% 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\Sous{\Cal S\text{ous}}  %used in vol 6
\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{mt47.tex}
\versiondate{8.4.13}
\copyrightdate{2002}

\def\chaptername{Geometric measure theory}
\def\sectionname{Introduction}

\def\varinnerprod#1#2{#1\dotproduct#2}

\newchapter{47}

I offer a chapter on geometric measure theory,
continuing from Chapter 26.   The greater part of it is
directed specifically at two topics:
a version of the Divergence Theorem (475N) and the elementary theory
of Newtonian capacity and potential (\S479).
I do not attempt to provide a balanced view of the subject, 
for which I must refer you to {\smc Mattila 95},
{\smc Evans \& Gariepy 92} and {\smc Federer 69}.
However \S472, at least, deals with something which must be central to
any approach, Besicovitch's Density Theorem for Radon measures on
$\BbbR^r$ (472D).   In \S473 I examine Lipschitz functions, and give crude
forms of some fundamental ineqalities relating integrals
$\int\|\grad f\|d\mu$ with other measures of the variation of a function
$f$ (473H, 473K).   
In \S474 I introduce perimeter measures $\lambda^{\partial}_E$
and outward-normal functions $\psi_E$ as those for which the Divergence
Theorem, in the form $\int_E\diverg\phi\,d\mu
=\int\varinnerprod{\phi}{\psi_E}\,d\lambda^{\partial}_E$, will be valid
(474E), and give the geometric description of $\psi_E(x)$ as the Federer
exterior normal to $E$ at $x$ (474R).   In \S475 I show that
$\lambda^{\partial}_E$ can be identified with normalized Hausdorff
$(r-1)$-dimensional measure on the essential boundary of $E$.

\S471 is devoted to Hausdorff measures on general metric spaces, extending
the ideas introduced in \S264 for Euclidean space, up to basic results on
densities (471P) and Howroyd's theorem (471S).
In \S476 I turn to a different topic, the problem of finding the subsets
of $\BbbR^r$ on which Lebesgue measure is most `concentrated' in some
sense.   I present a number of classical results, the deepest being
the Isoperimetric Theorem (476H):  among sets with a given measure,
those with the smallest perimeters are the balls.

The last three sections are different again.   Classical electrostatics led
to a vigorous theory of capacity and potential, based on the idea of
`harmonic function'.   It turns out that `Brownian motion' in
$\BbbR^r$ (\S477)
gives an alternative and very powerful approach to the subject.   I have
brought Brownian motion and Wiener measure to this chapter because I wish
to use them to illuminate the geometry of $\BbbR^r$;  but much of \S477
(in particular, the strong Markov property, 477G) is
necessarily devoted to adapting ideas developed in the more general
contexts of L\'evy and Gaussian processes, as described in \S\S455-456.   
In \S478 I give the
most elementary parts of the theory of harmonic and superharmonic
functions, building up to a definition of `harmonic
measures' based on Brownian motion (478P).   In \S479
I use these techniques to describe Newtonian capacity and its extension
Choquet-Newton capacity (479C) on Euclidean space of three or more
dimensions, and establish their basic properties (479E,
479F, 479N, 479P, 479U).

\discrpage


\frfilename{mt471.tex}
\versiondate{10.2.16}
\copyrightdate{2000}

\def\chaptername{Geometric measure theory}
\def\sectionname{Hausdorff measures}

\def\NN{\BbbN^{\Bbb N}}

\newsection{471}

I begin the chapter by returning to a class of measures which we have
not examined in depth since
Chapter 26.  The primary importance of these measures is in studying the
geometry of Euclidean space;  in \S265 I looked briefly at their use in
describing surface measures, which will reappear in \S475.   Hausdorff
measures are also one of the basic tools in
the study of fractals, but for such applications I must refer you to
{\smc Falconer 90} and {\smc Mattila 95}.
All I shall attempt to do here is to indicate some of the principal
ideas which are applicable to general metric spaces, and to look at some
special properties of Hausdorff measures related to the concerns of this
chapter and of \S261.

\leader{471A}{Definition} Let $(X,\rho)$ be a metric space and
$r\in\ooint{0,\infty}$.
For $\delta>0$ and $A\subseteq X$, set

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

\noindent\cmmnt{(As in \S264, take $\diam\emptyset=0$ and
$\inf\emptyset=\infty$.) }It will be useful to note that every
$\theta_{r\delta}$ is an outer measure.   Now set

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

\noindent for $A\subseteq X$;  $\theta_r$ also is an outer measure on
$X$\cmmnt{, as in
264B};  this is {\bf $r$-dimensional Hausdorff outer measure} on $X$.
Let $\mu_{Hr}$ be the measure defined by \Caratheodory's method from
$\theta_r$;  $\mu_{Hr}$ is {\bf $r$-dimensional Hausdorff measure} on
$X$.

\cmmnt{\medskip

\noindent{\bf Notation} It may help if I list some notation already used
elsewhere.   Suppose that $(X,\rho)$ is a metric space.    I write

\Centerline{$B(x,\delta)=\{y:\rho(y,x)\le\delta\}$,
\quad$U(x,\delta)=\{y:\rho(y,x)<\delta\}$}

\noindent for $x\in X$, $\delta\ge 0$;  recall that $U(x,\delta)$ is
open (2A3G).   For $x\in X$ and $A$, $A'\subseteq X$ I write

\Centerline{$\rho(x,A)=\inf_{y\in A}\rho(x,y)$,
\quad$\rho(A,A')=\inf_{y\in A,z\in A'}\rho(y,z)$;}

\noindent for definiteness, take $\inf\emptyset$ to be $\infty$, as
before.}%end of comment

\leader{471B}{Definition} Let $(X,\rho)$ be a metric space.   An outer
measure $\theta$ on $X$ is a {\bf metric outer measure} if
$\theta (A\cup B)=\theta A+\theta B$ whenever $A$,
$B\subseteq X$ and $\rho(A,B)>0$.

\leader{471C}{Proposition} Let $(X,\rho)$ be a metric space and $\theta$
a metric outer measure on $X$.  Let $\mu$ be the measure on $X$ defined
from $\theta$ by \Caratheodory's method.   Then $\mu$ is a topological
measure.

\proof{ (Compare 264E, part (b) of the proof.)  Let $G\subseteq X$ be
open, and $A$ any subset of $X$ such that $\theta A<\infty$.   Set

\Centerline{$A_n=\{x:x\in A,\,\rho(x,A\setminus G)\ge 2^{-n}\}$,}

\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
$\rho(y,z)<2^{-n+1}\le 2^{-m-1}$, while $\rho(x,z)$ must be at least
$2^{-m}$, so $\rho(x,y)\ge\rho(x,z)-\rho(y,z)\ge 2^{-m-1}$.   Thus
$\rho(B_m,B_n)>0$ whenever $n\ge m+2$.   It follows
that for any $k\ge 0$

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

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

\noindent Consequently $\sum_{n=0}^{\infty}\theta B_n<\infty$.

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

$$\eqalignno{\theta(A\cap G)+\theta(A\setminus G)
&\le\theta A_m+\sum_{n=m}^{\infty}\theta B_n+\theta(A\setminus G)\cr
&\le\epsilon+\theta A_m+\theta(A\setminus G)
=\epsilon+\theta(A_m\cup(A\setminus G))\cr
\noalign{\noindent (since $\rho(A_m,A\setminus G)\ge 2^{-m}$)}
&\le\epsilon+\theta A.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\theta(A\cap G)+\theta(A\setminus G)\le\theta A$.   As $A$ is
arbitrary, $G$ is measured by $\mu$;  as $G$ is arbitrary, $\mu$ is a
topological measure.
}%end of proof of 471C

\leader{471D}{Theorem} Let $(X,\rho)$ be a metric space and $r>0$.   Let
$\mu_{Hr}$ be $r$-dimensional Hausdorff measure on $X$, and $\Sigma$ its
domain;  write $\theta_r$ for $r$-dimensional Hausdorff outer measure on
$X$\cmmnt{, as defined in 471A}.

(a) $\mu_{Hr}$ is a topological measure.

(b) For every $A\subseteq X$ there is a G$_{\delta}$ set
$H\supseteq A$ such that $\mu_{Hr}H=\theta_rA$.

(c) $\theta_r$ is the outer measure defined from $\mu_{Hr}$\cmmnt{
(that is, $\theta_r$ is a regular outer measure)}.

(d) $\Sigma$ is closed under Souslin's operation.

(e) $\mu_{Hr}E=\sup\{\mu_{Hr}F:F\subseteq E$ is closed$\}$ whenever
$E\in\Sigma$ and $\mu_{Hr}E<\infty$.

(f) If $A\subseteq X$ and $\theta_rA<\infty$ then $A$ is separable and
the set of isolated points of $A$ is $\mu_{Hr}$-negligible.

(g) $\mu_{Hr}$ is atomless.

(h) If $\mu_{Hr}$ is totally finite it is a quasi-Radon measure.

\proof{{\bf (a)} The point is that $\theta_r$, as defined in 471A, is a
metric outer measure.   \Prf\ (Compare 264E, part (a) of the proof.)
Let $A$, $B$ be subsets of $X$ such that $\rho(A,B)>0$.   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 defining the $\theta_{r\delta_i}$ as in 471A.
Set $\delta=\min(\delta_1,\delta_2,\bover12\rho(A,B))>0$ and let
$\sequencen{D_n}$ be a sequence of sets of diameter at most $\delta$,
covering $A\cup B$, and such that $\sum_{n=0}^{\infty}(\diam D_n)^r
\le\theta_{r\delta}(A\cup B)+\epsilon$.   Set

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

\noindent Because $\rho(x,y)>\diam D_n$ whenever $x\in A$, $y\in B$
and $n\in\Bbb N$,
$K\cap L=\emptyset$;  and of course $A\subseteq\bigcup_{n\in K}D_k$,
$B\subseteq\bigcup_{n\in L}D_n$.   Consequently

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

\noindent As $\epsilon$ is arbitrary,
$\theta_r(A\cup B)\ge\theta_rA+\theta_rB$, and we therefore have
equality.   As $A$ and $B$ are
arbitrary, $\theta_r$ is a metric outer measure.\ \Qed

Now 471C tells us that $\mu_{Hr}$ must be a topological measure.

\medskip

{\bf (b)} (Compare 264Fa.)   If $\theta_rA=\infty$ this is trivial.
Otherwise, for each $n\in\Bbb N$, let $\sequence{i}{D_{ni}}$ be a
sequence of sets of diameter at most $2^{-n}$ such that
$A\subseteq\bigcup_{i\in\Bbb N}D_i$ and
$\sum_{i=0}^{\infty}(\diam D_{ni})^r\le\theta_{r,2^{-n}}(A)+2^{-n}$,
defining $\theta_{r,2^{-n}}$
as in 471A.   Let $\eta_{ni}\in\ocint{0,2^{-n}}$ be such that
$(2\eta_{ni}+\diam D_{ni})^r\le 2^{-n-i}+(\diam D_{ni})^r$, and set
$G_{ni}=\{x:\rho(x,D_{ni})<\eta_{ni}\}$, for all $n$, $i\in\Bbb N$;
then
$G_{ni}=\bigcup_{x\in D_{ni}}U(x,\eta_{ni})$ is an open set including
$D_{ni}$ and $(\diam G_{ni})^r\le 2^{-n-i}+(\diam D_{ni})^r$.   Set

\Centerline{$H=\bigcap_{n\in\Bbb N}\bigcup_{i\in\Bbb N}G_{ni}$,}

\noindent so that $H$ is a G$_{\delta}$ set including $A$.

For any $\delta>0$, there is an $n\in\Bbb N$ such that
$3\cdot 2^{-n}\le\delta$, so that
$\diam G_{mi}\le\diam D_{mi}+2\eta_{mi}\le\delta$ for every
$i\in\Bbb N$ and $m\ge n$, and

$$\eqalign{\theta_{r\delta}H
&\le\sum_{i=0}^{\infty}(\diam G_{mi})^r
\le\sum_{i=0}^{\infty}2^{-m-i}+(\diam D_{mi})^r\cr
&\le 2^{-m+1}+\theta_{r,2^{-m}}(A)+2^{-m}
\le 2^{-m+2}+\theta_rA\cr}$$

\noindent for every $m\ge n$.   Accordingly
$\theta_{r\delta}H\le\theta_rA$ for every $\delta>0$, so
$\theta_rH\le\theta_rA$.   Of course this means that
$\theta_rH=\theta_rA$;  and since, by (a), $\mu_{Hr}$ measures $H$, we
have $\mu_{Hr}H=\theta_rA$, as required.

\medskip

{\bf (c)} (Compare 264Fb.)   If $A\subseteq X$,

$$\eqalignno{\theta_rA&\ge\mu_{Hr}^*A\cr
\displaycause{by (b)}
&=\inf\{\theta_rE:A\subseteq E\in\Sigma\}
\ge\theta_rA.\cr}$$

\medskip

{\bf (d)} Use 431C.

\medskip

{\bf (e)} By (b), there is a Borel set $H\supseteq E$ such that
$\mu_{Hr}H=\mu_{Hr}E$, and now there is a Borel set
$H'\supseteq H\setminus E$ such that
$\mu_{Hr}H'=\mu_{Hr}(H\setminus E)=0$, so that
$G=H\setminus H'$ is a Borel set included in $E$ and
$\mu_{Hr}G=\mu_{Hr}E$.   Now $G$ is a Baire set (4A3Kb), so is Souslin-F
(421L), and $\mu_{Hr}G=\sup_{F\subseteq G\text{ is closed}}\mu_{Hr}F$,
by 431E.

\medskip

{\bf (f)} For every $n\in\Bbb N$, there must be a sequence
$\sequence{i}{D_{ni}}$ of sets of diameter at most
$2^{-n}$ covering $A$;  now if $D\subseteq A$ is a countable set which
meets $D_{ni}$ whenever $i$, $n\in\Bbb N$ and
$A\cap D_{ni}\ne\emptyset$, $D$ will be dense in $A$.   If $A_0$ is
the set of isolated points in $A$, it is still separable (4A2P(a-iv));
but as the only dense subset of $A_0$ is itself, it is countable.
Since $\theta_{r\delta}\{x\}=(\diam\{x\})^r=0$ for every $\delta>0$,
$\mu_{Hr}\{x\}=0$ for every $x\in X$, and $A_0$ is negligible.

\medskip

{\bf (g)} In fact, if $A\subseteq X$ and $\theta_rA>0$, there are
disjoint $A_0$, $A_1\subseteq A$ such that $\theta_rA_i>0$ for both $i$.
\Prf\ (i) Suppose first that $A$ is not separable.   For each
$n\in\Bbb N$, let $D_n\subseteq A$ be a maximal set such that
$\rho(x,y)\ge 2^{-n}$ for all distinct $x$, $y\in D_n$;  then
$\bigcup_{n\in\Bbb N}D_n$ is dense in $A$, so there is some $n\in\Bbb N$
such that $D_n$ is uncountable;  if we take $A_1$, $A_2$ to be disjoint
uncountable subsets of $D_n$, then $\theta_rA_1=\theta_rA_2=\infty$.
(ii) If $A$ is separable, then set $\Cal G=\{G:G\subseteq X$ is open,
$\theta_r(A\cap G)=0\}$.   Because $A$ is hereditarily Lindel\"of
(4A2P(a-iii)), there is a countable subset $\Cal G_0$ of $\Cal G$ such
that $A\cap\bigcup\Cal G=A\cap\bigcup\Cal G_0$ (4A2H(c-i)), so
$A\cap\bigcup\Cal G$ is negligible and $A\setminus\bigcup\Cal G$ has at
least two points $x_0$, $x_1$.   If we set
$A_i=A\cap U(x_i,\bover12\rho(x_i,x_{1-i}))$ for each $i$,
these are disjoint subsets of $A$ of non-zero outer measure.\ \Qed

\medskip

{\bf (h)} If $\mu_{Hr}$ is totally finite, then it is inner regular with
respect to the closed sets, by (e).   Also, because $X$ must be
separable, by (f), therefore hereditarily Lindel\"of, $\mu_{Hr}$ must be
$\tau$-additive (414O).   Finally, $\mu_{Hr}$ is complete just
because it is defined by \Caratheodory's method.   So $\mu_{Hr}$ is a
quasi-Radon measure.
}%end of proof of 471D

\leader{471E}{Corollary} If $(X,\rho)$ is a metric space, $r>0$ and
$Y\subseteq X$ then $r$-dimensional Hausdorff measure $\mu^{(Y)}_{Hr}$
on $Y$ extends the subspace measure $(\mu^{(X)}_{Hr})_Y$ on $Y$ induced
by $r$-dimensional Hausdorff measure $\mu^{(X)}_{Hr}$ on $X$;  and if
either $Y$ is measured by $\mu^{(X)}_{Hr}$ or $Y$ has finite
$r$-dimensional Hausdorff outer measure in $X$, then
$\mu^{(Y)}_{Hr}=(\mu^{(X)}_{Hr})_Y$.

\proof{ Write $\theta_r^{(X)}$ and $\theta_r^{(Y)}$ for the two
$r$-dimensional Hausdorff outer measures.

If $A\subseteq Y$ and $\sequencen{D_n}$ is any sequence of subsets of
$X$ covering $A$, then $\sequencen{D_n\cap Y}$ is a sequence of subsets
of $Y$ covering $A$, and $\sum_{n=0}^{\infty}(\diam(D_n\cap Y))^r
\le\sum_{n=0}^{\infty}(\diam D_n)^r$;  moreover, when calculating
$\diam(D_n\cap Y)$, it doesn't matter whether we use the metric $\rho$
on $X$ or the subspace metric $\rho\restrp Y\times Y$ on $Y$.   What
this means is that, for any $\delta>0$, $\theta_{r\delta}A$ is the same
whether calculated in $Y$ or in $X$, so that
$\theta^{(Y)}_rA=\sup_{\delta>0}\theta_{r\delta}A=\theta_r^{(X)}A$.

Thus $\theta^{(Y)}_r=\theta^{(X)}_r\restrp\Cal PY$.   Also, by 471Db,
$\theta^{(X)}_r$ is a regular outer measure.   So 214Hb gives the
results.
}%end of proof of 471E

\leader{471F}{Corollary} Let $(X,\rho)$ be an analytic metric
space\cmmnt{ (that is, a metric space in which the
topology is analytic in the sense of \S423)}, and write $\mu_{Hr}$ for
$r$-dimensional Hausdorff measure on $X$.
Suppose that $\nu$ is a locally finite indefinite-integral measure over
$\mu_{Hr}$.   Then $\nu$ is a Radon measure.

\proof{ Since $\dom\nu\supseteq\dom\mu_{Hr}$, $\nu$ is a topological measure.
Because $X$ is separable, therefore hereditarily Lindel\"of, $\nu$ is
$\sigma$-finite and $\tau$-additive, therefore locally determined and
effectively locally finite.   Next, it is inner regular with respect to
the closed sets.   \Prf\ Let $f$ be a Radon-Nikod\'ym derivative of
$\nu$.   If $\nu E>0$, there is an $E'\subseteq E$ such that

\Centerline{$0<\nu E'=\biggerint f\times\chi E'\,d\mu_{Hr}<\infty$.}

\noindent There is a $\mu_{Hr}$-simple function $g$ such that
$g\le f\times\chi E'\,\,\mu_{Hr}$-a.e.\ and $\int g\,d\mu_{Hr}>0$;
setting $H=E'\cap\{x:g(x)>0\}$, $\nu_{Hr}H<\infty$.
Now there is a closed set $F\subseteq H$ such that $\mu_{Hr}F>0$, by
471De, and in this case $\nu F\ge\int_Fg\,d\mu_{Hr}>0$.   By 412B, this
is enough to show that $\nu$ is inner regular with respect to the closed
sets.\ \Qed

Since $\nu$ is complete (234I), it is a
quasi-Radon measure, therefore a Radon measure (434Jf, 434Jb).
}%end of proof of 471F

\leader{471G}{Increasing Sets Lemma}\cmmnt{ ({\smc Davies 70})} Let
$(X,\rho)$ be a metric space and $r>0$.

(a) Suppose that $\delta>0$ and that $\sequencen{A_n}$ is a
non-decreasing sequence of subsets of $X$ with union $A$.   Then
$\theta_{r,6\delta}(A)
\le(5^r+2)\sup_{n\in\Bbb N}\theta_{r\delta}A_n$.

(b) Suppose that $\delta>0$ and that $\sequencen{A_n}$ is a
non-decreasing sequence of subsets of $X$ with union $A$.   Then
$\theta_{r\delta}A=\sup_{n\in\Bbb N}\theta_{r\delta}A_n$.

\proof{{\bf (a)} If $\sup_{n\in\Bbb N}\theta_{r\delta}A_n=\infty$ this
is trivial;  suppose otherwise.

\medskip

\quad{\bf (i)} Take any
$\gamma>\gamma'>\sup_{n\in\Bbb N}\theta_{r\delta}A_n$.   For each
$i\in\Bbb N$, let $\zeta_i\in\ocint{0,\bover14\delta}$ be such that
$(\alpha+\zeta_i)^r\le\alpha^r+2^{-i-1}(\gamma-\gamma')$ whenever
$0\le\alpha\le\delta$.   For each
$n\in\Bbb N$, there is a sequence $\sequence{i}{C_{ni}}$ of
sets covering $A_n$ such that $\diam C_{ni}\le\delta$ for every $i$ and
$\sum_{i=0}^{\infty}(\diam C_{ni})^r<\gamma'$;  let
$\sequence{i}{\gamma_{ni}}$ be such that
$\diam C_{ni}\le\gamma_{ni}\le\delta$ and $\gamma_{ni}>0$ for
every $i$ and $\sum_{i=0}^{\infty}\gamma_{ni}^r\le\gamma'$.   Since
$\sum_{i=0}^{\infty}\gamma_{ni}^r$ is finite,
$\lim_{i\to\infty}\gamma_{ni}=0$.   Because $\gamma_{ni}>0$ for every $i$,
we may rearrange the sequences $\sequence{i}{C_{ni}}$,
$\sequence{i}{\gamma_{ni}}$ in such a way that
$\gamma_{ni}\ge\gamma_{n,i+1}$ for each $i$.
%(We may have singleton $C_{ni}$.)

In this case, $\lim_{i\to\infty}\sup_{n\in\Bbb N}\gamma_{ni}=0$.
\Prf\

\Centerline{$(i+1)\gamma_{ni}^r\le\sum_{j=0}^i\gamma_{nj}^r\le\gamma$}

\noindent for every $n$, $i\in\Bbb N$.\ \Qed

\medskip

\quad{\bf (ii)} By Ramsey's theorem (4A1G, with $n=2$), there is an
infinite set $I\subseteq\Bbb N$ such that

\inset{for all $i$, $j\in\Bbb N$ there is an $s\in\Bbb N$ such that
either $C_{mi}\cap C_{nj}=\emptyset$ whenever $m$, $n\in I$ and
$s\le m<n$ or $C_{mi}\cap C_{nj}\ne\emptyset$ whenever $m$, $n\in I$ and
$s\le m<n$,}

\inset{for each $i\in\Bbb N$,
$\alpha_i=\lim_{n\in I,n\to\infty}\gamma_{ni}$ is defined in $\Bbb R$.}

\noindent (Apply 4A1Fb with the families

$$\eqalign{\Cal J_{ij}&=\{J:J\in[\Bbb N]^{\omega},
\text{ either }C_{mi}\cap C_{nj}=\emptyset
  \text{ whenever }m,\,n\in J\text{ and }m<n\cr
&\mskip150mu\text{ or }C_{mi}\cap C_{nj}\ne\emptyset
  \text{ whenever }m,\,n\in J\text{ and }m<n\}\cr
\Cal J'_{iq}&=\{J:J\in[\Bbb N]^{\omega},
\text{ either }\gamma_{ni}\le q\text{ for every }n\in J\cr
&\mskip200mu\text{ or }\gamma_{ni}\ge q\text{ for every }n\in J\}\cr}$$

\noindent for $i$, $j\in\Bbb N$ and $q\in\Bbb Q$.)

Of course $\alpha_j\le\alpha_i\le\delta$ whenever
$i\le j$, because $\gamma_{nj}\le\gamma_{ni}\le\delta$ for
every $n$.   Set $D_{ni}=\{x:\rho(x,C_{ni})\le 2\alpha_i+2\zeta_i\}$ for
all $n$, $i\in\Bbb N$, and
$D_i=\bigcup_{s\in\Bbb N}\bigcap_{n\in I\setminus s}D_{ni}$ for
$i\in\Bbb N$.   (I am identifying each $s\in\Bbb N$ with the set of its
predecessors.)   Note that if $i\in\Bbb N$ and $x$, $y\in D_i$, then
there is an $s\in\Bbb N$ such that $x$, $y$ both belong to $D_{ni}$ for
every $n\in I\setminus s$, so $\rho(x,y)\le\diam D_{ni}$ for every
$n\in I\setminus s$ and
$\rho(x,y)\le\liminf_{n\in I,n\to\infty}\diam D_{ni}$.   Accordingly
$\diam D_i\le\liminf_{n\in I,n\to\infty}\diam D_{ni}$.

\medskip

\quad{\bf (iii)} Set

\Centerline{$L=\{(i,j):i,\,j\in\Bbb N,\,
  \Forall s\in\Bbb N\,\Exists m,\,n\in I,\,s\le m<n$
  and $C_{mi}\cap C_{nj}\ne\emptyset\}$.}

\noindent If $(i,j)\in L$ then there is an $s\in\Bbb N$ such that
$C_{mi}\subseteq D_{\min(i,j)}$ whenever $m\in I$ and
$m\ge s$.   \Prf\ By the choice of $I$, we know that there is an
$s_0\in\Bbb N$ such
that $C_{mi}\cap C_{nj}\ne\emptyset$ whenever $m$, $n\in I$ and
$s_0\le m<n$.   Let $s_1\ge s_0$ be such that

\Centerline{$\gamma_{mi}\le\alpha_i+\min(\zeta_i,\zeta_j)$,
\quad$\gamma_{mj}\le\alpha_j+\min(\zeta_i,\zeta_j)$}

\noindent whenever $m\in I$ and $m\ge s_1$.   Take $m_0\in I$ such that
$m_0\ge s_1$, and set $s=m_0+1$.   Let $m\in I$ be such that $m\ge s$.

\medskip

\qquad\grheada\ Suppose that $i\le j$ and $x\in C_{mi}$.   Take any
$n\in I$ such that $m\le n$.   Then there is an $n'\in I$ such that
$n<n'$.   We know that
$C_{mi}\cap C_{n'j}$ and $C_{ni}\cap C_{n'j}$ are both non-empty.   So

\Centerline{$\rho(x,C_{ni})\le\diam C_{mi}+\diam C_{n'j}
\le\gamma_{mi}+\gamma_{n'j}
\le\alpha_i+\zeta_i+\alpha_j+\zeta_i\le 2\alpha_i+2\zeta_i$}

\noindent and $x\in D_{ni}$.   This is true for all $n\in I$ such that
$n\ge m$, so $x\in D_i$.   As $x$ is arbitrary, $C_{mi}\subseteq D_i$.

\medskip

\qquad\grheadb\ Suppose that $j\le i$ and $x\in C_{mi}$.   Take any
$n\in I$ such that $n>m$.   Then $C_{mi}\cap C_{nj}$ is not empty, so

\Centerline{$\rho(x,C_{nj})
\le\diam C_{mi}\le\gamma_i\le\alpha_i+\zeta_j\le\alpha_j+\zeta_j$}

\noindent and $x\in D_{nj}$.   As $x$ and $n$ are arbitrary,
$C_{mi}\subseteq D_j$.

Thus $C_{mi}\subseteq D_{\min(i,j)}$ in both cases.\ \Qed

\medskip

\quad{\bf (iv)} Set

\Centerline{$D=\bigcup_{i\in\Bbb N}D_i$,
\quad$J=\{i:i\in\Bbb N,\,\Exists s\in\Bbb N$, $C_{ni}\subseteq D$
whenever $n\in I$ and $n\ge s\}$.}

\noindent If $i\in\Bbb N\setminus J$ and $j\in\Bbb N$, then (iii) tells
us that $(i,j)\notin L$, so there is some $s\in\Bbb N$ such that
$C_{mi}\cap C_{nj}=\emptyset$ whenever $m$,
$n\in I$ and $s\le m<n$.

\medskip

\quad{\bf (v)} For $l\in\Bbb N$,
$\mu^*_{Hr}(A_l\setminus D)\le 2\gamma$.   \Prf\ Let $\epsilon>0$.
Then there is a
$k\in\Bbb N$ such that $\gamma_{ni}\le\epsilon$ whenever $n\in\Bbb N$
and $i>k$.   Next, there is an $s\in\Bbb N$ such that

\Centerline{$C_{ni}\subseteq D$ whenever $i\le k$, $i\in J$, $n\in I$
and $s\le n$,}

\Centerline{$C_{mi}\cap C_{nj}=\emptyset$ whenever $i$, $j\le k$,
$i\notin J$, $m$, $n\in I$ and $s\le m<n$.}

\noindent Take $m$, $n\in I$ such that $\max(l,s)\le m<n$.   Then

$$\eqalign{A_l\setminus D
&=\bigcup_{i\in\Bbb N}A_l\cap C_{mi}\setminus D\cr
&\subseteq\bigcup_{i\le k}(A_n\cap C_{mi}\setminus D)
  \cup\bigcup_{i>k}C_{mi}\cr
&\subseteq\bigcup_{i\le k,i\notin J}(A_n\cap C_{mi})
  \cup\bigcup_{i>k}C_{mi}\cr
&\subseteq\bigcup_{i\le k,i\notin J,j\le k}(C_{mi}\cap C_{nj})
  \cup\bigcup_{j>k}C_{nj}
  \cup\bigcup_{i>k}C_{mi}\cr
&=\bigcup_{j>k}C_{nj}\cup\bigcup_{i>k}C_{mi}.\cr}$$

\noindent Since $\diam C_{nj}\le\gamma_j\le\epsilon$ and
$\diam C_{mi}\le\gamma_i\le\epsilon$ for all $i$, $j>k$,

\Centerline{$\theta_{r\epsilon}(A_l\setminus D)
\le\sum_{i=k+1}^{\infty}\gamma_{ni}^r
  +\sum_{i=k+1}^{\infty}\gamma_{mi}^r
\le 2\gamma$.}

\noindent This is true for every $\epsilon>0$, so
$\mu^*_{Hr}(A_l\setminus D)\le 2\gamma$, as claimed.\ \Qed

\medskip

\quad{\bf (vi)} This is true for each $l\in\Bbb N$.   But this means
that $\mu_{Hr}^*(A\setminus D)\le 2\gamma$ (132Ae).
Now $\theta_{r,6\delta}D\le 5^r\gamma$.   \Prf\ For each $i\in\Bbb N$,

$$\eqalign{\diam D_i
&\le\liminf_{n\in I,n\to\infty}\diam D_{ni}
\le\liminf_{n\in I,n\to\infty}\diam C_{ni}+4\alpha_i+4\zeta_i\cr
&\le\lim_{n\in I,n\to\infty}\gamma_{ni}+4\alpha_i+4\zeta_i
=5\alpha_i+4\zeta_i\le 6\delta.\cr}$$

\noindent Next, for any $k\in\Bbb N$,

\Centerline{$\sum_{i=0}^k\alpha_i^r
=\lim_{n\in I,n\to\infty}\sum_{i=0}^k\gamma_{ni}^r
\le\gamma'$,}

\noindent so

$$\eqalignno{\sum_{i=0}^k(\diam D_i)^r
&\le 5^r\sum_{i=0}^k(\alpha_i+\zeta_i)^r
\le 5^r(\sum_{i=0}^k\alpha_i^r+2^{-i-1}(\gamma-\gamma'))
\Displaycause{by the choice of the $\zeta_i$}
\le 5^r(\gamma'+\sum_{i=0}^k2^{-i-1}(\gamma-\gamma'))
\le 5^r\gamma.\cr}$$

\noindent Letting $k\to\infty$,

\Centerline{$\theta_{r,6\delta}D\le\sum_{i=0}^{\infty}(\diam D_i)^r
\le 5^r\gamma$.   \Qed}

Putting these together,

\Centerline{$\theta_{r,6\delta}A
\le\theta_{r,6\delta}D+\theta_{r,6\delta}(A\setminus D)
\le\theta_{r,6\delta}D+\mu^*_{Hr}(A\setminus D)
\le(5^r+2)\gamma$.}

\noindent As $\gamma$ is arbitrary, we have the preliminary result (a).

\medskip

{\bf (b)} Now let us turn to the sharp form (b).   Once again, we may
suppose that $\sup_{n\in\Bbb N}\theta_{r\delta}A_n$ is finite.

\medskip

\quad{\bf (i)} Take $\gamma$ such that
$\sup_{n\in\Bbb N}\theta_{r\delta}A_n<\gamma$.   As in (a)(i)
above, we can find a family
$\langle C_{ni}\rangle_{n,i\in\Bbb N}$ such that

\Centerline{$A_n\subseteq\bigcup_{i\in\Bbb N}C_{ni}$,}

\Centerline{$\diam C_{ni}\le\delta$ for every $i\in\Bbb N$,}

\Centerline{$\sum_{i=0}^{\infty}(\diam C_{ni})^r\le\gamma$}

\noindent for each $n$, and

\Centerline{$\lim_{i\to\infty}\sup_{n\in\Bbb N}\diam C_{ni}=0$.}

\noindent Replacing each $C_{ni}$ by its closure if necessary, we may
suppose that every $C_{ni}$ is a Borel set.

Let $Q\subseteq X$ be a countable set which meets $C_{ni}$ whenever $n$,
$i\in\Bbb N$ and $C_{ni}$ is not empty.   This time, let
$I\subseteq\Bbb N$ be an infinite set such that

\Centerline{$\alpha_i=\lim_{n\in I,n\to\infty}\diam C_{ni}$ is defined
in $[0,\delta]$ for every $i\in\Bbb N$,}

\Centerline{$\lim_{n\in I,n\to\infty}\rho(z,C_{ni})$ is defined in
$[0,\infty]$ for every $i\in\Bbb N$ and every $z\in Q$.}

\noindent (Take $\rho(z,\emptyset)=\infty$ if any of the $C_{ni}$ are
empty.)   It will be helpful to note straight away that the limit
$\lim_{n\in I,n\to\infty}\rho(x,C_{ni})$ is defined in $[0,\infty]$ for
every $i\in\Bbb N$ and $x\in\overline{Q}$.   \Prf\ If
$\lim_{n\in I,n\to\infty}\rho(y,C_{ni})=\infty$ for some $y\in Q$, then
$\lim_{n\in I,n\to\infty}\rho(x,C_{ni})\penalty-100=\infty$,
and we can stop.
Otherwise, for any $\epsilon>0$, there are a $z\in Q$ such that
$\rho(x,z)\le\epsilon$ and an $s\in\Bbb N$ such that $C_{mi}$ is not
empty and $|\rho(z,C_{mi})-\rho(z,C_{ni})|\le\epsilon$ whenever $m$,
$n\in I\setminus s$;  in which case
$|\rho(x,C_{mi})-\rho(x,C_{ni})|\le 3\epsilon$ whenever $m$,
$n\in I\setminus s$.   As $\epsilon$ is arbitrary,
$\lim_{n\in I,n\to\infty}\rho(x,C_{ni})$ is defined in $\Bbb R$.\ \Qed

Let $\Cal F$ be a non-principal ultrafilter on $\Bbb N$ containing $I$,
and for $i\in\Bbb N$ set

\Centerline{$D_i=\{x:\lim_{n\to\Cal F}\rho(x,C_{ni})=0\}$.}

\noindent Set $D=\bigcup_{i\in\Bbb N}\overline{D}_i$.   (Actually it is
easy to check that every $D_i$ is closed.)

\medskip

\quad{\bf (ii)} Set

\Centerline{$A^*=\bigcup_{m\in\Bbb N}\bigcap_{n\in I\setminus m}
  \bigcup_{i\in\Bbb N}C_{ni}\setminus D$;}

\noindent note that $A^*$ is a Borel set.   For $k$, $m\in\Bbb N$, set

\Centerline{$A^*_{km}
=\bigcap_{n\in I\setminus m}\bigcup_{i\ge k}C_{ni}$.}

\noindent For fixed $k$, $\sequence{m}{A^*_{km}}$ is a non-decreasing
sequence of sets.   Also its union includes $A^*$.   \Prf\ Take
$x\in A^*$.

\medskip

\qquad\grheada\Quer\ If $x\notin\overline{Q}$, let $\epsilon>0$ be such
that $Q\cap B(x,\epsilon)=\emptyset$.   Let $l\in\Bbb N$ be such that
$\diam C_{ni}\le\epsilon$ whenever $n\in\Bbb N$ and $i\ge l$;   then
$x\notin C_{ni}$ whenever $n\in\Bbb N$ and $i\ge l$.   Let $m\in\Bbb N$
be such that $x\in\bigcup_{i\in\Bbb N}C_{ni}$ whenever $n\in I$ and
$n\ge m$.   Then $x\in\bigcup_{i<l}C_{ni}$ whenever $n\in I$ and
$n\ge m$.   But this means that there must be some $i<l$ such that
$\{n:x\in C_{ni}\}\in\Cal F$ and $x\in D_i\subseteq D$;  which is
impossible.\ \Bang

\medskip

\qquad\grheadb\ Thus $x\in\overline{Q}$, so
$\lim_{n\in I,n\to\infty}\rho(x,C_{ni})$ is defined for each $i$ (see
(i) above), and must be greater than $0$, since $x\notin D_i$.   In
particular, there is an $s\in\Bbb N$ such that $x\notin C_{ni}$ whenever
$i<k$ and $n\in I\setminus s$;   there is also an $m\in\Bbb N$ such that
$x\in\bigcap_{n\in I\setminus m}\bigcup_{i\in\Bbb N}C_{ni}$;  so that
$x\in A^*_{k,\max(s,m)}$.   As $x$ is arbitrary,
$A^*\subseteq\bigcup_{m\in\Bbb N}A^*_{km}$.\ \Qed

\medskip

\quad{\bf (iii)} $\mu_{Hr}A^*$ is finite.   \Prf\ Take any $\epsilon>0$.
Let $k\in\Bbb N$ be such that $\diam C_{ni}\le\epsilon$ whenever
$i\ge k$ and $n\in\Bbb N$.   For any $m\in I$,
$\theta_{r\epsilon}A^*_{km}
\le\sum_{i=k}^{\infty}(\diam C_{mi})^r\le\gamma$.   By (a),

$$\eqalignno{\theta_{r,6\epsilon}A^*
&\le(5^r+2)\sup_{m\in\Bbb N}\theta_{r\epsilon}A^*_{km}\cr
&=(5^r+2)\sup_{m\in I}\theta_{r\epsilon}A^*_{km}
\le(5^r+2)\gamma.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\mu_{Hr}A^*\le(5^r+2)\gamma<\infty$.\ \Qed

\medskip

\quad{\bf (iv)} Actually,
$\mu_{Hr}A^*\le\gamma-\sum_{i=0}^{\infty}\alpha_i^r$.  \Prf\Quer\
Suppose, if possible, otherwise.   Take $\beta$ such that
$\gamma-\sum_{i=0}^{\infty}\alpha_i^r<\beta<\mu_{Hr}A^*$.
For $x\in A^*$ and $k\in\Bbb N$, set
$f_k(x)=\min\{n:n\in I$, $x\in A^*_{kn}\}$;   then $\sequence{k}{f_k}$
is a non-decreasing sequence of Borel measurable functions from $A^*$ to
$\Bbb N$.   Choose $\sequence{k}{s_k}$ inductively so that

\Centerline{$\mu_{Hr}\{x:x\in A^*$, $f_j(x)\le s_j$ for every $j\le k\}
>\beta$}

\noindent for every $k\in\Bbb N$.   Set
$\tilde A=\{x:x\in A^*$, $f_j(x)\le s_j$ for every $j\in\Bbb N\}$;
because $\mu_{Hr}A^*$ is finite, $\mu_{Hr}\tilde A\ge\beta$.   Take
$\epsilon>0$ such that
$\theta_{r\epsilon}\tilde A>\gamma-\sum_{i=0}^{\infty}\alpha_i^r$.
Let $k\in\Bbb N$ be such that
$\theta_{r\epsilon}\tilde A+\sum_{i=0}^{k-1}\alpha_i^r>\gamma$ and
$\diam C_{ni}\le\epsilon$ whenever $n\in\Bbb N$ and $i\ge k$.
Take $n\in I$ such that $n\ge s_j$ for every $j\le k$ and
$\theta_{r\epsilon}\tilde A+\sum_{i=0}^{k-1}(\diam C_{ni})^r>\gamma$.
If $x\in\tilde A$, then

\Centerline{$f_k(x)\le s_k\le n$,
\quad$x\in A^*_{kn}\subseteq\bigcup_{i\ge k}C_{ni}$,}

\noindent so
$\theta_{r\epsilon}\tilde A\le\sum_{i=k}^{\infty}(\diam C_{ni})^r$;
but this means that $\sum_{i=0}^{\infty}(\diam C_{ni})^r>\gamma$,
contrary to the choice of the $C_{ni}$.\ \Bang\Qed

\medskip

\quad{\bf (v)} Now observe that

\Centerline{$A
\subseteq\bigcup_{m\in\Bbb N}\bigcap_{n\ge m}\bigcup_{i\in\Bbb N}C_{ni}
\subseteq A^*\cup D$.}

\noindent Moreover, for any $i\in\Bbb N$,
$\diam\overline{D}_i\le\alpha_i\le\delta$.   \Prf\ If $x$, $y\in D_i$
then for every $\epsilon>0$

\Centerline{$\rho(x,C_{ni})\le\epsilon$,
\quad$\rho(y,C_{ni})\le\epsilon$,
\quad$\diam C_{ni}\le\alpha_i+\epsilon$}

\noindent for all but finitely many $n\in I$.   So
$\rho(x,y)\le\alpha_i+3\epsilon$.   As $x$, $y$ and $\epsilon$ are
arbitrary, $\diam\overline{D}_i=\diam D_i\le\alpha_i$.   Of course
$\alpha_i\le\delta$ because $\diam C_{ni}\le\delta$ for every $n$.\ \Qed

Now

\Centerline{$\theta_{r\delta}D
\le\sum_{i=0}^{\infty}(\diam\overline{D}_i)^r
\le\sum_{i=0}^{\infty}\alpha_i^r$.}

\noindent Putting this together with (iv),

\Centerline{$\theta_{r\delta}A
\le\theta_{r\delta}D+\theta_{r\delta}A^*
\le\theta_{r\delta}D+\mu_{Hr}A^*
\le\gamma$.}

\noindent As $\gamma$ is arbitrary,

\Centerline{$\theta_{r\delta}A
\le\sup_{n\in\Bbb N}\theta_{r\delta}A_n$;}

\noindent as $\theta_{r\delta}$ is an outer measure, we have equality.
}%end of proof of 471G

\leader{471H}{Corollary}\dvArevised{2016}
Let $(X,\rho)$ be a metric space, and $r>0$.   For $A\subseteq X$, set

\Centerline{$\theta_{r\infty}A
=\inf\{\sum_{n=0}^{\infty}(\diam D_n)^r:\sequencen{D_n}$ is a sequence
of subsets of $X$ covering $A\}$.}

\noindent Then $\theta_{r\infty}$ is an outer regular
Choquet capacity on $X$.

\proof{{\bf (a)} Of course $0\le\theta_{r\infty}A\le\theta_{r\infty}B$
whenever $A\subseteq B\subseteq X$.

\medskip

{\bf (b)} Suppose that $\sequencen{A_n}$ is a non-decreasing sequence of
subsets of $A$ with union $A$.   By (a),
$\gamma=\lim_{n\to\infty}\theta_{r\infty}A_n$ is defined
and less than or equal to $\theta_{r\infty}A$.   If $\gamma=\infty$, of
course it is equal to $\theta_{r\infty}A$.   Otherwise,
take $\beta=(\gamma+1)^{1/r}$.   For $n$, $k\in\Bbb N$ there is a
sequence
$\sequence{i}{D_{nki}}$ of sets, covering $A_n$, such that
$\sum_{i=0}^{\infty}(\diam D_{nki})^r\le\gamma+2^{-k}$.   But in this
case $\diam D_{nki}\le\beta$ for all
$n$, $k$ and $i$, so the $D_{nki}$ witness that
$\theta_{r\beta}A_n\le\gamma$.   By 471Gb,
$\gamma\ge\theta_{r\beta}A\ge\theta_{r\infty}A$ and again we have
$\gamma=\theta_{r\infty}A$.

\medskip

{\bf (c)} Let $A\subseteq X$ be any set, and suppose that
$\gamma>\theta_{r\infty}A$.   Let $\sequencen{D_n}$ be a sequence of
sets, covering $A$, such that
$\sum_{n=0}^{\infty}(\diam D_n)^r<\gamma$.   Let
$\sequencen{\epsilon_n}$ be a sequence of strictly positive real numbers
such that
$\sum_{n=0}^{\infty}(\diam D_n+2\epsilon_n)^r\le\gamma$.   Set
$G_n=\{x:\rho(x,D_n)<\epsilon_n\}$ for each $n$;  then $G_n$ is open and
$\diam G_n\le\diam D_n+2\epsilon_n$.
So $G=\bigcup_{n\in\Bbb N}G_n$ is an open set including $A$, and
$\sequencen{G_n}$ witnesses that $\theta_{r\infty}G\le\gamma$.   As $A$
and $\gamma$ are arbitrary,
the condition of 432Jb is satisfied and $\theta_{r\infty}$ is an outer
regular Choquet capacity.
}%end of proof of 471H

\medskip

\noindent{\bf Remark} $\theta_{r\infty}$ is {\bf $r$-dimensional
Hausdorff capacity} on $X$.

\leader{471I}{Theorem} Let $(X,\rho)$ be a metric space, and
$r>0$.   Write $\mu_{Hr}$ for $r$-dimensional Hausdorff measure on
$X$.   If $A\subseteq X$ is analytic, then $\mu_{Hr}A$ is defined and
equal to $\sup\{\mu_{Hr}K:K\subseteq A$ is compact$\}$.

\proof{{\bf (a)} Before embarking on the main line of the proof, it will
be convenient to set out a preliminary result.   For $\delta>0$,
$n\in\Bbb N$, $B\subseteq X$ set

\Centerline{$\theta^{(n)}_{r\delta}(B)
=\inf\{\sum_{i=0}^n(\diam D_i)^r:
B\subseteq\bigcup_{i\le n}D_i,\,\diam D_i\le\delta$
for every $i\le n\}$,}

\noindent taking $\inf\emptyset=\infty$ as usual.   Then
$\theta_{r\delta}B\le\theta^{(n)}_{r\delta}(B)$ for
every $n$.   Now the point is that $\theta^{(n)}_{r\delta}(B)
=\sup\{\theta^{(n)}_{r\delta}(I):I\subseteq B$ is finite$\}$.   \Prf\
Set $\gamma=\sup_{I\in[B]^{<\omega}}\theta^{(n)}_{r\delta}(I)$.   Of
course $\gamma\le\theta^{(n)}_{r\delta}(B)$.   If $\gamma=\infty$ there
is nothing more to say.   Otherwise, take any $\gamma'>\gamma$.   For
each $I\in[B]^{<\omega}$, we have a function $f_I:I\to\{0,\ldots,n\}$
such that $\sum_{i\in J}\rho(x_i,y_i)^r\le\gamma'$ whenever
$J\subseteq\{0,\ldots,n\}$ and $x_i$, $y_i\in I$ and
$f_I(x_i)=f_I(y_i)=i$ for every $i\in J$, while $\rho(x,y)\le\delta$
whenever $x$, $y\in I$ and $f_I(x)=f_I(y)$.   Let $\Cal F$ be an
ultrafilter on $[B]^{<\omega}$ such that
$\{I:x\in I\in[B]^{<\omega}\}\in\Cal F$ for every $x\in B$ (4A1Ia).
Then for every $x\in B$ there is an $f(x)\in\{0,\ldots,n\}$ such that
$\{I:x\in I\in[B]^{<\omega},\,f_I(x)=f(x)\}\in\Cal F$.   Set
$D_i=f^{-1}[\{i\}]$ for $i\le n$.   If
$x$, $y\in B$ and $f(x)=f(y)$, there is an $I\in[B]^{<\omega}$
containing both $x$ and $y$ such that $f_I(x)=f(x)=f(y)=f_I(y)$, so that
$\rho(x,y)\le\delta$;  thus $\diam D_i\le\delta$ for each $i$.   If
$J\subseteq\{0,\ldots,n\}$ and
for each $i\in J$ we take $x_i$, $y_i\in D_i$, then there is an
$I\in[B]^{<\omega}$ such that $f_I(x_i)=f_I(y_i)=i$ for every $i\in J$,
so $\sum_{i\in J}\rho(x_i,y_i)^r\le\gamma'$.   This means that
$\sum_{i\le n}(\diam D_i)^r\le\gamma'$, so that
$\theta^{(n)}_{r\delta}(B)\le\gamma'$.   As $\gamma'$ is arbitrary,
$\theta^{(n)}_{r\delta}(B)\le\gamma$, as claimed.\ \Qed

\medskip

{\bf (b)} Now let us turn to the set $A$.
Because $A$ is Souslin-F (422Ha), $\mu_{Hr}$ measures $A$ (471Da,
471Dd).   Set
$\gamma=\sup\{\mu_{Hr}K:K\subseteq A$ is compact$\}$.

\Quer\ Suppose, if possible, that $\mu_{Hr}A>\gamma$.   Take
$\gamma'\in\ooint{\gamma,\mu_{Hr}A}$.  Let $\delta>0$ be such that
$\gamma'<\theta_{r\delta}A$.   Let $f:\NN\to A$ be a continuous
surjection.   For
$\sigma\in S=\bigcup_{n\in\Bbb N}\BbbN^n$, set

\Centerline{$F_{\sigma}=\{\phi:\phi\in\NN$, $\phi(i)\le\sigma(i)$ for
every $i<\#(\sigma)\}$,}

\noindent so that $f[F_{\emptyset}]=A$.   Now choose $\psi\in\NN$ and
a sequence $\sequencen{I_n}$ of finite subsets of $\NN$ inductively, as
follows.   Given that $I_j\subseteq F_{\psi\restr n}$ for every $j<n$
and that $\theta_{r\delta}(f[F_{\psi\restr n}])>\gamma'$, then
$\theta^{(n)}_{r\delta}(f[F_{\psi\restr n}])>\gamma'$, so by (a) above
there is a finite subset $I_n$ of $F_{\psi\restr n}$ such that
$\theta^{(n)}_{r\delta}(f[I_n])\ge\gamma'$.   Next,

$$\eqalignno{\lim_{i\to\infty}\theta_{r\delta}
  f[F_{(\psi\restr n)^{\smallfrown}\fraction{i}}]
&=\theta_{r\delta}
  (\bigcup_{i\in\Bbb N}f[F_{(\psi\restr n)^{\smallfrown}\fraction{i}}])\cr
\displaycause{by 471G}
&=\theta_{r\delta}f[F_{\psi\restr n}]>\gamma',\cr}$$

\noindent so we can take $\psi(n)$ such
that $\bigcup_{j\le n}I_j\subseteq F_{\psi\restr n+1}$ and
$\theta_{r\delta}f[F_{\psi\restr n+1}]>\gamma'$, and continue.

At the end of the induction, set $K=\{\phi:\phi\le\psi\}$.   Then $f[K]$
is a compact subset of $A$, and
$I_n\subseteq K$ for every $n\in\Bbb N$, so

\Centerline{$\theta^{(n)}_{r\delta}(f[K])
\ge\theta^{(n)}_{r\delta}(f[I_n])\ge\gamma'$}

\noindent for every $n\in\Bbb N$.   On the
other hand, $\mu_{Hr}(f[K])\le\gamma$, so there is a sequence
$\sequence{i}{D_i}$ of sets, covering $f[K]$, all of diameter less than
$\delta$, such that $\sum_{i=0}^{\infty}(\diam D_i)^r<\gamma'$.
Enlarging the $D_i$ slightly if need be, we may suppose that they are
all open.   But in this case there is some finite $n$ such that
$K\subseteq\bigcup_{i\le n}D_i$, and
$\theta^{(n)}_{r\delta}(K)\le\sum_{i=0}^n(\diam D_i)^r<\gamma'$;  which
is impossible.\ \Bang

This contradiction shows that $\mu_{Hr}A=\gamma$, as required.
}%end of proof of 471I

\leader{471J}{Proposition} Let $(X,\rho)$ and $(Y,\sigma)$ be metric
spaces, and $f:X\to Y$ a $\gamma$-Lipschitz function, where $\gamma\ge 0$.
If $r>0$ and $\theta^{(X)}_r$,
$\theta^{(Y)}_r$ are the $r$-dimensional Hausdorff outer measures on $X$
and $Y$ respectively, then
$\theta^{(Y)}_rf[A]\le\gamma^r\theta^{(X)}_rA$ for every $A\subseteq X$.

\proof{ (Compare 264G.)  Let $\delta>0$.   Set
$\eta=\delta/(1+\gamma)$ and consider
$\theta^{(X)}_{r\eta}:\Cal PX\to[0,\infty]$, defined as in 471A.   We
know that $\theta^{(X)}_rA\ge\theta^{(X)}_{r\eta}A$, so there is a
sequence $\sequencen{D_n}$ of sets, all of
diameter at most $\eta$, covering $A$, with
$\sum_{n=0}^{\infty}(\diam D_n)^r\le\theta^{(X)}_rA+\delta$.
Now $f[A]\subseteq\bigcup_{n\in\Bbb N}f[D_n]$ and

\Centerline{$\diam f[D_n]
\le\gamma\diam D_n\le\gamma\eta\le\delta$}

\noindent for every $n$.   Consequently

\Centerline{$\theta^{(Y)}_{r\delta}(f[A])
\le\sum_{n=0}^{\infty}(\diam f[D_n])^r
\le\sum_{n=0}^{\infty}\gamma^r(\diam D_n)^r
\le\gamma^r(\theta_r^{(X)}A+\delta)$,}

\noindent and

\Centerline{$\theta^{(Y)}_r(f[A])
=\lim_{\delta\downarrow 0}\theta^{(Y)}_{r\delta}(f[A])
\le\gamma^r\theta_r^{(X)}A$,}

\noindent as claimed.
}%end of proof of 471J

\leader{471K}{Lemma} Let $(X,\rho)$ be a metric space, and $r>0$.   Let
$\mu_{Hr}$ be $r$-dimensional Hausdorff measure on $X$.   If
$A\subseteq X$, then $\mu_{Hr}A=0$ iff for every $\epsilon>0$ there is a
countable family $\Cal D$ of sets, covering $A$, such that
$\sum_{D\in\Cal D}(\diam D)^r\le\epsilon$.

\proof{ If $\mu_{Hr}A=0$ and $\epsilon>0$, then, in the language of
471A, $\theta_{r1}A\le\epsilon$, so there is a sequence
$\sequencen{D_n}$ of sets covering $A$ such that
$\sum_{n=0}^{\infty}(\diam D_n)^r\le\epsilon$.

If the condition is satisfied, then for any $\epsilon$, $\delta>0$ there
is a countable family $\Cal D$ of sets, covering $A$, such that
$\sum_{D\in\Cal D}(\diam D)^r\le\min(\epsilon,\delta^r)$.   If $\Cal D$
is infinite, enumerate it as $\sequencen{D_n}$;  if it is finite,
enumerate it as $\langle D_n\rangle_{n<m}$ and set $D_n=\emptyset$ for
$n\ge m$.   Now $A\subseteq\bigcup_{n\in\Bbb N}D_n$ and
$\diam D_n\le\delta$ for every $n\in\Bbb N$, so
$\theta_{r\delta}A\le\sum_{n=0}^{\infty}(\diam D_n)^r\le\epsilon$.
As $\epsilon$ is arbitrary, $\theta_{r\delta}A=0$;  as $\delta$ is
arbitrary, $\theta_rA=0$;  it follows at once that $\mu_{Hr}A$ is
defined and is zero (113Xa).
}%end of proof of 471K

\leader{471L}{Proposition} Let $(X,\rho)$ be a metric space and
$0<r<s$.   If $A\subseteq X$ is such that $\mu^*_{Hr}A$ is finite, then
$\mu_{Hs}A=0$.

\proof{ Let $\epsilon>0$.   Let $\delta>0$ be such that
$\delta^{s-r}(1+\mu^*_{Hr}A)\le\epsilon$.   Then there is a sequence
$\sequencen{A_n}$ of sets of diameter at most $\delta$ such that
$A\subseteq\bigcup_{n\in\Bbb N}A_n$ and
$\sum_{n=0}^{\infty}(\diam A_n)^r\le 1+\mu^*_{Hr}A$.   But now, by the
choice of $\delta$, $\sum_{n=0}^{\infty}(\diam A_n)^s\le\epsilon$.   As
$\epsilon$ is arbitrary, $\mu_{Hs}A=0$, by 471K.
}%end of proof of 471L

\leader{471M}{}\cmmnt{ There is a generalization of the density
theorems of \S\S223 and 261 for general Hausdorff measures, which (as
one expects) depends on a kind of Vitali theorem.   I will use the
following notation for the next few paragraphs.

\medskip

\noindent}{\bf Definition} If $(X,\rho)$ is a metric space and
$A\subseteq X$, write $A^{\sim}$ for
$\{x:x\in X,\,\rho(x,A)\le 2\diam A\}$, where
$\rho(x,A)=\inf_{y\in A}\rho(x,y)$.   (\cmmnt{Following the
conventions of 471A, }$\emptyset^{\sim}=\emptyset$.)

\leader{471N}{Lemma} Let $(X,\rho)$ be a metric space.
Let $\Cal F$ be a family of subsets of $X$ such that
$\{\diam F:F\in\Cal F\}$ is bounded.   Set

\Centerline{$Y
=\bigcap_{\delta>0}\bigcup\{F:F\in\Cal F,\,\diam F\le\delta\}$.}

\noindent Then there is a disjoint family $\Cal I\subseteq\Cal F$ such
that

(i) $\bigcup\Cal F\subseteq\bigcup_{F\in\Cal I}F^{\sim}$;

(ii) $Y
\subseteq\overline{\bigcup\Cal J}
  \cup\bigcup_{F\in\Cal I\setminus\Cal J}F^{\sim}$
for every $\Cal J\subseteq\Cal I$.

\proof{{\bf (a)} Let $\gamma$ be an upper bound for
$\{\diam F:F\in\Cal F\}$.   Choose
$\sequencen{\Cal I_n}$, $\sequencen{\Cal J_n}$ inductively, as follows.
$\Cal I_0=\emptyset$.   Given $\Cal I_n$, set
$\Cal F_n'=\{F:F\in\Cal F,\,\diam F\ge 2^{-n}\gamma,\,
F\cap\bigcup\Cal I_n=\emptyset\}$, and let $\Cal J_n\subseteq\Cal F'_n$
be a maximal disjoint set;  now set $\Cal I_{n+1}=\Cal I_n\cup\Cal J_n$,
and continue.

At the end of the induction, set

\Centerline{$\Cal I'=\bigcup_{n\in\Bbb N}\Cal I_n$,
\quad$\Cal I
=\Cal I'\cup\{\{x\}:x\in F\setminus\bigcup\Cal I',\,\{x\}\in\Cal F\}$.}

\noindent The construction ensures that every $\Cal I_n$ is a disjoint
subset of $\Cal F$, so $\Cal I'$ and $\Cal I$ are also disjoint
subfamilies of $\Cal F$.

\medskip

{\bf (b)} \Quer\ Suppose, if possible, that there is a point $x$ in
$\bigcup\Cal F\setminus\bigcup_{F\in\Cal I}F^{\sim}$.   Let $F\in\Cal F$
be such that $x\in F$.   Since $x\notin\bigcup\Cal I'$ and
$\{x\}\notin\Cal I$, $\{x\}\notin\Cal F$, and $\diam F>0$;  let
$n\in\Bbb N$ be such that $2^{-n}\gamma\le\diam F\le 2^{-n+1}\gamma$.
If $F\notin\Cal F'_n$, there is a $D\in\Cal I_n$ such that
$F\cap D\ne\emptyset$;  otherwise, since $\Cal J_n$ is maximal and
$F\notin\Cal J_n$, there is a $D\in\Cal J_n$ such that
$F\cap D\ne\emptyset$.   In either case, we have a $D\in\Cal I$ such
that $F\cap D\ne\emptyset$ and $\diam F\le 2\diam D$.   But in this case
$\rho(x,D)\le\diam F\le 2\diam D$ and $x\in D^{\sim}$, which is
impossible.\ \Bang

\medskip

{\bf (c)} \Quer\ Suppose, if possible, that there are a point $x\in Y$
and a set $\Cal J\subseteq\Cal I$ such that
$x\notin\overline{\bigcup\Cal J}
  \cup\bigcup_{F\in\Cal I\setminus\Cal J}F^{\sim}$.   Then there is an
$F\in\Cal F$ such that $x\in F$ and
$\diam F<\rho(x,\bigcup\Cal J)$, so that
$F\cap\bigcup\Cal J=\emptyset$.   As in (b), $F$ cannot be $\{x\}$, and
there must be an $n\in\Bbb N$ such that
$2^{-n}\gamma<\diam F\le 2^{-n+1}\gamma$.   As in (b), there must be a
$D\in\Cal I_{n+1}$ such that $F\cap D\ne\emptyset$, so that
$x\in D^{\sim}$;  and as $D$ cannot belong to $\Cal J$, we again have a
contradiction.\ \Bang
}%end of proof of 471N

\leader{471O}{Lemma} Let $(X,\rho)$ be a metric space, and $r>0$.
Suppose that $A$, $\Cal F$ are such that

\inset{(i) $\Cal F$ is a family of closed subsets of $X$ such that
$\sum_{n=0}^{\infty}(\diam F_n)^r$ is finite for every disjoint sequence
$\sequencen{F_n}$ in $\Cal F$,

(ii) for every $x\in A$, $\delta>0$ there is an $F\in\Cal F$ such that
$x\in F$ and $0<\diam F\le\delta$.}

\noindent Then there is a countable disjoint family
$\Cal I\subseteq\Cal F$ such
that $A\setminus\bigcup\Cal I$ has zero $r$-dimensional Hausdorff
measure.

\proof{ Replacing $\Cal F$ by $\{F:F\in\Cal F,\,0<\diam F\le 1\}$ if
necessary, we may suppose that $\sup_{F\in\Cal F}\diam F$ is finite and
that $\diam F>0$ for every $F\in\Cal F$.   Take a disjoint family
$\Cal I\subseteq\Cal F$ as in 471N.   If $\Cal I$ is finite, then
$A\subseteq Y\subseteq\bigcup\Cal I$, where $Y$ is defined as in 471N,
so we can stop.   Otherwise, hypothesis (i) tells us that
$\{F:F\in\Cal I,\,\diam F\ge\delta\}$ is finite for every $\delta>0$, so
$\Cal I$ is countable;  enumerate it as $\sequencen{F_n}$;  we must have
$\sum_{n=0}^{\infty}(\diam F_n)^r<\infty$.   Since
$\diam F_n^{\sim}\le 5\diam F_n$ for every $n$,
$\sum_{n=0}^{\infty}(\diam F_n^{\sim})^r$ is finite, and
$\inf_{n\in\Bbb N}\sum_{i=n}^{\infty}(\diam F_i^{\sim})^r=0$.   But now
observe that the construction ensures that
$A\setminus\bigcup\Cal I\subseteq\bigcup_{i\ge n}F_i^{\sim}$ for every
$n\in\Bbb N$.   By 471K, $\mu_{Hr}(A\setminus\bigcup\Cal I)=0$, as
required.
}%end of proof of 471O

\leader{471P}{Theorem} Let $(X,\rho)$ be a metric space, and $r>0$.
Let $\mu_{Hr}$ be $r$-dimensional Hausdorff measure on $X$.   Suppose
that $A\subseteq X$ and $\mu_{Hr}^*A<\infty$.

(a) $\lim_{\delta\downarrow 0}
  \sup\{\Bover{\mu_{Hr}^*(A\cap D)}{(\diam D)^r}:
    x\in D,\,0<\diam D\le\delta\}
=1$ for $\mu_{Hr}$-almost every $x\in A$.

(b) $\limsup_{\delta\downarrow 0}
  \Bover{\mu_{Hr}^*(A\cap B(x,\delta))}{\delta^r}\ge 1$ for
$\mu_{Hr}$-almost every $x\in A$.   So

\Centerline{$2^{-r}
\le\limsup_{\delta\downarrow 0}
  \Bover{\mu_{Hr}^*(A\cap B(x,\delta))}{(\diam B(x,\delta))^r}
\le 1$}

\noindent for $\mu_{Hr}$-almost every $x\in A$.

(c) If $A$ is measured by $\mu_{Hr}$, then

\Centerline{$\lim_{\delta\downarrow 0}
  \sup\{\Bover{\mu_{Hr}^*(A\cap D)}{(\diam D)^r}:
    x\in D,\,0<\diam D\le\delta\}
=0$}

\noindent for $\mu_{Hr}$-almost every $x\in X\setminus A$.

\proof{{\bf (a)(i)} Note first that as the quantities

\Centerline{$\sup\{\Bover{\mu_{Hr}^*(A\cap D)}{(\diam D)^r}:
x\in D,\,0<\diam D\le\delta\}$}

\noindent decrease with $\delta$, the limit is defined in $[0,\infty]$
for every $x\in X$.   Moreover, since $\diam D=\diam\overline{D}$ and
$\mu_{Hr}^*(A\cap\overline{D})\ge\mu_{Hr}^*(A\cap D)$ for every $D$,

$$\eqalign{&\sup\{\Bover{\mu_{Hr}^*(A\cap D)}{(\diam D)^r}:x
\in D\subseteq X,\,0<\diam D\le\delta\}\cr
&\qquad\qquad\qquad
=\sup\{\Bover{\mu_{Hr}^*(A\cap F)}{(\diam F)^r}:
    F\subseteq X\text{ is closed},\,x\in F,\,0<\diam F\le\delta\}\cr}$$

\noindent for every $x$ and $\delta$.

\medskip

\quad{\bf (ii)} Fix $\epsilon$ for the moment, and set

\Centerline{$A_{\epsilon}=\{x:x\in A,\,\lim_{\delta\downarrow 0}
  \sup\{\Bover{\mu_{Hr}^*(A\cap D)}{(\diam D)^r}:
    x\in D,\,0<\diam D\le\delta\}
>1+\epsilon\}$.}

\noindent Then $\theta_{r\eta}(A)
\le\mu_{Hr}^*A-\Bover{\epsilon}{1+\epsilon}\mu^*_{Hr}A_{\epsilon}$ for
every
$\eta>0$, where $\theta_{r\eta}$ is defined in 471A.   \Prf\ Let
$\Cal F$ be the family

\Centerline{$\{F:F\subseteq X$ is closed, $0<\diam F\le\eta$,
$(1+\epsilon)(\diam F)^r\le\mu_{Hr}^*(A\cap F)\}$.}

\noindent Then every member of $A_{\epsilon}$ belongs to sets in
$\Cal F$ of arbitrarily small diameter.   Also, if $\sequencen{F_n}$ is
any disjoint sequence in $\Cal F$,

\Centerline{$\sum_{n=0}^{\infty}(\diam F_n)^r
\le\sum_{n=0}^{\infty}\mu^*_{Hr}(A\cap F_n)
\le\mu^*_{Hr}A<\infty$}

\noindent because every $F_n$, being closed, is measured by $\mu_{Hr}$.
(If you like, $F_n\cap A$ is measured by the subspace measure on $A$ for
every $n$.)   So 471O tells us that there is a countable disjoint family
$\Cal I\subseteq\Cal F$ such that $A_{\epsilon}\setminus\bigcup\Cal I$
is negligible, and
$\mu_{Hr}^*A_{\epsilon}=\mu_{Hr}^*(A_{\epsilon}\cap\bigcup\Cal I)$.

Because $\theta_{r\eta}$ is an outer measure and
$\theta_{r\eta}\le\mu_{Hr}^*$,

$$\eqalignno{\theta_{r\eta}A
&\le\theta_{r\eta}(A\cap\bigcup\Cal I)
  +\theta_{r\eta}(A\setminus\bigcup\Cal I)
\le\sum_{F\in\Cal I}(\diam F)^r+\mu_{Hr}^*(A\setminus\bigcup\Cal I)\cr
\displaycause{because $\Cal I$ is countable}
&\le\Bover1{1+\epsilon}\mu_{Hr}^*(A\cap\bigcup\Cal I)
  +\mu_{Hr}^*(A\setminus\bigcup\Cal I)
=\mu_{Hr}^*A
  -\Bover{\epsilon}{1+\epsilon}\mu_{Hr}^*(A\cap\bigcup\Cal I)\cr
&\le\mu_{Hr}^*A-\Bover{\epsilon}{1+\epsilon}
      \mu_{Hr}^*(A_{\epsilon}\cap\bigcup\Cal I)
=\mu_{Hr}^*A
  -\Bover{\epsilon}{1+\epsilon}\mu_{Hr}^*A_{\epsilon},\cr}$$

\noindent as claimed.\ \Qed

\medskip

\quad{\bf (iii)} Taking the supremum as $\eta\downarrow 0$,
$\mu_{Hr}^*A
\le\mu_{Hr}^*A-\Bover{\epsilon}{1+\epsilon}\mu_{Hr}^*A_{\epsilon}$ and
$\mu_{Hr}A_{\epsilon}=0$.

\woddheader{}{0}{0}{0}{40pt}

This is true for any $\epsilon>0$.   But

\Centerline{$\{x:x\in A,\,\lim_{\delta\downarrow 0}
  \sup\{\Bover{\mu_{Hr}^*(A\cap D)}{(\diam D)^r}:
    x\in D,\,0<\diam D\le\delta\}
>1\}$}

\noindent is just $\bigcup_{n\in\Bbb N}A_{2^{-n}}$, so is negligible.

\medskip

\quad{\bf (iv)} Next, for $0<\epsilon\le 1$, set

$$\eqalign{A'_{\epsilon}
=\{x:x\in A,\,\mu_{Hr}^*(A\cap D)&\le(1-\epsilon)(\diam D)^r\cr
&\text{ whenever }x\in D\text{ and }0<\diam D\le\epsilon\}.\cr}$$

\noindent Then $A'_{\epsilon}$ is negligible.   \Prf\ Let
$\sequencen{D_n}$ be any sequence of sets of diameter at most $\epsilon$
covering $A'_{\epsilon}$.   Set
$K=\{n:D_n\cap A'_{\epsilon}\ne\emptyset\}$.   Then

$$\eqalign{\mu_{Hr}^*A'_{\epsilon}
&\le\sum_{n\in K}\mu_{Hr}^*(A\cap D_n)\cr
&\le(1-\epsilon)\sum_{n\in K}(\diam D_n)^r
\le(1-\epsilon)\sum_{n=0}^{\infty}(\diam D_n)^r.\cr}$$

\noindent As $\sequencen{D_n}$ is arbitrary,

\Centerline{$\mu_{Hr}^*A'_{\epsilon}
\le(1-\epsilon)\theta_{r\epsilon}A'_{\epsilon}
\le(1-\epsilon)\mu_{Hr}^*A'_{\epsilon}$,}

\noindent and $\mu_{Hr}^*A'_{\epsilon}$ (being finite) must be zero.\
\Qed

This means that

\Centerline{$\{x:x\in A,\,\lim_{\delta\downarrow 0}
  \sup\{\Bover{\mu_{Hr}^*(A\cap D)}{(\diam D)^r}:
    x\in D,\,0<\diam D\le\delta\}<1\}
\subseteq\bigcup_{n\in\Bbb N}A'_{2^{-n}}$}

\noindent is also negligible, and we have the result.

\medskip

{\bf (b)} We need a slight modification of the argument in (a)(iv).
This time, for $0<\epsilon\le 1$, set

\Centerline{$\tilde A'_{\epsilon}
=\{x:x\in A,\,\mu_{Hr}^*(A\cap B(x,\delta))\le(1-\epsilon)\delta^r$
whenever $0<\delta\le\epsilon\}$.}

\noindent Then $\mu_{Hr}^*\tilde A'_{\epsilon}\le\epsilon$.   \Prf\ Note
first that, as $\mu_{Hr}\{x\}=0$ for every $x$,
$\mu_{Hr}^*(A\cap B(x,\delta))\le(1-\epsilon)\delta^r$ whenever
$x\in\tilde A'_{\epsilon}$ and $0\le\delta\le\epsilon$.   Let
$\sequencen{D_n}$ be a sequence of sets of diameter at most $\epsilon$
covering $\tilde A'_{\epsilon}$.   Set
$K=\{n:D_n\cap\tilde A'_{\epsilon}\ne\emptyset\}$, and for $n\in K$
choose $x_n\in D_n\cap\tilde A'_{\epsilon}$ and set
$\delta_n=\diam D_n$.   Then
$D_n\subseteq B(x_n,\delta_n)$ and $\delta_n\le\epsilon$ for each $n$,
so $\tilde A'_{\epsilon}\subseteq\bigcup_{n\in K}B(x_n,\delta_n)$ and

$$\eqalign{\mu_{Hr}^*\tilde A'_{\epsilon}
&\le\sum_{n\in K}\mu_{Hr}^*(\tilde A'_{\epsilon}\cap B(x_n,\delta_n))\cr
&\le\sum_{n\in K}(1-\epsilon)\delta_n^r
\le(1-\epsilon)\sum_{n=0}^{\infty}(\diam D_n)^r.\cr}$$

\noindent As $\sequencen{D_n}$ is arbitrary,
$\mu_{Hr}^*\tilde A'_{\epsilon}
\le(1-\epsilon)\mu_{Hr}^*\tilde A'_{\epsilon}$ and
$\tilde A'_{\epsilon}$ must be negligible.\ \Qed

Now

\Centerline{$\{x:x\in A,\,\limsup_{\delta\downarrow 0}
  \Bover{\mu_{Hr}^*(A\cap B(x,\delta))}{\delta^r}<1\}
=\bigcup_{n\in\Bbb N}\tilde A'_{2^{-n}}$}

\noindent is negligible.   As for the second formula, we need note only
that $\diam B(x,\delta)\le 2\delta$ for every $x\in X$,
$\delta>0$ to obtain the first inequality, and apply (a) to obtain the
second.

\medskip

{\bf (c)} Let $\epsilon>0$.   This time, write $\tilde A_{\epsilon}$ for

\Centerline{$\{x:x\in X,\,\lim_{\delta\downarrow 0}
  \sup\{\Bover{\mu_{Hr}^*(A\cap D)}{(\diam D)^r}:
    x\in D,\,0<\diam D\le\delta\}
>\epsilon\}$.}

\noindent Let $E\subseteq A$ be a closed set such that
$\mu(A\setminus E)\le\epsilon^2$ (471De).   For $\eta>0$, let
$\Cal F_{\eta}$ be the family

\Centerline{$\{F:F\subseteq X\setminus E$ is closed, $0<\diam F\le\eta$,
$\mu_{Hr}(A\cap F)\ge\epsilon(\diam F)^r\}$.}

\noindent Just as in (a) above, every point in
$\tilde A_{\epsilon}\setminus E$
belongs to members of $\Cal F_{\eta}$ of arbitrarily small diameter.
If $\familyiI{F_i}$ is a countable disjoint family in $\Cal F_{\eta}$,

\Centerline{$\sum_{i\in I}(\diam F_i)^r
\le\Bover1{\epsilon}\mu_{Hr}(A\setminus E)\le\epsilon$}

\noindent is finite.   There is therefore a countable disjoint family
$\Cal I_{\eta}\subseteq\Cal F_{\eta}$ such that
$\mu_{Hr}((\tilde A_{\epsilon}\setminus E)
  \setminus\bigcup\Cal I_{\eta})=0$.
If $\theta_{r\eta}$ is the outer measure defined in 471A, we have

$$\eqalign{\theta_{r\eta}(\tilde A_{\epsilon}\setminus A)
&\le\theta_{r\eta}(\bigcup\Cal I_{\eta})
  +\theta_{r\eta}(\tilde A_{\epsilon}
      \setminus(E\cup\bigcup\Cal I_{\eta}))\cr
&\le\sum_{F\in\Cal I_{\eta}}(\diam F)^r
  +\mu_{Hr}^*(\tilde A_{\epsilon}\setminus(E\cup\bigcup\Cal I_{\eta}))
\le\epsilon.\cr}$$

\noindent As $\eta$ is arbitrary,
$\mu_{Hr}^*(\tilde A_{\epsilon}\setminus A)\le\epsilon$.   But now

\Centerline{$\{x:x\in X\setminus A,\,\lim_{\delta\downarrow 0}
  \sup\{\Bover{\mu_{Hr}^*(A\cap D)}{(\diam D)^r}:
    x\in D,\,0<\diam D\le\delta\}
>0\}$}

\noindent is $\bigcap_{n\in\Bbb N}\tilde A_{2^{-n}}\setminus A$, and is
negligible.
}%end of proof of 471P

\leader{471Q}{}\cmmnt{ I now come to a remarkable fact about Hausdorff
measures on analytic spaces:  their Borel versions are semi-finite
(471S).   We need some new machinery.

\medskip

\noindent}{\bf Lemma} Let $(X,\rho)$ be a metric space, and $r>0$,
$\delta>0$.   Suppose that $\theta_{r\delta}X$, as defined in 471A, is
finite.

(a) There is a non-negative additive functional $\nu$ on $\Cal PX$ such
that $\nu X=5^{-r}\theta_{r\delta}X$ and
$\nu A\le(\diam A)^r$ whenever $A\subseteq X$ and
$\diam A\le\bover15\delta$.

(b) If $X$ is compact, there is a Radon measure $\mu$ on $X$ such that
$\mu X=5^{-r}\theta_{r\delta}X$ and $\mu G\le(\diam G)^r$ whenever
$G\subseteq X$ is open and $\diam G\le\bover15\delta$.

\proof{{\bf (a)} I use 391E.   If $\theta_{r\delta}X=0$ the result is
trivial.   Otherwise, set $\gamma=5^r/\theta_{r\delta}X$ and define
$\phi:\Cal PX\to[0,1]$ by setting $\phi A=\min(1,\gamma(\diam A)^r)$ if
$\diam A\le\bover15\delta$, $1$ for other $A\subseteq X$.   Now

\doubleinset{
whenever $\familyiI{A_i}$ is a finite family of subsets of $X$,
$m\in\Bbb N$ and $\sum_{i\in I}\chi A_i\ge m\chi X$,
then $\sum_{i\in I}\phi A_i\ge m$.}

\noindent\Prf\   Discarding any
$A_i$ for which $\phi A_i=1$, if necessary, we may suppose that
$\diam A_i\le\bover15\delta$ and $\phi A_i=\gamma(\diam A_i)^r$ for
every $i$.   Choose
$\langle I_j\rangle_{j\le m}$, $\langle J_j\rangle_{j<m}$ inductively,
as follows.   $I_0=I$.   Given that $j<m$ and that $I_j\subseteq I$ is
such that $\sum_{i\in I_j}\chi A_i\ge(m-j)\chi X$, apply 471N to
$\{A_i:i\in I_j\}$ to find $J_j\subseteq I_j$ such that
$\family{i}{J_j}{A_i}$ is disjoint and
$\bigcup_{i\in I_j}A_i\subseteq\bigcup_{i\in J_j}A_j^{\sim}$.   Set
$I_{j+1}=I_j\setminus J_j$.   Observe that
$\sum_{i\in J_j}\chi A_j\le\chi X$, so
$\sum_{i\in I_{j+1}}\chi A_i\ge(m-j-1)\chi X$ and the induction
proceeds.

Now note that, for each $j<m$, $\family{i}{J_j}{A_i^{\sim}}$ is a cover
of $\bigcup_{i\in I_j}A_i=X$ by sets of diameter at most $\delta$.   So
$\sum_{i\in J_j}(\diam A_i^{\sim})^r\ge\theta_{r\delta}X$ for each
$j$, and $\sum_{i\in I}(\diam A_i^{\sim})^r\ge m\theta_{r\delta}X$.
Accordingly

$$\eqalign{\sum_{i\in I}\phi A_i
&=\gamma\sum_{i\in I}(\diam A_i)^r
\ge 5^{-r}\gamma\sum_{i\in I}(\diam A_i^{\sim})^r\cr
&\ge 5^{-r}m\gamma\theta_{r\delta}X
=m.  \text{ \Qed}\cr}$$

By 391E, there is an additive functional $\nu_0:\Cal PX\to[0,1]$
such that $\nu_0X=1$ and $\nu_0A\le\phi A$ for every $A\subseteq X$.
Setting $\nu=5^{-r}\theta_{r\delta}X\nu_0$, we have the result.

\medskip

{\bf (b)} Now suppose that $X$ is compact.   By 416K, there is a Radon
measure $\mu$ on $X$ such that $\mu K\ge\nu K$ for every compact
$K\subseteq X$ and $\mu G\le\nu G$ for every open $G\subseteq X$.
Because $X$ itself is compact, $\mu X=\nu X=5^{-r}\theta_{r\delta}X$.
If $G$ is open and $\diam G\le\bover15\delta$,

\Centerline{$\mu G\le\nu G\le(\diam G)^r$,}

\noindent as required.
}%end of proof of 471Q

\leader{471R}{Lemma}\cmmnt{ ({\smc Howroyd 95})} Let $(X,\rho)$ be a
compact metric space and $r>0$.   Let $\mu_{Hr}$ be $r$-dimensional
Hausdorff measure on $X$.   If $\mu_{Hr}X>0$, there is a Borel set
$H\subseteq X$ such that $0<\mu_{Hr}H<\infty$.

\proof{{\bf (a)} Let $\delta>0$ be such that $\theta_{r,5\delta}(X)>0$,
where $\theta_{r,5\delta}$ is defined as in 471A.   Then there is a
family $\Cal V$ of open subsets of $X$ such that (i) $\diam V\le\delta$
for every $V\in\Cal V$ (ii) $\{V:V\in\Cal V,\,\diam V\ge\epsilon\}$ is
finite for every $\epsilon>0$ (iii) whenever $A\subseteq X$ and
$0<\diam A<\bover14\delta$ there is a $V\in\Cal V$ such that
$A\subseteq V$ and $\diam V\le 8\diam A$.   \Prf\ For each $k\in\Bbb N$,
let $I_k$ be a finite subset of $X$ such that
$X=\bigcup_{x\in I_k}B(x,2^{-k-2}\delta)$;  now set
$\Cal V=\{U(x,2^{-k-1}\delta):k\in\Bbb N,\,x\in I_k\}$.   Then $\Cal V$
is a family of open
sets and (i) and (ii) are satisfied.   If $A\subseteq X$ and
$0<\diam A<\bover14\delta$, let $k\in\Bbb N$ be such that
$2^{-k-3}\delta\le\diam A<2^{-k-2}\delta$.   Take $x\in I_k$ such that
$B(x,2^{-k-2}\delta)\cap A\ne\emptyset$;  then
$A\subseteq U(x,2^{-k-1}\delta)\in\Cal V$ and
$\diam U(x,2^{-k-1}\delta)\le 2^{-k}\delta\le 8\diam A$.\ \Qed

In particular, $\{V:V\in\Cal V,\,\diam V\le\epsilon\}$ covers $X$ for
every $\epsilon>0$.

\medskip

{\bf (b)} Set

\Centerline{$P=\{\mu:\mu$ is a Radon measure on $X$,
$\mu V\le(\diam V)^r$ for every $V\in\Cal V\}$.}

\noindent $P$ is non-empty (it contains the zero measure, for instance).
Now if $G\subseteq X$ is open,
$\mu\mapsto\mu G$ is lower semi-continuous for the narrow topology
(437Jd), so $P$ is a closed set in the narrow topology on the set of
Radon measures on
$X$, which may be identified with a subset of $C(X)^*$ with its
weak* topology (437Kc).   Moreover, since there is a finite subfamily of
$\Cal V$ covering
$X$, $\gamma=\sup\{\mu X:\mu\in P\}$ is finite, and $P$ is compact
(437Pb/437Rf).   Because $\mu\mapsto\mu X$ is continuous,
$P_0=\{\mu:\mu\in P,\,\mu X=\gamma\}$ is non-empty.   Of course $P$ and
$P_0$ are both convex, and $P_0$, like $P$, is compact.   By
the \Krein-Mil'man theorem (4A4Gb), applied in $C(X)^*$, $P$ has an
extreme point $\nu$ say.

Note next that $\theta_{r,5\delta}(X)$ is certainly finite, again
because $X$ is compact.   By 471Qb, $\gamma>0$, and $\nu$ is
non-trivial.   For any $\epsilon>0$, there is a finite cover of $X$ by
sets in $\Cal V$ of diameter at most $\epsilon$, which have measure at
most $\epsilon^r$ (for $\nu$);  so $\nu$ is atomless.   In particular,
$\nu\{x\}=0$ for every $x\in X$.

\medskip

{\bf (c)} For $\epsilon>0$, set

\Centerline{$G_{\epsilon}
=\bigcup\{V:V\in\Cal V,\,0<\diam V\le\epsilon$ and
  $\nu V\ge\bover12(\diam V)^r\}$.}

\noindent Then $G_{\epsilon}$ is $\nu$-conegligible.  \Prf\Quer\
Otherwise, $\nu(X\setminus G_{\epsilon})>0$.   Because
$\Cal V'_{\epsilon}=\{V:V\in\Cal V,\,\diam V>\epsilon\}$ is finite,
there is a Borel set $E\subseteq X\setminus G_{\epsilon}$ such that
$\nu E>0$ and, for every $V\in\Cal V'_{\epsilon}$, either $E\subseteq V$
or $E\cap V=\emptyset$.   Because $\nu$ is atomless, there is a
measurable set
$E_0\subseteq E$ such that $\nu E_0=\bover12\nu E$ (215D);  set
$E_1=E\setminus E_0$.

Define Radon measures $\nu_0$, $\nu_1$ on $X$ by setting

\Centerline{$\nu_i(F)=2\nu(F\cap E_i)+\nu(F\setminus E)$}

\noindent whenever $\nu$ measures $F\setminus E_{1-i}$, for each $i$
(use 416S if you feel the need to check that this defines a Radon
measure on the definitions of this book).   If $V\in\Cal V$, then, by
the choice of $E$,

\inset{{\it either} $E\subseteq V$ and $\nu_iV=\nu V\le(\diam V)^r$

{\it or} $E\cap V=\emptyset$ and $\nu_iV=\nu V\le(\diam V)^r$

{\it or} $0<\diam V\le\epsilon$ and $\nu V<\bover12(\diam V)^r$, in
which case $\nu_iV\le 2\nu V\le(\diam V)^r$

{\it or} $\diam V=0$ and $\nu_iV=\nu V=0=(\diam V)^r$.}

\noindent So both $\nu_i$ belong to $P$ and therefore to $P_0$, since
$\nu_iX=\nu X=\gamma$.   But $\nu=\bover12(\nu_0+\nu_1)$ and
$\nu_0\ne\nu_1$, so this is impossible, because $\nu$ is supposed to be
an extreme point of $P_0$.\ \Bang\Qed

\medskip

{\bf (d)} Accordingly, setting $H=\bigcap_{n\in\Bbb N}G_{2^{-n}}$,
$\nu H=\nu X=\gamma$.   Now examine $\mu_{Hr}H$.

\medskip

\quad{\bf (i)} $\mu_{Hr}H\ge 8^{-r}\gamma$.   \Prf\ Let
$\sequencen{A_n}$ be a sequence of sets covering $H$ with
$\diam A_n\le\bover18\delta$ for every $n$.   Set $K=\{n:\diam A_n>0\}$,
$H'=H\cap\bigcup_{n\in K}A_n$;  then $H\setminus H'$ is countable, so
$\nu H'=\nu H$.   For each $n\in K$, let $V_n\in\Cal V$ be such that
$A_n\subseteq V_n$ and $\diam V_n\le 8\diam A_n$ ((a) above).   Then

$$\eqalignno{\sum_{n=0}^{\infty}(\diam A_n)^r
&=\sum_{n\in K}(\diam A_n)^r
\ge 8^{-r}\sum_{n\in K}(\diam V_n)^r\cr
&\ge 8^{-r}\sum_{n\in K}\nu V_n
\ge 8^{-r}\nu H'
=8^{-r}\gamma.\cr}$$

\noindent As $\sequencen{A_n}$ is arbitrary,

\Centerline{$8^{-r}\gamma\le\theta_{r,\delta/8}(H)\le\mu_{Hr}^*H
=\mu_{Hr}H$.  \Qed}

\medskip

\quad{\bf (ii)} $\mu_{Hr}H\le 2\gamma$.   \Prf\ Let $\eta>0$.   Set
$\Cal F=\{\overline{V}:V\in\Cal V,\,0<\diam V\le\eta,\,
\nu V\ge\bover12(\diam V)^r\}$.   Then $\Cal F$ is a family of closed
subsets of $X$, and (by the definition of $G_{\epsilon}$) every member
of $H$ belongs to members of $\Cal F$ of arbitrarily small diameter.
Also $\nu F\ge\bover12(\diam F)^r$ for every $F\in\Cal F$, so

\Centerline{$\sum_{n=0}^{\infty}(\diam F_n)^r
\le 2\sum_{n=0}^{\infty}\nu F_n<\infty$}

\noindent for any disjoint sequence $\sequencen{F_n}$ in $\Cal F$.
By 471O, there is a countable disjoint family $\Cal I\subseteq\Cal F$
such that $\mu_{Hr}(H\setminus\bigcup\Cal I)=0$.   Accordingly

\Centerline{$\theta_{r\eta}(H)
\le\sum_{F\in\Cal I}(\diam F)^r+\theta_{r\eta}(H\setminus\bigcup\Cal I)
\le\sum_{F\in\Cal I}2\nu F
\le 2\gamma$.}

\noindent As $\eta$ is arbitrary, $\mu_{Hr}H=\mu_{Hr}^*H\le 2\gamma$.\
\Qed

\medskip

{\bf (e)} But this means that we have found a Borel set $H$ with
$0<\mu_{Hr}H<\infty$, as required.
}%end of proof of 471R

\leader{471S}{Theorem}\cmmnt{ ({\smc Howroyd 95})} Let $(X,\rho)$ be
an analytic metric space, and $r>0$.   Let
$\mu_{Hr}$ be $r$-dimensional Hausdorff measure on $X$, and $\Cal B$ the
Borel $\sigma$-algebra of $X$.   Then the Borel measure
$\mu_{Hr}\restr\Cal B$ is semi-finite and tight\cmmnt{ (that is, inner
regular with respect to the closed compact sets)}.

\proof{ Suppose that $E\in\Cal B$ and $\mu_{Hr}E>0$.   Since
$E$ is analytic (423Eb), 471I above tells us that there is a compact set
$K\subseteq E$ such that $\mu_{Hr}K>0$.   Next, by 471R, there is
a Borel set $H\subseteq K$ such that $0<\mu_{Hr}H<\infty$.   (Strictly
speaking, $\mu_{Hr}H$ here should be calculated as the $r$-dimensional
Hausdorff measure of $H$ defined by the subspace metric
$\rho\restr K\times K$ on $K$.   By 471E we do not need to distinguish
between this and the $r$-dimensional measure calculated from $\rho$
itself.)   By 471I again (applied to the subspace metric on $H$), there
is a compact set $L\subseteq H$ such that $\mu_{Hr}L>0$.

Thus $E$ includes a non-negligible compact set of finite measure.   As
$E$ is arbitrary, this is enough to show both that
$\mu_{Hr}\restr\Cal B$ is semi-finite and that it is tight.
}%end of proof of 471S

\leader{471T}{Proposition} Let $(X,\rho)$ be a metric space, and $r>0$.

(a) If $X$ is analytic and
$\mu_{Hr}X>0$, then for every $s\in\ooint{0,r}$
there is a non-zero Radon measure $\mu$ on $X$ such that
$\biggeriint\Bover1{\rho(x,y)^s}\mu(dx)\mu(dy)<\infty$.

(b) If there is a non-zero topological measure $\mu$ on $X$ such that
$\biggeriint\Bover1{\rho(x,y)^r}\mu(dx)\mu(dy)$ is finite, then
$\mu_{Hr}X=\infty$.

\proof{{\bf (a)} By 471S, there is a compact set $K\subseteq X$ such that
$\mu_{Hr}K>0$.   Set $\delta=5\diam K$ and define $\theta_{r\delta}$ as in
471A.  Then $\theta_{r\delta}K>0$, by 471K, and
$\theta_{r\delta}K\le(\diam K)^r<\infty$.   By 471Qb, there is
a Radon measure $\nu$ on $K$ such that $\nu K>0$ and $\nu G\le(\diam G)^r$
whenever $G\subseteq K$ is relatively
open;  consequently $\nu^*A\le(\diam A)^r$ for
every $A\subseteq K$.   Now, for any $y\in X$,

$$\eqalign{\int_K\Bover1{\rho(x,y)^s}\nu(dx)
&=\int_0^{\infty}\nu\{x:x\in K,\,\Bover1{\rho(x,y)^s}\ge t\}dt
=\int_0^{\infty}\nu\{x:x\in K,\,\rho(x,y)\le\Bover1{t^{1/s}}\}dt\cr
&=\int_0^{\infty}\nu(K\cap B(y,\Bover1{t^{1/s}}))dt
\le\int_0^{\infty}(\diam(K\cap B(y,\Bover1{t^{1/s}})))^rdt\cr
&\le\int_0^{\infty}(\min(\diam K,\Bover2{t^{1/s}}))^rdt
\le 2^r\int_0^{\infty}\min((\diam K)^r,\Bover1{t^{r/s}})dt
<\infty\cr}$$

\noindent because $r>s$.   It follows at once that
$\int_K\int_K\Bover1{\rho(x,y)^s}\nu(dx)\nu(dy)$ is finite.   Taking $\mu$
to be the extension of $\nu$ to a Radon measure on $X$ for which
$X\setminus K$ is negligible, we have an appropriate $\mu$.

\medskip

{\bf (b)(i)} We can suppose that $X$ is separable (471Df).
Since the integrand
is strictly positive, $\mu$ must be $\sigma$-finite, so that
there is no difficulty with the repeated integral.   Replacing $\mu$ by
$\mu\LLcorner F$ for some set $F$ of non-zero finite measure, we can
suppose that $\mu$ is totally finite;  and replacing $\mu$ by a scalar
multiple of itself, we can suppose that it is a probability measure.

\medskip

\quad{\bf (ii)} Let $\epsilon>0$.   Let $H$ be the conegligible set
$\{y:\biggerint\Bover1{\rho(x,y)^r}\mu(dx)<\infty\}$.
For any $y\in X$, $\mu\{y\}=0$, so

\Centerline{$\lim_{\delta\downarrow 0}
\int_{B(y,\delta)}\Bover1{\rho(x,y)^r}\mu(dx)=0$}

\noindent for every $y\in H$.   For each $\delta>0$,

\Centerline{$(x,y)\mapsto\Bover{\chi B(y,\delta)(x)}{\rho(x,y)^r}:
X\times X\to[0,\infty]$}

\noindent is Borel measurable, so

\Centerline{$y\mapsto\int_{B(y,\delta)}\Bover1{\rho(x,y)^r}\mu(dx):
X\to[0,\infty]$}

\noindent is Borel measurable (252P, applied to the restriction of $\mu$ to
the Borel $\sigma$-algebra of $X$).   There is therefore a $\delta>0$ such
that $E=\{y:y\in H$,
$\biggerint_{B(y,\delta)}\Bover1{\rho(x,y)^r}\mu(dy)\le\epsilon\}$ has
measure
$\mu E\ge\bover12$.   Note that if $C\subseteq X$ has diameter less than or
equal to $\delta$ and meets $E$ then
$\mu\overline{C}\le\epsilon(\diam C)^r$.   \Prf\ Set $\gamma=\diam C$ and
take $y\in C\cap E$.   If $C=\{y\}$ then $\mu\overline{C}=0$.   Otherwise,

\Centerline{$\mu\overline{C}
\le\mu B(y,\gamma)
\le\gamma^r\int_{B(y,\delta)}\Bover1{\rho(x,y)^r}\mu(dx)
\le\gamma^r\epsilon$.  \Qed}

Now suppose that $E\subseteq\bigcup_{i\in I}C_i$ where
$\diam C_i\le\delta$ for every $i$, and each $C_i$ is either empty or meets
$E$.   Then

\Centerline{$\Bover12\le\mu E\le\sum_{i=0}^{\infty}\mu\overline{C}_i
\le\sum_{i=0}^{\infty}\epsilon(\diam C_i)^r$.}

\noindent As $\sequence{i}{C_i}$ is arbitrary,
$\epsilon\mu_{Hr}E\ge\Bover12$ and $\mu_{Hr}X\ge\Bover1{2\epsilon}$.
As $\epsilon$ is arbitrary, $\mu_{Hr}X=\infty$.
}%end of proof of 471T

\cmmnt{\medskip

\noindent{\bf Remark} See 479Cb below.
}

\exercises{\leader{471X}{Basic exercises (a)}
%\spheader 471Xa
Define a metric $\rho$ on $X=\{0,1\}^{\Bbb N}$ by
setting $\rho(x,y)=2^{-n}$ if $x\restr n=y\restr n$ and $x(n)\ne y(n)$.
Show that the usual measure $\mu$ on $X$ is one-dimensional Hausdorff
measure.   \Hint{$\diam F\ge\mu F$ for every closed set $F\subseteq X$.}
%471A

\spheader 471Xb Suppose that $g:\Bbb R\to\Bbb R$ is continuous and
non-decreasing,
and that $\nu$ is the corresponding Lebesgue-Stieltjes measure
(114Xa).   Define $\rho(x,y)=|x-y|+\sqrt{|g(x)-g(y)|}$ for
$x$, $y\in\Bbb R$.
Show that $\rho$ is a metric on $\Bbb R$ defining the usual topology.
Show that $\nu$ is $2$-dimensional Hausdorff measure for the metric $\rho$.
%471A

\spheader 471Xc\dvAnew{2016}
Let $r\ge 1$ be an integer, and give $\BbbR^r$ the metric
$((\xi_1,\ldots,\xi_r),(\eta_1,\ldots,\eta_r))
\mapsto\max_{i\le r}|\xi_i-\eta_i|$.   Show that Lebesgue measure on
$\BbbR^r$ is Hausdorff $r$-dimensional measure for this metric.
%471D

\spheader 471Xd Let $(X,\rho)$ be a metric space and $r>0$;  let
$\mu_{Hr}$, $\theta_{r\infty}$ be $r$-dimensional Hausdorff measure and
capacity on $X$.   (i) Show that, for $A\subseteq X$, $\mu_{Hr}A=0$ iff
$\theta_{r\infty}A=0$.   (ii)  Suppose that $E\subseteq X$ and
$\delta>0$ are such that
$\delta\mu_{Hr}E<\theta_{r\infty}E$.   Show that there is a closed set
$F\subseteq E$ such that $\mu_{Hr}F>0$ and
$\delta\mu_{Hr}(F\cap G)\le(\diam G)^r$ whenever $\mu_{Hr}$ measures
$G$.   \Hint{show that
$\{G^{\ssbullet}:\theta_{r\infty}G<\delta\mu_{Hr}G\}$ cannot
be order-dense in the measure algebra of $\mu_{Hr}$.   This is a version
of `Frostman's Lemma'.}   (iii) Let $\Cal C$ be the family of closed
subsets of $X$, with its Vietoris topology.
Show that $\theta_{r\infty}\restr\Cal C$ is upper semi-continuous.
%471H

\spheader 471Xe\dvAnew{2016}
Show that all the outer measures $\theta_{r\delta}$
described in 471A are outer regular Choquet capacities.
%471H out of order query

\spheader 471Xf Let $(X,\rho)$ be an analytic metric space, $(Y,\sigma)$
a metric space, and $f:X\to Y$ a Lipschitz function.   Show that if $r>0$
and $A\subseteq X$ is measured by Hausdorff $r$-dimensional measure on
$X$, with finite measure, then $f[A]$ is measured by Hausdorff
$r$-dimensional measure on $Y$.
%471J

\spheader 471Xg Let $(X,\rho)$ be a metric space and $r>0$.   Show that
a set $A\subseteq X$ is negligible for Hausdorff $r$-dimensional measure
on $X$ iff there is a sequence $\sequencen{A_n}$ of subsets of $X$ such
that $\sum_{n=0}^{\infty}(\diam A_n)^r$ is finite and
$A\subseteq\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}A_m$.
%471K

\spheader 471Xh Let $(X,\rho)$ be a metric space.   (i) Show that there
is a unique $\dim_H(X)\in[0,\infty]$ such that the $r$-dimensional
Hausdorff measure of $X$ is infinite if $0<r<\dim_H(X)$, zero if
$r>\dim_H(X)$.   ($\dim_H(X)$ is the {\bf Hausdorff dimension} of $X$.)
(ii) Show that if $\sequencen{A_n}$ is any sequence of subsets of $X$,
then $\dim_H(\bigcup_{n\in\Bbb N}A_n)=\sup_{n\in\Bbb N}\dim_H(A_n)$.
%471L

\spheader 471Xi Let $(X,\rho)$ be a metric space, and $\mu$ any
topological measure on $X$.   Suppose that $E\subseteq X$ and that
$\mu E$ is defined and finite.   (i) Show that
$(x,\delta)\mapsto\mu(E\cap B(x,\delta)):
X\times\coint{0,\infty}\to\Bbb R$ is upper semi-continuous.  (ii) Show
that $x\mapsto\limsup_{\delta\downarrow 0}
\Bover1{\delta^r}\mu(E\cap B(x,\delta)):X\to[0,\infty]$ is Borel
measurable, for every $r\ge 0$.   (iii) Show that if $X$ is separable,
then
$\mu B(x,\delta)>0$ for every $\delta>0$, for $\mu$-almost every $x\in X$.
%471P

\spheader 471Xj Give $\Bbb R$ its usual metric.   Let $C\subseteq\Bbb R$
be the Cantor set, and $r=\ln 2/\ln 3$.   Show that

\Centerline{$\liminf_{\delta\downarrow 0}
  \Bover{\mu_{Hr}(C\cap B(x,\delta))}{(\diam B(x,\delta))^r}
\le 2^{-r}$}

\noindent for every $x\in\Bbb R$.
%471P

\spheader 471Xk Let $(X,\rho)$ be a metric space and $r>0$.   Let
$\mu_{Hr}$ be $r$-dimensional Hausdorff measure on $X$ and
$\tilde\mu_{Hr}$ its c.l.d.\ version (213D-213E).   Show that
$\tilde\mu_{Hr}$ is inner regular with respect to the closed sets, and
that $\tilde\mu_{Hr}A=\mu_{Hr}A$ for every analytic set $A\subseteq X$.
%471D, 471Q

\leader{471Y}{Further exercises (a)}
%\spheader 471Ya
The next few exercises (down to 471Yd) will be based on the
following.    Let $(X,\rho)$ be a metric space and
$\psi:\Cal PX\to[0,\infty]$ a function such that $\psi\emptyset=0$ and
$\psi A\le\psi A'$ whenever $A\subseteq A'\subseteq X$.   Set

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

\noindent for $\delta>0$, and
$\theta_{\psi}A=\sup_{\delta>0}\theta_{\psi\delta}A$ for $A\subseteq X$.
Show that $\theta_{\psi}$ is a metric outer measure.   Let
$\mu_{\psi}$ be the measure defined from $\theta_{\psi}$ by
\Caratheodory's method.
%471A

\spheader 471Yb Suppose that
$\psi A=\inf\{\psi E:E$ is a Borel set including $A\}$ for every
$A\subseteq X$.   Show that $\theta_{\psi}=\mu_{\psi}^*$ and that
$\mu_{\psi}E=\sup\{\mu_{\psi}F:F\subseteq E$ is closed$\}$ whenever
$\mu_{\psi}E<\infty$.
%471D, 471Ya

\spheader 471Yc Suppose that $X$ is separable and that $\beta>0$ is
such that $\psi A^{\sim}\le\beta\psi A$ for every $A\subseteq X$, where
$A^{\sim}$ is defined in 471M.   (i) Suppose that $A\subseteq X$ and
$\Cal F$ is a family of closed subsets of $X$ such that
$\sum_{n=0}^{\infty}\psi F_n$ is finite for every disjoint sequence
$\sequencen{F_n}$ in $\Cal F$ and for every $x\in A$, $\delta>0$ there
is an $F\in\Cal F$ such that $x\in F$ and $0<\diam F\le\delta$.   Show
that there is a disjoint family $\Cal I\subseteq\Cal F$ such that
$\mu_{\psi}(A\setminus\bigcup\Cal I)=0$.   (ii) Suppose that $\delta>0$
and that $\theta_{\psi\delta}(X)<\infty$.   Show that there is a
non-negative additive functional $\nu$ on $\Cal PX$ such that
$\nu X=\bover1{\beta}\theta_{\psi\delta}(X)$ and $\nu A\le\psi A$
whenever $A\subseteq X$ and $\diam A\le\bover15\delta$.
(iii) Now suppose that for every $x\in X$ and $\epsilon>0$ there is a
$\delta>0$ such that $\psi B(x,\delta)\le\epsilon$.
Show that if $X$ is compact and $\mu_{\psi}X>0$ there is a
compact set $K\subseteq X$ such that $0<\mu_{\psi}K<\infty$.
%471O, 471Q, 471R, 471Ya  mt43bits mt47bits

\spheader 471Yd State and prove a version of 471P appropriate to this
context.
%471P, 471Ya, 471Yc

\spheader 471Ye Give an example of a set $A\subseteq\BbbR^2$ which is
measured by Hausdorff 1-dimensional measure on $\BbbR^2$ but is such
that its projection onto the first coordinate is not measured by
Hausdorff 1-dimensional measure on $\Bbb R$.
%471Xf, 471J, 439H

\spheader 471Yf\dvAnew{2016} Let $\rho$ be a metric on $\Bbb R$ inducing
the usual topology.   Show that the corresponding Hausdorff dimension of
$\Bbb R$ is at least $1$.
%471Xh 471L mt47bits

\spheader 471Yg Show that the space $(X,\rho)$ of 471Xa can be
isometrically embedded
as a subset of a metric space $(Y,\sigma)$ in such a way that (i)
$\diam B(y,\delta)=2\delta$ for every $y\in Y$ and $\delta\ge 0$ (ii)
$Y\setminus X$ is countable (iii) if $\mu_{H1}$ is
one-dimensional Hausdorff measure on $Y$, then
$\mu_{H1}B(y,\delta)\in\{0,\delta\}$ for every $y\in Y$ and $\delta\ge 0$,
so that

\Centerline{$\lim_{\delta\downarrow 0}
  \Bover{\mu_{H1}B(y,\delta)}{\diam B(x,\delta)}
\in\{0,\bover12\}$}

\noindent for every $y\in Y$.
%471P  Y = (X\times\{0\})\cup(A\times\Bbb Q) , max metric

\spheader 471Yh Let $\sequencen{k_n}$ be a sequence in
$\Bbb N\setminus \{0,1,2,3\}$ such that
$\sum_{n=0}^{\infty}\Bover1{k_n}<\infty$.   Set
$X=\prod_{n\in\Bbb N}k_n$.
% (regarding each $k_n$ as the set of its predecessors).   %471G
Set $m_0=1$, $m_{n+1}=k_0k_1\ldots k_n$ for
$n\in\Bbb N$.   Define a metric $\rho$ on $X$ by saying that

$$\eqalign{\rho(x,y)
&=1/2m_n\text{ if }n=\min\{i:x(i)\ne y(i)\}
  \text{ and }\min(x(n),y(n))=0,\cr
&=1/m_n\text{ if }n=\min\{i:x(i)\ne y(i)\}
  \text{ and }\min(x(n),y(n))>0.\cr}$$

\noindent Let $\nu$ be the product measure on $X$ obtained by giving
each factor $k_n$ the uniform probability measure in which each
singleton set has measure $1/k_n$.   (i) Show that if $A\subseteq X$
then $\nu^*A\le\diam A$.   (ii) Show that $\nu$ is one-dimensional
Hausdorff measure on $X$.   (iii) Set
$E=\bigcup_{n\in\Bbb N}\{x:x\in X,\,x(n)=0\}$.   Show that $\nu E<1$.
(iv) Show that

\Centerline{$\limsup_{\delta\downarrow 0}
\Bover{\nu(E\cap B(x,\delta))}{\nu B(x,\delta)}\ge\Bover12$}

\noindent for every $x\in X$.   (v) Show that there is a family $\Cal F$
of closed balls in $X$ such that every point of $X$ is the centre of
arbitrarily small members of $\Cal F$, but $\nu(\bigcup\Cal I)<1$ for
any disjoint subfamily $\Cal I$ of $\Cal F$.
%471P   mt47bits

\spheader 471Yi Let $(X,\rho)$ be a metric space and $0<r<s$.
Suppose that there is an analytic set $A\subseteq X$ such that
$\mu_{Hs}A>0$.
Show that there is a Borel surjection $f:X\to\Bbb R$ such that
$\mu_{Hr}f^{-1}[\{\alpha\}]=\infty$ for every $\alpha\in\Bbb R$.
%471S 471H mt47bits

\spheader 471Yj\dvArevised{2016}
Let $\rho$ be the metric on $\{0,1\}^{\Bbb N}$ defined
in 471Xa.   (i)
Show that for any integer $k\ge 1$ there are a $\gamma_k>0$ and a bijection
$f:[0,1]^k\to\{0,1\}^{\Bbb N}$ such that whenever $0<r\le 1$,
$\mu_{H,rk}$ is Hausdorff $rk$-dimensional
measure on $[0,1]^k$ (for its usual metric) and
$\tilde\mu_{Hr}$ is Hausdorff $r$-dimensional measure on
$\{0,1\}^{\Bbb N}$, then
$\mu_{H,rk}^*A\le\gamma_k\tilde\mu^*_{Hr}f[A]\le\gamma_k^2\mu_{H,rk}^*A$
for every $A\subseteq[0,1]^k$.   (ii) Show that in this case
$\mu_{H,rk}$ and the image measure $\tilde\mu_{Hr}f^{-1}$ have the same
measurable sets, the same negligible sets and the same sets of finite
measure.
%+ mt47bits maybe \gamma=2^{3k/2} will do

\spheader 471Yk\dvArevised{2016}
Let $(X,\rho)$ be a metric space, and $r>0$.   Give
$X\times\Bbb R$ the metric $\sigma$ where
$\sigma((x,\alpha),(y,\beta))=\max(\rho(x,y),|\alpha-\beta|)$.   Write
$\mu_L$, $\mu_r$ and
$\mu_{r+1}$ for Lebesgue measure on $\Bbb R$, $r$-dimensional Hausdorff
measure on $(X,\rho)$ and $(r+1)$-dimensional Hausdorff measure on
$(X\times\Bbb R,\sigma)$
respectively.   Let $\lambda$ be the c.l.d.\ product of $\mu_r$ and
$\mu_L$.
(i) Show that if $W\subseteq X\times\Bbb R$ then
$\overline{\int}\mu_r^*W^{-1}[\{\alpha\}]d\alpha\le\mu_{r+1}^*W$.
(ii) Show that if $I\subseteq\Bbb R$ is a bounded interval,
$A\subseteq X$ and $\mu_r^*A$ is finite, then
$\mu^*_{r+1}(A\times I)=\mu_r^*A\cdot\mu_LI$.
(iii) Give an example in which there is a compact set
$K\subseteq X\times\Bbb R$
such that $\mu_{r+1}K=1$ and $\lambda K=0$.
(iv) Show that if $\mu_r$ is $\sigma$-finite then $\mu_{r+1}=\lambda$.
\Hint{{\smc Federer 69}, 2.10.45.}
%mt47bits
}%end of exercises

\leader{471Z}{Problems (a)} Let $\mu^{(2)}_{H1}$, $\mu^{(1)}_{H,1/2}$
be one-dimensional Hausdorff measure on $\Bbb R^2$ and
$\bover12$-dimensional Hausdorff measure on $\Bbb R$ respectively, for
their usual metrics.
Are the measure spaces $(\Bbb R^2,\mu^{(2)}_{H1})$ and
$(\Bbb R,\mu^{(1)}_{H,1/2})$ isomorphic?   (See 471Yj.)

\spheader 471Zb Let $\rho$ be a metric on $\Bbb R^2$ inducing the usual
topology, and $\mu^{(\rho)}_{H2}$ the corresponding $2$-dimensional
Hausdorff measure.   Is it necessarily the case that 
$\mu^{(\rho)}_{H2}(\Bbb R^2)>0$?
(See 471Yf.)

\endnotes{
\Notesheader{471} In the exposition above, I have worked throughout with
simple $r$-dimensional measures for $r>0$.   As noted in 264Db, there
are formulae in which it is helpful to interpret $\mu_{H0}$ as counting
measure.   More interestingly, when we come to use Hausdorff measures to
give us information about the geometric structure of an object (e.g., in
the observation that the Cantor set has $\ln 2/\ln 3$-dimensional
Hausdorff measure $1$, in 264J), it is sometimes useful to refine the
technique by using other functionals than $A\mapsto(\diam A)^r$ in the
basic formulae of 264A or 471A.   The most natural generalization is to
functionals of the form $\psi A=h(\diam A)$ where
$h:[0,\infty]\to[0,\infty]$ is a non-decreasing function (264Yo).   But
it is easy to see that many of the arguments are valid in much greater
generality, as in 471Ya-471Yc.   %471Ya 471Yb 471Yc
For more in these directions see {\smc Rogers 70} and {\smc Federer 69}.

In the context of this book, the most conspicuous peculiarity of
Hausdorff measures is that they are often very far from being
semi-finite.   (This is trivial for non-separable spaces, by 471Df.
That Hausdorff one-dimensional measure on a subset of $\BbbR^2$ can be
purely infinite is not I think obvious;  I gave an example in 439H.)
The response I ordinarily recommend in such cases is to take the c.l.d.\
version.   But then of course we need to know just what effect this will
have.   In geometric applications, one usually begins by checking that
the sets one is interested in have $\sigma$-finite measure, and that
therefore no problems arise;  but it is a striking fact that Hausdorff
measures behave relatively well on analytic sets, even when not
$\sigma$-finite, provided we ask exactly the right questions (471I,
471S, 471Xk).

The geometric applications of Hausdorff measures, naturally, tend to
rely heavily on density theorems;  it is therefore useful to know that
we have effective versions of Vitali's theorem available in this context
(471N-471O), leading to a general density theorem (471P) similar to that
in 261D;  see also 472D below.   I note that 471P is useful only
after we have
been able to concentrate our attention on a set of finite measure.   And
traps remain.   For instance, the formulae of 261C-261D cannot be
transferred to the present context without re-evaluation (471Yh).

This section, and indeed the chapter as a whole, is devoted to calculations
involving metrics, which is why the phrase `metric space' is constantly
repeated while the word `metrizable' does not appear.   But of course
topological ideas are omnipresent.   See 471Yf for an interesting
elementary fact with an obvious implied challenge (471Zb).   
There is a less elementary fact in 471Yj, which shows that 
much of the measure space structure, if not the
geometry, of Hausdorff $r$-dimensional measure on $\BbbR^k$ is determined
by the ratio $r/k$.   (See 471Za.)
}%end of notes

\discrpage

\frfilename{mt472.tex}
\versiondate{22.3.11}
\copyrightdate{2000}

\def\chaptername{Geometric measure theory}
\def\sectionname{Besicovitch's Density Theorem}

\newsection{472}

The first step in the program of the next few sections
is to set out some very remarkable properties of
Euclidean space.   We find that in $\BbbR^r$, for geometric reasons
(472A), we have versions of Vitali's theorem (472B-472C) and Lebesgue's
Density Theorem (472D) for arbitrary Radon measures.   I add a version
of the Hardy-Littlewood Maximal Theorem (472F).

Throughout the section, $r\ge 1$ will be a fixed integer.   As usual, I
write $B(x,\delta)$ for the closed ball with centre $x$ and radius
$\delta$.   $\|\,\|$ will represent the Euclidean norm, and
$x\dotproduct y$ the scalar product of $x$ and $y$, so that
$x\dotproduct y=\sum_{i=1}^r\xi_i\eta_i$ if $x=(\xi_1,\ldots,\xi_r)$ and
$y=(\eta_1,\ldots,\eta_r)$.

\leader{472A}{Besicovitch's Covering Lemma} Suppose that $\epsilon>0$ is
such that $(5^r+1)(1-\epsilon-\epsilon^2)^r>(5+\epsilon)^r$.   Let
$x_0,\ldots,x_n\in\BbbR^r$, $\delta_0,\ldots,\delta_n>0$ be such that

\Centerline{$\|x_i-x_j\|>\delta_i$,
\quad$\delta_j\le(1+\epsilon)\delta_i$}

\noindent whenever $i<j\le n$.   Then

\Centerline{$\#(\{i:i\le n,\,\|x_i-x_n\|\le\delta_i+\delta_n\})
\le 5^r$.}

\proof{ Set $I=\{i:i\le n,\,\|x_i-x_n\|\le\delta_n+\delta_i\}$.

\medskip

{\bf (a)} It will simplify the formulae of the main argument if we
suppose for the time being that $\delta_n=1$;  in this case
$1\le(1+\epsilon)\delta_i$, so that $\delta_i\ge\Bover1{1+\epsilon}$
for every $i\le n$, while we still have $\delta_i<\|x_i-x_n\|$ for every
$i<n$, and $\|x_i-x_n\|\le 1+\delta_i$ for every $i\in I$.

For $i\in I$, define $x'_i$ by saying that

\inset{-- if $\|x_i-x_n\|\le 2+\epsilon$, $x_i'=x_i$;}

\inset{-- if $\|x_i-x_n\|>2+\epsilon$, $x'_i$ is to be that point of the
closed line segment from $x_n$ to $x_i$ which is at distance
$2+\epsilon$ from $x_n$.}

\medskip

{\bf (b)} The point is that $\|x'_i-x'_j\|>1-\epsilon-\epsilon^2$
whenever $i$, $j$ are distinct members of $I$.   \Prf\ We may suppose
that $i<j$.

\medskip

\quad{\bf case 1} Suppose that $\|x_i-x_n\|\le 2+\epsilon$ and
$\|x_j-x_n\|\le 2+\epsilon$.   In this case

\Centerline{$\|x'_i-x'_j\|
=\|x_i-x_j\|\ge\delta_i\ge\Bover1{1+\epsilon}\ge 1-\epsilon$.}

\medskip

\quad{\bf case 2} Suppose that $\|x_i-x_n\|\ge 2+\epsilon
\ge\|x_j-x_n\|$.   In this case

$$\eqalign{\|x'_i-x'_j\|
&=\|x'_i-x_j\|
\ge\|x_i-x_j\|-\|x_i-x'_i\|\cr
&\ge\delta_i-\|x_i-x_n\|+2+\epsilon
\ge\delta_i-\delta_i-1+2+\epsilon
=1+\epsilon.\cr}$$

\medskip

\quad{\bf case 3} Suppose that $\|x_i-x_n\|\le 2+\epsilon
\le\|x_j-x_n\|$.   Then

$$\eqalignno{\|x'_i-x'_j\|
&=\|x_i-x'_j\|
\ge\|x_i-x_j\|-\|x_j-x'_j\|
>\delta_i-\|x_j-x_n\|+2+\epsilon\cr
&\ge\delta_i-\delta_j-1+2+\epsilon
\ge\delta_i-\delta_i(1+\epsilon)+1+\epsilon
\ge 1+\epsilon-\epsilon(2+\epsilon)\cr
\noalign{\noindent (because $\delta_i<\|x_i-x_n\|\le 2+\epsilon$)}
&=1-\epsilon-\epsilon^2.\cr}$$

\medskip

\quad{\bf case 4} Suppose that $2+\epsilon\le\|x_j-x_n\|
\le\|x_i-x_n\|$. Let $y$ be the point on the line segment between $x_i$
and $x_n$ which is the same distance from $x_n$ as $x_j$.   In this case

\Centerline{$\|y-x_j\|
\ge\|x_i-x_j\|-\|x_i-y\|
\ge\delta_i-\|x_i-x_n\|+\|x_j-x_n\|
\ge\|x_j-x_n\|-1$.}

\noindent Because the triangles $(x_n,y,x_j)$ and $(x_n,x'_i,x'_j)$ are
similar,

\Centerline{$\|x'_i-x'_j\|=\Bover{2+\epsilon}{\|x_j-x_n\|}\|y-x_j\|
\ge(2+\epsilon)\Bover{\|x_j-x_n\|-1}{\|x_j-x_n\|}\ge 1+\epsilon$}

\noindent because $\|x_j-x_n\|\ge 2+\epsilon$.

\medskip

\quad{\bf case 5} Suppose that $2+\epsilon\le\|x_i-x_n\|\le\|x_j-x_n\|$.
This time, let $y$ be the point on the line segment from $x_n$ to $x_j$
which is the same distance from $x_n$ as $x_i$ is.   We now have

$$\eqalign{\|y-x_i\|
&\ge\|x_i-x_j\|-\|x_j-y\|
>\delta_i-\|x_j-x_n\|+\|x_i-x_n\|\cr
&\ge\delta_j-\epsilon\delta_i-(\delta_j+1)+\|x_i-x_n\|\cr
&=\|x_i-x_n\|-1-\epsilon\delta_i
\ge\|x_i-x_n\|(1-\epsilon)-1,\cr}$$

\noindent so that

$$\eqalign{\|x'_i-x'_j\|
&=\Bover{2+\epsilon}{\|x_i-x_n\|}\|y-x_i\|
>(2+\epsilon)\Bover{\|x_i-x_n\|(1-\epsilon)-1}{\|x_i-x_n\|}\cr
&\ge(2+\epsilon)\Bover{(2+\epsilon)(1-\epsilon)-1}{2+\epsilon}
=1-\epsilon-\epsilon^2.\cr}$$

So we have the required inequality in all cases.\ \Qed

\medskip

{\bf (c)} Now consider the balls
$B(x'_i,\bover{1-\epsilon-\epsilon^2}2)$ for $i\in I$.   These are
disjoint, all have Lebesgue measure
$2^{-r}\beta_r(1-\epsilon-\epsilon^2)^r$ where $\beta_r$ is the measure
of the unit ball $B(\tbf{0},1)$, and are all included in the ball
$B(x_n,2+\epsilon+\bover{1-\epsilon}2)$, which has measure
$2^{-r}\beta_r(5+\epsilon)^r$.   So we must have

\Centerline{$2^{-r}\beta_r(1-\epsilon-\epsilon^2)^r\#(I)
\le 2^{-r}\beta_r(5+\epsilon)^r$.}

\noindent But $\epsilon$ was declared to be so small that this implies
that $\#(I)\le 5^r$, as claimed.

\medskip

{\bf (d)} This proves the lemma in the case $\delta_n=1$.   For the
general case, replace each $x_i$ by $\delta_n^{-1}x_i$ and each
$\delta_i$ by $\delta_i/\delta_n$;  the change of scale does not affect
the hypotheses or the set $I$.
}%end of proof of 472A

\leader{472B}{Theorem} Let $A\subseteq\BbbR^r$ be a bounded set,
and $\Cal I$ a family of non-trivial closed balls in $\BbbR^r$ such
that every point of $A$ is the centre of a member of $\Cal I$.   Then
there is a family $\langle\Cal I_k\rangle_{k<5^r}$ of countable
subsets of $\Cal I$ such that each $\Cal I_k$ is disjoint and
$\bigcup_{k<5^r}\Cal I_k$ covers $A$.

\proof{{\bf (a)} For each $x\in A$ let $\delta_x>0$ be such that
$B(x,\delta_x)\in\Cal I$.   If either $A$ is empty or $\sup_{x\in
A}\delta_x=\infty$, the result is trivial.   (In the latter case, take
$x\in A$ such that $\delta_x\ge\diam A$ and set
$\Cal I_0=\{B(x,\delta_x)\}$, $\Cal I_k=\emptyset$ for $k>0$.)   So let us
suppose henceforth that $\{\delta_x:x\in A\}$ is bounded in $\Bbb R$.
In this case, $C=\bigcup_{x\in A}B(x,\delta_x)$ is bounded in $\Bbb R^r$.

Fix $\epsilon>0$ such that
$(5^r+1)(1-\epsilon-\epsilon^2)^r>(5+\epsilon)^r$.

\medskip

{\bf (b)} Choose inductively a sequence $\sequencen{B_n}$ in
$\Cal I\cup\{\emptyset\}$ as follows.
Given $\langle B_i\rangle_{i<n}$, then
if $A\subseteq\bigcup_{i<n}B_i$ set $B_n=\emptyset$.   Otherwise, set
$\alpha_n=\sup\{\delta_x:x\in A\setminus\bigcup_{i<n}B_i\}$, choose
$x_n\in A\setminus\bigcup_{i<n}B_i$ such that
$(1+\epsilon)\delta_{x_n}\ge\alpha_n$, set $B_n=B(x_n,\delta_{x_n})$ and
continue.

Now whenever $n\in\Bbb N$, $I_n=\{i:i<n,\,B_i\cap B_n\ne\emptyset\}$ has
fewer than $5^r$ members.   \Prf\ We may suppose that $B_n\ne\emptyset$,
in which case $B_i=B(x_i,\delta_{x_i})$ for every $i\le n$, and the
$x_i$, $\delta_{x_i}$ are such that, whenever $i<j\le n$,

\Centerline{$x_j\notin B_i$, i.e., $\|x_i-x_j\|>\delta_{x_i}$,}

\Centerline{$\delta_{x_j}\le\alpha_i\le(1+\epsilon)\delta_{x_i}$.}

\noindent But now 472A gives the result at once.
\Qed

\medskip

{\bf (c)} We may therefore define a function
$f:\Bbb N\to\{0,1,\ldots,5^r-1\}$ by setting

\Centerline{$f(n)=\min\{k:0\le k<5^r,\,f(i)\ne k$ for every $i\in
I_n\}$}

\noindent for every $n\in\Bbb N$.   Set $\Cal I_k=\{B_i:i\in\Bbb
N,\,f(i)=k,\,B_i\ne\emptyset\}$  for each $k<5^r$.   By the choice of
$f$, $i\notin I_j$, so that $B_i\cap B_j=\emptyset$, whenever $i<j$ and
$f(i)=f(j)$;  thus every $\Cal I_k$ is disjoint.   Since $B_i\subseteq
C$ for every $i$, $\sum\{\mu B_i:f(i)=k\}\le\mu^*C$ for every $k<5^r$,
and $\sum_{i=0}^{\infty}\mu B_i\le 5^r\mu^*C$ is finite.

\medskip

{\bf (d)} \Quer\ Suppose, if possible, that

\Centerline{$A\not\subseteq\bigcup_{k<5^r}\bigcup\Cal
I_k=\bigcup_{n\in\Bbb N}B_n$.}

\noindent Take $x\in A\setminus\bigcup_{n\in\Bbb N}B_n$.   Then, first,
$A\not\subseteq\bigcup_{i<n}B_i$ for every $n$, so that $\alpha_n$ is
defined;  next, $\alpha_n\ge\delta_x$, so that
$(1+\epsilon)\delta_{x_n}\ge\delta_x$ for every $n$.   But this means
that $\mu B_n\ge\beta_r\bigl(\Bover{\delta_x}{1+\epsilon}\bigr)^r$
for every $n$, and $\sum_{n=0}^{\infty}\mu B_n=\infty$;  which is
impossible.\ \Bang

\medskip

{\bf (e)} Thus $A\subseteq\bigcup_{k<5^r}\bigcup\Cal I_k$, as required.
}%end of proof of 472B

\leader{472C}{Theorem} Let $\lambda$ be a Radon measure on
$\BbbR^r$, $A$ a subset of $\BbbR^r$ and $\Cal I$ a family of
non-trivial closed
balls in $\BbbR^r$ such that every point of $A$ is the centre of
arbitrarily small members of $\Cal I$.   Then

(a) there is a countable disjoint $\Cal I_0\subseteq\Cal I$ such that
$\lambda(A\setminus\bigcup\Cal I_0)=0$;

(b) for every $\epsilon>0$ there is a countable
$\Cal I_1\subseteq\Cal I$ such that $A\subseteq\bigcup\Cal I_1$ and
$\sum_{B\in\Cal I_1}\lambda B\le\lambda^*A+\epsilon$.

\proof{{\bf (a)(i)} The first step is to show that if $A'\subseteq A$ is
bounded then there is a finite disjoint set $\Cal J\subseteq\Cal I$ such
that $\lambda^*(A'\cap\bigcup\Cal J)\ge 6^{-r}\lambda^*A'$.   \Prf\ If
$\lambda^*A'=0$ take $\Cal J=\emptyset$.   Otherwise, by 472B, there is
a
family $\langle\Cal J_k\rangle_{k<5^r}$ of disjoint countable subsets of
$\Cal I$ such that $\bigcup_{k<5^r}\Cal J_k$ covers $A'$.   Accordingly

\Centerline{$\lambda^*A'
\le\sum_{k=0}^{5^r-1}\lambda^*(A'\cap\bigcup\Cal J_k)$}

\noindent and there is some $k<5^r$ such that
$\lambda^*(A'\cap\bigcup\Cal J_k)\ge 5^{-r}\lambda^*A'$.   Let
$\sequence{i}{B_i}$ be a sequence running
over $\Cal J_k$;  then

\Centerline{$\lim_{n\to\infty}\lambda^*(A'\cap\bigcup_{i\le n}B_i)
=\lambda^*(A'\cap\bigcup\Cal J_k)\ge 5^{-r}\lambda^*A'$,}

\noindent so there is some $n\in\Bbb N$ such that
$\lambda^*(A'\cap\bigcup_{i\le n}B_i)\ge 6^{-r}\lambda^*A'$, and we can
take
$\Cal J=\{B_i:i\le n\}$.\ \Qed

\medskip

\quad{\bf (ii)} Now choose $\sequencen{\Cal K_n}$ inductively, as
follows.   Start by fixing on a sequence $\sequencen{m_n}$ running over
$\Bbb N$ with cofinal repetitions.
Take $\Cal K_0=\emptyset$.   Given that $\Cal K_n$ is a finite disjoint
subset of $\Cal I$, set
$\Cal I'=\{B:B\in\Cal I,\,B\cap\bigcup\Cal K_n=\emptyset\}$,
$A_n=A\cap B(\tbf{0},m_n)\setminus\bigcup\Cal K_n$.
Because every point of $A$ is the centre of arbitrarily small members of
$\Cal I$, and $\bigcup\Cal K_n$ is closed, every member of $A_n$ is the
centre of (arbitrarily small) members of $\Cal I'$, and (i) tells us
that there is a finite disjoint set $\Cal J_n\subseteq\Cal I'$ such that
$\lambda^*(A_n\cap\bigcup\Cal J_n)\ge 6^{-r}\lambda^*A_n$.   Set
$\Cal K_{n+1}=\Cal K\cup\Cal J_n$, and continue.   At the end of the
induction, set $\Cal I_0=\bigcup_{n\in\Bbb N}\Cal K_n$;  because
$\sequencen{\Cal K_n}$ is non-decreasing and every $\Cal K_n$ is
disjoint, $\Cal I_0$ is disjoint, and of course
$\Cal I_0\subseteq\Cal I$.

The effect of this construction is to ensure that

$$\eqalignno{\lambda^*(A\cap B(\tbf{0},m_n)\setminus\bigcup\Cal K_{n+1})
&=\lambda^*(A_n\setminus\bigcup\Cal J_n)
=\lambda^*A_n-\lambda^*(A_n\cap\bigcup\Cal J_n)\cr
\noalign{\noindent (because $\bigcup\Cal J_n$ is a closed set, therefore
measured by $\lambda$)}
&\le(1-6^{-r})\lambda^*A_n\cr
&=(1-6^{-r})
  \lambda^*(A\cap B(\tbf{0},m_n)\setminus\bigcup\Cal K_n)\cr}$$

\noindent for every $n$.   So, for any $m\in\Bbb N$,

\Centerline{$\lambda^*(A\cap B(\tbf{0},m)\setminus\bigcup\Cal K_n)
\le\lambda^*(A\cap B(\tbf{0},m))
  (1-6^{-r})^{\#(\{j:j<n,m_j=m\})}
\to 0$}

\noindent as $n\to\infty$, and
$\lambda^*(A\cap B(\tbf{0},m)\setminus\bigcup\Cal I_0)=0$.   As $m$ is
arbitrary, $\lambda^*(A\setminus\bigcup\Cal I_0)=0$, as required.

\medskip

{\bf (b)(i)} Let $E\supseteq A$ be such that $\lambda E=\lambda^*A$, and
$H\supseteq E$ an open set such that
$\lambda H\le\lambda E+\bover12\epsilon$
(256Bb).   Set $\Cal I'=\{B:B\in\Cal I,\,B\subseteq H\}$.   Then every
point of $A$ is the centre of arbitrarily small members of $\Cal I'$, so
by (a) there is a disjoint family $\Cal I_0\subseteq\Cal I'$ such that
$\lambda(A\setminus\bigcup\Cal I_0)=0$.   Of course

\Centerline{$\sum_{B\in\Cal I_0}\lambda B
=\lambda(\bigcup\Cal I_0)\le\lambda H\le\lambda^*A+\bover12\epsilon$.}

\medskip

\quad{\bf (ii)} For $m\in\Bbb N$ set
$A_m=A\cap B(\tbf{0},m)\setminus\bigcup\Cal I_0$.   Then there is a
$\Cal J_m\subseteq\Cal I$, covering $A_m$, such that
$\sum_{B\in\Cal J_m}\lambda B\le 2^{-m-2}\epsilon$.   \Prf\  There is an
open set $G\supseteq A_m$
such that $\lambda G\le 5^{-r}2^{-m-2}\epsilon$.   Now
$\Cal I''=\{B:B\in\Cal I,\,B\subseteq G\}$ covers $A_m$, so there is a
family $\langle\Cal J_{mk}\rangle_{k<5^r}$ of disjoint countable
subfamilies of
$\Cal I''$ such that $\Cal J_m=\bigcup_{k<5^r}\Cal J_{mk}$ covers $A_m$.
For each $k$,

\Centerline{$\sum_{B\in\Cal J_{mk}}\lambda B=\lambda(\bigcup\Cal J_{mk})
\le\lambda G$,}

\noindent so

\Centerline{$\sum_{B\in\Cal J_m}\lambda B\le 5^r\lambda G
\le 2^{-m-2}\epsilon$.
\Qed}

\medskip

\quad{\bf (iii)} Setting
$\Cal I_1=\Cal I_0\cup\bigcup_{m\in\Bbb N}\Cal J_m$ we have a cover of
$A$ by members of $\Cal I$, and

$$\eqalign{\sum_{B\in\Cal I_1}\lambda B
&\le\sum_{B\in\Cal I_0}\lambda B
  +\sum_{m=0}^{\infty}\sum_{B\in\Cal J_m}\lambda B\cr
&\le\lambda^*A+\Bover12\epsilon+\sum_{m=0}^{\infty}2^{-m-2}\epsilon
=\lambda^*A+\epsilon.\cr}$$
}%end of proof of 472C

\leader{472D}{Besicovitch's Density Theorem} Let $\lambda$ be any Radon
measure on $\BbbR^r$.   Then, for any locally $\lambda$-integrable
real-valued function $f$,

(a) $f(y)
=\lim_{\delta\downarrow 0}\Bover1{\lambda B(y,\delta)}
  \int_{B(y,\delta)}fd\lambda$,

(b) $\lim_{\delta\downarrow 0}\Bover1{\lambda B(y,\delta)}
  \int_{B(y,\delta)}|f(x)-f(y)|\lambda(dx)=0$

\noindent for $\lambda$-almost every $y\in\BbbR^r$.

\cmmnt{\medskip

\noindent{\bf Remark} The theorem asserts that, for $\lambda$-almost
every $y$, limits of the form
$\lim_{\delta\downarrow 0}\Bover1{\lambda B(y,\delta)}\ldots$ are
defined;  \cmmnt{in my usage,} this includes the assertion that
$\lambda B(y,\delta)\ne 0$ for all sufficiently small $\delta>0$.
}

\proof{ (Compare 261C and 261E.)

\medskip

{\bf (a)} Let $Z$ be the support of $\lambda$ (411Nd);  then $Z$ is
$\lambda$-conegligible and $\lambda B(y,\delta)>0$ whenever $y\in Z$ and
$\delta>0$.   For $q<q'$ in $\Bbb Q$ and $n\in\Bbb N$ set

\Centerline{$A_{nqq'}=\{y:y\in Z\cap\dom f,\,\|y\|<n,\,f(y)\le q,\,
  \limsup_{\delta\downarrow 0}\Bover1{\lambda B(y,\delta)}
  \biggerint_{B(y,\delta)}fd\lambda>q'\}$.}

\noindent Then $\lambda A_{nqq'}=0$.   \Prf\ Let $\epsilon>0$.   Then
there is an
$\eta\in\ocint{0,\epsilon}$ such that $\int_F|f|d\lambda\le\epsilon$
whenever $F\subseteq B(\tbf{0},n)$ and $\lambda F\le\eta$ (225A).   Let
$E$ be a measurable envelope of $A_{nqq'}$ included in
$\{y:y\in Z\cap\dom f,\,f(y)\le q,\,\|y\|<n\}$, and take an open set
$G\supseteq E$ such that $G\subseteq B(\tbf{0},n)$ and
$\lambda(G\setminus E)\le\eta$ (256Bb again).
Let $\Cal I$ be the family of
non-singleton closed balls $B\subseteq G$ such that
$\int_Bf\ge q'\lambda B$.   Then every point of $A_{nqq'}$ is the centre of arbitrarily small members of
$\Cal I$, so there is a disjoint family $\Cal I_0\subseteq\Cal I$ such
that $\lambda(A_{nqq'}\setminus\bigcup\Cal I_0)=0$ (472C).   Now
$\lambda(E\setminus\bigcup\Cal I_0)=0$ and
$\lambda((\bigcup\Cal I_0)\setminus E)\le\eta\le\epsilon$, so

$$\eqalign{q'\lambda E
&\le q'\lambda(\bigcup\Cal I_0)+\epsilon|q'|
=\sum_{B\in\Cal I_0}q'\lambda B+\epsilon|q'|\cr
&\le\sum_{B\in\Cal I_0}\int_Bfd\lambda+\epsilon|q'|
=\int_{\bigcup\Cal I_0}fd\lambda+\epsilon|q'|\cr
&\le\int_Efd\lambda+\epsilon(1+|q'|)
\le q\lambda E+\epsilon(1+|q'|),\cr}$$

\noindent and

\Centerline{$(q'-q)\lambda^*A_{nqq'}=(q'-q)\lambda E
\le(1+|q'|)\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\lambda^*A_{nqq'}=0$.\ \Qed

As $n$, $q$ and $q'$ are arbitrary,

\Centerline{$\limsup_{\delta\downarrow 0}\Bover1{\lambda B(y,\delta)}
 \biggerint_{B(y,\delta)}f\le f(y)$}

\noindent for $\lambda$-almost every $y\in Z$, therefore for
$\lambda$-almost every $y\in\BbbR^r$.   Similarly, or applying the
same argument to $-f$,

\Centerline{$\liminf_{\delta\downarrow 0}\Bover1{\lambda B(y,\delta)}
 \biggerint_{B(y,\delta)}f\ge f(y)$}

\noindent for $\lambda$-almost every $y$, and

\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\lambda B(y,\delta)}
 \biggerint_{B(y,\delta)}f$ exists $=f(y)$}

\noindent for $\lambda$-almost every $y$.

\medskip

{\bf (b)} Now, for each $q\in\Bbb Q$ set
$g_q(x)=|f(x)-q|$ for $x\in\dom f$.   By (a), we have a
$\lambda$-conegligible set $D$ such that

\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\lambda B(y,\delta)}
  \biggerint_{B(y,\delta)}g_qd\lambda=g_q(y)$}

\noindent for every $y\in D$ and $q\in\Bbb Q$.   Now, if
$y\in D$ and $\epsilon>0$, there is a $q\in\Bbb Q$ such that
$|f(y)-q|\le\epsilon$, and a $\delta_0>0$ such that

\Centerline{$|\Bover1{\lambda B(y,\delta)}
  \biggerint_{B(y,\delta)}g_qd\lambda-g_q(y)|
\le\epsilon$}

\noindent whenever $0<\delta\le\delta_0$.   But in this case

$$\eqalign{\Bover1{\lambda B(y,\delta)}
&\int_{B(y,\delta)}|f(x)-f(y)|\lambda(dx)\cr
&\le\Bover1{\lambda B(y,\delta)}\int_{B(y,\delta)}|f(x)-q|\lambda(dx)
  +\Bover1{\lambda B(y,\delta)}\int_{B(y,\delta)}|q-f(y)|\lambda(dx)\cr
&\le 3\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\lambda B(y,\delta)}
  \biggerint_{B(y,\delta)}|f(x)-f(y)|\lambda(dx)
=0$;}

\noindent as this is true for every $y\in D$, the theorem is proved.
}%end of proof of 472D

\leader{*472E}{Proposition} Let $\lambda$, $\lambda'$ be Radon
measures on $\BbbR^r$, and $G\subseteq\BbbR^r$ an open set.   Let $Z$
be the support of $\lambda$, and for $x\in Z\cap G$ set

\Centerline{$M(x)=\sup\{\Bover{\lambda'B}{\lambda B}:B\subseteq G$ is a
non-trivial ball with centre $x\}$.}

\noindent Then

\Centerline{$\lambda\{x:x\in Z,\,M(x)\ge t\}\le\Bover{5^r}t\lambda'G$}

\noindent for every $t>0$.

\proof{ The function
$M:Z\to[0,\infty]$ is lower semi-continuous.   \Prf\ If $M(x)>t\ge 0$,
there is a $\delta>0$ such that $B(x,\delta)\subseteq G$ and
$\lambda'B(x,\delta)>t\lambda B(x,\delta)$.   Because $\lambda$ is a
Radon measure, there is an open set $V\supseteq B(x,\delta)$ such that
$V\subseteq G$ and
$\lambda'B(x,\delta)>t\lambda V$;  because $B(x,\delta)$ is compact,
there is an $\eta>0$ such that $B(x,\delta+2\eta)\subseteq V$.   Now if
$y\in Z$ and $\|y-x\|\le\eta$,

\Centerline{$B(x,\delta)\subseteq B(y,\delta+\eta)\subseteq V$,}

\noindent so $\lambda'B(y,\delta+\eta)>t\lambda B(y,\delta+\eta)$ and
$M(y)>t$.\ \Qed

In particular, $H_t=\{x:x\in Z\cap G,\,M(x)>t\}$ is always measured by
$\lambda$.   Now, given $t>0$, let $\Cal I$ be the set of non-trivial
closed balls $B\subseteq G$ such that $\lambda'B>t\lambda B$.   By 472B,
there is a family $\ofamily{k}{5^r}{\Cal I_k}$ of countable disjoint
subsets of $\Cal I$ such that $\bigcup_{k<5^r}\Cal I_k$ covers $H_t$.
So

\Centerline{$\lambda H_t
\le\sum_{k=0}^{5^r-1}\sum_{B\in\Cal I_k}\lambda B
\le\Bover1t\sum_{k=0}^{5^r-1}\sum_{B\in\Cal I_k}\lambda'B
\le\Bover{5^r}t\lambda'G$,}

\noindent as claimed.
}%end of proof of 472E

\leader{*472F}{Theorem} Let $\lambda$ be a Radon measure on
$\BbbR^r$, and $f\in\eusm L^p(\lambda)$ any function, where
$1<p<\infty$.   Let $Z$ be the support of $\lambda$, and for $x\in Z$
set $f^*(x)=\sup_{\delta>0}\Bover1{\lambda B(x,\delta)}
  \int_{B(x,\delta)}|f|d\lambda$.   Then $f^*$ is lower semi-continuous,
and $\|f^*\|_p\le 2\bigl(\Bover{5^rp}{p-1}\bigr)^{1/p}\|f\|_p$.

\proof{{\bf (a)} Replacing $f$ by $|f|$ if necessary, we may suppose
that $f\ge 0$.   $Z$ is $\lambda$-conegligible, so that $f^*$
is defined $\lambda$-almost everywhere.   Next, $f^*$ is lower
semi-continuous.   \Prf\ I repeat an idea from the proof of 472E.
If $f^*(x)>t\ge 0$, there is a $\delta>0$ such that
$\int_{B(x,\delta)}|f|d\lambda>t\lambda B(x,\delta)$.   Because
$\lambda$ is a Radon measure, there is an open set
$V\supseteq B(x,\delta)$ such that
$\int_{B(x,\delta)}|f|d\lambda>t\lambda V$;  because $B(x,\delta)$ is
compact, there is an $\eta>0$ such that $B(x,\delta+2\eta)\subseteq V$;
and now $f^*(y)>t$ for every $y\in Z\cap B(x,\eta)$.\ \Qed

\medskip

{\bf (b)} For $t>0$, set
$H_t=\{x:x\in Z,\,f^*(x)>t\}$ and $F_t=\{x:x\in\dom f,\,f(x)\ge t\}$.
Then

\Centerline{$\lambda H_t
\le\Bover{2\cdot 5^r}{t}\biggerint_{F_{t/2}}fd\lambda$.}

\noindent\Prf\ Set $g=f\times\chi F_{t/2}$.   Because
$(\bover{t}2)^p\lambda F_{t/2}\le\|f\|_p^p$ is finite, $\lambda F_{t/2}$
is finite,  $\chi F_{t/2}\in\eusm L^q(\lambda)$ (where
$\bover1p+\bover1q=1$) and $g\in\eusm L^1(\lambda)$ (244Eb).
Let $\lambda'$ be the indefinite-integral measure defined by $g$ over
$\lambda$ (234J);
then $\lambda'$ is totally finite, and is a Radon measure (416Sa).   Set

\Centerline{$M(x)=\sup\{\Bover{\lambda'B}{\lambda B}:B\subseteq\BbbR^r$
is a non-trivial ball with centre $x\}$}

\noindent for $x\in Z$.   Then $f^*(x)\le M(x)+\bover{t}2$ for every
$x\in Z$, just because

\Centerline{$\biggerint_Bfd\lambda
\le\Bover{t}2\lambda B+\int_Bg\,d\lambda
=\Bover{t}2\lambda B+\lambda'B$}

\noindent for every closed ball $B$.   Accordingly

$$\eqalignno{\lambda H_t
&\le\lambda\{x:M(x)>\Bover{t}2\}
\le\Bover{2\cdot 5^r}{t}\lambda'\BbbR^r\cr
\displaycause{by 472E}
&=\Bover{2\cdot 5^r}t\biggerint_{F_{t/2}}fd\lambda. \text{ \Qed}\cr}$$

\medskip

{\bf (c)} As in part (c) of the proof of 286A, we now have

$$\eqalign{\int(f^*)^pd\lambda
&=\int_0^{\infty}\lambda\{x:f^*(x)^p>t\}dt
=p\int_0^{\infty}t^{p-1}\lambda\{x:f^*(x)>t\}dt\cr
&\le 2\cdot 5^rp\int_0^{\infty}t^{p-2}\int_{F_{t/2}}fd\lambda dt
=2\cdot 5^rp\int_{\Bbb R^r}f(x)\int_0^{2f(x)}t^{p-2}dt\lambda(dx)\cr
&=2\cdot 5^rp\int_{\Bbb R^r}\Bover{2^{p-1}}{p-1}f(x)^p\lambda(dx)
=\Bover{2^p5^rp}{p-1}\int f^pd\lambda.\cr}$$

\noindent Taking $p$th roots, we have the result.
}%end of proof of 472F

\exercises{\leader{472X}{Basic exercises (a)}
%\spheader 472Xa
Show that if $\lambda$, $\lambda'$ are Radon measures on $\Bbb R^r$
which agree on closed balls, they are equal.   (Cf.\ 466Xj.)
%472C

\spheader 472Xb Let $\lambda$ be a Radon measure on $\BbbR^r$.   Let
$A\subseteq\BbbR^r$ be a non-empty set, and
$\epsilon>0$.   Show that there is a sequence $\sequencen{B_n}$ of
closed balls in $\BbbR^r$, all of radius at most $\epsilon$ and with
centres in $A$, such that $A\subseteq\bigcup_{n\in\Bbb N}B_n$ and
$\sum_{n=0}^{\infty}\lambda B_n\le\lambda^*A+\epsilon$.
%472C

\spheader 472Xc Let $\lambda$ be a non-zero Radon measure on $\BbbR^r$
and $Z$ its support.
Show that we have a lower density $\phi$ (definition: 341C) for the
subspace measure $\lambda_Z$ defined by
setting $\phi E=\{x:x\in Z$, $\lim_{\delta\downarrow 0}
\Bover{\lambda(E\cap B(x,\delta))}{\lambda B(x,\delta)}=1\}$ whenever
$\lambda_Z$ measures $E$.
%472D

\spheader 472Xd\dvAnew{2012}
Let $\lambda$ be a Radon measure on $\BbbR^r$, and
$f$ a locally $\lambda$-integrable function.   Show that
$E=\{y:g(y)=\lim_{\delta\downarrow 0}\Bover1{\lambda B(y,\delta)}
\int_{B(y,\delta)}f\,d\lambda$ is defined in $\Bbb R\}$ is a Borel set, and
that $g:E\to\Bbb R$ is Borel measurable.
%472D

\leader{472Y}{Further exercises (a)}%
%\spheader 472Ya
(i) Let $\Cal I$ be a finite family of intervals (open,
closed or half-open) in $\Bbb R$.   Show that there are subfamilies
$\Cal I_0$, $\Cal I_1\subseteq\Cal I$, both disjoint, such that
$\Cal I_0\cup\Cal I_1$ covers $\bigcup\Cal I$.   \Hint{induce on
$\#(\Cal I)$.}    Show that this remains true if any totally ordered set
is put in place of $\Bbb R$.   (ii) Show that if $\Cal I$ is any family
of non-empty intervals in $\Bbb R$ such that none contains the centre
of any other, then $\Cal I$ is expressible as $\Cal I_0\cup\Cal I_1$
where both $\Cal I_0$ and $\Cal I_1$ are disjoint.
%472B

\spheader 472Yb Let $m=m(r)$ be the largest number such that there are
$u_1,\ldots,u_m\in\BbbR^r$ such that $\|u_i\|=1$ for every $i$ and
$\|u_i-u_j\|\ge 1$ for all $i\ne j$.   Let $A\subseteq\BbbR^r$ be a
bounded set and $x\mapsto\delta_x:A\to\ooint{0,\infty}$ a bounded
function;  set
$B_x=B(x,\delta_x)$ for $x\in A$.   (i) Show that $m<3^r$.
(ii) Show that there is an $\epsilon\in\ocint{0,\bover1{10}}$ such that
whenever $\|u_0\|=\ldots=\|u_m\|=1$ there are distinct $i$,
$j\le m$ such that $u_i\dotproduct u_j>\bover12(1+\epsilon)$.
(iii) Suppose that $u$, $v\in\BbbR^r$ are such that
$\bover13\le\|u\|\le 1$, $\|v\|\le 1+\epsilon$ and $\|u-v\|>1$.   Show
that the angle $u\widehat{\tbf{0}}v$ has cosine at most
$\bover12(1+\epsilon)$.   \Hint{maximise $\Bover{a^2+b^2-c^2}{2ab}$
subject to $\bover13\le a\le 1$, $b\le 1+\epsilon$ and $c\ge 1$.}   (iv)
Suppose that $\sequencen{x_n}$ is a
sequence in $A$ such that $x_n\notin B_{x_i}$ for $i<n$ and
$(1+\epsilon)\delta_{x_n}
\ge\sup\{\delta_x:x\in A\setminus\bigcup_{i<n}B_{x_i}\}$ for every $n$.
Show that $A\subseteq\bigcup_{n\in\Bbb N}B_{x_n}$.
(v) Take $y\in\BbbR^r$.   Show that there is at most one $n$ such that
$\|y-x_n\|\le\bover13\delta_{x_n}$.
(vi) Show that if $i<j$,
$\bover13\delta_{x_i}\le\|y-x_i\|\le\delta_{x_i}$ and
$\|y-x_j\|\le\delta_j$ then the cosine of the angle
$x_i\widehat{y}x_j$ is at most $\bover12(1+\epsilon)$.
(vii) Show that $\#(\{i:y\in B_{x_i}\})\le m+1$.

Hence show that if $\Cal I$ is any family of non-trivial closed balls
such that every point of $A$ is the centre of some member of $\Cal I$,
then there is a countable $\Cal I_0\subseteq\Cal I$, covering $A$, such
that no point of $\BbbR^r$ belongs to more than $3^r$ members of
$\Cal I_0$.
%472B

\spheader 472Yc Use 472Yb to prove an alternative version of 472B, but
with the constant $9^r+1$ in place of $5^r$.
%472Yb 472B

\spheader 472Yd\dvAnew{2010}
Let $A\subseteq\BbbR^r$ be a bounded set, and $\Cal I$ a
family of non-trivial closed balls in $\BbbR^r$ such that whenever $x\in A$
and $\epsilon>0$ there is a ball $B(y,\delta)\in\Cal I$ such that
$\|x-y\|\le\epsilon\delta$.   Show that there is a family
$\langle\Cal I_k\rangle_{k<5^r}$ of
subsets of $\Cal I$ such that each $\Cal I_k$ is disjoint and
$\bigcup_{k<5^r}\Cal I_k$ covers $A$.
%472B  query out of order

\spheader 472Ye\dvAnew{2010} Give an example of a strictly positive
Radon probability measure $\mu$ on a
compact metric space $(X,\rho)$ for which there is a Borel set
$E\subseteq X$ such that

\Centerline{$\liminf_{\delta\downarrow 0}
  \Bover{\mu(E\cap B(x,\delta))}{\mu B(x,\delta)}=0$,
\quad$\limsup_{\delta\downarrow 0}
  \Bover{\mu(E\cap B(x,\delta))}{\mu B(x,\delta)}=1$}

\noindent for every $x\in X$.
%472D mt47bits

\spheader 472Yf Let $\lambda$ be a Radon measure on $\BbbR^r$, and $f$ a
$\lambda$-integrable real-valued function.   Show that
$\sup_{\delta>0}\Bover1{\lambda B(x,\delta)}
  \int_{B(x,\delta)}|f|d\lambda$  is defined and finite for
$\lambda$-almost every $x\in\BbbR^r$.
%472F

\spheader 472Yg Let $\lambda$, $\lambda'$ be Radon measures on
$\Bbb R^r$.   (i) Show that $g(x)=\lim_{\delta\downarrow 0}
\Bover{\lambda'B(x,\delta)}{\lambda B(x,\delta)}$ is defined in
$\Bbb R$ for $\lambda$-almost every $x$.   (ii) Setting
$\lambda_0=\sup_{n\in\Bbb N}\lambda'\wedge n\lambda$ in the cone of Radon
measures on $\Bbb R^r$ (437Yi),
show that $g$ is a Radon-Nikod\'ym derivative of $\lambda_0$ with respect
to $\lambda$.
\Hint{show that if $\lambda$ and $\lambda'$ are mutually singular then
$g=0\,\,\lambda$-a.e.}
%472E
}%end of exercises

\endnotes{
\Notesheader{472}
I gave primacy to the `weak' Vitali's theorem in 261B because I
think it is easier than the `strong' form in 472C, it uses the same
ideas as the original one-dimensional theorem in 221A, and it is
adequate for the needs of Volume 2.   Any proper study of general
measures on $\BbbR^r$, however, will depend on the ideas in 472A-472C.
You will see that in 472B, as in other forms of Vitali's theorem, there
is a key step in which a sequence is chosen greedily.   This time we
must look much more carefully at the geometry of $\BbbR^r$ because we
can no longer rely on a measure to tell us what is happening.   (Though
you will observe that I still use the elementary properties of Euclidean
volume in the argument of 472A.)   Once we have reached 472C, however,
we are in a position to repeat all the arguments of
261C-261E %261C 261D 261E
in much greater generality (472D), and, as a bonus, can refine
261F (472Xb).   For more in this direction see {\smc Mattila 95} and
{\smc Federer 69}, \S2.8.

It is natural to ask whether the constant `$5^r$' in 472B is best
possible.   The argument of 472A is derived from {\smc Sullivan 94},
where a more thorough analysis is given.   It seems that even for $r=2$
the best constant is unknown.   (For $r=1$, the best constant is $2$;
see 472Ya.)   Note that even for finite families $\Cal I$ we should have
to find the colouring number of a graph (counting two balls as linked if
they intersect), so it may well be a truly difficult problem.   The
method in 472B amounts to using the greedy colouring algorithm after
ordering the balls by size, and one does not expect such approaches to
give exact colouring numbers.   Of
course the questions addressed here depend only on the existence of {\it
some} function of $r$ to do the job.

An alternative argument runs through a kind of pointwise version of 472A
(472Yb-472Yc).   It gives a worse constant but is attractive in other
ways.   For many of the applications of 472C, the result of 472Yb is
already sufficient.

The constant $2\bigl(\Bover{5^rp}{p-1}\bigr)^{1/p}$ in 472F makes no
pretence to be `best', or even `good'.   The only reason for giving a
formula at all is to emphasize the remarkable fact that it does not
depend on the measure $\lambda$.   The theorems of this section are
based on the metric geometry of Euclidean space, not on any special
properties of Lebesgue measure.   The constants {\it do} depend on the
dimension, so that even in Hilbert space (for instance) we cannot expect
any corresponding results.
}%end of notes

\discrpage


\frfilename{mt473.tex}
\versiondate{25.7.11/7.8.20}
\copyrightdate{2000}

\def\chaptername{Geometric measure theory}
\def\sectionname{Poincar\'e's inequality}

\newsection{473}

In this section I embark on the main work of the first half of the
chapter, leading up to
the Divergence Theorem in \S475.   I follow the method in
{\smc Evans \& Gariepy 92}.   The first step is to add some minor results on
differentiable and Lipschitz functions to those already set out in \S262
(473B-473C).   Then we need to know something about convolution products
(473D), extending ideas in \S\S256 and 444;  in particular, it will be
convenient to have a fixed sequence $\sequencen{\tilde h_n}$ of
smoothing functions with some useful special properties (473E).

The new ideas of the section begin with the Gagliardo-Nirenberg-Sobolev
inequality, relating $\|f\|_{r/(r-1)}$ to $\int\|\grad f\|$.   In its
simplest form (473H) it applies only to functions with compact support;
we need to work much harder to get results which we can use to estimate
$\int_B|f|^{r/(r-1)}$ in terms of $\int_B\|\grad f\|$ and $\int_B|f|$ for
balls $B$ (473I, 473K).

\vleader{48pt}{473A}{Notation} For the next three sections,
$r\ge 2$ will be a fixed integer.   For $x\in\BbbR^r$ and $\delta\ge 0$,
$B(x,\delta)=\{y:\|y-x\|\le\delta\}$ will be the closed ball with centre
$x$ and radius $\delta$.
I will write $\partial B(x,\delta)$ for the boundary of $B(x,\delta)$,
the sphere  $\{y:\|y-x\|=\delta\}$.
$S_{r-1}=\partial B(\tbf{0},1)$ will be the unit sphere.
As in Chapter 26, I will use Greek letters to represent coordinates of
vectors, so that $x=(\xi_1,\ldots,\xi_r)$, etc.

$\mu$ will always be Lebesgue measure on $\BbbR^r$.
$\beta_r=\mu B(\tbf{0},1)$ will be the $r$-dimensional volume of the
unit ball, that is,

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

\noindent (252Q).
$\nu$ will be normalized Hausdorff $(r-1)$-dimensional measure on
$\BbbR^r$, that is,
$\nu=2^{-r+1}\beta_{r-1}\mu_{H,r-1}$, where $\mu_{H,r-1}$ is
$(r-1)$-dimensional Hausdorff measure on $\BbbR^r$ as described in \S264.
Recall from 265F and 265H that
$\nu S_{r-1}=2\pi\beta_{r-2}=r\beta_r$ (counting $\beta_0$ as $1$).

\leader{473B}{Differentiable functions (a)} Recall from \S262 that a
function $\phi$ from a subset of $\BbbR^r$ to $\BbbR^s$ (where
$s\ge 1$) is differentiable at $x\in\BbbR^r$, with
derivative an $s\times r$ matrix $T$, if for every $\epsilon>0$ there
is a $\delta>0$ such that $\|\phi(y)-\phi(x)-T(y-x)\|\le\epsilon\|y-x\|$
whenever $\|y-x\|\le\delta$;  this includes the assertion that
$B(x,\delta)\subseteq\dom\phi$.   In this case, the coefficients of $T$
are the partial derivatives $\Pd{\phi_j}{\xi_i}(x)$ at $x$, where
$\phi_1,\ldots,\phi_s$ are the coordinate functions of $\phi$, and
$\Pd{}{\xi_i}$ represents partial differentiation with respect to the
$i$th coordinate\cmmnt{ (262Ic)}.

\spheader 473Bb When $s=1$\cmmnt{, so that we have a real-valued
function $f$
defined on a subset of $\BbbR^r$}, I will write $(\grad f)(x)$ for the
derivative of $f$ at $x$, the {\bf gradient} of $f$.   \cmmnt{If we
strictly adhere to the language of (a), $\grad f$ is a $1\times r$
matrix $\Matrix{\Pd{f}{\xi_1}&\ldots&\Pd{f}{\xi_r}}$;  but it is
convenient to
treat it as a vector, so that $\grad f(x)$ (when defined) belongs to
$\BbbR^r$, and we can speak of $y\dotproduct\grad f(x)$ rather
than $(\grad f(x))(y)$, etc.}

\spheader 473Bc {\bf Chain rule for functions of many
variables}\cmmnt{ I find that I have not written out the
following basic fact.}   Let $\phi:A\to\BbbR^s$ and
$\psi:B\to\BbbR^p$ be functions, where $A\subseteq\BbbR^r$ and
$B\subseteq\BbbR^s$.   If $x\in A$ is such that $\phi$ is
differentiable at $x$, with derivative $S$, and $\psi$ is differentiable
at $\phi(x)$, with derivative $T$, then the composition $\psi\phi$ is
differentiable at $x$, with derivative $TS$.

\prooflet{\Prf\ Recall that if we regard $S$ and $T$ as linear
operators, they have finite norms (262H).   Given $\epsilon>0$, let
$\eta>0$ be such that  $\eta\|T\|+\eta(\|S\|+\eta)\le\epsilon$.   Let
$\delta_1$, $\delta_2>0$ be such that
$\phi(y)$ is defined and $\|\phi(y)-\phi(x)-S(y-x)\|\le\eta\|y-x\|$
whenever $\|y-x\|\le\delta_1$, and $\psi(z)$ is defined and
$\|\psi(z)-\psi\phi(x)-T(z-\phi(x))\|\le\eta\|z-\phi(x)\|$ whenever
$\|z-\phi(x)\|\le\delta_2$.   Set
$\delta=\min(\delta_1,\Bover{\delta_2}{\eta+\|S\|})>0$.   If
$\|y-x\|\le\delta$, then $\phi(y)$ is defined and

\Centerline{$\|\phi(y)-\phi(x)\|\le\|S(y-x)\|+\|\phi(y)-\phi(x)-S(y-x)\|
\le(\|S\|+\eta)\|y-x\|\le\delta_2$,}

\noindent so $\psi\phi(y)$ is defined and

$$\eqalign{\|\psi\phi(y)&-\psi\phi(x)-TS(y-x)\|\cr
&\le\|\psi\phi(y)-\psi\phi(x)-T(\phi(y)-\phi(x))\|
     +\|T\|\|\phi(y)-\phi(x)-S(y-x)\|\cr
&\le\eta\|\phi(y)-\phi(x)\|+\|T\|\eta\|y-x\|\cr
&\le\eta(\|S\|+\eta)\|y-x\|+\|T\|\eta\|y-x\|
\le\epsilon\|y-x\|;\cr}$$

\noindent as $\epsilon$ is arbitrary, $\psi\phi$ is differentiable at
$x$ with derivative $TS$.\ \Qed}%end of prooflet

\spheader 473Bd It follows that if $f$ and $g$ are real-valued functions
defined on a
subset of $\BbbR^r$, and $x\in\dom f\cap\dom g$ is such that
$(\grad f)(x)$ and $(\grad g)(x)$ are both defined, then
$\grad(f\times g)(x)$ is defined and equal to
$f(x)\grad g(x)+g(x)\grad f(x)$.   \prooflet{\Prf\ Set
$\phi(y)=\Matrix{f(y)\\g(y)}$ for $y\in\dom f\cap\dom g$;  then $\phi$
is differentiable at $x$ with derivative the $2\times r$ matrix
$\Matrix{\grad f(x)\\ \grad g(x)}$ (262Ib).   Set
$\psi(z)=\zeta_1\zeta_2$ for $z=(\zeta_1,\zeta_2)\in\BbbR^2$;  then
$\psi$ is differentiable everywhere, with derivative the $1\times 2$
matrix $\Matrix{\zeta_2&\zeta_1}$.   So $f\times g=\psi\phi$ is
differentiable at $x$ with derivative

\Centerline{$\Matrix{g(x)&f(x)}\Matrix{\grad f(x)\\ \grad g(x)}
=g(x)\grad f(x)+f(x)\grad g(x)$.  \Qed}
}%end of prooflet

\spheader 473Be Let $D$ be a subset of $\BbbR^r$ and $\phi:D\to\BbbR^s$
any function.   Set $D_0=\{x:x\in D,\,\phi$ is differentiable at $x\}$.
Then the derivative of $\phi$, regarded as a function from $D_0$ to
$\BbbR^{rs}$, is (Lebesgue) measurable.
\prooflet{\Prf\ Use 262P;  the point is that, writing $T(x)$ for the
derivative of $\phi$ at $x$, $T(x)$ is surely a derivative of
$\phi\restr D_0$, relative to $D_0$, at every point of $D_0$.\ \QeD\
}%end of prooflet
\cmmnt{(See also 473Ya.)}

\spheader 473Bf If $G\subseteq\BbbR^r$ is an open set, a function
$\phi:G\to\BbbR^s$ is {\bf smooth} if
it is differentiable arbitrarily often;  that is, if all its repeated
partial derivatives

\Centerline{$\Bover{\partial^m\phi_j}{\partial
\xi_{i_1}\ldots\partial\xi_{i_m}}$}

\noindent are defined and continuous everywhere on $G$.
I will write $\eusm D$ for the family of real-valued functions
from $\BbbR^r$ to $\Bbb R$ which are smooth and have compact support.

\leader{473C}{Lipschitz functions (a)} If $f$ and
$g$ are bounded real-valued Lipschitz functions, defined on any subsets
of $\BbbR^r$, then $f\times g$, defined on $\dom f\cap\dom g$, is
Lipschitz.   \prooflet{\Prf\ Let $\gamma_f$, $M_f$, $\gamma_g$ and $M_g$
be such that $|f(x)|\le M_f$ and $|f(x)-f(y)|\le\gamma_f\|x-y\|$ for all
$x$, $y\in\dom f$, while $|g(x)|\le M_g$ and
$|g(x)-g(y)|\le\gamma_g\|x-y\|$ for all $x$, $y\in\dom g$.   Then for
any $x$, $y\in\dom f\cap\dom g$,

$$\eqalign{|f(x)g(x)-f(y)g(y)|
&\le|f(x)||g(x)-g(y)|+|g(y)||f(x)-f(y)|\cr
&\le (M_f\gamma_g+M_g\gamma_f)\|x-y\|.\cr}$$

\noindent So $M_f\gamma_g+M_g\gamma_f$ is a Lipschitz constant for
$f\times g$.\ \Qed}

\leaveitout{It follows that if $f$ and $g$ are bounded real-valued Lipschitz functions defined on $\BbbR^r$, and $g$ has compact support, then $f\times g$ is Lipschitz.}

\spheader 473Cb Suppose that $F_1$, $F_2\subseteq\BbbR^r$ are closed
sets with convex union $C$.   Let $f:C\to\Bbb R$ be a function such that
$f\restr F_1$ and $f\restr F_2$ are both Lipschitz.   Then $f$ is
Lipschitz.   \prooflet{\Prf\ For each $j$, let $\gamma_j$ be a Lipschitz
constant for $f\restr F_j$, and set $\gamma=\max(\gamma_1,\gamma_2)$, so
that $\gamma$ is a Lipschitz constant for both $f\restr F_1$ and
$f\restr F_2$.   Take any $x$, $y\in C$.   If both belong to the same
$F_j$, then
$|f(x)-f(y)|\le\gamma\|x-y\|$.   If $x\in F_j$ and $y\notin F_j$, then
$y$ must belong to $F_{3-j}$, and $(1-t)x+ty\in F_1\cup F_2$ for every
$t\in[0,1]$, because $C$ is convex.   Set
$t_0=\sup\{t:t\in[0,1],\,(1-t)x+ty\in F_j\}$,
$z=(1-t_0)x+t_0y$;  then $z\in F_1\cap F_2$, because both are closed, so

\Centerline{$|f(x)-f(y)|\le|f(x)-f(z)|+|f(z)-f(y)|
\le\gamma\|x-z\|+\gamma\|z-y\|=\gamma\|x-y\|$.}

\noindent As $x$ and $y$ are arbitrary, $\gamma$ is a Lipschitz constant
for $f$.\ \Qed}

\spheader 473Cc Suppose that $f:\BbbR^r\to\Bbb R$ is Lipschitz.
\cmmnt{Recall that by Rademacher's theorem (262Q),} $\grad f$ is
defined almost everywhere.   \cmmnt{All the partial
derivatives of $f$ are (Lebesgue) measurable, by 473Be, so} $\grad f$ is
(Lebesgue) measurable on its domain.   If $\gamma$ is a Lipschitz
constant for $f$, $\|\grad f(x)\|\le\gamma$ whenever $\grad f(x)$ is
defined.   \prooflet{\Prf\ If $z\in\BbbR^r$, then

\Centerline{$\lim_{t\downarrow 0}
  \Bover1t|f(x+tz)-f(x)-tz\dotproduct\grad f(x)|=0$,}

\noindent so

\Centerline{$|z\dotproduct\grad f(x)|
=\lim_{t\downarrow 0}\Bover1t|f(x+tz)-f(x)|
\le\gamma\|z\|$;}

\noindent as $z$ is arbitrary, $\|\grad f(x)\|\le\gamma$.\ \Qed}

\spheader 473Cd Conversely, if $f:\BbbR^r\to\Bbb R$ is differentiable
and $\|\grad f(x)\|\le\gamma$ for every $x$, then $\gamma$ is a
Lipschitz constant for $f$.   \prooflet{\Prf\ Take $x$, $y\in\BbbR^r$.   Set $g(t)=f((1-t)x+ty)$ for $t\in\Bbb R$.   The function
$t\mapsto (1-t)x+ty:\Bbb R\to\BbbR^r$ is everywhere differentiable, with
constant derivative $y-x$, so by 473Bc $g$ is differentiable, with
derivative $g'(t)=(y-x)\dotproduct\grad f((1-t)x+ty)$ for every $t$;
in particular, $|g'(t)|\le\gamma\|y-x\|$ for every $t$.   Now, by the Mean Value Theorem, there is a $t\in[0,1]$ such that
$g(1)-g(0)=g'(t)$, so that $|f(y)-f(x)|=|g'(t)|\le\gamma\|y-x\|$.   As $x$ and $y$ are
arbitrary, $f$ is $\gamma$-Lipschitz.\ \Qed}

\spheader 473Ce Note that if
$f\in\eusm D$ then\cmmnt{ all its partial derivatives are continuous
functions with compact support, so are bounded (436Ia), and} $f$ is
Lipschitz as well as bounded\cmmnt{, by (d) here}.

\spheader 473Cf{\bf (i)} If $D\subseteq\BbbR^r$ is bounded and
$f:D\to\Bbb R$ is Lipschitz, then there is a Lipschitz function
$g:\BbbR^r\to\Bbb R$, with compact support,
extending $f$.   \prooflet{\Prf\ By 262Bb there is a Lipschitz function
$f_1:\BbbR^r\to\Bbb R$ which extends $f$.   Let $\gamma>0$ be such that
$D\subseteq B(\tbf{0},\gamma)$ and $\gamma$ is a Lipschitz constant for
$f_1$;  set $M=|f_1(0)|+\gamma^2$;  then $|f_1(x)|\le M$ for every
$x\in D$, so if we set
$f_2(x)=\med(-M,f_1(x),M)$ for $x\in\BbbR^r$, $f_2$ is a bounded
Lipschitz function extending $f$.   Set
$f_3(x)=\med(0,1+\gamma-\|x\|,1)$ for $x\in\BbbR^r$;  then
$f_3$ is a bounded Lipschitz function with compact support.   By (a),
$g=f_3\times f_2$ is Lipschitz, and $g:\BbbR^r\to\Bbb R$ is a function
with compact support extending $f$.\ \Qed}

\medskip

\quad{\bf (ii)}
It follows that if $D\subseteq\BbbR^r$ is bounded and $f:D\to\Bbb R^s$ is
Lipschitz, then there is a Lipschitz function $g:\BbbR^r\to\BbbR^s$, with
compact support,
extending $f$.   \prooflet{\Prf\ By 262Ba, we need only apply (i) to each
coordinate of $f$.\ \Qed}

\leader{473D}{Smoothing by \dvrocolon{convolution}}\cmmnt{ We shall
need a miscellany of facts, many of them special cases of results in
\S\S255 and 444, concerning convolutions on $\BbbR^r$.   Recall that I
write $(f*g)(x)=\int f(y)g(x-y)\mu(dy)$ whenever $f$ and $g$ are
real-valued functions defined almost everywhere in $\BbbR^r$ and the
integral is defined, and that $f*g=g*f$ (255Fb, 444Og).   Now we
have the following.

\medskip

\noindent}{\bf Lemma} Suppose that $f$ and $g$ are Lebesgue measurable
real-valued functions defined $\mu$-almost everywhere in $\BbbR^r$.

(a) If $f$ is integrable and $g$ is essentially bounded, then their
convolution $f*g$ is defined everywhere in $\BbbR^r$ and uniformly
continuous, and $\|f*g\|_{\infty}\le\|f\|_1\esssup|g|$.

(b) If $f$ is locally integrable and $g$ is bounded and has compact
support, then $f*g$ is defined everywhere in $\BbbR^r$ and is
continuous.

(c) If $f$ and $g$ are defined everywhere in $\BbbR^r$ and
$x\in\BbbR^r\setminus(\{y:f(y)\ne 0\}+\{z:g(z)\ne 0\})$, then $(f*g)(x)$
is defined and equal to $0$.

(d) If $f$ is integrable and $g$ is bounded, Lipschitz and defined
everywhere, then $f*\grad g$ and $\grad(f*g)$ are defined everywhere and
equal, where $f*\grad g=(f*\Pd{g}{\xi_1},\ldots,f*\Pd{g}{\xi_r})$.
Moreover, $f*g$ is Lipschitz.

(e) If $f$ is locally integrable, and $g\in\eusm D$,
then $f*g$ is defined everywhere and is smooth.

(f) If $f$ is essentially bounded and $g\in\eusm D$, then all the
derivatives of $f*g$ are bounded, and $f*g$ is Lipschitz.

(g) If $f$ is integrable and $\phi:\BbbR^r\to\BbbR^r$ is a bounded
measurable function with components $\phi_1,\ldots,\phi_r$, and we write
$(f*\phi)(x)=((f*\phi_1)(x),\ldots,(f*\phi_r)(x))$, then
$\|(f*\phi)(x)\|\le\|f\|_1\sup_{y\in\BbbR^r}\|\phi(y)\|$ for every
$x\in\BbbR^r$.

\proof{{\bf (a)} See 255K.

\medskip

{\bf (b)} Suppose that $g(y)=0$ when $\|y\|\ge n$.   Given
$x\in\BbbR^r$, set $\tilde f=f\times\chi B(x,n+1)$.   Then $\tilde f*g$
is defined everywhere and continuous, by (a), while
$(f*g)(z)=(\tilde f*g)(z)$ whenever $z\in B(x,1)$;  so $f*g$ is defined
everywhere in $B(x,1)$ and is continuous at $x$.

\medskip

{\bf (c)} We have only to note that $f(y)g(x-y)=0$ for every $y$.

\medskip

{\bf (d)} Let $\gamma$ be a Lipschitz constant for $g$.   We know that
$\grad g$ is defined almost everywhere, is measurable, and that
$\|\grad g(x)\|\le\gamma$ whenever it is defined (473Cc);  so
$(f*\grad g)(x)$ is defined for every $x$, by (a) here.   Fix
$x\in\BbbR^r$.   If $y$, $z\in\BbbR^r$ set

\Centerline{$\theta(y,z)
=\Bover1{\|z\|}\bigl(g(x-y+z)-g(x-y)
  -z\dotproduct\grad g(x-y)\bigr)$}

\noindent whenever this is defined.   Then $|\theta(y,z)|\le 2\gamma$
whenever it is defined.   Now suppose that $\sequencen{z_n}$ is a
sequence in $\BbbR^r\setminus\{\tbf{0}\}$ converging to $\tbf{0}$.
Then $\lim_{n\to\infty}\theta(y,z_n)=0$ whenever $\grad g(x-y)$ is
defined, which almost everywhere.   So
$\lim_{n\to\infty}\int f(y)\theta(y,z_n)\mu(dy)=0$, by Lebesgue's
Dominated Convergence Theorem.   But this means that

\Centerline{$\Bover1{\|z_n\|}
  \bigl((f*g)(x+z_n)-(f*g)(x)-((f*\grad g)(x))\dotproduct z_n\bigr)
\to 0$}

\noindent as $n\to\infty$.   As $\sequencen{z_n}$ is arbitrary,
$\grad(f*g)(x)$ is defined and is equal to $(f*\grad g)(x)$.

Now $\grad g$ is bounded, because $g$ is Lipschitz, so
$\grad(f*g)=f*\grad g$ also is bounded, by (a), and $f*g$ must be
Lipschitz (473Cd).

\medskip

{\bf (e)} By (b), $f*g$ is defined everywhere and is continuous.   Now,
for any $i\le r$, $\Pd{}{\xi_i}(f*g)=f*\Pd{g}{\xi_i}$ everywhere.
\Prf\ Let $n\in\Bbb N$ be such that $g(y)=0$ if $\|y\|\ge n$.   Given
$x\in\BbbR^r$, set $\tilde f=f\times\chi B(x,n+1)$.   Then
$(f*g)(z)=(\tilde f*g)(z)$ for every $z\in B(x,1)$, so that

$$\eqalignno{\Pd{(f*g)}{\xi_i}(x)
&=\Pd{(\tilde f*g)}{\xi_i}(x)
=(\tilde f*\Pd{g}{\xi_i})(x)\cr
\displaycause{by (d)}
&=(f*\Pd{g}{\xi_i})(x)\cr}$$

\noindent (because of course $\Pd{g}{\xi_i}$ is also zero outside
$B(\tbf{0},n)$).\ \QeD\   Inducing on $k$,

$$\bover{\partial^k}{\partial\xi_{i_1}\ldots\partial\xi_{i_k}}(f*g)(x)
=(f*\bover{\partial^kg}{\partial\xi_{i_1}\ldots\partial\xi_{i_k}})(x)$$

\noindent for every $x\in\BbbR^r$ and every $i_1,\ldots,i_k$;  so we
have the result.

\medskip

{\bf (f)} The point is just that all the partial derivatives of $g$,
being smooth functions with compact support, are integrable, and that

\Centerline{$|\Pd{}{\xi_i}(f*g)(x)|
=|(f*\Pd{g}{\xi_i})(x)|
\le\|f\|_{\infty}\|\Pd{g}{\xi_i}\|_1$}

\noindent for every $x$ and every $i\le r$.   Inducing on the order of $D$,
we see that $D(f*g)=f*Dg$ and
$\|D(f*g)\|_{\infty}\le\|f\|_{\infty}\|Dg\|_1$, so that $D(f*g)$ is
bounded, for any partial differential operator $D$.   In particular,
$\grad(f*g)$ is bounded, so that $f*g$ is Lipschitz, by 473Cd.

\medskip

{\bf (g)} If $x$, $z\in\BbbR^r$, then

$$\eqalign{z\dotproduct(f*\phi)(x)
&=\sum_{i=1}^r\zeta_i(f*\phi_i)(x)
=\int f(y)\sum_{i=1}^r\zeta_i\phi_i(x-y)\mu(dy)\cr
&\le\int|f(y)||\sum_{i=1}^r\zeta_i\phi_i(x-y)|\mu(dy)\cr
&\le\int|f(y)|\|z\|\|\phi(x-y)\|\mu(dy)
\le\|z\|\|f\|_1\sup_{y\in\BbbR^r}\|\phi(y)\|.\cr}$$

\noindent As $z$ is arbitrary,
$\|(f*\phi)(x)\|\le\|f\|_1\sup_{y\in\BbbR^r}\|\phi(y)\|$.
}%end of proof of 473D

\leader{473E}{Lemma} (a) Define $h:\Bbb R\to[0,1]$ by
setting $h(t)=\exp(\Bover{1}{t^2-1})$ for $|t|<1$, $0$ for
$|t|\ge 1$.   Then $h$ is smooth, and $h'(t)\le 0$ for $t\ge 0$.

(b) For $n\ge 1$, define $\tilde h_n:\BbbR^r\to\Bbb R$ by setting

\Centerline{$\alpha_n=\int h((n+1)^2\|x\|^2)\mu(dx)$,
\quad$\tilde h_n(x)=\Bover1{\alpha_n}h((n+1)^2\|x\|^2)$}

\noindent for every $x\in\BbbR^r$.   Then
$\tilde h_n\in\eusm D$, $\tilde h_n(x)\ge 0$ for every $x$,
$\tilde h_n(x)=0$ if
$\|x\|\ge\Bover1{n+1}$, and $\int\tilde h_nd\mu=1$.

(c) If $f\in\eusm L^0(\mu)$, then
$\lim_{n\to\infty}(f*\tilde h_n)(x)=f(x)$ for every $x\in\dom f$ at
which $f$ is continuous.

(d) If $f:\BbbR^r\to\Bbb R$ is uniformly continuous (in particular, if
it is either Lipschitz or a continuous function with compact support),
then $\lim_{n\to\infty}\|f-f*\tilde h_n\|_{\infty}=0$.

(e) If $f\in\eusm L^0(\mu)$ is locally integrable, then
$f(x)=\lim_{n\to\infty}(f*\tilde h_n)(x)$ for $\mu$-almost every
$x\in\BbbR^r$.

(f) If $f\in\eusm L^p(\mu)$, where $1\le p<\infty$, then
$\lim_{n\to\infty}\|f-f*\tilde h_n\|_p=0$.

\proof{{\bf (a)} Set $h_0(t)=\exp(-\bover1t)$ for $t>0$, $0$ for
$t\le 0$.
A simple induction on $n$ shows that the $n$th derivative $h_0^{(n)}$ of
$h_0$ is of the form

$$\eqalign{h_0^{(n)}(t)&=q_n(\Bover1t)\exp(-\Bover1t)\text{ for }t>0\cr
&=0\text{ for }t\le 0,\cr}$$

\allowmorestretch{468}{\noindent where each $q_n$ is a polynomial of degree $2n$;  the
inductive hypothesis depends on the fact that
$\lim_{s\to\infty}q(s)e^{-s}=0$ for every polynomial $q$.   So $h_0$ is
smooth.   Now $h(t)=h_0(1-t^2)$ so $h$ also is smooth.   If $0\le t<1$
then}

\Centerline{$h'(t)=-\exp(\Bover1{t^2-1})\cdot\Bover{2t}{(t^2-1)^2}<0$;}

\noindent if $t>1$ then $h'(t)=0$;  since $h'$ is continuous,
$h'(t)\le 0$ for every $t\ge 0$.

\medskip

{\bf (b)} We need only observe that

\Centerline{$x\mapsto(n+1)^2\|x\|^2=(n+1)^2\sum_{i=1}^r\xi_i^2$}

\noindent is smooth
and that the composition of smooth functions is smooth (using 473Bc).

\medskip

{\bf (c)} If $f$ is continuous at $x$ and $\epsilon>0$, let
$n_0\in\Bbb N$ be such that $|f(y)-f(x)|\le\epsilon$ whenever
$y\in\dom f$ and  $\|y-x\|\le\Bover1{n_0+1}$.   Then for any $n\ge n_0$,

$$\eqalignno{|(f*\tilde h_n)(x)-f(x)|
&=|\int f(x-y)\tilde h_n(y)\mu(dy)-\int f(x)\tilde h_n(y)\mu(dy)|\cr
&\le\int|f(x-y)-f(x)|\tilde h_n(y)\mu(dy)
\le\int\epsilon\tilde h_n(y)\mu(dy)
=\epsilon.\cr}$$

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

\medskip

{\bf (d)} Repeat the argument of (c), but `uniformly in $x$';  that is,
given $\epsilon>0$, take $n_0$ such that $|f(y)-f(x)|\le\epsilon$
whenever $x$, $y\in\BbbR^r$ and $\|y-x\|\le\Bover1{n_0+1}$, and see that
$|(f*\tilde h_n)(x)-f(x)|\le\epsilon$ for every $n\ge n_0$ and every
$x$.

\medskip

{\bf (e)} We know from 472Db or 261E that, for almost every $x\in\BbbR^r$,

\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\mu B(x,\delta)}
  \int_{B(x,\delta)}|f(y)-f(x)|\mu(dy)=0$.}

\noindent Take any such $x$.   Set $\gamma=f(x)$,
Set $g(y)=|f(y)-\gamma|$ for every $y\in\dom f$.   Let $\epsilon>0$.   Then there is some $\delta>0$ such that
$\Bover{q(t)}{\beta_rt^r}\le\epsilon$ whenever $0<t\le\delta$, where

\Centerline{$q(t)=\int_{B(x,t)}g\,d\mu
=\int_0^t\int_{\partial B(x,s)}g(y)\nu(dy)dt$}

\noindent by 265G, so $q'(t)=\int_{\partial B(y,t)}g\,d\nu$ for almost
every $t\in[0,\delta]$, by 222E.   If $n+1\ge\Bover1{\delta}$, then

$$\eqalignno{(g*\tilde h_n)(x)
&=\int g(y)\tilde h_n(x-y)\mu(dy)
=\int_{B(x,\delta)}g(y)\tilde h_n(x-y)\mu(dy)\cr
&=\Bover1{\alpha_n}\int_0^{\delta}\int_{\partial B(x,t)}
  g(y)h((n+1)^2t^2)\nu(dy)dt\cr
\displaycause{265G again}
&=\Bover1{\alpha_n}\int_0^{\delta}h((n+1)^2t^2)q'(t)dt\cr
&=-\Bover1{\alpha_n}\int_0^{\delta}2(n+1)^2th'((n+1)^2t^2)q(t)dt\cr
\displaycause{integrating by parts (225F), because
$q(0)=h((n+1)^2\delta^2)=0$ and both $q$ and $h$ are absolutely
continuous}
&\le-\Bover{\epsilon}{\alpha_n}\int_0^{\delta}
  2(n+1)^2th'((n+1)^2t^2)\beta_rt^rdt\cr
\displaycause{because $0\le q(t)\le\epsilon\beta_rt^r$ and
$h'((n+1)^2t^2)\le 0$ for $0\le t\le\delta$}
&=\epsilon\cr}$$

\noindent (applying the same calculations with $\chi\BbbR^r$ in place of
$g$).   But now, since $(\gamma\chi\BbbR^r*\tilde h_n)(x)=\gamma$ for
every $n$,

\Centerline{$|(f*\tilde h_n)(x)-\gamma|
=|\int(f(y)-\gamma)\tilde h_n(x-y)\mu(dy)|
\le\int|f(y)-\gamma|\tilde h_n(x-y)\mu(dy)
\le\epsilon$}

\noindent whenever $n+1\ge\Bover1{\delta}$.  As $\epsilon$ is arbitrary,
$f(x)=\gamma=\lim_{n\to\infty}(f*\tilde h_n)(x)$;  and this is true for
$\mu$-almost every $x$.

\medskip

{\bf (f)} Apply 444T to the indefinite-integral measure
$\tilde h_n\mu$ over $\mu$ defined by $\tilde h_n$;
use 444Pa for the identification of
$(\tilde h_n\mu)*f$ with $\tilde h_n*f=f*\tilde h_n$.
}%end of proof of 473E

\leader{473F}{Lemma} For any measure space $(X,\Sigma,\lambda)$ and any
non-negative $f_1,\ldots,f_k\in\eusm L^0(\lambda)$,

\Centerline{$\int\prod_{i=1}^kf_i^{1/k}d\lambda
\le\prod_{i=1}^k\bigl(\int f_id\lambda\bigr)^{1/k}$.}

\proof{ Induce on $k$.   Note that we can suppose that every
$f_i$ is integrable;  for if any
$\int f_i$ is zero, then $f_i=0$ a.e.\ and the result is trivial;  and
if all the $\int f_i$ are greater than zero and any of them is infinite,
the result is again trivial.

The induction starts with the trivial case $k=1$.   For the inductive
step to $k\ge 2$, we have

$$\eqalignno{\int\prod_{i=1}^kf_i^{1/k}d\lambda
&\le\|\prod_{i=1}^{k-1}f_i^{1/k}\|_{k/(k-1)}\|f_k^{1/k}\|_k\cr
\displaycause{by H\"older's inequality, 244E}
&=\bigl(\int\prod_{i=1}^{k-1}f_i^{1/(k-1)}d\lambda\bigr)^{(k-1)/k}
   \bigl(\int f_kd\lambda\bigr)^{1/k}\cr
&\le\bigl(\prod_{i=1}^{k-1}(\int f_id\lambda)^{1/(k-1)}\bigr)^{(k-1)/k}
   \bigl(\int f_kd\lambda\bigr)^{1/k}\cr
\displaycause{by the inductive hypothesis}
&=\prod_{i=1}^k(\int f_id\lambda)^{1/k},\cr}$$

\noindent as required.
}%end of proof of 473F

\leader{473G}{Lemma} Let $(X,\Sigma,\lambda)$ be a $\sigma$-finite
measure space and
$k\ge 2$ an integer.   Write $\lambda_k$ for the product measure on
$X^k$.   For
$x=(\xi_1,\ldots,\xi_k)\in X^k$, $t\in X$ and $1\le i\le k$ set
$S_i(x,t)=(\xi'_1,\ldots,\xi'_k)$ where $\xi'_i=t$ and $\xi'_j=\xi_j$
for $j\ne i$.   Then if $h\in\eusm L^1(\lambda_k)$ is non-negative, and
we set $h_i(x)=\int h(S_i(x,t))\lambda(dt)$ whenever this is defined in
$\Bbb R$, we have

\Centerline{$\int(\prod_{i=1}^kh_i)^{1/(k-1)}d\lambda_k
\le(\int h\,d\lambda_k)^{k/(k-1)}$.}

\proof{ Induce on $k$.

\medskip

{\bf (a)} If $k=2$, we have

$$\eqalign{\int h_1\times h_2d\lambda_k
&=\iint\bigl(\int h(\tau_1,\xi_2)d\tau_1\bigr)
  \bigl(\int h(\xi_1,\tau_2)d\tau_2\bigr)d\xi_1d\xi_2\cr
&=\iiiint h(\tau_1,\xi_2)h(\xi_1,\tau_2)d\tau_1d\tau_2d\xi_1d\xi_2\cr
&=\iint h(\tau_1,\xi_2)d\tau_1d\xi_2
  \cdot\iint h(\xi_1,\tau_2)d\tau_2d\xi_1
=\bigl(\int h\,d\lambda_2\bigr)^2\cr}$$

\noindent by Fubini's theorem (252B) used repeatedly, because (by 253D)
\discrcenter{390pt}{$(\tau_1,\tau_2,\xi_1,\xi_2)\mapsto
h(\xi_1,\tau_2)h(\tau_1,\xi_2)$ }is $\lambda_4$-integrable.
(See 251W for a sketch of the manipulations
needed to apply 252B, as stated, to the integrals above.)

\medskip

{\bf (b)} For the inductive step to $k\ge 3$, argue as follows.
For $y\in X^{k-1}$, set $g(y)=\int h(y,t)dt$
whenever this is defined in $\Bbb R$, identifying $X^k$ with
$X^{k-1}\times X$, so that $g(y)=h_k(y,t)$ whenever either is defined.
If $1\le i<k$, we can consider $S_i(y,t)$ for $y\in X^{k-1}$ and
$t\in X$, and we have

\Centerline{$\int g(S_i(y,t))dt
=\iint h(S_i(y,t),u)dudt
=\int h_i(y,t)dt$}

\noindent for almost every $y\in X^{k-1}$.   So

$$\eqalignno{\int(\prod_{i=1}^kh_i)^{1/(k-1)}d\lambda_k
&=\iint(\prod_{i=1}^{k-1}h_i(y,t))^{1/(k-1)}g(y)^{1/(k-1)}
  dt\,\lambda_{k-1}(dy)\cr
&=\int g(y)^{1/(k-1)}\int(\prod_{i=1}^{k-1}h_i(y,t))^{1/(k-1)}
  dt\,\lambda_{k-1}(dy)\cr
&\le\int g(y)^{1/(k-1)}
  \prod_{i=1}^{k-1}\bigl(\int h_i(y,t)dt\bigr)^{1/(k-1)}
  \lambda_{k-1}(dy)\cr
\displaycause{473F}
&=\int g(y)^{1/(k-1)}
  \prod_{i=1}^{k-1}g_i(y)^{1/(k-1)}\lambda_{k-1}(dy)\cr
\displaycause{where $g_i$ is defined from $g$ in the same way as $h_i$
is defined from $h$}
&\le\bigl(\int g(y)\lambda_{k-1}(dy)\bigr)^{1/(k-1)}
  \bigl(\int\prod_{i=1}^{k-1}g_i(y)^{1/(k-2)}
      \lambda_{k-1}(dy)\bigr)^{(k-2)/(k-1)}\cr
\displaycause{by H\"older's inequality again, this time with
$\Bover1{k-1}+\Bover{k-2}{k-1}=1$}
&\le\bigl(\int g(y)\lambda_{k-1}(dy)\bigr)^{1/(k-1)}
  \cdot\int g(y)\lambda_{k-1}(dy)\cr
\displaycause{by the inductive hypothesis}
&=\bigl(\int g(y)\lambda_{k-1}(dy)\bigr)^{k/(k-1)}
=\bigl(\int h(x)\lambda_k(dx)\bigr)^{k/(k-1)},\cr}$$

\noindent and the induction proceeds.
}%end of proof of 473G

\leader{473H}{Gagliardo-Nirenberg-Sobolev inequality} Suppose that
$f:\BbbR^r\to\Bbb R$ is a Lipschitz function with compact support.
Then $\|f\|_{r/(r-1)}\le\int\|\grad f\|d\mu$.

\proof{ By 473Cc, $\grad f$ is measurable and bounded, so $\|\grad f\|$
also is;  since it must have compact support, it is integrable.

For $1\le i\le r$,
$x=(\xi_1,\ldots,\xi_r)\in\BbbR^r$ and $t\in\Bbb R$ write
$S_i(x,t)=(\xi'_1,\ldots,\xi'_r)$ where $\xi'_i=t$ and $\xi'_j=\xi_j$
for $j\ne i$.   Set
$h_i(x)=\int_{-\infty}^{\infty}\|\grad f(S_i(x,t))\|dt$ when this is
defined,
which will be the case for almost every $x$.   Now, whenever $h_i(x)$ is
defined,

\Centerline{$|f(x)|=|f(S_i(x,\xi_i))|
=|\int_{-\infty}^{\xi_i}\pd{}{t}f(S_i(x,t))dt|
\le h_i(x)$.}

\noindent (Use 225E and the fact that a Lipschitz function on any
bounded interval in $\Bbb R$ is absolutely continuous.)
So $|f|\leae h_i$ for every $i\le r$.   Accordingly

\Centerline{$\int|f(x)|^{r/(r-1)}\mu(dx)
\le\int\prod_{i=1}^rh_i(x)^{1/(r-1)}\mu(dx)
\le\bigl(\int\|\grad f(x)\|\mu(dx)\bigr)^{r/(r-1)}$}

\noindent by 473G.   Raising both sides to the power $(r-1)/r$ we
have the result.
}%end of proof of 473H

\leader{473I}{Lemma} For any Lipschitz function
$f:B(\tbf{0},1)\to\Bbb R$,

\Centerline{$\int_{B(\tbf{0},1)}|f|^{r/(r-1)}d\mu
\le \bigl(2^{r+4}\sqrt r
  \int_{B(\tbf{0},1)}\|\grad f\|+|f|d\mu\bigr)^{r/(r-1)}$.}

\proof{{\bf (a)} Set $g(x)=\max(0,2\|x\|^2-1)$ for $x\in B(\tbf{0},1)$.
Then $\grad g$ is defined at
every point $x$ such that $\|x\|<1$ and $\|x\|\ne\Bover1{\sqrt{2}}$, and at all such points $\pd{g}{\xi_i}$ is either $0$ or $4\xi_i$ for each $i$, so that $\|\grad g(x)\|\le 4\|x\|\le 4$.   Hence
(or otherwise) $g$ is
Lipschitz.   So $f_1=f\times g$ is Lipschitz (473Ca).

By Rademacher's theorem again, $\grad f_1$ is defined almost everywhere
in $B(\tbf{0},1)$.   Now

$$\eqalignno{\int_{B(\tbf{0},1)}\|\grad f_1\|d\mu
&=\int_{B(\tbf{0},1)}\|f(x)\grad g(x)+g(x)\grad f(x)\|\mu(dx)\cr
\displaycause{473Bd}
&\le\int_{B(\tbf{0},1)}4|f|+\|\grad f\|d\mu.\cr}$$

\medskip

{\bf (b)} It will be convenient to have an elementary fact out in the
open.   Set $\phi(x)=\Bover{x}{\|x\|^2}$ for
$x\in\BbbR^r\setminus\{\tbf{0}\}$;  note that $\phi^2(x)=x$.   Then
$\phi\restr\{x:\|x\|\ge\delta\}$ is Lipschitz, for any $\delta>0$.
\Prf\ If $\|x\|=\alpha\ge\delta$, $\|y\|=\beta\ge\delta$, then  we have

$$\eqalign{\|\phi(x)-\phi(y)\|^2
&=\Bover1{\alpha^4}\|x\|^2-\Bover2{\alpha^2\beta^2}x\dotproduct y
  +\Bover1{\beta^4}\|y\|^2\cr
&=\Bover1{\alpha^2\beta^2}
  \bigl(\|y\|^2-2x\dotproduct y+\|x\|^2\bigr)
\le\Bover1{\delta^4}\|x-y\|^2,\cr}$$

\noindent so $\Bover1{\delta^2}$ is a Lipschitz constant for
$\phi\restrp\BbbR^r\setminus B(\tbf{0},\delta)$.\ \Qed

\medskip

{\bf (c)} Set $f_2(x)=f(x)$ if $\|x\|\le 1$, $f_1\phi(x)$ if
$\|x\|\ge 1$.   Then $f_2$ is
well-defined (because $f_1(x)=f(x)$ if $\|x\|=1$), is zero outside
$B(\tbf{0},\sqrt{2})$ (because $g(x)=0$ if $\|x\|\le\Bover1{\sqrt 2}$),
and is Lipschitz.   \Prf\ By 473Cb, it will be enough to show that
$f_2\restr F$ is Lipschitz, where $F=\{x:\|x\|\ge 1\}$.   But (b) shows
that $\phi\restr F$ is $1$-Lipschitz, so any Lipschitz constant
for $f_1$ is also a Lipschitz constant for
$f_2\restr F$.\ \Qed

If $\|x\|>1$, then, for any $i\le r$,

$$\eqalign{\Pd{f_2}{\xi_i}(x)
&=\sum_{j=1}^r\Pd{f_1}{\xi_j}(\phi(x))
  \cdot\Pd{}{\xi_i}(\Bover{\xi_j}{\|x\|^2})
=\Pd{f_1}{\xi_i}(\phi(x))\cdot\Bover1{\|x\|^2}
   -2\sum_{j=1}^r\Pd{f_1}{\xi_j}(\phi(x))
      \cdot\Bover{\xi_i\xi_j}{\|x\|^4}\cr
&=\Pd{f_1}{\xi_i}(\phi(x))\cdot\Bover1{\|x\|^2}
   -\Bover{2\xi_i}{\|x\|^4}x\dotproduct\grad f(\phi(x))\cr}$$

\noindent wherever the right-hand side is defined, that is, wherever all
the partial derivatives $\Pd{f_1}{\xi_j}(\phi(x))$ are defined.   But
$H=B(\tbf{0},1)\setminus\dom(\grad f_1)$ is negligible, and does not
meet $\{x:\|x\|<\Bover1{\sqrt 2}\}$, so $\phi\restr H$ is Lipschitz and
$\phi[H]=\phi^{-1}[H]$ is
negligible (262D);  while $\grad f_1(\phi(x))$ is defined whenever
$\|x\|>1$
and $x\notin\phi^{-1}[H]$.   So the formula here is valid for almost
every $x\in F$, and

$$\eqalign{|\Pd{f_2}{\xi_i}(x)|
&\le\|\grad f_1(\phi(x))\|\cdot\Bover1{\|x\|^2}
   +\Bover{2|\xi_i|}{\|x\|^4}\|\grad f_1(\phi(x))\|\|x\|\cr
&=\|\grad f_1(\phi(x))\|\Bover{\|x\|+2|\xi_i|}{\|x\|^3}
\le 3\|\grad f_1(\phi(x))\|\cr}$$

\noindent for almost every $x\in F$.   But (since we know that
$\grad f_2$ is
defined almost everywhere, by Rademacher's theorem, as usual) we have

\Centerline{$\|\grad f_2(x)\|
\le 3\sqrt r\|\grad f_1(\phi(x))\|$}

\noindent for almost every $x\in F$.

\medskip

{\bf (d)} We are now in a position to estimate

$$\eqalignno{\int_F\|\grad f_2\|d\mu
&=\int_{B(\tbf{0},\sqrt 2)}\|\grad f_2\|d\mu
  -\int_{B(\tbf{0},1)}\|\grad f_2\|d\mu\cr
\displaycause{because $f_2(x)=0$ if $\|x\|\ge\sqrt 2$}
&=\int_1^{\sqrt 2}\int_{\partial B(\tbf{0},t)}
  \|\grad f_2(x)\|\nu(dx)dt\cr
\displaycause{265G, as usual}
&\le 3\sqrt r\int_1^{\sqrt 2}\int_{\partial B(\tbf{0},t)}
   \|\grad f_1(\Bover1{t^2}x)\|\nu(dx)dt\cr
\displaycause{by (b) above}
&\le 3\sqrt r\int_1^{\sqrt 2}\int_{\partial B(\tbf{0},1/t)}
   t^{2r-2}\|\grad f_1(y)\|\nu(dy)dt\cr}$$

\noindent substituting $x=t^2y$ in the inner integral;  the point
being that as the function $y\mapsto t^2y$ changes all distances by a
scalar multiple $t^2$, it must transform Hausdorff $(r-1)$-dimensional
measure by a multiple $t^{2r-2}$.   But now, substituting $s=\bover1t$
in the outer integral, we have

$$\eqalign{\int_F\|\grad f_2\|d\mu
&\le 3\sqrt r\int_{1/\sqrt 2}^1\Bover1{s^{2r}}
  \int_{\partial B(\tbf{0},s)}\|\grad f_1(y)\|\nu(dy)ds\cr
&\le 2^r\cdot 3\sqrt r\int_{1/\sqrt 2}^1
    \int_{\partial B(\tbf{0},s)}\|\grad f_1(y)\|\nu(dy)ds\cr
&=2^r\cdot 3\sqrt r\int_{B(\tbf{0},1)}\|\grad f_1\|d\mu\cr
&\le 2^{r+2}\sqrt r\int_{B(\tbf{0},1)}4|f|+\|\grad f\|d\mu\cr}$$

\noindent by (a) above.

\medskip

{\bf (e)} Accordingly

$$\eqalign{\int_{\Bbb R^r}\|\grad f_2\|d\mu
&=\int_{B(\tbf{0},1)}\|\grad f\|d\mu+\int_F\|\grad f_2\|d\mu\cr
&\le 2^{r+4}\sqrt r\int_{B(\tbf{0},1)}|f|+\|\grad f\|d\mu.\cr}$$

\noindent But now we can apply 473H to see that

$$\eqalign{\int_{B(\tbf{0},1)}|f|^{r/(r-1)}d\mu
&\le\int|f_2|^{r/r-1}d\mu
\le(\int\|\grad f_2\|d\mu)^{r/(r-1)}\cr
&\le\bigl(2^{r+4}\sqrt r
  \int_{B(\tbf{0},1)}|f|+\|\grad f\|d\mu\bigr)^{r/(r-1)},\cr}$$

\noindent as claimed.
}%end of proof of 473I

\leader{473J}{Lemma} Let $f:\BbbR^r\to\Bbb R$ be a Lipschitz function.
Then

\Centerline{$\int_{B(y,\delta)}|f(x)-f(z)|\mu(dx)
\le \Bover{2^r}r\delta^r\int_{B(y,\delta)}
  \|\grad f(x)\|\|x-z\|^{1-r}\mu(dx)$}

\noindent whenever $y\in\BbbR^r$, $\delta>0$ and $z\in B(y,\delta)$.
%OK for other convex sets of finite diameter?

\proof{{\bf (a)} To begin with, suppose that $f$ is smooth.   In this
case, for any $x$, $z\in B(y,\delta)$,

$$\eqalign{|f(x)-f(z)|
&=\bigl|\int_0^1\Bover{d}{dt}f(z+t(x-z))dt\bigr|\cr
&=\bigl|\int_0^1(x-z)\dotproduct\grad f(z+t(x-z))dt\bigr|\cr
&\le\|x-z\|\int_0^1\|\grad f(z+t(x-z))\|dt.\cr}$$

\noindent So, for $\eta>0$,

$$\eqalignno{\int_{B(y,\delta)\cap\partial B(z,\eta)}
  &|f(x)-f(z)|\nu(dx)\cr
&\le\eta\int_0^1\int_{B(y,\delta)\cap\partial B(z,\eta)}
  \|\grad f(z+t(x-z))\|\nu(dx)dt\cr
\displaycause{$\grad f$ is continuous and bounded, and the subspace
measure induced by $\nu$ on $\partial B(z,\eta)$ is a
(quasi-\nobreak)Radon measure (471E, 471Dh), so
its product with Lebesgue measure also is (417T), and there is no
difficulty with the change in order of integration}
&\le\eta\int_0^1\Bover1{t^{r-1}}
  \int_{B(y,\delta)\cap\partial B(z,t\eta)}
  \|\grad f(w)\|\nu(dw)dt\cr
\displaycause{because if $\phi(x)=z+t(x-z)$, then
$\nu\phi^{-1}[E]=\Bover1{t^{r-1}}\nu E$ whenever $\nu$ measures $E$ and
$t>0$, while $\phi(x)\in B(y,\delta)$ whenever $x\in B(y,\delta)$}
&=\eta^r\int_0^1\int_{B(y,\delta)\cap\partial B(z,t\eta)}
  \|\grad f(w)\|\|w-z\|^{1-r}\nu(dw)dt\cr
&=\eta^{r-1}\int_0^{\eta}\int_{B(y,\delta)\cap\partial B(z,s)}
  \|\grad f(w)\|\|w-z\|^{1-r}\nu(dw)ds\cr
\displaycause{substituting $s=t\eta$}
&=\eta^{r-1}\int_{B(y,\delta)\cap B(z,\eta)}
  \|\grad f(w)\|\|w-z\|^{1-r}\mu(dw).\cr}$$

\noindent So

$$\eqalign{\int_{B(y,\delta)}|f(x)-f(z)|\mu(dx)
&=\int_0^{2\delta}\int_{B(y,\delta)\cap\partial B(z,\eta)}
  |f(x)-f(z)|\nu(dx)d\eta\cr
&\le\int_0^{2\delta}\eta^{r-1}\int_{B(y,\delta)\cap B(z,\eta)}
  \|\grad f(w)\|\|w-z\|^{1-r}\mu(dw)d\eta\cr
&\le\int_0^{2\delta}\eta^{r-1}\int_{B(y,\delta)}
  \|\grad f(w)\|\|w-z\|^{1-r}\mu(dw)d\eta\cr
&=\bover{2^r}r\delta^r\int_{B(y,\delta)}
  \|\grad f(w)\|\|w-z\|^{1-r}\mu(dw).\cr}$$

\medskip

{\bf (b)} Now turn to the general case in which $f$ is not necessarily
differentiable everywhere, but is known to be Lipschitz and bounded.   We need to know that
$\int_{B(y,\delta)}\|x-z\|^{1-r}\mu(dx)$ is finite;  this is because

$$\eqalignno{\int_{B(y,\delta)}\|x-z\|^{1-r}\mu(dx)
&\le\int_{B(z,2\delta)}\|x-z\|^{1-r}\mu(dx)
=\int_0^{2\delta}t^{1-r}\nu(\partial B(z,t))dt\cr
&=\int_0^{2\delta}t^{1-r}r\beta_rt^{r-1}dt
=2\delta r\beta_r.\cr}$$

\noindent Take the sequence $\sequencen{\tilde h_n}$ from 473E.
Then $\sequencen{f*\tilde h_n}$ converges uniformly to $f$ (473Ed),
while $\sequencen{\grad(f*\tilde h_n)}=\sequencen{\tilde h_n*\grad f}$
(473Dd) is uniformly bounded (473Cc, 473Dg) and
converges almost everywhere to $\grad f$ (473Ee).   But this means
that, setting $f_n=f*\tilde h_n$,

$$\eqalignno{\int_{B(y,\delta)}|f(x)-f(z)|\mu(dx)
&=\lim_{n\to\infty}\int_{B(y,\delta)}|f_n(x)-f_n(z)|\mu(dx)\cr
&\le\lim_{n\to\infty}\Bover{2^r}r\delta^r\int_{B(y,\delta)}
  \|\grad f_n(x)\|\|x-z\|^{1-r}\mu(dx)\cr
\displaycause{because every $f_n$ is smooth, by 473De}
&=\Bover{2^r}r\delta^r\int_{B(y,\delta)}
  \|\grad f(x)\|\|x-z\|^{1-r}\mu(dx)\cr}$$

\noindent by Lebesgue's Dominated Convergence Theorem.

\medskip

{\bf (c)} Finally, if $f$ is not bounded on the whole of $\BbbR^r$,
it is surely bounded on $B(y,\delta)$, so we can apply (b) to the function
$x\mapsto\med(-M,f(x),M)$ for a suitable $M\ge 0$ to get the result as
stated.
}%end of proof of 473J

\vleader{48pt}{473K}{Poincar\'e's inequality for balls} Let $B\subseteq\BbbR^r$
be a non-trivial closed ball, and $f:B\to\Bbb R$ a Lipschitz function.
Set $\gamma=\Bover1{\mu B}\int_Bfd\mu$.   Then

\Centerline{$\bigl(\int_B|f-\gamma|^{r/(r-1)}d\mu\bigr)^{(r-1)/r}
\le c\int_B\|\grad f\|d\mu$,}

\noindent where $c=2^{r+4}\sqrt r(1+2^{r+1})$.

\proof{{\bf (a)} To begin with (down to the end of (b)) suppose that
$B$ is the unit ball $B(\tbf{0},1)$.
Then, for any $x\in B$,

$$\eqalign{|f(x)-\gamma|
&=\Bover1{\mu B}
  \bigl|\int_Bf(x)-f(z)\mu(dz)\bigr|\cr
&\le\Bover1{\mu B}\int_B|f(x)-f(z)|\mu(dz)\cr
&\le\Bover{2^r}r\cdot\Bover1{\mu B}
  \int_B\|\grad f(z)\|\|x-z\|^{1-r}\mu(dz),\cr}$$

\noindent by 473J.   Also, for any $z\in B$,

$$\eqalignno{\int_{B(z,2)}\|x-z\|^{1-r}\mu(dx)
&=\int_0^2\int_{\partial B(z,t)}\|x-z\|^{1-r}\nu(dx)dt\cr
&=\int_0^2t^{1-r}\nu(\partial B(z,t))dt
=\int_0^2t^{1-r}t^{r-1}\nu S_{r-1}dt
=2r\beta_r.\cr}$$

\noindent So

$$\eqalign{\int_B|f(x)-\gamma|\mu(dx)
&\le\Bover{2^r}r\cdot\Bover1{\mu B}
  \int_B\int_B\|\grad f(z)\|\|x-z\|^{1-r}\mu(dz)\mu(dx)\cr
&=\Bover{2^r}{r\beta_r}\int_B\int_B
  \|\grad f(z)\|\|x-z\|^{1-r}\mu(dx)\mu(dz)\cr
&\le\Bover{2^r}{r\beta_r}
  \int_B\|\grad f(z)\|\int_{B(z,2)}
  \|x-z\|^{1-r}\mu(dx)\mu(dz)\cr
&\le 2^{r+1}\int_B\|\grad f(z)\|\mu(dz).\cr}$$

\medskip

{\bf (b)} Now apply 473I to $g=f(x)-\gamma$.   We have

$$\eqalignno{\int_B|f-\gamma|^{r/(r-1)}d\mu
&\le\bigl(2^{r+4}\sqrt r
  \int_B\|\grad f\|+|g|d\mu\bigr)^{r/(r-1)}\cr
&\le\bigl(2^{r+4}\sqrt r(1+2^{r+1})
  \int_B\|\grad f\|d\mu\bigr)^{r/(r-1)}\cr
\displaycause{by (a)}
&=\bigl(c\int_B\|\grad f\|d\mu\bigr)^{r/(r-1)}.\cr}$$

\medskip

{\bf (c)} For the general case, express $B$ as $B(y,\delta)$, and set
$h(x)=f(y+\delta x)$ for
$x\in B(\tbf{0},1)$.   Then $\grad h(x)=\delta\grad f(y+\delta x)$ for
almost every $x\in B(\tbf{0},1)$.
Now

\Centerline{$\int_{B(\tbf{0},1)}h\,d\mu
=\Bover1{\delta^r}\int_{B(y,\delta)}fd\mu$,}

\noindent so

\Centerline{$\Bover1{\mu B(\tbf{0},1)}\int_{B(\tbf{0},1)}h\,d\mu
=\Bover1{\mu B(y,\delta)}\int_{B(y,\delta)}fd\mu
=\gamma$.}

\noindent We therefore have

$$\eqalignno{\int_{B(y,\delta)}|f-\gamma|^{r/(r-1)}d\mu
&=\delta^r\int_{B(\tbf{0},1)}|h-\gamma|^{r/(r-1)}d\mu\cr
&\le\delta^r
  \bigl(c\int_{B(\tbf{0},1)}\|\grad h\|d\mu\bigr)^{r/(r-1)}\cr
\displaycause{by (a)-(b) above}
&=\delta^r\bigl(\Bover{\delta c}{\delta^r}
      \int_{B(y,\delta)}\|\grad f\|d\mu\bigr)^{r/(r-1)}\cr
&=\bigl(c\int_{B(y,\delta)}\|\grad f\|d\mu\bigr)^{r/(r-1)}.\cr}$$

\noindent Raising both sides to the power $(r-1)/r$ we have the result
as stated.
}%end of proof of 473K

\cmmnt{\medskip

\noindent{\bf Remark} As will be plain from the way in which the proof
here is constructed, there is no suggestion that the formula offered for
$c$ gives anything near the best possible value.
}%end of comment

\leader{473L}{Corollary} Let $B\subseteq\BbbR^r$ be a non-trivial
closed ball, and $f:B\to[0,1]$ a Lipschitz function.   Set

\Centerline{$F_0=\{x:x\in B,\,f(x)\le\bover14\}$,
\quad$F_1=\{x:x\in B,\,f(x)\ge\bover34\}$.}

\noindent Then

\Centerline{$\bigl(\min(\mu F_0,\mu F_1)\bigr)^{(r-1)/r}
\le 4c\int_B\|\grad f\|d\mu$,}

\noindent where
$c=2^{r+4}\sqrt r(1+2^{r+1})$.

\proof{ Setting $\gamma=\Bover1{\mu B}\int_Bfd\mu$,

$$\eqalign{\int_B|f-\gamma|^{r/(r-1)}d\mu
&\ge\Bover1{4^{r/(r-1)}}\mu F_0\text{ if }\gamma\ge\Bover12,\cr
&\ge\Bover1{4^{r/(r-1)}}\mu F_1\text{ if }\gamma\le\Bover12.\cr}$$

\noindent So 473K tells us that

\Centerline{$\Bover14\bigl(\min(\mu F_0,\mu F_1)\bigr)^{(r-1)/r}
\le c\int_B\|\grad f\|d\mu$,}

\noindent as required.
}%end of proof of 473L

\leader{473M}{The case $r=1$}\cmmnt{ The general rubric for this
section declares that $r$ is taken to be at least $2$, which is clearly
necessary for the formula in 473K to be appropriate.   For the sake of
an application in the next section, however, I mention the elementary
corresponding result when $r=1$.}   In this case, $B$ is just a closed
interval, and \cmmnt{$\grad f$ is the ordinary derivative of $f$;
interpreting $(\int_B|f-\gamma|^{r/(r-1)})^{(r-1)/r}$ as
$\|f\times\chi B-\gamma\chi B\|_{r/(r-1)}$, it is natural to look at}

\Centerline{$\|f\times\chi B-\gamma\chi B\|_{\infty}
=\sup_{x\in B}|f(x)-\gamma|
\le\sup_{x,y\in B}|f(x)-f(y)|
\le\int_B|f'|d\mu$,}

\noindent giving a version of 473K for $r=1$.   \cmmnt{We see that
the formula for $c$ remains valid in the case $r=1$, with a good deal to
spare.}   As for 473L, if $\int_B|f'|<\bover12$ then at least one
of $F_0$, $F_1$ must be empty.

\exercises{\leader{473X}{Basic exercises (a)}
%\spheader 473Xa
Set $f(x)=\max(0,-\ln\|x\|)$, $f_k(x)=\min(k,f(x))$ for
$x\in\BbbR^2\setminus\{0\}$, $k\in\Bbb N$.   Show that
$\lim_{k\to\infty}\|f-f_k\|_2
=\lim_{k\to\infty}\|\grad f-\grad f_k\|_1=0$, so that all the
inequalities 473H-473L %473H 473I 473J 473K 473L
are valid for $f$.
%473L

\spheader 473Xb Let $k\in[1,r]$ be an integer, and set
$m=\Bover{(r-1)!}{(k-1)!(r-k)!}$.   Let $e_1,\ldots,e_r$ be the standard orthonormal basis of $\BbbR^r$ and $\Cal J$ the family of subsets of $\{1,\ldots,r\}$ with $k$ members.   For $J\in\Cal J$ let
$V_J$ be the linear span of $\{e_i:i\in J\}$, $\pi_J:\BbbR^r\to V_J$ the orthogonal projection and $\nu_J$ the normalized $k$-dimensional Hausdorff measure on $V_J$.   Show that if
$A\subseteq\BbbR^r$ then
$(\mu^*A)^m\le\prod_{J\in\Cal J}\nu_J^*\pi_J[A]$.
\Hint{start with $A\subseteq[0,1]^r$ and note that $([0,1]^r)^m$ can be identified with $\prod_{J\in\Cal J}[0,1]^J$.}
%not much to do with anything

\leader{473Y}{Further exercises (a)}
%\spheader 473Ya
Let $D\subseteq\BbbR^r$ be any set and
$\phi:D\to\BbbR^s$ any function.   Show that $D_0=\{x:x\in D,\,\phi$ is
differentiable at $x\}$ is a Borel subset of $\BbbR^r$, and that the
derivative of $\phi$ is a Borel measurable function.
(Compare 225J.)

}%end of exercises

\endnotes{
\Notesheader{473} The point of all the inequalities
473H-473L %473H 473I 473J 473K 473L
is that they bound some measure of variance of a function $f$ by the
integral of $\|\grad f\|$.   If $r=2$, indeed, we are looking at
$\|f\|_2$ (473H) or $\int_B|f|^2$ (473I) or something essentially equal
to the variance of probability theory (473K).   In higher dimensions we
need to look at $\|\,\|_{r/(r-1)}$ in place of $\|\,\|_2$, and when
$r=1$ we can interpret the inequalities in terms of the supremum norm
$\|\,\|_{\infty}$ (473M).   In all cases we want to develop inequalities
which will enable us to discuss a function in terms of its first
derivative.   In one dimension, this is the familiar Fundamental Theorem
of Calculus (Chapter 22).   We find there a
straightforward criterion (`absolute continuity') to determine whether a
given function of one variable
is an indefinite integral, and that if so it is the indefinite
integral of its own derivative.   Even in two dimensions, this
simplicity disappears.   The essential problem is that a function can be
the indefinite integral of an integrable gradient function without being
bounded (473Xa).   The principal results of this section are stated for
Lipschitz functions, but in fact they apply much more widely.   The
argument suggested in 473Xa involves approximating the unbounded
function $f$ by Lipschitz functions $f_k$ in a sharp enough sense to
make it possible to read off all the inequalities for $f$ from the
corresponding inequalities for the $f_k$.   This idea leads naturally to
the concept of `Sobolev space', which I leave on one side for the
moment;  see {\smc Evans \& Gariepy 92}, chap.\ 4, for details.
}%end of notes

\discrpage


\frfilename{mt474.tex}
\versiondate{17.11.12}
\copyrightdate{2001}

\def\varinnerprod#1#2{#1\dotproduct#2}

\def\chaptername{Geometric measure theory}
\def\sectionname{The distributional perimeter}

\newsection{474}

The next step is a dramatic excursion, defining (for appropriate sets
$E$) a perimeter {\it measure} for which a version of the Divergence
Theorem is true (474E).   I begin the section with elementary notes on
the divergence of a vector field
(474B-474C).   I then define `locally finite perimeter' (474D),
`perimeter measure' and `outward normal' (474F) and `reduced boundary'
(474G).   The definitions rely on the Riesz representation theorem, and
we have to work very hard to relate them to any geometrically natural
idea of `boundary'.   Even half-spaces (474I) demand some attention.
From Poincar\'e's inequality\cmmnt{ (473K)} we can prove isoperimetric
inequalities for perimeter measures (474L).   With some effort we can
locate the reduced boundary as a subset of the topological
boundary\cmmnt{ (474Xc)}, and obtain asymptotic inequalities on the
perimeter measures of small balls (474N).   With much more effort we can
find a geometric description of outward normal functions in terms of
`Federer exterior normals' (474R), and get a tight asymptotic
description of the perimeter measures of small balls (474S).   I end
with the Compactness Theorem for sets of bounded perimeter (474T).

%taken from {\smc Evans \& Gariepy 92}

\cmmnt{
\leader{474A}{Notation} I had better repeat some of the notation from
\S473.   $r\ge 2$ is a fixed integer.   $\mu$ is Lebesgue measure on
$\BbbR^r$, and $\beta_r=\mu B(\tbf{0},1)$ is the volume of the unit
ball.   $S_{r-1}=\partial B(\tbf{0},1)$ is the unit sphere.   $\nu$ is
normalized $(r-1)$-dimensional Hausdorff measure on $\BbbR^r$.
We shall sometimes need to look at Lebesgue measure on $\BbbR^{r-1}$,
which I will denote $\mu_{r-1}$.   As in \S473, I will use Greek letters
to represent coordinates, so that $x=(\xi_1,\ldots,\xi_r)$ for
$x\in\BbbR^r$, etc., and $\beta_r$ will be the $r$-dimensional volume of
the unit ball in $\BbbR^r$.

$\eusm D$ is the set of smooth functions $f:\BbbR^r\to\Bbb R$ with
compact support;
$\eusm D_r$ the set of smooth functions $\phi:\BbbR^r\to\BbbR^r$ with
compact support, that is, the set of functions
$\phi=(\phi_1,\ldots,\phi_r):\BbbR^r\to\BbbR^r$ such that
$\phi_i\in\eusm D$ for every $i$.

I continue to use the sequence $\sequencen{\tilde h_n}$ from 473E;
these functions all belong to $\eusm D$, are non-negative everywhere and
zero outside $B(\tbf{0},\bover1{n+1})$, are even, and have integral $1$.
}%end of comment

\leader{474B}{The divergence of a vector field (a)} For a function
$\phi$ from a subset of $\BbbR^r$ to $\BbbR^r$, write
$\diverg\phi=\sum_{i=1}^r\pd{\phi_i}{\xi_i}$, where
$\phi=(\phi_1,\ldots,\phi_r)$;  for definiteness, let us take the domain
of $\diverg\phi$ to be the set of points at which $\phi$ is
differentiable\cmmnt{ (in the strict sense of 262Fa)}.   \cmmnt{Note
that} $\diverg\phi\in\eusm D$ for every $\phi\in\eusm D_r$.
\cmmnt{We need the following elementary facts.}

\spheader 474Bb  If $f:\BbbR^r\to\Bbb R$ and $\phi:\BbbR^r\to\BbbR^r$
are functions, then
$\diverg(f\times\phi)=\varinnerprod{\phi}{\grad f}+f\times\diverg\phi$
at any point at which $f$ and $\phi$ are both differentiable.
\prooflet{(Use 473Bc;  compare 473Bd.)}

\spheader 474Bc If $\phi:\BbbR^r\to\BbbR^r$ is a Lipschitz function with
compact support, then $\int\diverg\phi\,d\mu=0$.
\prooflet{\Prf\ $\diverg\phi$ is defined almost everywhere (by
Rademacher's theorem, 262Q), measurable (473Be), bounded (473Cc) and
with compact support, so

\Centerline{$\int\diverg\phi\,d\mu
=\sum_{i=1}^r\int\Pd{\phi_i}{\xi_i}d\mu$}

\noindent is defined in $\Bbb R$.   For each $i\le r$,
Fubini's theorem tells us that we can replace integration with respect
to $\mu$ by a repeated integral, in which the inner integral is

\Centerline{$\int_{-\infty}^{\infty}
\Pd{\phi_i}{\xi_i}(\xi_1,\ldots,\xi_r)d\xi_i
=0$}

\noindent because $\phi_i(\xi_1,\ldots,\xi_r)=0$ whenever $|\xi_i|$ is
large enough.   So $\int\Pd{\phi_i}{\xi_i}d\mu$ also is zero.   Summing
over $i$, we have the result.\ \Qed}

\spheader 474Bd If $\phi:\BbbR^r\to\BbbR^r$ and $f:\BbbR^r\to\Bbb R$ are
Lipschitz functions, one of which has compact support, then
$f\times\phi$ is Lipschitz.   \prooflet{\Prf\
Take $n\in\Bbb N$ such that
$f(x)\phi(x)=\tbf{0}$ for $\|x\|>n$, and $\gamma\ge 0$ such that
$|f(x)-f(y)|\le\gamma\|x-y\|$ and $\|\phi(x)-\phi(y)\|\le\gamma\|x-y\|$ for
all $x$, $y\in\BbbR^r$, while also
$|f(x)|\le\gamma$ whenever $\|x\|\le n+1$ and
$\|\phi(x)\|\le\gamma$ whenever $\|x\|\le n+1$.   If $x$, $y\in\BbbR^r$
then

\inset{----- if $\|x\|\le n+1$ and $\|y\|\le n+1$,

\Centerline{$\|f(x)\phi(x)-f(y)\phi(y)\|
\le|f(x)|\|\phi(x)-\phi(y)\|+\|\phi(y)\||f(x)-f(y)|
\le 2\gamma^2\|x-y\|$;}

----- if $\|x\|\le n$ and $\|y\|>n+1$,

\Centerline{$|f(x)\phi(x)-f(y)\phi(y)\|
=|f(x)|\|\phi(x)\|\le\gamma^2
\le\gamma^2\|x-y\|$;}

----- if $\|x\|>n$ and $\|y\|>n$, $|f(x)\phi(x)-f(y)\phi(y)\|=0$.}

\noindent So $2\gamma^2$ is a Lipschitz constant for $f\times\phi$.\ \Qed}

It follows that

\Centerline{$\int\varinnerprod{\phi}{\grad f}\,d\mu
  +\int f\times\diverg\phi\,d\mu=0$.}

\prooflet{\noindent\Prf\ $f$ and $\phi$ and $f\times\phi$
are all differentiable almost everywhere.   So

\Centerline{$\int\varinnerprod{\phi}{\grad f}\,d\mu
  +\int f\times\diverg\phi\,d\mu
=\int\diverg(f\times\phi)d\mu
=0$}

\noindent by (b) and (c) above.\ \Qed}

\spheader 474Be If $f\in\eusm L^{\infty}(\mu)$, $g\in\eusm L^1(\mu)$ is
even\cmmnt{ (that is, $g(-x)$ is defined and equal to $g(x)$ for
every $x\in\dom g$),} and $\phi:\BbbR^r\to\BbbR^r$ is a Lipschitz
function with compact support, then
$\int(f*g)\times\diverg\phi=\int f\times\diverg(g*\phi)$\cmmnt{, where
$g*\phi=(g*\phi_1,\ldots,g*\phi_r)$}.   \prooflet{\Prf\ For each $i$,

$$\eqalignno{\int(f*g)\times\Pd{\phi_i}{\xi_i}d\mu
&=\iint f(x)g(y)\Pd{\phi_i}{\xi_i}(x+y)\mu(dy)\mu(dx)\cr
\displaycause{255G/444Od}
&=\iint f(x)g(-y)\Pd{\phi_i}{\xi_i}(x+y)\mu(dy)\mu(dx)\cr
\displaycause{because $g$ is even}
&=\int f\times(g*\Pd{\phi_i}{\xi_i})d\mu
=\int f\times\Pd{}{\xi_i}(g*\phi_i)d\mu\cr}$$

\noindent as in 473Dd.   Now take the sum over $i$ of both sides.\
\Qed}

\leader{474C}{Invariance under isometries:  Proposition} Suppose that
$T:\BbbR^r\to\BbbR^r$ is an isometry, and that $\phi$ is a function
from a subset of $\BbbR^r$ to $\BbbR^r$.   Then

\Centerline{$\diverg(T^{-1}\phi T)=(\diverg\phi)T$.}

\proof{ Set $z=T(\tbf{0})$.   By 4A4Jb, the isometry $x\mapsto T(x)-z$
is linear and preserves inner products, so there is an orthogonal matrix
$S$ such that $T(x)=z+S(x)$ for every $x\in\BbbR^r$.   Now suppose that
$x\in\BbbR^r$ is such that $(\diverg\phi)(T(x))$ is defined.   Then
$T(y)-T(x)-S(y-x)=0$ for every $y$, so $T$ is differentiable at $x$,
with derivative $S$, and $\phi T$ is differentiable at $x$, with
derivative $DS$, where $D$ is the derivative of $\phi$ at $T(x)$, by
473Bc.   Also $T^{-1}(y)=S^{-1}(y-z)$ for every $y$, so $T^{-1}$ is
differentiable at $\phi(T(x))$ with derivative $S^{-1}$, and $T^{-1}\phi
T$ is differentiable at $x$, with derivative $S^{-1}DS$.   Now if $D$ is
$\langle\delta_{ij}\rangle_{1\le i,j\le r}$ and $S$ is
$\langle\sigma_{ij}\rangle_{1\le i,j\le r}$ and $S^{-1}DS$ is
$\langle\tau_{ij}\rangle_{1\le i,j\le r}$, then $S^{-1}$ is the
transpose $\langle\sigma_{ji}\rangle_{1\le i,j\le r}$ of $S$, because
$S$ is orthogonal, so

$$\eqalign{\diverg(T^{-1}\phi T)(x)
&=\sum_{i=1}^r\tau_{ii}
=\sum_{i=1}^r\sum_{j=1}^r\sigma_{ji}\sum_{k=1}^r
  \delta_{jk}\sigma_{ki}\cr
&=\sum_{j=1}^r\sum_{k=1}^r\delta_{jk}
  \sum_{i=1}^r\sigma_{ji}\sigma_{ki}
=\sum_{j=1}^r\delta_{jj}
=\diverg\phi(T(x))\cr}$$

\noindent because $\sum_{i=1}^r\sigma_{ji}\sigma_{jk}=1$ if $j=k$ and
$0$ otherwise.   If $\diverg(T^{-1}\phi T)(x)$ is defined, then (because
$T^{-1}$ also is an isometry)

\Centerline{$(\diverg\phi)(T(x))=\diverg(TT^{-1}\phi TT^{-1})(T(x))
=\diverg(T^{-1}\phi T)(T^{-1}T(x))=\diverg(T^{-1}\phi T)(x)$.}

\noindent So the functions $\diverg(T^{-1}\phi T)$ and $(\diverg\phi)T$
are identical.
}%end of proof of 474C

\leader{474D}{Locally finite perimeter:  Definition} Let
$E\subseteq\BbbR^r$ be a Lebesgue measurable set.    Its {\bf perimeter}
$\per E$ is

\Centerline{$\sup\{|\int_E\diverg\phi\,d\mu|:
  \phi:\BbbR^r\to B(\tbf{0},1)$ is a Lipschitz function with compact
support$\}$}

\noindent (allowing $\infty$).   A set $E\subseteq\BbbR^r$
has {\bf locally finite perimeter} if it is Lebesgue measurable and

\Centerline{$\sup\{|\int_E\diverg\phi\,d\mu|:
  \phi:\BbbR^r\to\BbbR^r$ is a Lipschitz function,
  $\|\phi\|\le\chi B(\tbf{0},n)\}$}

\noindent is finite for every $n\in\Bbb N$.   Of course a Lebesgue
measurable set with finite perimeter also has locally finite perimiter.
%Cacciopoli set = finite perimeter (Tamanini & Giacomelli)

\leader{474E}{Theorem} Suppose that $E\subseteq\BbbR^r$ has locally
finite perimeter.

(i) There are a Radon measure $\lambda^{\partial}_E$ on $\BbbR^r$ and a
Borel measurable function $\psi:\BbbR^r\to S_{r-1}$ such that

\Centerline{$\int_E\diverg\phi\,d\mu
=\int\varinnerprod{\phi}{\psi}\,d\lambda^{\partial}_E$}

\noindent for every Lipschitz function $\phi:\BbbR^r\to\BbbR^r$ with
compact support.

(ii) This formula uniquely determines $\lambda^{\partial}_E$, which can
also be defined by saying that

\Centerline{$\lambda^{\partial}_E(G)
=\sup\{|\int_E\diverg\phi\,d\mu|:
  \phi:\BbbR^r\to\BbbR^r$ is Lipschitz, $\|\phi\|\le\chi G\}$}

\noindent whenever $G\subseteq\BbbR^r$ is open.

(iii) If $\hat\psi$ is another function defined
$\lambda^{\partial}_E$-a.e.\ and satisfying the formula in (i),
then $\hat\psi$ and $\psi$ are equal
$\lambda^{\partial}_E$-almost everywhere.

\proof{{\bf (a)(i)} For each $l\in\Bbb N$, set

\Centerline{$\gamma_l=\sup\{|\int_E\diverg\phi\,d\mu|:
  \phi:\BbbR^r\to\BbbR^r$ is Lipschitz,
$\|\phi\|\le\chi B(\tbf{0},l)\}$.}

\noindent If $f:\BbbR^r\to\Bbb R$ is a Lipschitz function and
$f(x)=0$ for $\|x\|\ge l$, then
$|\int_E\Pd{f}{\xi_i}d\mu|\le\gamma_l\|f\|_{\infty}$ for every $i\le r$.
\Prf\ It is enough
to consider the case $\|f\|_{\infty}=1$, since the result is trivial if
$\|f\|_{\infty}=0$, and otherwise we can look at an appropriate scalar
multiple of $f$.   In this case, set $\phi(x)=f(x)e_i$ for every $x$, where
$e_i$ is
the unit vector $(0,\ldots,0,1,0,\ldots,0)$ with a $1$ in the $i$th place.
Then $\phi$ is Lipschitz and $\|\phi\|=|f|\le\chi B(\tbf{0},l)$, so

\Centerline{$|\int_E\Pd{f}{\xi_i}d\mu|
=|\int_E\diverg\phi\,d\mu|\le\gamma_l$.\ \Qed}

\medskip

\quad{\bf (ii)}
Write $C_k$ for the space of continuous functions with compact
support from $\BbbR^r$ to $\Bbb R$.   By 473Dc and 473De,
$f*\tilde h_n\in\eusm D$ for every $f\in C_k$ and $n\in\Bbb N$.
Now the point is that

\Centerline{$L_i(f)
=\lim_{n\to\infty}\int_E\Pd{}{\xi_i}(f*\tilde h_n)d\mu$}

\noindent is defined whenever $f\in C_k$ and $i\le r$.   \Prf\ Applying
473Ed, we see that
$\|f-f*\tilde h_n\|_{\infty}\to 0$ as $n\to\infty$.   Let $l$ be such that
$f(x)=0$ for $\|x\|\ge l$.   Then
$\|(f*\tilde h_m)-(f*\tilde h_n)\|_{\infty}\to 0$ as $m$,
$n\to\infty$, while all the $f*\tilde h_m$ are zero outside
$B(\tbf{0},l+1)$ (473Dc), so that

\Centerline{$\bigl|\int_E\Pd{}{\xi_i}(f*\tilde h_m)d\mu
  -\int_E\Pd{}{\xi_i}(f*\tilde h_n)d\mu\bigr|
\le\gamma_{l+1}\|(f*\tilde h_m)-(f*\tilde h_n)\|_{\infty}
\to 0$}

\noindent as $m$, $n\to\infty$.   Thus
$\sequencen{\int_E\Pd{}{\xi_i}(f*\tilde h_n)d\mu}$ is a Cauchy sequence
and must have a limit.\ \Qed

\medskip

\quad{\bf (iii)} If $f\in C_k$ is Lipschitz and zero outside
$B(\tbf{0},l)$, then

\Centerline{$|\int_E\Pd{f}{\xi_i}d\mu
  -\int_E\Pd{}{\xi_i}(f*\tilde h_n)d\mu|
\le\gamma_{l+1}\|f-f*\tilde h_n\|_{\infty}\to 0$}

\noindent as $n\to\infty$, and
$L_i(f)=\int_E\Pd{f}{\xi_i}d\mu$.   Consequently
$|L_i(f)|\le\gamma_l\|f\|_{\infty}$.

\medskip

{\bf (b)} Because all the functionals
$f\mapsto\int_E\Pd{}{\xi_i}(f*\tilde h_n)d\mu$ are linear, $L_i$ is
linear.   Moreover, by the last remark in (a-iii), it is order-bounded when
regarded as a linear functional on the Riesz space $C_k$, so is
expressible as a difference $L_i^+-L_i^-$ of positive linear functionals
(356B).

By the Riesz Representation Theorem (436J), we have Radon measures
$\lambda_i^+$, $\lambda_i^-$ on $\BbbR^r$ such that
$L_i^+(f)=\int fd\lambda_i^+$, $L_i^-(f)=\int fd\lambda_i^-$ for every
$f\in C_k$.   Let $\hat\lambda$ be the sum
$\sum_{i=1}^r\lambda_i^++\lambda_i^-$, so
that $\hat\lambda$ is a Radon measure (416De) and every $\lambda_i^+$,
$\lambda_i^-$ is an indefinite-integral measure over $\hat\lambda$
(416Sb).

For each $i\le r$, let $g_i^+$, $g_i^-$ be Radon-Nikod\'ym derivatives
of $\lambda_i^+$, $\lambda_i^-$ with respect to $\hat\lambda$.
Adjusting them on a $\hat\lambda$-negligible set if necessary, we may
suppose that they are all bounded non-negative Borel measurable
functions from $\BbbR^r$ to $\Bbb R$.   (Recall from 256C that
$\hat\lambda$ must be the completion of its restriction to the Borel
$\sigma$-algebra.)   Set $g_i=g_i^+-g_i^-$ for each $i$.   Then

$$\eqalign{\int_E\Pd{f}{\xi_i}d\mu
&=L_i^+(f)-L_i^-(f)
=\int fd\lambda_i^+-\int fd\lambda_i^-\cr
&=\int f\times g_i^+d\hat\lambda-\int f\times g_i^-d\hat\lambda
=\int f\times g_id\hat\lambda\cr}$$

\noindent for every Lipschitz function $f$ with compact support (235K).
Set $g=\sqrt{\sum_{i=1}^rg_i^2}$.   For $i\le r$, set
$\psi_i(x)=\Bover{g_i(x)}{g(x)}$ when $g(x)\ne 0$, $\Bover1{\sqrt{r}}$
when $g(x)=0$, so that
$\psi=(\psi_1,\ldots,\psi_r):\BbbR^r\to S_{r-1}$ is Borel measurable.
Let $\lambda^{\partial}_E$ be the indefinite-integral measure over
$\hat\lambda$
defined by $g$;  then $\lambda^{\partial}_E$ is a Radon measure on
$\BbbR^r$ (256E/416Sa).

\medskip

{\bf (c)} Now take any Lipschitz function $\phi:\BbbR^r\to\BbbR^r$ with
compact support.   Express it as
$(\phi_1,\ldots,\phi_r)$ where $\phi_i:\BbbR^r\to\Bbb R$ is a Lipschitz
function with compact support for each $i$.   Then

$$\eqalignno{\int_E\diverg\phi\,d\mu
&=\sum_{i=1}^r\int_E\Pd{\phi_i}{\xi_i}d\mu
=\sum_{i=1}^rL_i(\phi_i)\cr
&=\sum_{i=1}^rL_i^+(\phi_i)-\sum_{i=1}^rL_i^-(\phi_i)
=\sum_{i=1}^r\int\phi_id\hat\lambda^+_i
  -\sum_{i=1}^r\int\phi_id\hat\lambda^-_i\cr
&=\sum_{i=1}^r\int\phi_i\times g^+_id\hat\lambda
  -\sum_{i=1}^r\int\phi_i\times g^-_id\hat\lambda\cr
\displaycause{by 235K again}
&=\sum_{i=1}^r\int\phi_i\times g_id\hat\lambda
=\sum_{i=1}^r\int\phi_i\times\psi_id\lambda^{\partial}_E\cr
\displaycause{235K once more, because $\psi_i\times g=g_i$}
&=\int\varinnerprod{\phi}{\psi}\,d\lambda^{\partial}_E.\cr}$$

\noindent So we have $\lambda^{\partial}_E$ and $\psi$ satisfying (i).

\medskip

{\bf (d)} Now suppose that $G\subseteq\BbbR^r$ is open.   If
$\phi:\BbbR^r\to\BbbR^r$ is a Lipschitz function with compact support
and $\|\phi\|\le\chi G$, then

\Centerline{$|\int_E\diverg\phi\,d\mu|
=|\int\varinnerprod{\phi}{\psi}\,d\lambda^{\partial}_E|
\le\int\|\phi\|d\lambda^{\partial}_E
\le\lambda^{\partial}_E(G)$.}

\noindent On the other hand, if $\gamma<\lambda^{\partial}_E(G)$, let
$G_0\subseteq G$ be a bounded open set such that
$\gamma<\lambda^{\partial}_E(G_0)$, and set
$\epsilon=\bover13(\lambda^{\partial}_E(G_0)-\gamma)$.   Let
$K\subseteq G_0$ be a compact set such that
$\lambda^{\partial}_E(G_0\setminus K)\le\epsilon$.   Let $\delta>0$ be
such that $\|x-y\|\ge 2\delta$ whenever $y\in K$ and
$x\in\BbbR^r\setminus G_0$, and set
$H=\{x:\inf_{y\in K}\|x-y\|<\delta\}$.
Now there are $f_1,\ldots,f_r\in C_k$ such that

\Centerline{$\sum_{i=1}^rf_i^2\le\chi H$,
\quad$\int\sum_{i=1}^r
  f_i\times\psi_id\lambda^{\partial}_E
\ge\gamma$.}

\noindent\Prf\ For each $i\le r$, we can find a sequence
$\sequence{m}{g_{mi}}$ in $C_k$ such that
$\int|g_{mi}-(\psi_i\times\chi K)|d\lambda^{\partial}_E\le 2^{-m}$ for
every $m\in\Bbb N$ (416I);
multiplying the $g_{mi}$ by a function which takes the value $1$ on $K$
and $0$ outside $H$ if necessary, we can suppose that $g_{mi}(x)=0$ for
$x\notin H$.   Set

\Centerline{$f_{mi}
=\Bover{g_{mi}}{\max(1,\sqrt{\sumop_{j=1}^rg_{mj}^2})}\in C_k$}

\noindent for each $m$ and $i$.   Now
$\lim_{m\to\infty}f_{mi}(x)=\psi_i(x)$ for every $i\le r$ whenever
$\lim_{m\to\infty}g_{mi}(x)=\psi_i(x)$ for every $i\le r$, which is the
case for $\lambda^{\partial}_E$-almost every $x\in K$.   Also
$\sum_{i=1}^rf_{mi}^2\le\chi H$ for every $m$, so
$|\sum_{i=1}^rf_{mi}\times\psi_i|\le\chi H$ for every $m$, while

\Centerline{$\lim_{m\to\infty}\sum_{i=1}^r\int_K
  f_{mi}\times\psi_id\lambda^{\partial}_E
=\sum_{i=1}^r\int_K\psi_i^2d\lambda^{\partial}_E
=\lambda^{\partial}_E(K)$.}

\noindent At the same time,

\Centerline{$|\sum_{i=1}^r\int_{\Bbb R^r\setminus K}
  f_{mi}\times\psi_id\lambda^{\partial}_E|
\le\lambda^{\partial}_E(H\setminus K)\le\epsilon$}

\noindent for every $m$, so

\Centerline{$\sum_{i=1}^r\int f_{mi}\times\psi_id\lambda^{\partial}_E
\ge\lambda^{\partial}_E(G_0)-3\epsilon=\gamma$}

\noindent for all $m$ large enough, and we may take $f_i=f_{mi}$ for
such an $m$.\ \Qed

Now, for $n\in\Bbb N$, set

\Centerline{$\phi_n=(f_1*\tilde h_n,\ldots,f_r*\tilde h_n)
\in\eusm D_r$.}

\noindent For all $n$ large enough, we shall have
$\|x-y\|\ge\Bover1{n+1}$ for every $x\in\BbbR^r\setminus G_0$ and
$y\in H$, so that $\phi_n(x)=0$ if $x\notin G_0$.   By 473Dg,

\Centerline{$\|\phi_n(x)\|
\le\sup_{y\in\Bbb R}\sqrt{\sumop_{i=1}^rf_i(y)^2}\le 1$}

\noindent for every $x$ and $n$, so that $\|\phi_n\|\le\chi G_0$ for all
$n$ large enough.   Next,
$\lim_{n\to\infty}\phi_n(x)=(f_1(x),\ldots,
\ifdim\pagewidth=390pt\penalty-100\fi
f_r(x))$ for every $x\in\BbbR^r$ (473Ed), so

\Centerline{$\int_E\diverg\phi_nd\mu
=\int\varinnerprod{\phi_n}{\psi}\,d\lambda^{\partial}_E
\to\int\sum_{i=1}^rf_i\times\psi_id\lambda^{\partial}_E
\ge\gamma$}

\noindent as $n\to\infty$, by Lebesgue's Dominated Convergence Theorem.
As $\gamma$ is arbitrary,

$$\eqalign{\lambda^{\partial}_E(G)
&\le\sup\{\int_E\diverg\phi\,d\mu:
  \phi\in\eusm D_r,\,\|\phi\|\le\chi G\}\cr
&\le\sup\{\int_E\diverg\phi\,d\mu:
  \phi\text{ is Lipschitz},\,\|\phi\|\le\chi G\}\cr}$$

\noindent and we have equality.

\medskip

{\bf (e)} Thus $\lambda^{\partial}_E$ must satisfy (ii).   By 416Eb, it
is uniquely defined.   Now suppose that $\hat\psi$ is another function
from a $\lambda^{\partial}_E$-conegligible set to $\BbbR^r$ and
satisfies (i).   Then

\Centerline{$\int\varinnerprod{\phi}{\psi}\,d\lambda^{\partial}_E
=\int\varinnerprod{\phi}{\hat\psi}\,d\lambda^{\partial}_E$}

\noindent for every Lipschitz function $\phi:\BbbR^r\to\BbbR^r$ with
compact support.   Take any $i\le r$ and any compact set
$K\subseteq\BbbR^r$.   For $m\in\Bbb N$, set
$f_m(x)=\max(0,1-2^m\inf_{y\in K}\|y-x\|)$ for $x\in\BbbR^r$, so that
$\sequence{m}{f_m}$ is a sequence of Lipschitz functions with compact
support and $\lim_{m\to\infty}f_m=\chi K$.   Set

\Centerline{$\phi_{m}=(0,\ldots,f_m,\ldots,0)$,}

\noindent where the non-zero term is in the $i$th position.   Then

$$\eqalignno{\int_K\psi_id\lambda^{\partial}_E
&=\lim_{m\to\infty}\int f_m\times\psi_id\lambda^{\partial}_E
=\lim_{m\to\infty}
 \int\varinnerprod{\phi_{m}}{\psi}\,d\lambda^{\partial}_E\cr
&=\lim_{m\to\infty}
  \int\varinnerprod{\phi_{m}}{\hat\psi}\,d\lambda^{\partial}_E
=\int_K\hat\psi_id\lambda^{\partial}_E.\cr}$$

\noindent By the Monotone Class Theorem (136C), or otherwise,
$\int_F\hat\psi_id\lambda^{\partial}_E
=\int_F\psi_id\lambda^{\partial}_E$ for every bounded Borel set $F$, so
that $\hat\psi_i=\psi_i\,\,\lambda^{\partial}_E$-a.e.;  as $i$ is
arbitrary, $\psi=\hat\psi\,\,\lambda^{\partial}_E$-a.e.   This completes
the proof.
}%end of proof of 474E

\medskip

\leader{474F}{Definitions} In the context of 474E, I will call
$\lambda^{\partial}_E$ the {\bf perimeter measure} of $E$, and if $\psi$
is a function from a $\lambda^{\partial}_E$-conegligible subset of
$\BbbR^r$ to $S_{r-1}$ which has the property in (i) of the theorem, I
will call it an {\bf outward-normal} function for $E$.

\cmmnt{The words `perimeter' and `outward normal' are intended to
suggest geometric interpretations;  much of this section and the next
will be devoted to validating this suggestion.}

Observe that if $E$ has locally finite perimeter, then
$\per E=\lambda^{\partial}_E(\BbbR^r)$.   The definitions in 474D-474E also
make it clear that if $E$, $F\subseteq\Bbb R$ are Lebesgue measurable and
$\mu(E\symmdiff F)=0$, then $\per E=\per F$ and $E$ has locally finite
perimeter iff $F$ has;  and in this case
$\lambda^{\partial}_E=\lambda^{\partial}_F$ and an outward-normal function
for $E$ is an outward-normal function for $F$.

\leader{474G}{The reduced boundary} Let $E\subseteq\BbbR^r$ be a set
with locally finite
perimeter;  let $\lambda^{\partial}_E$ be its perimeter measure and
$\psi$ an outward-normal function for $E$.   The {\bf reduced boundary}
$\partial^{\$}E$ is the set of those $y\in\BbbR^r$ such that, for some
$z\in S_{r-1}$,

\Centerline{$\lim_{\delta\downarrow 0}
\Bover1{\lambda^{\partial}_EB(y,\delta)}
\int_{B(y,\delta)}\|\psi(x)-z\|\,\lambda^{\partial}_E(dx)=0$.}

\noindent\cmmnt{(When requiring that the limit be defined, I mean to
insist that $\lambda^{\partial}_EB(y,\delta)$ should be non-zero for
every $\delta>0$, that is, that $y$ belongs to the support of
$\lambda^{\partial}_E$.  {\bf Warning!}  Some authors use the phrase
`reduced boundary' for a slightly larger set.)  }Note that, writing
$\psi=(\psi_1,\ldots,\psi_r)$ and $z=(\zeta_1,\ldots,\zeta_r)$, we must
have

\Centerline{$\zeta_i
=\lim_{\delta\downarrow 0}\Bover1{\lambda^{\partial}_EB(y,\delta)}
  \int_{B(y,\delta)}\psi_id\lambda^{\partial}_E$,}

\noindent so that $z$ is uniquely defined;  call it $\psi_E(y)$.
\cmmnt{Of course }$\partial^{\$}E$ and $\psi_E$ are determined
entirely by the set $E$\cmmnt{, because
$\lambda^{\partial}_E$ is uniquely determined and $\psi$ is determined
up to a $\lambda^{\partial}_E$-negligible set (474E)}.

\cmmnt{By Besicovitch's Density Theorem (472Db),}

\Centerline{$\lim_{\delta\downarrow 0}
  \Bover1{\lambda^{\partial}_EB(x,\delta)}
  \int_{B(x,\delta)}|\psi_i(x)-\psi_i(y)|\lambda^{\partial}_E(dx)
=0$}

\noindent for every $i\le r$, for $\lambda^{\partial}_E$-almost every
$y\in\BbbR^r$;  and for any such $y$, $\psi_E(y)$ is defined and equal
to $\psi(y)$.   \cmmnt{Thus} $\partial^{\$}E$ is
$\lambda^{\partial}_E$-conegligible and $\psi_E$ is an outward-normal
function for $E$.   I will call $\psi_E:\partial^{\$}E\to S_{r-1}$ the
{\bf canonical outward-normal function} of $E$.

Once again, we see that if $E$, $F\subseteq\BbbR^r$ are sets with locally
finite perimeter and $E\symmdiff F$ is Lebesgue negligible, then
they have the same reduced boundary and the same canonical outward-normal
function.

\leader{474H}{Invariance under isometries:  Proposition} Let
$E\subseteq\BbbR^r$ be a set with locally finite perimeter.   Let
$\lambda^{\partial}_E$ be its perimeter measure, and $\psi_E$ its canonical
outward-normal function.   If $T:\BbbR^r\to\BbbR^r$ is any
isometry, then $T[E]$ has locally finite perimeter,
$\lambda^{\partial}_{T[E]}$ is the image measure
$\lambda^{\partial}_ET^{-1}$,
the reduced boundary $\partial^{\$}T[E]$ is $T[\partial^{\$}E]$,
and the canonical
outward-normal function of $T[E]$ is $S\psi_ET^{-1}$,
where $S$ is the derivative of $T$.

\proof{{\bf (a)} As noted in 474C, the derivative of $T$ is constant,
and is an orthogonal matrix.    Suppose that $n\in\Bbb N$.
Let $\phi:\BbbR^r\to\BbbR^r$ be a Lipschitz function such that
$\|\phi\|\le\chi B(\tbf{0},n)$.   Then

$$\eqalignno{|\int_{T[E]}\diverg\phi\,d\mu|
&=|\int_E(\diverg\phi)T\,d\mu|\cr
\displaycause{263D, because $|\det S|=1$}
&=|\int_E\diverg(T^{-1}\phi T)d\mu|\cr
\displaycause{474C}
&=|\int_E\diverg(S^{-1}\phi T)d\mu|\cr
\displaycause{because $S^{-1}\phi T$ and $T^{-1}\phi T$ differ by a
constant, and must have the same derivative}
&\le\lambda^{\partial}_E(T^{-1}[B(\tbf{0},n)])\cr}$$

\noindent because $S^{-1}\phi T$ is a Lipschitz function and

\Centerline{$\|S^{-1}\phi T\|=\|\phi T\|\le\chi T^{-1}[B(\tbf{0},n)]$.}

\noindent Since $T^{-1}[B(\tbf{0},n)]$ is bounded,
$\lambda^{\partial}_E(T^{-1}[B(\tbf{0},n)])$ is finite
for every $n$, and $T[E]$ has locally finite perimeter.

\medskip

{\bf (b)} We can therefore speak of its perimeter measure
$\lambda^{\partial}_{T[E]}$.   Let $G\subseteq\BbbR^r$ be an open set.
If $\phi:\BbbR^r\to\BbbR^r$ is a Lipschitz function and
$\|\phi\|\le\chi T[G]$, then

\Centerline{$|\int_{T[E]}\diverg\phi\,d\mu|
=|\int_E\diverg(S^{-1}\phi T)d\mu|
\le\lambda^{\partial}_E(G)$}

\noindent because $S^{-1}\phi T$ is a Lipschitz function dominated by
$\chi G$.   As $\phi$ is arbitrary,
$\lambda^{\partial}_{T[E]}(T[G])\le\lambda^{\partial}_E(G)$.   Applying
the same argument in reverse, with $T^{-1}$ in the place of $T$, we see
that $\lambda^{\partial}_E(G)\le\lambda^{\partial}_{T[E]}(T[G])$, so the
two are equal.   This means that the Radon measures
$\lambda^{\partial}_{T[E]}$ and $\lambda^{\partial}_ET^{-1}$ (418I)
agree on open sets, and must be identical (416Eb again).

\medskip

{\bf (c)} Now consider $S\psi_ET^{-1}$.   Since $\psi_E$ is defined
$\lambda^{\partial}_E$-almost everywhere and takes values in $S_{r-1}$,
$\psi_E T^{-1}$ and $S\psi_E T^{-1}$ are defined
$\lambda^{\partial}_{T[E]}$-almost everywhere and take values in
$S_{r-1}$.   If $\phi:\BbbR^r\to\BbbR^r$ is a Lipschitz function with
compact support,

$$\eqalignno{\int_{T[E]}\diverg\phi\,d\mu
&=\int_E(\diverg\phi)T\,d\mu
=\int_E\diverg(T^{-1}\phi T)d\mu\cr
&=\int_E\diverg(S^{-1}\phi T)d\mu
=\int\varinnerprod{(S^{-1}\phi T)}{\psi_E}\,d\lambda^{\partial}_E\cr
&=\int\varinnerprod{(\phi T)}{(S\psi_E)}d\lambda^{\partial}_E\cr
\displaycause{because $S$ is orthogonal}
&=\int\varinnerprod{\phi}{(S\psi_E T^{-1})}
  d(\lambda^{\partial}_ET^{-1})\cr
\displaycause{235G}
&=\int\varinnerprod{\phi}{(S\psi_E T^{-1})}
  d\lambda^{\partial}_{T[E]}.\cr}$$

\noindent Accordingly $S\psi_ET^{-1}$ is an outward-normal function for
$T[E]$.    Write $\psi_{T[E]}$ for the canonical outward-normal function
of $T[E]$.

\medskip

{\bf (d)} Take $y\in\Bbb R$ and consider

$$\eqalign{\Bover1{\lambda^{\partial}_{T[E]}B(y,\delta)}
   &\int_{B(y,\delta)}\|S\psi_ET^{-1}(x)-S\psi_ET^{-1}(y)\|
   \lambda^{\partial}_{T[E]}(dx)\cr
&=\Bover1{\lambda^{\partial}_EB(T^{-1}(y),\delta)}
   \int_{B(T^{-1}(y),\delta)}\|S\psi_E(x)-S\psi_ET^{-1}(y)\|
   \lambda^{\partial}_E(dx)\cr
&=\Bover1{\lambda^{\partial}_EB(T^{-1}(y),\delta)}
   \int_{B(T^{-1}(y),\delta)}\|\psi_E(x)-\psi_ET^{-1}(y)\|
   \lambda^{\partial}_E(dx)\cr}$$

\noindent for any $\delta>0$ for which

\Centerline{$\lambda^{\partial}_{T[E]}B(y,\delta)
=\lambda^{\partial}_ET^{-1}[B(y,\delta)]
=\lambda^{\partial}_EB(T^{-1}(y),\delta)$}

\noindent is non-zero.   We see that

\Centerline{$\lim_{\delta\downarrow 0}
   \Bover1{\lambda^{\partial}_{T[E]}B(y,\delta)}
   \int_{B(y,\delta)}\|S\psi_ET^{-1}(x)-S\psi_ET^{-1}(y)\|
   \lambda^{\partial}_{T[E]}(dx)$}

\noindent is defined and equal to $0$ whenever

\Centerline{$\Bover1{\lambda^{\partial}_EB(T^{-1}(y),\delta)}
   \int_{B(T^{-1}(y),\delta)}\|\psi_E(x)-\psi_ET^{-1}(y)\|
   \lambda^{\partial}_E(dx)$}

\noindent is defined and equal to $0$, that is,
$T^{-1}(y)\in\partial^{\$}E$.   In this case,
$y\in\partial^{\$}T[E]$ and $S\psi_ET^{-1}(y)=\psi_{T[E]}(y)$.
So $\partial^{\$}T[E]\supseteq T[\partial^{\$}E]$
and $S\psi_ET^{-1}$ extends $\psi_{T[E]}$.

Applying the argument to $T^{-1}$, we see that
$S^{-1}\psi_{T[E]}T$ extends $\psi_E$, that is,
$\psi_{T[E]}$ extends $S\psi_ET^{-1}$.   So
$S\psi_ET^{-1}$ is exactly the canonical outward-normal function of
$T[E]$, and its domain $T[\partial^{\$}E]$ is $\partial^{\$}T[E]$.
}%end of proof of 474H

\leader{474I}{\dvrocolon{Half-spaces}}\cmmnt{ It will be useful, and
perhaps instructive, to check the most elementary special case.

\medskip

\noindent}{\bf Proposition} Let $H\subseteq\BbbR^r$ be a half-space
$\{x:\varinnerprod{x}{v}\le\alpha\}$, where $v\in S^{r-1}$.   Then $H$
has locally finite perimeter;  its perimeter measure
$\lambda^{\partial}_H$ is defined by saying

\Centerline{$\lambda^{\partial}_H(F)=\nu(F\cap\partial H)$}

\noindent whenever $F\subseteq\BbbR^r$ is such that $\nu$ measures
$F\cap\partial H$, and the constant function with value $v$ is an
outward-normal function for $H$.

\proof{{\bf (a)} Suppose, to begin with, that $v$ is the unit vector
$(0,\ldots,0,1)$ and that $\alpha=0$, so that $H=\{x:\xi_r\le 0\}$.   Let
$\phi:\BbbR^r\to\BbbR^r$ be a Lipschitz function with compact support.
Then for any $i<r$

\Centerline{$\int_{H}\Pd{\phi_i}{\xi_i}\mu(dx)=0$}

\noindent because we can regard this as a multiple integral in which the
inner integral is with respect to $\xi_i$ and is therefore always zero.
On the other hand, integrating with respect to the $r$th coordinate
first,

$$\eqalignno{\int_H\Pd{\phi_r}{\xi_r}\mu(dx)
&=\int_{\Bbb R^{r-1}}\int_{-\infty}^0
  \Pd{\phi_r}{\xi_r}(z,t)dt\,\mu_{r-1}(dz)\cr
&=\int_{\Bbb R^{r-1}}\phi_r(z,0)\mu_{r-1}(dz)
=\int_{\partial H}\phi_r(x)\nu(dx)\cr
\displaycause{identifying $\nu$ on $\BbbR^{r-1}\times\{0\}$ with
$\mu_{r-1}$ on $\BbbR^{r-1}$}
&=\int_{\partial H}\varinnerprod{\phi}{v}\,d\nu
=\int\varinnerprod{\phi}{v}\,d\lambda\cr}$$

\noindent where $\lambda$ is the indefinite-integral measure over $\nu$
defined by the function $\chi(\partial H)$.   Note that (by 234La)
$\lambda$ can also be regarded as $\nu_{\partial H}\iota^{-1}$, where
$\nu_{\partial H}$ is the subspace measure on $\partial H$ and
$\iota:\partial H\to\BbbR^r$ is the identity map.   Now
$\nu_{\partial H}$ can be identified with Lebesgue measure on
$\BbbR^{r-1}$, by 265B or otherwise, so in particular is a Radon
measure, and $\lambda$ also is a Radon measure, by 418I again or otherwise.

This means that $\lambda$ and the constant function with value $v$
satisfy the conditions of 474E, and must be the perimeter measure of $H$
and an outward-normal function.

\medskip

{\bf (b)} For the general case, let $S$ be an orthogonal matrix such
that $S(0,\ldots,0,1)=v$, and set $T(x)=S(x)+\alpha v$ for every $x$, so
that $H=T[\{x:\xi_r\le 0\}]$.   By 474H, the perimeter measure of $H$ is
$\lambda T^{-1}$ and the constant function with value $v$ is an
outward-normal function for $H$.   Now the Radon measure
$\lambda^{\partial}_H=\lambda T^{-1}$ is defined by saying that

\Centerline{$\lambda^{\partial}_HF
=\lambda T^{-1}[F]
=\nu(T^{-1}[F]\cap\{x:\xi_r=0\})
=\nu(F\cap T[\{x:\xi_r=0\}])
=\nu(F\cap\partial H)$}

\noindent whenever $\nu(F\cap\partial H)$ is defined, because $\nu$
(being a scalar multiple of a Hausdorff measure) must be invariant under
the isometry $T$.
}%end of proof of 474I

\leader{474J}{Lemma} Let $E\subseteq\BbbR^r$ be a set with locally
finite perimeter.   Let $\lambda^{\partial}_E$ be the perimeter measure
of $E$, and $\psi_E$ its canonical outward-normal function.   Then
$\BbbR^r\setminus E$ also has locally finite perimeter;  its perimeter
measure is $\lambda^{\partial}_E$, its reduced boundary is
$\partial^{\$}E$, and its canonical outward-normal function is
$-\psi_E$.

\proof{ Of course $\BbbR^r\setminus E$ is Lebesgue measurable.
By 474Bc,

\Centerline{$\int_{\Bbb R^r\setminus E}\diverg\phi\,d\mu
=-\int_E\diverg\phi\,d\mu
=\int\varinnerprod{\phi}{(-\psi_E)}\,d\lambda^{\partial}_E$}

\noindent for every Lipschitz function $\phi:\BbbR^r\to\BbbR^r$ with
compact support.   The uniqueness assertions in 474E tell us that
$\BbbR^r\setminus E$ has locally
finite perimeter, that its perimeter measure is $\lambda^{\partial}_E$,
and that $-\psi_E$ is an outward-normal function for
$\BbbR^r\setminus E$.   Referring to the definition of `reduced
boundary' in 474G, we see at once that
$\partial^{\$}(\BbbR^r\setminus E)=\partial^{\$}E$ and that
$\psi_{\Bbb R^r\setminus E}=-\psi_E$.
}%end of proof of 474J

\leader{474K}{Lemma} Let $E\subseteq\BbbR^r$ be a set with locally
finite perimeter;  let $\lambda^{\partial}_E$ be its perimeter measure,
and $\psi$ an outward-normal function for $E$.   Let
$\phi:\BbbR^r\to\BbbR^r$ be a Lipschitz function with compact support,
and $g\in\eusm D$ an even function.   Then

\Centerline{$\int\varinnerprod{\phi}{\grad(g*\chi E)}d\mu
  +\int\varinnerprod{(g*\phi)}{\psi}\,d\lambda^{\partial}_E
=0$.}

\proof{

$$\eqalignno{\int\varinnerprod{\phi}{\grad(g*\chi E)}d\mu
&=-\int(g*\chi E)\times\diverg\phi\,d\mu\cr
\displaycause{474Bd, using 473Dd to see that $g*\chi E$ is
Lipschitz}
&=-\int\chi E\times\diverg(g*\phi)d\mu\cr
\displaycause{474Be}
&=-\int\varinnerprod{(g*\phi)}{\psi}\,d\lambda^{\partial}_E\cr}$$

\noindent (because $g*\phi$ is smooth and has compact support, so is
Lipschitz), as required.
}%end of proof of 474K

\leader{474L}{Two isoperimetric inequalities:  Theorem} Let
$E\subseteq\BbbR^r$ be a set with locally finite perimeter, and
$\lambda^{\partial}_E$ its perimeter measure.

(a) If $E$ is bounded, then $(\mu E)^{(r-1)/r}\le\per E$.

(b) If $B\subseteq\BbbR^r$ is a closed ball, then

\Centerline{$\min(\mu(B\cap E),\mu(B\setminus E))^{(r-1)/r}
\le 2c\lambda^{\partial}_E(\interior B)$,}

\noindent where $c=2^{r+4}\sqrt r(1+2^{r+1})$.

\proof{{\bf (a)} Let $\epsilon>0$.   By 473Ef, there is an $n\in\Bbb N$
such that
$\|f-\chi E\|_{r/(r-1)}\le\epsilon$, where $f=\chi E*\tilde h_n$.   Note
that $f$ is smooth (473De again) and has compact support, because $E$ is
bounded.   Let $\eta>0$ be such that

\Centerline{$\int\|\grad f\|d\mu
\le\int\Bover{\|\grad f\|^2}{\sqrt{\eta+\|\grad f\|^2}}
  \,d\mu+\epsilon$,}

\noindent and set $\phi=\Bover{\grad f}{\sqrt{\eta+\|\grad f\|^2}}$.
Then $\phi\in\eusm D_r$ and $\|\phi(x)\|\le 1$ for every $x\in\BbbR^r$.
Now we can estimate

$$\eqalignno{\int\|\grad f\|d\mu
&\le\int\varinnerprod{\phi}{\grad f}\,d\mu+\epsilon\cr
&=-\int\varinnerprod{(\tilde h_n*\phi)}{\psi}\,d\lambda^{\partial}_E
  +\epsilon\cr
\displaycause{where $\psi$ is an outward-normal function for $E$, by
474K}
&\le\int\|\tilde h_n*\phi\|d\lambda^{\partial}_E+\epsilon
\le\per E+\epsilon\cr}$$

\noindent because $\|(\tilde h_n*\phi)(x)\|\le 1$ for every
$x\in\BbbR^r$, by 473Dg.   Accordingly

$$\eqalignno{(\mu E)^{(r-1)/r}
&=\|\chi E\|_{r/(r-1)}
\le\|f\|_{r/(r-1)}+\epsilon
\le\int\|\grad f\|d\mu+\epsilon\cr
\displaycause{473H}
&\le\per E+2\epsilon.\cr}$$

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

\medskip

{\bf (b)(i)} Set
$\alpha=\min(\mu(B\cap E),\mu(B\setminus E))$.   If $\alpha=0$, the
result is trivial;  so suppose that $\alpha>0$.   Take
any $\epsilon\in\ocint{0,\alpha}$.   Let $B_1$ be a closed ball, with
the same centre as $B$ and strictly smaller non-zero radius, such that
$\mu(B\setminus B_1)\le\epsilon$;  then
$\alpha-\epsilon\le\min(\mu(B_1\cap E),\mu(B_1\setminus E))$.   For
$f\in\eusm L^{r/(r-1)}(\mu)$ set

\Centerline{$\gamma_0(f)=\Bover1{\mu B_1}\int_{B_1}fd\mu$,
\quad$\gamma_1(f)=\|(f\times\chi B_1)-\gamma_0(f)\chi B_1\|_{r/(r-1)}$;}

\noindent then both $\gamma_0$ and $\gamma_1$ are continuous functions
on $\eusm L^{r/(r-1)}(\mu)$ if we give it its usual pseudometric
$(f,g)\mapsto\|f-g\|_{r/(r-1)}$.   Now
$\gamma_1(\chi(E\cap B))\ge\bover12(\alpha-\epsilon)^{(r-1)/r}$.   \Prf\
We have

\Centerline{$\gamma_0(\chi(E\cap B))=\Bover{\mu(B_1\cap E)}{\mu B_1}$,}

$$\eqalign{\gamma_1(\chi(E\cap B))^{r/(r-1)}
&=\int_{B_1}|\chi(E\cap B)
  -\gamma_0(\chi(E\cap B))|^{r/(r-1)}\cr
&=\mu(B_1\cap E)\bigl(1-\Bover{\mu(B_1\cap E)}{\mu B_1}\bigr)^{r/(r-1)}
   +\mu(B_1\setminus E)
     \bigl(\Bover{\mu(B_1\cap E)}{\mu B_1}\bigr)^{r/(r-1)}\cr
&=\mu(B_1\cap E)
    \bigl(\Bover{\mu(B_1\setminus E)}{\mu B_1}\bigr)^{r/(r-1)}
   +\mu(B_1\setminus E)
    \bigl(\Bover{\mu(B_1\cap E)}{\mu B_1}\bigr)^{r/(r-1)}.\cr}$$

\noindent Either $\mu(B_1\cap E)\ge\bover12\mu B_1$ or
$\mu(B_1\setminus E)\ge\bover12\mu B_1$;  suppose the former.   Then

\Centerline{$\gamma_1(\chi(E\cap B))^{r/(r-1)}
\ge\Bover1{2^{r/(r-1)}}\mu(B_1\setminus E)
\ge\Bover1{2^{r/(r-1)}}(\alpha-\epsilon)$}

\noindent and
$\gamma_1(\chi(E\cap B))\ge\Bover12(\alpha-\epsilon)^{(r-1)/r}$.
Exchanging $B_1\cap E$ and $B_1\setminus E$ we have the same result if
$\mu(B_1\cap E)\ge\bover12\mu B_1$.\ \Qed

\medskip

\quad{\bf (ii)} Express $B$ as $B(y,\delta)$ and $B_1$ as
$B(y,\delta_1)$.   Take $n_0\ge\Bover2{\delta-\delta_1}$.  Because
$\gamma_1$ is
$\|\,\|_{r/(r-1)}$-continuous, there is an $n\ge n_0$ such that
$\gamma_1(f)\ge\bover12(\alpha-\epsilon)^{(r-1)/r}-\epsilon$, where
$f=\tilde h_n*\chi(E\cap B)$ (473Ef);  as in part (a) of the proof,
$f\in\eusm D$.   Let $\eta>0$ be such that

\Centerline{$\int_{B_1}
  \Bover{\|\grad f\|^2}{\sqrt{\eta+\|\grad f\|^2}}d\mu
\ge\int_{B_1}\|\grad f\|d\mu-\epsilon$.}

\noindent Let $m\ge n_0$ be such that
$\int\varinnerprod{\phi}{\grad f}\,d\mu
\ge\int_{B_1}\|\grad f\|d\mu-2\epsilon$, where

\Centerline{$\phi=\tilde h_m
  *(\Bover{\grad f}{\sqrt{\eta+\|\grad f\|^2}}\times\chi B_1)$.}

\noindent Note that $\phi(x)=0$ if
$\|x-y\|\ge\bover12(\delta+\delta_1)$, so that $(\tilde h_n*\phi)(x)=0$
if $x\notin\interior B$.   By 473Dg, $\|\phi(x)\|\le 1$ for every $x$
and $\|(\tilde h_n*\phi)(x)\|\le 1$ for every $x$, so
$\|\tilde h_n*\phi\|\le\chi(\interior B)$.

Now we have

$$\eqalignno{\int\varinnerprod{\phi}{\grad f}\,d\mu
&=\int\varinnerprod{\phi}{\grad(\tilde h_n*\chi(E\cap B))}\,d\mu\cr
&=-\int(\tilde h_n*\chi(E\cap B))\times\diverg\phi\,d\mu\cr
\displaycause{474Bd}
&=-\int_{E\cap B}\diverg(\tilde h_n*\phi)d\mu\cr
\displaycause{474Be}
&=-\int_E\diverg(\tilde h_n*\phi)d\mu
\le\lambda^{\partial}_E(\interior B)\cr}$$

\noindent (474E).

\medskip

\quad{\bf (iii)} Accordingly

$$\eqalignno{\Bover12(\alpha-\epsilon)^{(r-1)/r}-\epsilon
&\le\gamma_1(f)
\le c\int_{B_1}\|\grad f\|d\mu\cr
\displaycause{473K}
&\le c\bigl(\int\varinnerprod{\phi}{\grad f}\,d\mu+2\epsilon\bigr)
\le c(\lambda^{\partial}_E(\interior B)+2\epsilon).\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\alpha^{(r-1)/r}\le 2c\lambda^{\partial}_E(\interior B)$, as claimed.
}%end of proof of 474L

\leader{474M}{Lemma} Suppose that $E\subseteq\BbbR^r$ has locally finite
perimeter, with perimeter measure $\lambda^{\partial}_E$ and an
outward-normal function $\psi$.
Then for any $y\in\BbbR^r$ and any Lipschitz function
$\phi:\BbbR^r\to\BbbR^r$,

\Centerline{$\int_{E\cap B(y,\delta)}\diverg\phi\,d\mu
=\int_{B(y,\delta)}\varinnerprod{\phi}{\psi}\,d\lambda^{\partial}_E
+\int_{E\cap\partial B(y,\delta)}
  \varinnerprod{\phi(x)}{\Bover1{\delta}(x-y)}\,\nu(dx)$}

\noindent for almost every $\delta>0$.

\proof{{\bf (a)} For $t>0$, set

\Centerline{$w(t)
=\int_{E\cap\partial B(y,t)}\varinnerprod{\phi(x)}
  {\Bover1t(x-y)}\,\nu(dx)$}

\noindent when this is defined.   By 265G, applied to functions of the
form

$$x\mapsto
\oldcases{\varinnerprod{\phi(x)}{\Bover{x-y}{\|x-y\|}}
  &if $x\in E$ and $0<\|x-y\|\le\alpha$\cr
0&otherwise,\cr}$$

\noindent $w$ is defined almost everywhere in $\ooint{0,\infty}$ and is
measurable (for Lebesgue measure on $\Bbb R$).

Let $\delta>0$ be any point in the Lebesgue set of $w$ (223D).   Then

\Centerline{$\lim_{t\downarrow 0}\Bover1t\int_{\delta}^{\delta+t}
  |w(s)-w(\delta)|ds
\le 2\lim_{t\downarrow 0}\Bover1{2t}\int_{\delta-t}^{\delta+t}
  |w(s)-w(\delta)|
=0$.}

\noindent Let $\epsilon>0$.   Then there is an $\eta>0$ such that

\Centerline{$\Bover1{\eta}\int_{\delta}^{\delta+\eta}
  |w(s)-w(\delta)|ds
  \le\epsilon$,
\quad$\int_{B(y,\delta+\eta)\setminus B(y,\delta)}
  \|\phi\|d\lambda^{\partial}_E\le\epsilon$,}

\Centerline{$\int_{B(y,\delta+\eta)\setminus B(y,\delta)}
  \|\diverg\phi\|d\mu
\le\epsilon$.}

\medskip

{\bf (b)} Set

$$\eqalign{g(x)&=1\text{ if }\|x-y\|\le\delta,\cr
&=1-\Bover1{\eta}(\|x-y\|-\delta)
  \text{ if }\delta\le\|x-y\|\le\delta+\eta,\cr
&=0\text{ if }\|x-y\|\ge\delta+\eta.\cr}$$

\noindent Then $g$ is continuous, and $\grad g(x)=\tbf{0}$ if
$\|x-y\|<\delta$
or $\|x-y\|>\delta+\eta$;  while if $\delta<\|x-y\|<\delta+\eta$,
$\grad g(x)=-\Bover{x-y}{\eta\|x-y\|}$.   This means that

$$\eqalign{\int_E\varinnerprod{\phi}{\grad g}\,d\mu
&=-\Bover1{\eta}\int_{\delta}^{\delta+\eta}\int_{E\cap\partial B(y,t)}
  \varinnerprod{\Bover1t(x-y)}{\phi(x)}\,\nu(dx)dt\cr
&=-\Bover1{\eta}\int_{\delta}^{\delta+\eta}w(t)dt.\cr}$$

\noindent By the choice of $\eta$,

\Centerline{$|\int_E\varinnerprod{\phi}{\grad g}\,d\mu+w(\delta)|
\le\epsilon$.}

\medskip

{\bf (c)} By 474E and 474Bb we have

$$\eqalignno{\int\varinnerprod{(g\times\phi)}{\psi}
  \,d\lambda^{\partial}_E
&=\int_E\diverg(g\times\phi)d\mu\cr
\displaycause{of course $g\times\phi$ is Lipschitz, by 473Ca and 262Ba}
&=\int_E\varinnerprod{\phi}{\grad g}\,d\mu
   +\int_Eg\times\diverg\phi\,d\mu.\cr}$$

\noindent Next, by the choice of $\eta$,

\Centerline{$|\int\varinnerprod{((g\times\phi)}{\psi}\,
  d\lambda^{\partial}_E-\int_{B(y,\delta)}\varinnerprod{\phi}{\psi}
  \,d\lambda^{\partial}_E|
\le\int_{B(y,\delta+\eta)\setminus B(y,\delta)}
  \|\phi\|d\lambda^{\partial}_E
\le\epsilon$,}

\noindent while

\Centerline{$|\int_E\varinnerprod{\phi}{\grad g}\,d\mu
+\int_{E\cap\partial B(y,\delta)}
    \varinnerprod{\phi(x)}{\Bover1{\delta}(x-y)}\nu(dx)|
=|\int_E\varinnerprod{\phi}{\grad g}\,d\mu+w(\delta)|
\le\epsilon$}

\noindent and

$$\eqalign{\bigl|\int_Eg\times\diverg\phi\,d\mu
  &-\int_{E\cap B(y,\delta)}\diverg\phi\,d\mu\bigr|\cr
&\le\int_{B(y,\delta+\eta)\setminus B(y,\delta)}\|\diverg\phi\|d\mu
\le\epsilon.\cr}$$

\noindent Putting these together, we have

\Centerline{$|\int_{E\cap B(y,\delta)}\diverg\phi\,d\mu
-\int_{B(y,\delta)}\varinnerprod{\phi}{\psi}\,d\lambda^{\partial}_E
-\int_{E\cap\partial B(y,\delta)}
  \varinnerprod{\phi(x)}{\Bover1{\delta}(x-y)}\nu(dx)|
\le 3\epsilon$.}

\noindent As $\epsilon$ is arbitrary, this gives the result.
}%end of proof of 474M

\vleader{60pt}{474N}{Lemma} Let $E\subseteq\BbbR^r$ be a set with locally
finite perimeter, and $\lambda^{\partial}_E$ its perimeter measure.
Then, for any $y\in\partial^{\$}E$,

\quad(i) $\liminf_{\delta\downarrow 0}
  \Bover{\mu(B(y,\delta)\cap E)}{\delta^r}
\ge\Bover1{(3r)^r}$;

\quad(ii) $\liminf_{\delta\downarrow 0}
  \Bover{\mu(B(y,\delta)\setminus E)}{\delta^r}
\ge\Bover1{(3r)^r}$;

\quad(iii) $\liminf_{\delta\downarrow 0}
  \Bover{\lambda^{\partial}_EB(y,\delta)}{\delta^{r-1}}
\ge\Bover1{2c(3r)^{r-1}}$,

\noindent where $c=2^{r+4}\sqrt r(1+2^{r+1})$;

\quad(iv) $\limsup_{\delta\downarrow 0}
  \Bover{\lambda^{\partial}_EB(y,\delta)}{\delta^{r-1}}
\le 4\pi\beta_{r-2}$.

\proof{{\bf (a)} Let $\psi_E$ be the canonical outward-normal function
of $E$ (474G).   Take $y\in\partial^{\$}E$.   Set

\Centerline{$\Phi=\{\phi:\phi$ is a Lipschitz function from
$\BbbR^r$ to $B(\tbf{0},1)\}$.}

\noindent Because the space $L^1(\mu)$ is separable in its usual (norm)
topology (244I), so is
$\{(\diverg\phi\times\chi B(y,1))^{\ssbullet}:\phi\in\Phi\}$
(4A2P(a-iv)), and there must be
a countable set $\Phi_0\subseteq\Phi$ such that

\inset{whenever $\phi\in\Phi$ and $m\in\Bbb N$ there is a
$\hat\phi\in\Phi_0$ such that
\discrcenter{390pt}
{$\int_{B(y,1)}|\diverg\phi-\diverg\hat\phi|d\mu\le 2^{-m}$.}}

\noindent Now, for each $\phi\in\Phi_0$,

$$\eqalign{|\int_{E\cap B(y,\delta)}\diverg\phi\,d\mu|
&=|\int_{B(y,\delta)}\varinnerprod{\phi}{\psi_E}\,d\lambda^{\partial}_E
+\int_{E\cap\partial B(y,\delta)}
  \varinnerprod{\phi(x)}{\Bover1{\delta}(x-y)}\nu(dx)|\cr
&\le\lambda^{\partial}_EB(y,\delta)
+\nu(E\cap\partial B(y,\delta))\cr}$$

\noindent for almost every $\delta>0$, by 474M.   But this means that,
for almost every $\delta\in\ocint{0,1}$,

$$\eqalign{\per(E\cap B(y,\delta))
&=\sup_{\phi\in\Phi}|\int_{E\cap B(y,\delta)}\diverg\phi\,d\mu|\cr
&=\sup_{\phi\in\Phi_0}|\int_{E\cap B(y,\delta)}\diverg\phi\,d\mu|
\le\lambda^{\partial}_EB(y,\delta)
  +\nu(E\cap\partial B(y,\delta)).\cr}$$

\medskip

{\bf (b)} It follows that, for some $\delta_0>0$,

\Centerline{$\per(E\cap B(y,\delta))
\le 3\nu(E\cap\partial B(y,\delta))$}

\noindent for almost every $\delta\in\ocint{0,\delta_0}$.   \Prf\
Applying 474M with $\phi(x)=\psi_E(y)$ for every $x$, we have

\Centerline{$0
=\int_{B(y,\delta)}\varinnerprod{\psi_E(y)}{\psi_E(x)}
     \lambda^{\partial}_E(dx)
  +\int_{E\cap\partial B(y,\delta)}
    \varinnerprod{\psi_E(y)}{\Bover1{\delta}(x-y)}\nu(dx)$}

\noindent for almost every $\delta\in[0,1]$.   But by the definition of
$\psi_E(y)$,

\Centerline{$\lim_{\delta\downarrow 0}
  \Bover1{\lambda^{\partial}_EB(y,\delta)}\int_{B(y,\delta)}
     \varinnerprod{\psi_E(y)}{\psi_E}\,d\lambda^{\partial}_E
=\lim_{\delta\downarrow 0}
  \Bover1{\lambda^{\partial}_EB(y,\delta)}\int_{B(y,\delta)}
     \varinnerprod{\psi_E(y)}{\psi_E(y)}\,d\lambda^{\partial}_E
=1$.}

\noindent So there is some $\delta_0>0$ such that, for almost every
$\delta\in\ocint{0,\delta_0}$,

$$\eqalignno{\lambda^{\partial}_EB(y,\delta)
&\le 2\int_{B(y,\delta)}
  \varinnerprod{\psi_E(y)}{\psi_E}\,d\lambda^{\partial}_E\cr
&=-2\int_{E\cap\partial B(y,\delta)}
   \varinnerprod{\psi_E(y)}{\Bover1{\delta}(x-y)}\nu(dx)
\le 2\nu(E\cap\partial B(y,\delta)).&(\dagger)\cr}$$

\noindent But this means that, for almost every such $\delta$,

\Centerline{$\per(E\cap B(y,\delta))
\le\lambda^{\partial}_EB(y,\delta)
  +\nu(E\cap\partial B(y,\delta))
\le 3\nu(E\cap\partial B(y,\delta))$.   \Qed}

\medskip

{\bf (c)} Set $g(t)=\mu(E\cap B(y,t))$ for $t\ge 0$.   By 265G,
$g(t)=\int_0^{t}\nu(E\cap\partial B(y,s))ds$ for every $t$, so $g$ is
absolutely continuous on $[0,1]$ and $g'(t)=\nu(E\cap\partial B(y,t))$
for almost every $t$.   Now we can estimate

$$\eqalignno{g(t)^{(r-1)/r}
&=\mu(E\cap B(y,t))^{(r-1)/r}
\le\per(E\cap B(y,t))\cr
\displaycause{474La}
&\le 3\nu(E\cap\partial B(y,t))
=3g'(t)\cr}$$

\noindent for almost every $t\in[0,\delta_0]$.   So

\Centerline{$\Bover{d}{dt}\bigl(g(t)^{1/r}\bigr)
=\Bover1rg(t)^{(1-r)/r}g'(t)\ge\Bover1{3r}$}

\noindent for almost every $t\in[0,\delta_0]$;  since
$t\mapsto g(t)^{1/r}$ is non-decreasing, $g(t)^{1/r}\ge\Bover{t}{3r}$
(222C) and $g(t)\ge(3r)^{-r}t^r$ for every $t\in[0,\delta_0]$.

\medskip

{\bf (d)} Accordingly

\Centerline{$\liminf_{\delta\downarrow 0}
  \Bover{\mu(B(y,\delta)\cap E)}{\delta^r}
\ge\inf_{0<\delta\le\delta_0}
  \Bover{\mu(B(y,\delta)\cap E)}{\delta^r}
\ge\Bover1{(3r)^r}$.}

\noindent This proves (i).

\medskip

{\bf (e)} Because
$\lambda^{\partial}_{\Bbb R^r\setminus E}=\lambda^{\partial}_E$ and
$-\psi_E$ is the canonical outward-normal function of
$\BbbR^r\setminus E$ (474J), $y$ also belongs to
$\partial^{\$}(\BbbR^r\setminus E)$, so
the second formula of this lemma follows from the first.

\medskip

{\bf (f)} By 474Lb,

\Centerline{$\lambda^{\partial}_EB(y,\delta)
\ge\Bover1{2c}\min\bigl(\mu(B(y,\delta)\cap E),
  \mu(B(y,\delta)\setminus E)\bigr)^{(r-1)/r}$}

\noindent for every $\delta\ge 0$.   So

$$\eqalign{\liminf_{\delta\downarrow 0}
  \bover{\lambda^{\partial}_EB(y,\delta)}{\delta^{r-1}}
&\ge\Bover1{2c}\liminf_{\delta\downarrow 0}
  \min\bigl(\bover{\mu(B(y,\delta)\cap E)}{\delta^r},
  \bover{\mu(B(y,\delta)\setminus E)}{\delta^r}\bigr)^{(r-1)/r}\cr
&\ge\Bover1{2c}\min\bigl(\liminf_{\delta\downarrow 0}
  \bover{\mu(B(y,\delta)\cap E)}{\delta^r},
  \liminf_{\delta\downarrow 0}\bover{\mu(B(y,\delta)\setminus E)}{\delta^r}
  \bigr)^{(r-1)/r}\cr
&\ge\Bover1{2c}\bigl(\Bover1{(3r)^r}\bigr)^{(r-1)/r}
=\Bover1{2c(3r)^{r-1}}.\cr}$$

\noindent Thus (iii) is true.

\medskip

{\bf (g)} Returning to the inequality ($\dagger$) in the proof of (b)
above, we have a $\delta_0>0$ such that

\Centerline{$\lambda^{\partial}_EB(y,\delta)
\le 2\nu(E\cap\partial B(y,\delta))
\le 2\nu(\partial B(y,\delta))=4\pi\beta_{r-2}\delta^{r-1}$}

\noindent (265F) for almost every $\delta\in\ocint{0,\delta_0}$.   But this means that, for any $\delta\in\coint{0,\delta_0}$,

\Centerline{$\lambda^{\partial}_EB(y,\delta)
\le\inf_{t>\delta}\lambda^{\partial}_EB(y,t)
\le\inf_{t>\delta}4\pi\beta_{r-2}t^{r-1}
=4\pi\beta_{r-2}\delta^{r-1}$,}

\noindent and (iv) is true.
}%end of proof of 474N

\leader{474O}{Definition} Let $A\subseteq\BbbR^r$ be any set, and
$y\in\BbbR^r$.   A {\bf Federer exterior normal to
$A$ at $y$} is a $v\in S_{r-1}$ such that,

\Centerline{$\lim_{\delta\downarrow 0}
  \Bover{\mu^*((H\symmdiff A)\cap B(y,\delta))}{\mu B(y,\delta)}=0$,}

\noindent where $H$ is the half-space
$\{x:\varinnerprod{(x-y)}{v}\le 0\}$.

\leader{474P}{Lemma} If $A\subseteq\BbbR^r$ and $y\in\BbbR^r$, there can
be at most one Federer exterior normal to $A$ at $y$.

\proof{ Suppose that $v$, $v'\in S_{r-1}$ are two Federer exterior
normals to $E$ at $y$.   Set

\Centerline{$H=\{x:\varinnerprod{(x-y)}{v}\le 0\}$,
\quad$H'=\{x:\varinnerprod{(x-y)}{v'}\le 0\}$.}


\noindent Then

\Centerline{$\lim_{\delta\downarrow 0}
  \Bover{\mu((H\symmdiff H')\cap B(y,\delta))}{\mu B(y,\delta)}
\le\lim_{\delta\downarrow 0}
  \Bover{\mu^*((H\symmdiff A)\cap B(y,\delta))}{\mu B(y,\delta)}
 +\lim_{\delta\downarrow 0}
  \Bover{\mu^*((H'\symmdiff A)\cap B(y,\delta))}{\mu B(y,\delta)}
=0$.}

\noindent But for any $\delta>0$,

\Centerline{$(H\symmdiff H')\cap B(y,\delta)
=y+\delta((H_0\symmdiff H'_0)\cap B(\tbf{0},1))$,}

\noindent where

\Centerline{$H_0=\{x:\varinnerprod{x}{v}\le 0\}$,
\quad$H'_0=\{x:\varinnerprod{x}{v'}\le 0\}$.}

\noindent So

\Centerline{$0
=\lim_{\delta\downarrow 0}
  \Bover{\mu((H\symmdiff H')\cap B(y,\delta))}{\mu B(y,\delta)}
=\Bover{\mu((H_0\symmdiff H_0')\cap B(\tbf{0},1))}{\mu B(\tbf{0},1)}
=\Bover{\mu((H_0\symmdiff H_0')\cap B(\tbf{0},n))}{\mu B(\tbf{0},n)}$}

\noindent for every $n\ge 1$, and $\mu(H_0\symmdiff H_0')=0$.   Since
$\mu$ is strictly positive, and $H_0$ and $H_0'$ are both the closures
of their interiors, they must be identical;  and it follows that $v=v'$.
}%end of proof of 474P

\leader{474Q}{Lemma}\dvAnew{2012} Set $c'=2^{r+3}\sqrt{r-1}(1+2^r)$.
Suppose that $c^*$, $\epsilon$ and $\delta$ are such that

\Centerline{$c^*\ge 0$,
\quad$\delta>0$,\quad$0<\epsilon<\Bover1{\sqrt{2}}$,
\quad$c^*\epsilon^3<\bover14\beta_{r-1}$,
\quad$4c'\epsilon\le\bover18\beta_{r-1}$.}

\noindent Set
$V_{\delta}=\{z:z\in\BbbR^{r-1}$, $\|z\|\le\delta\}$ and
$C_{\delta}=V_{\delta}\times[-\delta,\delta]$, regarded as a cylinder
in $\BbbR^r$.   Let $f\in\eusm D$ be such that

\Centerline{$\int_{C_{\delta}}
  \|\grad_{r-1}f\|+\max(\Pd{f}{\xi_r},0)d\mu
\le c^*\epsilon^3\delta^{r-1}$,}

\noindent where $\grad_{r-1}f
=(\Pd{f}{\xi_1},\ldots,\Pd{f}{\xi_{r-1}},0)$.  Set

\Centerline{$F=\{x:x\in C_{\delta}$, $f(x)\ge\Bover34\}$,
\quad$F'=\{x:x\in C_{\delta}$, $f(x)\le\Bover14\}$.}

\noindent and for $\gamma\in\Bbb R$ set
$H_{\gamma}=\{x:x\in\BbbR^r$, $\xi_r\le\gamma\}$.   Then there is a
$\gamma\in\Bbb R$ such that

\Centerline{$\mu(F\symmdiff(H_{\gamma}\cap C_{\delta}))
\le 9\mu(C_{\delta}\setminus(F\cup F'))
   +(c^*\beta_{r-1}+16c')\epsilon\delta^r$.}

\proof{{\bf (a)} For $t\in[-\delta,\delta]$ set

\Centerline{$f_t(z)=f(z,t)$ for $z\in\BbbR^{r-1}$,}

\Centerline{$F_t=\{z:z\in V_{\delta}$, $f_t(z)\ge\Bover34\},$
\quad$F'_t=\{z:z\in V_{\delta}$, $f_t(z)\le\Bover14\}$;}

\noindent set

\Centerline{$\gamma
=\sup\bigl(\{-\delta\}\cup\{t:t\in[-\delta,\delta]$,
  $\mu_{r-1}F_t\ge\Bover34\mu_{r-1}V_{\delta}\}\bigr)$,}

\Centerline{$G
=\{t:t\in[-\delta,\delta]$,
  $\int_{V_{\delta}}\|\grad f_t\|d\mu_{r-1}\ge\epsilon^2\delta^{r-2}\}$.}

\noindent Note that $(\grad_{r-1}f)(z,t)=((\grad f_t)(z),0)$, so we
have

\Centerline{$\int_{-\delta}^{\delta}\int_{V_{\delta}}
   \|\grad f_t\|d\mu_{r-1}dt
=\int_{C_{\delta}}\|\grad_{r-1}f\|d\mu
\le c^*\epsilon^3\delta^{r-1}$}

\noindent and $\mu_1G\le c^*\epsilon\delta$, where $\mu_1$ is Lebesgue
measure on $\Bbb R$.

\medskip

{\bf (b)} If $t\in[-\delta,\delta]\setminus G$, then

\Centerline{$\min(\mu_{r-1}F'_t,\mu_{r-1}F_t)
\le 4c'\epsilon\delta^{r-1}$.}

\noindent\Prf\ If $r>2$,

$$\eqalignno{\min\bigl(\mu_{r-1}F'_t,
  \mu_{r-1}F_t)\bigr)^{(r-2)/(r-1)}
&\le 4c'\int_{V_{\delta}}\|\grad f_t\|d\mu_{r-1}\cr
\displaycause{473L}
&\le 4c'\epsilon^2\delta^{r-2}\cr}$$

\noindent because $t\notin G$, so that

$$\eqalign{\min(\mu_{r-1}F't,\mu_{r-1}F_t)
&\le (4c'\epsilon^2)^{(r-1)/(r-2)}\delta^{r-1}\cr
&\le 4c'\epsilon\delta^{r-1}\cr}$$

\noindent because $4c'\ge 1$ and $\Bover{2(r-1)}{r-2}\ge 1$ and
$\epsilon\le 1$.   If $r=2$, then

\Centerline{$\int_{V_{\delta}}\|\grad f_t\|d\mu_{r-1}
\le\epsilon^2<\Bover12$,}

\noindent so at least one of $F'_t$,
$F_t$ is empty, as noted in 473M, and
$\min(\mu_{r-1}F'_t,\mu_{r-1}F_t)=0$.\ \Qed

\medskip

{\bf (c)} If $-\delta\le s<t\le\delta$, then

\Centerline{$\int_{-\delta}^{\delta}
  \max(\Pd{f}{\xi_r}(z,\xi),0)d\xi
\ge\int_{s}^{t}\Pd{f}{\xi_r}(z,\xi)d\xi
=f(z,t)-f(z,s)\ge\Bover12$}

\noindent for every $z\in F'_s\cap F_t$.   Accordingly

\Centerline{$\Bover12\mu_{r-1}(F'_s\cap F_t)
\le\int_{C_{\delta}}\max(\Pd{f}{\xi_r},0)d\mu
\le c^*\epsilon^3\delta^{r-1}
<\Bover14\beta_{r-1}\delta^{r-1}
=\Bover14\mu_{r-1}V_{\delta}$}

\noindent and $\mu_{r-1}(F'_s\cap F_t)<\bover12\mu_{r-1}V_{\delta}$.
It follows that if $-\delta\le s<\gamma$, so that there is a $t>s$ such
that $\mu_{r-1}F_t\ge\bover34\mu_{r-1}V_{\delta}$, then
$\mu_{r-1}F'_s<\bover34\mu_{r-1}V_{\delta}$.

\medskip

{\bf (d)} Now

\Centerline{$\mu((F\symmdiff(H_{\gamma}\cap C_{\delta}))
\le 9\mu(C_{\delta}\setminus(F\cup F'))
   +\epsilon\delta^r(c^*\beta_{r-1}+16c')$.}

\noindent\Prf\ Set

\Centerline{$\tilde G
=\{t:-\delta\le t\le\delta$,
  $\mu_{r-1}(F_t\cup F'_t)
    \le\Bover78\mu_{r-1}V_{\delta}\}$,}

\Centerline{$\hat G=\{t:-\delta\le t\le\delta$,
  $\mu_{r-1}F_t
    \le 4c'\epsilon\delta^{r-1}\}$,}

\Centerline{$\hat G'=\{t:-\delta\le t\le\delta$,
  $\mu_{r-1}F'_t
    \le 4c'\epsilon\delta^{r-1}\}$.}

\noindent Then

\Centerline{$\Bover18\mu_{r-1}V_{\delta}\cdot\mu_1\tilde G
\le\mu(C_{\delta}\setminus(F\cup F'))$,}

\Centerline{$\mu(F\cap(V_{\delta}\times\hat G))
\le 8c'\epsilon\delta^r$,}

\Centerline{$\mu(F'\cap(V_{\delta}\times\hat G'))
\le 8c'\epsilon\delta^r$.}

\noindent So if we set

$$\eqalign{W
=(C_{\delta}\setminus(F\cup F'))
  &\cup(V_{\delta}
    \times(\tilde G\cup G\cup\{\gamma\}))\cr
&\cup(F\cap(V_{\delta}\times\hat G))
  \cup(F'\cap(V_{\delta}\times\hat G')),\cr}$$

\noindent we shall have

$$\eqalign{\mu W
&\le\mu(C_{\delta}\setminus(F\cup F'))
   +\mu_1\tilde G\cdot\mu_{r-1}V_{\delta}
   +\mu_1G\cdot\mu_{r-1}V_{\delta}
   +16c'\epsilon\delta^r\cr
&\le 9\mu(C_{\delta}\setminus(F\cup F'))
   +c^*\epsilon\delta\mu_{r-1}V_{\delta}
   +16c'\epsilon\delta^r\cr
&=9\mu(C_{\delta}\setminus(F\cup F'))
   +\epsilon\delta^r(c^*\beta_{r-1}+16c')\cr
}$$

\noindent (using the estimate of $\mu_1G$ in (a)).

\Quer\ Suppose, if possible, that there is a point
$(z,t)\in(F\symmdiff(H_{\gamma}\cap C_{\delta}))\setminus W$.
Since $t\notin G$, (b) tells us that

\Centerline{$\min(\mu_{r-1}F'_t,\mu_{r-1}F_t)
\le 4c'\epsilon\delta^{r-1}
\le\Bover18\mu_{r-1}V_{\delta}$.}

\noindent So
$t\in\hat G\cup\hat G'$.    Also, since $t\notin\tilde G$,
$\mu_{r-1}F_t+\mu_{r-1}F'_t\ge\Bover78\mu_{r-1}V_{\delta}$;  so
(since $t\ne\gamma$) either
$\mu_{r-1}F_t\ge\Bover34\mu_{r-1}V_{\delta}$ and
$t<\gamma$, or
$\mu_{r-1}F'_t\ge\Bover34\mu_{r-1}V_{\delta}$ and
$t>\gamma$ (by (c)).   Now

$$\eqalignno{t<\gamma
&\Longrightarrow\mu_{r-1}F_t
  \ge\Bover34\mu_{r-1}V_{\delta}\cr
&\Longrightarrow\mu_{r-1}F'_t
   \le 4c'\epsilon\delta^{r-1}\cr
&\Longrightarrow t\in\hat G'\cr
&\Longrightarrow (z,t)\notin F'\cr
\displaycause{because $(z,t)\notin F'\cap(V_{\delta}\times\hat G')$}
&\Longrightarrow (z,t)\in F\cr
\displaycause{because $(z,t)\notin C_{\delta}\setminus(F\cup F')$}
&\Longrightarrow (z,t)\in F\cap H_{\gamma},\cr}$$

\noindent which is impossible.   And similarly

$$\eqalignno{t>\gamma
&\Longrightarrow\mu_{r-1}F'
  \ge\Bover34\mu_{r-1}V_{\delta}\cr
&\Longrightarrow\mu_{r-1}F_t
   \le 4c'\epsilon\delta^{r-1}\cr
&\Longrightarrow t\in\hat G\cr
&\Longrightarrow (z,t)\notin F\cr
&\Longrightarrow (z,t)\notin F\cup H_{\gamma},\cr}$$

\noindent which is equally impossible.\ \Bang

Thus
$F\symmdiff(H_{\gamma}\cap C_{\delta})\subseteq W$
has measure at most

\Centerline{$9\mu(C_{\delta}\setminus(F\cup F'))
   +\epsilon\delta^r(c^*\beta_{r-1}+16c')$,}

\noindent as claimed.\ \Qed
}%end of proof of 474Q

\leader{474R}{Theorem} Let $E\subseteq\BbbR^r$ be a set with locally
finite perimeter, $\psi_E$
its canonical outward-normal function, and $y$
any point of its reduced boundary $\partial^{\$}E$.   Then $\psi_E(y)$
is the Federer exterior normal to $E$ at $y$.

\proof{ Write $\lambda^{\partial}_E$ for the perimeter measure of $E$, as
usual.

\medskip

{\bf (a)} To begin with (down to the end of (c-ii) below) suppose
that $y=\tbf{0}$ and that $\psi_E(y)=(0,\ldots,0,1)=v$ say.    Set

\Centerline{$c=2^{r+4}\sqrt r(1+2^{r+1})$,
\quad$c'=2^{r+3}\sqrt{r-1}(1+2^r)$,}

\Centerline{$c_1=1+\max\bigl(4\pi\beta_{r-2},
   2c(3r)^{r-1}\bigr)$,}

\noindent (counting $\beta_0$ as $1$, if $r=2$),

\Centerline{$c^*=\sqrt2(2\sqrt2)^{r-1}c_1$,
\quad$c_1^*=10+\Bover12(c^*+\Bover{16c'}{\beta_{r-1}})$.}

\noindent As in 474Q, set

\Centerline{$V_{\delta}=\{z:z\in\BbbR^{r-1}$, $\|z\|\le\delta\}$,
\quad$C_{\delta}=V_{\delta}\times[-\delta,\delta]$.
\quad$H_{\gamma}=\{x:\xi_r\le\gamma\}$}

\noindent for $\delta>0$ and $\gamma\in\Bbb R$, and
$\grad_{r-1}f=(\Pd{f}{\xi_1},\ldots,\Pd{f}{\xi_{r-1}},0)$ for
$f\in\eusm D$.

\medskip

{\bf (b)(i)} Take any $\epsilon>0$ such that

\Centerline{$\epsilon<\Bover1{\sqrt 2}$,
\quad$c^*\epsilon^3<\Bover14\beta_{r-1}$,
\quad$2^{r+1}c'\epsilon<\Bover18\beta_{r-1}$.}

\noindent Then there is a $\delta_0\in\ocint{0,1}$ such that

\Centerline{$\Bover1{\lambda^{\partial}_EB(\tbf{0},\delta)}
  \int_{B(\tbf{0},\delta)}\|\psi_E(x)-v\|\lambda^{\partial}_E(dx)
\le\epsilon^3$,}

\Centerline{$\Bover1{c_1}\delta^{r-1}
\le\lambda^{\partial}_EB(\tbf{0},\delta)
\le c_1\delta^{r-1}$}

\noindent for every $\delta\in\ocint{0,2\delta_0\sqrt2}$ (using
474N(iii) and 474N(iv) for the inequalities bounding
$\lambda^{\partial}_EB(\tbf{0},\delta)$).

\medskip

\quad{\bf (ii)} Suppose that $0<\delta\le\delta_0$.
Note first that

$$\eqalignno{\int_{C_{2\delta}}\|v-\psi_E\|d\lambda^{\partial}_E
&\le\int_{B(\tbf{0},2\delta\sqrt2)}\|v-\psi_E\|d\lambda^{\partial}_E
\le\epsilon^3\lambda^{\partial}_EB(\tbf{0},2\delta\sqrt2)\cr
&\le c_1\epsilon^3(2\delta\sqrt2)^{r-1}
=\Bover{c^*}{\sqrt2}\epsilon^3\delta^{r-1}.\cr}$$

\medskip

\quad{\bf (iii)}
$\lim_{n\to\infty}\tilde h_n*\chi E\eae\chi E$ (473Ee), so there is an
$n\ge\bover1{\delta}$ such that
$\int_{C_{\delta}}|\tilde h_n*\chi E-\chi E|d\mu
\le\bover14\epsilon\mu C_{\delta}$.   Setting

\Centerline{$f=\tilde h_n*\chi E$,
\quad$F=\{x:x\in C_{\delta}$, $f(x)\ge\bover34\}$,
\quad$F'=\{x:x\in C_{\delta}$, $f(x)\le\bover14\}$,}

\noindent we have $f\in\eusm D$ (473De once more) and

\Centerline{$\mu(C_{\delta}\Bsetminus(F\cup F'))\le\epsilon\mu C_{\delta}$,
\quad$\mu(F\symmdiff(E\cap C_{\delta}))\le\epsilon\mu C_{\delta}$.}

\medskip

\quad{\bf (iv)}

\Centerline{$\int_{C_{\delta}}
  \|\grad_{r-1}f\|+\max(\Pd{f}{\xi_r},0)d\mu
\le c^*\epsilon^3\delta^{r-1}$.}

\noindent\Prf\Quer\ Suppose, if possible, otherwise.   Note that because
$\mu C_{\delta}=2\beta_{r-1}\delta^r$,
$\lim_{\delta'\uparrow\delta}\mu C_{\delta'}=\mu C_{\delta}$, so there
is some $\delta'<\delta$ such that

\Centerline{$\int_{C_{\delta'}}
  \|\grad_{r-1}f\|+\max(\Pd{f}{\xi_r},0)d\mu
>c^*\epsilon^3\delta^{r-1}$.}

\noindent For $1\le i\le r$ and $x\in\BbbR^r$, set

$$\eqalignno{\theta_i(x)
&=\bover{\pd{f}{\xi_i}(x)}{\|\grad_{r-1}f(x)\|}
  \text{ if }i<r,\,x\in C_{\delta'}
  \text{ and }\grad_{r-1}(x)\ne 0,\cr
&=1\text{ if }i=r,\,x\in C_{\delta'}
  \text{ and }\Pd{f}{\xi_r}(x)\ge 0,\cr
&=0\text{ otherwise}.\cr}$$

\noindent Then all the $\theta_i$ are
$\mu$-integrable.   Setting $\theta=(\theta_1,\ldots,\theta_r)$,

\Centerline{$\int\varinnerprod{\theta}{\grad f}d\mu
=\int_{C_{\delta'}}
  \|\grad_{r-1}f\|+\max(\Pd{f}{\xi_r},0)d\mu
>c^*\epsilon^3\delta^{r-1}$.}

\noindent By 473Ef,
$\sequence{k}{\|\theta_i-\theta_i*\tilde h_k\|_1}\to 0$ for each $i$;
since $\grad f$ is bounded,

\Centerline{$\int
  \varinnerprod{(\tilde h_k*\theta)}{\grad f}\,d\mu
>c^*\epsilon^3\delta^{r-1}$}

\noindent for any $k$ large enough.  If we ensure also that
$\Bover1{k+1}\le\delta-\delta'$, and set $\phi=\tilde h_k*\theta$, we
shall get a function $\phi\in\eusm D$, with
$\|\phi(x)\|\le\sqrt2\chi C_{\delta}$ for every $x$ (by 473Dc and 473Dg),
such that

\Centerline{$\int\varinnerprod{\phi}{\grad f}d\mu
>c^*\epsilon^3\delta^{r-1}$.}

\noindent Moreover, referring to the definition of $*$ in 473Dd and
473Dg,

\Centerline{$\varinnerprod{(\tilde h_n*\phi)(x)}{v}
=(\tilde h_n*(\tilde h_k*\theta_r))(x)\ge 0$}

\noindent for every $x$, because $\tilde h_n$, $\tilde h_k$ and
$\theta_r$ are all non-negative.

Now

$$\eqalignno{c^*\epsilon^3\delta^{r-1}
&<\int\varinnerprod{\phi}{\grad f}d\mu
=\int\varinnerprod{\phi}{\grad(\tilde h_n*\chi E)}d\mu\cr
&=-\int\varinnerprod{(\tilde h_n*\phi)}{\psi_E}
  \,d\lambda^{\partial}_E\cr
\displaycause{474K}
&\le\int\varinnerprod{(\tilde h_n*\phi)}{(v-\psi_E)}
  \,d\lambda^{\partial}_E
\le\sqrt{2}\int_{C_{2\delta}}\|v-\psi_E\|\,d\lambda^{\partial}_E\cr
\displaycause{because $\|(\tilde h_n*\phi)(x)\|\le\sqrt2$ for every $x$,
by 473Dg again, and $(\tilde h_n*\phi)(x)=0$ if
$x\notin C_{\delta}+C_{1/(n+1)}\subseteq C_{2\delta}$}
&\le c^*\epsilon^3\delta^{r-1};\cr}$$

\noindent which is absurd.\ \Bang\Qed

\medskip

\quad{\bf (v)} By 474Q, there is a $\gamma\in\Bbb R$ such that

$$\eqalign{\mu(F\symmdiff(H_{\gamma}\cap C_{\delta}))
&\le 9\mu(C_{\delta}\setminus(F\cup F'))
   +(c^*\beta_{r-1}+16c')\epsilon\delta^r\cr
&\le 9\epsilon\mu C_{\delta}
   +\Bover1{2\beta_{r-1}}(c^*\beta_{r-1}+16c')\epsilon\mu C_{\delta}
=(c_1^*-1)\epsilon\mu C_{\delta},\cr}$$

\noindent and

\Centerline{$\mu((E\symmdiff H_{\gamma})\cap C_{\delta})
\le\mu(F\symmdiff(E\cap C_{\delta}))
    +\mu(F\symmdiff(H_{\gamma}\cap C_{\delta}))
\le c_1^*\epsilon\mu C_{\delta}$.}

\medskip

\quad{\bf (vi)} As $\epsilon$ is arbitrary, we see that

\Centerline{$\lim_{\delta\downarrow 0}\inf_{\gamma\in\Bbb R}
   \Bover1{\mu C_{\delta}}
   \mu((E\symmdiff H_{\gamma})\cap C_{\delta})=0$.}

\medskip

{\bf (c)} Again take $\epsilon\in\ocint{0,1}$.

\medskip

\quad{\bf (i)} By (b) above and 474N(i)-(ii) there is a
$\delta_1>0$ such that whenever $0<\delta\le\delta_1$ then

\Centerline{$\mu(B(\tbf{0},\delta)\cap E)
\ge\Bover1{2(3r)^r\beta_r}\mu B(\tbf{0},\delta)$,}

\Centerline{$\mu(B(\tbf{0},\delta)\setminus E)
\ge\Bover1{2(3r)^r\beta_r}\mu B(\tbf{0},\delta)$}

\noindent and there is a $\gamma\in\Bbb R$ such that

\Centerline{$\mu((E\symmdiff H_{\gamma})\cap C_{\delta})
<\min(\epsilon,\Bover{\epsilon^r}{4\beta_{r-1}(3r)^r})\mu C_{\delta}$.}

\noindent In this case, $|\gamma|\le\epsilon\delta$.
\Prf\Quer\ Suppose, if possible, that $\gamma<-\epsilon\delta$.
Then

$$\eqalign{\mu(B(\tbf{0},\epsilon\delta)\cap E)
&\le\mu(E\cap C_{\delta}\setminus H_{\gamma})\cr
&<\Bover{\epsilon^r}{4\beta_{r-1}(3r)^r}\mu C_{\delta}
=\Bover1{2\beta_r(3r)^r}\mu B(\tbf{0},\epsilon\delta)\cr}$$

\noindent which is impossible.\ \BanG\  In the same way,
\Quer\ if $\gamma>\epsilon\delta$,

$$\eqalign{\mu(B(\tbf{0},\epsilon\delta)\setminus E)
&\le\mu((C_{\delta}\setminus H_{\gamma})\setminus E)\cr
&<\Bover{\epsilon^r}{4\beta_{r-1}(3r)^r}\mu C_{\delta}
=\Bover1{2\beta_r(3r)^r}\mu B(\tbf{0},\epsilon\delta).
\text{ \Bang}\Qed\cr}$$

\medskip

\quad{\bf (ii)} It follows that

$$\eqalign{\mu((E\symmdiff H_0)\cap C_{\delta})
&\le\mu((E\symmdiff H_{\gamma})\cap C_{\delta})
   +\mu((H_{\gamma}\symmdiff H_0)\cap C_{\delta})\cr
&\le\epsilon\mu C_{\delta}+\epsilon\delta\mu_{r-1}V_{\delta}
=\Bover32\epsilon\mu C_{\delta}.\cr}$$

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\lim_{\delta\downarrow 0}
  \Bover{\mu((E\symmdiff H_0)\cap C_{\delta})}{\mu C_{\delta}}
=0$,}

\noindent and

\Centerline{$\lim_{\delta\downarrow 0}
  \Bover{\mu((E\symmdiff H_0)\cap B_{\delta})}{\mu B_{\delta}}
\le\Bover{2\beta_{r-1}}{\beta_r}\lim_{\delta\downarrow 0}
  \Bover{\mu((E\symmdiff H_0)\cap C_{\delta})}{\mu C_{\delta}}
=0$.}

\medskip

{\bf (d)} Thus $v$ is a Federer exterior normal to $E$ at $\tbf{0}$ if
$\psi_E(\tbf{0})=v$.   For the general case, let $S$ be an orthogonal
matrix such that
$S\psi_E(y)=v$, and set $T(x)=S(x-y)$ for every $x$.   
The point is of course that

\Centerline{$\tbf{0}=T(y)\in T[\partial^{\$}E]=\partial^{\$}T[E]$,
\quad$v=S\psi_ET^{-1}(\tbf{0})=\psi_{T[E]}(\tbf{0})$}

\noindent (474H).   So if we set

\Centerline{$H
=\{x:\varinnerprod{(x-y)}{\psi_E(y)}\le 0\}
=\{x:\varinnerprod{T(x)}{v}\le 0\}
=T^{-1}[H_0]$,}

\noindent then

\Centerline{$\Bover{\mu((H\symmdiff E)\cap B(y,\delta))}
   {\mu B(y,\delta)}
=\Bover{\mu((H_0\symmdiff T[E])\cap B(\tbf{0},\delta))}
   {\mu B(\tbf{0},\delta)}
\to 0$}

\noindent as $\delta\downarrow 0$, and $\psi_E(y)$ is a Federer exterior
normal to $E$ at $y$, as required.
}%end of proof of 474R

\leader{474S}{Corollary} Let $E\subseteq\BbbR^r$ be a set with locally
finite perimeter, and $\lambda^{\partial}_E$ its perimeter measure.
Let $y$ be any point of the reduced boundary of $E$.   Then

\Centerline{$\lim_{\delta\downarrow 0}
\Bover{\lambda^{\partial}_EB(y,\delta)}{\beta_{r-1}\delta^{r-1}}=1$.}

\proof{{\bf (a)} Set $v=\psi_E(y)$ and
$H=\{x:\varinnerprod{(x-y)}{v}\le 0\}$, as in 474R.   Now

\Centerline{$\int_{H\cap\partial B(y,\delta)}
  \varinnerprod{v}{\Bover1{\delta}(x-y)}\nu(dx)
=-\beta_{r-1}\delta^{r-1}$}

\noindent for almost every $\delta>0$.   \Prf\ Set $\phi(x)=v$ for every
$x\in\BbbR^r$.   By 474I, $\phi$ is an outward-normal function for $H$,
so 474M tells us that, for almost every $\delta>0$,

$$\eqalignno{\int_{H\cap\partial B(y,\delta)}
  \varinnerprod{v}{\Bover1{\delta}(x-y)}\,\nu(dx)
&=\int_{H\cap B(y,\delta)}\diverg\phi\,d\mu
   -\int_{B(y,\delta)}\varinnerprod{v}{v}\,d\lambda^{\partial}_H\cr
&=-\lambda^{\partial}_HB(y,\delta)
=-\nu(B(y,\delta)\cap\partial H)\cr
\displaycause{using the identification of $\lambda^{\partial}_H$ in
474I}
&=-\beta_{r-1}\delta^{r-1}\cr}$$

\noindent(identifying $\nu$ on the hyperplane $\partial H$ with Lebesgue
measure on $\BbbR^{r-1}$, as usual).\ \Qed

\medskip

{\bf (b)} Now, given $\epsilon>0$, there is a $\delta_0>0$ such that
whenever $0<\delta\le\delta_0$ there is an $\eta$ such that
$\delta\le\eta\le\delta(1+\epsilon)$ and
$|\lambda^{\partial}_EB(y,\eta)-\beta_{r-1}\eta^{r-1}|
\le\epsilon\eta^{r-1}$.   \Prf\ Let $\zeta>0$ be such that

\Centerline{$\zeta(1+\Bover{5\pi}{r}\beta_{r-2})(1+\epsilon)^r
\le\epsilon^2$.}

\noindent By 474N(iv) and 474R and the definition of $\psi_E$, there is
a $\delta_0>0$ such that

\Centerline{$\lambda^{\partial}_EB(y,\delta)
\le 5\pi\beta_{r-2}\delta^{r-1}$,}

\Centerline{$\mu((E\symmdiff H)\cap B(y,\delta))\le\zeta\delta^r$,}

\Centerline{$\int_{B(y,\delta)}
  \|\psi_E(x)-v\|\lambda^{\partial}_E(dx)
\le\zeta\lambda^{\partial}_EB(y,\delta)$}

\noindent whenever $0<\delta\le(1+\epsilon)\delta_0$.   Take
$0<\delta\le\delta_0$.   Then, for almost every $\eta>0$, we have

\Centerline{$\int_{B(y,\eta)}
  \varinnerprod{v}{\psi_E(x)}\,\lambda^{\partial}_E(dx)
  +\int_{E\cap\partial B(y,\eta)}
    \varinnerprod{v}{\Bover1{\eta}(x-y)}\,\nu(dx)
=0$}

\noindent by 474M, applied with $\phi$ the constant function with value
$v$.   Putting this together
with (a), we see that, for almost every
$\eta\in\ocint{0,(1+\epsilon)\delta_0}$,

$$\eqalignno{|\lambda^{\partial}_EB(y,\eta)&-\beta_{r-1}\eta^{r-1}|
=|\int_{B(y,\eta)}\varinnerprod{v}{v}\,d\lambda^{\partial}_E
   -\beta_{r-1}\eta^{r-1}|\cr
&\le|\int_{B(y,\eta)}\varinnerprod{v}{(v-\psi_E)}\,d\lambda^{\partial}_E|
   +|\int_{B(y,\eta)}\varinnerprod{v}{\psi_E}\,d\lambda^{\partial}_E
       -\beta_{r-1}\eta^{r-1}|\cr
&\le\int_{B(y,\eta)}\|\psi_E-v\|d\lambda^{\partial}_E\cr
&\hskip3em
   +|\int_{B(y,\eta)}
      \varinnerprod{v}{\psi_E(x)}\,\lambda^{\partial}_E(dx)
     +\int_{H\cap\partial B(y,\eta)}
       \varinnerprod{v}{\Bover1{\eta}(x-y)}\,\nu(dx)|\cr
\displaycause{using (a) above}
&\le\zeta\lambda^{\partial}_EB(y,\eta)\cr
&\hskip3em
   +|\int_{H\cap\partial B(y,\eta)}
       \varinnerprod{v}{\Bover1{\eta}(x-y)}\,\nu(dx)
      -\int_{E\cap\partial B(y,\eta)}
       \varinnerprod{v}{\Bover1{\eta}(x-y)}\,\nu(dx)|\cr
&\le 5\pi\beta_{r-2}\zeta\eta^{r-1}
   +\nu((E\symmdiff H)\cap\partial B(y,\eta)).\cr}$$

\woddheader{474S}{0}{0}{0}{30pt}

Integrating with respect to $\eta$, we have

$$\eqalignno{\int_0^{\delta(1+\epsilon)}
  |\lambda^{\partial}_EB(y,\eta)-\beta_{r-1}\eta^{r-1}|d\eta
&\le \Bover{5\pi}{r}\beta_{r-2}\zeta\delta^r(1+\epsilon)^r
   +\mu((E\symmdiff H)\cap B(y,\delta(1+\epsilon)))\cr
\displaycause{using 265G, as usual}
&\le \Bover{5\pi}{r}\beta_{r-2}\zeta\delta^r(1+\epsilon)^r
   +\zeta\delta^r(1+\epsilon)^r
\le\epsilon^2\delta^r\cr}$$

\noindent by the choice of $\zeta$.   But this means that there must be
some $\eta\in[\delta,\delta(1+\epsilon)]$ such that

\Centerline{$|\lambda^{\partial}_EB(y,\eta)-\beta_{r-1}\eta^{r-1}|
\le\epsilon\delta^{r-1}
\le\epsilon\eta^{r-1}$. \Qed}

\medskip

{\bf (c)} Now we see that

\Centerline{$\lambda^{\partial}_EB(y,\delta)
\le\lambda^{\partial}_EB(y,\eta)
\le(\beta_{r-1}+\epsilon)\eta^{r-1}
\le(\beta_{r-1}+\epsilon)(1+\epsilon)^{r-1}\delta^{r-1}$.}

\noindent But by the same argument we have an
$\hat\eta\in[\Bover{\delta}{1+\epsilon},\delta]$ such that
$|\lambda^{\partial}_EB(y,\hat\eta)-\beta_{r-1}\hat\eta^{r-1}|
\le\epsilon\hat\eta^{r-1}$, so that

\Centerline{$\lambda^{\partial}_EB(y,\delta)
\ge\lambda^{\partial}_EB(y,\hat\eta)
\ge(\beta_{r-1}-\epsilon)\hat\eta^{r-1}
\ge(\beta_{r-1}-\epsilon)(1+\epsilon)^{1-r}\delta^{r-1}$.}

Thus, for every $\delta\in\ocint{0,\delta_0}$,

\Centerline{$(\beta_{r-1}-\epsilon)(1+\epsilon)^{1-r}\delta^{r-1}
\le\lambda^{\partial}_EB(y,\delta)
\le(\beta_{r-1}+\epsilon)(1+\epsilon)^{r-1}\delta^{r-1}$.}

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\lim_{\delta\downarrow 0}
  \Bover{\lambda^{\partial}_EB(y,\delta)}{\delta^{r-1}}
=\beta_{r-1}$,}

\noindent as claimed.
}%end of proof of 474S

\leader{474T}{The Compactness Theorem} Let $\Sigma$ be the algebra of
Lebesgue measurable subsets of $\BbbR^r$, and give it the topology
$\frak T_m$ of convergence in measure defined by the pseudometrics
$\rho_H(E,F)=\mu((E\symmdiff F)\cap H)$ for measurable sets $H$ of
finite measure\cmmnt{ (cf.\ \S\S245 and 323)}.   Then

(a) $\per:\Sigma\to[0,\infty]$ is lower semi-continuous;

(b) for any $\gamma<\infty$, $\{E:E\in\Sigma$, $\per E\le\gamma\}$ is
compact.

\proof{{\bf (a)} Let $\sequencen{E_n}$ be any $\frak T_m$-convergent
sequence in $\Sigma$ with limit $E\in\Sigma$.   If
$\phi:\BbbR^r\to B(\tbf{0},1)$ is a Lipschitz function with compact
support, then $\diverg\phi$ is integrable, so
$F\mapsto\int_F\diverg\phi\,d\mu$ is truly continuous (225A), and

\Centerline{$|\int_E\diverg\phi\,d\mu|
=\lim_{n\to\infty}|\int_{E_n}\diverg\phi\,d\mu|
\le\sup_{n\in\Bbb N}\per E_n$.}

\noindent As $\phi$ is arbitrary, $\per E\le\sup_{n\in\Bbb N}\per E_n$.
This means that $\{E:\per E\le\gamma\}$ is sequentially closed,
therefore closed (4A2Ld), for any $\gamma$, and $\per$ is lower
semi-continuous.

\medskip

%\allowmorestretch{468}{
{\bf (b)} Let us say that a
`dyadic cube' is a set expressible in the form
$\prod_{1\le i\le r}\coint{2^{-n}k_i,2^{-n}(k_i+1)}$ where $n$,
$k_1,\ldots,k_r\in\Bbb Z$.   Set $\Cal A=\{E:\per E\le\gamma\}$.   
%}

\medskip

\quad{\bf (i)} For $E\in\Cal A$, $n\in\Bbb N$ and
$\epsilon\in\ocint{0,1}$ let $G(E,n,\epsilon)$ be the union of all the
dyadic cubes $D$ with side length $2^{-n}$ such that
$\epsilon\mu D\le\mu(E\cap D)\le(1-\epsilon)\mu D$.   Then
$\mu G(E,n,\epsilon)\le\Bover{c_1}{2^n\epsilon}\gamma$, where
$c_1=2^{r+5}(1+2^{r+1})(1+\sqrt r)^{r+1}$.

\Prf\ Express $G(E,n,\epsilon)$ as a disjoint union
$\bigcup_{i\in I}D_i$ where each $D_i$ is a dyadic cube of side length
$2^{-n}$ and
$\min(\mu(D_i\cap E),\mu(D_i\setminus E))\ge\epsilon\mu D_i$.   Let
$x_i$ be the centre of $D_i$ and $B_i$ the ball
$B(x_i,2^{-n-1}\sqrt r)$, so that $D_i\subseteq B_i$ and
$\mu B_i=\beta_r(\bover12\sqrt r)^r\mu D_i$.   For any $x\in\BbbR^r$,
the ball $B(x,2^{-n-1}\sqrt r)$ is included in a closed cube with side
length $2^{-n}\sqrt r$, so can contain at the very most $(1+\sqrt r)^r$
different $x_i$, because different $x_i$ differ by at least $2^{-n}$ in
some coordinate.   Turning this round,
$\sum_{i\in I}\chi B_i\le(1+\sqrt r)^r\chi(\BbbR^r)$.

Set $c=2^{r+4}\sqrt r(1+2^{r+1})$.   Then, for each $i\in I$,

$$\eqalignno{2c\lambda^{\partial}_EB_i
&\ge\min(\mu(B_i\cap E),\mu(B_i\setminus E))^{(r-1)/r}\cr
\displaycause{474Lb}
&\ge\min(\mu(D_i\cap E),\mu(D_i\setminus E))^{(r-1)/r}
\ge(\epsilon\mu D_i)^{(r-1)/r}
\ge 2^{-n(r-1)}\epsilon.\cr}$$

\noindent So

$$\eqalignno{\mu G(E,n,\epsilon)
&=2^{-nr}\#(I)
\le\Bover{2c}{2^n\epsilon}\sum_{i\in I}\lambda^{\partial}_EB_i\cr
&\le\Bover{2c(1+\sqrt r)^r}{2^n\epsilon}
  \lambda^{\partial}_E(\BbbR^r)
\le\Bover{c_1}{2^n\epsilon}\gamma.
  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (ii)} Now let $\sequencen{E_n}$ be any sequence in $\Cal A$.
Then we can find a subsequence $\sequencen{E'_n}$ such that whenever
$n\in\Bbb N$, $D$ is a dyadic cube of side length $2^{-n}$ meeting
$B(\tbf{0},n)$, and $i$, $j\ge n$, then
$|\mu(D\cap E'_i)-\mu(D\cap E'_j)|\le\Bover1{(n+1)^{r+2}}\mu D$.   Now

\Centerline{$\mu((E'_n\symmdiff E'_{n+1})\cap B(\tbf{0},n))
\le\Bover{3\beta_r(n+\sqrt r)^r}{(n+1)^{r+2}}
   +2^{-n}(n+1)^{r+1}c_1\gamma$}

\noindent whenever $n\ge 1$.   \Prf\ Let $\Cal E$ be
the set of dyadic cubes of side length $2^{-n}$ meeting $B(\tbf{0},n)$;
then every member of $\Cal E$ is included in
$B(\tbf{0},n+2^{-n}\sqrt r)$, so
$\mu(\bigcup\Cal E)\le\beta_r(n+\sqrt r)^r$.   Let $\Cal E_1$ be the
collection of those dyadic cubes of side length $2^{-n}$ included in
$G(E'_n,n,\Bover1{(n+1)^{r+2}})$.   If $D\in\Cal E\setminus\Cal E_1$,
either $\mu(E'_n\cap D)\le\Bover1{(n+1)^{r+2}}\mu D$ and
$\mu(E'_{n+1}\cap D)\le\Bover2{(n+1)^{r+2}}\mu D$ and
$\mu((E'_n\symmdiff E'_{n+1})\cap D)\le\Bover3{(n+1)^{r+2}}\mu D$, or
$\mu(D\setminus E'_n)\le\Bover1{(n+1)^{r+2}}\mu D$ and
$\mu(D\setminus E'_{n+1})\le\Bover2{(n+1)^{r+2}}\mu D$ and
$\mu((E'_n\symmdiff E'_{n+1})\cap D)\le\Bover3{(n+1)^{r+2}}\mu D$.
So

$$\eqalign{\mu((E'_n\symmdiff E'_{n+1})\cap B(\tbf{0},n))
&\le\sum_{D\in\Cal E}\mu((E'_n\symmdiff E'_{n+1})\cap D)\cr
&\le\sum_{D\in\Cal E\setminus\Cal E_1}
   \mu((E'_n\symmdiff E'_{n+1})\cap D)
   +\mu(\bigcup\Cal E_1)\cr
&\le\Bover3{(n+1)^{r+2}}\mu(\bigcup\Cal E)
   +\mu G(E'_n,n,\Bover1{(n+1)^{r+2}})\cr
&\le\Bover{3\beta_r(n+\sqrt r)^r}{(n+1)^{r+2}}
   +2^{-n}(n+1)^{r+2}c_1\gamma,
\cr}$$

\noindent as claimed.\ \Qed

\medskip

\quad{\bf (iii)} This means that
$\sum_{i=0}^{\infty}\mu((E'_i\symmdiff E'_{i+1})\cap B(\tbf{0},n))$ is
finite for each $n\in\Bbb N$, so that if we set
$E=\bigcup_{i\in\Bbb N}\bigcap_{j\ge i}E'_j$, then

\Centerline{$\mu((E\symmdiff E'_i)\cap B(\tbf{0},n))
\le\sum_{j=i}^{\infty}\mu((E'_j\symmdiff E'_{j+1})\cap B(\tbf{0},n))
\to 0$}

\noindent as $i\to\infty$ for every $n\in\Bbb N$.   It follows that
$\lim_{i\to\infty}\rho_H(E,E'_i)=0$ whenever $\mu H<\infty$ (see the
proofs of 245Eb and 323Gb).   Thus we have a subsequence
$\sequence{i}{E'_i}$ of the original sequence $\sequence{i}{E_i}$ which
is convergent for the topology $\frak T_m$ of convergence in measure.
By (a), its limit belongs to $\Cal A$.   But since $\frak T_m$ is
pseudometrizable (245Eb/323Gb), this is enough to show that $A$ is
compact for $\frak T_m$ (4A2Lf).
}%end of proof of 474T

\exercises{\leader{474X}{Basic exercises (a)}
%\spheader 474Xa
Show that for any $E\subseteq\BbbR^r$ with locally
finite perimeter, its reduced boundary is a Borel set and its canonical
outward-normal function is Borel measurable.
%474G 472Xd

\sqheader 474Xb Show that if $E\subseteq\Bbb R^r$ has finite perimeter
then either $E$ or its complement has finite measure.
%474L

\spheader 474Xc(i) Show that if $E\subseteq\BbbR^r$ has locally finite
perimeter, then $\partial^{\$}E\subseteq\partial E$.
\Hint{474N(i)-(ii).}   (ii) Show that if $H\subseteq\BbbR^r$ is a
half-space, as in 474I, then $\partial^{\$}E=\partial E$.
%474N 474H

\spheader 474Xd Let $E\subseteq\BbbR^r$ be a set with locally finite
perimeter, and $y\in\partial^{\$}E$.   Show that
$\lim_{\delta\downarrow 0}
\Bover{\mu(E\cap B(y,\delta))}{\mu B(y,\delta)}=\Bover12$.
%474R

\spheader 474Xe In the proof of 474S, use 265E to show that

\Centerline{$\int_{H\cap\partial B(y,\delta)}
  \varinnerprod{v}{\Bover1{\delta}(x-y)}\nu(dx)
=-\beta_{r-1}\delta^{r-1}$}

\noindent for {\it every} $\delta>0$.
%474S


\leader{474Y}{Further exercises (a)}
%\spheader 474Ya
In 474E, explain how to interpret the pair $(\psi,\lambda^{\partial}_E)$
as a vector measure (definition: 394O\formerly{3{}93O})
$\theta_E:\Cal B\to\BbbR^r$,
where $\Cal B$ is the Borel $\sigma$-algebra of $\BbbR^r$,
in such a way that we
have $\int_E\diverg\phi\,d\mu=\int\varinnerprod{\phi}{d\theta_E}$
for Lipschitz functions $\phi$ with compact support.
}%end of exercises

\endnotes{
\Notesheader{474} When we come to the Divergence Theorem itself in the
next section, it will be nothing but a repetition of Theorem 474E with
the perimeter measure and the outward-normal function properly
identified.   The idea of the indirect approach here is to start by
defining the pair $(\psi_E,\lambda^{\partial}_E)$ as a kind of
`distributional derivative' of the set $E$.   I take the space to match
the details with the language of the rest of this treatise, but really
474E amounts to nothing more than the Riesz representation theorem;
since the functional $\phi\mapsto\int_E\diverg\phi\,d\mu$ is linear, and
we restrict attention to sets $E$ for which it is continuous in an
appropriate sense (and can therefore be extended to arbitrary continuous
functions $\phi$ with compact support), it must be representable by a
(vector) measure, as in 474Ya.   For the process to be interesting, we
have to be able to identify at least some of the appropriate sets $E$
with their perimeter measures and outward-normal functions.
Half-spaces are straightforward enough (474I), and 474R tells us what
the outward-normal functions have to be;  but for a proper description
of the family of sets with locally finite perimeter we must wait until
the next section.   I see no quick way to show from the results here
that (for instance) the union of two sets with finite perimeter again
has finite perimeter.   And I notice that I have not even shown that balls
have finite perimeters.   After 475M things should be much clearer.

I have tried to find the shortest path to the Divergence Theorem itself,
and have not attempted to give `best' results in the intermediate
material.   In particular, in the isoperimetric inequality 474La, I show
only that the measure of a set $E$ is controlled by the magnitude of its
perimeter measure.   Simple scaling arguments show that if there is any
such control, then it must be of the form
$\gamma(\mu E)^{(r-1)/r}\le\per E$;  the
identification of the best constant $\gamma$ as $r\beta_r^{1/r}$, giving
equality for balls, is the real prize, to which I shall come in 476H.
Similarly, there is a dramatic jump from the crude estimates in 474N
to the exact limits in 474Xd and 474S.
When we say that a set $E$ has a Federer exterior normal at a point $y$,
we are clearly saying that there is an `approximate' tangent plane at that
point, as measured by ordinary volume $\mu$.
474S strengthens this by saying that, when measured by the
perimeter measure, the boundary of $E$ looks like a hyperplane through $y$
with normalised $(r-1)$-dimensional measure.   In 475G below we shall
come to a partial explanation of this.

The laborious arguments of 474C and 
474H are doing no more than establish the
geometric invariance of the concepts here, which ought, one would think, to
be obvious.   The trouble is that I have given
definitions of inner product and divergence and Lebesgue measure
in terms of the standard coordinate system of $\BbbR^r$.
If these were not invariant under isometries they would be far less
interesting.   But even if we are confident that there must be a result
corresponding to 474H, I think a little thought is required to identify the
exact formulae involved in the transformation.

I leave the Compactness Theorem (474T) to the end of the section because
it is off the line I have chosen to the Divergence Theorem (though it
can be used to make the proof of 474R more transparent;
see {\smc Evans \& Gariepy 92}, 5.7.2).
I have expressed 474T in terms of the topology of convergence in measure
on the algebra of Lebesgue measurable sets.   But since the perimeter of
a measurable set $E$ is not altered if we change $E$ by a negligible
set (474F),
`perimeter' can equally well be regarded as a function defined on
the measure algebra, in which case 474T becomes a theorem about the
usual topology of the measure algebra of Lebesgue measure, as described
in \S323.
}%end of notes

\discrpage

\frfilename{mt475.tex}
\versiondate{24.1.13}
\copyrightdate{2000}

\def\chaptername{Geometric measure theory}
\def\sectionname{The essential boundary}

\newsection{475}

The principal aim of this section is to translate Theorem 474E into
geometric terms.   We have already identified the Federer exterior
normal as an outward-normal function\cmmnt{ (474R)}, so we need to
find a description of perimeter measures.   Most remarkably, these turn
out, in every case considered in 474E,
to be just normalized Hausdorff measures (475G).   This
description needs the
concept of `essential boundary' (475B).   In order to complete the
programme, we need to be able to determine which sets have `locally
finite perimeter';  there is a natural criterion in the same language
(475L).   We now have all the machinery for a direct statement of the
Divergence Theorem (for Lipschitz functions) which depends on nothing
more advanced than the definition of Hausdorff measure (475N).   (The
definitions, at least, of `Federer exterior normal' and `essential
boundary' are elementary.)

This concludes the main work of the first part of this chapter.   But
since we are now
within reach of a reasonably direct proof of a fundamental fact about
the $(r-1)$-dimensional Hausdorff measure of the boundaries of subsets
of
$\BbbR^r$ (475Q), I continue up to Cauchy's Perimeter Theorem and the
Isoperimetric Theorem for convex sets (475S, 475T).


\cmmnt{\leader{475A}{Notation} As in the last two sections, $r$ will
be an integer (greater than or equal to $2$, unless explicitly permitted
to take the value $1$).   $\mu$ will be Lebesgue measure on $\BbbR^r$;
I will sometimes write
$\mu_{r-1}$ for Lebesgue measure on $\BbbR^{r-1}$ and $\mu_1$ for
Lebesgue measure on $\Bbb R$.   $\beta_r=\mu B(\tbf{0},1)$ will be the
measure of the unit ball in $\BbbR^r$, and
$S_{r-1}=\partial B(\tbf{0},1)$ will be the unit sphere.   $\nu$ will be
normalized
$(r-1)$-dimensional Hausdorff measure on $\BbbR^r$ (265A), that is,
$\nu=2^{-r+1}\beta_{r-1}\mu_{H,r-1}$, where $\mu_{H,r-1}$ is
$(r-1)$-dimensional Hausdorff measure on $\BbbR^r$.   Recall that
$\nu S_{r-1}=r\beta_r$ (265H).
I will take it for granted that $x\in\BbbR^r$ has coordinates
$(\xi_1,\ldots,\xi_r)$.

If $E\subseteq\BbbR^r$ has locally finite perimeter (474D),
$\lambda^{\partial}_E$ will be its perimeter measure (474F),
$\partial^{\$}E$ its reduced boundary (474G) and $\psi_E$ its
canonical outward-normal function (474G).
}%end of comment

\vleader{40pt}{475B}{The essential boundary}\cmmnt{ (In this paragraph I
allow $r=1$.)}   Let $A\subseteq\BbbR^r$ be any
set.   The {\bf essential closure} of $A$ is the set

\Centerline{$\clstar A=\{x:\limsup_{\delta\downarrow 0}
\Bover{\mu^*(B(x,\delta)\cap A)}{\mu B(x,\delta)}>0\}$}

\noindent\cmmnt{(see 266B).}
Similarly, the {\bf essential interior} of $A$ is the set

\Centerline{$\intstar A=\{x:\liminf_{\delta\downarrow 0}
\Bover{\mu_*(B(x,\delta)\cap A)}{\mu B(x,\delta)}=1\}$.}

\noindent\cmmnt{(If $A$ is Lebesgue measurable, this is the lower
Lebesgue density of $A$,
as defined in 341E;  see also 223Yf.)  }Finally,
the {\bf essential boundary} $\partstar A$ of $A$
is\cmmnt{ the difference} $\clstar A\setminus\intstar A$.

\cmmnt{Note that if $E\subseteq\BbbR^r$ is Lebesgue measurable
then $\BbbR^r\setminus\partstar E$ is the Lebesgue set of the function
$\chi E$, as defined in 261E.}

\leader{475C}{Lemma}\cmmnt{ (In this lemma I allow $r=1$.)} Let $A$,
$A'\subseteq\BbbR^r$.

(a)
\Centerline{$\interior A\subseteq\intstar A
\subseteq\clstar A\subseteq\overline{A}$,
\quad$\partstar A\subseteq\partial A$,}

\Centerline{$\clstar A=\BbbR^r\setminus\intstar(\BbbR^r\setminus A)$,
\quad$\partstar(\BbbR^r\setminus A)=\partstar A$.}

(b) If $A\setminus A'$ is negligible, then
$\clstar A\subseteq\clstar A'$ and $\intstar A\subseteq\intstar A'$;
\cmmnt{in particular,} if $A$ itself is negligible, $\clstar A$,
$\intstar A$ and $\partstar A$ are all empty.

(c) $\intstar A$, $\clstar A$ and $\partstar A$ are Borel sets.

(d) $\clstar(A\cup A')=\clstar A\cup\clstar A'$ and
$\intstar(A\cap A')=\intstar A\cap\intstar A'$, so
$\partstar(A\cup A')$, $\partstar(A\cap A')$ and
$\partstar(A\symmdiff A')$ are all included in
$\partstar A\cup\partstar A'$.

(e) $\clstar A\cap\intstar A'\subseteq\clstar(A\cap A')$,
$\partstar A\cap\intstar A'\subseteq\partstar(A\cap A')$ and
$\partstar A\setminus\clstar A'\subseteq\partstar(A\cup A')$.

(f) $\partstar(A\cap A')
  \subseteq(\clstar A'\cap\partial A)\cup(\partstar A'\cap\interior A)$.

(g) If $E\subseteq\BbbR^r$ is Lebesgue measurable, then
$E\symmdiff\intstar E$, $E\symmdiff\clstar E$ and $\partstar E$ are
Lebesgue negligible.

(h) $A$ is Lebesgue measurable iff $\partstar A$ is Lebesgue negligible.

\proof{{\bf (a)} It is obvious that

\Centerline{$\interior A\subseteq\intstar A
\subseteq\clstar A\subseteq\overline{A}$,}

\noindent so that $\partstar A\subseteq\partial A$.   Since

\Centerline{$\Bover{\mu^*(B(x,\delta)\cap A)}{\mu B(x,\delta)}
+\Bover{\mu_*(B(x,\delta)\setminus A)}{\mu B(x,\delta)}
=1$}

\noindent for every $x\in\BbbR^r$ and every $\delta>0$ (413Ec),
$\BbbR^r\setminus\intstar A=\clstar(\BbbR^r\setminus A)$.   It follows
that

$$\eqalign{\partstar(\BbbR^r\setminus A)
&=\clstar(\BbbR^r\setminus A)
  \symmdiff\intstar(\BbbR^r\setminus A)\cr
&=(\BbbR^r\setminus\intstar A)\symmdiff(\BbbR^r\setminus\clstar A)
=\intstar A\symmdiff\clstar A
=\partstar A.\cr}$$

\medskip

{\bf (b)} If $A\setminus A'$ is negligible, then

\Centerline{$\mu_*(B(x,\delta)\cap A)\le\mu_*(B(x,\delta)\cap A')$,
\quad$\mu^*(B(x,\delta)\cap A)\le\mu^*(B(x,\delta)\cap A')$}

\noindent for all $x$ and $\delta$, so $\intstar A\subseteq\intstar A'$
and $\clstar A\subseteq\clstar A'$.

\medskip

{\bf (c)} The point is just that
$(x,\delta)\mapsto\mu^*(A\cap B(x,\delta))$ is continuous.   \Prf\ For
any $x$, $y\in\BbbR^r$ and $\delta$, $\eta>0$ we have

$$\eqalign{|\mu^*(A\cap B(y,\eta))-\mu^*(A\cap B(x,\delta))|
&\le\mu(B(y,\eta)\symmdiff B(x,\delta))\cr
&=2\mu(B(x,\delta)\cup B(y,\eta))-\mu B(x,\delta)-\mu B(y,\eta)\cr
&\le\beta_r\bigl(2(\max(\delta,\eta)+\|x-y\|)^r-\delta^r-\eta^r\bigr)
\to 0\cr}$$


\noindent as $(y,\eta)\to(x,\delta)$.\ \QeD\  So

\Centerline{$x\mapsto
\limsup_{\delta\downarrow 0}
  \Bover{\mu^*(A\cap B(x,\delta))}{\mu B(x,\delta)}
=\inf_{\alpha\in\Bbb Q,\alpha>0}\sup_{\gamma\in\Bbb Q,0<\gamma\le\alpha}
  \Bover1{\beta_r\gamma^r}\mu^*(A\cap B(x,\gamma))$}

\noindent is Borel measurable, and

\Centerline{$\clstar A=\{x:\limsup_{\delta\downarrow 0}
  \Bover{\mu^*(A\cap B(x,\delta))}{\mu B(x,\delta)}>0\}$}

\noindent is a Borel set.

Accordingly $\intstar A=\BbbR^r\setminus\clstar(\BbbR^r\setminus A)$ and
$\partstar A$ are also Borel sets.

\medskip

{\bf (d)} For any $x\in\BbbR^r$,

$$\eqalign{\limsup_{\delta\downarrow 0}
  \Bover{\mu^*((A\cup A')\cap B(x,\delta))}{\mu B(x,\delta)}
&\le\limsup_{\delta\downarrow 0}
  \Bover{\mu^*(A\cap B(x,\delta))}{\mu B(x,\delta)}
   +\Bover{\mu^*(A'\cap B(x,\delta))}{\mu B(x,\delta)}\cr
&\le\limsup_{\delta\downarrow 0}
  \Bover{\mu(A\cap B(x,\delta))}{\mu B(x,\delta)}
   +\limsup_{\delta\downarrow 0}
  \Bover{\mu(A'\cap B(x,\delta))}{\mu B(x,\delta)},\cr}$$

\noindent so $\clstar(A\cup A')\subseteq\clstar A\cup\clstar A'$.
By (b), $\clstar A\cup\clstar A'\subseteq\clstar(A\cup A')$, so we have
equality.   Accordingly

\Centerline{$\intstar(A\cap A')
=\BbbR^r\setminus\clstar((\BbbR^r\setminus A)\cup(\BbbR^r\setminus A'))
=\intstar A\cap\intstar A'$.}

\noindent Since
$\intstar(A\cup A')\supseteq\intstar A\cup\intstar A'$,
$\partstar(A\cup A')\subseteq\partstar A\cup\partstar A'$.   Now

\Centerline{$\partstar(A\cap A')
=\partstar(\BbbR^r\setminus(A\cap A'))
\subseteq\partstar(\BbbR^r\setminus A)
  \cup\partstar(\BbbR^r\setminus A'))
=\partstar A\cup\partstar A'$}

\noindent and

\Centerline{$\partstar(A\symmdiff A')
\subseteq\partstar(A\cap(\BbbR^r\setminus A'))
  \cup\partstar(A'\cap(\BbbR^r\setminus A))
\subseteq\partstar A\cup\partstar A'$.}

\medskip

{\bf (e)} If $x\in\clstar A\cap\intstar A'$, then
$x\notin\clstar(\BbbR^r\setminus A')$ so
$x\notin\clstar(A\setminus A')$.   But from (d) we know that
$x\in\clstar(A\cap A')\cup\clstar(A\setminus A')$, so
$x\in\clstar(A\cap A')$.

Now

$$\eqalign{\partstar A\cap\intstar A'
&=(\clstar A\cap\intstar A')\setminus\intstar A\cr
&\subseteq\clstar(A\cap A')\setminus\intstar(A\cap A')
=\partstar(A\cap A'),\cr
\cr
\partstar A\setminus\clstar A'
&=\partstar(\BbbR^r\setminus A)\cap\intstar(\BbbR^r\setminus A')\cr
&\subseteq\partstar((\BbbR^r\setminus A)\cap(\BbbR^r\setminus A'))\cr
&=\partstar(\BbbR^r\setminus(A\cup A'))
=\partstar(A\cup A').\cr}$$

\medskip

{\bf (f)} If $x\in\partstar(A\cap A')\cap\partial A$, then of course
$x\in\clstar(A\cap A')\subseteq\clstar A$, so
$x\in\clstar A\cap\partial A$.   If
$x\in\partstar(A\cap A')\setminus\partial A$, then surely
$x\in\overline{A}$, so $x\in\interior A$.   But this means that

\Centerline{$\mu_*(B(x,\delta)\cap A')=\mu_*(B(x,\delta)\cap A\cap A')$,
\quad$\mu^*(B(x,\delta)\cap A')=\mu^*(B(x,\delta)\cap A\cap A')$}

\noindent for all $\delta$ small enough, so $x\in\partstar A'$ and
$x\in\partstar A'\cap\interior A$.

\medskip

{\bf (g)} Applying 472Da to $\mu$ and $\chi E$, we see that

$$\eqalign{\partstar E
&\subseteq(E\symmdiff\clstar E)\cup(E\symmdiff\intstar E)\cr
&\subseteq\{x:\chi E(x)\ne\lim_{\delta\downarrow 0}
  \Bover{\mu(E\cap B(x,\delta))}{\mu B(x,\delta)}\}\cr}$$

\noindent are all $\mu$-negligible.

\medskip

{\bf (h)} If $A$ is Lebesgue measurable then (g) tells us that
$\partstar A$ is negligible.   If $\partstar A$ is negligible, let $E$
be a measurable envelope
of $A$.   Then $\mu(E\cap B(x,\delta))=\mu^*(A\cap B(x,\delta))$ for all
$x$ and $\delta$, so $\clstar E=\clstar A$.   Similarly, if $F$ is a
measurable envelope
of $\BbbR^r\setminus A$, then $\clstar F=\clstar(\BbbR^r\setminus
A)=\BbbR^r\setminus\intstar A$ (using (a)).   Now (g) tells us that

\Centerline{$\mu(E\cap F)=\mu(\clstar E\cap\clstar F)=\mu(\clstar
A\setminus\intstar A)=0$.}

\noindent But now $A\setminus E$ and $E\setminus A\subseteq(E\cap
F)\cup((\BbbR^r\setminus A)\setminus F)$ are Lebesgue negligible, so $A$
is Lebesgue measurable.
}%end of proof of 475C

\leader{475D}{Lemma} Let $E\subseteq\BbbR^r$ be a set with locally
finite
perimeter\cmmnt{, and $\partial^{\$}E$ its reduced boundary}.   Then
$\partial^{\$}E\subseteq\partstar E$ and
$\nu(\partstar E\setminus\partial^{\$}E)=0$.

\proof{{\bf (a)} By 474N(i), $\partial^{\$}E\subseteq\clstar E$;  by
474N(ii), $\partial^{\$}E\cap\intstar E=\emptyset$;  so
$\partial^{\$}E\subseteq\partstar E$.

\medskip

{\bf (b)} For any $y\in\partstar E$,

\Centerline{$\limsup_{\delta\downarrow 0}\Bover1{\delta^{r-1}}
  \lambda^{\partial}_EB(y,\delta)>0$.}

\noindent\Prf\ We have an $\epsilon\in\ocint{0,\bover12}$ such that

\Centerline{$\liminf_{\delta\downarrow 0}
  \Bover{\mu(E\cap B(y,\delta))}{\mu B(y,\delta)}<1-\epsilon$,
\quad$\limsup_{\delta\downarrow 0}
  \Bover{\mu(E\cap B(y,\delta))}{\mu B(y,\delta)}>\epsilon$.}

\allowmorestretch{468}{
\noindent Since the function
$\delta\mapsto\Bover{\mu(E\cap B(y,\delta))}{\mu B(y,\delta)}$ is
continuous, there is a sequence $\sequencen{\delta_n}$ in
$\ooint{0,\infty}$ such that $\lim_{n\to\infty}\delta_n=0$ and
}

\Centerline{$\epsilon\mu B(y,\delta_n)\le\mu(E\cap B(y,\delta_n))
\le(1-\epsilon)\mu B(y,\delta_n)$}

\noindent for every $n$.   Now from 474Lb we have

$$\eqalign{(\epsilon\beta_r)^{(r-1)/r}\delta_n^{r-1}
&=(\epsilon\mu B(y,\delta_n))^{(r-1)/r}\cr
&\le\min(\mu(B(y,\delta_n)\cap E),
   \mu(B(y,\delta_n)\setminus E))^{(r-1)/r}
\le 2c\lambda^{\partial}_EB(y,\delta_n)\cr}$$

\noindent for every $n$, where $c>0$ is the constant there.
But this means that

\Centerline{$\limsup_{\delta\downarrow 0}\Bover1{\delta^{r-1}}
  \lambda^{\partial}_EB(y,\delta)
\ge\limsup_{n\to\infty}\Bover1{\delta_n^{r-1}}
  \lambda^{\partial}_EB(y,\delta_n)
\ge\Bover1{2c}(\epsilon\beta_r)^{(r-1)/r}
>0$.  \Qed}

\medskip

{\bf (c)} Let $\epsilon>0$.   Set

\Centerline{$F_{\epsilon}
=\{y:y\in\BbbR^r\setminus\partial^{\$}E,\,
\limsup_{\delta\downarrow 0}\Bover1{\delta^{r-1}}
\lambda^{\partial}_EB(y,\delta)>\epsilon\}$.}

\noindent Because $\partial^{\$}E$ is
$\lambda^{\partial}_E$-conegligible (474G),
$\lambda^{\partial}_EF_{\epsilon}=0$.   So there is an open set
$G\supseteq F_{\epsilon}$ such that $\lambda^{\partial}_EG\le\epsilon^2$
(256Bb/412Wb).   Let $\delta>0$.   Let $\Cal I$ be the family of all
those
non-singleton closed balls $B\subseteq G$ such that $\diam B\le\delta$
and $\lambda^{\partial}_EB\ge 2^{-r+1}\epsilon(\diam B)^{r-1}$.   Then
every point of $F_{\epsilon}$ is the centre of arbitrarily small members
of $\Cal I$.   By Besicovitch's Covering Lemma (472B), there is a family
$\ofamily{k}{5^r}{\Cal I_k}$ of disjoint countable subsets of $\Cal I$
such that $\Cal I^*=\bigcup_{k<5^r}\bigcup\Cal I_k$ covers
$F_{\epsilon}$.   Now

\Centerline{$\sum_{B\in\Cal I^*}(\diam B)^{r-1}
\le\sum_{k<5^r}\sum_{B\in\Cal I_k}\Bover{2^{r-1}}{\epsilon}
  \lambda^{\partial}_EB
\le\Bover{5^r2^{r-1}}{\epsilon}\lambda^{\partial}_EG
\le 5^r2^{r-1}\epsilon$.}

\noindent As $\delta$ is arbitrary, $\mu_{H,r-1}^*F_{\epsilon}$ is at
most $5^r2^{r-1}\epsilon$ (264Fb/471Dc) and
$\nu^*F_{\epsilon}\le 5^r\beta_{r-1}\epsilon$.   As $\epsilon$ is
arbitrary,

\Centerline{$\partstar E\setminus\partial^{\$}E
\subseteq\{y:y\in\BbbR^r\setminus\partial^{\$}E,\,
\limsup_{\delta\downarrow 0}\Bover1{\delta^{r-1}}
\lambda^{\partial}_EB(y,\delta)>0\}$}

\noindent is $\nu$-negligible, as claimed.
}%end of proof of 475D

\leader{475E}{Lemma} Let $E\subseteq\BbbR^r$ be a set with locally
finite perimeter.

(a) If $A\subseteq\partial^{\$}E$, then
$\nu^*A\le(\lambda^{\partial}_E)^*A$.

(b) If $A\subseteq\BbbR^r$ and $\nu A=0$, then
$\lambda^{\partial}_EA=0$.

\proof{{\bf (a)} Given $\epsilon$, $\delta>0$ let $\Cal I$ be the family
of non-trivial closed balls $B\subseteq\BbbR^r$ of diameter at most
$\delta$ such that $\beta_{r-1}(\Bover12\diam B)^{r-1}
\le(1+\epsilon)\lambda^{\partial}_EB$.   By 474S, every point of $A$ is
the centre of arbitrarily small members of $\Cal I$.   By 472Cb, there is
a countable family $\Cal I_1\subseteq I$ such that
$A\subseteq\bigcup\Cal I_1$ and
$\sum_{B\in\Cal I_1}\lambda^{\partial}_EB
\le(\lambda^{\partial}_E)^*A+\epsilon$.   But this means that

\Centerline{$\sum_{B\in\Cal I_1}(\diam B)^{r-1}
\le(1+\epsilon)\Bover{2^{r-1}}{\beta_{r-1}}
  ((\lambda^{\partial}_E)^*A+\epsilon)$.}

\noindent As $\delta$ is arbitrary,

\Centerline{$\nu^*A
=\Bover{\beta_{r-1}}{2^{r-1}}\mu_{H,r-1}^*A
\le(1+\epsilon)((\lambda^{\partial}_E)^*A+\epsilon)$.}

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

\medskip

{\bf (b)} For $n\in\Bbb N$, set

\Centerline{$A_n=\{x:x\in A$, $\lambda^{\partial}_EB(x,\delta)
\le 2\beta_{r-1}\delta^{r-1}$ whenever $0<\delta\le 2^{-n}\}$.}

\noindent Now, given $\epsilon>0$, there is a sequence
$\sequence{i}{D_i}$ of sets covering $A_n$ such that
$\diam D_i\le 2^{-n}$ for every $i$ and
$\sum_{i=0}^{\infty}(\diam D_i)^{r-1}\le\epsilon$.
Passing over the trivial case $A_n=\emptyset$, we may suppose that for
each $i\in\Bbb N$ there is an $x_i\in A_n\cap D_i$, so that
$D_i\subseteq B(x_i,\diam D_i)$ and

$$\eqalign{(\lambda^{\partial}_E)^*A_n
&\le\sum_{i=0}^{\infty}(\lambda^{\partial}_E)^*D_i
\le\sum_{i=0}^{\infty}\lambda^{\partial}_EB(x_i,\diam D_i)\cr
&\le\sum_{i=0}^{\infty}2\beta_{r-1}(\diam D_i)^{r-1}
\le 2\beta_{r-1}\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\lambda^{\partial}_EA_n=0$.
And this is true for every $n$.   As
$\bigcup_{n\in\Bbb N}A_n\supseteq A\cap\partial^{\$}E$ (474S again),
$A\setminus\bigcup_{n\in\Bbb N}A_n$
is $\lambda^{\partial}_E$-negligible (474G), and so is $A$.
}%end of proof of 475E

\leader{475F}{Lemma} Let $E\subseteq\BbbR^r$ be a set with locally
finite perimeter, and $\epsilon>0$.   Then $\lambda^{\partial}_E$ is
inner regular with respect to the family
$\Cal E=\{F:F\subseteq\BbbR^r$ is Borel,
$\lambda^{\partial}_EF\le(1+\epsilon)\nu F\}$.

\proof{{\bf (a)} We need some elementary bits of geometry.

\medskip

\quad{\bf (i)} If $x\in\BbbR^r$, $\delta>0$, $\alpha\ge 0$ and
$v\in S_{r-1}$, then

\Centerline{$\mu\{z:z\in B(x,\delta)$,
$|\varinnerprod{(z-x)}{v}|\le\alpha\}
\le 2\alpha\beta_{r-1}\delta^{r-1}$.}

\noindent\Prf\ Translating and rotating, if necessary, we can reduce to
the case $x=\tbf{0}$, $v=(0,\ldots,1)$.   In this case we are looking at

\Centerline{$\{z:\|z\|\le\delta,\,|\zeta_r|\le\alpha\}
\subseteq\{u:u\in\BbbR^{r-1},\,\|u\|\le\delta\}\times[-\alpha,\alpha]$}

\noindent which has measure $2\alpha\beta_{r-1}\delta^{r-1}$.\ \Qed

\medskip

\quad{\bf (ii)} If $x\in\BbbR^r$, $\delta>0$, $0<\eta\le\bover12$,
$v\in S_{r-1}$, $H=\{z:\varinnerprod{z}{v}\le\alpha\}$ and

\Centerline{$|\mu(H\cap B(x,\delta))-\Bover12\mu B(x,\delta)|
<2^{-r+1}\beta_{r-1}\eta\delta^r$,}

\noindent then $|\varinnerprod{x}{v}-\alpha|\le\eta\delta$.   \Prf\
Again translating and rotating if necessary, we may suppose that
$x=\tbf{0}$ and $v=(0,\ldots,1)$.   Set $H_0=\{z:\zeta_r\le 0\}$.
\Quer\ If $\alpha>\eta\delta$, then $H\cap B(\tbf{0},\delta)$ includes

\Centerline{$(H_0\cap B(\tbf{0},\delta))
\cup(\{u:u\in\BbbR^{r-1},\,\|u\|\le\Bover12\delta\}\times[0,\alpha'])$}

\noindent where $\alpha'=\min(|\alpha|,\Bover{\sqrt3}2\delta)>\eta\delta)$,
so

$$\eqalign{\mu(H\cap B(\tbf{0},\delta))-\Bover12\mu B(\tbf{0},\delta)
&=\mu(H\cap B(\tbf{0},\delta))-\mu(H_0\cap B(\tbf{0},\delta))\cr
&\ge 2^{-r+1}\beta_{r-1}\delta^{r-1}\alpha'
>2^{-r+1}\beta_{r-1}\delta^r\eta,\cr}$$

\noindent contrary to hypothesis.\ \BanG\  Similarly, \Quer\ if
$\alpha<-\eta\delta$, then $H\cap B(\tbf{0},\delta)$ is included in

\Centerline{$H_0\cap B(\tbf{0},\delta)
\setminus(\{u:u\in\BbbR^{r-1},\,
\|u\|\le\Bover12\delta\}\times\ocint{\alpha,0})$,}

\noindent so

$$\eqalign{\Bover12\mu B(\tbf{0},\delta)-\mu(H\cap B(\tbf{0},\delta))
&=\mu(H_0\cap B(\tbf{0},\delta))-\mu(H\cap B(\tbf{0},\delta))\cr
&\ge 2^{-r+1}\beta_{r-1}\delta^{r-1}\alpha'
>2^{-r+1}\beta_{r-1}\delta^r\eta,\cr}$$

\noindent which is equally impossible.\ \BanG\   So
$|\alpha|\le\eta\delta$.\ \Qed

\medskip

{\bf (b)} Let $F$ be such that $\lambda^{\partial}_EF>0$.   Let $\eta$,
$\zeta>0$ be such that

\Centerline{$\eta<1$,
\quad$\Bover{(1+\eta)^2}{(1-\eta)^{r-1}}\le 1+\epsilon$,
\quad$2(1+2^r)\beta_r\zeta<2^{-r}\beta_{r-1}\eta$.}

\noindent Because $\partial^{\$}E$ is
$\lambda^{\partial}_E$-conegligible (474G again),
$\lambda^{\partial}_E(F\cap\partial^{\$}E)>0$.   Because
$\lambda^{\partial}_E$ is a Radon measure (474E) and
$\psi_E:\partial^{\$}E\to S_{r-1}$ is
$\dom(\lambda^{\partial}_E)$-measurable (474E(i), 474G), there is a
compact set
$K_1\subseteq F\cap\partial^{\$}E$ such that $\lambda^{\partial}_EK_1>0$
and $\psi_E\restr K_1$ is continuous, by Lusin's theorem (418J).   For
$x\in\partial^{\$}E$, set
$H_x=\{z:\varinnerprod{(z-x)}{\psi_E(x)}\le 0\}$.   The function

\Centerline{$(x,\delta)\mapsto\mu((E\symmdiff H_x)\cap B(x,\delta)):
K_1\times\ooint{0,\infty}\to\Bbb R$}

\noindent is Borel measurable.   \Prf\ Take a Borel set $E'$ such that
$\mu(E\symmdiff E')=0$.   Then

\Centerline{$\{(x,\delta,z):x\in K_1$,
$z\in(E'\symmdiff H_x)\cap B(x,\delta)\}$}

\noindent is a Borel set in
$\BbbR^r\times\ooint{0,\infty}\times\BbbR^r$, so its sectional measure
is a Borel measurable function, by 252P.\ \Qed

For each $x\in K_1$,

\Centerline{$\lim_{n\to\infty}
  \Bover{\mu((E\symmdiff H_x)\cap B(x,2^{-n}))}{\mu B(x,2^{-n})}=0$}

\noindent (474R).   So there is an $n_0\in\Bbb N$ such that
$\lambda^{\partial}_EF_1>0$, where $F_1$ is the Borel set

\Centerline{$\{x:x\in K_1$, $\mu((E\symmdiff H_x)\cap B(x,2^{-n}))
\le\zeta\mu B(x,2^{-n})$ for every $n\ge n_0\}$.}

\noindent Let $K_2\subseteq F_1$ be a compact set such that
$\lambda^{\partial}_EK_2>0$.

For each $n\in\Bbb N$, the function

\Centerline{$x\mapsto\lambda^{\partial}_EB(x,2^{-n}))
=\lambda^{\partial}_E(x+B(\tbf{0},2^{-n}))$}

\noindent is Borel measurable (444Fe).   Let $y\in K_2$ be such that
$\lambda^{\partial}_E(K_2\cap B(y,\delta))>0$ for every $\delta>0$
(cf.\ 411Nd).   Set $v=\psi_E(y)$.   Let $n>n_0$ be so large that
$2\beta_{r-1}\|\psi_E(x)-v\|\le\beta_r\zeta$ whenever $x\in K_1$ and
$\|x-y\|\le 2^{-n}$.   Set $K_3=K_2\cap B(y,2^{-n-1})$, so that
$\lambda^{\partial}_EK_3>0$.

\medskip

{\bf (c)} We have $|\varinnerprod{(x-z)}{v}|\le\eta\|x-z\|$ whenever
$x$, $z\in K_3$.   \Prf\ If $x=z$ this is trivial.
Otherwise, let $k\ge n$ be such that $2^{-k-1}\le\|x-z\|\le 2^{-k}$, and
set $\delta=2^{-k}$.   Set

\Centerline{$H'_x=\{w:\varinnerprod{(w-x)}{v}\le 0\}$,
\quad$H'_z=\{w:\varinnerprod{(w-z)}{v}\le 0\}$.}

\noindent Since
$|\varinnerprod{(w-x)}{v}-\varinnerprod{(w-x)}{\psi_E(x)}|
\le 2\delta\|\psi_E(x)-v\|$ whenever $w\in B(x,2\delta)$,

\Centerline{$(H_x\symmdiff H'_x)\cap B(x,2\delta)
\subseteq\{w:w\in B(x,2\delta),
  \,|\varinnerprod{(w-x)}{v}|\le 2\delta\|\psi_E(x)-v\|\}$}

\noindent has measure at most

\Centerline{$4\delta\|\psi_E(x)-v\|\beta_{r-1}(2\delta)^{r-1}
\le 2\delta\beta_r\zeta(2\delta)^{r-1}
=\zeta\mu B(x,2\delta)$,}

\noindent using (a-i) for the first inequality.   So

$$\eqalign{\mu((E\symmdiff H'_x)\cap B(x,2\delta))
&\le\mu((E\symmdiff H_x)\cap B(x,2\delta))
+\mu((H_x\symmdiff H'_x)\cap B(x,2\delta))\cr
&\le 2\zeta\mu B(x,2\delta)\cr}$$

\noindent because $k>n_0$ and $x\in F_1$.   Similarly,
$\mu((E\symmdiff H'_z)\cap B(z,\delta))\le 2\zeta\mu B(z,\delta)$.   Now
observe that because $\|x-z\|\le\delta$,
$B(z,\delta)\subseteq B(x,2\delta)$,

\Centerline{$\mu((E\symmdiff H'_x)\cap B(z,\delta))
\le 2\zeta\mu B(x,2\delta)=2^{r+1}\zeta\mu B(z,\delta)$,}

\noindent and

$$\eqalign{\mu((H'_x\symmdiff H'_z)\cap B(z,\delta))
&\le\mu((E\symmdiff H'_x)\cap B(z,\delta))
  +\mu((E\symmdiff H'_z)\cap B(z,\delta))\cr
&\le(2+2^{r+1})\zeta\mu B(z,\delta).\cr}$$

\noindent Since $\mu(H'_z\cap B(z,\delta))=\Bover12\mu B(z,\delta)$,

\Centerline{$|\mu(H'_x\cap B(z,\delta))-\Bover12\mu B(z,\delta)|
\le 2(1+2^r)\zeta\mu B(z,\delta)<2^{-r}\beta_{r-1}\eta\delta^r$,}

\noindent and (using (a-ii) above)

\Centerline{$|\varinnerprod{(x-z)}{v}|\le\Bover12\eta\delta
\le\eta\|x-v\|$.  \Qed}

\medskip

{\bf (d)} Let $V$ be the hyperplane $\{w:\varinnerprod{w}{v}=0\}$, and
let $T:K_3\to V$ be the orthogonal projection, that is,
$Tx=x-(\varinnerprod{x}{v})v$ for every $x\in K_3$.   Then (c)
tells us that if $x$, $z\in K_3$,

\Centerline{$\|Tx-Tz\|\ge\|x-z\|-|\varinnerprod{(x-z)}{v}|
\ge(1-\eta)\|x-z\|$.}

\noindent Because $\eta<1$, $T$ is injective.   Consider the compact set
$T[K_3]$.   The inverse $T^{-1}$ of $T$ is $\Bover1{1-\eta}$-Lipschitz,
and $\nu K_3>0$ (by 475Eb), so
$\nu(T[K_3])\ge(1-\eta)^{r-1}\nu K_3>0$ (264G/471J).
Let $G\supseteq T[K_3]$ be an open set such that
$\nu(G\cap V)\le(1+\eta)\nu(T[K_3])$.   (I am using the fact that the
subspace measure $\nu_V$ induced by $\nu$ on $V$ is a copy of Lebesgue
measure on $\BbbR^{r-1}$, so is a Radon measure.)   Let $\Cal I$ be the
family of non-trivial closed balls $B\subseteq G$ such that
$(1-\eta)^{r-1}\lambda^{\partial}_ET^{-1}[B]
\le(1+\eta)\nu(B\cap V)$.   Then every point of $T[K_3]$ is the centre
of arbitrarily small members of
$\Cal I$.   \Prf\ If $x\in K_3$ and $\delta_0>0$, there is a
$\delta\in\ocint{0,\delta_0}$ such that $B(Tx,\delta)\subseteq G$ and
$\lambda^{\partial}_EB(x,\delta)\le(1+\eta)\beta_{r-1}\delta^{r-1}$
(474S once more).   Now consider $B=B(Tx,(1-\eta)\delta)$.  Then
$T^{-1}[B]\subseteq B(x,\delta)$, so

\Centerline{$\lambda^{\partial}_ET^{-1}[B]
\le\lambda^{\partial}_EB(x,\delta)
\le(1+\eta)\beta_{r-1}\delta^{r-1}
=\Bover{1+\eta}{(1-\eta)^{r-1}}\nu(B\cap V)$.  \Qed}

\noindent By 261B/472Ca, applied in $V\cong\BbbR^{r-1}$, there is a
countable disjoint family $\Cal I_0\subseteq\Cal I$ such that
$\nu(T[K_3]\setminus\bigcup\Cal I_0)=0$.

Now $\nu(K_3\setminus\bigcup_{B\in\Cal I_0}T^{-1}[B])=0$, because
$T^{-1}$ is Lipschitz, so
$\lambda^{\partial}_E(K_3\setminus\bigcup_{B\in\Cal I_0}T^{-1}[B])=0$
(475Eb again), and

$$\eqalignno{\lambda^{\partial}_EK_3
&\le\sum_{B\in\Cal I_0}\lambda^{\partial}_ET^{-1}[B]
\le\Bover{1+\eta}{(1-\eta)^{r-1}}\sum_{B\in\Cal I_0}\nu(B\cap V)\cr
&\le\Bover{1+\eta}{(1-\eta)^{r-1}}\nu(G\cap V)
\le\Bover{(1+\eta)^2}{(1-\eta)^{r-1}}\nu(T[K_3])
\le\Bover{(1+\eta)^2}{(1-\eta)^{r-1}}\nu K_3\cr
\displaycause{because $T$ is $1$-Lipschitz}
&\le(1+\epsilon)\nu K_3\cr}$$

\noindent by the choice of $\eta$.   Thus $K_3\in\Cal E$.

\medskip

{\bf (e)} This shows that every $\lambda^{\partial}_E$-non-negligible set
measured by
$\lambda^{\partial}_E$ includes a $\lambda^{\partial}_E$-non-negligible
member of $\Cal E$.
As $\Cal E$ is closed under disjoint unions, $\lambda^{\partial}_E$ is
inner regular with respect to $\Cal E$ (412Aa).
}%end of proof of 475F

\leader{475G}{Theorem} Let $E\subseteq\BbbR^r$ be a set with locally
finite perimeter.   Then
$\lambda^{\partial}_E=\nu\LLcorner\partstar E$\cmmnt{, that is, for
$F\subseteq\BbbR^r$,
$\lambda^{\partial}_EF=\nu(F\cap\partstar E)$ whenever either is
defined}.

\proof{{\bf (a)} Suppose first that $F$ is a Borel set included in the
reduced boundary $\partial^{\$}E$ of $E$.   Then
$\nu F\le\lambda^{\partial}_EF$, by 475Ea.   On the other hand, for any
$\epsilon>0$ and $\gamma<\lambda^{\partial}_EF$, there is an
$F_1\subseteq F$ such that

\Centerline{$\gamma\le\lambda^{\partial}_EF_1\le(1+\epsilon)\nu F_1
\le(1+\epsilon)\nu F$,}

\noindent by 475F;  so we must have $\lambda^{\partial}_EF=\nu F$.

\medskip

{\bf (b)} Now suppose that $F$ is measured by $\lambda^{\partial}_E$.
Because $\partial^{\$}E$ is $\lambda^{\partial}_E$-conegligible, and
$\lambda^{\partial}_E$ is a $\sigma$-finite Radon measure, there is a
Borel set $F'\subseteq F\cap\partial^{\$}E$ such that
$\lambda^{\partial}_E(F\setminus F')=0$.   Now
$\nu F'=\lambda^{\partial}_EF'$, by (a), and
$\nu(F\cap\partial^{\$}E\setminus F')=0$, by 475Ea, and
$\nu(\partstar E\setminus\partial^{\$}E)=0$, by 475D;  so
$\nu(F\cap\partstar E)$ is defined and equal to
$\lambda^{\partial}_EF'=\lambda^{\partial}_EF$.

\medskip

{\bf (c)} Let $\sequencen{K_n}$ be a non-decreasing sequence of compact
subsets of $\partial^{\$}E$ such that $\bigcup_{n\in\Bbb N}K_n$ is
$\lambda^{\partial}_E$-conegligible.
By (b), $\nu(\partstar E\setminus\bigcup_{n\in\Bbb N}K_n)=0$, while
$\nu K_n=\lambda^{\partial}_EK_n$ is finite for each $n$.
For each $n$, the subspace measure $\nu_{K_n}$ on $K_n$ is a multiple of
Hausdorff $(r-1)$-dimensional measure on $K_n$ (471E), so is a Radon
measure (471Dh, 471F),
as is $(\lambda^{\partial}_E)_{K_n}$;  since, by (b), $\nu_{K_n}$
and $(\lambda^{\partial}_E)_{K_n}$ agree on the Borel subsets of $K_n$,
they are actually identical.   So if $F\subseteq\BbbR^r$ is such that
$\nu$ measures $F\cap\partstar E$, $\lambda^{\partial}_E$ will measure
$F\cap K_n$ for every $n$, and therefore will measure $F$;  so that in
this case also $\lambda^{\partial}_EF=\nu(F\cap\partstar E)$.
}%end of proof of 475G

\leader{475H}{Proposition} Let $V\subseteq\BbbR^r$ be a hyperplane, and
$T:\BbbR^r\to V$ the orthogonal projection.   Suppose that
$A\subseteq\BbbR^r$ is such that $\nu A$ is defined and finite,
and for $u\in V$ set

$$\eqalign{f(u)&=\#(A\cap T^{-1}[\{u\}])\text{ if this is finite},\cr
&=\infty\text{ otherwise}.\cr}$$

\noindent Then $\int_Vf(u)\nu(du)$ is defined and at most $\nu A$.

\proof{{\bf (a)} Because $\nu$ is invariant under isometries, we can
suppose that $V=\{x:\xi_r=0\}$, so that
$Tx=(\xi_1,\ldots,\xi_{r-1},0)$ for $x=(\xi_1,\ldots,\xi_r)$.   For
$m$, $n\in\Bbb N$ and $u\in V$ set

\Centerline{$f_{mn}(u)
=\#(\{k:k\in\Bbb Z$, $|k|\le 4^m$,
$A\cap(\{u\}\times\coint{2^{-m}k,2^{-m}(k+1-2^{-n})})\ne\emptyset\})$;}

\noindent so that $f(u)=\lim_{m\to\infty}\lim_{n\to\infty}f_{mn}(u)$ for
every $u\in V$.

\medskip

{\bf (b)} Suppose for the moment that $A$ is actually a Borel set.
Then $T[A\cap(\BbbR^r\times\coint{\alpha,\beta})]$ is always analytic
(423Eb, 423Bb),
therefore measured by $\nu$ (432A), and every $f_{mn}$ is measurable.
Next, given $\epsilon>0$ and $m$, $n\in\Bbb N$, there is a sequence
$\sequence{i}{F_i}$ of closed sets of diameter at most $2^{-m-n}$,
covering $A$, such that
$\sum_{i=0}^{\infty}2^{-r+1}\beta_{r-1}(\diam F_i)^{r-1}
\le\nu A+\epsilon$.   Now each $T[F_i]$ is a compact set of diameter at
most $\diam F_i$, so
$\nu(T[F_i])\le 2^{-r+1}\beta_{r-1}(\diam F_i)^{r-1}$ (264H);  and if we
set $g=\sum_{i=0}^{\infty}\chi T[F_i]$,
$f_{mn}\le g$ everywhere on $V$, so
$\int f_{mn}d\nu\le\int g\,d\nu\le\nu A+\epsilon$.   As $\epsilon$ is
arbitary, $\int f_{mn}d\nu\le\nu A$.  Accordingly

$$\eqalignno{\int fd\nu
&=\lim_{m\to\infty}\lim_{n\to\infty}\int f_{mn}d\nu\cr
\displaycause{because the limits are monotonic}
&\le\nu A+\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\int fd\nu\le\nu A$.

\medskip

{\bf (c)} In general, there are Borel sets $E$, $F$ such that
$E\setminus F\subseteq A\subseteq E$ and $\nu F=0$, by 264Fc/471Db.
By (b),

\Centerline{$\int\#(E\cap T^{-1}[\{u\}])\nu(du)\le\nu E$,
\quad$\int\#(F\cap T^{-1}[\{u\}])\nu(du)\le\nu F=0$,}

\noindent so
$\int\#(A\cap T^{-1}[\{u\}])\nu(du)$ is defined and equal to
$\int\#(E\cap T^{-1}[\{u\}])\nu(du)\le\nu A$.
}%end of proof of 475H

\leader{475I}{Lemma}\cmmnt{ (In this lemma I allow $r=1$.)}   Let
$\Cal K$ be the family of compact sets $K\subseteq\BbbR^r$ such that
$K=\clstar K$.   Then $\mu$ is inner regular with respect to $\Cal K$.

\proof{
%\allowmorestretch{468}{
{\bf (a)} Write $\Cal D$ for the set of dyadic (half-open) cubes in
$\BbbR^r$, that is, sets expressible in the form
$\prod_{i<r}[2^{-m}k_i$,\discretionary{}{}{}$2^{-m}(k_i+1)[$
where $m$,
$k_0,\ldots,k_{r-1}\in\Bbb Z$.   For $m\in\Bbb N$ and $x\in\BbbR^r$
write $C(x,m)$ for the dyadic cube with side of length $2^{-m}$ which
contains $x$.   Then, for any $A\subseteq\BbbR^r$,
%}

\Centerline{$\lim_{m\to\infty}2^{mr}\mu_*(A\cap C(x,m))=1$}

\noindent for every $x\in\intstar A$.   \Prf\
$C(x,m)\subseteq B(x,2^{-m}\sqrt r)$ for each $m$, so

$$2^{mr}\mu^*(C(x,m)\setminus A)
\le\beta_rr^{r/2}\Bover{\mu^*(B(x,2^{-m}\sqrt r)\setminus A)}
  {\mu B(x,2^{-m}\sqrt r)}
\to 0$$

\noindent as $m\to\infty$.\ \Qed

\medskip

{\bf (b)} Now, given a Lebesgue measurable set $E\subseteq\BbbR^r$ and
$\gamma<\mu E$, choose $\sequencen{E_n}$,
$\sequencen{\gamma_n}$, $\sequencen{m_n}$ and $\sequencen{K_n}$
inductively, as follows.   Start with $E_0=E$ and $\gamma_0=\gamma$.
Given that $\gamma_n<\mu E_n$, let $K_n\subseteq E_n$ be a compact set
of measure greater than $\gamma_n$.   Now, by (a), there is an
$m_n\ge n$ such that $\mu^*E_{n+1}>\gamma_n$, where

\Centerline{$E_{n+1}=\{x:x\in K_n,\,
\mu(K_n\cap C(x,m_n))\ge\Bover23\mu C(x,m_n)\}$;}

\noindent note that $E_{n+1}$ is of the form $K_n\cap\bigcup\Cal D_0$ for
some set $\Cal D_0$ of half-open cubes of side $2^{-m_n}$, so that
$E_{n+1}$ is measurable and

\Centerline{$\mu(E_{n+1}\cap C(x,m_n))
=\mu(K_n\cap C(x,m_n))\ge\Bover23\mu C(x,m_n)$}

\noindent for every $x\in E_{n+1}$.   Now set

\Centerline{$\gamma_{n+1}
=\max(\gamma_n,\mu E_{n+1}-\Bover13\cdot 2^{-m_nr})$,}

\noindent and continue.

At the end of the induction, set
$K=\bigcap_{n\in\Bbb N}K_n=\bigcap_{n\in\Bbb N}E_n$.   Then $K$ is
compact, $K\subseteq E$ and

\Centerline{$\mu K=\lim_{n\to\infty}\mu K_n
=\lim_{n\to\infty}\gamma_n\ge\gamma$.}

\noindent If $x\in K$ and $n\in\Bbb N$, then

$$\eqalign{\mu(K\cap B(x,2^{-m_n}\sqrt r))
&\ge\mu(K\cap C(x,m_n))
\ge\mu(E_{n+1}\cap C(x,m_n))-\mu E_{n+1}+\mu K\cr
&\ge\Bover23\mu C(x,m_n)-\mu E_{n+1}+\gamma_{n+1}\cr
&\ge\Bover13\cdot 2^{-m_nr}
=\Bover1{3\beta_rr^{r/2}}\mu B(x,2^{-m_n}\sqrt r),\cr}$$

\noindent so

\Centerline{$\limsup_{\delta\downarrow 0}
  \Bover{\mu(K\cap B(x,\delta))}{\mu B(x,\delta)}
\ge\Bover1{3\beta_rr^{r/2}}>0$}

\noindent and $x\in\clstar K$.

\medskip

{\bf (c)} Thus $K\subseteq\clstar K$.   Since certainly 
$\clstar K\subseteq\overline{K}=K$, we have $K=\clstar K$.
}%end of proof of 475I

\vleader{60pt}{475J}{Lemma} Let $E$ be a Lebesgue measurable subset of
$\BbbR^r$, identified with $\BbbR^{r-1}\times\Bbb R$.
For $u\in\BbbR^{r-1}$, set $G_u=\{t:(u,t)\in\intstar E\}$,
$H_u=\{t:(u,t)\in\intstar(\BbbR^r\setminus E)\}$ and
$D_u=\{t:(u,t)\in\partstar E\}$, so that $G_u$, $H_u$ and $D_u$ are
disjoint and cover $\Bbb R$.

(a) There is a
$\mu_{r-1}$-conegligible set $Z\subseteq\BbbR^{r-1}$ such that whenever
$u\in Z$, $t<t'$ in $\Bbb R$, $t\in G_u$ and
$t'\in H_u$, there is an $s\in D_u\cap\ooint{t,t'}$.

(b) There is a
$\mu_{r-1}$-conegligible set $Z_1\subseteq\BbbR^{r-1}$ such that whenever
$u\in Z_1$, $t$, $t'\in\Bbb R$, $t\in G_u$ and
$t'\in H_u$, there is a member of $D_u$ between $t$ and $t'$.

(c) If $E$ has locally finite perimeter, there is a conegligible set
$Z_2\subseteq Z_1$ such that,
for every $u\in Z_2$, $D_u\cap[-n,n]$ is finite for every $n\in\Bbb N$,
$G_u$ and $H_u$ are open, and $D_u=\partial G_u=\partial H_u$, so that
the constituent intervals of $\Bbb R\setminus D_u$ lie alternately in
$G_u$ and $H_u$.

\proof{{\bf (a)(i)} For $q\in\Bbb Q$, set
$f_q(u)=\sup(G_u\cap\ooint{-\infty,q})$, taking $-\infty$ for
$\sup\emptyset$.   Then
$f_q:\BbbR^{r-1}\to[-\infty,q]$ is Lebesgue measurable.   \Prf\ For
$\alpha<q$,

\Centerline{$\{u:f_q(u)>\alpha\}=\{u:$ there is some
$t\in\ooint{\alpha,q}$ such that $(u,t)\in\intstar E\}$.}

\noindent Because $\intstar E$ is a Borel set (475Cc),
$\{u:f_q(u)>\alpha\}$ is
analytic (423Eb, 423Bc), therefore measurable (432A again).\ \Qed

\medskip

\quad{\bf (ii)} For any $q\in\Bbb Q$,
$W_q=\{u:f_q(u)<q,\,f_q(u)\in G_u\}$ is negligible.
\Prf\Quer\ Suppose, if possible, otherwise.   Let $n\in\Bbb N$ be such
that $\{u:u\in W_q,\,f_q(u)>-n\}$ is not negligible.
If we think of $[-\infty,q]$ as a compact metrizable space, 418J again
tells us that there is a Borel set
$F\subseteq\BbbR^{r-1}$ such that $f_q\restr F$ is continuous and
$F_1=\{u:u\in F\cap W_q$, $f_q(u)>-n\}$ is not
negligible.   Note that $F_1$ is measurable, being the projection of the
Borel set $\{(u,f_q(u)):u\in F,\,-n<f_q(u)<q\}\cap\intstar E$.   By 475I,
there is a non-negligible compact set $K\subseteq F_1$ such that
$K\subseteq\clstar K$, interpreting $\clstar K$ in $\BbbR^{r-1}$.
Because $g\restr K$ is continuous, it attains its maximum at $u\in K$
say.

Set $x=(u,f_q(u))$.   Then, whenever
$0<\delta\le\min(n+f_q(u),q-f_q(u))$,

$$\eqalignno{B(x,2\delta)\setminus\intstar E
&\supseteq\{(w,t):w\in K\cap V(u,\delta),\,|t-f_q(u)|\le\delta,\,
                   f_q(w)<t<q\}\cr
\displaycause{writing $V(u,\delta)$ for
   $\{w:w\in\BbbR^{r-1},\,\|w-u\|\le\delta\}$}
&\supseteq\{(w,t):w\in K\cap V(u,\delta),\,
  f_q(u)<t<f_q(u)+\delta\}\cr}$$

\noindent because $f_q(w)\le f_q(u)$ for every $w\in K$.
So, for such $\delta$,

$$\eqalign{\mu(B(x,2\delta)\setminus\intstar E)
&\ge\delta\mu_{r-1}(K\cap V(u,\delta))\cr
&=\Bover{\beta_{r-1}}{2^r\beta_r}
  \cdot\Bover{\mu_{r-1}(K\cap V(u,\delta))}{\mu_{r-1}V(u,\delta)}
  \cdot\mu B(x,2\delta),\cr}$$

\noindent and

$$\eqalign{\limsup_{\delta\downarrow 0}
  \Bover{\mu(B(x,\delta)\setminus E)}{\mu B(x,\delta)}
&=\limsup_{\delta\downarrow 0}
  \Bover{\mu(B(x,2\delta)\setminus\intstar E)}{\mu B(x,2\delta)}\cr
&\ge\Bover{\beta_{r-1}}{2^r\beta_r}\limsup_{\delta\downarrow 0}
  \Bover{\mu_{r-1}(K\cap V(u,\delta))}{\mu_{r-1}V(u,\delta)}
>0\cr}$$

\noindent because $u\in\clstar K$;  but $x=(u,f_q(u))$ is supposed to
belong to $\intstar E$.\ \Bang\Qed

\medskip

\quad{\bf (iii)} Similarly, setting

\Centerline{$f'_q(u)
=\inf(H_u\cap\ooint{u,\infty})$,
\quad$W'_q=\{u:q<f'_q(u)<\infty,\,f'_q(u)\in H_u\}$,}

\noindent every $W'_q$ is negligible.   So
$Z=\BbbR^{r-1}\setminus\bigcup_{q\in\Bbb Q}(W_q\cup W'_q)$ is
$\mu_{r-1}$-conegligible.

Now if $u\in Z$, $t\in G_u$ and
$t'\in H_u$, where $t<t'$, there is some
$s\in\ooint{t,t'}\cap D_u$.   \Prf\Quer\
Suppose, if possible, otherwise.   Since, by hypothesis, neither $t$
nor $t'$ belongs to $D_u$, $D_u\cap[t,t']=\emptyset$.   Note that,
because $u\in Z$,

\Centerline{$s=\inf(G_u\cap\ooint{s,\infty})$ for every $s\in G_u$,
\quad$s=\sup(H_u\cap\ooint{-\infty,s})$ for every $s\in H_u$.}

\noindent Choose $\sequencen{s_n}$, $\sequencen{s'_n}$ and
$\sequencen{\delta_n}$ inductively, as follows.   Set $s_0=t$,
$s'_0=t'$.   Given that $t\le s_n<s'_n\le t'$ and $s_n\in G_u$ and
$s'_n\in H_u$, where $n$ is even,
set $s'_{n+1}=\sup(G_u\cap[s_n,s'_n])$.   Then either $s'_{n+1}=s'_n$,
so $s'_{n+1}\in H_u$, or $s'_{n+1}<s'_n$ and
$\ocint{s'_{n+1},s'_n}\cap G_u=\emptyset$, so $s'_{n+1}\notin G_u$ and
again $s'_{n+1}\in H_u$.   Let $\delta_n>0$ be such that
$\delta_n\le 2^{-n}$ and
\penalty-1000
$\mu(E\cap B((u,s'_{n+1}),\delta_n))<\bover12\beta_r\delta_n^r$.
Because the function $s\mapsto\mu(E\cap B((u,s),\delta_n))$ is
continuous (443C, or otherwise), there is an
$s_{n+1}\in G_u\cap\coint{s_n,s'_{n+1}}$ such that
$\mu(E\cap B((u,s),\delta_n))\le\bover12\beta_r\delta_n^r$ whenever
$s\in[s_{n+1},s'_{n+1}]$.

This is the inductive step from even $n$.   If $n$ is odd and
$s_n\in G_u$, $s'_n\in H_u$ and $s_n<s'_n$, set
$s_{n+1}=\inf(H_u\cap[s_n,s'_n])$.   This time we find that $s_{n+1}$
must belong to $G_u$.   Let $\delta_n\in\ocint{0,2^{-n}}$ be such that
$\mu(E\cap B((u,s_{n+1}),\delta_n))>\bover12\beta_r\delta_n^r$, and let
$s'_{n+1}\in H_u\cap\ocint{s_{n+1},s'_n}$ be such that
$\mu(E\cap B((u,s),\delta_n))\ge\bover12\beta_r\delta_n^r$ whenever
$s\in[s_{n+1},s'_{n+1}]$.

The construction provides us with a non-increasing sequence
$\sequencen{[s_n,s'_n]}$ of closed intervals, so there must be some $s$
in their intersection.   In this case

\Centerline{$\mu(E\cap B((u,s),\delta_n)\le\bover12\beta_r\delta_n^r$ if
$n$ is even,}

\Centerline{$\mu(E\cap B((u,s),\delta_n)\ge\bover12\beta_r\delta_n^r$ if
$n$ is odd.}

\noindent Since $\lim_{n\to\infty}\delta_n=0$,

\Centerline{$\liminf_{\delta\downarrow 0}
  \Bover{\mu(E\cap B((u,s),\delta))}{\mu B((u,s),\delta)}\le\Bover12$,
\quad$\limsup_{\delta\downarrow 0}
  \Bover{\mu(E\cap B((u,s),\delta))}{\mu B((u,s),\delta)}\ge\Bover12$,}

\noindent and $s\in D_u$, while of course $t\le s\le t'$.\ \Bang\Qed

\medskip

\quad{\bf (iv)} Thus the conegligible set $Z$ has the property required
by (a).

\medskip

{\bf (b)} Applying
(a) to $\BbbR^r\setminus E$, there is a conegligible set
$Z'\subseteq\Bbb R^{r-1}$ such that if $u\in Z'$, $t\in H_u$,
$t'\in G_u$ and $t<t'$, then $D_u$ meets $\ooint{t,t'}$.
So we can use $Z\cap Z'$.

\medskip

{\bf (c)} Now suppose that $E$ has locally finite perimeter.
We know that
$\nu(\partstar E\cap B(\tbf{0},n))=\lambda^{\partial}_EB(\tbf{0,}n)$ is
finite for every $n\in\Bbb N$ (475G).   By 475H,

\Centerline{$\int_{\|u\|\le n}\#(D_u\cap[-n,n])\mu_{r-1}(du)
\le\nu(\partstar E\cap B(\tbf{0},2n))<\infty$}

\noindent for every $n\in\Bbb N$;  but this means that $D_u\cap[-n,n]$
must be finite for almost every $u$ such that $\|u\|\le n$, for every
$n$, and therefore that

\Centerline{$Z_1'=\{u:u\in Z_1,\,
  D_u\cap[-n,n]$ is finite for every $n\}$}

\noindent is conegligible.   For $u\in Z'_1$, $\Bbb R\setminus D_u$ is an
open set, so is made up of a disjoint sequence of intervals with
endpoints in $D_u\cup\{-\infty,\infty\}$ (see 2A2I);  and because
$u\in Z_1$, each of these intervals is included in either $G_u$ or
$H_u$.   Now

\Centerline{$A=\{u:$ there are successive constituent intervals of
$\Bbb R\setminus D_u$ both included in $G_u\}$}

\noindent is negligible.   \Prf\Quer\ Otherwise, there are rationals
$q<q'$ such that

\Centerline{$F
=\{u:\ooint{q,q'}\setminus G_u$ contains exactly one point$\}$}

\noindent is not negligible.   Note that, writing
$T:\BbbR^r\to\BbbR^{r-1}$ for the orthogonal projection,

$$\eqalign{F&=T[(\BbbR^{r-1}\times\ooint{q,q'})\setminus\intstar E]\cr
&\qquad\setminus\bigcup_{q''\in\Bbb Q,q<q''<q'}
  T[(\BbbR^{r-1}\times\ooint{q,q''})\setminus\intstar E]
  \cap T[(\BbbR^{r-1}\times\ooint{q'',q'})\setminus\intstar E]\cr}$$

\noindent is measurable.   Take any $u\in F\cap\intstar F$, and let $t$
be the unique point in $\ooint{q,q'}\setminus G_u$.   Then whenever
$0<\delta\le\min(t-q,q'-t)$ we shall have

\Centerline{$\mu(B((u,t),\delta)\setminus E)
=\mu(B((u,t),\delta)\setminus\intstar E)
\le 2\delta\mu_{r-1}(V(u,\delta)\setminus F)$,}

\noindent because if $w\in F$ then $(w,s)\in\intstar E$ for almost
every $s\in[t-\delta,t+\delta]$.   So

\Centerline{$\limsup_{\delta\downarrow 0}
  \Bover{\mu(B((u,t),\delta)\setminus E)}{\mu B((u,t),\delta)}
\le\Bover{2\beta_{r-1}}{\beta_r}\limsup_{\delta\downarrow 0}
   \Bover{\mu_{r-1}(V(u,\delta)\setminus F)}{\mu_{r-1}V(u,\delta)}
=0$,}

\noindent and $(u,t)\in\intstar E$.\ \Bang\Qed

Similarly,

\Centerline{$A'=\{u:$ there are successive constituent intervals of
$\Bbb R\setminus D_u$ both included in $H_u\}$}

\noindent is negligible.   So $Z_2=Z'_1\setminus(A\cup A')$
is a conegligible set of the kind we need.
}%end of proof of 475J

\leader{475K}{Lemma} Suppose that $h:\BbbR^r\to[-1,1]$ is a Lipschitz
function with compact support;  let $n\in\Bbb N$ be such that $h(x)=0$
for $\|x\|\ge n$.  Suppose that $E\subseteq\BbbR^r$ is a Lebesgue
measurable set.   Then

\Centerline{$|\int_E\Pd{h}{\xi_r}d\mu|
\le 2\bigl(\beta_{r-1}n^{r-1}
  +\nu(\partstar E\cap B(\tbf{0},n))\bigr)$.}

\proof{ By Rademacher's theorem (262Q), $\Pd{h}{\xi_r}$ is defined
almost everywhere;  as it is
measurable and bounded, and is zero outside $B(\tbf{0},n)$, the
integral is well-defined.   If $\nu(\partstar E\cap B(\tbf{0},n))$ is
infinite, the result is trivial;  so henceforth let us suppose that
$\nu(\partstar E\cap B(\tbf{0},n))<\infty$.   Identify $\BbbR^r$ with
$\BbbR^{r-1}\times\Bbb R$.   For $u\in\BbbR^{r-1}$, set

$$\eqalign{f(u)&=\#(\{t:(u,t)\in\partstar E\cap B(\tbf{0},n)\})
  \text{ if this is finite},\cr
&=\infty\text{ otherwise}.\cr}$$

\noindent By 475H,
$\int fd\mu_{r-1}\le\nu(\partstar E\cap B(\tbf{0},n))$.   By 475Jb,
there is a
$\mu_{r-1}$-conegligible set $Z\subseteq\BbbR^{r-1}$ such that
$f(u)$ is finite for every $u\in Z$ and

\inset{\noindent whenever $u\in Z$ and $(u,t)\in\intstar E$ and
$(u,t')\in\BbbR^r\setminus\clstar E$, there is an $s$ lying between $t$
and $t'$ such that $(u,s)\in\partstar E$.}

\noindent For $u\in Z$, set
$D'_u=\{t:(u,t)\in B(\tbf{0},n)\setminus\partstar E\}$,
and define $g_u:D'_u\to\{0,1\}$ by setting

$$\eqalign{g_u(t)
&=1\text{ if }(u,t)\in B(\tbf{0},n)\cap\intstar E,\cr
&=0\text{ if }(u,t)\in B(\tbf{0},n)\setminus\clstar E.\cr}$$

\noindent Now if $t$, $t'\in D'_u$ and $g(t)\ne g(t')$, there is a point
$s$ between $t$ and $t'$ such that $(u,s)\in\partstar E$;  so the
variation $\Var_{D'_u}g_u$ of $g_u$ (224A) is at most $f(u)$.   Setting
$h_u(t)=h(u,t)$ for $u\in\BbbR^{r-1}$ and $t\in\Bbb R$, we now have

$$\eqalignno{\int_{-\infty}^{\infty}\Pd{h}{\xi_r}(u,t)
  \chi(B(\tbf{0},n)\cap\intstar E)(u,t)dt
&=\int_{D'_u}h'_u(t)g_u(t)dt\cr
\displaycause{because $h'_u(t)=0$ if $(u,t)\notin B(\tbf{0},n)$}
&\le(1+\Var_{D'_u}g_u)\sup_{a\le b}|\int_a^bh'_u(t)|dt\cr
\displaycause{by 224J, recalling that $D'_u$ is either empty or a
bounded interval with finitely many points deleted}
&\le (1+f(u))\sup_{a\le b}|h_u(b)-h_u(a)|\cr
\displaycause{because $h_u$ is Lipschitz, therefore absolutely
continuous on any bounded interval}
&\le 2(1+f(u)).\cr}$$

\noindent Integrating over $u$, we now have

$$\eqalignno{|\int_E\Pd{h}{\xi_r}d\mu|
&=|\int_{B(\tbf{0},n)\cap\intstar E}\Pd{h}{\xi_r}d\mu|\cr
\displaycause{475Cg}
&=\int_{V(\tbf{0},n)}\int_{-\infty}^{\infty}\Pd{h}{\xi_r}(u,t)
  \chi(B(\tbf{0},n)\cap\intstar E)(u,t)dt\,\mu_{r-1}(du)\cr
\displaycause{where $V(\tbf{0},n)=\{u:u\in\BbbR^{r-1},\,\|u\|\le n\}$}
&\le\int_{V(\tbf{0},n)}2(1+f(u))\mu_{r-1}(du)\cr
&\le 2\bigl(\beta_{r-1}n^{r-1}
  +\nu(\partstar E\cap B(\tbf{0},n))\bigr).\cr}$$
}%end of proof of 475K

\leader{475L}{Theorem} Suppose that $E\subseteq\BbbR^r$.   Then $E$ has
locally finite perimeter iff $\nu(\partstar E\cap B(\tbf{0},n))$ is
finite for every $n\in\Bbb N$.

\proof{ If $E$ has locally finite perimeter, then
$\nu(\partstar E\cap B(\tbf{0},n))=\lambda^{\partial}_EB(\tbf{0},n)$ is
finite for every $n$, by 475G.   So let us suppose that
$\nu(\partstar E\cap B(\tbf{0},n))$ is finite for every
$n\in\Bbb N$.   Then $\mu(\partstar E\cap B(\tbf{0},n))=0$ for every
$n$ (471L), $\mu(\partstar E)=0$ and $E$ is Lebesgue measurable (475Ch).

If $n\in\Bbb N$, then

$$\eqalign{\sup\{|\int_E\diverg\phi\,d\mu|:
  \phi:\BbbR^r\to\BbbR^r&\text{ is Lipschitz},\,
  \|\phi\|\le\chi B(\tbf{0},n)\}\cr
&\le 2r\bigl(\beta_{r-1}n^{r-1}
   +\nu(\partstar E\cap B(\tbf{0},n))\bigr)\cr}$$

\noindent is finite.   \Prf\ If
$\phi=(\phi_1,\ldots,\phi_r):\BbbR^r\to\BbbR^r$ is a Lipschitz
function and $\|\phi\|\le\chi B(\tbf{0},n)$, then 475K tells us that

\Centerline{$|\int_E\Pd{\phi_r}{\xi_r}d\mu|
\le 2\bigl(\beta_{r-1}n^{r-1}
  +\nu(\partstar E\cap B(\tbf{0},n))\bigr)$.}

\noindent But of course it is equally true that

\Centerline{$|\int_E\Pd{\phi_j}{\xi_j}d\mu|
\le 2\bigl(\beta_{r-1}n^{r-1}
  +\nu(\partstar E\cap B(\tbf{0},n))\bigr)$}

\noindent for every other $j\le r$;  adding, we have the result.\ \Qed

Since $n$ is arbitrary, $E$ has locally finite perimeter.
}%end of proof of 475L

\leader{475M}{Corollary} (a) The family of sets with locally finite
perimeter is a subalgebra of the algebra of Lebesgue measurable subsets
of $\BbbR^r$.

(b) A set $E\subseteq\BbbR^r$ is Lebesgue measurable and
has finite perimeter iff
$\nu(\partstar E)<\infty$, and in this case $\nu(\partstar E)$ is the
perimeter of $E$.

(c) If $E\subseteq\BbbR^r$ has finite measure, then
$\per E=\liminf_{\alpha\to\infty}\per(E\cap B(\tbf{0},\alpha))$.

%?? shouldn't it be lim_{\alpha\to\infty} ? yes, see 475Xk

\proof{{\bf (a)} Recall that the definition in 474D insists that sets
with locally finite perimeter should be Lebesgue measurable.
If $E\subseteq\BbbR^r$ has locally finite perimeter, then so
has $\BbbR^r\setminus E$, by 474J.   If $E$, $F\subseteq\BbbR^r$ have
locally finite perimeter, then

\Centerline{$\nu(\partstar(E\cup F)\cap B(\tbf{0},n))
\le\nu(\partstar E\cap B(\tbf{0},n))
  +\nu(\partstar F\cap B(\tbf{0},n))$}

\noindent is finite for every $n\in\Bbb N$, by 475L and 475Cd;  by 475L
in the other direction, $E\cup F$ has locally finite perimeter.

\medskip

{\bf (b)} If $E$ is Lebesgue measurable and has
finite perimeter (on the definition in 474D), then
$\nu(\partstar E)=\lambda^{\partial}_E\BbbR^r$ is the perimeter of $E$,
by 475G.
If $\nu(\partstar E)<\infty$, then $\mu(\partstar E)=0$ and
$E$ is measurable (471L and 475Ch);  now
$E$ has locally finite perimeter, by 475L, and 475G again tells us that
$\nu(\partstar E)=\lambda^{\partial}_E\BbbR^r$ is the perimeter of $E$.

\medskip

{\bf (c)} Now suppose that $\mu E<\infty$.   For any $\alpha\ge 0$,

\Centerline{$\partstar(E\cap B(\tbf{0},\alpha))
\subseteq(\partstar E\cap B(\tbf{0},\alpha))
\cup(\clstar E\cap\partial B(\tbf{0},\alpha))
\subseteq\partstar E\cup(\clstar E\cap\partial B(\tbf{0},\alpha))$}

\noindent by 475Cf.    Now we know also that

\Centerline{$\int_0^{\infty}
   \nu(\clstar E\cap\partial B(\tbf{0},\alpha))d\alpha
=\mu(\clstar E)=\mu E<\infty$}

\noindent (265G), so
$\liminf_{\alpha\to\infty}\nu(\clstar E\cap\partial B(\tbf{0},\alpha))
=0$.   This means that

$$\eqalign{\liminf_{\alpha\to\infty}\per(E\cap B(\tbf{0},\alpha))
&=\liminf_{\alpha\to\infty}\nu(\partstar(E\cap B(\tbf{0},\alpha)))\cr
&\le\liminf_{\alpha\to\infty}\nu(\partstar E)
  +\nu(\clstar E\cap\partial B(\tbf{0},\alpha))
=\nu(\partstar E).\cr}$$

\noindent In the other direction,

\Centerline{$\partstar E\cap\interior B(\tbf{0},\alpha)
=\partstar E\cap\intstar B(\tbf{0},\alpha)
\subseteq\partstar(E\cap B(\tbf{0},\alpha))$}

\noindent for every $\alpha$, by 475Ce, so

$$\eqalign{\per E
&=\nu(\partstar E)
=\lim_{\alpha\to\infty}
  \nu(\partstar E\cap\interior B(\tbf{0},\alpha))\cr
&\le\liminf_{\alpha\to\infty}\nu(\partstar(E\cap B(\tbf{0},\alpha)))
=\liminf_{\alpha\to\infty}\per(E\cap B(\tbf{0},\alpha))\cr}$$

\noindent and we must have equality.
}%end of proof of 475M

\cmmnt{\medskip

\noindent{\bf Remark} See 475Xk.
}

\leader{475N}{The Divergence Theorem}
%de Giorgi-Federer
Let $E\subseteq\BbbR^r$ be such
that $\nu(\partstar E\cap B(\tbf{0},n))$ is finite for every
$n\in\Bbb N$.

(a) $E$ is Lebesgue measurable.

(b) For $\nu$-almost every $x\in\partstar E$, there is a Federer
exterior normal $v_x$ of $E$ at $x$.

(c) For every Lipschitz function $\phi:\BbbR^r\to\BbbR^r$ with compact
support,

\Centerline{$\int_E\diverg\phi\,d\mu
=\int_{\partstar E}\varinnerprod{\phi(x)}{v_x}\,\nu(dx)$.}

\proof{ By 475L, $E$ has locally finite perimeter, and in particular is
Lebesgue measurable.   By 474R, there is a
Federer exterior normal $v_x=\psi_E(x)$ of $E$ at $x$ for every
$x\in\partial^{\$}E$;  by 475D, $\nu$-almost every point in
$\partstar E$ is of this kind.   By 474E-474F,

\Centerline{$\int_E\diverg\phi\,d\mu
=\int\varinnerprod{\phi}{\psi_E}\,d\lambda^{\partial}_E
=\int_{\partial^{\$}E}
   \varinnerprod{\phi}{\psi_E}\,d\lambda^{\partial}_E$,}

\noindent and this is also equal to
$\int_{\partstar E}
   \varinnerprod{\phi}{\psi_E}\,d\lambda^{\partial}_E$, because
$\partial^{\$}E\subseteq\partstar E$ and $\partial^{\$}E$ is
$\lambda^{\partial}_E$-conegligible.   But $\lambda^{\partial}_E$ and
$\nu$ induce the same subspace measures on $\partstar E$, by 475G, so

\Centerline{$\int_E\diverg\phi\,d\mu
=\int_{\partstar E}\varinnerprod{\phi}{\psi_E}\,d\nu
=\int_{\partstar E}\varinnerprod{\phi(x)}{v_x}\,\nu(dx)$,}

\noindent as claimed.
}%end of proof of 475N

\leader{475O}{}\cmmnt{ At the price of some technicalities which are
themselves instructive, we can now proceed to some basic properties of
the essential boundary.

\medskip

\noindent}{\bf Lemma} Let $E\subseteq\BbbR^r$ be a set with locally
finite perimeter, and $\psi_E$ its canonical outward-normal function.
Let $v$ be the unit vector $(0,\ldots,0,1)$.   Identify $\BbbR^r$ with
$\BbbR^{r-1}\times\Bbb R$.   Then we have sequences $\sequencen{F_n}$,
$\sequencen{g_n}$ and $\sequencen{g'_n}$ such that

(i) for each $n\in\Bbb N$, $F_n$ is a Lebesgue measurable subset of
$\BbbR^{r-1}$, and $g_n$, $g'_n:F_n\to[-\infty,\infty]$ are Lebesgue
measurable functions such that $g_n(u)<g'_n(u)$ for every $u\in F_n$;

(ii) if $m$, $n\in\Bbb N$ are distinct and $u\in F_m\cap F_n$, then
$[g_m(u),g'_m(u)]\cap[g_n(u),g'_n(u)]=\emptyset$;

(iii) $\sum_{n=0}^{\infty}\int_{F_n}g'_n-g_nd\mu_{r-1}=\mu E$;

(iv) for any continuous function $h:\BbbR^r\to\Bbb R$ with compact
support,

\Centerline{$\int_{\partstar E}
  h(x)\varinnerprod{v}{\psi_E(x)}\,\nu(dx)
=$$\sum_{n=0}^{\infty}$$\int_{F_n}h(u,g'_n(u))-h(u,g_n(u))
  \mu_{r-1}(du)$,}

\noindent where we interpret $h(u,\infty)$ and $h(u,-\infty)$ as $0$ if
necessary;

(v) for $\mu_{r-1}$-almost every $u\in\BbbR^{r-1}$,

$$\eqalign{\{t:(u,t)\in\partstar E\}
&=\{g_n(u):n\in\Bbb N,\,u\in F_n,\,g_n(u)\ne-\infty\}\cr
&\hskip10em
  \cup\{g'_n(u):n\in\Bbb N,\,u\in F_n,\,g'_n(u)\ne\infty\}.\cr}$$

\proof{{\bf (a)} Take a conegligible set $Z_2\subseteq\BbbR^{r-1}$ as in
475Jc.   Let $Z\subseteq Z_2$ be a conegligible Borel set.
For $u\in Z$ set $D_u=\{t:(u,t)\in\partstar E\}$.

\medskip

{\bf (b)} For each $q\in\Bbb Q$, set
$F'_q=\{u:u\in Z,\,(u,q)\in\intstar E\}$, so that $F'_q$ is a Borel set,
and for $u\in F'_q$ set

\Centerline{$f_q(u)=\sup(D_u\cap\ooint{-\infty,q})$,
\quad$f'_q(u)=\inf(D_u\cap\ooint{-\infty,q})$,}

\noindent allowing $-\infty$ as $\sup\emptyset$ and $\infty$ as
$\inf\emptyset$.   Observe that (because $D_u\cap[q-1,q+1]$ is finite)
$f_q(u)<q<f'_q(u)$.   Now $f_q$ and $f'_q$ are measurable, by the
argument of part (a-i) of the proof of 475J.   Enumerate $\Bbb Q$ as
$\sequencen{q_n}$, and set

\Centerline{$F_n
=F'_{q_n}\setminus\bigcup_{m<n}\{u:u\in F'_{q_m},
  \,f_{q_m}(u)<q_n<f'_{q_m}(u)\}$}

\noindent for $n\in\Bbb N$.   Set $g_n(u)=f_{q_n}(u)$,
$g'_n(u)=f'_{q_n}(u)$ for $u\in F_n$.

The effect of this construction is to ensure that, for any $u\in Z$,
$u\in F_n$ iff $q_n$ is the first rational lying in one of those
constituent intervals $I$ of $\Bbb R\setminus D_u$ such that
$\{u\}\times I\subseteq\intstar E$, and that now $g_n(u)$ and $g'_n(u)$
are the endpoints of that interval, allowing $\pm\infty$ as endpoints.

\medskip

{\bf (c)} Now let us look at the items (i)-(v) of the statement of this
lemma.    We have already achieved (i).   If $u\in F_m\cap F_n$, then,
in the language of 475J,
$\ooint{g_m(u),g'_m(u)}$ is one of the constituent intervals of $G_u$,
and $\ooint{g_n(u),g'_n(u)}$ is another;  since these must be separated by
one of the constituent intervals of $H_u$, their closures are disjoint.
Thus (ii) is true.   For any $u\in Z$,

\Centerline{$\sum_{n\in\Bbb N,u\in F_n}g'_n(u)-g_n(u)
=\mu_1\{t:(u,t)\in\intstar E\}$,}

\noindent so

\Centerline{$\sum_{n=0}^{\infty}$$\int_{F_n}g'_n-g_nd\mu_{r-1}
=\int_Z\mu_1\{t:(u,t)\in\intstar E\}\mu_{r-1}(du)
=\mu(\intstar E)
=\mu E$.}

\noindent So (iii) is true.   Also, for $u\in Z$,

$$\eqalign{\{t:(u,t)\in\partstar E\}
&=D_u\cr
&=\{g_n(u):n\in\Bbb N,\,u\in F_n,\,g_n(u)\ne-\infty\}\cr
&\hskip10em
  \cup\{g'_n(u):n\in\Bbb N,\,u\in F_n,\,g'_n(u)\ne\infty\},\cr}$$

\noindent so (v) is true.

\medskip

{\bf (d)} As for (iv), suppose first that $h:\BbbR^r\to\Bbb R$ is a
Lipschitz function with compact support.   Set $\phi(x)=(0,\ldots,h(x))$
for $x\in\BbbR^r$.   Then

$$\eqalignno{\int_{\partstar E}
  h(x)\varinnerprod{v}{\psi_E(x)}\,\nu(dx)
&=\int h(x)\varinnerprod{v}{\psi_E(x)}\,\lambda^{\partial}_E(dx)\cr
\displaycause{475G}
&=\int\varinnerprod{\phi}{\psi_E}\,d\lambda^{\partial}_E
=\int_E\diverg\phi\,d\mu\cr
\displaycause{474E}
&=\int_E\Pd{h}{\xi_r}\,d\mu
=\int_{\intstar E}\Pd{h}{\xi_r}\,d\mu\cr
&=\int\sum_{n\in\Bbb N,u\in F_n}
  \int_{g_n(u)}^{g'_n(u)}\Pd{h}{\xi_r}(u,t)dt\,\mu_{r-1}(du)\cr
&=\int\sum_{n\in\Bbb N,u\in F_n}
  h(u,g'_n(u))-h(u,g_n(u))\mu_{r-1}(du)\cr
\displaycause{with the convention that
$h(u,\pm\infty)=\lim_{t\to\pm\infty}h(u,t)=0$}
&=\sum_{n=0}^{\infty}\int_{F_n}
  h(u,g'_n(u))-h(u,g_n(u))\mu_{r-1}(du)\cr}$$

\noindent because if $|h|\le m\chi B(\tbf{0},m)$ then

$$\eqalign{\int\sum_{n\in\Bbb N,u\in F_n}
  |h(u,g'_n(u))&-h(u,g_n(u))|\mu_{r-1}(du)\cr
&\le 2m\int
   \#(\{t:(u,t)\in B(\tbf{0},m)\cap\partstar E\})\mu_{r-1}(du)\cr
&\le 2m\mskip2mu\nu(B(\tbf{0},m)\cap\partstar E)\cr}$$

\noindent (475H) is finite.

For a general continuous function $h$ of compact support, consider the
convolutions $h_k=h*\tilde h_k$ for large $k$, where $\tilde h_k$ is
defined in 473E.   If $|h|\le m\chi B(\tbf{0},m)$ then
$|h_k|\le m\chi B(\tbf{0},m+1)$ for every $k$, so that

\Centerline{$\sum_{n\in\Bbb N,u\in F_n}|h_k(g'_n(u))-h_k(g_n(u))|
\le 2m\#(\{t:(u,t)\in B(\tbf{0},m+1)\cap\partstar E\})$}

\noindent for every $u\in\Bbb R^r$, $k\in\Bbb N$.   Since $h_k\to h$
uniformly (473Ed),

$$\eqalign{\int_{\partstar E}
  h(x)\varinnerprod{v}{\psi_E(x)}\,\nu(dx)
&=\lim_{k\to\infty}\int_{\partstar E}
  h_k(x)\varinnerprod{v}{\psi_E(x)}\,\nu(dx)\cr
&=\lim_{k\to\infty}\int\sum_{n\in\Bbb N,u\in F_n}
  h_k(u,g'_n(u))-h_k(u,g_n(u))\mu_{r-1}(du)\cr
&=\int\sum_{n\in\Bbb N,u\in F_n}h(u,g'_n(u))-h(u,g_n(u))
  \mu_{r-1}(du)\cr
&=\sum_{n=0}^{\infty}\int_{F_n}h(u,g'_n(u))-h(u,g_n(u))
  \mu_{r-1}(du),\cr}$$

\noindent as required.
}%end of proof of 475O

\leader{475P}{Lemma} Let $v\in S_{r-1}$ be any unit vector, and
$V\subseteq\BbbR^r$ the hyperplane $\{x:\varinnerprod{x}{v}=0\}$.   Let
$T:\BbbR^r\to V$ be the orthogonal projection.   If $E\subseteq\BbbR^r$
is any set with locally finite perimeter and canonical outward-normal
function $\psi_E$, then

\Centerline{$\int_{\partstar E}|\varinnerprod{v}{\psi_E}|d\nu
=\int_V\#(\partstar E\cap T^{-1}[\{u\}])\nu(du)$,}

\noindent interpreting $\#(\partstar E\cap T^{-1}[\{u\}])$ as $\infty$
if $\partstar E\cap T^{-1}[\{u\}]$ is infinite.

\proof{ As usual, we may suppose that the structure $(E,v)$ is rotated
until $v=(0,\ldots,1)$, so that we can identify $T(\xi_1,\ldots,\xi_r)$
with $(\xi_1,\ldots,\xi_{r-1})\in\BbbR^{r-1}$, and
$\int_V\#(\partstar E\cap T^{-1}[\{u\}])\nu(du)$ with
$\int_{\Bbb R^{r-1}}\#(D_u)\mu_{r-1}(du)$, where
$D_u=\{t:(u,t)\in\partstar E\}$.   For each $m\in\Bbb N$ and
$u\in\BbbR^{r-1}$, set
$D^{(m)}_u=\{t:(u,t)\in\partstar E,\,\|(u,t)\|<m\}$;  note that
$\int\#(D^{(m)}_u)\mu_{r-1}(du)\le\nu(\partstar E\cap B(\tbf{0},m))$ is
defined and finite (475H).   It follows that the integral
$\int_V\#(\partstar E\cap T^{-1}[\{u\}])\nu(du)$ is defined in
$[0,\infty]$.

\woddheader{475P}{4}{2}{2}{30pt}

{\bf (a)} Write $\Phi$ for the set of continuous functions
$h:\BbbR^r\to[-1,1]$ with compact support.   Then

\Centerline{$\int_{\partstar E}|\varinnerprod{v}{\psi_E}|d\nu
=\sup_{h\in\Phi}\int_{\partstar E}
   h(x)\varinnerprod{v}{\psi_E(x)}\,\nu(dx)$.}

\noindent\Prf\ Of course

\Centerline{$\int_{\partstar E}|\varinnerprod{v}{\psi_E}|d\nu
\ge\int_{\partstar E}
   h(x)\varinnerprod{v}{\psi_E(x)}\,\nu(dx)$}

\noindent for any $h\in\Phi$.   On the other hand, if

\Centerline{$\gamma
<\int_{\partstar E}|\varinnerprod{v}{\psi_E}|d\nu
=\int|\varinnerprod{v}{\psi_E}|d\lambda^{\partial}_E$}

\noindent (475G), then, because $\lambda^{\partial}_E$ is a Radon
measure, there is an $n\in\Bbb N$ such that

\Centerline{$\gamma<\int_{B(\tbf{0},n)}|\varinnerprod{v}{\psi_E}|
  d\lambda^{\partial}_E
=\int h_0(x)\varinnerprod{v}{\psi_E(x)}\,\lambda^{\partial}_E(dx)$}

\noindent where

$$\eqalign{h_0(x)
&=\bover{\varinnerprod{v}{\psi_E(x)}}{|\varinnerprod{v}{\psi_E(x)}|}
\text{ if }x\in B(\tbf{0},n)
  \text{ and }\varinnerprod{v}{\psi_E(x)}\ne 0,\cr
&=0\text{ otherwise}.\cr}$$

\noindent Again because
$\lambda^{\partial}_E$ is a Radon measure, there is a
continuous function $h_1:\BbbR^r\to\Bbb R$ with compact support such
that

\Centerline{$\int|h_1-h_0|d\lambda^{\partial}_E
\le\int_{B(\tbf{0},n)}|\varinnerprod{v}{\psi_E}|
  d\lambda^{\partial}_E-\gamma$}

\noindent (416I);  of course we may suppose that $h_1$, like $h$, takes
values in $[-1,1]$, so that $h_1\in\Phi$.   Now

\Centerline{$\int_{\partstar E}h_1(x)
  \varinnerprod{v}{\psi_E(x)}\nu(dx)
=\int h_1(x)\varinnerprod{v}{\psi_E(x)}\,\lambda^{\partial}_E(dx)
\ge\gamma$.}

\noindent As $\gamma$ is arbitrary,

\Centerline{$\sup_{h\in\Phi}
  \int h(x)\varinnerprod{v}{\psi_E(x)}\,\nu(dx)
=\int|\varinnerprod{v}{\psi_E}|d\nu$}

\noindent as required.\ \Qed

\medskip

{\bf (b)} Now take $\sequencen{F_n}$, $\sequencen{g_n}$ and
$\sequencen{g'_n}$ as in 475O.   Then

\Centerline{$\int\#(D_u)\mu_{r-1}(du)
=\sup_{h\in\Phi}$$\sum_{n=0}^{\infty}$$\int_{F_n}
  h(u,g'_n(u))-h(u,g_n(u))\mu_{r-1}(du)$.}

\noindent (As in 475O(iv), interpret $h(u,\pm\infty)$ as $0$ if
necessary.)   \Prf\ By 475O(ii), $g'_m(u)\ne g_n(u)$ whenever
$m$, $n\in\Bbb N$ and $u\in F_m\cap F_n$,
while for almost all $u\in\BbbR^{r-1}$

\Centerline{$D_u
=(\{g_n(u):n\in\Bbb N,\,u\in F_n\}\setminus\{-\infty\})
  \cup(\{g'_n(u):n\in\Bbb N,\,u\in F_n\}\setminus\{\infty\})$.}

\noindent So if $h\in\Phi$,

$$\eqalign{\sum_{n\in\Bbb N,u\in F_n}h(u,g'_n(u))-h(u,g_n(u))
&\le\#(\{g_n(u):u\in F_n,\,g_n(u)\ne-\infty\})\cr
&\hskip5em +\#(\{g'_n(u):u\in F_n,\,g'_n(u)\ne\infty\})\cr
&=\#(D_u)\cr}$$

\noindent for almost every $u\in\BbbR^{r-1}$, and

\Centerline{$\sum_{n=0}^{\infty}$$\int_{F_n}
   h(u,g'_n(u))-h(u,g_n(u))\mu_{r-1}(du)
\le\int\#(D_u)\mu_{r-1}(du)$.}

\allowmorestretch{468}{
In the other direction, given $\gamma<\int\#(D_u)\mu_{r-1}(du)$,
there is an $m\in\Bbb N$ such that
$\gamma<\int\#(D^{(m)}_u)\mu_{r-1}(du)$.   Setting
}

\Centerline{$H_n=\{u:u\in F_n,\,g_n(u)\in D^{(m)}_u\}$,
\quad$H'_n=\{u:u\in F_n,\,g'_n(u)\in D^{(m)}_u\}$}

\noindent for $n\in\Bbb N$, we have

\Centerline{$\gamma
<\sum_{n=0}^{\infty}\mu_{r-1}H_n+\sum_{n=0}^{\infty}\mu_{r-1}H'_n
\le\int\#(D^{(m)}_u\mu_{r-1}(du)
<\infty$.}

\noindent By 418J once more, we can find $n\in\Bbb N$ and
compact sets
$K_i\subseteq H_i$, $K'_i\subseteq H'_i$ such that $g_i\restr K_i$ and
$g'_i\restr K'_i$ are continuous for every $i$ and

$$\eqalign{\gamma
&\le\sum_{i=0}^n(\mu_{r-1}K_i+\mu_{r-1}K'_i)
   -\sum_{i=n+1}^{\infty}(\mu_{r-1}H_i+\mu_{r-1}H'_i)\cr
&\qquad\qquad-\sum_{i=0}^n(\mu_{r-1}(H_i\setminus K_i)
       +\mu_{r-1}(H'_i\setminus K'_i)).\cr}$$

\noindent Set

\Centerline{$K=\bigcup_{i\le n}\{(u,g_i(u)):u\in K_i\}$,
\quad$K'=\bigcup_{i\le n}\{(u,g'_i(u)):u\in K'_i\}$,}

\noindent so that $K$ and $K'$ are disjoint compact subsets of
$\interior B(\tbf{0},m)$.   Let $h:\BbbR^r\to\Bbb R$ be a continuous
function such that

$$\eqalign{h(x)
&=1\text{ for }x\in K',\cr
&=-1\text{ for }x\in K,\cr
&=0\text{ for }x\in\BbbR^r\setminus\interior B(\tbf{0},m)\cr}$$

\noindent (4A2F(d-ix));  we can suppose that $-1\le h(x)\le 1$ for
every $x$.   Then $h\in\Phi$, and

$$\eqalignno{\sum_{i=0}^{\infty}\int_{F_i}h(u,g'_i(u))&-h(u,g_i(u))
  \mu_{r-1}(du)\cr
&=\sum_{i=0}^{\infty}\int_{H'_i}h(u,g'_i(u))\mu_{r-1}(du)
  -\sum_{i=0}^{\infty}\int_{H_i}h(u,g_i(u))\mu_{r-1}(du)\cr
\displaycause{because $h$ is zero outside $\interior B(\tbf{0},m)$}
&\ge\sum_{i=0}^n\int_{K'_i}h(u,g'_i(u))\mu_{r-1}(du)
  -\sum_{i=0}^n\int_{K_i}h(u,g_i(u))\mu_{r-1}(du)\cr
&\qquad\qquad-\sum_{i=0}^n(\mu(H'_i\setminus K'_i)+\mu(H_i\setminus
K_i))
  -\sum_{i=n+1}^{\infty}(\mu H'_i+\mu H_i)\cr
&\ge\gamma.\cr}$$

\noindent This shows that

\Centerline{$\int\#(D_u)\mu_{r-1}(du)
\le\sup_{h\in\Phi}$$\sum_{n=0}^{\infty}$$\int_{F_n}
   h(u,g'_n(u))-h(u,g_n(u))\mu_{r-1}(du)$,}

\noindent and we have equality.\ \Qed

\medskip

{\bf (c)} Putting (a) and (b) together with 475O(iv), we have the
result.
}%end of proof of 475P

\leader{475Q}{Theorem}\discrversionA{\footnote{New in 2009:
the final formula in (a)
and part (b).}}{} (a) Let $E\subseteq\BbbR^r$ be a set with finite
perimeter.   For $v\in S_{r-1}$ write $V_v$ for
$\{x:\varinnerprod{x}{v}=0\}$, and let $T_v:\BbbR^r\to V_v$ be the
orthogonal projection.   Then

$$\eqalignno{\per E
&=\nu(\partstar E)
=\Bover1{2\beta_{r-1}}\int_{S_{r-1}}\int_{V_v}
  \#(\partstar E\cap T_v^{-1}[\{u\}])\nu(du)\nu(dv)\cr
&=\lim_{\delta\downarrow 0}\Bover1{2\beta_{r-1}\delta}
  \int_{S_{r-1}}\mu(E\symmdiff(E+\delta v))\nu(dv).\cr}$$

(b) Suppose that $E\subseteq\BbbR^r$ is Lebesgue measurable.   Set

\Centerline{$\gamma
=\sup_{x\in\BbbR^r\setminus\{0\}}\Bover1{\|x\|}\mu(E\symmdiff(E+x))$.}

\noindent Then $\gamma\le\per E\le\Bover{r\beta_r\gamma}{2\beta_{r-1}}$.

\proof{{\bf (a)(i)} I start with an elementary fact:  there is a constant
$c$ such that
$\int|\varinnerprod{w}{v}|\lambda^{\partial}_{S_{r-1}}(dv)=c$ for every
$w\in S_{r-1}$;
this is because whenever $w$, $w'\in S_{r-1}$ there is an orthogonal
linear transformation taking $w$ to $w'$, and this transformation is an
automorphism of the structure
$(\BbbR^r,\nu,S_{r-1},\lambda^{\partial}_{S_{r-1}})$ (474H).   (In (iii)
below I will come to the calculation of $c$.)

\medskip

\quad{\bf (ii)} Now, writing $\psi_E$ for the canonical outward-normal
function of $E$, we have

$$\eqalignno{\int_{S_{r-1}}\int_{V_v}
  \#(\partstar E\cap T_v^{-1}[\{u\}])\nu(du)\nu(dv)
&=\int_{S_{r-1}}\int_{\partstar E}
   |\varinnerprod{\psi_E(x)}{v}|\nu(dx)\nu(dv)\cr
\displaycause{475P}
&=\iint|\varinnerprod{\psi_E(x)}{v}|\lambda^{\partial}_E(dx)
  \lambda^{\partial}_{S_{r-1}}(dv)\cr
\displaycause{by 235K, recalling that $\lambda^{\partial}_{S_{r-1}}$ and
$\lambda^{\partial}_E$ are just
indefinite-integral measures over $\nu$, while $S_{r-1}$ and
$\partstar E$ are Borel sets}
&=\iint|\varinnerprod{\psi_E(x)}{v}|\lambda^{\partial}_{S_{r-1}}(dv)
  \lambda^{\partial}_E(dx)\cr
\displaycause{because $\varinnerprod{}{}$ is continuous, so
$(x,v)\mapsto\varinnerprod{\psi_E(x)}{v}$ is measurable, while
$\lambda^{\partial}_{S_{r-1}}$
and $\lambda^{\partial}_E$ are totally finite, so we can use Fubini's
theorem}
&=c\per E\cr
\displaycause{by (i) above}
&=c\nu(\partstar E).\cr}$$

\medskip

\quad{\bf (iii)}
We have still to identify the constant $c$.   But observe that
the argument above applies whenever $\nu(\partstar E)$ is finite, and in
particular applies to $E=B(\tbf{0},1)$.   In this case,
$\partstar B(\tbf{0},1)=S_{r-1}$, and for any $v\in S_{r-1}$, $u\in V_v$
we have

$$\eqalign{\#(\partstar B(\tbf{0},1)\cap T^{-1}[\{u\}])
&=2\text{ if }\|u\|<1,\cr
&=1\text{ if }\|u\|=1,\cr
&=0\text{ if }\|u\|>1.\cr}$$

\noindent Since we can identify $\nu$ on $V_v$ as a copy of Lebesgue
measure $\mu_{r-1}$,

\Centerline{$\int_{V_v}
  \#(\partstar B(\tbf{0},1)\cap T^{-1}[\{u\}])\nu(du)
=2\nu\{u:u\in V_v,\,\|u\|<1\}
=2\beta_{r-1}$.}

\noindent This is true for every $v\in V_u$, so from (ii) we get

\Centerline{$2\beta_{r-1}\nu S_{r-1}
=\int_{S_{r-1}}\int_{V_v}
  \#(\partstar E\cap T_v^{-1}[\{u\}])\nu(du)\nu(dv)
=c\nu S_{r-1}$,}

\noindent and $c=2\beta_{r-1}$.   Substituting this into the result of
(ii), we get

\Centerline{$\per E=\nu(\partstar E)
=\Bover1{2\beta_{r-1}}\int_{S_{r-1}}\int_{V_v}
  \#(\partstar E\cap T_v^{-1}[\{u\}])\nu(du)\nu(dv)$.}

\medskip

\quad{\bf (iv)} Before continuing with the main argument, it will help
to set out another elementary fact, this time about
translates of certain
simple subsets of $\Bbb R$.   Suppose that $(G,H,D)$ is a
partition of $\Bbb R$ such that $G$ and $H$ are open,
$D=\partial G=\partial H$ is the common boundary of $G$ and $H$,
and $D$ is locally finite, that is,
$D\cap[-n,n]$ is finite for every $n\in\Bbb N$.   Then

\Centerline{$\sup_{\delta>0}\Bover1{\delta}\mu_1(G\symmdiff(G+\delta))
=\lim_{\delta\downarrow 0}\Bover1{\delta}\mu_1(G\symmdiff(G+\delta))
=\#(D)$}

\noindent if you will allow me to identify `$\infty$' with the cardinal
$\omega$.   \Prf\ If we look at the components of $G$, these are intervals
with endpoints in $D$;  and because $\partial G=\partial H$, distinct
components of $G$ have disjoint closures.   Set
$f(\delta)=\mu_1(G\symmdiff(G+\delta))$ for $\delta>0$.   If
$D$ is infinite, then for any $n\in\Bbb N$ we have disjoint
bounded components
$I_0,\ldots,I_n$ of $G$;  for any $\delta$ small enough,
$G\cap(I_j+\delta)\subseteq I_j$ for every $j\le n$ (because $D$ is locally
finite);  so that

\Centerline{$f(\delta)\ge\sum_{j=0}^n\mu(I_j\symmdiff(I_j+\delta))
=2(n+1)\delta$,}

\noindent and
$\liminf_{\delta\downarrow 0}\Bover1{\delta}f(\delta)\ge 2(n+1)$.
As $n$ is arbitrary,

\Centerline{$\sup_{\delta>0}\Bover1{\delta}f(\delta)
=\lim_{\delta\downarrow 0}\Bover1{\delta}f(\delta)
=\infty$.}

If $D$ is empty, then $G$ is either empty or $\Bbb R$, and

\Centerline{$\sup_{\delta>0}\Bover1{\delta}f(\delta)
=\lim_{\delta\downarrow 0}\Bover1{\delta}f(\delta)
=0$.}

\noindent If $D$ is finite and not empty, let $I_0,\ldots,I_n$ be the
components of $G$.   Then, for all small $\delta>0$, we have

\Centerline{$f(\delta)
=\sum_{j=0}^n\mu(I_j\symmdiff(I_j+\delta))=2(n+1)\delta=\delta\#(D)$,}

\noindent so again

\Centerline{$\sup_{\delta>0}\Bover1{\delta}f(\delta)
=\lim_{\delta\downarrow 0}\Bover1{\delta}f(\delta)
=\#(D)$.  \Qed}

\medskip

\quad{\bf (v)} Returning to the proof in hand, we find that if
$v\in S_{r-1}$ is such that
$\int_{V_v}\#(\partstar E\cap T_v^{-1}[\{u\}])\nu(du)$ is finite, then
the integral is equal to

\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\delta}
  \mu(E\symmdiff(E+\delta v))
=\sup_{\delta>0}\Bover1{\delta}\mu(E\symmdiff(E+\delta v))$.}

\Prf\ It is enough to consider the case in which $v$ is the unit
vector $(0,\ldots,0,1)$, so that we can identify $V_v$ with $\BbbR^{r-1}$
and $\BbbR^r$ with $\BbbR^{r-1}\times\Bbb R$, as in 475J;  in this case,
$E\cap T_v^{-1}[\{u\}]$ turns into $E[\{u\}]$.   Let
$Z_2\subseteq\BbbR^{r-1}$ and $G_u$, $H_u$ and $D_u$, for $u\in\BbbR^{r-1}$,
be as in 475Jc.   In this case, for any $\delta>0$,

$$\eqalign{\mu(E\symmdiff(E+\delta v))
&=\mu(\intstar E\symmdiff(\intstar E+\delta v))\cr
&=\int_{\BbbR^{r-1}}
   \mu_1((\intstar E\symmdiff(\intstar E+\delta v))[\{u\}])
   \mu_{r-1}(du)\cr
&=\int_{\BbbR^{r-1}}\mu_1(G_u\symmdiff(G_u+\delta))\mu_{r-1}(du)\cr
&=\int_{Z_2}\mu_1(G_u\symmdiff(G_u+\delta))\mu_{r-1}(du)\cr}$$

\noindent because $Z_2$ is conegligible.   By (iv),

\Centerline{$\lim_{\delta\downarrow 0}
   \Bover1{\delta}\mu_1(G_u\symmdiff(G_u+\delta))
=\sup_{\delta>0}\Bover1{\delta}\mu_1(G_u\symmdiff(G_u+\delta))
=\#(D_u)$}

\noindent for any $u\in Z_2$.   Applying Lebesgue's Dominated
Convergence Theorem to arbitrary sequences
$\sequencen{\delta_n}\downarrow 0$, we see that

\Centerline{$\lim_{\delta\downarrow 0}
   \Bover1{\delta}\mu(E\symmdiff(E+\delta v))
=\int_{Z_2}\#(D_u)\nu(du)
=\int_{\BbbR^{r-1}}\#(\partstar E\cap T_v^{-1}[\{u\}])\nu(du)$,}

\noindent as required.   To see that

\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\delta}
  \mu(E\symmdiff(E+\delta v))
=\sup_{\delta>0}\Bover1{\delta}\mu(E\symmdiff(E+\delta v))$,}

\noindent set $g(\delta)=\mu(E\symmdiff(E+\delta v))$ for $\delta>0$.
Then for $\delta$, $\delta'>0$ we have

$$\eqalign{g(\delta+\delta')
&=\mu(E\symmdiff(E+(\delta+\delta')v))
\le\mu(E\symmdiff(E+\delta v))
   +\mu((E+\delta v)\symmdiff(E+(\delta+\delta'v)))\cr
&=g(\delta)
   +\mu(\delta v+(E\symmdiff(E+\delta'v)))
=g(\delta)+g(\delta').\cr}$$

\noindent Consequently $g(\delta)\le ng(\bover1n\delta)$ whenever
$\delta>0$ and $n\ge 1$, so

\Centerline{$\Bover1{\delta}g(\delta)
\le\liminf_{n\to\infty}\Bover{n}{\delta}g(\bover1n\delta)
=\lim_{\delta'\downarrow 0}\Bover1{\delta'}g(\delta')$}

\noindent for every $\delta>0$.\ \Qed

\medskip

\quad{\bf (vi)} Putting (v) together with (i)-(iii) above,

\Centerline{$\per E
=\Bover1{2\beta_{r-1}}\int_{S_{r-1}}
  \lim_{\delta\downarrow 0}\Bover1{\delta}
  \mu(E\symmdiff(E+\delta v))\nu(dv)$.}

\noindent To see that we can exchange the limit and the integral,
observe that we can again use the dominated convergence theorem, because

\Centerline{$\int_{S_{r-1}}\sup_{\delta>0}\Bover1{\delta}
  \mu(E\symmdiff(E+\delta v))=2\beta_{r-1}\per E$}

\noindent is finite.   So

\Centerline{$\per E
=\lim_{\delta\downarrow 0}\Bover1{2\beta_{r-1}\delta}
  \int_{S_{r-1}}\mu(E\symmdiff(E+\delta v))\nu(dv)$.}

\medskip

{\bf (b)(i)} $\gamma\le\per E$.   \Prf\
We can suppose that $E$ has finite perimeter.

\medskip

\qquad\grheada\ To begin with, suppose that $x=(0,\ldots,0,\delta)$ where
$\delta>0$.
As in part (a-v) of this proof, set $v=(0,\ldots,0,1)$, identify
$\BbbR^r$ with $\BbbR^{r-1}\times\Bbb R$, and define $G_u$,
$D_u\subseteq\Bbb R$, for $u\in\BbbR^{r-1}$, and
$Z_2\subseteq\BbbR^{r-1}$ as in 475Jc.   Suppose that $u\in Z_2$.
Then $G_u=(\intstar E)[\{u\}]$ is an open set and its
constituent intervals have endpoints in $D_u=(\partstar E)[\{u\}]$.
It follows that for any $t$ in

\Centerline{$G_u\symmdiff(G_u+\delta)
=(\intstar E\symmdiff(\intstar E+\delta v))[\{u\}]$,}

\noindent there must be an
$s\in D_u\cap[t-\delta,\delta]$, and $t\in D_u+[0,\delta]$.
Accordingly

\Centerline{$(\intstar E\symmdiff(\intstar E+\delta v))
     \cap(Z_2\times\Bbb R)
  \subseteq\partstar E+[0,\delta v]$,}

\noindent writing $[0,\delta v]$ for $\{tv:t\in[0,\delta]\}$.   So

\Centerline{$\mu(E\symmdiff(E+\delta v))
=\mu(\intstar E\symmdiff(\intstar E+\delta v))
\le\mu^*(\partstar E+[0,\delta v])$.}

Take any $\epsilon>0$.   We have a sequence $\sequencen{A_n}$ of sets,
all of
diameter at most $\epsilon$, covering $\partstar E$, and such that
$2^{-r+1}\beta_{r-1}\sum_{n=0}^{\infty}(\diam A_n)^{r-1}
\le\epsilon+\nu(\partstar E)$;
we can suppose that every $A_n$ is closed.
Taking $T:\BbbR^r\to\BbbR^{r-1}$ to be the natural projection,
$T[A_n]$ is compact and has
diameter at most $\diam A_n$, so that
$\mu_{r-1}T[A_n]\le 2^{-r+1}\beta_{r-1}(\diam A_n)^{r-1}$ (264H again).
For each $n\in\Bbb N$, the vertical sections of
$A_n+[0,\delta v]$ have diameter at most $\epsilon+\delta$.   So

\Centerline{$\mu(A_n+[0,\delta v])
\le 2^{-r+1}\beta_{r-1}(\diam A_n)^{r-1}(\epsilon+\delta)$.}

\noindent Consequently,

$$\eqalign{\mu(E\symmdiff(E+\delta v))
&\le\mu^*(\partstar E+[0,\delta v])
\le\sum_{n=0}^{\infty}\mu(A_n+[0,\delta v])\cr
&\le\sum_{n=0}^{\infty}
   2^{-r+1}\beta_{r-1}(\diam A_n)^{r-1}(\epsilon+\delta)
\le(\epsilon+\delta)(\epsilon+\nu(\partstar E)).\cr}$$

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\mu(E\symmdiff(E+x))
=\mu(E\symmdiff(E+\delta v))\le\delta\nu(\partstar E)
=\|x\|\nu(\partstar E)$.}

\medskip

\qquad\grheadb\ Of course the same must be true for all other non-zero
$x\in\BbbR^r$, so $\gamma\le\nu(\partstar E)=\per E$.\ \Qed

\medskip

\quad{\bf (ii)} For the other inequality, we need look only at the case
in which $\gamma$ is finite.

\medskip

\qquad\grheada\ In this case, $E$ has finite perimeter.   \Prf\
Let $\phi:\BbbR^r\to B(\tbf{0},1)$ be a Lipschitz function with compact
support.   Take $i$ such that $1\le i\le r$, and consider

$$\eqalignno{\bigl|\int_E\Pd{\phi_i}{\xi_i}d\mu\bigr|
&=\bigl|\int_E\lim_{n\to\infty}2^n(\phi_i(x+2^{-n}e_i)-\phi_i(x))\mu(dx)
   \bigr|\cr
\displaycause{where $\phi=(\phi_1,\ldots,\phi_r)$}
&=\bigl|\lim_{n\to\infty}\int_E2^n(\phi_i(x+2^{-n}e_i)-\phi_i(x))\mu(dx)
   \bigr|\cr
\displaycause{by Lebesgue's Dominated Convergence Theorem, because
$\phi_i$ is Lipschitz and has bounded support}
&=\lim_{n\to\infty}2^n\bigl|\int_E(\phi_i(x+2^{-n}e_i)-\phi_i(x))\mu(dx)
   \bigr|\cr
&=\lim_{n\to\infty}2^n\bigl|\int_{E+2^{-n}e_i}\phi_id\mu
  -\int_E\phi_id\mu\bigr|\cr
&\le\limsup_{n\to\infty}2^n\mu((E+2^{-n}e_i)\symmdiff E)\cr
\displaycause{because $\|\phi_i\|_{\infty}\le 1$}
&\le\gamma.\cr}$$

\noindent Summing over $i$,
$|\int_E\diverg\phi\,d\mu|\le r\gamma$.   As $\phi$ is arbitrary,
$\per E\le r\gamma$ is finite.\ \Qed

\medskip

\qquad\grheadb\ By (a), we have

$$\eqalign{\per E
&=\lim_{\delta\downarrow 0}\Bover1{2\beta_{r-1}\delta}\int_{S_{r-1}}
   \mu(E\symmdiff(E+\delta v))\nu(dv)\cr
&\le\Bover1{2\beta_{r-1}}\gamma\nu S_{r-1}
=\Bover{r\beta_r\gamma}{2\beta_{r-1}},\cr}$$

\noindent as required.
}%end of proof of 475Q

\leader{475R}{Convex sets in \dvrocolon{$\BbbR^r$}}\cmmnt{ For the
next result it will help to have some elementary facts about convex
sets in finite-dimensional spaces out in the open.

\medskip

\noindent}{\bf Lemma}\cmmnt{ (In this lemma I allow $r=1$.)} Let
$C\subseteq\BbbR^r$ be a convex set.

(a) If $x\in C$ and $y\in\interior C$, then $ty+(1-t)x\in\interior C$
for every $t\in\ocint{0,1}$.

(b) $\overline{C}$ and $\interior C$ are convex.

(c) If $\interior C\ne\emptyset$ then
$\overline{C}=\overline{\interior C}$.

(d) If $\interior C=\emptyset$ then $C$ lies within some hyperplane.

(e) $\interior\overline{C}=\interior C$.

\proof{{\bf (a)} Setting $\phi(z)=x+t(z-x)$ for $z\in\BbbR^r$,
$\phi:\BbbR^r\to\BbbR^r$ is a homeomorphism and $\phi[C]\subseteq C$, so

\Centerline{$\phi(y)\in\phi[\interior C]
=\interior\phi[C]\subseteq\interior C$.}

\medskip

{\bf (b)} It follows at once from (a) that $\interior C$ is convex;
$\overline{C}$ is convex because $(x,y)\mapsto tx+(1-t)y$ is continuous
for every $t\in[0,1]$.

\medskip

{\bf (c)} From (a) we see also that if $\interior C\ne\emptyset$ then
$C\subseteq\overline{\interior C}$, so that
$\overline{C}\subseteq\overline{\interior C}$ and
$\overline{C}=\overline{\interior C}$.

\medskip

{\bf (d)} It is enough to consider the case in which $\tbf{0}\in C$,
since if $C=\emptyset$ the result is trivial.   \Quer\ If
$x_1,\ldots,x_r$ are linearly independent elements of $C$, set
$x=\bover1{r+1}\sum_{i=1}^rx_i$;  then

\Centerline{$x+\sum_{i=1}^r\alpha_ix_i
=\sum_{i=1}^r(\alpha_i+\Bover1{r+1})x_i\in C$}

\noindent whenever $\sum_{i=1}^r|\alpha_i|\le\Bover1{r+1}$.   Also,
writing $e_1,\ldots,e_r$ for the standard orthonormal basis of
$\BbbR^r$, we can express $e_i$ as $\sum_{j=1}^r\alpha_{ij}x_j$ for each
$j$;  setting $M=(r+1)\max_{i\le r}\sum_{j=1}^r|\alpha_{ij}|$, we have
$x\pm\Bover1Me_i\in C$ for every $i\le r$, so that $x+y\in C$ whenever
$\|y\|\le\Bover1{M\sqrt r}$, and $x\in\interior C$.\ \Bang

So the linear subspace of $\BbbR^r$ spanned by $C$ has dimension at most
$r-1$.

\medskip

{\bf (e)} If $\interior C=\BbbR^r$ the result is trivial.   If
$\interior C$ is empty, then (d) shows that $C$ is included
in a hyperplane, so that $\interior\overline{C}$ is empty.   Otherwise,
if $x\in\BbbR^r\setminus\interior C$, there is a non-zero $e\in\BbbR^r$
such that $\varinnerprod{e}{y}\le\varinnerprod{e}{x}$ for every
$y\in\interior C$ (4A4Db, or otherwise).   Now, by (c),
$\varinnerprod{e}{y}\le\varinnerprod{e}{x}$ for every
$y\in\overline{C}$, so $x\notin\interior\overline{C}$.   This shows that
$\interior\overline{C}\subseteq\interior C$, so that the two are equal.
}%end of proof of 475R

\leader{475S}{Corollary:  Cauchy's Perimeter Theorem} Let
$C\subseteq\BbbR^r$ be a bounded convex set with non-empty interior.
For $v\in S_{r-1}$ write $V_v$ for $\{x:\varinnerprod{x}{v}=0\}$, and
let $T_v:\BbbR^r\to V_v$ be the orthogonal projection.   Then

\Centerline{$\nu(\partial C)=\Bover1{\beta_{r-1}}\int_{S_{r-1}}
  \nu(T_v[C])\nu(dv)$.}

\proof{{\bf (a)} The first thing to note is that
$\partstar C=\partial C$.   \Prf\ Of course
$\partstar C\subseteq\partial C$ (475Ca).   If
$x\in\partial C$, there is a half-space $V$ containing $x$ and disjoint
from $\interior C$ (4A4Db again, because $\interior C$ is convex), so that

\Centerline{$\limsup_{\delta\downarrow 0}
  \Bover{\mu(B(x,\delta)\setminus C)}{\mu B(x,\delta)}
\ge\limsup_{\delta\downarrow 0}
  \Bover{\mu(B(x,\delta)\setminus\interior V)}{\mu B(x,\delta)}
=\Bover12$,}

\noindent and $x\notin\intstar C$.   On the other hand, if
$x_0\in\interior C$ and $\eta>0$ are
such that $B(x_0,\eta)\subseteq\interior C$, then for any
$\delta\in\ocint{0,1}$ we
can set $t=\Bover{\delta}{\eta+\|x_0-x\|}$, and then

\Centerline{$B(x+t(x_0-x),t\eta)
\subseteq B(x,\delta)\cap((1-t)x+tB(x_0,\eta))
\subseteq B(x,\delta)\cap C$,}

\noindent so that

\Centerline{$\mu(B(x,\delta)\cap C)
\ge\beta_rt^r\eta^r
=\bigl(\Bover{\eta}{\|x_0-x\|+\eta}\bigr)^r\mu B(x,\delta)$;}

\noindent as $\delta$ is arbitrary, $x\in\clstar C$.   This shows that
$\partial C\subseteq\partstar C$ so that $\partstar C=\partial C$.\ \Qed

\medskip

{\bf (b)} We have a function
$\phi:\BbbR^r\to\overline{C}$ defined by taking $\phi(x)$ to be the
unique point
of $C$ closest to $x$, for every $x\in\BbbR^r$ (3A5Md).   This function
is $1$-Lipschitz.
\Prf\ Take any $x$, $y\in\BbbR^r$ and set $e=\phi(x)-\phi(y)$.   We know
that $\phi(x)-\epsilon e\in\overline{C}$, so that
$\|x-\phi(x)-\epsilon e\|\ge\|x+\phi(x)\|$,  for $0\le\epsilon\le 1$;
it follows that $\varinnerprod{(x-\phi(x))}{e}\ge 0$.   Similarly,
$\varinnerprod{(y-\phi(y))}{(-e)}\ge 0$.   Accordingly
$\varinnerprod{(x-y)}{e}\ge\varinnerprod{e}{e}$ and $\|x-y\|\ge\|e\|$.
As $x$ and $y$ are arbitrary, $\phi$ is $1$-Lipschitz.
\Qed

Now suppose that $C'\supseteq C$ is a closed bounded convex set.
Then $\nu(\partial C')\ge\nu(\partial C)$.   \Prf\ Let $\phi$ be the
function defined just above.   By 264G/471J again,
$\nu^*(\phi[\partial C'])\le\nu(\partial C')$.   But if
$x\in\partial C$, there is an $e\in\BbbR^r\setminus\{0\}$ such that
$\varinnerprod{x}{e}\ge\varinnerprod{y}{e}$ for every $y\in C$.   Then
$\phi(x+\alpha e)=x$ for every $\alpha\ge 0$.   Because $C'$ is closed
and bounded, and $x\in C\subseteq C'$, there is a greatest $\alpha\ge 0$
such that $x+\alpha e\in C'$, and in this case
$x+\alpha e\in\partial C'$;
since $\phi(x+\alpha e)=x$, $x\in\phi[\partial C']$.   As $x$ is
arbitrary, $\partial C\subseteq\phi[\partial C']$, and

\Centerline{$\nu(\partial C)
\le\nu^*(\phi[\partial C'])\le\nu(\partial C')$.   \Qed}

Since we can certainly find a closed convex set $C'\supseteq C$ such
that $\nu(\partial C')$ is finite (e.g., any sufficiently large ball or
cube), $\nu(\partial C)<\infty$.   It follows at once that
$\mu(\partial C)=0$ (471L once more).

\medskip

{\bf (c)} The argument so far applies, of course, to every $r\ge 1$ and
every bounded convex set with non-empty interior in $\BbbR^r$.   Moving
to the intended case $r\ge 2$, and fixing $v\in S_{r-1}$ for the moment,
we see that, because $T_v$ is an open map (if we give $V_v$ its subspace
topology), $T_v[C]$ is again a bounded convex set with non-empty
(relative) interior.   Since the subspace measure induced by $\nu$ on
$V_v$ is just a copy of Lebesgue measure, (b) tells us that
$\nu T_v[C]=\nu(\interior_{V_v}T_v[C])$, where here I write
$\interior_{V_v}\mskip-3muT_v[C]$ for the interior of $T_v[C]$ in the subspace topology
of $V_v$.   Now the point is that
$\interior_{V_v}\mskip-3muT_v[C]\subseteq T_v[\interior C]$.
\Prf\ $\interior C$ (taken in $\BbbR^r$) is
dense in $C$ (475Rc), so $W=T_v[\interior C]$ is a relatively
open convex set
which is dense in $T_v[C]$;  now $W=\interior_{V_v}\overline{W}$ (475Re,
applied in $V_v\cong\BbbR^{r-1}$),
so $W\supseteq\interior_{V_v}\mskip-3muT_v[C]$.\ \Qed

It follows that $\#(\partial C\cap T_v^{-1}[\{u\}])=2$ for every
$u\in\interior_{V_v}\mskip-3muT_v[C]$.   \Prf\ $T_v^{-1}[\{u\}]$ is a straight line
meeting $\interior C$ in $y_0$ say.   Because $\overline{C}$ is a
bounded convex set, it meets $T_v^{-1}[\{u\}]$ in a bounded convex set,
which must be a non-trivial closed line segment with endpoints $y_1$,
$y_2$ say.   Now certainly neither $y_1$ nor $y_2$ can be in the
interior of $C$.   Moreover, the open line segments between $y_1$ and
$y_0$, and between $y_2$ and $y_0$, are covered by $\interior C$, by
475Ra;  so
$T_v^{-1}[\{u\}]\cap\partial C=\{y_1,y_2\}$ has just two members.\ \Qed

\medskip

{\bf (d)} This is true for every $v\in S_{r-1}$.   But this means that
we can apply 475Q to see that

$$\eqalign{\nu(\partial C)
&=\nu(\partstar C)
=\Bover1{2\beta_{r-1}}\int_{S_{r-1}}\int_{V_v}
  \#(\partstar C\cap T_v^{-1}[\{u\}])\nu(du)\nu(dv)\cr
&=\Bover1{2\beta_{r-1}}\int_{S_{r-1}}\int_{T_v[C]}
  \#(\partstar C\cap T_v^{-1}[\{u\}])\nu(du)\nu(dv)\cr
&=\Bover1{2\beta_{r-1}}\int_{S_{r-1}}\int_{\interior_{V_v}\mskip-3muT_v[C]}
  \#(\partstar C\cap T_v^{-1}[\{u\}])\nu(du)\nu(dv)\cr
&=\Bover1{\beta_{r-1}}\int_{S_{r-1}}
   \nu(\interior_{V_v}\mskip-3muT_v[C])\nu(dv)
=\Bover1{\beta_{r-1}}\int_{S_{r-1}}\nu(T_v[C])\nu(dv),\cr}$$

\noindent as required.
}%end of proof of 475S

\leader{475T}{Corollary:  the Convex Isoperimetric Theorem} If
$C\subseteq\BbbR^r$ is a bounded convex set, then
$\nu(\partial C)\le r\beta_r(\bover12\diam C)^{r-1}$.

\proof{{\bf (a)} If $C$ is included in some $(r-1)$-dimensional affine
subspace, then

\Centerline{$\nu(\partial C)=\nu\overline{C}
\le\beta_{r-1}(\Bover12\diam C)^{r-1}$}

\noindent by 264H once more.   For completeness, I should check that
$\beta_{r-1}\le r\beta_r$.   \Prf\ Comparing 265F with 265H, or working
from the formulae in 252Q, we have $r\beta_r=2\pi\beta_{r-2}$.   On the
other hand, by the argument of 252Q,

\Centerline{$\beta_{r-1}
=\beta_{r-2}\int_{-\pi/2}^{\pi/2}\cos^{r-1}t\,dt
\le\pi\beta_{r-2}$,}

\noindent so (not coincidentally) we have a factor of two to
spare.\ \Qed

\medskip

{\bf (b)} Otherwise, $C$ has non-empty interior (475Rd), and for any
orthogonal projection $T$ of $\BbbR^r$ onto an $(r-1)$-dimensional
linear
subspace, $\diam T[C]\le\diam C$, so
$\nu(T[C])\le\beta_{r-1}(\Bover12\diam C)^{r-1}$.    Now 475S tells us
that

\Centerline{$\nu(\partial C)\le(\Bover12\diam C)^{r-1}\nu S_{r-1}
=r\beta_r(\Bover12\diam C)^{r-1}$.}
}%end of proof of 475T

\cmmnt{\medskip

\noindent{\bf Remark} Compare 476H below.
}

\exercises{\leader{475X}{Basic exercises (a)}
%\spheader 475Xa
Show that if $C\subseteq\BbbR^r$ is convex, then either
$\mu C=0$ and $\partstar C=\emptyset$, or $\partstar C=\partial C$.
%475B

\spheader 475Xb
Let $A$, $A'\subseteq\BbbR^r$ be any sets.   Show that

\Centerline{$(\partstar A\cap\intstar A')
   \cup(\partstar A'\cap\intstar A)
\subseteq\partstar(A\cap A')
\subseteq(\partstar A\cap\clstar A')\cup(\partstar A'\cap\clstar A)$.}
%475C

\spheader 475Xc Let $A\subseteq\BbbR^r$ be any set, and $B$ a
non-trivial closed ball.   Show that

\Centerline{$\partstar(A\cap B)
  \symmdiff((B\cap\partstar A)\cup(A\cap\partial B))
\subseteq A\cap\partial B\setminus\partstar A$.}
%475C

\sqheader 475Xd Let $E$, $F\subseteq\BbbR^r$ be measurable sets, and $v$
the Federer exterior normal to $E$ at $x\in\intstar F$.   Show that $v$ is
the Federer exterior normal to $E\cap F$ at $x$.
%475C

\spheader 475Xe Let $\frak T$ be the density topology on $\BbbR^r$
(414P) defined from lower Lebesgue density (341E).   Show that, for any
$A\subseteq\BbbR^r$, $A\cup\clstar A$ is the $\frak T$-closure of $A$
and $\intstar A$ is the $\frak T$-interior of the $\frak T$-closure of
$A$.
%475C

\spheader 475Xf Let $A\subseteq\BbbR^r$ be any set.   Show that
$A\setminus\clstar A$ and $\intstar A\setminus A$ are Lebesgue
negligible.
%475C

\sqheader 475Xg Let $E\subseteq\BbbR^r$ be such that $\nu(\partstar E)$
and $\mu E$ are both finite.   Show that, taking $v_x$ to be the Federer
exterior normal to $E$ at any point $x$ where this is defined,

\Centerline{$\int_E\diverg\phi\,d\mu
=\int_{\partstar E}\varinnerprod{\phi(x)}{v_x}\,\nu(dx)$}

\noindent for every bounded Lipschitz function $\phi:\BbbR^r\to\BbbR^r$.
%475L

\sqheader 475Xh Let $\sequencen{E_n}$ be a sequence of measurable
subsets of $\BbbR^r$ such that (i) there is a measurable set $E$ such
that $\lim_{n\to\infty}\mu((E_n\symmdiff E)\cap B(\tbf{0},m))=0$ for
every $m\in\Bbb N$ (ii)
$\sup_{n\in\Bbb N}\nu(\partstar E_n\cap B(\tbf{0},m))$ is finite for
every $m\in\Bbb N$.   Show that $E$ has locally finite
perimeter.
\Hint{$\int_E\diverg\phi\,d\mu
=\lim_{n\to\infty}\int_{E_n}\diverg\phi\,d\mu$
for every Lipschitz function $\phi$ with compact support.}
%475L

\spheader 475Xi Give an example of bounded convex sets $E$ and $F$ such
that
$\partial^{\$}(E\cup F)\not\subseteq\partial^{\$}E\cup\partial^{\$}F$.
%475M

\spheader 475Xj(i) Show that if $A$, $B\subseteq\Bbb R^r$ then
$\partstar(A\cap B)\cap\partstar(A\cup B)
\subseteq\partstar A\cap\partstar B$.    (ii) Show that if $E$,
$F\subseteq\Bbb R$ are Lebesgue measurable, then
$\per(E\cap F)+\per(E\cup F)\le\per E+\per F$.
%475M

\sqheader 475Xk Let $E\subseteq\BbbR^r$ be a set with finite Lebesgue
measure and finite perimeter.   (i) Show that if $H\subseteq\BbbR^r$ is
a half-space, then $\per(E\cap H)\le\per E$.   \Hint{475Ja.}   (ii) Show
that if $C\subseteq\BbbR^r$ is convex, then $\per(E\cap C)\le\per E$.
\Hint{by the Hahn-Banach theorem, $C$ is a limit of polytopes;  use
474Ta.}   (iii) Show that in 475Mc we have
$\per E=\lim_{\alpha\to\infty}\per(E\cap B(\tbf{0},\alpha))$.
%475M

\spheader 475Xl\dvAnew{2009}
Let $E\subseteq\BbbR^r$ be a set with finite measure and finite
perimeter, and
$f:\BbbR^r\to\Bbb R$ a Lipschitz function.   Show that for any unit vector
$v\in\BbbR^r$,
$|\int_E\varinnerprod{v}{\grad f}\,d\mu|\le\|f\|_{\infty}\per E$.
%475M out of order query

\spheader 475Xm\dvAnew{2009} For measurable $E\subseteq\BbbR^r$ set
$p(E)
=\sup_{x\in\BbbR^r\setminus\{0\}}\Bover1{\|x\|}\mu(E\symmdiff(E+x))$.
(i) Show that for any measurable $E$,
$p(E)=\limsup_{x\to\tbf{0}}\Bover1{\|x\|}\mu(E\symmdiff(E+x))$.
(ii) Show that for every $\epsilon>0$ there is an
$E\subseteq\BbbR^r$ such that $\per E=1$ and $p(E)\ge 1-\epsilon$.
(iii) Show that if $E\subseteq\BbbR^r$ is a non-trivial
ball then $\per E=\Bover{r\beta_r}{2\beta_{r-1}}p(E)$.
(iv) Show that if $E\subseteq\BbbR^r$ is a cube then
$\per E=\sqrt{r}p(E)$.   \Hint{475Yf.}
%475Q

\spheader 475Xn\dvAnew{2013} Suppose that $E\subseteq\BbbR^r$ is a bounded
set with finite perimeter, and $\phi$, $\psi:\BbbR^r\to\Bbb R$ two
differentiable functions such that $\grad\phi$ and $\grad\psi$ are
Lipschitz.   Show that

\Centerline{$\int_E\phi\times\nabla^2\psi-\psi\times\nabla^2\phi\,d\mu
=\int_{\partstar E}
 \varinnerprod{(\phi\times\grad\psi-\psi\times\grad\phi)}{v_x}\,\nu(dx)$}

\noindent where $v_x$ is the Federer exterior normal to $E$ at $x$ when
this is defined.   (This is {\bf Green's second identity}.)

\leader{475Y}{Further exercises (a)}
%\spheader 475Ya
Show that if $A\subseteq\BbbR^r$ is Lebesgue
negligible, then there is a Borel set $E\subseteq\BbbR^r$ such that
$A\subseteq\partstar E$.
%475C

\spheader 475Yb\dvAnew{2013} Let $(X,\rho)$ be a metric space and $\mu$ a
strictly positive locally finite topological measure on $X$.   Show that we
can define operations $\clstar$, $\intstar$ and $\partstar$ on $\Cal PX$
for which parts (a)-(f) of 475C will be true.
%475C (easy, but not much use for anything that I know of)

\spheader 475Yc Let $B$ be a ball in $\BbbR^r$ with centre
$y$, and $v$, $v'$ two unit vectors in $\BbbR^r$.   Set

\Centerline{$H=\{x:x\in\BbbR^r,\,\varinnerprod{(x-y)}{v}\le 0\}$,
\quad$H'=\{x:x\in\BbbR^r,\,\varinnerprod{(x-y)}{v'}\le 0\}$.}

\noindent Show that $\mu((H\symmdiff H')\cap B)
=\Bover1{\pi}\arccos(\varinnerprod{v}{v'})\mu B$.
%475F

\allowmorestretch{468}{
\spheader 475Yd Show that $\mu$ is inner regular with respect to the
family of compact sets $K\subseteq\BbbR^r$ such that
$\limsup_{\delta\downarrow 0}
\Bover{\mu(B(x,\delta)\cap K)}{\mu B(x,\delta)}\ge\Bover12$ for every
$x\in K$.
}
%475I 47bits

\spheader 475Ye Let $\sequencen{f_n}$ be a
sequence of functions from $\BbbR^r$ to $\Bbb R$ which is uniformly
bounded and {\bf uniformly Lipschitz}
in the sense that there is some $\gamma\ge 0$
such that every $f_n$ is $\gamma$-Lipschitz.   Suppose that
$f=\lim_{n\to\infty}f_n$ is defined everywhere in $\BbbR^r$.   (i) Show
that if $E\subseteq\BbbR^r$ has finite measure, then
$\int_E\varinnerprod{z}{\grad f}d\mu
=\lim_{n\to\infty}\int_E\varinnerprod{z}{\grad f_n}d\mu$ for every
$z\in\BbbR^r$.   \Hint{look at $E$ of finite perimeter first.}   
(ii) Show that
for any convex function $\phi:\BbbR^r\to\coint{0,\infty}$,
$\int\phi(\grad f)d\mu\le\liminf_{n\to\infty}\int\phi(\grad f_n)d\mu$.
%233Yg %461C %475N out of order query

\spheader 475Yf Let $E\subseteq\BbbR^r$ be a set with locally
finite perimeter.   Show that

\Centerline{$\sup_{x\in\BbbR^r\setminus\{0\}}
  \Bover1{\|x\|}\mu(E\symmdiff(E+x))
=\sup_{\|v\|=1}\int_{\partstar E}|\varinnerprod{v}{v_x}|\nu(dx)$,}

\noindent where $v_x$ is the Federer exterior normal of $E$ at $x$ when
this is defined.
%475Q 475Xm

\spheader 475Yg\dvAnew{2011}
Let $E\subseteq\BbbR^r$ be Lebesgue measurable.
(i) Show that $\intstar E$ is an F$_{\sigma\delta}$
(= $\pmb{\Pi}^0_3$) set, that is, is expressible as the intersection of a
sequence of F$_{\sigma}$ sets.   (ii)(cf.\ {\smc Andretta \& Camerlo 13})
Show that if $E$ is not negligible and $\clstar E$ has empty interior,
then $\intstar E$ is not G$_{\delta\sigma}$
(= $\pmb{\Sigma}^0_3$), that is, cannot be expressed as the union of
sequence of G$_{\delta}$ sets.
%475B out of order query
%n11516

\leaveitout{\query\spheader 475Y? Let $F\subseteq[0,1]$ be the `Cantor
$\bover14$-set', that is, the set
$\{\sum_{i=1}^{\infty}4^{-i}\epsilon_i:\epsilon_i\in\{0,3\}$ for every
$i\}$, so that $F=\bover14F\cup(\bover34+\bover14F)$, and for each
$k\in\Bbb N\,\,F$ is the union of $2^k$ copies of $4^{-k}F$.   Set
$K=F^2\subseteq\BbbR^2$.   Show that the one-dimensional Hausdorff
measure of $K$ is $\sqrt 2$.   (Compare 264J.)   Show that, in the
language of 475Q, $\nu(T_v[K])=0$ for almost every $v\in S_1$.
}%end of leaveitout
}%end of exercises

\endnotes{
\Notesheader{475} The successful identification of the
distributionally-defined notion of `perimeter', as described in \S474,
with the geometrically accessible concept of Hausdorff measure of an
appropriate boundary, is of course the key to any proper understanding
of the results of the last section as well as this one.   The very word
`perimeter' would be unfair if the perimeter of $E\cup F$ were unrelated
to the perimeters of $E$ and $F$;  and from this point of view the
reduced boundary is less suitable than the essential boundary (475Cd,
475Xi).   If we re-examine 474M, we see that it is saying, in effect,
that for many balls $B$ the boundary $\partstar(E\cap B)$ is nearly
$(B\cap\partstar E)\cup(E\cap\partial B)$, and that an outward-normal
function for $E\cap B$ can be assembled from outward-normal functions
for $E$ and $B$.   But looking at 475Xc-475Xd we see that this is
entirely natural;  we need only ensure that $\nu(F\cap\partial B)=0$ for
a $\mu$-negligible set $F$ defined from $E$;  and the `almost every
$\delta$' in the statement of 474M is fully enough to arrange this.
On the other hand, 475Xg seems to be very hard to prove without using
the identification between $\nu(\partstar E)$ and $\per E$.

Concerning 475Q, I ought to emphasize that it is {\it not} generally
true that

\Centerline{$\nu F
=\Bover1{2\beta_{r-1}}\int_{S_{r-1}}\int_{V_v}
  \#(F\cap T_v^{-1}[\{u\}])\nu(du)\nu(dv)$}

\noindent even for $r=2$ and compact sets $F$ with $\nu F<\infty$.
%(See 474Xb\query.)
We are here approaching one of the many fundamental
concepts of geometric measure theory which I am ignoring.   The key word
is `rectifiability';
for `rectifiable' sets a wide variety of concepts of $k$-dimensional
measure coincide, including the integral-geometric form above, and
$\partstar E$ is rectifiable whenever $E$ has locally finite perimeter
({\smc Evans \& Gariepy 92}, 5.7.3).
For the general theory of rectifiable sets, see the last quarter of
{\smc Mattila 95}, or Chapter 3 of {\smc Federer 69}.

I have already noted that the largest volumes for sets of given diameter
or perimeter are provided by balls (see 264H and the notes to \S474).
The isoperimetric theorem for convex sets (475T) is of the same form:
once again, the best constant (here, the largest perimeter for a convex
set of given diameter, or the smallest diameter for a convex set of
given perimeter) is provided by balls.

475Qb gives an alternative characterization of `set of finite perimeter',
with bounds on the perimeter which are sometimes useful.
}%end of notes

\discrpage

\frfilename{mt476.tex}
\versiondate{29.7.21}
\copyrightdate{2001}

\def\varinnerprod#1#2{#1\dotproduct#2}

\def\chaptername{Geometric measure theory}
\def\sectionname{Concentration of measure}

\newsection{476}

Among the myriad special properties of Lebesgue measure, a particularly
interesting one is `concentration of measure'.   For a set of given
measure in the plane, it is natural to feel that it is most
`concentrated' if it is a disk.   There are many ways of defining
`concentration', and I examine three of them in this section (476F, 476G
and 476H);  all lead us to Euclidean balls as the `most concentrated'
shapes.   On the sphere the same criteria lead us to caps
(476K\cmmnt{, 476Xe}).

All the main theorems of this section will be based on the fact that
semi-continuous functions on compact spaces attain their bounds.   The
compact spaces in question will be spaces of subsets, and I begin with
some general facts concerning the topologies introduced in 4A2T
(476A-476B).
The particular geometric properties of Euclidean
space which make all these results possible are described in 476D-476E,
where I describe concentrating operators based on reflections.   The
actual theorems 476F-476H %476F 476G 476H
and 476K can now almost be mass-produced.

\leader{476A}{Proposition} Let $X$ be a topological space,
$\Cal C$ the
family of closed subsets of $X$, $\Cal K\subseteq\Cal C$ the family of
closed compact sets and $\mu$ a topological measure on $X$.

(a)(i) If $\mu$ is outer regular with respect to the open sets then
$\mu\restr\Cal C:\Cal C\to\coint{0,\infty}$ is upper semi-continuous
with respect to the Vietoris topology on $\Cal C$.

\quad(ii) If $\mu$ is locally finite and inner regular with respect to the
closed sets then
$\mu\restr\Cal K$ is upper semi-continuous
with respect to the Vietoris topology.

\quad(iii) If $\mu$ is inner regular with respect to the
closed sets and $f$ is a non-negative $\mu$-integrable real-valued function
then $F\mapsto\int_Ffd\mu:\Cal C\to\Bbb R$ is upper semi-continuous with
respect to the Vietoris topology.

(b) Suppose that $\mu$ is tight.

\quad(i) If $\mu$ is totally finite then
$\mu\restr\Cal C$ is upper semi-continuous
with respect to the Fell topology on $\Cal C$.

\quad(ii) If $f$ is a non-negative $\mu$-integrable real-valued function
then $F\mapsto\int_Ffd\mu:\Cal C\to\Bbb R$ is upper semi-continuous with
respect to the Fell topology.

(c) Suppose that $X$ is metrizable, and that $\rho$ is a metric on $X$
defining its topology;  let $\tilde\rho$ be the Hausdorff metric on
$\Cal C\setminus\{\emptyset\}$.

\quad(i) If $\mu$ is totally finite, then
$\mu\restr\Cal C\setminus\{\emptyset\}$ is upper semi-continuous with
respect to $\tilde\rho$.

\quad(ii) If $\mu$ is locally finite, then
$\mu\restr\Cal K\setminus\{\emptyset\}$ is upper semi-continuous with
respect to $\tilde\rho$.

\quad(iii) If $f$ is a non-negative
$\mu$-integrable real-valued function, then
$F\mapsto\int_Ffd\mu:\Cal C\setminus\{\emptyset\}\to\Bbb R$ is upper
semi-continuous with respect to $\tilde\rho$.

\proof{{\bf (a)(i)} Suppose that $F\in\Cal C$ and that $\mu F<\alpha$.
Because $\mu$ is outer regular with respect to the
open sets, there is an open set
$G\supseteq F$ such that $\mu G<\alpha$.
Now $\Cal V=\{E:E\in\Cal C,\,E\subseteq G\}$ is an open set
for the Vietoris topology containing $F$,
and $\mu E<\alpha$ for every $E\in\Cal V$.   As $F$ and
$\alpha$ are arbitrary, $\mu\restr\Cal C$ is upper semi-continuous for the
Vietoris topology.

\medskip

\quad{\bf (ii)} Given that $K\in\Cal K$ and $\mu K<\alpha$, then,
because $\mu$ is locally finite, there is an open set $G$ of
finite measure including $K$ (cf.\ 411Ga).   Now there is a closed set
$F\subseteq G\setminus K$ such that $\mu F>\mu G-\alpha$, so that
$\Cal V=\{L:L\in\Cal K$, $L\subseteq G\setminus F\}$ is a relatively open
subset of $\Cal K$ for the Vietoris topology containing $K$, and
$\mu L<\alpha$ for every $L\in\Cal V$.

\medskip

\quad{\bf (iii)} Apply (i) to the indefinite-integral measure over $\mu$
defined by $f$;  by 412Q this is still inner regular with respect to the
closed sets.

\medskip

{\bf (b)} If $F\in\Cal C$ and $\mu F<\alpha$, let
$K\subseteq X\setminus F$ be a compact set such that
$\mu K>\mu X-\alpha$.   Then
$\Cal V=\{E:E\in\Cal C$, $E\cap K=\emptyset\}$ is a neighbourhood of $F$
and $\mu E<\alpha$ for every $E\in\Cal V$.   This proves (i).
Now (ii) follows as in (a-iii) above.

\medskip

{\bf (c)(i)} If $F\in\Cal C\setminus\{\emptyset\}$ and
$\mu F<\alpha$, then for each $n\in\Bbb N$ set
$F_n=\{x:\rho(x,F)\le 2^{-n}\}$.   Since $\sequencen{F_n}$ is a
non-increasing sequence of closed sets with intersection $F$, and $\mu$
is totally finite, there is an $n$ such that $\mu F_n<\alpha$.   If now
we take $E\in\Cal C\setminus\{\emptyset\}$ such that
$\tilde\rho(E,F)\le 2^{-n}$, then $E\subseteq F_n$ so $\mu E<\alpha$.
As $F$ and $\alpha$ are arbitrary,
$\mu\restr\Cal C\setminus\{\emptyset\}$ is upper semi-continuous.

\medskip

\quad{\bf (ii)} If $K\in\Cal K\setminus\{\emptyset\}$ and $\mu K<\alpha$,
let $G\supseteq K$ be an open set of finite measure, as in (a-ii) above.
The function $x\mapsto\rho(x,X\setminus G)$ is continuous and
strictly positive on $K$, so has a non-zero lower bound on $K$, and
there is some $m\in\Bbb N$ such that $\rho(x,X\setminus G)>2^{-m}$ for
every $x\in K$.   If, as in (i) just above, we set
$F_n=\{x:\rho(x,K)\le 2^{-n}\}$ for each $n$, $F_m\subseteq G$ has finite
measure.   So, as in (i), we have an $n\ge m$ such that $\mu F_n<\alpha$,
and we can continue as before.

\medskip

\quad{\bf (iii)} Once again this follows at once from (i).
}%end of proof of 476A

\vleader{60pt}{476B}{Lemma} Let $(X,\rho)$ be a metric space, and $\Cal C$
the family of closed subsets of $X$, with its Fell topology.   For
$\epsilon>0$, set
$U(A,\epsilon)=\{x:x\in X$, $\rho(x,A)<\epsilon\}$ if
$A\subseteq X$ is not
empty;  set $U(\emptyset,\epsilon)=\emptyset$.   Then for any
$\tau$-additive topological measure $\mu$ on $X$, the function

\Centerline{$(F,\epsilon)\mapsto\mu U(F,\epsilon):
  \Cal C\times\ooint{0,\infty}\to[0,\infty]$}

\noindent is lower semi-continuous.

\proof{ Set
$Q=\{(F,\epsilon):F\in\Cal C$, $\epsilon>0$, $\mu U(F,\epsilon)>\gamma\}$,
where $\gamma\in\Bbb R$.   Take any $(F_0,\epsilon_0)\in Q$.
Note first that
$\mu U(F_0,\epsilon_0)=\sup_{\epsilon<\epsilon_0}\mu U(F,\epsilon)$, so
there is a $\delta\in\ooint{0,\bover12\epsilon_0}$ such that
$\mu U(F_0,\epsilon_0-2\delta)>\gamma$.
Next, $\{U(x,\epsilon_0-2\delta):x\in F_0\}$ is an open cover of
$U(F_0,\epsilon_0-2\delta)$;  because $\mu$ is $\tau$-additive, there is a
finite set $I\subseteq F_0$ such that
$\mu(\bigcup_{x\in I}U(x,\epsilon_0-2\delta))>\gamma$.   Consider

\Centerline{$\Cal V
=\{F:F\in\Cal C$, $F\cap U(x,\delta)\ne\emptyset$ for every $x\in I\}$.}

\noindent By the definition of the Fell topology, $\Cal V$ is open.
So $\Cal V\times\ooint{\epsilon_0-\delta,\infty}$ is an open
neighbourhood of $(F_0,\epsilon)$ in $\Cal C\times\Bbb R$.   If
$F\in\Cal V$ and $\epsilon>\epsilon_0-\delta$, then

\Centerline{$U(F,\epsilon)\supseteq\bigcup_{x\in I}U(x,\epsilon-\delta)
\supseteq\bigcup_{x\in I}U(x,\epsilon_0-2\delta)$}

\noindent has measure greater than $\gamma$ and $(F,\epsilon)\in Q$.   As
$(F_0,\epsilon_0)$ is arbitrary, $Q$ is open;  as $\gamma$ is arbitrary,
$(F,\epsilon)\mapsto\mu U(F,\epsilon)$ is lower semi-continuous.
}%end of proof of 476B

\cmmnt{\medskip

\noindent{\bf Remark} Recall that all `ordinary' topological measures on
metric spaces are $\tau$-additive;  see 438J.
}

\leader{476C}{Proposition} Let $(X,\rho)$ be a non-empty compact metric
space, and suppose that its isometry group $G$ acts transitively on $X$.
Then $X$ has a unique $G$-invariant Radon probability measure $\mu$,
which is strictly positive.

\proof{ By 441G, $G$, with its topology of pointwise convergence, is a
compact topological group, and the action of $G$ on $X$ is continuous.
So 443Ud gives the result.
}%end of proof of 476C

\leader{476D}{Concentration by partial reflection}\cmmnt{ The
following construction will be
used repeatedly in the rest of the section.

\medskip

} {\bf (a)}   Let $X$ be an inner
product space.   \cmmnt{(In this section, $X$ will be usually be
$\BbbR^r$, but in 493G below it
will be helpful to be able to speak of abstract Hilbert spaces.)}   For
any unit vector $e\in X$ and any $\alpha\in\Bbb R$, write
$R=R_{e\alpha}:X\to X$ for the reflection in the hyperplane
$V=V_{e\alpha}=\{x:x\in X,\,\innerprod{x}{e}=\alpha\}$, so that
$R(x)=x+2(\alpha-\innerprod{x}{e})e$ for every $x\in X$.
Next, for any
$A\subseteq X$, \cmmnt{we can} define \cmmnt{a set}
$\psi(A)=\psi_{e\alpha}(A)$ by setting

$$\eqalign{\psi(A)
&=\{x:x\in A,\,\innerprod{x}{e}\ge\alpha\}
\cup\{x:x\in A,\innerprod{x}{e}<\alpha,\,R(x)\in A\}\cr
&\hskip10em\cup
\{x:x\in \BbbR^r\setminus A,\,\innerprod{x}{e}\ge\alpha,\,
R(x)\in A\}\cr
&=(W\cap(A\cup R[A]))
  \cup(A\cap R[A]),\cr}$$

\noindent where $W=W_{e\alpha}$ is the half-space
$\{x:\innerprod{x}{e}\ge\alpha\}$.   \cmmnt{Geometrically, we
construct $\psi(A)$ by moving those points of $A$ on the
`wrong' side of the hyperplane
$V$ to their reflections, provided those points are not
already occupied.   We have the following facts.}

\spheader 476Db(i) If $A\subseteq B\subseteq X$, $\psi(A)\subseteq\psi(B)$.
(ii) For any $A\subseteq X$, $\psi(R[A])=\psi(A)$.
(iii) If $F\subseteq X$ is closed, then $\psi(F)$ is
closed.   \prooflet{\Prf\ Use the second formula in (a) for
$\psi(F)$.\ \Qed}

\spheader 476Dc\cmmnt{ We need a fragment of elementary geometry.} If 
$x\in X\setminus W$ and $y\in W$ then $\|x-R(y)\|\le\|x-y\|$.   \prooflet{\Prf\
Write $Y$ for the linear subspace of $X$ generated by $e$, $Y^{\perp}$ for its
orthogonal complement and $P$ for the orthogonal projection of $X$ onto
$Y^{\perp}$.   Then $P(e)=0$, $PR(y)=P(y)$,

\Centerline{$\innerprod{x}{e}<\alpha\le\innerprod{y}{e}$, 
\quad$\innerprod{R(y)}{e}=\innerprod{y}{e}+2(\alpha-\innerprod{y}{e})
=2\alpha-\innerprod{y}{e}$,}

\Centerline{$|\innerprod{x-y}{e}|
=|\innerprod{x}{e}-\alpha|+|\innerprod{y}{e}-\alpha|
\ge|\innerprod{x}{e}-2\alpha+\innerprod{y}{e}|
=|\innerprod{x-R(y)}{e}|$,}

\noindent and

$$\eqalignno{\|x-y\|^2
&=\innerprod{x-y}{e}^2+\|P(x-y)\|^2
\Displaycause{4A4Jf}
\ge\innerprod{x-R(y)}{e}^2+\|P(x-R(y))\|^2
=\|x-R(y)\|^2.  \text{ \Qed}\cr}$$}

\spheader 476Dd For non-empty $A\subseteq X$ and $\epsilon>0$, set
$U(A,\epsilon)=\{x:\rho(x,A)<\epsilon\}$, where $\rho$ is the
standard metric on $X$.   \cmmnt{Now}
$U(\psi(A),\epsilon)\subseteq\psi(U(A,\epsilon))$.
\prooflet{\Prf

\medskip

\quad{\bf (i)} If $x\in U(W\cap A,\epsilon)\setminus W$, there is a
$y\in W\cap A$ such that $\|x-y\|<\epsilon$;  by (c), $\|x-R(y)\|<\epsilon$
so $x\in U(R[A],\epsilon)=R[U(A,\epsilon)]$ (because $R:X\to X$ is an isometry)
and

\Centerline{$x\in U(A,\epsilon)\cap R[U(A,\epsilon)]
\subseteq\psi(U(A,\epsilon))$;}

\noindent thus $U(W\cap A,\epsilon)\setminus W\subseteq\psi(U(A,\epsilon))$;
as also 

\Centerline{$U(W\cap A,\epsilon)\cap W
\subseteq U(A,\epsilon)\cap W\subseteq\psi(U(A,\epsilon))$,}

\noindent we have $U(W\cap A,\epsilon)\subseteq\psi(U(A,\epsilon))$.

\medskip

\quad{\bf (ii)} Now 

$$\eqalignno{U(\psi(A),\epsilon)&
=U(A\cap R[A],\epsilon)\cup U(A\cap W,\epsilon)\cup U(R[A]\cap W,\epsilon)\cr&
\subseteq(U(A,\epsilon)\cap U(R[A],\epsilon))\cup\psi(U(A,\epsilon))
    \cup\psi(U(R[A],\epsilon))\cr&
=(U(A,\epsilon)\cap R[U(A,\epsilon)])\cup\psi(U(A,\epsilon))
    \cup\psi(R[U(A,\epsilon)])\cr&
=(U(A,\epsilon)\cap R[U(A,\epsilon)])\cup\psi(U(A,\epsilon))
\Displaycause{using (b-i)}
=\psi(U(A,\epsilon)).  \text{ \Qed}\cr}$$
}

\vleader{60pt}{476E}{Lemma} 
Let $X$ be an inner product space, $e\in X$ a unit vector
and $\alpha\in\Bbb R$.   Let $R=R_{e\alpha}:X\to X$ be the reflection
operator, and $\psi=\psi_{e\alpha}:\Cal PX\to\Cal PX$ the associated
transformation, as described in 476D.   For $x\in A\subseteq X$, define

$$\eqalign{\phi_A(x)
&=x\text{ if }\innerprod{x}{e}\ge\alpha,\cr
&=x\text{ if }\innerprod{x}{e}<\alpha
  \text{ and }R(x)\in A,\cr
&=R(x)\text{ if }\innerprod{x}{e}<\alpha
  \text{ and }R(x)\notin A.\cr}$$

\noindent Let $\nu$ be a topological measure on $X$ which is
$R$-invariant\cmmnt{, that is, $\nu$ coincides with the image measure
$\nu R^{-1}$}.

(a) For any $A\subseteq X$, $\phi_A:A\to\psi(A)$ is a bijection.
If $\alpha<0$, then $\|\phi_A(x)\|\le\|x\|$ for every $x\in A$, with
$\|\phi_A(x)\|<\|x\|$ iff $\innerprod{x}{e}<\alpha$ and
$R(x)\notin A$.

(b)(i) If $E\subseteq X$ is measured by $\nu$, then $\psi(E)$ is measured by 
$\nu$, $\nu\psi(E)=\nu E$ and $\phi_E$ is a measure space
isomorphism for the subspace measures on $E$ and $\psi(E)$ induced by $\nu$.

\quad(ii) For any $A\subseteq X$, $\nu^*\psi(A)\le\nu^*A\le 2\nu^*\psi(A)$.

(c) If $\alpha<0$ and $E\subseteq X$ is measured by $\nu$, then
$\int_E\|x\|\nu(dx)\ge\int_{\psi(E)}\|x\|\nu(dx)$, with equality iff
$\{x:x\in E,\,\innerprod{x}{e}<\alpha,\,R(x)\notin E\}$ is negligible.

(d) Suppose that $X$ is separable.   Let $\lambda$ be the c.l.d.\
product measure of $\nu$ with itself on $X\times X$.   If
$E\subseteq X$ is measured by $\nu$, then

\Centerline{$\int_{E\times E}\|x-y\|\lambda(d(x,y))
\ge\int_{\psi(E)\times\psi(E)}\|x-y\|\lambda(d(x,y))$.}

(e) Now suppose that $X=\BbbR^r$.   Then
$\nu(\partstar\psi(A))\le\nu(\partstar A)$ for every
$A\subseteq\BbbR^r$, where $\partstar A$ is the essential boundary of
$A$\cmmnt{ (definition:  475B)}.

\proof{{\bf (a)} That $\phi_A:A\to\psi(A)$ is a bijection is immediate
from the definitions of $\psi$ and $\phi_A$.   If $\alpha<0$, then for
any $x\in A$ either $\phi_A(x)=x$ or $\innerprod{x}{e}<\alpha$ and
$R(x)\notin A$.   In the latter case

$$\eqalignno{\|\phi_A(x)\|^2
&=\|R(x)\|^2=\|x+2\gamma e\|^2\cr
\displaycause{where $\gamma=\alpha-\innerprod{x}{e}>0$}
&=\|x\|^2+4\gamma\innerprod{x}{e}+4\gamma^2
=\|x\|^2+4\gamma\alpha
<\|x\|^2,\cr}$$

\noindent so $\|\phi_A(x)\|<\|x\|$.

\medskip

{\bf (b)(i)} If we set

\Centerline{$E_1=\{x:x\in E,\,\innerprod{x}{e}\ge\alpha\}$,}

\Centerline{$E_2=\{x:x\in E,\innerprod{x}{e}<\alpha,\,
R(x)\in E\}$,}

\Centerline{$E_3=\{x:x\in E,\,\innerprod{x}{e}<\alpha,\,
R(x)\notin E\}$,}

\Centerline{$E_4=\{x:x\in\BbbR^r\setminus E,\,
\innerprod{x}{e}>\alpha,\,R(x)\in E\}$,}

\noindent then $E_1$, $E_2$, $E_3$ and $E_4$ are disjoint and measured by
$\nu$, $E=E_1\cup E_2\cup E_3$, $\psi(E)=E_1\cup E_2\cup E_4$ and
$\phi_E\restr E_3=R\restr E_3$ is a measure space isomorphism for the
subspace measures on $E_3$ and $E_4$.

\medskip

\quad{\bf (ii)} There is an $E\supseteq A$ such that $\nu$ measures $E$ and
$\nu^*A=\nu E$.   Now $\psi(A)\subseteq\psi(E)$, so 

\Centerline{$\nu^*\psi(A)\le\nu\psi(E)=\nu E=\nu^*A$.}

\noindent On the other hand, if $x\in A$ then at least one of $x$, $R(x)$
belongs to $\psi(A)$, so $A\subseteq\psi(A)\cup R[\psi(A)]$ and

\Centerline{$\nu^*A\le\nu^*\psi(A)+\nu^*R[\psi(A)]=2\nu^*\psi(A)$.}


\medskip

{\bf (c)} By (a),

\Centerline{$\int_E\|x\|\nu(dx)
\ge\int_E\|\phi_E(x)\|\nu(dx)
=\int_{\psi(E)}\|x\|\nu(dx)$}

\noindent by 235Gc, because $\phi_E:E\to\psi(E)$ is \imp, with equality
only when

\Centerline{$\{x:\|x\|>\|\phi_E(x)\|\}
=\{x:x\in E,\,\innerprod{x}{e}<\alpha,\,R(x)\notin E\}$}

\noindent is negligible.

\medskip

{\bf (d)} Note first that if $\Lambda$ is the domain of $\lambda$ then
$\Lambda$ includes the Borel algebra of $X\times X$ (because $X$ is
second-countable, so this is just the $\sigma$-algebra generated by
products of Borel sets, by 4A3Ga);  so that $(x,y)\mapsto\|x-y\|$ is
$\Lambda$-measurable, and the integrals are defined in $[0,\infty]$.   Now
consider the sets

\Centerline{$W_1=\{(x,y):x\in E,\,y\in E,\,R(x)\notin E,\,
R(y)\in E,\,\innerprod{x}{e}<\alpha,\,\innerprod{y}{e}<\alpha\}$,}

\Centerline{$W'_1=\{(x,y):x\in E,\,y\in E,\,R(x)\notin E,\,
R(y)\in E,\,\innerprod{x}{e}<\alpha,\,\innerprod{y}{e}>\alpha\}$,}

\Centerline{$W_2=\{(x,y):x\in E,\,y\in E,\,R(x)\in E,\,
R(y)\notin E,\,\innerprod{x}{e}<\alpha,\,\innerprod{y}{e}<\alpha\}$,}

\Centerline{$W'_2=\{(x,y):x\in E,\,y\in E,\,R(x)\in E,\,
R(y)\notin E,\,\innerprod{x}{e}>\alpha,\,\innerprod{y}{e}<\alpha\}$.}

\noindent Then $(x,y)\mapsto(x,R(y)):W'_1\to W_1$ is a measure space
isomorphism for the subspace measures induced on $W_1$ and $W'_1$ by
$\lambda$, so

$$\eqalign{\int_{W_1}\|\phi_E(x)-\phi_E(y)\|\lambda(d(x,y))
&=\int_{W_1}\|R(x)-y\|\lambda(d(x,y))\cr
&=\int_{W'_1}\|R(x)-R(y)\|\lambda(d(x,y))\cr
&=\int_{W'_1}\|x-y\|\lambda(d(x,y)).\cr}$$

\noindent Similarly,

$$\eqalign{\int_{W'_1}\|\phi_E(x)-\phi_E(y)\|\lambda(d(x,y))
&=\int_{W'_1}\|R(x)-y\|\lambda(d(x,y))\cr
&=\int_{W_1}\|R(x)-R(y)\|\lambda(d(x,y))\cr
&=\int_{W_1}\|x-y\|\lambda(d(x,y)).\cr}$$

\noindent So we get

\Centerline{$\int_{W_1\cup W'_1}
  \|\phi_E(x)-\phi_E(y)\|\lambda(d(x,y))
=\int_{W_1\cup W'_1}\|x-y\|\lambda(d(x,y))$.}

\noindent In the same way, $(x,y)\mapsto(R(x),y)$ is an isomorphism of
the subspace measures on $W_2$ and $W'_2$, and we have

\Centerline{$\int_{W_2\cup W'_2}
  \|\phi_E(x)-\phi_E(y)\|\lambda(d(x,y))
=\int_{W_2\cup W'_2}\|x-y\|\lambda(d(x,y))$.}

\noindent On the other hand, for all
$(x,y)\in(E\times E)\setminus(W_1\cup W'_1\cup W_2\cup W'_2)$, we have
$\|\phi_E(x)-\phi_E(y)\|\le\|x-y\|$.   (Either $x$ and $y$ are both left
fixed by $\phi_E$, or both are moved, or one is on the reflecting
hyperplane, or one is moved to the same side of the reflecting
hyperplane as the other.)   So we get

$$\eqalign{\int_{E\times E}\|x-y\|\lambda(d(x,y))
&\ge\int_{E\times E}\|\phi_E(x)-\phi_E(y)\|\lambda(d(x,y))\cr
&=\int_{\psi(E)\times\psi(E)}\|x-y\|\lambda(d(x,y))\cr}$$

\noindent because $(x,y)\mapsto(\phi_E(x),\phi_E(y))$ is an \imp\
transformation for the subspace measures on $E\times E$ and
$\psi(E)\times\psi(E)$.

\medskip

{\bf (e)(i)} Because $R$ is both an isometry and a
measure space automorphism for Lebesgue measure $\mu$ on $\BbbR^r$, 
$\clstar R[A]=R[\clstar A]$ and
$\intstar R[A]=R[\intstar A]$, where $\clstar A$ and $\intstar A$ are
the essential closure and the essential interior of $A$, as in 475B.
Recall that $\clstar A$, $\intstar A$ and $\partstar A$ are all Borel
sets (475Cc), so that $\partstar A$ and $\partstar\psi(A)$ are
measured by $\nu$.

\medskip

\quad{\bf (ii)} Suppose that $\varinnerprod{x}{e}=\alpha$.   Then
$x\in\partstar\psi(A)$
iff $x\in\partstar A$.   \Prf\ For any $\delta>0$, it is easy to check that
$B(x,\delta)\cap\psi(A)=\psi(B(x,\delta)\cap A)$, while
$\mu^*(B(x,\delta)\cap A)$ lies between $\mu^*\psi(B(x,\delta)\cap A)$ and
$2\mu^*\psi(B(x,\delta)\cap A)$ by (b-ii) above;  so

\Centerline{$\mu^*(B(x,\delta)\cap\psi(A))
\le\mu^*(B(x,\delta)\cap A)
\le 2\mu^*(B(x,\delta)\cap\psi(A))$}

\noindent for every $\delta>0$ and
$x\in\clstar A$ iff $x\in\clstar\psi(A)$.   If
$x=R(x)\in\intstar A$, then $x\in\intstar R[A]$ so
$x\in\intstar(A\cap R[A])$ (475Cd) and $x\in\intstar\psi(A)$.   If
$x\in\intstar\psi(A)$ then
$x\in\intstar(R[\psi(A)]\cap\psi(A))\subseteq\intstar A$.\ \Qed

\medskip

\quad{\bf (iii)} If $x\in\partstar\psi(A)\setminus\partstar A$ then
$R(x)\in\partstar A\setminus\partstar\psi(A)$.   \Prf\ By (ii),
$\varinnerprod{x}{e}\ne\alpha$.

\qquad{\bf case 1} Suppose that $\varinnerprod{x}{e}>\alpha$.   Setting
$\delta=\varinnerprod{x}{e}-\alpha$, we see that
$U(x,\delta)\cap\psi(A)=U(x,\delta)\cap(A\cup R[A])$, while
$U(R(x),\delta)\cap\psi(A)=U(R(x),\delta)\cap A\cap R[A]$.   Since
$x\notin\intstar\psi(A)$, $x\notin\intstar(A\cup R[A])$ and
$x\notin\intstar A$;  since $x$ also does not belong to $\partstar A$,
$x\notin\clstar A$.   However,
$x\in\clstar(A\cup R[A])=\clstar A\cup\clstar R[A]$ (475Cd), so
$x\in\clstar R[A]$ and $R(x)\in\clstar A$.   Next,
$x\notin\intstar R[A]$, so $R(x)\notin\intstar A$ and
$R(x)\in\partstar A$.   Since $x\notin\clstar A$,
$R(x)\notin\clstar R[A]$, while $\varinnerprod{R(x)}{e}<\alpha$;  so
$R(x)\notin\clstar\psi(A)$ and
$R(x)\in\partstar A\setminus\partstar\psi(A)$.

\qquad{\bf case 2} Suppose that $\varinnerprod{x}{e}<\alpha$.    This
time, set $\delta=\alpha-\varinnerprod{x}{e}$, so that
$U(x,\delta)\cap\psi(A)=U(x,\delta)\cap A\cap R[A]$ and
$U(R(x),\delta)\cap\psi(A)=U(R(x),\delta)\cap(A\cup R[A])$.   As
$x\in\clstar\psi(A)$, $x\in\clstar(A\cap R[A])$ and $R(x)\in\clstar A$.
Also $x\in\clstar A$;  as $x\notin\partstar A$, $x\in\intstar A$,
$R(x)\in\intstar R[A]$ and $R(x)\in\intstar\psi(A)$, so that
$R(x)\notin\partstar\psi(A)$.   Finally, we know that $x\in\intstar A$
but $x\notin\intstar(A\cap R[A])$ (because $x\notin\intstar\psi(A)$);  it
follows that $x\notin\intstar R[A]$ so $R(x)\notin\intstar A$ and
$R(x)\in\partstar A\setminus\partstar\psi(A)$.

Thus all possibilities are covered and we have the result.\ \Qed

\medskip

\quad{\bf (iv)} What this means is that if we set
$E=\partstar\psi(A)\setminus\partstar A$
then $R[E]\subseteq\partstar A\setminus\partstar\psi(A)$.   So

\Centerline{$\nu\partstar\psi(A)
=\nu E+\nu(\partstar\psi(A)\cap\partstar A)
=\nu R[E]+\nu(\partstar\psi(A)\cap\partstar A)
\le\nu\partstar A$,}

\noindent as required.
}%end of proof of 476E

\leader{476F}{Theorem} Let $r\ge 1$ be an integer, and let $\mu$ be
Lebesgue measure on $\BbbR^r$.   For non-empty $A\subseteq\BbbR^r$ and
$\epsilon>0$, write $U(A,\epsilon)$ for $\{x:\rho(x,A)<\epsilon\}$,
where $\rho$ is the Euclidean metric on $\BbbR^r$.   If $\mu^*A$ is
finite, then $\mu U(A,\epsilon)\ge\mu U(B_A,\epsilon)$, where $B_A$ is
the closed ball with centre $\tbf{0}$ and measure $\mu^*A$.

\proof{{\bf (a)} To begin with, suppose that $A$ is bounded.   Set
$\gamma=\mu^*A$ and $\beta=\mu U(A,\epsilon)$.   If $\gamma=0$ then
(because $A\ne\emptyset$)

\Centerline{$\mu U(A,\epsilon)\ge\mu U(\{\tbf{0}\},\epsilon)
=\mu U(B_A,\epsilon)$,}

\noindent and we can stop.   So let us suppose henceforth that
$\gamma>0$.   Let $M\ge 0$ be such that $A\subseteq B(\tbf{0},M)$, and
consider the family

\Centerline{$\Cal F
=\{F:F\in\Cal C$, $F\subseteq B(\tbf{0},M)$,
$\mu F\ge\gamma$, $\mu U(F,\epsilon)\le\beta\}$,}

\noindent where $\Cal C$ is the family of closed subsets of
$\BbbR^r$ with its Fell topology.   Because
$U(\overline{A},\epsilon)=U(A,\epsilon)$,
$\overline{A}\in\Cal F$ and $\Cal F$ is non-empty.   By the definition of
the Fell topology, $\{F:F\subseteq B(\tbf{0},M)\}$ is closed;
by 476A(b-ii) (applied to the functional
$F\mapsto\int_F\chi B(\tbf{0},M)d\mu$)
and 476B, $\Cal F$ is closed in $\Cal C$, therefore compact,
by 4A2T(b-iii).   Next, the function

\Centerline{$F\mapsto\int_F\max(0,M-\|x\|)\mu(dx):
\Cal C\to\coint{0,\infty}$}

\noindent is upper semi-continuous, by 476A(b-ii) again.
It therefore attains its
supremum on $\Cal F$ at some $F_0\in\Cal F$ (4A2Gl).   Let
$F_1\subseteq F_0$ be a closed self-supporting set of the same measure
as $F_0$;  then $U(F_1,\epsilon)\subseteq U(F_0,\epsilon)$ and
$\mu F_1=\mu F_0$, so $F_1\in\Cal F$;  also

\Centerline{$\int_{F_1}M-\|x\|\mu(dx)
=\int_{F_0}M-\|x\|\mu(dx)
\ge\int_{F}M-\|x\|\mu(dx)$}

\noindent for every $F\in\Cal F$.

\medskip

{\bf (b)} Now $F_1$ is a ball with centre $\tbf{0}$.   \Prf\Quer\
Suppose, if possible, otherwise.   Then there are $x_1\in F_1$ and
$x_0\in\BbbR^r\setminus F_1$ such that $\|x_0\|<\|x_1\|$.
Set $e=\Bover1{\|x_0-x_1\|}(x_0-x_1)$, so that $e$ is a unit vector, and
$\alpha=\Bover12\varinnerprod{e}{(x_0+x_1)}$;  then

\Centerline{$\alpha
=\Bover1{2\|x_0-x_1\|}\varinnerprod{(x_0-x_1)}{(x_0+x_1)}
=\Bover1{2\|x_0-x_1\|}(\|x_0\|^2-\|x_1\|^2)<0$,}

\Centerline{$\varinnerprod{e}{(x_0-x_1)}>0$,
\quad$\varinnerprod{e}{x_0}>\varinnerprod{e}{x_1}$,
\quad$\varinnerprod{e}{x_0}>\alpha>\varinnerprod{e}{x_1}$.}

\noindent Define $R=R_{e\alpha}:\BbbR^r\to\BbbR^r$ and
$\psi=\psi_{e\alpha}$ as in 476D;  note that $R(x_1)=x_0$.   Set $F=\psi(F_1)$.   Then $F$ is
closed (476Db) and $\mu F=\mu F_1\ge\mu^*A$ (476Eb).   Also
$U(F,\epsilon)\subseteq\psi(U(F_1,\epsilon))$ (476Dd), so

\Centerline{$\mu U(F,\epsilon)\le\mu(\psi(U(F_1,\epsilon))
=\mu U(F_1,\epsilon)\le\beta$}

\noindent and $F\in\Cal F$.   It follows that
$\int_FM-\|x\|\mu(dx)\le\int_{F_1}M-\|x\|\mu(dx)$;  as $\mu F=\mu F_1$,
$\int_F\|x\|\mu(dx)\ge\int_{F_1}\|x\|\mu(dx)$.   By 476Ec,
$G=\{x:x\in F_1,\,\varinnerprod{x}{e}<\alpha,\,R(x)\notin F_1\}$ is
negligible.   But $G$ contains $x_1$ and is relatively open in $F_1$,
and $F_1$ is supposed to be self-supporting;  so this is impossible.\
\Bang\Qed

\medskip

{\bf (c)} Since $\mu F_1\ge\gamma$, $F_1\supseteq B_A$, and

\Centerline{$\mu U(B_A,\epsilon)\le\mu U(F_1,\epsilon)\le\beta
=\mu U(A,\epsilon)$.}

\noindent So we have the required result for bounded $A$.   In general,
given an unbounded set $A$ of finite measure, let $\delta$ be the radius
of $B_A$;  then

$$\eqalign{\mu U(B_A,\epsilon)
&=\mu B(\tbf{0},\delta+\epsilon)
=\sup_{\alpha<\delta}\mu B(\tbf{0},\alpha+\epsilon)\cr
&\le\sup_{A'\subseteq A\text{ is bounded}}\mu U(B_{A'},\epsilon)
\le\sup_{A'\subseteq A\text{ is bounded}}\mu U(A',\epsilon)
=\mu U(A,\epsilon)\cr}$$

\noindent because $\{U(A',\epsilon):A'\subseteq A$ is bounded$\}$ is an
upwards-directed family of open sets with union $U(A,\epsilon)$, and
$\mu$ is $\tau$-additive.   So the theorem is true for unbounded $A$ as
well.
}%end of proof of 476F

\leader{476G}{Theorem} Let $r\ge 1$ be an integer, and let $\mu$ be
Lebesgue measure on $\BbbR^r$;  write $\lambda$ for the product measure
on $\BbbR^r\times\BbbR^r$.   For any measurable set $E\subseteq\BbbR^r$
of finite measure, write $B_E$ for the closed ball with centre $\tbf{0}$
and the same measure as $E$.   Then

\Centerline{$\int_{E\times E}\|x-y\|\lambda(d(x,y))
\ge\int_{B_E\times B_E}\|x-y\|\lambda(d(x,y))$.}

\proof{{\bf (a)} Suppose for the time being (down to the end of (c) below)
that $E$ is compact and not
empty, and that $\epsilon>0$.   Let $M\ge 0$ be such that $\|x\|\le M$
for every $x\in E$.   For a non-empty set $A\subseteq\BbbR^r$ set
$U(A,\epsilon)=\{x:\rho(x,A)<\epsilon\}$, where $\rho$ is Euclidean
distance on $\BbbR^r$.   Set
$\beta=\int_{U(E,\epsilon)\times U(E,\epsilon)}\|x-y\|\lambda(d(x,y))$.
Let $\Cal F$ be the family of non-empty closed subsets $F$ of the ball
$B(\tbf{0},M)=\{x:\|x\|\le M\}$ such that $\mu F\ge\mu E$ and
$\int_{U(F,\epsilon)\times U(F,\epsilon)}
\|x-y\|\lambda(d(x,y))\le\beta$.
Then $\Cal F$ is compact for the Fell topology on the family
$\Cal C$ of closed subsets of $\BbbR^r$.   \Prf\ We know from 4A2T(b-iii)
that $\Cal C$ is compact, and from 476A(b-ii) that
$\{F:\int_F\chi B(\tbf{0},M)d\mu\ge\mu E\}$ is closed;
also $\{F:F\subseteq B(\tbf{0},M)\}$ is closed.
Let $\sigma$ be the metric on $\BbbR^r\times\BbbR^r$
defined by setting $\sigma((x,y),(x',y'))=\max(\|x-x'\|,\|y-y'\|)$, and
$\nu$ the indefinite-integral measure over $\lambda$ defined by
the function $(x,y)\mapsto\|x-y\|$.   Then

\Centerline{$U(F,\epsilon)\times U(F,\epsilon)
=\{(x,y):\sigma((x,y),F\times F)<\epsilon\}=U(F\times F,\epsilon;\sigma)$}

\noindent for $F\in\Cal C$ and $\epsilon>0$.   Now, writing $\Cal C_2$ for
the family of closed sets in $\BbbR^r\times\BbbR^r$ with its Fell topology,
we know that

\inset{$F\mapsto F\times F:\Cal C\to\Cal C_2$
is continuous, by 4A2Td,

$E\mapsto\nu U(E,\epsilon;\sigma):\Cal C_2\to\Bbb R$ is
lower semi-continuous, by 476B;}

\noindent so $F\mapsto\nu(U(F,\epsilon)\times U(F,\epsilon))$ is
lower semi-continuous, and $\{F:
\int_{U(F,\epsilon)\times U(F,\epsilon)}\|x-y\|\lambda(d(x,y))\le\beta\}$
is closed.   Putting these together, $\Cal F$ is a closed subset of
$\Cal C$ and is compact.\ \Qed

\medskip

{\bf (b)} Since $E\in\Cal F$, $\Cal F$ is not empty.   By 476A(b-ii),
there is an $F_0\in\Cal F$ such that
$\int_{F_0}M-\|x\|\mu(dx)\ge\int_FM-\|x\|\mu(dx)$ for every
$F\in\Cal F$.   Let $F_1\subseteq F_0$ be a closed self-supporting set
of the same measure;  then $U(F_1,\epsilon)\subseteq U(F_0,\epsilon)$,
so $\int_{U(F_1,\epsilon)\times U(F_1,\epsilon)}\|x-y\|\lambda(d(x,y))
\le\beta$ and $F_1\in\Cal F$;  also

\Centerline{$\int_{F_1}M-\|x\|\mu(dx)
=\int_{F_0}M-\|x\|\mu(dx)\ge\int_FM-\|x\|\mu(dx)$}

\noindent for every $F\in\Cal F$.

Now $F_1$ is a ball with centre $\tbf{0}$.   \Prf\Quer\ Suppose, if
possible, otherwise.   Then (just as in the proof of 476F) there are
$x_1\in F_1$ and $x_0\in\BbbR^r\setminus F_1$ such that
$\|x_1\|>\|x_0\|$.   Once again, set $e=\Bover1{\|x_0-x_1\|}(x_0-x_1)$ and
$\alpha=\Bover12\varinnerprod{e}{(x_0+x_1)}<0$.
Define $R=R_{e\alpha}:\BbbR^r\to\BbbR^r$ and $\psi=\psi_{e\alpha}$ as in
476D.   Set $F=\psi(F_1)$.   Then $F$ is closed and
$\mu F=\mu F_1\ge\mu E$ and
$U(F,\epsilon)\subseteq\psi(U(F_1,\epsilon))$.   So

$$\eqalignno{\int_{U(F,\epsilon)\times U(F,\epsilon)}
  \|x-y\|\lambda(d(x,y))
&\le\int_{\psi(U(F_1,\epsilon))\times\psi(U(F_1,\epsilon))}
  \|x-y\|\lambda(d(x,y))\cr
&\le\int_{U(F_1,\epsilon)\times U(F_1,\epsilon)}
  \|x-y\|\lambda(d(x,y))\cr
\displaycause{476Ed}
&\le\beta.\cr}$$

\noindent This means that $F\in\Cal F$.   Accordingly
$\int_{F_1}M-\|x\|\mu(dx)\ge\int_FM-\|x\|\mu(dx)$;  since
$\mu F=\mu F_1$, $\int_{F_1}\|x\|\mu(dx)\le\int_F\|x\|\mu(dx)$.
By 476Ec,
$G=\{x:x\in F_1,\,\varinnerprod{x}{e}<\alpha,\,R(x)\notin F_1\}$
must be negligible.   But $G$ contains $x_1$ and is relatively open in
$F_1$, and $F_1$ is supposed to be self-supporting;  so this is
impossible.\ \Bang\Qed

\medskip

{\bf (c)} Since $\mu F_1\ge\mu E$, $F_1\supseteq B_E$, and

$$\eqalign{\int_{B_E\times B_E}\|x-y\|\lambda(d(x,y))
&\le\int_{U(F_1,\epsilon)\times U(F_1,\epsilon)}
  \|x-y\|\lambda(d(x,y))\cr
&\le\beta
=\int_{U(E,\epsilon)\times U(E,\epsilon)}
  \|x-y\|\lambda(d(x,y)).\cr}$$

\noindent At this point, recall that $\epsilon$ was arbitrary.   Since
$E$ is compact,

$$\eqalign{\int_{E\times E}\|x-y\|\lambda(d(x,y))
&=\inf_{\epsilon>0}\int_{U(E,\epsilon)\times U(E,\epsilon)}
  \|x-y\|\lambda(d(x,y))\cr
&\ge\int_{B_E\times B_E}\|x-y\|\lambda(d(x,y)).\cr}$$

\medskip

{\bf (d)} Thus the result is proved for non-empty compact sets $E$.   In
general, given a measurable set $E$ of finite measure, then if $E$ is
negligible the result is trivial;  and otherwise, writing $\delta$ for
the radius of $B_E$,

$$\eqalign{\int_{B_E\times B_E}\|x-y\|\lambda(d(x,y))
&=\sup_{\alpha<\delta}\int_{B(\tbf{0},\alpha)\times B(\tbf{0},\alpha)}
  \|x-y\|\lambda(d(x,y))\cr
&\le\sup_{K\subseteq E\text{ is compact}}
  \int_{B_K\times B_K}\|x-y\|\lambda(d(x,y))\cr
&\le\sup_{K\subseteq E\text{ is compact}}
  \int_{K\times K}\|x-y\|\lambda(d(x,y))\cr
&=\int_{E\times E}\|x-y\|\lambda(d(x,y)),\cr}$$

\noindent so the proof is complete.
}%end of proof of 476G

\leader{476H}{The Isoperimetric Theorem} Let $r\ge 1$ be an integer, and
let $\mu$ be Lebesgue measure on $\BbbR^r$.   If $E\subseteq\BbbR^r$ is
a measurable set of finite measure, then $\per E\ge\per B_E$, where
$B_E$ is the closed ball with centre $\tbf{0}$ and the same measure as
$E$\cmmnt{, while $\per E$ is the perimeter of $E$ as defined in
474D}.

\proof{{\bf (a)} Suppose to begin with that $E\subseteq B(\tbf{0},M)$,
where $M\ge 0$, and that $\per E<\infty$.   Let $\Cal F$ be the family
of measurable sets $F\subseteq\BbbR^r$ such that
$F\setminus B(\tbf{0},M)$ is negligible, $\mu F\ge\mu E$ and
$\per F\le\per E$, with
the topology of convergence in measure (474T).   Then

$$\eqalign{\Cal F
&=\{F:\per F\le\per E,\,\mu(F\cap B(\tbf{0},M))\ge\mu E,\cr
&\hskip10em
\mu(F\cap B(\tbf{0},\alpha))\le\mu(F\cap B(\tbf{0},M))
  \text{ for every }\alpha\ge 0\}\cr}$$

\noindent is a closed subset of $\{F:\per F\le\per E\}$, which is
compact (474Tb), so $\Cal F$ is compact.   For $F\in\Cal F$ set
$h(F)=\int_F\|x\|\mu(dx)$;  then
$|h(F)-h(F')|\le M\mu((F\symmdiff F')\cap B(\tbf{0},M))$ for all $F$,
$F'\in\Cal F$, so $h$ is continuous.   There is therefore an
$F_0\in\Cal F$ such that $h(F_0)\le h(F)$ for every $F\in\Cal F$.   Set
$F_1=\clstar F_0$, so that $F_1\symmdiff F_0$ is negligible (475Cg),
$\per F_1=\per F_0$ (474F),
$F_1\in\Cal F$, $F_1\subseteq B(\tbf{0},M)$ and $h(F_1)=h(F_0)$.

\medskip

{\bf (b)} Writing $\delta=\sup_{x\in F_1}\|x\|$, we have
$U(\tbf{0},\delta)\subseteq F_1$.   \Prf\Quer\ Otherwise, there are
$x_0\in\BbbR^r\setminus F_1$ and $x_1\in F_1$ such that
$\|x_0\|<\|x_1\|$.   Set $e=\Bover1{\|x_0-x_1\|}(x_0-x_1)$,
$\alpha=\Bover12\varinnerprod{e}{(x_0+x_1)}<0$,
$R=R_{e\alpha}:\BbbR^r\to\BbbR^r$ and $\psi=\psi_{e\alpha}$.
As before, $\varinnerprod{e}{x_1}<\alpha<\varinnerprod{e}{x_0}$,
and $R(x_0)=x_1$ so $R[B(x_0,\eta)]=B(x_1,\eta)$ for every $\eta>0$.
Set $F=\psi(F_1)$ and let $\phi=\phi_{F_1}:F_1\to F$ be the function
described in 476E.   Then $\|\phi(x)\|\le\|x\|$ for every $x\in F_1$
(476Ea).   In particular, $F=\phi[F_1]\subseteq B(\tbf{0},M)$.   Now
$F$ is measurable and $\mu F=\mu F_1\ge\mu E$, by 476Eb.   Also, writing
$\nu$ for normalized $(r-1)$-dimensional Hausdorff measure on $\BbbR^r$,

\Centerline{$\per F=\nu(\partstar F)\le\nu(\partstar F_1)
=\per F_1\le\per E$,}

\noindent by 475Mb and 476Ee.   So $F\in\Cal F$, and

\Centerline{$\int_F\|x\|\mu(dx)\ge\int_{F_0}\|x\|\mu(dx)
=\int_{F_1}\|x\|\mu(dx)$.}

\noindent By 476Ec,
$G=\{x:x\in F_1,\,\varinnerprod{x}{e}<\alpha,\,R(x)\notin F_1\}$ is
negligible.   But now consider $G\cap U(x_1,\eta)$ for small $\eta>0$.
Since $x_1$ belongs to $F_1=\clstar F_0=\clstar F_1$, but $x_0$ does
not,

\Centerline{$\limsup_{\eta\downarrow 0}
\Bover{\mu(F_1\cap B(x_1,\eta))}{\mu B(x_1,\eta)}
>0
=\limsup_{\eta\downarrow 0}
\Bover{\mu(F_1\cap B(x_0,\eta))}{\mu B(x_0,\eta)}$.}

\noindent There must therefore be some $\eta>0$ such that
$\eta<\bover12\|x_1-x_0\|$ and
$\mu(F_1\cap B(x_0,\eta))<\mu(F_1\cap B(x_1,\eta))$.   In this case,
however,
$G\supseteq F_1\cap B(x_1,\eta)\setminus R[F_1\cap B(x_0,\eta)]$ has
measure at least $\mu(F_1\cap B(x_1,\eta))-\mu(F_1\cap B(x_0,\eta))>0$,
which is impossible.\ \Bang\Qed

\medskip

{\bf (c)} Thus
$U(\tbf{0},\delta)\subseteq F_1\subseteq B(\tbf{0},\delta)$ and
$\per F_1=\per B(\tbf{0},\delta)$.   Since $\mu F_1\ge\mu E$, the radius
of $B_E$ is at most $\delta$ and

\Centerline{$\per B_E\le\per B(\tbf{0},\delta)=\per F_1\le\per E$.}

\medskip

{\bf (d)} Thus the result is proved when $E$ is bounded and has finite
perimeter.   Of course it is trivial when $E$ has infinite perimeter.
Now suppose that $E$ is any measurable set with finite measure and
finite perimeter.   Set $E_{\alpha}=E\cap B(\tbf{0},\alpha)$ for
$\alpha\ge 0$;  then
$\per E=\liminf_{\alpha\to\infty}\per E_{\alpha}$ (475Mc, 475Xk).
%probably lim_{\alpha\to\infty}
By (a)-(c), $\per E_{\alpha}\ge\per B_{E_{\alpha}}$ for every $\alpha\ge 0$;  
since $\per B_{E_{\alpha}}\to\per B_E$ as $\alpha\to\infty$,
$\per E\ge\per B_E$ in this case also.
}%end of proof of 476H

\leader{476I}{Spheres in inner product spaces} For the rest of the
section I will use the following notation.
Let $X$ be a (real) inner product space.\cmmnt{ Then} $S_X$ will be
the unit
sphere $\{x:x\in X,\,\|x\|=1\}$.   Let $H_X$ be the isometry group of
$S_X$ with its topology of pointwise convergence\cmmnt{ (441G)}.

A {\bf cap} in $S_X$ will be a set of the form
$\{x:x\in S_X,\,\innerprod{x}{e}\ge\alpha\}$ where $e\in S_X$ and
$-1\le\alpha\le 1$.

When $X$ is finite-dimensional, it is isomorphic\cmmnt{, as inner
product
space,} to $\BbbR^r$, where $r=\dim X$\cmmnt{ (4A4Je)}.   If $r\ge 1$,
$S_X$\cmmnt{ is non-empty and
compact, so} has a unique $H_X$-invariant Radon probability measure
$\nu_X$, which is strictly positive\cmmnt{ (476C)}.   If $r\ge 1$ is an
integer, \cmmnt{we know that} the $(r-1)$-dimensional Hausdorff
measure of the
sphere $S_{\Bbb R^r}$ is finite and non-zero\cmmnt{ (265F)}.
\cmmnt{Since Hausdorff
measures are invariant under isometries (471J), and are
quasi-Radon measures when totally finite (471Dh),} $(r-1)$-dimensional
Hausdorff
measure on $S_{\Bbb R^r}$ is a multiple of the normalized invariant
measure $\nu_{\Bbb R^r}$\cmmnt{, by 476C}.   The same is therefore true
in any
$r$-dimensional inner product space.

\leader{476J}{Lemma} Let $X$ be a real inner product space and
$f\in H_X$.
Then $\innerprod{f(x)}{f(y)}=\innerprod{x}{y}$ for all $x$, $y\in S_X$.
\cmmnt{Consequently} $f(\alpha x+\beta y)=\alpha f(x)+\beta f(y)$
whenever $x$, $y\in S_X$ and $\alpha$, $\beta\in\Bbb R$ are such that
$\alpha x+\beta y\in S_X$.

\proof{{\bf (a)} We have

\Centerline{$\rho(x,y)^2=\innerprod{x-y}{x-y}
=\innerprod{x}{x}-2\innerprod{x}{y}+\innerprod{y}{y}
=2-2\innerprod{x}{y}$,}

\noindent so

\Centerline{$\innerprod{x}{y}=1-\Bover12\rho(x,y)^2
=1-\Bover12\rho(f(x),f(y))^2=\innerprod{f(x)}{f(y)}$.}

\wheader{476J}{6}{2}{2}{72pt}

{\bf (b)}

$$\eqalign{\|f(\alpha x&+\beta y)-\alpha f(x)-\beta f(y)\|^2\cr
&=\innerprod{f(\alpha x+\beta y)-\alpha f(x)-\beta f(y)}
   {f(\alpha x+\beta y)-\alpha f(x)-\beta f(y)}\cr
&=1+\alpha^2+\beta^2-2\alpha\innerprod{f(\alpha x+\beta y)}{f(x)}\cr&
\mskip100mu
  -2\beta\innerprod{f(\alpha x+\beta y)}{f(y)}
  +2\alpha\beta\innerprod{f(x)}{f(y)}\cr
&=1+\alpha^2+\beta^2-2\alpha\innerprod{\alpha x+\beta y}{x}
  -2\beta\innerprod{\alpha x+\beta y}{y}
  +2\alpha\beta\innerprod{x}{y}\cr
&=\|(\alpha x+\beta y)-\alpha x-\beta y\|^2
=0.\cr}$$

}%end of proof of 476J

\leader{476K}{}\cmmnt{ I give a theorem on concentration of measure on
the sphere corresponding to 476F.

\medskip

\noindent}{\bf Theorem} Let $X$ be a finite-dimensional inner product
space of dimension at least $2$, $S_X$ its unit sphere and $\nu_X$ the
invariant Radon probability measure on $S_X$.   For a non-empty set
$A\subseteq S_X$ and $\epsilon>0$, write
$U(A,\epsilon)=\{x:\rho(x,A)<\epsilon\}$, where $\rho$ is
the usual metric of $X$.   Then there is a cap $C\subseteq S_X$ such
that $\nu_XC=\nu_X^*A$, and
$\nu_X(S_X\cap U(A,\epsilon))\ge\nu_X(S_X\cap U(C,\epsilon))$
for any such $C$ and every $\epsilon>0$.

\proof{ In order to apply the results of 476D-476E directly, and
simplify some of the formulae slightly, it will be helpful to write
$\nu$ for the Radon measure on $X$ defined by setting
$\nu E=\nu_X(E\cap S_X)$ whenever this is defined.   By 214Cd, $\nu^*$
agrees with $\nu_X^*$ on $\Cal PS_X$.

\medskip

{\bf (a)} The first step is to check that there is a cap $C$ of $S_X$
such that $\nu C=\nu^*A$.   \Prf\ Take any $e_0\in S_X$, and set
$C_{\alpha}=\{x:x\in S_X,\,\innerprod{x}{e_0}\ge\alpha\}$ for
$\alpha\in[-1,1]$.
$\nu C_{\alpha}$ is defined for every $\alpha\in\Bbb R$
because every $C_{\alpha}$ is closed and $\nu$ is a topological
measure.   Now examine the formulae of 265F.   We can identify $X$
with $\BbbR^{r+1}$ where $r+1=\dim X$;  do this in such a way that $e_0$
corresponds to the unit vector $(0,\ldots,0,1)$.   We have a
parametrization
$\phi_r:D_r\to S_X$, where $D_r$ is a Borel subset of $\BbbR^r$ with
interior $\ooint{-\pi,\pi}\times\ooint{0,\pi}^{r-1}$ and $\phi_r$ is
differentiable with continuous derivative.   Moreover, if
$x=(\xi_1,\ldots,\xi_r)\in D_r$, then
$\varinnerprod{\phi_r(x)}{e_0}=\cos\xi_r$, and the Jacobian $J_r$ of
$\phi_r$ is bounded by $1$ and never zero on $\interior D_r$.   Finally,
the boundary $\partial D_r$ is negligible.   What this means is that
$\nu_rC_{\alpha}=\int_{E_{\alpha}}J_rd\mu_r$, where $\mu_r$ is Lebesgue
measure on $\BbbR^r$, $\nu_r$ is normalized Hausdorff $r$-dimensional
measure on $\BbbR^{r+1}$, and
$E_{\alpha}=\{x:x\in D_r,\,\cos\xi_r\ge\alpha\}$.   So if
$-1\le\alpha\le\beta\le 1$ then

\Centerline{$\nu_rC_{\alpha}-\nu_rC_{\beta}
\le\mu_r(E_{\alpha}\setminus E_{\beta})
\le 2\pi^{r-1}(\arccos\alpha-\arccos\beta)$;}

\noindent because $\arccos$ is continuous, so is
$\alpha\mapsto\nu_rC_{\alpha}$.   Also, if $\alpha<\beta$, then
$E_{\alpha}\setminus E_{\beta}$ is non-negligible, so
$\int_{E_{\alpha}\setminus E_{\beta}}J_rd\mu_r\ne 0$ and
$\nu_rC_{\alpha}>\nu_rC_{\beta}$.

This shows that $\alpha\mapsto\nu_rC_{\alpha}$ is continuous and
strictly decreasing;  since $\nu_r$ is just a multiple of $\nu$ on
$S_X$, the same is true of $\alpha\mapsto\nu C_{\alpha}$.

Since $\nu C_{-1}=\nu S_X=1$ and $\nu C_1=\nu\{e_0\}=0$, the
Intermediate
Value Theorem tells us that there is a unique $\alpha$ such that
$\nu C_{\alpha}=\nu^*A$, and we can set $C=C_{\alpha}$.\ \Qed

\medskip

{\bf (b)} Now take any non-empty set $A\subseteq S_X$ and any
$\epsilon>0$, and set $\gamma=\nu^*A$, $\beta=\nu U(A,\epsilon)$.
Let $C$ be a cap such that $\nu^*A=\nu C$;  let $e_0$ be the centre of
$C$.   Consider the family

\Centerline{$\Cal F
=\{F:F\in\Cal C$, $F\subseteq S_X$,
$\nu F\ge\gamma$, $\nu U(F,\epsilon)\le\beta\}$,}

\noindent where $\Cal C$ is the family of closed subsets of $X$ with its
Fell topology.
Because $U(\overline{A},\epsilon)=U(A,\epsilon)$,
$\overline{A}\in\Cal F$ and $\Cal F$ is non-empty.   By
476A(b-i) and 476B, $\Cal F$ is closed in
$\Cal C$, therefore compact, by 4A2T(b-iii) once more.   Next, the function

\Centerline{$F\mapsto\int_F\max(0,1+\innerprod{x}{e_0})\nu(dx):
\Cal C\to\coint{0,\infty}$}

\noindent is upper semi-continuous, by 476A(b-ii).
It therefore attains its
supremum on $\Cal F$ at some $F_0\in\Cal F$.   Let $F_1\subseteq F_0$ be
a self-supporting closed set with the same measure as $F_0$;  then
$F_1\in\Cal F$ and $\int_F(1+\innerprod{x}{e_0})\nu (dx)
\le\int_{F_1}(1+\innerprod{x}{e_0})\nu(dx)$ for every $F\in\Cal F$.

\medskip

{\bf (c)} $F_1$ is a cap with centre $e_0$.   \Prf\Quer\ Otherwise,
there are $x_0\in S_X\setminus F_1$ and
$x_1\in F_1$ such that $\innerprod{x_0}{e_0}>\innerprod{x_1}{e_0}$.
Set $e=\Bover1{\|x_0-x_1\|}(x_0-x_1)$.   Then $e\in S_X$ and
$\innerprod{e}{e_0}>0$.   Set $R=R_{e0}$ and $\psi=\psi_{e0}$ as defined
in 476D;  write $F$ for $\psi(F_1)$.   Note that
$\innerprod{x_0+x_1}{x_0-x_1}=\|x_0\|^2-\|x_1\|^2=0$, so
$\innerprod{x_0+x_1}{e}=0$ and $R(x_1)=x_0$.   Also $R[S_X]=S_X$, so
$\nu$ is $R$-invariant, because $\nu$ is a multiple of Hausdorff
$(r-1)$-dimensional measure on $S_X$ and must be invariant under
isometries of $S_X$.

We have $\nu F=\nu F_1\ge\gamma$, by 476Eb, and

\Centerline{$\nu U(F,\epsilon)\le\nu\psi(U(F_1,\epsilon))
=\nu U(F_1,\epsilon)\le\nu U(F_0,\epsilon)\le\beta$}

\noindent by 476Dd.   So $F\in\Cal F$.   But consider the standard
bijection $\phi=\phi_{F_1}:F_1\to F$ as defined in 476E.
We have

\Centerline{$\int_{F_1}(1+\innerprod{\phi(x)}{e_0})\nu(dx)
=\int_F(1+\innerprod{x}{e_0})\nu(dx)
\le\int_{F_1}(1+\innerprod{x}{e_0})\nu(dx)$.}

\noindent If we examine the definition of $\phi$, we see that
$\phi(x)\ne x$ only when $\innerprod{x}{e}<0$ and $\phi(x)=R(x)$, so
that in this case $\phi(x)-x$ is a positive multiple of $e$ and
$\innerprod{\phi(x)}{e_0}>\innerprod{x}{e_0}$.   So
$G=\{x:x\in F_1,\,\innerprod{x}{e}<0,\,R(x)\notin F_1\}$ must be
$\nu$-negligible.   But $G$ includes a relative neighbourhood of $x_1$
in $F_1$ and $F_1$ is supposed to be self-supporting for $\nu$, so this
is impossible.\ \Bang\Qed

\medskip

{\bf (d)} Now $\nu^*A=\gamma\le\nu F_1$, so $C\subseteq F_1$ and

\Centerline{$\nu U(A,\epsilon)=\beta
\ge\nu U(F_1,\epsilon)\ge\nu U(C,\epsilon)$,}

\noindent as claimed.
}%end of proof of 476K

\leader{476L}{Corollary} For any $\epsilon>0$, there is an $r_0\ge 1$
such that whenever $X$ is a finite-dimensional inner product space of
dimension at least $r_0$, $A_1$, $A_2\subseteq S_X$ and
$\min(\nu_X^*A_1,\nu_X^*A_2)\ge\epsilon$, then there are $x\in A_1$,
$y\in A_2$ such that $\|x-y\|\le\epsilon$.

\proof{ Take $r_0\ge 2$ such that $r_0\epsilon^3>2$.   Suppose that
$\dim X=r\ge r_0$.   Fix $e_0\in S_X$.   We need an estimate of
$\nu_XC_{\epsilon/2}$, where
$C_{\epsilon/2}=\{x:x\in S_X,\,\innerprod{x}{e_0}\ge\epsilon/2\}$ as in
476K.   To get this, let $e_1,\ldots,e_{r-1}$ be such that
$e_0,\ldots,e_{r-1}$ is an orthonormal basis of $X$ (4A4Kc).   For each
$i<r$, there is an $f\in H_X$ such that $f(e_i)=e_0$, so that
$\innerprod{x}{e_i}=\innerprod{f(x)}{e_0}$ for every $x$ (476J),
and

\Centerline{$\int\innerprod{x}{e_i}^2\nu_X(dx)
=\int\innerprod{f(x)}{e_0}^2\nu_X(dx)
=\int\innerprod{x}{e_0}^2\nu_X(dx)$,}

\noindent because $f:S_X\to S_X$ is \imp\ for $\nu_X$.

Accordingly

$$\eqalign{\nu_XC_{\epsilon/2}
&=\Bover12\nu_X\{x:x\in S_X,\,|\innerprod{x}{e_0}|\ge\epsilon/2\}
\le\Bover2{\epsilon^2}\int_{S_X}\innerprod{x}{e_0}^2\nu_X(dx)\cr
&<r\epsilon\int_{S_X}\innerprod{x}{e_0}^2\nu_X(dx)
=\epsilon\sum_{i=0}^{r-1}\int_{S_X}\innerprod{x}{e_i}^2\nu_X(dx)\cr
&=\epsilon\int_{S_X}\sum_{i=0}^{r-1}\innerprod{x}{e_i}^2\nu_X(dx)
=\epsilon
\le\nu_X^*A_1.\cr}$$

\noindent So, taking $C$ to be the cap of $S_X$ with centre $e_0$ and
measure $\nu_X^*A_1$, $C=C_{\alpha}$ where $\alpha<\bover12\epsilon$, and

\Centerline{$\nu_X(S_X\cap U(A_1,\bover12\epsilon))
\ge\nu_X(S_X\cap U(C_{\alpha},\bover12\epsilon))
\ge\nu_XC_{\alpha-\epsilon/2}>\Bover12$}

\noindent by 476K.
Similarly, $\nu_X(S_X\cap U(A_2,\bover12\epsilon))>\bover12$
and there must be some
$z\in S_X\cap U(A_1,\bover12\epsilon)\cap U(A_2,\bover12\epsilon)$.
Take $x\in A_1$ and $y\in A_2$ such that $\|x-z\|<\bover12\epsilon$ and
$\|y-z\|<\bover12\epsilon$;  then $\|x-y\|\le\epsilon$, as required.
}%end of proof of 476L

\exercises{\leader{476X}{Basic exercises (a)}
%\spheader 476Xa
Let $X$ be a topological space, $\Cal C$ the set of
closed subsets of $X$, $\mu$ a topological measure on $X$ and $f$ a
$\mu$-integrable real-valued function;  set $\phi(F)=\int_Ffd\mu$ for
$F\in\Cal C$.   (i) Show that if {\it either} $\mu$ is inner regular with
respect to the closed sets and $\Cal C$ is given its Vietoris topology
{\it or} $\mu$ is tight and $\Cal C$ is given its Fell topology,
then $\phi$ is Borel measurable.   (ii) Show that if $X$ is
metrizable and $\Cal C\setminus\{\emptyset\}$ is given an appropriate
Hausdorff metric, then $\phi\restr\Cal C\setminus\{\emptyset\}$ is Borel
measurable.
%476A

\spheader 476Xb In the context of 476D, show that
$\diam\psi_{e\alpha}(A)\le\diam A$ for all $A$, $e$ and $\alpha$.
%476D

\sqheader 476Xc Find an argument along the lines of those in 476F and
476G to prove 264H.
%?? calculation of volume of ball
\Hint{476Xb.}
%476G

\sqheader 476Xd Let $X$ be an inner product space and $S_X$ its unit
sphere.   Show that every isometry $f:S_X\to S_X$ extends uniquely to an
isometry $T_f:X\to X$ which is a linear operator.   \Hint{first check
the cases in which $\dim X\le 2$.}   Show that $f$ is surjective iff
$T_f$ is, so that we have a natural isomorphism between the isometry
group of $S_X$ and the group of invertible isometric linear operators.
Show that this isomorphism is a homeomorphism for the topologies of
pointwise convergence.
%476J

\spheader 476Xe Let $X$ be a finite-dimensional inner product space,
$\nu_X$ the invariant Radon probability measure on the sphere $S_X$, and
$E\in\dom\nu_X$;  let $C\subseteq S_X$ be a cap with the same measure as
$E$, and let $\lambda$ be the product measure of $\nu_X$ with itself on
$S_X\times S_X$.    Show that
$\int_{C\times C}\|x-y\|\lambda(d(x,y))
\le\int_{E\times E}\|x-y\|\lambda(d(x,y))$.
%476G, 476K

\spheader 476Xf Let $X$ be a finite-dimensional inner product space and
$\nu_X$ the invariant Radon probability measure on the sphere $S_X$.
(i) Without appealing to the formulae in \S265, show that
$\nu_X(S_X\cap H)=0$ whenever $H\subseteq X$ is a proper affine
subspace.   \Hint{induce on
$\dim H$.}   (ii) Use this to prove that if $e\in S_X$ then
$\alpha\mapsto\nu_X\{x:\innerprod{x}{e}\ge\alpha\}$ is continuous.
%476K

\leader{476Y}{Further exercises (a)}
%\spheader 476Ya
Let $X$ be a compact metric space and $G$ its isometry
group.   Suppose that $H\subseteq G$ is a subgroup such that the action
of $H$ on $X$ is transitive.   Show that $X$ has a unique $H$-invariant
Radon probability measure which is also $G$-invariant.
%476C mt47bits

\spheader 476Yb
Let $r\ge 1$ be an integer, and $g\in C_0(\BbbR^r)$ a non-negative
$\gamma$-Lipschitz function, where $\gamma\ge 0$.   Let $F$ be the
set of non-negative $\gamma$-Lipschitz functions $f\in C(\BbbR^r)$ such
that $f$ has the same decreasing rearrangement as $g$ with respect to
Lebesgue measure $\mu$ on $\BbbR^r$ (\S373)
and $\int\phi(\grad f)d\mu\le\int\phi(\grad g)d\mu$ for every convex function
$\phi:\BbbR^r\to\coint{0,\infty}$.   (i) Show that $F$ is
compact for the topology of pointwise convergence.   %\Hint{475Ye.}
(ii) Show that there
is an $f\in F$ such that $f(x)\ge f(y)$ whenever $\|x\|\le\|y\|$.
\Hint{parts (a) and (b-i) of the proof of 479V.}
}%end of exercises

\endnotes{
\Notesheader{476}
The main theorems here (476F-476H, %476F 476G 476H
476K), like 264H, are all
`classical';  they go back to the roots of geometric measure theory, and
the contribution of the twentieth century was to extend the classes
of sets for which balls or caps provide the bounding examples.   It is
very striking that they can all be proved with the same
tools (see 476Xc).   Of course I should remark that the Compactness
Theorem (474T) lies at a much deeper level than the rest of the ideas
here.   (The proof of 474T relies on the distributional definition of
`perimeter' in 474D, while the arguments of 476Ee and 476H work with the
Hausdorff measures of essential boundaries;  so that we can join these
ideas together only after proving all the principal theorems of
\S\S472-475.)   So while `Steiner symmetrization' (264H) and
`concentration by partial reflection' (476D) are natural companions,
476H is essentially harder than the other results.

In all the theorems here, as in 264H, I have been content to show that a
ball or a cap is an optimum for whatever inequality is being considered.
I have not examined the question of whether, and in what sense, 
the optimum is unique.   It seems that this requires deeper analysis.
}%end of notes

\discrpage

%The formula $r_0\epsilon^3\ge 2$ in the proof of 476L can be greatly
%improved;  see {\smc Milman \& Schechtman 86}, 2.2.

\frfilename{mt477.tex}
\versiondate{4.1.08/2.1.10}
\copyrightdate{2008}

\def\hp{\mathop{\text{hp}}\nolimits}

\def\chaptername{Geometric measure theory}
\def\sectionname{Brownian motion}

\newsection{477}

I presented \S455 with an extraordinary omission:  the leading example of a
L\'evy process, and the inspiration for the whole project, was
relegated to an anonymous example (455Xg).   In this section I will take
the subject up again.   The theorem that the sum of independent normally
distributed random variables is again normally distributed (274B), when
translated into the language of this volume, tells us that we have a family
$\langle\lambda_t\rangle_{t>0}$ of centered normal distributions such that
$\lambda_{s+t}=\lambda_s*\lambda_t$ for all $s$, $t>0$.   Consequently we
have a corresponding example of a L\'evy process on $\Bbb R$, and this is
the process which we call `Brownian motion' (477A).   This is special in
innumerable ways, but one of them is central:  we can represent it in such
a way that sample paths are continuous (477B), that is, as a Radon measure
on the space of continuous paths starting at $0$.   In this form, it also
appears as a limit, for the narrow topology, of interpolations of
random walks (477C).

For the geometric ideas of \S479, we need Brownian motion in three
dimensions;  the $r$-dimensional theory of 477D-477G gives no new
difficulties.   The simplest expression of Brownian motion in $\BbbR^r$ is
just to take a product measure (477Da), but in order to apply the results
of \S455, and match the construction with the ideas of \S456, a fair bit of
explanation is necessary.   The geometric properties of Brownian motion
begin with the invariant transformations of 477E.   As for all L\'evy
processes, we have a strong Markov property, and Theorem 455U translates
easily into the new formulation (477G), as does the theory of hitting
times (477I).   I conclude with a classic result on maximal values
which will be useful later (477J), and with proofs that
almost all Brownian paths are nowhere differentiable (477K) and have
zero two-dimensional Hausdorff measure (477L).

\leader{477A}{Brownian motion:  Theorem} There are a probability space
$(\Omega,\Sigma,\mu)$ and a family $\langle X_t\rangle_{t\ge 0}$ of
real-valued random variables on $\Omega$ such that

(i) $X_0=0$ almost everywhere;

(ii) whenever $0\le s<t$ then $X_t-X_s$ is normally
distributed with expectation $0$ and variance $t-s$;

(iii) $\langle X_t\rangle_{t\ge 0}$ has independent increments.

\ifwithproofs\medskip

\noindent{\bf First proof} In 455P, take $U=\Bbb R$ and
$\lambda_t$, for $t>0$, to be the distribution of a normal random
variable with expectation $0$ and variance $t$;  that is, the distribution
with probability density function
$x\mapsto\Bover1{\sqrt{2\pi t}}e^{-x^2/2t}$.   By 272T\formerly{2{}72S},
$\lambda_{s+t}=\lambda_s*\lambda_t$ for all $s$, $t>0$.
If $\epsilon>0$, then

\def\shortstrut{\hbox{\vrule height7.5pt depth2.5pt width0pt}}

\Centerline{$\lim_{t\downarrow 0}\lambda_t\ooint{-\epsilon,\epsilon}
=\lim_{t\downarrow 0}
\lambda_1\bigl]-\Bover{\epsilon}{\sqrt t},\Bover{\epsilon}{\sqrt t}\bigr[
=1$,}

\noindent so $\langle\lambda_t\rangle_{t>0}$ satisfies the conditions of
455P.   Accordingly we have a probability measure $\hat\mu$ on
$\Omega=\BbbR^{\coint{0,\infty}}$ for which, setting
$X_t(\omega)=\omega(t)$, $\langle X_t\rangle_{t\ge 0}$ has the required
properties, as noted in 455Q-455R.

\medskip

\noindent{\bf Second proof} Let $\mu_L$ be Lebesgue measure on $\Bbb R$,
and for $t\ge 0$ set $u_t=\chi[0,t]^{\ssbullet}$ in $L^2(\mu_L)$, so that
$\innerprod{u_s}{u_t}=\min(s,t)$ for $s$, $t\ge 0$.   By 456C, there is a
centered Gaussian distribution $\mu$ on $\BbbR^{\coint{0,\infty}}$ with
covariance matrix $\langle\min(s,t)\rangle_{s,t\ge 0}$.   Set
$X_t(x)=x(t)$ for $x\in\BbbR^{\coint{0,\infty}}$.   Then $X_0$ has
expectation and variance both $0$, that is, $X_0=0$ a.e.   If
$0\le s<t$, then $X_t-X_s$ is a linear combination of $X_s$ and $X_t$, so
is normally distributed with expectation $0$, and its variance is

\Centerline{$\Expn(X_t-X_s)^2
=\Expn(X_t)^2-2\Expn(X_t\times X_s)+\Expn(X_s)^2
=t-2s+s=t-s$.}

\noindent If $0\le t_0<\ldots<t_n$ and $Y_i=X_{t_{i+1}}-X_{t_i}$ for
$i<n$, then $\langle Y_i\rangle_{i<n}$ has a centered Gaussian
distribution, by 456Ba.   Also, if $i<j<n$, then

$$\eqalign{\Expn(Y_i\times Y_j)
&=\Expn(X_{t_{i+1}}\times X_{t_{j+1}})
  -\Expn(X_{t_{i+1}}\times X_{t_j})
  -\Expn(X_{t_i}\times X_{t_{j+1}})
  +\Expn(X_{t_i}\times X_{t_j})\cr
&=t_{i+1}-t_{i+1}-t_i+t_i
=0.\cr}$$

\noindent So 456E assures us that $\langle Y_i\rangle_{i<n}$ is
independent.

Thus $\langle X_t\rangle_{t\ge 0}$ satisfies the conditions required.
\else\fi %end of proof of 477A

\leader{477B}{}\cmmnt{ These constructions of Brownian motion are
sufficient to show that there is a process, satisfying the defining
conditions (i)-(iii), which can be studied with the tools of measure
theory.   From 455H we see that we have a Radon measure on the space
of \callal\ functions representing the process, and from 455P that we
have the option of moving to the \cadlag\ functions, with a corresponding
description of the strong Markov property in terms of \imp\ functions, as
in 455U.   But there is no hint yet of the most important property of
Brownian motion, that `sample paths are continuous'.   With some simple
inequalities from Chapter 27 and the ideas of 454Q-454S, %454Q 454R 454S
we can find a proof of this, as follows.

\medskip

\noindent}{\bf Theorem} Let
$\langle X_t\rangle_{t\ge 0}$ be as in 477A, and $\hat\mu$ the
distribution of the process $\langle X_t\rangle_{t\ge 0}$.
Let $C(\coint{0,\infty})_0$ be
the set of continuous functions
$\omega:\coint{0,\infty}\to\Bbb R$ such that $\omega(0)=0$.
Then $C(\coint{0,\infty})_0$ has full outer measure for
$\hat\mu$, and the subspace measure $\mu_W$ on $C(\coint{0,\infty})_0$
induced by $\hat\mu$
is a Radon measure when $C(\coint{0,\infty})_0$ is given the topology
$\frak T_c$ of uniform convergence on compact sets.

\proof{{\bf (a)} The main part of the argument here (down to
the end of (e)) is devoted to showing that
$\hat\mu^*C(\coint{0,\infty})=1$;
the result will then follow easily from 454Sb.

\medskip

{\bf (b)} \Quer\ Suppose, if possible, that
$\hat\mu^*C(\coint{0,\infty})<1$.
Then there is a non-negligible Baire set
$H\subseteq\BbbR^{\coint{0,\infty}}\setminus C(\coint{0,\infty})$.
There is a countable set $D\subseteq\coint{0,\infty}$ such that $H$ is
determined by coordinates in $D$ (4A3Nb);  we may suppose that $D$
includes $\Bbb Q\cap\coint{0,\infty}$.

\medskip

{\bf (c)} (The key.)  Let $q$, $q'$ be rational numbers such that
$0\le q<q'$, and $\epsilon>0$.   Then

\Centerline{$\Pr(\sup_{t\in D\cap[q,q']}|X_t-X_q|>\epsilon)
\le\Bover{18\sqrt{q'-q}}{\epsilon\sqrt{2\pi}}
   e^{-\epsilon^2/18(q'-q)}$.}

\noindent\Prf\ If $q=t_0<t_1<\ldots<t_n=q'$, set
$Y_i=X_{t_i}-X_{t_{i-1}}$ for $1\le i\le n$, so that
$X_{t_m}-X_q=\sum_{i=1}^mY_i$ for $1\le m\le n$, and $Y_1,\ldots,Y_n$
are independent.   By Etemadi's lemma (272V\formerly{2{}72U}),

$$\eqalignno{\Pr(\sup_{i\le n}|X_{t_i}-X_q|>\epsilon)
&\le 3\max_{i\le n}\Pr(|X_{t_i}-X_q|>\Bover13\epsilon)\cr
&=3\max_{1\le i\le n}\Pr(\Bover1{\sqrt{t_i-q}}|X_{t_i}-X_q|
                         >\Bover{\epsilon}{3\sqrt{t_i-q}})\cr
&=6\max_{1\le i\le n}\Bover1{\sqrt{2\pi}}
  \int_{\epsilon/3\sqrt{t_i-q}}^{\infty}e^{-x^2/2}dx\cr
\displaycause{because $\Bover1{\sqrt{t_i-q}}(X_{t_i}-X_q)$ is
standard normal}
&=\Bover6{\sqrt{2\pi}}
  \int_{\epsilon/3\sqrt{q'-q}}^{\infty}e^{-x^2/2}dx\cr
&\le\Bover{18\sqrt{q'-q}}{\epsilon\sqrt{2\pi}}
   e^{-\epsilon^2/18(q'-q)}\cr}$$

\noindent by 274Ma.   Thus if $I\subseteq[q,q']$ is any finite set
containing $q$ and $q'$,

\Centerline{$\Pr(\sup_{t\in I}|X_t-X_q|>\epsilon)
\le\Bover{18\sqrt{q'-q}}{\epsilon\sqrt{2\pi}}
   e^{-\epsilon^2/18(q'-q)}$.}

\noindent Taking $\sequencen{I_n}$ to be a non-decreasing sequence of
finite sets with union $D\cap[q,q']$, starting from $I_0=\{q,q'\}$, we
get

$$\eqalign{\Pr(\sup_{t\in D\cap[q,q']}|X_t-X_q|>\epsilon)
&=\lim_{n\to\infty}\Pr(\sup_{t\in I_n}|X_t-X_q|>\epsilon)\cr
&\le\Bover{18\sqrt{q'-q}}{\epsilon\sqrt{2\pi}}
   e^{-\epsilon^2/18(q'-q)},\cr}$$

\noindent as required.\ \Qed

\medskip

{\bf (d)} If $\epsilon>0$ and $n\ge 1$, then

$$\eqalign{\Pr(\text{there are }t,\,u\in D\cap[0,n]&\text{ such that }
|t-u|\le\Bover1{n^2}\text{ and }|X_t-X_u|>3\epsilon)\cr
&\le\Bover{18n^2}{\epsilon\sqrt{2\pi}}e^{-n^2\epsilon^2/18}.\cr}$$

\noindent\Prf\ Divide $[0,n]$ into $n^3$ intervals $[q_i,q_{i+1}]$ of
length $1/n^2$.   For each of these,

\Centerline{$\Pr(\sup_{t\in D\cap[q_i,q_{i+1}]}|X_t-X_{q_i}|>\epsilon)
\le\Bover{18}{n\epsilon\sqrt{2\pi}}e^{-n^2\epsilon^2/18}$.}

\noindent So

\Centerline{$\Pr(\text{there are }i<n^3,\,t\in D\cap[q_i,q_{i+1}]
\text{ such that }|X_t-X_{q_i}|>\epsilon)$}

\noindent is at most
$\Bover{18n^2}{\epsilon\sqrt{2\pi}}e^{-n^2\epsilon^2/18}$.

But if
$t$, $u\in[0,n]$ and $|t-u|\le 1/n^2$ and $|X_t-X_u|>3\epsilon$,
there must be an $i<n^3$ such that both $t$ and $u$ belong to
$[q_i,q_{i+2}]$, so that either there is a $t'\in D\cap[q_i,q_{i+1}]$
such that $|X_{t'}-X_{q_i}|>\epsilon$ or there is a
$t'\in D\cap[q_{i+1},q_{i+2}]$ such that
$|X_{t'}-X_{q_{i+1}}|>\epsilon$.   So

$$\eqalign{\Pr(&\text{there are }t,\,u\in D\cap[0,n]\text{ such that }
|t-u|\le\Bover1{n^2}\text{ and }|X_t-X_u|> 3\epsilon)\cr
&\quad\le\Pr(\text{there are }i<n^3,\,t\in D\cap[q_i,q_{i+1}]
\text{ such that }\text{ and }|X_t-X_{q_i}|>\epsilon)\cr
&\quad\le\Bover{18n^2}{\epsilon\sqrt{2\pi}}e^{-n^2\epsilon^2/18},\cr}$$

\noindent as required.\ \Qed

\medskip

{\bf (e)} So if we take $G_{\epsilon n}$ to be the Baire set

$$\eqalign{\{\omega:\omega\in\BbbR^{\coint{0,\infty}},
  \text{ there are }t,\,
  u\in D\cap[0,n]&\text{ such that }|t-u|\le\Bover1{n^2}\cr
&\text{ and }|\omega(t)-\omega(u)|> 3\epsilon\},\cr}$$

\noindent we have

\Centerline{$\hat\mu G_{\epsilon n}
\le\Bover{18n^2}{\epsilon\sqrt{2\pi}}e^{-n^2\epsilon^2/18}$,}

\ifdim\pagewidth>467pt\fontdimen3\tenrm=2pt\fi
\ifdim\pagewidth>467pt\fontdimen4\tenrm=1.67pt\fi
\noindent and $\lim_{n\to\infty}\hat\mu G_{\epsilon n}=0$.   We can
therefore find a strictly increasing sequence $\sequence{k}{n_k}$
in $\Bbb N$ such that $\sum_{k=1}^{\infty}\hat\mu(G_{1/k,n_k})<\hat\mu H$,
so that there is
an $\omega\in H\setminus\bigcup_{k\ge 1}G_{1/k,n_k}$.
\fontdimen3\tenrm=1.67pt
\fontdimen4\tenrm=1.11pt

What this means is that if $k\ge 1$ and $t$, $u\in D\cap[0,n_k]$ are
such that $|t-u|\le\Bover1{n_k^2}$, then
$|\omega(t)-\omega(u)|\le\Bover3k$.   Since
$n_k\to\infty$ as $k\to\infty$, there is a continuous function
$\omega':\coint{0,\infty}\to\Bbb R$ such
that $\omega'\restr D=\omega\restr D$.
But $H$ is determined by coordinates in
$D$, so $\omega'$ belongs to $H\cap C(\coint{0,\infty})$,
which is supposed to be empty.\ \Bang

\medskip

{\bf (f)} Thus $\hat\mu^*C(\coint{0,\infty})=1$.
Since $\hat\mu\{\omega:\omega(0)=0\}=1$,
$C(\coint{0,\infty})\setminus C(\coint{0,\infty})_0$
is $\hat\mu$-negligible and
$C(\coint{0,\infty})_0$ has full outer measure for $\hat\mu$.
By 454Sb, the subspace
measure $\hat\mu_C$ on $C(\coint{0,\infty})$ induced by $\hat\mu$ is a
Radon measure for $\frak T_c$;  now $C(\coint{0,\infty})_0$ is
$\hat\mu_C$-conegligible.   The subspace measure $\mu_W$ on
$C(\coint{0,\infty})_0$ induced by $\hat\mu$ is also the subspace measure
induced by $\hat\mu_C$, so is a Radon measure for the topology on
$C(\coint{0,\infty})_0$ induced by $\frak T_c$.
}%end of proof of 477B

\cmmnt{\medskip

\noindent{\bf Remark} We can put this together with the ideas of 455H.
Following the First Proof of 477A, and using 455Pc, we see that
there is a unique Radon
measure $\tilde\mu$ on $\BbbR^{\coint{0,\infty}}$ (for the topology
$\frak T_p$ of
pointwise convergence) extending $\hat\mu$.   The identity map
$\iota:C(\coint{0,\infty})_0\to\BbbR^{\coint{0,\infty}}$ is continuous for
$\frak T_c$ and $\frak T_p$, so the image measure $\mu_W\iota^{-1}$ is a
Radon measure on $\BbbR^{\coint{0,\infty}}$ (418I).   If
$E\subseteq\BbbR^{\coint{0,\infty}}$ is a Baire set, then

\Centerline{$\mu_W\iota^{-1}[E]=\mu_W(E\cap C(\coint{0,\infty})_0)
=\hat\mu E$,}

\noindent so $\mu_W\iota^{-1}$ agrees with $\tilde\mu$ on Baire sets,
and the two must be equal.
Now $C(\coint{0,\infty})_0$ is $\tilde\mu$-conegligible,
just because its complement has empty inverse image under $\iota$.
So $\mu_W$ is also the subspace measure on $C(\coint{0,\infty})_0$
induced by $\tilde\mu$.

Equally, since of course $C(\coint{0,\infty})_0$ is a subspace of the set
$\Cdlg$ of \cadlag\ functions from $\coint{0,\infty}$ to $\Bbb R$,
$\mu_W$ is the subspace measure induced by the measure $\ddot\mu$ of
Theorem 455O.
}%end of comment

\leader{*477C}{}\cmmnt{ I star the next theorem because it is very hard
work and will not be relied on later.
Nevertheless I think the statement, at
least, should be part of your general picture of Brownian motion.

\medskip

\noindent}{\bf Theorem} For $\alpha>0$, define
$f_{\alpha}:\BbbR^{\Bbb N}\to\Omega=C(\coint{0,\infty})_0$ by setting
$f_{\alpha}(z)(t)
=\sqrt\alpha(\sum_{i<n}z(i)+\Bover1{\alpha}(t-n\alpha)z(n))$
when $z\in\BbbR^{\Bbb N}$, $n\in\Bbb N$ and
$n\alpha\le t\le(n+1)\alpha$.   Give $\Omega$ its topology $\frak T_c$ of
uniform convergence on compact sets, and $\BbbR^{\Bbb N}$ its product
topology;  then $f_{\alpha}$ is continuous.
For a Radon probability measure $\nu$ on $\Bbb R$,
let $\mu_{\nu\alpha}$ be the image Radon measure
$\nu^{\Bbb N}f_{\alpha}^{-1}$ on $\Omega$, where $\nu^{\Bbb N}$ is the
product measure on $\BbbR^{\Bbb N}$.   Let $\mu_W$ be the Radon measure of
477B, and $U$ a neighbourhood of $\mu_W$ in the space
$P_{\text{R}}(\Omega)$ of Radon probability measures on $\Omega$
for the narrow topology\cmmnt{ (437Jd)}.   Then there is a
$\delta>0$ such
that $\mu_{\nu\alpha}\in U$ whenever $\alpha\in\ocint{0,\delta}$ and $\nu$
is a Radon probability measure on $\Bbb R$ with mean $0=\int x\,\nu(dx)$
and variance $1=\int x^2\nu(dx)$ and

%\vskip2pt

\hfill$\biggerint_{\{x:|x|\ge\delta/\sqrt{\alpha}\}}x^2\nu(dx)\le\delta$.
\hfill($\dagger$)

\cmmnt{\medskip

\noindent{\bf Remark} The idea is that, for a given $\alpha$ and $\nu$, we
consider a random walk with independent identically distributed steps,
with expectation 0 and variance $\alpha$, at time intervals of $\alpha$,
and then interpolate to get a continuous function on $\coint{0,\infty}$;
and that if the step-lengths are small the result should look like Brownian
motion.   Moreover, this ought not to depend on the distribution $\nu$;
but in order to apply the Central Limit Theorem in a sufficiently uniform
way, we need the extra regularity condition ($\dagger$).
On first reading you may well prefer to fix on a
particular distribution $\nu$ with mean $0$ and expectation $1$ (e.g., the
distribution which gives measure $\bover12$ to each of $\{1\}$ and
$\{-1\}$), so that ($\dagger$) is satisfied whenever $\alpha$ is small
enough compared with $\delta$.}

\proof{ For $\delta>0$ I will write $Q(\delta)$ for the set
of pairs $(\nu,\alpha)$ such that $\nu$ is a
Radon probability measure on $\Bbb R$ with mean $0$ and variance $1$,
$0<\alpha\le\delta$ and
$\int_{\{x:|x|\ge\delta/\sqrt{\alpha}\}}x^2\nu(dx)\le\delta$.
Note that $Q(\delta')\subseteq Q(\delta)$ when $\delta'\le\delta$.

\medskip

{\bf (a)(i)} If $\gamma$, $\epsilon>0$ there is a $\delta>0$ such
that whenever $(\nu,\alpha)\in Q(\delta)$, $s$, $t\ge 0$
are multiples of $\alpha$
such that $t-s\ge\gamma$, and $I\subseteq\Bbb R$ is an interval (open,
closed or half-open), then

\Centerline{$
|\mu_{\nu\alpha}\{\omega:\omega\in\Omega,\,\omega(t)-\omega(s)\in I\}
-\Bover1{\sqrt{2\pi(t-s)}}\int_Ie^{-x^2/2(t-s)}dx|\le\epsilon$.}

\noindent\Prf\ For $\delta>0$, $x\in\Bbb R$ set
$\psi_{\delta}(x)=x^2$ if $|x|>\delta$, $0$ if $|x|\le\delta$.
Let $\eta>0$ be such that whenever $Y_1,\ldots,Y_k$ are
independent random variables with finite variance and zero expectation,
$\sum_{i=1}^k\Var(Y_i)=1$ and
$\sum_{i=1}^k\Expn(\psi_{\eta}(Y_i))\le\eta$, then

\Centerline{$|\Pr(\sum_{i=1}^kY_i\le\beta)
  -\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\beta}e^{-x^2/2}dx|
\le\Bover{\epsilon}2$}

\noindent for every $\beta\in\Bbb R$ (274F);  observe that in this case

$$\eqalignno{|\Pr(\sum_{i=1}^k&Y_i<\beta)
  -\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\beta}e^{-x^2/2}dx|\cr
&=\lim_{\beta'\uparrow\beta}|\Pr(\sum_{i=1}^kY_i\le\beta')
  -\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\beta'}e^{-x^2/2}dx|
\le\Bover{\epsilon}2\cr}$$

\noindent for every $\beta\in\Bbb R$, so that

\Centerline{$|\Pr(\sum_{i=1}^kY_i\in J)
  -\Bover1{\sqrt{2\pi}}\int_Je^{-x^2/2}dx|
\le\epsilon$}

\noindent for every interval $J\subseteq\Bbb R$.

Set $\delta=\min(\eta,\eta\sqrt{\gamma})$.
If $I\subseteq\Bbb R$ is an interval,
$(\nu,\alpha)\in Q(\delta)$ and $s$, $t$ are multiples of $\alpha$ such
that $t-s\ge\gamma$, set $j=\Bover{s}{\alpha}$,
$k=\Bover{t-s}{\alpha}$ and $J=\sqrt{t-s}I=\sqrt{\bover{k}{\alpha}}I$.
Then

$$\eqalign{\mu_{\nu\alpha}\{\omega:\omega(t)-\omega(s)\in I\}
&=\nu^{\Bbb N}\{z:f_{\alpha}(z)(t)-f_{\alpha}(z)(s)\in I\}\cr
&=\nu^{\Bbb N}\{z:\sqrt{\alpha}\sum_{i=j}^{j+k-1}z(i)\in I\}
=\Pr(\sum_{i=0}^{k-1}Y_i\in J)\cr}$$

\noindent where $Y_i(z)=\Bover1{\sqrt{k}}z(j+i)$.   For each $i$,
the mean and variance of $Y_i$ are $0$ and $\bover1k$, because the mean
and expectation of $\nu$ are $0$ and $1$.   Next,

$$\eqalign{\sum_{i=0}^{k-1}\Expn(\psi_{\eta}(Y_i))
&=k\int_{\{x:|x|>\eta\sqrt{k}\}}\Bover1kx^2\nu(dx)
\le\int_{\{x:|x|>\eta\sqrt{\gamma/\alpha}\}}x^2\nu(dx)\cr
&\le\int_{\{x:|x|>\delta/\sqrt{\alpha}\}}x^2\nu(dx)
\le\delta
\le\eta,\cr}$$

\noindent so by the choice of $\eta$,

$$\eqalign{
&|\mu_{\nu\alpha}\{\omega:\omega\in\Omega,\,\omega(t)-\omega(s)\in I\}
  -\Bover1{\sqrt{2\pi(t-s)}}\int_Ie^{-x^2/2(t-s)}dx|\cr
&\mskip230mu
=|\Pr(\sum_{i=0}^{k-1}Y_i\in J)
  -\Bover1{\sqrt{2\pi}}\int_Je^{-x^2/2}dx|
\le\epsilon.  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (ii)} If $\gamma$, $\epsilon>0$, there is a $\delta>0$
such that

\Centerline{$\mu_{\nu\alpha}
  \{\omega:\diam(\omega[\,[\beta,\beta+\gamma]\,])
>12\epsilon\}
\le\Bover{3\sqrt{\gamma}}{\epsilon}e^{-\epsilon^2/2\gamma}$}

\noindent whenever $(\nu,\alpha)\in Q(\delta)$ and
$\beta\ge 0$.   \Prf\ Let $\eta>0$ be such that

\Centerline{$6(\eta+\Bover{\sqrt{\gamma+\eta}}{\epsilon\sqrt{2\pi}}
   e^{-\epsilon^2/2(\gamma+\eta)})
\le\Bover{3\sqrt{\gamma}}{\epsilon}e^{-\epsilon^2/2\gamma}$,}

\noindent and let $\delta_0>0$ be such that

\Centerline{$
|\mu_{\nu\alpha}\{\omega:\omega\in\Omega,\,\omega(t)-\omega(s)\in I\}
-\Bover1{\sqrt{2\pi(t-s)}}\int_Ie^{-x^2/2(t-s)}dx|\le\eta$}

\noindent whenever $I\subseteq\Bbb R$ is an interval,
$(\nu,\alpha)\in Q(\delta_0)$
and $s$ and $t$ are
multiples of $\alpha$ such that $t-s\ge\bover14\gamma$.
Set $\delta=\min(\bover14\gamma,\bover12\eta,\delta_0)$.

Fix $(\nu,\alpha)\in Q(\delta)$.   Applying the last formula with
$I=[-\epsilon,\epsilon]$ and then taking complements,

$$\eqalign{\mu_{\nu\alpha}\{\omega:|\omega(t)-\omega(s)|>\epsilon\}
&\le\eta+\Bover{2}{\sqrt{2\pi(t-s)}}
  \int_{\epsilon}^{\infty}e^{-x^2/2(t-s)}dx\cr
&=\eta+\Bover{2}{\sqrt{2\pi}}
  \int_{\epsilon/\sqrt{t-s}}^{\infty}e^{-x^2/2}dx\cr
&\le\eta+\Bover{2}{\sqrt{2\pi}}
  \int_{\epsilon/\sqrt{\gamma+\eta}}^{\infty}e^{-x^2/2}dx
\le\eta
  +\Bover{\sqrt{\gamma+\eta}}
  {\epsilon}e^{-\epsilon^2/2(\gamma+\eta)}\cr}$$

\noindent whenever $s$, $t$ are multiples of $\alpha$ such that
$\bover14\gamma\le t-s\le\gamma+\eta$, using
274Ma for the last step, as in part (c) of the proof of 477B.
Now if $s$, $t$ are multiples of $\alpha$ such
that $s\le t\le\gamma+\eta$, either $t-s\ge\bover14\gamma$ and

\Centerline{$\mu_{\nu\alpha}\{\omega:|\omega(t)-\omega(s)|>2\epsilon\}
\le\eta+\Bover{\sqrt{\gamma+\eta}}
  {\epsilon}e^{-\epsilon^2/2(\gamma+\eta)}$,}

\noindent or $t\le s+\bover14\gamma$ and
there is a multiple $u$ of $\alpha$ such that
$t+\bover14\gamma\le u\le t+\bover12\gamma$, in which case

$$\eqalign{\mu_{\nu\alpha}\{\omega:|\omega(t)-\omega(s)|>2\epsilon\}
&\le\mu_{\nu\alpha}\{\omega:|\omega(u)-\omega(s)|>\epsilon\}
  +\mu_{\nu\alpha}\{\omega:|\omega(u)-\omega(t)|>\epsilon\}\cr
&\le 2(\eta+\Bover{\sqrt{\gamma+\eta}}
  {\epsilon}e^{-\epsilon^2/2(\gamma+\eta)}).\cr}$$

Let $j$, $k$ be such that $\beta-\bover12\eta<j\alpha\le\beta$ and
$\beta+\gamma\le k\alpha<\beta+\gamma+\bover12\eta$.
We have

$$\eqalignno{
\mu_{\nu\alpha}\{\omega&:\diam(\omega[\,[\beta,\beta+\gamma]\,])
  >12\epsilon\}\cr
&=\nu^{\Bbb N}\{z:\diam(f_{\alpha}(z)[\,[\beta,\beta+\gamma]\,])
   >12\epsilon\}\cr
&\le\nu^{\Bbb N}\{z:\sup_{t\in[\beta,\beta+\gamma]}
   |f_{\alpha}(z)(t)-f_{\alpha}(z)(j\alpha)|
   >6\epsilon\}\cr
&\le\nu^{\Bbb N}\{z:\text{there is an }l\text{ such that }j<l\le k
  \text{ and }
  |f_{\alpha}(z)(l\alpha)-f_{\alpha}(z)(j\alpha)|>6\epsilon\}\cr
\displaycause{because $f_{\alpha}(z)$ is linear between its
determining values at multiples of $\alpha$}
&=\nu^{\Bbb N}\{z:\text{there is an }l\text{ such that }j<l\le k
  \text{ and }
  |\sum_{i=j}^{l-1}z(i)|>\Bover{6\epsilon}{\sqrt{\alpha}}\}\cr
&\le 3\sup_{j<l\le k}
  \nu^{\Bbb N}\{z:|\sum_{i=j}^{l-1}z(i)|
  >\Bover{2\epsilon}{\sqrt{\alpha}}\}\cr
\displaycause{Etemadi's lemma, 272V}
&=3\sup_{j<l\le k}
  \mu_{\nu\alpha}\{\omega:|\omega(l\alpha)-\omega(j\alpha)|
  >2\epsilon\}\cr
&\le 6(\eta+\Bover{\sqrt{\gamma+\eta}}{\epsilon}
  e^{-\epsilon^2/2(\gamma+\eta)})
\le\Bover{3\sqrt{\gamma}}{\epsilon}e^{-\epsilon^2/2\gamma},\cr}$$

\noindent as required.\ \Qed

\medskip

\quad{\bf (iii)}
If $\gamma$, $\epsilon>0$ there is a $\delta>0$ such that

\Centerline{$\mu_{\nu\alpha}\{\omega:\text{there are }s,\,t\in[0,\gamma]
   \text{ such that }|t-s|\le\delta
   \text{ and }|\omega(t)-\omega(s)|>\epsilon\}\le\epsilon$}

\noindent whenever $(\nu,\alpha)\in Q(\delta)$.
\Prf\ Set $\eta=\epsilon/12$, and let $k\ge 1$ be such that
$\Bover{6\gamma k}{\eta}e^{-k^2\eta^2/2}\le\epsilon$;  set
$m=\lfloor 2k^2\gamma\rfloor$.
By (ii), there is a $\delta\in\ocint{0,\bover1{2k^2}}$
such that

\Centerline{$\mu_{\nu\alpha}\{\omega:
  \diam(\omega[\,[\beta,\beta+\Bover1{k^2}]\,])>12\eta\}
\le\Bover{3}{k\eta}e^{-k^2\eta^2/2}$}

\noindent whenever $(\nu,\alpha)\in Q(\delta)$ and
$\beta\ge 0$.   Now, for such $\nu$ and $\alpha$,

$$\eqalign{&\mu_{\nu\alpha}\{\omega:\text{there are }s,\,t\in[0,\gamma]
   \text{ such that }|t-s|\le\delta
   \text{ and }|\omega(t)-\omega(s)|>\epsilon\}\cr
&\mskip150mu
\le\mu_{\nu\alpha}(\bigcup_{i<m}
  \{\omega:\diam(\omega[\,[\Bover{i}{2k^2},\Bover{i+2}{2k^2}]\,]>12\eta\})
  \cr
&\mskip150mu
\le\Bover{3m}{k\eta}e^{-k^2\eta^2/2}
\le\Bover{6\gamma k}{\eta}e^{-k^2\eta^2/2}
\le\epsilon,\cr}$$

\noindent as required.\ \Qed

\medskip

{\bf (b)} Suppose that $0=t_0<t_1<\ldots<t_n$ and that $E_0,\ldots,E_{n-1}$
are intervals in $\Bbb R$;  set
$E=\{\omega:\omega\in\Omega$, $\omega(t_{i+1})-\omega(t_i)\in E_i$ for
$i<n\}$.   Then for every $\epsilon>0$ there is a $\delta>0$ such
that $\mu_WE\le 3\epsilon+\mu_{\nu\alpha}E$ whenever
$(\nu,\alpha)\in Q(\delta)$.   \Prf\ Of course

\Centerline{$\mu_WE
=\prod_{i<n}\Bover1{\sqrt{2\pi(t_{i+1}-t_i)}}
  \int_{E_i}e^{-x^2/2(t_{i+1}-t_i)}dx$.}

\noindent For $\eta>0$ and $i<n$, let $F_{i\eta}$ be the interval
$\{x:[x-2\eta,x+2\eta]\subseteq E_i\}$.
Set $\gamma=\bover12\min_{i<n}(t_{i+1}-t_i)$;  let
$\eta\in\ocint{0,\gamma}$ be such that

\Centerline{$\prod_{i<n}\Bover1{\sqrt{2\pi\gamma_i}}
  \int_{F_{i\eta}}e^{-x^2/2\gamma_i}dx
\ge\mu_WE-\epsilon$}

\noindent whenever $|\gamma_i-(t_{i+1}-t_i)|\le\eta$ for every $i<n$.
Next, by (a-i) and (a-iii), there is a $\delta\in\ocint{0,\bover12\eta}$
such that

$$\eqalign{\prod_{i<n}\Bover1{\sqrt{2\pi(s_{i+1}-s_i)}}
   &\int_{F_{i\eta}}e^{-x^2/2(s_{i+1}-s_i)}dx\cr
&\le\epsilon+\prod_{i<n}\mu_{\nu\alpha}\{\omega:
    \omega(s_{i+1})-\omega(s_i)\in F_{i\eta}\},\cr}$$

$$\eqalign{\mu_{\nu\alpha}\{\omega:
  \text{there are }&s,\,t\in[0,t_n+\eta]\cr
&\text{ such that }|s-t|\le\delta
  \text{ and }|\omega(s)-\omega(t)|>\eta\}
\le\epsilon\cr}$$

\noindent whenever $(\nu,\alpha)\in Q(\delta)$ and
$s_0,\ldots,s_n$ are multiples of $\alpha$ such that
$s_{i+1}-s_i\ge\gamma$ for every $i\le n$.   Take any
$(\nu,\alpha)\in Q(\delta)$,
and for each $i\le n$ let $s_i$ be a multiple of $\alpha$ such
that $t_i\le s_i\le t_i+\alpha$.   Then

$$\eqalign{\{\omega:
   \omega&(s_{i+1})-\omega(s_i)\in F_{i\eta}\text{ for every }i<n\}
\setminus E\cr
&=\bigcup_{i<n}\{\omega:\omega(s_{i+1})-\omega(s_i)\in F_{i\eta},\,
\omega(t_{i+1})-\omega(t_i)\notin E_i\}\cr
&\subseteq\bigcup_{i<n}\{\omega:
|(\omega(s_{i+1})-\omega(s_i))-(\omega(t_{i+1}-\omega(t_i))|>2\eta\}\cr
&\subseteq\bigcup_{i\le n}\{\omega:|\omega(s_i)-\omega(t_i)|>\eta\}\cr
&\subseteq\{\omega:\text{there are }s,\,t\in[0,t_n+\eta]\cr
&\mskip100mu\text{ such that }
|s-t|\le\delta\text{ and }|\omega(s_i)-\omega(t_i)|>\eta\},\cr}$$

\noindent so

\Centerline{$\mu_{\nu\alpha}\{\omega:
  \omega(s_{i+1})-\omega(s_i)\in F_{i\eta}$ for every $i<n\}
\le\epsilon+\mu_{\nu\alpha}E$.}

\noindent Next, if $s_i=k_i\alpha$ for each $i$,

$$\eqalignno{\mu_{\nu\alpha}\{\omega:\omega&(s_{i+1})-\omega(s_i)
  \in F_{i\eta}\text{ for every }i<n\}\cr
&=\nu^{\Bbb N}\{z:f_{\alpha}(z)(s_{i+1})-f_{\alpha}(z)(s_i)
  \in F_{i\eta}
  \text{ for every }i<n\}\cr
&=\nu^{\Bbb N}\{z:\sqrt{\alpha}\sum_{j=k_i}^{k_{i+1}-1}z(j)\in F_{i\eta}
  \text{ for every }i<n\}\cr
&=\prod_{i<n}\nu^{\Bbb N}\{z:\sqrt{\alpha}\sum_{j=k_i}^{k_{i-1}}z(j)
  \in F_{i\eta}\}\cr
&=\prod_{i<n}\mu_{\nu\alpha}\{\omega:\omega(s_{i+1})-\omega(s_i)
  \in F_{i\eta}\}.\cr}$$

\noindent So

$$\eqalignno{\mu_WE
&\le\epsilon+\prod_{i<n}\Bover1{\sqrt{2\pi(s_{i+1}-s_i)}}
   \int_{F_{i\eta}}e^{-x^2/2(s_{i+1}-s_i)}dx\cr
\displaycause{because $|(s_{i+1}-s_i)-(t_{i+1}-t_i)|\le\alpha\le\eta$
for $i<n$}
&\le 2\epsilon
  +\prod_{i<n}\mu_{\nu\alpha}\{\omega:\omega(s_{i+1})-\omega(s_i)
      \in F_{i\eta}\}\cr
\displaycause{because $s_{i+1}-s_i\ge\gamma$ for every $i<n$}
&=2\epsilon+\mu_{\nu\alpha}\{\omega:\omega(s_{i+1})-\omega(s_i)
  \in F_{i\eta}\text{ for every }i<n\}\cr
&\le 3\epsilon+\mu_{\nu\alpha}E,\cr}$$

\noindent as required.\ \Qed

\medskip

{\bf (c)(i)} For $k\in\Bbb N$ let $\delta_k>0$ be such that
$\mu_{\nu\alpha}G_k\le 2^{-k}$ whenever $(\nu,\alpha)\in Q(\delta_k)$,
where

\Centerline{$G_k=\{\omega:$ there are $s$, $t\in[0,k]$ such that
   $|t-s|\le\delta_k$ and $|\omega(t)-\omega(s)|>2^{-k}\}$;}

\noindent such exists by (a-iii) above.   For $k$, $n\in\Bbb N$ set
$H_{kn}=\bigcup_{i\le n}G_{k+i}$.   If $k\in\Bbb N$ and
$\sequencen{\omega'_n}$ is a sequence such that
$\omega'_n\in\Omega\setminus H_{kn}$ for every $n\in\Bbb N$,
$\{\omega'_n:n\in\Bbb N\}$ is relatively compact in $\Omega$.   \Prf\
If $\gamma\ge 0$ and $\epsilon>0$, there is an $n\in\Bbb N$
such that $2^{-k-n}\le\epsilon$ and $k+n\ge\gamma$;  now for $m\ge n$,
$\omega'_m\notin G_{k+n}$ so
$|\omega'_m(t)-\omega'_m(s)|\le\epsilon$ whenever
$s$, $t\in[0,\gamma]$ and $|s-t|\le\delta_{k+n}$.   Of course there is a
$\delta\in\ocint{0,\delta_{k+n}}$ such that
$|\omega'_m(s)-\omega'_m(t)|\le\epsilon$ whenever $m<k+n$ and $s$,
$t\in[0,\gamma]$ are such that $|s-t|\le\delta$.   Since $\omega'_n(0)=0$
for every $n$, the conditions of
4A2U(e-ii) are satisfied, and $\{\omega'_n:n\in\Bbb N\}$ is relatively
compact in $C(\coint{0,\infty})$, therefore in its closed subset $\Omega$.\
\Qed

Now if we have a compact set
$K\subseteq\Omega$, an open set $G\subseteq\Omega$ including $K$, and
$k\in\Bbb N$, there are an $n\in\Bbb N$ and a finite set
$I\subseteq\coint{0,\infty}$ such that $\omega'\in G\cup H_{kn}$ whenever
$\omega\in K$, $\omega'\in\Omega$ and $|\omega'(s)-\omega(s)|\le 2^{-n}$
for every $s\in I$.   \Prf\Quer\ Otherwise, let $\sequence{i}{q_i}$
enumerate $\Bbb Q\cap\coint{0,\infty}$.   For each $n\in\Bbb N$ we have
$\omega_n\in K$ and $\omega'_n\in\Omega\setminus(G\cup H_{kn})$ such that
$|\omega'_n(q_i)-\omega_n(q_i)|\le 2^{-n}$ for every $i\le n$.
Since the topology $\frak T_c$ on $\Omega$ is metrizable (4A2U(e-i)),
and both
$\{\omega_n:n\in\Bbb N\}$ and $\{\omega'_n:n\in\Bbb N\}$ are relatively
compact, there is a strictly increasing sequence $\sequence{i}{n_i}$ such
that $\omega=\lim_{i\to\infty}\omega_{n_i}$ and
$\omega'=\lim_{i\to\infty}\omega'_{n_i}$ are both defined (use 4A2Lf
twice).   Since $|\omega'(q)-\omega(q)|
=\lim_{i\to\infty}|\omega'_{n_i}(q)-\omega_{n_i}(q)|$ is zero for
every $q\in\Bbb Q\cap\coint{0,\infty}$, $\omega=\omega'$;  but
$\omega\in K$ and $\omega'\notin G$, so this is impossible.\ \Bang\Qed

\medskip

\quad{\bf (ii)} Suppose that $G\subseteq\Omega$ is open and
$\gamma<\mu_WE$.   Then there is a $\delta>0$ such that
$\mu_{\nu\alpha}G>\gamma$ whenever $(\nu,\alpha)\in Q_{\delta}$.
\Prf\ Let $K\subseteq G$ be a compact set such that
$\mu_WK>\gamma$.   Let $k\in\Bbb N$, $\epsilon>0$ be such that
$\mu_WK\ge\gamma+\epsilon+2^{-k+1}$.
By (i), there are an $n\in\Bbb N$ and
a finite set $I\subseteq\coint{0,\infty}$ such that
$\omega'\in G\cup H_{kn}$ whenever $\omega'\in\Omega$, $\omega\in K$ and
$|\omega'(t)-\omega(t)|\le 2^{-n}$ for every $t\in I$;  of course we can
suppose that $0\in I$ and that $\#(I)\ge 2$.   Enumerate $I$ in increasing
order as $\langle t_i\rangle_{i\le m}$.   For $z\in\BbbZ^m$, set

\Centerline{$E_z
=\{\omega:\omega\in\Omega$,
 $\lfloor 2^nm(\omega(t_{i+1}-\omega(t_i))\rfloor=z(i)$ for every $i<m\}$;}

\noindent set $D=\{z:z\in\BbbZ^m$, $E_z\cap K\ne\emptyset\}$ and
$F=\bigcup_{z\in D}E_z$.   If $z\in D$ and $\omega'\in E_z$, there is
an $\omega\in K\cap E_z$, in which case

\Centerline{$|(\omega'(t_{i+1})-\omega'(t_i))-(\omega(t_{i+1}-\omega(t_i))|
\le\Bover{2^{-n}}m$ for every $i<m$,}

\Centerline{$|\omega'(t_i)-\omega(t_i)|\le 2^{-n}$ for every $i\le m$}

\noindent and $\omega'\in G\cup H_{kn}$.   Thus $F\subseteq G\cup H_{kn}$.
As $K$ is compact, $\{\omega(t_i):\omega\in K\}$ is bounded for every $i$
and $D$ is finite.
By (b) there is a $\delta>0$ such that $\delta\le\delta_{k+i}$
for every $i\le n$ and

\Centerline{$\mu_WE_z\le\Bover{\epsilon}{1+\#(D)}+\mu_{\nu\alpha}E_z$}

\noindent whenever $z\in D$ and $(\nu,\alpha)\in Q(\delta)$.
Now, for such $\nu$ and $\alpha$,

$$\eqalignno{\epsilon+2^{-k+1}+\gamma
&\le\mu_WK
\le\mu_WF
=\sum_{z\in D}\mu_WE_z
\le\epsilon+\sum_{z\in D}\mu_{\nu\alpha}E_z\cr
&=\epsilon+\mu_{\nu\alpha}F
\le\epsilon+\mu_{\nu\alpha}G+\sum_{i=0}^n\mu_{\nu\alpha}G_{k+i}\cr
&\le\epsilon+\mu_{\nu\alpha}G+\sum_{i=0}^n2^{-k-i}
<\epsilon+2^{-k+1}+\mu_{\nu\alpha}G\cr}$$

\noindent and $\mu_{\nu\alpha}G>\gamma$, as required.\ \Qed

\medskip

\quad{\bf (iii)} So if $U$ is a neighbourhood of $\mu_W$ for the narrow
topology on $P_{\text{R}}(\Omega)$,
there is a $\delta>0$ such that $\mu_{\nu\alpha}\in U$
whenever $(\nu,\alpha)\in Q(\delta)$.   \Prf\ There are
open sets $G_0,\ldots,G_n$ and $\gamma_0,\ldots,\gamma_n$ such that
$\gamma_i<\mu_WG_i$ for each $i<n$ and $U$ includes
$\{\mu:\mu\in P_{\text{R}}(\Omega)$, $\mu G_i>\gamma_i$ for every $i<n\}$.
But from (ii) we see that for each $i\le n$ there will be a
$\delta'_i>0$ such that $\mu_{\nu\alpha}G_i>\gamma_i$ for every
$i$ whenever $(\nu,\alpha)\in Q(\delta'_i)$;  so setting
$\delta=\min_{i\le n}\delta'_i$ we get the result.\ \Qed

And this is just the conclusion declared in the
statement of the theorem, rephrased in the
language developed in the course of the proof.
}%end of proof of 477C

\leader{477D}{Multidimensional Brownian motion}\cmmnt{ In \S\S478-479
we shall need the theory of Brownian motion in
$r$-dimensional space.   I sketch the relevant details.}
Fix an integer $r\ge 1$.

\spheader 477Da Let $\mu_{W1}$ be the Radon probability
measure on $\Omega_1=C(\coint{0,\infty})_0$ described
in 477B;  I will call it {\bf one-dimensional Wiener measure}.
We can identify the power $\Omega_1^r$ with
$\Omega=C(\coint{0,\infty};\BbbR^r)_0$, the space of continuous functions
$\omega:\coint{0,\infty}\to\BbbR^r$ such that $\omega(0)=0$,
with the topology of uniform convergence on compact sets;  note that
$\Omega_1$ is Polish\cmmnt{ (4A2U(e-i))}, so $\Omega_1^r$ also is.
Because $\Omega_1$ is separable and metrizable, the c.l.d.\
product measure $\mu_{W1}^r$ measures every Borel
set\cmmnt{ (4A3Dc, 4A3E)}, while it is inner regular with respect to
the compact sets\cmmnt{ (412Sb)}, so
it is a Radon measure.   I will say that
$\mu_W=\mu_{W1}^r$, interpreted as a measure on
$C(\coint{0,\infty};\BbbR^r)_0$, is {\bf $r$-dimensional Wiener measure}.

\cmmnt{As observed in 477B,}
$\mu_{W1}$ is the subspace measure on $\Omega_1$ induced by the
distribution $\hat\mu$ of the process
$\langle X_t\rangle_{t\ge 0}$ in 477A.   \cmmnt{So }$\mu_W$ here,
regarded as a
measure on $C(\coint{0,\infty})_0^r$, is the subspace measure induced by
the measure $\hat\mu^r$ on
$(\BbbR^{\coint{0,\infty}})^r
  \cong\BbbR^{\coint{0,\infty}\times r}$\cmmnt{ (254La)}.

\spheader 477Db For $\omega\in\Omega$, $t\ge 0$ and $i<r$,
set $X_t^{(i)}(\omega)=\omega(t)(i)$.   Then
$\langle X_t^{(i)}\rangle_{t\ge 0,i<r}$ is a centered
Gaussian process, with covariance matrix

$$\eqalign{\Expn(X_s^{(i)}\times X_t^{(j)})
&=0\text{ if }i\ne j,\cr
&=\min(s,t)\text{ if }i=j.\cr}$$

\prooflet{\noindent\Prf\ Taking $\mu$, $\hat\mu$ and $\hat\mu^r$ as in (a),
$\hat\mu^r$, like $\hat\mu$, is a centered Gaussian
distribution (456Be);  but it is easy to check from the formula in 454J(i)
that $\hat\mu^r$ can be identified with the distribution of the family
$\langle X_t^{(i)}\rangle_{t\ge 0,i<r}$.   So
$\langle X_t^{(i)}\rangle_{t\ge 0,i<r}$ is a centered Gaussian process.
As for the covariance matrix, if $i\ne j$ then $X_s^{(i)}$ and $X_t^{(j)}$
are determined by different factors in the product $\Omega=\Omega_1^r$,
so must be independent;  while if $i=j$ then $(X^{(i)}_s,X^{(i)}_t)$ have
the same joint distribution as $(X_s,X_t)$ in 477A.\ \Qed}

\spheader 477Dc\cmmnt{ We shall need a variety of characterizations of
the Radon measure $\mu_W$.

\medskip

\quad}{\bf (i)} $\mu_W$ is the only Radon probability measure on $\Omega$
such that the process $\langle X_t^{(i)}\rangle_{t\ge 0,i<r}$ described in
(b) is a Gaussian process with the covariance matrix there.
\prooflet{\Prf\ Suppose $\nu$ is another measure with these properties.
The distribution of $\langle X_t^{(i)}\rangle_{t\ge 0,i<r}$ (with respect
to $\nu$) must be a centered Gaussian process on
$\BbbR^{r\times\coint{0,\infty}}\cong(\BbbR^{\coint{0,\infty}})^r$, and
because it has the same covariance matrix it must be equal to $\hat\mu^r$,
by 456Bb.   But this says just that $\langle X_t^{(i)}\rangle_{t\ge 0,i<r}$
has the same joint distribution with respect to $\mu_W$ and $\nu$.   By
454N, $\nu=\mu_W$.\ \Qed}

\medskip

\quad{\bf (ii)}
Another way of looking at the family
$\langle X^{(i)}_t\rangle_{i<r,t\ge 0}$ is to write $X_t(\omega)=\omega(t)$
for $t\ge 0$, so that $\langle X_t\rangle_{t\ge 0}$ is now a family of
$\BbbR^r$-valued random variables defined on $\Omega$.
We can describe its distribution in terms matching those of 455Q and 477A,
which become

\inset{(i) $X_0=0$ everywhere\cmmnt{ (on $\Omega$, that is)};

(ii) whenever $0\le s<t$ then $\Bover1{\sqrt{t-s}}(X_t-X_s)$
has the standard Gaussian distribution $\mu_G^r$\cmmnt{ (that is,
$\omega\mapsto\Bover1{\sqrt{t-s}}(\omega(t)-\omega(s))$ is \imp\ for
$\mu_W$ and $\mu_G^r$)};

(iii) whenever $0\le t_1<\ldots<t_n$, then $X_{t_2}-X_{t_1},\ldots,
X_{t_n}-X_{t_{n-1}}$ are independent\cmmnt{ (that is, taking $\Tau_i$ to
be the $\sigma$-algebra
$\{\{\omega:\omega(t_{i+1})-\omega(t_i)\in E\}:E\subseteq\BbbR^r$
is a Borel set$\}$, $\Tau_1,\ldots,\Tau_{n-1}$ are independent)}.
}

\noindent Note that
these properties also determine the Radon measure $\mu_W$.
\prooflet{\Prf\ Once again, suppose $\nu$ is a Radon probability measure on
$\Omega$ for which (ii) and (iii) are true.   We wish to show that $\mu_W$
and $\nu$ give the same distribution to
$\langle X^{(i)}_t\rangle_{i<r,t\ge 0}$.   If $0=t_0<t_1<\ldots<t_n$, we
know that $\mu_W$ and $\nu$ give the same distribution to each of the
differences $Y_j=X_{t_{j+1}}-X_{t_j}$ (or, if you prefer, to each of the
families $\ofamily{i}{r}{Y^{(i)}_j}$, where
$Y^{(i)}_j=X^{(i)}_{t_{j+1}}-X^{(i)}_{t_j}$);  moreover,
if $\Sigma_j$ is the $\sigma$-algebra generated by $\{Y^{(i)}_j:i<r\}$ for
each $j$, then $\mu_W$ and $\nu$ agree that $\ofamily{j}{n}{\Sigma_j}$ is
independent.   So $\mu_WE=\nu E$ whenever $E$ is of the form
$\bigcap_{j<n}E_j$ where $E_j\in\Sigma_j$ for each $j<n$.   By the Monotone
Class Theorem, $\mu_W$ and $\nu$ agree on the $\sigma$-algebra
$\Sigma$ generated by
sets of this type, which is the $\sigma$-algebra generated by
$\{Y^{(i)}_j:i<r$, $j<n\}$.   But as $X_{t_j}=\sum_{i<j}Y_i$ for every
$j\le n$, every $X^{(i)}_{t_j}$ is $\Sigma$-measurable, and $\mu_W$ and
$\nu$ give the same distribution to
$\langle X^{(i)}_{t_j}\rangle_{i<r,j\le n}$.  As this is true whenever
$0=t_0<\ldots<t_n$, $\mu_W$ and $\nu$ give the same distribution to the
whole family $\langle X^{(i)}_t\rangle_{i<r,t\ge 0}$, and must be equal.\
\Qed}

\cmmnt{
\spheader 477Dd In order to apply Theorem 455U, we need to go a little
deeper, in order to relate the product-measure definition of $\mu_W$ to the
construction in 455P.   I will use the ideas of part (b) of the proof of
455R.   Consider the process $\langle X^{(i)}_t\rangle_{t\ge 0,i<r}$
and the associated distribution $\hat\mu^r$ on
$(\BbbR^{\coint{0,\infty}})^r\cong(\BbbR^r)^{\coint{0,\infty}}$.
Setting $X_t=\ofamily{i}{r}{X^{(i)}_t}$, $\langle X_t\rangle_{t\ge 0}$ is
an $\BbbR^r$-valued process satisfying the conditions of 455Q with
$U=\BbbR^r$.   \Prf\ $X_0=0$ a.e.\  because every $X^{(i)}_0$ is zero a.e.
If $0\le s<t$ then $X_t-X_s=\ofamily{i}{r}{X^{(i)}_t-X^{(i)}_s}$ has the
same distribution as $X_{t-s}$ because $X^{(i)}_t-X^{(i)}_s$ has the same
distribution as $X^{(i)}_{t-s}$ for each $i$ and
$\ofamily{i}{r}{X^{(i)}_t-X^{(i)}_s}$,
$\ofamily{i}{r}{X^{(i)}_{t-s}}$ are both independent.   If
$0\le t_0<t_1<\ldots<t_n$ then
$\langle X^{(i)}_{t_{j+1}}-X^{(i)}_{t_j}\rangle_{i<r,j<n}$ is independent
so $\ofamily{j}{n}{X_{t_{j+1}}-X_{t_j}}$ is independent (using 272K, or
otherwise).   Finally, when $t\downarrow 0$,
$X_t\to 0$ in measure because $X^{(i)}_t\to 0$ in
measure for each $i$.\ \Qed

For $t>0$, let $\lambda_t$ be the distribution of
$X_t$.   Then $\lambda_t$ is the centered Gaussian distribution
on $\BbbR^r$ with
covariance matrix $\langle\sigma_{ij}\rangle_{i,j<r}$ where
$\sigma_{ij}=t$
if $i=j$ and $0$ if $i\ne j$ (456Ba, with $T(\omega)=\omega(t)$ for
$\omega\in\BbbR^{\coint{0,\infty}\times r}
\cong(\BbbR^r)^{\coint{0,\infty}}$).
By 455R, the process of 455P can be applied to
$\langle\lambda_t\rangle_{t>0}$ to give us a measure
$\hat\nu$ on $(\BbbR^r)^{\coint{0,\infty}}$,
the completion of a Baire measure, such that

$$\eqalign{\hat\nu\{\omega:\omega(t_i)\in F_i\text{ for every }i\le n\}
&=\Pr(X_{t_i}\in F_i\text{ for every }i\le n)\cr
&=\hat\mu^r\{\omega:\omega(t_i)\in F_i\text{ for every }i\le n\}\cr}$$

\noindent whenever $F_0,\ldots,F_n\subseteq\BbbR^r$ are Borel sets and
$t_0,\ldots,t_n\in\coint{0,\infty}$.
Since sets of this kind generate the Baire $\sigma$-algebra of
$(\BbbR^r)^{\coint{0,\infty}}$,
$\hat\nu$ must be equal to $\hat\mu^r$, that is, $\hat\mu^r$
is the result of applying 455P to $\langle\lambda_t\rangle_{t>0}$.

By 455H, $\hat\nu$
has a unique extension to a measure $\tilde\nu$ on
$(\BbbR^r)^{\coint{0,\infty}}$ which is a Radon measure for the product
topology.   But if we write
$\iota:C(\coint{0,\infty};\BbbR^r)_0\to(\BbbR^r)^{\coint{0,\infty}}$ for
the identity map, the image measure $\mu_W\iota^{-1}$ is a Radon measure on
$(\BbbR^r)^{\coint{0,\infty}}$ for the
product topology and extends $\hat\mu^r$, so must be equal to $\tilde\nu$.
Thus $\tilde\nu C(\coint{0,\infty};\BbbR^r)_0=1$.
Of course $C(\coint{0,\infty};\BbbR^r)_0$ is included in the space of
\callal\ functions from $\coint{0,\infty}$ to $\BbbR^r$, so that we have a
strengthening of the results in \S455.   Similarly, writing $\ddot\nu$ for
the subspace measure induced by $\hat\nu$ or $\hat\mu^r$
on the space $\Cdlg$ of \cadlag\ functions from
$\coint{0,\infty}$ to $\BbbR^r$, $\mu_W$ is the subspace measure
on $C(\coint{0,\infty};\BbbR^r)_0$ induced by $\ddot\nu$.

By 4A3Qa, every Baire subset of $\Cdlg$ is the intersection of $\Cdlg$ with
a Baire subset of $(\BbbR^r)^{\coint{0,\infty}}$, and is therefore measured
by $\ddot\nu$.   In particular,
$C(\coint{0,\infty};\BbbR^r)$ and $C(\coint{0,\infty};\BbbR^r)_0$ are
measured by $\ddot\nu$ (4A3Qd).
}%end of comment

\leader{477E}{Invariant transformations of Wiener measure:  Proposition}
Let $r\ge 1$ be an integer, and $\mu_W$ Wiener measure on
$\Omega=C(\coint{0,\infty};\BbbR^r)_0$.
Let $\hat\mu^r$ be the product measure on
$(\BbbR^{\coint{0,\infty}})^r$ as described in 477D.

(a) Suppose that
$f:(\BbbR^{\coint{0,\infty}})^r\to(\BbbR^{\coint{0,\infty}})^r$
is \imp\ for $\hat\mu^r$, and that $\Omega_0\subseteq\Omega$ is a
$\mu_W$-conegligible set such that $f[\Omega_0]\subseteq\Omega_0$.
Then $f\restr\Omega_0$ is \imp\ for the subspace measure induced by
$\mu_W$ on $\Omega_0$.

(b) Suppose that
$\hat T:\BbbR^{r\times\coint{0,\infty}}\to\BbbR^{r\times\coint{0,\infty}}$
is a linear operator such that, for $i$, $j<r$ and $s$, $t\ge 0$,

$$\eqalign{\int(\hat T\omega)(i,s)(\hat T\omega)(j,t)\hat\mu^r(d\omega)
&=\min(s,t)\text{ if }i=j,\cr
&=0\text{ if }i\ne j.\cr}$$

\noindent Then, identifying $\BbbR^{r\times\coint{0,\infty}}$
with $(\BbbR^{\coint{0,\infty}})^r$,
$\hat T$ is \imp\ for $\hat\mu^r$.

(c) Suppose that $t\ge 0$.   Define $S_t:\Omega\to\Omega$ by setting
$(S_t\omega)(s)=\omega(s+t)-\omega(s)$ for $s\ge 0$ and $\omega\in\Omega$.
Then $S_t$ is \imp\ for $\mu_W$.

(d) Let $T:\BbbR^r\to\BbbR^r$ be an orthogonal
transformation.   Define $\tilde T:\Omega\to\Omega$ by setting
$(\tilde T\omega)(t)=T(\omega(t))$ for $t\ge 0$ and $\omega\in\Omega$.
Then $\tilde T$ is an automorphism of $(\Omega,\mu_W)$.

(e) Suppose that $\alpha>0$.   Define $U_{\alpha}:\Omega\to\Omega$ by
setting $U_{\alpha}(\omega)(t)=\Bover1{\sqrt{\alpha}}\omega(\alpha t)$ for
$t\ge 0$ and $\omega\in\Omega$.   Then $U_{\alpha}$ is an automorphism of
$(\Omega,\mu_W)$.

(f) Set

\Centerline{$\Omega_0=\{\omega:\omega\in\Omega$,
$\lim_{t\to\infty}\Bover1t\omega(t)=0\}$,}

\noindent and define $R:\Omega_0\to\Omega_0$ by setting

$$\eqalign{(R\omega)(t)
&=t\omega(\Bover1t)\text{ if }t>0,\cr
&=0\text{ if }t=0.\cr}$$

\noindent Then $\Omega_0$ is $\mu_W$-conegligible and $R$ is an
automorphism of $\Omega_0$ with its subspace measure.

(g) Suppose that $1\le r'\le r$, and that $\mu_W^{(r')}$ is Wiener measure
on $C(\coint{0,\infty};\BbbR^{r'})_0$.   Define
$P:\Omega\to C(\coint{0,\infty};\BbbR^{r'})_0$ by setting
$(P\omega)(t)(i)=\omega(t)(i)$ for $t\ge 0$, $i<r'$ and $\omega\in\Omega$.
Then $\mu_W^{(r')}$ is the image measure $\mu_WP^{-1}$.

\proof{ The following arguments will unscrupulously identify
$C(\coint{0,\infty};\BbbR^r)_0$ with $C(\coint{0,\infty})_0^r$, and
$\BbbR^{r\times\coint{0,\infty}}$ with $(\BbbR^r)^{\coint{0,\infty}}$ and
$(\BbbR^{\coint{0,\infty}})^r$.

\medskip

{\bf (a)} Because $\mu_W$ is the subspace measure on $\Omega$ induced by
$\hat\mu^r$ (477Da), the subspace measure $\nu$
on $\Omega_0$ induced by $\mu_W$
is also the subspace measure on $\Omega_0$ induced by $\hat\mu^r$ (214Ce).
If $E\subseteq\Omega_0$ is measured by $\nu$, there is an
$F\in\dom\hat\mu^r$ such that $E=F\cap\Omega_0$, and now

\Centerline{$\nu E=\hat\mu^rF=\hat\mu^rf^{-1}[F]=\nu(\Omega_0\cap f^{-1}[F])
=\nu(f\restr\Omega_0)^{-1}[E]$.}

\noindent As $E$ is arbitrary, $f\restr\Omega_0$ is \imp\ for $\nu$.

\medskip

{\bf (b)} By 456Ba, $\hat\mu^r\hat T^{-1}$ is a centered Gaussian
distribution on $\BbbR^{r\times\coint{0,\infty}}$.   The hypothesis
asserts that its covariance matrix is the same as that of $\hat\mu^r$
(477Db), so that $\hat\mu^r=\hat\mu^r\hat T^{-1}$ (456Bb), that is,
$\hat T$ is \imp\ for $\hat\mu^r$.

\medskip

{\bf (c)} Define $\hat S_t:\BbbR^{r\times\coint{0,\infty}}
  \to\BbbR^{r\times\coint{0,\infty}}$ by setting
$(\hat S_t\omega)(i,s)=\omega(i,s+t)-\omega(i,s)$, this time for
$\omega\in\BbbR^{r\times\coint{0,\infty}}$, $i<r$ and $s\ge 0$.
Then $\hat S_t$ is linear, and for $s$, $u\in\coint{0,\infty}$, $i$, $j<r$

$$\eqalign{\int(\hat S_t\omega)&(i,s)(\hat S_t\omega)(j,u)
   \mu^r(d\omega)\cr
&=\int(\omega(i,s+t)-\omega(t))(\omega(j,u+t)-\omega(j,t))\hat\mu^r(d\omega)
   \cr
&=0\text{ if }i\ne j,\cr
&=\min(s+t,u+t)-\min(s+t,t)\cr
&\mskip100mu-\min(t,u+t)+\min(t,t)
=\min(s,u)\text{ if }i=j.\cr}$$

\noindent By (b), $\hat S_t$ is \imp\ for $\hat\mu^r$.   Now
$\hat S_t[\Omega]\subseteq\Omega$, so $S_t=\hat S_t\restr\Omega$ is \imp\
for $\mu W$, by (a).

\medskip

{\bf (d)} If we define
$\hat T:(\BbbR^r)^{\coint{0,\infty}}\to(\BbbR^r)^{\coint{0,\infty}}$
by setting
$(\hat T)(\omega)(t)=T(\omega(t))$ for $x\in(\BbbR^r)^{\coint{0,\infty}}$ and
$t\ge 0$, then $\hat T$ is linear.   Suppose that $T$ is defined by the
matrix $\langle\alpha_{ij}\rangle_{i,j<r}$.
For $\omega\in\BbbR^{r\times\coint{0,\infty}}$, $t\in\coint{0,\infty}$
and $i<r$,

\Centerline{$(\hat T\omega)(i,t)
=\sum_{k=0}^{r-1}\alpha_{ik}\omega(k,t)$.}

\noindent So, for $i$, $j<r$ and $s$, $t\ge 0$,

$$\eqalign{\int(\hat T\omega)(i,s)(\hat T\omega)(j,t)\hat\mu^r(d\omega)
&=\sum_{k=0}^{r-1}\sum_{l=0}^{r-1}\alpha_{ik}\alpha_{jl}
  \int\omega(k,s)\omega(l,t)\hat\mu^r(d\omega)\cr
&=\sum_{k=0}^{r-1}\alpha_{ik}\alpha_{jk}\min(s,t)\cr
&=\min(s,t)\text{ if }i=j,\cr
&=0\text{ if }i\ne j.\cr}$$

\noindent So $\hat T$ is $\hat\mu^r$-\imp.   If we think of $\hat T$ as
operating on $(\BbbR^r)^{\coint{0,\infty}}$, then $\hat T(\omega)=T\omega$
is continuous for every $\omega\in C(\coint{0,\infty};\BbbR^r)$, so
$\hat T[\Omega]\subseteq\Omega$ and $\tilde T=\hat T\restr\Omega$ is
$\mu_W$-\imp.

Now the same argument applies to $T^{-1}$, so
$\tilde T^{-1}=(T^{-1})\sptilde$ also is \imp, and $\tilde T$ is an
automorphism of $(\Omega,\mu_W)$.

\medskip

{\bf (e)} This time, we have
$\hat U_{\alpha}:\BbbR^{r\times\coint{0,\infty}}
\to\BbbR^{r\times\coint{0,\infty}}$ defined by the formula
$(\hat U_{\alpha}\omega)(i,t)=\Bover1{\sqrt{\alpha}}\omega(i,\alpha t)$
for $i<r$, $t\ge 0$ and $\omega\in\BbbR^{r\times\coint{0,\infty}}$.
Once again, $\hat U_{\alpha}$ is linear.   This time,


$$\eqalign{\int(\hat U_{\alpha}\omega)(i,s)(\hat U_{\alpha}\omega)(j,t)
  \hat\mu^r(d\omega)
&=\Bover1{\alpha}\int\omega(i,\alpha s)\omega(j,\alpha t)\hat\mu^r(d\omega)
   \cr
&=0\text{ if }i\ne j,\cr
&=\Bover1{\alpha}\min(\alpha s,\alpha t)=\min(s,t)\text{ if }i=j.\cr}$$

\noindent As before, it follows that $\hat U_{\alpha}$ is $\hat\mu^r$-\imp,
so that $U_{\alpha}=\hat U_{\alpha}\restr\Omega$ is $\mu_W$-\imp.
In the same way as in
(d),  $U_{\alpha}^{-1}=U_{1/\alpha}$ is $\mu_W$-\imp, so
$U_{\alpha}$ is an automorphism of $(\Omega,\mu_W)$.

\medskip

{\bf (f)} Define
$\hat R:\BbbR^{r\times\coint{0,\infty}}\to\BbbR^{r\times\coint{0,\infty}}$
by setting

$$\eqalign{\hat R(\omega)(i,t)
&=t\omega(i,\Bover1t)\text{ if }i<r\text{ and }t>0,\cr
&=\omega(i,0)\text{ if }i<r\text{ and }t=0.\cr}$$

\noindent Then, if $i$, $j<r$ and $s$, $t>0$,

$$\eqalign{\int(\hat R\omega)(i,s)(\hat R\omega)(j,t)\hat\mu^r(d\omega)
&=st\int\omega(i,\Bover1s)\omega(j,\Bover1t)\hat\mu^r(d\omega)\cr
&=0\text{ if }i\ne j,\cr
&=st\min(\Bover1s,\Bover1t)=\min(s,t)\text{ if }i=j.\cr}$$

\noindent If $s=0$, then
$(\hat R\omega)(i,s)=\omega(i,s)=0$ for almost every $\omega$, so
that $\int(\hat R\omega)(i,s)(\hat R\omega)(j,t)\hat\mu^r(d\omega)=0$;
and similarly if $t=0$.
So $\hat R$ is $\hat\mu^r$-\imp.

At this point I think we need a new argument.   Consider the set

\Centerline{$E
=\{\omega:\omega\in(\BbbR^r)^{\coint{0,\infty}}$,
$\lim_{q\in\Bbb Q,q\downarrow 0}\omega(q)=0\}$.}

\noindent Then $E$ is a Baire set in
$(\BbbR^r)^{\coint{0,\infty}}\cong\BbbR^{r\times\coint{0,\infty}}$.
Since $E\supseteq\Omega$, $\hat\mu^rE=1$.   Consequently
$\hat\mu^r\hat R^{-1}[E]=1$.   But, for
$\omega\in(\BbbR^r)^{\coint{0,\infty}}$,

\Centerline{$\omega\in\hat R^{-1}[E]
\iff 0
=\lim_{q\in\Bbb Q,q\downarrow 0}(\hat R\omega)(q)
=\lim_{q\in\Bbb Q,q\downarrow 0}q\omega(\Bover1q)
=\lim_{q\in\Bbb Q,q\to\infty}\Bover1q\omega(q)$.}

\noindent So

$$\eqalignno{\Omega_0
&=\{\omega:\omega\in\Omega,\,\lim_{t\to\infty}\Bover1t\omega(t)=0\}
=\{\omega:\omega\in\Omega,\,
  \lim_{q\in\Bbb Q,q\to\infty}\Bover1q\omega(q)=0\}\cr
\displaycause{because every member of $\Omega$ is continuous}
&=\Omega\cap\hat R^{-1}[E]\cr}$$

\noindent is $\mu_W$-conegligible.   Next, for $\omega\in\Omega_0$,
$\hat R\omega$ is continuous on $\ooint{0,\infty}$ and

$$\eqalign{0
&=\omega(0)
=\lim_{t\to\infty}\Bover1t\omega(t)=\lim_{t\downarrow 0}\omega(t)\cr
&=(\hat R\omega)(0)
=\lim_{t\to\infty}(\hat R\omega)(\Bover1t)
=\lim_{t\downarrow 0}t(\hat R\omega)(\Bover1t)\cr
&=\lim_{t\downarrow 0}(\hat R\omega)(t)
=\lim_{t\to\infty}\Bover1t(\hat R\omega)(t),\cr}$$

\noindent so $\hat R\omega\in\Omega_0$.   By (a), $R=\hat R\restr\Omega_0$
is \imp\ for the subspace measure $\nu=(\mu_W)_{\Omega_0}$;
being an involution, it is an automorphism of $(\Omega_0,\nu)$.

\medskip

{\bf (g)} If we identify $\mu_W$ and $\mu_W^{(r')}$ with
$\mu_{W1}^r$ and $\mu_{W1}^{r'}$, as in 477Da, this is elementary.
}%end of proof of 477E

\leader{477F}{Proposition} Let $r\ge 1$ be an integer.   Then
Wiener measure on $\Omega=C(\coint{0,\infty};\BbbR^r)_0$ is strictly
positive for the topology $\frak T_c$ of uniform convergence on compact
sets.

\proof{{\bf (a)} Let $\Cal I$ be a partition of $\coint{0,\infty}$ into
bounded intervals (open, closed or half-open).   As usual, set
$X_t(\omega)=\omega(t)$ for $t\in\coint{0,\infty}$ and
$\omega\in(\BbbR^r)^{\coint{0,\infty}}$.   Define
$\langle Y_t\rangle_{t\ge 0}$, $\langle Z_t\rangle_{t\ge 0}$ as follows.
If $t\in I\in\Cal I$, $a=\inf I$ and $b=\sup I$, then

$$\eqalign{Y_t
&=X_a+\Bover{t-a}{b-a}(X_b-X_a)\text{ if }a<b,\cr
&=X_a=X_t=X_b\text{ if }a=b,\cr
Z_t&=X_t-Y_t.\cr}$$

\medskip

\quad{\bf (i)} With respect to the centered Gaussian distribution
$\hat\mu^r$ on
$(\BbbR^{\coint{0,\infty}})^r\cong(\BbbR^r)^{\coint{0,\infty}}$,
$(\langle Y_t\rangle_{t\ge 0},\langle Z_t\rangle_{t\ge 0})$
is a centered Gaussian process.   \Prf\ The map
$\omega\mapsto(\langle Y_t(\omega)\rangle_{t\ge 0},
\langle Z_t(\omega)\rangle_{t\ge 0})$ is linear and continuous, so we can
apply 456Ba (strictly speaking, we apply this to the family
$(\langle Y^{(i)}_t(\omega)\rangle_{i<r,t\ge 0},
\langle Z^{(i)}_t(\omega)\rangle_{i<r,t\ge 0})$ regarded as linear operators
from $\BbbR^{r\times\coint{0,\infty}}$ to its square).\ \Qed

\medskip

\quad{\bf (ii)} If $s$, $t\in\coint{0,\infty}$ then
$\Expn(Y_s\times Y_t)=\Expn(X_s\times Y_t)$.   \Prf\ Let $I$, $J$ be the
members of $\Cal I$ containing $s$, $t$ respectively;  set $a=\min I$,
$b=\max I$, $c=\min J$ and $d=\max J$.

\qquad{\bf case 1} If $a=b$ then $Y_s=X_s$ and we can stop.

\qquad{\bf case 2} If $a<b\le c$, then

\Centerline{$\Expn(X_a\times X_c)=\Expn(X_a\times X_d)=a$,
\quad$\Expn(X_a\times Y_t)=a$,}

\Centerline{$\Expn(X_b\times X_c)=\Expn(X_b\times X_d)=b$,
\quad$\Expn(X_b\times Y_t)=b$,}

\Centerline{$\Expn(X_s\times X_c)=\Expn(X_s\times X_d)=s$,
\quad$\Expn(X_s\times Y_t)=s$,}

\Centerline{$\Expn(Y_s\times Y_t)
=a+\Bover{s-a}{b-a}(b-a)=s=\Expn(X_s\times Y_t)$.}

\qquad{\bf case 3} If $d\le a<b$, then

\Centerline{$c=\Expn(X_s\times X_c)
=\Expn(X_a\times X_c)=\Expn(X_b\times X_c)=\Expn(Y_s\times X_c)$,}

\Centerline{$d=\Expn(X_s\times X_d)
=\Expn(X_a\times X_d)=\Expn(X_b\times X_d)=\Expn(Y_s\times X_d)$;}

\noindent since $Y_t$ is a convex combination of $X_c$ and $X_d$,
$\Expn(Y_s\times Y_t)=\Expn(X_s\times Y_t)$.

\qquad{\bf case 4} If $a=c<b=d$, then

\Centerline{$\Expn(X_a\times Y_t)
=\Expn(X_a\times X_a)+\Bover{t-a}{b-a}\,\Expn(X_a\times(X_b-X_a))=a$,}

\Centerline{$\Expn(X_b\times Y_t)
=\Expn(X_b\times X_a)+\Bover{t-a}{b-a}\,\Expn(X_b\times(X_b-X_a))
=a+\Bover{t-a}{b-a}(b-a)=t$,}

\Centerline{$\Expn(X_s\times Y_t)
=\Expn(X_s\times X_a)+\Bover{t-a}{b-a}\,\Expn(X_s\times(X_b-X_a))
=a+\Bover{t-a}{b-a}(s-a)$,}

$$\eqalign{\Expn(Y_s\times Y_t)
&=\Expn(X_a\times Y_t)+\Bover{s-a}{b-a}\,\Expn((X_b-X_a)\times Y_t)\cr
&=a+\Bover{s-a}{b-a}(t-a)=\Expn(X_s\times Y_t).  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (iii)} Accordingly $\Expn(Z_s\times Y_t)=0$ for all $s$,
$t\ge 0$.   It follows that if $\Sigma_1$, $\Sigma_2$ are the
$\sigma$-algebras of subsets of $(\BbbR^r)^{\coint{0,\infty}}$ defined by
$\{Y_t:t\ge 0\}$ and $\{Z_t:t\ge 0\}$ respectively, $\Sigma_1$ and
$\Sigma_2$ are $\hat\mu^r$-independent (456Eb).

\medskip

{\bf (b)} Observe next that if
$x_1,\ldots,x_n\in\BbbR^r$, $0<t_1\le\ldots\le t_n$ and $\delta>0$, then
$\hat\mu^r\{\omega:\|\omega(t_i)-x_i\|\le\delta$
for $1\le i\le n\}$ is greater than $0$.   \Prf\ Set $x_0=0$ in
$\BbbR^r$ and $t_0=0$.   For each $i<n$, the distribution of
$\Bover1{\sqrt{t_{i+1}-t_i}}(X_{t_{i+1}}-X_{t_i})$ is the standard Gaussian
distribution $\mu_G^r$
on $\BbbR^r$, which has strictly positive probability density
function with respect to Lebesgue measure.   So

$$\eqalign{\Pr(\|(X_{t_{i+1}}-X_{t_i})-(x_{i+1}-x_i)\|\le\Bover1n\delta)
&=\mu_G^r\{x:\|\sqrt{t_{i+1}-t_i}x-x_{i+1}+x_i\|\le\Bover1n\delta\}\cr
&>0.\cr}$$

\noindent Next, $X_{t_1}-X_{t_0},\ldots,X_{t_n}-X_{t_{n-1}}$ are
independent, so

$$\eqalign{0
&<\prod_{i<n}\Pr(\|(X_{t_{i+1}}-X_{t_i})-(x_{i+1}-x_i)\|\le\Bover1n\delta)
  \cr
&=\Pr(\|(X_{t_{i+1}}-X_{t_i})-(x_{i+1}-x_i)\|\le\Bover1n\delta
   \text{ for every }i<n)\cr
&\le\Pr(\|X_{t_i}-x_i\|\le\delta\text{ for every }i\le n).
   \text{ \Qed}\cr}$$

\medskip

{\bf (c)} Let $G\subseteq\Omega$ be a non-empty $\frak T_c$-open set.
Then there are
$\omega_0\in\Omega$, $m\in\Bbb N$ and $\delta>0$ such that $G$ includes

\Centerline{$V=\{\omega:\omega\in\Omega$,
$\|\omega(t)-\omega_0(t)\|\le 6\delta$ for every $t\in[0,m]\}$.}

\noindent For $n\in\Bbb N$, let $F_n$ be

$$\eqalign{\{\omega:\omega&\in(\BbbR^r)^{\coint{0,\infty}},
\|\omega(q)-\omega(q')\|\le\delta\cr
&\mskip50mu
\text{ whenever }q,\,q'\in\Bbb Q\cap[0,m]\text{ and }
  |q-q'|\le 2^{-n}\}.\cr}$$

\noindent Then $\Omega\subseteq\bigcup_{n\in\Bbb N}F_n$ so there is an
$n\ge 0$ such that $\omega_0\in F_n$ and
$\hat\mu^rF_n>0$.   Let $\Cal I$ be
$\{\coint{2^{-n}k,2^{-n}(k+1)}:k\in\Bbb N\}$, and let
$\langle Y_t\rangle_{t\ge 0}$, $\langle Z_t\rangle_{t\ge 0}$ be the
families of random variables defined from $\Cal I$ by the method of (a)
above, with corresponding independent $\sigma$-algebras $\Sigma_1$,
$\Sigma_2$.   If $\omega\in F_n$, then $\|Z_t(\omega)\|\le\delta$
for every $t\in\Bbb Q\cap[0,m]$.
\Prf\ If $t\in\coint{a,b}\in\Cal I$, then
$b-a=2^{-n}$ so $\|X_t(\omega)-X_a(\omega)\|$,
$\|X_t(\omega)-X_b(\omega)\|$ and therefore
$\|Z_t(\omega)\|=\|X_t(\omega)-Y_t(\omega)\|$ are all at most $\delta$.\
\QeD\  Set

\Centerline{$F=\{\omega:\omega\in(\BbbR^r)^{\coint{0,\infty}}$,
$\|Z_t(\omega)\|\le\delta$ for every
$t\in\Bbb Q\cap[0,m]\}$;}

\noindent then $F\in\Sigma_2$ and $\hat\mu^rF>0$.

Next, set

\Centerline{$E=\{\omega:\omega\in(\BbbR^r)^{\coint{0,\infty}}$,
$\|\omega(2^{-n}k)-\omega_0(2^{-n}k)\|\le\delta$ for $1\le k\le 2^nm\}$.}

\noindent By (b), $\hat\mu^rE>0$.   But since
$Y_{2^{-n}k}(\omega)=X_{2^{-n}k}(\omega)=\omega(2^{-n}k)$ whenever
$k\le 2^nm$, $E\in\Sigma_1$.    Accordingly
$\hat\mu^r(E\cap F)=\hat\mu^rE\cdot\hat\mu^rF>0$.   But
$E\cap F\cap\Omega\subseteq V$.   \Prf\ If $\omega\in E\cap F\cap\Omega$,
then $t\mapsto X_t(\omega)$, $t\mapsto Y_t(\omega)$ and
$t\mapsto Z_t(\omega)$ are all continuous, so $\|Z_t(\omega)\|\le\delta$
for every $t\in[0,m]$.
If $t\in[0,m]$, let $k<2^nm$ be such that
$2^{-n}k\le t\le 2^{-n}(k+1)$.   Then

$$\eqalign{\|\omega(t)-\omega_0(t)\|
&\le\|\omega(t)-\omega(2^{-n}k)\|
   +\|\omega(2^{-n}k)-\omega_0(2^{-n}k)\|
   +\|\omega_0(2^{-n}k)-\omega_0(t)\|\cr
&\le\|Z_t(\omega)\|+\|Y_t(\omega)-Y_{2^{-n}k}(\omega)\|
   +2\delta\cr
&\le\delta+\|\omega(2^{-n}(k+1))-\omega(2^{-n}k)\|
   +2\delta\cr
&\le\|\omega(2^{-n}(k+1))-\omega_0(2^{-n}(k+1))\|
   +\|\omega_0(2^{-n}(k+1))-\omega_0(2^{-n}k)\|\cr
&\mskip200mu
   +\|\omega_0(2^{-n}k)-\omega(2^{-n}k)\|
   +3\delta\cr
&\le 6\delta.\cr}$$

\noindent As $t$ is arbitrary, $\omega\in V$.\ \Qed

Accordingly

\Centerline{$\mu_WG
\ge\mu_W(E\cap F\cap\Omega)=\hat\mu^r(E\cap F)>0$.}

\noindent As $G$ is arbitrary, $\mu_W$ is strictly positive.
}%end of proof of 477F


\leader{477G}{The strong Markov \dvrocolon{property}}\cmmnt{ With the
identification in 477Dd, we are ready for
one of the most important properties of Brownian motion.

\medskip

\noindent}{\bf Theorem} Suppose that $r\ge 1$,
$\mu_W$ is Wiener measure on $\Omega=C(\coint{0,\infty};\BbbR^r)_0$ and
$\Sigma$ is its domain.   For $t\ge 0$ let $\Sigma_t$ be

\Centerline{$\{F:F\in\Sigma$, $\omega'\in F$ whenever $\omega\in F$,
$\omega'\in\Omega$ and $\omega'\restr[0,t]=\omega\restr[0,t]\}$,}

\Centerline{$\Sigma^+_t=\bigcap_{s>t}\Sigma_s$,}

\noindent and let
$\tau:\Omega\to[0,\infty]$ be a stopping time adapted to the family
$\langle\Sigma^+_t\rangle_{t\ge 0}$.
Define $\phi_{\tau}:\Omega\times\Omega\to\Omega$ by saying that

$$\eqalign{\phi_{\tau}(\omega,\omega')(t)
&=\omega(t)\text{ if }t\le\tau(\omega),\cr
&=\omega(\tau(\omega))+\omega'(t-\tau(\omega))
   \text{ if }t\ge\tau(\omega).\cr}$$

\noindent Then $\phi_{\tau}$ is \imp\ for $\mu_W\times\mu_W$ and $\mu_W$.

\proof{{\bf (a)}
At this point I apply the general theory of \S455 in something like
its full strength.   As in 477Dd, let
$\langle\lambda_t\rangle_{t>0}$ be the
standard family of Gaussian distributions on $\BbbR^r$, $\ddot\nu$
the corresponding measure on the space $\Cdlg$ of \cadlag\ functions from
$\coint{0,\infty}$ to $\BbbR^r$, and $\ddot\Sigma$ its domain;  then
$\ddot\nu\Omega=1$ and $\mu_W$ is the subspace measure on
$\Omega$ induced by $\ddot\nu$.   As in 455U, let $\ddot\Sigma_t$ be

\Centerline{$\{F:F\in\ddot\Sigma$, $\omega'\in F$ whenever $\omega\in F$,
   $\omega'\in\Cdlg$ and $\omega'\restr[0,t]=\omega\restr[0,t]\}$,}

\noindent and $\ddot\Sigma^+_t=\bigcap_{s>t}\ddot\Sigma_s$, for $t>0$.

\medskip

{\bf (b)} For $\omega\in\Cdlg$ set

\Centerline{$\ddot\tau(\omega)
=\inf\{t:$ there is an $\omega'\in\Omega$ such that
$\omega'\restr[0,t]=\omega\restr[0,t]$ and $\tau(\omega')\le t\}$,}

\noindent counting $\inf\emptyset$ as $\infty$.   Then $\ddot\tau$ is a
stopping time adapted to $\langle\ddot\Sigma^+_t\rangle_{t\ge 0}$.
\Prf\ For $t\ge 0$, set

\Centerline{$F_t=\{\omega:\omega\in\Omega$, $\tau(\omega)<t\}\in\Sigma_t$}

\noindent (455Lb), and

\Centerline{$F'_t=\{\omega:\omega\in\Cdlg$, there is an $\omega'\in F_t$
such that $\omega\restr[0,t]=\omega'\restr[0,t]\}$.}

\noindent Since $F_t\in\Sigma_t$, $F'_t\cap\Omega=F_t$;  as $\Omega$ is
$\ddot\nu$-conegligible and $\ddot\nu$ is complete, $F'_t\in\ddot\Sigma$;
now of course $F'_t\in\ddot\Sigma_t$.   If $\omega\in F'_t$, let
$\omega'\in F_t$ be such that $\omega\restr[0,t]=\omega'\restr[0,t]$;  then
$\omega'$ witnesses that $\ddot\tau(\omega)\le\tau(\omega')<t$.   If
$\omega\in\Cdlg$ and $\ddot\tau(\omega)<t$, let $q<t$ and
$\omega'\in\Omega$ be such that $q$ is rational,
$\omega\restr[0,q]=\omega'\restr[0,q]$ and
$\tau(\omega')<q$;  then

\Centerline{$\omega\in F'_q\in\ddot\Sigma_q\subseteq\ddot\Sigma_t$.}

\noindent This shows that

\Centerline{$\{\omega:\omega\in\Cdlg$, $\ddot\tau(\omega)<t\}
=\bigcup_{q\in[0,t]\cap\Bbb Q}F'_q
\in\ddot\Sigma_t$.}

\noindent By 455Lb in the other direction, $\ddot\tau$ is a stopping time
adapted to $\langle\ddot\Sigma^+_t\rangle_{t\ge 0}$.\ \Qed

\medskip

{\bf (c)} By 455U, $\ddot\phi:\Cdlg\times\Cdlg\to\Cdlg$ is \imp\ for
$\ddot\nu\times\ddot\nu$ and $\ddot\nu$, where

$$\eqalign{\ddot\phi(\omega,\omega')(t)
&=\omega(t)\text{ if }t<\ddot\tau(\omega),\cr
&=\omega(\ddot\tau(\omega))+\omega'(t-\ddot\tau(\omega))
   \text{ if }t\ge\ddot\tau(\omega).\cr}$$

\noindent Now $\phi_{\tau}=\ddot\phi\restr\Omega\times\Omega$
and $\mu_W\times\mu_W$ is the subspace measure on
$\Omega\times\Omega$ induced by $\ddot\nu\times\ddot\nu$ (251Q),
so $\phi_{\tau}$ also is \imp.
}%end of proof of 477G

%see mt47bits, 477M \query

\leader{477H}{Some families of \dvrocolon{$\sigma$-algebras}}\cmmnt{ The
$\sigma$-algebras considered in Theorem 477G can be looked at in other ways
which are sometimes useful.

\medskip

\noindent}{\bf Proposition} Let $r\ge 1$
be an integer, $\mu_W\,\,r$-dimensional Wiener measure on
$\Omega=C(\coint{0,\infty};\BbbR^r)_0$ and $\Sigma$ its domain.   Set
$X^{(i)}_t(\omega)=\omega(t)(i)$ for $t\ge 0$ and $i<r$.   For
$I\subseteq\coint{0,\infty}$, let $\Tau_I$ be the $\sigma$-algebra
of subsets of $\Omega$
generated by $\{X^{(i)}_s-X^{(i)}_t:s$, $t\in I$, $i<r\}$, and
$\hat\Tau_I$ the $\sigma$-algebra
$\{E\symmdiff F:E\in\Tau_I$, $\mu_WF=0\}$.

(a) $\Tau_{\coint{0,\infty}}$ is the Borel $\sigma$-algebra of $\Omega$
either for the topology of pointwise convergence inherited from
$(\BbbR^r)^{\coint{0,\infty}}$ or $\BbbR^{r\times\coint{0,\infty}}$, or for
the topology of uniform convergence on compact sets.

(b) If $\Cal I$ is a family of subsets of $\coint{0,\infty}$ such that
for all distinct $I$, $J\in\Cal I$ either $\sup I\le\inf J$ or
$\sup J\le\inf I$ (counting $\inf\emptyset$ as $\infty$ and $\sup\emptyset$
as $0$), then $\family{I}{\Cal I}{\hat\Tau_I}$ is an
independent family of $\sigma$-algebras.

(c) For $t\ge 0$, let $\Sigma_t$ be
the $\sigma$-algebra of sets $F\in\Sigma$
such that $\omega'\in F$ whenever $\omega\in F$, $\omega'\in\Omega$ and
$\omega'\restr[0,t]=\omega\restr[0,t]$, and
$\Sigma^+_t=\bigcap_{s>t}\Sigma_s$.   Write $\hat\Tau_{[0,t]}^+$ for
$\bigcap_{s>t}\hat\Tau_{[0,s]}$.   Then, for any $t\ge 0$,

%\Centerline{$\Tau_{[0,t]}\subseteq\Sigma_t\subseteq\Sigma^+_t
%\subseteq\hat\Tau_{[0,t]}^+=\hat\Tau_{[0,t]}$.}

\Centerline{$\Tau_{[0,t]}\subseteq\Sigma_t\subseteq\Sigma^+_t
\subseteq\hat\Tau_{[0,t]}^+=\hat\Tau_{[0,t]}=\hat\Tau_{\coint{0,t}}$.}

(d) On the tail $\sigma$-algebra
$\bigcap_{t\ge 0}\hat\Tau_{\coint{t,\infty}}$, $\mu_W$
takes only the values $0$ and $1$.

\proof{{\bf (a)} Write $\Cal B(\Omega,\frak T_p)$,
$\Cal B(\Omega,\frak T_c)$ for the Borel algebras under the topologies
$\frak T_p$, $\frak T_c$ of pointwise convergence and uniform convergence
on compact sets.   Then
$\Tau_{\coint{0,\infty}}\subseteq\Cal B(\Omega,\frak T_p)$ because the
functionals $X^{(i)}_t$ are all $\frak T_p$-continuous, and
$\Cal B(\Omega,\frak T_p)\subseteq\Cal B(\Omega,\frak T_c)$ because
$\frak T_p\subseteq\frak T_c$.

Now $\Tau_{\coint{0,\infty}}$ includes a base for $\frak T_c$.
\Prf\ Suppose that $\omega\in\Omega$,
$n\in\Bbb N$ and
$V=\{\omega':\omega'\in\Omega,\,
  \sup_{t\in[0,n]}\|\omega'(t)-\omega(t)\|<2^{-n}\}$.   Then
(because every member of $\Omega$ is continuous)

\Centerline{$V=\bigcup_{m\in\Bbb N}\bigcap_{q\in\Bbb Q\cap[0,n]}
  \{\omega':\sum_{i=0}^{r-1}|X^{(i)}_q(\omega')-X^{(i)}_q(\omega)|^2
  \le 2^{-2n}-2^{-m}\}$}

\noindent belongs to $\Tau_{\coint{0,\infty}}$;  but such sets form a base
for $\frak T_c$.\ \QeD\

Since $\frak T_c$ is separable and metrizable,
$\Cal B(\Omega,\frak T_c)\subseteq\Tau_{\coint{0,\infty}}$ (4A3Da).

\medskip

{\bf (b)(i)} Suppose to begin with that $\Cal I$ is finite and
every member of $\Cal I$ is finite.
If we enumerate $\{0\}\cup\bigcup\Cal I$ in ascending order as
$\langle t_j\rangle_{j\le n}$,
$\langle X^{(i)}_{t_{j+1}}-X^{(i)}_{t_j}\rangle_{j<n,i<r}$ is an
independent family of real-valued random variables.   Taking
$J_I=\{t_j:j<n$, $t_j\in I$, $t_{j+1}\in I\}$ for $I\in\Cal I$,
$\{X^{(i)}_{t_{j+1}}-X^{(i)}_{t_j}:j\in J_I\}$ generates $\Tau_I$ for each
$I\in\Cal I$, because of the separation property of the members of
$\Cal I$, and $\family{I}{\Cal I}{J_I}$ is disjoint.   By 272K,
$\family{I}{\Cal I}{\Tau_I}$ is independent.

\medskip

\quad{\bf (ii)} Now suppose only that $\Cal I$ is finite and not empty.
For $I\in\Cal I$,
set $\Tau'_I=\bigcup_{J\subseteq I\text{ is finite}}\Tau_J$ for
$I\in\Cal I$;  then $\Tau'_I$ is an algebra of sets, and $\Tau_I$ is the
$\sigma$-algebra generated by $\Tau'_I$.   If $E_I\in\Tau'_I$ for
$I\in\Cal I$, there are $J_I\in[I]^{<\omega}$ such that $E_I\in\Tau_{J_I}$
for $I\in\Cal I$, so
$\mu_W(\bigcap_{I\in\Cal I}E_I)=\prod_{I\in\Cal I}\mu_W(E_I)$, by
($\alpha$).   Inducing on $n$, and using the Monotone Class Theorem
for the inductive step, we see that
$\mu_W(\bigcap_{I\in\Cal I}E_I)=\prod_{I\in\Cal I}\mu_W(E_I)$
whenever $E_I\in\Tau_I$ for every $I\in\Cal I$ and
$\#(\{I:E_I\notin\Tau'_I\})\le n$.   At the end of the induction, with
$n=\#(\Cal I)$, we have
$\mu_W(\bigcap_{I\in\Cal I}E_I)=\prod_{I\in\Cal I}\mu_W(E_I)$
whenever $E_I\in\Tau_I$ for every $I\in\Cal I$;  that is,
$\family{I}{\Cal I}{\Tau_I}$ is independent.

\medskip

\quad{\bf (iii)} Thus $\family{I}{\Cal J}{\Tau_I}$ is independent for every
non-empty finite $\Cal J\subseteq\Cal I$, and $\family{I}{\Cal I}{\Tau_I}$
is independent (272Bb).

\medskip

{\bf (c)(i)} If $s$, $s'\le t$ and $i<r$
then $X^{(i)}_s$, $X^{(i)}_{s'}$ and $X^{(i)}_s-X^{(i)}_{s'}$ are
$\Sigma_t$-measurable, so $\Tau_{[0,t]}\subseteq\Sigma_t$.
Of course $\Sigma_t\subseteq\Sigma^+_t$.

\medskip

\quad{\bf (ii)} $\Sigma_t\subseteq\hat\Tau_{[0,t]}$.   \Prf\ Suppose that
$F\in\Sigma_t$.   Set $D=[0,t]\cap(\Bbb Q\cup\{t\})$, and set
$g(\omega)=\omega\restr D$ for $\omega\in\Omega$;  then
$g:\Omega\to(\BbbR^r)^D$ is continuous (when $\Omega$ is given the topology
of pointwise convergence inherited from $(\BbbR^r)^{\coint{0,\infty}}$,
for definiteness), and $F=g^{-1}[g[F]]$.   Now the Borel $\sigma$-algebra
of $(\BbbR^r)^D\cong\BbbR^{r\times D}$ is the $\sigma$-algebra generated
by the functionals $\omega\mapsto\omega(t)(i):(\BbbR^r)^D\to\Bbb R$ for
$t\in D$ and $i<r$ (4A3D(c-i)), and for such $t$ and $i$,
$\omega\mapsto g(\omega)(t)(i)=X_t^{(i)}(\omega)$ is
$\Tau_{[0,t]}$-measurable;  so $g$ is $\Tau_{[0,t]}$-measurable.   Now
there is a sequence $\sequencen{K_n}$ of compact subsets of $F$ such that
$\sup_{n\in\Bbb N}\mu_WK_n=\mu_WF$.    In this case,
$g[K_n]\subseteq(\BbbR^r)^D$ is compact and $K'_n=g^{-1}[g[K_n]]$ belongs
to $\Tau_{[0,t]}$, for each $n$.   So $F'=\bigcup_{n\in\Bbb N}K'_n$ belongs
to $\Tau_{[0,t]}$, and $\mu_W(F\setminus F')=0$.

Similarly, applying the same argument to $\Omega\setminus F$, we have an
$F''\in\Tau_{[0,t]}$ such that $F''\supseteq F$ and
$\mu_W(F''\setminus F)=0$.   So $F\in\hat\Tau_{[0,t]}$.\ \Qed

Consequently $\Sigma_t^+\subseteq\hat\Tau_{[0,t]}^+$.

\medskip

\quad{\bf (iii)}\grheada\
Let $\Cal A$ be the family of those sets $G\in\Sigma$
such that $\chi G$ has a conditional expectation on
$\hat\Tau_{[0,t]}^+$ which is $\Tau_{[0,t]}$-measurable.
Then $\Cal A$ is a Dynkin class (definition:  136A).
If $E\in\Tau_{[0,t]}$ and $F\in\Tau_{\coint{s,\infty}}$ where $s>t$,
then $(\mu_WF)\chi\Omega$ is a conditional
expectation of $\chi F$ on $\hat\Tau_{[0,t]}^+$, because
$\hat\Tau_{[0,t]}^+\subseteq\hat\Tau_{[0,s]}$ and
$\Tau_{\coint{s,\infty}}$ are independent.   As
$E\in\hat\Tau_{[0,t]}^+$,
$(\mu_WF)\chi E$ is a conditional expectation of $\chi(E\cap F)$ on
$\hat\Tau_{[0,t]}^+$ (233Eg), and $E\cap F\in\Cal A$.   Since
$\Cal E=\{E\cap F:E\in\Sigma_t$,
$F\in\bigcup_{s>t}\Tau_{\coint{s,\infty}}\}$ is closed
under finite intersections, the Monotone Class Theorem, in the form 136B,
shows that $\Cal A$ includes the $\sigma$-algebra $\Tau$ generated by
$\Cal E$;  note that $\Tau$ includes
$\Tau_{[0,t]}\cup\Tau_{\coint{s,\infty}}$ whenever
$s>t$.   Now $X^{(i)}_u-X^{(i)}_s$ is $\Tau$-measurable whenever
$0\le s\le u$.   \Prf\ If $u\le t$, $X^{(i)}_u-X^{(i)}_s$ is
$\Tau_{[0,t]}$-measurable, therefore $\Tau$-measurable;  if $t<s$,
$X^{(i)}_u-X^{(i)}_s$ is
$\Tau_{\coint{s,\infty}}$-measurable, therefore $\Tau$-measurable.
If $s\le t<u$,
let $\sequencen{t_n}$ be a sequence in $\ocint{t,u}$ with
limit $t$.   Then

\Centerline{$X^{(i)}_u-X^{(i)}_s
=\lim_{n\to\infty}(X^{(i)}_u-X^{(i)}_{t_n})+(X^{(i)}_t-X^{(i)}_s)$}

\noindent is $\Tau$-measurable.\ \Qed

\medskip

\qquad\grheadb\ This means that $\Tau$ includes $\Tau_{\coint{0,\infty}}$.
It follows that $\Cal A=\Sigma$, because for any $G\in\Sigma$ there is a
$G'\in\Tau_{\coint{0,\infty}}$ such that $G\symmdiff G'$ is negligible,
and now $\chi G$ and $\chi G'$ have the same conditional expectations.
So $\hat\Tau_{[0,t]}^+=\hat\Tau_{[0,t]}$.   \Prf\ Of course
$\hat\Tau_{[0,t]}^+\supseteq\hat\Tau_{[0,t]}$.
If $H\in\hat\Tau_{[0,t]}^+$ there is a $\Tau_{[0,t]}$-measurable function
$g$ which is a conditional expectation of
$\chi H$ on $\hat\Tau_{[0,t]}^+$.   But in this case $g\eae\chi H$,
so, setting $E=\{\omega:g(\omega)=1\}\in\Tau_{[0,t]}$, $E\symmdiff H$ is
negligible and $H\in\hat\Tau_{[0,t]}$.   Thus
$\hat\Tau_{[0,t]}^+\subseteq\hat\Tau_{[0,t]}$.\ \Qed

\medskip

\qquad\grheadc\ Observe next that $X^{(i)}_t$ is
$\Tau_{\coint{0,t}}$-measurable for $i<r$.    \Prf\ If $t=0$ then
$X^{(i)}_t$ is the constant function with value $0$.
Otherwise, there is a strictly
increasing sequence $\sequencen{s_n}$ in $\coint{0,t}$ with limit $t$, so
that $X^{(i)}_t=\lim_{n\to\infty}X^{(i)}_{s_n}$ is the limit of a sequence
of $\Tau_{\coint{0,t}}$-measurable functions and is itself
$\Tau_{\coint{0,t}}$-measurable.\ \QeD\   But this means that
$\Tau_{[0,t]}=\Tau_{\coint{0,t}}$, so
$\hat\Tau_{[0,t]}^+\subseteq\hat\Tau_{[0,t]}=\hat\Tau_{\coint{0,t}}$.
In the other direction, of course
$\Tau_{\coint{0,t}}\subseteq\Tau_{[0,t]}^+$ and
$\hat\Tau_{\coint{0,t}}\subseteq\hat\Tau_{[0,t]}^+$, so we have equality.

\medskip

{\bf (d)} Set $\Tau'=\bigcup_{t\ge 0}\Tau_{[0,t]}$.
If $E\in\bigcap_{t\ge 0}\hat\Tau_{\coint{t,\infty}}$, then
$\mu_W(E\cap F)=\mu_WE\cdot\mu_WF$ for every $F\in\Tau'$.   By the
Monotone Class Theorem again, $\mu_W(E\cap F)=\mu_WE\cdot\mu_WF$ for every
$F$ in the $\sigma$-algebra
generated by $\Tau'$, which is $\Cal B(\Omega)$, by (a).
Now $\mu_W\LLcorner E$ (definition:  234M\formerly{2{}34E}) and $(\mu_WE)\mu_W$
are Radon measures on $\Omega$ (416Sa) which agree on $\Cal B(\Omega)$, so
must be identical.   In particular,

\Centerline{$\mu_WE=(\mu_W\LLcorner E)(E)=(\mu_WE)^2$}

\noindent and $\mu_WE$ must be either $0$ or $1$.
}%end of proof of 477H

\leader{477I}{Hitting times}\cmmnt{ In 455M I introduced `hitting times'.
I give a paragraph now to these in the special case of Brownian motion;
such stopping times will dominate
the applications of the theory in \S\S478-479.} Take $r\ge 1$, and let
$\mu_W$ be Wiener measure on $\Omega=C(\coint{0,\infty};\BbbR^r)_0$ and
$\Sigma$ its domain;
for $t\ge 0$ define $\Sigma^+_t$ and $\Tau_{[0,t]}$ as in 477G and 477H.
Give $\Omega$ its topology of uniform convergence on compact sets.

\spheader 477Ia Suppose that $A\subseteq\BbbR^r$.
For $\omega\in\Omega$ set
$\tau(\omega)=\inf\{t:t\in\coint{0,\infty}$, $\omega(t)\in A\}$, counting
$\inf\emptyset$ as $\infty$.
I will call $\tau$ the {\bf Brownian hitting time}
to $A$, or the {\bf Brownian exit time} from $\BbbR^r\setminus A$.
I will say that the {\bf Brownian hitting probability} of $A$, or
the {\bf Brownian exit probability} of $\BbbR^r\setminus A$, is
$\hp(A)=\mu_W\{\omega:\tau(\omega)<\infty\}$ if this is defined.
More generally, I will write

\Centerline{$\hp^*(A)=\mu_W^*\{\omega:\tau(\omega)<\infty\}
=\mu_W^*\{\omega:\omega^{-1}[A]\ne\emptyset\}$,}

\noindent the {\bf outer Brownian hitting probability},
for any $A\subseteq\BbbR^r$.

\spheader 477Ib If $A\subseteq\BbbR^r$ is analytic,
the Brownian hitting time to $A$
is a stopping time adapted to the family
$\langle\Sigma^+_t\rangle_{t\ge 0}$.   \prooflet{\Prf\ Let $\Cdlg$ be the
space of \cadlag\ functions from $\coint{0,\infty}$ to $\BbbR^r$, and
define $\ddot\Sigma$ as in the proof
of 477G;  let $\ddot\tau$ be the hitting time on $\Cdlg$ defined by $A$.
By 455Ma, $\ddot\tau$ is $\ddot\Sigma$-measurable, so
$\tau=\ddot\tau\restr\Omega$ is $\Sigma$-measurable.   Now (as in 455Mb)
$\{\omega:\tau(\omega)<t\}\in\Sigma_t$ for every $t$, so
$\tau$ is adapted to $\langle\Sigma^+_t\rangle_{t\ge 0}$.\ \Qed}

In particular, there is a well-defined Brownian hitting probability of $A$.

\spheader 477Ic Let $F\subseteq\BbbR^r$ be a closed set, and $\tau$ the
Brownian hitting time to $F$.

\medskip

\quad{\bf (i)} If $\tau(\omega)<\infty$, then

\Centerline{$\tau(\omega)
=\inf\omega^{-1}[F]=\min\omega^{-1}[F]$}

\noindent because $\omega$ is continuous.   If $0\notin F$ and
$\tau(\omega)<\infty$, then
$\omega(\tau(\omega))\in\partial F$.

\medskip

\quad{\bf (ii)} $\tau$ is lower semi-continuous.
\prooflet{\Prf\ For any $t\in\coint{0,\infty}$,

\Centerline{$\{\omega:\tau(\omega)>t\}
=\{\omega:\omega(s)\notin F$ for every $s\le t\}$}

\noindent is open in $\Omega$.\ \QeD}

\medskip

\quad{\bf (iii)} $\tau$ is adapted to $\langle\Tau_{[0,t]}\rangle_{t\ge 0}$.
\prooflet{\Prf\
Let $\sequencen{G_n}$ be a non-increasing sequence of open sets including
$F$ such that $F=\bigcap_{n\in\Bbb N}\overline{G}_n$.
Then, for $\omega\in\Omega$ and $t>0$,

$$\eqalignno{\tau(\omega)\le t
&\iff\omega[\,[0,t]\,]\cap F\ne\emptyset\cr
\displaycause{because $\omega$ is continuous}
&\iff\omega[\,[0,t]\,]\cap G_n\ne\emptyset\text{ for every }n\in\Bbb N\cr
\displaycause{because $\omega[\,[0,t]\,]$ is compact}
&\iff\text{ for every }n\in\Bbb N\text{ there is a rational }q\le t
  \text{ such that }\omega(q)\in G_n.\cr}$$

\noindent So

\Centerline{$\{\omega:\tau(\omega)\le t\}
=\bigcap_{n\in\Bbb N}\bigcup_{q\in\Bbb Q\cap[0,t]}
   \{\omega:\omega(q)\in G_n\}\in\Tau_{[0,t]}$.}

\noindent Of course $\{\omega:\tau(\omega)=0\}$ is either $\Omega$ (if
$0\in F$) or $\emptyset$ (if $0\notin F$), so belongs to
$\Tau_{[0,0]}$.\ \Qed}

In the language of 477G, we have $\Tau_{[0,t]}\subseteq\Sigma_t$ for every
$t\ge 0$\cmmnt{ (477Hc)}, so $\tau$ must also be adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$.

\spheader 477Id If $A\subseteq\BbbR^r$ is any set, then

\Centerline{$\hp^*(A)
=\min\{\hp(B):B\supseteq A$ is an analytic set$\}
=\min\{\hp(E):E\supseteq A$ is a G$_{\delta}$ set$\}$.}

\prooflet{\noindent\Prf\ Of course

$$\eqalign{\hp^*(A)
&\le\inf\{\hp(B):B\supseteq A\text{ is an analytic set}\}\cr
&=\min\{\hp(B):B\supseteq A\text{ is an analytic set}\}\cr
&\le\inf\{\hp(E):E\supseteq A\text{ is a G}_{\delta}\text{ set}\}
=\min\{\hp(E):E\supseteq A\text{ is a G}_{\delta}\text{ set}\}\cr}$$

\noindent just because $\hp^*$ is an order-preserving function.   If
$\gamma>\hp^*(A)$, there is a compact $K\subseteq\Omega$ such that
$\omega^{-1}[A]=\emptyset$ for every $\omega\in K$ and
$\mu_WK\ge 1-\gamma$.   Now
$F=\{\omega(t):\omega\in K$, $t\in\coint{0,\infty}\}$ is a K$_{\sigma}$ set
not meeting $A$, so $E=\BbbR^r\setminus F$ is a G$_{\delta}$ set including
$A$.   Since $\omega^{-1}[E]$ is empty for every $\omega\in K$,
$\hp(E)\le\mu_W(\Omega\setminus K)\le\gamma$.   As $\gamma$ is arbitrary,

\Centerline{$\inf\{\hp(E):E\supseteq A$ is a G$_{\delta}$ set$\}
\le\hp^*(A)$}

\noindent and we have equality throughout.\ \Qed}

\spheader 477Ie If $A\subseteq\BbbR^r$ is analytic, then
$\hp(A)=\sup\{\hp(K):K\subseteq A$ is compact$\}$.   \prooflet{\Prf\
Suppose that $\gamma<\hp(A)$.
Set $E=\{(\omega,t):\omega\in\Omega$, $t\ge 0$, $\omega(t)\in A\}$.
Then $E$ is analytic and $\hp(A)=\mu_W\pi_1[E]$, where
$\pi_1(\omega,t)=\omega$ for $(\omega,t)\in E$.   Let $\lambda$ be the
subspace measure $(\mu_W)_{\pi_1[E]}$.   By 433D, there is a Radon measure
$\lambda'$ on $E$ such that $\lambda=\lambda'\pi_1^{-1}$.
Then $\lambda'E=\hp(A)>\gamma$, so there is a compact set
$L\subseteq E$ such that $\lambda'L\ge\gamma$.   Set
$K=\{\omega(t):(\omega,t)\in L\}$;  then $K\subseteq\BbbR^r$ is compact,
and

\Centerline{$\hp(K)=\mu_W\{\omega:\omega(t)\in K$ for some $t\ge 0\}
\ge\mu_W\pi_1[L]\ge\lambda'L\ge\gamma$.}

\noindent As $\gamma$ is arbitrary,
$\hp(A)\le\sup\{\hp(K):K\subseteq A$ is compact$\}$;  the reverse
inequality is trivial.\ \Qed}

\cmmnt{\medskip

\noindent{\bf Remark} 477Id-477Ie are characteristic of Choquet capacities
(432J-432L);  see 478Xe below.}

\leader{477J}{}\cmmnt{ As an example of the use of 477G, I give
a classical result on one-dimensional Brownian motion.

\medskip

\noindent}{\bf Proposition} Let $\mu_W$ be Wiener measure
on $\Omega=C(\coint{0,\infty})_0$.   Set
$X_t(\omega)=\omega(t)$ for $\omega\in\Omega$.
Then

\Centerline{$\Pr(\max_{s\le t}X_s\ge\alpha)
=2\Pr(X_t\ge\alpha)
=\Bover2{\sqrt{2\pi}}\int_{\alpha/\sqrt t}^{\infty}e^{-u^2/2}du$}

\noindent whenever $t>0$ and $\alpha\ge 0$.

\proof{ Let $\tau$ be the Brownian hitting time to
$F=\{x:x\in\Bbb R$, $x\ge\alpha\}$;  because $F$ is closed,
$\tau$ is a stopping time adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$, as in 477Ic.   Let
$\phi_{\tau}:\Omega\times\Omega\to\Omega$ be the corresponding \imp\
function as in 477G, and set $E=\{\omega:\tau(\omega)<t\}$.
Note that as $\omega(\tau(\omega))=\alpha$ whenever $\tau(\omega)$ is
finite, $\Pr(\tau=t)\le\Pr(X_t=\alpha)=0$, and

\Centerline{$\mu_WE=\Pr(\tau\le t)=\Pr(\max_{s\le t}X_s\ge\alpha)$.}

\noindent Now

$$\eqalignno{\Pr(X_t\ge\alpha)
&=\mu_W\{\omega:\omega(t)\ge\alpha\}
=\mu_W^2\{(\omega,\omega'):\phi_{\tau}(\omega,\omega')(t)\ge\alpha\}\cr
&=\mu_W^2\{(\omega,\omega'):\tau(\omega)\le t,\,
   \phi_{\tau}(\omega,\omega')(t)\ge\alpha\}\cr
\displaycause{because if $\tau(\omega)>t$ then
$\phi_{\tau}(\omega,\omega')(t)=\omega(t)<\alpha$}
&=\mu_W^2\{(\omega,\omega'):\tau(\omega)<t,\,
   \phi_{\tau}(\omega,\omega')(t)\ge\alpha\}\cr
\displaycause{because $\{\omega:\tau(\omega)=t\}$ is negligible}
&=\mu_W^2\{(\omega,\omega'):\tau(\omega)<t,\,
   \omega(\tau(\omega))+\omega'(t-\tau(\omega))\ge\alpha\}\cr
&=\mu_W^2\{(\omega,\omega'):\tau(\omega)<t,\,
   \omega'(t-\tau(\omega))\ge 0\}\cr
&=\int_E\mu_W\{\omega':\omega'(t-\tau(\omega))\ge 0\}\mu_W(d\omega)\cr
&=\Bover12\mu_WE
=\Bover12\Pr(\max_{s\le t}X_s\ge\alpha).\cr}$$

To compute the value, observe that $X_t$ has the same distribution as
$\sqrt tZ$ where $Z$ is a standard normal random variable, so that

\Centerline{$\Pr(X_t\ge\alpha)
=\Pr(Z\ge\bover{\alpha}{\sqrt t})
=\Bover1{\sqrt{2\pi}}\int_{\alpha/\sqrt{t}}^{\infty}e^{-u^2/2}du$.}
}%end of proof of 477J

\leader{477K}{Typical Brownian \dvrocolon{paths}}\cmmnt{ A vast amount is
known concerning the nature of `typical' members of
$\Omega$;   that is to say, a great many
interesting $\mu_W$-conegligible sets have been found.   Here I will
give only a couple of basic results;  the first because it is essential to any
picture of Brownian motion, and the second because it is relevant to a
question in \S479.   Others are in 478M, 478Yi and 479R.

\medskip

\noindent}{\bf Proposition} Let $\mu_W$
be\cmmnt{ one-dimensional} Wiener
measure on $\Omega=C(\coint{0,\infty})_0$.
Then $\mu_W$-almost every element of $\Omega$ is nowhere differentiable.

\proof{ Note first that if $\eta>0$ and $Z$ is a standard normal random
variable, then $\Pr(|Z|\le\eta)\le\eta$, because
the maximum value of the probability density function of $Z$
is $\Bover1{\sqrt{2\pi}}\le\Bover12$.   For $m$, $n$, $k\in\Bbb N$, set

\Centerline{$F_m=\{\omega:\omega\in\Omega$ and
there is a $t\in\coint{0,m}$ such that
$\limsup_{s\downarrow 0}\Bover{|\omega(s)-\omega(t)|}{s-t}<m\}$,}

$$\eqalign{E_{mnk}
&=\{\omega:\omega\in\Omega,\,
        |\omega(2^{-n}(k+2))-\omega(2^{-n}(k+1))|\le 3\cdot 2^{-n}m,\cr
&\mskip200mu
        |\omega(2^{-n}(k+3))-\omega(2^{-n}(k+2))|\le 5\cdot 2^{-n}m,\cr
&\mskip200mu
        |\omega(2^{-n}(k+4))-\omega(2^{-n}(k+3))|\le 7\cdot 2^{-n}m\},\cr}$$
	
\Centerline{$E_{mn}=\bigcup_{k<2^nm}E_{mnk}$.}

\noindent Now we can estimate the measure of $E_{mnk}$, because for any
$\alpha$, $t\ge 0$, $2^{n/2}(X_{t+2^{-n}}-X_t)$ has a standard normal
distribution (taking $X_t(\omega)=\omega(t)$, as usual), so

\Centerline{$\Pr(|X_{t+2^{-n}}-X_t|\le\alpha)\le 2^{n/2}\alpha$;}

\noindent since $E_{mnk}$ is the intersection of three independent sets of
this type,

\Centerline{$\mu_WE_{mnk}
\le 2^{n/2}\cdot 3\cdot 2^{-n}m\cdot2^{n/2}\cdot 5
   \cdot 2^{-n}m\cdot2^{n/2}\cdot 7\cdot 2^{-n}m
=105\,m^32^{-3n/2}$.}

\noindent Accordingly

\Centerline{$\mu_WE_{mn}\le\sum_{k<2^nm}\mu_WE_{mnk}\le 105\,m^42^{-n/2}$.}

Next, observe that
$F_m\subseteq\bigcup_{l\in\Bbb N}\bigcap_{n\ge l}E_{mn}$.   \Prf\
If $\omega\in F_m$, let $t\in\coint{0,m}$ be such that
$\limsup_{s\downarrow 0}\Bover{|\omega(s)-\omega(t)|}{s-t}<m$,
and $l\in\Bbb N$ such that
$|\omega(s)-\omega(t)|\le m(s-t)$ whenever $t<s\le t+4\cdot 2^{-l}$.
Take any $n\ge l$.   Then there is a $k<2^nm$ such that
$2^{-n}k\le t<2^{-n}(k+1)$.   In this case,

\Centerline{$|\omega(2^{-n}(k+j))-\omega(t)|\le 2^{-n}jm$}

\noindent for $1\le j\le 4$,

\Centerline{$|\omega(2^{-n}(k+j+1))-\omega(2^{-n}(k+j))|\le (2j+1)2^{-n}m$}

\noindent for $1\le j\le 3$, and $\omega\in E_{mnk}\subseteq E_{mn}$.\ \Qed

Since

\Centerline{$\mu_W(\bigcup_{l\in\Bbb N}\bigcap_{n\ge l}E_{mn})
\le\liminf_{n\to\infty}\mu_WE_{mn}=0$,}

\noindent $F_m$ is negligible.   So $F=\bigcup_{m\in\Bbb N}F_m$ is
negligible.   But $F$ includes any member of $\omega$ which is
differentiable at any point of $\ooint{0,\infty}$, and more.   So almost
every path is nowhere differentiable.
}%end of proof of 477K

\leader{477L}{Theorem} Let $r\ge 1$ be an integer,
and $\mu_W$ Wiener measure on $\Omega=C(\coint{0,\infty};\BbbR^r)_0$;
for $s>0$ let $\mu_{Hs}$ be $s$-dimensional Hausdorff measure on
$\BbbR^r$.

(a)\cmmnt{ ({\smc Taylor 53})}
$\{\omega(t):t\in\coint{0,\infty}\}$ is $\mu_{H2}$-negligible
for $\mu_W$-almost every $\omega$.

(b) Now suppose that $r\ge 2$.   For
$\omega\in\Omega$, let $F_{\omega}$ be the compact set
$\{\omega(t):t\in[0,1]\}$.
Then for $\mu_W$-almost every $\omega\in\Omega$,
$\mu_{Hs}F_{\omega}=\infty$ for every $s\in\ooint{0,2}$.

\proof{{\bf (a)(i)} For $0\le s\le t$ and $\omega\in\Omega$ set
$K_{st}(\omega)=\{\omega(u):s\le u\le t\}$ and
$d_{st}(\omega)=\diam K_{st}(\omega)$.   Note that
$d_{st}:\Omega\to\coint{0,\infty}$ is continuous
(for the topology of uniform convergence on compact sets, of course).

\medskip

\qquad\grheada\ If $0\le s\le t$ then $\Expn(d_{st}^2)\le 8r(t-s)$.
\Prf\ As $\langle X_{u+s}-X_s\rangle_{u\ge 0}$ and
$\langle X_u\rangle_{u\ge 0}$ have the same distribution, $d_{st}$ has
the same distribution as $d_{0,t-s}$, and we may suppose that $s=0$.
[If you prefer:  if $S_s:\Omega\to\Omega$ is the shift operator of 477Ec,
$K_{st}(\omega)=\omega(s)+K_{0,t-s}(S_s\omega)$, so
$d_{st}(\omega)=d_{0,t-s}(S_s\omega)$, while $S_s$ is \imp.]
In this case,

\Centerline{$d_{0t}(\omega)^2
\le 4\max_{s\in[0,t]}\|\omega(s)\|^2
\le 4\sum_{j=0}^{r-1}\max_{s\in[0,t]}\omega(s)(j)^2$.}

\noindent For each $j<r$,

$$\eqalignno{\int\max_{s\in[0,t]}\omega(s)(j)^2\mu_W(d\omega)
&=\int_0^{\infty}\mu_W\{\omega:
   \max_{s\in[0,t]}\omega(s)(j)^2\ge\beta\}d\beta\cr
&\le\int_0^{\infty}\mu_W\{\omega:\max_{s\in[0,t]}\omega(s)(j)
   \ge\sqrt\beta\}\cr
&\mskip100mu
   +\mu_W\{\omega:\min_{s\in[0,t]}\omega(s)(j)
      \le-\sqrt\beta\}
   d\beta\cr
&=2\int_0^{\infty}\mu_W\{\omega:\max_{s\in[0,t]}\omega(s)(j)
   \ge\sqrt\beta\}d\beta\cr
\displaycause{because $\mu_W$ is invariant under reflections
in $\BbbR^r$, see 477Ed}
&=4\int_0^{\infty}\mu_W\{\omega:\omega(t)(j)\ge\sqrt\beta\}d\beta\cr
\displaycause{by 477J, applied to the $j$th coordinate projection of
$\Omega$ onto $C(\coint{0,\infty})_0$, which is \imp, by 477Da or
477Ed and 477Eg}
&=2\int_0^{\infty}\mu_W\{\omega:\omega(t)(j)^2\ge\beta\}d\beta\cr
\displaycause{again because $\mu_W$ is symmetric}
&=2\int_{\Omega}\omega(t)(j)^2\mu_W(d\omega)
=2\Expn(tZ^2)\cr
\displaycause{where $Z$ is a standard normal random variable}
&=2t.\cr}$$

\noindent Summing,

\Centerline{$\Expn(d_{0t}^2)
\le 4\sum_{j=0}^{r-1}\int_{\Omega}\max_{s\in[0,t]}\omega(s)(j)^2
    \mu_W(d\omega)
\le 8rt$.  \Qed}

\medskip

\qquad\grheadb\ For any $\epsilon>0$, $\Pr(d_{01}\le\epsilon)>0$.   \Prf\
$\{\omega:d_{01}(\omega)\le\epsilon\}$ is a neighbourhood of $0$ for the
topology of uniform convergence on compact sets, so has non-zero measure,
by 477F.\ \Qed

\medskip

\quad{\bf (ii)}
For a non-empty finite set $I\subseteq\coint{0,\infty}$ and
$\omega\in\Omega$ set

\Centerline{$g_I(\omega)
=\sum_{j=0}^{n-1}d_{t_{j-1},t_j}(\omega)^2$}

\noindent where $\langle t_j\rangle_{j\le n}$ enumerates $I$ in increasing
order.   For $0\le s\le t$ and $\omega\in\Omega$ set

\Centerline{$h_{st}(\omega)
=\inf_{\{s,t\}\subseteq I\subseteq[s,t]\text{ is finite}}g_I(\omega)$;}

\noindent then $h_{st}$ is $\Tau_{[s,t]}$-measurable, in the language of
477H.   \Prf\ The point is that
$I\mapsto g_I(\omega)$ is a continuous function of the members of $I$,
at least if we restrict attention to sets $I$ of a fixed size.
So if $D$ is any countable dense subset of $[s,t]$ containing $s$ and $t$,

\Centerline{$h_{st}
=\inf_{\{s,t\}\subseteq I\subseteq D\text{ is finite}}g_I$.}

\noindent On the other hand, if $I\subseteq D$ is enumerated as
$\langle t_i\rangle_{i\le n}$,

\Centerline{$g_I(\omega)
=\sum_{i=0}^{n-1}\max_{u,u'\in D\cap[t_i,t_{i+1}]}
  \|\omega(u)-\omega(u')\|^2$,}

\noindent so $g_I$ is $\Tau_{[s,t]}$-measurable.\ \Qed

\medskip

\quad{\bf (iii)} We need the following facts about the $h_{st}$.

\medskip

\qquad\grheada\ If $0\le s\le t$, then the distribution of $h_{st}$ is the
same as the distribution of $h_{0,t-s}$, again because
$\langle X_{s+u}-X_s\rangle_{u\ge 0}$ has the same distribution as
$\langle X_u\rangle_{u\ge 0}$.   [In the language suggested in the proof of
(i-$\alpha$),
we have $g_{s+I}(\omega)=g_I(S_s(\omega))$ for any $\omega\in\Omega$
and non-empty finite $I\subseteq\coint{0,\infty}$, so
$h_{st}(\omega)=h_{0,t-s}(S_s\omega)$.]

\medskip

\qquad\grheadb\ $h_{0t}$ has finite expectation.   \Prf\
$h_{0t}\le g_{\{0,t\}}=d_{0t}^2$, so we can use (i).\ \Qed

\medskip

\qquad\grheadc\ $h_{su}\le h_{st}+h_{tu}$ if $s\le t\le u$.   \Prf\ If
$\{s,t\}\subseteq I\subseteq[s,t]$ and $\{t,u\}\subseteq J\subseteq[t,u]$
then $\{s,u\}\subseteq I\cup J\subseteq[s,u]$ and
$g_{I\cup J}=g_I+g_J$.\ \QeD\

\medskip

\qquad\grheadd\ If $s\le t\le u$ then $h_{st}$ and $h_{tu}$ are
independent, because
$\Tau_{[s,t]}$ and $\Tau_{[t,u]}$ are independent (477H(b-i)).

\medskip

\qquad\grheade\ The
distribution of $h_{0t}$ is the same as the distribution of $th_{01}$
whenever $t\ge 0$.
\Prf\ The case $t=0$ is trivial.   For $t>0$, define
$U_t:\Omega\to\Omega$ by saying that
$U_t(\omega)(s)=\Bover1{\sqrt{t}}\omega(ts)$, as in 477Ee.   Then

\Centerline{$K_{su}(U_t(\omega))=\Bover1{\sqrt{t}}K_{ts,tu}(\omega)$,
\quad$d_{su}(U_t(\omega))=\Bover1{\sqrt{t}}d_{ts,tu}(\omega)$,}

\Centerline{$g_I(U_t(\omega))=\Bover1tg_{tI}(\omega)$,
\quad$h_{su}(U_t(\omega))=\Bover1th_{ts,tu}(\omega)$,}

\noindent whenever $s\le u$, $\{s,u\}\subseteq I\subseteq[s,u]$
and $\omega\in\Omega$, and

$$\eqalignno{\mu_W\{\omega:th_{01}(\omega)\ge\alpha\}
&=\mu_W\{\omega:th_{01}(U_t(\omega))\ge\alpha\}\cr
\displaycause{because $U_t$ is an automorphism of $(\Omega,\mu_W)$}
&=\mu_W\{\omega:h_{0t}(\omega)\ge\alpha\}\cr}$$

\noindent for every $\alpha\in\Bbb R$.\ \Qed

\medskip

\qquad{\bf($\pmb{\zeta}$)} Consequently

\Centerline{$\Expn(h_{st})=\Expn(h_{0,t-s})=(t-s)\Expn(h_{01})$}

\noindent whenever $s\le t$, and

\Centerline{$\Expn(h_{st})+\Expn(h_{tu})=\Expn(h_{su})$}

\noindent whenever $s\le t\le u$.   Since $h_{st}+h_{tu}\ge h_{su}$, by
(iii), we must have $h_{st}+h_{tu}\eae h_{su}$.

\medskip

\qquad{\bf($\pmb{\eta}$)} For any $\eta>0$, $\Pr(h_{01}\le 4\eta^2)>0$.
\Prf\ By
(i-$\beta$),
 $\Pr(d_{01}\le 2\eta)>0$, and $h_{01}\le g_{\{0,1\}}=d_{01}^2$.\ \Qed

\medskip

\quad{\bf (iv)} For $t\ge 0$ let $\varphi_t$ be the characteristic function of
$h_{0t}$, that is, $\varphi_t(\alpha)=\Expn(\exp(i\alpha h_{0t}))$ for
$\alpha\in\Bbb R$ (285Ab).   Working through the facts listed above, we see
that

$$\eqalignno{\varphi_t(1)
&=\Expn(\exp(ih_{0t}))
=\Expn(\exp(ith_{01}))\cr
\displaycause{by (iii-$\epsilon$)}
&=\varphi_1(t),\cr
\varphi_1(s)\varphi_1(t)
&=\varphi_s(1)\varphi_t(1)
=\Expn(\exp(ih_{0s}))\Expn(\exp(ih_{0t}))\cr
&=\Expn(\exp(ih_{0s}))\Expn(\exp(ih_{s,s+t}))\cr
\displaycause{by (iii-$\alpha$)}
&=\Expn(\exp(ih_{0s})\exp(ih_{s,s+t}))\cr
\displaycause{because $h_{0s}$ and $h_{s,s+t}$ are independent, by (c-iv)}
&=\Expn(\exp(i(h_{0s}+h_{s,s+t})))
=\Expn(ih_{0,s+t})\cr
\displaycause{by (iii-$\zeta$)}
&=\varphi_1(s+t),\cr}$$

\noindent for all $s$, $t\ge 0$;  while $\varphi_1$ is differentiable, because
$h_{01}$ has finite expectation ((iii-$\alpha$) above and 285Fd).
It follows that there is a
$\gamma\in\Bbb R$ such that $\varphi_1(t)=e^{i\gamma t}$ for every
$t\in\Bbb R$.   \Prf\ Set $\gamma=\Bover1i\varphi_1'(0)=\Expn(h_{01})$
(285Fd) and $\psi(t)=e^{-i\gamma t}\varphi_1(t)$ for $t\in\Bbb R$.
If $t>0$, then

\Centerline{$\varphi'_1(t)
=\lim_{s\downarrow 0}\Bover1s(\varphi_1(t+s)-\varphi_1(t))
=\varphi_1(t)\lim_{s\downarrow 0}\Bover1s(\varphi_1(s)-1)
=i\varphi_1(t)\gamma$}

\noindent and $\psi'(t)=0$.
Since $\psi$ is continuous on $\coint{0,\infty}$ (285Fb), it must be
constant, and $\varphi_1(t)=e^{i\gamma t}\psi(0)=e^{i\gamma t}$ for every
$t\ge 0$.   As for negative $t$, we have

\Centerline{$\varphi_1(t)=\overline{\varphi_1(-t)}=\overline{e^{-i\gamma t}}
=e^{i\gamma t}$}

\noindent for $t\le 0$, by 285Fc.\ \Qed

\medskip

\quad{\bf (v)}
Thus we see that $h_{01}$ has the same characteristic function
as the Dirac measure concentrated at $\gamma$, and this must therefore be
the distribution of $h_{01}$ (285M);
that is, $h_{01}\eae\gamma$.   Now (iii-$\eta$) tells us that $\gamma=0$.

Since $h_{0t}$ has the same distribution as $th_{01}$, $h_{0t}\eae 0$ for
every $t\ge 0$.   But now observe that if $t\ge 0$, $\omega\in\Omega$ and
$h_{0t}(\omega)=0$, then for any $\eta>0$
there is a finite $I\subseteq[0,t]$, containing
$0$ and $t$, such that $g_I(\omega)\le\eta^2$.
This means that $K_{0t}(\omega)$ can be covered by finitely many
sets $K_{t_j,t_{j+1}}(\omega)$ with
$\sum_{j=0}^{n-1}\diam K_{t_j,t_{j+1}}(\omega)^2\le\eta^2$.   All the
diameters here must of course be less than or equal to $\eta$.   As
$\eta$ is arbitrary, $\mu_{H2}K_{0t}(\omega)=0$.

For each $t\ge 0$, this is true for almost every $\omega$.   But this means
that, for almost every $\omega$, $\mu_{H2}K_{0n}(\omega)=0$ for every $n$,
and $\mu_{H2}\{\omega(t):t\ge 0\}=0$, as claimed.

\medskip

{\bf (b)(i)} To begin with, take a fixed $s\in\ooint{0,2}$.
Let $\mu_{L1}$ be Lebesgue measure on $[0,1]$.
For each $\omega\in\Omega$, let $\zeta_{\omega}$ be the image
measure $\mu_{L1}(\omega\restr[0,1])^{-1}$ on $F_{\omega}$.   Then

$$\eqalignno{\int_{\Omega}\int_{F_{\omega}}\int_{F_{\omega}}
   &\Bover1{\|x-y\|^s}\zeta_{\omega}(dx)\zeta_{\omega}(dy)
    \mu_W(d\omega)\cr
&=\int_{\Omega}\int_0^1\int_0^1
   \Bover1{\|\omega(t)-\omega(u)\|^s}dt\,du\,\mu_W(d\omega)\cr
&=\int_0^1\int_0^1\int_{\Omega}
   \Bover1{\|\omega(t)-\omega(u)\|^s}\mu_W(d\omega)\,dt\,du\cr
\displaycause{of course
$(\omega,t,u)\mapsto\Bover1{\|\omega(t)-\omega(u)\|^s}$ is continuous and
non-negative, so there is no difficulty with the change in order of
integration}
&=2\int_0^1\int_u^1\int_{\Omega}
   \Bover1{\|\omega(t)-\omega(u)\|^s}\mu_W(d\omega)\,dt\,du\cr
&=2\int_0^1\int_u^1\int_{\Omega}
   \Bover1{\|\omega(t-u)\|^s}\mu_W(d\omega)\,dt\,du\cr
\displaycause{because $X_t-X_u$ has the same distribution as $X_{t-u}$, as
in (a-i-$\alpha$)}
&=2\int_0^1\int_0^{1-u}\int_{\Omega}
   \Bover1{\|\omega(t)\|^s}\mu_W(d\omega)\,dt\,du
\le 2\int_0^1\int_{\Omega}
   \Bover1{\|\omega(t)\|^s}\mu_W(d\omega)\,dt\cr
&=2\int_0^1\int_{\BbbR^r}\Bover1{(\sqrt{2\pi t})^r}
   \Bover1{\|x\|^s}e^{-\|x\|^2/2t}\mu(dx)\,dt\cr
\displaycause{here $\mu$ is Lebesgue measure on $\BbbR^r$}
&=\Bover2{(\sqrt{2\pi})^r}\int_0^1\Bover1{t^{r/2}}\int_0^{\infty}
   \Bover1{\alpha^s}\cdot r\beta_r\alpha^{r-1}e^{-\alpha^2/2t}
   d\alpha\,dt\cr
&=\Bover{2r\beta_r}{(\sqrt{2\pi})^r}\int_0^1\Bover1{t^{r/2}}
   \int_0^{\infty}
   \Bover{(\beta\sqrt t)^{r-1}}{(\beta\sqrt t)^s}e^{-\beta^2/2}
   \sqrt{t}d\beta\,dt\cr
&=\Bover{2r\beta_r}{(\sqrt{2\pi})^r}\int_0^1
   \Bover1{t^{s/2}}dt
   \int_0^{\infty}\beta^{r-s-1}e^{-\beta^2/2}d\beta
<\infty\cr}$$

\noindent because $\Bover{s}2<1$ and $r-s>0$.
So $\biggerint_{F_{\omega}}\biggerint_{F_{\omega}}
   \Bover1{\|x-y\|^s}\zeta_{\omega}(dx)\zeta_{\omega}(dy)$
is finite for almost every
$\omega$.   Since $\zeta_{\omega}$ is always a probability measure
with support included in $F_{\omega}$,
$\mu_{Hs}F_{\omega}=\infty$ for all such $\omega$ (471Tb).

\medskip

\quad{\bf (ii)} Setting $s_n=2-2^{-n}$ for each $n$,
we see that, for almost
every $\omega\in\Omega$, $\mu_{Hs_n}F_{\omega}=\infty$ for every $n$.
But for
any such $\omega$, $\mu_{Hs}F_{\omega}=\infty$ for every $s\in\ooint{0,2}$,
by 471L.
}%end of proof of 477L

\exercises{\leader{477X}{Basic exercises (a)}
%\spheader 477Xa
Use 272Yc\formerly{2{}72Ye} to simplify the formulae in the proof of 477B.
%477B

\spheader 477Xb Let $\mu_W$ be Wiener measure on
$\Omega=C(\coint{0,\infty})_0$,
and set $X_t(\omega)=\omega(t)$ for $t\ge 0$ and $\omega\in\Omega$.
Let $f$ be a real-valued tempered function on $\Bbb R$ (definition:
284D).   For $x\in\Bbb R$ and $0<t<b$, let
$\nu_x^{(t,b)}$ be the distribution of a normally distributed random
variable with mean $x$ and variance $b-t$, so that
$g(x,t)=\int f(y)\nu_x^{(t,b)}(dy)$ can be regarded as
the expectation of $f(X_b)$ given that $X_t=x$.   (i) Show that
$g$ satisfies the {\bf backwards heat equation}
$2\Bover{\partial g}{\partial t}+\Bover{\partial^2g}{\partial x^2}=0$.
(ii) Interpret this in terms of the disintegration
$\nu_x^{(t,b)}=\int\nu_z^{(u,b)}\nu_x^{(t,u)}(dz)$ as $u\downarrow t$.
%477B

\spheader 477Xc(i) Show that the measure $\hat\mu^r$ of 477Da can be
constructed directly by applying 455A with
$(X_t,\Cal B_t)=(\BbbR^r,\Cal B(\BbbR^r))$ for every $t\ge 0$ and suitable
Gaussian distributions $\nu_x^{(s,t)}$ on $\BbbR^r$.   (ii) Show that the
measure $\hat\mu^r$ can be constructed by applying 455A to
$T=r\times\coint{0,\infty}$ with its lexicographic ordering and suitable
Gaussian distributions $\nu_x^{(s,t)}$ on $\Bbb R$.
%477D

\spheader 477Xd Let $r\ge 1$ be an integer.
(i) Show that there is a centered Gaussian process
$\langle Y_t\rangle_{t\in[0,1]}=\langle Y^{(i)}_t\rangle_{t\in[0,1],i<r}$
such that
$\Expn(Y^{(i)}_s\times Y^{(j)}_t)=0$ if $i\ne j$, $\min(s,t)-st$ otherwise.
(ii) Show that if $\langle X_t\rangle_{t\ge 0}$ is ordinary $r$-dimensional
Brownian motion, then $\family{t}{[0,1]}{Y_t}$ has the same distribution as
$\family{t}{[0,1]}{X_t-tX_1}$.
(iii) Show that the process $\family{t}{[0,1]}{Y_t}$
(the {\bf Brownian bridge})
can be represented by a Radon probability
measure $\mu_{\text{bridge}}$ on the space $C([0,1];\BbbR^r)_{00}$ of
continuous functions from $[0,1]$ to $\BbbR^r$ taking the value $0$ at both
ends of the interval.   (iv) For
$\omega\in C([0,1];\BbbR^r)_{00}$ define
$\Reverse{\omega}\in C([0,1];\BbbR^r)_{00}$ by setting
$\Reverse{\omega}(t)=\omega(1-t)$ for $t\in[0,1]$.   Show that
$\omega\mapsto\Reverse{\omega}$ is an automorphism of
$(C([0,1];\BbbR^r)_{00},\mu_{\text{bridge}})$.
%477D

\spheader 477Xe Let $(\Omega,\Sigma,\mu)$ be a complete
probability space and
$\langle X_t\rangle_{t\ge 0}$ a family of real-valued random variables on
$\Omega$ with independent increments.   For $I\subseteq\coint{0,\infty}$
let $\Tau_I$ be the $\sigma$-algebra generated by
$\{X_s-X_t:s$, $t\in I\}$.
Let $\Cal I$ be a family of subsets of $\coint{0,\infty}$ such that
for all distinct $I$, $J\in\Cal I$ either $\sup I\le\inf J$ or
$\sup J\le\inf I$.   Show that $\family{I}{\Cal I}{\Tau_I}$ is an
independent family of $\sigma$-algebras.
%477H

\spheader 477Xf Suppose that $H\subseteq\BbbR^r$ is an F$_{\sigma}$ set
and that $\tau:\Omega\to[0,\infty]$ is the Brownian hitting time to $H$,
as defined in 477I.   Show that $\tau$ is Borel measurable.
%477I

\spheader 477Xg Let $\mu_W$ be one-dimensional Wiener measure, and $\tau$
the hitting time to $\{1\}$.   Show that the distribution of
$\tau$ has probability density function
$x\mapsto\Bover1{x\sqrt{2\pi x}}e^{-1/2x}$ for $x>0$.
%477J

\spheader 477Xh Let $\mu_W$ be one-dimensional Wiener measure on
$\Omega=C(\coint{0,\infty})_0$.   Show that, for $\mu_W$-almost every
$\omega\in\Omega$, the total variation $\Var_{[s,t]}(\omega)$ is infinite
whenever $0\le s<t$.
%477K

\leader{477Y}{Further exercises (a)}
%\spheader 477Ya
Write $D_n$ for
$\{2^{-n}i:i\in\Bbb N\}$ and $D=\bigcup_{n\in\Bbb N}D_n$,
the set of dyadic rationals in
$\coint{0,\infty}$.   For $d\in D$ define $f_d\in C(\coint{0,\infty})$
as follows.   If $n\in\Bbb N$, $f_n(t)=0$ if $t\le n$, $t-n$ if
$n\le t\le n+1$, $1$ if $t\ge n+1$.   If $d=2^{-n}k$ where $n\ge 1$ and
$k\in\Bbb N$ is
odd, $f_d(t)=\max(0,\Bover1{\sqrt{2^{n+1}}}(1-2^n|t-d|))$ for $t\ge 0$.
Now let $\family{d}D{Z_d}$ be an independent family of standard normal
distributions, and set $\omega_n(t)=\sum_{d\in D_n}f_d(t)Z_d$ for $t\ge 0$,
so that each $\omega_n$ is a random continuous function on
$\coint{0,\infty}$.   Show that for any $n\in\Bbb N$ and $\epsilon>0$,

\Centerline{$\Pr(\sup_{t\in[0,n]}|\omega_{n+1}(t)-\omega_n(t)|\ge\epsilon)
\le 2^nn\cdot\Bover{2}{\sqrt{2\pi}}
  \biggerint_{2\epsilon\sqrt{2^n}}^{\infty}e^{-x^2/2}dx$,}

\noindent and hence that $\sequencen{\omega_n}$ converges almost surely to a
continuous function.   Explain how to interpret this as a
construction of Wiener measure on $\Omega=C(\coint{0,\infty})_0$,
as the image measure $\mu_G^Dg^{-1}$ where $g:\BbbR^D\to\Omega$ is almost
continuous (for the topology $\frak T_c$ on $\Omega$) and $\mu_G^D$ is the
product of copies of the standard normal distribution $\mu_G$.
%477B

\spheader 477Yb\dvAformerly{4{}56Ye}
Fix $p\in\ooint{0,1}\setminus\{\bover12\}$.   (i) For $\alpha\in\Bbb R$ set
$h_{\alpha}(t)=|t-\alpha|^{p-\bover12}-|t|^{p-\bover12}$ when
this is defined.   Show that $h_{\alpha}\in\eusm L^2(\mu_L)$,
where $\mu_L$ is Lebesgue measure,
and that $\|h_{\alpha}\|^2_2=|\alpha|^{2p}\|h_1\|^2_2$ and
$\|h_{\alpha}-h_{\beta}\|_2=\|h_{\alpha-\beta}\|_2$ for all $\alpha$,
$\beta\in\Bbb R$.
(ii) Show that there is a centered Gaussian process
$\family{\alpha}{\Bbb R}{X_{\alpha}}$ such that
$\Expn(X_{\alpha}\times X_{\beta})
=|\alpha|^{2p}+|\beta|^{2p}-|\alpha-\beta|^{2p}$ for all $\alpha$,
$\beta\in\Bbb R$.
(iii) Show that such a process can be represented by a Radon measure on
$C(\Bbb R)$.   \Hint{477Ya.}   (This is {\bf fractional Brownian motion}.)
%477B

\spheader 477Yc Let $\mu_W$ be Wiener measure on
$\Omega=C(\coint{0,\infty};\BbbR^r)_0$, where $r\ge 1$,
and set $X_t(\omega)=\omega(t)$ for $t\ge 0$ and $\omega\in\Omega$.
Let $f$ be a real-valued tempered function on $\BbbR^r$ (definition:
284Wa).   For $x\in\BbbR^r$ and $0<t<b$, let
$\nu_x^{(t,b)}$ be the distribution of $x+X_{b-t}$, so that
$g(x,t)=\int f(y)\nu_x^{(t,b)}(dy)$ can be regarded as
the expectation of $f(X_b)$ given that $X_t=x$.   Show that
$g$ satisfies the backwards heat equation
$2\Bover{\partial g}{\partial t}+\sum_{i=0}^{r-1}\Pdd{g}{\xi_i}=0$.
%477Xb 477B

\spheader 477Yd Let $r\ge 1$ be an integer,
and $\nu$ a Radon probability measure on $\BbbR^r$ such that
$x\mapsto\varinnerprod{a}{x}$ has expectation $0$ and variance
$\|a\|^2$ for every $a\in\BbbR^r$.
Let $\Omega$ be $C(\coint{0,\infty};\BbbR^r)_0$, and
for $\alpha>0$ define $f_{\alpha}:(\BbbR^r)^{\Bbb N}\to\Omega$ by setting
$f_{\alpha}(x)(t)
=\sqrt{\alpha}(\sum_{i<n}z(i)+\Bover1{\alpha}(t-n\alpha)z(n))$ when
$z\in(\BbbR^r)^{\Bbb N}$, $n\in\Bbb N$ and $n\alpha\le t\le(n+1)\alpha$;
let $\mu_{\alpha}$ be the image Radon measure $\nu^{\Bbb N}f_{\alpha}^{-1}$
on $\Omega$.   Show that Wiener measure $\mu_W$ is the limit
$\lim_{\alpha\downarrow 0}\mu_{\alpha}$ for the narrow topology.
%477C calculate characteristic function of X_t

\spheader 477Ye Let $\mu_W$ be Wiener measure on $C(\coint{0,\infty})_0$,
and $\gamma>\bover12$.   Show that
$\lim_{t\to\infty}\Bover1{t^{\gamma}}\omega(t)
=\lim_{t\downarrow 0}\Bover1{t^{1-\gamma}}\omega(t)=0$
for $\mu_W$-almost every $\omega$.
%477E


\spheader 477Yf Write out a proof of 477G which works directly from the
Gaussian-distribution
characterization of Wiener measure, without appealing to results from
\S455 other than 455L.   (I think you will need to start with stopping
times taking finitely and countably many values, as in 455C;  but
you will find great simplifications.)
%477G

\spheader 477Yg Let $\hat\mu$ be the Gaussian distribution on
$\BbbR^{\coint{0,\infty}}$ corresponding to Brownian motion, as in
477A.   For $t\ge 0$ let
$\Sigma_t$ be the family of Baire subsets of $\BbbR^{\coint{0,\infty}}$
determined by coordinates in $[0,t]$, and
$\Sigma^+_t=\bigcap_{s>t}\Sigma_s$.
For $\omega\in\BbbR^{\coint{0,\infty}}$ set
$\tau(\omega)=\inf\{q:q\in\Bbb Q$, $\omega(q)\ge 1\}$, counting
$\inf\emptyset$ as $\infty$.   Show that $\tau$ is a stopping time adapted
to $\langle\Sigma^+_t\rangle_{t\ge 0}$.
For $\omega$, $\omega'\in\BbbR^{\coint{0,\infty}}$
define $\phi_{\tau}(\omega,\omega')\in\BbbR^{\coint{0,\infty}}$ by setting
$\phi_{\tau}(\omega,\omega')(t)=\omega(t)$ if $t\le\tau(\omega)$,
$\omega(\tau(\omega))+\omega'(t-\tau(\omega))$ if $t>\tau(\omega)$.
Show that $\phi_{\tau}$ is not \imp\ for the product measure
$\hat\mu^2$ and $\hat\mu$.    \Hint{show that
$\{\omega:\tau(\omega)\in D\}$ is negligible for every countable set $D$.}
%477G

\spheader 477Yh Let $\mu_W$ be one-dimensional Wiener measure
on $\Omega=C(\coint{0,\infty})_0$.   (i) Show
by induction on $k$ that $\Pr($there are $t_0<t_1<\ldots<t_k\le t$ such
that $X_{t_j}=(-1)^j$ for every $j\le k)
=\Pr($there is an $s\le t$ such that
$X_s=2k+1)$ for any $t\ge 0$.   \Hint{477J.}  (ii) For $\omega\in\Omega$,
$k\in\Bbb N$ define $\tau_k(\omega)$ by saying that
$\tau_0(\omega)=\inf\{t:|\omega(t)|=1\}$,
$\tau_{k+1}(\omega)=\inf\{t:t>\tau_k(\omega)$,
$\omega(t)=-\omega(\tau_k(\omega))\}$.   Show that $\tau_k$ is a stopping
time adapted to the family $\langle\Sigma_t\rangle_{t\ge 0}$ of 477H,
and is finite $\mu_W$-a.e.   (iii) Set
$E_k=\{\omega:\tau_k(\omega)\le 1<\tau_{k+1}(\omega)\}$,
$p_k=\mu_WE_k$, $F_k=\{\omega:$ there is an $s\le 1$ such that
$\omega(s)=2k+1\}$, $q_k=\mu_WF_k$.   Show that

\Centerline{$q_k
=\Bover12p_k+\sum_{j=k+1}^{\infty}p_j
=\Bover2{\sqrt{2\pi}}\int_{2k+1}^{\infty}e^{-x^2/2}dx$.}

\noindent (iv) Show that
$\Pr(\tau_0\le 1)=\sum_{k=0}^{\infty}p_k=2\sum_{k=0}^{\infty}(-1)^kq_k$.
%477J

\spheader 477Yi Let $\mu_W$ be Wiener measure on
$\Omega=C(\coint{0,\infty};\BbbR^r)_0$, where $r\ge 1$.
Show that, for $\mu_W$-almost every $\omega\in\Omega$,
$\{\Bover{\omega(t)}{\|\omega(t)\|}:t\ge t_0$, $\omega(t)\ne 0\}$
is dense in $\partial B(\tbf{0},1)$ for every $t_0\ge 0$.
%477I mt47bits out of order query

\spheader 477Yj(i) Show that, for any $r\ge 1$,
the topology $\frak T_c$ of uniform
convergence on compact sets is a complete linear space topology on
$\Omega=C(\coint{0,\infty};\BbbR^r)_0$.   (ii) Show that Wiener measure on
$\Omega$ is a centered Gaussian measure in the sense of 466N.
\Hint{466Ye.}
%477D out of order query
}%end of exercises

\endnotes{
\Notesheader{477} The `first proof' of 477A calls on a result which is
twenty-five %checked 2010
pages into \S455, and you will probably be glad to be assured
that all it really needs is 455A, fragments from the proof
of 455P, and part (a) of the proof of 455R.
So it is not enormously harder than the `second proof', based on the
elementary theory of Gaussian processes.

With this theorem, we have two routes to the first target:  setting up a
measure space with a family of random variables representing Brownian
motion.   I repeat that this is a secondary issue.   Brownian
motion begins with
the family of joint distributions of finite indexed sets
$(X_{t_0},\ldots,X_{t_n})$ satisfying the properties listed in 477A.
It is one of the triumphs of Kolmogorov's theory of probability that these
distributions can be represented by a family of
real-valued functions on a set with a countably additive measure;  but they
would still be of the highest importance if they could not.   In order to
show that it can be done, we
can use either the time-dependent approach based on conditional
expectations, as in 455A, 455P and the `first proof' of 477A, or the
timeless Gaussian-distribution approach through 456C, as in the `second
proof' of 477A.   Both, of course, depend on Kolmogorov's theorem 454D.
They have different advantages, and it will be very useful to be able
to call on the intuitions of both.   The `first proof' leads us naturally
into the theory of L\'evy processes, in which other families of
distributions replace normal distributions.

To get to the continuity of
sample paths, we need to do quite a bit more, and the proof of 477B is one
way of filling the gap.   At this point it becomes tempting to abandon both
proofs of 477A and start again with the method of 477Ya, the
`L\'evy-Ciesielski' construction, {\it not} using
Kolmogorov's theorem.   But if we do this, we shall have to devise a new
argument to prove the strong Markov property 477G, rather than quoting
455U.   Of course the special properties of Gaussian processes mean that a
direct proof of 477G is still quite a bit easier than the general results
of \S455 (477Yf).   I make no claim that one
approach is `better' than another;  they all throw light on the result.

What I here call `Wiener measure' (477B) is a particular realization of
Brownian motion.
It is so convincing that it is tempting to regard it as `the' real basis
of Brownian motion.   I do not mean to assert this in any way which
will bind me in future.   But (as a measure theorist, rather than a
probabilist) I think that the specific measures of 477B and 477D
are worth as much attention as any.
One reason for not insisting that the space
$C(\coint{0,\infty})_0$ is the only right place to start is that we may at
any moment wish to move to something smaller, as in 477Ef.   The approach
here gives a very direct language in which to express
theorems of the form `almost
every Brownian path is $\ldots$' (477K, 477L), and every such theorem
carries an implicit suggestion that we could move to a conegligible set
and a subspace measure.

In 477C I sketch an alternative characterization of one-dimensional Wiener
measure.   Five pages seem to be rather a lot for a proof of
something which surely has to be true, if we can get the hypotheses right;
but I do not see a genuinely
shorter route, and I think in fact that the indigestibility of the argument
as presented is due to compression more than to pedantry.
At least I have tried to put the key step into part (a-i) of the proof.
We have to use the Central Limit Theorem;  we have to use a
finite-approximation version of it, rather than a limit version;  the ideas
of this proof do not demand Lindeberg's formulation, but this is what we
have to hand in Volume 2;  and if we are going to consider interpolations
for general random walks, we need something to force a sufficient degree
of equicontinuity, and ($\dagger$) is what comes naturally from the result
in 274F.   It is surely obvious that I have been half-hearted in the
generality of the theorem as given.   There can be no reason for insisting
on steps being at uniform time intervals, or on stationary processes, or
even on variances being exactly correct, provided that everything averages
out nicely in the limit.   The idea does require that steps be
independent, but after that we just need hypotheses adequate for the
application of the Central Limit Theorem.

Clearly we can also look for $r$-dimensional versions of the theorem.   I
have not done so because they would inevitably demand vector-valued
versions of the Central Limit Theorem, and while a combination of
the ideas of
\S\S274 and 456 would take us a long way, it would not belong to this
section.   However I give 477Yd as an example which can be dealt with
without much general theory.

Already in the elementary results
477Eb-477Eg % 477Eb 477Ec 477Ed 477Ee 477Ef 477Eg
we see that Wiener measure is a remarkable construction.
It is a general principle that the more symmetries an object has, the more
important it is;  this one has a surprising symmetry (477Ef), which is even
better.   I take it as confirmation that we have a good representation,
that all these symmetries can be represented by actual \imp\ functions,
rather than leaving them as manipulations of distributions.

477F is a natural result, and a further confirmation that in
$C(\coint{0,\infty};\BbbR^r)_0$ we have got hold
of an appropriate space of functions.   The proof I give depends on an
aspect of the structure developed in 477Ya.

The next really important result is the `strong Markov property' (477G).
This is clearly a central property of Brownian motion.   It may not
be quite so clear what the formulation here is trying to say.   As in 477E,
I am expressing the result in terms of an \imp\ function.   This makes no
sense unless we have a probability space $\Omega$ in which we can put two
elements $\omega$, $\omega'$ together to form a third;  so we are more or
less forced to look at a space of paths.   But not all spaces of paths will
do.   For an indication of what can happen if we work with
the wrong realization, see 477Yg.

In 477H we have two kinds of zero-one law.   One, 477Hd, is explicit; the
tail $\sigma$-algebra $\bigcap_{t\ge 0}\hat\Tau_{\coint{t,\infty}}$ behaves
like the tail of an independent sequence of  $\sigma$-algebras (272O).
But the formula `$\bigcap_{s>t}\Sigma_s\subseteq\hat\Tau_{[0,t]}$' in
477Hc can
be thought of as a relative zero-one law.   There are many events (e.g.,
$\{\omega:\liminf_{s\downarrow t}\bover1{s-t}\omega(s)\ge 0\}$)
which belong to $\bigcap_{s>t}\Sigma_s$ and not to $\Tau_{[0,t]}$ or
$\Sigma_t$, but all collapse to events in $\Tau_{[0,t]}$
if we rearrange them on appropriate negligible sets.   This is really a
special case of 455T.

The formulae in the first application of the strong Markov property (477J)
demand a bit of concentrated attention, but
I think that the key step at the end of the proof (moving from $\mu_W^2$ to
$\int\ldots d\mu_W$) faithfully represents the intuition:  once we have
reached the level $\alpha$, we have an even chance of rising farther.
For the discrete case, 272Yc is a version of the same idea.
From the distribution of the
hitting time to $\{\alpha\}$ we can deduce the distribution of the
hitting time to $\{-1,1\}$ (477Yh);   but I do not know of a corresponding
exact result for the hitting time to the unit circle for two-dimensional
Brownian motion.

I expect you have been shown a continuous
function which is nowhere differentiable.   In 477K we see that
`almost every'
function is of this type.   What a hundred and fifty years ago seemed to be
an exotic counter-example now presents itself as a representative of the
typical case.
The very crude estimates in the proof of 477K are supposed to furnish a
straightforward proof of the result, without asking for anything which
might lead to refinements.   Of course there is much more to be said,
starting with 477Xh.   In 477L we have an interesting result which will
be useful in \S479, when
I return to geometric properties of Brownian paths.


}%end of notes

\discrpage

\frfilename{mt478.tex}
\versiondate{4.6.09/7.8.20}

\def\energy{\mathop{\text{energy}}\nolimits}
\def\hp{\mathop{\text{hp}}\nolimits}
\def\varinnerprod#1#2{#1\dotproduct#2}
\def\zz{}

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\bigbaselineskip=36pt

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\ifdim\pagetotalplus>\pagegoalminus\break
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\def\chaptername{Geometric measure theory}
\def\sectionname{Harmonic functions}

\newsection{478}

In this section and the next I will attempt an introduction to potential
theory.   This is an enormous subject and my choice of results is
necessarily somewhat arbitrary.   My principal aim is to give the most
elementary properties of Newtonian capacity, which will appear in \S479.
It seems that this necessarily involves a basic understanding of harmonic
and superharmonic functions.   I approach these by the `probabilistic'
route, using Brownian motion as described in \S477.

The first few paragraphs, down to 478J, do not
in fact involve Brownian motion;
they rely on multidimensional advanced calculus and on the Divergence
Theorem.   \cmmnt{(The latter is applied only to continuously
differentiable functions and domains of very simple types, so
we need far less than the quoted result in 475N.)}   Defining `harmonic
function' in terms of average values over concentric
spherical shells (478B), the
first step is to identify this with the definition in terms of the
Laplacian differential operator (478E).   An essential result is a formula
for a harmonic function inside a sphere in terms of its values on the
boundary and the `Poisson kernel' (478Ib),
and we also need to understand the effects of smoothing
by convolution with appropriate functions (478J\cmmnt{, following
473D-473E}).
I turn to Brownian motion with Dynkin's formula (478K), relating the
expected value of $f(X_{\tau})$ for
a stopped Brownian process $X_{\tau}$
to an integral in terms of $\nabla^2f$.   This is already enough to deal
with the asymptotic behaviour
of Brownian paths, which depends in a striking way on the dimension of the
space (478M).

We can now approach Dirichlet's problem.   If we have a bounded open set
$G\subseteq\BbbR^r$, there is a family $\family{x}{G}{\mu_x}$ of
probability measures such that whenever
$f:\overline{G}\to\Bbb R$ is continuous and $f\restr G$ is harmonic, then
$f(x)=\int fd\mu_x$ for every $x\in G$ (478Pc).   So this
family of `harmonic measures' gives a formula continuously extending a
function on $\partial G$ to a harmonic function on $G$, whenever
such an extension exists (478S).   The method used expresses $\mu_x$ in
terms of
the distribution of points at which Brownian paths starting at $x$ strike
$\partial G$, and relies on Dynkin's formula through Theorem 478O.   The
strong Markov property of Brownian motion now enables us to relate harmonic
measures associated with different sets (478R).

\leader{478A}{Notation} $r\ge 1$ will be an integer;
if you find it easier to focus on one dimensionality at a time, you should
start with $r=3$, because $r=1$ is too easy and $r=2$ is exceptional.
$\mu$ will be Lebesgue measure on $\BbbR^r$, and
$\|\,\|$ the Euclidean norm on $\BbbR^r$;  $\nu$ will be normalized
$(r-1)$-dimensional Hausdorff measure on $\BbbR^r$.
In the elementary case $r=1$, $\nu$ will be counting measure on $\Bbb R$.

\cmmnt{$\beta_r$ will be the volume of the unit ball in $\BbbR^r$,
that is,}

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

\cmmnt{\noindent Recall that}

$$\eqalign{\nu(\partial B(\tbf{0},1))=r\beta_r
&=\Bover{2}{(k-1)!}\pi^k\text{ if }r=2k\text{ is even},\cr
&=\Bover{2^{2k+1}k!}{(2k)!}\pi^k\text{ if }r=2k+1\text{ is odd}
  \dvro{.}{}\cr}$$

\cmmnt{\noindent (265F/265H).}

In the formulae below, there are repeated expressions of the form
$\Bover1{\|x-y\|^{r-1}}$, $\Bover1{\|x-y\|^{r-2}}$;  in these, it will
often be
convenient to interpret $\bover10$ as $\infty$\cmmnt{, so that we have
$[0,\infty]$-valued functions defined everywhere}.

It will be convenient to do some
calculations in the one-point compactification $\BbbR^r\cup\{\infty\}$ of
$\BbbR^r$\cmmnt{ (3A3O)}.
For a set $A\subseteq\BbbR^r$\cmmnt{, write $\overline{A}^{\infty}$
and $\partial^{\infty}A$ for its closure and boundary taken in
$\BbbR^r\cup\{\infty\}$;  that is,}

\Centerline{$\overline{A}^{\infty}=\overline{A}$,
\quad$\partial^{\infty}A=\partial A$}

\noindent if $A$ is bounded, and

\Centerline{$\overline{A}^{\infty}=\overline{A}\cup\{\infty\}$,
\quad$\partial^{\infty}A=\partial A\cup\{\infty\}$}

\noindent if $A$ is unbounded.
\cmmnt{Note that }$\overline{A}^{\infty}$ and
$\partial^{\infty}A$ are always compact.
In this context I will take $x+\infty=\infty$ for every $x\in\BbbR^r$.

$\mu_W$ will be $r$-dimensional Wiener measure on
$\Omega=C(\coint{0,\infty};\BbbR^r)_0$, the space of continuous functions
$\omega$ from
$\coint{0,\infty}$ to $\BbbR^r$ such that $\omega(0)=0$\cmmnt{ (477D)},
endowed with the topology of uniform convergence on compact sets;  $\Sigma$
will be the domain of $\mu_W$.
\cmmnt{The probabilistic notations $\Expn$ and $\Pr$ will always be
with respect to $\mu_W$ or some directly associated probability.}
$\mu_W^2$ will be the product measure on $\Omega\times\Omega$.
I will write $X_t(\omega)=\omega(t)$ for $t\in\coint{0,\infty}$ and
$\omega\in\Omega$, and if $\tau:\Omega\to[0,\infty]$ is a function,
I will write
$X_{\tau}(\omega)=\omega(\tau(\omega))$ whenever $\omega\in\Omega$ and
$\tau(\omega)$ is finite.

\cmmnt{As in 477Hc,} I will write $\Sigma_t$ for
the $\sigma$-algebra of sets $F\in\Sigma$
such that $\omega'\in F$ whenever $\omega\in F$, $\omega'\in\Omega$ and
$\omega'\restr[0,t]=\omega\restr[0,t]$, and
$\Sigma^+_t=\bigcap_{s>t}\Sigma_s$.   $\Tau_{[0,t]}$ will be the
$\sigma$-algebra of subsets of $\Omega$ generated by $\{X_s:s\le t\}$.

\leader{478B}{Harmonic and superharmonic functions} Let $G\subseteq\BbbR^r$
be an open set and $f:G\to[-\infty,\infty]$ a function.

\spheader 478Ba $f$ is {\bf superharmonic} if
$\Bover1{\nu(\partial B(x,\delta))}\int_{\partial B(x,\delta)}fd\nu$ is
defined in $[-\infty,\infty]$ and
less than or equal to $f(x)$ whenever $x\in G$, $\delta>0$ and
$B(x,\delta)\subseteq G$.

\spheader 478Bb $f$ is {\bf subharmonic} if $-f$ is superharmonic, that is,
$\Bover1{\nu(\partial B(x,\delta))}\int_{\partial B(x,\delta)}fd\nu$ is
defined in $[-\infty,\infty]$ and
greater than or equal to $f(x)$ whenever $x\in G$, $\delta>0$ and
$B(x,\delta)\subseteq G$.

\spheader 478Bc $f$ is {\bf harmonic} if it is both superharmonic and
subharmonic, that is,
$\Bover1{\nu(\partial B(x,\delta))}\int_{\partial B(x,\delta)}fd\nu$
is defined and equal to $f(x)$ whenever $x\in G$, $\delta>0$ and
$B(x,\delta)\subseteq G$.

\leader{478C}{Elementary facts} Let $G\subseteq\BbbR^r$ be an open set.

\spheader 478Ca If $f:G\to[-\infty,\infty]$ is a function,
then $f$ is superharmonic iff $-f$ is subharmonic.

\spheader 478Cb If $f$, $g:G\to\coint{-\infty,\infty}$
are superharmonic functions,
then $f+g$ is superharmonic.   \prooflet{\Prf\ If $x\in G$, $\delta>0$ and
$B(x,\delta)\subseteq G$, then $\int_{\partial B(x,\delta)}fd\nu$ and
$\int_{\partial B(x,\delta)}g\,d\nu$ are defined in
$\coint{-\infty,\infty}$, so $\int_{\partial B(x,\delta)}f+g\,d\nu$ is
defined and is

\Centerline{$\int_{\partial B(x,\delta)}fd\nu+
\int_{\partial B(x,\delta)}g\,d\nu
\le\nu(\partial B(x,\delta))\,(f(x)+g(x))$.  \Qed}}

\spheader 478Cc If $f$, $g:G\to[-\infty,\infty]$ are superharmonic
functions, then $f\wedge g$ is superharmonic.
\prooflet{\Prf\ If $x\in G$, $\delta>0$ and
$B(x,\delta)\subseteq G$, then $\int_{\partial B(x,\delta)}fd\nu$ and
$\int_{\partial B(x,\delta)}g\,d\nu$ are defined in
$[-\infty,\infty]$, so $\int_{\partial B(x,\delta)}f\wedge g\,d\nu$ is
defined and is at most

\Centerline{$\min\bigl(\int_{\partial B(x,\delta)}fd\nu,
\int_{\partial B(x,\delta)}g\,d\nu\bigr)
\le\nu(\partial B(x,\delta))\,\min(f(x),g(x))$.  \Qed}}

\woddheader{478Cd}{0}{0}{0}{48pt}

\spheader 478Cd Let $f:G\to\Bbb R$ be a harmonic function which is
locally integrable with respect to Lebesgue measure on $G$\cmmnt{ (that
is, every point of $G$ has a neighbourhood $V$ such that $\int_Vfd\mu$ is
defined and finite)}.   Then

\Centerline{$f(x)=\Bover1{\mu B(x,\delta)}\int_{B(x,\delta)}fd\mu$}

\noindent whenever $x\in G$, $\delta>0$ and $B(x,\delta)\subseteq G$.
\prooflet{\Prf\ By 265G,

\Centerline{$\int_{B(x,\delta)}fd\mu
=\int_0^t\int_{\partial B(x,t)}fd\nu\,dt
=\int_0^t\nu(\partial B(x,\delta))f(x)dt
=\beta_r\delta^rf(x)$.   \Qed}

\noindent}So $f$ is continuous.   \prooflet{\Prf\ If $x\in G$, take
$\delta>0$ such that $B(x,2\delta)\subseteq G$, and set $f_1(y)=f(y)$ for
$y\in B(x,2\delta)$, $0$ for $y\in\BbbR^r\setminus B(x,2\delta)$.   Set
$g=\Bover1{\mu B(\tbf{0},\delta)}\chi B(\tbf{0},\delta)$.
Then $f_1$ is integrable, so the convolution $f_1*g$ is continuous (444Rc).
Also, for any $y\in B(x,\delta)$,

$$\eqalign{(f_1*g)(y)
&=\int f_1(z)g(y-z)\mu(dz)
=\int_{B(y,\delta)}\Bover{f_1(z)}{\mu B(y,\delta)}\mu(dz)\cr
&=\Bover1{\mu B(y,\delta)}\int_{B(y,\delta)}f(z)\mu(dz)
=f(y),}$$

\noindent so $f$ is continuous at $x$.\ \Qed}

%do we really need to assume "locally integrable"?  could it be a
%consequence of "measurable harmonic real-valued"? query

\leader{478D}{Maximal \dvrocolon{principle}}\cmmnt{ One of the
fundamental properties of harmonic functions will hardly be used in the
exposition here, but I had better give it a suitably prominent place.

\medskip

\noindent}{\bf Proposition} Let $G\subseteq\BbbR^r$ be a
non-empty open set.  Suppose that
$g:\overline{G}^{\infty}\to\ocint{-\infty,\infty}$ is
lower semi-continuous,
$g(y)\ge 0$ for every $y\in\partial^{\infty}G$, and $g\restr G$ is
superharmonic.   Then $g(x)\ge 0$ for every $x\in G$.

\proof{ \Quer\ Otherwise, set
$\gamma=\inf_{x\in G}g(x)=\inf\{g(y):y\in\overline{G}^{\infty}\}$.
Because $\overline{G}^{\infty}$ is compact and
$g$ is lower semi-continuous, $K=\{x:x\in G$, $g(x)=\gamma\}$ is
non-empty and compact (4A2B(d-viii)).
Let $x\in K$ be such that $\|x\|$ is maximal, and $\delta>0$
such that $B(x,\delta)\subseteq G$.   Then
$\Bover1{\nu(\partial B(x,\delta))}\int_{\partial B(x,\delta)}g\,d\nu
\le g(x)$.   But $g(y)\ge g(x)$ for
every $y\in\partial B(x,\delta)$ and

\Centerline{$\{y:y\in\partial B(x,\delta)$, $g(y)>g(x)\}
\supseteq\{y:y\in\partial B(x,\delta)$, $(y-x)\dotproduct x\ge 0\}$}

\noindent is not $\nu$-negligible, so this is impossible.\ \Bang
}%end of proof of 478D

\leader{478E}{Theorem} Let $G\subseteq\BbbR^r$ be an open set and
$f:G\to\Bbb R$ a function with continuous second derivative.
Write $\nabla^2f$ for its Laplacian
$\diverg\grad f=\sum_{i=1}^r\Pdd{f}{\xi_i}$.

(a) $f$ is superharmonic iff $\nabla^2f\le 0$ everywhere in $G$.

(b) $f$ is subharmonic iff $\nabla^2f\ge 0$ everywhere in $G$.

(c) $f$ is harmonic iff $\nabla^2f=0$ everywhere in $G$.

\proof{{\bf (a)(i)} For $x\in G$ set

\Centerline{$R_x=\rho(x,\BbbR^r\setminus G)
=\inf_{y\in\BbbR^r\setminus G}\|x-y\|$,}

\noindent counting $\inf\emptyset$ as $\infty$;  for $0<\gamma<R_x$ set

\Centerline{$g_x(\gamma)
=\Bover1{\gamma^{r-1}}\int_{\partial B(x,\gamma)}f(y)\nu(dy)
=\int_{\partial B(\tbf{0},1)}f(x+\gamma z)\nu(dz)$.}

\noindent Because $f$ is
continuously differentiable, $g'_x(\gamma)$ is defined and equal to
$\int_{\partial B(\tbf{0},1)}\Pd{}{\gamma}f(x+\gamma z)\nu(dz)$ for
$\gamma\in\ooint{0,R_x}$.

Set $\phi=\grad f$, so that $\nabla^2f=\diverg\phi$.
Each ball $B(x,\gamma)$ has
finite perimeter;  its essential boundary is its ordinary boundary;
the Federer exterior normal
$v_y$ at $y$ is $\Bover1{\gamma}(y-x)$;  and if $y=x+\gamma z$, where
$\|z\|=1$, then
$\phi(y)\dotproduct v_y$ is $\Pd{}{\gamma}f(x+\gamma z)$.
So the Divergence Theorem (475N) tells us that

$$\eqalignno{\int_{B(x,\gamma)}\nabla^2fd\mu
&=\int_{\partial B(x,\gamma)}\phi(y)\dotproduct v_y\nu(dy)\cr
&=\gamma^{r-1}\int_{\partial B(\tbf{0},1)}
  \Pd{}{\gamma}f(x+\gamma z)\nu(dz)
=\gamma^{r-1}g'_x(\gamma).\cr}$$

\medskip

\quad{\bf (ii)} If $\nabla^2f\le 0$ everywhere in $G$, and
$B(x,\gamma)\subseteq G$, then $g'_x(t)\le 0$ for $0<t\le \gamma$, so

\Centerline{$g_x(\gamma)\le\lim_{t\downarrow 0}g_x(t)=r\beta_rf(x)$;}

\noindent as $x$ and $\gamma$ are arbitrary, $f$ is superharmonic.

\medskip

\quad{\bf (iii)} If $f$ is superharmonic, and $x\in G$, then

\Centerline{$g_x(\gamma)\le r\beta_rf(x)=\lim_{t\downarrow 0}g_x(t)$}

\noindent for every $\gamma\in\ooint{0,R_x}$.   So there must be arbitrarily
small $\gamma >0$ such that $g'_x(\gamma)\le 0$ and
$\int_{B(x,\gamma)}\nabla^2fd\mu\le 0$;  as $\nabla^2f$ is continuous,
$(\nabla^2f)(x)\le 0$.

\medskip

{\bf (b)-(c)} are now immediate.
}%end of proof of 478E

\leader{478F}{Basic examples} (a) For any $y$, $z\in\BbbR^r$,

\Centerline{$x\mapsto\Bover1{\|x-z\|^{r-2}}$,
\quad$x\mapsto\Bover{(x-z)\dotproduct y}{\|x-z\|^r}$,}

\Centerline{$x\mapsto\Bover{\|y-z\|^2-\|x-y\|^2}{\|x-z\|^r}
=2\Bover{(x-z)\dotproduct(y-z)}{\|x-z\|^r}-\Bover1{\|x-z\|^{r-2}}$}

\noindent are harmonic on $\BbbR^r\setminus\{z\}$.

(b) For any $z\in\BbbR^2$,

\Centerline{$x\mapsto\ln\|x-z\|$}

\noindent is harmonic on $\Bbb R^2\setminus\{z\}$.


\proof{ The Laplacians are easiest to calculate when $z=0$, of course,
but in any case you only have to get the algebra right to apply 478Ec.
}%end of proof of 478F

\cmmnt{\medskip

\noindent{\bf Remark} The function
$x\mapsto\Bover{\|y-z\|^2-\|x-y\|^2}{\|x-z\|^r}$ is the {\bf Poisson
kernel};  see 478I below.}

\leader{478G}{}\cmmnt{ We shall
need a pair of exact integrals involving the functions here, with an easy
corollary.

\medskip

\noindent}{\bf Lemma}
(a) $\Bover1{\nu(\partial B(\tbf{0},\delta))}\biggerint_{\partial B(\tbf{0},\delta)}
\Bover1{\|x-z\|^{r-2}}\nu(dz)
=\Bover1{\max(\delta,\|x\|)^{r-2}}$ whenever $x\in\BbbR^r$
and $\delta>0$.

(b) $\Bover1{\nu(\partial B(\tbf{0},\delta))}\biggerint_{\partial B(\tbf{0},\delta)}
  \Bover{|\delta^2-\|x\|^2|}{\|x-z\|^r}\nu(dz)
=\Bover1{\max(\delta,\|x\|)^{r-2}}$ whenever $x\in\BbbR^r$,
$\delta>0$ and $\|x\|\ne\delta$.

(c) $\biggerint_{B(\tbf{0},\delta)}\Bover1{\|x-z\|^{r-2}}\mu(dz)
\le\Bover12r\beta_r\delta^2$
whenever $x\in\BbbR^r$ and $\delta>0$.

\proof{{\bf (a)(i)} The first thing to note is that there is a function
$g:\coint{0,\infty}\setminus\{\delta\}\to\coint{0,\infty}$ such that
$g(\|x\|)=\Bover1{\nu(\partial B(\tbf{0},\delta))}
\biggerint_{\partial B(\tbf{0},\delta)}\Bover1{\|x-z\|^{r-2}}\nu(dz)$
whenever $\|x\|\ne\delta$.   \Prf\ If $\|x\|=\|y\|$, then
there is an orthogonal transformation $T:\BbbR^r\to\BbbR^r$ such that
$Tx=y$, so that

$$\eqalignno{\int_{\partial B(\tbf{0},\delta)}\Bover1{\|y-z\|^{r-2}}\nu(dz)
&=\int_{\partial B(\tbf{0},\delta)}\Bover1{\|y-Tz\|^{r-2}}\nu(dz)\cr
\displaycause{because $T$ is an automorphism of
$(\BbbR^r,B(\tbf{0},\delta),\nu)$}
&=\int_{\partial B(\tbf{0},\delta)}\Bover1{\|Tx-Tz\|^{r-2}}\nu(dz)
=\int_{\partial B(\tbf{0},\delta)}\Bover1{\|x-z\|^{r-2}}\nu(dz).
\text{ \Qed}\cr}$$

\woddheader{478G}{4}{2}{2}{60pt}

\quad{\bf (ii)} Now suppose that $0<\gamma<\delta$.   Then

$$\eqalignno{g(\gamma)
&=\Bover1{\nu(\partial B(\tbf{0},\gamma))}\int_{\partial B(\tbf{0},\gamma)}
  g(\gamma)\nu(dx)\cr
&=\Bover1{\nu(\partial B(\tbf{0},\gamma))}\int_{\partial B(\tbf{0},\gamma)}
   \Bover1{\nu(\partial B(\tbf{0},\delta))}\int_{\partial B(\tbf{0},\delta)}
   \Bover1{\|x-z\|^{r-2}}\nu(dz)\nu(dx)\cr
&=\Bover1{\nu(\partial B(\tbf{0},\delta))}\int_{\partial B(\tbf{0},\delta)}
   \Bover1{\nu(\partial B(\tbf{0},\gamma))}
   \int_{\partial B(\tbf{0},\gamma)}\Bover1{\|x-z\|^{r-2}}\nu(dx)\nu(dz)\cr
&=\Bover1{\nu(\partial B(\tbf{0},\delta))}\int_{\partial B(\tbf{0},\delta)}
   \Bover1{\|z\|^{r-2}}\nu(dz)\cr
\displaycause{because the function $x\mapsto\Bover1{\|x-z\|^2}$ is harmonic
in $\BbbR^r\setminus\{z\}$, by 478Fa}
&=\Bover1{\delta^{r-2}}.\cr}$$

\medskip

\quad{\bf (iii)} Next, if $\gamma>\delta$,

$$\eqalignno{g(\gamma)
&=\Bover1{\nu(\partial B(\tbf{0},\delta))}\int_{\partial B(\tbf{0},\delta)}
   \Bover1{\nu(\partial B(\tbf{0},\gamma))}
   \int_{\partial B(\tbf{0},\gamma)}\Bover1{\|x-z\|^{r-2}}\nu(dx)\nu(dz)\cr
\displaycause{as in (ii)}
&=\Bover1{\nu(\partial B(\tbf{0},\delta))}\int_{\partial B(\tbf{0},\delta)}
   \Bover1{\gamma^{r-2}}\nu(dz)\cr
\displaycause{by (ii), with $\gamma$ and $\delta$ interchanged}
&=\Bover1{\gamma^{r-2}}.\cr}$$

\medskip

\quad{\bf (iv)} So we have the result if $\|x\|\ne\delta$.   If
$\|x\|=\delta$ and $r\ge 2$,
set $x_n=(1+2^{-n})x$ for each $n\in\Bbb N$.   If
$z\in\partial B(\tbf{0},\delta)$,
$\sequencen{\|x_n-z\|}$ is a decreasing sequence with limit $\|x-z\|$, so
$\sequencen{\Bover1{\|x_n-z\|^{r-2}}}$ is a non-decreasing sequence with
limit $\Bover1{\|x-z\|^{r-2}}$.   By B.Levi's theorem,

$$\eqalign{\Bover1{\nu(\partial B(\tbf{0},\delta))}\int_{\partial B(\tbf{0},\delta)}
  \Bover1{\|x-z\|^{r-2}}\nu(dz)
&=\lim_{n\to\infty}\Bover1{\nu(\partial B(\tbf{0},\delta))}
  \int_{\partial B(\tbf{0},\delta)}\Bover1{\|x_n-z\|^{r-2}}\nu(dz)\cr
&=\lim_{n\to\infty}\Bover1{\|x_n\|^{r-2}}
=\Bover1{\|x\|^{r-2}}
=\Bover1{\delta^{r-2}}.\cr}$$

\noindent Finally, if $r=1$ and $\|x\|=|x|=\delta$, we are just trying to
take the average of $|x-\delta|$ and  $|x-(-\delta)|$, which will be
$\delta=\Bover1{\delta^{r-2}}$.

\medskip

{\bf (b)} We can follow the same general line.

\medskip

\quad{\bf (i)}
Define $f:\BbbR^r\setminus\partial B(\tbf{0},\delta)\to\Bbb R$ by setting
$f(x)=\Bover1{\nu(\partial B(\tbf{0},\delta))}\int_{\partial B(\tbf{0},\delta)}
\Bover{\|x\|^2-\delta^2}{\|x-z\|^r}\nu(dz)$ when
$\|x\|\ne\delta$.   Then $f$ is harmonic.   \Prf\ If $x$ and $\gamma>0$ are
such that $B(x,\gamma)\subseteq\BbbR^r\setminus\partial B(\tbf{0},\delta)$, then

$$\eqalignno{\Bover1{\nu(\partial B(x,\gamma))}
   &\int_{\partial B(x,\gamma)}f(y)\nu(dy)\cr
&=\Bover1{\nu(\partial B(x,\gamma))}\int_{\partial B(x,\gamma)}
   \Bover1{\nu(\partial B(\tbf{0},\delta))}\int_{\partial B(\tbf{0},\delta)}
   \Bover{\|y\|^2-\delta^2}{\|y-z\|^r}
   \nu(dz)\nu(dy)\cr
&=\Bover1{\nu(\partial B(\tbf{0},\delta))}\int_{\partial B(\tbf{0},\delta)}
   \Bover1{\nu(\partial B(x,\gamma))}\int_{\partial B(x,\gamma)}
   \Bover{\|y\|^2-\delta^2}{\|y-z\|^r}
   \nu(dy)\nu(dz)\cr
&=\Bover1{\nu(\partial B(\tbf{0},\delta))}\int_{\partial B(\tbf{0},\delta)}
   \Bover{\|x\|^2-\delta^2}{\|x-z\|^r}\nu(dz)\cr
\displaycause{because the functions
$y\mapsto\Bover{\|y\|^2-\delta^2}{\|y-z\|^r}$ are harmonic when
$\|z\|=\delta$, by 478Fa}
&=f(x).  \text{\Qed}\cr}$$

\noindent
Since $f$ is smooth, $\nabla^2f=0$ everywhere off $\partial B(\tbf{0},\delta)$,
by 478Ec.

\medskip

\quad{\bf (ii)} As before, we have a function
$h:\coint{0,\infty}\setminus\{\delta\}\to\coint{0,\infty}$ such that
$f(x)=h(\|x\|)$ whenever $\|x\|\ne\delta$.   If $0<\gamma<\delta$ then

$$\eqalign{h(\gamma)
&=\Bover1{\nu(\partial B(\tbf{0},\gamma))}
  \int_{\partial B(\tbf{0},\gamma)}h(\gamma)\nu(dy)
=\Bover1{\nu(\partial B(\tbf{0},\gamma))}
  \int_{\partial B(\tbf{0},\gamma)}f(y)\nu(dy)\cr
&=f(0)
=\Bover1{\nu(\partial B(\tbf{0},\delta))}\int_{\partial B(\tbf{0},\delta)}
  \Bover{-\delta^2}{\|z\|^r}\nu(dz)
=-\Bover1{\delta^{r-2}},\cr}$$

\noindent and

\Centerline{$\Bover1{\nu(\partial B(\tbf{0},\delta))}\int_{\partial B(\tbf{0},\delta)}
  \Bover{|\delta^2-\|x\|^2|}{\|x-z\|^r}\nu(dz)
=-h(\gamma)=\Bover1{\delta^{r-2}}$}

\noindent if $\|x\|=\gamma<\delta$.

\medskip

\quad{\bf (iii)} For $\gamma>\delta$ I start with an elementary estimate.
If $\|x\|=\gamma>\delta$ then
$\Bover{\|x\|^2-\delta^2}{\|x-z\|^r}$ lies between
$\Bover{\gamma^2-\delta^2}{(\gamma+\delta)^r}$ and
$\Bover{\gamma^2-\delta^2}{(\gamma-\delta)^r}$ for every
$z\in\partial B(x,\delta)$, so that $\gamma^{r-2}f(x)$ lies
between
$\Bover{1-(\delta/\gamma)^2}{(1+(\delta/\gamma))^r}$
and $\Bover{1-(\delta/\gamma)^2}{(1-(\delta/\gamma))^r}$,
and is approximately $1$ if $\gamma$ is large.

\medskip

\quad{\bf (iv)} Now we can use the Divergence Theorem
again, as follows.   If $\delta<\gamma<\beta$ consider the region
$E=B(\tbf{0},\beta)\setminus B(\tbf{0},\gamma)$ and the function $\phi=\grad f$.
As $f$ is smooth, $\phi$ is defined everywhere off
$\partial B(\tbf{0},\delta)$, and $\phi(x)=\Bover{h'(\|x\|)}{\|x\|}x$
at every $x\in\BbbR^r\setminus\partial B(\tbf{0},\delta)$.
The essential boundary of $E$ is
$\partial E=\partial B(\tbf{0},\gamma)\cup\partial B(\tbf{0},\beta)$;  the Federer
exterior normal at $x\in\partial B(\tbf{0},\gamma)$ is $v_x=-\Bover1{\gamma}x$
and at $x\in\partial B(\tbf{0},\beta)$ it is $v_x=\Bover1{\beta}x$;
and $\diverg\phi=\nabla^2f$ is zero everywhere on $E$.
So 475N tells us that

$$\eqalign{0
&=\int_{\partial B(\tbf{0},\beta)}\phi(x)\dotproduct v_x\nu(dx)
   +\int_{\partial B(\tbf{0},\gamma)}\phi(x)\dotproduct v_x\nu(dx)\cr
&=\int_{\partial B(\tbf{0},\beta)}h'(\beta)\nu(dx)
   -\int_{\partial B(\tbf{0},\gamma)}h'(\gamma)\nu(dx)\cr
&=r\beta_r\beta^{r-1}h'(\beta)
   -r\beta_r\gamma^{r-1}h'(\gamma).\cr}$$

\noindent
This shows that $h'(\gamma)$ is inversely proportional to $\gamma^{r-1}$.

\medskip

\quad{\bf (v)} If $r\ge 3$,
there are $\alpha$, $\beta\in\Bbb R$ such that
$h(\gamma)=\alpha+\Bover{\beta}{\gamma^{r-2}}$ for every $\gamma>\delta$.
But since (iii) shows us that
$\lim_{\gamma\to\infty}\gamma^{r-2}h(\gamma)=1$,
we must have $h(\gamma)=\Bover1{\gamma^{r-2}}$
for $\gamma>\delta$, as declared.   If $r=2$, then we can express $h$ in
the form $h(\gamma)=\alpha+\beta\ln\gamma$;  this time,
$\lim_{\gamma\to\infty}h(\gamma)=1$, so once more
$h(\gamma)=1=\Bover1{\gamma^{r-2}}$ for every $\gamma$.

\medskip

\quad{\bf (vi)} Finally, if $r=1$ and $|x|>\delta$, then, as in
(a-iv) above,
$\Bover1{\nu(\partial B(\tbf{0},\delta))}\biggerint_{\partial B(\tbf{0},\delta)}
  \Bover{|\delta^2-\|x\|^2|}{\|x-z\|^r}\nu(dz)$ is the average of
$\Bover{x^2-\delta^2}{|x-\delta|}=|x+\delta|$ and
$\Bover{x^2-\delta^2}{|x+\delta|}=|x-\delta|$, so is
$|x|=\Bover1{|x|^{r-2}}$.

\medskip

{\bf (c)} This follows easily from (a);

$$\eqalign{\int_{B(\tbf{0},\delta)}\Bover1{\|x-z\|^{r-2}}\mu(dz)
&=\int_0^{\delta}\int_{\partial B(\tbf{0},t)}\Bover1{\|x-z\|^{r-2}}\nu(dz)dt
   \cr
&=\int_0^{\delta}\Bover{r\beta_rt^{r-1}}
   {\max(t,\|x\|)^{r-2}}dt
\le\int_0^{\delta}r\beta_rt\,dt
=\Bover12r\beta_r\delta^2.\cr}$$
}%end of proof of 478G

\leader{478H}{Corollary} If $r\ge 2$, then
$x\mapsto\Bover1{\|x-z\|^{r-2}}:\BbbR^r\to[0,\infty]$ is superharmonic
for any $z\in\BbbR^r$.

\woddheader{478H}{0}{0}{0}{40pt}

\proof{ If $\delta>0$ and $x\in\BbbR^r$,

$$\eqalign{\Bover1{\nu\partial B(x,\delta)}
  \int_{\partial B(x,\delta)}\Bover1{\|y-z\|^{r-2}}\nu(dy)
&=\Bover1{\nu\partial B(\tbf{0},\delta)}
  \int_{\partial B(\tbf{0},\delta)}\Bover1{\|y+x-z\|^{r-2}}\nu(dy)\cr
&=\Bover1{\max(\delta,\|x-z\|)^{r-2}}
\le\Bover1{\|x-z\|^{r-2}}.\cr}$$
}%end of proof of 478H

\leader{478I}{}\cmmnt{ The Poisson kernel gives a basic method of
building and describing harmonic functions.

\medskip

\noindent}{\bf Theorem} Suppose that $y\in\BbbR^r$ and $\delta>0$;
let $S=\partial B(y,\delta)$\cmmnt{ be the sphere with centre $y$ and
radius $\delta$}.

(a) Let $\zeta$ be a totally finite Radon measure on $S$, and define $f$ on
$\BbbR^r\setminus S$ by setting

\Centerline{$f(x)
=\Bover1{r\beta_r\delta}\int_S
  \Bover{|\delta^2-\|x-y\|^2|}{\|x-z\|^r}\zeta(dz)$}

\noindent for $x\in\BbbR^r\setminus S$.
Then $f$ is continuous and harmonic.

(b) Let $g:S\to\Bbb R$ be a $\nu_S$-integrable function, where $\nu_S$ is
the subspace measure on $S$ induced by $\nu$, and define
$f:\BbbR^r\to\Bbb R$ by setting

$$\eqalign{f(x)
&=\Bover1{r\beta_r\delta}\int_S
  g(z)\Bover{|\delta^2-\|x-y\|^2|}{\|x-z\|^r}\nu(dz)
  \text{ if }x\in\BbbR^r\setminus S,\cr
&=g(x)\text{ if }x\in S.\cr}$$

\quad(i) $f$ is continuous and harmonic in $\BbbR^r\setminus S$.

\quad(ii) If $r\ge 2$, then

\Centerline{$\liminf_{z\in S,z\to z_0}g(x)=\liminf_{x\to z_0}f(x)$,
\quad$\limsup_{x\to z_0}f(x)=\limsup_{z\in S,z\to z_0}g(x)$}

\noindent for every $z_0\in S$.

\quad(iii) $f$
is continuous at any point of $S$ where $g$ is continuous, and if $g$ is
lower semi-continuous then $f$ also is.

\quad(iv) $\sup_{x\in\BbbR^r}f(x)=\sup_{z\in S}g(z)$ and
$\inf_{x\in\BbbR^r}f(x)=\inf_{z\in S}g(z)$.

\proof{{\bf (a)} $f$ is continuous just because
$x\mapsto\Bover{\delta^2-\|x-y\|^2}{\|x-z\|^r}$ is continuous for each
$z\in S$ and uniformly bounded for $z\in S$ and
$x$ running over any compact set not meeting $S$.

Suppose that $\|x-y\|<\delta$
and $\eta>0$ is such
that $B(x,\eta)\cap S=\emptyset$.   Then

$$\eqalignno{
\Bover1{\nu(\partial B(x,\eta))}\int_{\partial B(x,\eta)}fd\nu
&=\Bover1{\nu(\partial B(x,\eta))}\int_{\partial B(x,\eta)}
  \Bover1{r\beta_r\delta}
    \int_S\Bover{\delta^2-\|w-y\|^2}{\|w-z\|^r}\zeta(dz)\nu(dw)\cr
&=\Bover1{r\beta_r\delta}\int_S\Bover1{\nu(\partial B(x,\eta))}
    \int_{\partial B(x,\eta)}
    \Bover{\delta^2-\|w-y\|^2}{\|w-z\|^r}\nu(dw)\zeta(dz)\cr
&=\Bover1{r\beta_r\delta}\int_S
    \Bover{\delta^2-\|x-y\|^2}{\|x-z\|^r}\zeta(dz)\cr
\displaycause{because $w\mapsto\Bover{\delta^2-\|w-y\|^2}{\|w-z\|^r}$
is harmonic on $\BbbR^r\setminus\{z\}$ whenever $z\in S$, by 478Fa}
&=f(x).\cr}$$

\noindent As $x$ and $\eta$ are arbitrary, $f$ is harmonic on
$\interior B(y,\delta)$.   Similarly, it is harmonic on
$\BbbR^r\setminus B(y,\delta)$.
\medskip

{\bf (b)(i)} Applying (a) to the indefinite-integral measures over
the subspace measure $\nu_S$
defined by the positive and negative parts of $g$, we see that $f$ is
continuous and harmonic in $\BbbR^r\setminus S$.

\medskip

\quad{\bf (ii)}\grheada\ If $x\notin S$,

$$\eqalign{\int_S\Bover{|\delta^2-\|x-y\|^2|}{\|x-z\|^r}\nu(dz)
&=\int_{\partial B(\tbf{0},\delta)}
   \Bover{|\delta^2-\|x-y\|^2|}{\|x-y-z\|^r}\nu(dz)\cr
&=\Bover{\nu(\partial B(\tbf{0},\delta))}{\max(\delta,\|x-y\|)^{r-2}}\cr}$$

\noindent by 478Gb.   In particular, if
$x$ is close to, but not on, the sphere $S$,
$\int_S\Bover{|\delta^2-\|x-y\|^2|}{\|x-z\|^r}\nu(dz)$ is
approximately
$\Bover{\nu(\partial B(\tbf{0},\delta)}{\delta^{r-2}}=r\beta_r\delta$.

\medskip

\qquad\grheadb\ Set $M=\int_S|g|d\nu$, and take $z_0\in S$;   set
$\gamma=\limsup_{x\in S,x\to z_0}g(x)$.   If $\gamma=\infty$ then certainly
$\limsup_{x\to z_0}f(x)\le\gamma$.   Otherwise, take $\eta>0$.
Let $\alpha_0\in\ooint{0,\delta}$ be such that

\Centerline{$|\Bover1{r\beta_r\delta}\int_S
  \Bover{|\delta^2-\|x-y\|^2|}{\|x-z\|^r}\nu(dz)-1|\le\eta$}

\noindent whenever $0<|\delta-\|x-y\||\le\alpha_0$, and
$g(z)\le\gamma+\eta$
whenever $z\in S$ and $0<\|z-z_0\|\le 2\alpha_0$.
Let $\alpha\in\ocint{0,\alpha_0}$ be such that
$(2\delta+\alpha_0)(M+|\gamma|)\alpha\nu S
\le 2^rr\beta_r\delta\alpha_0^r\eta$.

If $\|x-z_0\|\le\alpha$ and $\|x-y\|\ne \delta$,
then $|\delta -\|x-y\||\le\|x-z_0\|\le\alpha_0$ and
$|\delta^2-\|x-y\|^2|\le\|x-z_0\|(2\delta +\alpha_0)$, so

$$\eqalignno{f(x)-\gamma
&\le\eta|\gamma|+\Bover1{r\beta_r\delta}\int_S
  (g(z)-\gamma)\Bover{|\delta^2-\|x-y\|^2|}{\|x-z\|^r}\nu(dz)\cr
&\le\eta|\gamma|
  +\Bover1{r\beta_r\delta}\int_{S\cap B(z_0,2\alpha_0)}
    (g(z)-\gamma)\Bover{|\delta^2-\|x-y\|^2|}{\|x-z\|^r}\nu(dz)\cr
&\mskip100mu
  +\Bover1{r\beta_r\delta}\int_{S\setminus B(z_0,2\alpha_0)}
    (|g(z)|+|\gamma|)\Bover{|\delta^2-\|x-y\|^2|}{\|x-z\|^r}\nu(dz)\cr
&\le\eta|\gamma|
  +\Bover{\eta}{r\beta_r\delta}\int_S
    \Bover{|\delta^2-\|x-y\|^2|}{\|x-z\|^r}\nu(dz)\cr
&\mskip100mu
  +\Bover1{r\beta_r\delta}\int_{S\setminus B(z_0,2\alpha_0)}
    (|g(z)|+|\gamma|)\Bover{\|x-z_0\|(2\delta +\alpha_0)}{2^r\alpha_0^r}
    \nu(dz)\cr
\displaycause{because $r\ge 2$, so $\nu\{z_0\}=0$}
&\le\eta|\gamma|
  +\eta(1+\eta)+(M+|\gamma|)
      \Bover{\alpha(2\delta +\alpha_0)}{2^rr\beta_r\delta\alpha_0^r}\nu S
  \cr
&\le (|\gamma|+1+\eta+1)\eta.\cr}$$

\noindent Also, of course, $f(x)-\gamma=g(x)-\gamma\le\eta$ if
$0<\|x-z_0\|\le\alpha$ and $\|x-y\|=\delta$.   As $\eta$
is arbitrary, $\limsup_{x\to z_0}f(x)\le\gamma$.
In the other direction, $\limsup_{x\to z_0}f(x)\ge\gamma$ just because $f$
extends $g$.

\medskip

\qquad\grheadc\
Similarly, or applying ($\beta$) to $-g$,
$\liminf_{x\to z_0}f(x)=\liminf_{x\in S,x\to z_0}g(x)$.

\medskip

\quad{\bf (iii)}\grheada\ If $r\ge 2$, it follows at once from (ii)
that if $g$ is continuous at $z_0\in S$, so is $f$, and
that if $g$ is lower semi-continuous (so that
$g(z_0)\le\liminf_{x\in S,x\to z_0}g(x)$ for every $z_0\in S$) then $f$
also is lower semi-continuous.

\medskip

\qquad\grheadb\ If $r=1$, then $S=\{y-\delta,y+\delta\}$ and
$\nu\{y-\delta\}=\nu\{y+\delta\}=1$, so

$$\eqalign{f(x)
&=\Bover1{2\delta}\bigl(
   \Bover{|\delta^2-(x-y)^2|}{|x-(y-\delta)|}g(y-\delta)
     +\Bover{|\delta^2-(x-y)^2|}{|x-(y+\delta)|}g(y+\delta)\bigr)\cr
&=\Bover1{2\delta}\bigl(|x-(y+\delta)|g(y-\delta)
           +|x-(y-\delta)|g(y+\delta)\bigr)\cr}$$

\noindent for $x\in\Bbb R\setminus S$, and
$\lim_{x\mapsto y\pm\delta}f(x)=g(y\pm\delta)$, so $f$ is continuous.

\medskip

\quad{\bf (iv)} If $g(z)\le\alpha<\infty$ for every $z\in S$, then

$$\eqalign{f(x)
&=\Bover1{r\beta_r\delta}\int_S
   g(z)\Bover{|\delta^2-\|x-y\|^2|}{\|x-z\|^r}\nu(dz)
\le\Bover{\alpha}{r\beta_r\delta}\int_{\partial B(\tbf{0},\delta)}
   \Bover{|\delta^2-\|x-y\|^2|}{\|x-y-z\|^r}\nu(dz)\cr
&=\Bover{\alpha}{r\beta_r\delta}
   \cdot\Bover{\nu(\partial B(\tbf{0},\delta))}{\max(\delta,\|x-y\|)^{r-2}}
=\Bover{\alpha\delta^{r-2}}{\max(\delta,\|x-y\|)^{r-2}}
\le\alpha\cr}$$

\noindent for every $x\in\BbbR^r\setminus S$.   So
$\sup_{x\in\BbbR^r}f(x)=\sup_{z\in S}g(z)$;  similarly,
$\inf_{x\in\BbbR^r}f(x)=\inf_{z\in S}g(z)$.
}%end of proof of 478I

\leader{478J}{Convolutions and smoothing:  Proposition} (a) Suppose that
$f:\BbbR^r\to[0,\infty]$ is Lebesgue measurable, and $G\subseteq\BbbR^r$
an open set such that $f\restr G$ is superharmonic.   Let
$h:\BbbR^r\to[0,\infty]$ be a Lebesgue integrable function, and $f*h$ the
convolution of $f$ and $h$.   If
$H\subseteq G$ is an open set such that $H-\{z:h(z)\ne 0\}\subseteq G$,
then $(f*h)\restr H$ is superharmonic.

(b) Suppose, in (a), that
$h(y)=h(z)$ whenever $\|y\|=\|z\|$ and that
$\int_{\BbbR^r}h\,d\mu\le 1$.   If $x\in G$ and $\gamma>0$ are such that
$B(x,\gamma)\subseteq G$ and $h(y)=0$ whenever $\|y\|\ge\gamma$,
then $(f*h)(x)\le f(x)$.

(c) Let $f:\BbbR^r\to[0,\infty]$ be a lower semi-continuous
function, and $\sequencen{\tilde h_n}$ the sequence of 473E.
If $G\subseteq\BbbR^r$ is an open set such that $f\restr G$ is
superharmonic, then $f(x)=\lim_{n\to\infty}(f*\tilde h_n)(x)$ for every
$x\in G$.

(d)\dvAnew{2020}
Let $G\subseteq\BbbR^r$ be an open set, $K\subseteq G$ a compact set
and $g:G\to\Bbb R$ a smooth function.   Then there is a smooth function
$f:G\to\Bbb R$ with compact support included in $G$
such that $f$ agrees with $g$ on an open set including $K$.

\proof{{\bf (a)} If $x\in H$ and $\delta>0$ are such that
$B(x,\delta)\subseteq H$, then

$$\eqalignno{\int_{\partial B(x,\delta)}(f*h)(y)\nu(dy)
&=\int_{\partial B(x,\delta)}\int_{\BbbR^r}f(y-z)h(z)\mu(dz)\nu(dy)\cr
&=\int_{\BbbR^r}h(z)\int_{\partial B(x,\delta)}f(y-z)\nu(dy)\mu(dz)\cr
&=\int_{\BbbR^r}h(z)\int_{\partial B(x-z,\delta)}f(y)\nu(dy)\mu(dz)\cr
&\le r\beta_r\delta^{r-1}\int_{\BbbR^r}h(z)f(x-z)\mu(dz)\cr
\displaycause{because if $h(z)\ne 0$ then $B(x-z,\delta)=B(x,\delta)-z$ is
included in $G$}
&=\nu(\partial B(x,\delta))\cdot(f*h)(x).\cr}$$

\medskip

{\bf (b)} Let $g:\coint{0,\infty}\to[0,\infty]$ be such that
$h(y)=g(\|y\|)$ for every $y$.   Then

$$\eqalign{(f*h)(x)
&=\int_{\BbbR^r}f(y)h(x-y)\mu(dy)
=\int_0^{\gamma}\int_{\partial B(x,t)}f(y)g(t)\nu(dy)dt\cr
&\le\int_0^{\gamma}r\beta_rt^{r-1}f(x)g(t)dt
=f(x)\int_{\BbbR^r}h\,d\mu
\le f(x).\cr}$$

\medskip

{\bf (c)} By (b), $(f*\tilde h_n)(x)\le f(x)$ for every sufficiently
large $n$, so $\limsup_{n\to\infty}(f*\tilde h_n)(x)\le f(x)$.
In the other
direction, if $x\in G$ and $\alpha<f(x)$, there is a $\delta>0$ such
that $B(x,\delta)\subseteq G$ and
$f(y)\ge\alpha$ for every $y\in B(x,\delta)$.   Now there is an
$m\in\Bbb N$ such that $\tilde h_n(y)=0$ whenever $n\ge m$ and
$\|y\|\ge\delta$;  so that

$$\eqalign{(f*\tilde h_n)(x)
&=\int f(y)\tilde h_n(x-y)\mu(dy)
=\int_{B(x,\delta)}f(y)\tilde h_n(x-y)\mu(dy)\cr
&\ge\alpha\int_{B(x,\delta)}\tilde h_n(x-y)\mu(dy)
=\alpha\cr}$$

\noindent whenever $n\ge m$.   As $\alpha$ is arbitrary,
$f(x)=\lim_{n\to\infty}(f*\tilde h_n)(x)$.

\medskip

{\bf (d)} If $K$ is empty we can take $f$ to be the constant function with value
$0$, so suppose
otherwise.   Let $\delta>0$ be such that $B(x,4\delta)\subseteq G$ for every
$x\in K$, and set $H_k=\{x:\rho(x,K)<k\delta\}$ for $k\in\Bbb N$.
Let $n$ be such that
$\bover1{n+1}\le\delta$, and set $f=g\times(\chi H_2*\tilde h_n)$.   By 473De,
$\chi H_2*\tilde h_n$ and therefore $f$ are smooth.  If
$x\in H_1$ then $(\chi H_2*\tilde h_n)(x)=1$ so $f(x)=g(x)$, while if
$x\in G\setminus H_3$ then $f(x)=(\chi H_2*\tilde h_n)(x)=0$ so the support of
$f$ is included in the compact set $\overline{H}_3\subseteq G$.
}%end of proof of 478J

\leader{478K}{Dynkin's formula:  Lemma}
Let $\mu_W$ be $r$-dimensional Wiener measure on
$\Omega=C(\coint{0,\infty};\BbbR^r)_0$;  set
$X_t(\omega)=\omega(t)$ for $\omega\in\Omega$ and $t\ge 0$.
Let $f:\BbbR^r\to\Bbb R$ be a three-times-differentiable
function such that $f$ and its
first three derivatives are continuous and bounded.

(a) $\Expn(f(X_t))=f(0)+\Bover12\Expn(\int_0^t(\nabla^2f)(X_s)ds)$ for
every $t\ge 0$.

(b) If $\tau:\Omega\to\coint{0,\infty}$ is a stopping time adapted to
$\langle\Sigma^+_t\rangle_{t\ge 0}$ and $\Expn(\tau)$ is finite, then

\Centerline{
$\Expn(f(X_{\tau}))=f(0)+\Bover12\Expn(\int_0^{\tau}(\nabla^2f)(X_s)ds)$.}

\proof{{\bf (a)(i)} We need a special case of the multidimensional
Taylor's theorem.   If $f:\BbbR^r\to\Bbb R$ is three times differentiable
and $x=(\xi_1,\ldots,\xi_r)$, $y=(\eta_1,\ldots,\eta_r)\in\BbbR^r$,
then there is a $z$ in the line segment $[x,y]$ such that

$$\eqalign{f(y)&=f(x)+\sum_{i=1}^r(\eta_i-\xi_i)\Pd{f}{\xi_i}(x)
             +\Bover12\sum_{i=1}^r\sum_{j=1}^r(\eta_i-\xi_i)(\eta_j-\xi_j)
	        \Pde{f}{\xi_i}{\xi_j}(x)\cr
&\mskip100mu
+\Bover16\sum_{i=1}^r\sum_{j=1}^r\sum_{k=1}^r
	        (\eta_i-\xi_i)(\eta_j-\xi_j)(\eta_k-\xi_k)
      \Bover{\partial^3f}{\partial\xi_i\partial\xi_j\partial\xi_k}(z).\cr}$$

\noindent\Prf\ Set
$g(\beta)=f(\beta y+(1-\beta)x)$ for $\beta\in\Bbb R$.   Then $g$ is three
times differentiable, with

$$\eqalign{g'(\beta)
&=\sum_{k=1}^r(\eta_k-\xi_k)\Pd{f}{\xi_k}(\beta y+(1-\beta)x),\cr
g''(\beta)
&=\sum_{j=1}^r\sum_{k=1}^r(\eta_j-\xi_j)(\eta_k-\xi_k)
         \Pde{f}{\xi_j}{\xi_k}(\beta y+(1-\beta)x),\cr
g'''(\beta)	
&=\sum_{i=1}^r\sum_{j=1}^r\sum_{k=1}^r
	        (\eta_i-\xi_i)(\eta_j-\xi_j)(\eta_k-\xi_k)
      \Bover{\partial^3f}{\partial\xi_i\partial\xi_j\partial\xi_k}
        (\beta y+(1-\beta)x).\cr}$$

\noindent Now by Taylor's theorem with remainder, in one dimension, there
is a $\beta\in\ooint{0,1}$ such that

\Centerline{$g(1)=g(0)+g'(0)+\Bover12g''(0)+\Bover16g'''(\beta)$}

\noindent and all we have to do is to
set $z=\beta y+(1-\beta)x$ and substitute in the values for
$g(1),\ldots,g'''(\beta)$.\ \Qed

\medskip

\quad{\bf (ii)} Let $M\ge 0$ be such that
$\|\Bover{\partial^3f}{\partial\xi_i\partial\xi_j\partial\xi_k}\|_{\infty}
\le M$ whenever $1\le i$, $j$, $k\le r$.   Let $K$ be
$\Expn((\sum_{i=1}^r|Z_i|)^3)$ when $Z_1,\ldots,Z_r$ are independent
real-valued
random variables with standard normal distribution.   (To see that this is
finite, observe that

$$\eqalignno{\Expn((\sum_{i=1}^r|Z_i|)^3)
&\le\Expn(r^3\max_{i\le r}|Z_i|^3)
\le r^3\,\Expn(\sum_{i=1}^r|Z_i|^3)
=r^4\,\Expn(|Z|^3)\cr
\displaycause{where $Z$ is a random variable with standard normal
distribution}
&=\Bover{2r^4}{\sqrt{2\pi}}\int_0^{\infty}t^3e^{-t^2/2}dt
<\infty.)\cr}$$

\noindent For any $x$, $y\in\BbbR^r$ we have

$$\eqalign{\bigl|f(y)-f(x)-\sum_{i=1}^r(\eta_i-\xi_i)\Pd{f}{\xi_i}(x)
             &+\Bover12\sum_{i=1}^r\sum_{j=1}^r(\eta_i-\xi_i)(\eta_j-\xi_j)
	        \Pde{f}{\xi_i}{\xi_j}(x)\bigr|\cr
&\le\Bover{M}6(\sum_{i=1}^r|\eta_i-\xi_i|)^3.\cr}$$
		
\noindent If
$0\le s\le t$ and $\omega\in\Omega$, then

$$\eqalign{\bigl|f(\omega(t))-f(\omega(s))
  &-\sum_{i=1}^r\Pd{f}{\xi_i}(\omega(s))(\omega_i(t)-\omega_i(s))\cr
&\mskip100mu
   -\Bover12\sum_{i=1}^r\sum_{j=1}^r\Pde{f}{\xi_i}{\xi_j}(\omega(s))
      (\omega_i(t)-\omega_i(s))(\omega_j(t)-\omega_j(s))\bigr|\cr
&\le\Bover{M}6(\sum_{i=1}^r|\omega_i(t)-\omega_i(s)|)^3,\cr}$$

\noindent writing $\omega_1,\ldots,\omega_r\in C(\coint{0,\infty})_0$
for the coordinates of
$\omega\in\Omega$.   Integrating with respect to $\omega$, we have

$$\eqalignno{|\Expn(f(X_t)-&f(X_s)-\Bover12(t-s)(\nabla^2f)(X_s))|\cr
&=\bigl|\Expn(f(X_t)-f(X_s))
  -\sum_{i=1}^r\Expn(\Pd{f}{\xi_i}(X_s))\Expn(X^{(i)}_t-X^{(i)}_s)\cr
&\mskip100mu
  -\Bover12\sum_{i=1}^r\Expn(\Pdd{f}{\xi_i}(X_s))
      \Expn(X^{(i)}_t-X^{(i)}_s)^2\cr
&\mskip100mu
  -\Bover12\sum_{i=1}^r\sum_{j\ne i}\Expn(\Pde{f}{\xi_i}{\xi_j})(X_s)
      \Expn(X^{(i)}_t-X^{(i)}_s)\Expn(X^{(j)}_t-X^{(j)}_s)\bigr|\cr
\displaycause{writing $X^{(i)}_t(\omega)=\omega_i(t)$ for
$1\le i\le r$, and recalling that $\Expn(X^{(i)}_t-X^{(i)}_s)=0$ for
every $i$, while $\Expn(X^{(i)}_t-X^{(i)}_s)^2=t-s$}
&=\bigl|\Expn(f(X_t)-f(X_s))
  -\sum_{i=1}^r\Expn((X^{(i)}_t-X^{(i)}_s)\Pd{f}{\xi_i}(X_s))\cr
&\mskip100mu
  -\Bover12\sum_{i=1}^r\Expn((X^{(i)}_t-X^{(i)}_s)^2\Pdd{f}{\xi_i}(X_s))\cr
&\mskip100mu
  -\Bover12\sum_{i=1}^r\sum_{j\ne i}
     \Expn((X^{(i)}_t-X^{(i)}_s)(X^{(j)}_t-X^{(j)}_s)
        \Pde{f}{\xi_i}{\xi_j}(X_s))\bigr|\cr
\displaycause{because for any $i\le r$ the random variables
$\Pd{f}{\xi_i}(X_s)$ and $X^{(i)}_t-X^{(i)}_s$ are independent, while
for any distinct $i$, $j\le r$ the random variables
$\Pde{f}{\xi_i}{\xi_j}(X_s)$, $X^{(i)}_t-X^{(i)}_s$ and
$X^{(j)}_t-X^{(j)}_s$ are independent}
&=\bigl|\Expn\bigl(f(X_t)-f(X_s)
  -\sum_{i=1}^r(X^{(i)}_t-X^{(i)}_s)\Pd{f}{\xi_i}(X_s)\cr
&\mskip100mu  -\Bover12\sum_{i=1}^r\sum_{j=1}^r
     (X^{(i)}_t-X^{(i)}_s)(X^{(j)}_t-X^{(j)}_s)
        \Pde{f}{\xi_i}{\xi_j}(X_s)\bigr)\bigr|\cr
&\le\Bover{M}6\Expn((\sum_{i=1}^r|X^{(i)}_t-X^{(i)}_s|)^3)
\le\Bover{MK}6(t-s)^{3/2}\cr}$$

\noindent because the $X^{(i)}_t-X^{(i)}_s$ are independent random
variables all with the same distribution as $\sqrt{t-s}\,Z$ where $Z$ is
standard normal.

\woddheader{478K}{4}{2}{2}{36pt}

{\bf (iii)} Now fix $t\ge 0$ and $n\ge 1$;  set $s_k=\Bover{k}{n}t$ for
$k\le n$.   Set

\Centerline{$g_n(\omega)
=\sum_{k=0}^{n-1}(s_{k+1}-s_k)(\nabla^2f)(\omega(s_k))$}

\noindent for $\omega\in\Omega$.   Then

$$\eqalign{
\bigl|\int\bigl(f(\omega(t))-&f(0)
  -\Bover12g_n(\omega)\bigr)\mu_W(d\omega)\bigr|\cr
&=\bigl|\sum_{k=0}^{n-1}\Expn(f(X_{s_{k+1}})-f(X_{s_k})
  -\Bover12(s_{k+1}-s_k)(\nabla^2f)(X_{s_k}))\bigr|\cr
&\le\sum_{k=0}^{n-1}\Bover{MK}6(\Bover{t}n)^{3/2}
=\Bover{MKt\sqrt t}{6\sqrt n}.\cr}$$

\noindent On the other hand,

\Centerline{$\lim_{n\to\infty}g_n(\omega)
=\int_0^t(\nabla^2f)(\omega(s))ds$}

\noindent for every $\omega$ (the Riemann integral
$\int_0^t(\nabla^2f)(\omega(s))ds$ is defined because
$s\mapsto(\nabla^2f)(\omega(s))$ is continuous), and
$|g_n(\omega)|\le t\|\nabla^2f\|_{\infty}<\infty$ for every $\omega$, so by
Lebesgue's Dominated Convergence Theorem

$$\eqalign{\Expn(f(X_t)-f(0))
&=\Bover12\lim_{n\to\infty}\int g_n(\omega)\mu_W(d\omega)
=\Bover12\int\lim_{n\to\infty}g_n(\omega)\mu_W(d\omega)\cr
&=\Bover12\int\int_0^t(\nabla^2f)(\omega(s))ds\,\mu_W(d\omega)
=\Bover12\Expn\bigl(\int_0^t(\nabla^2f)(X_s)ds\bigr)\cr}$$

\noindent as claimed.

\medskip

{\bf (b)(i)}
Consider first the case in which $\tau$ takes values in a
finite set $I\subseteq\coint{0,\infty}$.
In this case we can induce on $\#(I)$.   If $I=\{t_0\}$ then

\Centerline{
$\Expn(f(X_{\tau}))
=\Expn(f(X_{t_0}))
=f(0)+\Bover12\Expn(\int_0^{t_0}(\nabla^2f)(X_s)ds)
=f(0)+\Bover12\Expn(\int_0^{\tau}(\nabla^2f)(X_s)ds)$}

\noindent by (a).   For the inductive step to $\#(I)>1$, set $t_0=\min I$,
$E=\{\omega:\tau(\omega)=t_0\}$ and

$$\eqalign{\phi(\omega,\omega')(t)
&=\omega(t)\text{ if }t\le t_0,\cr
&=\omega(t_0)+\omega'(t-t_0)\text{ if }t\ge t_0,\cr}$$

\noindent for $\omega$, $\omega'\in\Omega$, so that $\phi$ is \imp\ (477G).
Set

\Centerline{$\sigma_{\omega}(\omega')
=\tau(\phi(\omega,\omega'))-t_0$}

\noindent for $\omega$, $\omega'\in\Omega$.   If
$\omega\in\Omega\setminus E$,
$\sigma_{\omega}$ is a stopping time adapted to
$\langle\Sigma_t^+\rangle_{t\ge 0}$, taking fewer than $\#(I)$ values.
\Prf\ Suppose that $t>0$ and $F=\{\omega':\sigma_{\omega}(\omega')<t\}$.
If $\omega'\in F$, $\tilde\omega'\in\Omega$ and
$\tilde\omega'\restr[0,t]=\omega'\restr[0,t]$,
then $\tau(\phi(\omega,\omega'))<t+t_0$, while
$\phi(\omega,\tilde\omega')(s)=\phi(\omega,\omega')(s)$ whenever
$s\le t+t_0$;  so that

\Centerline{$\sigma_{\omega}(\tilde\omega')+t_0
=\tau(\phi(\omega,\tilde\omega'))<t+t_0$}

\noindent and $\tilde\omega'\in F$.   Thus $F\in\Sigma_t$;  as $t$ is
arbitrary, $\sigma_{\omega}$ is adapted to
$\langle\Sigma_t^+\rangle_{t\ge 0}$ (455Lb).
Also every value of $\sigma_{\omega}$ belongs to
$\{t-t_0:t\in I$, $t>t_0\}$ which is smaller than $I$.\ \Qed

Writing $\int\ldots d\omega$ and
$\int\ldots d\omega'$ for integration with respect to $\mu_W$,

\global\pagegoalminus=\the\pagegoal
\global\pagetotalplus=\the\pagetotal
\global\advance\pagegoalminus by -\pageshrink
\global\advance\pagetotalplus by \bigbaselineskip

$$\eqalignno{\Expn\bigl(\int_0^{\tau}&(\nabla^2f)(X_s)ds\bigr)\cr
\global\advance\pagetotalplus by \bigbaselineskip
&=\int_{\Omega}\int_0^{\tau(\omega)}(\nabla^2f)(\omega(s))
  ds\,d\omega\cr
\global\advance\pagetotalplus by \bigbaselineskip
&=\int_{\Omega}\int_{\Omega}\int_0^{\tau(\phi(\omega,\omega'))}
  (\nabla^2f)(\phi(\omega,\omega')(s))
  ds\,d\omega'd\omega\cr
\checkbreak
&=\int_E\int_{\Omega}\int_0^{t_0}
    (\nabla^2f)(\omega(s))ds\,d\omega'd\omega\cr
\global\advance\pagetotalplus by \bigbaselineskip
&\mskip100mu  +\int_{\Omega\setminus E}\int_{\Omega}
    \int_0^{t_0+\sigma_{\omega}(\omega')}
    (\nabla^2f)(\phi(\omega,\omega')(s))
    ds\,d\omega'd\omega\cr
\checkbreak
&=\int_E\int_0^{t_0}(\nabla^2f)(\omega(s))ds\,d\omega
  +\int_{\Omega\setminus E}\int_{\Omega}\int_0^{t_0}
    (\nabla^2f)(\phi(\omega,\omega')(s))
    ds\,d\omega'd\omega\cr
\global\advance\pagetotalplus by \bigbaselineskip
&\mskip100mu  +\int_{\Omega\setminus E}\int_{\Omega}
    \int_{t_0}^{t_0+\sigma_{\omega}(\omega')}
    (\nabla^2f)(\phi(\omega,\omega')(s))
    ds\,d\omega'd\omega\cr
&=\int_{\Omega}\int_0^{t_0}(\nabla^2f)(\omega(s))ds\,d\omega\cr
\global\advance\pagetotalplus by \bigbaselineskip
&\mskip100mu
  +\int_{\Omega\setminus E}\int_{\Omega}
    \int_{t_0}^{t_0+\sigma_{\omega}(\omega')}
    (\nabla^2f)(\omega(t_0)+\omega'(s-t_0))
    ds\,d\omega'd\omega\cr
\checkbreak
&=\int_{\Omega}\int_0^{t_0}(\nabla^2f)(\omega(s))ds\,d\omega
  +\int_{\Omega\setminus E}\int_{\Omega}\int_0^{\sigma_{\omega}(\omega')}
    (\nabla^2f)(\omega(t_0)+\omega'(s))
    ds\,d\omega'd\omega\cr
\checkbreak
&=2\int_{\Omega}f(\omega(t_0))-f(0)d\omega\cr
\global\advance\pagetotalplus by \bigbaselineskip
&\mskip100mu
  +2\int_{\Omega\setminus E}\int_{\Omega}
    f(\omega(t_0)+\omega'(\sigma_{\omega}(\omega')))-f(\omega(t_0))
    d\omega'd\omega\cr
\displaycause{\global\advance\pagetotalplus by \baselineskip
applying the inductive hypothesis to the function
$x\mapsto f(\omega(t_0)+x)$ and the stopping time $\sigma_{\omega}$}
\checkbreak
&=2\int_E f(\omega(t_0))-f(0)d\omega
  +2\int_{\Omega\setminus E}\int_{\Omega}
    f(\omega(t_0)+\omega'(\sigma_{\omega}(\omega')))-f(0)
    d\omega'd\omega\cr
\checkbreak
&=2\int_E\int_{\Omega}
   f\bigl(\phi(\omega,\omega')(\tau(\phi(\omega,\omega')))\bigr)
     -f(0)d\omega'd\omega\cr
\global\advance\pagetotalplus by \bigbaselineskip
&\mskip100mu  +2\int_{\Omega\setminus E}\int_{\Omega}
    f\bigl(\phi(\omega,\omega')(\tau(\phi(\omega,\omega')))\bigr)
      -f(0)d\omega'd\omega\cr
\checkbreak
&=2\int_{\Omega}\int_{\Omega}
    f\bigl(\phi(\omega,\omega')(\tau(\phi(\omega,\omega')))\bigr)
      -f(0)d\omega'd\omega\cr
\checkbreak
&=2\int_{\Omega}f(\omega(\tau(\omega)))-f(0)d\omega
=2\bigl(\Expn(f(X_{\tau}))-f(0)\bigr).\cr}$$

\noindent Turning this around, we have the formula we want, so the
induction proceeds.

\medskip

\quad{\bf (ii)} Now suppose that every value of
$\tau$ belongs to an infinite set of
the form $\{t_n:n\in\Bbb N\}\cup\{\infty\}$ where $\sequencen{t_n}$ is
a strictly increasing sequence in $\coint{0,\infty}$.
In this case, for $n\in\Bbb N$, define

\Centerline{$\tau_n(\omega)=\min(\tau(\omega),t_n)$}

\noindent for $\omega\in\Omega$, so that $\tau_n$ takes
values in the finite set $\{t_0,\ldots,t_n\}$, and

$$\eqalign{\{\omega:\tau_n(\omega)<t\}
&=\{\omega:\tau(\omega)<t\}\in\Sigma_t\text{ if }t\le t_n,\cr
&=\Omega\in\Sigma_t\text{ if }t>t_n.\cr}$$
Now $\tau$ is finite a.e., so
$\tau\eae\lim_{n\to\infty}\tau_n$;  it follows that

\Centerline{$f(X_{\tau}(\omega))
=f(\omega(\tau(\omega)))
=\lim_{n\to\infty}f(\omega(\tau_n(\omega)))
=\lim_{n\to\infty}f(X_{\tau_n}(\omega))$}

\noindent for almost
every $\omega$;  because $f$ is bounded,

\Centerline{$\Expn(f(X_{\tau}))=\lim_{n\to\infty}\Expn(f(X_{\tau_n}))$.}

\noindent On the other side,

\Centerline{$\int_0^{\tau(\omega)}(\nabla^2f)(\omega(s))ds
=\lim_{n\to\infty}\int_0^{\tau_n(\omega)}(\nabla^2f)(\omega(s))ds$}

\noindent for almost
every $\omega$.   At this point, recall that we are supposing
that $\tau$ has finite expectation and that $\nabla^2f$ is bounded.
So

\Centerline{$|\int_0^{\tau_n(\omega)}(\nabla^2f)(\omega(s))ds|
\le\int_0^{\tau(\omega)}|(\nabla^2f)(\omega(s))ds|
\le\|\nabla^2f\|_{\infty}\tau(\omega)$}

\noindent for every $\omega$, and the dominated convergence theorem assures
us that

$$\eqalign{\Expn\bigl(\int_0^{\tau}(\nabla^2f)(X_s)ds\bigr)
&=\int_{\Omega}\int_0^{\tau(\omega)}(\nabla^2f)(\omega(s))ds\,d\omega\cr
&=\lim_{n\to\infty}
  \int_{\Omega}\int_0^{\tau_n(\omega)}(\nabla^2f)(\omega(s))ds\,d\omega\cr
&=\lim_{n\to\infty}\Expn\bigl(\int_0^{\tau_n}(\nabla^2f)(X_s)ds\bigr).
\cr}$$

\noindent Accordingly

$$\eqalignno{\Expn(f(X_{\tau}))
&=\lim_{n\to\infty}\Expn(f(X_{\tau_n}))
=\lim_{n\to\infty}f(0)
  +\Bover12\Expn\bigl(\int_0^{\tau_n}(\nabla^2f)(X_s)ds\bigr)\cr
\displaycause{by (i)}
&=f(0)+\Bover12\Expn\bigl(\int_0^{\tau}(\nabla^2f)(X_s)ds\bigr),\cr}$$

\noindent as required.

\medskip

\quad{\bf (iii)} Suppose just that $\tau$ has finite
expectation.   This time, for $n\in\Bbb N$, define a stopping time $\tau_n$
by saying that

$$\eqalign{\tau_n(\omega)&=2^{-n}k\text{ if }k\ge 1
  \text{ and }2^{-n}(k-1)\le\tau(\omega)<2^{-n}k,\cr
&=\infty\text{ if }\tau(\omega)=\infty.\cr}$$

\noindent If $t>0$, set $t'=2^{-n}k$ where $2^{-n}k<t\le 2^{-n}(k+1)$;
then

\Centerline{$\{\omega:\tau_n(\omega)<t\}
=\{\omega:\tau(\omega)<t'\}\in\Sigma_{t'}\subseteq\Sigma_t$.}

\noindent So $\tau_n$ is a stopping time adapted to
$\langle\Sigma_t^+\rangle_{t\ge 0}$;  as $\tau_n\le 2^{-n}+\tau$,
$\Expn(\tau_n)<\infty$.   Again we have
$\tau(\omega)=\lim_{n\to\infty}\tau_n(\omega)$ for every
$\omega$.   The arguments of (ii) now tell us that, as before,

\Centerline{$f(X_{\tau}(\omega))=\lim_{n\to\infty}f(X_{\tau_n}(\omega))$}

\noindent(because $f$ is continuous),

\Centerline{$\int_0^{\tau(\omega)}(\nabla^2f)(\omega(s))ds
=\lim_{n\to\infty}\int_0^{\tau_n(\omega)}(\nabla^2f)(\omega(s))ds$}

\noindent for almost every $\omega$, so that

\Centerline{$\Expn(f(X_{\tau}))=\lim_{n\to\infty}\Expn(f(X_{\tau_n}))$,}

\Centerline{$\Expn(\int_0^{\tau}(\nabla^2f)(X_s)ds)
=\lim_{n\to\infty}\Expn(\int_0^{\tau_n}(\nabla^2f)(X_s)ds)$.}

\noindent (This time, of course, we need to check that

\Centerline{$|\int_0^{\tau_n(\omega)}(\nabla^2f)(\omega(s))ds|
\le\|\nabla^2f\|_{\infty}\tau_0(\omega)$}

\noindent for almost every $\omega$, to confirm that we have dominated
convergence.)   So once again the desired formula can be got by taking the
limit of a sequence of equalities we already know.
}%end of proof of 478K

\leader{478L}{Theorem} Let $\mu_W$ be $r$-dimensional Wiener measure on
$\Omega=C(\coint{0,\infty};\BbbR^r)_0$,
$f:\BbbR^r\to[0,\infty]$ a lower semi-continuous superharmonic
function, and $\tau:\Omega\to[0,\infty]$ a stopping time adapted to
$\langle\Sigma_t^+\rangle_{t\ge 0}$.   Set
$H=\{\omega:\omega\in\Omega$, $\tau(\omega)<\infty\}$.   Then

\Centerline{$f(x)
  \ge\int_Hf(x+\omega(\tau(\omega)))\mu_W(d\omega)$}

\noindent for every $x\in\BbbR^r$.

\proof{{\bf (a)} To begin with, suppose that $f$ is
real-valued and bounded.
Let $\sequence{m}{\tilde h_m}$ be the sequence of 473E/478J, and
for $m\in\Bbb N$ set $f_m=f*\tilde h_m$.   Then each $f_m$ is non-negative,
smooth with
bounded derivatives of all orders (473De) and superharmonic (478Ja), so
$\nabla^2f_m\le 0$ (478Ea).   Set $\tau_n(\omega)=\min(n,\tau(\omega))$ for
$n\in\Bbb N$ and $\omega\in\Omega$;  then each $\tau_n$ is a stopping time,
adapted to $\langle\Sigma^+_t\rangle_{t\ge 0}$, with finite expectation.
In the language of 478K,

\Centerline{$\Expn(f_m(x+X_{\tau_n}))
=f_m(x)+\Bover12\Expn\bigl(\int_0^{\tau_n}(\nabla^2f_m)(x+X_s)ds\bigr)
\le f_m(x)$}

\noindent whenever $m$, $n\in\Bbb N$.   Letting $n\to\infty$,

$$\eqalignno{\int_Hf_m(x+\omega(\tau(\omega)))\mu_W(d\omega)
&=\lim_{n\to\infty}\int_Hf_m(x+\omega(\tau_n(\omega)))\mu_W(\omega)\cr
\displaycause{because $f_m$, and every $\omega\in\Omega$, are continuous,
and $\tau(\omega)=\lim_{n\to\infty}\tau_n(\omega)$ for every $\omega$}
&\le\liminf_{n\to\infty}
   \int_{\BbbR^r}f_m(x+\omega(\tau_n(\omega)))\mu_W(d\omega)\cr
\displaycause{because $f_m$ is non-negative}
&\le f_m(x)\cr}$$

\noindent by Fatou's Lemma.   Now $f=\lim_{m\to\infty}f_m$ (478Jc), so

$$\eqalign{\int_Hf(x+\omega(\tau(\omega)))\mu_W(\omega)
&\le\liminf_{m\to\infty}
    \int_Hf_m(x+\omega(\tau(\omega)))\mu_W(\omega)\cr
&\le\liminf_{m\to\infty}f_m(x)
=f(x),\cr}$$

\noindent which is what we need to know.

\medskip

{\bf (b)} For the general case, set $g_k=f\wedge k\chi\BbbR^r$ for each
$k\in\Bbb N$.   Then $g_k$ is non-negative,
lower semi-continuous, superharmonic (478Cc) and bounded.   So

$$\eqalign{\int_Hf(x+\omega(\tau(\omega)))\mu_W(d\omega)
&=\lim_{k\to\infty}\int_Hg_k(x+\omega(\tau(\omega)))\mu_W(d\omega)\cr
&\le\lim_{k\to\infty}g_k(x)
=f(x).\cr}$$
}%end of proof of 478L

\leader{478M}{Proposition} (a) If $r=1$, then $\{\omega(t):t\ge 0\}=\Bbb R$
for almost every $\omega\in\Omega$.

(b) If $r\le 2$, then
$\{\omega(t):t\ge 0\}$ is dense in $\BbbR^2$ for almost every
$\omega\in\Omega$.

(c) If $r\ge 2$, then
for every $z\in\BbbR^2$, $z\notin\{\omega(t):t>0\}$ for almost
every $\omega\in\Omega$.

(d) If $r\ge 3$, then
$\lim_{t\to\infty}\|\omega(t)\|=\infty$ for
almost every $\omega\in\Omega$.

\proof{{\bf (a)} Suppose that
$\alpha$, $\beta>0$ and that $\tau$ is the Brownian exit time from
$\ooint{-\alpha,\beta}$;  then $\tau$ is a stopping time adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$ (477Ic).   Now
$\tau$ is almost everywhere finite and
$\Pr(X_{\tau}=\beta)=\Bover{\alpha}{\alpha+\beta}$.   \Prf\ Since
$\Pr(|X_t|\le\max(\alpha,\beta))\to 0$ as $t\to\infty$, $\tau$ is finite
a.e., and $\Pr(X_{\tau}=\beta)+\Pr(X_{\tau}=-\alpha)=1$.   Set
$\tau_n(\omega)=\min(n,\tau(\omega))$ for each $n$, and let $f:\Bbb R\to\Bbb R$
be a smooth function with compact support such that $f(x)=x$ for
$x\in[-\alpha-1,\beta+1]$ (478Jd).   Now 478K tells us that

\Centerline{$\Expn(X_{\tau_n})=\Expn(f(X_{\tau_n}))
=f(0)+\Bover12\Expn(\int_0^{\tau_n}(\nabla^2f)(X_s)ds)=f(0)=0$.}

\noindent Since $\sequencen{X_{\tau_n}}$ is a uniformly bounded sequence
converging almost everywhere to $X_{\tau}$,

\Centerline{$\beta\Pr(X_{\tau}=\beta)-\alpha\Pr(X_{\tau}=-\alpha)
=\Expn(X_{\tau})=\lim_{n\to\infty}\Expn(X_{\tau_n})=0$,}

\noindent and $\Pr(X_{\tau}=\beta)=\Bover{\alpha}{\alpha+\beta}$.\ \Qed

Letting $\alpha\to\infty$, we see that
$\Pr(\Exists t\ge 0,\,X_t=\beta)=1$.   Similarly, $-\alpha$ lies on
almost every sample path.

Thus almost every sample path must pass through every point of $\Bbb Z$;
since sample paths are continuous, they almost all cover $\Bbb R$.

\medskip

{\bf (b)} For $r=1$ this is covered by (a);  take $r=2$.
Suppose that $z\in\Bbb R^2$ and that
$\delta>0$.   Then almost every sample path meets $B(z,\delta)$.
\Prf\ If $\delta\ge\|z\|$ this is trivial.
Otherwise, take $R>\|z\|$ and let $\tau$ be
the Brownian exit time from $G=\interior B(z,R)\setminus B(z,\delta)$.
We have $\Pr(\|X_t\|\le R+\|z\|)\to 0$ as $t\to\infty$
(because $\Pr(\|X_t\|\le\alpha)\le\Pr(|Z|\le\Bover{\alpha}{\sqrt{t}})$
where
$Z$ is a standard normal random variable), so $\tau$ is finite a.e.
Once again, set $\tau_n(\omega)=\min(n,\tau(\omega))$ for $n\in\Bbb N$ and
$\omega\in\Omega$;  this time,
take a smooth function $f:\BbbR^2\to\Bbb R$ with compact support
such that $f(x)=\ln\|x-z\|$ for $x\in B(z,2R)\setminus B(z,\bover12\delta)$.
Then

\Centerline{$\Expn(f(X_{\tau_n}))=f(0)=\ln\|z\|$}

\noindent(use 478Fb), so

$$\eqalign{\ln R\cdot\Pr(X_{\tau}\in\partial B(z,R))
  &+\ln\delta\cdot\Pr(X_{\tau}\in\partial B(z,\delta))\cr
&=\Expn(f(X_{\tau}))
=\lim_{n\to\infty}\Expn(f(X_{\tau_n}))
=\ln\|z\|\cr}$$

\noindent and

\Centerline{$\Pr(X_{\tau}\in\partial B(z,\delta))
=\Bover{\ln R-\ln\|z\|}{\ln R-\ln\delta}$.}

\noindent Letting $R\to\infty$, we see that
$\Pr(\Exists t\ge 0,\,\omega(t)\in B(z,\delta))=1$;  that is, almost every
sample path meets $B(z,\delta)$.\ \Qed

Letting $B(z,\delta)$ run over a
sequence of balls constituting a network for the topology of $\BbbR^2$,
we see that almost every path meets every non-empty open set and is dense
in $\BbbR^2$.

\medskip

{\bf (c)(i)} Consider first the case $r=2$.

\medskip

\qquad\grheada\ Suppose that $z\ne 0$.   In this case,
take $\delta$, $R$ such that
$0<\delta<\|z\|<R$ and let $\tau$ be
the Brownian exit time from $G=\interior B(z,R)\setminus B(z,\delta)$,
as in the proof of (b).   As before, we have

\Centerline{$\Pr(X_{\tau}\in\partial B(z,\delta))
=\Bover{\ln R-\ln\|z\|}{\ln R-\ln\delta}$.}

\noindent This time, looking at the limit as $\delta\downarrow 0$, we see
that

\Centerline{$\{\omega:$ there is a $t\ge 0$ such that $\omega(t)=z$ but
$\|\omega(s)-z\|<R$ for every $s\le t\}$}

\noindent is negligible.   Taking the union of these sets over large
integer $R$, we see that

\Centerline{$\{\omega:$ there is a $t\ge 0$ such that $\omega(t)=z\}$}

\noindent is negligible, as required.

\medskip

\qquad\grheadb\ As for $z=0$, take any $\epsilon>0$.   Then

$$\eqalignno{
\mu_W\{\omega:&\text{ there is some }t\ge\epsilon
  \text{ such that }\omega(t)=0\}\cr
&=\mu_W^2\{(\omega,\omega'):
  \text{ there is some }t\ge 0
  \text{ such that }\omega'(t)=-\omega(\epsilon)\}\cr
&=\mu_W^2\{(\omega,\omega'):\omega(\epsilon)\ne 0
  \text{ and there is some }t\ge 0
  \text{ such that }\omega'(t)=-\omega(\epsilon)\}\cr
\displaycause{because the distribution of $X_{\epsilon}$ is atomless, so
$\{\omega:\omega(\epsilon)=0\}$ is negligible}
&=0\cr}$$

\noindent by ($\alpha$).   Taking the union over rational $\epsilon>0$,
$\{\omega:\text{ there is some }t>0\text{ such that }\omega(t)=0\}$ is
negligible.

\medskip

\quad{\bf (ii)} If $r>2$, set $Tx=(\xi_1,\xi_2)$ for
$x=(\xi_1,\ldots,\xi_r)\in\BbbR^r$.   Then
$\omega\mapsto T\omega:
C(\coint{0,\infty};\BbbR^r)_0\to C(\coint{0,\infty};\BbbR^2)_0$ is
\imp\ for $r$-dimensional Wiener measure $\mu_{Wr}$ on
$C(\coint{0,\infty};\BbbR^r)_0$ and
two-dimensional Wiener measure $\mu_{W2}$ on
$C(\coint{0,\infty};\BbbR^2)_0$, by 477D(c-i) or otherwise.
So

\Centerline{$\{\omega:z\in\omega[\,\ooint{0,\infty}\,]\}
\subseteq\{\omega:Tz\in(T\omega)[\,\ooint{0,\infty}\,]\}$}

\noindent is negligible.

\medskip

{\bf (d)(i)} Fix $\gamma\in\coint{0,\infty}$ and $\epsilon>0$ for the
moment.
Set $g(x)=\int\Bover1{\|y\|^{r-2}}\tilde h_0(x-y)\mu(dy)$ for
$x\in\BbbR^r$, where $\tilde h_0$ is the function of 473E;
then $g$ is smooth (473De), strictly positive and superharmonic (478Ja).
In addition, we have the following.

\medskip

\qquad\grheada\ All the derivatives of $g$ are bounded.   \Prf\
As shown in the proof of 473De, $\Pd{g}{\xi_i}(x)
=\int\Bover1{\|y\|^{r-2}}\Pd{}{\xi_i}\tilde h_0(x-y)\mu(dy)$ for
$1\le i\le r$ and $x\in\BbbR^r$.
Inducing on the order of $D$, and using 478Gc at the last step,
we see that

$$\eqalign{(Dg)(x)
&=\int\Bover1{\|y\|^{r-2}}(D\tilde h_0)(x-y)\mu(dy)
=\int\Bover1{\|x-y\|^{r-2}}(D\tilde h_0)(y)\mu(dy)\cr
&=\int_{B(\tbf{0},1)}\Bover1{\|x-y\|^{r-2}}(D\tilde h_0)(y)\mu(dy)\cr
&\le\|D\tilde h_0\|_{\infty}\int_{B(\tbf{0},1)}\Bover1{\|x-y\|^{r-2}}\mu(dy)
\le\Bover12r\beta_r\|D\tilde h_0\|_{\infty}\cr}$$

\noindent for
any partial differential operator $D$ and any $x\in\BbbR^r$.\ \Qed

\medskip

\qquad\grheadb\ $\lim_{\|x\|\to\infty}g(x)=0$, because

\Centerline{$g(x)
\le\|\tilde h_0\|_{\infty}\int_{B(\tbf{0},1)}\Bover1{\|x-y\|^{r-2}}\mu(dy)
\le\|\tilde h_0\|_{\infty}\Bover{\beta_r}{(\|x\|-1)^{r-2}}$}

\noindent whenever $\|x\|>1$.

\medskip

\qquad\grheadc\ $g(x)=g(y)$ whenever $\|x\|=\|y\|$.   \Prf\ Let
$T:\BbbR^r\to\BbbR^r$ be an orthogonal transformation such that $Tx=y$.
Then

$$\eqalignno{g(x)
&=\int\Bover1{\|x-z\|^{r-2}}\tilde h_0(z)\mu(dz)
=\int\Bover1{\|T(x-z)\|^{r-2}}\tilde h_0(Tz)\mu(dz)\cr
\displaycause{because $\tilde h_0T=\tilde h_0$}
&=\int\Bover1{\|y-Tz\|^{r-2}}\tilde h_0(Tz)\mu(dz)
=\int\Bover1{\|y-z\|^{r-2}}\tilde h_0(z)\mu(dz)\cr
\displaycause{because $T$ is an automorphism of $(\BbbR^r,\mu)$}
&=g(y).  \text{  \Qed}\cr}$$

\medskip

\quad{\bf (ii)} Let $\beta>0$ be the common value of $g(y)$ for
$\|y\|=\gamma$.  Take $x\in\BbbR^r$ such that $\|x\|>\gamma$,
and $n\in\Bbb N$.   Define

\Centerline{$\tau(\omega)=\min(\{n\}\cup\{t:\|x+\omega(t)\|\le\gamma\})$}

\noindent for $\omega\in\Omega$.   Then $\tau$ is a bounded stopping time
adapted to $\langle\Sigma_t\rangle_{t\ge 0}$, so

$$\eqalign{\beta\Pr(\tau<n)
&\le\Expn(g(x+X_{\tau}))\cr
&=g(x)+\Bover12\Expn(\int_0^{\tau}(\nabla^2g)(x+X_s)ds)
\le g(x).\cr}$$

\noindent Letting $n\to\infty$, we see that

\Centerline{$\mu_W\{\omega:\|x+\omega(t)\|\le\gamma$ for some $t\ge 0\}
\le\Bover1{\beta}g(x)$.}

\medskip

\quad{\bf (iii)} Now let $n>\gamma$ be an integer such that
$\Bover1{\beta}g(x)\le\epsilon$
whenever $\|x\|\ge n$.   As in (a) and (b-i) above,
$\lim_{t\to\infty}\Pr(\|X_t\|\le n)=0$;  take $m\in\Bbb N$ such that
$\Pr(\|X_m\|\le n)\le\epsilon$.   Let $\sigma$ be the stopping time with
constant value $m$, with $\phi_{\sigma}:\Omega\times\Omega\to\Omega$ the
corresponding \imp\ function (477G).   Set
$F=\{\omega:\|\omega(m)\|>n\}$.   Now

$$\eqalignno{\Pr(\|X_t\|\le\gamma&\text{ for some }t\ge m)\cr
&=\mu_W^2\{(\omega,\omega'):\|\phi_{\sigma}(\omega,\omega')(t)\|\le\gamma
  \text{ for some }t\ge m\}\cr
\displaycause{where $\mu_W^2$ is the product measure on
$\Omega\times\Omega$}
&=\mu_W^2\{(\omega,\omega'):\|\omega(m)+\omega'(t-m)\|\le\gamma
  \text{ for some }t\ge m\}\cr
&\le\mu_W^2\{(\omega,\omega'):\|\omega(m)\|\le n\text{ or }
  \|\omega(m)\|\ge n\cr
&\mskip200mu
  \text{and }\|\omega(m)+\omega'(t)\|\le\gamma\text{ for some }t\ge 0\}\cr
&\le\mu\{\omega:\|\omega(m)\|\le n\}\cr
&\mskip50mu
  +\int_F\mu_W\{\omega':\|\omega(m)+\omega'(t)\|\le\gamma\text{ for some }
     t\ge 0\}\mu_W(d\omega)\cr
&\le\epsilon
  +\int_F\Bover1{\beta}g(\omega(m))\mu_W(d\omega)
\le\epsilon+\epsilon\mu_WF
\le 2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\Pr(\liminf_{t\to\infty}\|X_t\|<\gamma)=0$;  as $\gamma$ is arbitrary,
$\Pr(\lim_{t\to\infty}\|X_t\|=\infty)=1$.
}%end of proof of 478M

\cmmnt{\medskip

\noindent{\bf Remark} In 479R I will show that there is a
surprising difference between the cases $r=3$ and $r\ge 4$.
}

\vleader{48pt}{478N}{Wandering paths}
Let $G\subseteq\BbbR^r$ be an open set, and for $x\in G$ set

\Centerline{$F_x(G)=\{\omega:$ either $\tau_x(\omega)<\infty$ or
$\lim_{t\to\infty}\|\omega(t)\|=\infty\}$}

\noindent where $\tau_x$ is the
Brownian exit time from $G-x$.   I will say that
$G$ has {\bf few wandering paths} if $F_x(G)$ is conegligible for every
$x\in G$.   In this case we can
be sure that, if $x\in G$, then for almost every $\omega$ {\it either}
$\lim_{t\to\infty}\|\omega(t)\|=\infty$ {\it or} $\omega(t)\notin G-x$ for
some $t$.   So
we can speak of $X_{\tau_x}(\omega)=\omega(\tau_x(\omega))$, taking this to
be $\infty$ if $\omega\in F_x(G)$ and
$\tau_x(\omega)=\infty$;  and $\omega$ will be
continuous on $[0,\tau_x(\omega)]$ for every $\omega\in F_x(G)$.
\cmmnt{We find that }$X_{\tau_x}:\Omega\to\partial^{\infty}(G-x)$
is Borel
measurable.   \prooflet{\Prf\ $\tau_x$ is the Brownian hitting time to the
closed set $\BbbR^r\setminus(G-x)$, so is a stopping time adapted to
$\langle\Tau_{[0,t]}\rangle_{t\ge 0}$ (477Ic).   Let
$\Cal B(\Omega)$ be the
Borel $\sigma$-algebra of $\Omega$ for the topology of uniform convergence
on compact sets;  then $\Tau_{[0,t]}\subseteq\Cal B(\Omega)$ for every
$t\ge 0$.   The function

\Centerline{$(t,\omega)\mapsto X_t(\omega):
\coint{0,\infty}\times\Omega\to\BbbR^r$}

\noindent is continuous, therefore
$\Cal B(\coint{0,\infty})\tensorhat\Cal B(\Omega)$-measurable
(4A3D(c-i));  so $X_{\tau_x}$ is $\Cal B(\Omega)$-measurable (455Ld).\
\Qed}

From 478M, we see that if $r\ge 3$
then any open set in $\BbbR^r$ will have
few wandering paths, while if $r\le 2$ then $G$ will have few wandering
paths whenever it is not dense in $\BbbR^r$.
Note that if $G\subseteq\BbbR^r$ is open, $H$ is a component of $G$, and
$x\in H$, then the exit times from $H-x$ and $G-x$ are the same\cmmnt{,
just because sample paths are continuous}, and $F_x(G)=F_x(H)$.
It follows\cmmnt{ at once}
that if $G$ has more than one component then it has few wandering
paths.

\leader{478O}{Theorem} Let $G\subseteq\BbbR^r$ be an open set with few
wandering paths and
$f:\overline{G}^{\infty}\to\Bbb R$ a bounded lower semi-continuous function
such that $f\restr G$ is superharmonic.   Take $x\in G$ and let
$\tau:\Omega\to[0,\infty]$ be the
Brownian exit time from $G-x$\cmmnt{ (477Ia)}.
Then $f(x)\ge\Expn(f(x+X_{\tau}))$.

\proof{ It will be enough to deal with the case $f\ge 0$.

\medskip

{\bf (a)} Extend $f$ to a function
$\tilde f:\BbbR^r\cup\{\infty\}\to\Bbb R$ by setting $\tilde f(x)=0$
for $x\notin\overline{G}^{\infty}$.   Since $f$ is bounded, so is
$\tilde f$.
Let $\sequencen{\tilde h_n}$ be the sequence of 473E/478Jc, and for
$n\in\Bbb N$ set $f_n=(\tilde f\restr\BbbR^r)*\tilde h_n$.
Also, for $n\in\Bbb N$, set

\Centerline{$G_n
=\{y:y\in G$, $\|y\|<n$, $\rho(y,\BbbR^r\setminus G)>\Bover1{n+1}\}$}

\noindent (interpreting $\rho(y,\emptyset)$ as $\infty$ if $G=\BbbR^r$),
and let $\tau_n$ be the Brownian exit time from $G_n-x$.

\medskip

{\bf (b)} For $y\in G$, $f_n(y)\le f(y)$ for all sufficiently large
$n$ and $f(y)=\lim_{n\to\infty}f_n(y)$ (478Jb).   Also $f_n\restr G_n$ is
superharmonic (478Ja).   Each $f_n$ is smooth with bounded
derivatives of all orders (473De), and $(\nabla^2f_n)(y)\le 0$ for
$y\in G_n$ (478Ea).

If $m\ge n$,

\Centerline{$\Expn(f_m(x+X_{\tau_n}))
=f_m(x)+\Bover12\Expn(\int_0^{\tau_n}(\nabla^2f_m)(x+X_s)ds)
\le f_m(x)$}

\noindent (478K).   Consequently

$$\eqalign{\Expn(f(x+X_{\tau_n}))
&\le\liminf_{m\to\infty}\Expn(f_m(x+X_{\tau_n}))\cr
&\le\liminf_{m\to\infty}f_m(x)
=f(x).\cr}$$

{\bf (c)} For every $\omega\in\Omega$,
$\sequencen{\tau_n(\omega)}$ is a non-decreasing sequence with limit
$\tau(\omega)$.   \Prf\ Since
$\tau_n(\omega)\le\tau_{n+1}(\omega)\le\tau(\omega)$ for every $n$,
$t=\lim_{n\in\Bbb N}\tau_n(\omega)$ is defined in $[0,\infty]$.   If
$t=\infty$ then surely $t=\tau(\omega)$.   Otherwise,
$\omega(t)=\lim_{n\to\infty}\omega(\tau_n(\omega))\notin G-x$, so again
$t=\tau(\omega)$.\ \Qed

Consequently

\Centerline{$f(x+\omega(\tau(\omega)))
\le\liminf_{n\to\infty}f(x+\omega(\tau_n(\omega)))$}

\noindent for almost every $\omega$.   \Prf\ In the language of
478N, we can suppose that $\omega\in F_x(G)$, so that
$\omega(\tau(\omega))=\lim_{n\to\infty}\omega(\tau_n(\omega))$
in $\BbbR^r\cup\{\infty\}$, and
we can use the fact that $f$ is lower semi-continuous.\ \QeD\
So

\Centerline{$\Expn(f(x+X_{\tau}))
\le\liminf_{n\to\infty}\Expn(f(x+X_{\tau_n}))\le f(x)$}

\noindent as required.
}%end of proof of 478O

\leader{478P}{Harmonic measures (a)}
Let $A\subseteq\BbbR^r$ be an analytic
set and $x\in\BbbR^r$.   Let $\tau:\Omega\to[0,\infty]$ be the Brownian
hitting time to $A-x$\cmmnt{ (477I)}.   Then $\tau$ is
$\Sigma$-measurable, where $\Sigma$ is the domain of
$\mu_W$\cmmnt{ (455Ma)}.   Setting $H=\{\omega:\tau(\omega)<\infty\}$,
$X_{\tau}:H\to\BbbR^r$ is $\Sigma$-measurable.
\prooflet{\Prf\ By 4A3Qc,
$(\omega,t)\mapsto\omega(t)$ is
$\Sigma\tensorhat\Cal B(\coint{0,\infty})$-measurable, while
$\omega\mapsto(\omega,\tau(\omega))$ is
$(\Sigma,\Sigma\tensorhat\Cal B(\coint{0,\infty}))$-measurable.   So
$\omega\mapsto\omega(\tau(\omega))$ is $\Sigma$-measurable on $H$.\ \Qed}

Consider the function
$\omega\mapsto x+\omega(\tau(\omega)):H\to\BbbR^r$.
This induces a Radon image measure $\mu_x$ on $\BbbR^r$ defined by saying
that

\Centerline{$\mu_xF
=\mu_W\{\omega:\omega\in H$, $x+\omega(\tau(\omega))\in F\}
=\Pr(x+X_{\tau}\in F)$}

\noindent whenever this is defined.   \cmmnt{ Because
every $\omega\in\Omega$ is
continuous,} $X_{\tau}(\omega)\in\partial(A-x)$ for every $\omega\in H$,
and $\partial A$ is conegligible for $\mu_x$.   I will call $\mu_x$ the
{\bf harmonic measure for arrivals in $A$ from $x$}.   Of course
$\mu_x\BbbR^r$ is the Brownian hitting probability of $A$.

Note that if $F\subseteq\BbbR^r$ is closed and $x\in\BbbR^r\setminus F$,
then the Brownian hitting time to $F-x$ is the same as the Brownian hitting
time to $\partial F-x$,\cmmnt{ because all paths are continuous,}
so that the harmonic measure for arrivals in $F$
from $x$ coincides with the harmonic measure for arrivals in $\partial F$
from $x$.

\spheader 478Pb\cmmnt{ We now have an easy corollary of 478L.}   Let
$A\subseteq\BbbR^r$ be an analytic set, $x\in\BbbR^r$, and $\mu_x$
the harmonic measure for arrivals in $A$ from $x$.   If
$f:\BbbR^r\to[0,\infty]$ is a lower semi-continuous superharmonic function,
$f(x)\ge\int fd\mu_x$.   \prooflet{\Prf\ Let $\tau$ be the Brownian hitting
time to $A-x$, and $H=\{\omega:\tau(\omega)<\infty\}$.   Then

$$\eqalignno{\int fd\mu_x
&=\int_Hf(x+\omega(\tau(\omega)))d\mu_W\cr
\displaycause{because $\mu_x$ is the image measure of
the subspace measure $(\mu_W)_H$ under
$\omega\mapsto x+\omega(\tau(\omega))$}
&\le f(x)\cr}$$

\noindent by 478L.\ \Qed}

\spheader 478Pc We can re-interpret 478O in this language.   Let
$G\subseteq\BbbR^r$ be an open set with few wandering paths, and $x\in G$.
Let $\mu_x$ be the harmonic measure for arrivals in $\BbbR^r\setminus G$
from $x$.   In this case, taking $\tau$ to be the Brownian exit time from
$G-x$ and $H=\{\omega:\tau(\omega)<\infty\}$,\cmmnt{ we know that}
$\lim_{t\to\infty}\|\omega(t)\|=\infty$ for almost every
$\omega\in\Omega\setminus H$.   If
$f:\partial^{\infty}G\to[-\infty,\infty]$ is a function, then

$$\eqalignno{\Expn(f(x+X_{\tau}))
&=\int_Hf(x+X_{\tau}(\omega))\mu_W(d\omega)
   +f(\infty)\mu_W(\Omega\setminus H)\cr
\displaycause{counting $f(\infty)$ as zero if $G$ is bounded}
&=\int fd\mu_x+f(\infty)(1-\mu_x\BbbR^r)\cr}$$

\noindent if either integral is defined in
$[-\infty,\infty]$\cmmnt{ (235J\formerly{2{}35L})}.
In particular, if $f:\overline{G}^{\infty}\to\Bbb R$ is a bounded lower
semi-continuous function and $f\restr G$ is superharmonic, then
$f(x)\ge\int fd\mu_x+f(\infty)(1-\mu_x\BbbR^r)$\cmmnt{, by 478O}.
Similarly, if
$f:\overline{G}^{\infty}\to\Bbb R$ is continuous and $f\restr G$ is
harmonic, then $f(x)=\int fd\mu_x+f(\infty)(1-\mu_x\BbbR^r)$ for every
$x\in G$.

\spheader 478Pd Suppose that $\sequencen{A_n}$ is a non-decreasing sequence
of analytic subsets of $\BbbR^r$, with union $A$.   For $x\in\BbbR^r$, let
$\mu^{(n)}_x$, $\mu_x$ be the harmonic measures for arrivals in $A_n$, $A$
respectively from $x$.   Then $\mu_x$ is the limit
$\lim_{n\to\infty}\mu^{(n)}_x$ for the narrow topology on the space of
totally finite Radon measures on $\BbbR^r$\cmmnt{ (437Jd)}.
\prooflet{\Prf\ Let $\tau_n$, $\tau$ be the Brownian hitting times for
$A_n-x$, $A-x$ respectively.   Then $\sequencen{\tau_n}$ is a
non-increasing sequence with limit $\tau$.   Since every $\omega\in\Omega$
is continuous, $X_{\tau}(\omega)=\lim_{n\to\infty}X_{\tau_n}(\omega)$
whenever $\tau(\omega)<\infty$.   Set

\Centerline{$H_n=\{\omega:\tau_n(\omega)<\infty\}$ for each $n$,
$H=\{\omega:\tau(\omega)<\infty\}=\bigcup_{n\in\Bbb N}H_n$.}

\noindent If $f\in C_b(\BbbR^r)$, then
$f(x+X_{\tau}(\omega))=\lim_{n\to\infty}f(x+X_{\tau_n}(\omega))$ for every
$\omega\in H$, so

\Centerline{$\int fd\mu_x
=\int_Hf(x+X_{\tau})d\mu_W
=\lim_{n\to\infty}\int_{H_n}f(x+X_{\tau_n})d\mu_W
=\lim_{n\to\infty}\int fd\mu^{(n)}_x$.}

\noindent As $f$ is arbitrary, $\mu_x=\lim_{n\to\infty}\mu^{(n)}_x$
(437Kc).\ \Qed}

\leader{478Q}{}\cmmnt{ It is generally difficult to find formulae
describing
harmonic measures.   Theorem 478I, however, gives us a technique for
an important special case.

\wheader{478Q}{4}{2}{2}{60pt}

\noindent}{\bf Proposition} Let $S$ be the sphere $\partial B(y,\delta)$,
where $y\in\BbbR^r$ and $\delta>0$.   For $x\in\BbbR^r\setminus S$, let
$\zeta_x$ be the indefinite-integral measure over $\nu$ defined by the
function

$$\eqalign{z
&\mapsto\Bover{|\delta^2-\|x-y\|^2|}{r\beta_r\delta\|x-z\|^r}
  \text{ if }z\in S,\cr
&\mapsto 0\text{ if }z\in\BbbR^r\setminus S.\cr}$$

(a) If $x\in\interior B(y,\delta)$,
then the harmonic measure $\mu_x$ for arrivals in $S$ from $x$
is $\zeta_x$.

(b) In particular, the harmonic measure $\mu_y$ for arrivals in $S$ from
$y$ is $\Bover1{\nu S}\nu\LLcorner S$.

(c) Suppose that $r\ge 2$.   If $x\in\BbbR^r\setminus B(y,\delta)$,
then the harmonic measure $\mu_x$ for arrivals in $S$ from
$x$ is $\zeta_x$.\cmmnt{  In particular,}
$\mu_x\BbbR^r=\Bover{\delta^{r-2}}{\|x-y\|^{r-2}}$.

\proof{{\bf (a)} If $g\in C_b(\BbbR^r)$,
then 478Ib tells us that we have a continuous function $f_g$ which extends
$g$, is harmonic on $\BbbR^r\setminus S$ and is such that
$f_g(x)=\int g\,d\zeta_x$.   Now $G=\interior B(y,\delta)$ is bounded,
so it has
few wandering paths (478N) and the harmonic measure $\mu_x$ is defined,
with $f_g(x)=\int f_g\,d\mu_x$, by 478Pc.   But this means that

\Centerline{$\int g\,d\mu_x=\int f_gd\mu_x=f_g(x)=\int g\,d\zeta_x$.}

\noindent As $g$ is arbitrary, $\mu_x=\zeta_x$ (415I), as claimed.

\medskip

{\bf (b)} If $x=y$ then

\Centerline{$\Bover{\delta^2-\|x-y\|^2}{r\beta_r\delta\|x-z\|^r}
=\Bover{\delta^2}{r\beta_r\delta^{r+1}}
=\Bover1{\nu S}$}

\noindent if $z\in S$.   Since $\nu\LLcorner S$ is the indefinite-integral
measure over $\nu$ defined by $\chi S$ (234M\formerly{2{}34E}),
we have the result.

\medskip

{\bf (c)(i)} To see that $\zeta_x$ is the harmonic measure,
we can use the same argument as in (a), with decorations.
If $g\in C_b(\BbbR^r)$, then 478Ib gives us a bounded
continuous function $f_g$, harmonic on $H=\BbbR^r\setminus B(y,\delta)$,
such that $f_g$ agrees with $g$ on $S$, and $f_g(x)=\int g\,d\zeta_x$.
$S$ is conegligible for both $\mu_x$ and $\zeta_x$.

\medskip

\qquad\grheada\ If $r\ge 3$, then

\Centerline{$\limsup_{\|x\|\to\infty}|f_g(x)|
\le\Bover{\nu S}{r\beta_r\delta}
  \limsup_{\|x\|\to\infty}
    \Bover{\|x-y\|^2-\delta^2}{(\|x\|-\delta-\|y\|)^r}
=0$.}

\noindent So setting $f_g(\infty)=0$, $f_g:\overline{H}^{\infty}\to\Bbb R$
is continuous and bounded, and harmonic on $H$;  so that

\Centerline{$\int g\,d\zeta_x
=f_g(x)
=\int f_gd\mu_x+f_g(\infty)(1-\mu_x\BbbR^r)
=\int g\,d\mu_x$.}

\noindent As in (a), we conclude that $\zeta_x=\mu_x$.

\medskip

\qquad\grheadb\ If $r=2$, then by 478Mb we see that almost every
$\omega\in\Omega$ takes values in $B(y,\delta)-x$;  so $\mu_x\BbbR^r=1$.
Set $\underline{f}\vthsp_g(x)=f_g(x)$ for $x\in\BbbR^2$,
$\underline{f}\vthsp_g(\infty)=\liminf_{\|x\|\to\infty}f_g(x)$.   Then
$\underline{f}\vthsp_g$ is lower semi-continuous on
$\overline{H}^{\infty}$ and harmonic on $H$, so

\Centerline{$\int g\,d\zeta_x
=f_g(x)
=\underline{f}\vthsp_g(x)
\ge\int\underline{f}\vthsp_gd\mu_x
   +\underline{f}\vthsp_g(\infty)(1-\mu_x\BbbR^r)
=\int g\,d\mu_x$.}

\noindent Applying the same argument to $-g$, we see that
$\int g\,d\zeta_x\le\int g\,d\mu_x$, so in fact the integrals are equal,
and we have the result in this case also.

\medskip

\quad{\bf (ii)} Now

$$\eqalignno{\mu_x\BbbR^r
&=\zeta_xS
=\int_S\Bover{\|x-y\|^2-\delta^2}{r\beta_r\delta\|x-z\|^r}\nu(dz)\cr
&=\int_{\partial B(\tbf{0},\delta)}
  \Bover{\|x-y\|^2-\delta^2}{r\beta_r\delta\|x-y-z\|^r}\nu(dz)
=\Bover{\nu(\partial B(\tbf{0},\delta))}{r\beta_r\delta\|x-y\|^{r-2}}\cr
\displaycause{478Gb}
&=\Bover{\delta^{r-2}}{\|x-y\|^{r-2}}.\cr}$$
}%end of proof of 478Q

\leader{478R}{Theorem} Let $A$, $B\subseteq\BbbR^r$ be analytic sets
with $A\subseteq B$.
For $x\in\BbbR^r$, let $\mu^{(A)}_x$, $\mu^{(B)}_x$ be the harmonic
measures for arrivals in $A$, $B$ respectively from $x$.   Then, for any
$x\in\BbbR^r$,
$\family{y}{\BbbR^r}{\mu^{(A)}_y}$ is a disintegration of $\mu^{(A)}_x$
over $\mu^{(B)}_x$.

\proof{{\bf (a)} Let $\tau$ be the
Brownian hitting time to $B-x$, and $\tau'$ the hitting time
to $A-x$; then $\tau(\omega)\le\tau'(\omega)$ for every $\omega$.
If $\tau(\omega)<\infty$, set $f(\omega)=x+\omega(\tau(\omega))$, so that
$\mu_x^{(A)}$ is the image measure $(\mu_W)_Hf^{-1}$, where
$H=\{\omega:\tau(\omega)<\infty\}$.
Define $\phi_{\tau}:\Omega\times\Omega\to\Omega$ as in
477G, so that

$$\eqalign{\phi_{\tau}(\omega,\omega')(t)
&=\omega(t)\text{ if }t\le\tau(\omega),\cr
&=\omega(\tau(\omega))+\omega'(t-\tau(\omega))
   \text{ if }t\ge\tau(\omega).\cr}$$

\noindent Then we have $x+\phi_{\tau}(\omega,\omega')(t)\in A$ iff
$t\ge\tau(\omega)$ and
$f(\omega)+\omega'(t-\tau(\omega))\in A$;   so if we write
$\sigma_{\omega}$ for the Brownian hitting time to $A-f(\omega)$ when
$\omega\in H$,
$\tau'(\phi_{\tau}(\omega,\omega'))=\tau(\omega)+\sigma_{\omega}(\omega')$.

\medskip

{\bf (b)} Now suppose that $E\subseteq\BbbR^r$ is a Borel set.   Then

$$\eqalignno{\mu_x^{(A)}(E)
&=\mu_W\{\omega:\tau'(\omega)<\infty,
  \,x+\omega(\tau'(\omega))\in E\}\cr
&=\mu_W^2\{(\omega,\omega'):
  \tau'(\phi_{\tau}(\omega,\omega'))<\infty,\,
  x+\phi_{\tau}(\omega,\omega')(\tau'(\phi_{\tau}(\omega,\omega')))\in E\}
  \cr
&=\mu_W^2\{(\omega,\omega'):\tau(\omega)<\infty,\,
  \sigma_{\omega}(\omega')<\infty,\,
  f(\omega)+\omega'(\sigma_{\omega}(\omega'))\in E\}\cr
&=\int_H\mu_W\{\omega':\sigma_{\omega}(\omega')<\infty,\,
  f(\omega)+\omega'(\sigma_{\omega}(\omega'))\in E\}\mu_W(d\omega)\cr
&=\int_H\mu^{(A)}_{f(\omega)}(E)\mu_W(d\omega)
=\int\mu^{(A)}_y(E)\mu_x^{(B)}(dy).
}$$

\noindent The definition in 452E demands that this formula should be valid
whenever $E$ is measured by $\mu_x^{(A)}$;  but in general there will be
Borel sets $E'$, $E''$ such that $E'\subseteq E\subseteq E''$ and
$\mu^{(A)}_x(E')=\mu^{(A)}_x(E)=\mu^{(A)}_x(E'')$,
in which case we must have
$\mu^{(A)}_y(E')=\mu^{(A)}_y(E)=\mu^{(A)}_y(E'')$ for $\mu^{(B)}_x$-almost
every $y$, and again
$\mu_x^{(A)}(E)=\int\mu^{(A)}_y(E)\mu^{(B)}_x(dy)$.
}%end of proof of 478R

\leader{478S}{Corollary} Let $A\subseteq\BbbR^r$ be an analytic set,
and $f:\partial A\to\Bbb R$ a bounded universally measurable function.
For $x\in\BbbR^r\setminus\overline{A}$ set $g(x)=\int fd\mu_x$, where
$\mu_x$ is the harmonic measure for arrivals in $A$ from
$x$.   Then $g$ is harmonic.

\proof{ Suppose that $\delta>0$ is such that
$B(x,\delta)\cap\overline{A}=\emptyset$, and set
$S=\partial B(x,\delta)=\partial(\BbbR^r\setminus\interior B(x,\delta))$.
Then the harmonic measure for arrivals in
$\BbbR^r\setminus\interior B(x,\delta)$ from $x$ is
$\Bover1{\nu S}\nu\LLcorner S$ (478Qb).   So

$$\eqalignno{g(x)
&=\int fd\mu_x
=\int_S\Bover1{\nu S}\int fd\mu_y\nu(dy)\cr
\displaycause{478R, 452F}
&=\Bover1{\nu S}\int_Sg(y)\nu(dy).\cr}$$

\noindent As $x$ and $\delta$ are arbitrary, $g$ is harmonic.
}%end of proof of 478S

\leader{478T}{Corollary} Let $A\subseteq\BbbR^r$ be an analytic set,
and for $x\in\BbbR^r$ let $\mu_x$ be the harmonic measure for arrivals in
$A$ from $x$.   Then $x\mapsto\mu_x$ is continuous on
$\BbbR^r\setminus\overline{A}$ for the
total variation metric on the set of totally finite
Radon measures on $\BbbR^r$\cmmnt{ (definition:  437Qa)}.

\proof{ Take any $y\in\BbbR^r\setminus\overline{A}$.   Let $\delta>0$ be
such that $B(y,\delta)\cap A=\emptyset$, and set
$S=\partial B(y,\delta)$.   For $x\in\interior B(y,\delta)$,
let $\zeta_x$ be the harmonic measure for arrivals in
$\BbbR^r\setminus B(y,\delta)$ from $x$, so that $\zeta_x$ is the
indefinite-integral measure over $\nu$ defined by the function

$$\eqalign{z
&\mapsto\Bover{\delta^2-\|x-y\|^2}{r\beta_r\delta\|x-z\|^r}
  \text{ if }z\in S,\cr
&\mapsto 0\text{ if }z\in\BbbR^r\setminus S\cr}$$

\noindent (478Qa).   Then, for any $x\in\interior B(y,\delta)$,
$\family{z}{\BbbR^r}{\mu_z}$ is a disintegration of $\mu_x$
over $\zeta_x$.   So if $E\subseteq\BbbR^r$ is a Borel set,

\Centerline{$\mu_xE
=\int\mu_z(E)\,\zeta_x(dz)
=\Bover1{r\beta_r\delta}\int_S\mu_z(E)\,
   \Bover{\delta^2-\|x-y\|^2}{\|x-z\|^r}\nu(dz)$.}

\noindent But this means that

\Centerline{$|\mu_x(E)-\mu_y(E)|
\le\Bover{\nu S}{r\beta_r\delta}
  \sup_{z\in S}\bigl|\Bover{\delta^2-\|x-y\|^2}{\|x-z\|^r}
      -\Bover{\delta^2}{\|y-z\|^r}\bigr|$;}

\noindent as $E$ is arbitrary, the distance from $\mu_x$ to
$\mu_y$ is at most

\Centerline{$\delta^{r-2}
  \sup_{z\in S}\bigl|\Bover{\delta^2-\|x-y\|^2}{\|x-z\|^r}
      -\Bover{\delta^2}{\|y-z\|^r}\bigr|$,}

\noindent which is small if $x$ is close to $y$.
}%end of proof of 478T

\leader{478U}{}\cmmnt{ A variation on the technique of 478R enables us to
say something about Brownian paths starting from a point in the essential
closure of a set.

\medskip

\noindent}{\bf Proposition} Suppose that $A\subseteq\BbbR^r$ and that
$0$ belongs to the essential closure\cmmnt{ $\clstar A$} of $A$\cmmnt{ as
defined in 475B}.   Then the outer Brownian hitting probability
$\hp^*(A)$ of $A$\cmmnt{ (477Ia)} is $1$.

\proof{{\bf (a)} Take that $\alpha\in\ooint{0,1}$ such that
$\Bover{1-\alpha^2}{(1+\alpha)^r}=\Bover12$.
Suppose that $E\subseteq\BbbR^r$ is analytic, and that
$0<\delta_0<\ldots<\delta_n$ are such that
$\delta_i\le\alpha\delta_{i+1}$ for $i<n$.   For $i\le n$, let $\tau_i$
be the Brownian hitting time to $S_i=\partial B(\tbf{0},\delta_i)$.   Then

\Centerline{$\mu_W\{\omega:
   \omega(\tau_i(\omega))\notin E$ for every $i\le n\}
\le\prod_{i=0}^n(1-\Bover{\nu(E\cap S_i)}{2\nu S_i})$.}

\noindent\Prf\ Induce on $n$.   If $n=0$, then

$$\eqalignno{\mu_W\{\omega:\omega(\tau_0(\omega))\notin E\}
&=1-\mu_0^{(S_i)}(E)\cr
\displaycause{where $\mu_0^{(S_i)}$ is the harmonic measure for arrivals in
$S_i$ from $0$}
&=1-\Bover{\nu(E\cap S_i)}{\nu S_i}
\le 1-\Bover{\nu(E\cap S_i)}{2\nu S_i}\cr}$$

\noindent(478Qb).   For the inductive step to $n+1\ge 1$, let
$\phi:\Omega\times\Omega\to\Omega$ be the \imp\ function
corresponding to the stopping time $\tau_n$ as in 477G;  when
$\tau_n(\omega)$ is finite, set
$y(\omega)=\omega(\tau_n(\omega))\in S_n$.   Since
$\tau_i(\omega)<\tau_{i+1}(\omega)$ whenever $i\le n$ and
$\tau_i(\omega)$ is finite,

\Centerline{$\tau_i(\phi(\omega,\omega'))=\tau_i(\omega)$}

\noindent for $i\le n$ and $\omega$, $\omega'\in\Omega$.   As for
$\tau_{n+1}(\phi(\omega,\omega'))$, this is infinite if
$\tau_n(\omega)=\infty$, and otherwise is
$\sigma_{y(\omega)}(\omega')$, where $\sigma_y$
is the Brownian hitting time of $S_{n+1}-y$.   Now if $y\in S_n$, then

$$\eqalignno{\mu_y^{(S_{n+1})}(E)
&=\int_{E\cap S_{n+1}}
  \Bover{\delta_{n+1}^2-\delta_n^2}{r\beta_r\delta_{n+1}\|x-y\|^r}
  \nu(dx)\cr
\displaycause{478Qa}
&\ge\Bover{\delta_{n+1}^2-\delta_n^2}
   {r\beta_r\delta_{n+1}(\delta_{n+1}+\delta_n)^r}
   \nu(E\cap S_{n+1})\cr
&\ge\Bover{1-\alpha^2}
   {r\beta_r\delta_{n+1}^{r-1}(1+\alpha)^r}
   \nu(E\cap S_{n+1})
=\Bover{\nu(E\cap S_{n+1})}{2\nu S_{n+1}}.\cr}$$

\noindent Consequently

$$\eqalignno{
&\mu_W\{\omega:\omega(\tau_i(\omega))\notin E\text{ for every }i\le n+1\}
  \cr
&\mskip50mu=(\mu_W\times\mu_W)\{(\omega,\omega'):
  \phi(\omega,\omega')(\tau_i(\phi(\omega,\omega')))\notin E
  \text{ for every }i\le n+1\}\cr
&\mskip50mu=(\mu_W\times\mu_W)\{(\omega,\omega'):
  \omega(\tau_i(\omega))\notin E
  \text{ for every }i\le n,
  \,\omega'(\sigma_{y(\omega)}(\omega'))\notin E\}\cr
&\mskip50mu
=\int_V\mu_W\{\omega':\omega'(\sigma_{y(\omega)}(\omega'))\notin E\}
   \mu_W(d\omega)\cr
\displaycause{setting $V=\{\omega:\omega(\tau_i(\omega))\notin E$ for every
$i\le n\}$}
&\mskip50mu\le\mu_WV\cdot\sup_{y\in S_n}(1-\mu_y^{(S_{n+1})}E)\cr
&\mskip50mu\le\mu_WV\cdot(1-\Bover{\nu(E\cap S_{n+1})}{2\nu S_{n+1}})
\le\prod_{i=0}^{n+1}(1-\Bover{\nu(E\cap S_i)}{2\nu S_i})\cr}$$

\noindent by the inductive hypothesis.   So the induction continues.\ \Qed

\medskip

{\bf (b)} In particular, under the conditions of (a),
$\hp(E)\ge 1-\prod_{i=0}^n(1-\Bover{\nu(E\cap S_i)}{2\nu S_i})$.
Now suppose that $A\subseteq\BbbR^r$ and that $0\in\clstar A$.
Let $E\supseteq A$ be an analytic set such that $\hp(E)=\hp^*(A)$
(477Id).   Then $0\in\clstar E$;  set
$\gamma=\Bover13\limsup_{\delta\downarrow 0}
   \Bover{\mu(E\cap B(\tbf{0},\delta))}{\mu B(\tbf{0},\delta)}>0$.
For any $\delta>0$, there is a $\delta'\in\ocint{0,\delta}$ such that
$\Bover{\nu(E\cap\partial B(\tbf{0},\delta'))}{\nu\partial B(\tbf{0},\delta')}
\ge 2\gamma$.   \Prf\ Let $\beta\in\ocint{0,\delta}$ be such that
$\mu(E\cap B(\tbf{0},\beta))\ge 2\gamma\mu B(\tbf{0},\beta)$.   Then

\Centerline{$\int_0^{\beta}\nu(E\cap\partial B(\tbf{0},t))dt
\ge 2\gamma\int_0^{\beta}\nu\partial B(\tbf{0},t)dt$,}

\noindent so there must be a $\delta'\in\ocint{0,\beta}$ such that
$\nu(E\cap\partial B(\tbf{0},\delta'))\ge 2\gamma\nu\partial B(\tbf{0},\delta')$.\ \Qed

We can therefore find, for any $n\in\Bbb N$, $0<\delta_0<\ldots<\delta_n$
such that $\delta_i\le\alpha\delta_{i+1}$ for every $i<n$ (where $\alpha$
is chosen as in (a) above) and
$\nu(E\cap\partial B(\tbf{0},\delta_i))\ge 2\gamma\nu\partial B(\tbf{0},\delta_i)$ for
every $i$.   As noted at the beginning of this part of the proof, it
follows that $\hp(E)\ge 1-(1-\gamma)^{n+1}$.   As this is true for every
$n\in\Bbb N$, $\hp(E)=1$, so $\hp^*(A)=1$, as claimed.
}%end of proof of 478U

\leader{*478V}{Theorem} (a)
Let $G\subseteq\BbbR^r$ be an open set with few wandering paths and
$f:\overline{G}^{\infty}\to\Bbb R$ a continuous function such that
$f\restr G$ is harmonic.   For $x\in\BbbR^r$ let
 $\tau_x:\Omega\to[0,\infty]$ be the Brownian exit time from
$G-x$.   Set

$$\eqalign{g_{\tau_x}(\omega)
&=f(x+\omega(\tau_x(\omega)))\text{ if }\tau_x(\omega)<\infty,\cr
&=f(\infty)\text{ if }\lim_{t\to\infty}\|\omega(t)\|=\infty
\text{ and }\tau_x(\omega)=\infty.\cr}$$

\noindent Then $f(x)=\Expn(g_{\tau_x})$.

(b) Now suppose that $\sigma$ is a stopping time
adapted to $\langle\Sigma_t\rangle_{t\ge 0}$ such that
$\sigma(\omega)\le\tau_x(\omega)$ for every $\omega$.   Set

$$\eqalign{g_{\sigma}(\omega)
&=g_{\tau_x}(\omega)\text{ if }\sigma(\omega)=\tau_x(\omega)=\infty,\cr
&=f(x+\omega(\sigma(\omega)))\text{ otherwise}.\cr}$$

\noindent As in 455Lc, set
$\Sigma_{\sigma}=\{E:E\in\dom\mu_W$,
$E\cap\{\omega:\sigma(\omega)\le t\}\in\Sigma_t$ for every $t\ge 0\}$.
Then $g_{\sigma}$ is a conditional expectation of
$g_{\tau_x}$ on $\Sigma_{\sigma}$.

\proof{{\bf (a)(i)} Of course if $x\notin G$ then $\tau_x(\omega)=0$
and $g_{\tau_x}(\omega)=f(x)$ for
every $\omega$ and the result is trivial.   So we can suppose that
$x\in G$.   Note next that if there is any $\omega$ such that
$\lim_{t\to\infty}\|\omega(t)\|=\infty$ and $\tau_x(\omega)=\infty$, then
$G$ must be unbounded, so $f(\infty)$ will be defined.   Because
$G$ has few wandering paths, $g_{\tau_x}$ is defined almost everywhere.

\medskip

\quad{\bf (ii)} Let $m\in\Bbb N$ be such that
$\rho(x,\BbbR^r\setminus G)>\bover1{m+1}$ and $\|x\|<m$;
for $n\ge m$, set
$G_n=\{y:\|y\|<n$, $\rho(y,\BbbR^r\setminus G)>\bover1{n+1}\}$,
let $\tau_{xn}'$ be the
Brownian exit time from $G_n-x$ and set
$\tau_{xn}(\omega)=\min(n,\tau_{xn}'(\omega))$ for every $\omega$.
Note that by 477I(c-i), $x+\omega(\tau_{xn}(\omega))\in\overline{G}_n$
for every $\omega$.

By 477I(c-iii) and 455L(c-v), $\tau_{xn}$ is a stopping time adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$.   Let $\sequencen{\tilde h_n}$ be the
sequence of 473E, and for $n\ge m$ set $f_n=\tilde f*\tilde h_n$, where
$\tilde f$ is the extension of $f\restr\overline{G}$ to
$\BbbR^r$ which takes the value $0$
on $\BbbR^r\setminus\overline{G}$.   Then $f_n$ and all its derivatives are
smooth and bounded.   By 478Jb
(applied in turn to the functions $\tilde f+M\chi\BbbR^r$
and $-\tilde f+M\chi\BbbR^r$ where $M=\sup_{y\in G}|f(y)|$,
which of course are both superharmonic on $G$),
$f_n$ agrees with $f$ on $G_n$, so that $f_n\restr G_n$ is harmonic and
$\nabla^2f_n$ is zero on $G_n$ (478Ec).
Also, because both $f_n$ and $f$ are continuous, they agree on
$\overline{G}_n$ and
$f(x+\omega(\tau_{xn}(\omega)))
=f_n(x+\omega(\tau_{xn}(\omega)))$
for every $\omega$.

If $n\ge m$, $\omega\in\Omega$ and $0\le s<\tau_{xn}(\omega)$, then
$x+\omega(s)\in G_n$ so $(\nabla^2f_n)(x+\omega(s))=0$.   Dynkin's formula
(478K), applied to the function $y\mapsto f_n(x+y)$,
therefore tells us that
$f(x)=f_n(x)
=\int f_n(x+\omega(\tau_{xn}(\omega)))\mu_W(d\omega)$.

\medskip

\quad{\bf (iii)} If $\omega\in\Omega$ and $t<\tau_x(\omega)$,
then the compact set
$x+\omega[\,[0,t]\,]$ is included in the open set $G$ and there is an
$n\ge\max(m,t)$ such that it is included in $G_n$.   So
$\lim_{n\to\infty}\tau_{xn}(\omega)=\tau_x(\omega)$ and, because $f$ is
continuous on $\overline{G}^{\infty}$,

\Centerline{$g_{\tau_x}(\omega)
=\lim_{n\to\infty}f(x+\omega(\tau_{xn}(\omega)))
=\lim_{n\to\infty}f_n(x+\omega(\tau_{xn}(\omega)))$}

\noindent
for almost every $\omega$.   Since
$\|f_n\|_{\infty}\le\|f\|_{\infty}<\infty$ for every $n$,
Lebesgue's Dominated Convergence Theorem tells us that

\Centerline{$\Expn(g_{\tau_x})
=\lim_{n\to\infty}\int f_n(x+\omega(\tau_{xn}(\omega)))
  \mu_W(d\omega)
=f(x)$,}

\noindent as required.

\medskip

{\bf (b)(i)} If $\omega_0$, $\omega_1\in\Omega$, $\sigma(\omega_0)=t$ and
$\omega_1\restr[0,t]=\omega_0\restr[0,t]$, then
$\sigma(\omega_1)=t$.
\Prf\ The set $\{\omega:\sigma(\omega)\le t\}$ belongs
to $\Sigma_t$;  as it contains $\omega_0$, it contains $\omega_1$, and
$\sigma(\omega_1)\le t$.   But now $\omega_0$ agrees with $\omega_1$ on
$[0,\sigma(\omega_1)]$, so $\sigma(\omega_1)\ge\sigma(\omega)=t$.\ \Qed

If $H\in\Sigma_{\sigma}$ then $\omega_0\in H$ iff $\omega_1\in H$.
\Prf\ For every $t\ge 0$,
$H\cap\{\omega:\sigma(\omega)\le t\}$ belongs to
$\Sigma_t$, so contains $\omega_0$ iff it contains $\omega_1$.\ \Qed

\medskip

\quad{\bf (ii)} Of course
$E_{\infty}=\{\omega:\sigma(\omega)=\infty\}$ belongs to $\Sigma_{\sigma}$,
because it has empty intersection with every set
$\{\omega:\sigma(\omega)\le t\}$.

\leaveitout{Also
$E_1=\{\omega:\tau(\omega)=\sigma(\omega)<\infty\}$ belongs to
$\Sigma_{\sigma}$.   \Prf\ For any $t\ge 0$,

$$\eqalign{E_1\cap\{\omega:\sigma(\omega)\le t\}
&=\{\omega:\sigma(\omega)\le t\}\cap\{\omega:\tau(\omega)\le t\}\cr
&\mskip50mu
  \setminus\bigcup_{q\in[0,t]\cap\Bbb Q}
      (\{\omega:\sigma(\omega)\le q<\tau(\omega)\}
         \cup\{\omega:\tau(\omega)\le q<\sigma(\omega)\})\cr
&\in\Sigma_t.  \text{ \Qed}\cr}$$

\noindent So

\Centerline{$E_0=\{\omega:\sigma(\omega)=0<\tau(\omega)\}
=\{\omega:\sigma(\omega)\le 0\}\setminus E_1$}

\noindent belongs to $\Sigma_{\sigma}$.
}

\medskip

\quad{\bf (iii)} $g_{\sigma}$ is $\Sigma_{\sigma}$-measurable.   \Prf\
For $n\in\Bbb N$, $\omega\in\Omega$ and $t\ge 0$, set

$$\eqalign{h_n(t,\omega)
&=f(\omega(2^{-n}\lfloor 2^nt\rfloor))
  \text{ if }\omega(2^{-n}\lfloor 2^nt\rfloor)\in G,\cr
&=0\text{ otherwise}.\cr}$$

\noindent Then $h_n$ is
$(\Cal B(\coint{0,\infty})\times\Sigma)$-measurable,
so if we set $h(t,\omega)=\lim_{n\to\infty}h_n(t,\omega)$ when this is
defined, $h$ also will be
$(\Cal B(\coint{0,\infty})\times\Sigma)$-measurable, and
$\omega\mapsto h(\sigma(\omega),\omega)$ is $\Sigma$-measurable.
Now, because
$\sigma\le\tau$, $g_{\sigma}(\omega)=h(\sigma(\omega),\omega)$ for almost
every $\omega\in\Omega\setminus E_{\infty}$;
because $\mu_W$ is complete,
$g_{\sigma}$ is $\Sigma$-measurable.   But now observe that if $t\ge 0$ and
$\alpha\in\Bbb R$,
$\{\omega:\sigma(\omega)\le t$, $g_{\sigma}(\omega)\ge\alpha\}$ belongs to
$\Sigma$ and is determined by coordinates less than or equal to $t$, so
belongs to $\Sigma_t$.   As $t$ is arbitrary,
$\{\omega:g_{\sigma}(\omega)\ge\alpha\}\in\Sigma_{\sigma}$;  as
$\alpha$ is arbitrary, $g_{\sigma}$ is $\Sigma_{\sigma}$-measurable.\ \Qed

\medskip
	
\quad{\bf (iv)} As in 477G, define
$\phi_{\sigma}:\Omega\times\Omega\to\Omega$ by saying that

$$\eqalign{\phi_{\sigma}(\omega,\omega')(t)
&=\omega(t)\text{ if }t\le\sigma(\omega),\cr
&=\omega(\sigma(\omega))+\omega'(t-\sigma(\omega))
   \text{ if }t\ge\sigma(\omega).\cr}$$

\noindent Then 477G tells us that $\phi_{\sigma}$ is \imp.

\medskip

\quad{\bf (v)} $\tau_x(\phi_{\sigma}(\omega,\omega'))
=\sigma(\omega)+\tau_{x+\omega(\sigma(\omega))}(\omega')$
for all $\omega$, $\omega'\in\Omega$.   \Prf\ If
$\sigma(\omega)=\infty$ then $\phi_{\sigma}(\omega,\omega')=\omega$
and

\Centerline{$\tau_x(\phi_{\sigma}(\omega,\omega'))
=\tau_x(\omega)=\sigma(\omega)$.}

\noindent If $\sigma(\omega)=\tau_x(\omega)$ is finite then
$\omega(\sigma(\omega))\notin G-x$ and
$\phi_{\sigma}(\omega,\omega')\restr[0,\tau_x(\omega)]
=\omega\restr[0,\tau_x(\omega)]$, so

\Centerline{$\tau_x(\phi_{\sigma}(\omega,\omega'))
=\tau_x(\omega)=\sigma(\omega)
=\sigma(\omega)+\tau_{x+\omega(\sigma(\omega))}(\omega')$.}

\noindent If $\sigma(\omega)<\tau_x(\omega)$ then
$\omega(t)=\phi_{\sigma}(\omega,\omega')(t)$ belongs to $G-x$ for every
$t\le\sigma(\omega))$ and

$$\eqalign{\tau_x(\phi_{\sigma}(\omega,\omega'))
&=\inf\{t:t\ge\sigma(\omega),\,
   \omega(\sigma(\omega))+\omega'(t-\sigma(\omega))\notin G-x\}\cr
&=\sigma(\omega)+\inf\{t:\omega'(t)\notin G-x-\omega(\sigma(\omega))\}
=\sigma(\omega)+\tau_{x+\omega(\sigma(\omega))}(\omega').
\text{ \Qed}\cr}$$

\noindent Consequently, if $\omega\in\Omega$, $\sigma(\omega)<\infty$ and
$y=\omega(\sigma(\omega))$,

\Centerline{$\phi_{\sigma}(\omega,\omega')
  (\tau_x(\phi_{\sigma}(\omega,\omega')))
=\phi_{\sigma}(\omega,\omega')
  (\sigma(\omega)+\tau_{x+y}(\omega'))
=y+\omega'(\tau_{x+y}(\omega'))$}

\noindent whenever either $\tau_x(\phi_{\sigma}(\omega,\omega'))$ or
$\tau_{x+y}(\omega')$ is finite,

\Centerline{$g_{\tau_x}(\phi_{\sigma}(\omega,\omega'))
=f(x+\phi_{\sigma}(\omega,\omega')
  (\tau_x(\phi_{\sigma}(\omega,\omega'))))
=f(x+y+\omega'(\tau_{x+y}(\omega')))$}

\noindent for almost every $\omega'$.

\medskip

\quad{\bf (vi)} If $H\in\Sigma_{\sigma}$ then
$\phi_{\sigma}^{-1}[H]=H\times\Omega$.   \Prf\ If $\omega$,
$\omega'\in\Omega$ then
$\phi_{\sigma}(\omega,\omega')\restr[0,\sigma(\omega)]
=\omega\restr[0,\sigma(\omega)]$, so by (i) above
$\phi_{\sigma}(\omega,\omega')\in H$ iff $\omega\in H$.\ \Qed

If $H\cap E_{\infty}=\emptyset$, we now have

$$\eqalignno{\int_Hg_{\tau_x}
&=\int_{\phi_{\sigma}^{-1}[H]}g_{\tau_x}(\phi_{\sigma}(\omega,\omega'))
  d(\omega,\omega')\cr
&=\int_H\int f(x+\omega(\sigma(\omega))
   +\omega'(\tau_{x+\omega(\sigma(\omega))}(\omega)))
  d\omega'd\omega\cr
&=\int_Hf(x+\omega(\sigma(\omega)))d\omega\cr
\displaycause{by (a) above}
&=\int_Hg_{\sigma}.\cr}$$

\noindent Of course we also have $\int_Hg_{\tau_x}=\int_Hg_{\sigma}$ if
$H\subseteq E_{\infty}$.   So $\int_Hg_{\sigma}=\int_Hg_{\tau_x}$ for every
$H\in\Sigma_{\sigma}$, and
$g_{\sigma}$ is a conditional expectation of
$g_{\tau_x}$ on $\Sigma_{\sigma}$.
}%end of proof of 478V

\leaveitout{\medskip

\noindent{\bf Remarks} Will this now cover 478R?

Anyway we can do quite a bit better;  e.g., assuming only that
$f:G\to\Bbb R$ is harmonic, but that there is a bound on
$f(\omega(t))$ for $\omega\in\Omega$, $t<\tau(\omega)$.
}

\exercises{\leader{478X}{Basic exercises (a)}
%\spheader 478Xa
Let $G\subseteq\BbbR^r$ be an open set, and
$\sequencen{f_n}$ a sequence of superharmonic functions from $G$ to
$[0,\infty]$.   Show that
$\liminf_{n\to\infty}f_n$ is superharmonic.
%478B

\spheader 478Xb Let $G\subseteq\BbbR^r$ be an open set, and $f:G\to\Bbb R$
a continuous harmonic function.   Show that $f$ is smooth.   \Hint{put 478I
and 478D together.}
%478I
%won't "Lebesgue measurable" do in place of "continuous"?

\spheader 478Xc Let $G\subseteq\BbbR^r$ be an open set, and
$f:G\to[0,\infty]$ a lower semi-continuous superharmonic function.
Show that there is are sequences $\sequencen{G_n}$, $\sequencen{f_n}$ such
that (i) $\sequencen{G_n}$ is a non-decreasing sequence of open sets with
union $G$ (ii) for each $n\in\Bbb N$,
$f_n:G_n\to\coint{0,\infty}$ is a bounded smooth superharmonic
function and $f_n\le f\restr G_n$ (iii) $f(x)=\lim_{n\to\infty}f_n(x)$ for
every $x\in G$.
%478J

\sqheader 478Xd Let $\langle X_t\rangle_{t\ge 0}$ be Brownian motion in
$\BbbR^r$, and $\delta>0$.   Let $\tau$ be the Brownian hitting time to
$\{x:\|x\|\ge\delta\}$.   Show that $\Expn(\tau)=\Bover{\delta^2}{r}$.
\Hint{in 478K, take $f$ extending $x\mapsto\|x\|^2:B(0,\delta)\to\Bbb R$.}
%478K

\spheader 478Xe Show that
$\hp^*:\Cal P\BbbR^r\to[0,1]$ is an outer regular Choquet
capacity (definition:  432J) iff $r\ge 3$.
\Hint{if $r\ge 3$, $\mu_W$ is inner regular with respect to
$\{K:K\subseteq\Omega$,
$\lim_{t\to\infty}\inf_{\omega\in K}\|\omega(t)\|=\infty\}$.}
%477I %478M

\spheader 478Xf Show that an open subset of $\Bbb R$ has few wandering
paths iff it is not $\Bbb R$ itself.
%478N

\spheader 478Xg Suppose $r=2$.  (i)
Show that if $x\in\BbbR^2\setminus\{0\}$, then
the Brownian exit time from $\BbbR^2\setminus\{x\}$ is infinite a.e.
\Hint{use the
method of part (b-ii) of the proof of 478M to show that if $R>\|x\|$ and
$\delta>0$ is small enough then most sample paths meet $\partial B(x,R)$
before they meet $B(x,\delta)$.}
(ii) Show that if $G\subseteq\BbbR^2$ is an open set
with countable complement then $G$ does not have few wandering paths.
%478N

\spheader 478Xh Suppose that $r\ge 3$.   Let $\sequencen{A_n}$ be a
non-increasing sequence of analytic sets in $\BbbR^r$ such that $A_0$ is
bounded and $\bigcap_{n\in\Bbb N}\bar A_n=\bigcap_{n\in\Bbb N}A_n$,
and $x\in\BbbR^r$.   Let $\mu_x^{(n)}$, $\mu_x$ be the
harmonic measures for arrivals in $A_n$, $\bigcap_{m\in\Bbb N}A_m$
from $x$.   Show that
$\mu_x=\lim_{n\to\infty}\mu^{(n)}_x$ for the narrow topology on the set of
totally finite Radon measures on $\BbbR^r$.   \Hint{478Xe, 478Pd.}
%478Xe 478M 478P

\spheader 478Xi Let $A\subseteq\Bbb R$ be an analytic set, $x\in\Bbb R$ and
$\mu_x$ the harmonic measure for arrivals in $A$ from $x$.   For
$y\in\Bbb R$ let $\delta_y$ be the Dirac measure on $\Bbb R$ concentrated
at $y$.   Show that (i) if $A$ is empty, then $\mu_x$ is the zero measure;
(ii) if $A\ne\emptyset$ but $A\cap\coint{x,\infty}=\emptyset$ then
$\mu_x=\delta_{\sup A}$;
(iii) if $A\ne\emptyset$ but $A\cap\ocint{-\infty,x}=\emptyset$ then
$\mu_x=\delta_{\inf A}$;  (iv) if $A$ meets both $\ocint{-\infty,x}$ and
$\coint{x,\infty}$, and $y=\sup(A\cap\ocint{-\infty,x})$,
$z=\inf(A\cap\coint{x,\infty})$, then $\mu_x=\delta_x$ if $y=z=x$, and
otherwise $\mu_x=\Bover{z-x}{z-y}\delta_y+\Bover{x-y}{z-y}\delta_z$.
%478P

\spheader 478Xj Prove 478Qb by a symmetry argument not involving the
calculations of 478I.
%478Q

\sqheader 478Xk Let $G\subseteq\BbbR^r$ be an open set, and for $x\in G$
let $\mu_x$ be the
harmonic measure for arrivals in $\BbbR^r\setminus G$ from $x$.
Show that for any bounded universally measurable function
$f:\partial G\to\Bbb R$, the function
$x\mapsto\int fd\mu_x:G\to\Bbb R$ is continuous and harmonic.
%478R

\spheader 478Xl (i)
Suppose that $r=2$, and that $x$, $y$, $z\in\BbbR^r$ are
such that $\|x\|<1=\|z\|$ and $y=0$.   Identify $\BbbR^2$ with $\Bbb C$,
and express $x$, $z$ as $\gamma e^{i\theta}$ and $e^{it}$ respectively.
Show that, in the language of 272Yg,
$\Bover{\|y-z\|^2-\|x-y\|^2}{\|x-z\|^r}=A_{\gamma}(\theta-t)$.
(ii) Compare 478I(b-iii) with 272Yg(iii).
%478I out of order query

\leader{478Y}{Further exercises (a)}%
%\spheader 478Ya
(i) Show that there is a function $f:\Bbb R\to\Bbb Q$ which
is `harmonic' in the sense of 478B, but is not continuous.   \Hint{take $f$
to be a linear operator when $\Bbb R$ is regarded as a linear space over
$\Bbb Q$.}   (ii) Show that if the continuum hypothesis is true, there is a
surjective
function $f:\BbbR^2\to\{0,1\}$ which is `harmonic' in the sense of 478B.
%478B

\spheader 478Yb Let $G\subseteq\BbbR^r$ be a connected open set, and
$f:G\to[0,\infty]$ a superharmonic Lebesgue measurable function which is
not everywhere infinite.   Show that $f$ is locally integrable.
%478C mt47bits

\spheader 478Yc Let $G\subseteq\BbbR^2$ be an open set, and $f:G\to\Bbb C$ a
function which is analytic when regarded as a function of a complex
variable.   Show that $\Real f$ is harmonic.   \Hint{The
non-trivial part is the theorem that $f$ has continuous second partial
derivatives.}
%478F

\spheader 478Yd Define $\psi:\BbbR^r\setminus\{0\}\to\BbbR^r$ by setting
$\psi(x)=\Bover{x}{\|x\|^2}$.
For a $[-\infty,\infty]$-valued function $f$ defined on a subset of
$\BbbR^r$, set $f^*(x)=\Bover1{\|x\|^{r-2}}f(\psi(x))$ for
$x\in\psi^{-1}[\dom f]\setminus\{0\}$.
(This is the {\bf Kelvin transform} of $f$ relative
to the sphere $\partial B(\tbf{0},1)$.)   (i) Show that if $f$ is real-valued
and twice continuously differentiable, then
$(\nabla^2f^*)(x)=\Bover1{\|x\|^{r+2}}(\nabla^2f)(\psi(x))$
for $x\in\dom f^*$.
%Armitage 1.6.3;  Axler \& Ramey 95
(ii) Show that if $\dom f$ is open and $f$ is non-negative,
lower semi-continuous and superharmonic, then $f^*$ is
superharmonic.  %478Xc
%478J

\spheader 478Ye Let $G\subseteq\BbbR^2$ be an open set.   Show that $G$ has
few wandering paths iff there is an $x\in\BbbR^2$ such that
$\hp((\BbbR^2\setminus(G\cup\{x\}))-x)>0$.
%478N mt47bits

\spheader 478Yf Show that 478K remains true if we replace
`three-times-differentiable function such that $f$ and its first three
derivatives are continuous and bounded' with
`twice-differentiable function such that $f$ and its first two
derivatives are continuous and bounded'.
%478K

\spheader 478Yg Let $f:\BbbR^r\to\Bbb R$ be a twice-differentiable
function such that $f$ and its first two
derivatives are continuous and bounded.   Show that

\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\delta^2}
  \bigl(f(x)-\Bover1{\mu B(x,\delta)}\int_{B(x,\delta)}fd\mu\bigr)
=-\Bover{(\nabla^2f)(x)}{2(r+2)}$}

\noindent for every $x\in\BbbR^r$.
%478Yf 478K mt47bits

\spheader 478Yh Let $G\subseteq\BbbR^r$ be an open set and $f:G\to\Bbb R$ a
continuous function such that

\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\delta^2}
  \bigl(f(x)-\Bover1{\mu B(x,\delta)}\int_{B(x,\delta)}fd\mu\bigr)
=0$}

\noindent for every $x\in G$.   Show that $f$ is harmonic.
%478Yg 478K apply 478Yg to smoothed versions of $f$

\spheader 478Yi Let $f:\BbbR^r\to[0,\infty]$ be a lower semi-continuous
superharmonic function.   Show
that $f\omega$ is continuous for $\mu_W$-almost every $\omega\in\Omega$.
%478L mt47bits

\spheader 478Yj Suppose that $A\subseteq\BbbR^r$.
Show that $x\mapsto\hp^*(A-x)$ is lower
semi-continuous at every point of $\BbbR^r\setminus A$, and continuous at
every point of $\BbbR^r\setminus\overline{A}$.
%if x_n\to x  there is an analytic  A'\supseteq A  such that
%\hp(A'-x_n)=\hp^*(A-x_n)$ for every  n , so enough to consider analytic
%A;  now use 477Ie
%478T

\spheader 478Yk Suppose that $A\subseteq\BbbR^r$ is such that
$\inf_{\delta>0}\hp^*(A\cap B(\tbf{0},\delta))>0$.   Show that $\hp^*(A)=1$.
%478U

\spheader 478Yl Let $\mu_W$ be three-dimensional Wiener measure on
$\Omega=C(\coint{0,\infty};\BbbR^3)_0$, and $e$ a unit vector in
$\BbbR^3$.   Set $Y_t(\omega)=\Bover1{\|\omega(t)-e\|}$ for
$\omega\in\Omega$ and $t\in\coint{0,\infty}$ such that $\omega(t)\ne e$.
(i) Show that if $R>1$ then

$$\eqalign{\Expn(Y_t)
&=\Bover1{(\sqrt{2\pi t})^3}
  \int\Bover{\exp(-\|x\|^2/2t)}{\|x-e\|}\mu(dx)\cr
&\le\Bover1{(\sqrt{2\pi t})^3}
  \int_{B(\tbf{0},1)}\Bover1{\|x\|}\mu(dx)
+\Bover1{(\sqrt{2\pi t})^3}
  \int_{B(e,R)\setminus B(e,1)}e^{-\|x\|^2/2t}\mu(dx)
+\Bover1R
\to\Bover1R\cr}$$

\noindent as $t\to\infty$, so that $\lim_{t\to\infty}\Expn(Y_t)=0$ and
$\langle Y_t\rangle_{t\ge 0}$ is not a martingale.    (ii) Show that if
$n\ge 1$ and $\tau_n$ is the Brownian hitting time to $B(e,2^{-n})$,
then $\langle Y_{t\wedge\tau_n}\rangle_{t\ge 0}$ is a martingale, where
$t\wedge\tau_n$ is the stopping time $\omega\mapsto\min(t,\tau_n(\omega))$.
(iii) Show that $\sequencen{\tau_n(\omega)}$ is a strictly increasing
sequence with limit $\infty$ for almost every $\omega$.
($\langle Y_t\rangle_{t\ge 0}$ is a `local martingale'.)
%478Pc Bain 5.2 out of order query

\spheader 478Ym\dvAnew{2019} Taking $r=2$, set $S=\{(\xi,1):\xi\in\Bbb R\}$
and let $\zeta$ be the
indefinite-integral measure over $\nu$ defined by the function

$$\eqalign{(\xi,\eta)
&\mapsto\Bover1{\pi(1+\xi^2)}\text{ if }\eta=1,\cr
&\mapsto 0\text{ otherwise}.\cr}$$

\noindent Show that $\zeta$ is the harmonic measure for arrivals in $S$ from
$(0,0)$.
%478Q 652O
}%end of exercises

\endnotes{
\Notesheader{478} I find that books are still being published on the
subject of potential theory which ignore Brownian motion.   To my eye,
Newtonian potential, at least (and this is generally acknowledged to be the
core of the subject), is an essentially geometric concept, and
random walks are an indispensable tool for understanding it.
So I am giving these priority, at the
cost of myself ignoring Green's functions.

The definitions in 478B are already unconventional;  most authors take it
for granted that harmonic functions should be finite-valued and continuous
(see 478Cd).   All the work of this section refers to measurable functions.
But there are things which can be said about non-measurable functions
satisfying the definitions here (478Ya).   Let me draw your attention to
478Fa and 478H.   If we want to say that $x\mapsto\Bover1{\|x\|^{r-2}}$ is
harmonic, we have to be careful not to define it at $0$.   If (for
$r\ge 3$) we allow $\Bover1{0^{r-2}}=\infty$, we get a superharmonic
function.
If (for $r=1$) we allow $\Bover1{0^{-1}}=0$, we get a subharmonic function.
The slightly paradoxical phenomenon of 478Yl is another manifestation of
this.

I hope that using the operations $\overline{\phantom{x}}^{\infty}$ and
$\partial^{\infty}$ does not make things more difficult.   The point is
that by compactifying $\BbbR^r$ we get an efficient way of talking about
$\lim_{\|x\|\to\infty}f(x)$ when we need to.   This is particularly
effective for Brownian paths, since in three and more dimensions almost all
paths go off to infinity (478Md).   In two dimensions the situation is more
complex (478Mb-478Mc),
and we have to consider the possibility that a path $\omega$ in an
open set may be `wandering', in the sense that it neither strikes the
boundary nor goes to infinity, and $\lim_{t\to\infty}f(\omega(t))$ may fail
to exist even for the best-behaved functions $f$.   Of course this already
happens in one dimension, but only when $G=\Bbb R$, and classical potential
theory (though not, I think, Brownian potential theory) is nearly trivial
in the one-dimensional case.

Many readers will also find that setting $r=3$ and $r\beta_r=4\pi$ will
make the formulae easier to digest.   I allow for variations in $r$ partly
in order to cover the case $r=2$ (in this section, though not in the
next, many of the ideas translate directly into the one- and
two-dimensional cases), and partly because it is not always easy to guess
at a formula for $r\ge 4$ from the formula for $r=3$.   There is
little extra work to be done, given that \S\S472-475 cover the general
case.

I call 478K a `lemma' because I have made no attempt to look for weakest
adequate hypotheses;  of course we don't really need third derivatives
(478Yf).
The `theorems' are 478L and 478O, where the hypotheses seem to mark
natural boundaries of the arguments given.   In 478O %maybe elsewhere
I use a language which is both unusual and slightly contorted, in
order to do as much as possible without splitting the cases $r\le 2$ from
the rest.   Of course any result involving the notion of `few wandering
paths' really has two forms;  one when $r\ge 3$, so that there is no
restriction on the open set and we are genuinely making use of the
one-point
compactification of $\BbbR^r$, and one when $r\le 2$, in which essentially
all our paths are bounded.

Theorem 478O leads directly to a solution of Dirichlet's problem, in the
sense that, for an open set $G$ with few wandering paths, we have a
family of measures enabling us to
calculate the values within $G$ of a continuous function on
$\overline{G}^{\infty}$ which satisfies Laplace's equation inside $G$
(478Pc).   We do not get a satisfactory existence theorem;  we can
use harmonic measures to generate
many harmonic functions on $G$ (478S), but
we do not get good information on their behaviour near $\partial G$, and
are left guessing at which continuous functions on $\partial^{\infty}G$
will be extended continuously.   The method does, however, make it clear
that what matters is the geometry of the boundary;  we need to know
whether, starting from a point near the boundary, a random walk will hit
the boundary soon.   So at least we can see from 478M that (if $r\ge 2$) an
isolated point of $\partial G$ will be at worst an irrelevant distraction.
For $r\ge 3$ the next section will give some useful information
(479P {\it
et seq.}), though I shall not have space for a proper analysis.

The idea of 478Vb is that we have a particularly dramatic kind of
martingale.   Writing $S$ for the set of stopping times $\sigma\le\tau$,
it is easy to see that the family
$\family{\sigma}{S}{g_{\sigma}}$ is a martingale in the sense that
$\Sigma_{\sigma}\subseteq\Sigma_{\sigma'}$
and $g_{\sigma}$ is a conditional expectation of $g_{\sigma'}$ on
$\Sigma_{\sigma}$  whenever $\sigma\le\sigma'$ in $S$.
}

\discrpage


%variation of $\mu_x^{(G)}$ as $G$ varies;  narrow convergence for
%increasing unions?
\frfilename{mt479.tex}
\versiondate{15.2.10/28.4.10}

\def\hp{\mathop{\text{hp}}\nolimits}
\def\rdtf{rapidly decreasing test function}
\def\varinnerprod#1#2{#1\dotproduct#2}
\def\zz{}


\def\chaptername{Geometric measure theory}
\def\sectionname{Newtonian capacity}

\newsection{479}
I end the chapter with a sketch of fragments of the theory of Newtonian
capacity.   I introduce equilibrium measures as integrals
of harmonic measures (479B);  this gives a quick definition of capacity
(479C), with a substantial number of basic properties (479D, 479E),
including its extendability to a Choquet capacity (479Ed).
I give sufficient fragments of the theory of Newtonian potentials (479F,
479J) and harmonic analysis (479G, 479I) to support the classical
definitions of capacity and equilibrium measures in terms of potential and
energy (479K, 479N).   The method demands some Fourier analysis extending
that of Chapter 28 (479H).
479P is a portmanteau theorem on generalized equilibrium measures and
potentials with an exact description of the latter in terms of outer
Brownian hitting probabilities.
I continue with notes on
capacity and Hausdorff measure (479Q), self-intersecting Brownian
paths (479R) and an example of a discontinuous equilibrium potential
(479S).   Yet another definition of capacity, for
compact sets, can be formulated in terms of gradients of potential
functions (479U);  this leads to a simple inequality relating
capacity to Lebesgue measure (479V).   The section ends with an alternative
description of capacity in terms of a measure on the family of closed
subsets of $\BbbR^r$ (479W).

\leader{479A}{Notation} In this section, unless otherwise stated,
$r$ will be a fixed integer greater than or equal to $3$.
\cmmnt{As in \S478, }$\mu$ will be Lebesgue measure on $\BbbR^r$,
and $\beta_r$ the measure of\cmmnt{ the unit ball} $B(\tbf{0},1)$;
$\nu$ will be normalized $(r-1)$-dimensional Hausdorff
measure on $\BbbR^r$\cmmnt{, so that
$\nu(\partial B(\tbf{0},1))=r\beta_r$}.

Recall that if $\zeta$ is a measure on a space $X$, and
$E\in\dom\zeta$, then $\zeta\LLcorner E$ is defined by saying that
$(\zeta\LLcorner E)(F)=\zeta(E\cap F)$ whenever $F\subseteq X$ and
$\zeta$ measures $E\cap F$\cmmnt{ (234M\formerly{2{}34E})}.
If $\zeta$ is a Radon measure, so is $\zeta\LLcorner E$\cmmnt{ (416Sa)}.

\cmmnt{As in \S478, }$\Omega$ will be $C(\coint{0,\infty};\BbbR^r)_0$,
with the topology of
uniform convergence on compact sets;
$\mu_W$ will be Wiener measure on $\Omega$.
Recall that the Brownian hitting probability $\hp(D)$
of a set $D\subseteq\BbbR^r$
is $\mu_W\{\omega:\omega^{-1}[D]\ne\emptyset\}$ if this is
defined, and that for any $D\subseteq\BbbR^r$ the outer
Brownian hitting probability is
$\hp^*(D)=\mu_W^*\{\omega:\omega^{-1}[D]\ne\emptyset\}$\cmmnt{ (477Ia)}.
%this is $\min\{\hp(E):E\supseteq D$ is G$_{\delta}\}$.

If $x\in\BbbR^r$ and $A\subseteq\BbbR^r$ is an analytic set, $\mu_x^{(A)}$
will be the harmonic measure for arrivals in $A$ from $x$\cmmnt{ (478P)};
note that $\mu_x^{(A)}(\BbbR^r)=\mu_x^{(A)}(\partial A)=\hp(A-x)$.

I will write $\rhotv$ for the total variation metric on the space
$M^+_{\text{R}}(\BbbR^r)$ of totally finite Radon measures on $\BbbR^r$, so
that

\Centerline{$\rhotv(\lambda,\zeta)
=\sup_{E,F\subseteq\BbbR^r\text{ are Borel}}
 \lambda E-\zeta E-\lambda F+\zeta F$}

\noindent for $\lambda$,
$\zeta\in M^+_{\text{R}}(\BbbR^r)$\cmmnt{ (437Qa)}.

\leader{479B}{Theorem} Let $A\subseteq\BbbR^r$ be a bounded analytic set.

\quad(i) There is a Radon measure $\lambda_A$ on $\BbbR^r$, with support
included in $\partial A$, defined by saying that
\discretionary{}{}{}$\family{x}{\partial B(\tbf{0},\gamma)}
{\Bover1{r\beta_r\gamma}\mu_x^{(A)}}$ is a
disintegration of $\lambda_A$ over the subspace measure
$\nu_{\partial B(\tbf{0},\gamma)}$ whenever $\gamma>0$ and
$\overline{A}\subseteq\interior B(\tbf{0},\gamma)$.

\quad(ii) $\lambda_A$ is the limit
$\lim_{\|x\|\to\infty}\|x\|^{r-2}\mu_x^{(A)}$ for
the total variation metric on $M^+_{\text{R}}(\BbbR^r)$.

\leaveitout{Whenever $\gamma>0$ is such that
$\overline{A}\subseteq\interior B(\tbf{0},\gamma)$, $M>1$ and $\|x\|=M\gamma$,
then

\Centerline{$|\lambda_AE-\|x\|^{r-2}\mu_x^{(A)}E|
\le\gamma^{r-2}
  \bigl(|\Bover{M^r}{(M-1)^r}-1|+\Bover{M^{r-2}}{(M-1)^r}\bigr)$}

\noindent for every Borel set $E\subseteq\BbbR^r$.}

\proof{{\bf (a)} Suppose that $\gamma>0$ is such that
$\overline{A}\subseteq\interior B(\tbf{0},\gamma)$.   By 478T,
$x\mapsto\mu^{(A)}_x(E):\partial B(\tbf{0},\gamma)\to\coint{0,\infty}$ is
continuous for every Borel set $E\subseteq\BbbR^r$,
and $\mu^{(A)}_x(\BbbR^r\setminus\overline{A})=0$ for every $x$, so
$\{\mu^{(A)}_x:x\in\partial B(\tbf{0},\gamma)\}$ is uniformly tight.
By 452Da, we have a unique
totally finite Radon measure $\zeta_{\gamma}$ such that
$\family{x}{\partial B(\tbf{0},\gamma)}{\Bover1{r\beta_r\gamma}\mu_x^{(A)}}$ is a
disintegration of $\zeta_{\gamma}$ over the subspace measure
$\nu_{\partial B(\tbf{0},\gamma)}$.   Since $\BbbR^r\setminus\partial A$ is
$\mu_x^{(A)}$-negligible for every $x\in\partial B(\tbf{0},\gamma)$ (478Pa),
it is also $\zeta_{\gamma}$-negligible.

\medskip

{\bf (b)} Now suppose that $\overline{A}\subseteq\interior B(\tbf{0},\gamma)$ and
that $\|x\|=M\gamma$, where $M>1$.   Then 478R tells us that
$\family{y}{\BbbR^r}{\mu^{(A)}_y}$ is a disintegration of
$\mu_x^{(A)}$ over $\mu_x^{(B(\tbf{0},\gamma))}$.  So, for any Borel set
$E\subseteq\BbbR^r$,

$$\eqalignno{|\zeta_{\gamma}E-\|x\|^{r-2}\mu_x^{(A)}E|
&=\Bover1{r\beta_r\gamma}\bigl|\int_S\mu_y^{(A)}E\,\nu(dy)
  -\|x\|^{r-2}\int_{\BbbR^r}\mu_y^{(A)}E\,\mu_x^{(B(\tbf{0},\gamma))}(dy)\bigr|
  \cr
\displaycause{where $S=\partial B(\tbf{0},\gamma)$}
&=\Bover1{r\beta_r\gamma}\bigl|\int_S\mu_y^{(A)}E\,\nu(dy)
  -\|x\|^{r-2}\int_S\mu_y^{(A)}E\,\mu_x^{(S)}(dy)\bigr|\cr
\displaycause{478Pa}
&=\Bover1{r\beta_r\gamma}\bigl|\int_S\mu_y^{(A)}E\,\nu(dy)
  -\|x\|^{r-2}\int_S\Bover{\|x\|^2-\gamma^2}{r\beta_r\gamma\|x-y\|^r}
     \mu_y^{(A)}E\,\nu(dy)\bigr|\cr
\displaycause{478Q}
&\le\Bover1{r\beta_r\gamma}\int_S
  |1-\|x\|^{r-2}\,\Bover{\|x\|^2-\gamma^2}{\|x-y\|^r}|
  \mu_y^{(A)}E\nu(dy)\cr
&\le\Bover{\nu S}{r\beta_r\gamma}
  \sup_{y\in S}|1-\|x\|^{r-2}\,\Bover{\|x\|^2-\gamma^2}{\|x-y\|^r}|\cr
&\le\gamma^{r-2}
  \sup_{y\in S}(|1-\Bover{\|x\|^r}{\|x-y\|^r}|
     +\Bover{\gamma^2\|x\|^{r-2}}{\|x-y\|^r})
  \cr
&=\gamma^{r-2}
  \bigl(|\Bover{M^r\gamma^r}{(M\gamma-\gamma)^r}-1|
     +\Bover{\gamma^rM^{r-2}}{\gamma^r(M-1)^r}\bigr)
  \cr
&=\gamma^{r-2}\bigl(|\Bover{M^{r-2}}{(M-1)^r}-1|
  +\Bover{M^r}{(M-1)^r}\bigr).\cr}$$

\medskip

{\bf (c)} Accordingly

\Centerline{$\rhotv(\zeta_{\gamma},\|x\|^{r-2}\mu_x^{(A)})
\le 2\gamma^{r-2}
   \bigl(|\Bover{M^r}{(M-1)^r}-1|+\Bover{M^{r-2}}{(M-1)^r}\bigr)$}

\noindent whenever $\gamma>0$,
$\overline{A}\subseteq\interior B(\tbf{0},\gamma)$ and
$\|x\|=M\gamma>\gamma$;  so that

\Centerline{$\zeta_{\gamma}=\lim_{\|x\|\to\infty}\|x\|^{r-2}\mu_x^{(A)}$}

\noindent for the total variation metric whenever $\gamma>0$ and
$\overline{A}\subseteq\interior B(\tbf{0},\gamma)$.   We can therefore write
$\lambda_A$ for the limit, and both (i) and (ii) will be true, since I have
already checked that $\supp(\zeta_{\gamma})\subseteq\partial A$ for all
large $\gamma$.
}%end of proof of 479B

\leader{479C}{Definitions (a)(i)} In the context of 479B, I will call
$\lambda_A$ the {\bf equilibrium measure} of the bounded analytic set
$A$, and $\lambda_A\BbbR^r\cmmnt{\mskip5mu=\lambda_A(\partial A)}$ the
{\bf Newtonian capacity} $\capacity A$ of $A$.

\medskip

\quad{\bf (ii)} For any $D\subseteq\BbbR^r$, its {\bf Choquet-Newton
capacity} will be

\Centerline{$c(D)
=\inf_{G\supseteq D\text{ is open}}\sup_{K\subseteq G\text{ is compact}}
\capacity K$.}

\noindent\cmmnt{(I will confirm in 479Ed below
that $c$ is in fact a capacity as defined in \S432.) }Sets with zero
Choquet-Newton capacity are called {\bf polar}.

\spheader 479Cb If $\zeta$ is a Radon
measure on $\BbbR^r$, the {\bf Newtonian potential} associated with
$\zeta$ is
the function $W_{\zeta}:\BbbR^r\to[0,\infty]$ defined by the formula

\Centerline{$W_{\zeta}(x)=\int_{\BbbR^r}\Bover1{\|y-x\|^{r-2}}\zeta(dy)$}

\noindent for $x\in\BbbR^r$.   The {\bf energy} of $\zeta$ is now

\Centerline{$\energy(\zeta)=\int W_{\zeta}d\zeta
=\int_{\BbbR^r}\int_{\BbbR^r}\Bover1{\|x-y\|^{r-2}}\zeta(dy)\zeta(dx)$.}

\noindent If $A$ is a bounded analytic subset of $\BbbR^r$, the potential
$\tilde W_A=W_{\lambda_A}$ is the {\bf equilibrium potential} of $A$.

\cmmnt{(In 479P below I will describe constructions of equilibrium
measures and
potentials for arbitrary subsets $D$ of $\BbbR^r$ such that $c(D)$ is
finite.)}

\spheader 479Cc If $\zeta$ is a Radon measure on $\BbbR^r$,
I will write $U_{\zeta}$ for the
{\bf ($r-1$)-potential} of $\zeta$, defined by saying that
$U_{\zeta}(x)
=\biggerint_{\BbbR^r}\Bover1{\|x-y\|^{r-1}}\zeta(dy)\in[0,\infty]$ for
$x\in\BbbR^r$.

\leader{479D}{}\cmmnt{ The machinery in Theorem 479B gives an efficient
method of approaching
several fundamental properties of equilibrium measures.   I start with
some elementary calculations.

\medskip

\noindent}{\bf Proposition}
(a) For any $\gamma>0$ and $z\in\BbbR^r$,
the Newtonian capacity of
$B(z,\gamma)$ is $\gamma^{r-2}$, the equilibrium measure
of $B(z,\gamma)$ is
$\Bover1{r\beta_r\gamma}\nu\LLcorner\partial B(z,\gamma)$, and the
equilibrium potential of $B(z,\gamma)$ is given by

\Centerline{$\tilde W_{B(z,\gamma)}(x)
=\min(1,\Bover{\gamma^{r-2}}{\|x-z\|^{r-2}})$}

\noindent for every $x\in\BbbR^r$.

(b) Let $A\subseteq\BbbR^r$ be a bounded analytic
set with equilibrium measure $\lambda_A$ and equilibrium potential
$\tilde W_A$.

\quad(i) $\tilde W_A(x)\le 1$ for every $x\in\BbbR^r$.

\quad(ii) If $B\subseteq A$ is another analytic set,
$\tilde W_B\le\tilde W_A$.

\quad(iii) $\tilde W_A(x)=1$ for every $x\in\interior A$.

(c) Let $A$, $B\subseteq\BbbR^r$ be bounded analytic sets.

\quad(i)\cmmnt{ Defining $+$ and $\le$ as in 234G\formerly{1{}12Xe} and
234P,} $\lambda_{A\cup B}\le\lambda_A+\lambda_B$.

\quad(ii) $\lambda_AB\le\capacity B$.

\proof{{\bf (a)} For $x\in\BbbR^r\setminus B(z,\gamma)$,
$\mu_x^{(B(z,\gamma))}$ is the indefinite-integral measure over
$\nu\LLcorner\partial B(z,\gamma)$ defined by the function
$y\mapsto\Bover{\|x-z\|^2-\gamma^2}{r\beta_r\gamma\|x-y\|^r}$ (478Qc).
So $\|x\|^{r-2}\mu_x^{(B(z,\gamma))}$ is the indefinite-integral measure
defined by

\Centerline{$y\mapsto f_x(y)
=\Bover{\|x\|^{r-2}(\|x-z\|^2-\gamma^2)}{r\beta_r\gamma\|x-y\|^r}$.}

\noindent As $\|x\|\to\infty$, $f_x(y)\to\Bover1{r\beta_r\gamma}$
uniformly for $y\in\partial B(z,\gamma)$, so
$\lambda_{B(\tbf{0},\gamma)}
=\lim_{\|x\|\to\infty}\|x\|^{r-2}\mu_x^{(B(z,\gamma))}$ is
$\Bover1{r\beta_r\gamma}\nu\LLcorner\partial B(z,\gamma)$.
Consequently the capacity of $B(z,\gamma)$ is
$\Bover1{r\beta_r\gamma}\nu(\partial B(z,\gamma))=\gamma^{r-2}$, and the
equilibrium potential is

$$\eqalignno{\tilde W_{B(z,\gamma)}(x)
&=\Bover1{r\beta_r\gamma}\int_{\partial B(z,\gamma)}
   \Bover1{\|y-x\|^{r-2}}\nu(dy)\cr
&=\Bover1{r\beta_r\gamma}\int_{\partial B(\tbf{0},\gamma)}
   \Bover1{\|y+z-x\|^{r-2}}\nu(dy)
=\Bover1{r\beta_r\gamma}
  \cdot\Bover{\nu(\partial B(\tbf{0},\gamma))}{\max(\gamma,\|x-z\|)^{r-2}}\cr
\displaycause{478Ga}
&=\min(1,\Bover{\gamma^{r-2}}{\|x-z\|^{r-2}}).\cr}$$

\medskip

{\bf (b)(i)} Let $\gamma>0$ be such that
$\overline{A}\subseteq\interior B(\tbf{0},\gamma)$.   Then

$$\eqalignno{\tilde W_A(x)
&=\int_{\BbbR^r}\Bover1{\|x-y\|^{r-2}}\lambda_A(dy)
=\Bover1{r\beta_r\gamma}\int_{\partial B(\tbf{0},\gamma)}\int_{\BbbR^r}
  \Bover1{\|x-y\|^{r-2}}\mu_z^{(A)}(dy)\nu(dz)\cr
\displaycause{452F}
&\le\Bover1{r\beta_r\gamma}\int_{\partial B(\tbf{0},\gamma)}
  \Bover1{\|x-z\|^{r-2}}\nu(dz)\cr
\displaycause{478Pb, 478H}
&=\Bover{\nu(\partial B(\tbf{0},\gamma))}{r\beta_r\gamma}
  \Bover1{\max(\gamma,\|x\|)^{r-2}}\cr
\displaycause{478Ga}
&\le 1.\cr}$$

\medskip

\quad{\bf (ii)} Let $\gamma>0$ be such that
$\overline{A}\subseteq\interior B(\tbf{0},\gamma)$.   Then

$$\eqalignno{\tilde W_B(x)
&=\int_{\BbbR^r}\Bover1{\|x-y\|^{r-2}}\lambda_B(dy)\cr
&=\Bover1{r\beta_r\gamma}\int_{\partial B(\tbf{0},\gamma)}
   \int_{\BbbR^r}\Bover1{\|x-y\|^{r-2}}\mu_z^{(B)}(dy)\nu(dz)\cr
&=\Bover1{r\beta_r\gamma}\int_{\partial B(\tbf{0},\gamma)}
   \int_{\BbbR^r}\int_{\BbbR^r}\Bover1{\|x-y\|^{r-2}}
   \mu_w^{(B)}(dy)\mu_z^{(A)}(dw)\nu(dz)\cr
\displaycause{because $\family{w}{\BbbR^r}{\mu_w^{(B)}}$ is a
disintegration of $\mu_z^{(B)}$ over $\mu_z^{(A)}$ for every $z$, by
478R}
&\le\Bover1{r\beta_r\gamma}\int_{\partial B(\tbf{0},\gamma)}
   \int_{\BbbR^r}\Bover1{\|x-w\|^{r-2}}\mu_z^{(A)}(dw)\nu(dz)\cr
\displaycause{by 478Pb, because $y\mapsto\Bover1{\|x-y\|^{r-2}}$ is
continuous and superharmonic}
&=\tilde W_A(x).\cr}$$

\medskip

\quad{\bf (iii)} If $x\in\interior A$, there is a $\gamma>0$ such that
$B(x,\gamma)\subseteq A$;  now, putting (a) and (ii) above together,

\Centerline{$\tilde W_A(x)\ge\tilde W_{B(x,\gamma)}(x)=1$.}

\noindent Since we know from (i) that $\tilde W_A(x)\le 1$, we have
equality.

\medskip

{\bf (c)(i)} Suppose that $K\subseteq\BbbR^r$ is
compact  and that $x\in\BbbR^r$.
Let $\tau_A$, $\tau_B$ and $\tau_{A\cup B}$ be the Brownian
hitting times to $A-x$, $B-x$ and $(A\cup B)-x$ respectively.   Then
$\tau_{A\cup B}=\tau_A\wedge\tau_B$.   Now

$$\eqalign{\mu^{(A\cup B)}_x(K)
&=\mu_W\{\omega:\tau_{A\cup B}(\omega)<\infty,\,
   x+\omega(\tau_{A\cup B}(\omega))\in K\}\cr
&\le\mu_W\{\omega:\tau_{A\cup B}(\omega)=\tau_A(\omega)<\infty,\,
      x+\omega(\tau_A(\omega))\in K\}\cr
&\mskip150mu
  +\mu_W\{\omega:\tau_{A\cup B}(\omega)=\tau_B(\omega)<\infty,\,
      x+\omega(\tau_B(\omega))\in K\}\cr
&\le\mu^{(A)}_x(K)+\mu^{(B)}_x(K).\cr}$$

\noindent Multiplying by $\|x\|^{r-2}$ and letting $\|x\|\to\infty$,

\Centerline{$\lambda_{A\cup B}(K)\le\lambda_AK+\lambda_BK
=(\lambda_A+\lambda_B)(K)$}

\noindent for every $K$, which is the criterion of 416E(a-ii).

\medskip

\quad{\bf (ii)} For any $x\in\BbbR^r$,

$$\eqalignno{\mu^{(A)}_x(B)
&=\mu_W\{\omega:\tau_A(\omega)<\infty,\,x+\omega(\tau_A(\omega))\in B\}\cr
\displaycause{where $\tau_A(\omega)$ is the Brownian hitting time
to $A-x$}
&\le\mu_W\{\omega:\omega^{-1}[B-x]\ne\emptyset\}
=\mu^{(B)}_x(\BbbR^r).\cr}$$

\noindent Multiplying by $\|x\|^{r-2}$ and taking the limit as
$\|x\|\to\infty$, $\lambda_AB\le\capacity B$.
}%end of proof of 479D

\leader{479E}{Theorem} (a) Newtonian capacity $\capacity$ is
submodular\cmmnt{ (definition:  413Qb)}.

(b) Suppose that $\sequencen{A_n}$ is a
non-decreasing sequence of analytic subsets of
$\BbbR^r$ with bounded union $A$.

\quad(i) The equilibrium measure $\lambda_A$ is the limit
$\lim_{n\to\infty}\lambda_{A_n}$ for the narrow topology on the space
$M^+_{\text{R}}(\BbbR^r)$ of
totally finite Radon measures on $\BbbR^r$.

\quad(ii) $\capacity A=\lim_{n\to\infty}\capacity A_n$.

\quad(iii) The equilibrium potential
$\tilde W_A$ is
$\lim_{n\to\infty}\tilde W_{A_n}=\sup_{n\in\Bbb N}\tilde W_{A_n}$.

(c) Suppose that $\sequencen{A_n}$ is a non-increasing sequence of bounded
analytic subsets of $\BbbR^r$ such that
$\bigcap_{n\in\Bbb N}A_n=\bigcap_{n\in\Bbb N}\overline{A_n}=A$ say.

\quad(i) $\lambda_A$ is the limit
$\lim_{n\to\infty}\lambda_{A_n}$ for the narrow topology on
$M^+_{\text{R}}(\BbbR^r)$.

\quad(ii) $\capacity A=\lim_{n\to\infty}\capacity A_n$.

(d)(i) Choquet-Newton capacity $c:\Cal P\BbbR^r\to[0,\infty]$ is the unique
outer regular Choquet capacity on $\BbbR^r$ extending $\capacity$.

\quad(ii) $c$ is submodular.

\quad(iii) $c(A)=\sup\{\capacity K:K\subseteq A\text{ is compact}\}$
for every analytic set $A\subseteq\BbbR^r$.

\proof{{\bf (a)} Let $A$, $B\subseteq\BbbR^r$ be bounded analytic sets.
If $x\in\BbbR^r$, then

\Centerline{$\hp((A\cup B)-x)+\hp((A\cap B)-x)
\le\hp(A-x)+\hp(B-x)$.}

\noindent\Prf\ For $C\subseteq\BbbR^r$ set

\Centerline{$H_C=\{\omega:\omega\in\Omega$,
there is some $t\ge 0$ such that $x+\omega(t)\in C\}$,}

\noindent so that if $C$ is an analytic set,
$\hp(C-x)=\mu_WH_C$.   Then

\Centerline{$H_{A\cup B}=H_A\cup H_B$,
\quad$H_{A\cap B}\subseteq H_A\cap H_B$,}

\noindent so

$$\eqalign{\hp((A\cup B)-x)+\hp((A\cap B)-x)
&=\mu_WH_{A\cup B}+\mu_WH_{A\cap B}\cr
&\le\mu_W(H_A\cup H_B)+\mu_W(H_A\cap H_B)\cr
&=\mu_WH_A+\mu_WH_B\cr
&=\hp(A-x)+\hp(B-x).  \text{ \Qed}\cr}$$

\noindent Consequently

$$\eqalign{\capacity(A\cup B)+\capacity(A\cap B)
&=\lambda_{A\cup B}(\BbbR^r)+\lambda_{A\cap B}(\BbbR^r)\cr
&=\lim_{\|x\|\to\infty}\|x\|^{r-2}
  (\hp((A\cup B)-x)+\hp((A\cap B)-x)\cr
&\le\lim_{\|x\|\to\infty}\|x\|^{r-2}
  (\hp(A-x)+\hp(B-x))\cr
&=\capacity A+\capacity B.\cr}$$

\noindent As $A$ and $B$ are arbitrary, $\capacity$ is submodular.

\medskip

{\bf (b)(i)} Let $f:\BbbR^r\to\Bbb R$ be any bounded continuous function.
For any $x\in\BbbR^r$,
$\int fd\mu_x^{(A)}=\lim_{n\to\infty}\int fd\mu_x^{(A_n)}$.
\Prf\ Let $\tau$, $\tau_n$ be the Brownian hitting times to $A-x$, $A_n-x$
respectively.   Observe that $\sequencen{\tau_n(\omega)}$ is non-increasing
and

\Centerline{$\tau(\omega)=\inf\{t:x+\omega(t)\in\bigcup_{n\in\Bbb N}A_n\}
=\lim_{n\to\infty}\tau_n(\omega)$}

\noindent for every $\omega\in\Omega$.
Set $H=\{\omega:\tau(\omega)<\infty\}$,
$H_n=\{\omega:\tau_n(\omega)<\infty\}$.   Then $\sequencen{H_n}$ is a
non-decreasing sequence with union $H$, and for $\omega\in H$

\Centerline{$f(x+\tau(\omega))=\lim_{n\to\infty}f(x+\tau_n(\omega))$}

\noindent because $f$ and $\omega$ are continuous.   Accordingly

$$\eqalign{\int fd\mu^{(A)}_x
&=\int_Hf(x+\omega(\tau(\omega)))\cr
&=\lim_{n\to\infty}\int_{H_n}f(x+\omega(\tau_n(\omega)))
=\lim_{n\to\infty}\int fd\mu^{(A_n)}_x. \text{ \Qed}\cr}$$

\woddheader{479E}{0}{0}{0}{48pt}

Taking $\gamma>0$ such that $\overline{A}\subseteq\interior B(\tbf{0},\gamma)$,

$$\eqalignno{\int fd\lambda_A
&=\Bover1{r\beta_r\gamma}
   \int_{\partial B(\tbf{0},\gamma)}\int fd\mu_x^{(A)}\nu(dx)\cr
\displaycause{452F}
&=\Bover1{r\beta_r\gamma}\int_{\partial B(\tbf{0},\gamma)}
   \lim_{n\to\infty}\int fd\mu_x^{(A_n)}\nu(dx)\cr\cr
&=\lim_{n\to\infty}\Bover1{r\beta_r\gamma}
   \int_{\partial B(\tbf{0},\gamma)}\int fd\mu_x^{(A_n)}\nu(dx)
=\lim_{n\to\infty}\int fd\lambda_{A_n}.\cr}$$

\noindent As $f$ is arbitrary, $\lambda_A=\lim_{n\to\infty}\lambda_{A_n}$
for the narrow topology (437Kc).

\medskip

\quad{\bf (ii)} Taking $f=\chi\BbbR^r$ in (i), we see that
$\capacity A=\lim_{n\to\infty}\capacity A_n$.

\medskip

\quad{\bf (iii)} For any $x\in\BbbR^r$,

\Centerline{$\tilde W_A(x)
=\int\Bover1{\|y-x\|^{r-2}}\lambda_A(dy)
\le\liminf_{n\to\infty}\int\Bover1{\|y-x\|^{r-2}}\lambda_{A_n}(dy)$}

\noindent because $y\mapsto\Bover1{\|y-x\|^{r-2}}$ is non-negative and
continuous (437Jg).   As $\tilde W_{A_n}(x)\le\tilde W_A(x)$
for every $n$ (479D(b-ii)),
$\tilde W_A(x)=\lim_{n\to\infty}\tilde W_{A_n}(x)
=\sup_{n\in\Bbb N}\tilde W_{A_n}(x)$.

\medskip

{\bf (c)} Most of the ideas of (b) still work.   Again take
$f\in C_b(\BbbR^r)$.   Then
$\int fd\mu_x^{(A)}=\lim_{n\to\infty}\int fd\mu_x^{(A_n)}$
for any $x\in\BbbR^r$.
\Prf\ As before, let $\tau$, $\tau_n$ be the Brownian hitting times to
$A-x$, $A_n-x$
respectively.   This time, $\sequencen{\tau_n}$ is non-decreasing.
Let $\Omega'$ be the conegligible subset of $\Omega$ consisting of those
functions $\omega$ such that $\lim_{t\to\infty}\|\omega(t)\|=\infty$.
If $\omega\in\Omega'$ and $t=\lim_{n\to\infty}\tau_n(\omega)$ is finite,
then for every $n\in\Bbb N$ there is a $t_n\le t+2^{-n}$ such that
$x+\omega(t_n)\in A_n$.   Let $s\in[0,t]$ be a cluster point of
$\sequencen{t_n}$;  then $x+\omega(s)$ is a cluster point of
$\sequencen{x+\omega(t_n)}$, so belongs to
$\bigcap_{n\in\Bbb N}\overline{A}_n=A$, and $\tau(\omega)\le s\le t$.
Since $\tau(\omega)\ge\tau(\omega_n)$ for every $n$,
we have $\tau(\omega)=\lim_{n\to\infty}\tau(\omega_n)$.

Setting $H=\{\omega:\tau(\omega)<\infty\}$ and
$H_n=\{\omega:\tau_n(\omega)<\infty\}$, $\sequencen{H_n}$ is a
non-increasing sequence with intersection $H$, and for $\omega\in H$

\Centerline{$f(x+\tau(\omega))=\lim_{n\to\infty}f(x+\tau_n(\omega))$.}

\noindent So once again

$$\eqalign{\int fd\mu^{(A)}_x
&=\int_Hf(x+\omega(\tau(\omega)))\mu_W(d\omega)\cr
&=\lim_{n\to\infty}\int_{H_n}f(x+\omega(\tau_n(\omega)))\mu_W(d\omega)
=\lim_{n\to\infty}\int fd\mu^{(A_n)}_x. \text{ \Qed}\cr}$$

\noindent The rest of the argument follows (b-i) and (b-ii) unchanged.

\medskip

{\bf (d)(i)} I seek to apply 432Lb.

\medskip

\qquad\grheada\ Let $\Cal K$ be the family of compact
subsets of $\BbbR^r$ and set $c_1=\capacity\restr\Cal K$.   By
(a), $c_1$ is submodular.
If $G\subseteq\BbbR^r$ is a bounded open set, then it is
expressible as the union of a
non-decreasing sequence of compact sets, so by (b-ii) we
have $\capacity G=\sup\{c_1(L):L\in\Cal K$, $L\subseteq G\}$;  and if
$K\in\Cal K$, there is a non-increasing sequence $\sequencen{G_n}$ of
bounded open sets such that
$K=\bigcap_{n\in\Bbb N}G_n=\bigcap_{n\in\Bbb N}\overline{G}_n$, and now
$c_1(K)=\lim_{n\to\infty}\capacity G_n$, by (c-ii).   But this means that

\Centerline{$c_1(K)
\le\inf_{G\supseteq K\text{ is open}}
  \sup_{L\subseteq G\text{ is compact}}c_1(L)
\le\inf_{n\in\Bbb N}\capacity G_n
=c_1(K)$.}

\noindent So all the conditions of 432Lb are satisfied, and $c$, as defined
in 479C(a-ii), is the unique extension of $c_1$
to an outer regular Choquet capacity on $\BbbR^r$.

\medskip

\qquad\grheadb\ Now $c(A)=\capacity A$ for every bounded analytic set
$A\subseteq\BbbR^r$.   \Prf\

$$\eqalignno{c(A)
&=\sup_{K\subseteq A\text{ is compact}}c(K)\cr
\displaycause{432K}
&=\sup_{K\subseteq A\text{ is compact}}\capacity K
\le\capacity A
\le\inf_{G\supseteq A\text{ is bounded and open}}\capacity G\cr
&=\mathop{\text{inf}\hbox{\vrule depth2pt width0pt}}
  _{G\supseteq A\text{ is open}}
   \,\,\sup_{L\subseteq G\text{ is compact}}c(L)
=c(A).  \text{ \Qed}\cr}$$

\noindent So $c$ extends $\capacity$, as claimed, and must be the unique
outer regular Choquet capacity doing so.

\medskip

\quad{\bf (ii)-(iii)} By 432Lb, $c$ is submodular;  and (iii) is covered by
the argument in (i-$\beta$).
}%end of proof of 479E

\leader{479F}{}\cmmnt{ I now wish to describe an entirely different
characterization of the capacity and equilibrium measure of a compact set,
which demands a substantial investment in harmonic analysis (down to
479I) and an excursion into Fourier analysis (479H).   I begin with general
remarks about Newtonian potentials.

\medskip

\noindent}{\bf Theorem} Let $\zeta$ be a totally finite Radon measure on
$\BbbR^r$, and set $G=\BbbR^r\setminus\supp\zeta$\cmmnt{, where
$\supp\zeta$ is the support of $\zeta$ (411Nd)}.   Let $W_{\zeta}$ be the
Newtonian potential associated with $\zeta$.

(a) $W_{\zeta}:\BbbR^r\to[0,\infty]$ is lower semi-continuous,
and $W_{\zeta}\restr G:G\to\coint{0,\infty}$ is continuous.

(b) $W_{\zeta}$ is superharmonic, and $W_{\zeta}\restr G$ is harmonic.

(c) $W_{\zeta}$ is locally $\mu$-integrable;  in particular, it is finite
$\mu$-a.e.

(d) If $\zeta$ has compact support, then
$\zeta\BbbR^r=\lim_{\|x\|\to\infty}\|x\|^{r-2}W_{\zeta}(x)$.

(e) If $W_{\zeta}\restr\supp\zeta$ is continuous then $W_{\zeta}$ is
continuous.

(f) If $K$ is a compact set such that $W_{\zeta}\restr K$ is continuous
and finite-valued then $W_{\zeta\LLcorner K}$ is continuous.

(g) If $W_{\zeta}$ is finite $\zeta$-a.e.\ and
$f:\BbbR^r\to[0,\infty]$ is a lower semi-continuous superharmonic
function such that $f\ge W_{\zeta}\,\,\zeta$-a.e., then $f\ge W_{\zeta}$.
%index as `domination principle'

(h) If $\zeta'$ is another Radon measure on $\BbbR^r$ and
$\zeta'\le\zeta$, then $W_{\zeta'}\le W_{\zeta}$ and
$\energy(\zeta')\le\energy(\zeta)$.

%If $h:\BbbR^r\to[0,\infty]$ a Lebesgue integrable function, then
%$W_{\zeta*h\mu}=W_{\zeta}*h$, where $h\mu$ is the indefinite-integral
%measure over $\mu$ defined by $h$ and
%the convolution $\zeta*h\mu$ is defined as in 444A or 257A.
%see 479Q

\proof{{\bf (a)}
If $\sequencen{x_n}$ is a convergent sequence in
$\BbbR^r$ with limit $x$, then
$\Bover1{\|y-x\|^{r-2}}=\lim_{n\to\infty}\Bover1{\|y-x_n\|^{r-2}}$
for every $y$
(counting $\Bover10$ as $\infty$, as usual), so that
$W_{\zeta}(x)\le\liminf_{n\to\infty}W_{\zeta}(x_n)$, by Fatou's Lemma.   As $x$ and
$\sequencen{x_n}$ are arbitrary, $W_{\zeta}$ is lower semi-continuous.

If $x\in G$, then
$\Bover1{\|y-x_n\|^{r-2}}\le\Bover2{\rho(x,\supp\zeta)^{r-2}}$ for all $n$
large enough and all $y\in\supp\zeta$, so Lebesgue's Dominated Convergence Theorem
tells us that $W_{\zeta}(x)=\lim_{n\to\infty}W_{\zeta}(x_n)$ and that
$W_{\zeta}$ is continuous at $x$, as well as finite-valued there.

\medskip

{\bf (b)} If $x\in\BbbR^r$ and $\delta>0$, then

$$\eqalign{\Bover1{\nu(\partial B(x,\delta))}\int_{\partial B(x,\delta)}
  W_{\zeta}d\nu
&=\Bover1{\nu(\partial B(x,\delta))}\int_{\partial B(x,\delta)}
  \int_{\BbbR^r}\Bover1{\|z-y\|^{r-2}}\zeta(dz)\nu(dy)\cr
&=\int_{\BbbR^r}\Bover1{\nu(\partial B(x,\delta))}
  \int_{\partial B(x,\delta)}\Bover1{\|z-y\|^{r-2}}\nu(dy)\zeta(dz)\cr
&=\int_{\BbbR^r}\Bover1{\nu(\partial B(\tbf{0},\delta))}
  \int_{\partial B(\tbf{0},\delta)}\Bover1{\|z-x-y\|^{r-2}}\nu(dy)\zeta(dz)\cr
&\ge\int_{\BbbR^r}\Bover1{\|z-x\|^{r-2}}\zeta(dz)
=W_{\zeta}(x)\cr}$$

\noindent (478Ga) with equality if $B(x,\delta)$ does not meet
$\supp\zeta$, since then $z-x\notin B(\tbf{0},\delta)$ and

\Centerline{$\Bover1{\nu(\partial B(x,\delta))}
  \int_{\partial B(x,\delta)}\Bover1{\|z-x-y\|^{r-2}}\nu(dy)
=\Bover1{\|z-x\|^{r-2}}$}

\noindent for $\zeta$-almost every $z$.

\medskip

{\bf (c)} For any $\gamma>0$ and $y\in\BbbR^r$,
$\biggerint_{B(\tbf{0},\gamma)}\Bover1{\|y-x\|^{r-2}}\mu(dx)
\le\Bover12r\beta_r\gamma^2$ (478Gc), so

$$\eqalign{\int_{B(\tbf{0},\gamma)}W_{\zeta}d\mu
&=\int_{B(\tbf{0},\gamma)}\int_{\BbbR^r}\Bover1{\|y-x\|^{r-2}}\zeta(dy)\mu(dx)\cr
&=\int_{\BbbR^r}\int_{B(\tbf{0},\gamma)}\Bover1{\|y-x\|^{r-2}}\mu(dx)\zeta(dy)
\le\Bover12r\beta_r\gamma^2\zeta\BbbR^r\cr}$$

\noindent is finite.

\medskip

{\bf (d)} If $\zeta$ has compact support, there is an $\gamma>0$ such that
$\supp\zeta\subseteq B(\tbf{0},\gamma)$.
In this case, for $\|x\|>\gamma$, we have

\Centerline{$\Bover{\|x\|^{r-2}}{(\|x\|+\gamma)^{r-2}}\zeta\BbbR^r
\le\int_{\BbbR^r}\Bover{\|x\|^{r-2}}{\|x-y\|^{r-2}}\zeta(dy)
=W_{\zeta}(x)\|x\|^{r-2}
\le\Bover{\|x\|^{r-2}}{(\|x\|-\gamma)^{r-2}}\zeta\BbbR^r$}

\noindent so all the terms converge to $\zeta\BbbR^r$ as
$\|x\|\to\infty$.

\medskip

{\bf (e)} Since $W_{\zeta}$ is lower semi-continuous, it will be
enough to show that $H=\{x:W_{\zeta}(x)<\gamma\}$ is open for every
$\gamma\in\Bbb R$.   Take $x_0\in H$.
If $x_0\notin\supp\zeta$ then $W_{\zeta}$ is continuous at $x_0$,
by 479Fa, and
$H$ is certainly a neighbourhood of $x_0$.   If $x_0\in\supp\zeta$
take $\eta\in\ooint{0,2^{-r}(\gamma-W_{\zeta}(x_0))}$.
Because $W_{\zeta}(x_0)$ is finite, $\zeta\{x_0\}=0$;
because $W_{\zeta}\restr\supp\zeta$ is continuous, there is a $\delta>0$
such that
$\biggerint_{B(x_0,\delta)}\Bover1{\|x_0-y\|^{r-2}}\zeta(dy)\le\eta$ and
$|W_{\zeta}(x)-W_{\zeta}(x_0)|\le\eta$ whenever
$x\in B(x_0,\delta)\cap\supp\zeta$.   Let
$\delta'\in\ooint{0,\delta}$ be such that

\Centerline{$\bigl|\Bover1{\|x-y\|^{r-2}}-\Bover1{\|x_0-y\|^{r-2}}\bigr|
   \le\Bover{\eta}{\zeta\BbbR^r}$}

\noindent whenever $\|x-x_0\|\le\delta'$ and $\|y-x_0\|\ge\delta$;
then

\Centerline{$
|\int_{\BbbR^r\setminus B(x_0,\delta)}\Bover1{\|x-y\|^{r-2}}\zeta(dy)
  -\int_{\BbbR^r\setminus B(x_0,\delta)}\Bover1{\|x_0-y\|^{r-2}}\zeta(dy)|
\le\eta$}

\noindent whenever $x\in B(x_0,\delta')$.

Take $x\in B(x_0,\bover12\delta')$, and let
$z\in B(x_0,\delta)\cap\supp\zeta$ be such that
$\|x-z\|=\rho(x,B(x_0,\delta)\cap\supp\zeta)$.
We have $\|x-z\|\le\|x-x_0\|$ so $\|z-x_0\|\le 2\|x-x_0\|\le\delta'$.
If $y\in B(x_0,\delta)\cap\supp\zeta$, then
$\|y-z\|\le\|x-y\|+\|x-z\|\le 2\|x-y\|$;  so

$$\eqalign{\int_{B(x_0,\delta)}\Bover1{\|x-y\|^{r-2}}\zeta(dy)
&\le 2^{r-2}\int_{B(x_0,\delta)}\Bover1{\|z-y\|^{r-2}}\zeta(dy)\cr
&=2^{r-2}(W_{\zeta}(z)
  -\int_{\BbbR^r\setminus B(x_0,\delta)}\Bover1{\|z-y\|^{r-2}}\zeta(dy))\cr
&\le 2^{r-2}(2\eta+W_{\zeta}(x_0)
  -\int_{\BbbR^r\setminus B(x_0,\delta)}
     \Bover1{\|x_0-y\|^{r-2}}\zeta(dy))\cr
&=2^{r-1}\eta+2\int_{B(x_0,\delta)}\Bover1{\|x_0-y\|^{r-2}}\zeta(dy)\cr
&\le 2^{r-1}\eta+2\eta\le(2^r-1)\eta,\cr}$$

$$\eqalign{W_{\zeta}(x)
&\le (2^r-1)\eta+\int_{\BbbR^r\setminus B(x_0,\delta)}
  \Bover1{\|x-y\|^{r-2}}\zeta(dy)\cr
&\le 2^r\eta+\int_{\BbbR^r\setminus B(x_0,\delta)}
  \Bover1{\|x_0-y\|^{r-2}}\zeta(dy)
\le 2^r\eta+W_{\zeta}(x_0)
<\gamma.\cr}$$

\noindent Thus $B(x_0,\bover12\delta')\subseteq H$ and again $H$ is a
neighbourhood of $x_0$.   As $x_0$ is arbitrary, $H$ is open;  as
$\gamma$ is arbitrary, $W_{\zeta}$ is continuous.

\medskip

{\bf (f)} Setting $H=\BbbR^r\setminus K$,
$\zeta=\zeta\LLcorner K+\zeta\LLcorner H$, so
$W_{\zeta}=W_{\zeta\LLcorner K}+W_{\zeta\LLcorner H}$
(234Hc\formerly{2{}12Xh}).   Now
$W_{\zeta\LLcorner K}$ and $W_{\zeta\LLcorner H}$ are both lower
semi-continuous and non-negative,
so if $W_{\zeta}\restr K$ is continuous and finite-valued
then $W_{\zeta\LLcorner K}\restr K$
is continuous (4A2B(d-ix)).   Since $\supp(\zeta\LLcorner K)\subseteq K$,
(e) tells us that $W_{\zeta\LLcorner K}$ is continuous.

\medskip

{\bf (g)} \Quer\ Suppose that $f(x_0)<W_{\zeta}(x_0)$.   Since
$\{x:f(x)\ge W_{\zeta}(x)$, $W_{\zeta}(x)<\infty\}$ is
$\zeta$-conegligible, and $W_{\zeta}$ is measurable therefore
$\zeta$-almost continuous (418J), there is a compact set $K$ such that
$W_{\zeta}(x)<\infty$ and $W_{\zeta}(x)\le f(x)$ for every $x\in K$,
$W_{\zeta}\restr K$ is continuous and
$\biggerint_K\Bover1{\|x_0-y\|^{r-2}}\zeta(dy)>f(x_0)$.   Set
$\zeta'=\zeta\LLcorner K$.   By (f), $W_{\zeta'}$ is
continuous;  $W_{\zeta'}(x_0)>f(x_0)$;  and
$f(x)\ge W_{\zeta}(x)\ge W_{\zeta'}(x)$ for every
$x\in K\supseteq\supp\zeta'$.

Set $g=f-W_{\zeta'}$ and $\alpha=\inf_{x\in\BbbR^r}g(x)<0$.
Because $f$ is lower semi-continuous and $W_{\zeta'}$ is continuous,
$g$ is lower semi-continuous;  because $\zeta'$ has compact
support, $\lim_{\|x\|\to\infty}W_{\zeta'}(x)=0$ ((d) above) and
$\liminf_{\|x\|\to\infty}g(x)\ge 0$;  so
$L=\{x:g(x)=\alpha\}$ is non-empty and compact.   Note that
$L$ is disjoint from $K$.
Let $x_1\in L$ be a point of maximum norm.   Then $x_1\notin K$, while
$W_{\zeta'}\restr\BbbR^r\setminus K$ is harmonic ((b) above).
Let $\delta>0$
be such that $B(x_1,\delta)\cap K=\emptyset$.   Then we have

\Centerline{$\Bover1{\nu(\partial B(x_1,\delta))}
   \int_{\partial B(x_1,\delta)}g\,d\nu
>\alpha$}

\noindent because $g(x)\ge\alpha$ for every $x$ and
$g(x)>\alpha$ whenever $x\ne x_1$ and
$(x-x_1)\dotproduct x_1\ge 0$.   But we also have

$$\eqalign{\Bover1{\nu(\partial B(x_1,\delta))}
   \int_{\partial B(x_1,\delta)}g\,d\nu
&=\Bover1{\nu(\partial B(x_1,\delta))}
   \int_{\partial B(x_1,\delta)}f\,d\nu
   -\Bover1{\nu(\partial B(x_1,\delta))}
     \int_{\partial B(x_1,\delta)}W_{\zeta'}\,d\nu\cr
&\le f(x_1)-W_{\zeta'}(x_1)
=\alpha,\cr}$$

\noindent which is impossible.\ \Bang

\medskip

{\bf (h)} By 234Qc, $W_{\zeta'}\le W_{\zeta}$;  so

\Centerline{$\energy(\zeta')=\int W_{\zeta'}d\zeta'
\le\int W_{\zeta}d\zeta'\le\int W_{\zeta}d\zeta=\energy(\zeta)$.}

}%end of proof of 479F

\leader{479G}{}\cmmnt{ At this point
I embark on an extended parenthesis, down to 479I, covering
some essential material from harmonic analysis and Fourier analysis.
The methods here apply equally to the cases $r=1$ and $r=2$.

\medskip

\noindent}{\bf Lemma} (In this result, $r$ may be any integer greater than
or equal to $1$.)   For $\alpha\in\Bbb R$, set
$k_{\alpha}(x)=\Bover1{\|x\|^{\alpha}}$ for $x\in\BbbR^r\setminus\{0\}$.
If $\alpha<r$, $\beta<r$ and $\alpha+\beta>r$, then
$k_{\alpha+\beta-r}$ is a constant multiple of the convolution
$k_{\alpha}*k_{\beta}$\cmmnt{ (definition:  255E, 444O)}.

\proof{{\bf (a)} First note that

\Centerline{$\int_{B(\tbf{0},1)}k_{\alpha}(x)\mu(dx)
=r\beta_r\int_0^1\Bover{t^{r-1}}{t^{\alpha}}dt
=\Bover{r\beta_r}{r-\alpha}$}

\noindent is finite.   Consequently $k_{\alpha}$ is expressible as the sum
of an integrable function and a bounded function, and in particular is
locally integrable.

\medskip

{\bf (b)} For $x\in\BbbR^r$, set
$f(x)=\int_{\BbbR^r}k_{\alpha}(x-y)k_{\beta}(x-y)\mu(dy)\in[0,\infty]$.
If $e\in\BbbR^r$ is a unit vector, then $f(e)$ is finite.
\Prf\ For any $y\in\BbbR^r$,
at least one of $\|e-y\|$, $\|e+y\|$ is greater than or equal to $1$, so
$k_{\alpha}(e-y)k_{\beta}(e+y)\le k_{\alpha}(e-y)+k_{\beta}(e+y)$.
Consequently

\Centerline{$\int_{B(\tbf{0},2)}k_{\alpha}(e-y)k_{\alpha}(e+y)\mu(dy)
\le\int_{B(\tbf{0},2)}k_{\alpha}(e-y)+k_{\alpha}(e+y)\mu(dy)$}

\noindent is finite.   On the other hand, if $\|y\|\ge 2$,
$\|e-y\|$ and $\|e-y\|$ are both at least $\bover12\|y\|$, so

$$\eqalign{\int_{\BbbR^r\setminus B(\tbf{0},2)}
  k_{\alpha}(e-y)k_{\beta}(e+y)\mu(dy)
&\le\int_{\BbbR^r\setminus B(\tbf{0},2)}\Bover{2^{\alpha}}{\|y\|^{\alpha}}
  \cdot\Bover{2^{\beta}}{\|y\|^{\beta}}\mu(dy)\cr
&=2^{\alpha+\beta}r\beta_r\int_2^{\infty}
   \Bover{t^{r-1}}{t^{\alpha+\beta}}dt
=\Bover{2^rr\beta_r}{\alpha+\beta-r}\cr}$$

\noindent is finite.   Putting these together,
$f(e)=\int_{\BbbR^r}k_{\alpha}(e-y)k_{\beta}(e-y)\mu(dy)$ is finite.\
\Qed

\medskip

{\bf (c)} If $e$, $e'\in\BbbR^r$ are unit vectors, then $f(e)=f(e')$.
\Prf\ Let $T:\BbbR^r\to\BbbR^r$ be an orthogonal transformation such that
$Te=e'$.   Then

$$\eqalignno{f(e')
&=\int_{\BbbR^r}k_{\alpha}(Te-y)k_{\beta}(Te-y)\mu(dy)
=\int_{\BbbR^r}k_{\alpha}(Te-Ty)k_{\beta}(Te-Ty)\mu(dy)\cr
\displaycause{because $T$ is an automorphism of $(\BbbR^r,\mu)$}
&=\int_{\BbbR^r}k_{\alpha}(e-y)k_{\beta}(e-y)\mu(dy)\cr
\displaycause{because $k_{\alpha}(x)$, $k_{\beta}(x)$ are functions of
$\|x\|$}
&=f(e).  \text{ \Qed}\cr}$$

Let $c$ be the constant value of $f(e)$ for $\|e\|=1$.

\woddheader{479G}{4}{2}{2}{72pt}

{\bf (d)} If $x\in\BbbR^r\setminus\{0\}$,
$f(x)=\Bover{c}{\|x\|^{\alpha+\beta-r}}$.   \Prf\ Set $t=\|x\|$,
$e=\bover1tx$.   Then

$$\eqalignno{f(x)
&=\int_{\BbbR^r}k_{\alpha}(te-y)k_{\beta}(te-y)\mu(dy)
=\int_{\BbbR^r}t^rk_{\alpha}(te-tz)k_{\beta}(te-tz)\mu(dz)\cr
\displaycause{substituting $y=tz$}
&=t^{r-\alpha-\beta}\int_{\BbbR^r}k_{\alpha}(e-z)k_{\beta}(e-z)\mu(dz)
=\Bover{c}{\|x\|^{\alpha+\beta-r}}. \text{ \Qed}\cr}$$

\medskip

{\bf (e)} If $x\in\BbbR^r\setminus\{0\}$,
$(k_{\alpha}*k_{\beta})(x)=2^{\alpha+\beta-r}ck_{\alpha+\beta-r}(x)$.
\Prf\ Set $z=\bover12x$.   Then

$$\eqalign{(k_{\alpha}*k_{\beta})(x)
&=\int_{\BbbR^r}\Bover1{\|x-y\|^{\alpha}\|y\|^{\beta}}\mu(dy)
=\int_{\BbbR^r}\Bover1{\|x-y-z\|^{\alpha}\|y+z\|^{\beta}}\mu(dy)\cr
&=\int_{\BbbR^r}\Bover1{\|z-y\|^{\alpha}\|z+y\|^{\beta}}\mu(dy)
=\Bover{c}{\|z\|^{\alpha+\beta-r}}
=2^{\alpha+\beta-r}ck_{\alpha+\beta-r}(x).  \text{ \Qed}\cr}$$

\medskip

{\bf (f)} Of course it is of no importance what happens at $0$, but for
completeness:  $k_{\alpha+\beta-r}$ is declared to be undefined there, and
$\biggerint_{\BbbR^r}\Bover1{\|y\|^{\alpha}\|y\|^{\beta}}\mu(dy)$ is
infinite for
any $\alpha$ and $\beta$, so $k_{\alpha}*k_{\beta}$ also is undefined at
$0$ on the convention of 255E or 444O.   Thus we have
$k_{\alpha}*k_{\beta}=2^{\alpha+\beta-r}ck_{\alpha+\beta-r}$ in the strict
sense.
}%end of proof of 479G

\medskip

\noindent{\bf Remark}\cmmnt{ The functions $k_{\alpha}$ are called
{\bf Riesz kernels}.   It will be helpful later to have a name
for the constant arising here in a special case.}   If $r\ge 3$,
I will take $c_r>0$
to be the constant such that $c_rk_{r-2}=k_{r-1}*k_{r-1}$.

\leader{479H}{}\cmmnt{ Now for some Fourier analysis which wasn't
quite reached in Chapter 28.   In the following, I will define the Fourier
transform $\varhatf$ and the inverse Fourier transform
$\varcheckf$, for $\mu$-measurable complex-valued functions $f$ defined
$\mu$-almost everywhere in $\BbbR^r$, as in 283W and 284W, and
$\varhat{\zeta}$, for a totally finite Radon measure $\zeta$ on $\BbbR^r$,
by the formula offered in 285Ya for probability measures.   The convolution
$\zeta*f$ of a measure and a function will be defined as in 444H.   Thus
the basic formulae are

\Centerline{$\varhatf(y)=\Bover1{(\sqrt{2\pi})^r}
  \int_{\BbbR^r}e^{-iy\dotproduct x}f(x)\mu(dx)$,
\quad$\varcheckf(y)=\Bover1{(\sqrt{2\pi})^r}
  \int_{\BbbR^r}e^{iy\dotproduct x}f(x)\mu(dx)$}

\noindent for $\mu$-integrable $f$,

\Centerline{$\varhat{\zeta}(y)=\Bover1{(\sqrt{2\pi})^r}
  \int_{\BbbR^r}e^{-iy\dotproduct x}\zeta(dx)$,
\quad$(\zeta*f)(x)
  =\int_{\BbbR^r}f(x-y)\zeta(dy)$.}

\medskip

\noindent}{\bf Theorem} (In this result, $r$ may be any integer greater
than or equal to $1$.)   Let $\zeta$ be a totally finite Radon measure on
$\BbbR^r$\dvro{ and $\varhat{\zeta}$ its Fourier transform}{}.

(a) If $f\in\eusm L^1_{\Bbb C}(\mu)$, then $\zeta*f$ is
$\mu$-integrable and
$(\zeta*f)\varsphat=(\sqrt{2\pi})^r\varhat{\zeta}\times\varhatf$.

(b) If $\zeta$ has compact support and
$h:\BbbR^r\to\Bbb C$ is a rapidly decreasing test
function\cmmnt{ (284Wa)},
then $\zeta*h$ and $h\times\varhat{\zeta}$ are rapidly decreasing test
functions.

(c) Suppose that $f$ is a tempered function on $\BbbR^r$\cmmnt{ (284Wa)}.
If {\it either} $\zeta$ has compact support {\it or} $f$ is expressible as
the sum of a $\mu$-integrable function and a bounded function, then
$\zeta*f$ is defined $\mu$-almost everywhere and is a tempered function.

(d) Suppose that $f$, $g$ are tempered functions on
$\BbbR^r$ such that $g$ represents the Fourier transform
of $f$\cmmnt{ (284Wd)}.   If {\it either} $\zeta$ has compact support
{\it or} $f$ is expressible as the sum of a bounded function and a
$\mu$-integrable function, then
$(\sqrt{2\pi})^r\varhat{\zeta}\times g$ represents the Fourier transform of
$\zeta*f$.

\proof{{\bf (a)(i)} To begin with, suppose that $f$ is real-valued and
non-negative.   As in \S444, I will write $f\mu$ for the
indefinite-integral measure defined by $f$ over $\mu$.
By 444K, $\zeta*f$ is $\mu$-integrable and $(\zeta*f)\mu=\zeta*f\mu$.

As the formula used here for $\varhat{\zeta}$ does not
quite match that of 445C, whatever parametrization we use for the
characters of the topological group $\BbbR^r$, I had better not try to
quote Chapter 44 when discussing Fourier transforms.
Going back to first principles,

$$\eqalignno{(\zeta*f)\varsphat(y)
&=\Bover1{(\sqrt{2\pi})^r}\int e^{-iy\dotproduct x}(\zeta*f)(x)\mu(dx)
=\Bover1{(\sqrt{2\pi})^r}\int e^{-iy\dotproduct x}(\zeta*f)\mu(dx)\cr
&=\Bover1{(\sqrt{2\pi})^r}\int e^{-iy\dotproduct x}(\zeta*f\mu)(dx)
=\Bover1{(\sqrt{2\pi})^r}\iint e^{-iy\dotproduct(x+z)}
  \zeta(dz)(f\mu)(dx)\cr
\displaycause{444C}
&=\Bover1{(\sqrt{2\pi})^r}\int e^{-iy\dotproduct z}\zeta(dz)
   \int e^{-iy\dotproduct x}(f\mu)(dx)\cr
&=\varhat{\zeta}(y)\int e^{-iy\dotproduct x}f(x)\mu(dx)
=(\sqrt{2\pi})^r\varhat{\zeta}(y)\varhatf(y)\cr}$$

\noindent for every $y\in\BbbR^r$.

\medskip

\quad{\bf (ii)} For general integrable complex-valued functions $f$,
apply (i) to the positive and negative parts of the real and imaginary
parts of $f$.

\medskip

{\bf (b)(i)} Because $h$ is continuous, so is $\zeta*h$ (444Ib).
If $j<r$, then, as in 123D,

\Centerline{$(\Pd{}{\xi_j}(\zeta*h))(x)
=\Pd{}{\xi_j}\int h(x-y)\zeta(dy)
=\int\Pd{}{\xi_j}h(x-y)\zeta(dy)
=(\zeta*\Pd{h}{\xi_j})(x)$}

\noindent because $\Pd{h}{\xi_j}$ is bounded.
Since $\Pd{h}{\xi_j}$ is again a \rdtf,
we can repeat this process
to see that $\zeta*h$ is smooth.   Next, let $\gamma>0$ be such that
the support of $\zeta$ is included in $B(\tbf{0},\gamma)$.
If $k\in\Bbb N$, then
$M=\sup_{x\in\BbbR^r}(\gamma^k+\|x\|^k)|h(x)|$ is finite, and

\Centerline{$\|x\|^k|h(y)|\le(\|y\|+\gamma)^k|h(y)|
\le 2^kM$}

\noindent whenever $\|x-y\|\le\gamma$.   So

\Centerline{$\|x\|^k|(\zeta*h)(x)|
\le\zeta\BbbR^r\cdot\|x\|^k\sup_{\|y-x\|\le\gamma}|h(y)|
\le 2^kM\zeta\BbbR^r$}

\noindent for every $x\in\BbbR^r$.   Applying this to all the partial
derivatives of $h$, we see that $\zeta*h$ is a \rdtf.

\medskip

\quad{\bf (ii)} Again suppose that $j<r$.
Because $\zeta$ has compact support, $\int\|x\|\zeta(dx)$ is finite, so
$\Pd{}{\eta_j}\varhat{\zeta}(y)$ is defined and equal to
$-i\int\xi_je^{-iy\dotproduct x}\zeta(dx)$ for every $y\in\BbbR^r$
(cf.\ 285Fd).   More generally, whenever $j_1,\ldots,j_m<r$,

\Centerline{$
\Bover{\partial^m}{\partial\eta_{j_1}\ldots\partial\eta_{j_m}}
  \varhat{\zeta}(y)
=(-i)^m\int\xi_{j_1}\ldots\xi_{j_m}e^{-iy\dotproduct x}\zeta(dx)$,}

\noindent so all the partial derivatives of $\varhat{\zeta}$
are defined everywhere and bounded.   It follows that $\varhat{\zeta}$ is
smooth and $h\times\varhat{\zeta}$ is a \rdtf.

\medskip

{\bf (c)(i)} To begin with, suppose that $f$ is real and non-negative, and
that $\zeta$ has compact support.
Set $f_n=f\times\chi B(\tbf{0},n)$ for each $n\in\Bbb N$.   Then $f_n$ is
integrable, so $\zeta*f_n$ is defined $\mu$-a.e.;  also
$\sequencen{\zeta*f_n}$ is non-decreasing, and
$(\zeta*f)(x)=\sup_{n\in\Bbb N}(\zeta*f_n)(x)$ whenever the latter is
defined and finite.

Let $\gamma\ge 1$ be such that the support of $\zeta$ is included in
$B(\tbf{0},\gamma)$, and let $k\in\Bbb N$ be such that
$\biggerint_{\BbbR^r}\Bover1{1+\|x\|^k}f(x)\mu(dx)$ is finite.
If $y\in B(\tbf{0},\gamma)$ and $x\in\BbbR^r$, then
$\|x\|\le 2\max(\gamma,\|x+y\|)$, so
$\Bover1{1+\|x+y\|^k}\le\Bover{M}{1+\|x\|^k}$, where $M=1+2^k\gamma^k$.
It follows that

$$\eqalign{\int_{\BbbR^r}\Bover1{1+\|x\|^k}(\zeta*f_n)(x)\mu(dx)
&=\int_{\BbbR^r}\int_{B(\tbf{0},\gamma)}\Bover1{1+\|x\|^k}
  f_n(x-y)\zeta(dy)\mu(dx)\cr
&=\int_{B(\tbf{0},\gamma)}\int_{\BbbR^r}\Bover1{1+\|x\|^k}
  f_n(x-y)\mu(dx)\zeta(dy)\cr
&=\int_{B(\tbf{0},\gamma)}\int_{\BbbR^r}\Bover1{1+\|x+y\|^k}
  f_n(x)\mu(dx)\zeta(dy)\cr
&\le M\int_{B(\tbf{0},\gamma)}\int_{\BbbR^r}\Bover1{1+\|x\|^k}
  f(x)\mu(dx)\zeta(dy)}$$

\noindent for every $n\in\Bbb N$.   Consequently
$\biggerint_{\BbbR^r}\Bover1{1+\|x\|^k}(\zeta*f)(x)\mu(dx)$ is defined and
finite.

\medskip

\quad{\bf (ii)} Now suppose that $f$ is expressible as $f_1+f_{\infty}$,
where $f_1$ is $\mu$-integrable, $f_{\infty}$ is bounded and both are
real-valued and non-negative.
Adjusting $f_1$ and $f_2$ on a $\mu$-negligible set if necessary, we can
suppose that $f_{\infty}$ is Borel measurable and defined everywhere on
$\BbbR^r$.
By (a), $\zeta*f_1$ is defined and $\mu$-integrable.   Next,
$\zeta*f_{\infty}$
is defined everywhere, is bounded, and is Borel measurable (444Ia).    So
$\zeta*f\eae\zeta*f_1+\zeta*f_{\infty}$ is the sum of
a $\mu$-integrable function and a bounded Borel measurable function, and is
tempered.

\medskip

\quad{\bf (iii)} These arguments deal with the case in which $f\ge 0$.
For the general case, apply (i) or (ii) to the four parts of $f$,
as in (a-ii).

\medskip

{\bf (d)(i)} Suppose to begin with that $\zeta$ has compact support.
Let $h$ be a \rdtf.
Set $\Reverse{h}(x)=h(-x)$ for every $x\in\BbbR^r$.   Then $\Reverse{h}$
is a \rdtf, and

\Centerline{$(\zeta*\Reverse{h})(-x)
=\int\Reverse{h}(-x-y)\zeta(dy)
=\int h(x+y)\zeta(dy)$}

\noindent for every $x\in\BbbR^r$.   Accordingly

$$\eqalignno{\int(\zeta*f)\times h\,d\mu
&=\iint h(x)f(x-y)\zeta(dy)\mu(dx)\cr
&=\iint h(x)f(x-y)\mu(dx)\zeta(dy)\cr
\displaycause{because $\zeta*|f|$ is tempered, so
$\iint|h(x)f(x-y)|\zeta(dy)\mu(dx)=\int|h|\times(\zeta*|f|)d\mu$ is finite}
&=\iint h(x+y)f(x)\mu(dx)\zeta(dy)\cr
&=\iint h(x+y)f(x)\zeta(dy)\mu(dx)\cr
\displaycause{because $\iint|h(x+y)f(x)|\mu(dx)\zeta(dy)
=\iint|h(x)f(x-y)|\mu(dx)\zeta(dy)$ is finite}
&=\int f\times(\zeta*\Reverse{h})^{\scriptscriptstyle\leftrightarrow}
  d\mu
=\int g\times((\zeta*\Reverse{h})^{\scriptscriptstyle\leftrightarrow})
  \varspcheck d\mu\cr
\displaycause{because $\zeta*\Reverse{h}$ and
$(\zeta*\Reverse{h})^{\scriptscriptstyle\leftrightarrow}$ are \rdtf s,
by (b)}
&=\int g\times(\zeta*\Reverse{h})\varsphat d\mu
=(\sqrt{2\pi})^r\int g\times\varhat{\zeta}\times(\Reverse{h})\varsphat
  d\mu\cr
\displaycause{by (a)}
&=(\sqrt{2\pi})^r\int g\times\varhat{\zeta}\times\varcheck{h}\,d\mu.\cr}$$

\noindent As $h$ is arbitrary, $(\sqrt{2\pi})^rg\times\varhat{\zeta}$
represents the Fourier transform of $\zeta*f$.

\medskip

\quad{\bf (ii)} Now suppose that $f$ is expressible as $f_1+f_{\infty}$
where $f_1$ is $\mu$-integrable and $f_{\infty}$ is bounded.   By (c),
$\zeta*|f|$ is defined almost everywhere and is a tempered function.
Set $\zeta_n=\zeta\LLcorner B(\tbf{0},n)$ for each $n$.   Then
$(\sqrt{2\pi})^rg\times\varhat{\zeta}_n$ represents the Fourier transform
of $\zeta_n*f$, for each $n$.   Now $\sequencen{\varhat{\zeta}_n}$
converges uniformly to $\varhat{\zeta}$, and $\sequencen{\zeta_n*f}$
converges to $\zeta*f$ at every point at which $\zeta*|f|$ is defined and
finite, which is $\mu$-almost everywhere.   So if $h$ is a \rdtf,

$$\eqalignno{\int h\times(\sqrt{2\pi})^rg\times\varhat{\zeta}
&=\lim_{n\to\infty}\int h\times(\sqrt{2\pi})^rg\times\varhat{\zeta}_n\cr
\displaycause{the convergence is dominated by the integrable function
$(\sqrt{2\pi})^r\zeta\BbbR^r\cdot|h\times g|$}
&=\lim_{n\to\infty}\int\varhat{h}\times(\zeta_n*f)
=\int\varhat{h}\times(\zeta*f)\cr}$$

\noindent(this convergence being dominated by the integrable function
$|\varhat{h}|\times(\zeta*|f|)$).   As $h$ is arbitrary,
$(\sqrt{2\pi})^rg\times\varhat{\zeta}$
represents the Fourier transform of $\zeta*f$.
}%end of proof of 479H

\vleader{60pt}{479I}{Proposition} (In this result,
$r$ may be any integer greater than or equal to $1$.)

(a) Suppose that $0<\alpha<r$.

\quad(i) There is a tempered function representing the
Fourier transform of $k_{\alpha}$.

\quad(ii) There is a measurable function $g_0$, defined almost everywhere
on $\coint{0,\infty}$, such that $y\mapsto g_0(\|y\|)$ represents the
Fourier transform of $k_{\alpha}$.

\quad(iii) In (ii),

\Centerline{$2^{\alpha/2}\Gamma(\bover{\alpha}2)
   \int_0^{\infty}t^{r-1}g_0(t)e^{-\epsilon t^2}dt
=2^{(r-\alpha)/2}\Gamma(\bover{r-\alpha}2)
   \int_0^{\infty}t^{\alpha-1}e^{-\epsilon t^2}dt$}

\noindent for every $\epsilon>0$.

\quad(iv) $2^{\alpha/2}\Gamma(\bover{\alpha}2)g_0(t)
=2^{(r-\alpha)/2}\Gamma(\bover{r-\alpha}2)t^{\alpha-r}$ for
almost every $t>0$.

\quad(v) $2^{(r-\alpha)/2}\Gamma(\bover{r-\alpha}2)k_{r-\alpha}$
represents the Fourier transform
of $2^{\alpha/2}\Gamma(\bover{\alpha}2)k_{\alpha}$.

(b) Suppose that
$\zeta_1$, $\zeta_2$ are totally finite Radon measures on $\BbbR^r$, and
$0<\alpha<r$.   If $\zeta_1*k_{\alpha}=\zeta_2*k_{\alpha}\,\,\mu$-a.e.,
then $\zeta_1=\zeta_2$.

\proof{{\bf (a)(i)} Set $\beta=\bover12(\alpha+r)$.
Then $k_{\beta}$ is expressible as $f_1+f_2$ where
$f_1$ is integrable and $f_2$ is square-integrable.
\Prf\

\Centerline{$\int_{B(\tbf{0},1)}k_{\beta}d\mu
=r\beta_r\int_0^1\Bover{t^{r-1}}{t^{\beta}}dt$}

\noindent is finite because $\beta<r$;

\Centerline{$\int_{\BbbR^r\setminus B(\tbf{0},1)}k_{\beta}^2d\mu
=r\beta_r\int_1^{\infty}\Bover{t^{r-1}}{t^{2\beta}}dt$}

\noindent is finite because $2\beta>r$.   So we can take
$f_1=k_{\alpha}\times\chi B(\tbf{0},1)$ and $f_2=k_{\alpha}-f_1$.\ \Qed

479G tells us that there is a constant $c$ such that

\Centerline{$k_{\alpha}
=ck_{\beta}*k_{\beta}
=c(f_1*f_1+2f_1*f_2+f_2*f_2)$.}

\noindent Now $f_1*f_1$ is integrable and $f_1*f_2$ is square-integrable
(444Ra), so both have Fourier transforms represented by tempered functions;
while the continuous function $f_2*f_2$ also has a Fourier transform
represented by an integrable function (284Wi).   Assembling these,
$k_{\alpha}$ has a Fourier transform represented by a tempered function.

\medskip

\quad{\bf (ii)} We can
therefore represent the Fourier transform of $k_{\alpha}$ by
the function $g$, where

\Centerline{$g(y)=\lim_{n\to\infty}\Bover1{(\sqrt{2\pi})^r}
  \int e^{-iy\dotproduct x}e^{-\|x\|^2/n}k_{\alpha}(x)\mu(dx)$}

\noindent is defined $\mu$-almost everywhere
(284M/284Wg).   Now suppose that $T:\BbbR^r\to\BbbR^r$ is any orthogonal
transformation, and that $y\in\dom g$.   Then

$$\eqalignno{g(y)
&=\lim_{n\to\infty}\Bover1{(\sqrt{2\pi})^r}
  \int e^{-iy\dotproduct x}e^{-\|x\|^2/n}k_{\alpha}(x)\mu(dx)\cr
&=\lim_{n\to\infty}\Bover1{(\sqrt{2\pi})^r}
  \int e^{-iy\dotproduct T\trs x}e^{-\|T\trs x\|^2/n}k_{\alpha}(T\trs x)
  \mu(dx)\cr
\displaycause{because the transpose $T\trs$ of $T$
acts as an automorphism of $(\BbbR^r,\mu)$}
&=\lim_{n\to\infty}\Bover1{(\sqrt{2\pi})^r}
  \int e^{-iTy\dotproduct x}e^{-\|x\|^2/n}k_{\alpha}(x)\mu(dx),\cr}$$

\noindent and $g(Ty)$ is defined and equal to $g(y)$.   So we can set
$g_0(t)=g(y)$ whenever $y\in\dom g$ and $\|y\|=t$, and we shall have
$y\mapsto g_0(\|y\|)$ representing the Fourier transform of $k_{\alpha}$.

\medskip

\quad{\bf (iii)} If $\epsilon>0$, then $x\mapsto e^{-\epsilon\|x\|^2}$ is a
\rdtf, and its Fourier transform is the function
$x\mapsto\Bover1{(\sqrt{2\epsilon})^r}e^{-\|x\|^2/4\epsilon}$
(283N/283Wi\formerly{2{}83We}).   We therefore have

\Centerline{$\int g_0(\|y\|)e^{-\epsilon\|y\|^2}\mu(dy)
=\Bover1{(\sqrt{2\epsilon})^r}\int k_{\alpha}(x)e^{-\|x\|^2/4\epsilon}
   \mu(dx)$,}

\noindent that is,

\Centerline{$r\beta_r\int_0^{\infty}t^{r-1}g_0(t)e^{-\epsilon t^2}dt
=\Bover{r\beta_r}{(\sqrt{2\epsilon})^r}\int_0^{\infty}
  \Bover{t^{r-1}}{t^{\alpha}}e^{-t^2/4\epsilon}dt$;}

\noindent simplifying,

$$\eqalignno{\int_0^{\infty}t^{r-1}g_0(t)e^{-\epsilon t^2}dt
&=\Bover1{(\sqrt{2\epsilon})^r}\int_0^{\infty}
  t^{r-1-\alpha}e^{-t^2/4\epsilon}dt\cr
&=\Bover{2^{r-\alpha}}{2\cdot 2^{r/2}\epsilon^{\alpha/2}}
  \int_0^{\infty}u^{(r-\alpha-2)/2}e^{-u}du\cr
\displaycause{substituting $u=t^2/4\epsilon$}
&=\Bover{2^{r-\alpha}}{2\cdot 2^{r/2}\epsilon^{\alpha/2}}
  \Gamma(\Bover{r-\alpha}2).\cr}$$

\noindent On the other hand,

\Centerline{$\int_0^{\infty}t^{\alpha-1}e^{-\epsilon t^2}dt
=\int_0^{\infty}
  \Bover{(\sqrt u)^{\alpha-2}}{2\epsilon^{\alpha/2}}
  e^{-u}du
=\Bover1{2\epsilon^{\alpha/2}}\Gamma(\Bover{\alpha}2)$.}

\noindent Putting these together,

\Centerline{$2^{\alpha/2}\Gamma(\bover{\alpha}2)
   \int_0^{\infty}t^{r-1}g_0(t)e^{-\epsilon t^2}dt
=2^{(r-\alpha)/2}\Gamma(\bover{r-\alpha}2)
   \int_0^{\infty}t^{(r-1)-(r-\alpha)}e^{-\epsilon t^2}dt$}

\noindent for every $\epsilon>0$.

\medskip

\quad{\bf (iv)} Set

\Centerline{$g_1(t)=t^{r-1}e^{-t^2}
  \bigl(2^{\alpha/2}\Gamma(\bover{\alpha}2)g_0(t)
    -2^{(r-\alpha)/2}\Gamma(\bover{r-\alpha}2)t^{\alpha-r}\bigr)$}

\noindent for $t>0$.   Then $g_1$ is integrable and
$\int_0^{\infty}g_1(t)e^{-\epsilon t^2}dt=0$ for every $\epsilon\ge 0$.
It follows that $g_1=0$ a.e.   \Prf\ Consider the linear span $A$ of the
functions $t\mapsto e^{-\epsilon t^2}$ for $\epsilon\ge 0$.   This is a
subalgebra of $C_b(\coint{0,\infty})$ containing the constant functions and
separating the points of $\coint{0,\infty}$.   It follows that for every
$\gamma\ge 0$, $\delta>0$ and $h\in C_b(\coint{0,\infty})$, there
is an $f\in A$ such that $|f(t)-h(t)|\le\delta$ for $t\in[0,\gamma]$ and
$\|f\|_{\infty}\le\|h\|_{\infty}$ (281E).   Since
$\int_0^{\infty}g_1\times f=0$, we must have

\Centerline{$|\int_0^{\infty}g_1\times h|
\le\delta\|g_1\|_1+2\|h\|_{\infty}\int_{\gamma}^{\infty}|g_1(t)|dt$.}

\noindent As $\delta$ and $\gamma$ are arbitrary,
$\int_0^{\infty}g_1\times h=0$;  as $h$ is arbitrary, $\int_0^ag_1=0$
for every $a\ge 0$, and $g_1$ must be zero almost
everywhere (222D).\ \Qed

Accordingly
$2^{\alpha/2}\Gamma(\bover{\alpha}2)g_0(t)
  =2^{(r-\alpha)/2}\Gamma(\bover{r-\alpha}2)t^{\alpha-r}$
for almost every $t\ge 0$.

\medskip

\quad{\bf (v)} Now

\Centerline{$y
\mapsto 2^{(r-\alpha)/2}\Gamma(\bover{r-\alpha}2)k_{r-\alpha}(y)
\eae 2^{\alpha/2}\Gamma(\bover{\alpha}2)g_0(\|y\|)$}

\noindent represents the Fourier transform of
$2^{\alpha/2}\Gamma(\bover{\alpha}2)k_{\alpha}$.

\medskip

{\bf (b)} By (a), the Fourier transform of $k_{\alpha}$ is represented by a
tempered function $g$ which is non-zero $\mu$-a.e.
As $k_{\alpha}$ is the sum of an integrable function and a bounded
function, 479Hd tells us that the Fourier
transform of $\zeta_1*k_{\alpha}$ is represented by
$(\sqrt{2\pi})^r\varhat{\zeta}_1\times g$;  and similarly for $\zeta_2$.
As $\zeta_1*k_{\alpha}\eae\zeta_2*k_{\alpha}$,
$\varhat{\zeta}_1\times g\eae\varhat{\zeta}_2\times g$ (284Ib) and
$\varhat{\zeta}_1\eae\varhat{\zeta}_2$.   Since $\varhat{\zeta}_1$ and
$\varhat{\zeta}_2$ are both continuous (285Fb), they are equal everywhere;
in particular,

\Centerline{$\zeta_1\BbbR^r=\varhat{\zeta}_1(0)
=\varhat{\zeta}_2(0)=\zeta_2\BbbR^r$.}

\noindent If $\zeta_1=\zeta_2$ is the zero measure, we can stop.
Otherwise, they can be expressed as $\gamma\zeta'_1$ and $\gamma\zeta'_2$
where $\zeta'_1$ and $\zeta'_2$ are probability measures and $\gamma>0$.
In this case, $\zeta'_1$ and $\zeta'_2$ have the same characteristic
function (285D) and must be equal (285M);  so $\zeta_1=\zeta_2$, as
claimed.
}%end of proof of 479I

\cmmnt{\medskip

\noindent{\bf Remark} The functions $\zeta*k_{\alpha}$ are called {\bf
Riesz potentials}.
}

\leader{479J}{}\cmmnt{ Now I return to the study of
Newtonian potential when $r\ge 3$.

\medskip

\noindent}{\bf Lemma} (a) Let $\zeta$ be a totally finite Radon measure on
$\BbbR^r$.   Let $U_{\zeta}$ be the $(r-1)$-potential of $\zeta$ and
$W_{\zeta}$ the Newtonian potential of $\zeta$;  let
$k_{r-1}$ and $k_{r-2}$ be the Riesz kernels.
Then $U_{\zeta}\eae\zeta*k_{r-1}$ and $W_{\zeta}\eae\zeta*k_{r-2}$.

(b) Let $\zeta$,
$\zeta_1$ and $\zeta_2$ be totally finite Radon measures on $\BbbR^r$.

\quad(i) $\int_{\BbbR^r}W_{\zeta_1}d\zeta_2
=\int_{\BbbR^r}W_{\zeta_2}d\zeta_1
=\Bover1{c_r}\int_{\BbbR^r}U_{\zeta_1}\times U_{\zeta_2}d\mu$\cmmnt{,
where $c_r$ is the constant of 479G}.

\quad(ii) The energy $\energy(\zeta)$ of $\zeta$
is $\Bover1{c_r}\|U_{\zeta}\|_2^2$, counting
$\|U_{\zeta}\|_2$ as $\infty$ if $U_{\zeta}\notin\eusm L^2(\mu)$.

\quad(iii) If $\zeta=\zeta_1+\zeta_2$
then $U_{\zeta}=U_{\zeta_1}+U_{\zeta_2}$ and
$W_{\zeta}=W_{\zeta_1}+W_{\zeta_2}$;  similarly,
$U_{\alpha\zeta}=\alpha U_{\zeta}$ and $W_{\alpha\zeta}=\alpha W_{\zeta}$
for $\alpha\ge 0$.

\quad(iv) If $U_{\zeta_1}=U_{\zeta_2}\,\,\mu$-a.e., then $\zeta_1=\zeta_2$.

\quad(v) If $W_{\zeta_1}=W_{\zeta_2}\,\,\mu$-a.e., then $\zeta_1=\zeta_2$.

\quad(vi) $\zeta\BbbR^r
=\lim_{\gamma\to\infty}\Bover1{r\beta_r\gamma}
  \int_{\partial B(0,\gamma)}W_{\zeta}d\nu$.

(c) Let $M^+_{\text{R}}(\BbbR^r)$
be the set of totally finite Radon measures on
$\BbbR^r$, with its narrow topology.   Then
$\energy:M^+_{\text{R}}(\BbbR^r)\to[0,\infty]$ is lower semi-continuous.

\proof{{\bf (a)} As $k_{r-1}$ and $k_{r-2}$
are both expressible as sums
of integrable functions and bounded functions, $\zeta*k_{r-1}$ and
$\zeta*k_{r-2}$ are both defined a.e.\ (479Hc);  and now we have only to
read the definitions to see that $U_{\zeta}$ and $W_{\zeta}$ are these
convolutions with the technical adjustment that they are permitted to take
the value $\infty$.

\medskip

{\bf (b)(i)} For any $x$, $y\in\BbbR^r$,

$$\eqalign{\Bover1{\|x-y\|^{r-2}}
&=k_{r-2}(x-y)
=\Bover1{c_r}(k_{r-1}*k_{r-1})(x-y)\cr
&=\Bover1{c_r}\int_{\BbbR^r}\Bover1{\|x-y-z\|^{r-1}\|z\|^{r-1}}\mu(dz)\cr
&=\Bover1{c_r}\int_{\BbbR^r}\Bover1{\|x-z\|^{r-1}\|z-y\|^{r-1}}\mu(dz)
=\Bover1{c_r}\int_{\BbbR^r}\Bover1{\|x-z\|^{r-1}\|y-z\|^{r-1}}\mu(dz).
  \cr}$$

\noindent So

$$\eqalignno{\int_{\BbbR^r}W_{\zeta_1}d\zeta_2
&=\int_{\BbbR^r}\int_{\BbbR^r}\Bover1{\|x-y\|^{r-2}}\zeta_1(dx)\zeta_2(dy)
  \cr
&=\Bover1{c_r}\int_{\BbbR^r}\int_{\BbbR^r}\int_{\BbbR^r}
  \Bover1{\|x-z\|^{r-1}\|y-z\|^{r-1}}\mu(dz)\zeta_1(dx)\zeta_2(dy)\cr
&=\Bover1{c_r}\int_{\BbbR^r}\int_{\BbbR^r}\int_{\BbbR^r}
  \Bover1{\|x-z\|^{r-1}\|y-z\|^{r-1}}\zeta_1(dx)\zeta_2(dy)\mu(dz)\cr
&=\Bover1{c_r}\int_{\BbbR^r}U_{\zeta_1}(z)U_{\zeta_2}(z)\mu(dz)
=\int_{\BbbR^r}U_{\zeta_1}\times U_{\zeta_2}d\mu.\cr}$$

\noindent Hence (or otherwise)

\Centerline{$\int_{\BbbR^r}W_{\zeta_2}d\zeta_1
=\Bover1{c_r}\int_{\BbbR^r}U_{\zeta_2}\times U_{\zeta_1}d\mu
=\int_{\BbbR^r}W_{\zeta_1}d\zeta_2$.}

\medskip

\quad{\bf (ii)} Take $\zeta_1=\zeta_2=\zeta$ in (i).

\medskip

\quad{\bf (iii)} This is immediate from
234Hc.

\medskip

\quad{\bf (iv)-(v)} Put (a) and 479Ib together.

\medskip

\quad{\bf (vi)} For any $\gamma>0$,

$$\eqalignno{\Bover1{r\beta_r\gamma}
  \int_{\partial B(0,\gamma)}W_{\zeta}d\nu
&=\int W_{\zeta}d\lambda_{B(0,\gamma)}\cr
\displaycause{479Da}
&=\int\tilde W_{B(0,\gamma)}d\zeta\cr
\displaycause{(i) above}
&=\int\min(1,\Bover{\gamma^{r-2}}{\|x-z\|^{r-2}})\zeta(dx)\cr
\displaycause{479Da again}
&\to\zeta\BbbR^r\cr}$$

\noindent as $\gamma\to\infty$.

\medskip

{\bf (c)} The map
$\zeta\mapsto\zeta\times\zeta:M^+_{\text{R}}(\BbbR^r)
\to M^+_{\text{R}}(\BbbR^r\times\BbbR^r)$ is
continuous, by 437Ma.   Next, the function
$\lambda\mapsto\biggerint_{\BbbR^r}\Bover1{\|x-y\|^{r-2}}\lambda(d(x,y)):
M^+_{\text{R}}(\BbbR^r\times\BbbR^r)\to[0,\infty]$ is
lower semi-continuous, by 437Jg again.
So $\energy$ is the composition of a lower semi-continuous function with
a continuous function, and is lower semi-continuous (4A2B(d-ii)).
}%end of proof of 479J

\leader{479K}{Lemma} Let $K\subseteq\BbbR^r$ be a compact set, with
equilibrium measure $\lambda_K$.
Then $\lambda_KK=\capacity K=\energy(\lambda_K)$, and if
$\zeta$ is any Radon measure on $\BbbR^r$ such that
$\zeta K\ge\capacity K\ge\energy(\zeta)$, $\zeta=\lambda_K$.

\proof{{\bf (a)} We know that $\lambda_KK=\lambda_K\BbbR^r=\capacity K$
(479C(a-i)).   So if $K$ has zero capacity then $\lambda_K$ is the zero
measure
and $\energy(\lambda_K)=0$;  also the only Radon measure on $\BbbR^r$
with zero energy is
$\lambda_K$, and we can stop.   So henceforth let us suppose that
$\capacity K>0$.

Set

\Centerline{$e=\inf\{\energy(\zeta):\zeta$ is a Radon
probability measure on $\BbbR^r$ such that $\zeta K\ge\capacity K\}$.}

\noindent Because $\tilde W_K(x)\le 1$ for every $x\in\BbbR^r$
(479D(b-i)),

\Centerline{$e\le\energy(\lambda_K)=\int\tilde W_Kd\lambda_K
\le\lambda_K\BbbR^r=\capacity K$}

\noindent is finite.

\medskip

{\bf (b)} Consider the set $Q$ of Radon measures $\zeta$ on $\BbbR^r$ such
that $\zeta K=\zeta\BbbR^r=\capacity K$.   With its narrow topology,
$Q$ is homeomorphic to the set of Radon measures on $K$ of magnitude
$\capacity K$, which is compact (437R(f-ii)).
Since $\energy:Q\to[0,\infty]$ is
lower semi-continuous (479Jc), there is a $\lambda\in Q$ with
energy $e$ (4A2B(d-viii)).

In fact there is exactly one such member of $Q$.   \Prf\
Suppose that $\zeta$ is any other member of $Q$ with energy $e$.
Write $u_{\zeta}$ for the
equivalence class of $U_{\zeta}$ in $L^2$.   Then
$\bover12(\zeta+\lambda)$ belongs to $Q$ and
$U_{\bover12(\zeta+\lambda)}=\bover12(U_{\zeta}+U_{\lambda})$
(479J(b-iii)).   So, defining $c_r$ as in 479G,

$$\eqalignno{e+\Bover1{c_r}\|u_{\zeta}-u_{\lambda}\|_2^2
&\le\energy(\Bover12(\zeta+\lambda))
  +\Bover1{4c_r}\innerprod{u_{\zeta}-u_{\lambda}}{u_{\zeta}-u_{\lambda}}\cr
&=\Bover1{4c_r}\innerprod{u_{\zeta}+u_{\lambda}}{u_{\zeta}+u_{\lambda}}
  +\Bover1{4c_r}\innerprod{u_{\zeta}-u_{\lambda}}{u_{\zeta}-u_{\lambda}}\cr
\displaycause{479J(b-ii)}
&=\Bover1{2c_r}(\|u_{\zeta}\|_2^2+\|u_{\lambda}\|_2^2)
=e.\cr}$$

\noindent It follows that $\|u_{\zeta}-u_{\lambda}\|_2=0$ and
$U_{\zeta}\eae U_{\lambda}$.   Consequently $\zeta=\lambda$
(479J(b-iv)).\ \Qed

\medskip

{\bf (c)(i)} If $\zeta$ is any Radon measure on $\BbbR^r$ with finite
energy, then $\int W_{\zeta}d\lambda\ge\Bover{e\,\zeta K}{\capacity K}$.
\Prf\ If $\zeta K=0$ this is trivial.   Otherwise, set
$\zeta'=\Bover{\capacity K}{\zeta K}\zeta\LLcorner K$.   Then
$\zeta'$ has finite energy (479Fh) and belongs to $Q$, so for any
$\alpha\in[0,1]$ we have $\alpha\zeta'+(1-\alpha)\lambda\in Q$, and

$$\eqalign{c_re
&\le c_r\energy(\alpha\zeta'+(1-\alpha)\lambda)
=\|\alpha u_{\zeta'}+(1-\alpha)u_{\lambda}\|_2^2\cr
&=\alpha^2\|u_{\zeta'}\|_2^2
  +2\alpha(1-\alpha)\innerprod{u_{\zeta'}}{u_{\lambda}}
  +(1-\alpha)^2\|u_{\lambda}\|_2^2\cr
&=\alpha^2\|u_{\zeta'}\|_2^2
  +2\alpha(1-\alpha)\innerprod{u_{\zeta'}}{u_{\lambda}}
  +(1-\alpha)^2c_re\cr
&=c_re
  +2\alpha(\innerprod{u_{\zeta'}}{u_{\lambda}}-c_re)
  +\alpha^2(\|u_{\zeta'}\|_2^2
    -2\innerprod{u_{\zeta'}}{u_{\lambda}}
    +c_re).\cr}$$

\noindent It follows that
$\innerprod{u_{\zeta'}}{u_{\lambda}}-c_re\ge 0$ and

$$\eqalignno{\int W_{\zeta}d\lambda
&\ge\int W_{\zeta\LLcorner K}d\lambda\cr
\displaycause{479Fh}
&=\Bover{\zeta K}{\capacity K}\int W_{\zeta'}d\lambda
=\Bover{\zeta K}{c_r\capacity K}\int U_{\zeta'}\times U_{\lambda}d\mu\cr
\displaycause{479J(b-i)}
&=\Bover{\zeta K}{c_r\capacity K}\innerprod{u_{\zeta'}}{u_{\zeta}}
\ge\Bover{\zeta K}{c_r\capacity K}c_re
=\Bover{e\,\zeta K}{\capacity K},\cr}$$

\noindent as claimed.\ \Qed

\medskip

\quad{\bf (ii)} If $\zeta$ is any Radon measure on $\BbbR^r$ with finite
energy, then $W_{\lambda}(x)\ge\Bover{e}{\capacity K}$ for $\zeta$-almost
every $x\in K$.   \Prf\Quer\ Otherwise, set
$E=\{x:x\in K$, $W_{\lambda}(x)<\Bover{e}{\capacity K}\}$,
and consider $\zeta'=\zeta\LLcorner E$.   Then

$$\eqalignno{\int W_{\zeta'}d\lambda
&=\int W_{\lambda}d\zeta'\cr
\displaycause{479J(b-i)}
&<\Bover{e}{\capacity K}\zeta'E
\le\Bover{e\zeta'K}{\capacity K},\cr}$$

\noindent contradicting (i).\ \Bang\Qed

\medskip

\quad{\bf (iii)} $W_{\lambda}(x)=\Bover{e}{\capacity K}$ for
$\lambda$-almost every $x\in K$.   \Prf\
Since $\lambda$ has finite energy,
(ii) tells us that $W_{\lambda}(x)\ge\Bover{e}{\capacity K}$
for $\lambda$-almost every $x\in K$.   Since

\Centerline{$\int_KW_{\lambda}d\lambda\le\int W_{\lambda}d\lambda
=e=\Bover{e}{\capacity K}\lambda K$,}

\noindent we must have $W_{\lambda}(x)=\Bover{e}{\capacity K}$ for
$\lambda$-almost every $x\in K$.\ \Qed

\medskip

\quad{\bf (iv)} Since $\lambda K=\lambda\BbbR^r$, 479Fg,
with $f$ the constant function with value $\Bover{e}{\capacity K}$,
tells us that $W_{\lambda}(x)\le\Bover{e}{\capacity K}$ for every
$x\in\BbbR^r$.

\medskip

{\bf (d)} For $x\in\BbbR^r\setminus K$,
$W_{\lambda}(x)\le\Bover{e}{\capacity K}\hp(K-x)$.   \Prf\
Set $G=\BbbR^r\setminus K$, and
let $\tau$ be the Brownian exit time from $G-x$.   Define
$f:\overline{G}^{\infty}\to[0,1]$ by setting

$$\eqalign{f(y)
&=0\text{ if }y\in\partial G=\partial K,\cr
&=\Bover{e}{\capacity K}-W_{\lambda}(y)\text{ if }y\in G,\cr
&=\Bover{e}{\capacity K}\text{ if }y=\infty.\cr}$$

\noindent Because $W_{\lambda}\restr G$ is continuous
and harmonic (479Fa), so is $f\restr G$.
Because $\lambda$ has compact support,
$\lim_{y\to\infty}W_{\lambda}(y)=0$ (479Fd), so $f$ is continuous
at $\infty$;
because $W_{\lambda}(y)\le\Bover{e}{\capacity K}$ for every $y$, $f$ is
lower semi-continuous.   So

$$\eqalignno{\Bover{e}{\capacity K}-W_{\lambda}(x)
&=f(x)
\ge\Expn(f(x+X_{\tau}))\cr
\displaycause{478O, because $r\ge 3$ and $\BbbR^r$ has few wandering paths}
&=\Bover{e}{\capacity K}\Pr(\tau=\infty)
=\Bover{e}{\capacity K}(1-\Pr(\tau<\infty)).\cr}$$

\noindent Thus $W_{\lambda}(x)$ is at most
$\Bover{e}{\capacity K}\Pr(\tau<\infty)$.    But $\Pr(\tau<\infty)$ is just
the Brownian hitting probability $\hp(K-x)$.\ \Qed

\medskip

{\bf (e)} $e=\energy(\lambda_K)$.   \Prf\

$$\eqalignno{e
&\le\energy(\lambda_K)
\le\capacity K\cr
\displaycause{(a) above}
&=\lambda K
=\lambda\BbbR^r
=\lim_{\|x\|\to\infty}\|x\|^{r-2}W_{\lambda}(x)\cr
\displaycause{479Fd}
&\le\Bover{e}{\capacity K}
   \liminf_{\|x\|\to\infty}\|x\|^{r-2}\hp(K-x)\cr
\displaycause{by (d) of this proof}
&=\Bover{e}{\capacity K}\cdot\capacity K\cr
\displaycause{479B(ii)}
&=e. \text{ \Qed}\cr}$$

\medskip

{\bf (f)} From this we see at once
that $\lambda=\lambda_K$ and $\capacity K=\energy(\lambda_K)$.
Now suppose that
$\zeta$ is a Radon measure on $\BbbR^r$ such that
$\zeta K\ge\capacity K\ge\energy(\zeta)$.
Set $\zeta'=\Bover{\capacity K}{\zeta K}\zeta\LLcorner K$;  then
$\zeta'\in Q$, while $\zeta'\le\zeta$, so

\Centerline{$e=\capacity K\ge\energy(\zeta)\ge\energy(\zeta')$}

\noindent by 479Fh.   It follows that $\zeta'=\lambda$ and
$\lambda\le\zeta$.   Accordingly $W_{\lambda}\le W_{\zeta}$ (479Fh again),

$$\eqalignno{\energy(\zeta)
&=\int W_{\zeta}d\zeta
\ge\int W_{\lambda}d\zeta
\ge\int_KW_{\lambda}d\zeta
\ge\zeta K\cr
\displaycause{(c-ii) above}
&\ge\capacity K
\ge\energy(\zeta),\cr}$$

\noindent and we have equality throughout.   Since $\lambda$ is non-zero
and the kernel $(x,y)\mapsto\Bover1{\|x-y\|^{r-2}}$ is strictly positive,
$W_{\lambda}$ is strictly positive.
It follows that $\zeta(\BbbR^r\setminus K)=0$ and $\zeta\in Q$;
consequently $\zeta=\lambda=\lambda_K$, as required.
}%end of proof of 479K

\leader{479L}{}\cmmnt{ I shall wish later to quote a couple of
the facts which
appeared in the course of the proof above, and I think it will be safer to
list them now.

\medskip

\noindent}{\bf Corollary} Let $K\subseteq\BbbR^r$ be a compact set with
equilibrium potential $\tilde W_K$.

(a) If $\zeta$ is any Radon measure on $\BbbR^r$ with finite energy,
then $\tilde W_K(x)=1$ for $\zeta$-almost every $x\in K$.

(b) If $\zeta$ is a Radon measure on $\BbbR^r$ such that $W_{\zeta}\le 1$
everywhere on $K$, $\zeta K\le\capacity K$.

(c) $\tilde W_K(x)\le\hp(K-x)$ for every $x\in\BbbR^r\setminus K$.

\proof{{\bf (a)(i)} Suppose first that $\capacity K>0$.
Working through the proof of 479K, we discover, in parts
(e)-(f) of the proof, that
$e=\capacity K$ and $\lambda=\lambda_K$,
so we just have to put (c-ii) of the proof together with 479D(b-i).

\medskip

\quad{\bf (ii)} If $\capacity K=0$, let $B$ be a non-trivial closed
ball disjoint from $A$, and consider $L=K\cup B$.   Then
$\capacity B=\capacity L$ (479Ea) and
$\lambda_LK=0$, by 479D(c-ii), so

\Centerline{$\lambda_LB=\lambda_LL=\capacity L=\energy(\lambda_L)
=\capacity B$}

\noindent and $\lambda_L=\lambda_B$ (479K).   Now
$\tilde W_L=1\,\,\zeta$-a.e.\ on $L$, while

\Centerline{$\tilde W_L(x)=\tilde W_B(x)<1$}

\noindent for every $x\in\BbbR^r\setminus B$ (479Da), and in particular
for every $x\in K$;  so $K$ must be $\zeta$-negligible.

\medskip

{\bf (b)} Set $\zeta'=\zeta\LLcorner K$;  then $W_{\zeta'}\le W_{\zeta}$,
so $\energy(\zeta')=\int_KW_{\zeta'}d\zeta'\le\zeta'K$ is finite.
By (a), $\tilde W_K\ge 1\,\,\zeta'$-a.e., so

$$\eqalignno{\zeta K
&=\zeta'K
\le\int\tilde W_Kd\zeta'
=\int W_{\zeta'}d\lambda_K\cr
\displaycause{479J(b-i)}
&\le\lambda_K\BbbR^r
=\capacity K.\cr}$$

\medskip

{\bf (c)} If $\capacity K=0$ then $\lambda_K$ is the zero measure and the
result is trivial.   Otherwise, again look at the proof of 479K;  in part
(d), we saw that $W_{\lambda}(x)\le\Bover{e}{\capacity K}\hp(K-x)$;  but we
now know that $e=\capacity K$ and $\lambda=\lambda_K$, so we get
$\tilde W_K(x)\le\hp(K-x)$, as claimed.
}%end of proof of 479L

\leader{479M}{}\cmmnt{ In 479Ed we saw that there is a natural extension
of Newtonian capacity to a Choquet capacity defined on every subset of
$\BbbR^r$.   However the importance of Newtonian capacity lies as much in
the equilibrium measures and potentials as in the simple quantity of
capacity itself, and the methods of 479B-479E %479B 479C 479D 479E
do not seem to yield these by
any direct method.   With the new ideas of 479K-479L, we can now
approach the problem of defining equilibrium measures for unbounded
analytic sets of finite capacity.

\wheader{479M}{4}{2}{2}{60pt}

\noindent}{\bf Lemma} Let $A\subseteq\BbbR^r$ be an analytic set with
finite Choquet-Newton capacity $c(A)$.

(a) $\lim_{\gamma\to\infty}c(A\setminus B(\tbf{0},\gamma))=0$.

(b) $\lambda_A=\lim_{\gamma\to\infty}\lambda_{A\cap B(\tbf{0},\gamma)}$
is defined for the total variation metric on the space
$M^+_{\text{R}}(\BbbR^r)$ of totally finite Radon measures on $\BbbR^r$.

(c)(i) $\lambda_A\BbbR^r=c(A)$.

\quad(ii) $\supp(\lambda_A)\subseteq\partial A$.

\quad(iii) If $B\subseteq\BbbR^r$ is another analytic set such that
$c(B)<\infty$, then $\lambda_{A\cup B}\le\lambda_A+\lambda_B$.

(d)(i) $\tilde W_A=W_{\lambda_A}$ is the limit
$\lim_{\gamma\to\infty}\tilde W_{A\cap B(\tbf{0},\gamma)}
=\sup_{\gamma\ge 0}\tilde W_{A\cap B(\tbf{0},\gamma)}$.

\quad(ii) $\tilde W_A(x)\le 1$ for every $x\in\BbbR^r$.

\quad(iii) If $\zeta$ is any Radon measure on $\BbbR^r$ with finite energy,
$\tilde W_A(x)=1$ for $\zeta$-almost every $x\in A$.

\quad(iv) $\energy(\lambda_A)=c(A)$.

\proof{{\bf (a)} \Quer\ Otherwise, set

\Centerline{$\alpha=\lim_{\gamma\to\infty}c(A\setminus B(\tbf{0},\gamma))
=\inf_{\gamma>0}c(A\setminus B(\tbf{0},\gamma))>0$.}

\noindent Set $\epsilon=\bover18\alpha$
and $\delta=\bover32\sqrt{\alpha}$.
Let $\gamma$ be such that
$c(A\setminus B(\tbf{0},\gamma))\le\alpha+\epsilon$, and let
$K\subseteq A\setminus B(\tbf{0},\gamma)$ be a compact set such that
$\capacity K\ge\alpha-\epsilon$ (479E(d-iii)).   Let $\gamma'$ be such that
$K\subseteq B(\tbf{0},\gamma')$, and
let $L\subseteq A\setminus B(\tbf{0},\gamma'+\delta)$ be a compact set such that
$\capacity L\ge\alpha-\epsilon$.

Set $\zeta=\bover23(\lambda_K+\lambda_L)$.   Then
$W_{\zeta}=\bover23(\tilde W_K+\tilde W_L)$.   If $x\in K$, then
$\|x-y\|\ge\delta$ for every $y\in L$, so

\Centerline{$\tilde W_L(x)
\le\Bover1{\delta^2}\lambda_LL\le\Bover{\alpha+\epsilon}{\delta^2}
=\Bover12$;}

\noindent similarly, $\tilde W_K(x)\le\Bover12$ for every $x\in L$.
So $W_{\zeta}(x)\le 1$ for every $x\in K\cup L$, and therefore for every
$x\in\BbbR^r$, by 479Fg.   But this means that

$$\eqalignno{c(A\setminus B(\tbf{0},\gamma))
&\ge\capacity(K\cup L)
\ge\zeta(K\cup L)\cr
\displaycause{479Lb}
&=\Bover23(\capacity K+\capacity L)
\ge\Bover43(\alpha-\epsilon)
>\alpha+\epsilon,\cr}$$

\noindent which is impossible.\ \Bang

\medskip

{\bf (b)} For $\gamma\ge 0$ set
$\alpha_{\gamma}=c(A\setminus B(\tbf{0},\gamma))$
and $\zeta_{\gamma}=\lambda_{A\cap B(\tbf{0},\gamma)}$.
If $0\le\gamma\le\gamma'$ and $E\subseteq\BbbR^r$ is Borel, then
$|\zeta_{\gamma}E-\zeta_{\gamma'}E|\le\alpha_{\gamma}$.
\Prf\

$$\eqalignno{\zeta_{\gamma'}E
&\le\zeta_{\gamma}E+\lambda_{E\cap B(\tbf{0},\gamma')\setminus B(\tbf{0},\gamma)}(K)\cr
\displaycause{479D(c-i)}
&\le\zeta_{\gamma}E+c(A\cap B(\tbf{0},\gamma')\setminus B(\tbf{0},\gamma))
\le\zeta_{\gamma}E+c(A\setminus B(\tbf{0},\gamma))
=\zeta_{\gamma}E+\alpha_{\gamma}.\cr}$$

\noindent On the other side we now have

$$\eqalign{\zeta_{\gamma}E
&=c(A\cap B(\tbf{0},\gamma))-\zeta_{\gamma}(\BbbR^r\setminus E)\cr
&\le c(A\cap B(\tbf{0},\gamma'))-\zeta_{\gamma'}(\BbbR^r\setminus E)
   +\alpha_{\gamma}
=\zeta_{\gamma'}E+\alpha_{\gamma}.\cr}$$

\noindent So
$|\zeta_{\gamma}E-\zeta_{\gamma'}E|\le\alpha_{\gamma}$.\ \QeD\   It follows
at once that $\rhotv(\zeta_{\gamma},\zeta_{\gamma'})\le 2\alpha_{\gamma}$.

Since $\lim_{\gamma\to\infty}\alpha_{\gamma}=0$, by (a),
$\sequencen{\zeta_n}$ is a Cauchy sequence for $\rhotv$.
As noted in 437Q(a-iii), $M^+_{\text{R}}(\BbbR^r)$ is complete, so
$\lim_{\gamma\to\infty}\lambda_{A\cap B(\tbf{0},\gamma)}
=\lim_{n\to\infty}\zeta_n$ is defined, and we have our measure $\lambda_A$.

\medskip

{\bf (c)(i)} Now

\Centerline{$\lambda_A\BbbR^r
=\lim_{n\to\infty}\lambda_{A\cap B(\tbf{0},n)}(\BbbR^r)
=\lim_{n\to\infty}c(A\cap B(\tbf{0},n))=c(A)$.}

\medskip

\quad{\bf (ii)} For any $\gamma\ge 0$,

$$\eqalignno{\lambda_{A\cap B(\tbf{0},\gamma)}(\BbbR^r\setminus\partial A)
&\le\lambda_{A\cap B(\tbf{0},\gamma)}(\partial B(\tbf{0},\gamma))\cr
\displaycause{because the support of
$\lambda_{A\cap B(\tbf{0},\gamma)}$ is included in
$\partial(A\cap B(\tbf{0},\gamma))\subseteq\partial A\cup\partial B(\tbf{0},\gamma)$}
&\le|\lambda_A(\partial B(\tbf{0},\gamma))
  -\lambda_{A\cap B(\tbf{0},\gamma)}(\partial B(\tbf{0},\gamma))|
  +\lambda_A(\partial B(\tbf{0},\gamma))\cr
&\le\rhotv(\lambda_A,\lambda_{A\cap B(\tbf{0},\gamma)})
  +\lambda_A(\partial B(\tbf{0},\gamma)).\cr}$$

\noindent So

$$\eqalign{\lambda_A(\BbbR^r\setminus\partial A)
&=\lim_{\gamma\to\infty}
  \lambda_{A\cap B(\tbf{0},\gamma)}(\BbbR^r\setminus\partial A)\cr
&\le\lim_{\gamma\to\infty}\rhotv(\lambda_A,\lambda_{A\cap B(\tbf{0},\gamma)})
  +\lim_{\gamma\to\infty}\lambda_A(\partial B(\tbf{0},\gamma))
=0.\cr}$$

\medskip

\quad{\bf (iii)} For any compact set $K\subseteq\BbbR^r$,

$$\eqalignno{\lambda_{A\cup B}(K)
&=\lim_{\gamma\to\infty}\lambda_{(A\cup B)\cap B(\tbf{0},\gamma)}(K)
\le\lim_{\gamma\to\infty}
  \lambda_{A\cap B(\tbf{0},\gamma)}(K)+\lambda_{B\cap B(\tbf{0},\gamma)}(K)\cr
\displaycause{479D(c-i)}
&=\lambda_A(K)+\lambda_B(K)
=(\lambda_A+\lambda_B)(K).\cr}$$

\noindent By 416Ea, $\lambda_{A\cup B}\le\lambda_A+\lambda_B$.

\medskip

{\bf (d)(i)} By 479D(b-ii), the supremum and the limit are the same.
Suppose that $x\in\BbbR^r$ and $\epsilon>0$.
Start with $\gamma>\|x\|$.   Since $\tilde W_{A\cap B(\tbf{0},\gamma)}(x)$ is
finite, there is a $\delta\in\ooint{0,\gamma-\|x\|}$ such that
$\biggerint_{B(\tbf{0},\delta)}\Bover1{\|x-y\|^{r-2}}
  \lambda_{A\cap B(\tbf{0},\gamma)}(dy)\le\epsilon$.
If $\gamma'\ge\gamma\ge 0$, then

\Centerline{$\lambda_{A\cap B(\tbf{0},\gamma')}
\le\lambda_{A\cap B(\tbf{0},\gamma)}
  +\lambda_{A\cap B(\tbf{0},\gamma')\setminus B(\tbf{0},\gamma)}$,}

\noindent so

$$\eqalignno{\int_{B(\tbf{0},\delta)}\Bover1{\|x-y\|^{r-2}}
   \lambda_{A\cap B(\tbf{0},\gamma')}(dy)
&\le\int_{B(\tbf{0},\delta)}\Bover1{\|x-y\|^{r-2}}
  \lambda_{A\cap B(\tbf{0},\gamma)}(dy)\cr
&\mskip75mu
  +\int_{B(\tbf{0},\delta)}\Bover1{\|x-y\|^{r-2}}
   \lambda_{A\cap B(\tbf{0},\gamma')\setminus B(\tbf{0},\gamma)}(dy)\cr
\displaycause{234Hc, 234Qc}
% if \mu\le\nu then \int fd\mu\le\int fd\nu whenever f\ge 0 and
% latter is defined in [0,\infty]
&=\int_{B(\tbf{0},\delta)}\Bover1{\|x-y\|^{r-2}}
   \lambda_{A\cap B(\tbf{0},\gamma)}(dy)\cr
\displaycause{because $\interior B(\tbf{0},\gamma)$ is
$\lambda_{A\cap B(\tbf{0},\gamma')\setminus B(\tbf{0},\gamma)}$-negligible}
&\le\epsilon.\cr}$$

\noindent So, setting $M=\Bover1{\delta^{r-2}}$,

\Centerline{$|\tilde W_{A\cap B(\tbf{0},\gamma')}(x)
  -\int\min(M,\Bover1{\|x-y\|^{r-2}})\lambda_{A\cap B(\tbf{0},\gamma')}(dy)|
\le\epsilon$.}

\noindent Using (c-ii) and (c-iii) to apply
the same argument with $A$ in place of $A\cap B(\tbf{0},\gamma')$, we get

\Centerline{$|\tilde W_A(x)
  -\int\min(M,\Bover1{\|x-y\|^{r-2}})\lambda_A(dy)|
\le\epsilon$.}

\noindent On the other hand,

\Centerline{$\int\min(M,\Bover1{\|x-y\|^{r-2}})\lambda_A(dy)
=\lim_{\gamma'\to\infty}\int\min(M,\Bover1{\|x-y\|^{r-2}})
  \lambda_{A\cap B(\tbf{0},\gamma')}(dy)$}

\noindent (437Q(a-ii)), so

\Centerline{$\limsup_{\gamma'\to\infty}|\tilde W_{A\cap B(\tbf{0},\gamma')}(x)
  -\tilde W_A(x)|\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$W_A(x)=\lim_{\gamma'\to\infty}W_{A\cap B(\tbf{0},\gamma')}(x)$, as claimed.

\medskip

\quad{\bf (ii)} It follows at once that $\tilde W_A\le 1$ everywhere.

\medskip

\quad{\bf (iii)} Write $E=\{x:x\in A,\,\tilde W_A(x)<1\}$, and let $\zeta$
be a Radon measure on $\BbbR^r$ of finite energy.   \Quer\ If
$\zeta E>0$, there is a compact set $K\subseteq E$ such that
$\zeta K>0$.   Now there is a $\gamma>0$ such that
$K\subseteq B(\tbf{0},\gamma)$, in which case

\Centerline{$\tilde W_K(x)\le\tilde W_{A\cap B(\tbf{0},\gamma)}(x)<1$}

\noindent for every $x\in K$, and $\zeta K=0$, by 479La.\ \BanG\  So
$\zeta E=0$, as required.

\medskip

\quad{\bf (iv)} By (ii) and (c-i),

\Centerline{$\energy(\lambda_A)=\int\tilde W_Ad\lambda_A
\le\lambda_A\BbbR^r=c(A)$.}

\noindent In the other direction, for any $\gamma\ge 0$,

$$\eqalignno{\energy(\lambda_A)
&=\int\tilde W_Ad\lambda_A
\ge\int\tilde W_{A\cap B(\tbf{0},\gamma)}d\lambda_A
=\int\tilde W_Ad\lambda_{A\cap B(\tbf{0},\gamma)}\cr
\displaycause{479J(b-i)}
&\ge\int\tilde W_{A\cap B(\tbf{0},\gamma)}d\lambda_{A\cap B(\tbf{0},\gamma)}
=c(A\cap B(\tbf{0},\gamma));\cr}$$

\noindent taking the limit as $\gamma\to\infty$,
$\energy(\lambda_A)\ge c(A)$ and we have equality.
}%end of proof of 479M

\leader{479N}{}\cmmnt{ We are ready to match the definitions
in 479C to some alternative definitions of capacity.

\medskip

\noindent}{\bf Theorem} Let $A\subseteq\BbbR^r$ be an analytic set with
finite Choquet-Newton capacity $c(A)$.

(a) Writing $W_{\zeta}$ for the Newtonian potential of a Radon measure
$\zeta$ on $\BbbR^r$,

\Centerline{$c(A)
=\sup\{\zeta A:\zeta\text{ is a Radon measure on }\BbbR^r,\,
  W_{\zeta}(x)\le 1\text{ for every }x\in\BbbR^r\}$;}

\noindent if $A$ is closed, the supremum is attained.

(b) $c(A)
=\inf\{\energy(\zeta):\zeta\text{ is a Radon measure on }\BbbR^r,\,
  \zeta A\ge c(A)\}$;
if $A$ is closed, the infimum is attained.

(c) If $A\ne\emptyset$, $c(A)
=\sup\{\Bover1{\energy(\zeta)}:\zeta$ is a Radon measure on
$\BbbR^r$ such that $\zeta A=1\}$, counting $\Bover1{\infty}$ as zero;
if $A$ is closed, the supremum is attained.

\proof{ Note first that if there is
a Radon measure $\zeta$ on $\BbbR^r$, with finite energy, such that
$\zeta A>0$, then $c(A)>0$.   \Prf\ By 479M(d-iii),
$\tilde W_A=1\,\,\zeta$-a.e.\ on $A$.   So $\tilde W_A$ cannot be
identically $0$, and $0<\lambda_A\BbbR^r=c(A)$, by 479M(c-i).\ \Qed

\medskip

{\bf (a)(i)} We know from 479E(d-i) and 479D(b-i) that

$$\eqalign{c(A)
&=\sup\{\capacity K:K\subseteq A\text{ is compact}\}
=\sup\{\lambda_KK:K\subseteq A\text{ is compact}\}\cr
&=\sup\{\lambda_KA:K\subseteq A\text{ is compact}\}
\le\sup\{\zeta A:W_{\zeta}\le\chi\BbbR^r\}.\cr}$$

\medskip

\quad{\bf (ii)} If $\zeta$ is a Radon measure on $\BbbR^r$ and
$W_{\zeta}\le\chi\BbbR^r$, then

$$\eqalignno{\zeta A
&=\sup_{K\subseteq A\text{ is compact}}\zeta K
\le\sup_{K\subseteq A\text{ is compact}}\capacity K\cr
\displaycause{479Lb}
&=c(A).\cr}$$

\noindent Thus $\sup\{\zeta A:W_{\zeta}\le\chi\BbbR^r\}\le c(A)$ and we
have equality.

\medskip

\quad{\bf (iii)} If $A$ is closed, then by 479M(c-ii)

\Centerline{$\lambda_AA=\lambda_A(\partial A)=\lambda_A\BbbR^r
=c(A)$}

\noindent so $\lambda_A$ witnesses that the supremum is attained.

\medskip

{\bf (b)(i)} \Quer\ Suppose, if possible, that there is a Radon
measure $\zeta$ on $\BbbR^r$ such
that $\zeta A\ge c(A)>\energy(\zeta)$.   Let $\alpha\in\ooint{0,1}$
be such that $\alpha^4c(A)\ge\energy(\zeta)$.   Since

\Centerline{$\zeta A=\sup\{\zeta K:K\subseteq A$ is compact$\}$,
\quad$c(A)=\sup\{\capacity K:K\subseteq A$ is compact$\}$,}

\noindent there is a compact $K\subseteq A$ such that
$\zeta K\ge\alpha\zeta A$ and $\capacity K>\alpha^2c(A)$.
Set $\zeta'=\Bover{\capacity K}{\zeta K}\zeta$.   Then

$$\eqalign{\energy(\zeta')
&=\bigl(\Bover{\capacity K}{\zeta K}\bigr)^2\energy(\zeta)
\le\bigl(\Bover{c(A)}{\alpha\zeta A}\bigr)^2\alpha^4c(A)\cr
&\le\alpha^2c(A)
<\capacity K
=\zeta'K;\cr}$$

\noindent which is impossible, by 479K.\ \Bang

So $c(A)\le\inf\{\energy(\zeta):\zeta A\ge c(A)\}$.

\medskip

\quad{\bf (ii)} Take any $\epsilon>0$.
Then there is a compact set $K\subseteq A$ such that
$(1+\epsilon)\capacity K\ge c(A)$.   Set
$\zeta=(1+\epsilon)\lambda_K$;  then

\Centerline{$\zeta A\ge c(A)$,
\quad$\energy(\zeta)=(1+\epsilon)^2\energy(\lambda_K)
=(1+\epsilon)^2\capacity K\le(1+\epsilon)^2c(A)$.}

\noindent As $\epsilon$ is arbitrary,
$c(A)\ge\inf\{\energy(\zeta):\zeta A\ge c(A)\}$
and we have equality.

\medskip

\quad{\bf (iii)} If $A$ is closed, then

\Centerline{$\lambda_AA=\lambda_A(\partial A)=\lambda_A\BbbR^r=c(A)$}

\noindent by 479M(c-i) and (c-ii), while $\energy(\lambda_A)=c(A)$ by
479M(d-iv).   So $\lambda_A$ witnesses that
$c(A)=\min\{\energy(\zeta):\zeta A\ge c(A)\}$.

\medskip

{\bf (c)(i)} Suppose that
$\zeta$ is a Radon measure on $\BbbR^r$
such that $\zeta A=1$.   If $\energy(\zeta)=\infty$ then of course
$\Bover1{\energy(\zeta)}\le c(A)$.   Otherwise, $c(A)>0$, as
remarked at the beginning of this part of the proof.   Set
$\zeta'=c(A)\zeta$.   By (b),

\Centerline{$c(A)\le\energy(\zeta')=c(A)^2\energy(\zeta)$,}

\noindent so $\Bover1{\energy(\zeta)}\le c(A)$.

Thus $\sup\{\Bover1{\energy(\zeta)}:\zeta A=\zeta\BbbR^r=1\}\le c(A)$.

\medskip

\quad{\bf (ii)} If $c(A)=0$ then the supremum is attained by any
Radon measure $\zeta$ such that $\zeta A=1$, so we can stop.
If $c(A)>0$, then for any $\alpha\in\ooint{0,1}$ there is a compact set
$K\subseteq A$ such that $\capacity K\ge\alpha c(A)$.   Set
$\zeta=\Bover1{\capacity K}\lambda_K$;  then

\Centerline{$\zeta K=\zeta\BbbR^r=\zeta A=1$}

\noindent and

\Centerline{$\Bover1{\energy(\zeta)}
=\Bover{(\capacity K)^2}{\energy(\lambda_K)}
=\capacity K\ge\alpha c(A)$.}

\noindent As $\alpha$ is arbitrary,
$c(A)\le\sup\{\Bover1{\energy(\zeta)}:\zeta A=\zeta\BbbR^r=1\}$ and
we have equality.

\medskip

\quad{\bf (iii)} If $A$ is closed and $c(A)>0$, then
$\zeta=\Bover1{c(A)}\lambda_A$ witnesses that the supremum is
attained, as in (b) above.
}%end of proof of 479N

\leader{479O}{Polar \dvrocolon{sets}}\cmmnt{ To make the final step, to
arbitrary sets with finite Choquet-Newton capacity, we seem to
need an alternative description of polar sets.

\medskip

\noindent}{\bf Proposition} For a set $D\subseteq\BbbR^r$,
the following are equiveridical:

(i) $D$ is polar\cmmnt{, that is, $c(D)=0$};

(ii) there is a totally finite Radon measure $\zeta$ on $\BbbR^r$ such that
its Newtonian potential $W_{\zeta}$ is infinite at every point of $D$;

(iii) there is an analytic set $E\supseteq D$ such that
$\zeta E=0$ whenever $\zeta$ is a Radon measure on $\BbbR^r$ with
finite energy.

\proof{{\bf (i)$\Rightarrow$(ii)} If (i) is true, then for each
$n\in\Bbb N$ there is a bounded open set
$G_n\supseteq D\cap B(\tbf{0},n)$ such that
$c(G_n)\le 2^{-n}$.   Try $\zeta=\sum_{n=0}^{\infty}\lambda_{G_n}$,
defining the sum as in 234G.   Then
$\zeta\BbbR^r=\sum_{n=0}^{\infty}c(G_n)$ is finite, and
$W_{\zeta}=\sum_{n=0}^{\infty}\tilde W_{G_n}$ (234Hc).
If $x\in D\cap B(\tbf{0},n)$, then $\tilde W_{G_m}(x)=1$ for every $m\ge n$
(479D(b-iii)), so $W_{\zeta}(x)=\infty$.   Thus $\zeta$ witnesses that (ii)
is true.

\medskip

{\bf (ii)$\Rightarrow$(iii)} Suppose that $\lambda$ is a totally finite
Radon measure such that $W_{\lambda}(x)=\infty$ for every $x\in D$.
Set $E=\{x:W_{\lambda}(x)=\infty\}$;  then $E$ is a G$_{\delta}$ set,
because $W_{\lambda}$ is lower semi-continuous (479Fa).
\Quer\ If there is a Radon
measure $\zeta$ on $\BbbR^r$, with finite energy, such that $\zeta E>0$,
let $K\subseteq E$ be a compact set such that $\zeta K>0$.   Set
$\zeta_1=\Bover1{\zeta K}\zeta\LLcorner K$;  then $\zeta_1$ has finite
energy and $\zeta_1K=1$, so $\capacity K\ge\Bover1{\energy(\zeta_1)}>0$,
by 479Nc.

Let $G\supseteq K$ be a bounded open set;  set
$\lambda_1=\lambda\LLcorner G$ and
$\lambda_2=\lambda\LLcorner(\BbbR^r\setminus G)$, so that
$\lambda=\lambda_1+\lambda_2$ and $W_{\lambda}=W_{\lambda_1}+W_{\lambda_2}$
(234Hc).   Since $W_{\lambda_2}(x)$ is finite for $x\in G$ (479Fa),
$W_{\lambda_1}(x)=\infty$ for every $x\in K$.   Let $\epsilon>0$ be such
that $\epsilon\lambda_1\BbbR^r<\capacity K$.   Then
$\epsilon W_{\lambda_1}$ is a lower
semi-continuous superharmonic function greater than or equal to
$\tilde W_K$ on $K\supseteq\supp(\lambda_K)$, so
$\epsilon W_{\lambda_1}\ge\tilde W_K$ everywhere (479Fg).   But this
means that

$$\eqalignno{\epsilon\lambda_1\BbbR^r
&=\epsilon\lim_{\|x\|\to\infty}\|x\|^{r-2}W_{\lambda_1}(x)\cr
\displaycause{479Fd}
&\ge\lim_{\|x\|\to\infty}\|x\|^{r-2}\tilde W_K(x)
=\lambda_K\BbbR^r
=\capacity K
>\epsilon\lambda_1\BbbR^r,\cr}$$

\noindent which is absurd.\ \Bang

So $E$ witnesses that (iii) is true.

\medskip

{\bf (iii)$\Rightarrow$(i)} Suppose that $E\supseteq D$ is analytic and
that $\zeta E=0$ whenever $\energy(\zeta)$ is finite.   If
$K\subseteq E$ is compact and $\zeta$ is a Radon probability measure on
$\BbbR^r$ such that $\zeta K=1$, then $\energy(\zeta)$ must be infinite;
by 479Nc, $\capacity K=0$.   As $K$ is arbitrary, $c(E)=0$ and
$c(D)=0$.
}%end of proof of 479O

\leader{479P}{}\cmmnt{ At last I come to my final extension of the
notions of equilibrium measure and potential, together with a direct
expression of the latter in terms of Brownian hitting probabilities.

\medskip

\noindent}{\bf Theorem} Let $D\subseteq\BbbR^r$ be a set with finite
Choquet-Newton capacity $c(D)$.

(a) There is a totally finite Radon
measure $\lambda_D$ on $\BbbR^r$ such that
$\lambda_D=\lambda_A$, as defined in 479Mb, whenever
$A\supseteq D$ is analytic and $c(A)=c(D)$.

(b) Write
$\tilde W_D=W_{\lambda_D}$ for the equilibrium potential
corresponding to the equilibrium measure $\lambda_D$.   Then
$\tilde W_D(x)=\hp^*((D\setminus\{x\})-x)$ for every $x\in\BbbR^r$.

(c)(i)($\alpha$) $\lambda_D\BbbR^r=c(D)$;

\qquad($\beta$) if $\zeta$ is any Radon measure on $\BbbR^r$ with finite energy,
$\tilde W_D(x)=1$ for $\zeta$-almost every $x\in D$;

\qquad($\gamma$) $\energy(\lambda_D)=c(D)$;

\qquad($\delta$) if $D'\subseteq D$ and $c(D')=c(D)$, then
$\lambda_{D'}=\lambda_D$.

\quad(ii) $\supp(\lambda_D)\subseteq\partial D$.

\quad(iii) For any $D'\subseteq\BbbR^r$ such that $c(D')<\infty$,

\qquad($\alpha$) $\lambda_D^*(D')\le c(D')$;

\qquad($\beta$) $\lambda_{D\cup D'}\le\lambda_D+\lambda_{D'}$;

\qquad($\gamma$) $\tilde W_{D\cap D'}+\tilde W_{D\cup D'}
\le\tilde W_D+\tilde W_{D'}$;

\qquad($\delta$) $\rhotv(\lambda_D,\lambda_{D'})\le 2c(D\symmdiff D')$.

\quad(iv) If $\sequencen{D_n}$ is a non-decreasing sequence of sets with
union $D$, then

\qquad($\alpha$) $\tilde W_D=\lim_{n\to\infty}\tilde W_{D_n}
=\sup_{n\in\Bbb N}\tilde W_{D_n}$;

\qquad($\beta$) $\sequencen{\lambda_{D_n}}\to\lambda_D$ for the narrow
topology on $M^+_{\text{R}}(\BbbR^r)$.

\quad(v) $c(D)
=\inf\{\zeta\BbbR^r:\zeta\text{ is a Radon measure on }\BbbR^r,\,
   W_{\zeta}\ge\chi D\}$

\quad$\phantom{\text{(v) }c(D)}
=\inf\{\energy(\zeta):\zeta\text{ is a Radon measure on }\BbbR^r,\,
   W_{\zeta}\ge\chi D\}$.

\quad(vi) Writing $\clstar D$ for the essential closure of $D$,
$c(\clstar D)\le c(D)$ and $\tilde W_{\clstar D}\le\tilde W_D$.

\quad(vii) Suppose that $f:D\to\BbbR^r$ is $\gamma$-Lipschitz, where
$\gamma\ge 0$.   Then $c(f[D])\le\gamma^{r-2}c(D)$.

\proof{{\bf (a)(i)} If $A$, $B\subseteq\BbbR^r$ are analytic sets,
$c(B)<\infty$ and $A\subseteq B$, then

$$\eqalignno{\tilde W_A
&=\sup_{n\in\Bbb N}\tilde W_{A\cap B(\tbf{0},n)}\cr
\displaycause{479M(d-i)}
&\le\sup_{n\in\Bbb N}\tilde W_{B\cap B(\tbf{0},n)}\cr
\displaycause{479D(b-ii)}
&=\tilde W_B.\cr}$$

\noindent If $c(A)=c(B)$, then $\lambda_A=\lambda_B$.   \Prf\

$$\eqalignno{c(A)
&=\energy(\lambda_A)\cr
\displaycause{479M(d-iv)}
&=\int\tilde W_Ad\lambda_A
\le\int\tilde W_Bd\lambda_A
=\int\tilde W_Ad\lambda_B\cr
\displaycause{479J(b-i)}
&\le\int\tilde W_Bd\lambda_A
=\energy(\lambda_B)
=c(B)
=\lambda_B\BbbR^r}$$

\noindent (479M(c-i)).   So we must have equality throughout, and
$\tilde W_A=\tilde W_B\,\,\lambda_B$-a.e.   By 479Fg,
$\tilde W_A\ge\tilde W_B$ everywhere and

\Centerline{$W_{\lambda_B}=\tilde W_B=\tilde W_A=W_{\lambda_A}$.}

\noindent By 479J(b-v), $\lambda_B=\lambda_A$. \Qed

\quad{\bf (ii)} Now consider the given set $D$.
By 479E(d-i), there is an analytic set $A\supseteq D$ such
that $c(A)=c(D)$.   If $B$ is another such set, then
$c(A\cap B)=c(A)=c(B)$, so $\lambda_{A\cap B}=\lambda_A=\lambda_B$.
We therefore have a common measure which we can take to be
$\lambda_D$.   Of course this agrees with 479Mb if $D$ itself is analytic,
and with 479B if $D$ is bounded and analytic.

\medskip

{\bf (b)} Write $h_D(x)$ for $\hp^*((D\setminus\{x\})-x)$.

\medskip

\quad{\bf (i)} To begin with, suppose that
$D=K$ is compact and that $x\notin K$, so that $h_D(x)=h_K(x)=\hp(K-x)$.

\medskip

\qquad\grheada\ $h_K(x)\ge\tilde W_K(x)$.   \Prf\ 479Lc.\ \Qed

\medskip

\qquad\grheadb\ In fact $h_K(x)=\tilde W_K(x)$.  \Prf\ Let $\epsilon>0$.
Set $E=\{y:y\in K$, $\tilde W_K(y)<1\}$.   Because $\tilde W_K$ is lower
semi-continuous, $E$ is an F$_{\sigma}$ set, therefore analytic;
by 479M(d-iii), $E$ satisfies condition (iii) of 479O, and is polar.
By 479O(ii), there is a totally finite Radon measure $\zeta$
on $\BbbR^r$ such that $W_{\zeta}(y)=\infty$ for every $y\in E$.
Let $H$ be a bounded open set, including $K$, such that
$x\notin\overline{H}$;  set $\zeta_1=\zeta\LLcorner H$ and
$\zeta_2=\zeta\LLcorner(\BbbR^r\setminus H)$.
Then $\zeta=\zeta_1+\zeta_2$, so
$W_{\zeta}=W_{\zeta_1}+W_{\zeta_2}$ (479J(b-iii)).
Since $H$ is open and $\zeta_2$-negligible, $W_{\zeta_2}(y)$ is
finite for every $y\in K$ (479Fa), and $W_{\zeta_1}(y)=\infty$ for every
$y\in E$;  while $W_{\zeta_1}(x)$ is finite because the support of
$\zeta_1$ is included in $\overline{H}$.

There is therefore an $\eta>0$ such that $\eta W_{\zeta_1}(x)\le\epsilon$.
Consider $\lambda=\lambda_K+\eta\zeta_1$.   We have
$W_{\lambda}(y)\ge 1$ for every $y\in K$, while $W_{\lambda}$ is
superharmonic and lower semi-continuous (479Fa, 479Fb);  as
the support of $\lambda$ is included in the compact set $\overline{H}$,
$\lim_{\|y\|\to\infty}W_{\lambda}(y)=0$ (479Fd).   Consequently

$$\eqalignno{h_K(x)
&=\mu_x^{(K)}(K)
\le\int W_{\lambda}d\mu_x^{(K)}
\le W_{\lambda}(x)\cr
\displaycause{478Pc, with $G=\BbbR^r\setminus K$}
&\le\tilde W_K(x)+\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $h_K(x)\le\tilde W_K(x)$ and we
have equality.\ \Qed

\medskip

\quad{\bf (ii)} If $D=A$ is analytic,
note that $\capacity\{x\}=0$ (479Da, or otherwise), so
$c(A\setminus\{x\})=c(A)$, because $c$ is monotonic and submodular,
therefore subadditive (479E(d-ii)).   Now we know that

\Centerline{$h_A(x)
=\sup\{\hp(K-x):K\subseteq A\setminus\{x\}$ is compact$\}$}

\noindent (477Ie) and

\Centerline{$c(A\setminus\{x\})
=\sup\{\capacity K:K\subseteq A\setminus\{x\}$ is compact$\}$}

\noindent (479E(d-iii)).   So there is a non-decreasing sequence
$\sequencen{K_n}$ of compact subsets of $A\setminus\{x\}$ such that

\Centerline{$h_A(x)
=\sup_{n\in\Bbb N}\hp(K_n-x)$,
\quad$c(A\setminus\{x\})
=\sup_{n\in\Bbb N}\capacity K_n$.}

\noindent Set $E=\bigcup_{n\in\Bbb N}K_n$;  then $E\subseteq A$ and
$c(E)=c(A)$, so $\lambda_E=\lambda_A$ ((a-i) above) and
$\tilde W_E=\tilde W_A$.   Accordingly

$$\eqalignno{h_A(x)
&=\sup_{n\in\Bbb N}\hp(K_n-x)
=\sup_{n\in\Bbb N}\tilde W_{K_n}(x)\cr
\displaycause{(a-i) above}
&=\sup_{m,n\in\Bbb N}\tilde W_{K_n\cap B(\tbf{0},m)}(x)
=\sup_{m\in\Bbb N}\tilde W_{E\cap B(\tbf{0},m)}(x)\cr
\displaycause{apply 479E(b-iii) twice}
&=\tilde W_E(x)
=\tilde W_A(x).\cr}$$

\medskip

\quad{\bf (iii)} For the general case, note first that
$h_D\le\tilde W_D$.   \Prf\
There is a G$_{\delta}$ set $E\supseteq D$ such that $c(E)=c(D)$, so
$\lambda_E=\lambda_D$.   Now, for any $x\in\BbbR^r$,

\Centerline{$h_D(x)\le h_E(x)=\tilde W_E(x)=\tilde W_D(x)$,}

\noindent using (ii) for the central equality.\ \Qed

Equally, $h_D\ge\tilde W_D$.   \Prf\ If $x\in\BbbR^r$,
there is a G$_{\delta}$ set $H\supseteq(D\setminus\{x\})-x$ such that

\Centerline{$h_D(x)=\hp^*((D\setminus\{x\})-x)=\hp H$}

\noindent (477Id).   Set $A=(H+x)\cup\{x\}$;  then $A\supseteq D$ and

\Centerline{$h_D(x)=h_A(x)=\tilde W_A(x)\ge\tilde W_{A\cap E}(x)
=\tilde W_D(x)$,}

\noindent using (a-i) again for the inequality.\ \Qed

So $h_D=\tilde W_D$, as claimed.

\medskip

{\bf (c)} Fix an analytic set $A\supseteq D$ such that $c(A)=c(D)$;
replacing $A$ by $A\cap\overline{D}$ if necessary, we may suppose that
$A\subseteq\overline{D}$.   We have
$\lambda_D=\lambda_A$ and $\tilde W_D=\tilde W_A$.

\medskip

\quad{\bf (i)}\grheada\

\Centerline{$\lambda_D\BbbR^r=\lambda_A\BbbR^r=c(A)=c(D)$}

\noindent by 479M(c-i).

\medskip

\qquad\grheadb-\grheadc\ 479M(d-iii) tells us that
$\tilde W_D(x)=\tilde W_A(x)=1$ for $\zeta$-almost every $x\in A$, and
therefore for $\zeta$-almost every $x\in D$.
At the same time,

\Centerline{$\energy(\lambda_D)=\energy(\lambda_A)=c(A)=c(D)$}

\noindent by 479M(d-iv).

\medskip

\qquad\grheadd\ Of course $A\supseteq D'$ and $c(A)=c(D')$, so
$\lambda_{D'}=\lambda_A=\lambda_D$.

\medskip

\quad{\bf (ii)}
$\overline{A}=\overline{D}$ and $\interior A\supseteq\interior D$, so
$\partial A\subseteq\partial D$ and

\Centerline{$\lambda_D(\BbbR^r\setminus\partial D)
=\lambda_A(\BbbR^r\setminus\partial D)
\le\lambda_A(\BbbR^r\setminus\partial A)
=0$}

\noindent by 479M(c-ii).   As $\partial D$ is closed, it includes
$\supp(\lambda_D)$.

\medskip

\quad{\bf (iii)} Let $A'\supseteq D'$ be an analytic set such that
$c(A')=c(D')$.

\qquad\grheada\

$$\eqalignno{\lambda^*_D(D')
&\le\lambda_D(A')
=\lambda_A(A')
\le\sup_{m\in\Bbb N}\lambda_{A\cap B(\tbf{0},m)}(A')\cr
\displaycause{479Mb}
&=\sup_{m,n\in\Bbb N}
    \lambda_{A\cap B(\tbf{0},m)}(A'\cap B(\tbf{0},n))
\le\sup_{n\in\Bbb N}c(A'\cap B(\tbf{0},n))\cr
\displaycause{479D(c-ii)}
&=c(A')\cr
\displaycause{because $c$ is a capacity}
&=c(D').\cr}$$

\medskip

\qquad\grheadb\ Because $c$ is subadditive, we know that $c(D\cup D')$ is
finite.   Let $B\supseteq D\cup D'$ be an analytic set such that
$c(B)=c(D\cup D')$.   Then

$$\eqalignno{\lambda_{D\cup D'}
&=\lambda_{B\cap(A\cup A')}
\le\lambda_{B\cap A}+\lambda_{B\cap A'}\cr
\displaycause{479M(c-iii)}
&=\lambda_D+\lambda_{D'}.\cr}$$

\medskip

\qquad\grheadc\ This is immediate from (b) and the general fact that
$\zeta^*(U\cap V)+\zeta^*(U\cup V)\le\zeta^*U+\zeta^*V$ for any measure
$\zeta$ and any sets $U$ and $V$ (132Xk).

\medskip

\qquad\grheadd\ As usual, it will be enough to show that
$|\lambda_DE-\lambda_{D'}E|\le c(D\symmdiff D')$ for every Borel set
$E\subseteq\BbbR^r$;  by symmetry, all we need to check is that
$\lambda_{D'}E\le\lambda_DE+c(D\symmdiff D')$ for every Borel set $E$.
\Prf

\medskip

\qquad\quad{\bf case 1} Suppose that $D$ and $D'$ are both bounded Borel
sets.   Take $x\in\BbbR^r$, and let $\tau$, $\tau':\Omega\to[0,\infty]$
be the Brownian arrival times to $D-x$, $D'-x$ respectively.   Then

$$\eqalign{\mu_x^{(D')}E
&=\mu_W\{\omega:\tau'(\omega)<\infty,\,x+\omega(\tau'(\omega))\in E\}\cr
&\le\mu_W\{\omega:\tau(\omega)<\infty,\,x+\omega(\tau(\omega))\in E\}
   +\mu_W\{\omega:\tau(\omega)\ne\tau'(\omega)\}\cr
&\le\mu_x^{(D)}E+\mu_W\{\omega:\text{there is some }t\ge 0
   \text{ such that }x+\omega(t)\in D\symmdiff D'\}\cr
&=\mu_x^{(D)}E+\mu_x^{(D\symmdiff D')}\BbbR^r.\cr}$$

\noindent So

$$\eqalign{\lambda_{D'}E
&=\lim_{\|x\|\to\infty}\|x\|^{r-2}\mu_x^{(D')}E\cr
&\le\lim_{\|x\|\to\infty}\|x\|^{r-2}\mu_x^{(D)}E
   +\lim_{\|x\|\to\infty}\|x\|^{r-2}\mu_x^{(D\symmdiff D')}\BbbR^r
=\lambda_DE+c(D\symmdiff D').\cr}$$

\medskip

\qquad\quad{\bf case 2} Suppose that $D$, $D'$ are Borel sets, not
necessarity bounded.   Set $D_n=D\cap B(\tbf{0},n)$, $D'_n=D'\cap B(\tbf{0},n)$.
Then

$$\eqalignno{\lambda_{D'}E
&=\lim_{n\to\infty}\lambda_{D'_n}E\cr
\displaycause{479Mb}
&\le\lim_{n\to\infty}\lambda_{D_n}E
   +\lim_{n\to\infty}c(D_n\symmdiff D'_n)\cr
\displaycause{by case 1}
&=\lambda_DE
   +\lim_{n\to\infty}c((D\symmdiff D')\cap B(\tbf{0},n))
=\lambda_DE+c(D\symmdiff D')\cr}$$

\noindent because $c$ is a capacity.

\medskip

\qquad\quad{\bf case 3} In general, let $G\supseteq D$, $G'\supseteq D'$
and $H\supseteq D\symmdiff D'$ be G$_{\delta}$ sets such that
$c(G)=c(D)$, $c(G')=c(D')$ and $c(H)=c(D\symmdiff D')$.   Set

\Centerline{$G_1=G\cap(G'\cup H)$,
\quad$G'_1=G'\cap(G\cup H)$;}

\noindent these are Borel sets, while $D\subseteq G_1\subseteq G$,
$D'\subseteq G'_1\subseteq G'$ and $G_1\symmdiff G_2\subseteq H$.
So

$$\eqalignno{\lambda_{D'}E
&=\lambda_{G'_1}E
\le\lambda_{G_1}E+c(G\symmdiff G_1)\cr
\displaycause{by case 2}
&\le\lambda_DE+c(H)
=\lambda_DE+c(D\symmdiff D')\cr}$$

\noindent and we have the result in this case also.   So we're done.\ \Qed

\medskip

\quad{\bf (iv)}\grheada\ This follows immediately from (b) above.

\medskip

\qquad\grheadb\
Consider first the case in which every $D_n$ is analytic.
Returning to the proof of 479M, or putting 479Ma together
with (iii-$\delta$) here, we see that for any $m$, $n\in\Bbb N$ we shall
have

\Centerline{$\rhotv(\lambda_{D_n},\lambda_{D_n\cap B(\tbf{0},m)})
\le 2c(D_n\setminus B(\tbf{0},m))\le 2c(D\setminus B(\tbf{0},m))=2\alpha_m$}

\noindent say, and that $\lim_{m\to\infty}\alpha_m=0$.   So if
$G\subseteq\BbbR^r$ is any open set,

$$\eqalignno{\lambda_DG
&=\lim_{m\to\infty}\lambda_{D\cap B(\tbf{0},m)}G
\le\lim_{m\to\infty}\liminf_{n\to\infty}\lambda_{D_n\cap B(\tbf{0},m)}G\cr
\displaycause{479E(c-i)}
&\le\lim_{m\to\infty}\liminf_{n\to\infty}\lambda_{D_n}G+2\alpha_m
=\liminf_{n\to\infty}\lambda_{D_n}G.\cr}$$

\noindent Since we know also that

\Centerline{$\lambda_D\BbbR^r=c(D)=\lim_{n\to\infty}c(D_n)
=\lim_{n\to\infty}\lambda_{D_n}\BbbR^r$,}

\noindent $\sequencen{\lambda_{D_n}}\to\lambda_D$ for the narrow topology.

For the general case, take analytic sets $A_n\supseteq D_n$, $A\supseteq D$
such that $c(A_n)=c(D_n)$ for every $n$ and $c(A)=c(D)$.   Set
$A'_n=A\cap\bigcap_{m\ge n}A_m$ for each $n$, $A'=\bigcup_{n\in\Bbb N}A_n$;
then

\Centerline{$\sequencen{\lambda_{D_n}}
=\sequencen{\lambda_{A'_n}}\to\lambda_{A'}=\lambda_D$}

\noindent for the narrow topology.

\medskip

{\bf (v)} Let $Q$ be the set of Radon measures $\zeta$ on $\BbbR^r$ such
that $W_{\zeta}\ge\chi D$.

\medskip

\qquad\grheada\ I show first that
$\inf_{\zeta\in Q}\zeta\BbbR^r$ and $\inf_{\zeta\in Q}\energy(\zeta)\}$
are both less than or equal to $c(D)$.
\Prf\ Let $\epsilon>0$.   Because $c$ is outer regular (479E(d-i)),
there is an open set $G\supseteq D$ such that $c(G)\le c(D)+\epsilon$.
Set $\zeta=\lambda_G$.   Then

\Centerline{$W_{\zeta}=\tilde W_G\ge\chi G\ge\chi D$}

\noindent (479D(b-iii)), so $\zeta\in Q$, while

\Centerline{$\zeta\BbbR^r=\energy(\zeta)=c(G)\le c(D)+\epsilon$.  \Qed}

\medskip

\qquad\grheadb\ Now suppose that $\zeta\in Q$.   Then
$c(D)\le\min(\zeta\BbbR^r,\energy(\zeta))$.   \Prf\
Take any $\gamma<c(D)$ and $\epsilon>0$.
Let $A\supseteq D$ be an analytic set such that
$c(A)=c(D)$;  replacing $A$ by $\{x:x\in A$, $W_{\zeta}(x)\ge 1\}$ if
necessary, we can suppose that $W_{\zeta}\ge\chi A$.
For each $n\in\Bbb N$, let $\zeta_n$ be the totally finite measure
$(1+\epsilon)\zeta\LLcorner B(\tbf{0},n)$.
Then $\sequencen{W_{\zeta_n}}$ is non-decreasing
and has supremum $(1+\epsilon)W_{\zeta}$ (479M(d-i)),
so $A=\bigcup_{n\in\Bbb N}A_n$,
where $A_n=\{x:x\in A$, $W_{\zeta_n}(x)\ge 1\}$.   There are an
$n\in\Bbb N$ such that $c(A_n)>\gamma$ and a compact
$K\subseteq A_n$ such that $\capacity K\ge\gamma$ (432K).   Now
$W_{\zeta_n}\ge\tilde W_K\,\,\lambda_K$-a.e., so
$W_{\zeta_n}\ge\tilde W_K$ everywhere (479Fg) and

$$\eqalignno{\gamma
&\le\capacity K
=\int\tilde W_Kd\lambda_K
\le\int W_{\zeta_n}d\lambda_K
=\int\tilde W_Kd\zeta_n\cr
\displaycause{479J(b-i)}
&\le\int W_{\zeta_n}d\zeta_n
\le(1+\epsilon)\int W_{\zeta}d\zeta_n
\le(1+\epsilon)^2\int W_{\zeta}d\zeta
=(1+\epsilon)^2\energy(\zeta).\cr}$$

\noindent Moreover, 479J(c-vi), applied to $\zeta_n$ and $\lambda_K$,
tells us that

\Centerline{$\zeta\BbbR^r\ge\zeta_n\BbbR^r\ge\lambda_K\BbbR^r
=\capacity K\ge\gamma$.}

\noindent As $\gamma$ and $\epsilon$ are arbitrary,
$c(D)\le\min(\energy(\zeta),\zeta\BbbR^r)$, as claimed.\ \Qed

\medskip

\qquad\grheadc\ Putting these together, we see that
$c(D)=\inf_{\zeta\in Q}\zeta\BbbR^r=\inf_{\zeta\in Q}\energy(\zeta)$.

\medskip

\quad{\bf (vi)} If $x\in\clstar A$, then
$0\in\clstar((A\setminus\{x\})-x)$ and
$\tilde W_A(x)=\hp^*((A\setminus\{x\})-x)=1$ for every $x\in\clstar E$, by
478U and (b) above.   Now

$$\eqalignno{c(\clstar D)
&\le c(\clstar A)
\le\energy(\lambda_A)\cr
\displaycause{(v) above}
&=c(A)\cr
\displaycause{479M(d-iv)}
&=c(D).\cr}$$

\medskip

\quad{\bf (vii)}\grheada\ Consider first the case $D=A$, so
that $f[D]=f[A]$ is analytic.   We can suppose that $c(f[A])>0$,
in which case $A\ne\emptyset$ and $\gamma>0$.   Take any
$\epsilon>0$.   By 479Nc there is a Radon measure $\zeta$ on
$\BbbR^r$ such that $\zeta f[A]=1$ and
$c(f[A])\le\Bover{1+\epsilon}{\energy(\zeta)}$.   Applying 433D to
the subspace measure $\zeta_{f[A]}$, we see that there is a Radon
probability measure $\zeta'$ on $A$ such that $\zeta_{f[A]}$ is the image
measure $\zeta'f^{-1}$;  let $\lambda$ be the Radon probability measure on
$\BbbR^r$ extending $\zeta'$.   Then

$$\eqalignno{\energy(\zeta)
&=\int_{\BbbR^r}\int_{\BbbR^r}\Bover1{\|x-y\|^{r-2}}\zeta(dx)\zeta(dy)
\ge\int_{f[A]}\int_{f[A]}\Bover1{\|x-y\|^{r-2}}\zeta(dx)\zeta(dy)\cr
&=\int_A\int_{f[A]}\Bover1{\|x-f(v)\|^{r-2}}\zeta(dx)\zeta'(dv)
=\int_A\int_A\Bover1{\|f(u)-f(v)\|^{r-2}}\zeta'(du)\zeta'(dv)\cr
\displaycause{applying 235J\formerly{2{}35L} twice}
&\ge\int_A\int_A\Bover1{\gamma^{r-2}\|u-v\|^{r-2}}\zeta'(du)\zeta'(dv)\cr
&=\Bover1{\gamma^{r-2}}\int_{\BbbR^r}\int_{\BbbR^r}
   \Bover1{\|u-v\|^{r-2}}\lambda(du)\lambda(dv)
=\Bover1{\gamma^{r-2}}\energy(\lambda).\cr}$$

\noindent By 479Nc in the other direction,

\Centerline{$c(A)\ge\Bover1{\energy(\lambda)}
\ge\Bover1{\gamma^{r-2}\energy(\zeta)}
\ge\Bover1{(1+\epsilon)\gamma^{r-2}}c(f[A])$.}

\noindent As $\epsilon$ is arbitrary, $c(f[A])\le\gamma^{r-2}c(A)$.

\medskip

\qquad\grheadb\ In general, since $f:D\to\BbbR^r$ is certainly uniformly
continuous, it has a continuous extension $g:\overline{D}\to\BbbR^r$
(3A4G), which is still $\gamma$-Lipschitz.   Now ($\alpha$) tells us that

\Centerline{$c(f[D])\le c(g[A])
\le\gamma^{r-2}c(A)=\gamma^{r-2}c(D)$,}

\noindent as required.
}%end of proof of 479P

\leader{479Q}{Hausdorff measure:  Theorem}
For $s\in\ooint{0,\infty}$ let $\mu_{Hs}$ be
Hausdorff $s$-dimensional measure on $\BbbR^r$.
Let $D$ be any subset of $\BbbR^r$.

(a) If the Choquet-Newton capacity $c(D)$ is non-zero,
then $\mu^*_{H,r-2}D=\infty$.

(b) If $s>r-2$ and $\mu^*_{Hs}D>0$, then $c(D)>0$.
%Mattila 8.9

\proof{{\bf (a)} Let $E\supseteq D$ be a G$_{\delta}$ set such that
$\mu_{H,r-2}E=\mu^*_{H,r-2}D$ (471Db).   Then $c(E)>0$.
Let $K\subseteq E$ be a compact set such that $\capacity K>0$.   Then

\Centerline{$\capacity K
=\int_K\int_K\Bover1{\|x-y\|^{r-2}}\lambda_K(dx)\lambda_K(dy)$}

\noindent is finite and not $0$;  applying 471Tb to the subspace measure on
$K$,

\Centerline{$\infty=\mu_{H,r-2}K=\mu_{H,r-2}E=\mu^*_{H,r-2}D$.}

\medskip

{\bf (b)} Let $E\supseteq D$ be a G$_{\delta}$ set such that
$c(E)=c(D)$ (479E(d-i)).   Then $\mu_{Hs}E>0$.
By 471Ta, there is a non-zero Radon measure $\zeta_0$ on $E$ such
that
$\biggerint_E\biggerint_E\Bover1{\|x-y\|^{r-2}}\zeta_0(dx)\zeta_0(dy)$
is finite.
Let $K\subseteq E$ be a compact set such that $\zeta_0K>0$, and let $\zeta$
be the Radon measure on $\BbbR^r$ such that $\zeta H=\zeta_0(K\cap H)$ for
every Borel set $H\subseteq\BbbR^r$;  then

\Centerline{$\energy(\zeta)
=\int_K\int_K\Bover1{\|x-y\|^{r-2}}\zeta_0(dx)\zeta_0(dy)$}

\noindent is finite, while $K$ is $\zeta$-conegligible.
By 479Nc (or, more directly, by the first remark in the proof of 479N),
$\capacity K>0$, so that $c(D)=c(E)\ge\capacity K>0$.
}%end of proof of 479Q

\leader{479R}{}\cmmnt{ I come to the promised difference between Brownian
motion in $\BbbR^3$ and in higher dimensions, following 478M.

\medskip

\noindent}{\bf Proposition} (a) Suppose that $r=3$.
Then almost every $\omega\in\Omega$ is not injective.

(b) If $r\ge 4$, then almost every $\omega\in\Omega$ is injective.

\proof{ In this proof, I will take $X_t(\omega)=\omega(t)$ for
$\omega\in\Omega$ and $t\ge 0$.

\medskip

{\bf (a)(i)} For $\omega\in\Omega$ set
$F_{\omega}=\{\omega(t):t\in[0,1]\}$.   Then $\capacity F_{\omega}>0$ for
$\mu_W$-almost every $\omega$.   \Prf\ By 477Lb,
$\mu_{H,3/2}F_{\omega}=\infty$ for almost every $\omega$.   For any such
$\omega$, there is a non-zero
Radon measure $\zeta_0$ on $F_{\omega}$ such that
$\int_{F_{\omega}}\int_{F_{\omega}}\Bover1{\|x-y\|}\zeta_0(dx)\zeta_0(dy)$
is finite (471Ta).   Let $\zeta$ be the Radon measure on $\BbbR^r$,
extending $\zeta_0$, for which $F_{\omega}$ is conegligible.   Then
$\zeta(F_{\omega})>0$ and $\energy(\zeta)<\infty$.   (This is where we need
to know that $r=3$.)   So $\capacity F_{\omega}>0$ (479Nc).\ \Qed

\medskip

\quad{\bf (ii)} Consider
$E_0=\{\omega:$ there are $s\le 1$ and
$t\ge 2$ such that $\omega(s)=\omega(t)\}$.   (This is an F$_{\sigma}$ set,
so is measurable.)   Take $\tau$ to be the stopping time with constant
value $2$ and $\phi_{\tau}:\Omega\times\Omega\to\Omega$ the corresponding
\imp\ function as in 477G;  set
$H=\{\omega:\omega\in\Omega$, $\omega(2)\notin F_{\omega}\}$.   Then

$$\eqalignno{\mu_WE_0
&=\int_{\Omega}\mu_W\{\omega':\phi_{\tau}(\omega,\omega')\in E_0\}
  \mu_W(d\omega)\cr
&=\int_{\Omega}\mu_W\{\omega':\text{ there is some }t\ge 0\text{ such that }
  \omega(2)+\omega'(t)\in F_{\omega}\}\mu_W(d\omega)\cr
&=\mu_W(\Omega\setminus H)
  +\int_H\tilde W_{F_{\omega}}(\omega(2))\mu_W(d\omega)\cr
\displaycause{479Pb}
&>0\cr}$$

\noindent because $\tilde W_{F_{\omega}}(\omega(2))>0$ whenever
$\omega\in H$ and $\capacity F_{\omega}>0$, which is so for almost every
$\omega\in H$.

\medskip

\quad{\bf (iii)} Now, setting

\Centerline{$E_n=\{\omega:$ there are $s\in[n,n+1]$ and
$t\ge n+2$ such that $\omega(s)=\omega(t)\}$,}

\noindent we have $\mu_WE_n=\mu_WE_0$ for every $n$, because
$\langle X_{s+n}-X_n\rangle_{s\ge 0}$ has the same distribution as
$\langle X_s\rangle_{s\ge 0}$.   So if
$E=\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}E_m$, $\mu_WE>0$.   But $E$ belongs
to the tail $\sigma$-algebra
$\bigcap_{t\ge 0}\Tau_{\coint{t,\infty}}$,
so has measure either $0$ or $1$ (477Hd), and must be conegligible.
Since every $\omega\in E$ is self-intersecting, we see that almost every
Brownian path is self-intersecting.

\medskip

{\bf (b)(i)} Suppose that $q$, $q'\in\Bbb Q$ are such that
$0\le q<q'$.   This time, set
$F_{\omega}=\{\omega(t):t\in[0,q]\}$.   For almost every $\omega$,
$F_{\omega}$ has
zero two-dimensional Hausdorff measure (477La), so has zero
$(r-2)$-dimensional Hausdorff measure (because $r\ge 4$),
and therefore has zero capacity (479Qa).   Also

\Centerline{$\mu_W\{\omega:\omega(q')\in F_{\omega}\}
  =(\mu_W\times\mu_W)\{(\omega,\omega'):
     \omega'(q'-q)\in F_{\omega}-\omega(q)\}=0$}

\noindent because the distribution of $X_{q'-q}$ is absolutely continuous
with respect to Lebesgue measure and $\mu F_{\omega}=0$ for $\mu_W$-almost
every $\omega$.   But this means that

$$\eqalign{\mu_W\{\omega:&\text{ there is a }t\ge q'
  \text{ such that }\omega(t)\in F_{\omega}\}\cr
&=(\mu_W\times\mu_W)\{(\omega,\omega'):\text{there is a }t\ge 0
  \text{ such that }\omega'(t)\in F_{\omega}-\omega(q')\}\cr
&=\int_{\Omega}\tilde W_{F_{\omega}}(\omega(q'))\mu(d\omega)
=0,\cr}$$

\noindent that is,

\Centerline{$\{\omega:$ there are $s\le q$, $t\ge q'$ such that
$\omega(s)=\omega(t)\}$}

\noindent is negligible.   As $q$ and $q'$ are arbitrary, almost every
sample path is injective.
}%end of proof of 479R

\leader{479S}{}\cmmnt{ A famous classical problem concerned, in effect,
the continuity of potential functions, in particular the continuity of
functions of the form $\tilde W_K$.   I think
that even with the modern theory
as sketched above, this is not quite trivial, so I spell out an example.

\medskip

\noindent}{\bf Example} Suppose that $e\in\BbbR^r$ is a unit vector.   Then
there is a sequence $\sequencen{\delta_n}$ of strictly positive real
numbers such that the equilibrium potential
$\tilde W_K$ is discontinuous at $e$ whenever
$K\subseteq B(\tbf{0},1)$ is compact, $e\in\overline{\interior K}$ and
$\|x-te\|\le\delta_n$ whenever $n\in\Bbb N$, $t\in[1-2^{-n},1]$, $x\in K$
and $\|x\|=t$.

\proof{ For $n\in\Bbb N$, let $K_n$ be the line segment
$\{te:1-2^{-n}\le t\le 1-2^{-n-1}\}$.   Then the one-dimensional Hausdorff
measure of $K_n$ is finite, so $\capacity K_n=0$ (479Qa).
By 479E(c-ii), $\lim_{\delta\downarrow 0}\capacity(K_n+B(\tbf{0},\delta))=0$;
let $\delta_n\in\ooint{0,2^{-n-2}}$
be such that $\capacity(K_n+B(\tbf{0},\delta_n))\le 2^{-3n-6}$.
Setting $L_n=K_n+B(\tbf{0},\delta_n)$, the distance from $e$ to $L_n$ is at least
$2^{-n-2}$.   By 479Pb,

\Centerline{$\hp(L_n-e)
=\tilde W_{L_n}(e)
\le 4^{n+2}\lambda_{L_n}(\BbbR^r)
=4^{n+2}\capacity L_n
\le 2^{-n-2}$.}

Suppose that
$K\subseteq B(\tbf{0},1)$ is compact, $e\in\overline{\interior K}$ and
$\|x-te\|\le\delta_n$ whenever $n\in\Bbb N$, $t\in[1-2^{-n},1]$, $x\in K$
and $\|x\|=t$.   Then
$K\subseteq\bigcup_{n\in\Bbb N}L_n\cup\{e\}$.   Using the full strength of
479Pb,

\Centerline{$\tilde W_K(e)=\hp((K\setminus\{e\})-e)
\le\hp(\bigcup_{n\in\Bbb N}L_n-e)
\le\sum_{n=0}^{\infty}\hp(L_n-e)
\le\Bover12$.}

\noindent On the other hand, $\tilde W_K(x)=1$ for every
$x\in\interior K$ (479D(b-iii)), so $\tilde W_K$ is not continuous at
$e$.
}%end of proof of 479S

\leader{*479T}{}\cmmnt{ This concludes the main argument of the section,
which you may feel is quite enough.   However, there is an important
alternative method of calculating the capacity of a compact set,
based on gradients of potential functions (479U), and a couple of further
results are reasonably accessible (479V-479W) which reflect other concerns
of this volume.

\medskip

\noindent}{\bf Lemma} (a) If $g:\BbbR^r\to\Bbb R$ is a smooth function with
compact support,

\Centerline{$\int_{\BbbR^r}\Bover1{\|x-y\|^{r-2}}\nabla^2g\,d\mu
=-r(r-2)\beta_rg(x)$}

\noindent for every $x\in\BbbR^r$.
%this ought to work for general $r$

(b) Let $g$, $h:\BbbR^r\to\Bbb R$ be smooth functions with compact support.
Then

\Centerline{$\int_{\BbbR^r}h\times\nabla^2g\,d\mu
=\int_{\BbbR^r}g\times\nabla^2h
=-\int_{\BbbR^r}\varinnerprod{\grad h}{\grad g}\,d\mu$.}

(c) Let $\zeta$ be a totally finite Radon measure on $\BbbR^r$, and
$W_{\zeta}:\BbbR^r\to[0,\infty]$ the associated Newtonian potential.   Then
$\int_{\BbbR^r}W_{\zeta}\times\nabla^2g\,d\mu
=-r(r-2)\beta_r\int_{\BbbR^r}g\,d\zeta$ for every smooth
function $g:\BbbR^r\to\Bbb R$ with compact support.

(d) Let $\zeta$ be a totally finite Radon measure on $\BbbR^r$ such that
$W_{\zeta}$ is finite-valued everywhere and Lipschitz.   Then
$\int_{\BbbR^r}\varinnerprod{\grad f}{\grad W_{\zeta}}d\mu
=r(r-2)\beta_r\int_{\BbbR^r}f\,d\zeta$ for every Lipschitz function
$f:\BbbR^r\to\Bbb R$ with compact support.

%maybe can make this work with Lipschitz $f\in C_0(\BbbR^r)$
%what about $\energy(\zeta)$ and $\int\|\grad W_{\zeta}\|^2$?

(e) Let $K\subseteq\BbbR^r$ be a compact set, and $\epsilon>0$.   Then
there is a Radon measure $\zeta$ on $\BbbR^r$, with support included in
$K+B(\tbf{0},\epsilon)$, such that $W_{\zeta}$ is a smooth function with compact
support, $W_{\zeta}\ge\chi K$, $\zeta\BbbR^r\le\capacity K+\epsilon$ and

\Centerline{$\int_{\BbbR^r}\|\grad W_{\zeta}\|^2d\mu
=r(r-2)\beta_r\energy(\zeta)\le r(r-2)\beta_r\zeta\BbbR^r$.}

\proof{{\bf (a)(i)} Consider first the case $x=0$.
Setting $f(y)=\Bover1{\|y\|^{r-2}}$ for $y\ne 0$, we have
$\grad f(y)=-\Bover{r-2}{\|y\|^r}y$ and $(\nabla^2f)(y)=0$ for $y\ne 0$
(478Fa);   also $f$ is locally integrable, by 478Ga.
So $\int_{\BbbR^r}f\times\nabla^2g\,d\mu$ is well-defined.

Let $R>0$ be such that $g$ is zero outside $B(\tbf{0},R)$, and set
$M=\|\grad g\|_{\infty}$;
take $\epsilon\in\ooint{0,R}$.    Then

$$\eqalignno{\int_{\BbbR^r\setminus B(\tbf{0},\epsilon)}f\times\nabla^2g\,d\mu
&=\int_{B(\tbf{0},R)\setminus B(\tbf{0},\epsilon)}
  f\times\nabla^2g-g\times\nabla^2f\,d\mu\cr
&=\int_{B(\tbf{0},R)\setminus B(\tbf{0},\epsilon)}
  \diverg(f\times\grad g-g\times\grad f)d\mu\cr
\displaycause{use 474Bb}
&=\int_{\partial B(\tbf{0},R)}
  (\Bover1{\|y\|^{r-2}}\grad g(y)+\Bover{(r-2)g(y)}{\|y\|^r}y)
  \dotproduct\Bover{y}{\|y\|}\nu(dy)\cr
&\mskip100mu
 -\int_{\partial B(\tbf{0},\epsilon)}
  (\Bover1{\|y\|^{r-2}}\grad g(y)+\Bover{(r-2)g(y)}{\|y\|^r}y)
  \dotproduct\Bover{y}{\|y\|}\nu(dy)\cr
\displaycause{475Nc}
&=-\int_{\partial B(\tbf{0},\epsilon)}
  (\Bover1{\|y\|^{r-1}}y\dotproduct\grad g(y)
     +\Bover{(r-2)g(y)}{\|y\|^{r-1}})\nu(dy)\cr
&=-\Bover1{\epsilon^{r-1}}\int_{\partial B(\tbf{0},\epsilon)}
  (y\dotproduct\grad g(y)+(r-2)g(y))\nu(dy).
  \cr}$$

\noindent Now we have

\Centerline{$|\int_{\partial B(\tbf{0},\epsilon)}
  y\dotproduct\grad g(y)\nu(dy)|
\le\epsilon M\nu(\partial B(\tbf{0},\epsilon))\le r\beta_r\epsilon^rM$,}

\noindent so

$$\eqalign{
|\int_{\BbbR^r\setminus B(\tbf{0},\epsilon)}f\times&\nabla^2g\,d\mu
   +r(r-2)\beta_rg(0)|\cr
&\le r\beta_r\epsilon M
 +\Bover1{\epsilon^{r-1}}|r\beta_r\epsilon^{r-1}(r-2)g(0)
   -\int_{\partial B(\tbf{0},\epsilon)}(r-2)g(y)\nu(dy)|\cr
&\le r\beta_r\epsilon M
 +\Bover{r-2}{\epsilon^{r-1}}\int_{\partial B(\tbf{0},\epsilon)}
    |g(0)-g(y)|\nu(dy)\cr
&\le r\beta_r\epsilon M
 +r(r-2)\beta_r\sup_{y\in\partial B(\tbf{0},\epsilon)}|g(0)-g(y)|
\to 0\cr}$$

\noindent as $\epsilon\downarrow 0$;  that is,

\Centerline{$\int_{\BbbR^r}f\times\nabla^2g\,d\mu=-r(r-2)\beta_rg(0)$.}

\medskip

\quad{\bf (ii)} For the general case, apply (i) to the function
$y\mapsto g(x+y)$.

\medskip

{\bf (b)} Take $R>0$ so large that both $g$ and $h$ are zero outside
$B(\tbf{0},R)$, and $M\ge\max(\|\nabla^2g\|_{\infty},\|\nabla^2h\|_{\infty})$.

\medskip

\quad{\bf (i)} We have

$$\eqalignno{\int_{\BbbR^r}\int_{\BbbR^r}&\Bover1{\|x-y\|^{r-2}}
  |(\nabla^2g)(x)(\nabla^2h)(y)|\mu(dx)\mu(dy)\cr
&\le M^2\int_{B(\tbf{0},R)}\int_{B(\tbf{0},R)}\Bover1{\|x-y\|^{r-2}}\mu(dx)\mu(dy)
\le M^2\int_{B(\tbf{0},R)}\Bover12r\beta_rR^2\mu(dy)\cr
\displaycause{478Gc}
&<\infty.\cr}$$

\noindent So

$$\eqalignno{-r(r-2)\beta_r\int_{\BbbR^r}h\times\nabla^2g\,d\mu
&=\int_{\BbbR^r}\int_{\BbbR^r}\Bover1{\|x-y\|^{r-2}}
   (\nabla^2h)(x)(\nabla^2g)(y)\mu(dx)\mu(dy)\cr
\displaycause{by (a)}
&=\int_{\BbbR^r}\int_{\BbbR^r}\Bover1{\|x-y\|^{r-2}}
   (\nabla^2g)(y)(\nabla^2h)(x)\mu(dy)\mu(dx)\cr
&=-r(r-2)\beta_r\int_{\BbbR^r}g\times\nabla^2h\,d\mu.\cr}$$

\noindent Thus $\int_{\BbbR^r}g\times\nabla^2h\,d\mu
=\int_{\BbbR^r}h\times\nabla^2g\,d\mu$.

\medskip

\quad{\bf (ii)} By 473Bd, $\grad(g\times h)=g\times\grad h+h\times\grad g$,
so 474Bb tells us that
$\nabla^2(g\times h)
=2\varinnerprod{\grad g}{\grad h}+g\times\nabla^2h+h\times\nabla^2g$, and

$$\eqalign{\int_{\BbbR^r}\nabla^2(g\times h)d\mu
&=\int_{B(\tbf{0},R)}\nabla^2(g\times h)d\mu\cr
&=\int_{\partial B(\tbf{0},R)}\varinnerprod{\grad(g\times h)}{\Bover{x}{\|x\|}}
   \nu(dx)
=0,\cr}$$

\noindent so

\Centerline{$\int_{\BbbR^r}\varinnerprod{\grad g}{\grad h}\,d\mu
=-\Bover12\int_{\BbbR^r}g\times\nabla^2h+h\times\nabla^2g\,d\mu
=-\int_{\BbbR^r}g\times\nabla^2h\,d\mu$.}

\medskip

{\bf (c)} If $g(x)=0$ for $\|x\|\ge R$ and
$|(\nabla^2g)(x)|\le M$ for every $x$, then

\Centerline{$\int_{\BbbR^r}\Bover1{\|x-y\|^{r-2}}|(\nabla^2g)(x)|\mu(dx)
\le M\int_{B(\tbf{0},R)}\Bover1{\|x-y\|^{r-2}}\mu(dx)
\le\Bover12Mr\beta_rR^2$}

\noindent for every $y$ (478Gc again).
We can therefore apply (a) and
integrate with respect to $\zeta$ to see that

$$\eqalign{-r(r-2)\beta_r\int_{\BbbR^r}g\,d\zeta
&=\int_{\BbbR^r}\int_{\BbbR^r}
  \Bover1{\|x-y\|^{r-2}}(\nabla^2g)(x)\mu(dx)\zeta(dy)\cr
&=\int_{\BbbR^r}\int_{\BbbR^r}
  \Bover1{\|x-y\|^{r-2}}(\nabla^2g)(x)\zeta(dy)\mu(dx) \cr
&=\int_{\BbbR^r}W_{\zeta}(x)(\nabla^2g)(x)\mu(dx),\cr}$$

\noindent as required.

\medskip

{\bf (d)} Let $\sequencen{\tilde h_n}$ be the
smoothing sequence of 473E.

\medskip

\quad{\bf (i)} Suppose to begin with that $f$ is smooth.    For
$n\in\Bbb N$ set $g_n=\tilde h_n*W_{\zeta}$.   As $W_{\zeta}$ is
continuous, $\lim_{n\to\infty}g_n=W_{\zeta}$ (473Ec);  as
$\|W_{\zeta}\|_{\infty}\le 1$, $\|g_n\|_{\infty}\le 1$ for every $n$
(473Da).   Because $f$ has compact support,

\Centerline{$\lim_{n\to\infty}\int_{\BbbR^r}g_n\times\nabla^2f\,d\mu
=\int_{\BbbR^r}W_{\zeta}\times\nabla^2f\,d\mu$}

\noindent by the dominated convergence theorem.   Next,
$\grad g_n=\tilde h_n*\grad W_{\zeta}$ for each $n$ (473Dd).   As
$\grad W_{\zeta}$ is essentially bounded (473Cc), all its coordinates are
locally integrable, so
$\grad W_{\zeta}\eae\lim_{n\to\infty}\grad g_n$ (473Ee).   We therefore
have

$$\eqalignno{\int_{\BbbR^r}\varinnerprod{\grad f}{\grad W_{\zeta}}d\mu
&=\lim_{n\to\infty}\int_{\BbbR^r}
   \varinnerprod{\grad f}{\grad g_n}d\mu\cr
&=-\lim_{n\to\infty}\int_{\BbbR^r}g_n\times\nabla^2f\,d\mu\cr
\displaycause{(b) above}
&=-\int_{\BbbR^r}W_{\zeta}\times\nabla^2f\,d\mu
=r(r-2)\beta_r\int_{\BbbR^r}f\,d\zeta\cr}$$

\noindent by (c).

\medskip

\quad{\bf (ii)} For the general case, smooth on the other side, setting
$f_n=\tilde h_n*f$ for every $n$.   This time, $f_n\to f$ uniformly
(473Ed), so
$\int_{\BbbR^r}fd\zeta=\lim_{n\to\infty}\int_{\BbbR^r}f_nd\zeta$.
On the other hand, if $f$ is $M$-Lipschitz,
$\grad f_n=\tilde h_n*\grad f$ converges $\mu$-a.e.\ to
$\grad f$, and $\|\grad f_n\|_{\infty}$ is at most $M$ for every $n$;
also there is a bounded set outside which all
the $f_n$ and $\grad f_n$ are zero, and $\|\grad W_{\zeta}\|$ is bounded.
So

\Centerline{$\int_{\BbbR^r}\varinnerprod{\grad f}{\grad W_{\zeta}}d\mu
=\lim_{n\to\infty}\int_{\BbbR^r}
  \varinnerprod{\grad f_n}{\grad W_{\zeta}}d\mu$.}

\noindent
Applying (i) to each $f_n$ and taking the limit, we get the equality we
seek.

\medskip

{\bf (e)(i)} There is a compact set $L\subseteq K+B(\tbf{0},\bover{\epsilon}2)$
such that $K\subseteq\interior L$ and $\capacity L\le\capacity K+\epsilon$
(479Ed).   Let $n\in\Bbb N$ be such that
$\Bover1{n+1}\le\Bover{\epsilon}2$
and $K+B(\tbf{0},\bover1{n+1})\subseteq L$.   Set
$h=\lambda_L*\tilde h_n$, where $\tilde h_n$ is the
function of 473E, as before;  let $\zeta=h\mu$ be the corresponding
indefinite-integral measure over $\mu$.   Because $\tilde h_n$ is zero
outside $B(\tbf{0},\bover1{n+1})$
and the support of $\lambda_L$ is included in $L$,
the support of $\zeta$ is included in
$L+B(\tbf{0},\bover1{n+1})\subseteq K+B(\tbf{0},\epsilon)$.

\medskip

\quad{\bf (ii)} By 444Pa, we have

\Centerline{$W_{\zeta}=\zeta*k_{r-2}=(h\mu)*k_{r-2}=h*k_{r-2}$}

\noindent where $k_{r-2}$ is the Riesz kernel (479G).
Now $W_{\zeta}=\tilde W_L*\tilde h_n$.   \Prf\ For $m\in\Bbb N$, set
$f_m=k_{r-2}\times\chi B(\tbf{0},m)$, so that $f_m$ is $\mu$-integrable.
Observe that

$$\eqalign{\tilde W_L(x)
&=\int_{\BbbR^r} k_{r-2}(x-y)\lambda_L(dy)
=\int_{\BbbR^r}\lim_{m\to\infty}f_m(x-y)\lambda_L(dy)\cr
&=\lim_{m\to\infty}\int_{\BbbR^r} f_m(x-y)\lambda_L(dy)
=\lim_{m\to\infty}(\lambda_L*f_m)(x)\cr}$$

\noindent for each $x$;  moreover, because $\sequence{m}{f_m}$ is
non-decreasing, so is $\sequence{m}{\lambda_L*f_m}$.
For each $m$,

$$\eqalignno{h*f_m
&=(h\mu)*f_m
=(\lambda_L*\tilde h_n)\mu*f_m
=(\lambda_L*\tilde h_n\mu)*f_m\cr
\displaycause{444K}
&=\lambda_L*(\tilde h_n\mu*f_m)\cr
\displaycause{444Ic}
&=\lambda_L*(\tilde h_n*f_m)
=\lambda_L*(f_m*\tilde h_n)
=\lambda_L*(f_m\mu*\tilde h_n)\cr
&=(\lambda_L*f_m\mu)*\tilde h_n
=(\lambda_L*f_m)\mu*\tilde h_n
=(\lambda_L*f_m)*\tilde h_n.\cr}$$

\noindent Now, for each $x$,

$$\eqalign{W_{\zeta}(x)
&=\int_{\BbbR^r} h(x-y)k_{r-2}(y)\mu(dy)
=\lim_{m\to\infty}\int_{\BbbR^r} h(x-y)f_m(y)\mu(dy)\cr
&=\lim_{m\to\infty}(h*f_m)(x)
=\lim_{m\to\infty}((\lambda_L*f_m)*\tilde h_n)(x)\cr
&=\lim_{m\to\infty}\int_{\BbbR^r}(\lambda_L*f_m)(y)\tilde h_n(x-y)\mu(dy)
  \cr
&=\int_{\BbbR^r}\lim_{m\to\infty}(\lambda_L*f_m)(y)
   \tilde h_n(x-y)\mu(dy)\cr
&=\int_{\BbbR^r}\tilde W_L(y)\tilde h_n(x-y)\mu(dy)
=(\tilde W_L*\tilde h_n)(x).  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (iii)} Since $\tilde W_L(x)=1$ whenever $x\in\interior L$
(479D(b-iii)), and
$x+y\in\interior L$ whenever $x\in K$ and $\tilde h_n(y)\ne 0$,
$W_{\zeta}(x)=1$ for every $x\in K$.   Because both $\tilde W_L$ and
$\tilde h_n$ have compact support, so does $W_{\zeta}$;
because $\tilde h_n$ is smooth, so is $W_{\zeta}$ (473De).

\medskip

\quad{\bf (iv)} Now

$$\eqalignno{\int_{\BbbR^r}\|\grad W_{\zeta}\|^2d\mu
&=-\int_{\BbbR^r}W_{\zeta}\times\nabla^2W_{\zeta}\,d\mu\cr
\displaycause{(b) above}
&=r(r-2)\beta_r\int_{\BbbR^r}W_{\zeta}\,d\zeta\cr
\displaycause{(c) above}
&=r(r-2)\beta_r\energy(\zeta)
\le r(r-2)\beta_r\zeta\BbbR^r\cr}$$

\noindent because
$\|W_{\zeta}\|_{\infty}\le\|\tilde W_L\|_{\infty}\|\tilde h_n\|_1\le 1$.

\medskip

\quad{\bf (v)} Finally,

$$\eqalign{\zeta\BbbR^r=(h\mu)\BbbR^r
&=(\lambda_L*\tilde h_n\mu)\BbbR^r
=\lambda_L\BbbR^r\cdot(\tilde h_n\mu)\BbbR^r\cr
&=\lambda_L\BbbR^r
=\capacity L
\le\epsilon+\capacity K.\cr}$$
}%end of proof of 479T

\leader{*479U}{Theorem} Let $K\subseteq\BbbR^r$ be compact,
and let $\Phi$ be
the set of Lipschitz functions $g:\BbbR^r\to\Bbb R$ such that
$g(x)\ge 1$ for every $x\in K$ and $\lim_{\|x\|\to\infty}g(x)=0$.   Then

$$\eqalign{r(r-2)\beta_r\capacity K
&=\inf\{\int_{\BbbR^r}\|\grad g\|^2d\mu:g\in\Phi\text{ is smooth and has
compact support}\}\cr
&=\inf\{\int_{\BbbR^r}\|\grad g\|^2d\mu:g\in\Phi\}.\cr}$$

\proof{{\bf (a)} By 479Te,

\Centerline{$\inf\{\int_{\BbbR^r}\|\grad g\|^2d\mu:g\in\Phi$
  is smooth and has compact support$\}
\le r(r-2)\beta_r\capacity K$.}

\medskip

{\bf (b)} Now suppose that $g\in\Phi$ is a smooth function with compact
support.   Then
$r(r-2)\beta_r\capacity K\le\int_{\BbbR^r}\|\grad g\|^2d\mu$.   \Prf\
Take any $\epsilon\in\ooint{0,1}$.
Then there is a $\delta>0$ such that $g(x)\ge 1-\epsilon$ for every
$x\in K+B(\tbf{0},\delta)$.   By 479Te, there is a Radon measure $\zeta$ on
$\BbbR^r$, with support included in $K+B(\tbf{0},\delta)$, such that $W_{\zeta}$
is smooth and has compact support, $W_{\zeta}\ge\chi K$,
$\zeta\BbbR^r\le\capacity K+\epsilon$ and

\Centerline{$\int_{\BbbR^r}\|\grad W_{\zeta}\|^2d\mu
=r(r-2)\beta_r\energy(\zeta)\le r(r-2)\beta_r\zeta\BbbR^r$.}

\noindent In this case,

$$\eqalignno{\int_{\BbbR^r}\varinnerprod{\grad g}{\grad W_{\zeta}}d\mu
&=r(r-2)\beta_r\int_{\BbbR^r}g\,d\zeta\cr
\displaycause{479Td}
&\ge(1-\epsilon)r(r-2)\beta_r\zeta\BbbR^r
\ge(1-\epsilon)\int_{\BbbR^r}\|\grad W_{\zeta}\|^2.\cr}$$

\noindent Setting $v=(1-\epsilon)\grad W_{\zeta}$, we have

\Centerline{$\int_{\BbbR^r}\varinnerprod{v}{\grad g}d\mu
\ge\int_{\BbbR^r}\|v\|^2d\mu$.}

\noindent But this means that

$$\eqalignno{\int_{\BbbR^r}\|\grad g\|^2d\mu
&=2\int_{\BbbR^r}\varinnerprod{v}{\grad g}d\mu-\int_{\BbbR^r}\|v\|^2d\mu
   +\int_{\BbbR^r}\|v-\grad g\|^2d\mu\cr
&\ge\int_{\BbbR^r}\|v\|^2d\mu
\ge(1-\epsilon)^2\int_{\BbbR^r}\|\grad W_{\zeta}\|^2d\mu\cr
&=(1-\epsilon)^2r(r-2)\beta_r\int_{\BbbR^r}W_{\zeta}d\zeta
\ge(1-\epsilon)^2r(r-2)\beta_r\int_{\BbbR^r}\tilde W_Kd\zeta\cr
\displaycause{because $W_{\zeta}\ge\tilde W_K$ on $K$, so
$W_{\zeta}\ge\tilde W_K$ everywhere, by 479Fg}
&=(1-\epsilon)^2r(r-2)\beta_r\int_{\BbbR^r}W_{\zeta}d\lambda_K\cr
\displaycause{479J(b-i)}
&\ge(1-\epsilon)^2r(r-2)\beta_r\int_{\BbbR^r}\tilde W_Kd\lambda_K
=(1-\epsilon)^2r(r-2)\beta_r\capacity K.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$r(r-2)\beta_r\capacity K\le\int_{\BbbR^r}\|\grad g\|^2d\mu$.\ \Qed

\medskip

{\bf (c)} If $g\in\Phi$ has compact support, then
$\int_{\BbbR^r}\|\grad g\|^2d\mu\ge r(r-2)\beta_r\capacity K$.   \Prf\
Let $R>0$ be such that $g$ is zero outside $B(\tbf{0},R)$.   Let $M\ge 0$ be such
that $g$ is $M$-Lipschitz;   then
$\|\grad g(x)\|\le M$ for every $x\in\dom\grad g$ (473Cc).
Take any $\epsilon>0$.   As in 479T, let $\sequencen{\tilde h_n}$ be
the smoothing sequence of 473E.
For $n\in\Bbb N$, set $g_n=(1+\epsilon)\tilde h_n*g$.   Then
$\grad g_n=(1+\epsilon)\tilde h_n*\grad g$ (473Dd) and
$\|\grad g_n\|_{\infty}\le M(1+\epsilon)$ (473Da).   In the limit,
$(1+\epsilon)\grad g\eae\lim_{n\to\infty}\grad g_n$ (473Ee).

There is an $m\in\Bbb N$ such that $(1+\epsilon)g(x)\ge 1$ for every
$x\in K+B(\tbf{0},\bover1{m+1})$;  now if $n\ge m$,

\Centerline{$g_n(x)\ge(1+\epsilon)\inf_{\|y\|\le 1/(n+1)}g(x-y)\ge 1$}

\noindent for every $x\in K$.   So

$$\eqalignno{(1+\epsilon)^2\int_{\BbbR^r}\|\grad g\|^2d\mu
&=(1+\epsilon)^2\int_{B(\tbf{0},R+1)}\|\grad g\|^2d\mu\cr
&=\lim_{n\to\infty}\int_{B(\tbf{0},R+1)}\|\grad g_n\|^2d\mu\cr
\displaycause{by the dominated convergence theorem}
&=\lim_{n\to\infty}\int_{\BbbR^r}\|\grad g_n\|^2d\mu\cr
\displaycause{because every $g_n$ is zero outside $B(\tbf{0},R+1)$}
&\ge r(r-2)\beta_r\capacity K\cr}$$

\noindent (applying (b) to $g_n$ for $n\ge m$).
As $\epsilon$ is arbitrary,
$r(r-2)\beta_r\capacity K\le\int_{\BbbR^r}\|\grad g\|^2d\mu$.\ \Qed

\medskip

{\bf (d)} If $g\in\Phi$, then
$\int_{\BbbR^r}\|\grad g\|^2d\mu\ge r(r-2)\beta_r\capacity K$.   \Prf\
Let $\epsilon>0$.   Set
$g_1(x)=\max(0,(1+\epsilon)g(x)-\epsilon)$ for $x\in\BbbR^r$.
Then $g_1\in\Phi$ has compact support, and
$\|g_1(x)-g_1(y)\|\le(1+\epsilon)\|g(x)-g(y)\|$ for all $x$,
$y\in\Bbb R^r$, so
$\|\grad g_1(x)\|\le(1+\epsilon)\|\grad g(x)\|$ whenever both gradients are
defined.   Accordingly

\Centerline{$(1+\epsilon)^2\int_{\BbbR^r}\|\grad g\|^2d\mu
\ge\int_{\BbbR^r}\|\grad g_1\|^2d\mu
\ge r(r-2)\beta_r\capacity K$}

\noindent by (c).   As $\epsilon$ is arbitrary, we have the result.\ \Qed

\medskip

{\bf (e)} Putting (a) and (d) together, the theorem is proved.
}%end of proof of 479U

\leader{*479V}{}\cmmnt{ We are ready for another theorem
along the lines of 476H, this time relating capacity and Lebesgue
measure.

%what about capacity and r-1-dimensional measure?

\medskip

\noindent}{\bf Theorem} Let $D\subseteq\BbbR^r$ be a set of finite outer
Lebesgue measure, and $B_D$ the closed ball with centre $0$ and
the same outer measure as $D$.   Then
the Choquet-Newton capacity $c(D)$ of $D$ is at least $\capacity B_D$.

\proof{{\bf (a)} We need an elementary fact about gradients.   Suppose that
$f$, $g:\BbbR^r\to\Bbb R$ and $x\in\BbbR^r$ are such that $\grad f$,
$\grad g$, $\grad(f\vee g)$ and $\grad(f\wedge g)$ are all defined at $x$.
Then $\{\grad(f\vee g)(x),\grad(f\wedge g)(x)\}=\{\grad f(x),\grad g(x)\}$.
\Prf\ (i) If $f(x)>g(x)$ then (because $f$ and $g$ are both continuous at
$x$) we have $\grad(f\vee g)(x)=\grad f(x)$,
$\grad(f\wedge g)(x)=\grad g(x)$ and the result is immediate.   (ii) The
same argument applies if $f(x)<g(x)$.   (iii) If $f(x)=g(x)$, consider
$h=|f-g|=(f\vee g)-(f\wedge g)$.   Then $\grad h(x)$ is defined, and
$h(x)=0\le h(y)$ for every $y$.   So all the partial derivatives of $h$
have to be zero at $x$, and $\grad h(x)=0$, that is,
$\lim_{y\to x}\Bover1{\|y-x\|}h(y)=0$.   It follows at once that
$\grad f(x)=\grad g(x)$, and therefore both are equal to
$\grad(f\vee g)(x)$ and $\grad(f\wedge g)(x)$.   So again we have the
result.\ \Qed

\medskip

{\bf (b)} Now for a further clause to add to Lemma 476E.   Suppose that
$e\in S_{r-1}=\partial B(\tbf{0},1)$ and $\alpha\in\Bbb R$;
let $R=R_{e\alpha}$ be the
reflection in the plane $\{x:\varinnerprod{x}{e}=\alpha\}$, and
$\psi=\psi_{e\alpha}:\Cal P\BbbR^r\to\Cal P\BbbR^r$ the partial-reflection
operator of 476D-476E, that is,

\Centerline{$\psi(D)=(W\cap(D\cup R[D]))\cup(W'\cap D\cap R[D])$}

\noindent for $D\subseteq\BbbR^r$, where
$W=\{x:\varinnerprod{x}{e}\ge\alpha\}$ and
$W'=\{x:\varinnerprod{x}{e}\le\alpha\}$.   Then $c(\psi(D))\le c(D)$ for
every $D\subseteq\BbbR^r$.

\medskip

\Prf{\bf (i)} Suppose first that $D=K$ is compact.   Take any
$\gamma>r(r-2)\beta_r\capacity K$.   By 479U, there is a Lipschitz function
$f:\BbbR^r\to\Bbb R$ such that $f(x)\ge 1$ for every $x\in K$,
$\lim_{\|x\|\to\infty}f(x)=0$ and
$\int_{\BbbR^r}\|\grad f\|^2d\mu\le\gamma$.   Set $g=fR$.   Of course
$g$ is Lipschitz and
$\int_{\BbbR^r}\|\grad g\|^2d\mu=\int_{\BbbR^r}\|\grad f\|^2d\mu$.
Now $f\vee g$ and $f\wedge g$ are also Lipschitz, so for almost every
$x\in\BbbR^r$ all the gradients $\grad f(x)$, $\grad g(x)$,
$\grad(f\vee g)(x)$ and $\grad(f\wedge g)(x)$ are defined;  by (a),
$\|\grad f\|^2+\|\grad g\|^2
\eae\|\grad(f\vee g)\|^2+\|\grad(f\wedge g)\|^2$.

Now consider the function $h$ defined by saying that

$$\eqalign{h(x)
&=(f\vee g)(x)\text{ if }x\in W,\cr
&=(f\wedge g)(x)\text{ if }x\in W'.\cr}$$

\noindent $h$ is Lipschitz and $\lim_{\|x\|\to\infty}h(x)=0$;  also
$h(x)\ge 1$ for every $x\in\psi(K)$.   For $x\in W\setminus W'$,
$h(x)=(f\vee g)(x)$ and $h(Rx)=(f\wedge g)(x)$, so
$\{\grad h(x),\grad(hR)(x)\}=\{\grad(f\vee g)(x),\grad(f\wedge g)(x)\}$ if
the gradients are defined;  for $x\in W'\setminus W$,
$h(x)=(f\wedge g)(x)$ and $h(Rx)=(f\vee g)(x)$, so
$\{\grad h(x),\grad(hR)(x)\}=\{\grad(f\vee g)(x),\grad(f\wedge g)(x)\}$ if
the gradients are defined.   Accordingly

$$\eqalign{\|\grad h\|^2+\|\grad(hR)\|^2
&\eae\|\grad(f\vee g)\|^2+\|\grad(f\wedge g)\|^2\cr
&\eae\|\grad f\|^2+\|\grad g\|^2.\cr}$$

\noindent By 479U again,

$$\eqalign{r(r-2)\beta_r\capacity(\psi(K))
&\le\int_{\BbbR^r}\|\grad h\|^2d\mu
=\Bover12\int_{\BbbR^r}(\|\grad h\|^2+\|\grad(hR)\|^2)d\mu\cr
&=\Bover12\int_{\BbbR^r}(\|\grad f\|^2+\|\grad g\|^2)d\mu
\le\gamma.\cr}$$

\noindent As $\gamma$ is arbitrary, $\capacity(\psi(K))\le\capacity K$.

\medskip

\quad{\bf (ii)} Now suppose that $D=G$ is open.   Then there is a
non-decreasing sequence $\sequencen{K_n}$ of compact sets with union $G$,
and $\sequencen{\psi(K_n)}$ is a non-decreasing sequence with union
$\psi(G)$.   So

\Centerline{$c(\psi(G))=\sup_{n\in\Bbb N}\capacity(\psi(K_n))
\le\sup_{n\in\Bbb N}\capacity K_n=c(G)$.}

\medskip

\quad{\bf (iii)} Finally, for arbitrary $D\subseteq\BbbR^r$, take any
$\gamma>c(D)$.   Then
there is an open set $G$ such that $D\subseteq G$ and $c(G)\le\gamma$
(because $c$ is outer regular, see 479E(d-i)).   In this case,
$\psi(D)\subseteq\psi(G)$, so

\Centerline{$c(\psi(D))\le c(\psi(G))\le c(G)\le\gamma$.}

\noindent As $\gamma$ is arbitrary, $c(\psi(D))\le c(D)$ and we are done.\
\Qed

\medskip

{\bf (c)} Now suppose that $E$ is a bounded Lebesgue measurable subset of
$\BbbR^r$ with finite perimeter.

\medskip

\quad{\bf (i)} Let $M\ge 0$ be such that
$E\subseteq B(\tbf{0},M)$.   Consider

$$\eqalign{\Cal E
&=\{F:F\subseteq B(\tbf{0},M)\text{ is Lebesgue measurable},\cr
&\mskip100mu
\mu F=\mu E,\,\per F\le\per E,\,c(F)\le c(E)\}.\cr}$$

\noindent Then $\Cal E$ is compact for the topology $\frak T_m$ of
convergence in measure as described in 474T.   \Prf\ By 474T,

\Centerline{$\Cal E_1=\{F:F$ is Lebesgue measurable,
$\per F\le\per E\}$}

\noindent is compact.   So if $\sequencen{F_n}$ is any sequence in
$\Cal E$, it has a subsequence $\sequencen{F'_n}$ which is
$\frak T_m$-convergent to $F\in\Cal E_1$ say (4A2Le;  recall that, as noted
in the proof of 474T, $\frak T_m$ is pseudometrizable).
Taking a further subsequence
if necessary, we can suppose that
$\mu((F\symmdiff F'_n)\cap B(\tbf{0},M))\le 2^{-n}$ for every
$n\in\Bbb N$.   Set $F'=\bigcup_{m\in\Bbb N}\bigcap_{n\ge m}F'_n$.
Because every $F'_n$ is included in $B(\tbf{0},M)$,
$F'$ is a $\frak T_m$-limit of $\sequencen{F'_n}$.   So
$\mu F'=\lim_{n\to\infty}\mu F'_n=\mu E$, and

\Centerline{$c(F')
=\lim_{m\to\infty}c(\bigcap_{n\ge m}F'_n)\le c(E)$.}

\noindent Finally, $F'\symmdiff F$ is negligible, so
$\partstar F'=\partstar F$ and $\per(F')=\per F\le\per E$.   Thus
$F'\in\Cal E$.   As $\sequencen{F_n}$ is arbitrary,
$\Cal E$ is relatively compact, by 4A2Le in the opposite direction.\ \Qed

\medskip

\quad{\bf (ii)} Because $B(\tbf{0},M)$ is bounded,
the function $F\mapsto\int_F\|x\|\mu(dx):\Cal E\to\coint{0,\infty}$ is
continuous, and must attain its infimum at $H$ say.   Let $B_E$ be the ball
with centre $0$ and the same measure as $E$.   Then
$B_E\subseteq\clstar H$.   \Prf\ (Compare part (b) of the proof of
476H.)  \Quer\ Otherwise, take $z\in B_E\setminus\clstar H$.
Then

\Centerline{$\lim_{\delta>0}
  \Bover{\mu(B(z,\delta)\setminus H)}{\mu B(z,\delta)}=1$,
\quad$\lim_{\delta>0}
  \Bover{\mu(B(z,\delta)\setminus B_E)}{\mu B(z,\delta)}\le\Bover12$,}

\noindent so there is a $\delta>0$ such that
$\mu(B(z,\delta)\setminus B_E)<\mu(B(z,\delta)\setminus H)$ and
$\mu(B_E\setminus H)>0$.   Because

\Centerline{$\mu(\clstar H)=\mu H=\mu E=\mu B_E$,}

\noindent $\clstar H\setminus B_E$ is also non-negligible.
Take $x_1\in\clstar H\setminus B_E$ and $x_0\in B_E\setminus\clstar H$.
Then $\delta_0=\|x_1\|-\|x_0\|$ is greater than $0$.
Since

\Centerline{$\limsup_{\delta\downarrow 0}
  \Bover{\mu(H\cap B(x_1,\delta))}{\mu B(x_1,\delta)}
>0=\lim_{\delta\downarrow 0}
  \Bover{\mu(H\cap B(x_0,\delta))}{\mu B(x_0,\delta)}$,}

\noindent there is a $\delta\in\ooint{0,\bover12\delta_0}$ such that
$\mu(H\cap B(x_1,\delta))>\mu(H\cap B(x_0,\delta))$.   Now let $e$ be the
unit vector $\Bover1{\|x_0-x_1\|}(x_0-x_1)$, and set
$\alpha=\varinnerprod{e}{\bover12(x_0+x_1)}$.   Consider the reflection
$R=R_{e\alpha}$ and the operator $\psi=\psi_{e\alpha}$;  set $H_1=\psi(H)$
and let $\phi=\phi_{H_1}:H\to H_1$ be the function of 476E.   As
$\alpha<0$, $\|\phi(x)\|\le\|x\|$ for every $x\in H$;   moreover,
$R[B(x_1,\delta)]=B(x_0,\delta)$, so

\Centerline{$\{x:\|\phi(x)\|<\|x\|\}
\supseteq\{x:x\in B(x_1,\delta)\cap H$, $Rx\notin H\}$}

\noindent is not negligible.   So
$\int_{H_1}\|x\|\mu(dx)<\int_H\|x\|\mu(dx)$.   On the other hand, we
surely have $H_1\subseteq B(\tbf{0},M)$, $\mu H_1=\mu H=\mu E$ and
$\per H_1\le\per H\le\per E$ (476Ee);  and, finally,
$c(H_1)\le c(H)\le c(E)$, by (b) of this proof.   Thus $H_1\in\Cal E$ and
the functional $F\mapsto\int_F\|x\|\mu(dx)$ is not minimized at $H$.\
\Bang\Qed

\medskip

\quad{\bf (iii)} Accordingly

$$\eqalignno{\capacity B_E
&\le c(\clstar H)\le c(H)\cr
\displaycause{479P(c-vi)}
&\le c(E).\cr}$$

\medskip

{\bf (d)} Thus $\capacity B_E\le c(E)$ whenever $E\subseteq\BbbR^r$ is
Lebesgue
measurable, bounded and has finite perimeter.   Consequently
$\capacity B_K\le\capacity K$ for every compact set $K\subseteq\BbbR^r$.
\Prf\ If $\epsilon>0$, there is an open set $G\supseteq K$ such that
$c(G)\le\capacity K+\epsilon$.   Now there is a set $E$, a finite union
of balls,
such that $K\subseteq E\subseteq G$.   In this case, $E$ has finite
perimeter and is bounded, while of course $B_E\supseteq B_K$.   So

\Centerline{$\capacity B_K
\le\capacity B_E\le c(E)\le c(G)\le\capacity K+\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\capacity B_K\le\capacity K$.\ \Qed

It follows that $\capacity B_E\le c(E)$ for every measurable set
$E\subseteq\BbbR^r$ of finite measure.   \Prf\ If $K\subseteq E$
is compact, then $\capacity B_K\le\capacity K\le c(E)$.   But as
$\mu E=\sup\{\mu K:K\subseteq E$ is compact$\}$,
$\diam B_E=\sup\{\diam B_K:K\subseteq E$ is compact$\}$;
because capacity is a continuous function of radius (479Da),

$$\eqalign{\capacity B_E)
&=\sup\{\capacity B_K:K\subseteq E\text{ is compact}\}\cr
&\le\sup\{\capacity K:K\subseteq E\text{ is compact }\}
\le c(E).  \text{ \Qed}\cr}$$

Finally, if $D$ is any set of finite outer measure, there is a G$_{\delta}$
set $E\supseteq D$ such that $c(E)=c(D)$ and $\mu E=\mu^*D$, so that

\Centerline{$\capacity B_D=\capacity B_E\le c(E)=c(D)$,}

\noindent and we have the general result claimed.
}%end of proof of 479V

\leader{*479W}{}\cmmnt{ I conclude with an alternative representation of
Choquet-Newton capacity $c$ in terms of a measure on the space of closed
subsets of $\BbbR^r$.

\medskip

\noindent}{\bf Theorem}
Let $\Cal C^+$ be the family of non-empty closed subsets of $\BbbR^r$,
with its Fell topology\cmmnt{ (4A2T)}.   Then there is a unique Radon
measure $\theta$ on $\Cal C^+$ such that
$\theta^*\{C:C\in\Cal C^+$, $D\cap C\ne\emptyset\}$ is the Choquet-Newton
capacity $c(D)$ of $D$ for every $D\subseteq\BbbR^r$.

\proof{{\bf (a)} Recall that the Fell topology on
$\Cal C=\Cal C^+\cup\{\emptyset\}$ is compact (4A2T(b-iii)) and metrizable
(4A2Tf), so $\Cal C^+$ is locally compact and Polish.
For $D\subseteq\BbbR^r$, set
$\Psi D=\{C:C\in\Cal C^+$, $C\cap D\ne\emptyset\}$.   Of course
$\Psi(\bigcup\Cal A)=\bigcup_{D\in\Cal A}\Psi D$ for every family
$\Cal A$ of subsets of $\BbbR^r$.

\medskip

{\bf (b)} Let $\Omega'$ be the set of those $\omega\in\Omega$ such that
$\lim_{t\to\infty}\|\omega(t)\|=\infty$;  because $r\ge 3$, $\Omega'$ is
conegligible in $\Omega$ (478Md).   If $\omega\in\Omega'$, then
$\omega[\,\coint{0,\infty}\,]$ is closed.
For $x\in\BbbR^r$ and $\omega\in\Omega'$, set
$h_x(\omega)=x+\omega[\,\coint{0,\infty}\,]\in\Cal C^+$.   Then
$h_x:\Omega'\to\Cal C^+$ is Borel measurable.   \Prf\ ($\alpha$) If
$G\subseteq\BbbR^r$ is open, then

\Centerline{$\{\omega:\omega\in\Omega'$, $h_x(\omega)\cap G\ne\emptyset\}
=\bigcup_{t\ge 0}\{\omega:x+\omega(t)\in G\}$}

\noindent is relatively open in $\Omega'$.
($\beta$) If $K\subseteq\BbbR^r$ is compact,

\Centerline{$\{(\omega,t):x+\omega(t)\in K\}$}

\noindent is closed in $\Omega\times\coint{0,\infty}$, so its projection
$\{\omega:x+\omega[\,\coint{0,\infty}\,]\cap K\ne\emptyset\}$ is
F$_{\sigma}$, and
$\{\omega:\omega\in\Omega'$, $h_x(\omega)\cap K=\emptyset\}$ is a
G$_{\delta}$ set in $\Omega'$.   ($\gamma$)
Because $\Cal C^+$ is hereditarily Lindel\"of, this is enough to prove that
$h_x$ is Borel measurable (4A3Db).\ \Qed

\medskip

{\bf (c)} Let $\Tau$ be the ring of subsets of $\Cal C^+$ generated by
sets of the form $\Psi E$ where $E\subseteq\BbbR^r$ is bounded and is
either compact or open.   Then we have an additive functional
$\phi:\Tau\to\coint{0,\infty}$ such that
$\phi(\Psi K)=\capacity K$ for every compact set $K\subseteq\BbbR^r$.
\Prf\ For $x\in\BbbR^r$ let $h_x:\Omega'\to\Cal C^+$ be as in (b).
Then we have a corresponding scaled Radon
image measure $\phi_x=\|x\|^{r-2}(\mu_W)_{\Omega'}h_x^{-1}$ on $\Cal C^+$
(418I), defined by setting
$\phi_xH=\|x\|^{r-2}\mu_W\{\omega:x+\omega[\,\coint{0,\infty}\,]\in H\}$
whenever this is defined.   If $E\subseteq\BbbR^r$ is either compact or
open, then

\Centerline{$\{\omega:x+\omega[\,\coint{0,\infty}\,]\cap E\ne\emptyset\}$}

\noindent is F$_{\sigma}$ or open, respectively, so $\phi_x(\Psi E)$ is
defined;
accordingly $\phi_xH$ is defined for every $H\in\Tau$.   If
$\gamma>0$, $E\subseteq B(\tbf{0},\gamma)$ and
$\|x\|>\gamma$, then

\Centerline{$\phi_x^*(\Psi E)
\le\phi_x(\Psi(B(\tbf{0},\gamma)))
=\|x\|^{r-2}\hp(B(\tbf{0},\gamma)-x)
=\gamma^{r-2}$}

\noindent(478Qc).   So $\limsup_{\|x\|\to\infty}\phi_xH$ is finite for
every $H\in\Tau$.   Take an ultrafilter $\Cal F$ on $\BbbR^r$ containing
$\BbbR^r\setminus B(\tbf{0},\gamma)$ for every $\gamma>0$;  then
$\phi H=\lim_{x\to\Cal F}\phi_xH$ is defined in $\coint{0,\infty}$
for every $H\in\Tau$, and $\phi$ is additive.   If $E\subseteq\BbbR^r$ is
bounded and either compact or open, then

\Centerline{$\phi(\Psi E)
=\lim_{x\to\Cal F}\|x\|^{r-2}\hp(E-x)=c(E)$}

\noindent by 479B(ii).\ \Qed

\medskip

{\bf (d)} $\phi$ is inner regular with respect to the compact sets, in the
sense that $\phi H=\sup\{\phi L:L\in\Tau$ is compact, $L\subseteq H\}$
for every $H\in\Tau$.   \Prf\ Note first that the set

\Centerline{$\Cal H
=\{H:H\in\Tau$,
  $\phi H=\sup\{\phi L:L\in\Tau$ is compact, $L\subseteq H\}\}$}

\noindent is a sublattice of $\Tau$.   Suppose that $E$,
$H\subseteq\BbbR^r$ are bounded sets which are either compact or open, and
$\epsilon>0$.   Then there are a compact $K\subseteq E$ and a bounded open
$G\supseteq H$ such that $c(E)\le\epsilon+\capacity K$ and
$c(G)\le\epsilon+c(H)$ (479E).   Now $\Psi K$ is a closed
subset of $\Cal C$ included in $\Cal C^+$, so is a compact subset of
$\Cal C^+$, while $\Psi G$ is open;  thus
$L=\Psi K\setminus\Psi G$ is a compact subset of
$\Psi E\setminus\Psi H$, and of course $L\in\Tau$.   Now

$$\eqalign{\phi(\Psi E\setminus\Psi H)
&\le\phi L+\phi(\Psi E\setminus\Psi K)+\phi(\Psi G\setminus\Psi H)\cr
&=\phi L+\phi(\Psi E)-\phi(\Psi K)+\phi(\Psi G)-\phi(\Psi H)\cr
&=\phi L+c(E)-\capacity K+c(G)-c(H)
\le\phi L+2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\Psi E\setminus\Psi H$ belongs to
$\Cal H$.

Since any member of $\Tau$ is expressible as a finite union of finite
intersections of sets of this kind, $\Tau\subseteq\Cal H$, as required.\
\Qed

\medskip

{\bf (e)} Let $\Cal L$ be the family of compact subsets of $\Cal C^+$.
If $L\in\Cal L$, it is a closed subset of $\Cal C$ not containing
$\emptyset$, so there must be a compact set $K\subseteq\BbbR^r$ such that
$L\subseteq\Psi K$.   Thus every member of $\Cal L$ is covered by a member
of $\Tau$, and we have a functional $\phi_1:\Cal L\to\coint{0,\infty}$
defined by setting $\phi_1L=\inf\{\phi E:E\in\Tau$, $L\subseteq E\}$ for
$L\in\Cal L$.

I seek to apply 413J.   Of course $\emptyset\in\Cal L$ and $\Cal L$ is
closed under finite disjoint unions and countable intersections;  moreover,
if $\sequencen{L_n}$ is a non-increasing sequence in $\Cal L$ with empty
intersection, one of the $L_n$ must be empty, so
$\inf_{n\in\Bbb N}\phi_1L_n=0$.   Now turn to condition ($\alpha$) of the
theorem:

\Centerline{$\phi_1L_1
=\phi_1L_0+\sup\{\phi_1L:L\in\Cal L$, $L\subseteq L_1\setminus L_0\}$
whenever $L_0$, $L_1\in\Cal L$ and $L_0\subseteq L_1$.}

\medskip

\quad{\bf (i)} If $L_0$, $L\in\Cal L$ are disjoint, then
$\phi_1(L_0\cup L)\ge\phi_1L_0+\phi_1L$.   \Prf\ The topology $\frak S$ of
$\Cal C^+$ is generated by sets of the form $\Psi G$, where
$G\subseteq\BbbR^r$ is open, and by sets of the form
$\Cal C^+\setminus\Psi K$, where $K\subseteq\BbbR^r$ is compact.
It is therefore generated by

\Centerline{$\{\Psi G\setminus\Psi K:G\subseteq\BbbR^r$ is bounded and
open, $K\subseteq\BbbR^r$ is compact$\}
\subseteq\Tau$.}

\noindent So $\frak S\cap\Tau$ is a base for $\frak S$ and disjoint
compact sets in $\Cal C^+$ can be separated by members of $\Tau$
(4A2F(h-i));  let $E_0\in\Tau$ be such that
$L_0\subseteq E_0\subseteq\Cal C^+\setminus L$.   Now if $E\in\Tau$ and
$E\supseteq L_0\cup L$,

\Centerline{$\phi E=\phi(E\cap E_0)+\phi(E\setminus E_0)
\ge\phi_1L_0+\phi_1L$;}

\noindent as $E$ is arbitrary,
$\phi_1(L_0\cup L)\ge\phi_1L_0+\phi_1L$.\ \Qed

\medskip

\quad{\bf (ii)} If $L_0$, $L_1\in\Cal L$, $L_0\subseteq L_1$ and
$\epsilon>0$, there is an $L\in\Cal L$ such that
$L\subseteq L_1\cup L_0$ and $\phi_1L_1\le\phi_1L_0+\phi_1L+3\epsilon$.
\Prf\ Let $E_0$, $E_1\in\Tau$ be such that $L_0\subseteq E_0$,
$L_1\subseteq E_1$ and $\phi E_0\le\phi_1L_0+\epsilon$.
By (d), there is an
$L'\in\Cal L\cap\Tau$ such that $L'\subseteq E_1\setminus E_0$ and
$\phi L'\ge\phi(E_1\setminus E_0)-\epsilon$.   Set $L=L'\cap L_1$.
Then $L\in\Cal L$ and $L\subseteq L_1\setminus L_0$.   Let $E\in\Tau$ be
such that $L\subseteq E$ and $\phi E\le\phi_1L+\epsilon$.   Then
$L_1\subseteq E_0\cup E\cup((E_1\setminus E_0)\setminus L')$, so

\Centerline{$\phi_1L_1
\le\phi E_0+\phi E+\phi(E_1\setminus E_0)-\phi L'
\le\phi_1L_1+\phi L+3\epsilon$,}

\noindent as required.\ \Qed

Putting this together with (i), the final condition of 413J is satisfied.

\medskip

{\bf (f)} We therefore have a complete locally determined measure $\theta$
on $\Cal C^+$ extending $\phi_1$ and inner regular with respect to
$\Cal L$.   For $E\subseteq\Cal C^+$, $\theta$ measures $E$ iff $\theta$
measures $E\cap L$ for every $L\in\Cal L$ (412Ja);
so $\theta$ measures all
closed subsets of $\Cal C^+$, and is a topological measure.   Of course
$\theta$ is inner regular with respect to the compact sets.   If
$C\in\Cal C^+$, there is a bounded open set $G\subseteq\BbbR^r$ meeting
$C$, and now $\Psi G$ is an open set containing $C$ and included in the
compact set $\Psi\overline{G}$;  accordingly

\Centerline{$\theta(\Psi G)\le\theta(\Psi\overline{G})
=\phi_1(\Psi\overline{G})=\phi(\Psi\overline{G})
=\capacity\overline{G}$}

\noindent is finite.   Thus $\theta$ is locally finite and is a Radon
measure.

\medskip

{\bf (g)} As in (f), we have

\Centerline{$\theta(\Psi K)=\phi_1(\Psi K)=\phi(\Psi K)=\capacity K$}

\noindent for every compact $K\subseteq\BbbR^r$.   Next,
$\theta(\Psi G)=c(G)$ for every open $G\subseteq\BbbR^r$.   \Prf\
If $G$ is bounded,

$$\eqalign{\theta(\Psi G)
&=\sup\{\theta L:L\subseteq\Psi G\text{ is compact}\}\cr
&=\sup\{\phi_1L:L\subseteq\Psi G\text{ is compact}\}
\le\phi(\Psi G)
=c(G)\cr
&=\sup\{\capacity K:K\subseteq G\text{ is compact}\}\cr
&=\sup\{\theta(\Psi K):K\subseteq G\text{ is compact}\}
\le\theta(\Psi G).\cr}$$

\noindent If $G$ is unbounded, then there is a non-decreasing sequence
$\sequencen{G_n}$ of bounded open sets with union $G$, so

\Centerline{$\theta(\Psi G)
=\theta(\bigcup_{n\in\Bbb N}\Psi G_n)
=\sup_{n\in\Bbb N}\theta(\Psi G_n)
=\sup_{n\in\Bbb N}c(G_n)
=c(G)$.  \Qed}

\medskip

{\bf (h)} Now suppose that $D\subseteq\BbbR^r$ is any bounded set.   We
have

\Centerline{$\theta^*(\Psi D)
\le\inf\{\theta(\Psi G):G\supseteq D\text{ is open}\}
=\inf\{c(G):G\supseteq D\text{ is open}\}
=c(D)$.}

\noindent\Quer\ Suppose, if possible, that $\theta^*(\Psi D)<c(D)$.
Let $G\supseteq D$ be a bounded open set.   Then there is a compact
$L\subseteq\Psi G\setminus\Psi D$ such that
$\theta L>\theta(\Psi G)-c(D)$.   Set $F=\bigcup_{C\in L}C$;  then $F$
is closed (4A2T(e-iii)) and disjoint from $D$, so $G\setminus F$ is open,
$D\subseteq G\setminus F$ and $\Psi(G\setminus F)$ is disjoint from $L$.
But this means that

\Centerline{$c(D)\le c(G\setminus F)=\theta(\Psi(G\setminus F))
\le\theta(\Psi G)-\theta L<c(D)$,}

\noindent which is absurd.\ \BanG\  So $\theta^*(\Psi D)=c(D)$.

If $D\subseteq\BbbR^r$ is any set, then it is expressible as the union of a
non-decreasing sequence $\sequencen{D_n}$ of bounded sets, so

\Centerline{$c(D)
=\lim_{n\to\infty}c(D_n)
=\lim_{n\to\infty}\theta^*(\Psi D_n)
=\theta^*(\bigcup_{n\in\Bbb N}\Psi D_n)
=\theta^*(\Psi D)$.}

\noindent Thus $\theta$ has all the properties declared.

\medskip

{\bf (i)} To see that $\theta$ is unique, consider the base $\Cal V$ for
the topology of $\Cal C^+$ consisting of sets of the form
$\bigcap_{i\in I}\Psi G_i\setminus\Psi K$ where $\familyiI{G_i}$ is a
non-empty finite family of bounded open sets in $\BbbR^r$ and
$K\subseteq\BbbR^r$ is compact.   The conditions that $\theta$ must satisfy
determine its value on any set of the form
$\Psi(G\cup K)=\Psi G\cup\Psi K$ where $G\subseteq\BbbR^r$ is open and
$K\subseteq\BbbR^r$ is compact, and therefore determine its values on
$\Cal V$.   By 415H(iv), $\theta$ is fixed by these.
}%end of proof of 479W

\exercises{\leader{479X}{Basic exercises (a)}
%\spheader 479Xa
Let $\zeta$ be a Radon measure on $\BbbR^r$.  Show that

\Centerline{$\zeta\BbbR^r
=\lim_{\gamma\to\infty}\Bover2{r\beta_r\gamma^2}
  \int_{B(\tbf{0},\gamma)}W_{\zeta}d\mu$.}
%479J out of order query

\sqheader 479Xb(i) Show directly from 479B-479C that Choquet-Newton
capacity $c$ is
invariant under isometries of $\BbbR^r$.
(ii) Show that $c(\alpha D)=\alpha^{r-2}c(D)$
whenever $\alpha\ge 0$ and $D\subseteq\BbbR^r$.
%479C

\spheader 479Xc Suppose that $\zeta_1$ and $\zeta_2$ are totally finite
Radon measures on $\BbbR^r$.
Show that $W_{\zeta_1*\zeta_2}=\zeta_1*W_{\zeta_2}=\zeta_2*W_{\zeta_1}$.
%479J

\spheader 479Xd Show that there is a closed set $F\subseteq\Bbb R^r$
such that $\hp(F)<1$ but $c(F)=\infty$.   \Hint{look at the proof of
479Ma.}
%479M

\spheader 479Xe Let $K\subseteq\BbbR^r$ be compact.  Show that
$\interior\{x:\tilde W_K(x)<1\}$ is the unbounded component of
$\BbbR^r\setminus\supp\lambda_K$.   \Hint{setting $L=\supp\lambda_K$,
show that $\capacity(\supp\lambda_K)=\capacity K$ so that
$\tilde W_K=\tilde W_L$.}
%479N

\spheader 479Xf Let $A\subseteq\BbbR^r$ be an analytic set such that
$c(A)<\infty$.   Show that
$\tilde W_A=\sup\{W_{\zeta}:\zeta$ is a Radon measure on $\BbbR^r$,
$A$ is $\zeta$-conegligible, $W_{\zeta}\le 1\}$.
%479N 479M query out of order

\sqheader 479Xg Show that there
is a universally negligible set $D\subseteq B(\tbf{0},1)$ such that
$c(D)=1$.   \Hint{use the ideas of 439F to find $D$ such that
$\{\|x\|:x\in D\}$ is universally negligible and
$x\mapsto\|x\|:D\to[0,1]$ is injective, but
$\nu^*\{\Bover{x}{\|x\|}:x\in D$, $\|x\|\ge 1-\delta\}=r\beta_r$
for every $\delta\in\ooint{0,1}$;  compute $c(D)$ with the aid of
479D, 479P(c-iii-$\alpha$) and 479P(c-vii).}
		
\spheader 479Xh Suppose that $D\subseteq D'\subseteq\BbbR^r$ and
$c(D')<\infty$.
Show that $\int fd\lambda_D\le\int fd\lambda_{D'}$
for every lower semi-continuous superharmonic $f:\BbbR^r\to[0,\infty]$.
%479P

%if W_{\zeta}\le W_{\zeta'} does it follow that
%\int fd\zeta\le\int fd\zeta'  for non-neg lsc superharmonic f ?
%yes I think, see 479qS in mt47bits.tex
%so in fact  \zeta*f\le\zeta'*f

\spheader 479Xi Let $D\subseteq\BbbR^r$ be a bounded set.   Show that
$c(D)=\lim_{\|x\|\to\infty}\|x\|^{r-2}\hp^*(D-x)$.  \Hint{477Id.}
%479P

\spheader 479Xj Let $T:\BbbR^r\to\BbbR^r$ be an isometry, and
$D\subseteq\BbbR^r$ a set such that $c(D)<\infty$.   Show that
$\tilde W_{T[D]}Tx)=\tilde W_D(x)$ for every $x\in\BbbR^r$ and that
$\lambda_{T[D]}$ is the image measure $\lambda_DT^{-1}$.
%479P

\spheader 479Xk Let $D\subseteq\BbbR^r$ be a set such that
$c(D)<\infty$, and $\alpha>0$.   Show that
$\tilde W_{\alpha D}(x)=\tilde W_D(\bover1{\alpha}x)$
for every $x\in\BbbR^r$, and that
$\lambda_{\alpha D}=\alpha^{r-2}\lambda_DT^{-1}$,
where $T(x)=\alpha x$ for $x\in\BbbR^r$.
%479P 479Xj

\spheader 479Xl Show that
$c(D)
=\inf\{\zeta\BbbR^r:\zeta$ is a Radon measure on $\BbbR^r$,
$W_{\zeta}\ge\chi D\}
=\inf\{\energy(\zeta):\zeta$ is a Radon measure on $\BbbR^r$,
$W_{\zeta}\ge\chi D\}$
for any $D\subseteq\BbbR^r$.
%479P

\spheader 479Xm Let $K\subseteq\BbbR^r$ be a compact set, with complement
$G$, and $\Phi$ the set of
continuous harmonic functions $f:G\to[0,1]$ such that
$\lim_{\|x\|\to\infty}f(x)=0$.   Show that
$\tilde W_K\restr G$ is the greatest element of $\Phi$.
\Hint{479Pb, 478Pc.}
%479P

\spheader 479Xn(i) Show that if $G$ is a convex open set then
$\hp(G-x)=1$ for every $x\in\overline{G}$.   (ii) Show that if
$D\subseteq\BbbR^r$ is a convex bounded set with non-empty
interior, then $\tilde W_D$ is continuous.
%479P

\spheader 479Xo Show that if $D$, $D'\subseteq\BbbR^r$
and $c(D\cup D')<\infty$, then
$\tilde W_{D\cap D'}+\tilde W_{D\cup D'}\le\tilde W_D+\tilde W_{D'}$.
%479P

\spheader 479Xp Let $D\subseteq\BbbR^r$ be a set such that
$c(D)<\infty$, and set $\tilde D=\{x:\tilde W_D(x)=1\}$.   Show that
(i) $D\setminus\tilde D$ is polar (ii)
$\lambda_{\tilde D}=\lambda_D$.   \Hint{reduce to the case in which $D=A$
is analytic;  use 479Fg to show that $\tilde W_{\tilde A}\le\tilde W_A$;
use 479J(b-v).}
%479P

\spheader 479Xq Let $\Cal A$ be the set of subsets of $\BbbR^r$
with finite Choquet-Newton capacity, and
$\rho$ the pseudometric $(D,D')\mapsto 2c(D\cup D')-c(D)-c(D')$ (432Xj).
(i) Show that $\|U_{\lambda_D}-U_{\lambda_{D'}}\|^2_2\le 2c_r\rho(D,D')$
for $D$, $D'\in\Cal A$.
(ii) Show that $\rho(D,D')=0$ iff $\lambda_D=\lambda_{D'}$.
%479P query out of order

\spheader 479Xr Suppose that $D\subseteq\BbbR^r$.   Show that the
following are equiveridical:  (i) $D$ is polar;  (ii) there is some
$x\in\BbbR^r$ such that $\hp(D-x)=0$;  (iii)
$\hp((D\setminus\{x\})-x)=0$ for every $x\in\BbbR^r$.
%479P
%if (i), enlarge $D$ to analytic polar, then $\lambda_D=0$ so
%$\tilde W_D=0$ so (iii).   If (ii), again can enlarge to analytic,
%and $\tilde W_D=0$ so (iii) and (i).
%if (iii), $D\ne\BbbR^r$ so (ii).

\spheader 479Xs Suppose that $\sequencen{D_n}$ is a non-increasing sequence
of subsets of $\BbbR^r$ such that $\inf_{n\in\Bbb N}c(D_n)$ is finite and
$\bigcap_{n\in\Bbb N}D_n=\bigcap_{n\in\Bbb N}\overline{D_n}=F$ say.   Show
that $\sequencen{\lambda_{D_n}}\to\lambda_D$ for the narrow topology, and
that $\sequencen{c(D_n)}\to c(F)$.   (Compare 479Ye.)
%479P 479Ec

\spheader 479Xt For $\omega\in\Omega$ set
$\tau(\omega)=\sup\{t:\|\omega(t)\|\le 1\}$.   (i) Show that
$\tau:\Omega\to[0,\infty]$ is measurable.
(ii) Show that if $r\le 2$ then $\tau=\infty$ a.e.   (iii) Show that if
$r\ge 3$ then $\tau$ is not a stopping time.   (iv) Show that if
$3\le r\le 4$ then $\tau$ is finite a.e., but has infinite expectation.
(v) Show that if $r\ge 5$ then $\tau$ has finite expectation.
({\it Hint\/}:  show that if $r\ge 2$ then

\Centerline{$\Pr(\tau\ge t)
=\Bover1{(\sqrt{2\pi})^r}\int e^{-\|x\|^2/2}
\Bover1{\max(1,(\sqrt t)^{r-2}\|x\|^{r-2})}dx$.)}
%479R 479Da 479Pb {\smc Chung 95}, chap 11

\leader{479Y}{Further exercises (a)}%
%\spheader 479Ya
(i) Show that there is an open set $G\subseteq B(\tbf{0},1)$,
dense in $B(\tbf{0},1)$, such that $c(G)<1$.   (ii) Show that
$\capacity(\supp\lambda_G)=1$.
%479E

\spheader 479Yb In 479G, suppose that $0<\alpha<r$, $0<\beta<r$ and
$\alpha+\beta>r$.   Show that

$$k_{\alpha+\beta-r}
=\bover{\Gamma(r-\bover{\alpha+\beta}2)\Gamma(\bover{\alpha}2)
                  \Gamma(\bover{\beta}2)}
   {(\sqrt\pi)^r\Gamma(\bover{\alpha+\beta-r}2)\Gamma(\bover{r-\alpha}2)
                  \Gamma(\bover{r-\beta}2)}
   k_{\alpha}*k_{\beta}.$$
%479Ia

\spheader 479Yc Let $A\subseteq\BbbR^r$ be an analytic set with
$c(A)<\infty$.   (i) Show that for every $\gamma>0$ there is a Radon
measure $\zeta_{\gamma}$ on $\BbbR^r$ such that
$\family{x}{\partial B(\tbf{0},\gamma)}{\Bover1{r\beta_r\gamma}\mu^{(A)}_x}$ is a
disintegration of $\zeta_{\gamma}$ over the subspace measure
$\nu_{\partial B(\tbf{0},\gamma)}$.
(ii) Show that $\lim_{\gamma\to\infty}\zeta_{\gamma}=\lambda_A$ for the
total variation metric on $M^+_{\text{R}}(\BbbR^r)$.
%479M

\spheader 479Yd Set $c'(D)=\sup\{\zeta^*D:\zeta$ is a Radon measure on
$\BbbR^r$ such that $W_{\zeta}\le 1$ everywhere$\}$ for
$D\subseteq\BbbR^r$.   Show that $c'$ is a Choquet capacity on $\BbbR^r$,
extending Newtonian capacity for compact sets, which is different from
Choquet-Newton capacity.
%479N 479Xg 479P

\spheader 479Ye
Let $\sequencen{D_n}$ be a non-increasing sequence of
subsets of $\BbbR^r$ with finite Choquet-Newton capacity.
For each $n\in\Bbb N$, set $\tilde D_n=\{x:\tilde W_{D_n}(x)=1\}$, and set
$A=\bigcap_{n\in\Bbb N}\tilde D_n$.   Show that
$c(A)=\inf_{n\in\Bbb N}c(D_n)$ and that $\lambda_A$ is the limit of
$\sequencen{\lambda_{D_n}}$ for the narrow topology on
$M^+_{\text{R}}(\BbbR^r)$.
%479Xp 479P 478Yi

\spheader 479Yf Let $\Cal A$, $\rho$ be as in 479Xq, and let
$(\bar{\Cal A},\bar\rho)$ be the corresponding metric space, identifying
members of $\Cal A$
which are zero distance apart.   Show that $\bar{\Cal A}$ is complete.
%479Ye 479P

\spheader 479Yg
Let $E\subseteq\BbbR^r$ be a set of finite Lebesgue
measure, and $B_E$ the ball with centre $0$ and the same measure as $E$.
Show that $\energy(\mu\LLcorner E)\le\energy(\mu\LLcorner B_E)$.
%479V mt47bits

\spheader 479Yh Prove 479V from 479U and 476Yb.
%479V

\spheader 479Yi(i) Show that $c$ is alternating of all orders, that is,

\Centerline{$\sum_{J\subseteq I,\#(J)\text{ is even}}
   c(D\cup\bigcup_{i\in J}D_i)
\le\sum_{J\subseteq I,\#(J)\text{ is odd}}c(D\cup\bigcup_{i\in J}D_i)$}

\noindent whenever $I$ is a
non-empty finite set, $\familyiI{D_i}$ is a family of
subsets of $\BbbR^r$ and $D$ is another subset of
$\BbbR^r$.   (Cf.\ 132Yf.)
(ii) Show that if $c(D\cup\bigcup_{i\in I}D_i)<\infty$, then

\Centerline{$\sum_{J\subseteq I,\#(J)\text{ is even}}
   \tilde W_{D\cup\bigcupop_{i\in J}D_i}
\le\sum_{J\subseteq I,\#(J)\text{ is odd}}
   \tilde W_{D\cup\bigcupop_{i\in J}D_i}$.}
%479W

\spheader 479Yj Let us say that if $X$ is a Polish space, a set
$A\subseteq X$ is {\bf projectively universally measurable} if $W[A]$ is
universally measurable whenever $Y$ is a Polish space and
$W\subseteq X\times Y$ is analytic.   Show that we can replace
the word `analytic' by the phrase `projectively universally measurable' in
all the theorems of this section.
%479 notes

\spheader 479Yk Suppose that $A\subseteq\BbbR^r$ is analytic and non-empty,
and $x\in\BbbR^r$ is such that $\rho(x,A)=\delta>0$.
Show that $\energy(\mu^{(A)}_x)\le\Bover1{\delta^{r-2}}$.
%479C out of order query mt47bits

\def\clcap{\text{cl}_{\text{cap}}}
\spheader 479Yl Show that if $D\subseteq\BbbR^r$ and $c(D)<\infty$, then
$c(\{x:\tilde W_D(x)\ge\gamma\})\le\Bover1{\gamma}c(D)$ for every
$\gamma>0$.
%479P mt47bits

\spheader 479Ym
For a set $D\subseteq\BbbR^r$ with $c(D)<\infty$, set
$\clcap D=\{x:\tilde W_D(x)=1\}$.   Show that
$c(D)=c(\clcap D)=c(D\cup\clcap D)$.
%479P mt47bits 479Yl
}%end of exercises

\endnotes{
\Notesheader{479} Newtonian potential is another of the great concepts of
mathematics, and is one of the points at which physical problems and
intuitions have stimulated and illuminated the development of analysis.
As with all the best ideas of mathematics, there is more than one route to
it, and any proper understanding of it must include a matching of the
different approaches.   In the exposition here I start with a description
of equilibrium measures in terms of harmonic measures (479B), themselves
defined in 478P in terms of Brownian motion.   We are led quickly to
definitions of capacity and equilibrium potential (479C), with some
elementary properties (479D).   Moreover, some very striking further
results (479E, 479W) are already accessible.

However we are still rather
far from the original physical concept of `capacity' of a conductor.
If you have ever studied electrostatics, the ideas here may recall some
basic physical principles.   The kernel $x\mapsto\Bover1{\|x\|^{r-2}}$
represents the potential energy field of a point mass or charge;
the potential $W_{\zeta}$ represents the field due to a mass or
charge with distribution $\zeta$.   The capacity of a set $K$ is the
largest charge that can be put on $K$ without raising the potential of
any point above $1$ (479Na), and the infimum of the charges
which raise the potential of every point of $K$ to $1$ (479P(c-v)).
The result that $\lambda_K$ is supported by $\partial K$ (479B(i))
corresponds to the principle that the charge on a conductor always collects
on the surface of the conductor;  479D(b-iii)
corresponds to the principle that there is no electric field
inside a conductor.

At the same time, the
equilibrium measure is supposed to be the (unique) distribution of the
charge, which on physical grounds ought to be the distribution with least
energy, as in 479K.   To reach these ideas, it seems that we need to know
various non-trivial facts from classical analysis, which I set out in
479G-479I.  %479G 479H 479I
The deepest of these is in 479Ib:  for the Riesz kernels $k_{\alpha}$,
the convolution $\zeta*k_{\alpha}$ determines the totally finite Radon
measure $\zeta$.   I do not know of any way of establishing this except
through the Fourier analysis of 479H and the detailed calculations of 479G
and 479Ia.

The ideas here are connected in so many ways that there is no clear flow to
the logic, and we are more than usually in danger of using circular
arguments.   In my style of exposition, this complexity manifests itself in
an exceptional density of detailed back-references;  I hope that these
will enable you to check the proofs effectively.   On a larger scale,
the laborious zigzag
progression from the original notion of capacity of compact
sets, as in 479K and 479U, through bounded analytic sets (479B, 479E)
and general analytic sets (479M, 479N) to arbitrary sets (479P),
displays a choice of path to which there are surely many
alternatives.

Of course we cannot expect all the properties of
Newtonian capacity to have recognizable forms in such a general context
as that of 479P
(see 479Xg, which shows that we cannot hope to replicate the ideas of
479Na-479Nc), but the elementary results if 479D mostly extend (479Pc).
More importantly, we have a quite new characterization of
equilibrium potentials (479Pb).
With these techniques available, we can learn a good deal more about
Brownian motion.   479R is a curious
and striking fact to go with 477K, 477L and 478M.   It is not a surprise
that capacity and Hausdorff dimension should be linked, but it is notable
that the phase change is at dimension $r-2$ (479Q);  this goes naturally
with 479P(c-vii).   I know of no such dramatic difference between four and
five dimensions, but for some purposes 479Xt marks a significant change.

My treatment is an unconventional one, so perhaps I should indicate points
where you should expect other authors to diverge from it.
While the notions of Newtonian
capacity, equilibrium measure and equilibrium potential are solidly
established for compact sets in $\BbbR^r$ (at least up to scalar factors,
and for $r\ge 3$),
the extension to general bounded analytic sets is not I think standard.
(I try to signal this by writing $c(A)$ in place of $\capacity A$, after
479E, for sets which are not guaranteed to be compact, even when the
definition in 479C(a-i) is applicable.
The fact that this step gives very little extra trouble is a
demonstration of the power of the Brownian-motion approach.)   The
further extension of Newtonian
capacity, defined on compact sets, to a Choquet capacity, defined on every
subset of $\BbbR^r$ (479Ed), is surely not standard, which is why I
give the extension a different name.   (While Choquet certainly
considered the capacity which I here call `Choquet-Newton capacity', I
fear that the phrase has no real historical justification;
but I hope it will convey some of the right ideas.)
You may have noticed that I
give essentially nothing concerning differential equations, which have
traditionally been one of the central concerns of potential theory;  there
are hints in 479Xm and 479T.

A weakness in the formulae of 479B is that
they are not self-evidently translation-invariant.   Of course it is
easy to show that in fact we have an isometry-invariant construction
(479Xb), and this can also be seen from the descriptions of capacity
and equilibrium potentials in 479N and 479Pb.
Because the capacity $c$ is countably subadditive,
it is easy to build a dense open subset $G$ of $\BbbR^r$ such that
$c(G)$ is finite (see 479Ya), and for such a set we cannot ask that
$\lambda_G$ should be
describable as a limit of $\|x\|^{r-2}\mu_x^{(G)}$ as $\|x\|\to\infty$.
But if we start from 479B(i) rather than 479B(ii), we do have an averaged
form, with

\Centerline{$\lambda_A=\lim_{\gamma\to\infty}\Bover1{r\beta_r\gamma}
  \int_{\partial B(\tbf{0},\gamma)}\mu_x^{(A)}\nu(dx)$}

\noindent whenever $A$ is an analytic set and $c(A)<\infty$ (479Yc; see
also 479J(b-vi) and 479Xa).

The factor $(\sqrt{2\pi})^r$ in 479H repeats that of
283Wg\formerly{2{}83Wj}.   The
appearance of $\sqrt{2\pi}$ in 283M, but not 445G, is proof that
the conventions of Chapter 28 are not reconcilable with those of
\S445.

In 479O I describe one of the important notions of `small' set in
Euclidean space, to go with `negligible' and `meager'.   I have no space to
deal with it properly here,
but the applications in the proofs of 479P, 479R
and 479S will give an idea of its uses;  another is in 479Xm.   As another
example of the logical complexity of the patterns here, consider the
problem of either proving 479Pb without 479O, or extending 479O to cover
479Xr without passing through a version of 479P.

Quite a lot of the work here is caused by the need to accommodate
discontinuous equilibrium potentials (479S).   This has been an
important theme in general potential theory.   479Pb shows that
the problem is essentially geometric:  if a compact set $K$ has a
sufficiently narrow spike at $e$, then a Brownian path starting at $e$ can
easily fail to enter $K$ again.

As I have written the theory out, 479T-479U are rather separate from the
rest, being closer in spirit to the work of \S\S473-475.   They explore
some more of the basic principles of potential theory.   Note, in
particular, the formula of 479Ta, which amounts to saying that (under the
right conditions) a function $g$ is a multiple of $k_{r-2}*\nabla^2g$;
of course this can be thought of as a method of finding a
particular solution of the equation $\nabla^2g=f$;  equally, it gives an
approach to the problem of expressing a given $g$ as a potential
$W_{\zeta}$.   From 479Tb and 479Tc we see that in the sense of
distributional
derivatives we can think of $r(r-2)\beta_r\zeta$ as representing
the Laplacian $-\nabla^2W_{\zeta}$;  recall that as $W_{\zeta}$ is
superharmonic (479Fb), we expect $\nabla^2W_{\zeta}$ to be negative (478E).

I give a bit of space to 479V because it links the material here to that of
\S476, and this book is about such linkages, and because it supports
my thesis that capacity is a geometrical concept.
479W is characteristic of
Choquet capacities which are alternating of all orders (479Yi).
I spell it out
here because it calls on the Fell topology, which is important
elsewhere in this volume.

It is natural to ask which of the ideas here applying to analytic sets
can be extended to wider classes.   If you look back
to where analytic sets first entered the discussion, in the theory of
hitting times
(455M), you will see that we needed a class of universally measurable
sets which would be invariant under various operations, notably
projections (479Yj).   In Volume 5 we shall meet axiom systems in which
there are various interesting possibilities.

This section is firmly directed at Euclidean space of three or more
dimensions.
The harmonic and Fourier analysis of 479G-479I %479G 479H 479I
applies unchanged to dimensions $1$ and $2$;  so does 479Tb.
On the line, Brownian
hitting probabilities are trivial;  in the plane, they are very
different from hitting probabilities in higher dimensions, but still of
considerable interest.   Theorems 479B, 479E and 479W still work,
but `capacity', if defined by the formulation of 479Ca, is bounded by
$1$.   The geometric nature of
the results changes dramatically, and 479I cannot be applied in the
same way, since we no longer have $0<r-2$.
}%end of notes

\discrpage

\ifresultsonly\else\frfilename{mt47conc.tex}
\versiondate{16.2.10}
\copyrightdate{2010}

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

\def\chaptername{}
\def\sectionname{Concordance}

\newsection{}

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

\medskip

\leader{4{}79Xe}{} Exercise 4{}79Xe on Choquet-Newton capacity, referred to
in the 2008 edition of Volume 5, is now 479Xi.

\frnewpage
\fi
\references{
\frfilename{mt4r.tex}
\versiondate{9.2.13}
\copyrightdate{2002}

% 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}

\Loadcyrillic

\Centerline{\bf References for Volume 4}


\medskip%A

Aldaz J.M.\ [95] `On compactness and Loeb measures', Proc.\ Amer.\
Math.\ Soc.\ 123 (1995) 173-175.
\cmmnt{[439Xa.]}

Aldaz J.M.\ [97] `Borel extensions of Baire measures', Fund.\ Math.\ 154
(1997) 275-293.
\cmmnt{[434Yn.]}

Aldaz J.M.\ \& Render H.\ [00] `Borel measure extensions of measures
defined on sub-$\sigma$-algebras', Advances in Math.\ 150 (2000)
233-263.
\cmmnt{[432D.]}

Alexandroff P.S.\ \& Urysohn P.S.\ [1929] {\it  M\'emoire sur les espaces
topologiques compacts},
Verhandelingen Akad.\ Wetensch.\ Amsterdam 14 (1929) 1-96.
\cmmnt{[419L.]}

Anderson I.\ [87] {\it Combinatorics of Finite Sets.}  Oxford U.P.,
1987.
\cmmnt{[4A4N.]}

Andretta A.\ \& Camerlo R.\ [13] `The descriptive set theory of the
Lebesgue density theorem', Advances in Math.\ 234 (2013).
%{\tt http://arxiv.org/abs/1105.3355}.
\cmmnt{[475Yg.]}

Arkhangel'skii A.V.\ [92] {\it Topological Function Spaces.}  Kluwer,
1992.
\cmmnt{[\S462 {\it intro.}, \S467 {\it notes\/}.]}

\leaveitout{
Armitage D.H.\ \& Gardiner S.J.\ [01] {\it Classical Potential Theory.}
Springer, 2001.
478Yd}

Aronov B., Basu S., Pach J.\ \& Sharir M.\ [03] (eds.)
{\it Discrete and Computational Geometry;  the Goodman-Pollack
Festschrift,}  Springer, 2003.

Asanov M.O.\ \& Velichko N.V.\ [81] `Compact sets in $C_p(X)$', Comm.\
Math.\ Helv.\ 22 (1981) 255-266.
\cmmnt{[462Ya.]}

Austin T.\ [10a] `On the norm convergence of nonconventional ergodic
averages', Ergodic Theory and Dynamical Systems 30 (2010) 321-338.
%({\tt http://{\penalty-100}arxiv.org/abs/{\penalty-100}0805.0320}).
\cmmnt{[497M.]}

Austin T.\ [10b] `Deducing the multidimensional Szemer\'edi theorem from
the infinitary hypergraph removal lemma',
J.\ d'Analyse Math.\ 111 (2010) 131-150.
%({\tt http://arxiv.org{\bsp}abs/0808.2267}).
\cmmnt{[497M.]}

\leaveitout{
Avil\'es A., Plebanek G.\ \& Rodr\'{\i}guez J.\ [p11] `Measurability in
$C(2^{\kappa})$ and Kunen cardinals', 2011.
%material on  C(\{0,1\}^{\kappa}) ;  Baire and Borel \sigma-algebras
%in weak and pointwise topologies
}

\leaveitout{Axler S. \& Ramey W.\ [95]
`Harmonic polynomials and Dirichlet-type problems',
Proc.\ Amer.\  Math.\ Soc.\ 123 (1995) 3765-3773.
http://www.axler.net/HarmonicPolynomials.html
478Yd}

\medskip%B

Balogh Z.\ [96] `A small Dowker space in ZFC', Proc.\ Amer.\ Math.\
Soc.\ 124 (1996) 2555-2560.
\cmmnt{[439O.]}

Banach S.\ [1923] `Sur le probl\`eme de la mesure',
Fundamenta Math.\ 4 (1923) 7-33.
\cmmnt{[449O.]}

Becker H.\ \& Kechris A.S.\ [96] {\it The descriptive set theory of
Polish group actions.}  Cambridge U.P., 1996 (London Math.\ Soc.\
Lecture Note Series 232).  \cmmnt{[424H, 448P, \S448 {\it notes\/}.]}

Bellow A.\ {\it see} Ionescu Tulcea A.

Bellow A.\ \& K\"olzow D.\ [76] {\it Measure Theory, Oberwolfach 1975.}
Springer, 1976 (Lecture Notes in Mathematics 541).
%4r

Bergelson V.\ [85] `Sets of recurrence of $\Bbb Z^m$-actions and properties of
sets of differences in $\Bbb Z^m$',
J.\ London Math.\ Soc.\ (2) 31 (1985) 295-304.
%http://people.dm.unipi.it/dinasso/ULTRA2015/BUMN/Bergelson%20-%20Sets%20of%20recurrence%20of%20Zm-actions%20and%20properties%20of%20sets%20of%20differences%20in%20Zm%20(1985).pdf
\cmmnt{[491Yc.]}

Bergman G.M.\ [06] `Generating infinite symmetric groups', Bull.\ London
Math.\ Soc.\ 38 (2006) 429-440.
%http://math.berkeley.edu/~gbergman/papers/
%({\tt http:/{\bsp}arxiv.org{\bsp}abs{\bsp}math{\bsp}0401304v2}).
\cmmnt{[\S494 {\it notes}.]}

Billingsley P.\ [99] {\it Convergence of Probability Measures.}  Wiley,
1999.
\cmmnt{[\S4A3 {\it notes}.]}
%second edition ISBN 0-471-19745-9

Blackwell D.\ [56] `On a class of probability spaces',
pp.\ 1-6 in {\smc Neyman 56}.
\cmmnt{[419K, 452P.]}

Bochner S.\ [1933] `Monotone Funktionen, Stieltjessche Integrale, und
harmonische Analyse', Math.\ Ann.\ 108 (1933) 378-410.
\cmmnt{[445N.]}

\indexheader{Bogachev}
Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.
\cmmnt{[437Q, 437Xs, 437Yo, 437Yu, 457K.]}

Bollob\'as B.\ [79] {\it Graph Theory.}   Springer, 1979.
\cmmnt{[4A1G, 4A4N.]}

Bongiorno B., Piazza L.di \ \& Preiss D.\ [00]
`A constructive minimal integral which includes Lebesgue integrable
functions and derivatives', J.\ London Math.\ Soc.\ (2) 62 (2000) 117-126.
\cmmnt{[483Yi.]}

Borodulin-Nadzieja P.\ \& Plebanek G.\ [05] `On compactness of measures on
Polish spaces', Illinois J.\ Math.\ 49 (2005) 531-545.
\cmmnt{[451L.]}

Bourbaki N.\ [65] {\it Int\'egration,} chaps.~1-4.   Hermann, 1965
(Actualit\'es Scientifiques et Industrielles 1175).
\cmmnt{[\S416 {\it notes\/}, \S436 {\it notes\/}.]}

Bourbaki N.\ [66] {\it Espaces Vectoriels Topologiques}, chaps.~1-2,
$2^{\text{e}}$ \'ed.
Hermann, 1966 (Actualit\'es Scientifiques et Industrielles 1189).
\cmmnt{[4A2J.]}
%Store A QA 37.B6 no 7792 0070 5

Bourbaki N.\ [69] {\it Int\'egration,} chap.~9.   Hermann, 1969
(Actualit\'es Scientifiques et Industrielles 1343).
\cmmnt{[\S436 {\it notes.}]}

Bourbaki N.\ [87] {\it Topological Vector Spaces.}  Springer, 1987.
\cmmnt{[\S461 {\it notes\/}, \S4A4.]}

\indexheader{Bourgain}
Bourgain J., Fremlin D.H.\ \& Talagrand M.\ [78] `Pointwise compact
sets of Baire-measurable functions', Amer.\ J.\ Math.\ 100 (1978)
845-886.
\cmmnt{[462Ye.]}

\indexheader{Brodski\v{\i}}
Brodski\v{\i} M.L.\ [1949] `On some properties of sets of positive measure',
Uspehi Matem.\ Nauk (N.S.) 4 (1949) 136-138.
\cmmnt{[498B.]}

Brook R.B.\ [70] `A construction of the greatest ambit',
Math.\ Systems Theory 4 (1970) 243-248.
\cmmnt{[449D.]}

Burke D.K.\ \& Pol R.\ [05] `Note on separate continuity and the Namioka
property', Top.\ Appl.\ 152 (2005) 258-268.
\cmmnt{[\S463 {\it notes.}]}

Burke M.R., Macheras N.D.\ \& Strauss W.\ [p21] `The strong marginal lifting
problem for hyperstonian spaces', preprint (2021).
%bms5s.pdf

\medskip%C

Carrington D.C.\ [72] `The generalised Riemann-complete integral', PhD
thesis, Cambridge, 1972.
\cmmnt{[481L.]}

\v{C}ech E.\ [66] {\it Topological Spaces.}  Wiley, 1966.
\cmmnt{[\S4A2, 4A3S.]}

Chacon R.V. [69]
`Weakly mixing transformations which are not strongly mixing',
Proc.\ Amer.\ Math.\ Soc.\ 22 (1969) 559-562.
\cmmnt{[494F.]}

\indexheader{Choquet}
Choquet G.\ [55] `Theory of capacities', Ann.\ Inst.\ Fourier (Grenoble)
5 (1955) 131-295.
\cmmnt{[432K.]}

\indexheader{Chung}
Chung K.L.\ [95] {\it Green, Brown and Probability.}   World Scientific,
1995.
% 978-9810225339  HA33.4.C5  479Xt

\indexheader{Ciesielski}
Ciesielski K.\ \& Pawlikowski J.\ [03]
`Covering Property Axiom CPA$_{\text{cube}}$ and its consequences',
Fundamenta Math. 176 (2003) 63-75.
\cmmnt{[498C.]}

Cs\'asz\'ar \'A.\ [78] {\it General Topology.}  Adam Hilger, 1978.
\cmmnt{[\S4A2, \S4A5.]}

\medskip%D

Davies R.O.\ [70] `Increasing sequences of sets and Hausdorff measure', Proc.\ London Math.\ Soc.\ (3) 20 (1970) 222-236.
\cmmnt{[471G.]}

Davies R.O.\ [71] `Measures not approximable or not specifiable by means
of balls', Mathematika 18 (1971) 157-160.
\cmmnt{[\S466 {\it notes\/}.]}

Davis W.J., Ghoussoub N.\ \& Lindenstrauss J.\ [81] `A lattice-renorming
theorem and applications to vector-valued processes', Trans.\ Amer.\
Math.\ Soc.\ 263 (1981) 531-540.
\cmmnt{[\S467 {\it notes\/}.]}

\indexheader{de Caux}
de Caux P.\ [76] `A collectionwise normal, weakly $\theta$-refinable
Dowker space which is neither irreducible nor realcompact', Topology
Proc.\ 1 (1976) 67-77.
\cmmnt{[439O.]}

Dellacherie C.\ [80] `Un cours sur les ensembles analytiques', pp.\
183-316 in {\smc Rogers 80}.
\cmmnt{[\S432 {\it notes\/}.]}

de Maria J.L.\ \& Rodriguez-Salinas B.\ [91] `The space
$(\ell_{\infty}/c_0)$,weak) is not a Radon space', Proc.\ Amer.\ Math.\
Soc.\ 112 (1991) 1095-1100.
\cmmnt{[466I.]}

Deville R., Godefroy G.\ \& Zizler V.\ [93] {\it Smoothness and
Renormings in Banach Spaces.}  Pitman, 1993.
\cmmnt{[\S467 {\it intro.}, 467Ye, \S467 {\it notes\/}.]}

Diaconis, P.\ \& Freedman, D.\ [80] `Finite exchangeable sequences.'
Ann. of Probability 8 (1980) 745-764.
\cmmnt{[459Xd.] }

Droste M., Holland C.\ \& Ulbrich G.\ [08]
`On full groups of measure preserving and ergodic transformations with
uncountable cofinalities',
Bull.\ London Math. Soc.\ 40 (2008) 463-472.
%{\tt http://web.mac.com/chollan1{\bsp}iWeb/Site{\bsp}publications.html}.
\cmmnt{[494Q.]}

Dubins L.E.\ \& Freedman D.A.\ [79]
`Exchangeable processes need not be mixtures of independent,
identically distributed random variables',
Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 48 (1979) 115-132.
\cmmnt{[459K.]}

Dunford N.\ \& Schwartz J.T.\ [57] {\it Linear Operators I.}  Wiley,
1957 (reprinted 1988).
\cmmnt{[\S4A4.]}

Du Plessis, N.\ [70] {\it Introduction to Potential Theory.}  Oliver \&
Boyd, 1970.
%used for \S478

\medskip%E

Eggleston H.G.\ [54] `Two measure properties of Cartesian product sets',
Quarterly J.\ Math.\ (2) 5 (1954) 108-115.
\cmmnt{[498B.]}

Emerson W.\ \& Greenleaf F.\ [67] `Covering properties and F{\o}lner
conditions for locally compact groups',
Math.\ Zeitschrift 102 (1967) 370-384.
\cmmnt{[449J.]}

Enderton H.B.\ [77] {\it Elements of Set Theory.}   Academic, 1977.
\cmmnt{[4A1A.]}

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).
\cmmnt{[Vol.\ 4 {\it intro.}, 415Yb, \S434 {\it notes\/},
\S435 {\it notes\/}, 462L, \S4A2, 4A4B.]}

Evans L.C.\ \& Gariepy R.F.\ [92] {\it Measure Theory and Fine
Properties of Functions.}   CRC Press, 1992.
\cmmnt{[Chap.\ 47 {\it intro.}, \S473 {\it intro.}, \S473 {\it notes\/},
\S474 {\it notes\/}, \S475 {\it notes\/}.]}

\medskip%F

Falconer K.J.\ [90] {\it Fractal Geometry.}  Wiley, 1990.
\cmmnt{[\S471 {\it intro.}]}

Federer H.\ [69] {\it Geometric Measure Theory.}
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\cmmnt{[\S441 {\it notes\/}, 442Ya,
\S443 {\it notes\/}, Chap.\ 47 {\it intro.}, 471Yk,
\S471 {\it notes\/}, \S472 {\it notes\/}, \S475 {\it notes\/}.]}

Fernique X.\ [97] {\it Fonctions al\'eatoires gaussiennes,
vecteurs al\'eatoires gaussiens}, CRM Publications, 1997.
\cmmnt{[\S456 {\it notes\/}.]}

Folland G.B.\ [95] {\it A Course in Abstract Harmonic Analysis.}   CRC
Press, 1995
\cmmnt{[\S4A5, \S4A6.]}

\indexheader{F{\o}lner}
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Math.\ Scand.\ 3 (1955) 243-254.
\cmmnt{[\S449 {\it notes\/}.]}

\indexheader{Ford}
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\cmmnt{[4A4N.]}

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\cmmnt{[497L.]}

\indexheader{Fremlin}
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\cmmnt{[\S462 {\it notes\/}.]}

Fremlin D.H.\ [75a] `Pointwise compact sets of measurable functions',
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\cmmnt{[463K.]}

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Math.\ Proc.\ Cambridge Phil.\ Soc.\ 78 (1975) 95-106.
\cmmnt{[419C, 419D.]}

Fremlin D.H.\ [76] `Products of Radon measures:  a counterexample',
Canadian Math.\ Bull.\ 19 (1976) 285-289.
\cmmnt{[419E.]}

Fremlin D.H.\ [81] `Measurable functions and almost continuous
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\cmmnt{[451T.]}

Fremlin D.H.\ [84] {\it Consequences of Martin's Axiom.}
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\cmmnt{[\S434 {\it notes\/}, 439Yd, 4A2E.]}

Fremlin D.H.\ [93] `Real-valued-measurable cardinals', pp.\ 151-304
in {\smc Judah 93}.
\cmmnt{[\S438 {\it notes\/}.]}

Fremlin D.H.\ [95] `The generalized McShane integral',
Illinois J.\ Math.\ 39 (1995) 39-67.
\cmmnt{[481N.]}

Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces',
Fundamenta Math.\ 165 (2000) 67-94.
\cmmnt{[451V.]}

Fremlin D.H.\ [n05] `Strictly localizable Borel measures', note of 7.5.05
({\tt http:/{\bsp}www.\penalty-100essex.ac.uk{\bsp}maths/people{\bsp}fremlin/preprints.htm}).
\cmmnt{[434Yr.]}

Fremlin D.H.\ [n15] `The Sorgenfrey line is not a Prokhorov space',
note of 5.8.15
({\tt http:/{\bsp}www.\penalty-100essex.ac.uk{\bsp}maths/people{\bsp}fremlin/preprints.htm}).
\cmmnt{[439Yk.]}

Fremlin D.H.\ \& Grekas S.\ [95] `Products of completion regular
measures', Fund.\ Math.\ 147 (1995) 27-37.
\cmmnt{[434Q.]}

Fremlin D.H.\ \& Talagrand M.\ [78] `On the representation of increasing
linear functionals on Riesz spaces by measures',
Mathematika 25 (1978) 213-215.
\cmmnt{[439I.]}

Fremlin D.H.\ \& Talagrand M.\ [79] `A decomposition theorem for
additive set functions and applications to Pettis integral and ergodic
means', Math.\ Zeitschrift 168 (1979) 117-142.
\cmmnt{[\S464.]}  %{\it 464C}, 464I, 464M, 464N, \S464 {\it notes\/}

Fristedt B.\ \& Gray L.\ [97] {\it A Modern Approach to Probability
Theory.}  Birkh\"auser, 1997.
\cmmnt{[\S455 {\it notes\/}, \S495 {\it notes\/}.]}
%HA33.4.F7

Frol{\'\i}k Z.\ [61] `On analytic spaces', Bull.\ Acad.\ Polon.\ Sci.\
%8 (1961) 747-750.
9 (1961) 721-726.   %Paulo Monteiro, email 8.9.22
\cmmnt{[422F.]}

Frol\'\i k Z.\ [88] (ed) {\it General Topology and its Relations to
Modern Analysis and Algebra VI} (Proc.\ Sixth Prague Topological
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Furstenberg H.\ [77]
`Ergodic behavior of diagonal measures and a theorem of Szemer\'edi
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\cmmnt{[\S497 {\it intro.}, \S497 {\it notes\/}.]}

Furstenberg H.\ [81]
{\it Recurrence in Ergodic Theory and Combinatorial Number Theory.}
Princeton U.P., 1981.
\cmmnt{[\S497 {\it intro.}, 497N.]}
% QA614

Furstenberg H.\ \& Katznelson Y.\ [85]
`An ergodic Szemer\'edi theorem for IP-systems and combinatorial theory',
J.\ d'Analyse Math.\ 45 (1985) 117-168.
\cmmnt{[\S497 {\it intro.}]}

\medskip%G

Gaal S.A.\ [64] {\it Point Set Topology.}  Academic, 1964.
\cmmnt{[\S4A2.]}

Gardner R.J.\ [75] `The regularity of Borel measures and Borel
measure-compactness', Proc.\ London Math.\ Soc.\ (3) 30 (1975) 95-113.
\cmmnt{[438M.]}

Gardner R.J.\ \& Pfeffer W.F.\ [84] `Borel measures', pp. 961-1043 in
{\smc Kunen \& Vaughan 84}.
\cmmnt{[434G.]}

Giordano T.\ \& Pestov V.\ [02] `Some extremely amenable groups',
C.R.A.S.\ Paris (S\'er.\ I) 334 (2002) 273-278.
\cmmnt{[494I.]}

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\cmmnt{[\S493 {\it notes\/}.]}

Gordon R.A.\ [94] {\it The Integrals of Lebesgue, Denjoy, Perron and
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4).
\cmmnt{[481Q, \S483 {\it notes\/}.]}

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\cmmnt{[\S449 {\it notes.}]}

Grothendieck A. [92] {\it Topological Vector Spaces.}  Gordon \& Breach,
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\cmmnt{[\S462 {\it notes\/}.]}

\indexheader{Grzegorek}
Grzegorek E.\ [81] `On some results of Darst and Sierpi\'nski concerning
universal null and universally negligible sets', Bull.\ Acad.\ Polon.\
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\cmmnt{[439F, \S439 {\it notes\/}.]}

\medskip%H

\indexheader{Halmos}
Halmos P.R.\ [1944] `In general a measure-preserving transformation is
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\cmmnt{[494E.]}

Halmos P.R.\ [50] {\it Measure Theory.}   Springer, 1974.
\cmmnt{[\S441 {\it notes\/}, \S442 {\it notes\/}, 443M, \S443 {\it notes\/}.]}

Halmos P.R.\ [56] {\it Lectures on ergodic theory.}  Chelsea, 1956.
\cmmnt{[494A.]}

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Banach spaces', Serdica Math.\ J.\ 27 (2001) 1-66.
\cmmnt{[466D.]}

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Real Analysis Exchange 25 (1999) 653-659.
\cmmnt{[416Yh.]}

Hartman S.\ \& Mycielski J.\ [58] `On the imbedding of topological groups into connected topological groups', Colloq.\ Math.\ 5 (1958) 167-169.
\cmmnt{[493Ya.]}

Haydon R.\ [74] `On compactness in spaces of measures and
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\cmmnt{[438J.]}

Henry J.P.\ [69] `Prolongement des mesures de Radon', Ann.\ Inst.\
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\cmmnt{[416N.]}

Henstock R.\ [63] {\it Theory of Integration.}  Butterworths, 1963.
\cmmnt{[Chap.\ 48 {\it intro.}, 481J.]}

Henstock R.\ [91] {\it The General Theory of Integration.}   Oxford
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\cmmnt{[\S481 {\it notes\/}.]}

Herer W.\ \& Christensen J.P.R.\ [75] `On the existence of pathological submeasures and the construction of exotic topological groups', Math.\ Ann.\ 213 (1975) 203-210.
\cmmnt{[\S493 {\it notes\/}.]}

Herglotz G.\ [1911] `\"Uber Potenzreihen mit positivem reellen Teil im
Einheitskreis', Leipziger Berichte 63 (1911) 501-511.
\cmmnt{[445N.]}

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\cmmnt{[\S442 {\it notes\/}, \S444 {\it intro.}, \S444 {\it notes\/},
\S4A5, \S4A6.]}

Hewitt E.\ \& Savage L.J.\ [55]
`Symmetric measures on Cartesian products',
Trans.\ Amer.\ Math.\ Soc.\ 80 (1955) 470-501.
\cmmnt{[459Xe.]}

%Holick\'y P.\ \& Spurn\'y J.\ [03] `Perfect images of absolute Souslin
%and absolute Borel Tychonoff spaces', Topology and its Appl.\ 131 (2003)
%281-294.
%\cmmnt{[434Yr.]}

\indexheader{Howroyd}
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\cmmnt{[471R, 471S.]}

\medskip%I

\indexheader{Ionescu Tulcea}
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\cmmnt{[463C.]}

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lifting commuting with the left translations of an arbitrary locally
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\cmmnt{[\S447 {\it intro.}, 447I, \S447 {\it notes\/}.]}

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\cmmnt{[\S453 {\it intro.}]}

\indexheader{Ismail}
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\cmmnt{[\S434 {\it notes\/}.]}

\medskip%J

%Jackson S.\ \& Mauldin R.D.\ [99] `On the $\sigma$-class generated by
%open balls', Math.\ Proc.\ Cambridge Phil.\ Soc.\ 127 (1999) 99-108.

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\cmmnt{[\S4A4.]}

\indexheader{Jayne}
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\cmmnt{[422Yc.]}

Jayne J.E.\ \& Rogers C.A.\ [95] `Radon measures on Banach spaces with
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\cmmnt{[466H, \S466 {\it notes\/}.]}

\indexheader{Jech}
Jech T.\ [78] {\it Set Theory.}  Academic, 1978.
\cmmnt{[423S, \S4A1.]}

Jech T.\ [03] {\it Set Theory, Millennium Edition.}  Springer, 2003.
%ISBN 3540440852
\cmmnt{[\S4A1.]}

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\cmmnt{[434R.]}

Judah H.\ [93] (ed.) {\it Proceedings of the Bar-Ilan Conference on Set
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\cmmnt{[\S437 {\it notes\/}, 439K, \S439 {\it notes\/}.]}

Just W.\ \& Weese M.\ [96] {\it Discovering Modern Set Theory I.}  Amer.\ Math.\ Soc., 1996 (Graduate Studies in Mathematics 8).
\cmmnt{[4A1A.]}

Just W.\ \& Weese M.\ [97] {\it Discovering Modern Set Theory II.}  Amer.\ Math.\ Soc., 1997 (Graduate Studies in Mathematics 18).
\cmmnt{[\S4A1.]}

\medskip%K

Kampen, E.R.van [1935] `Locally bicompact abelian groups and their character groups',
Ann.\ of Math.\ (2) 36 (1935) 448-463.
\cmmnt{[445U.]}

\indexheader{Kaplansky}
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\cmmnt{[\S446 {\it notes.}]}

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\cmmnt{[Chap.\ 42 {\it intro.}, 423T, \S423 {\it notes\/},
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%Kechris A.S.\ [p09] {\it Global Aspects of Ergodic Group Actions.}
%Amer.\ Math.\ Soc.\

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\cmmnt{[457M.]}

Kelley J.L.\ [55] {\it General Topology.} Van Nostrand, 1955.
\cmmnt{[438R.]}

Kelley J.L.\ \& Srinivasan T.P.\ [71] `Pre-measures on lattices of sets',
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\cmmnt{[413Xr.]}

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Ergodic Theory and Dynamical Systems 2009 (doi:10.1017/S0143385709000078).
\cmmnt{[494O, \S494 {\it notes}.]}

Kolmogorov A.N.\ [1933] {\it Grundbegriffe der
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\cmmnt{[454D.]}

K\"othe G.\ [69] {\it Topological Vector Spaces I.}   Springer, 1969.
\cmmnt{[\S462 {\it notes\/}, \S4A4, 4A6C.]}

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\cmmnt{[451T.]}

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Sequences.}   Wiley, 1974.
\cmmnt{[491Z.]}

Kunen K.\ [80] {\it Set Theory.}  North-Holland, 1980.
\cmmnt{[\S4A1, 4A2E.]}

Kunen K.\ \& Vaughan J.E.\ [84] (eds.) {\it Handbook of Set-Theoretic
Topology.}   North-Holland, 1984.
%4r

Kuratowski K.\  [66] {\it Topology,} vol.\ I.   Academic, 1966.
\cmmnt{[423T, 462Ye, \S4A2, 4A3S.]}

Kurzweil J.\ [57] `Generalized ordinary differential equations and
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\cmmnt{[Chap.\ 48 {\it intro.}]}

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%{\tt http://www.math.uiuc.edu/\penalty-100$^{\sim}$ssolecki/papers.html}.
\cmmnt{[\S448 {\it notes\/}.]}

\medskip%L

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\cmmnt{[\S449 {\it notes\/}.]}

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\cmmnt{[\S4A1.]}

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\cmmnt{[\S495 {\it notes\/}.]}

\indexheader{Losert}
Losert V.\ [79] `A measure space without the strong lifting property',
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\cmmnt{[453N, \S453 {\it notes\/}.]}

\indexheader{Luke\v{s}}
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\cmmnt{[414Ye.]}

\medskip%M

\indexheader{Mackey}
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\cmmnt{[448S.]}

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\cmmnt{[454C.]}

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\cmmnt{[435C.]}

Mattila P.\ [95] {\it Geometry of Sets and Measures in Euclidean
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%QA311.M2 9520 6456 1

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\cmmnt{[418T.]}

Maurey B.\ [79] `Construction des suites sym\'etriques', C.R.A.S.\ (Paris) 288 (1979) 679-681.
\cmmnt{[492H.]}
  % one m in sym\'etrique

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\cmmnt{[481M.]}

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\cmmnt{[\S495 {\it notes\/}.]}

Milman V.D.\ \& Schechtman G.\ [86]
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\cmmnt{[492G, \S492 {\it notes\/}.]}

\leaveitout{Mohammed D.\ \& Saadi K.\ [p09#]
`Probl\`emes de mesurabilit\'es sur
l'espace des fonctions continues sur un compact', 2009.
\cmmnt{[466Y?.]}
%refereeing for Canad Math Bull;  rejected
}

Molt\'o A., Orihuela J., Troyanski S.\ \& Valdivia M.\ [09] {\it A
Non-linear Transfer Technique for Renorming.}   Springer, 2009 (Lecture
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\cmmnt{[\S467 {\it notes.}]}

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\cmmnt{[\S446 {\it notes.}]}

\indexheader{Moran}
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\cmmnt{[439P.]}

Moran W.\ [69] `Measures and mappings on topological spaces', Proc.\
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\cmmnt{[435Xk.]}

Moran W.\ [70] `Measures on metacompact spaces', Proc.\ London Math.\
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\cmmnt{[438J.]}

\indexheader{Moschovakis}
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%ISBN 5-900975-01-0

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\medskip%N

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%Naimark M.A. [70] {\it Normed Rings.}  Wolters-Noordhoff, 1970.


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\medskip%O
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\cmmnt{[434Yq.]}

\medskip%P

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% QA 403.P2 store A

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\cmmnt{[439S.]}
%ILL 31.7.15

Preiss D.\ \& Ti\v{s}er J.\ [91]
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\medskip%Q\medskip%R

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\cmmnt{[\S451 {\it notes\/}.]}

\indexheader{Rao B.V.}
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\cmmnt{[419F.]}

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\indexheader{Rao M.}
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{\tt http://www1.essex.ac.uk/maths/people{\bsp}fremlin{\bsp}rao.htm}.
%used for \S477

\indexheader{Ressel}
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\cmmnt{[417C, 437M.]}

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\cmmnt{[\S494 {\it notes\/}.]}

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\cmmnt{[\S471 {\it notes\/}.]}

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Rogers L.C.J.\ \& Williams D.\ [94] {\it Diffusions, Markov Processes
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\cmmnt{[\S455 {\it notes\/}.]}
%HA33.4.W5

Rokhlin V.A.\ [1948] `A ``general'' measure-preserving transformation is not
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`The generic isometry and measure preserving homeomorphism are conjugate
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%({\tt http://www.math.uic.edu{\bsp}$\sptilde$rosendal/}).
\cmmnt{[494Ye.]}

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\cmmnt{[494Z, \S494 {\it notes}.]}

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\cmmnt{[439O.]}

Rudin M.E.\ [84] `Dowker spaces', pp.\ 761-780 in
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\cmmnt{[439O.]}

Rudin W.\ [67] {\it Fourier Analysis on Groups.} Wiley, 1967 and 1990.
\cmmnt{[\S445 {\it intro\/}.]}

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\cmmnt{[\S4A4, \S4A6.]}

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\cmmnt{[451C.]}

\medskip%S

Sazonov V.V.\ [66] `On perfect measures', A.M.S.\ Translations (2) 48
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\cmmnt{[451F.]}

Schachermayer W.\ [77] `Eberlein-compacts et espaces de Radon',
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\cmmnt{[467P.]}

Schaefer H.H.\ [71] {\it Topological Vector Spaces.}  Springer, 1971.
\cmmnt{[\S4A4.]}

Schwartz L.\ [73] {\it Radon Measures on Arbitrary Topological Spaces
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\cmmnt{[\S434 {\it notes\/}.]}

Shelah S.\ \& Fremlin D.H.\ [93] `Pointwise compact and stable sets of
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\cmmnt{[\S465 {\it notes\/}.]}

\indexheader{Sierpi\'nski}
Sierpi\'nski W.\ [1945] `Sur une suite infinie de fonctions de classe 1 dont toute fonction d'accumulation est non mesurable', Fundamenta Math.\ 33 (1945) 104-105.
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\indexheader{Solecki}
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\cmmnt{[444Ye.]}

Solymosi J.\ [03]
`Note on a generalization of Roth's theorem', pp.\ 825-827 in
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\cmmnt{[497M.]}

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\cmmnt{[421D, \S421 {\it notes\/}.]}

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\cmmnt{[434Ya, \S439 {\it notes\/}.]}

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\cmmnt{[441C.]}

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\cmmnt{[457D.]}

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\cmmnt{[449Yi.]}

\indexheader{Szemer\'edi}
Szemer\'edi, E.\ [75]
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\cmmnt{[497L.]}
\indexheader{Talagrand} %? before \medskip to act correctly

\medskip%T

Talagrand M.\ [75] `Sur une conjecture de H.H.Corson', Bull.\ Sci.\
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\cmmnt{[467M.]}

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\cmmnt{[466I.]}

Talagrand M.\ [78b] `Comparaison des boreliens pour les topologies
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\cmmnt{[466Za.]}

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\cmmnt{[464C, 464D.]}

Talagrand M.\ [81] `La $\tau$-r\'egularit\'e de mesures gaussiennes',
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Talagrand M.\ [82] `Closed convex hull of set of measurable functions,
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\cmmnt{[465M.]}

Talagrand M.\ [84] {\it Pettis integral and measure theory.}
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\cmmnt{[463Ya, 463Yd, 463Za, \S465 {\it intro.}, 465B, 465N,
465R-465T, %465R 465S 465T
466I, \S466 {\it notes\/}.]}

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\cmmnt{[\S465 {\it intro.}, 465L, 465M, {\it 465Yd},
\S465 {\it notes\/}.]}

Talagrand M.\ [88] `On liftings and the regularization of stochastic
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\cmmnt{[465U.]}

Talagrand M.\ [89] `A ``regular oscillation'' property of stable sets of
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\cmmnt{[\S465 {\it notes\/}.]}

Talagrand M.\ [95]
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\cmmnt{[492D.]}

Talagrand M.\ [96] `The Glivenko-Cantelli problem, ten years later', J.\
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\cmmnt{[\S465 {\it notes\/}.]}
%465Xm

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\cmmnt{[459I, \S497.]}

Tao T.\ [l08] {\it Topics in Ergodic Theory,} lecture notes.
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\cmmnt{[494F.]}

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\cmmnt{[\S4A4.]}

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Tops{\o}e F.\ [70b] {\it Topology and Measure.}  Springer, 1970 (Lecture
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\cmmnt{[437J.]}
%used in chap 41

T\"ornquist A.\ [11] `On the pointwise implementation of near-actions',
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\cmmnt{[425D, 425Ya.]}

\medskip%U

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Uspenski\v\i\ V.V.\ [88] `Why compact groups are dyadic', pp.\ 601-610
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\cmmnt{[4A5T.]}

\medskip%V

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Veech W.A.\ [77] `Topological dynamics', Bull.\ Amer.\ Math.\ Soc.\ 83 (1977) 775-830.
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{\cyr Vinokurov V.G.} \& {\cyr Ma{kh}kamov B.M.\ [73] `O prostranstva{kh}
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\cmmnt{[451U.]}
% `Ta{sh}ken{t}skogo instituta' taken from ref in Fund Math

\medskip%W

Wage M.L.\ [80] `Products of Radon spaces', Russian Math. Surveys 35:3
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\cmmnt{[\S434 {\it notes\/}.]}

Wagon S.\ [85] {\it The Banach-Tarski Paradox.}  Cambridge U.P., 1985.
\cmmnt{[449Yj.]}

Weil A.\ [1940] {\it L'int\'egration dans les groupes topologiques et ses
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\cmmnt{[445N.]}

\indexheader{Wheeler}
Wheeler R.F.\ [83] `A survey of Baire measures and strict topologies',
Expositiones Math.\ 1 (1983) 97-190.
  % printer's error "2" for "1" at head of paper refers to part, not
%volume
\cmmnt{[\S435 {\it notes\/}, \S437 {\it notes\/}.]}

%\medskip%X\medskip%Y\medskip%Z

%Zindulka O.\ [p98] `Killing residual measures', preprint, 1998.

\frnewpage
}%end of references


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