta)} 
\int_{B(x,\delta)}|f(y)-f(x)|dy 
=\lim_{\delta\downarrow 0}\Bover1{\mu B(x,\delta)} 
\int_{G\cap B(x,\delta)}|f(y)-f(x)|dy 
=0$} 
      
\noindent for almost every $x\in G$.   As $n$ is arbitrary, 
 
\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\mu B(x,\delta)} 
\int_{B(x,\delta)}|f(y)-f(x)|dy=0$} 
      
\noindent for almost every $x\in\BbbR^r$. 
}%end of proof of 261E 
      
\medskip 
\noindent{\bf Remark} The set 
      
\Centerline{$\{x:x\in\dom f,\, 
  \lim_{\delta\downarrow 0}\Bover1{\mu B(x,\delta)} 
\int_{B(x,\delta)}|f(y)-f(x)|dy=0\}$} 
      
\noindent is sometimes called the {\bf Lebesgue set} of $f$. 
      
\leader{261F}{}\cmmnt{ Another very useful consequence of 261B is the 
following. 
      
\medskip 
      
\noindent}{\bf Proposition} Let $A\subseteq\BbbR^r$ be any set, and 
$\epsilon>0$.   Then there is a sequence $\sequencen{B_n}$ of closed 
balls in $\BbbR^r$, all of radius at most $\epsilon$, such that 
$A\subseteq\bigcup_{n\in\Bbb N}B_n$ and 
$\sum_{n=0}^{\infty}\mu B_n\le\mu^*A+\epsilon$.   Moreover, we may 
suppose that the balls in the 
sequence whose centres do not lie in $A$ have measures summing to at 
most $\epsilon$. 
      
\proof{{\bf (a)} The first step is the 
obvious remark that if $x\in\BbbR^r$, $\delta>0$ then the half-open 
cube $I=\coint{x,x+\delta\tbf{1}}$ is a subset of the ball 
$B(x,\delta\sqrt{r})$, 
which has measure $\gamma_r\delta^r=\gamma_r\mu I$, where 
$\gamma_r=\beta_rr^{r/2}$.   It follows that if $G\subseteq\BbbR^r$ is 
any open set, then $G$ can be covered by a sequence of balls of total 
measure at most $\gamma_r\mu G$.   \Prf\ If $G$ is empty, we can take 
all the balls to be singletons.   Otherwise, for each $k\in\Bbb N$, set 
      
\Centerline{$Q_k 
=\{z:z\in \Bbb Z^r,\,\coint{2^{-k}z,2^{-k}(z+\tbf{1})}\subseteq G\}$,} 
      
\Centerline{$E_k 
=\bigcup_{z\in Q_k}\coint{2^{-k}z,2^{-k}(z+\tbf{1}})$.} 
      
\noindent Then $\sequence{k}{E_k}$ is a non-decreasing sequence of sets 
with union $G$, and $E_0$ and each of the differences $E_{k+1}\setminus 
E_k$ is expressible as a disjoint union of half-open cubes.   Thus $G$ 
also is expressible as a disjoint union of a sequence $\sequencen{I_n}$ 
of half-open cubes.   Each $I_n$ is covered by a ball $B_n$ of measure 
$\gamma_r\mu I_n$;  so that $G\subseteq\bigcup_{n\in\Bbb N}B_n$ and 
      
\Centerline{$\sum_{n=0}^{\infty}\mu B_n 
\le\gamma_r\sum_{n=0}^{\infty}\mu I_n=\gamma_r\mu G$.   \Qed} 
      
\medskip 
      
{\bf (b)} It follows at once that if $\mu A=0$ then for any $\epsilon>0$ 
there is a sequence $\sequencen{B_n}$ of balls covering $A$ of measures 
summing to at most $\epsilon$, because there is certainly an open set 
including $A$ with measure at most $\epsilon/\gamma_r$. 
      
\medskip 
      
{\bf (c)} Now take any set $A$, and $\epsilon>0$.   Let $G\supseteq A$ 
be an open set with $\mu G\le\mu^*A+\bover12\epsilon$.   Let $\Cal I$ be 
the family of non-trivial closed balls included in $G$, of radius at 
most $\epsilon$ and with centres in $A$.   Then every point of $A$ 
belongs to arbitrarily small members of $\Cal I$, so there is a 
countable disjoint $\Cal I_0\subseteq\Cal I$ such that 
$\mu(A\setminus\bigcup\Cal I_0)=0$.   Let $\sequencen{B'_n}$ be a 
sequence of balls covering $A\setminus\bigcup\Cal I_0$ with 
$\sum_{n=0}^{\infty}\mu 
B'_n\le\min(\bover12\epsilon,\beta_r\epsilon^r)$;  these surely all have 
radius at most $\epsilon$.   Let $\sequencen{B_n}$ be a sequence 
amalgamating $\Cal I_0$ with $\sequencen{B'_n}$;  then 
$A\subseteq\bigcup_{n\in\Bbb N}B_n$, every $B_n$ has radius at most 
$\epsilon$ and 
      
\Centerline{$\sum_{n=0}^{\infty}\mu B_n 
=\sum_{B\in\Cal I_0}\mu B+\sum_{n=0}^{\infty}\mu B'_n 
\le\mu G+\Bover12\epsilon 
\le\mu A+\epsilon$,} 
      
\noindent while the $B_n$ whose centres do not lie in $A$ must come from 
the sequence $\sequencen{B'_n}$, so their measures sum to at most 
$\bover12\epsilon\le\epsilon$. 
}%end of proof of 261F 
      
\cmmnt{\medskip 
      
\noindent{\bf Remark} In fact we can (if $A$ is not empty) arrange that 
the centre of {\it every} $B_n$ belongs to $A$.   This is an easy 
consequence of Besicovitch's Covering Lemma (see \S472 in Volume 4). 
}%end of comment 
      
\exercises{ 
\leader{261X}{Basic exercises (a)} Show that 261C is valid for 
any locally integrable real-valued function $f$;  in particular, for any 
$f\in\eusm L^p(\mu_D)$ for any $p\ge 1$, writing $\mu_D$ for the 
subspace measure on $D$. 
%261C 
      
\spheader 261Xb Show that 261C, 261Dc, 261Dd and 261E are valid for 
complex-valued functions $f$. 
%261E 
      
\sqheader 261Xc Take three disks in the plane, each touching the other 
two, so that they enclose an open region $R$ with three cusps.   In $R$ 
let $D$ be a disk tangent to each of the three original disks, and 
$R_0$, $R_1$, $R_2$ the three components of $R\setminus D$.   In each 
$R_j$ let $D_j$ be a disk tangent to each of the disks bounding $R_j$, 
and $R_{j0}$, $R_{j1}$, $R_{j2}$ the three components of $R_j\setminus 
D_j$.   Continue, obtaining 27 regions at the next step, 81 regions at 
the next, and so on. 
      
Show that the total area of the residual regions converges to zero as 
the process continues indefinitely.   \Hint{compare with the process in 
the proof of 261B.} 
%261B 
      
\leader{261Y}{Further exercises (a)} 
%\spheader 261Ya 
Formulate an abstract definition of 
`Vitali cover', meaning a family of sets satisfying the conclusion of 
261B in some sense, and corresponding generalizations of 261C-261E, 
covering (at least) (b)-(d) below. 
      
\header{261Yb}{\bf (b)} For $x\in\BbbR^r$, $k\in\Bbb N$ let $C(x,k)$ be 
the half-open 
cube of the form $\coint{2^{-k}z,2^{-k}(z+\tbf{1})}$, with $z\in \Bbb 
Z^r$, containing $x$.   Show that if $f$ is an integrable function on 
$\BbbR^r$ then 
      
\Centerline{$\lim_{k\to\infty}2^{kr}\int_{C(x,k)}f=f(x)$} 
      
\noindent for almost every $x\in\BbbR^r$. 
      
\spheader 261Yc Let $f$ be a real-valued function which is integrable 
over $\BbbR^r$.   Show that 
      
\Centerline{$\lim_{\delta\downarrow 
0}\Bover{1}{\delta^r}\int_{\coint{x,x+\delta\tbf{1}}}f=f(x)$} 
      
\noindent for almost every $x\in\BbbR^r$. 
      
\spheader 261Yd Give $X=\{0,1\}^{\Bbb N}$ its usual measure $\nu$ 
(254J).   For $x\in X$, $k\in\Bbb N$ set 
$C(x,k)=\{y:y\in X,\,y(i)=x(i)$ for $i<k\}$.   Show that if $f$ is any 
real-valued function which is integrable over $X$ then 
$\lim_{k\to\infty}2^k\int_{C(x,k)}fd\nu=f(x)$, 
$\lim_{k\to\infty}2^k\int_{C(x,k)}|f(y)-f(x)|\nu(dy)=0$ for almost every 
$x\in X$. 
      
\spheader 261Ye Let $f$ be a real-valued function which is integrable 
over $\BbbR^r$, and $x$ a point in the Lebesgue set of $f$.   Show that 
for every $\epsilon>0$ there is a $\delta>0$ such that 
\discrcenter{390pt}{$|f(x)-\int f(x-y)g(\|y\|)dy|\le\epsilon$ }whenever 
$g:\coint{0,\infty}\to\coint{0,\infty}$ is a non-increasing function 
such that $\int_{\Bbb R^r}g(\|y\|)dy=1$ and 
$\int_{B(\tbf{0},\delta)}g(\|y\|)dy\ge 1-\delta$.   \Hint{223Yg.} 
      
\spheader 261Yf Let $\frak T$ be the family of those measurable sets 
$G\subseteq\BbbR^r$ such that 
\discrcenter{390pt}{$\lim_{\delta\downarrow 0} 
\Bover{\mu(G\cap B(x,\delta))}{\mu B(x,\delta)}=1$ }for every $x\in G$. 
Show that $\frak T$ is a topology on $\BbbR^r$, the {\bf density 
topology} of $\BbbR^r$.   Show that a function $f:\BbbR^r\to\Bbb R$ is 
measurable iff it is $\frak T$-continuous at almost every point of 
$\Bbb R^r$. 
      
\spheader 261Yg A set $A\subseteq\BbbR^r$ is said to be {\bf porous} at 
$x\in\BbbR^r$ if $\limsup_{y\to x}\Bover{\rho(y,A)}{\|y-x\|}>0$,  
writing $\rho(y,A)=\inf_{z\in A}\|y-z\|$ (or $\infty$ if $A$ is empty). 
(i) Show that if $A$ is porous at all its points then it is negligible. 
(ii) Show that in the construction of 261B the residual set 
$A\setminus\bigcup\Cal I_0$ will be porous at all its points. 
 
\spheader 261Yh Let $A\subseteq\BbbR^r$ be a bounded set and $\Cal I$ a 
non-empty family of non-trivial closed balls covering $A$.   Show that 
for any $\epsilon>0$ there are disjoint $B_0,\ldots,B_n\in\Cal I$ such 
that $\mu^*A\le(3+\epsilon)^r\sum_{k=0}^n\mu B_k$. 
      
\spheader 261Yi Let $(X,\rho)$ be a metric space and $A\subseteq X$ any 
set, $x\mapsto\delta_x:A\to\coint{0,\infty}$ any bounded function. 
Show that if $\gamma>3$ then there is an $A'\subseteq A$ such that (i) 
$\rho(x,y)>\delta_x+\delta_y$ for all distinct $x$, $y\in A'$ (ii) 
$\bigcup_{x\in A}B(x,\delta_x) 
\subseteq\bigcup_{x\in A'}B(x,\gamma\delta_x)$, writing $B(x,\alpha)$ 
for the closed ball $\{y:\rho(y,x)\le\alpha\}$. 
      
\spheader 261Yj(i) Let $\Cal C$ be the family of those measurable sets 
$C\subseteq\BbbR^r$ such that 
\discrcenter{390pt}{$\limsup_{\delta\downarrow 0} 
\Bover{\mu(C\cap B(x,\delta))}{\mu B(x,\delta)}>0$ }for every $x\in C$. 
Show that $\bigcup\Cal C_0\in\Cal C$ for every $\Cal C_0\subseteq\Cal C$.
\Hint{215B(iv).}   
(ii) Show that any union of non-trivial closed balls in 
$\BbbR^r$ is Lebesgue measurable.   
      
\spheader 261Yk Suppose that $A\subseteq\BbbR^r$ and that $\Cal I$ is a 
family of closed subsets of $\BbbR^r$ such that 
      
\inset{\noindent for every $x\in A$ there is an $\eta>0$ such that for 
every $\epsilon>0$ there is an $I\in\Cal I$ such that $x\in I$ and 
$0<\eta(\diam I)^r\le\mu I\le\epsilon$.} 
      
\noindent Show that there is a countable disjoint set 
$\Cal I_0\subseteq\Cal I$ such that 
$A\setminus\bigcup\Cal I_0$ is negligible. 
%261B 
 
\spheader 261Yl Let $\frak T'$ be the family of measurable sets 
$G\subseteq\BbbR^r$ such that whenever $x\in G$ and $\epsilon>0$ 
there is a $\delta>0$ such that $\mu(G\cap I)\ge(1-\epsilon)\mu I$ whenever 
$I$ is an interval containing $x$ and included in $B(x,\delta)$.    
Show that  
$\frak T'$ is a topology on $\BbbR^r$ intermediate between the density 
topology (261Yf) and the Euclidean topology. 
}%end of exercises 
      
\endnotes{ 
\Notesheader{261} 
In the proofs of 261B-261E above, I have done my best to 
follow the lines of the one-dimensional case;  this section amounts to a 
series of generalizations of the work of \S\S221 and 223. 
      
It will be clear that 
the idea of 261A/261B can be used on other shapes than balls.   To make 
it work in the form above, we need a family $\Cal I$ such that there is 
a constant $K$ for which 
      
\Centerline{$\mu I'\le K\mu I$} 
      
\noindent for every $I\in \Cal I$, where we write 
      
\Centerline{$I'=\{x:\inf_{y\in I}\|x-y\|\le\diam(I)\}$.} 
      
\noindent Evidently this will be true for many classes $\Cal I$ 
determined by the shapes of the sets involved;  for instance, if 
$E\subseteq\BbbR^r$ is any bounded set of strictly positive measure, 
the family $\Cal I=\{x+\delta E:x\in\BbbR^r,\,\delta>0\}$ will satisfy 
the condition. 
      
In 261Ya I challenge you to find an appropriate generalization of the 
arguments depending on the conclusion of 261B. 
      
Another way of using 261B is to say that because sets 
can be essentially covered by {\it disjoint} sequences of balls, it 
ought to be possible to use balls, rather than half-open intervals, in 
the definition of Lebesgue measure on $\BbbR^r$.   This is indeed so 
(261F).  The difficulty in using balls in the basic 
definition comes right at the start, in proving that if a ball is 
covered by finitely many balls then the sum of the volumes of the 
covering balls is at least the volume of the covered ball.    (There is 
a trick, using the compactness of closed balls and the openness of open 
balls, to extend such a proof to infinite covers.)   Of course you could 
regard this fact as `elementary', on the ground that Archimedes 
would have noticed if it weren't true, but nevertheless it would be 
something of a challenge to prove it, unless you were willing to wait 
for a version of Fubini's theorem, as some authors do. 
      
I have given the results in 261C-261D for arbitrary subsets $D$ of  
$\Bbb R^r$ not because I have any applications in mind in which 
non-measurable subsets are significant, but because I wish to make it 
possible to notice when measurability matters.   Of course it is 
necessary to interpret the integrals $\int_Dfd\mu$ in the way laid down 
in \S214.   The game is given away in part (c) of the proof of 261C, 
where I rely on the fact that if $f$ is integrable over $D$ then there 
is an integrable $\tilde f:\BbbR^r\to\Bbb R$ such that 
$\int_F\tilde f=\int_{D\cap F}f$ for every measurable 
$F\subseteq\BbbR^r$.   In 
effect, for all the questions dealt with here, we can replace $f$, $D$ 
by $\tilde f$, $\BbbR^r$. 
      
The idea of 261C is that, for almost every $x$, $f(x)$ is approximated 
by its mean value on small balls $B(x,\delta)$, ignoring the missing 
values on $B(x,\delta)\setminus(D\cap\dom f)$;  261E is a sharper 
version of the same idea. 
The formulae of 261C-261E mostly involve the expression 
$\mu B(x,\delta)$.   Of course this is just $\beta_r\delta^r$.   But I 
think that leaving it unexpanded is actually more illuminating, as well 
as avoiding sub- and superscripts, since it makes it clearer what these 
density theorems are really about.   In \S472 of Volume 4 I will revisit  
this material, showing that a surprisingly large proportion of the ideas  
can be applied to arbitrary Radon measures on $\BbbR^r$, even though 
Vitali's theorem (in the form stated here) is no longer valid. 
      
}%end of notes 
      
      
\discrpage 
      
\frfilename{mt262.tex}
\versiondate{11.8.15}
\copyrightdate{1995}

\def\dist{\mathop{\text{dist}}}

\def\chaptername{Change of variable in the integral}
\def\sectionname{Lipschitz and differentiable functions}

\newsection{262}

In preparation for the main work of this chapter in \S263, I devote a
section to two important classes of functions between Euclidean spaces.
What we really need is the essentially elementary material down to 262I,
together with the technical lemma 262M and its corollaries.   Theorem
262Q is not relied on in this volume, though I believe that it makes the
patterns which will develop more natural and comprehensible.

As in \S261, $r$ (and here also $s$) will be a strictly positive integer,
and `measurable', `negligible', `integrable' will refer to Lebesgue measure
unless otherwise stated.

\leader{262A}{Lipschitz functions} Suppose that
$\phi:D\to\BbbR^s$ is a function, where $D\subseteq\BbbR^r$.   
\cmmnt{We say that }$\phi$ is {\bf $\gamma$-Lipschitz}, where
$\gamma\in\coint{0,\infty}$, if

\Centerline{$\|\phi(x)-\phi(y)\|\le \gamma \|x-y\|$}

\noindent for all $x$, $y\in D$, writing
$\|x\|=\sqrt{\xi_1^2+\ldots+\xi_r^2}$ if
$x=(\xi_1,\ldots,\xi_r)\in\BbbR^r$,
$\|z\|=\sqrt{\zeta_1^2+\ldots+\zeta_s^2}$ if
$z=(\zeta_1,\ldots,\zeta_s)\in\BbbR^s$.   In this case, $\gamma$ is
{\bf a Lipschitz constant for $\phi$}.

A {\bf Lipschitz function} is a function $\phi$ which is
$\gamma$-Lipschitz for some $\gamma\ge 0$.
\cmmnt{Note that in this case $\phi$ has a least
Lipschitz constant (since if $A$ is the set of Lipschitz constants for
$\phi$, and $\gamma_0=\inf A$, then $\gamma_0$ is a Lipschitz constant
for $\phi$).}
\dvAnew{2015}Evidently a Lipschitz function is (uniformly) continuous.

\leader{262B}{}\cmmnt{ We need the following easy facts.

\medskip

\noindent}{\bf Lemma} Let $D\subseteq\BbbR^r$ be a set and
$\phi:D\to\BbbR^s$ a function.

(a)  $\phi$ is Lipschitz iff $\phi_i:D\to\Bbb R$
is Lipschitz for every $i$, writing
$\phi(x)=(\phi_1(x),\ldots,\phi_s(x))$ for every $x\in
D=\dom\phi\subseteq\BbbR^r$.

(b) In this case, there is a Lipschitz function
$\tilde\phi:\BbbR^r\to\BbbR^s$ extending $\phi$.

(c) If $r=s=1$ and $D=[a,b]$ is an interval,
then $\phi$ is Lipschitz iff it is absolutely continuous and has a
bounded derivative.

\proof{{\bf (a)} For any $x$, $y\in D$ and $i\le s$,

\Centerline{$|\phi_i(x)-\phi_i(y)|\le\|\phi(x)-\phi(y)\|
\le\sqrt{s}\sup_{j\le s}|\phi_j(x)-\phi_j(y)|$,}

\noindent so any Lipschitz constant for $\phi$ will be a Lipschitz
constant for every $\phi_i$, and if $\gamma_j$ is a Lipschitz constant
for $\phi_j$ for each $j$, then $\sqrt{s}\sup_{j\le s}\gamma_j$ will be
a Lipschitz constant for $\phi$.

\medskip

{\bf (b)} By (a), it is
enough to consider the case $s=1$, for if every $\phi_i$ has a Lipschitz
extension $\tilde\phi_i$, we can set
$\tilde\phi(x)=(\tilde\phi_1(x),\ldots,\tilde\phi_s(x))$ for every $x$
to obtain a Lipschitz extension of $\phi$.
Taking $s=1$, then, note that the case $D=\emptyset$ is trivial;  so
suppose that $D\ne\emptyset$.   Let $\gamma$ be a Lipschitz constant for
$\phi$, and write

\Centerline{$\tilde\phi(z)=\sup_{y\in D}\phi(y)-\gamma\|y-z\|$}

\noindent for every $z\in\BbbR^r$.   If $x\in D$, then, for any
$z\in\BbbR^r$ and $y\in D$,

\Centerline{$\phi(y)-\gamma\|y-z\|
\le\phi(x)+\gamma\|y-x\|-\gamma\|y-z\|
\le\phi(x)+\gamma\|z-x\|$,}

\noindent so that $\tilde\phi(z)\le\phi(x)+\gamma\|z-x\|$;  this shows,
in particular, that $\tilde\phi(z)<\infty$.   Also, if $z\in D$, we must
have

\Centerline{$\phi(z)-\gamma\|z-z\|\le\tilde\phi(z)
\le\phi(z)+\gamma\|z-z\|$,}

\noindent so that $\tilde\phi$ extends $\phi$.   Finally, if $w$,
$z\in\BbbR^r$ and $y\in D$,

\Centerline{$\phi(y)-\gamma\|y-w\|
\le\phi(y)-\gamma\|y-z\|+\gamma\|w-z\|
\le\tilde\phi(z)+\gamma\|w-z\|$;}

\noindent and taking the supremum over $y\in D$,

\Centerline{$\tilde\phi(w)\le\tilde\phi(z)+\gamma\|w-z\|$.}

\noindent As $w$ and $z$ are arbitrary, $\tilde\phi$ is Lipschitz.

\medskip

{\bf (c)(i)} Suppose that $\phi$ is $\gamma$-Lipschitz.  If $\epsilon>0$
and
$a\le a_1\le b_1\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=1}^nb_i-a_i\le\epsilon/(1+\gamma)$, then

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

\noindent As $\epsilon$ is arbitrary, $\phi$ is absolutely continuous.
If $x\in[a,b]$ and $\phi'(x)$ is defined, then

\Centerline{$|\phi'(x)|
=\lim_{y\to x}\Bover{|\phi(y)-\phi(x)|}{|y-x|}\le\gamma$,}

\noindent so $\phi'$ is bounded.

\medskip

\quad{\bf (ii)} Now suppose that $\phi$ is absolutely continuous and
that $|\phi'(x)|\le\gamma$ for every $x\in\dom\phi'$, where
$\gamma\ge 0$.   Then whenever $a\le x\le y\le b$,

\Centerline{$|\phi(y)-\phi(x)|=|\int_x^y\phi'|\le\int_x^y|\phi'|
\le\gamma(y-x)$}

\noindent (using 225E for the first equality).   As $x$ and $y$ are
arbitrary, $\phi$ is $\gamma$-Lipschitz.
}%end of proof of 262B

\cmmnt{
\leader{262C}{Remark} The argument for (b) above shows that if
$\phi:D\to\Bbb R$ is a Lipschitz function, where
$D\subseteq\Bbb R^r$, then $\phi$ has an extension to $\BbbR^r$ with the same Lipschitz
constants.   In fact it is the case that if $\phi:D\to\BbbR^s$ is a
Lipschitz function, then $\phi$ has
an extension to $\tilde\phi:\BbbR^r\to\BbbR^s$ with the same Lipschitz
constants;  this is `Kirszbraun's theorem' ({\smc Kirszbraun 1934}, or
{\smc Federer 69}, 2.10.43).
}%end of comment

\leader{262D}{Proposition} If $\phi:D\to\BbbR^r$ is a $\gamma$-Lipschitz
function, where $D\subseteq\BbbR^r$, then
$\mu^*\phi[A]\le\gamma^r\mu^*A$ for every $A\subseteq D$, where $\mu$ is
Lebesgue measure on $\BbbR^r$.   In particular,
$\phi[D\cap A]$ is negligible for every negligible set $A\subseteq\BbbR^r$.

\proof{ Let $\epsilon>0$.   By 261F, there is a sequence
$\sequencen{B_n}=\sequencen{B(x_n,\delta_n)}$ of closed balls in
$\Bbb R^r$, covering $A$, such that
$\sum_{n=0}^{\infty}\mu B_n\le\mu^* A+\epsilon$ and
$\sum_{n\in\Bbb N\setminus K}\mu B_n\le\epsilon$, where
$K=\{n:n\in\Bbb N,\,x_n\in A\}$.   Set

\Centerline{$L=\{n:n\in\Bbb N\setminus K,\,B_n\cap D\ne\emptyset\}$,}

\noindent and for $n\in L$ choose $y_n\in D\cap B_n$.   Now set

$$\eqalign{B'_n
&=B(\phi(x_n),\gamma\delta_n)\text{ if }n\in K,\cr
&=B(\phi(y_n),2\gamma\delta_n)\text{ if }n\in L,\cr
&=\emptyset\text{ if }n\in\Bbb N\setminus(K\cup L).\cr}$$

\noindent Then $\phi[B_n\cap D]\subseteq B'_n$ for every $n$, so
$\phi[D\cap A]\subseteq\bigcup_{n\in\Bbb N}B'_n$, and

$$\eqalign{\mu^*\phi[A\cap D]
&\le\sum_{n=0}^{\infty}\mu B'_n
=\gamma^r\sum_{n\in K}\mu B_n+2^r\gamma^r\sum_{n\in L}\mu B_n\cr
&\le\gamma^r(\mu^*A+\epsilon)+2^r\gamma^r\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\mu^*\phi[A\cap
D]\le\gamma^r\mu^*A$, as claimed.
}%end of proof of 262D

\vleader{72pt}{262E}{Corollary} Let $\phi:D\to\BbbR^r$ be an injective
Lipschitz function, where $D\subseteq\BbbR^r$, and $f$ a measurable
function from a subset of $\BbbR^r$ to $\Bbb R$.

(a) If $\phi^{-1}$ is defined almost everywhere in a subset $H$ of
$\BbbR^r$ and $f$ is defined almost everywhere in $\BbbR^r$, then
$f\phi^{-1}$ is defined almost everywhere in $H$.

(b) If $E\subseteq D$ is Lebesgue measurable then $\phi[E]$ is
measurable.

(c) If $D$ is measurable then $f\phi^{-1}$ is measurable.

\proof{ Set

\Centerline{$C=\dom(f\phi^{-1})
=\{y:y\in\phi[D],\,\phi^{-1}(y)\in\dom f\}=\phi[D\cap\dom f]$.}

\medskip

{\bf (a)} Because $f$ is defined almost everywhere,
$\phi[D\setminus\dom f]$
is negligible.   But now

\Centerline{$C=\phi[D]\setminus \phi[D\setminus\dom f]
=\dom\phi^{-1}\setminus\phi[D\setminus\dom f]$,}

\noindent so

\Centerline{$H\setminus C
\subseteq(H\setminus\dom\phi^{-1})\cup\phi[D\setminus\dom f]$}

\noindent is negligible.

\medskip

{\bf (b)} Now suppose that $E\subseteq D$ and that $E$ is measurable.
Let $\sequencen{F_n}$ be a sequence of closed bounded
subsets of $E$ such that $\mu(E\setminus\bigcup_{n\in\Bbb N}F_n)=0$
(134Fb).   Because $\phi$ is Lipschitz, it is continuous, so
$\phi[F_n]$ is compact, therefore closed, therefore measurable for every
$n$ (2A2E, 115G);  also $\phi[E\setminus\bigcup_{n\in\Bbb N}F_n]$
is negligible, by 262D, therefore measurable.   So

\Centerline{$\phi[E]
=\phi[E\setminus\bigcup_{n\in\Bbb N}F_n]
  \cup\bigcup_{n\in\Bbb N}\phi[F_n]
$}

\noindent is measurable.

\medskip

{\bf (c)} For any $a\in\Bbb R$, take a measurable set
$E\subseteq\BbbR^r$
such that $\{x:f(x)\ge a\}=E\cap\dom f$.   Then

\Centerline{$\{y:y\in C,\,f\phi^{-1}(y)\ge a\}
=C\cap\phi[D\cap E]$.}

\noindent But $\phi[D\cap E]$ is measurable, by (b), so
$\{y:f\phi^{-1}(y)\ge a\}$ is relatively measurable in $C$.   As $a$ is
arbitrary, $f\phi^{-1}$ is measurable.
}%end of proof of 262E

\leader{262F}{\dvrocolon{Differentiability}}\cmmnt{ I come now to the
class of functions whose properties will take up most of the rest of the
chapter.

\medskip

\noindent}{\bf Definitions}  Suppose that $\phi$ is a
function from a subset $D=\dom\phi$ of $\BbbR^r$ to $\BbbR^s$.

\spheader 262Fa $\phi$  is {\bf differentiable} at $x\in D$ if there is a
real $s\times r$ matrix $T$ such that

\Centerline{$\lim_{y\to x}
\Bover{\|\phi(y)-\phi(x)-T(y-x)\|}{\|y-x\|}=0$;}

\noindent in this case we may write $T=\phi'(x)$.

\spheader 262Fb I will say that $\phi$ is {\bf differentiable relative to
its domain} at
$x$, and that $T$ is {\bf a} derivative of $\phi$ at $x$, if
$x\in D$ and for every $\epsilon>0$ there is a $\delta>0$ such that
$\|\phi(y)-\phi(x)-T(y-x)\|\le\epsilon\|y-x\|$ for every
$y\in B(x,\delta)\cap D$.

\cmmnt{
\vleader{72pt}{262G}{Remarks (a)} The standard definition in 262Fa,
involving an all-sided limit
`$\lim_{y\to x}$', implicitly requires $\phi$ to be defined on some
non-trivial ball centered on $x$, so that we can calculate
$\phi(y)-\phi(x)-T(y-x)$ for all $y$ sufficiently near $x$.   It has the
advantage that the derivative $T=\phi'(x)$ is uniquely defined (because
if $\lim_{z\to \tbf{0}}\Bover{\|T_1z-T_2z\|}{\|z\|}=0$ then

\Centerline{$\Bover{\|(T_1-T_2)z\|}{\|z\|}=\lim_{\alpha\to
0}\Bover{\|T_1(\alpha
z)-T_2(\alpha z)\|}{\|\alpha z\|}=0$}

\noindent for every non-zero $z$, so $T_1-T_2$ must be the zero matrix).
For our purposes here, there is some advantage in relaxing this slightly
to the form
in 262Fb, so that we do not need to pay special attention to the
boundary of $\dom\phi$.   In particular we find that if
$T$ is a derivative of $\phi:D\to\BbbR^s$
relative to its domain at $x$, and $x\in D'\subseteq D$, then
$T$ is a derivative of $\phi\restr D'$,
relative to its domain, at $x$.

\header{262Gb}{\bf (b)} If you have not seen this concept of
`differentiability'
before, but have some familiarity with partial differentiation, it is
necessary to emphasize that the concept of `differentiable' function
(at least in the strict sense demanded by 262Fa) is strictly
stronger than the concept of `partially differentiable' function.
For purposes of computation, the most useful method of finding true
derivatives is through 262Id below.   For a simple
example of a function with a full set of partial derivatives, which is
not everywhere differentiable, consider $\phi:\BbbR^2\to\Bbb R$ defined
by

$$\eqalign{\phi(\xi_1,\xi_2)
&={{\xi_1\xi_2}\over{\xi_1^2+\xi_2^2}}
 \text{ if }\xi_1^2+\xi_2^2\ne 0,\cr
&=0\text{ if }\xi_1=\xi_2=0.\cr}$$

\noindent Then $\phi$ is not even continuous at $\tbf{0}$, although
both partial derivatives $\pd{\phi}{\xi_j}$ are defined everywhere.

\header{262Gc}{\bf (c)} In the definition above, I speak of a derivative
as being a
matrix.   Properly speaking,  the derivative of a function defined on a
subset of $\BbbR^r$ and taking values in $\BbbR^s$ should be thought
of as a bounded linear operator from $\BbbR^r$ to $\BbbR^s$;   the
formulation in terms of matrices is acceptable just because there is a
natural one-to-one correspondence between $s\times r$ real matrices and
linear operators from $\BbbR^r$ to $\BbbR^s$, and all these linear
operators are
bounded.   I use the `matrix' description because it makes certain
calculations more direct;  in particular, the relationship between
$\phi'$ and the partial derivatives of $\phi$ (262Ic), and the notion of
the determinant $\det\phi'(x)$, used throughout \S\S263 and 265.
}%end of comment

\leader{262H}{The norm of a matrix} Some of the calculations below will
rely on the notion of `norm' of a matrix.   The one I
will use\cmmnt{ (in fact, for our purposes here, any norm would do)}
is the `operator norm', defined by saying

\Centerline{$\|T\|=\sup\{\|Tx\|:x\in\BbbR^r,\,\|x\|\le 1\}$}

\noindent for any $s\times r$ matrix $T$.   \cmmnt{For the basic facts
concerning these norms, see 2A4F-2A4G.   The following will also be
useful.}

\header{262Ha}{\bf (a)} If\cmmnt{ all the coefficients of $T$ are
small, so is $\|T\|$;  in fact,
if} $T=\langle\tau_{ij}\rangle_{i\le s,j\le r}$\cmmnt{, and
$\|x\|\le 1$,} then
\cmmnt{$|\xi_j|\le 1$ for each $j$, so

\Centerline{$\|Tx\|
=\bigl(\sum_{i=1}^s(\sum_{j=1}^r\tau_{ij}\xi_j)^2\bigr)^{1/2}
\le \bigl(\sum_{i=1}^s(\sum_{j=1}^r|\tau_{ij}|)^2\bigr)^{1/2}
\le r\sqrt{s}\max_{i\le s,j\le r}|\tau_{ij}|$,}

\noindent and} $\|T\|\le r\sqrt{s}\max_{i\le s,j\le r}|\tau_{ij}|$.
\cmmnt{(This is a singularly crude inequality.   A better one is in
262Yb.   But it tells us, in particular, that $\|T\|$ is always finite.)
}

\header{262Hb}{\bf (b)}\dvro{ $|\tau_{ij}|\le\|T\|$ for all $i$, $j$.}
{ If $\|T\|$ is small, so are all the coefficients of $T$;  in fact,
writing $e_j$ for the $j$th unit vector of $\BbbR^r$, then the $i$th
coordinate of $Te_j$ is $\tau_{ij}$, so
$|\tau_{ij}|\le\|Te_j\|\le\|T\|$.}

\leader{262I}{Lemma} Let $\phi:D\to\BbbR^s$ be a function, where
$D\subseteq\BbbR^r$.   For $i\le s$ let $\phi_i:D\to\Bbb R$ be its $i$th
coordinate, so that $\phi(x)=(\phi_1(x),\ldots,\phi_s(x))$ for $x\in D$.

(a) If $\phi$ is differentiable relative to its domain
at $x\in D$, then $\phi$ is continuous at $x$.

(b) If $x\in D$, then $\phi$ is differentiable relative to its domain at
$x$ iff each $\phi_i$ is differentiable relative to its domain at $x$.

(c) If $\phi$ is differentiable at $x\in D$, then all the partial
derivatives $\pd{\phi_i}{\xi_j}$ of $\phi$ are defined at $x$, and the
derivative of $\phi$ at $x$ is the matrix
$\langle\pd{\phi_i}{\xi_j}(x)\rangle_{i\le s,j\le r}$.

(d) If all the partial derivatives $\pd{\phi_i}{\xi_j}$, for $i\le s$
and $j\le r$, are defined in a neighbourhood of $x\in D$ and are continuous at $x$, then $\phi$ is differentiable at $x$.

\proof{{\bf (a)} Let $T$ be a derivative of $\phi$ at $x$.   Applying
the definition 262Fb with
$\epsilon=1$, we see that there is a $\delta>0$ such that
\Centerline{$\|\phi(y)-\phi(x)-T(y-x)\|\le\|y-x\|$}

\noindent whenever $y\in D$ and $\|y-x\|\le\delta$.   Now

\Centerline{$\|\phi(y)-\phi(x)\|\le\|T(y-x)\|+\|y-x\|
\le(1+\|T\|)\|y-x\|$}

\noindent whenever $y\in D$ and $\|y-x\|\le\delta$, so $\phi$ is
continuous at $x$.

\medskip

{\bf (b)(i)} If $\phi$ is differentiable relative to its domain at
$x\in D$, let $T$ be a
derivative of $\phi$ at $x$.   For $i\le s$ let $T_i$ be
the $1\times r$ matrix consisting of the $i$th row of $T$.    Let
$\epsilon>0$.  Then we have a $\delta>0$ such that

$$\eqalign{|\phi_i(y)-\phi_i(x)-T_i(y-x)|
&\le\|\phi(y)-\phi(x)-T(y-x)\|\cr
&\le\epsilon\|y-x\|\cr}$$

\noindent whenever $y\in D$ and $\|y-x\|\le\delta$, so that $T_i$ is a
derivative of $\phi_i$ at $x$.

\medskip

\quad{\bf (ii)} If each $\phi_i$ is differentiable relative to its
domain at
$x$, with corresponding derivatives $T_i$, let $T$ be the $s\times r$
matrix with rows $T_1,\ldots,T_s$.   Given $\epsilon>0$, there is for
each $i\le s$ a $\delta_i>0$ such that

\Centerline{$|\phi_i(y)-\phi_i(x)-T_iy|\le\epsilon\|y-x\|$ whenever
$y\in D$, $\|y-x\|\le\delta_i$;}

\noindent set $\delta=\min_{i\le s}\delta_i>0$;  then if $y\in D$ and
$\|y-x\|\le\delta$, we shall have

\Centerline{$\|\phi(y)-\phi(x)-T(y-x)\|^2
=\sum_{i=1}^s|\phi_i(y)-\phi_i(x)-T_i(y-x)|^2
\le s\epsilon^2\|y-x\|^2$,}

\noindent so that

\Centerline{$\|\phi(y)-\phi(x)-T(y-x)\|
\le\epsilon\sqrt{s}\|y-x\|$.}

\noindent As $\epsilon$ is arbitrary, $T$ is a derivative of $\phi$ at
$x$.

\medskip

{\bf (c)} Set $T=\phi'(x)$.   We have

\Centerline{$\lim_{y\to x}\Bover{\|\phi(y)-\phi(x)-T(y-x)\|}{\|y-x\|}
=0$;}

\noindent fix $j\le r$, and consider $y=x+\eta e_j$, where
$e_j=(0,\ldots,0,1,0,\ldots,0)$ is the $j$th unit vector in $\BbbR^r$.
Then we must have

\Centerline{$\lim_{\eta\to 0}\Bover{\|\phi(x+\eta e_j)-\phi(x)-\eta
T(e_j)\|}{|\eta|}
=0$.}

\noindent Looking at the $i$th coordinate of
$\phi(x+\eta e_j)-\phi(x)-\eta T(e_j)$, we have


\Centerline{$|\phi_i(x+\eta e_j)-\phi_i(x)-\tau_{ij}\eta|
\le\|\phi(x+\eta e_j)-\phi(x)-\eta T(e_j)\|$,}

\noindent where $\tau_{ij}$ is the
$(i,j)$th coefficient of $T$;  so that

\Centerline{$\lim_{\eta\to 0}
\Bover{|\phi_i(x+\eta e_j)-\phi_i(x)-\tau_{ij}\eta|}{|\eta|}
=0$.}

\noindent  But this just says that the partial derivative
$\pd{\phi_i}{\xi_j}(x)$ exists and is equal to $\tau_{ij}$, as claimed.

\medskip

{\bf (d)} Now suppose that the partial derivatives $\pd{\phi_i}{\xi_j}$
are defined near $x$ and continuous at $x$.   Let $\epsilon>0$.   Let
$\delta>0$ be such that

\Centerline{$|\pd{\phi_i}{\xi_j}(y)-\tau_{ij}|\le\epsilon$}

\noindent whenever $\|y-x\|\le\delta$, writing
$\tau_{ij}=\pd{\phi_i}{\xi_j}(x)$.   Now suppose that
$\|y-x\|\le\delta$.   Set

\Centerline{$y=(\eta_1,\ldots,\eta_r)$,
\quad$x=(\xi_1,\ldots,\xi_r)$,}

\Centerline{$y_j=(\eta_1,\ldots,\eta_j,\xi_{j+1},\ldots,\xi_r)$ for
$0\le
j\le r$,}

\noindent so that $y_0=x$, $y_r=y$ and the line segment between
$y_{j-1}$ and $y_{j}$ lies wholly within $\delta$ of $x$ whenever
$1\le j\le r$, since if $z=(\zeta_1,\ldots,\zeta_r)$
lies on this segment then $\zeta_i$ lies
between $\xi_i$ and $\eta_i$ for every $i$.   By the ordinary mean value
theorem for differentiable real functions, applied to the function

\Centerline{$t\mapsto
\phi_i(\eta_1,\ldots,\eta_{j-1},t,\xi_{j+1},\ldots,\xi_r)$,}

\noindent there is for each $i\le s$, $j\le r$ a point $z_{ij}$ on the
line segment between $y_{j-1}$ and $y_j$ such that

\Centerline{$\phi_i(y_j)-\phi_i(y_{j-1})
=(\eta_j-\xi_j)\pd{\phi_i}{\xi_j}(z_{ij})$.}

\noindent But

\Centerline{$|\pd{\phi_i}{\xi_j}(z_{ij})-\tau_{ij}|\le\epsilon$,}
\noindent so

\Centerline{$|\phi_i(y_j)-\phi_i(y_{j-1})
-\tau_{ij}(\eta_j-\xi_j)|\le\epsilon|\eta_j-\xi_j|\le\epsilon\|y-x\|$.}

\noindent Summing over $j$,

\Centerline{$|\phi_i(y)-\phi_i(x)-\sum_{j=1}^r\tau_{ij}(\eta_j-\xi_j)|
\le r\epsilon\|y-x\|$}

\noindent for each $i$.   Summing the squares and taking the square
root,

\Centerline{$\|\phi(y)-\phi(x)-T(y-x)\|\le \epsilon r\sqrt{s}\|y-x\|$,}

\noindent where $T=\langle\tau_{ij}\rangle_{i\le s,j\le r}$.   And this
is true
whenever $\|y-x\|\le\delta$.   As $\epsilon$ is arbitrary, $\phi'(x)=T$
is defined.
}%end of proof of 262I

\cmmnt{
\leader{262J}{Remark} I am not sure if I ought to apologize for the
notation $\pd{}{\xi_j}$.   In such formulae as
$(\eta_j-\xi_j)\pd{\phi_i}{\xi_j}(z_{ij})$ above, the two appearances of
$\xi_j$ clash most violently.   But I do not think that any person of
good will is likely to be misled, provided that the labels $\xi_j$ (or
whatever symbols are used to represent the variables involved) are
adequately described when the domain of $\phi$ is first introduced (and
always remembering that in partial differentiation, we are not only
moving one variable -- a $\xi_j$ in the present context -- but holding
fixed some further list of variables, not listed in the notation).   I
believe that the traditional notation $\pd{}{\xi_j}$ has survived for
solid reasons, and I should like to offer a welcome to those who are
more comfortable with it than with any of the many alternatives which
have been proposed, but have never taken root.
}%end of comment

\cmmnt{
\leader{262K}{The Cantor function revisited} It is salutary to
re-examine the examples of 134H-134I in the light of the present
considerations.   Let $f:[0,1]\to[0,1]$ be the Cantor function (134H)
and set $g(x)=\bover12(x+f(x))$ for $x\in [0,1]$.   Then
$g:[0,1]\to[0,1]$ is a homeomorphism (134I);  set
$\phi=g^{-1}:[0,1]\to[0,1]$.   We see that if $0\le x\le y\le 1$ then
$g(y)-g(x)\ge\bover12(y-x)$;  equivalently, $\phi(y)-\phi(x)\le 2(y-x)$
whenever $0\le x\le y\le 1$, so that $\phi$ is a Lipschitz function,
therefore absolutely continuous (262Bc).   If
$D=\{x:\phi'(x)$ is defined$\}$, then
$[0,1]\setminus D$ is negligible (225Cb), so
$[0,1]\setminus\phi[D]=\phi[\,[0,1]\setminus D]$ is negligible (262Da).
I noted in 134I that there is a measurable function $h:[0,1]\to\Bbb R$
such that the composition $h\phi$ is not measurable;  now
$h(\phi\restr D)=(h\phi)\restr D$ cannot be measurable, even though
$\phi\restr D$ is differentiable.
}%end of comment

\leader{262L}{}\cmmnt{ It will be convenient to be able to call on the
following straightforward result.

\medskip

\noindent}{\bf Lemma} Suppose that $D\subseteq\BbbR^r$ and $x\in\BbbR^r$
are such that
$\lim_{\delta\downarrow 0}
 \Bover{\mu^*(D\cap B(x,\delta))}{\mu B(x,\delta)}=1$.   Then
$\lim_{z\to\tbf{0}}\Bover{\rho(x+z,D)}{\|z\|}=0$, where
$\rho(x+z,D)=\inf_{y\in D}\|x+z-y\|$.

\proof{ Let $\epsilon>0$.  Let $\delta_0>0$ be such that

\Centerline{$\mu^*(D\cap B(x,\delta))
>(1-(\Bover{\epsilon}{1+\epsilon})^r)\mu B(x,\delta)$}

\noindent whenever $0<\delta\le\delta_0$.   Take any $z$ such that
$0<\|z\|\le\delta_0/(1+\epsilon)$.   \Quer\ Suppose, if possible, that
$\rho(x+z,D)>\epsilon\|z\|$.   Then $B(x+z,\epsilon\|z\|)\subseteq
B(x,(1+\epsilon)\|z\|)\setminus D$, so

$$\eqalign{\mu^*(D\cap B(x,(1+\epsilon)\|z\|))
&\le\mu B(x,(1+\epsilon)\|z\|)-\mu B(x+z,\epsilon\|z\|)\cr
&=(1-(\Bover{\epsilon}{1+\epsilon})^r)\mu B(x,(1+\epsilon)\|z\|),\cr}$$

\noindent which is impossible, as $(1+\epsilon)\|z\|\le\delta_0$.\
\BanG\   Thus $\rho(x+z,D)\le\epsilon\|z\|$.   As $\epsilon$ is
arbitrary, this proves the result.
}%end of proof of 262L

\cmmnt{\medskip

\noindent{\bf Remark} There is a word for this;  see 261Yg.}

\leader{262M}{}\cmmnt{ I come now to the first result connecting
Lipschitz functions with differentiable functions.   I approach it
through a substantial lemma which will be the foundation of \S263.

\medskip

\noindent}{\bf Lemma} Let $\phi$ be a
function from a subset $D$ of $\BbbR^r$ to $\BbbR^s$ which is
differentiable at
each point of its domain.   For each $x\in D$ let $T(x)$ be a derivative
of $\phi$.   Let $M_{sr}$ be the set of $s\times r$ matrices and
$\zeta:A\to\ooint{0,\infty}$ a strictly positive function, where
$A\subseteq M_{sr}$ is a non-empty set containing $T(x)$ for every
$x\in D$.   Then we
can find sequences $\sequencen{D_n}$, $\sequencen{T_n}$ such that

(i) $\sequencen{D_n}$ is a partition of $D$ into sets which are
relatively measurable in $D$\cmmnt{, that is, are intersections of $D$ with measurable subsets of $\BbbR^r$};

(ii) $T_n\in A$ for every $n$;

(iii) $\|\phi(x)-\phi(y)-T_n(x-y)\|\le\zeta(T_n)\|x-y\|$ for every
$n\in\Bbb N$ and $x$, $y\in D_n$;

(iv) $\|T(x)-T_n\|\le\zeta(T_n)$ for every $x\in D_n$.

\proof{{\bf (a)} The first step is to note that there is a sequence
$\sequencen{S_n}$ in $A$ such that

\Centerline{$A
\subseteq\bigcup_{n\in\Bbb N}\{T:T\in M_{sr},\,\|T-S_n\|<\zeta(S_n)\}$.}

\noindent\Prf\ (Of course this is a standard result about
separable metric
spaces.) Write $Q$ for the set of matrices in $M_{sr}$ with rational
coefficients;  then there is a natural bijection between $Q$ and
$\Bbb Q^{sr}$, so $Q$ and $Q\times\Bbb N$ are countable.   Enumerate
$Q\times\Bbb N$ as $\sequencen{(R_n,k_n)}$.   For each $n\in\Bbb N$,
choose $S_n\in A$ by the rule

\quad --- if there is an $S\in A$ such that $\{T:\|T-R_n\|\le
2^{-k_n}\}\subseteq\{T:\|T-S\|<\zeta(S)\}$, take such an $S$ for $S_n$;

\quad --- otherwise, take $S_n$ to be any member of $A$.

\noindent I claim that this works.   For let $S\in A$.   Then
$\zeta(S)>0$;  take $k\in\Bbb N$ such that $2^{-k}<\zeta(S)$.   Take
$R^*\in Q$ such that $\|R^*-S\|<\min(\zeta(S)-2^{-k},2^{-k})$;  this
is possible because $\|R-S\|$ will be small whenever all the
coefficients
of $R$ are close enough to the corresponding coefficients of $S$
(262Ha), and we can find rational numbers to achieve this.   Let
$n\in\Bbb N$ be such that $R^*=R_n$ and $k=k_n$.   Then

\Centerline{$\{T:\|T-R_n\|\le 2^{-k_n}\}
\subseteq\{T:\|T-S\|<\zeta(S)\}$}

\noindent (because $\|T-S\|\le\|T-R_n\|+\|R_n-S\|$), so we must have
chosen $S_n$ by the first part of the rule above, and

\Centerline{$S\in\{T:\|T-R_n\|\le 2^{-k_n}\}
\subseteq\{T:\|T-S_n\|<\zeta(S_n)\}$.}

\noindent As $S$ is arbitrary, this proves the result.\ \Qed

\medskip

{\bf (b)} Enumerate $\Bbb Q^r\times\Bbb Q^r\times\Bbb N$ as
$\sequencen{(q_n,q'_n,m_n)}$.   For each $n\in\Bbb N$, set

$$\eqalign{H_n&=\{x:x\in[q_n,q'_n]\cap D,\,
\|\phi(y)-\phi(x)-S_{m_n}(y-x)\|\le\zeta(S_{m_n})\|y-x\|\cr
&\mskip370mu
    \text{for every }y\in[q_n,q'_n]\cap D\}\cr
&=[q_n,q'_n]\cap D\cap\bigcap_{y\in[q_n,q'_n]\cap D}
       \{x:x\in D,\cr
&\mskip270mu
    \|\phi(y)-\phi(x)-S_{m_n}(y-x)\|\le\zeta(S_{m_n})\|y-x\|\}.\cr}$$

\noindent Because $\phi$ is continuous, $H_n=D\cap\overline{H}_n$,
writing $\overline{H}_n$ for the closure of $H_n$, so $H_n$ is
relatively measurable in $D$.
Note that if $x$, $y\in H_n$, then $y\in D\cap[q_n,q'_n]$, so that

\Centerline{$\|\phi(y)-\phi(x)-S_{m_n}(y-x)\|\le\zeta(S_{m_n})\|y-x\|$.}


Set

\Centerline{$H_n'=\{x:x\in H_n,\,\|T(x)-S_{m_n}\|\le\zeta(S_{m_n})\}$.}

\medskip

{\bf (c)} $D=\bigcup_{n\in\Bbb N}H'_n$.   \Prf\ Let $x\in D$.   Then
$T(x)\in A$, so there is a $k\in\Bbb N$ such that
$\|T(x)-S_k\|<\zeta(S_k)$.   Let $\delta>0$ be such that

\Centerline{$\|\phi(y)-\phi(x)-T(x)(x-y)\|
\le(\zeta(S_k)-\|T(x)-S_k\|)\|x-y\|$}

\noindent whenever $y\in D$ and $\|y-x\|\le\delta$.   Then

$$\eqalign{\|\phi(y)-\phi(x)-S_k(x-y)\|
&\le(\zeta(S_k)-\|T(x)-S_k\|)\|x-y\|+\|T(x)-S_k\|\|x-y\|\cr
&\le\zeta(S_k)\|x-y\|\cr}$$

\noindent whenever $y\in D\cap B(x,\delta)$.   Let $q$, $q'\in\Bbb Q^r$
be such that $x\in[q,q']\subseteq B(x,\delta)$.   Let $n$ be such that
$q=q_n$, $q'=q'_n$ and $k=m_n$.   Then $x\in H'_n$.\ \Qed

\medskip

{\bf (d)} Write

\Centerline{$C_n=\{x:x\in H_n,\,\lim_{\delta\downarrow 0}
\Bover{\mu^*(H_n\cap B(x,\delta))}{\mu B(x,\delta)}=1\}$.}

\noindent Then $C_n\subseteq H'_n$.

\medskip

\Prf\ {\bf (i)} Take $x\in C_n$, and set $\tilde T=T(x)-S_{m_n}$.   I
have
to show that $\|\tilde T\|\le\zeta(S_{m_n})$.   Take $\epsilon>0$.   Let
$\delta_0>0$ be such that

\Centerline{$\|\phi(y)-\phi(x)-T(x)(y-x)\|\le\epsilon\|y-x\|$}

\noindent whenever $y\in D$ and $\|y-x\|\le\delta_0$.   Since

\Centerline{$\|\phi(y)-\phi(x)-S_{m_n}(y-x)\|\le\zeta(S_{m_n})\|y-x\|$}

\noindent whenever $y\in H_n$, we have

\Centerline{$\|\tilde T(y-x)\|\le(\epsilon+\zeta(S_{m_n}))\|y-x\|$}

\noindent whenever $y\in H_n$ and $\|y-x\|\le\delta_0$.

\medskip

\quad{\bf (ii)} By 262L, there is a $\delta_1>0$ such that
$(1+2\epsilon)\delta_1\le\delta_0$ and $\rho(x+z,H_n)\le\epsilon\|z\|$
whenever $0<\|z\|\le\delta_1$.   So if $\|z\|\le\delta_1$ there is a
$y\in
H_n$ such that $\|x+z-y\|\le 2\epsilon\|z\|$.   (If $z=0$ we can take
$y=x$.)   Now $\|x-y\|\le (1+2\epsilon)\|z\|\le\delta_0$, so

$$\eqalign{\|\tilde Tz\|
&\le\|\tilde T(y-x)\|+\|\tilde T(x+z-y)\|\cr
&\le(\epsilon+\zeta(S_{m_n}))\|y-x\|+\|\tilde T\|\|x+z-y\|\cr
&\le(\epsilon+\zeta(S_{m_n}))\|z\|
    +(\epsilon+\zeta(S_{m_n})+\|\tilde T\|)\|x+z-y\|\cr
&\le(\epsilon+\zeta(S_{m_n})+2\epsilon^2+2\epsilon\zeta(S_{m_n})
    +2\epsilon\|\tilde T\|)\|z\|.\cr}$$

\noindent And this is true whenever $0<\|z\|\le\delta_1$.
But multiplying this inequality by
suitable positive scalars we see that

\Centerline{$\|\tilde Tz\|
\le\bigl(\epsilon+\zeta(S_{m_n})+2\epsilon^2+2\epsilon\zeta(S_{m_n})
+2\epsilon\|\tilde T\|\bigr)\|z\|$}

\noindent for all $z\in\BbbR^r$, and

\Centerline{$\|\tilde T\|
\le\epsilon+\zeta(S_{m_n})+2\epsilon^2+2\epsilon\zeta(S_{m_n})
  +2\epsilon\|\tilde T\|$.}

\noindent As $\epsilon$ is
arbitrary, $\|\tilde T\|\le\zeta(S_{m_n})$, as claimed.\ \Qed

\medskip

{\bf (e)} By 261Da, $H_n\setminus C_n$ is negligible for every $n$, so
$H_n\setminus H'_n$ is negligible, and

\Centerline{$H'_n=D\cap(\overline{H}_n\setminus(H_n\setminus H'_n))$}

\noindent is relatively measurable in $D$.   Set

\Centerline{$D_n=H'_n\setminus\bigcup_{k<n}H'_k$,
\quad$T_n=S_{m_n}$}

\noindent for each $n$;  these serve.
}%end of proof of 262M


\leader{262N}{Corollary} Let $\phi$ be a function from a subset $D$ of
$\Bbb R^r$ to $\BbbR^s$, and suppose that $\phi$ is differentiable relative to
its domain at each point of $D$.   Then $D$ can be expressed as the
union of
a disjoint sequence $\sequencen{D_n}$ of relatively measurable subsets
of $D$ such that $\phi\restr D_n$ is Lipschitz for each $n\in\Bbb N$.

\proof{ In 262M, take $\zeta(T)=1$ for every $T\in A=M_{sr}$.   If $x$,
$y\in D_n$ then

$$\eqalign{\|\phi(x)-\phi(y)\|
&\le\|\phi(x)-\phi(y)-T_n(x-y)\|+\|T_n(x-y)\|\cr
&\le\|x-y\|+\|T_n\|\|x-y\|,\cr}$$

\noindent so $\phi\restr D_n$ is $(1+\|T_n\|)$-Lipschitz.
}%end of proof of 262N

\leader{262O}{Corollary} Suppose that $\phi$ is an injective function
from a measurable subset $D$ of $\BbbR^r$ to $\BbbR^r$, and that
$\phi$ is differentiable relative to its domain at every point of $D$.

(a) If $A\subseteq D$ is negligible, $\phi[A]$ is negligible.

(b) If $E\subseteq D$ is measurable, then $\phi[E]$ is
measurable.

(c) If $D$ is measurable and $f$ is a measurable function defined on a
subset of $\BbbR^r$, then $f\phi^{-1}$ is measurable.

(d) If $H\subseteq\BbbR^r$ and $\phi^{-1}$ is defined almost everywhere
in $H$, and if $f$ is a function defined almost everywhere in
$\Bbb R^r$, then $f\phi^{-1}$ is defined almost everywhere in $H$.

\proof{ Take a sequence $\sequencen{D_n}$ as in 262N,
and apply 262E to $\phi\restr D_n$ for each $n$.
}%end of proof of 262O

\leader{262P}{Corollary} Let $\phi$ be a function from a a subset $D$ of
$\BbbR^r$ to $\BbbR^s$, and suppose that $\phi$ is differentiable
relative to its domain, with a derivative $T(x)$, at each point $x\in
D$.   Then the function $x\mapsto T(x)$ is measurable in the sense that
$\tau_{ij}:D\to\Bbb R$ is measurable for all $i\le s$ and $j\le r$,
where $\tau_{ij}(x)$ is the
$(i,j)$th coefficient of the matrix $T(x)$ for all $i$, $j$ and $x$.

\proof{ For each $k\in\Bbb N$, apply 262M with $\zeta(T)=2^{-k}$ for
each $T\in A=M_{sr}$, obtaining sequences $\sequencen{D_{kn}}$ of
relatively measurable subsets of $D$ and $\sequencen{T_{kn}}$ in
$M_{sr}$.   Let
$\tau^{(kn)}_{ij}$ be the $(i,j)$th coefficient of $T_{kn}$.   Then we
have functions $f_{ijk}:D\to\Bbb R$ defined by setting

\Centerline{$f_{ijk}(x)=\tau^{(kn)}_{ij}$ if $x\in D_{kn}$.}

\noindent Because the $D_{kn}$ are relatively measurable, the $f_{ijk}$
are
measurable functions.   For $x\in D_{kn}$,
\Centerline{$|\tau_{ij}(x)-f_{ijk}(x)|\le\|T(x)-T_n\|\le 2^{-k}$,}

\noindent so $|\tau_{ij}(x)-f_{ijk}(x)|\le 2^{-k}$ for every $x\in D$,
and

\Centerline{$\tau_{ij}=\lim_{k\to\infty}f_{ijk}$}

\noindent is measurable, as claimed.
}%end of proof of 262P

\leader{*262Q}{}\cmmnt{ This concludes the part of the section
which is essential for the rest of the chapter.   However the main
results of \S263
will I think be better understood if you are aware of the fact that any
Lipschitz function is differentiable (relative to its domain) almost
everywhere in its domain.   I devote the next couple of pages to a proof
of this fact, which apart from its intrinsic interest is a useful
exercise.

\medskip

\noindent}{\bf Rademacher's theorem} Let $\phi$ be a Lipschitz function
from a subset
of $\BbbR^r$ to $\BbbR^s$.   Then $\phi$ is
differentiable relative to its domain almost everywhere in its domain.

\proof{{\bf (a)} By 262Ba and 262Ib, it will be enough to
deal with the
case $s=1$.   By 262Bb, there is a Lipschitz function
$\tilde\phi:\BbbR^r\to\Bbb R$ extending $\phi$;  now $\phi$ is
differentiable with respect to its domain at any point of $\dom\phi$ at
which $\tilde \phi$ is differentiable, so it will be enough if I can
show that $\tilde\phi$ is differentiable almost everywhere.   To make
the notation more agreeable to the eye, I will suppose that $\phi$
itself was defined everywhere in $\BbbR^r$.   Let $\gamma$ be a
Lipschitz constant for $\phi$.

The proof proceeds by induction on $r$.   If $r=1$, we have a Lipschitz
function $\phi:\Bbb R\to\Bbb R$;  now $\phi$ is absolutely
continuous in any bounded interval (262Bc), therefore differentiable
almost everywhere (225Cb).   Thus the
induction starts.   The rest of the proof is devoted to the inductive
step to $r>1$.

\medskip

{\bf (b)} The first step is to show that all the partial derivatives
$\pd{\phi}{\xi_j}$ are defined almost everywhere and are Borel
measurable.
\Prf\ Take $j\le r$.   For $q\in\Bbb Q\setminus\{0\}$ set

\Centerline{$\Delta_q(x)=\Bover{1}{q}(\phi(x+qe_j)-\phi(x))$,}

\noindent writing $e_j$ for the $j$th unit vector of $\BbbR^r$.
Because $\phi$ is continuous, so is $\Delta_q$, so that $\Delta_q$ is a
Borel
measurable function for each $q$.   Next, for any $x\in\Bbb
R^r$,

\Centerline{$D^+(x)=\limsup_{\delta\to 0}\Bover1{\delta}
(\phi(x+\delta e_j)-\phi(x))
=\lim_{n\to\infty}\sup_{q\in\Bbb Q,0<|q|\le 2^{-n}}\Delta_q(x)$,}

\noindent so that the set on which $D^+(x)$ is defined in $\Bbb R$ is
Borel and
$D^+$ is a Borel measurable function.
Similarly,

\Centerline{$D^-(x)=\liminf_{\delta\to 0}\Bover1{\delta}
(\phi(x+\delta e_j)-\phi(x))$}

\noindent is a Borel measurable function with Borel domain.   So

\Centerline{$E=\{x:\pd{\phi}{\xi_j}(x)\text{ exists in }\Bbb R\}
=\{x:D^+(x)=D^-(x)\in\Bbb R\}$}

\noindent is a Borel set, and $\pd{\phi}{\xi_j}$ is a Borel measurable
function.

On the other hand, if we identify $\BbbR^r$ with
$\BbbR^J\times\Bbb R$, taking $J$ to be
$\{1,\ldots,j-1,j+1,\ldots,r\}$, then we can think of
Lebesgue measure $\mu$
on $\BbbR^r$ as being the product of Lebesgue measure $\mu_J$ on
$\BbbR^J$ with Lebesgue measure $\mu_1$ on $\Bbb R$ (251N).   Now for
every $y\in\BbbR^J$ we have a function $\phi_y:\Bbb R\to\Bbb R$ defined
by writing

\Centerline{$\phi_y(\sigma)=\phi(y,\sigma)$,}

\noindent and $E$ becomes

\Centerline{$\{(y,\sigma):\phi_y'(\sigma)$ is defined$\}$,}

\noindent so that all the sections

\Centerline{$\{\sigma:(y,\sigma)\in E\}$}

\noindent are conegligible subsets of $\Bbb R$, because every
$\phi_y$ is Lipschitz, therefore differentiable almost everywhere, as
remarked in part (a) of the proof.   Since we know that $E$ is
measurable, it must be conegligible, by Fubini's theorem (apply 252D or
252F to the complement of $E$).
Thus $\pd{\phi}{\xi_j}$ is defined almost everywhere, as claimed.\ \Qed

Write

\Centerline{$H=\{x:x\in\BbbR^r,\,\pd{\phi}{\xi_j}(x)
\text{ exists for every }j\le r\}$,}

\noindent so that $H$ is a conegligible Borel set in $\BbbR^r$.

\medskip

{\bf (c)} For the rest of this proof, I fix on the natural
identification of $\BbbR^r$ with $\BbbR^{r-1}\times\Bbb R$,
identifying $(\xi_1,\ldots,\xi_r)$ with
$((\xi_1,\ldots,\xi_{r-1}),\xi_r)$.
For $x\in H$, let $T(x)$ be the $1\times r$ matrix
$(\pd{\phi}{\xi_1}(x),\ldots,\pd{\phi}{\xi_r}(x))$.

\medskip

{\bf (d)} Set

\Centerline{$H_1=\{x:x\in H,\,\lim_{u\to\tbf{0}\text{ in }\BbbR^{r-1}}
\Bover{|\phi(x+(u,0))-\phi(x)-T(x)(u,0)|}{\|u\|}=0\}$.}

\noindent I claim that $H_1$ is conegligible in $\BbbR^r$.
\Prf\ This is really the same idea as in (b).   For $x\in H$, $x\in H_1$
iff

\inset{for every $\epsilon>0$ there is a $\delta>0$ such that

\Centerline{$|\phi(x+(u,0))-\phi(x)-T(x)(u,0)|\le\epsilon\|u\|$}

\noindent whenever $\|u\|\le\delta$,}

\noindent that is, iff

\inset{for every $m\in\Bbb N$ there is an $n\in\Bbb N$ such that

\Centerline{$|\phi(x+(u,0))-\phi(x)-T(x)(u,0)|\le 2^{-m}\|u\|$}

\noindent whenever $u\in\Bbb Q^{r-1}$ and $\|u\|\le 2^{-n}$.}

\noindent But for any particular $m\in\Bbb N$ and $u\in\Bbb Q^{r-1}$ the
set

\Centerline{$\{x:|\phi(x+(u,0))-\phi(x)-T(x)(u,0)|\le 2^{-m}\|u\|\}$}

\noindent is measurable, indeed Borel, because all the functions
$x\mapsto\phi(x+(u,0))$, $x\mapsto\phi(x)$, $x\mapsto T(x)(u,0)$ are
Borel measurable.   So $H_1$ is of the form

\Centerline{$\bigcap_{m\in\Bbb N}\bigcup_{n\in\Bbb N}
\bigcap_{u\in\Bbb Q^{r-1},\|u\|\le 2^{-n}}E_{mnu}$}

\noindent where every $E_{mnu}$ is a measurable set, and $H_1$ is
therefore measurable.

Now however observe that for any $\sigma\in\Bbb R$, the function

\Centerline{$v\mapsto\phi_{\sigma}(v)=\phi(v,\sigma):
\BbbR^{r-1}\to\Bbb R$}

\noindent is Lipschitz, therefore (by the inductive hypothesis)
differentiable almost everywhere in $\BbbR^{r-1}$;  and that
$(v,\sigma)\in H_1$ iff $(v,\sigma)\in H$ and $\phi_{\sigma}'(v)$ is
defined.   Consequently
$\{v:(v,\sigma)\in H_1\}$ is conegligible whenever
$\{v:(v,\sigma)\in H\}$
is, that is, for almost every $\sigma\in \Bbb R$;  so that $H_1$, being
measurable, must be conegligible.\ \Qed

\medskip

{\bf (e)} Now, for $q$, $q'\in\Bbb Q$ and $n\in\Bbb N$, set

\Centerline{$F(q,q',n)
=\{x:x\in\BbbR^r$,
  $q\le\Bover{\phi(x+(\tbf{0},\eta))-\phi(x)}{\eta}\le q'$
  whenever $0<|\eta|\le 2^{-n}\}$,}

\Centerline{$F_*(q,q',n)
=\{x:x\in F(q,q',n),\,\lim_{\delta\downarrow 0}
 \Bover{\mu^*(F(q,q',n)\cap B(x,\delta))}{\mu B(x,\delta)}=1\}$.}

\noindent By 261Da, $F(q,q',n)\setminus F_*(q,q',n)$ is negligible for
all $q$, $q'$, $n$, so that

\Centerline{$H_2
=H_1\setminus\bigcup_{q,q'\in\Bbb Q,n\in\Bbb N}
  (F(q,q',n)\setminus F_*(q,q',n))$}

\noindent is conegligible.

\medskip

{\bf (f)} I claim that $\phi$ is differentiable at every point of $H_2$.
\Prf\ Take $x=(u,\sigma)\in H_2$.   Then $\alpha=\Pd{\phi}{\xi_r}(x)$
and $T=T(x)$ are defined.   Let $\gamma$ be a Lipschitz constant for
$\phi$.

Take $\epsilon>0$;  take $q$, $q'\in\Bbb Q$ such that
$\alpha-\epsilon\le q<\alpha<q'\le\alpha+\epsilon$.   There must be an
$n\in\Bbb N$ such that $x\in F(q,q',n)$;  consequently
$x\in F_*(q,q',n)$,
by the definition of $H_2$.   By 262L, there is a $\delta_0>0$ such that
$\rho(x+z,F(q,q',n))\le\epsilon\|z\|$ whenever $\|z\|\le\delta_0$.
Next, there is a $\delta_1>0$ such that
$|\phi(x+(v,0))-\phi(x)-T(v,0)|\le\epsilon\|v\|$ whenever
$v\in\BbbR^{r-1}$ and $\|v\|\le\delta_1$.   Set

\Centerline{$\delta=\min(\delta_0,\delta_1,2^{-n})/(1+2\epsilon)>0$.}

Suppose that $z=(v,\tau)\in\BbbR^r$ and that $\|z\|\le\delta$.   Because
$\|z\|\le\delta_0$ there is an $x'=(u',\sigma')\in F(q,q',n)$ such that
$\|x+z-x'\|\le2\epsilon\|z\|$;  set $x^*=(u',\sigma)$.   Now

\Centerline{$\max(\|u-u'\|,|\sigma-\sigma'|)\le
\|x-x'\|\le(1+2\epsilon)\|z\|\le\min(\delta_1,2^{-n})$}

\noindent and $x^*=x+(u'-u,0)$, so

\Centerline{$|\phi(x^*)-\phi(x)-T(x^*-x)|
\le\epsilon\|u'-u\|\le\epsilon(1+2\epsilon)\|z\|$.}

\noindent But also

\Centerline{$|\phi(x')-\phi(x^*)-T(x'-x^*)|
=|\phi(x')-\phi(x^*)-\alpha(\sigma'-\sigma)|
\le\epsilon|\sigma'-\sigma|\le\epsilon(1+ 2\epsilon)\|z\|$,}

\noindent because $x'\in F(q,q',n)$ and $|\sigma-\sigma'|\le 2^{-n}$,
so that (if $x'\ne x^*$)

\Centerline{$\alpha-\epsilon\le q
\le\Bover{\phi(x^*)-\phi(x')}{\sigma-\sigma'}\le q'\le\alpha+\epsilon$}

\noindent and

\Centerline{$\bigl|\Bover{\phi(x')-\phi(x^*)}
{\sigma'-\sigma}-\alpha\bigr|\le\epsilon$.}
\noindent Finally,

\Centerline{$|\phi(x+z)-\phi(x')|
\le\gamma\|x+z-x'\|\le2\gamma\epsilon\|z\|$,}

\Centerline{$|Tz-T(x'-x)|\le\|T\|\|x+z-x'\|\le2\epsilon\|T\|\|z\|$.}

Putting all these together,

$$\eqalign{|\phi(x+z)-\phi x-Tz|
&\le|\phi(x+z)-\phi(x')|
+|T(x'-x)-Tz|\cr
&\qquad
+|\phi(x')-\phi(x^*)-T(x'-x^*)|
+|\phi(x^*)-\phi(x)-T(x^*-x)|\cr
&\le2\gamma\epsilon\|z\|
+2\epsilon\|T\|\|z\|
+\epsilon(1+2\epsilon)\|z\|
+\epsilon(1+2\epsilon)\|z\|\cr
&=\epsilon(2\gamma+2\|T\|+2+4\epsilon)\|z\|.\cr}$$

\noindent And this is true whenever $\|z\|\le\delta$.   As $\epsilon$ is
arbitrary, $\phi$ is differentiable at $x$.\ \Qed

Thus $\{x:\phi$ is differentiable at $x\}$ includes $H_2$ and is
conegligible;  and the induction continues.
}%end of proof of 262Q

\exercises{
\leader{262X}{Basic exercises (a)} Let $\phi$ and $\psi$ be Lipschitz
functions from subsets of $\BbbR^r$ to $\BbbR^s$.   Show that
$\phi+\psi$ is a Lipschitz function from $\dom\phi\cap\dom\psi$ to
$\Bbb R^s$.
%262A

\header{262Xb}{\bf (b)} Let $\phi$ be a Lipschitz function from a subset
of $\BbbR^r$ to $\BbbR^s$, and $c\in\Bbb R$.   Show that $c\phi$ is a
Lipschitz function.
%262A

\header{262Xc}{\bf (c)} Suppose $\phi:D\to\BbbR^s$ and
$\psi:E\to\BbbR^q$ are Lipschitz functions,
where $D\subseteq\BbbR^r$ and $E\subseteq\BbbR^s$.   Show that the
composition $\psi\phi:D\cap\phi^{-1}[E]\to\BbbR^q$ is Lipschitz.
%262A

\header{262Xd}{\bf (d)} Suppose $\phi$, $\psi$ are functions from
subsets of $\BbbR^r$ to $\BbbR^s$, and suppose that
$x\in\dom\phi\cap\dom\psi$ is such that each function is differentiable
relative to its domain at $x$, with derivatives $S$, $T$ there.   Show
that $\phi+\psi$ is differentiable relative to its domain at $x$, and
that $S+T$ is a derivative of $\phi+\psi$ at $x$.
%262F

\header{262Xe}{\bf (e)} Suppose that $\phi$ is a function from a subset
of $\BbbR^r$ to $\BbbR^s$, and is differentiable relative to its
domain at $x\in\dom\phi$.   Show that $c\phi$ is differentiable relative
to its domain at $x$ for every $c\in\Bbb R$.
%262F

\sqheader 262Xf Suppose $\phi:D\to\BbbR^s$ and
$\psi:E\to\Bbb R^q$ are functions,
where $D\subseteq\BbbR^r$ and $E\subseteq\BbbR^s$;  suppose that
$\phi$ is differentiable relative to its domain at
$x\in D\cap\phi^{-1}[E]$, with an
$s\times r$ matrix $T$ a derivative there, and that $\psi$ is
differentiable
relative to its domain at $\phi(x)$, with a $q\times s$ matrix  $S$
a derivative there.   Show that the composition $\psi\phi$ is
differentiable relative to its
domain at $x$, and that the $q\times r$ matrix $ST$ is a derivative of
$\psi\phi$ at $x$.
%262F

\sqheader 262Xg Let $\phi:\BbbR^r\to\BbbR^s$ be a linear
operator, with associated matrix $T$.   Show that $\phi$ is
differentiable everywhere, with $\phi'(x)=T$ for every $x$.
%262F

\spheader 262Xh Let $G\subseteq\BbbR^r$ be a convex open set,
and $\phi:G\to\BbbR^s$ a function such that all the partial derivatives
$\pd{\phi_i}{\xi_j}$ are defined everywhere in $G$.   Show that $\phi$
is Lipschitz iff all the partial derivatives are bounded on $G$.
%262I

\spheader 262Xi Let $\phi:\BbbR^r\to\BbbR^s$ be a function.   Show that
$\phi$ is differentiable at $x\in\BbbR^r$ iff for every $m\in\Bbb N$
there are an $n\in\Bbb N$ and an $r\times s$ matrix $T$ with rational
coefficients
such that $\|\phi(y)-\phi(x)-T(y-x)\|\le 2^{-m}\|y-x\|$ whenever
$\|y-x\|\le 2^{-n}$.
%262I

\spheader 262Xj Suppose that $f$ is a real-valued function which is
integrable over $\BbbR^r$, and that $g:\BbbR^r\to\Bbb R$ is a bounded
differentiable function such that the partial derivative $\Pd{g}{\xi_j}$
is bounded, where $j\le r$.   Let $f*g$ be
the convolution of $f$ and $g$ (255E, 255L).
Show that $\Pd{}{\xi_j}(f*g)$
is defined everywhere and equal to $f*\Pd{g}{\xi_j}$.   \Hint{255Xd.}
%262+

\spheader 262Xk Let $(X,\Sigma,\mu)$ be a measure space,
$G\subseteq\BbbR^r$ an open set, and $f:X\times G\to\Bbb R$ a function.
Suppose that

\inset{(i) for every $x\in X$, $t\mapsto f(x,t):G\to\Bbb R$ is
differentiable;

(ii) there is an integrable function $g$ on $X$ such that
$|\pd{f}{\tau_j}(x,t)|\le g(x)$ whenever $x\in X$, $t\in G$ and
$j\le r$;

(iii) $\int|f(x,t)|\mu(dx)$ exists in $\Bbb R$ for every $t\in G$.}

\noindent Show that $t\mapsto\int f(x,t)\mu(dx):G\to\Bbb R$ is
differentiable.  \Hint{show first that, for a suitable $M$,
$|f(x,t)-f(x,t')|\le M|g(x)|\|t-t'\|$ for every  $t$, $t'\in G$ and
$x\in X$.}
%262+

\spheader 262Xl\dvAnew{2015} Let $f:[a,b]\to\Bbb R$ 
be an absolutely continuous
function, where $a\le b$, and $g:f[\,[a,b]\,]\to\Bbb R$ a Lipschitz
function.   Show that $gf$ is absolutely continuous.
%262Bc out of order query

\leader{262Y}{Further exercises (a)}
%\spheader 262Ya
Let $L$ be the space of all Lipschitz functions from
$\BbbR^r$ to $\BbbR^s$, and for $\phi\in L$ set

\Centerline{$\|\phi\|
=\|\phi(0)\|+\min\{\gamma:\gamma\in\coint{0,\infty}$,
$\|\phi(y)-\phi(x)\|\le\gamma\|y-x\|$ for every $x$, $y\in\BbbR^r\}$.}

\noindent Show that $(L,\|\,\,\|)$ is a Banach space.
%262A

\spheader 262Yb Show that if
$T=\langle\tau_{ij}\rangle_{i\le s,j\le r}$ is an $s\times r$ matrix
then the operator norm $\|T\|$, as defined in 262H, is at most
$\sqrt{\sum_{i=1}^s\sum_{j=1}^r\tau_{ij}^2}$.
%262H

\spheader 262Yc Let $\phi:D\to\Bbb R$ be any function, where
$D\subseteq\BbbR^r$.   Show that $H=\{x:x\in D,\,\phi$ is differentiable
relative to its domain at $x\}$ is relatively measurable in $D$, and
that $\pd{\phi}{\xi_j}\restr H$ is measurable for every $j\le r$.
%262I

\spheader 262Yd\dvAnew{2015}
Let $\phi:D\to\Bbb R$ be a function, where
$D\subseteq\BbbR^r$.
(i) Show that if $\phi$ is measurable then all its partial derivatives
are measurable.
(ii) Show that if $\phi$ is Borel
measurable then all its partial derivatives are Borel measurable.
%262Yc 262I 225Yg

\spheader 262Ye A function $\phi:\BbbR^r\to\Bbb R$ is {\bf smooth} if
all its partial derivatives
$\Bover{\partial\ldots\partial\phi}
{\partial\xi_i\partial\xi_j\ldots\partial\xi_l}$ are defined everywhere
in $\BbbR^r$ and are
continuous.   Show that if $f$ is integrable over $\BbbR^r$ and
$\phi:\BbbR^r\to\Bbb R$ is smooth and has bounded support then the
convolution $f*\phi$ is smooth.   \Hint{262Xj, 262Xk.}
%262Xk 262+

\spheader 262Yf For $\delta>0$ set
$\tilde\phi_{\delta}(x) =e^{1/(\delta^2-\|x\|^2)}$ if $\|x\|<\delta$,
$0$ if $\|x\|\ge\delta$;  set
$\alpha_{\delta}=\int\tilde\phi_{\delta}(x)dx$,
$\phi_{\delta}(x)=\alpha_{\delta}^{-1}\tilde\phi_{\delta}(x)$ for every
$x$.  (i) Show that $\phi_{\delta}:\BbbR^r\to\Bbb R$ is smooth and has
bounded support.   (ii) Show that if $f$ is integrable over $\BbbR^r$
then $\lim_{\delta\downarrow 0}\int|f(x)-(f*\phi_{\delta})(x)|dx=0$.
\Hint{start with continuous functions $f$ with bounded support, and use
242O.}
%262Ye 262+

\spheader 262Yg Show that if
$f$ is integrable over $\BbbR^r$ and $\epsilon>0$ there is a smooth
function $h$ with bounded support such that $\int|f-h|\le\epsilon$.
\Hint{{\it either} reduce to the case in which $f$ has bounded support
and use 262Yf {\it or} adapt the method of 242Xi.}
%262Yf 262+

\spheader 262Yh Suppose that $f$ is a real function which is integrable
over every bounded subset of $\BbbR^r$.   (i) Show that $f\times\phi$ is
integrable whenever $\phi:\BbbR^r\to\Bbb R$ is a smooth function with
bounded support.   (ii) Show that if $\int f\times\phi=0$ for every
smooth function with bounded support then $f=0$ a.e.   ({\it Hint\/}:
show that
$\int_{B(x,\delta)}f=0$ for every $x\in\BbbR^r$ and $\delta>0$, and use
261C.   Alternatively show that $\int_Ef=0$ first for $E=[b,c]$, then
for open sets $E$, then for arbitrary measurable sets $E$.)
%262Ye 262+

\spheader 262Yi Let $f$ be integrable over $\BbbR^r$, and for $\delta>0$
let $\phi_{\delta}:\BbbR^r\to\Bbb R$ be the function of 262Yf.   Show
that $\lim_{\delta\downarrow 0}(f*\phi_{\delta})(x)=f(x)$ for every $x$
in the Lebesgue set of $f$.   ({\it Hint\/}:  261Ye.)
%262Yf 262+

}%end of exercises

\endnotes{
\Notesheader{262} The emphasis of this section has turned out to be on
the connexions between the concepts of `Lipschitz function' and
`differentiable function'.   It is the delight of classical real
analysis
that such intimate relationships arise between concepts which belong to
different categories.   `Lipschitz functions' clearly belong to the
theory of metric spaces (I will return to this in \S264), while
`differentiable functions' belong to the theory of differentiable
manifolds, which is outside the scope of this volume.
I have written this section out carefully just in case there are readers
who have so far missed the theory of differentiable mappings between
multi-dimensional Euclidean spaces;  but it also gives me a chance to
work through the notion of `function differentiable relative to its
domain', which will make it possible in the next section to ride
smoothly past a variety of problems arising at boundaries.   The
difficulties I am concerned with arise in the first place with such
functions as the polar-coordinate transformation

\Centerline{$(\rho,\theta)\mapsto (\rho\cos\theta,\rho\sin\theta):
\{(0,0)\}\cup(\ooint{0,\infty}\times\ocint{-\pi,\pi})\to\BbbR^2$.}

\noindent In order to make this a bijection we have to do something
rather arbitrary, and the domain of the transformation cannot be an open
set.   On the definitions I am using, this function is differentiable
relative to its domain at every point of its domain, and we can apply
such results as 262O uninhibitedly.   You will observe that in this case
the non-interior points of the domain form a negligible set
$\{(0,0)\}\cup(\ooint{0,\infty}\times\{\pi\})$, so we can expect to be
able to ignore them;
and for most of the geometrically straightforward transformations that
the theory is applied to, judicious excision of negligible sets will
reduce problems to the case of honestly differentiable functions with
open domains.   But while open-domain theory will deal with a large
proportion of the most important examples, there is a danger that you
would be left with real misapprehensions concerning the scope of these
methods.

The essence of differentiability is that a differentiable function
$\phi$ is
approximable, near any given point of its domain, by an affine function.
The idea of 262M is to describe a widely effective method of dissecting
$D=\dom\phi$ into countably many pieces on each of which $\phi$ is
well controlled.   This will be applied in \S\S263 and 265 to investigate
the measure of $\phi[D]$;  but we already have several straightforward
consequences (262N-262P).

I have offered a number of results suggesting that (on the definitions I
have chosen) a derivative can be expected to share at least some
`descriptive' properties
with the original function;  see 222Yd, 225J, 225Yg, 262Yc, 262Yd.
For partial derivatives, there are complications concerning their domains
(419Yd, 431Yd) which do not arise with full derivatives (225J, 262Yc).
}%end of notes

\discrpage

\frfilename{mt263.tex}
\versiondate{4.4.13}
\copyrightdate{2000}

\def\diam{\mathop{\text{diam}}}
\def\dist{\mathop{\text{dist}}}

\def\chaptername{Change of variable in the integral}
\def\sectionname{Differentiable transformations in $\BbbR^r$}

\newsection{263}

\def\headlinesectionname{Differentiable transformations in
$\eightBbb R^r$}

This section is devoted to the proof of a single major theorem
(263D) concerning differentiable transformations between subsets of
$\BbbR^r$.   There will be a generalization of this result in \S265,
and those with some familiarity with the topic, or sufficient hardihood,
may wish to read \S264 before taking this section and \S265
together.   I end with a few simple corollaries and an extension of the
main result which can be made in the one-dimensional case (263J).

Throughout this section, as in the rest of the chapter, $\mu$ will
denote Lebesgue measure on $\BbbR^r$.

\leader{263A}{Linear \dvrocolon{transformations}}\cmmnt{ I begin with
the special
case of linear operators, which is not only the basis of the proof of
263D, but is also one of its most important
applications, and is indeed sufficient for many very striking results.

\medskip

\noindent}{\bf Theorem} Let $T$ be a real $r\times r$ matrix;  regard
$T$ as a
linear operator from $\BbbR^r$ to itself.   Let $J=|\det T|$ be the
modulus of its determinant.   Then

\Centerline{$\mu T[E]=J\mu E$}

\noindent for every measurable set $E\subseteq\BbbR^r$.   If
$T$ is a permutation (that is, if $J\ne 0$), then

\Centerline{$\mu F=J\mu(T^{-1}[F])$}

\noindent for every measurable $F\subseteq\BbbR^r$, and

\Centerline{$\int_F g\,d\mu=J\int_{T^{-1}[F]} gT\,d\mu$}

\noindent for every integrable function $g$ and measurable set $F$.

\proof{{\bf (a)} The first step is to show that $T[I]$ is
measurable for every half-open interval $I\subseteq\BbbR^r$.   \Prf\
Any non-empty half-open interval
$I=\coint{a,b}$ is a countable union of closed
intervals $I_n =[a,b-2^{-n}\tbf{1}]$, and each $I_n$ is compact (2A2F),
so that $T[I_n]$ is compact (2A2Eb), therefore closed (2A2Ec),
therefore measurable (115G), and $T[I]=\bigcup_{n\in\Bbb N}T[I_n]$
is measurable.\ \Qed

\medskip

{\bf (b)} Set $J^*=\mu T[\,\coint{\tbf{0},\tbf{1}}\,]$, where
$\tbf{0}=(0,\dots,0)$ and $\tbf{1}=(1,\ldots,1)$;  because
$T[\,\coint{\tbf{0},\tbf{1}}\,]$ is bounded, $J^*<\infty$.   (I will
eventually show that $J^*=J$.)   It is convenient to deal with the case
of singular $T$ first.   Recall that $T$, regarded as a linear
transformation from $\BbbR^r$ to itself, is either bijective or onto a
proper linear subspace.   In the latter case, take any $e\in\Bbb
R^r\setminus T[\BbbR^r]$;  then the sets

\Centerline{$T[\,\coint{\tbf{0},\tbf{1}}\,]+\gamma e$,}

\noindent as $\gamma$ runs over $[0,1]$, are disjoint and all of the
same measure $J^*$, because $\mu$ is translation-invariant (134A);
moreover, their union is bounded, so has finite outer measure.   As
there are infinitely many such $\gamma$, the common measure $J^*$ must
be zero.   Now observe that

\Centerline{$T[\BbbR^r]=\bigcup_{z\in \Bbb
Z^r}T[\,\coint{\tbf{0},\tbf{1}}\,]+Tz$,}

\noindent and

\Centerline{$\mu(T[\,\coint{\tbf{0},\tbf{1}}\,]+Tz)=J^*=0$}

\noindent for every $z\in\Bbb Z^r$, while $\Bbb Z^r$ is countable, so
$\mu T[\Bbb R^r]=0$.   At the same time, because
$T$ is singular, it has zero determinant, and $J=0$.
Accordingly

\Centerline{$\mu T[E]=0=J\mu E$}

\noindent for every measurable $E\subseteq\BbbR^r$, and we're done.

\medskip

{\bf (c)} Henceforth, therefore, let us assume that $T$ is
non-singular.
Note that it and its inverse are continuous, so that $T$ is a
homeomorphism, and $T[G]$ is open iff $G$ is open.

If $a\in\BbbR^r$ and $k\in\Bbb N$, then

\Centerline{$\mu T[\,\coint{a,a+2^{-k}\tbf{1}}\,]=2^{-kr}J^*$.}

\noindent   \Prf\ Set
$J^*_k=\mu T[\,\coint{\tbf{0},2^{-k}\tbf{1}}\,]$.   Now
$T[\,\coint{a,a+2^{-k}\tbf{1}}\,]
=T[\,\coint{\tbf{0},2^{-k}\tbf{1}}\,]+Ta$;  because $\mu$ is
translation-invariant, its measure also is $J^*_k$.   Next,
$\coint{\tbf{0},\tbf{1}}$ is expressible as a disjoint union of $2^{kr}$
sets of the form $\coint{a,a+2^{-k}\tbf{1}}$;  consequently,
$T[\,\coint{\tbf{0},\tbf{1}}\,]$ is expressible as a disjoint union of
$2^{kr}$ sets of the form $T[\,\coint{a,a+2^{-k}\tbf{1}}\,]$, and


\Centerline{$J^*=\mu T[\,\coint{\tbf{0},\tbf{1}}\,]=2^{kr}J^*_k$,}

\noindent that is,
$J^*_k=2^{-kr}J^*$, as claimed.\ \Qed

\medskip

{\bf (d)} Consequently $\mu T[G]=J^*\mu G$ for every open set
$G\subseteq\BbbR^r$.   \Prf\ For each $k\in\Bbb N$, set

\Centerline{$Q_k
=\{z:z\in\Bbb Z^r,\,\coint{2^{-k}z,2^{-k}z+2^{-k}\tbf{1}}\subseteq G\}$,}

\Centerline{$G_k=\bigcup_{z\in Q_k}
\coint{2^{-k}z,2^{-k}z+2^{-k}\tbf{1}}$.}

\noindent Then $G_k$ is a disjoint union of $\#(Q_k)$ sets of the form
$\coint{2^{-k}z,2^{-k}z+2^{-k}\tbf{1}}$, so $\mu G_k=2^{-kr}\#(Q_k)$;
also, $T[G_k]$ is a disjoint union of $\#(Q_k)$ sets of the form
$T[\,\coint{2^{-k}z,2^{-k}z+2^{-k}\tbf{1}}\,]$, so has measure
$2^{-kr}J^*\#(Q_k)=J^*\mu G_k$, using (c).

Observe next that $\sequence{k}{G_k}$ is a non-decreasing sequence with
union $G$, so that

\Centerline{$\mu T[G]=\lim_{k\to\infty}\mu T[G_k]
=\lim_{k\to\infty}J^*\mu G_k
=J^*\mu G$.   \Qed}

\medskip

{\bf (e)} It follows that $\mu^*T[A]=J^*\mu^* A$ for every
$A\subseteq\Bbb R^r$.   \Prf\ Given $A\subseteq\BbbR^r$ and
$\epsilon>0$, there are
open sets $G$, $H$ such that $G\supseteq A$, $H\supseteq T[A]$,
$\mu G\le\mu^* A+\epsilon$ and $\mu H\le\mu^*T[A]+\epsilon$ (134Fa).
Set $G_1=G\cap T^{-1}[H]$;  then $G_1$ is open because $T^{-1}[H]$
is.   Now $\mu T[G_1]=J^*\mu G_1$, so

$$\eqalign{\mu^* T[A]
&\le\mu T[G_1]
=J^*\mu G_1
\le J^*\mu^* A+J^*\epsilon\cr
&\le J^*\mu G_1+J^*\epsilon
=\mu T[G_1]+J^*\epsilon
\le\mu H+J^*\epsilon\cr
&\le\mu^*T[A]+\epsilon+J^*\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\mu^*T[A]=J^*\mu^* A$.\ \Qed

\medskip

{\bf (f)} Consequently $\mu T[E]$ exists and is equal to $J^*\mu E$
for every measurable $E\subseteq\BbbR^r$.   \Prf\ Let
$E\subseteq\BbbR^r$ be measurable, and take any
$A\subseteq\BbbR^r$.   Set $A'=T^{-1}[A]$.   Then

$$\eqalign{\mu^*(A\cap T[E])+\mu^*(A\setminus T[E])
&=\mu^*(T[A'\cap E])+\mu^*(T[A'\setminus E])\cr
&=J^*(\mu^*(A'\cap E)+\mu^*(A'\setminus E))\cr
&=J^*\mu^* A'
=\mu^*T[A']
=\mu^* A.\cr}$$

\noindent As $A$ is arbitrary, $T[E]$ is measurable, and now


\Centerline{$\mu T[E]=\mu^*T[E]=J^*\mu^* E=J^*\mu E$.  \Qed}

\medskip

{\bf (g)} We are at last ready for the calculation of $J^*$.
Recall that the matrix $T$ must be expressible as $PDQ$, where $P$ and
$Q$ are orthogonal matrices and $D$ is diagonal, with non-negative
diagonal entries (2A6C).  Now we must have

\Centerline{$T[\,\coint{\tbf{0},\tbf{1}}\,]
=P[D[Q[\,\coint{\tbf{0},\tbf{1}}\,]]]$,}

\noindent so, using (f),

\Centerline{$J^*=J^*_PJ^*_DJ^*_Q$,}

\noindent where $J^*_P=\mu P[\,\coint{\tbf{0},\tbf{1}}\,]$, etc.
Now
we find that $J^*_P=J^*_Q=1$.   \Prf\ Let $B=B(\tbf{0},1)$ be the unit
ball of $\BbbR^r$.    Because $B$ is closed, it is measurable;  because
it is
bounded, $\mu B<\infty$;  and because $B$ includes the non-empty
half-open interval $\coint{\tbf{0},r^{-1/2}\tbf{1}}$, $\mu B>0$.   Now
$P[B]=Q[B]=B$, because $P$ and $Q$ are orthogonal matrices;
so we have

\Centerline{$\mu B=\mu P[B]=J^*_P\mu B$,}

\noindent and $J^*_P$ must be $1$;  similarly, $J^*_Q=1$.\ \Qed

\medskip

{\bf (h)} So we have only to calculate $J^*_D$.   Suppose the
coefficients of $D$ are $\delta_1,\ldots,\delta_r\ge 0$, so that
$Dx=(\delta_1\xi_1,\ldots,\delta_r\xi_r)\penalty-100=d\times x$.   We have
been assuming since the beginning of (c) that $T$ is non-singular, so
no $\delta_i$ can be $0$.   Accordingly

\Centerline{$D[\,\coint{\tbf{0},\tbf{1}}\,]
=\coint{\tbf{0},d}$,}

\noindent and

\Centerline{$J^*_D=\mu\coint{\tbf{0},d}=\prod_{i=1}^r\delta_i
=\det D$.}

\noindent Now because $P$ and $Q$ are orthogonal, both have determinant
$\pm 1$, so $\det T=\pm\det D$ and $J^*=\pm\det T$;  because $J^*$ is
surely non-negative, $J^*=|\det T|=J$.

\medskip

{\bf (i)} Thus $\mu T[E]=J\mu E$ for every Lebesgue measurable
$E\subseteq\BbbR^r$.   As $T$ is non-singular, we may use the
above argument to show
that $T^{-1}[F]$ is measurable for every measurable $F$, and

\Centerline{$\mu F=\mu T[T^{-1}[F]]=J\mu T^{-1}[F]
=\int J\times\chi(T^{-1}[F])\,d\mu$,}

\noindent identifying $J$ with the constant function with value $J$.
By 235A,

\Centerline{$\int_F g\,d\mu=\int_{T^{-1}[F]}JgT\,d\mu
=J\int_{T^{-1}[F]}gT\,d\mu$}

\noindent for every integrable function $g$ and measurable set $F$.
}%end of proof of 263A

\cmmnt{
\leader{263B}{Remark} Perhaps I should have warned you that I
should be calling on the results of \S235.   But if they were fresh in
your mind the formulae of the statement of the theorem will have
recalled them, and if not then it is perhaps better to turn back to them
now rather than before reading the theorem, since they are used only in
the last sentence of the proof.

I have taken the argument above at a leisurely,
not to say pedestrian, pace.   The point is that while the
translation-invariance of Lebesgue measure, and its behaviour under
simple magnification of a single coordinate, are more or less built into
the definition, its behaviour under general rotations is not, since a
rotation takes half-open intervals into skew cuboids.   Of
course the calculation of the measure of such an object is not really
anything to do with the Lebesgue theory, and it will be clear that much
of the argument would apply equally to any geometrically reasonable
notion of $r$-dimensional volume.

We come now to the central result of the chapter.  We have already done
some of the detail work in 262M.   The next basic element is the
following lemma.
}%end of comment

\leader{263C}{Lemma} Let $T$ be a real $r\times r$ matrix;  set
$J=|\det T|$.
Then for any $\epsilon>0$ there is a $\zeta=\zeta(T,\epsilon)>0$ such
that

(i) $|\det S-\det T|\le\epsilon$ whenever $S$ is an $r\times r$ matrix
and $\|S-T\|\le\zeta$;

(ii) whenever $D\subseteq\BbbR^r$ is a bounded set and
$\phi:D\to\BbbR^r$ is a function such that
$\|\phi(x)-\phi(y)-T(x-y)\|\le\zeta\|x-y\|$ for all $x$, $y\in D$, then
$|\mu^*\phi[D]-J\mu^*D|\le\epsilon\mu^*D$.

\proof{{\bf (a)} Of course (i) is the easy part.   Because
$\det S$ is a continuous function of the coefficients of $S$, and the
coefficients of $S$ must be close to those of $T$ if $\|S-T\|$ is small
(262Hb), there is surely a $\zeta_0>0$ such that
$|\det S-\det T|\le\epsilon$ whenever $\|S-T\|\le\zeta_0$.

\medskip

{\bf (b)(i)} Write $B=B(\tbf{0},1)$ for the unit ball of $\BbbR^r$, and
consider $T[B]$.      We know that $\mu T[B]=J\mu B$ (263A).    Let
$G\supseteq T[B]$ be an open set such that $\mu G\le (J+\epsilon)\mu B$
(134Fa again).   Because $B$ is compact (2A2F again) so is $T[B]$, so
there is a $\zeta_1>0$ such that $T[B]+\zeta_1B\subseteq G$ (2A2Ed).
This means that $\mu^*(T[B]+\zeta_1B)\le(J+\epsilon)\mu B$.

\medskip

\quad{\bf (ii)}
Now suppose that $D\subseteq\BbbR^r$ is a bounded set, and that
$\phi:D\to\BbbR^r$ is a function such that
$\|\phi(x)-\phi(y)-T(x-y)\|\le\zeta_1\|x-y\|$ for all $x$, $y\in D$.
Then if $x\in D$ and $\delta>0$,

\Centerline{$\phi[D\cap B(x,\delta)]\subseteq \phi(x) + \delta T[B] +
\delta\zeta_1B$,}

\noindent because if $y\in D\cap B(x,\delta)$ then $T(y-x)\in\delta
T[B]$ and

$$\eqalign{\phi(y)
&= \phi(x) + T(y-x) + (\phi(y)-\phi(x)-T(y-x))\cr
&\in\phi(x) + \delta T[B]+\zeta_1\|y-x\|B
\subseteq\phi(x)+\delta T[B]+\zeta_1\delta B.\cr}$$

\noindent Accordingly

$$\eqalign{\mu^*\phi[D\cap B(x,\delta)]
&\le\mu^*(\delta T[B]+\delta\zeta_1B)
=\delta^r\mu^*(T[B]+\zeta_1B)\cr
&\le\delta^r(J+\epsilon)\mu B
=(J+\epsilon)\mu B(x,\delta).\cr}$$

Let $\eta>0$.   Then there is a sequence $\sequencen{B_n}$ of balls in
$\BbbR^r$ such that $D\subseteq\bigcup_{n\in\Bbb N}B_n$,
$\sum_{n=0}^{\infty}\mu B_n\le\mu^*D+\eta$ and the sum of the measures
of those $B_n$ whose centres do not lie in $D$ is at most $\eta$ (261F).
Let $K$ be the set of those $n$ such that the centre of $B_n$ lies in
$D$.   Then $\mu^*\phi[D\cap B_n]\le(J+\epsilon)\mu B_n$ for every
$n\in K$.    Also, of course, $\phi$ is $(\|T\|+\zeta_1)$-Lipschitz, so
$\mu^*\phi[D\cap B_n]\le(\|T\|+\zeta_1)^r\mu B_n$
for $n\in\Bbb N\setminus K$ (262D).   Now

$$\eqalign{\mu^*\phi[D]
&\le\sum_{n=0}^{\infty}\mu^*\phi[D\cap B_n]\cr
&\le\sum_{n\in K}(J+\epsilon)\mu B_n
   +\sum_{n\in\Bbb N\setminus K}(\|T\|+\zeta_1)^r\mu B_n\cr
&\le(J+\epsilon)(\mu^*D+\eta)+\eta(\|T\|+\zeta_1)^r.\cr}$$

\noindent As $\eta$ is arbitrary,

\Centerline{$\mu^*\phi[D]\le (J+\epsilon)\mu^*D$.}

\medskip

{\bf (c)} If $J=0$, we can stop here, setting
$\zeta=\min(\zeta_0,\zeta_1)$;  for then we surely have $|\det S-\det
T|\le\epsilon$ whenever $\|S-T\|\le\zeta$, while if $\phi:D\to\Bbb
R^r$ is such that $\|\phi(x)-\phi(y)-T(x-y)\|\le\zeta\|x-y\|$ for all
$x$, $y\in D$, then

\Centerline{$|\mu^*\phi[D]-J\mu^*D|=\mu^*\phi[D]\le\epsilon\mu^*D$.}

\noindent If $J\ne 0$, we have more to do.   Because $T$ has non-zero
determinant, it has an inverse $T^{-1}$, and $|\det T^{-1}|=J^{-1}$.
As in (b-i) above, there is a $\zeta_2>0$ such that
$\mu^*(T^{-1}[B]+\zeta_2B)\le(J^{-1}+\epsilon')\mu B$, where
$\epsilon'=\epsilon/J(J+\epsilon)$.   Repeating (b), we see that if
$C\subseteq\Bbb R^r$ is bounded and $\psi:C\to\BbbR^r$ is such that
$\|\psi(u)-\psi(v)-T^{-1}(u-v)\|\le\zeta_2\|u-v\|$ for all $u$,
$v\in C$, then $\mu^*\psi[C]\le(J^{-1}+\epsilon')\mu^*C$.

Now suppose that
$D\subseteq\BbbR^r$ is bounded and $\phi:D\to\BbbR^r$ is such that
$\|\phi(x)-\phi(y)-T(x-y)\|\le\zeta_2' \|x-y\|$ for all $x$, $y\in D$,
where $\zeta_2'=\min(\zeta_2,\|T^{-1}\|)/2\|T^{-1}\|^2>0$.   Then

\Centerline{$\|T^{-1}(\phi(x)-\phi(y))-(x-y)\|
\le\|T^{-1}\|\zeta_2'\|x-y\|\le\Bover12\|x-y\|$}

\noindent for all $x$, $y\in D$, so $\phi$ must be injective;  set
$C=\phi[D]$ and $\psi=\phi^{-1}:C\to D$.   Note that $C$ is bounded,
because

\Centerline{$\|\phi(x)-\phi(y)\|\le(\|T\|+\zeta_2')\|x-y\|$}

\noindent whenever $x$, $y\in D$.   Also

\Centerline{$\|T^{-1}(u-v)-(\psi(u)-\psi(v))\|
\le\|T^{-1}\|\zeta_2'\|\psi(u)-\psi(v)\|\le\Bover12\|\psi(u)-\psi(v)\|$}

\noindent for all $u$, $v\in C$.   But this means that

\Centerline{$\|\psi(u)-\psi(v)\|-\|T^{-1}\|\|u-v\|
\le\Bover12\|\psi(u)-\psi(v)\|$}

\noindent and $\|\psi(u)-\psi(v)\|\le 2\|T^{-1}\|\|u-v\|$ for all $u$,
$v\in C$, so that

\Centerline{$\|\psi(u)-\psi(v)-T^{-1}(u-v)\|
\le 2\zeta_2'\|T^{-1}\|^2\|u-v\|\le\zeta_2\|u-v\|$}

\noindent for all $u$, $v\in C$.

It follows that

\Centerline{$\mu^*D=\mu^*\psi[C]\le(J^{-1}+\epsilon')\mu^*C
=(J^{-1}+\epsilon')\mu^*\phi[D]$,}

\noindent and

\Centerline{$J\mu^*D\le (1+J\epsilon')\mu^*\phi[D]$.}

\medskip

{\bf (d)} So if we set $\zeta=\min(\zeta_0,\zeta_1,\zeta_2')>0$, and if
$D\subseteq\BbbR^r$, $\phi:D\to\BbbR^r$ are such that  $D$ is bounded
and
$\|\phi(x)-\phi(y)-T(x-y)\|\le\zeta\|x-y\|$ for all $x$, $y\in D$, we
shall have

\Centerline{$\mu^*\phi[D]\le(J+\epsilon)\mu^*D$,}

\Centerline{$\mu^*\phi[D]\ge J\mu^*D-J\epsilon'\mu^*\phi[D]
\ge J\mu^*D-J\epsilon'(J+\epsilon)\mu^*D
= J\mu^*D-\epsilon\mu^*D$,}

\noindent so we get the required formula

\Centerline{$|\mu^*\phi[D]-J\mu^*D|\le\epsilon\mu^*D$.}
}%end of proof of 263C

\leader{263D}{}\cmmnt{ We are ready for the theorem.

\medskip

\noindent}{\bf Theorem} Let $D\subseteq\BbbR^r$ be any set, and
$\phi:D\to\BbbR^r$ a function differentiable relative to its domain at
each point of $D$.   For each $x\in D$ let $T(x)$ be a derivative of
$\phi$ relative to $D$ at $x$, and set $J(x)=|\det T(x)|$.   Then

\quad (i) $J:D\to\coint{0,\infty}$ is a measurable function,

\quad (ii) $\mu^*\phi[D]\le\int_DJ\,d\mu$,

\noindent allowing $\infty$ as the value of the integral.  If $D$ is
measurable, then

\quad (iii) $\phi[D]$ is measurable.

\noindent If $D$ is measurable and $\phi$ is injective, then

\quad (iv) $\mu\phi[D]=\int_DJ\,d\mu$,

\quad (v) for every real-valued function $g$ defined on a subset of
$\phi[D]$,

\Centerline{$\int_{\phi[D]}g\,d\mu=\int_{D}J\times g\phi\,d\mu$}

\noindent if either integral is defined in $[-\infty,\infty]$, provided
we interpret $J(x)g(\phi(x))$ as zero when $J(x)=0$ and $g(\phi(x))$ is
undefined.

\proof{{\bf (a)} To see that $J$ is measurable, use 262P;  the function
$T\mapsto|\det T|$ is a continuous function of the coefficients of $T$,
and the coefficients of $T(x)$ are measurable functions of $x$, by 262P,
so $x\mapsto|\det T(x)|$ is measurable (121K).   We also know that if
$D$ is measurable, $\phi[D]$ will be measurable, by 262Ob.   Thus (i)
and (iii) are done.

\medskip

{\bf (b)} For the moment, assume
that $D$ is bounded, and fix $\epsilon>0$.    For $r\times r$ matrices
$T$, take $\zeta(T,\epsilon)>0$ as in 263C.   Take $\sequencen{D_n}$,
$\sequencen{T_n}$ as in 262M, so that $\sequencen{D_n}$ is a disjoint
cover of $D$ by sets which are relatively measurable in $D$, and each
$T_n$ is an $r\times r$ matrix such that

\Centerline{$\|T(x)-T_n\|\le\zeta(T_n,\epsilon)$ whenever $x\in D_n$,}

\Centerline{$\|\phi(x)-\phi(y)-T_n(x-y)\|\le\zeta(T_n,\epsilon)\|x-y\|$
for all $x$, $y\in D_n$.}

\noindent Then, setting $J_n=|\det T_n|$, we have

\Centerline{$|J(x)-J_n|\le\epsilon$ for every $x\in D_n$,}

\Centerline{$|\mu^*\phi[D_n]-J_n\mu^*D_n|\le\epsilon\mu^*D_n$,}

\noindent by the choice of $\zeta(T_n,\epsilon)$.   So we have

\Centerline{$\int_DJ\,d\mu
\le\sum_{n=0}^{\infty}J_n\mu^*D_n+\epsilon\mu^*D
\le\int_DJ\,d\mu+2\epsilon\mu^*D$;}

\noindent I am using here the fact that all the $D_n$ are relatively
measurable in $D$, so that, in particular,
$\mu^*D=\sum_{n=0}^{\infty}\mu^*D_n$.   Next,

\Centerline{$\mu^*\phi[D]\le\sum_{n=0}^{\infty}\mu^*\phi[D_n]
\le\sum_{n=0}^{\infty}J_n\mu^*D_n+\epsilon\mu^*D$.}

\noindent Putting these together,

\Centerline{$\mu^*\phi[D]\le\int_DJ\,d\mu+2\epsilon\mu^*D$.}

If $D$ is measurable and $\phi$ is injective, then all the $D_n$ are
measurable subsets of $\BbbR^r$, so all the $\phi[D_n]$ are measurable,
and they are also disjoint.   Accordingly

\Centerline{$\int_DJ\,d\mu
\le\sum_{n=0}^{\infty}J_n\mu D_n+\epsilon\mu D
\le\sum_{n=0}^{\infty}(\mu\phi[D_n]+\epsilon\mu D_n)+\epsilon\mu D
=\mu\phi[D]+2\epsilon\mu D$.}

Since $\epsilon$ is arbitrary, we get

\Centerline{$\mu^*\phi[D]\le\int_DJ\,d\mu$,}

\noindent and if $D$ is measurable and $\phi$ is injective,

\Centerline{$\int_DJ\,d\mu\le\mu\phi[D]$;}

\noindent thus we have (ii) and (iv), on the assumption that $D$ is
bounded.

\medskip

{\bf (c)} For a general set $D$, set $B_k=B(\tbf{0},k)$;  then

\Centerline{$\mu^*\phi[D]=\lim_{k\to\infty}\mu^*\phi[D\cap B_k]
\le\lim_{k\to\infty}\int_{D\cap B_k}J\,d\mu
=\int_DJ\,d\mu$,}

\noindent with equality if $\phi$ is injective and $D$ is measurable.

\medskip

{\bf (d)} For part (v), I seek to show that the
hypotheses of 235J are satisfied, taking $X=D$ and $Y=\phi[D]$.   \Prf\
Set $G=\{x:x\in D,\,J(x)>0\}$.

\medskip

\quad\grheada\ If $F\subseteq\phi[D]$ is measurable, then there are
Borel sets $F_1$, $F_2$ such
that $F_1\subseteq F\subseteq F_2$ and $\mu(F_2\setminus F_1)=0$.   Set
$E_j=\phi^{-1}[F_j]$ for each $j$, so that
$E_1\subseteq \phi^{-1}[F]\subseteq E_2$, and both the sets $E_j$ are
measurable,
because $\phi$ and $\dom\phi$ are measurable.   Now, applying (iv) to
$\phi\restr E_j$,

\Centerline{$\int_{E_j}J\,d\mu=\mu\phi[E_j]=\mu(F_j\cap\phi[D])=\mu F$}

\noindent for both $j$, so $\int_{E_2\setminus E_1}J\,d\mu=0$ and $J=0$
a.e.\ on $E_2\setminus E_1$.   Accordingly
$J\times\chi(\phi^{-1}[F])\eae J\times\chi E_1$, and
$\int J\times\chi(\phi^{-1}[F])d\mu$ exists and is equal to
$\int_{E_1}J\,d\mu=\mu F$.   At the same time,
$(\phi^{-1}[F]\cap G)\symmdiff(E_1\cap G)$ is negligible, so
$\phi^{-1}[F]\cap G$ is measurable.

\medskip

\quad\grheadb\ If $F\subseteq\phi[D]$ and $G\cap\phi^{-1}[F]$ is
measurable, then we know that $\mu\phi[D\setminus G]=\int_{D\setminus
G}J=0$ (by (iv) applied to $\phi\restr D\setminus G$),
so $F\setminus\phi[G]$ must be negligible;  while
$F\cap\phi[G]=\phi[G\cap\phi^{-1}[F]]$ also is measurable, by (iii).
Accordingly $F$ is measurable whenever $G\cap\phi^{-1}[F]$ is
measurable.

\medskip

Thus all the hypotheses of 235J are satisfied.\ \QeD\   Now (v) can be
read off from the conclusion of 235J.
}%end of proof of 263D

\cmmnt{
\leader{263E}{Remarks (a)} This is a version of the classical result on
change of
variable in a many-dimensional integral.   What I here call $J(x)$ is
the {\bf Jacobian} of $\phi$ at $x$;  it describes the change in volumes
of objects near $x$, following the rule already established in 263A for
functions with constant derivative.   The idea of the
proof is also the classical one:  to break the set $D$ up into small
enough pieces $D_m$ for us to be able to approximate $\phi$ by affine
operators $y\mapsto \phi(x) + T_{m}(y-x)$ on each.   The potential
irregularity of the set $D$, which in this theorem may be any set, is
compensated for by a corresponding freedom in
choosing the sets $D_m$.   In fact there is a further decomposition of
the sets $D_m$ hidden in part (b-ii) of the proof of 263C;  each $D_m$
is essentially covered by a disjoint family of balls, the measures of
whose images we can estimate with an adequate accuracy.   There is
always a danger of a negligible exceptional set, and we need the crude
inequalities of the proof of 262D to deal with it.

\header{263Eb}{\bf (b)} Throughout the work of this chapter, from 261B
to 263D, I have chosen
balls $B(x,\delta)$ as the basic shapes to work with.   I think it
should be clear that in fact any reasonable shapes would do just as
well.    In particular, the `balls'

\Centerline{$B_1(x,\delta)
=\{y:\sum_{i=1}^r|\eta_i-\xi_i|\le\delta\}$,\quad
$B_{\infty}(x,\delta)=\{y:|\eta_i-\xi_i|\le\delta\Forall
i\}$}

\noindent would serve perfectly.   There are many alternatives.   We
could use sets of the form $C(x,k)$, for $x\in\BbbR^r$ and
$k\in\Bbb N$, defined to be the half-open cube of the form
$\coint{2^{-k}z,2^{-k}(z+\tbf{1})}$ with $z\in\Bbb Z^r$ containing $x$,
instead;  or even $C'(x,\delta)=\coint{x,x+\delta\tbf{1}}$.   In all
such cases we have versions of the density theorems (261Yb-261Yc) which
support the remaining theory.


\header{263Ec}{\bf (c)} I have presented 263D as a theorem about
differentiable functions, because that is the normal form in which one
uses it in
elementary applications.   However, the proof depends essentially on the
fact that a differentiable function is a countable union of Lipschitz
functions, and 263D would follow at once from the same theorem
proved for Lipschitz functions only.   Now the fact is that the theorem
applies to {\it any} countable union of Lipschitz functions,
because a Lipschitz function is differentiable almost everywhere.   For
more advanced work (see {\smc Federer 69} or {\smc Evans \& Gariepy 92},
or Chapter 47 in Volume 4) it seems clear that Lipschitz
functions are the vital ones, so I spell out the result.
}%end of comment

\leader{*263F}{Corollary} Let $D\subseteq\BbbR^r$ be any set and
$\phi:D\to\BbbR^r$ a Lipschitz function.   Let $D_1$ be the set of
points at which $\phi$ has a derivative relative to $D$, and for each
$x\in D_1$ let $T(x)$ be such a derivative, with $J(x)=|\det T(x)|$.
Then

\quad (i) $D\setminus D_1$ is negligible;

\quad (ii) $J:D_1\to\coint{0,\infty}$ is measurable;

\quad (iii) $\mu^*\phi[D]\le\int_DJ(x)dx$.

\noindent If $D$ is measurable, then

\quad (iv) $\phi[D]$ is measurable.

\noindent If $D$ is measurable and $\phi$ is injective, then

\quad (v) $\mu\phi[D]=\int_DJ\,d\mu$,

\quad (vi) for every real-valued function $g$ defined on a subset of
$\phi[D]$,

\Centerline{$\int_{\phi[D]}g\,d\mu=\int_{D}J\times g\phi\,d\mu$}

\noindent if either integral is defined in $[-\infty,\infty]$, provided
we interpret $J(x)g(\phi(x))$ as zero when $J(x)=0$ and $g(\phi(x))$ is
undefined.

\proof{ This is now just a matter of putting 262Q and 263D
together, with a little help from 262D.   Use 262Q to show that
$D\setminus D_1$ is negligible, 262D to show that $\phi[D\setminus D_1]$
is negligible, and apply 263D to $\phi\restr D_1$.
}%end of proof of 263F

\dvro{\leader{263G}{Proposition}
$\int_{-\infty}^{\infty}e^{-t^2/2}dt=\sqrt{2\pi}$.}
{\leader{263G}{Polar coordinates in the plane} I offer an elementary
example with a useful consequence.   Define $\phi:\BbbR^2\to\BbbR^2$
by setting $\phi(\rho,\theta)=(\rho\cos\theta,\rho\sin\theta)$ for
$\rho$, $\theta\in\BbbR^2$.   Then $\phi'(\rho,\theta)
=\Matrix{\cos\theta&-\rho\sin\theta\\ \sin\theta&\rho\cos\theta}$, so
$J(\rho,\theta)=|\rho|$ for all $\rho$, $\theta$.   Of course $\phi$ is
not injective, but if we restrict it to the domain
$D=\{(0,0)\}\cup\{(\rho,\theta):\rho>0,\,-\pi<\theta\le\pi\}$ then
$\phi\restr D$ is a bijection between $D$ and $\BbbR^2$, and

\Centerline{$\int g\,d\xi_1d\xi_2=\int_Dg(\phi(\rho,\theta))\rho\,d\rho
d\theta$}

\noindent for every real-valued function $g$ which is integrable over
$\BbbR^2$.

Suppose, in particular, that we set

\Centerline{$g(x)=e^{-\|x\|^2/2}
=e^{-\xi_1^2/2}e^{-\xi_2^2/2}$}

\noindent for $x=(\xi_1,\xi_2)\in\Bbb R$.   Then

\Centerline{$\int g(x)dx=\int e^{-\xi_1^2/2}d\xi_1
\int e^{-\xi_2^2/2}d\xi_2$,}

\noindent as in 253D.   Setting $I=\int e^{-t^2/2}dt$, we have
$\int g=I^2$.   (To see that $I$ is well-defined in $\Bbb R$, note that
the integrand is continuous, therefore measurable, and that

\Centerline{$\int_{-1}^1e^{-t^2/2}dt\le 2$,}

\Centerline{$\int_{-\infty}^{-1}e^{-t^2/2}dt
=\int_1^{\infty}e^{-t^2/2}dt
\le\int_1^{\infty}e^{-t/2}dt
=\lim_{a\to\infty}\int_1^ae^{-t/2}dt
=\Bover12e^{-1/2}$}

\noindent are both finite.)   Now looking at the alternative expression
we have

$$\eqalignno{I^2
&=\int g(x)dx
=\int_Dg(\rho\cos\theta,\rho\sin\theta)\rho\,d(\rho,\theta)\cr
&=\int_De^{-\rho^2/2}\rho\,d(\rho,\theta)
=\int_0^{\infty}\int_{-\pi}^{\pi}\rho e^{-\rho^2/2}d\theta d\rho\cr
\noalign{\noindent (ignoring the point $(0,0)$, which has zero measure)}
&=\int_0^{\infty}2\pi\rho e^{-\rho^2/2}d\rho
=2\pi\lim_{a\to\infty}\int_0^a\rho e^{-\rho^2/2}d\rho\cr
&=2\pi\lim_{a\to\infty}(-e^{-a^2/2}+1)
=2\pi.\cr}$$

\noindent Consequently

\Centerline{$\int_{-\infty}^{\infty}e^{-t^2/2}dt=I=\sqrt{2\pi}$,}

\noindent which is one of the many facts every mathematician should
know, and in particular is vital for Chapter 27 below.
}%end of dvro

\leader{263H}{Corollary} If $k\in\Bbb N$ is odd,

\Centerline{$\int_{-\infty}^{\infty}x^ke^{-x^2/2}dx=0$;}

\noindent if $k=2l\in\Bbb N$ is even, then

\Centerline{$\int_{-\infty}^{\infty}x^ke^{-x^2/2}dx
=\Bover{(2l)!}{2^ll!}\sqrt{2\pi}$.}

\proof{{\bf (a)} To see that all the integrals are well-defined and
finite, observe that
$\lim_{x\to\pm\infty}x^ke^{-x^2/4}=0$, so that
$M_k=\sup_{x\in\Bbb R}|x^ke^{-x^2/4}|$ is finite, and

\Centerline{$\int_{-\infty}^{\infty}|x^ke^{-x^2/2}|dx
\le M_k\int_{-\infty}^{\infty}e^{-x^2/4}dx<\infty$.}

\medskip

{\bf (b)} If $k$ is odd, then substituting $y=-x$ we get

\Centerline{$\int_{-\infty}^{\infty}x^ke^{-x^2/2}dx
=-\int_{-\infty}^{\infty}y^ke^{-y^2/2}dy$,}

\noindent so that both integrals must be zero.

\medskip

{\bf (c)} For even $k$, proceed by induction.   Set
$I_l=\int_{-\infty}^{\infty}x^{2l}e^{-x^2/2}dx$.
$I_0=\sqrt{2\pi}=\bover{0!}{2^00!}\sqrt{2\pi}$ by 263G.   For the
inductive step to $l+1\ge 1$, integrate by parts to see that

\Centerline{$\int_{-a}^ax^{2l+1}\cdot xe^{-x^2/2}dx
=-a^{2l+1}e^{-a^2/2}+(-a)^{2l+1}e^{-a^2/2}
  +\int_{-a}^a(2l+1)x^{2l}e^{-x^2/2}dx$}

\noindent for every $a\ge 0$.   Letting $a\to\infty$,

\Centerline{$I_{l+1}=(2l+1)I_l$.}

\noindent Because

\Centerline{$\Bover{(2(l+1))!}{2^{l+1}(l+1)!}\sqrt{2\pi}
=(2l+1)\Bover{(2l)!}{2^ll!}\sqrt{2\pi}$,}

\noindent the induction proceeds.
}%end of proof of 263H

\leader{263I}{}\cmmnt{ The following is a version of 263D for non-injective
transformations.

\medskip

\noindent}{\bf Theorem}\dvAnew{2013}
Let $D\subseteq\BbbR^r$ be a measurable set, and
$\phi:D\to\BbbR^r$ a function differentiable relative to its domain at
each point of $D$.   For each $x\in D$ let $T(x)$ be a derivative of
$\phi$ relative to $D$ at $x$, and set $J(x)=|\det T(x)|$.

(a) Let $\nu$ be counting measure on $\BbbR^r$.   Then
$\int_{\BbbR^r}\nu(\phi^{-1}[\{y\}])dy$ and
$\int_DJ\,d\mu$ are defined in $[0,\infty]$ and equal.

(b) Let $g$ be a real-valued function defined on a subset of $\phi[D]$
such that $\int_Dg(\phi(x))\det T(x)dx$ is defined in $\Bbb R$,
interpreting $g(\phi(x))\det T(x)$ as zero when $\det T(x)=0$ and
$g(\phi(x))$ is undefined.   Set

\Centerline{$C=\{y:y\in\phi[D]$, $\phi^{-1}[\{y\}]$ is finite$\}$,
\quad$R(y)=\sum_{x\in\phi^{-1}[\{y\}]}\sgn\det T(x)$}

\noindent for $y\in C$, where $\sgn(0)=0$ and
$\sgn(\alpha)=\Bover{\alpha}{|\alpha|}$ for non-zero $\alpha$.
If we interpret $g(y)R(y)$ as zero when $g(y)=0$ and $R(y)$ is undefined,
then $\int_{\phi[D]}g\times R\,d\mu$ is defined and equal to
$\int_Dg(\phi(x))\det T(x)dx$.

\proof{{\bf (a)} By 263D(i), $J$ is measurable, so
$\int_DJ\,d\mu$ is defined in $[0,\infty]$ and
$D_0=\{x:x\in D$, $J(x)=0\}$ is measurable.   Applying 263D(ii) to
$\phi\restr D_0$, we see that $\phi[D_0]$ is negligible.

Applying 262M to $\phi\restr D\setminus D_0$,
the set $A$ of non-singular
$r\times r$-matrices and $\zeta(S)=\bover1{2\|S^{-1}\|}$ for $S\in A$,
we have a partition $\sequencen{E_n}$ of $D\setminus D_0$ into measurable
sets and a sequence $\sequencen{T_n}$ in $A$ such that

\Centerline{$\|\phi(x)-\phi(y)-T_n(x-y)\|\le\Bover1{2\|T_n^{-1}\|}\|x-y\|$,
\quad$\|T(x)-T_n\|\le\Bover1{2\|T_n^{-1}\|}$}

\noindent whenever
$n\in\Bbb N$ and $x$, $y\in E_n$.   In this case, for $x$, $y\in E_n$,

\Centerline{$\phi(x)=\phi(y)
\,\Longrightarrow\,\|x-y\|\le\|T_n^{-1}\|\|T_n(x-y)\|
\le\Bover12\|x-y\|
\,\Longrightarrow\, x=y$,}

\noindent so $\phi\restr E_n$ is injective, for each $n$.
Consequently
$\#(\phi^{-1}[\{y\}])=\#(\{n:y\in\phi[E_n]\})$ for
$y\in\phi[D]\setminus\phi[D_0]$, and

$$\eqalignno{\int_{\phi[D]}\nu(\phi^{-1}[\{y\}])dy
&=\int_{\phi[D]\setminus\phi[D_0]}\nu(\phi^{-1}[\{y\}])dy
=\sum_{n=0}^{\infty}\mu(\phi[E_n]\setminus\phi[D_0])\cr
\displaycause{applying 263D(iii) to $\phi\restr E_n$, we know that
$\phi[E_n]$ is measurable for each $n$}
&=\sum_{n=0}^{\infty}\mu\phi[E_n]
=\sum_{n=0}^{\infty}\int_{E_n}J\cr
\displaycause{applying 263D(iv) to $\phi\restr E_n$}
&=\int_{D\setminus D_0}J
=\int_DJ,\cr}$$

\noindent each sum or integral being defined in $[0,\infty]$ because the
next one is.

\medskip

{\bf (b)(i)} Setting $D'=\phi^{-1}[\dom g]$, we see that
$D\setminus(D_0\cup D')$ is negligible, so (using 263D(ii) again)

\Centerline{$\phi[D]\setminus\dom g=\phi[D]\setminus\phi[D']
\subseteq\phi[D_0]\cup\phi[D\setminus(D_0\cup D')]$}

\noindent is negligible.   Next, if we set
$C'=C\cup g^{-1}[\{0\}]$, $\phi[D]\setminus C'$ is negligible.
\Prf\ For each $m\in\Bbb N$, set
$F_m=\{y:y\in\dom g$, $|g(y)|\ge 2^{-m}\}$.   Then

\Centerline{$\phi^{-1}[F_m]\setminus D_0=\{x:x\in D$, $J(x)\ne 0$,
   $g(\phi(x))$ is defined and $|g(\phi(x))|\ge 2^{-m}\}$}

\noindent is measurable (because $J\times g\phi$ is measurable) and
(applying (a) to $\phi\restr\phi^{-1}[F_m]$)

\Centerline{$\int_{F_m}\nu(\phi^{-1}[\{y\}])dy
=\int_{\phi^{-1}[F_m]}J\,d\mu
\le 2^m\int_{\phi^{-1}[F_m]}|J\times g(\phi)|\,d\mu$}

\noindent is finite.   But this means that $\nu(\phi^{-1}[\{y\}])$ must be
finite for almost every $y\in F_m$, that is, that $F_m\setminus C$ is
negligible.   As $m$ is arbitrary, $\dom g\setminus C'$ and
$\phi[D]\setminus C'$ are negligible.\ \Qed

\medskip

\quad{\bf (ii)} Taking $\sequencen{E_n}$ and $\sequencen{T_n}$ as in (a),
set $\epsilon_n=\sgn\det T_n\in\{-1,1\}$ for each $n$.   Then
$\sgn\det T(x)=\epsilon_n$ whenever $n\in\Bbb N$ and $x\in E_n$.   \Prf\
For any $\alpha\in[0,1]$,

\Centerline{$\|(\alpha T(x)+(1-\alpha)T_n)-T_n\|
\le\|T(x)-T_n\|\le\Bover1{2\|T_n^{-1}\|}$,}

\noindent so (using 2A4Fd)
$\|(\alpha T(x)+(1-\alpha)T_n)-I_r\|\le\Bover12$,
where $I_r$ is the $r\times r$ identity matrix,
and $\alpha T(x)+(1-\alpha)T_n$ is non-singular
(since if $(\alpha T(x)+(1-\alpha)T_n)z=0$, then
$\|z\|\le\bover12\|z\|$).   Thus $\det(\alpha T(x)+(1-\alpha)T_n)$ is
non-zero for $0\le\alpha\le 1$.   But as
$\alpha\mapsto\det(\alpha T(x)+(1-\alpha)T_n)$ is continuous,
$\epsilon_n=\sgn\det T_n=\sgn\det T(x)$.\ \Qed

\medskip

\quad{\bf (iii)} Now

$$\eqalignno{\int_Dg(\phi(x))\det T(x)\,dx
&=\int_{D\setminus D_0}g(\phi(x))\det T(x)\,dx\cr
&=\sum_{n=0}^{\infty}\int_{E_n}g(\phi(x))\det T(x)\,dx\cr
\displaycause{in this series of formulae, each sum and integral is
well-defined because the preceding ones are}
&=\sum_{n=0}^{\infty}\epsilon_n\int_{E_n}g(\phi(x))J(x)\,dx
=\sum_{n=0}^{\infty}\epsilon_n\int_{\phi[E_n]}g\,dx\cr
&=\sum_{n=0}^{\infty}\epsilon_n\int_{\phi[E_n]\cap C'}g\,dx
=\sum_{n=0}^{\infty}\epsilon_n\int_{C'}g\times\chi(\phi[E_n])\,dx\cr}$$

\noindent(using 131Fa, if you like, for the last step).
Since we also have

\Centerline{$\infty
>\int_D|g(\phi(x))\det T(x)|\,dx
=\sum_{n=0}^{\infty}\int_{C'}|g|\times\chi(\phi[E_n])\,dx$}

\noindent (going through the same stages with the absolute values of
integrands), we have

$$\eqalignno{\int_Dg(\phi(x))\det T(x)\,dx
&=\sum_{n=0}^{\infty}\epsilon_n\int_{C'}g\times\chi(\phi[E_n])\,dx\cr
&=\int_{C'}g\times\sum_{n=0}^{\infty}\epsilon_n\chi(\phi[E_n])\,dx\cr
&=\int_{C'\setminus\phi[D_0]}
   g\times\sum_{n=0}^{\infty}\epsilon_n\chi(\phi[E_n])\,dx
=\int_{C'\setminus\phi[D_0]}g\times R\,d\mu\cr
\displaycause{because if $y\in C'\setminus\phi[D_0]$, either $g(y)=0$ or
$R(y)$ is defined and equal to
$\sum_{n=0}^{\infty}\epsilon_n\chi(\phi[E_n])(y)$}
&=\int_{\phi[D]}g\times R\,d\mu,\cr}$$

\noindent as claimed.
}%end of proof of 263I

\leader{263J}{The one-dimensional \dvrocolon{case}}\cmmnt{ The
restriction to injective functions $\phi$ in 263D(v) is unavoidable in
the context of the result there.   But in the substitutions of
elementary calculus it is not always essential.   In the hope of
clarifying the position I give a result here which covers many of the
standard tricks.

\medskip

\noindent}{\bf Proposition}
Let $I\subseteq\Bbb R$ be an interval with more
than one point, and $\phi:I\to\Bbb R$ a function which is absolutely
continuous on any closed bounded subinterval of $I$.   Write $u=\inf I$,
$u'=\sup I$ in
$[-\infty,\infty]$, and suppose that $v=\lim_{x\downarrow u}\phi(x)$ and
$v'=\lim_{x\uparrow u'}\phi(x)$ are defined in $[-\infty,\infty]$.   Let
$g$ be a real function such that
$\int_Ig(\phi(x))\phi'(x)dx$ is defined,
on the understanding that we interpret $g(\phi(x))\phi'(x)$ as $0$ when
$\phi'(x)=0$ and $g(\phi(x))$ is undefined.   Then
$\int_v^{v'}g$ is defined and equal to $\int_Ig(\phi(x))\phi'(x)dx$, where
here we interpret $\int_v^{v'}g$ as $-\int_{v'}^vg$ if
$v'<v$.

\proof{{\bf (a)} $\phi$ is differentiable almost everywhere
on $I$ and $\phi[A]$ is negligible for every negligible
$A\subseteq I$.   \Prf\ We can express $I$ as the union of a
sequence $\sequencen{I_n}$ of closed bounded intervals such that
$\phi\restr I_n$ is absolutely continuous for every $n$.   By
225Cb and 225G, $\phi\restr I_n$ is differentiable almost everywhere
on $I_n$ and $\phi[A]$ is negligible for every negligible
$A\subseteq I_n$, for each $n$.   So
$\phi$ is differentiable almost everywhere
on $\bigcup_{n\in\Bbb N}I_n=I$ and
$\phi[A]=\bigcup_{n\in\Bbb N}\phi[A\cap I_n]$ is negligible for every
negligible $A\subseteq I$.\ \Qed

Because $\phi\restr J$ is continuous for every closed interval
$J\subseteq I$, $\phi$ is continuous.   By the Intermediate Value Theorem,
$\phi[I]$ is an interval including $\ooint{\min(v,v'),\max(v,v')}$.

\medskip

{\bf (b)} Let $D\subseteq I$ be the domain of $\phi'$.   For $x\in D$,
we can think of $\phi'(x)$ as a $1\times 1$ matrix with determinant
$\phi'(x)$.   As $I\setminus D$ is negligible,
$\phi[I]\setminus\phi[D]\subseteq\phi[I\setminus D]$ is negligible.
Now $\int_Dg(\phi(x))\phi'(x)dx=\int_Ig(\phi(x))\phi'(x)dx$.
Applying 263I to $\phi\restr D$ and $g\restr\phi[D]$, we see that
$\int_{\phi[D]}g\times R$ is defined and equal to
$\int_Dg\phi\times\phi'$, where
$R(y)=\sum_{x\in D\cap\phi^{-1}[\{y\}]}\sgn\phi'(x)$ whenever
$y\in\phi[D]$ and $D\cap\phi^{-1}[\{y\}]$ is finite.

\medskip

{\bf (c)} (The key.) Set $D_0=\{x:x\in D$, $\phi'(x)=0\}$.   By 263D(ii),
applied to $\phi\restr D_0$, $\phi[D_0]$ is negligible.
Set

\Centerline{$C=\{y:y\in\phi[D]\cap\dom g\setminus(\phi[D_0]\cup\{v,v'\})$,
$\phi^{-1}[\{y\}]$ is finite, $g(y)\ne 0\}$.}

\noindent If $y\in C$ and $K=\phi^{-1}[\{y\}]$, then

$$\eqalign{R(y)
=\sum_{x\in K}\sgn\phi'(x)
&=1\text{ if }v<y<v',\cr
&=-1\text{ if }v'<y<v,\cr
&=0\text{ otherwise}.\cr}$$

\noindent\Prf\
If $J\subseteq I\setminus K$ is an interval, $\phi(z)\ne y$ for
$z\in J$;  since $\phi$ is continuous, the Intermediate Value Theorem
tells us that $\sgn(\phi(z)-y)$ is constant on $J$.
Also $\phi'(x)\ne 0$ for every $x\in K$, because $y\notin\phi[D_0]$.
A simple induction
on $\#(K\cap\ooint{-\infty,z})$ shows that
$\sgn(\phi(z)-y)=\sgn(v-y)+2\sum_{x\in K,x<z}\sgn\phi'(x)$ for every
$z\in I\setminus K$;  taking the limit as $z\uparrow u'$,
$\sum_{x\in K}\sgn\phi'(x)=\bover12(\sgn(v'-y)-\sgn(v-y))$.
(Here we may have to interpret $\sgn(\pm\infty)$ as $\pm 1$ in the
obvious way.)   This turns out to be just what we need to know.\ \Qed

\medskip

{\bf (d)} So now we have

$$\eqalign{\int_Ig\phi\times\phi'
&=\int_Dg\phi\times\phi'
=\int_{\phi[D]}g\times R\cr
&=\int_{\phi[D]}g\times R\times\chi C
=\int_v^{v'}g\times\chi C
=\int_v^{v'}g\cr}$$

\noindent because $\phi[D]\setminus(C\cup g^{-1}[\{0\}])$ is negligible.
}%end of proof of 263J

\exercises{
\leader{263X}{Basic exercises (a)}
%\spheader 263Xa
Let $(X,\Sigma,\mu)$ be any measure space, $f\in\eusm L^0(\mu)$ and
$p\in\coint{1,\infty}$.   Show that $f\in\eusm L^p(\mu)$ iff

\Centerline{$\gamma=p\int_0^{\infty}\alpha^{p-1}
\mu^*\{x:x\in\dom f,\,|f(x)|>\alpha\}d\alpha$}

\noindent is finite, and in this case $\|f\|_p=\gamma^{1/p}$.
\Hint{$\int|f|^p=\int_0^{\infty}\mu^*\{x:|f(x)|^p>\beta\}d\beta$, by
252O;  now substitute $\beta=\alpha^p$.}
%263D

\spheader 263Xb Let $f$ be an integrable function defined almost
everywhere in $\BbbR^r$.   Show that if $\alpha<r-1$ then
$\sum_{n=1}^{\infty}n^{\alpha}|f(nx)|$ is finite for almost every
$x\in\BbbR^r$.   \Hint{estimate
$\sum_{n=0}^{\infty}n^{\alpha}\int_B|f(nx)|dx$ for balls $B$
centered at the origin.}
%263D

\spheader 263Xc Let $A\subseteq\ooint{0,1}$ be a set such that
$\mu^*A=\mu^*([0,1]\setminus A)=1$, where $\mu$ is Lebesgue measure on
$\Bbb R$.   Set $D=A\cup\{-x:x\in\ooint{0,1}\setminus A\}
\subseteq[-1,1]$, and set $\phi(x)=|x|$ for $x\in D$.   Show that $\phi$
is injective, that $\phi$ is differentiable relative to its domain
everywhere in $D$, and that $\mu^*\phi[D]<\int_D|\phi'(x)|dx$.
%263D

\spheader 263Xd Let $\phi:D\to\BbbR^r$ be a function differentiable
relative to $D$ at each point of $D\subseteq\BbbR^r$, and suppose that
for each
$x\in D$ there is a non-singular derivative $T(x)$ of $\phi$ at $x$.
Show that $D$ is expressible as
$\bigcup_{k\in\Bbb N}D_k$ where $D_k=D\cap\overline{D}_k$ and
$\phi\restr D_k$ is injective for each $k$.
%263D

\sqheader 263Xe(i) Show that for any Lebesgue measurable
$E\subseteq\Bbb R$ and $t\in\Bbb R\setminus\{0\}$,
$\int_{tE}\bover1{|u|}du=\int_E\bover1{|u|}du$.   (ii) For $t\in\Bbb R$,
$u\in\Bbb R\setminus\{0\}$ set $\phi(t,u)=(\bover{t}{u},u)$.   Show that
$\int_{\phi[E]}\bover1{|tu|}d(t,u)=\int_E\bover1{|tu|}d(t,u)$ for any
Lebesgue measurable $E\subseteq\BbbR^2$.
%263D

\sqheader 263Xf Define $\phi:\BbbR^3\to\BbbR^3$ by
setting

\Centerline{$\phi(\rho,\theta,\alpha)
=(\rho\sin\theta\sin\alpha,\rho\cos\theta\sin\alpha,\rho\cos\alpha)$.}

\noindent Show that $\det\phi'(\rho,\theta,\alpha)=\rho^2\sin\alpha$.
%263G

\spheader 263Xg Show that if $k=2l+1$ is odd, then
$\int_0^{\infty}x^ke^{-x^2/2}dx=2^ll!$.   (Compare 252Xi.)
%263H

\leader{263Y}{Further exercises (a)}
%\spheader 263Ya
Define a measure $\nu$ on $\Bbb R$ by setting
$\nu E=\int_E\bover1{|x|}dx$ for Lebesgue measurable sets
$E\subseteq\Bbb R$.   For $f$, $g\in\eusm L^1(\nu)$ set
$(f*g)(x)=\int f(\bover{x}{t})g(t)\nu(dt)$ whenever this is defined in
$\Bbb R$.   (i) Show that $f*g=g*f\in\eusm L^1(\nu)$.   (ii) Show that
$\int h(x)(f*g)(x)\nu(dx)=\int h(xy)f(x)f(y)\nu(dx)\nu(dy)$ for every
$h\in\eusm L^{\infty}(\nu)$.   (iii) Show that $f*(g*h)=(f*g)*h$ for
every $h\in\eusm L^1(\nu)$.   \Hint{263Xe.}
%263Xe 263D

\spheader 263Yb Let $E\subseteq\BbbR^2$ be a measurable set such that
$\limsup_{\alpha\to\infty}
\bover1{\alpha^2}\mu_2(E\cap B(\tbf{0},\alpha))>0$, writing $\mu_2$ for
Lebesgue measure on
$\BbbR^2$.   Show that there is some $\theta\in\ocint{-\pi,\pi}$ such
that $\mu_1E_{\theta}=\infty$, where
$E_{\theta}=\{\rho:\rho\ge 0,\,(\rho\cos\theta,\rho\sin\theta)\in E\}$.
\Hint{show that $\mu_2(E\cap B(\tbf{0},\alpha))
\le\alpha\int_{-\pi}^{\pi}\mu_1E_{\theta})d\theta$.}
Generalize to higher dimensions and to functions other than $\chi E$.
%263G

\spheader 263Yc Let $E\subseteq\BbbR^r$ be a measurable set,
and $\phi:E\to\BbbR^r$ a function differentiable relative to its domain,
with a derivative $T(x)$, at each point $x$ of $E$;  set
$J(x)=|\det T(x)|$.
Show that for any integrable function $g$ defined on $\phi[E]$,

\Centerline{$\int g(y)\#(\phi^{-1}[\{y\}])dy
=\int_EJ(x)g(\phi(x))dx$.}
%263I

\spheader 263Yd Find a proof of 263J based on the ideas of \S225.
\Hint{225Xe.}
%263J

\spheader 263Ye Let $f:[a,b]\to\Bbb R$ be a function of bounded
variation, where $a<b$ in $\Bbb R$, with Lebesgue decomposition
$f=f_p+f_{cs}+f_{ac}$ as in 226Cd;  let $\mu$ be Lebesgue measure on
$\Bbb R$.   Show that the following are equiveridical:  (i) $f_{cs}$ is
constant;  (ii) $\mu f[\,[c,d]\,]\le\int_c^d|f'|d\mu$ whenever
$a\le c\le d\le b$;   (iii)
$\mu^*f[A]\le\int_A|f'|d\mu$ for every $A\subseteq[a,b]$;   (iv) $f[A]$
is negligible for every negligible set $A\subseteq[a,b]$.   \Hint{for
(iv)$\Rightarrow$(i) put 226Yd and 263D(ii) together to show that
$|f(d)-f(c)|\le\int_c^d|f'|d\mu+\Var_{[c,d]}f_p$ whenever
$a\le c\le d\le b$, and therefore that
$\Var_{[a,b]}f\le\Var_{[a,b]}f_p+\Var_{[a,b]}f_{ac}$.}
%263D

\spheader 263Yf\dvAnew{2013} Suppose that $r=2$ and that
$\phi:\BbbR^2\to\BbbR^2$ is continuously differentiable with non-singular
derivative $T$ at $\tbf{0}$.   (i) Show that there is an $\epsilon>0$
such that whenever
$\Gamma$ is a small circle with centre $\tbf{0}$ and radius at most
$\epsilon$ then
$\phi\restr\Gamma$ is a homeomorphism between $\Gamma$ and a simple closed
curve around $\tbf{0}$.   (ii) Show that if $\det T>0$, then
for such circles $\phi(x)$ runs anticlockwise around $\phi[\Gamma]$ as $x$
runs anticlockwise around $\Gamma$.   (iii) What happens if
$\det T<0$?
}%end of exercises

\endnotes{
\Notesheader{263} Yet again, approaching 263D, I find myself having to
choose between giving an accessible, relatively weak result and making
the extra effort to set out a theorem which is somewhere near the
natural boundary of what is achievable within the concepts being
developed in this volume;  and, as usual, I go for the more powerful
form.   There are three basic sources of difficulty:  (i) the fact that
we are dealing with more than one dimension;  (ii) the fact that we are
dealing with irregular domains;   (iii) the fact that we are dealing
with arbitrary integrable functions.   I do not think I need to
apologise for (iii) in a book on measure theory.   Concerning (ii), it
is quite true that the principal applications of these results are to
cases in which the transformation $\phi$ is differentiable everywhere,
with continuous derivative, and the set $D$ has negligible boundary;
and in these cases there are substantial simplifications available --
mostly because the sets $D_m$ of the proof of 263D can be taken to be
cubes.   Nevertheless, I think any form of the result which makes such
assumptions is deeply unsatisfactory at this level, being an awkward
compromise between ideas natural to the Riemann integral and those
natural to the Lebesgue integral.   Concerning (i), it might even have
been right to lay out the whole argument for the case $r=1$  before
proceeding to the general case, as I did in \S\S114-115, because the
one-dimensional case is already important and interesting;  and if you
find the work above difficult -- which it is -- and your immediate
interests are in one-dimensional integration by substitution, then I
think you might find it worth your time to reproduce the $r=1$ argument
yourself, up to a proof of 263J.   In fact the biggest difference is
in 263A, which becomes nearly trivial;  the work of 262M and 263C
becomes more readable, because all the matrices turn into scalars and we
can drop the word `determinant', but I do not think we can dispense
with any of the ideas, at least if we wish to obtain 263D as stated.
(But see 263Yd.)

I found myself insisting, in the last paragraph, that a distinction can
be made between `ideas natural to the Riemann integral and those
natural to the Lebesgue integral'.   We are approaching deep questions
here, like `what are books on measure theory for?', which I do not
think can be answered without some -- possibly unconscious --
reference to the question `what is mathematics for?'.   I do of
course want to present here some of the wonderful general theorems which
arise in the Lebesgue theory.   But more important than any specific
theorem is a general idea of what can be proved by these methods.
It is the essence of modern measure theory that continuity does not
matter, or, if you prefer, that measurable functions are in some sense
so nearly continuous that we do not have to add hypotheses of continuity
in our theorems.   Now this is in a sense a great liberation, and the
Lebesgue integral is now the standard one.   But you must not regard the
Riemann integral as outdated.   The intuitions on which it is founded --
for instance, that the surface of a solid body has zero volume -- remain
of great value in their proper context, which certainly includes the
study of differentiable functions with continuous derivatives.   What I
am saying here is that I believe we can use these intuitions best if we
maintain a division, a flexible and permeable one, of course, between
the ideas of the two theories;  and that when transferring a theorem
from one side of the boundary to the other we should do so
whole-heartedly, seeking to express the full power of the methods we are
using.

I have already said that the essential difference between the
one-dimensional and multi-dimensional cases lies in 263A, where the
Jacobian $J=|\det T|$ enters the argument.   Shorn of the technical
devices necessary to deal with arbitrary Lebesgue measurable sets, this
amounts to a calculation of the volume of the parallelepiped $T[I]$
where $I$ is the interval $\coint{\tbf{0},\tbf{1}}$.   I have dealt with
this by a little bit of algebra, saying that the result is essentially
obvious if $T$ is diagonal, whereas if $T$ is an isometry it follows
from the fact that the unit ball is left invariant;  and the algebra
comes in to express an arbitrary matrix as a product of diagonal and
orthogonal matrices.    It is also plain from 261F that Lebesgue
measure must be rotation-invariant as well as translation-invariant;
that is to
say, it is invariant under all isometries.   Another way of looking at
this will appear in the next section.

I feel myself that the centre of the argument for 263D is in the lemma
263C.   This is where we turn the exact result for linear operators into
an approximate result for almost-linear functions;  and the whole point
of differentiability is that a differentiable function is well
approximated, near any point of its domain, by a linear
operator.   The lemma involves two rather different ideas.   To show
that $\mu^*\phi[D]\le(J+\epsilon)\mu^*D$, we look first at balls and
then use Vitali's theorem to see that $D$ is economically covered by
balls, so that an upper bound for $\mu^*\phi[D]$ in terms of a sum
$\sum_{B\in\Cal I_0}\mu^*\phi[D\cap B]$ is adequate.    To obtain a
lower bound, we need to reverse the argument by looking at
$\psi=\phi^{-1}$, which involves checking first that $\phi$ is
invertible, and then that $\psi$ is appropriately linked to $T^{-1}$.
I have written out exact formulae for $\epsilon'$, $\zeta_2'$ and so on,
but this is only in case you do not trust your intuition;  the fact that
$\|\phi^{-1}(u)-\phi^{-1}(v)-T^{-1}(u-v)\|$ is small compared with
$\|u-v\|$ is pretty clearly a consequence of the hypothesis that
$\|\phi(x)-\phi(y)-T(x-y)\|$ is small compared with $\|x-y\|$.

The argument of 263D itself is now a matter of breaking the set $D$ up
into appropriate pieces on each of which $\phi$ is sufficiently nearly
linear for 263C to apply, so that

\Centerline{$\mu^*\phi[D]\le\sum_{m=0}^{\infty}\mu^*\phi[D_m]
\le\sum_{m=0}^{\infty}(J_m+\epsilon)\mu^*D_m$.}

\noindent With a little care (taken in 263C, with its condition (i)), we
can also ensure that the Jacobian $J$ is
well approximated by $J_m$ almost everywhere in $D_m$, so that
$\sum_{m=0}^{\infty}J_m\mu^*D_m\bumpeq\int_DJ(x)dx$.

These ideas, joined with the results of \S262, bring us to the point

\Centerline{$\int_EJ\,d\mu=\mu\phi[E]$}

\noindent when $\phi$ is injective and $E\subseteq D$ is measurable.
We need a final trick, involving Borel sets, to translate this into

\Centerline{$\int_{\phi^{-1}[F]}J\,d\mu=\mu F$}

\noindent whenever $F\subseteq\phi[D]$ is measurable, which is what is
needed for the application of 235J.

I hope that you long ago saw, and were delighted by, the device in 263G.
Once again, this is not really Lebesgue integration;  but I include it
just to show that the machinery of this chapter can be turned to deal
with the classical results, and that indeed we have a tiny profit from
our labour, in that no apology need be made for the boundary of the set
$D$ into which the polar coordinate system maps the plane.   I have
already given the actual
result as an exercise in 252Xi.   That involved (if you chase through
the references) a one-dimensional substitution (performed in 225Xh),
Fubini's theorem and an application of the formulae of \S235;  that is
to say, very much the same elements as those used above, though in a
different order.   I could present this with no mention of
differentiation in higher dimensions because the first change of
variable was in one dimension, and the second (involving the function
$x\mapsto\|x\|$, in 252Xi(i)) was of a particularly simple type, so that
a different method could be used to find the function $J$.

The function $R(y)=\sum_{x\in\phi^{-1}[\{y\}]}\sgn\det T(x)$ of 263Ib
belongs to rather deeper notions in differential geometry than I wish to
enlarge on here.   In the one-dimensional context it simply counts up- and
down-crossings of $y$ (see part (c) of the proof of 263J), because we can
think of each $T(x)$ as a scalar which is either positive or negative.
(It is relevant that in this case we can assume that $T(x)$ is non-singular
whenever $\phi(x)=y$.)   In higher dimensions, I suppose the first thing to
look for is a geometric interpretation of $\sgn\det T(x)$.   (See 263Ye.)
A geometric interpretation of the sum is something else again.
But it is worth noting that (subject to a natural interpretation of
`0$\times$undefined') we can
relate $\int_Dg\phi\times\det\phi'$ to $\int_{\phi[D]}g\times R$ where
$R$ is definable from $\phi$ without reference to $g$;  it counts folds
in the graph of $\phi$.

The abstract ideas to which this treatise is devoted do not, indeed,
lead us to many particular examples on which to practise the ideas of
this section.   The ones which do arise tend to be very straightforward,
as in 263G, 263Xa-263Xb and 263Xe.   I mention the last because it
provides a formula needed to discuss a new type of convolution (263Ya).
In effect, this depends on the multiplicative group
$\Bbb R\setminus\{0\}$ in place of the additive group $\Bbb R$ treated
in \S255.   The formula $\bover1x$ in the definition of $\nu$ is of
course the derivative of $\ln x$, and $\ln$ is an isomorphism between
$(\ooint{0,\infty},\cdot,\nu)$ and $(\Bbb R,+,\text{Lebesgue measure})$.
}%end of notes

\discrpage
\frfilename{mt264.tex}
\versiondate{12.5.03}
\copyrightdate{1994}

\def\chaptername{Change of variable in the integral}
\def\sectionname{Hausdorff measures}

\newsection{264}

The next topic I wish to approach is the question of `surface
measure';  a useful example to bear in mind throughout this section and
the next is the notion of area for regions on the sphere, but any other
smoothly curved two-dimensional surface in three-dimensional space will
serve equally well.   It is I think more than plausible that our
intuitive concepts of `area' for such surfaces should correspond to
appropriate measures.   But formalizing this intuition is
non-trivial, especially if we seek the generality that simple geometric
ideas lead us to;  I mean, not contenting ourselves with arguments that
depend on the special nature of the sphere, for instance, to describe
spherical surface area.   I divide the problem into two parts.   In this
section I will describe a construction which enables us to define the
$r$-dimensional measure of an $r$-dimensional surface -- among other
things -- in $s$-dimensional space.   In the next section I will set out
the basic theorems making it possible to calculate these measures
effectively in the leading cases.

\leader{264A}{Definitions} Let $s\ge 1$ be an integer, and $r>0$.
\cmmnt{(I am primarily concerned with integral $r$,
but will not insist on this until it becomes necessary, since there are
some very interesting ideas which involve non-integral `dimension'
$r$.)}   For any $A\subseteq \BbbR^s$, $\delta>0$ set

$$\eqalign{\theta_{r\delta}A
=\inf\{\sum_{n=0}^{\infty}(\diam A_n)^r:\sequencen{A_n}
&\text{ is a sequence of subsets of }\BbbR^s\text{ covering }A,\cr
&\qquad\qquad\qquad\quad\diam A_n\le\delta\text{ for every }
  n\in\Bbb N\}.\cr}$$

\noindent   It is convenient in this context to
say that $\diam\emptyset=0$.
Now set

\Centerline{$\theta_rA=\sup_{\delta>0}\theta_{r\delta}A$;}

\noindent $\theta_r$ is {\bf $r$-dimensional Hausdorff outer measure} on
$\BbbR^s$.

\leader{264B}{}\cmmnt{ Of course we must immediately check the
following:

\medskip

\noindent}{\bf Lemma} $\theta_r$, as defined in 264A, is always an outer
measure.

\proof{ You should be used to these arguments by now, but there is an
extra step in this one, so I spell out the details.

\medskip

{\bf (a)} Interpreting the diameter of the empty set as $0$, we have
$\theta_{r\delta}\emptyset=0$ for every $\delta>0$, so
$\theta_r\emptyset=0$.

\medskip

{\bf (b)} If $A\subseteq B\subseteq\BbbR^s$, then every sequence
covering $B$
also covers $A$, so $\theta_{r\delta}A\le\theta_{r\delta}B$ for every
$\delta$ and $\theta_rA\le\theta_rB$.

\medskip

{\bf (c)} Let $\sequencen{A_n}$ be a sequence of subsets of $\BbbR^s$
with union $A$, and take any  $a<\theta_rA$.   Then there is a $\delta>0$
such that $a\le\theta_{r\delta}A$.   Now
$\theta_{r\delta}A\le\sum_{n=0}^{\infty}\theta_{r\delta}(A_n)$.   \Prf\
Let $\epsilon>0$, and for each $n\in\Bbb N$ choose a sequence
$\sequence{m}{A_{nm}}$ of sets, covering $A_n$, with 
$\diam A_{nm}\le\delta$ for every $m$ and 
$\sum_{m=0}^{\infty}(\diam A_{nm})^r\le\theta_{r\delta}A_n+2^{-n}\epsilon$.
Then $\langle A_{nm}\rangle_{m,n\in\Bbb N}$ is a cover of $A$ by 
countably many sets of diameter at most $\delta$, so

\Centerline{$\theta_{r\delta}A
\le\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}(\diam A_{nm})^r
\le\sum_{n=0}^{\infty}\theta_{r\delta}A_n+2^{-n}\epsilon
=2\epsilon+\sum_{n=0}^{\infty}\theta_{r\delta}A_n$.}

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

Accordingly

\Centerline{$a\le\theta_{r\delta}A
\le\sum_{n=0}^{\infty}\theta_{r\delta}A_n
\le\sum_{n=0}^{\infty}\theta_rA_n$.}

\noindent As $a$ is arbitrary,

\Centerline{$\theta_rA\le\sum_{n=0}^{\infty}\theta_rA_n$;}

\noindent as $\sequencen{A_n}$ is arbitrary, $\theta_r$ is an outer
measure.
}%end of proof of 264B

\leader{264C}{Definition} If $s\ge 1$ is an integer, and $r>0$, then
{\bf Hausdorff $r$-dimensional measure} on $\BbbR^s$ is the measure
$\mu_{Hr}$ on $\BbbR^s$ defined by \Caratheodory's method from the outer
measure $\theta_r$ of 264A-264B.

\leader{264D}{Remarks}\cmmnt{ {\bf (a)} It is important to note that
the sets used in
the definition of the $\theta_{r\delta}$ need not be balls;  even in
$\BbbR^2$ not every set $A$ can be covered by a ball of the same
diameter as $A$.

\header{264Db}} {\bf (b)} In the definitions above I require $r>0$.   It
is sometimes appropriate to take $\mu_{H0}$ to be counting measure.
\cmmnt{This
is nearly the result of applying the formulae above with $r=0$, but
there can be difficulties if we interpret them over-literally.}

\header{264Dc}{\bf (c)} All Hausdorff measures must be
complete\cmmnt{, because
they are defined by \Caratheodory's method (212A)}.   For $r>0$, they
are atomless\cmmnt{ (264Yg)}.   \cmmnt{In terms of the other
criteria of \S211,
however, they are very ill-behaved;  for instance, if $r$, $s$ are
integers and $1\le r<s$, then $\mu_{Hr}$ on $\BbbR^s$ is not
semi-finite.   (I will give a proof of this in 439H in Volume 4.)
Nevertheless, they do have some striking properties which make them
reasonably tractable.}

\cmmnt{
\header{264Dd}{\bf (d)} In 264A, note that
$\theta_{r\delta}A\le\theta_{r\delta'}A$ when $0<\delta'\le\delta$;
consequently, for instance,
$\theta_rA=\lim_{n\to\infty}\theta_{r,2^{-n}}A$.   I have allowed
arbitrary sets $A_n$ in the covers, but it makes no difference if we
restrict our attention to covers consisting
of open sets or of closed sets (264Xc).
}%end of comment

\leader{264E}{Theorem} Let $s\ge 1$ be an integer, and $r\ge 0$;
let $\mu_{Hr}$ be Hausdorff $r$-dimensional measure on $\BbbR^s$, and
$\Sigma_{Hr}$ its domain.   Then every Borel subset of $\BbbR^s$ belongs
to $\Sigma_{Hr}$.

\proof{ This is trivial if $r=0$;  so suppose henceforth that $r>0$.

\medskip

{\bf (a)} The first step is to note that if $A$, $B$ are subsets of
$\Bbb R^s$
and $\eta>0$ is such that $\|x-y\|\ge\eta$ for all $x\in A$, $y\in B$,
then $\theta_r(A\cup B)=\theta_rA+\theta_rB$, where $\theta_r$ is
$r$-dimensional Hausdorff outer measure on $\BbbR^s$.   \Prf\ Of course
$\theta_r(A\cup B)\le\theta_rA+\theta_rB$, because $\theta_r$ is an
outer measure.   For the reverse inequality, we may suppose that
$\theta_r(A\cup B)<\infty$, so that $\theta_rA$ and $\theta_rB$ are both
finite.   Let $\epsilon>0$ and let $\delta_1$, $\delta_2>0$ be such that

\Centerline{$\theta_rA+\theta_rB\le
\theta_{r\delta_1}A+\theta_{r\delta_2}B+\epsilon$.}

\noindent Set $\delta=\min(\delta_1,\delta_2,\bover12\eta)>0$ and let
$\sequencen{A_n}$ be a sequence of sets of diameter at most $\delta$,
covering $A\cup B$, and such that
$\sum_{n=0}^{\infty}(\diam A_n)^r\le\theta_{r\delta}(A\cup B)+\epsilon$.
Set

\Centerline{$K=\{n:A_n\cap A\ne\emptyset\}$,
\quad$L=\{n:A_n\cap B\ne\emptyset\}$.}

\noindent Because

\Centerline{$\|x-y\|\ge\eta>\diam A_n$}

\noindent whenever $x\in A$, $y\in B$ and $n\in\Bbb N$, $K\cap
L=\emptyset$;  and of course $A\subseteq\bigcup_{n\in K}A_k$,
$B\subseteq\bigcup_{n\in L}A_n$.   Consequently

$$\eqalign{\theta_rA+\theta_rB
&\le\epsilon+\theta_{r\delta_1}A+\theta_{r\delta_2}B\cr
&\le\epsilon+\sum_{n\in K}(\diam A_n)^r+\sum_{n\in L}(\diam A_n)^r\cr
&\le\epsilon+\sum_{n=0}^{\infty}(\diam A_n)^r
\le2\epsilon+\theta_{r\delta}(A\cup B)
\le 2\epsilon+\theta_r(A\cup B).\cr}$$

\noindent As $\epsilon$ is arbitrary, $\theta_r(A\cup
B)\ge\theta_rA+\theta_rB$, as required.\ \Qed

\medskip

{\bf (b)} It follows that
$\theta_rA=\theta_r(A\cap G)+\theta_r(A\setminus G)$ whenever
$A\subseteq \BbbR^s$ and $G$ is open.
\Prf\ As usual, it is enough to consider the case $\theta_rA<\infty$
and to show that in this case
$\theta_r(A\cap G)+\theta_r(A\setminus G)\le\theta_rA$.   Set

\Centerline{$A_n=\{x:x\in A,\,\|x-y\|\ge 2^{-n}$ for every
$y\in A\setminus G\}$,}

\Centerline{$B_0=A_0$,
\quad$B_n=A_n\setminus A_{n-1}$ for $n>1$.}

\noindent Observe that $A_n\subseteq A_{n+1}$ for every $n$ and
$\bigcup_{n\in\Bbb N}A_n=\bigcup_{n\in\Bbb N}B_n=A\cap G$.   The point
is that if $m$, $n\in\Bbb N$ and $n\ge m+2$, and if $x\in B_m$ and
$y\in B_n$, then there is a $z\in A\setminus G$ such that
$\|y-z\|<2^{-n+1}\le 2^{-m-1}$, while $\|x-z\|$ must be at least
$2^{-m}$, so $\|x-y\|\ge\|x-z\|-\|y-z\|\ge 2^{-m-1}$.   It follows
that for any $k\ge 0$

\Centerline{$\sum_{m=0}^k\theta_rB_{2m}
=\theta_r(\bigcup_{m\le k}B_{2m})\le\theta_r(A\cap G)<\infty$,}

\Centerline{$\sum_{m=0}^k\theta_rB_{2m+1}
=\theta_r(\bigcup_{m\le k}B_{2m+1})\le\theta_r(A\cap G)<\infty$,}

\noindent (inducing on $k$, using (a) above for the inductive step).
Consequently $\sum_{n=0}^{\infty}\theta_rB_n<\infty$.

But now, given $\epsilon>0$, there is an $m$ such that
$\sum_{n=m}^{\infty}\theta_rB_m\le\epsilon$, so that

$$\eqalignno{\theta_r(A\cap G)+\theta_r(A\setminus G)
&\le\theta_rA_m+\sum_{n=m}^{\infty}\theta_rB_n+\theta_r(A\setminus G)\cr
&\le\epsilon+\theta_rA_m+\theta_r(A\setminus G)
=\epsilon+\theta_r(A_m\cup(A\setminus G))\cr
\noalign{\noindent (by (a) again, since $\|x-y\|\ge 2^{-m}$ for
$x\in A_m$, $y\in A\setminus G$)}
&\le\epsilon+\theta_rA.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\theta_r(A\cap G)+\theta_r(A\setminus G)\le\theta_rA$,
as required.\ \Qed

\medskip

{\bf (c)} Part (b) shows exactly that open sets belong to $\Sigma_{Hr}$.
It follows at once that the Borel $\sigma$-algebra of $\BbbR^s$ is
included in $\Sigma_{Hr}$, as claimed.
}%end of proof of 264E

\leader{264F}{Proposition} Let $s\ge 1$ be an integer, and $r>0$;
let $\theta_r$ be $r$-dimensional Hausdorff outer measure on $\BbbR^s$,
and write  $\mu_{Hr}$ for $r$-dimensional Hausdorff measure on
$\BbbR^s$, $\Sigma_{Hr}$ for its domain.   Then

(a) for every $A\subseteq \BbbR^s$ there is a Borel set $E\supseteq A$
such that $\mu_{Hr}E=\theta_rA$;

(b) $\theta_r=\mu^*_{Hr}$, the outer measure defined from $\mu_{Hr}$;

(c) if $E\in\Sigma_{Hr}$ is expressible as a countable union of sets of
finite measure,
there are Borel sets $E'$, $E''$ such that
$E'\subseteq E\subseteq E''$ and $\mu_{Hr}(E''\setminus E')=0$.

\proof{{\bf (a)} If $\theta_rA=\infty$ this is trivial -- take
$E=\BbbR^s$.
Otherwise, for each $n\in\Bbb N$ choose a sequence
$\sequence{m}{A_{nm}}$ of sets of diameter at most $2^{-n}$, covering
$A$, and such that
$\sum_{m=0}^{\infty}(\diam A_{nm})^r\le\theta_{r,2^{-n}}A+2^{-n}$.   Set
$F_{nm}=\overline{A}_{nm}$,   $E=\bigcap_{n\in\Bbb N}\bigcup_{m\in\Bbb
N}F_{nm}$;  then $E$ is a Borel set in $\BbbR^s$.   Of course

\Centerline{$A\subseteq\bigcap_{n\in\Bbb N}\bigcup_{m\in\Bbb N}A_{mn}
\subseteq\bigcap_{n\in\Bbb N}\bigcup_{m\in\Bbb N}F_{nm}=E$.}

For any $n\in\Bbb N$,

\Centerline{$\diam F_{nm}=\diam A_{nm}\le 2^{-n}$
for every $m\in\Bbb N$,}

\Centerline{$\sum_{m=0}^{\infty}(\diam F_{nm})^r
=\sum_{m=0}^{\infty}(\diam A_{nm})^r
\le\theta_{r,2^{-n}}A+2^{-n}$,}

\noindent so

\Centerline{$\theta_{r,2^{-n}}E\le\theta_{r,2^{-n}}A+2^{-n}$.}

\noindent Letting $n\to\infty$,

\Centerline{$\theta_rE=\lim_{n\to\infty}\theta_{r,2^{-n}}E
\le\lim_{n\to\infty}\theta_{r,2^{-n}}A+2^{-n}
=\theta_rA$;}

\noindent of course it follows that $\theta_rA=\theta_rE$, because
$A\subseteq E$.    Now by 264E we know that $E\in\Sigma_{Hr}$, so we can
write $\mu_{Hr}E$ in place of $\theta_rE$.

\medskip

{\bf (b)} This follows at once, because we have

\Centerline{$\mu_{Hr}^*A
=\inf\{\mu_{Hr}E:E\in\Sigma_{Hr},\,A\subseteq E\}
=\inf\{\theta_rE:E\in\Sigma_{Hr},\,A\subseteq E\}
\ge\theta_rA$}

\noindent for every $A\subseteq \BbbR^s$.   On the other hand, if
$A\subseteq\BbbR^s$, we have a Borel set $E\supseteq A$ such that
$\theta_rA=\mu_{Hr}E$, so that $\mu_{Hr}^*A\le\mu_{Hr}E=\theta_rA$.

\medskip

{\bf (c)(i)} Suppose first that $\mu_{Hr}E<\infty$.   By (a), there are
Borel sets $E''\supseteq E$, $H\supseteq E''\setminus E$ such that
$\mu_{Hr}E''=\theta_rE$,

\Centerline{$\mu_{Hr}H=\theta_r(E''\setminus E)
=\mu_{Hr}(E''\setminus E)
=\mu_{Hr}E''-\mu_{Hr}E
=\mu_{Hr}E''-\theta_rE
=0$.}

\noindent So setting $E'=E''\setminus H$, we obtain a Borel set
included in $E$, and

\Centerline{$\mu_{Hr}(E''\setminus E')\le\mu_{Hr}H=0$.}

\medskip

\quad{\bf (ii)} For the general case, express $E$ as $\bigcup_{n\in\Bbb
N}E_n$ where $\mu_{Hr}E_n<\infty$ for each $n$;  take Borel sets $E'_n$,
$E_n''$ such that $E_n'\subseteq E_n\subseteq E_n''$ and
$\mu_{Hr}(E_n''\setminus E_n')=0$ for each $n$;  and set
$E'=\bigcup_{n\in\Bbb N}E'_n$, $E''=\bigcup_{n\in\Bbb N}E_n''$.
}%end of proof of 264F

\leader{264G}{Lipschitz functions}\dvro{\bf: }{ The definition of
Hausdorff measure is exactly adapted to the following result,
corresponding to 262D.

\wheader{264G}{6}{2}{2}{48pt}

\noindent}{\bf Proposition} Let $m$, $s\ge 1$ be integers, and
$\phi:D\to\BbbR^s$ a $\gamma$-Lipschitz function, where $D$ is a subset
of $\BbbR^m$.   Then for any $A\subseteq D$ and $r\ge 0$,

\Centerline{$\mu_{Hr}^*(\phi[A])\le\gamma^r\mu_{Hr}^*A$}

\noindent for every $A\subseteq D$, writing $\mu_{Hr}$ for
$r$-dimensional Hausdorff outer measure on either $\BbbR^m$ or
$\BbbR^s$.

\proof{{\bf (a)} The case $r=0$ is trivial, since then $\gamma^r=1$ and
$\mu_{Hr}^*A=\mu_{H0}A=\#(A)$ if $A$ is finite, $\infty$
otherwise, while $\#(\phi[A])\le\#(A)$.

\medskip

{\bf (b)} If $r>0$, then take any $\delta>0$.   Set
$\eta=\delta/(1+\gamma)$ and consider $\theta_{r\eta}:\Cal
P\BbbR^m\to[0,\infty]$, defined as in 264A.   We know from 264Fb
that

\Centerline{$\mu_{Hr}^*A=\theta_rA\ge\theta_{r\eta}A$,}

\noindent so there is a sequence $\sequencen{A_n}$ of sets, all of
diameter at most $\eta$, covering $A$, with
$\sum_{n=0}^{\infty}(\diam A_n)^r\le\mu_{Hr}^*A+\delta$.
Now $\phi[A]\subseteq\bigcup_{n\in\Bbb N}\phi[A_n\cap D]$ and

\Centerline{$\diam \phi[A_n\cap D]
\le\gamma\diam A_n\le\gamma\eta\le\delta$}

\noindent for every $n$.   Consequently

\Centerline{$\theta_{r\delta}(\phi[A])\le\sum_{n=0}^{\infty}
(\diam \phi[A_n])^r
\le\sum_{n=0}^{\infty}\gamma^r(\diam A_n)^r
\le\gamma^r(\mu_{Hr}^*A+\delta)$,}

\noindent and

\Centerline{$\mu_{Hr}^*(\phi[A])
=\lim_{\delta\downarrow 0}\theta_{r\delta}(\phi[A])
\le\gamma^r\mu_{Hr}^*A$,}

\noindent as claimed.
}%end of proof of 264G

\leader{264H}{}\cmmnt{ The next step is to relate $r$-dimensional
Hausdorff
measure on $\BbbR^r$ to Lebesgue measure on $\BbbR^r$.   The basic
fact we need is the following, which is even more important for the idea
in its proof than for the result.

\wheader{264H}{6}{2}{2}{48pt}\noindent}{\bf Theorem}
%hold this line to get spacing right
Let $r\ge 1$ be an integer, and $A$ a bounded
subset of $\BbbR^r$;  write $\mu_r$ for Lebesgue measure on $\BbbR^r$
and $d=\diam A$.   Then

\Centerline{$\mu_r^*(A)\le\mu_rB(\tbf{0},\Bover{d}2)=2^{-r}\beta_rd^r$,}

\noindent where $B(\tbf{0},\bover{d}2)$ is the ball with centre
$\tbf{0}$ and
diameter $d$, so that $B(\tbf{0},1)$ is the unit ball in $\BbbR^r$, and
has measure

$$\eqalign{\beta_r
&=\Bover{1}{k!}\pi^k\text{ if }r=2k\text{ is even},\cr
&=\Bover{2^{2k+1}k!}{(2k+1)!}\pi^k\text{ if }r=2k+1\text{ is odd}.\cr}$$

\proof{{\bf (a)} For the calculation of $\beta_r$, see 252Q or
252Xi.

\medskip

{\bf (b)} The case $r=1$ is elementary, for in this case $A$ is included
in an interval of length $\diam A$, so that $\mu_1^*A\le\diam A$.   So
henceforth let us suppose that $r\ge 2$.

\medskip

{\bf (c)} For $1\le i\le r$ let $S_i:\BbbR^r\to\BbbR^r$ be reflection
in the $i$th coordinate, so that
$S_ix=(\xi_1,\ldots,
\ifdim\pagewidth>467pt\penalty-100\fi
\xi_{i-1},-\xi_i,\xi_{i+1},
\ifdim\pagewidth>467pt\penalty-50\fi
\ldots,
\ifdim\pagewidth>467pt\penalty-10\fi
\xi_r)$ for every
$x=(\xi_1,\ldots,\xi_r)\in \BbbR^r$.   Let us say that a set
$C\subseteq\BbbR^r$ is {\bf symmetric in coordinates in $J$}, where
$J\subseteq\{1,\ldots,r\}$, if $S_i[C]=C$ for $i\in J$.   Now the centre
of the argument is the following fact:  if $C\subseteq\Bbb R$ is a
bounded set which is symmetric in coordinates in $J$, where $J$ is a
proper subset of $\{1,\ldots,r\}$, and $j\in\{1,\ldots,r\}\setminus J$,
then there is a set $D$, symmetric in coordinates in $J\cup\{j\}$,
such that $\diam D\le\diam C$ and $\mu_r^*C\le\mu^*_rD$.


\Prf\ {\bf (i)} Because Lebesgue measure is invariant under permutation
of coordinates, it is enough to deal with the case $j=r$.   Start by
writing $F=\overline{C}$, so that $\diam F=\diam C$ and
$\mu_rF\ge\mu_r^*C$.   Note that because $S_i$ is a homeomorphism for
every $i$,

\Centerline{$S_i[F]=S_i[\overline{C}]=\overline{S_i[C]}=\overline{C}=F$}

\noindent for $i\in J$, and $F$ is symmetric in coordinates in $J$.

For $y=(\eta_1,\ldots,\eta_{r-1})\in\BbbR^{r-1}$, set

\Centerline{$F_y=\{\xi:(\eta_1,\ldots,\eta_{r-1},\xi)\in F\}$,\quad
$f(y)=\mu_1F_y$,}

\noindent where $\mu_1$ is Lebesgue measure on $\Bbb R$.
Set

\Centerline{$D=\{(y,\xi):y\in\BbbR^{r-1},\,|\xi|<\Bover12f(y)\}
\subseteq\BbbR^r$.}

\medskip

\quad{\bf (ii)} If $H\subseteq\BbbR^r$ is measurable and $H\supseteq D$,
then, writing $\mu_{r-1}$ for Lebesgue measure on $\BbbR^{r-1}$, we have

$$\eqalignno{\mu_rH
&=\int\mu_1\{\xi:(y,\xi)\in H\}\mu_{r-1}(dy)\cr
\noalign{\noindent (using 251N and 252D)}
&\ge\int\mu_1\{\xi:(y,\xi)\in D\}\mu_{r-1}(dy)
=\int f(y)\mu_{r-1}(dy)\cr
&=\int\mu_1\{\xi:(y,\xi)\in F\}\mu_{r-1}(dy)
=\mu_rF
\ge\mu^*_rC.\cr}$$

\noindent As $H$ is arbitrary, $\mu_r^*D\ge\mu_r^*C$.

\medskip

\quad{\bf (iii)} The next step is to check that $\diam D\le\diam C$.
If $x$, $x'\in D$, express them as $(y,\xi_r)$ and $(y',\xi_r')$.
Because $F$ is a bounded closed set in $\BbbR^r$, $F_y$ and $F_{y'}$ are
bounded closed subsets of $\Bbb R$.   Also both $f(y)$ and $f(y')$ must
be
greater than $0$, so that $F_y$, $F_{y'}$ are both non-empty.
Consequently

\Centerline{$\alpha=\inf F_y$,\quad $\beta=\sup F_y$,
\quad$\alpha'=\inf F_{y'}$,
\quad$\beta'=\sup F_{y'}$}

\noindent are all defined in $\Bbb R$, and $\alpha$, $\beta\in F_y$,
while
$\alpha'$ and $\beta'$ belong to $F_{y'}$.   We have

$$\eqalign{|\xi_r-\xi_r'|
&\le|\xi_r|+|\xi'_r|
<\Bover12f(y)+\Bover12f(y')\cr
&=\Bover12(\mu_1F_y+\mu_1F_{y'})
\le\Bover12(\beta-\alpha+\beta'-\alpha')
\le\max(\beta'-\alpha,\beta-\alpha').\cr}$$

\noindent So taking $(\xi,\xi')$ to be one of $(\alpha,\beta')$ or
$(\beta,\alpha')$, we can find $\xi\in F_y$, $\xi'\in F_{y'}$ such that
$|\xi-\xi'|\ge|\xi_r-\xi'_r|$.   Now $z=(y,\xi)$, $z'=(y',\xi')$ both
belong to $F$, so

\Centerline{$\|x-x'\|^2=\|y-y'\|^2+|\xi_r-\xi_r'|^2
\le\|y-y'\|^2+|\xi-\xi'|^2
=\|z-z'\|^2
\le(\diam F)^2$,}

\noindent and $\|x-x'\|\le\diam F$.   As $x$ and $x'$ are arbitrary,
$\diam D\le\diam F=\diam C$, as claimed.

\medskip

\quad{\bf (iv)} Evidently $S_r[D]=D$.   Moreover, if $i\in J$, then
(interpreting $S_i$ as an operator on $\BbbR^{r-1}$)

\Centerline{$F_{S_i(y)}=F_y$ for every $y\in\BbbR^{r-1}$,}

\noindent so $f(S_i(y))=f(y)$ and, for $\xi\in \Bbb R$,
$y\in\BbbR^{r-1}$,

\Centerline{$(y,\xi)\in D\,\iff\,|\xi|<\bover12 f(y)\,
\iff\,|\xi|<\bover12f(S_i(y))
\,\iff\,(S_i(y),\xi)\in D$,}

\noindent so that $S_i[D]=D$.   Thus $D$ is symmetric in coordinates in
$J\cup\{r\}$.\ \Qed

\medskip

{\bf (d)} The rest is easy.   Starting from any bounded
$A\subseteq\BbbR^r$, set $A_0=A$ and construct inductively
$A_1,\ldots,A_r$ such that

\Centerline{$d=\diam A=\diam A_0\ge\diam A_1\ge\ldots\ge\diam A_r$,}

\Centerline{$\mu_r^*A=\mu_r^*A_0\le\ldots\le\mu_r^*A_r$,}

\Centerline{$A_j$ is symmetric in coordinates in $\{1,\ldots,j\}$ for
every $j\le r$.}

\noindent At the end, we have $A_r$ symmetric in coordinates in
$\{1,\ldots,r\}$.   But this means that if $x\in A_r$ then

\Centerline{$-x=S_1S_2\ldots S_rx\in A_r$,}

\noindent so that

\Centerline{$\|x\|=\Bover12\|x-(-x)\|
\le\Bover12\diam A_r\le\Bover{d}2$.}

\noindent Thus $A_r\subseteq B(\tbf{0},\bover{d}2)$, and

\Centerline{$\mu_r^*A\le\mu_r^*A_r\le\mu_rB(\tbf{0},\Bover{d}2)$,}

\noindent as claimed.
}%end of proof of 264H


\leader{264I}{Theorem} Let $r\ge 1$ be an integer;  let $\mu$ be
Lebesgue measure on $\BbbR^r$, and let $\mu_{Hr}$ be
$r$-dimensional Hausdorff measure on $\BbbR^r$.   Then $\mu$ and
$\mu_{Hr}$ have the same measurable sets and

\Centerline{$\mu E=2^{-r}\beta_r\mu_{Hr}E$}

\noindent for every measurable set $E\subseteq\BbbR^r$, where
$\beta_r=\mu B(\tbf{0},1)$, so that the normalizing factor is

$$\eqalign{2^{-r}\beta_r
&=\Bover{1}{2^{2k}k!}\pi^k\text{ if }r=2k\text{ is even},\cr
&=\Bover{k!}{(2k+1)!}\pi^k\text{ if }r=2k+1\text{ is odd}.\cr}$$

\proof{{\bf (a)} Of course if $B=B(x,\alpha)$ is any ball of radius
$\alpha$,

\Centerline{$2^{-r}\beta_r(\diam B)^r
=\beta_r\alpha^r=\mu B$.}

\medskip

{\bf (b)} The point is that $\mu^*=2^{-r}\beta_r\mu_{Hr}^*$.   \Prf\
Let $A\subseteq\BbbR^r$.

\medskip

\quad{\bf (i)} Let $\delta$, $\epsilon>0$.   By
261F, there is a sequence $\sequencen{B_n}$ of  balls, all of diameter
at most $\delta$, such that $A\subseteq\bigcup_{n\in\Bbb N}B_n$ and
$\sum_{n=0}^{\infty}\mu B_n\le\mu^*A+\epsilon$.   Now, defining
$\theta_{r\delta}$ as in 264A,

\Centerline{$2^{-r}\beta_r\theta_{r\delta}(A)
\le 2^{-r}\beta_r\sum_{n=0}^{\infty}(\diam B_n)^r
=\sum_{n=0}^{\infty}\mu B_n
\le\mu^*A+\epsilon$.}

\noindent Letting $\delta\downarrow 0$,

\Centerline{$2^{-r}\beta_r\mu^*_{Hr}A\le\mu^*A+\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$2^{-r}\beta_r\mu_{Hr}^*A\le\mu^*A$.

\medskip

\quad{\bf (ii)} Let $\epsilon>0$.   Then there is a sequence
$\sequencen{A_n}$ of sets of diameter at most $1$ such that
$A\subseteq\bigcup_{n\in\Bbb N}A_n$ and $\sum_{n=0}^{\infty}(\diam
A_n)^r\le\theta_{r1}A+\epsilon$, so that

\Centerline{$\mu^*A\le\sum_{n=0}^{\infty}\mu^*A_n
\le\sum_{n=0}^{\infty}2^{-r}\beta_r(\diam A_n)^r
\le 2^{-r}\beta_r(\theta_{r1}A+\epsilon)
\le 2^{-r}\beta_r(\mu_{Hr}^*A+\epsilon)$}

\noindent by 264H.   As $\epsilon$ is arbitrary,
$\mu^*A\le 2^{-r}\beta_r\mu_{Hr}^*A$.\ \Qed

\medskip

{\bf (c)} Because $\mu$, $\mu_{Hr}$ are the measures defined from
their respective outer measures by \Caratheodory's method, it follows
at once that $\mu=2^{-r}\beta_r\mu_{Hr}$ in the strict sense required.
}%end of proof of 264I

\leader{*264J}{The Cantor \dvrocolon{set}}\cmmnt{ I remarked in 264A
that fractional
`dimensions' $r$ were of interest.   I have no space for these here, and
they are off the main lines of this volume, but I will give one result
for its intrinsic interest.

\medskip

\noindent}{\bf Proposition} Let $C$ be the Cantor set in $[0,1]$.   Set
$r=\ln 2/\ln 3$.   Then the $r$-dimensional Hausdorff measure of $C$ is
$1$.

\proof{{\bf (a)} Recall that $C=\bigcap_{n\in\Bbb N}C_n$, where each
$C_n$ consists of $2^n$ closed intervals of length $3^{-n}$, and
$C_{n+1}$ is obtained from $C_n$ by deleting the middle (open) third of
each interval of $C_n$.   (See 134G.)    Because $C$ is
closed, $\mu_{Hr}C$ is defined (264E).   Note that $3^r=2$.

\medskip

{\bf (b)} If $\delta>0$, take $n$ such that $3^{-n}\le\delta$;  then $C$
can be covered by $2^n$ intervals of diameter $3^{-n}$, so

\Centerline{$\theta_{r\delta}C\le 2^n(3^{-n})^r=1$.}

\noindent Consequently

\Centerline{$\mu_{Hr}C=\mu_{Hr}^*C
=\lim_{\delta\downarrow 0}\theta_{r\delta}C\le 1$.}

\medskip

{\bf (c)} We need the following elementary fact:  if $\alpha$, $\beta$,
$\gamma\ge 0$ and $\max(\alpha,\gamma)\le\beta$, then
$\alpha^r+\gamma^r\le(\alpha+\beta+\gamma)^r$.   \Prf\ Because
$0<r\le 1$,

\Centerline{$\xi\mapsto(\xi+\eta)^r-\xi^r
=r\biggerint_0^{\eta}(\xi+\zeta)^{r-1}d\zeta$}

\noindent is non-increasing for every $\eta\ge 0$.
Consequently

$$\eqalign{(\alpha+\beta+\gamma)^r-\alpha^r-\gamma^r
&\ge (\beta+\beta+\gamma)^r-\beta^r-\gamma^r\cr
&\ge(\beta+\beta+\beta)^r-\beta^r-\beta^r
=\beta^r(3^r-2)
=0,\cr}$$

\noindent as required.\ \Qed

\medskip

{\bf (d)} Now suppose that $I\subseteq\Bbb R$ is any interval, and
$m\in\Bbb N$;  write $j_m(I)$ for the number of the intervals composing
$C_m$ which are included in $I$.   Then $2^{-m}j_m(I)\le(\diam I)^r$.
\Prf\ If $I$ does not meet $C_m$, this is trivial.   Otherwise, induce
on

\Centerline{$l=\min\{i:I$ meets only one of the intervals composing
$C_{m-i}\}$.}

\noindent If $l=0$, so that $I$ meets only one of the intervals
composing $C_m$, then $j_m(I)\le 1$, and if $j_m(I)=1$ then
$\diam I\ge 3^{-m}$ so $(\diam I)^r\ge 2^{-m}$;  thus the induction
starts.   For
the inductive step to $l>1$, let $J$ be the interval of
$C_{m-l}$ which meets $I$, and $J'$, $J''$ the two intervals of
$C_{m-l+1}$ included in $J$, so that $I$ meets both $J'$ and $J''$, and


\Centerline{$j_m(I)=j_m(I\cap J)=j_m(I\cap J')+j_m(I\cap J'')$.}

\noindent By the inductive hypothesis,

\Centerline{$(\diam(I\cap J'))^r+(\diam(I\cap J''))^r
\ge 2^{-m}j_m(I\cap J')+2^{-m}j_m(I\cap J'')
=2^{-m}j_m(I)$.}

\noindent On the other hand, by (c),

$$\eqalign{(\diam(I\cap J'))^r+(\diam(I\cap J''))^r
&\le(\diam(I\cap J')+3^{-m+l-1}+\diam(I\cap J''))^r\cr
&=(\diam(I\cap J))^r
\le(\diam I)^r\cr}$$

\noindent because $J'$, $J''$ both have diameter at most $3^{-(m-l+1)}$,
the length of the interval between them.
Thus the induction continues.\ \Qed

\medskip

{\bf (e)} Now suppose that $\epsilon>0$.   Then there is a sequence
$\sequencen{A_n}$ of sets, covering $C$, such that
\discrcenter{468pt}{$\sum_{n=0}^{\infty}(\diam A_n)^r
<\mu_{Hr}C+\epsilon$.  }Take $\eta_n>0$ such that
$\sum_{n=0}^{\infty}(\diam A_n+\eta_n)^r\le\mu_{Hr}C+\epsilon$, and for
each $n$ take an open
interval $I_n\supseteq A_n$ of length at most $\diam A_n+\eta_n$ and
with neither endpoint belonging to $C$;  this is possible because $C$
does not include any non-trivial interval.   Now
$C\subseteq\bigcup_{n\in\Bbb N}I_n$;  because $C$ is compact, there is a
$k\in\Bbb N$ such that $C\subseteq\bigcup_{n\le k}I_n$.   Next, there is
an $m\in\Bbb N$ such that no endpoint of any $I_n$, for $n\le k$,
belongs to $C_m$.   Consequently each of the intervals composing $C_m$
must be included in some $I_n$, and
(in the terminology of (d) above) $\sum_{n=0}^kj_m(I_n)\ge 2^m$.
Accordingly

\Centerline{$1
\le\sum_{n=0}^k2^{-m}j_m(I_n)
\le \sum_{n=0}^k(\diam I_n)^r
\le \sum_{n=0}^{\infty}(\diam A_n+\eta_n)^r
\le\mu_{Hr}C+\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\mu_{Hr}C\ge 1$, as required.
}%end of proof of 264J

\leader{*264K}{General metric spaces}\cmmnt{ While this chapter deals
exclusively with Euclidean spaces, readers familiar with the general
theory
of metric spaces may find the nature of the theory clearer if they use
the language of metric spaces in the basic definitions and results.   I
therefore repeat the definition here, and spell out the corresponding
results in the exercises 264Yb-264Yl.

} Let $(X,\rho)$ be a metric space, and $r>0$.   For any $A\subseteq X$,
$\delta>0$ set

$$\eqalign{\theta_{r\delta}A
=\inf\{\sum_{n=0}^{\infty}(\diam A_n)^r:\sequencen{A_n}
&\text{ is a sequence of subsets of }X\text{ covering }A,\cr
&\qquad\qquad\qquad\quad\diam A_n\le\delta\text{ for every }
  n\in\Bbb N\},\cr}$$

\noindent interpreting the diameter of the empty set as $0$, and
$\inf\emptyset$ as $\infty$, so that $\theta_{r\delta}A=\infty$ if $A$
cannot be covered by a sequence of sets of diameter at most $\delta$.
Say
that $\theta_rA=\sup_{\delta>0}\theta_{r\delta}A$ is the
{\bf $r$-dimensional Hausdorff outer measure} of $A$, and take the
measure $\mu_{Hr}$ defined by \Caratheodory's method from this outer measure to be {\bf $r$-dimensional Hausdorff measure} on $X$.

\exercises{
\leader{264X}{Basic exercises $\pmb{>}$(a)}
%\spheader 264Xa
Show that all the functions $\theta_{r\delta}$ of 264A are outer
measures.   Show that in that context, $\theta_{r\delta}(A)=0$ iff
$\theta_r(A)=0$, for any $\delta>0$ and any $A\subseteq\BbbR^s$.

\spheader 264Xb Let $s\ge 1$ be an integer, and $\theta$ an
outer measure on $\BbbR^s$ such that $\theta(A\cup B)=\theta A+\theta B$
whenever $A$, $B$ are non-empty subsets of $\BbbR^s$ and
$\inf_{x\in A,y\in B}\|x-y\|>0$.   Show that every Borel subset of
$\BbbR^s$ is measured by the
measure defined from $\theta$ by \Caratheodory's method.

\sqheader 264Xc Let $s\ge 1$ be an integer and $r>0$;
define $\theta_{r\delta}$ as in 264A.   Show that for any
$A\subseteq\BbbR^s$, $\delta>0$,

$$\eqalign{\theta_{r\delta}A
&=\inf\{\sum_{n=0}^{\infty}(\diam F_n)^r:\sequencen{F_n}
 \text{ is a sequence of closed subsets of }X\cr
&\qquad\qquad\qquad\qquad\qquad\qquad\text{ covering }A,\diam
F_n\le\delta
  \text{ for every }n\in\Bbb N\}\cr
&=\inf\{\sum_{n=0}^{\infty}(\diam G_n)^r:\sequencen{G_n}
  \text{ is a sequence of open subsets of }X\cr
&\qquad\qquad\qquad\qquad\qquad\qquad\text{ covering }A,\diam
G_n\le\delta
  \text{ for every }n\in\Bbb N\}.\cr}$$

\sqheader 264Xd Let $s\ge 1$ be an integer and $r\ge 0$;
let $\mu_{Hr}$ be $r$-dimensional Hausdorff measure on $\BbbR^s$.
Show that for every $A\subseteq \BbbR^s$ there is a
G$_{\delta}$ set (that is, a set expressible as the intersection of a
sequence of open sets) $H\supseteq A$ such that $\mu_{Hr}H=\mu^*_{Hr}A$.
({\it Hint\/}:  use 264Xc.)

\sqheader 264Xe Let $s\ge 1$ be an integer, and $0\le r<r'$.   Show that
if $A\subseteq\BbbR^s$ and the $r$-dimensional
Hausdorff outer measure $\mu_{Hr}^*A$ of $A$ is finite, then
$\mu_{Hr'}^*A$ must be zero.

\spheader 264Xf(i) Suppose that $f:[a,b]\to\Bbb R$ has graph
$\Gamma_f\subseteq\BbbR^2$, where $a\le b$ in $\Bbb R$.   Show that the
outer measure $\mu_{H1}^*(\Gamma_f)$ of $\Gamma$ for one-dimensional
Hausdorff measure on $\BbbR^2$ is at most $b-a+\Var_{[a,b]}(f)$.
\Hint{if $f$ has finite variation, show that
$\diam(\Gamma_{f\restr\ooint{t,u}})
\le u-t+\Var_{\ooint{t,u}}(f)$;  then use 224E.}  (ii) Let
$f:[0,1]\to[0,1]$ be the Cantor function (134H).   Show that
$\mu_{H1}(\Gamma_f)=2$.   \Hint{264G.}
%264G

\spheader 264Xg\dvAnew{2009.} In 264A, show that

$$\eqalign{\theta_{r\delta}A
=\inf\{\sum_{n=0}^{\infty}(\diam A_n)^r:\sequencen{A_n}
&\text{ is a sequence of convex sets covering }A,\cr
&\qquad\qquad\qquad\quad\diam A_n\le\delta\text{ for every }
  n\in\Bbb N\}\cr}$$

\noindent for any $A\subseteq\BbbR^s$.
%264A

\leader{264Y}{Further exercises (a)}
%\spheader 264Ya
Let $\theta_{11}$ be the outer measure on $\BbbR^2$ defined in 264A,
with
$r=\delta=1$, and $\mu_{11}$ the measure derived from $\theta_{11}$ by
\Caratheodory's method, $\Sigma_{11}$ its domain.   Show that any set in
$\Sigma_{11}$ is either negligible or conegligible.

\spheader 264Yb Let $(X,\rho)$ be a metric space and
$r\ge 0$.   Show that if $A\subseteq X$ and $\mu_{Hr}^*A<\infty$,
then $A$ is separable.

\spheader 264Yc Let $(X,\rho)$ be a metric space, and $\theta$ an
outer measure on $X$ such that $\theta(A\cup B)=\theta A+\theta B$
whenever $A$, $B$ are non-empty subsets of $X$ and $\inf_{x\in A,y\in
B}\rho(x,y)>0$.   (Such an outer measure is called a {\bf metric outer
measure}.)    Show that every open subset of $X$ is measured by the
measure defined from $\theta$ by \Caratheodory's method.

\spheader 264Yd Let $(X,\rho)$ be a metric space and $r>0$;
define $\theta_{r\delta}$ as in 264K.    Show that for any
$A\subseteq X$,

$$\eqalign{\mu_{Hr}^*A
&=\sup_{\delta>0}\inf\{\sum_{n=0}^{\infty}(\diam F_n)^r:\sequencen{F_n}
  \text{ is a sequence of closed subsets of }X\cr
&\qquad\qquad\qquad\qquad\qquad\qquad\text{ covering }A,\,\diam
F_n\le\delta
  \text{ for every }n\in\Bbb N\}\cr
&=\sup_{\delta>0}\inf\{\sum_{n=0}^{\infty}(\diam G_n)^r:\sequencen{G_n}
  \text{ is a sequence of open subsets of }X\cr
&\qquad\qquad\qquad\qquad\qquad\qquad\text{ covering }A,\,\diam
G_n\le\delta
  \text{ for every }n\in\Bbb N\}.\cr}$$
\spheader 264Ye Let $(X,\rho)$ be a metric space and $r\ge 0$;
let $\mu_{Hr}$ be $r$-dimensional Hausdorff measure on $X$.
Show that for every $A\subseteq X$ there is a
G$_{\delta}$ set $H\supseteq A$ such that $\mu_{Hr}H=\mu^*_{Hr}A$ is the
$r$-dimensional Hausdorff outer measure of $A$.

\spheader 264Yf Let $(X,\rho)$ be a metric space and $r\ge 0$;
let $Y$ be any subset of $X$, and give $Y$ its induced metric $\rho_Y$.
(i) Show that the $r$-dimensional Hausdorff outer measure
$\mu^{(Y)*}_{Hr}$ on $Y$ is just the restriction to $\Cal PY$ of the
outer measure $\mu_{Hr}^*$ on $X$.   (ii) Show that if {\it either}
$\mu_{Hr}^*Y<\infty$ {\it or} $\mu_{Hr}$ measures $Y$ then
$r$-dimensional Hausdorff measure $\mu^{(Y)}_{Hr}$ on
$Y$ is just the subspace measure on $Y$ induced by the measure
$\mu_{Hr}$ on $X$.

\spheader 264Yg Let $(X,\rho)$ be a metric space
and $r>0$.   Show that $r$-dimensional Hausdorff measure on $X$ is
atomless.   ({\it Hint\/}: Let $E\in\dom\mu_{Hr}$.   (i) If
$E$ is not separable, there is an open set $G$ such that $E\cap G$ and
$E\setminus G$ are both non-separable, therefore both non-negligible.
(ii) If there is an $x\in E$ such that $\mu_{Hr}(E\cap B(x,\delta))>0$
for
every $\delta>0$, then one of these sets has non-negligible complement
in
$E$.    (iii) Otherwise, $\mu_{Hr}E=0$.)

\spheader 264Yh  Let $(X,\rho)$ be a metric space and $r\ge 0$;
let $\mu_{Hr}$ be
$r$-dimensional Hausdorff measure on $X$.   Show that if
$\mu_{Hr}E<\infty$
then $\mu_{Hr}E=\sup\{\mu_{Hr}F:F\subseteq E$ is closed and totally
bounded$\}$.   ({\it Hint\/}: given $\epsilon>0$, use 264Yd to find a
closed totally bounded set $F$ such that $\mu_{Hr}(F\setminus E)=0$
and $\mu_{Hr}(E\setminus F)\le\epsilon$, and now apply 264Ye to
$F\setminus E$.)

\spheader 264Yi  Let $(X,\rho)$ be a complete metric space and $r\ge 0$;
let $\mu_{Hr}$ be
$r$-dimensional Hausdorff measure on $X$.   Show that if
$\mu_{Hr}E<\infty$
then $\mu_{Hr}E=\sup\{\mu_{Hr}F:F\subseteq E$ is compact$\}$.

\spheader 264Yj Let $(X,\rho)$ and $(Y,\sigma)$ be metric spaces.   If
$D\subseteq X$ and $\phi:D\to Y$ is a function, then $\phi$ is {\bf
$\gamma$-Lipschitz} if
$\sigma(\phi(x),\phi(x'))\le\gamma\rho(x,x')$ for every $x$, $x'\in D$.
(i) Show that in this case, if $r\ge 0$,
$\mu_{Hr}^*(\phi[A])\le\gamma^r\mu_{Hr}^*A$ for every $A\subseteq D$,
writing $\mu_{Hr}^*$ for $r$-dimensional Hausdorff outer measure on
either $X$ or $Y$.   (ii) Show that if $X$ is complete and $\mu_{Hr}E$
is defined
and finite, then $\mu_{Hr}(\phi[E])$ is defined.   ({\it Hint\/}:
264Yi.)

\spheader 264Yk Let $(X,\rho)$ be a metric space, and for $r\ge 0$ let
$\mu_{Hr}$ be Hausdorff $r$-dimensional measure on $X$.   Show that
there is a unique $\Delta=\Delta(X)\in[0,\infty]$ such that
$\mu_{Hr}X=\infty$ if
$r\in\coint{0,\Delta}$, $0$ if $r\in\ooint{\Delta,\infty}$.

\spheader 264Yl Let $(X,\rho)$ be a metric space and $\phi:I\to X$ a
continuous function, where $I\subseteq\Bbb R$ is an interval.   Write
$\mu_{H1}$ for one-dimensional Hausdorff measure on $X$.   Show that

\Centerline{$\mu_{H1}(\phi[I])
\le\sup\{\sum_{i=1}^n\rho(\phi(t_i),\phi(t_{i-1})):t_0,\ldots,t_n\in
I,\,t_0\le\ldots\le t_n\}$,}

\noindent the length of the curve $\phi$, with equality if $\phi$ is
injective.


\spheader 264Ym Set $r=\ln 2/\ln 3$, as in 264J, and write
$\mu_{Hr}$ for $r$-dimensional Hausdorff measure on the Cantor set $C$.
Let $\lambda$ be the usual measure on $\{0,1\}^{\Bbb N}$ (254J).
Define $\phi:\{0,1\}^{\Bbb N}\to C$ by setting
$\phi(x)=\bover23\sum_{n=0}^{\infty}3^{-n}x(n)$ for
$x\in\{0,1\}^{\Bbb N}$.   Show that $\phi$ is an isomorphism between
$(\{0,1\}^{\Bbb N},\lambda)$ and $(C,\mu_{Hr})$, so that $\mu_{Hr}$ is the
subspace measure on $C$ induced by `Cantor measure' as defined in 256Hc.

\spheader 264Yn Set $r=\ln 2/\ln 3$ and write
$\mu_{Hr}$ for $r$-dimensional Hausdorff measure on the Cantor set $C$.
Let $f:[0,1]\to[0,1]$ be the Cantor function and let $\mu$ be
Lebesgue measure on $\Bbb R$.   Show that $\mu f[E]=\mu_{Hr}E$ for every
$E\in\dom\mu_{Hr}$ and $\mu_{Hr}(C\cap f^{-1}[F])=\mu F$ for every
Lebesgue measurable set $F\subseteq[0,1]$.

\spheader 264Yo Let $(X,\rho)$ be a metric space and
$h:\coint{0,\infty}\to\coint{0,\infty}$ a non-decreasing function.   For
$A\subseteq X$ set

$$\eqalign{\theta_hA
&=\sup_{\delta>0}\inf\{\sum_{n=0}^{\infty}h(\diam A_n):\sequencen{A_n}
  \text{ is a sequence of subsets of }X\cr
&\qquad\qquad\qquad\qquad\qquad\qquad\text{ covering }A,\,\diam
A_n\le\delta
  \text{ for every }n\in\Bbb N\},\cr}$$

\noindent interpreting $\diam\emptyset$ as $0$, $\inf\emptyset$ as
$\infty$ as usual.   Show that $\theta_h$ is an outer measure on $X$.
State and prove theorems corresponding to 264E and 264F.   Look through
264X and 264Y for further results which might be generalizable, perhaps
on the assumption that $h$ is continuous on the right.

\spheader 264Yp Let $(X,\rho)$ be a metric space.   Let us say that if
$a<b$ in $\Bbb R$ and $f:[a,b]\to X$ is a function, then $f$ is {\bf
absolutely continuous} if for every $\epsilon>0$ there is a $\delta>0$
such that $\sum_{i=1}^n\rho(f(a_i),f(b_i))\le\epsilon$ whenever
$a\le a_0\le b_0\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=0}^nb_i-a_i\le\delta$.   Show that $f:[a,b]\to X$ is absolutely
continuous iff it is continuous and of bounded variation (in the sense
of 224Ye) and $\mu_{H1}f[A]=0$ whenever $A\subseteq[a,b]$ is Lebesgue
negligible, where $\mu_{H1}$ is 1-dimensional Hausdorff measure on $X$.
(Compare 225M.)
Show that in this case $\mu_{H1}f[\,[a,b]\,]<\infty$.
%-  mt26bits

\spheader 264Yq Let $s\ge 1$ be an integer, and $r\in\coint{1,\infty}$.   
For $x$, $y\in\BbbR^s$ set $\rho(x,y)=\|x-y\|^{s/r}$.   (i) Show that 
$\rho$ is a metric on $\BbbR^s$ inducing the Euclidean topology.   (ii) 
Let $\mu_{Hr}$ be the associated $r$-dimensional Hausdorff measure.   
Show that $\mu_{Hr}B(\tbf{0},1)=2^s$.
%264I out of order query
}%end of exercises

\endnotes{
\Notesheader{264} In this section we have come to the next step in
`geometric measure theory'.  I am taking this very slowly, because there
are real difficulties in the subject, and for the purposes of this
volume we do not need to master very much of it.   The idea here is to
find a definition of $r$-dimensional Lebesgue measure which will be
`geometric' in the strict sense, that is, dependent only on the metric
structure of $\BbbR^r$, and therefore applicable to sets which have a
metric structure but no linear structure.   As has happened before, the
definition of Hausdorff measure from an outer measure gives no
problems -- the only new idea in 264A-264C is that of using a supremum
$\theta_r=\sup_{\delta>0}\theta_{r\delta}$ of outer measures -- and the
difficult part is proving that our new measure has any useful
properties.   Concerning the properties of Hausdorff measure, there are
two essential objectives;  first, to check that these measures, in
general, share a reasonable proportion of the properties of Lebesgue
measure;  and second, to justify the term `$r$-dimensional measure' by
relating Hausdorff $r$-dimensional measure on $\BbbR^r$ to Lebesgue
measure on $\BbbR^r$.

As for the properties of general Hausdorff measures, we have to go
rather carefully.   I do not give counter-examples here because they
involve concepts which belong to Volumes 4 and 5 rather than this
volume, but I must warn you to expect the worst.   However, we do at
least have open sets measurable, so that all Borel sets are measurable
(264E).   The outer measure of a set $A$ can be defined in terms of the
Borel sets including $A$ (264Fa), though not in general in terms of the
open sets including $A$;  but the measure of a measurable set  $E$ is
not necessarily the supremum of the measures of the Borel sets included
in $E$, unless $E$ has finite measure (264Fc).   We do find that the
outer measure $\theta_r$ defined in 264A is the outer measure defined
from $\mu_{Hr}$ (264Fb), so that the phrase
`$r$-dimensional Hausdorff outer measure' is unambiguous.   A crucial
property of Lebesgue measure is the fact that the measure of a
measurable set $E$ is the supremum of the measures of the compact
subsets
of $E$;  this is not generally shared by Hausdorff measures, but is
valid for sets $E$ of finite measure in complete spaces (264Yi).
Concerning subspaces, there are no problems with the outer measures, and
for sets of finite measure the subspace measures are also consistent
(264Yf).   Because Hausdorff measure is defined in metric terms, it
behaves regularly for Lipschitz maps (264G);  one of the most
natural classes of functions to consider when studying metric spaces is
that of $1$-Lipschitz functions, so that (in the language of 264G)
$\mu_{Hr}^*\phi[A]\le \mu_{Hr}^*A$ for every $A$.

The second essential feature of Hausdorff measure, its relation with
Lebesgue measure in the appropriate dimension, is Theorem 264I.
Because both Hausdorff measure and Lebesgue measure are
translation-invariant, this can be proved by relatively elementary
means, except for the evaluation of the normalizing constant;  all we
need to know is that $\mu\coint{0,1}^r=1$ and
$\mu_{Hr}\coint{0,1}^r$ are both finite and non-zero, and this is
straightforward.   (The arguments of part (a) of the proof of 261F are
relevant.)   For the purposes of this chapter, we do not I think have to
know the value of the constant;  but I cannot leave it unsettled, and
therefore give Theorem 264H, the {\bf isodiametric inequality}, to show
that it is just the Lebesgue measure of an $r$-dimensional ball
of diameter $1$, as one would hope and expect.   The critical step in
the argument of 264H is in part (c) of the proof.   This is called
`Steiner symmetrization';  the idea is that given a set $A$, we
transform
$A$ through a series of steps, at each stage lowering, or at least not
increasing, its diameter, and raising, or at least not decreasing, its
outer measure, progressively making $A$ more symmetric, until at the end
we have a set which is sufficiently constrained to be amenable.   The
particular symmetrization operation used in this proof is important
enough;  but the idea of progressive regularization of an object is one
of the most powerful methods in measure theory, and you should give all
your attention to mastering any example you encounter.   In my
experience, the idea is principally useful when seeking an inequality
involving disparate quantities -- in the present example, the diameter
and volume of a set.

Of course it is awkward having two measures on $\BbbR^r$, differing by
a constant multiple, and for the purposes of the next section it would
actually have been a little more convenient to follow {\smc Federer 69}
in using `normalized Hausdorff measure' $2^{-r}\beta_r\mu_{Hr}$.
(For non-integral $r$, we could take
$\beta_r=\pi^{r/2}/\Gamma(1+\bover{r}2)$, as suggested in
252Xi.)   However, I believe this to be a minority position, and
the striking example of Hausdorff measure on the Cantor set (264J,
264Ym-264Yn) looks much better in the non-normalized version.

Hausdorff ($\ln 2/\ln 3$)-dimensional measure on the Cantor set is of
course but one, perhaps the easiest, of a large class of examples.
Because the Hausdorff
$r$-dimensional outer measure of a set $A$, regarded as a function of
$r$, behaves dramatically (falling from $\infty$ to $0$) at a certain
critical
value $\Delta(A)$ (see 264Xe, 264Yk), it gives us a metric space
invariant
of $A$;  $\Delta(A)$ is the {\bf Hausdorff dimension} of $A$.
Evidently the Hausdorff dimension of $C$ is $\ln 2/\ln 3$, while that of
$r$-dimensional Euclidean space is $r$.
}%end of notes

\discrpage


\frfilename{mt265.tex}
\versiondate{3.9.13}
\copyrightdate{2000}

\def\chaptername{Change of variable in the integral}
\def\sectionname{Surface measures}

\newsection{265}

In this section I offer a new version of the arguments of \S263, this
time not with the intention of justifying integration-by-substitution,
but instead to give a practically effective method of computing the
Hausdorff $r$-dimensional measure of a smooth $r$-dimensional surface
in an $s$-dimensional space.   The basic case to bear in mind is $r=2$,
$s=3$, though any other combination which you can easily visualize
will also be a valuable aid to intuition.   I give a fundamental theorem
(265E) providing a formula from which we can hope to calculate the
$r$-dimensional measure of a surface in $s$-dimensional space which is
parametrized by a differentiable function, and work through some of
the calculations in the case of the $r$-sphere
(265F-265H). %265F 265G 265H

\leader{265A}{Normalized Hausdorff measure}\dvro{ In}{ As I remarked
at the end of
the last section, Hausdorff measure, as defined in 264A-264C, is not
quite the most appropriate measure for our work here;  so in}
this section I will use {\bf normalized Hausdorff measure}, meaning
$\nu_r=2^{-r}\beta_r\mu_{Hr}$, where $\mu_{Hr}$ is $r$-dimensional
Hausdorff measure\cmmnt{ (interpreted in whichever space is under
consideration)} and
$\beta_r=\mu_rB(\tbf{0},1)$ is the Lebesgue measure of any ball
of radius $1$ in $\BbbR^r$.   \cmmnt{It will be convenient to take
$\beta_0=1$.}   \dvro{This}{As shown in 264H-264I, this} normalization
makes $\nu_r$ on $\BbbR^r$ agree with Lebesgue measure $\mu_r$.
\cmmnt{Observe that of course}
$\nu_r^*=2^{-r}\beta_r\mu_{Hr}^*$\cmmnt{ (264Fb)}.

\leader{265B}{Linear \dvrocolon{subspaces}}\cmmnt{ Just as in \S263,
the first step is to deal with linear operators.

\medskip

\noindent}{\bf Theorem} Suppose that $r$, $s$ are integers with
$1\le r\le s$, and that $T$ is a real $s\times r$ matrix;  regard $T$
as
a linear operator from $\BbbR^r$ to $\BbbR^s$.   Set
$J=\sqrt{\det T\trs T}$,
where $T\trs $ is the transpose of $T$.   Write $\nu_r$ for normalized
$r$-dimensional Hausdorff measure on $\BbbR^s$, $\Tau_r$ for its
domain,
and $\mu_r$ for Lebesgue measure on $\BbbR^r$.   Then

\Centerline{$\nu_r T[E]=J\mu_rE$}

\noindent for every measurable set $E\subseteq\BbbR^r$.   If $T$ is
injective (that is, if $J\ne 0$), then

\Centerline{$\nu_rF=J\mu_rT^{-1}[F]$}

\noindent whenever $F\in\Tau_r$ and $F\subseteq T[\BbbR^r]$.

\proof{ The formula for $J$ assumes that $\det T\trs T$ is
non-negative, which is a fact not in evidence;  but the argument below
will establish it adequately soon.
\medskip
{\bf (a)} Let $V$ be the linear subspace of $\BbbR^s$ consisting of
vectors $y=(\eta_1,\ldots,\eta_s)$ such that $\eta_i=0$ whenever
$r<i\le
s$.   Let $R$ be the $r\times s$ matrix $\langle
\rho_{ij}\rangle_{i\le
r,j\le s}$, where $\rho_{ij}=1$ if $i=j\le r$, $0$ otherwise;  then
the
$s\times r$ matrix $R\trs $
may be regarded as a bijection from $\BbbR^r$ to $V$.   Let $W$ be an
$r$-dimensional linear subspace of $\BbbR^s$ including $T[\BbbR^r]$,
and let $P$ be an orthogonal $s\times s$ matrix such that $P[W]=V$.
Then $S=RPT$ is an $r\times r$ matrix.   We have $R\trs Ry=y$ for $y\in
V$,
so $R\trs RPT=PT$ and

\Centerline{$S\trs S=T\trs P\trs R\trs RPT=T\trs P\trs PT=T\trs T$;}

\noindent accordingly

\Centerline{$\det T\trs T=\det S\trs S=(\det S)^2\ge 0$}

\noindent and $J=|\det S|$.   At the same time,

\Centerline{$P\trs R\trs S=P\trs R\trs RPT=P\trs PT=T$.}

\noindent Observe that $J=0$ iff $S$ is not injective, that is, $T$ is
not injective.

\medskip

{\bf (b)} If we consider the $s\times r$ matrix $P\trs R\trs $ as a map from
$\BbbR^r$ to $\BbbR^s$, we see that $\phi=P\trs R\trs $ is an isometry between
$\BbbR^r$ and $W$, with inverse $\phi^{-1}=RP\restrp W$.   It follows
that $\phi$ is an isomorphism between the measure spaces
$(\Bbb R^r,\mu_{Hr}^{(r)})$ and $(W,\mu_{HrW}^{(s)})$, where
$\mu_{Hr}^{(r)}$ is
$r$-dimensional Hausdorff measure on $\BbbR^r$ and $\mu_{HrW}^{(s)}$
is
the subspace measure on $W$ induced by $r$-dimensional
Hausdorff measure $\mu_{Hr}^{(s)}$ on $\BbbR^s$.

\medskip

\Prf\ {\bf (i)} If $A\subseteq\BbbR^r$ and $A'\subseteq W$,

\Centerline{$\mu_{Hr}^{(s)*}(\phi[A])\le\mu_{Hr}^{(r)*}(A)$,
\quad $\mu_{Hr}^{(r)*}(\phi^{-1}[A'])\le\mu_{Hr}^{(s)*}(A')$,}

\noindent using 264G twice.   Thus
$\mu_{Hr}^{(s)*}(\phi[A])=\mu_{Hr}^{(r)*}(A)$ for
every $A\subseteq \BbbR^r$.

\medskip

\quad{\bf (ii)} Now because $W$ is closed, therefore in the domain of
$\mu_{Hr}^{(s)}$ (264E), the subspace measure $\mu_{HrW}^{(s)}$ is
just
the measure induced by $\mu_{Hr}^{(s)*}\restrp W$ by \Caratheodory's
method (214H(b-ii)).   Because $\phi$ is an isomorphism between
$(\Bbb R^r,\mu_{Hr}^{(r)*})$ and $(W,\mu_{Hr}^{(s)*}\restrp W)$, it
is an
isomorphism between $(\BbbR^r,\mu_{Hr}^{(r)})$ and
$(W,\mu_{HrW}^{(s)})$.
\Qed

\medskip

{\bf (c)}  It follows that $\phi$ is also an isomorphism between the
normalized versions $(\BbbR^r,\mu_r)$ and $(W,\nu_{rW})$, writing
$\nu_{rW}$ for the subspace measure on $W$ induced by $\nu_r$.

Now if $E\subseteq\BbbR^r$ is  Lebesgue measurable, we have
$\mu_rS[E]=J\mu_rE$, by 263A;  so that

\Centerline{$\nu_rT[E]=\nu_r(P\trs R\trs [S[E]])=\nu_r(\phi[S[E]])
=\mu_rS[E]=J\mu_rE$.}

\noindent If $T$ is injective, then $S=\phi^{-1}T$ must also be
injective, so that $J\ne 0$ and

\Centerline{$\nu_rF=\mu_r(\phi^{-1}[F])
=J\mu_r(S^{-1}[\phi^{-1}[F]])=J\mu_rT^{-1}[F]$}

\noindent whenever $F\in\Tau_r$ and $F\subseteq W=T[\BbbR^r]$.
}%end of proof of 265B

\leader{265C}{Corollary} Under the conditions of 265B,

\Centerline{$\nu_r^*T[A]=J\mu_r^*A$}

\noindent for every $A\subseteq\BbbR^r$.

\proof{{\bf (a)} If $E$ is  Lebesgue measurable and $A\subseteq E$,
then
$T[A]\subseteq T[E]$, so

\Centerline{$\nu_r^*T[A]\le \nu_rT[E]=J\mu_rE$;}

\noindent as $E$ is arbitrary, $\nu_r^*T[A]\le J\mu_r^*A$.

\medskip

{\bf (b)} If $J=0$ we can stop.   If $J\ne 0$ then $T$ is injective,
so
if $F\in\Tau_r$ and $T[A]\subseteq F$ we shall have

\Centerline{$J\mu_r^*A\le J\mu_rT^{-1}[F\cap T[\BbbR^r]]
=\nu_r(F\cap T[\BbbR^r])\le\nu_rF$;}

\noindent as $F$ is arbitrary, $J\mu_r^*A\le\nu_r^*T[A]$.
}%end of proof of 265C

\leader{265D}{}\cmmnt{ I now proceed to the lemma corresponding to
263C.

\medskip

\noindent}{\bf Lemma} Suppose that $1\le r\le s$ and that $T$ is an
$s\times r$ matrix;  set $J=\sqrt{\det T\trs T}$, and suppose that 
$J\ne 0$.
Then for any $\epsilon>0$ there is a $\zeta=\zeta(T,\epsilon)>0$ such
that

(i) $|\sqrt{\det S\trs S}-J|\le\epsilon$ whenever $S$ is an $s\times r$
matrix and $\|S-T\|\le\zeta$\cmmnt{, defining the norm of a matrix as in
262H};

(ii) whenever $D\subseteq\BbbR^r$ is a bounded set and $\phi:D\to\Bbb
R^s$ is a function such that
$\|\phi(x)-\phi(y)-T(x-y)\|\le\zeta\|x-y\|$
for all $x$, $y\in D$, then
$|\nu_r^*\phi[D]-J\mu_r^*D|\le\epsilon\mu_r^*D$.

\proof{{\bf (a)} Because $\det S\trs S$ is a continuous function of the
coefficients of $S$, 262Hb tells us that there must be a $\zeta_0>0$
such that $|J-\sqrt{\det S\trs S}|\le\epsilon$ whenever
$\|S-T\|\le\zeta_0$.

\medskip

{\bf (b)} Because $J\ne 0$, $T$ is injective, and there is an
$r\times s$ matrix $T^*$ such
that $T^*T$ is the identity $r\times r$ matrix.   Take $\zeta>0$ such
that $\zeta\le\zeta_0$,
$\zeta\|T^*\|<1$, $J(1+\zeta\|T^*\|)^r\le J+\epsilon$ and
$1-J^{-1}\epsilon\le(1-\zeta\|T^*\|)^r$.

Let $\phi:D\to\BbbR^s$ be such that
$\|\phi(x)-\phi(y)-T(x-y)\|\le\zeta\|x-y\|$ whenever $x$, $y\in D$.
Set $\psi=\phi T^*$, so that $\phi=\psi T$.   Then

\Centerline{$\|\psi(u)-\psi(v)\|\le(1+\zeta\|T^*\|)\|u-v\|$,
\quad$\|u-v\|\le(1-\zeta\|T^*\|)^{-1}\|\psi(u)-\psi(v)\|$}

\noindent whenever $u$, $v\in T[D]$.
\Prf\ Take $x$, $y\in D$ such that $u=Tx$, $v=Ty$;  of course
$x=T^*u$, $y=T^*v$.   Then

$$\eqalign{\|\psi(u)-\psi(v)\|
&=\|\phi(T^*u)-\phi(T^*v)\|
=\|\phi(x)-\phi(y)\|\cr
&\le\|T(x-y)\|+\zeta\|x-y\|\cr
&=\|u-v\|+\zeta\|T^*u-T^*v\|
\le\|u-v\|(1+\zeta\|T^*\|).\cr}$$

\noindent Next,

$$\eqalign{\|u-v\|
&=\|Tx-Ty\|
\le\|\phi(x)-\phi(y)\|+\zeta\|x-y\|\cr
&=\|\psi(u)-\psi(v)\|+\zeta\|T^*u-T^*v\|\cr
&\le\|\psi(u)-\psi(v)\|+\zeta\|T^*\|\|u-v\|,\cr}$$

\noindent so that $(1-\zeta\|T^*\|)\|u-v\|\le\|\psi(u)-\psi(v)\|$ and
$\|u-v\|\le(1-\zeta\|T^*\|)^{-1}\|\psi(u)-\psi(v)\|$.   \Qed

\medskip

{\bf (c)} Now from 264G and 265C we see that

\Centerline{$\nu^*_r\phi[D]
=\nu_r^*\psi[T[D]]
\le(1+\zeta\|T^*\|)^r\nu_r^*T[D]
=(1+\zeta\|T^*\|)^rJ\mu_r^*D
\le(J+\epsilon)\mu_r^*D$,}

\noindent and (provided $\epsilon\le J$)

$$\eqalignno{(J-\epsilon)\mu_r^*D
&=(1-J^{-1}\epsilon)\nu_r^*T[D]
\le(1-J^{-1}\epsilon)(1-\zeta\|T^*\|)^{-r}\nu_r^*\psi[T[D]]\cr
\displaycause{applying 264G to $\psi^{-1}:\psi[T[D]]\to T[D]$}
&\le\nu_r^*\psi[T[D]]
=\nu_r^*\phi[D].\cr}$$

\noindent (Of course, if $\epsilon\ge J$, then surely
$(J-\epsilon)\mu_r^*D\le\nu_r^*\phi[D]$.)   Thus
\Centerline{$(J-\epsilon)\mu_r^*D\le\nu_r^*\phi[D]
\le(J+\epsilon)\mu_r^*D$}

\noindent as required, and we have an appropriate $\zeta$.
}%end of proof of 265D

\leader{265E}{Theorem} Suppose that $1\le r\le s$;  write $\mu_r$ for
Lebesgue measure on $\BbbR^r$, $\nu_r$ for normalized Hausdorff
measure on $\BbbR^s$, and $\Tau_r$ for the domain of $\nu_r$.   Let
$D\subseteq\BbbR^r$ be any set, and
$\phi:D\to\BbbR^s$ a function differentiable relative to its domain at
each point of $D$.   For each $x\in D$ let $T(x)$ be a derivative of
$\phi$ at $x$ relative to $D$, and set $J(x)=\sqrt{\det T(x)\trs T(x)}$.
Set $D'=\{x:x\in D,\,J(x)>0\}$.   Then

\quad (i) $J:D\to\coint{0,\infty}$ is a measurable function;

\quad (ii) $\nu_r^*\phi[D]\le\int_DJ(x)\mu_r(dx)$,

\noindent allowing $\infty$ as the value of the integral;

\quad (iii) $\nu_r^*\phi[D\setminus D']=0$.

\noindent If $D$ is Lebesgue measurable, then

\quad (iv) $\phi[D]\in\Tau_r$.

\noindent If $D$ is measurable and $\phi$ is injective, then

\quad (v) $\nu_r\phi[D]=\int_DJ\,d\mu_r$;

\quad (vi) for any set $E\subseteq\phi[D]$, $E\in\Tau_r$
iff $\phi^{-1}[E]\cap D'$ is  Lebesgue measurable, and in this case

\Centerline{$\nu_rE
=\int_{\phi^{-1}[E]}J(x)\mu_r(dx)
=\int_DJ\times\chi(\phi^{-1}[E])d\mu_r$;}

\quad (vii) for every real-valued function $g$ defined on a subset of
$\phi[D]$,

\Centerline{$\int_{\phi[D]}g\,d\nu_r=\int_DJ\times g\phi\,d\mu_r$}

\noindent if either integral is defined in $[-\infty,\infty]$,
provided
we interpret $J(x)g(\phi(x))$ as zero when $J(x)=0$ and $g(\phi(x))$
is undefined.

\proof{ I seek to follow the line laid out in the proof of 263D.

\medskip

{\bf (a)} Just as in 263D, we know that $J:D\to\Bbb R$ is measurable,
since $J(x)$ is a continuous function of the coefficients of $T(x)$,
all of which are measurable, by 262P.   If $D$ is Lebesgue measurable,
then there is a sequence $\sequencen{F_n}$ of compact subsets of $D$ such
that $D\setminus\bigcup_{n\in\Bbb N}F_n$ is $\mu_r$-negligible.   Now
$\phi[F_n]$ is compact, therefore belongs to $\Tau_r$, for each
$n\in\Bbb N$.   As for $\phi[D\setminus\bigcup_{n\in\Bbb N}F_n]$, this
must be $\nu_r$-negligible by 264G, because $\phi$ is a countable union of
Lipschitz functions (262N).   So

\Centerline{$\phi[D]=\bigcup_{n\in\Bbb N}
  \phi[F_n]\cup\phi[D\setminus\bigcup_{n\in\Bbb N}F_n]\in\Tau_r$.}

\noindent This deals with (i) and (iv).

\medskip

{\bf (b)} For the moment, assume that $D$ is bounded and that $J(x)>0$
for every $x\in D$, and fix $\epsilon>0$.   Let $M^*_{sr}$ be the set
of $s\times r$ matrices $T$ such that $\det T\trs T\ne 0$, that is, the
corresponding map $T:\BbbR^r\to\BbbR^s$ is injective.   For  
$T\in M_{sr}^*$ take $\zeta(T,\epsilon)>0$ as in 265D.

Take $\sequencen{D_n}$, $\sequencen{T_n}$ as in 262M, with $A=M^*_{sr}$,
so that $\sequencen{D_n}$ is a partition of $D$ into sets which are
relatively measurable in $D$, and each $T_n$ is an $s\times r$ matrix
such that

\Centerline{$\|T(x)-T_n\|\le\zeta(T_n,\epsilon)$ whenever $x\in D_n$,}

\Centerline{$\|\phi(x)-\phi(y)-T_n(x-y)\|\le\zeta(T_n,\epsilon)\|x-y\|$
for all $x$, $y\in D_n$.}

\noindent Then, setting $J_n=\sqrt{\det T_n\trs T_n}$, we have

\Centerline{$|J(x)-J_n|\le\epsilon$ for every $x\in D_n$,}

\Centerline{$|\nu_r^*\phi[D_n]-J_n\mu_r^*D_n|\le\epsilon\mu_r^*D_n$,}

\noindent by the choice of $\zeta(T_n,\epsilon)$.   So

$$\eqalignno{\nu_r^*\phi[D]
&\le\sum_{n=0}^{\infty}\nu_r^*\phi[D_n]\cr
\displaycause{because $\phi[D]=\bigcup_{n\in\Bbb N}\phi[D_n]$}
&\le\sum_{n=0}^{\infty}J_n\mu_r^*D_n+\epsilon\mu_r^*D_n
\le\epsilon\mu_r^*D+\sum_{n=0}^{\infty}J_n\mu_r^*D_n\cr
\displaycause{because the $D_n$ are disjoint and relatively measurable
in $D$}
&=\epsilon\mu_r^*D+\int_D\sum_{n=0}^{\infty}J_n\chi D_nd\mu\cr
&\le\epsilon\mu_r^*D+\int_DJ(x)+\epsilon\mu_r(dx)
=2\epsilon\mu_r^*D+\int_DJ\,d\mu_r.\cr}$$

If $D$ is measurable and $\phi$ is injective, then all the $D_n$ are
Lebesgue measurable subsets of $\BbbR^r$, so all the $\phi[D_n]$ are
measured by $\nu_r$, and they are also disjoint.   Accordingly

$$\eqalign{\int_{D}J\,d\mu
&\le\sum_{n=0}^{\infty}J_n\mu_r D_n+\epsilon\mu_r D\cr
&\le\sum_{n=0}^{\infty}(\nu_r\phi[D_n]
   +\epsilon\mu_r D_n)+\epsilon\mu_rD
=\nu_r\phi[D]+2\epsilon\mu_rD.\cr}$$

\noindent Since $\epsilon$ is arbitrary, we get

\Centerline{$\nu_r^*\phi[D]\le\int_DJ\,d\mu_r$,}

\noindent and if $D$ is measurable and $\phi$ is injective,

\Centerline{$\int_DJ\,d\mu_r\le\nu_r\phi[D]$;}

\noindent thus we have (ii) and (v), on the assumption that $D$ is
bounded and $J>0$ everywhere on $D$.

\medskip

{\bf (c)} Just as in the proof of
263D, we can now relax the assumption that $D$ is
bounded by considering $B_k=B(\tbf{0},k)\subseteq\BbbR^r$;  provided
$J>0$ everywhere on $D$, we get

\Centerline{$\nu_r^*\phi[D]
=\lim_{k\to\infty}\nu_r^*\phi[D\cap B_k]
\le\lim_{k\to\infty}\int_{D\cap B_k}J\,d\mu_r
=\int_DJ\,d\mu_r$,}

\noindent with equality if $D$ is measurable and $\phi$ is injective.

\medskip

{\bf (d)} Now we find that $\nu_r^*\phi[D\setminus D']=0$.

\medskip

\Prf\grheada\ Let $\eta\in\ocint{0,1}$.   Define
$\psi_{\eta}:D\to\BbbR^{s+r}$ by setting $\psi_{\eta}(x)=(\phi(x),\eta x)$,
identifying $\BbbR^{s+r}$ with $\BbbR^s\times\BbbR^r$.   $\psi_{\eta}$
is differentiable relative to its domain at each point of $D$, with
derivative $\tilde T_{\eta}(x)$, being the $(s+r)\times r$ matrix in
which the top $s$ rows consist of the $s\times r$ matrix $T(x)$, and
the bottom $r$ rows are $\eta I_r$, writing $I_r$ for the $r\times r$
identity matrix.   (Use 262Ib.)   Now of course $\tilde T_{\eta}(x)$,
regarded as a map from $\BbbR^r$ to $\BbbR^{s+r}$, is injective, so

\Centerline{$\tilde J_{\eta}(x)
=\sqrt{\det\tilde T_{\eta}(x)\trs \tilde T_{\eta}(x)}
=\sqrt{\det(T(x)\trs T(x)+\eta^2I)}>0$.}

\noindent   We have $\lim_{\eta\downarrow 0}\tilde J_{\eta}(x)=J(x)=0$
for $x\in D\setminus D'$.

\medskip

\quad\grheadb\ Express $T(x)$ as
$\langle\tau_{ij}(x)\rangle_{i\le s,j\le r}$ for each $x\in D$.   Set

\Centerline{$C_m=\{x:x\in D,\,\|x\|\le m,\,|\tau_{ij}(x)|\le m$ for
all $i\le s,\,j\le r\}$}

\noindent for each $m\ge 1$.   For $x\in C_m$, all the coefficients of
$\tilde T_{\eta}(x)$ have moduli at most $m$;  consequently (giving
the crudest and most immediately available inequalities) all the
coefficients of $\tilde T_{\eta}(x)\trs \tilde T_{\eta}(x)$ have moduli at
most $(r+s)m^2$ and $\tilde J_{\eta}(x)\le\sqrt{r!(s+r)^r}m^r$.
Consequently we can use Lebesgue's Dominated Convergence Theorem to
see that

\Centerline{$\lim_{\eta\downarrow 0}\int_{C_m\setminus D'}\tilde
J_{\eta}d\mu_r=0$.}

\medskip

\quad\grheadc\ Let $\tilde\nu_r$ be normalized Hausdorff
$r$-dimensional measure on $\BbbR^{s+r}$.   Applying (b) of this proof
to $\psi_{\eta}\restr C_m\setminus D'$, we see that

\Centerline{$\tilde\nu_r^*\psi_{\eta}[C_m\setminus D']
\le\int_{C_m\setminus D'}\tilde J_{\eta}d\mu_r$.}

\noindent Now we have a natural map $P:\BbbR^{s+r}\to\BbbR^s$ given by
setting $P(\xi_1,\ldots,\xi_{s+r})=(\xi_1,\ldots,\xi_s)$, and $P$ is
$1$-Lipschitz, so by 264G once more we have (allowing for
the normalizing constants $2^{-r}\beta_r$)

\Centerline{$\nu_r^*P[A]\le\tilde\nu_r^*A$}

\noindent for every $A\subseteq\BbbR^{s+r}$.   In particular,

\Centerline{$\nu_r^*\phi[C_m\setminus D']
=\nu_r^*P[\psi_{\eta}[C_m\setminus D']]
\le\tilde\nu_r^*\psi_{\eta}[C_m\setminus D']
\le\int_{C_m\setminus D'}\tilde J_{\eta}d\mu_r
\to 0$}

\noindent as $\eta\downarrow 0$.   But this means that
$\nu_r^*\phi[C_m\setminus D']=0$.   As $D=\bigcup_{m\ge 1}C_m$,
$\nu_r^*\phi[D\setminus D']=0$, as claimed.   \Qed

\medskip

{\bf (e)} This proves (iii) of the theorem.   But of course this is
enough to give (ii) and (v), because (applying (b)-(c) to $\phi\restr D'$)
we must have

\Centerline{$\nu_r^*\phi[D]
=\nu_r^*\phi[D']
\le\int_{D'}J\,d\mu_r
=\int_DJ\,d\mu_r$,}

\noindent with equality if $D$ (and therefore also $D'$)
is measurable and $\phi$ is injective.

\medskip

{\bf (f)} So let us turn to part (vi).   Assume that $D$ is measurable
and that $\phi$ is injective.

\medskip

\quad\grheada\ Suppose that $E\subseteq\phi[D]$ belongs to $\Tau_r$.
Let

\Centerline{$H_k=\{x:x\in D,\,\|x\|\le k,\,J(x)\le k\}$}

\noindent for each $k$;  then each $H_k$ is Lebesgue measurable, so
(applying (iii) to $\phi\restr H_k$) $\phi[H_k]\in\Tau_r$, and

\Centerline{$\nu_r\phi[H_k]\le k\mu_rH_k<\infty$.}

\noindent Thus $\phi[D]$ can be covered by a sequence of
sets of finite measure for $\nu_r$, which of course are of finite
measure for $r$-dimensional Hausdorff measure on $\BbbR^s$.   By 264Fc,
there are Borel sets $E_1$, $E_2\subseteq\BbbR^s$ such that
$E_1\subseteq E\subseteq E_2$ and $\nu_r(E_2\setminus E_1)=0$.
Now $F_1=\phi^{-1}[E_1]$, $F_2=\phi^{-1}[E_2]$ are Lebesgue measurable
subsets of $D$, and

\Centerline{$\int_{F_2\setminus F_1}J\,d\mu_r=\nu_r\phi[F_2\setminus
F_1]=\nu_r(\phi[D]\cap E_2\setminus E_1)=0$.}

\noindent Accordingly $\mu_r(D'\cap(F_2\setminus F_1))=0$.   But as

\Centerline{$D'\cap F_1\subseteq D'\cap\phi^{-1}[E]
\subseteq D'\cap F_2$,}

\noindent it follows that $D'\cap\phi^{-1}[E]$ is measurable, and that

$$\eqalign{\int_{\phi^{-1}[E]}J\,d\mu_r
&=\int_{D'\cap\phi^{-1}[E]}J\,d\mu_r
=\int_{D'\cap F_1}J\,d\mu_r\cr
&=\int_{D\cap F_1}J\,d\mu_r
=\nu_r\phi[D\cap F_1]
=\nu_rE_1
=\nu_rE.\cr}$$

\noindent Moreover, $J\times\chi(\phi^{-1}[E])
=J\times\chi(D'\cap\phi^{-1}[E])$ is measurable, so we can write 
$\int J\times\chi(\phi^{-1}[E])$ in place of $\int_{\phi^{-1}[E]}J$.

\medskip

\quad\grheadb\ If $E\subseteq\phi[D]$ and $D'\cap\phi^{-1}[E]$ is
measurable, then of course

\Centerline{$E=\phi[D'\cap\phi^{-1}[E]]
  \cup\phi[(D\setminus D')\cap\phi^{-1}[E]]\in\Tau_r$,}

\noindent because $\phi[G]\in\Tau_r$ for every measurable $G\subseteq D$
and $\phi[D\setminus D']$ is $\nu_r$-negligible.

\medskip

{\bf (g)} Finally, (vii) follows at once from (vi), applying 235J to
$\mu_r$ and the subspace measure induced by $\nu_r$ on $\phi[D]$.
}%end of proof of 265E

\leader{265F}{The surface of a sphere}\cmmnt{ To show how these ideas
can be applied to one of the basic cases, I give the details of a method
of describing spherical surface measure in $s$-dimensional space.
Take $r\ge 1$ and $s=r+1$.}   Write $S_r$ for
$\{z:z\in\Bbb R^{r+1},\,\|z\|=1\}$, the $r$-sphere.
\dvro{Then the normalized $r$-dimensional Hausdorff measure of $S_r$ is
$2\pi\beta_{r-1}$, where $\beta_{r-1}$ is the volume of the unit ball of
$\BbbR^{r-1}$ (interpreting $\beta_0$ as $1$.)}
{Then we have a parametrization
$\phi_r$ of $S_r$ given by setting

$$\phi_r\Matrix{\xi_1\\ \xi_2\\ \ldots \\ \ldots \\ \ldots\\ \xi_r}
=\Matrix{\sin\xi_1\sin\xi_2\sin\xi_3\ldots\sin\xi_r\\
\cos\xi_1\sin\xi_2\sin\xi_3\ldots\sin\xi_r\\
\cos\xi_2\sin\xi_3\ldots\sin\xi_r\\
\ldots\\
\cos\xi_{r-2}\sin\xi_{r-1}\sin\xi_r\\
\cos\xi_{r-1}\sin\xi_r\\
\cos\xi_r}.$$

\noindent I choose this formulation because I wish to use an inductive
argument based on the fact that

\Centerline{$\phi_{r+1}\Matrix{x\\ \xi}
=\Matrix{\sin\xi\,\phi_r(x)\\ \cos\xi}$}
\noindent for $x\in\BbbR^r$, $\xi\in\Bbb R$.   Every $\phi_r$ is
differentiable, by 262Id.   If we set

$$\eqalign{D_r
=\{x:\xi_1&\in\ocint{-\pi,\pi},\,\xi_2,\ldots,\xi_r\in[0,\pi],\cr
&\text{ if }\xi_j\in\{0,\pi\}\text{ then }
  \xi_i=0\text{ for }i<j\},\cr}$$

\noindent then it is easy to check that $D_r$ is a Borel subset of
$\Bbb R^r$ and that $\phi_r\restr D_r$ is a bijection between $D_r$
and
$S_r$. Now let $T_r(x)$ be the $(r+1)\times r$ matrix $\phi_r'(x)$.
Then

$$T_{r+1}\Matrix{x\\ \xi}=\Matrix{\sin\xi\,T_r(x)&\cos\xi\,\phi_r(x)\\
                                  \tbf{0}&-\sin\xi}.$$

\noindent So

$$(T_{r+1}\Matrix{x\\ \xi})\trs T_{r+1}\Matrix{x\\ \xi}
=\Matrix{\sin^2\xi\,T_r(x)\trs T_r(x)&\sin\xi\cos\xi\,T_r(x)\trs \phi_r(x)\\
\cos\xi\sin\xi\,\phi_r(x)\trs T_r(x)&\cos^2\xi\phi_r(x)\trs \phi_r(x)+\sin^2\xi}
.$$

\noindent But of course $\phi_r(x)\trs \phi_r(x)=\|\phi_r(x)\|^2=1$ for every
$x$, and (differentiating with respect to each coordinate of $x$, if you
wish) $T_r(x)\trs \phi_r(x)=\tbf{0}$, $\phi_r(x)\trs T_r(x)=\tbf{0}$.   So we get

$$(T_{r+1}\Matrix{x\\ \xi})\trs T_{r+1}\Matrix{x\\ \xi}
=\Matrix{\sin^2\xi\,T_r(x)\trs T_r(x)&\tbf{0}\\
\tbf{0}&1},$$

\noindent and writing $J_r(x)=\sqrt{\det T_r(x)\trs T_r(x)}$,

$$J_{r+1}\Matrix{x\\ \xi}=|\sin^r\xi|J_r(x).$$

At this point we induce on $r$ to see that

\Centerline{$J_r(x)
=|\sin^{r-1}\xi_r\sin^{r-2}\xi_{r-1}\ldots\sin\xi_2|$}

\noindent (since of course the induction starts with the case $r=1$,
\discrcenter{468pt}{$\phi_1(x)=\Matrix{\sin x\\ \cos x}$,
\ifdim\pagewidth>467pt\quad\fi
$T_1(x)=\Matrix{\cos x\\ -\sin x}$,
\ifdim\pagewidth>467pt\quad\fi
$T_1(x)\trs T_1(x)=1$,
\ifdim\pagewidth>467pt\quad\fi
$J_1(x)=1$).}

To find the surface measure of $S_r$, we need to calculate

$$\eqalignno{\int_{D_r}J_rd\mu_r
&=\int_0^{\pi}\ldots\int_0^{\pi}\int_{-\pi}^{\pi}
\sin^{r-1}\xi_r\ldots\sin\xi_2
d\xi_1d\xi_2\ldots d\xi_r\cr
&=2\pi\prod_{k=2}^r\int_0^{\pi}\sin^{k-1}t\,dt
=2\pi\prod_{k=1}^{r-1}\int_{-\pi/2}^{\pi/2}\cos^kt\,dt\cr}$$

\noindent (substituting $\bover{\pi}{2}-t$ for $t$).   But in the
language of 252Q, this is just

\Centerline{$2\pi\prod_{k=1}^{r-1}I_k=2\pi\beta_{r-1}$,}

\noindent where $\beta_{r-1}$ is the volume of the unit ball of
$\BbbR^{r-1}$\cmmnt{ (interpreting $\beta_0$ as $1$, if you like)}.
}%end of dvro

\leader{265G}{}\cmmnt{ The surface area of a sphere can also be
calculated through the following result.

\medskip

\noindent}{\bf Theorem} Let $\mu_{r+1}$ be Lebesgue measure on
$\BbbR^{r+1}$, and $\nu_r$ normalized $r$-dimensional Hausdorff
measure
on $\BbbR^{r+1}$.   If $f$ is a locally $\mu_{r+1}$-integrable
real-valued function, $y\in\BbbR^{r+1}$ and $\delta>0$,

\Centerline{$\int_{B(y,\delta)}fd\mu_{r+1}
=\int_0^{\delta}\int_{\partial B(y,t)}fd\nu_rdt$,}

\noindent where I write $\partial B(y,t)$ for the sphere
$\{x:\|x-y\|=t\}$\cmmnt{ and the integral $\int\ldots dt$ is to be
taken with respect to Lebesgue measure on $\Bbb R$}.

\proof{ Take any differentiable function
$\phi:\BbbR^r\to S_r$ with a
Borel set $F\subseteq\BbbR^r$ such that $\phi\restr F$ is a
bijection between $F$ and $S_r$;  such a pair $(\phi,F)$ is
described in 265F.   Define
$\psi:\BbbR^r\times\Bbb R\to\BbbR^{r+1}$ by setting
$\psi(z,t)=y+t\phi(z)$;  then $\psi$ is differentiable and
$\psi\restr F\times\ocint{0,\delta}$ is a bijection between
$F\times\ocint{0,\delta}$ and $B(y,\delta)\setminus\{y\}$.   For
$t\in\ocint{0,\delta}$, $z\in\BbbR^r$ set $\psi_t(z)=\psi(z,t)$;
then $\psi_t\restr F$ is a bijection between $F$ and the sphere
$\{x:\|x-y\|=t\}=\partial B(y,t)$.

The derivative of $\phi$ at $z$ is an $(r+1)\times r$ matrix $T_1(z)$
say, and the derivative $T_t(z)$ of $\psi_t$ at $z$ is just $tT_1(z)$;
also the derivative of $\psi$ at $(z,t)$ is the the $(r+1)\times(r+1)$
matrix $T(z,t)=\Matrix{tT_1(z)&\phi(z)}$, where $\phi(z)$ is
interpreted
as a column vector.   If we set

\Centerline{$J_t(z)=\sqrt{\det T_t(z)\trs T_t(z)}$,
\quad$J(z,t)=|\det T(z,t)|$,}

\noindent then

$$\eqalign{J(z,t)^2
=\det T(z,t)\trs T(z,t)
&=\det\Matrix{tT_1(z)\trs \\ \phi(z)\trs }\Matrix{tT_1(z)&\phi(z)}\cr
&=\det\Matrix{t^2T_1(z)\trs T_1(z)&\tbf{0}\\ \tbf{0}&1}
=J_t(z)^2,\cr}$$

\noindent because when we come to calculate the $(i,r+1)$-coefficient
of
$T(z,t)\trs T(z,t)$, for $1\le i\le r$, it is

\Centerline{$\sum_{j=1}^{r+1}t\Pd{\phi_j}{\zeta_i}(z)\phi_j(z)
=\Bover{t}2\Pd{}{\zeta_i}(\sum_{j=1}^{r+1}\phi_j(z)^2)
=0$,}

\noindent where $\phi_j$ is the $j$th coordinate of $\phi$;  while the
$(r+1,r+1)$-coefficient of $T(z,t)\trs T(z,t)$ is just
$\sum_{j=1}^{r+1}\phi_j(z)^2=1$.
So in fact $J(z,t)=J_t(z)$ for all $z\in\BbbR^r$, $t>0$.

Now, given $f\in\eusm L^1(\mu_{r+1})$, we can calculate

$$\eqalignno{\int_{B(y,\delta)}fd\mu_{r+1}
&=\int_{B(y,\delta)\setminus\{y\}}fd\mu_{r+1}\cr
&=\int_{F\times\ocint{0,\delta}}f(\psi(z,t))J(z,t)\mu_{r+1}(d(z,t))\cr
\displaycause{by 263D}
&=\int_0^{\delta}\int_Ff(\psi_t(z))J_t(z)\mu_r(dz)dt\cr
\displaycause{where $\mu_r$ is Lebesgue measure on $\BbbR^r$,
by Fubini's theorem, 252B}
&=\int_0^{\delta}\int_{\partial B(y,t)}fd\nu_rdt\cr}$$

\noindent by 265E(vii).
}%end of proof of 265G

\leader{265H}{Corollary} If $\nu_r$ is normalized $r$-dimensional
Hausdorff measure on $\BbbR^{r+1}$, then $\nu_rS_r=(r+1)\beta_{r+1}$.

\proof{ In 265G, take $y=\tbf{0}$, $\delta=1$, and
$f=\chi B(\tbf{0},1)$;  then

\Centerline{$\beta_{r+1}=\int fd\mu_{r+1}
=\int_0^1\nu_r(\partial B(\tbf{0},t))dt=\int_0^1t^r\nu_rS_rdt
=\Bover1{r+1}\nu_rS_r$,}

\noindent this time
applying 264G to the maps $x\mapsto tx$, $x\mapsto\bover1tx$
from $\BbbR^{r+1}$ to itself to see that
$\nu_r(\partial B(\tbf{0},t))=t^r\nu_rS_r$ for $t>0$.
}%end of proof of 265H

\exercises{
\leader{265X}{Basic exercises (a)} Let $r\ge 1$, and let
$S_r(\alpha)=\{z:z\in\BbbR^{r+1},\,\|z\|=\alpha\}$ be the
$r$-sphere of radius $\alpha$.   Show that
$\nu_rS_r(\alpha)=2\pi\beta_{r-1}\alpha^r=(r+1)\beta_{r+1}\alpha^r$
for
every $\alpha\ge 0$.

\sqheader 265Xb Let $r\ge 1$, and for $a\in[-1,1]$ set
$C_a=\{z:z\in\BbbR^{r+1},\,\|z\|=1,\,\zeta_{r+1}\ge a\}$, writing
$z=(\zeta_1,\ldots,\zeta_{r+1})$ as usual.   (i) Show that

\Centerline{$\nu_rC_a=r\beta_r\int_0^{\arccos a}\sin^{r-1}t\,dt$.}

\noindent (ii)\dvAnew{2013} 
Compute the integral in the cases $r=2$, $r=4$.

\sqheader 265Xc Again write
$C_a=\{z:z\in S_r,\,\zeta_{r+1}\ge a\}$, where $S_r\subseteq\BbbR^{r+1}$
is the unit sphere.   Show that, for any $a\in\ocint{0,1}$,
$\nu_rC_a\le\Bover{\nu_rS_r}{2(r+1)a^2}$.   \Hint{calculate
$\sum_{i=1}^{r+1}\int_{S_r}\|\xi_i\|^2\nu_r(dx)$.}
%265F

\sqheader 265Xd Let $\phi:\ooint{0,1}\to\BbbR^r$ be an injective
differentiable function.   Show that the `length' or
one-dimensional Hausdorff measure of $\phi[\,\ooint{0,1}\,]$ is just
$\int_0^1\|\phi'(t)\|dt$.

\spheader 265Xe(i) Show that if $I$ is the identity $r\times r$ matrix
and $z\in\BbbR^r$, then $\det(I+zz\trs )=1+\|z\|^2$.   \Hint{induce on
$r$.}
(ii) Write $U_{r-1}$ for the open unit ball in $\BbbR^{r-1}$, where
$r\ge 2$.   Define $\phi:U_{r-1}\times\Bbb R\to S_r$ by setting

$$\phi\Matrix{x\\ \xi}=\Matrix{x\\ \theta(x)\cos\xi \\
\theta(x)\sin\xi},$$

\noindent where $\theta(x)=\sqrt{1-\|x\|^2}$.   Show that

$$\phi'\Matrix{x\\ \xi}\trs \phi'\Matrix{x\\ \xi}
=\Matrix{I+\Bover1{\theta(x)^2}xx\trs  & \tbf{0}\\
        \tbf{0} & \theta(x)^2},$$

\noindent so that $J\Matrix{x\\ \xi}=1$ for all $x\in U_{r-1}$,
$\xi\in\Bbb R$.   (iii) Hence show that the normalized $r$-dimensional
Hausdorff measure of $\{y:y\in S_r,\,\sum_{i=1}^{r-1}\eta_i^2<1\}$ is
just $2\pi\beta_{r-1}$, where $\beta_{r-1}$ is the Lebesgue measure of
$U_{r-1}$.   (iv) By considering $\psi z=\Matrix{z\\ 0\\ 0}$ for
$z\in S_{r-2}$, or otherwise, show that the normalized $r$-dimensional
Hausdorff measure of $S_r$ is $2\pi\beta_{r-1}$.   (v) This time setting
$C_a=\{z:z\in\BbbR^{r+1},\,\|z\|=1,\,\zeta_1\ge a\}$, show that
$\nu_rC_a=2\pi\mu_{r-1}\{x:x\in\BbbR^{r-1},\,\|x\|\le 1,\,\xi_1\ge a\}$
for every $a\in[-1,1]$.
%265F

\spheader 265Xf\dvAnew{2013} Suppose that $r\ge 2$.   Identifying
$\BbbR^r$ with $\BbbR^{r-1}\times\Bbb R$, let $C_r$ be the cylinder
$B_{r-1}\times[-1,1]\supseteq B_r$, and 
$\partial C_r=(B_{r-1}\times\{-1,1\})\cup(S_{r-2}\times[-1,1])$ 
its boundary.   Show that 

\Centerline{$\Bover{\mu_rB_r}{\mu_rC_r}
=\Bover{\nu_{r-1}S_{r-1}}{\nu_{r-1}(\partial C_r)}$.}

\noindent(The case $r=3$ is due to Archimedes.)
%265F

\leader{265Y}{Further exercises (a)} 
%\spheader 265Ya
Take $a<b$ in $\Bbb R$.   (i)
Show that $\phi:[a,b]\to\BbbR^r$ is absolutely continuous in the
sense of 264Yp iff all its coordinates
$\phi_i:[a,b]\to\Bbb R$, for $i\le r$, are absolutely continuous in
the sense of \S225.   (ii) Let $\phi:[a,b]\to\BbbR^r$
be a continuous function, and set $F=\{x:x\in\ooint{a,b},\,\phi$ is
differentiable at $x\}$.    Show that $\phi$ is absolutely continuous
iff $\int_F\|\phi'(x)\|dx$ is finite and
$\nu_1(\phi[[a,b]\setminus F])=0$, where $\nu_1$ is (normalized)
Hausdorff one-dimensional measure on $\BbbR^r$.   ({\it Hint\/}:
225K.)   (iii) Show that if $\phi:[a,b]\to\BbbR^r$ is
absolutely continuous then $\nu_1^*(\phi[D])\le\int_D\|\phi'(x)\|dx$
for
every $D\subseteq[a,b]$, with equality if $D$ is measurable and
$\phi\restr D$ is injective.

\spheader 265Yb Suppose that $a\le b$ in $\Bbb R$, and that
$f:[a,b]\to\Bbb R$ is a continuous function of bounded variation with
graph $\Gamma_f$.   Show that the one-dimensional Hausdorff measure
of $\Gamma_f$ is $\Var_{[a,b]}(f)+\int_a^b(\sqrt{1+(f')^2}-|f'|)$.
%mt26bits
}%end of exercises

\endnotes{
\Notesheader{265} The proof of 265B seems to call on most of the
second
half of the alphabet.   The idea is supposed to be straightforward
enough.   Because $T[\BbbR^r]$ has dimension at most $r$, it can be
rotated by an
orthogonal transformation $P$ into a subspace of the canonical
$r$-dimensional subspace $V$, which is a natural copy of $\BbbR^r$;
the matrix $R$ represents the copying process from $V$ to $\BbbR^r$, and
$\phi$ or $P\trs R\trs $ is a copy of $\BbbR^r$ onto a subspace including
$T[\Bbb R^r]$.   All this copying back and forth is designed to turn
$T$ into a linear operator $S:\BbbR^r\to\BbbR^r$ to which we can apply
263A, and part (b)
of the proof is the check that we are copying the measures as well as
the linear structures.

In 265D-265E I have tried to follow 263C-263D as closely as possible.
In fact only one new idea is needed.   When $s=r$, we have a special
argument
available to show that $\mu_r^*\phi[D]\le J\mu_r^*D+\epsilon\mu_r^*D$
(in the language of 263C) which applies whether or not $J=0$.   When
$s>r$, this approach fails, because we can no longer approximate
$\nu_rT[B]$ by
$\nu_rG$ where $G\supseteq T[B]$ is open.   (See part (b-i) of the proof
of 263C.)   I therefore turn to a different argument, valid only when
$J>0$, and accordingly have to find a separate method to show that
$\{\phi(x):x\in D,\,J(x)=0\}$ is
$\nu_r$-negligible.   Since we are working without restrictions on the
dimensions $r$, $s$ except that $r\le s$, we can use the trick of
approximating $\phi:D\to\BbbR^s$ by $\psi_{\eta}:D\to\BbbR^{s+r}$, as
in part (d) of the proof of 265E.

I give three methods by which the area of the $r$-sphere can be
calculated;  a bare-hands approach (265F), the surrounding-cylinder
method (265Xe) and an important repeated-integral theorem (265G).
The
first two provide formulae for the area of a cap (265Xb, 265Xe(v)).
The surrounding-cylinder method is attractive because the Jacobian
comes
out to be $1$, that is, we have an \imp\ function.   I note that
despite
having developed a technique which
allows irregular domains, I am still forced by the singularity in the
function $\theta$ of 265Xe to take the sphere in two bites.   Theorem
265G is a special case of the Coarea Theorem
({\smc Evans \& Gariepy 92}, \S3.4;  {\smc Federer 69}, 3.2.12).

For the next steps in the geometric theory of measures on Euclidean
space, see Chapter 47 in Volume 4.

}%end of notes

\frnewpage

\frfilename{mt266.tex}
\versiondate{28.1.09}
\copyrightdate{2004}

\def\chaptername{Change of variable in the integral}
\def\sectionname{The Brunn-Minkowski inequality}

\newsection{*266\dvAnew{2004}}

We now have most of the essential ingredients for a proof of the
Brunn-Minkowski inequality (266C) in a strong form.   I do not at
present expect to use it in this treatise, but it is one of the
basic results of geometric measure theory and from where we now stand is
not difficult, so I include it here.   The preliminary results on
arithmetic and geometric means (266A) and essential closures (266B) are
of great importance for other reasons.

\leader{266A}{Arithmetic and geometric \dvrocolon{means}}\cmmnt{ We
shall need the following standard result.

\medskip

\noindent}{\bf Proposition} If
$u_0,\ldots,u_n,p_0,\ldots,p_n\in\coint{0,\infty}$ and
$\sum_{i=0}^np_i=1$, then $\prod_{i=0}^nu_i^{p_i}\le\sum_{i=0}^np_iu_i$.

\proof{ Induce on $n$.   For $n=0$, $p_0=1$ the result is trivial.   If
$n=1$, then if $u_1=0$ the result is trivial (even if, as is standard in
this book, we interpret $0^0$ as $1$).   Otherwise, set
$t=\Bover{u_0}{u_1}$;  then

\Centerline{$t^{p_0}\le p_0t+1-p_0=p_0t+p_1$}

\noindent(as in part (a) of the proof of 244E), so

\Centerline{$u_0^{p_0}u_1^{p_1}=t^{p_0}u_1\le p_0tu_1+p_1u_1
=p_0u_0+p_1u_1$.}

\noindent For the inductive step to $n\ge 2$, if $p_0=\ldots=p_{n-1}=0$
the result is trivial.   Otherwise, set $q=p_0+\ldots+p_{n-1}=1-p_n$;
then

$$\eqalignno{\prod_{i=0}^nu_i^{p_i}
&=(\prod_{i=0}^{n-1}u_i^{p_i/q})^qu_n^{p_n}
\le(\sum_{i=0}^{n-1}\Bover{p_i}qu_i)^qu_n^{p_n}\cr
\displaycause{by the inductive hypothesis}
&\le q(\sum_{i=0}^{n-1}\Bover{p_i}qu_i)+p_nu_n\cr
\displaycause{by the two-term case just examined}
&=\sum_{i=0}^np_iu_i,\cr}$$

\noindent and the induction continues.
}%end of proof of 266A

\leader{266B}{Proposition} For any set $D\subseteq\BbbR^r$ set

\Centerline{$\clstar D=\{x:\limsup_{\delta\downarrow 0}
\Bover{\mu^*(D\cap B(x,\delta))}{\mu B(x,\delta)}>0\}$,}

\noindent where $\mu$ is Lebesgue measure on $\BbbR^r$.

(a) $D\setminus\clstar D$ is negligible.

(b) $\clstar D\subseteq\overline{D}$.

(c) $\clstar D$ is a Borel set.

(d) $\mu(\clstar D)=\mu^*D$.

(e) If $C\subseteq\Bbb R$ then
$\overline{C}+\clstar D\subseteq\clstar(C+D)$, writing $C+D$ for
$\{x+y:x\in C$, $y\in D\}$.

\proof{{\bf (a)} 261Da.

\medskip

{\bf (b)} If $x\in\BbbR^r\setminus\overline{D}$ then
$D\cap B(x,\delta)=\emptyset$ for all small $\delta$.

\medskip

{\bf (c)} The point is just that
$(x,\delta)\mapsto\mu^*(D\cap B(x,\delta))$ is continuous.   \Prf\ For
any $x$, $y\in\BbbR^r$ and $\delta$, $\eta\ge 0$ we have

$$\eqalignno{|\mu^*(D\cap B(y,\eta))-\mu^*(D\cap B(x,\delta))|
&\le\mu(B(y,\eta)\symmdiff B(x,\delta))\cr
&=2\mu(B(x,\delta)\cup B(y,\eta))-\mu B(x,\delta)-\mu B(y,\eta)\cr
&\le\beta_r\bigl(2(\max(\delta,\eta)+\|x-y\|)^r-\delta^r-\eta^r\bigr)\cr
\displaycause{where $\beta_r=\mu B(\tbf{0},1)$}
&\to 0\cr}$$

\noindent as $(y,\eta)\to(x,\delta)$.\ \QeD\  So

\Centerline{$x\mapsto
\limsup_{\delta\downarrow 0}
  \Bover{\mu^*(D\cap B(x,\delta))}{\mu B(x,\delta)}
=\inf_{\alpha\in\Bbb Q,\alpha>0}\sup_{\beta\in\Bbb Q,0<\beta\le\alpha}
  \Bover1{\beta_r\beta^r}\mu^*(D\cap B(x,\beta))$}

\noindent is Borel measurable, and

\Centerline{$\clstar D=\{x:\limsup_{\delta\downarrow 0}
  \Bover{\mu^*(D\cap B(x,\delta))}{\mu B(x,\delta)}>0\}$}

\noindent is a Borel set.

\medskip

{\bf (d)} By (c), $\mu(\clstar D)$ is defined;  by (a),
$\mu(\clstar D)\ge\mu^*D$.   On the other hand, let $E$ be a measurable
envelope of $D$ (132Ee);  then 261Db tells us that

\Centerline{$\limsup_{\delta\downarrow 0}
  \Bover{\mu^*(D\cap B(x,\delta))}{\mu B(x,\delta)}
\le\limsup_{\delta\downarrow 0}
  \Bover{\mu(E\cap B(x,\delta))}{\mu B(x,\delta)}
=0$}

\noindent for almost every $x\in\BbbR^r\setminus E$, so
$\clstar D\setminus E$ is negligible and

\Centerline{$\mu(\clstar D)\le\mu E=\mu^*D$.}

\medskip

{\bf (e)} If
$x\in\overline{C}$ and $y\in\clstar D$, set

\Centerline{$\gamma=\Bover13\limsup_{\delta\downarrow 0}
  \Bover{\mu^*(D\cap B(y,\delta))}{\mu B(y,\delta)}>0$.}

\noindent For any $\eta>0$, there is a $\delta\in\ocint{0,\eta}$ such
that $\mu^*(D\cap B(y,\delta))\ge 2\gamma\mu B(x,\delta)$.   Let
$\delta_1\in\coint{0,\delta}$ be such that
$\delta^r-\delta_1^r\le\gamma\delta^r$.   Then there is an $x'\in C$
such that $\|x-x'\|\le\delta-\delta_1$.  In this case,

$$\eqalign{\mu^*((C+D)\cap B(x+y,\delta))
&\ge\mu^*((x'+D)\cap B(x'+y,\delta_1))
=\mu^*(D\cap B(y,\delta_1))\cr
&\ge\mu^*(D\cap B(y,\delta))-\mu B(y,\delta)+\mu B(y,\delta_1)\cr
&\ge 2\beta_r\gamma\delta^r-\beta_r\delta^r+\beta_r\delta_1^r
\ge\beta_r\gamma\delta^r.\cr}$$

\noindent As $\eta$ is arbitrary,

\Centerline{$\limsup_{\delta\downarrow 0}
  \Bover{\mu^*((C+D)\cap B(x+y,\delta))}{\mu B(y,\delta)}
\ge\gamma$}

\noindent and $x+y\in\clstar(C+D)$;  as $x$ and $y$ are arbitrary,
$\overline{C}+\clstar D\subseteq\clstar(C+D)$.
}%end of proof of 266B

\cmmnt{\medskip

\noindent{\bf Remark} In this context, $\clstar D$ is called the {\bf
essential closure} of $D$.
}

\leader{266C}{Theorem} Let $A$, $B\subseteq\BbbR^r$ be non-empty sets,
where $r\ge 1$ is an integer.   If $\mu$ is Lebesgue measure on
$\BbbR^r$, and $A+B=\{x+y:x\in A$, $y\in B\}$, then
$\mu^*(A+B)^{1/r}\ge(\mu^*A)^{1/r}+(\mu^*B)^{1/r}$.

\proof{{\bf (a)} Consider first the case in which $A=\coint{a,a'}$ and
$B=\coint{b,b'}$ are half-open intervals.   In this case
$A+B=\coint{a+b,a'+b'}$;  writing $a=(\alpha_1,\ldots,\alpha_r)$, etc.,
as in \S115, set

\Centerline{$u_i
=\Bover{\alpha'_i-\alpha_i}{\alpha'_i+\beta'_i-\alpha_i-\beta_i}$,
\quad$v_i
=\Bover{\beta'_i-\beta_i}{\alpha'_i+\beta'_i-\alpha_i-\beta_i}$}

\noindent for each $i$.   Then we have

$$\eqalignno{(\mu A)^{1/r}+(\mu B)^{1/r}
&=\prod_{i=0}^r(\alpha'_i-\alpha_i)^{1/r}
  +\prod_{i=0}^r(\beta'_i-\beta_i)^{1/r}\cr
&=\mu(A+B)^{1/r}(\prod_{i=1}^ru_i^{1/r}+\prod_{i=1}^rv_i^{1/r})\cr
&\le\mu(A+B)^{1/r}(\Bover1r\sum_{i=1}^ru_i+\Bover1r\sum_{i=1}^rv_i)\cr
\displaycause{266A}
&=\mu(A+B)^{1/r}.\cr}$$

\medskip

{\bf (b)} Now I show by induction on $m+n$ that if
$A=\bigcup_{j=0}^mA_j$ and $B=\bigcup_{j=0}^nB_j$, where
$\langle A_j\rangle_{j\le m}$ and $\langle B_j\rangle_{j\le n}$ are both
disjoint families of non-empty half-open intervals, then
$\mu(A+B)^{1/r}\ge(\mu A)^{1/r}+(\mu B)^{1/r}$.   \Prf\ The induction
starts with the case $m=n=0$, dealt with in (a).   For the inductive
step to $m+n=l\ge 1$, one of $m$, $n$ is non-zero;  the argument is the
same in both cases;  suppose the former.
Since $A_0\cap A_1=\emptyset$, there must be some $j\le r$ and
$\alpha\in\Bbb R$ such that $A_0$ and $A_1$ are separated by the
hyperplane $\{x:\xi_j=\alpha\}$.   Set $A'=\{x:x\in A$, $\xi_j<\alpha\}$
and $A''=\{x:x\in A$, $\xi_j\ge\alpha\}$;  then both $A'$ and $A''$ are
non-empty and can be expressed as the union of at most $m-1$ disjoint
half-open intervals.   Set $\gamma=\Bover{\mu A'}{\mu A}\in\ooint{0,1}$.
The function $\beta\mapsto\mu\{x:x\in B$, $\xi_j<\beta\}$ is continuous,
so there is a $\beta\in\Bbb R$ such that $\mu B'=\gamma\mu B$, where
$B'=\{x:x\in B$, $\xi_j<\beta\}$;  set $B''=B\setminus B$.   Then $B'$
and $B''$ can be expressed as unions of at most $n$ half-open intervals.
By the inductive hypothesis,

\Centerline{$\mu(A'+B')^{1/r}\ge(\mu A')^{1/r}+(\mu B')^{1/r}$,
\quad$\mu(A''+B'')^{1/r}\ge(\mu A'')^{1/r}+(\mu B'')^{1/r}$.}

\noindent Now $A'+B'\subseteq\{x:\xi_j<\alpha+\beta\}$, while
$A''+B''\subseteq\{x:\xi_j\ge\alpha+\beta\}$.   So

$$\eqalign{\mu(A+B)
&\ge\mu(A'+B')+\mu(A''+B'')\cr
&\ge\bigl((\mu A')^{1/r}+(\mu B')^{1/r}\bigr)^r
  +\bigl((\mu A'')^{1/r}+(\mu B'')^{1/r}\bigr)^r\cr
&=\bigl((\gamma\mu A)^{1/r}+(\gamma\mu B)^{1/r}\bigr)^r
  +\bigl(((1-\gamma)\mu A)^{1/r}+((1-\gamma)\mu B)^{1/r}\bigr)^r\cr
&=(\bigl(\mu A)^{1/r}+(\mu B)^{1/r}\bigr)^r.\cr}$$

\noindent Taking $r$th roots,
$\mu(A+B)^{1/r}\ge(\mu A)^{1/r}+(\mu B)^{1/r}$ and the induction
proceeds.\ \Qed

\medskip

{\bf (c)} Now suppose that $A$ and $B$ are compact non-empty subsets of
$\BbbR^r$.   Then $\mu(A+B)^{1/r}\ge(\mu A)^{1/r}+(\mu B)^{1/r}$.
\Prf\ $A+B$ is compact (because $A\times B\subseteq\BbbR^r\times\BbbR^r$
is compact, being closed and bounded, and addition is continuous, so we
can use 2A2Eb).   Let $\epsilon>0$.   Let $G\supseteq A+B$ be an open set
such that $\mu G\le\mu(A+B)+\epsilon$ (134Fa);  then there is a
$\delta>0$ such that $B(x,2\delta)\subseteq G$ for every $x\in A+B$
(2A2Ed).   Let $n\in\Bbb N$ be such that
$2^{-n}\sqrt{r}\le\delta$, and let $A_1$ be the union of all the
half-open
intervals of the form $\coint{2^{-n}z,2^{-n}z+2^{-n}e}$ which meet $A$,
where $z\in\BbbZ^r$ and $e=(1,1,\ldots,1)$.  Then $A_1$ is a finite
disjoint union of half-open intervals, $A\subseteq A_1$ and every point
of $A_1$ is within a distance $\delta$ of some point of $A$.
Similarly, we can find a set $B_1$, a finite disjoint union of half-open
intervals, including $B$ and such that every point of $B_1$ is within
$\delta$ of some point of $B$.   But this means that every point of
$A_1+B_1$ is within a distance $2\delta$ of some point of $A+B$, and
belongs to $G$.   Accordingly

$$\eqalignno{(\mu(A+B)+\epsilon)^{1/r}
&\ge(\mu G)^{1/r}
\ge\mu(A_1+B_1)^{1/r}
\ge(\mu A_1)^{1/r}+(\mu B_1)^{1/r}\cr
\displaycause{by (b)}
&\ge(\mu A)^{1/r}+(\mu B)^{1/r}.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\mu(A+B)^{1/r}\ge(\mu A)^{1/r}+(\mu B)^{1/r}$.\ \Qed

\medskip

{\bf (d)} Next suppose that $A$, $B\subseteq\BbbR^r$ are Lebesgue
measurable.   Then

$$\eqalignno{(\mu A)^{1/r}+(\mu B)^{1/r}
&=\sup\{(\mu K)^{1/r}+(\mu L)^{1/r}:
  K\subseteq A\text{ and }L\subseteq B\text{ are compact}\}\cr
\displaycause{134Fb}
&\le\sup\{\mu(K+L)^{1/r}:
  K\subseteq A\text{ and }L\subseteq B\text{ are compact}\}\cr
\displaycause{by (c)}
&\le\mu^*(A+B)^{1/r}.\cr}$$

\medskip

{\bf (e)} For the penultimate step, 
suppose that $A$, $B\subseteq\BbbR^r$ have
non-zero outer Lebesgue measure.   Consider $\clstar A$, $\clstar B$ and
$\clstar(A+B)$ as defined in 266B.   Then $\clstar A$ and $\clstar B$ are
non-empty and their sum is included in
$\clstar(A+B)$, by 266Bb and 266Be.   So we have

$$\eqalignno{(\mu^*A)^{1/r}+(\mu^*B)^{1/r}
&=\mu(\clstar A)^{1/r}+\mu(\clstar B)^{1/r}\cr
\displaycause{266Bd}
&\le\mu^*(\clstar A+\clstar B)^{1/r}\cr
\displaycause{by (d) here}
&\le\mu(\clstar(A+B))^{1/r}
=\mu^*(A+B)^{1/r}.\cr}$$

\medskip

{\bf (f)} Finally, for arbitrary non-empty sets $A$,
$B\subseteq\BbbR^r$, note that if (for instance) $A$ is negligible then
we can take any $x\in A$ and see that

\Centerline{$\mu^*(A+B)^{1/r}\ge\mu^*(x+B)^{1/r}=(\mu B)^{1/r}
=(\mu^*A)^{1/r}+(\mu^*B)^{1/r}$,}

\noindent and the result is similarly trivial if $B$ is negligible.   So
all cases are covered.
}%end of proof of 266C

\exercises{\leader{266X}{Basic exercises (a)}
Let $D$, $D'$ be subsets of $\BbbR^r$.   Show that (i)
$\clstar(D\cup D')=\clstar D\cup\clstar D'$ (ii) $\clstar D=\clstar D'$
iff $D$ and $D'$ have a common measurable envelope
(iii) $\clstar D\setminus\clstar(\BbbR^r\setminus D')
\subseteq\clstar(D\cap D')$ (iv) $D$ is Lebesgue measurable iff
$\clstar D\cap\clstar(\BbbR^r\setminus D)$ is Lebesgue negligible (v)
$D\cup\clstar D$ is a measurable envelope of $D$
(vi) $\clstar(\clstar D)=\clstar D$.
%266B

\spheader 266Xb Show that, for a measurable set $E\subseteq\Bbb R$,
$\clstar E$ is just the set of real numbers which are not density points of
$\Bbb R\setminus E$.
%266B

\spheader 266Xc In 266C, show that if $A$ and $B$ are similar convex
sets in the same orientation then $A+B$ is a convex set similar to both
and $\mu(A+B)^{1/r}=(\mu A)^{1/r}+(\mu B)^{1/r}$.
%266C

\spheader 266Xd\dvAnew{2020} 
Show that if $r\ge 1$, $\mu$ is Lebesgue measure on $\BbbR^r$
and $A_0,\ldots,A_n$ are non-empty subsets of $\BbbR^r$, then
$\mu^*(A_0+\ldots+A_n)^{1/r}\ge\sum_{i=0}^n(\mu^*A_i)^{1/r}$.
%266C

\spheader 266Xe\dvAnew{2020} 
In 266C, show that if $p\in[0,1]$ then (subject to an
appropriate interpretation of $\infty^0$)
$\mu^*(pA+(1-p)B)\ge(\mu^*A)^p(\mu^*B)^{1-p}$.
%266C
 
%\leader{266Y}{Further exercises (a)}

}%end of exercises

\endnotes{
\Notesheader{266}
The proof of 266C is taken from {\smc Federer 69}.
%Is there a Hausdorff-measure version?
There is a slightly specious generality in the form given here.   If the
sets $A$ and $B$ are at all irregular, then $\mu^*(A+B)^{1/r}$ is likely
to be much greater than $(\mu^*A)^{1/r}+(\mu^*B)^{1/r}$.   The critical
case, in which $A$ and $B$ are similar convex sets, is much easier
(266Xc).   The theorem is therefore most useful when $A$ and $B$ are
non-similar convex sets and we get a non-trivial estimate which may be
hard to establish by other means.   For this case we do not need 266B.
Theorem 266C is an instructive example of the way in which the
dimension $r$ enters formulae when we seek results applying to general
Euclidean spaces.   There will be many more when I return to geometric
measure theory in Chapter 47 of Volume 4.
}%end of notes

\discrpage



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

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

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

\centerline{\bf References for Volume 2}

\medskip

%A

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

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

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

\medskip%B

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

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

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

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

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

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

\medskip%C

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

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

\medskip%D

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

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

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

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

\medskip%E

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

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

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

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

\medskip%F

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

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

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

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

\medskip%G\medskip%H

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

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

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

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

Hanner O.\ [56] `On the uniform convexity of $L^p$ and $l^p$',
Arkiv f\"or Matematik 3 (1956) 239-244.
\cmmnt{[244O.]}

Henle J.M.\ [86] {\it An Outline of Set Theory.}  Springer, 1986.
\cmmnt{[\S2A1.]}

Hoeffding W.\ [63]
`Probability inequalities for sums of bounded random variables',
J.\ Amer.\ Statistical Association 58 (1963) 13-30.
\cmmnt{[272W.]}

Hunt R.A.\ [67] `On the convergence of Fourier series', pp.\ 235-255
in {\smc Haimo 67}.  \cmmnt{[\S286 {\it notes}.]}

\medskip%I\medskip%J

Jorsb{\o}e O.G.\ \& Mejlbro L.\ [82] {\it The Carleson-Hunt Theorem on
Fourier Series.}  Springer, 1982 (Lecture Notes in Mathematics 911).
\cmmnt{[\S286 {\it notes}.]}

Judah H.\ [93] (ed.) {Proceedings of the Bar-Ilan Conference on Set
Theory and the Reals, 1991.} Amer.\ Math.\ Soc.\ (Israel Mathematical
Conference Proceedings 6), 1993.

\medskip%K

\indexheader{Kelley}
Kelley J.L.\ [55] {\it General Topology.} Van Nostrand, 1955.
\cmmnt{[2A5F.]}

Kelley J.L.\ \& Namioka I.\ [76] {\it Linear Topological Spaces.}
Springer, 1976.   \cmmnt{[2A5C.]}

\indexheader{Kirszbraun}
Kirszbraun M.D.\ [1934] `\"Uber die zusammenziehenden und Lipschitzian
Transformationen', Fund.\ Math.\ 22 (1934) 77-108.  \cmmnt{[262C.]}

Kolmogorov A.N.\ [1926] `Une s\'erie de Fourier-Lebesgue divergente
partout', C.\ R.\ Acad.\ Sci.\ Paris 183 (1926) 1327-1328.
\cmmnt{[\S282 {\it notes}.]}

Koml\'os J.\ [67] `A generalization of a problem of Steinhaus', Acta
Math.\ Acad.\ Sci.\ Hung.\ 18 (1967) 217-229.  \cmmnt{[276H.]}

K\"orner T.W.\ [88] {\it Fourier Analysis.}  Cambridge U.P., 1988.
\cmmnt{[\S282 {\it notes}.]}

K\"othe G.\ [69] {\it Topological Vector Spaces I.}   Springer, 1969.
\cmmnt{[2A5C, 2A5E, 2A5J.]}

Krivine J.-L.\ [71] {\it Introduction to Axiomatic Set Theory.}
D.~Reidel, 1971.  \cmmnt{[\S2A1.]}

\medskip%L

Lacey M.T.\ [05] {\it Carleson's Theorem: Proof, Complements, Variations.}
{\tt http://arxiv.org/pdf/math{\bsp}0307008v4{\penalty-100}.pdf}.
%expanded from 
% Publ.\ Mat.\ 48 (2004) 251-307
\cmmnt{[\S286 {\it notes}.]}

Lacey M.T.\ \& Thiele C.M.\ [00] `A proof of boundedness of the Carleson
operator', Math.\ Research Letters 7 (2000) 1-10.
\cmmnt{[\S286 {\it intro.}, 286H.]}

Lebesgue H.\ [72] {\it Oeuvres Scientifiques.}   L'Enseignement
Math\'ematique, Institut de Math\'ematiques, Univ.\ de Gen\`eve, 1972.
\cmmnt{[Chap.\ 27 {\it intro.}]}

Liapounoff A.\ [1901] `Nouvelle forme du th\'eor\`eme sur la limite
de probabilit\'e', M\'em.\ Acad.\ Imp.\ Sci.\ St-P\'etersbourg 12(5)
(1901) 1-24. \cmmnt{[274Xh.]}

Lighthill M.J.\ [59] {\it Introduction to Fourier Analysis and
Generalised Functions.}  Cambridge U.P., 1959.
\cmmnt{[\S284 {\it notes}.]}

Lindeberg J.W.\ [1922] `Eine neue Herleitung des Exponentialgesetzes in
der Wahrscheinlichkeitsrechnung', Math.\ Zeitschrift 15 (1922) 211-225.
\cmmnt{[274H, \S274 {\it notes}.]}

Lipschutz S.\ [64] {\it Set Theory and Related Topics.}  McGraw-Hill,
1964 (Schaum's Outline Series).  \cmmnt{[\S2A1.]}

Lo\`eve M.\ [77] {\it Probability Theory I.} Springer, 1977.
\cmmnt{[Chap.\ 27 {\it intro.}, 274H.]}

Luxemburg W.A.J.\ \& Zaanen A.C.\ [71] {\it Riesz Spaces
I.} North-Holland, 1971.
\cmmnt{[241F.]}


\medskip%M

Mozzochi C.J.\ [71] {\it On the Pointwise Convergence of Fourier
Series.}   Springer, 1971 (Lecture Notes in Mathematics 199).
\cmmnt{[\S286 {\it notes}.]}

\medskip%N

Naor A.\ [04] `Proof of the uniform convexity lemma',
{\tt http://www.cims.nyu.edu/\penalty-100$^{\sim}$naor/homepage files/\ifnum\stylenumber=12\break\fi{\vthsp}inequality\discretionary{}{}{}.pdf},
26.2.04.
\cmmnt{[244O.]}
%anon04.PDF

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

R\'enyi A.\ [70] {\it Probability Theory.}  North-Holland, 1970.
\cmmnt{[274H.]}

Roitman J.\ [90] {\it  An Introduction to Set Theory.}  Wiley, 1990.
\cmmnt{[\S2A1.]}

Roselli P.\ \& Willem M.\ [02] `A convexity inequality',
Amer.\ Math.\ Monthly 109 (2002) 64-70.
\cmmnt{[244Ym.]}

\medskip%S

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

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

Schipp F.\ [78] `On Carleson's method', pp.\ 679-695 in
{\smc Alexits 78}.   \cmmnt{[\S286 {\it notes}.]}

Semadeni Z.\ [71] {\it Banach spaces of continuous functions I.}
Polish Scientific Publishers, 1971.  \cmmnt{[\S253 {\it notes}.]}

Shiryayev A.N.\ [84] {\it Probability.} Springer, 1984. \cmmnt{[285N.]}

Steele J.M.\ [86] `An Efron-Stein inequality of nonsymmetric statistics',
Annals of Statistics 14 (1986) 753-758.
\cmmnt{[274Ya.]}

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

Zaanen A.C.\ [83] {\it Riesz Spaces II.} North-Holland, 1983.
\cmmnt{[241F.]}

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

\frnewpage

}%end of references

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

\end



