% The following TeXfile comprises part of

%     MEASURE THEORY, by D.H.Fremlin
%          (University of Essex, Colchester, England;
%                   e-mail  fremdh@essex.ac.uk)

% Please contact me with any difficulties not covered by the remarks
% below.

% The file is set up for processing by plain TeX in the presence of
% files  amstex.tex ,  amssym.tex  and  amsppt.sty  suitably supported by the
% appropriate fonts.   If these are unavailable, a readable version
% can be obtained by removing the comment symbol  %  from the line

%        %   \amstexsupportedfalse

% below.   Some systems do not allow access to files  amstex.tex  and
% amsppt.sty , but instead employ a command "amstex <filename.tex>";
% for such systems it is sometimes right to delete or disable the
% commands

%        \input amstex
%        \documentstyle{amsppt}

% (I have little experience with this and should welcome comments.)

% Various formats of output are obtainable.   The principal variations
% are

%    1. "Standard" output uses the command
%               \smallprint
% and gives output adapted to A4 paper (21 cm x 29.7 cm).

%    2.  Using the command
%               \USsmallprint
% gives output adapted to 8.5" x 11" .

%    3.  Using the command
%               \noproofs
% produces an abridged version, in which proofs are omitted.

%    4.   Using the command
%               \resultsonly
% produces a super-abridged version, omitting comments and exercises
% as well as proofs.

% These commands are grouped together at the end of the introductory
% definitions;  they may be activated by deleting the initial comment
% symbols  % .

% Any diagrams included will be called for by inputting postscript files.
% (You can look for these by hunting for   \picture  in the code below.)
% To get these you will need the relevant  .ps  files with the expected
% names  ( \picture{mt133g}{...}  demands the file  mt133g.ps , for instance).


%(If you can read TeX fluently, you may wish to fast-forward to a line
%beginning   %file starts here   or   \frfilename
%if one of these is present.)

%TeXfiles with this heading are designed for processing with AmSTeX,
%accessed through TeXfiles  amstex.tex (with amssym.tex  and
% amsppt.sty  and possibly  amsppt.sti)  during compilation.   
%If these are unavailable, a readable
%version can (I hope) be produced by deleting the  %  before
% \amstexsupportedfalse  below.

%Some special difficulties arise with the importation of  .ps  files
%in diagrams.   Look for  \picture  below for notes on these.

%The following \count registers are used for special purposes:
%   \count1=\stylenumber  to specify style
%   \count2=\footnotenumber  to count footnotes

%AmSTeX declaration
\newif\ifamstexsupported
     \amstexsupportedtrue
%     \amstexsupportedfalse               %use when AMSTeX unavailable
\newif\ifDHFamstexloaded        %for detection of a non-standard AMSTeX
     \DHFamstexloadedfalse    
\newif\ifDHFamspptloaded        
     \DHFamspptloadedfalse    
\newif\iffourteensavailable
     \fourteensavailabletrue
    \fourteensavailablefalse
\newif\ifathome
     \athometrue     %to cope with a quirk     of my son's Ghostscript
     \athomefalse
\newif\iffirstpage
     \firstpagetrue

%Style assignments
\countdef\stylenumber=1
\countdef\footnotenumber=2
\def\fullsize{\magnification=\magstep1
     \def\pagewidth{390pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=1}
\def\smallprint{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{468pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=2}
\def\printersize{\magnification=\magstep1
     \baselineskip24pt
     \overfullrule0pt
     \def\pagewidth{330pt}
     \def\pageheight{8.5truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=3}
\def\refsize{\magnification=\magstep1
     \overfullrule0pt
     \def\pagewidth{390pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=4}
\def\rrversion{\magnification=\magstep1
     \overfullrule0pt
     \def\pagewidth{330pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=5}
\def\bigsize{\magnification=\magstep2
     \overfullrule0pt
     \def\pagewidth{325pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=6}
\def\bbigsize{\magnification=\magstep3
     \overfullrule0pt
     \def\pagewidth{271pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=7}
\def\USsmallprint{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{482pt}
     \def\pageheight{8.7truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=8}
\def\halfmagsize{\magnification=\magstephalf
     \def\pagewidth{427pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=9}
           % register  stylenumber  is used to control the output format

%addresses, letterheads
\def\deptaddress{\hskip 3truein Department of Mathematical Sciences\par
   \noindent e-mail: {\tt fremdh}\hskip 2.9truein\today\par
   \medskip}
\long\def\univaddress#1{\hskip 3truein Department of Mathematical Sciences\par
   \hskip 3truein University of Essex\par
   \hskip 3truein Colchester CO4 3SQ\par
   \hskip 3truein England\par
   %\hskip 3truein van Vleck Hall\par
   %\hskip 3truein 480 Lincoln Drive\par
   %\hskip 3truein Madison\par
   %\hskip 3truein WI 53706-1388, U.S.A.\par
   \hskip 3.2truein#1\par
   \noindent e-mail: {\tt david\@fremlin.org}
       %{\tt fremlin\@math.wisc.edu}
   \bigskip}
\long\def\homeaddress#1{\hskip 3truein 25 Ireton Road\par
   \hskip 3truein Colchester CO3 3AT\par
   \hskip 3truein England\par
   \hskip 3.2truein#1\par
   \noindent e-mail: {\tt david\@fremlin.org}
   \bigskip}
\long\def\letterhead#1#2#3{\univaddress{#1}
   {\parindent=0pt
    #2
   }
   \medskip

   \noindent Dear #3,
     }
\def\shortaddress{\noindent Departmentof Mathematical Sciences, University of Essex,
   Colchester CO4 3SQ, England\par
   \noindent e-mail: david\@fremlin.org}
\def\valediction#1#2{\bigskip
   \hskip 3truein #1,
   \vskip 1truein
   \hskip 3.3truein #2
   \bigskip}

%picture importation
   %It is clear that there is no general standard.   The syntax I
   %use when calling for a  .ps  file is
   %    \def\Caption{<...>}
   %    \picture{<filename>}{<dim>}
   %which is supposed to import  <filename>.ps  at height  <dim> ,
   %centered, and adding a one-line centered caption  <...> .
   %The basic macro listed in  \def\picture  is that which
   %used to work at Essex.   I now generally use the macro in
   % \def\picturewithpsfig , which requires a previous input
   % psfig.tex  or  psfig.sty ;
   %if you interpolate  \atUEssex  early in your file it will
   %give you the latter.
   %   I find that  psfig  is erratic in its centering, especially
   %under magnification, so I have written a routine
   % \centerpicturewithpsfig  (which may be called from
   % \centerpicture ) with the syntax
   %    \centerpicture{<filename>}{<dim>}{<dim>}
   %the first dimension being width and the second height.

   % A second package is called for by  \usegraphicx .   This requires
   % miniltx.tex  and  graphicx.sty ;  miniltx.tex  is currently (April 2016)
   % available from
   % http://mirror.ox.ac.uk/sites/ctan.org/macros/plain/graphics/miniltx.tex
   % The advantage of this is that it seems to allow manipulation of the
   % dvi>pdf stage using  dvipdfmx .

\def\picture#1#2{\bigskip
  \bigskip
  \immediate\write0{importing diagram #1.ps at height #2
         with dvitops}
  \vbox{\vskip #2
    \special{dvitops: import #1.ps \pagewidth #2}
    \Centerline{\Caption}
    }
   \medskip
   \def\Caption{}}
%an alternative version, without caption
\def\centerpicture#1#2#3{\bigskip
  \bigskip
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with dvitops}
  \vbox{\vskip #3
    \special{dvitops: import #1.ps \pagewidth #3}
  }}
%to put pictures on same line
\def\sideshiftedpicture#1#2#3#4{\bigskip
   \bigskip
   \immediate\write0{importing diagram #1.ps at height #4
        and width #3 with dvitops}
   \vbox{\special{dvitops:  import #1.ps #3 #4}}
   \hskip #2 plus 0pt minus 0pt
   }
\def\startsideshiftedpicture{}
\newcount\scaleup
\newcount\scaledown
\scaleup=1
\scaledown=1
\def\scalesideshiftedpicture#1#2#3#4{
  \dimen4=#4
  \dimen3=#3
  \multiply\dimen4 by\scaleup
  \divide\dimen4 by \scaledown
  \multiply\dimen3 by\scaleup
  \divide\dimen3 by \scaledown
  \sideshiftedpicture{#1}{#2}{\dimen3}{\dimen4}}

\def\picturewithpsfig#1#2{\bigskip
   \immediate\write0{importing diagram #1.ps at height #2 with psfig}
   \vbox{\hsize\pagewidth
   \centerline{\psfig{figure=#1.ps,height=#2}}
   \medskip
   \centerline{\Caption}}
   \medskip
   \def\Caption{}}
\def\centerpicturewithpsfig#1#2#3{
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with psfig}
  \dimen 3 = #2
  \dimen 4 = #3
  \dimen 2 = \pagewidth
  \advance\dimen2 by -\dimen3
  \divide\dimen2 by 2
  \discrversionB{\shrinkonestep}{}
  \discrversionH{\shrinkonestep}{}
  \discrversionF{\enlargeonestep}{}
  \discrversionG{\enlargeonestep\enlargeonestep}{}
  \dimen 6 = \dimen4
  \divide\dimen6 by 3
  \vskip\dimen6
  \ifdim\dimen2>0\hskip\dimen2\fi
     \vbox{\psfig{file=#1.ps,width=\dimen3,height=\dimen4}}
  \discrversionG{\vskip-\dimen6}{}
  }
\def\enlargeonestep{\dimen5 = 5truein
   \dimen6 = 7truein
   \ifdim\dimen3<\dimen5
      \ifdim\dimen4<\dimen6
         \divide\dimen3 by 5
         \multiply\dimen3 by 6
         \divide\dimen4 by 5
         \multiply\dimen4 by 6
      \fi\fi}
\def\shrinkonestep{\divide\dimen3 by 6
         \multiply\dimen3 by 5
         \divide\dimen4 by 6
         \multiply\dimen4 by 5}
\def\sideshiftedpicturewithpsfig#1#2#3#4{
   \immediate\write0{importing diagram #1.ps at height #4
      and width #3 with psfig}
   \dimen 4 = #4
   \dimen 3 = #3
   \discrversionB{\shrinkonestep}{}
   \discrversionH{\shrinkonestep}{}
   \discrversionF{\enlargeonestep}{}
   \discrversionG{\enlargeonestep\enlargeonestep}{}
%   \vbox{\psfig{figure=#1.ps,height=\dimen4,width=\dimen3}}
   \psfig{figure=#1.ps,height=\dimen4,width=\dimen3}
   \hskip #2 plus 0pt minus 0pt}

\def\picturewithgraphicx#1#2{\bigskip
   \immediate\write0{importing diagram #1.ps at height #2 with graphicx}
   \vbox{\hsize\pagewidth
   \centerline{\includegraphics[height=#2]{#1.ps}}
   \medskip
   \centerline{\Caption}}
   \medskip
   \def\Caption{}}
\def\centerpicturewithgraphicx#1#2#3{
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with graphicx}
  \dimen 3 = #2
  \dimen 4 = #3
  \dimen 2 = \pagewidth
  \advance\dimen2 by -\dimen3
  \divide\dimen2 by 2
  \discrversionB{\shrinkonestep}{}
  \discrversionH{\shrinkonestep}{}
  \discrversionF{\enlargeonestep}{}
  \discrversionG{\enlargeonestep\enlargeonestep}{}
  \dimen 6 = \dimen4
  \divide\dimen6 by 3
  \vskip\dimen6
  \ifdim\dimen2>0\hskip\dimen2\fi
     \vbox{\includegraphics[width=\dimen3,height=\dimen4]{#1.ps}}
  \discrversionG{\vskip-\dimen6}{}
  }
\def\enlargeonestep{\dimen5 = 5truein
   \dimen6 = 7truein
   \ifdim\dimen3<\dimen5
      \ifdim\dimen4<\dimen6
         \divide\dimen3 by 5
         \multiply\dimen3 by 6
         \divide\dimen4 by 5
         \multiply\dimen4 by 6
      \fi\fi}
\def\shrinkonestep{\divide\dimen3 by 6
         \multiply\dimen3 by 5
         \divide\dimen4 by 6
         \multiply\dimen4 by 5}
\def\sideshiftedpicturewithgraphicx#1#2#3#4{
   \immediate\write0{importing diagram #1.ps at height #4
      and width #3 with graphicx}
   \dimen 4 = #4
   \dimen 3 = #3
   \discrversionB{\shrinkonestep}{}
   \discrversionH{\shrinkonestep}{}
   \discrversionF{\enlargeonestep}{}
   \discrversionG{\enlargeonestep\enlargeonestep}{}
%   \vbox{\includegraphics[height=\dimen4,width=\dimen3]{#1.ps}}
   \includegraphics[height=\dimen4,width=\dimen3]{#1.ps}
   \hskip #2 plus 0pt minus 0pt}

%alternative picture importation code;  I've been told about this
  %but have not used it successfully myself
%\def\Caption{}
%\input epsf
%\epsfverbosetrue
%\def\picture#1#2{\bigskip
%    \bigskip
%    \immediate\write0{importing diagram #1.ps at height #2 with epsf}
%    \epsfysize=#2
%    \centerline{\epsffile{#1.ps}}
%    \medskip
%    \Centerline{\Caption}
%    \medskip
%    }

%in addition it may be necessary to amend the imported  .ps  file
%by replacing the   %%BoundingBox   header by  %%BoundingBox:

\def\atUEssex{
   \input psfig.sty
   \def\picture##1##2{\picturewithpsfig{##1}{##2}}
   \def\sideshiftedpicture##1##2##3##4{
     \sideshiftedpicturewithpsfig{##1}{##2}{##3}{##4}}
   \def\centerpicture##1##2##3{
       \centerpicturewithpsfig{##1}{##2}{##3}}
   \def\startsideshiftedpicture{
       \sideshiftedpicture{empty}{1pt}{1pt}{1pt}}
   }
\def\usegraphicx{
   \input miniltx
   \def\Gin@driver{dvips.def}
   \input graphicx.sty
   \def\picture##1##2{\picturewithgraphicx{##1}{##2}}
   \def\sideshiftedpicture##1##2##3##4{
     \sideshiftedpicturewithgraphicx{##1}{##2}{##3}{##4}}
   \def\centerpicture##1##2##3{
       \centerpicturewithgraphicx{##1}{##2}{##3}}
   \def\startsideshiftedpicture{
       \sideshiftedpicture{empty}{1pt}{1pt}{1pt}}
}


%special fonts
\def\biggerscriptfonts{\scriptfont0=\eightrm
     \scriptfont1=\eighti
     \scriptfont2=\eightsy
     \scriptfont3=\eightex
     \scriptfont\bffam=\eightbf
     \scriptfont\msbfam=\eightmsb
     \scriptfont\eufmfam=\eighteufm
     }
\def\Loadcyrillic{\ifamstexsupported
      \input cyracc.def
      \font\tencyr=wncyr10
      \def\cyr{\tencyr\cyracc} \def\itcyr{\tenitcyr\cyracc}
    \else\def\cyr{\smc}\def\cdprime{u\i}\def\Cdprime{UI}\fi}
%http://wkweb3.cableinet.co.uk/linux/LDP/HOWTO/Cyrillic-HOWTO-7.html
%cyrillic.tex

\def\Loadfourteens{\iffourteensavailable
       \font\fourteenrm=cmr14
       \font\fourteenbf=cmbx14
       \font\fourteenit=cmsl14
   \else\font\fourteenrm=cmr10 scaled \magstep2
      \font\fourteenbf=cmbx10 scaled \magstep2
      \font\fourteenit=cmsl10 scaled \magstep2
   \fi}
\def\smallerfonts{\eightpoint
   \textfont\msbfam=\eightmsb
   \textfont\eufmfam=\eighteufm
   \textfont\bffam=\eightbf
   \textfont\eurmfam=\eighteurm
   \textfont\eusmfam=\eighteusm
   }
\font\twelveex=cmex10 scaled \magstep1


%Copyright declarations
\newcount\oldcopyrightdate
\newcount\newcopyrightdate
\oldcopyrightdate=0
\def\copyrightdate#1{\newcopyrightdate=#1
       \ifnum\newcopyrightdate=\oldcopyrightdate\else
       \oldcopyrightdate=#1
       \oldfootnote{}{\eightrm \copyright\ #1 D.~H.~Fremlin}
       \immediate\write0{Copyright #1}
       \fi}

%hyphenation
\hyphenation{equi-ver-id-ical}

%miscellaneous definitions
\def\Aut{\mathop{\text{Aut}}}
\def\backstep#1{\hskip -0.#1em plus 0 pt minus 0 pt}
\def\Bang{\vthsp\hbox{\bf X\hskip0.07em\llap X\hskip0.07em\llap X\rm}}
\def\BanG{\Bang{\hskip 1.111pt plus 3.333pt minus 0pt}}
%\def\BbbN{\mathchoice{\hbox{$\Bbb N$\hskip0.02em}}
%  {\hbox{$\Bbb N$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb N$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb N$\hskip0.04em}}}
%\def\BbbQ{\mathchoice{\hbox{$\Bbb Q$\hskip0.02em}}
%  {\hbox{$\Bbb Q$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb Q$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb Q$\hskip0.04em}}}
%\def\BbbR{\mathchoice{\hbox{$\Bbb R$\hskip0.02em}}
%  {\hbox{$\Bbb R$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb R$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb R$\hskip0.04em}}}
%  %do not use with $\Bbb R^X$
%\def\BbbZ{\mathchoice{\hbox{$\Bbb Z$\hskip0.02em}}
%  {\hbox{$\Bbb Z$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb Z$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb Z$\hskip0.04em}}}
\def\BbbN{\openBbb{N}}
\def\BbbQ{\openBbb{Q}}
\def\BbbR{\openBbb{R}}
  %do not use with $\Bbb R^X$  do not use in $_{\Bbb R^r}$
\def\BbbZ{\openBbb{Z}}
\def\Bbbone{\hbox{1\hskip0.13em\llap{1}}}
\def\biggerint{\textfont3=\twelveex\intop\nolimits}
  % you can get this size into a displayed formula with
  %  $$\hbox{$\biggerint_{\Cal S}\pmb{u}\,d\pmb{v}$}$$
  % but it may then seem cramped
  % or try \hbox{$\biggerint_{\Cal S}\,$}
\def\biggeriint{\textfont3=\twelveex\iint}
  % remarkable fact:  this seems to infect all \int within a given
  % $..$ ; but  $\biggerint$$\int$ seems to leave the $\int$ normal
  % really  \textfont3=\twelveex  is more reliable, since  \biggerint
  % as written gets turned off rather easily (e.g. by  \comment,
  % or by an extra  {...})
\def\bover#1#2{{{#1}\over{#2}}}
\def\Bover#1#2{\backstep4\hbox{
     \biggerscriptfonts${\shortstrut#1}\over{\shortstrut#2}$}}
  % when there are superscripts in the denominator, as in \Bover1{2^m}
  % or  \Bover1{\sqrt{2\pi}^r} , these can sometimes come out with better
  % horizontal alignment if you use  \Bover1{{\sshortstrut 2}^m}  or
  % \Bover1{{\sqrt{2\pi}}^r}
\def\bsover#1#2{\textstyle\bover{#1}{#2}}
\def\bsp{/\penalty-100}
  % \penalty seems to behave oddly in a \tt string in \fullsize mode
\def\Caption{}
\def\Center#1{

   \Centerline{#1}

   \noindent}
\def\Centerline#1{\vskip 5pt
     {\def\int{\text{}$$\textfont3=\twelveex\intop\nolimits}
      \def\dashint{\text{}$$\textfont3=\twelveex
         \mskip2mu\vrule height2.65pt width4.5pt depth-2.35pt\mskip-11mu
	 \intop\nolimits\mskip-2mu}
      \def\idashint{\text{}$$\int\mskip-7mu\dashint}	 
      \def\dashiint{\text{}$$\dashint\mskip-6mu\int}	 
      \def\dashidashint{\text{}$$\dashint\mskip-6mu\dashint}	 
      \def\underlineint{
           \text{}$$\textfont3=\twelveex\underline{\intop\nolimits}}
      \def\overlineint{
           \text{}$$\textfont3=\twelveex\overline{\intop\nolimits}}
      \def\iint{\text{}$$\textfont3=\twelveex
           \intop\nolimits\hskip-0.5em\intop\nolimits}
      \def\sum{\text{}$$\sumop}
      \def\prod{\text{}$$\prodop}
      \def\bigcap{\text{}$$\bigcapop}
      \def\bigcup{\text{}$$\bigcupop}
   \centerline{#1}}\vskip 5pt}
   \mathchardef\bigcapop="1354
   \mathchardef\bigcupop="1353
   \mathchardef\prodop="1351
   \mathchardef\sumop="1350
   %in  \Centerline{$\overline{}$, $\cmmnt{}$} must use \sumop etc
\def\cf{\mathop{\text{cf}}}
\def\circumflex#1{{\accent"5E #1}}
\def\closeplus{\hbox{+}}
\long\def\cmmnt#1{\ifresultsonly\else{#1}\fi}
\def\coint#1{\left[#1\right[}
\def\curl{\mathop{\text{curl}}}
\def\dashint{\mathchoice{-\mskip-20mu\intop\nolimits}
   {{\vrule height2.75pt width4.5pt depth-2.45pt\mskip-10mu\intop\nolimits}}
      {-\mskip-16.5mu\intop\nolimits} %not yet used
      {-\mskip-16.5mu\intop\nolimits}} %not yet used
\def\idashint{\mathchoice{\int\mskip-6mu\dashint}
      {\int\mskip-5mu\dashint}
      {\int\mskip-5mu\dashint} %not yet used
      {\int\mskip-5mu\dashint}} %not yet used
\def\dashiint{\mathchoice{\dashint\mskip-10mu\int}
      {\dashint\mskip-6mu\int}
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\dashidashint{\mathchoice{\dashint\mskip-10mu\dashint}
      {\dashint\mskip-6mu\dashint}
      {\dashint\mskip-6mu\dashint} %not yet used
      {\dashint\mskip-6mu\dashint}} %not yet used
\def\dashidashiint{\mathchoice{\dashint\mskip-8mu\dashint\mskip-10mu\int}
      {\dashint\mskip-4mu\dashint\mskip-8mu\int} 
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\dashiidashint{\mathchoice{\dashint\mskip-10mu\int\mskip-8mu\dashint}
      {\dashint\mskip-4mu\dashint\mskip-8mu\int} 
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\idashidashint{\mathchoice{\int\mskip-8mu\dashint\mskip-8mu\dashint}
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\diam{\mathop{\text{diam}}}
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
           \else\ifnum\stylenumber=3{\frnewpage}
           \else\ifnum\stylenumber=5{\frnewpage}
           \else\bigskip\fi\fi\fi}
\long\def\discrversionA#1#2{\ifnum\stylenumber=1#1\else#2\fi}
\long\def\discrversionB#1#2{\ifnum\stylenumber=2#1\else#2\fi}
\long\def\discrversionC#1#2{\ifnum\stylenumber=3#1\else#2\fi}
\long\def\discrversionD#1#2{\ifnum\stylenumber=4#1\else#2\fi}
\long\def\discrversionE#1#2{\ifnum\stylenumber=5#1\else#2\fi}
\long\def\discrversionF#1#2{\ifnum\stylenumber=6#1\else#2\fi}
\long\def\discrversionG#1#2{\ifnum\stylenumber=7#1\else#2\fi}
\long\def\discrversionH#1#2{\ifnum\stylenumber=8#1\else#2\fi}
           % facility for special interpolations in particular output
           % versions
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\diverg{\mathop{\text{div}}}
\def\dom{\mathop{\text{dom}}}
\def\eae{=_{\text{a.e.}}}

\def\esssup{\mathop{\text{ess sup}}}
\mathchardef\Exists="3239

\def\family#1#2#3{\langle#3\rangle_{#1\in#2}}
\def\familyi#1#2{\family{i}{#1}{#2}}
\def\familyiI#1{\familyi{I}{#1}}

\def\frfilename#1{\ifnum\stylenumber=1{\noindent filename #1}
               \else\ifnum\stylenumber=6{\noindent{\fiverm filename #1}}
               \else\ifnum\stylenumber=7{\noindent{\fiverm filename #1}}
               \else\fi\fi\fi}
\def\Footnote#1#2{\footnote"#1"{#2}}
\mathchardef\Forall="3238
\def\frakc{\frak c\phantom{.}}
\def\frnewpage{\vskip \pageheight plus 0pt minus\pageheight
           \eject
           \gdef\topparagraph{}
           \gdef\bottomparagraph{}
           }
\def\frsmallcirc{{\hskip 1pt plus 0pt minus 0pt}
     {\raise 1pt\hbox{$\smallcirc$}}{\hskip 1pt plus 0pt minus 0pt}}

\def\geae{\ge_{\text{a.e.}}}
\def\grad{\mathop{\text{grad}}}
\def\grheada{{\bf ($\pmb{\alpha}$)}}
\def\grheadb{{\bf ($\pmb{\beta}$)}}
\def\grheadc{{\bf ($\pmb{\gamma}$)}}
\def\grheadd{{\bf ($\pmb{\delta}$)}}
\def\grheade{{\bf ($\pmb{\epsilon}$)}}
\def\grheadz{{\bf ($\pmb{\zeta}$)}}
\def\grv#1{{\accent"12 #1}}

\def\hfillall{\hskip\pagewidth plus0pt minus\pagewidth}

\def\Imag{\mathop{\Cal I\text{m}}}
\long\def\inset#1{{\narrower\narrower{\vskip 1pt plus 1pt minus 0pt}
     #1
     {\vskip 1pt plus 1pt minus 0pt}}}
\def\Int#1#2{\displaystyle{\int_{#1}^{#2}}}
\def\interior{\mathop{\text{int}}}

\def\leae{\le_{\text{a.e.}}}
\long\def\leaveitout#1{}
\def\LLcorner{\mathchoice{\,\hbox{\vrule height7pt width0.5pt depth0pt
       \vrule height0.5pt width5pt depth0pt}\,}
   {\,\hbox{\vrule height7pt width0.5pt depth0pt
       \vrule height0.5pt width5pt depth0pt}\,}
   {\,\hbox{\vrule height5pt width0.3pt depth0pt
       \vrule height0.3pt width4pt depth0pt}\,}
   {\,\hbox{\vrule height4pt width0.25pt depth0pt
       \vrule height0.25pt width3pt depth0pt}\,}
  }
\def\Ln{\mathop{\text{Ln}}}
\def\Loadfont#1{\font#1}
\def\loibr{\hbox{\hskip0.1em]\hskip0.05em}}
\def\Matrix#1{\pmatrix#1\endpmatrix}
\def\Mu{\text{M}}
\def\noamslogo{\def\amslogo{}}
\def\normalsubgroup{\vartriangleleft}
\def\Nu{\text{N}}
\def\ocint#1{\left]#1\right]}
\def\ofamily#1#2#3{\langle#3\rangle_{#1<#2}}
\let\oldcases=\cases
\let\oldfootnote=\footnote
\def\On{\mathop{\text{On}}}
\def\ooint#1{\left]#1\right[}
\def\openBbb#1{\Bbb#1\mskip1mu}
\def\pd#1#2{\bover{\partial#1}{\partial#2}}
\def\Pd#1#2{\Bover{\partial#1}{\partial#2}}
\def\pdd#1#2{{{\partial^2#1}\over{\partial#2^2}}}
\def\Pdd#1#2{\Bover{\partial^2#1}{\partial#2^2}}
\def\pde#1#2#3{{{\partial^2#1}\over{\partial#2\partial#3}}}
\def\Pde#1#2#3{\Bover{\partial^2#1}{\partial#2\partial#3}}
\def\Prf{\vthsp\hbox{\bf P\hskip0.07em\llap P\hskip0.07em\llap P\rm}}
\def\pushbottom#1{\hbox{\vrule height0pt depth#1 width0pt}}
\def\Qed{\vthsp\hbox{\bf Q\hskip0.07em\llap Q\hskip0.07em\llap Q\rm}}
\def\QeD{\Qed{\hskip 1.111pt plus 3.333pt minus 0pt}}
\def\Quer{\vthsp\hbox{\bf ?\hskip0.05em\llap ?\hskip0.05em\llap
   ?\thinspace\rm}}
\def\query{\discrversionA{\immediate\write0{query}
    \global\advance\footnotenumber by 1
    \oldfootnote{$^{\the\footnotenumber}$}{query}}{}}
\def\Real{\mathop{\Cal R\text{e}}}
\def\restr{\mathchoice
  {\hbox{$\restriction$}\mskip 1mu}
  {\hbox{$\restriction$}\mskip 1mu}
  {\hbox{$\scriptstyle\restriction$}\mskip 1mu}
  {\hbox{$\scriptscriptstyle\restriction$}\mskip 1mu}}
\def\restrp{\mathchoice
  {\hbox{$\restriction$}\mskip 2mu}
  {\hbox{$\restriction$}\mskip 2mu}
  {\mskip-1mu\hbox{$\scriptstyle\restriction$}\mskip 2mu}
  {\mskip-1mu\hbox{$\scriptscriptstyle\restriction$}\mskip 1mu}}
    %use in \restrp V \restrp W \restrp Y
    % \restrp\Cal P \restrp\Bbb R \restrp\frak
    % \restrp\Omega \restrp\Tau \restrp\Upsilon
    % \restrp[ \restrp\widehat{}
\def\Rho{\text{P}}
\def\roibr{\hbox{\hskip0.05em[\hskip0.1em}}
\def\RoverC{\hbox{\biggerscriptfonts${{\Bbb R}\atop{\Bbb C}}$}}
\def\sequence#1#2{\family{#1}{\Bbb N}{#2}}
\def\sequencen#1{\sequence{n}{#1}}
\def\setover#1#2{\overset{#1}\to{#2}}
\def\shortstrut{\hbox{\vrule height7.5pt depth2.5pt width0pt}}
\def\sshortstrut{\hbox{\vrule height5.8pt depth2.5pt width0pt}}
\def\smallcirc{{\scriptstyle\circ}}
\def\ssbullet{{\scriptscriptstyle\bullet}}
\def\symmdiff{\triangle}
\def\thicken#1{#1\hskip0.03em\llap #1\hskip0.03em\llap
   #1\hskip0.03em\llap #1}
\def\Tau{{\text{T}}}
\def\tbf#1{\text{\bf #1}}
\def\terror#1#2#3{#1:  `#2' for `#3'}
\def\today{\number\day.\number\month.\number\year}
\def\UEMDRR{University of Essex Mathematics Department Research Report}
\def\UEMDWP{University of Essex Mathematics Department Working Paper}
\def\versiondate#1{
    \immediate\write0{Version of #1}
    \def\vsdate{#1}
               \ifnum\stylenumber=1\hfill Version of #1
                              \medskip
               \else\ifnum\stylenumber=2\hfill Version of #1
                              \medskip
               \else\ifnum\stylenumber=6\hfill{\fiverm Version of #1}
                              \medskip
               \else\ifnum\stylenumber=7\hfill{\fiverm Version of #1}
                              \medskip
               \else\ifnum\stylenumber=8\hfill Version of #1
                              \medskip
               \else\fi\fi\fi\fi\fi}
\def\versionhead{\ifnum\stylenumber=3 \hfill PRINTER'S VERSION

                        \bigskip
                   \else\ifnum\stylenumber=4 \hfill REFEREE'S VERSION

                        \bigskip
                   \else \fi\fi}
\def\versionref{\ifnum\stylenumber=1\workingref
           \else\ifnum\stylenumber=2\workingref
           \else\ifnum\stylenumber=8\workingref
           \else\printerref\fi\fi\fi}
\def\vthsp{\hskip0.001em}

%Missing AmSTeX can be coped with by using the following lines:

\ifamstexsupported\else
   \immediate\write0{AmSTeX-unsupported version of  fremtex.tex }
   \fourteensavailablefalse
   \def\@{@}
   \def\\{\cr}
   \def\AmSTeX{AmS\TeX}
   \def\Bbb#1{\underline{\hbox{\bf #1}}}
   \def\biggerscriptfonts{}
   \def\bumpeq{=_{\text{approx}}}
   \def\Cal#1{{\cal#1}}
   \def\curlyeqprec{\preceq}
   \def\eightit{\it}
   \def\eightrm{\rm}
   \def\eightsmc{\smc}
   \def\eurm#1{\underline{\cal#1}}
   \def\Footnote#1#2{\footnote#1#2}
   \def\frac#1#2{{#1}\over{#2}}
   \def\frak#1{{\hbox{\bf #1\hskip0.1em\llap #1}}}
   \def\frnewpage{\vfill\eject}
   \def\iint{\int\!\!\!\int}
   \def\iiint{\int\!\!\!\int\!\!\!\int}
   \def\iiiint{\int\!\!\!\int\!\!\!\int\!\!\!\int}
   \def\llcorner{\lfloor}
   \def\loadeurm{}
   \def\loadeusm{}
%   \def\loadeusmx#1{\def\eusm#1{\underline{\cal#1}}}
   \def\ltimes{|\hskip-0.5em\times}
   \def\eusm#1{\underline{\cal#1}}
   \def\ltimes{|\times}
   \def\Loadfont#1{}
   \def\Matrix#1{\pmatrix{#1}}
   \def\noamslogo{}
   \def\nrightarrow{\not\rightarrow}
   \def\preccurlyeq{\preceq}
   \def\query{\discrversionA{\immediate\write0{query}
                             \footnote{$^*$}{query}}{}}
   \def\restriction{|}
   \def\rtimes{\times|}
   \def\setover#1#2{#2^{#1}}
   \def\shortstrut{\strut}
   \def\smallerfonts{\eightrm}
   \def\smallfrown{\wedge}
   \def\smc{}
   \def\spcheck{\check{\phantom{x}}}
   \def\sphat{\hat{\phantom{x}}}
   \def\square{\mathchoice\sqr64\sqr64\sqr43\sqr33}
   \def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt
          \hbox{\vrule width.#2pt height#1pt \kern#1pt
                  \vrule width.#2pt}
          \hrule height.#2pt}}}}
   \def\pmb#1{\underline{#1}}
   \def\text#1{\hbox{#1}}
   \def\therefore{\ssbullet\raise.9ex\hbox{$\ssbullet$}\ssbullet}
   \def\twelvebf{\bf}
   \def\vartriangleleft{\triangleleft}
   \def\Vdash{|\vdash}
   \def\vDash{|=}
   \fi

%End of lines for missing AmSTeX

\ifamstexsupported\input amstex
%              \documentstyle{amsppt}
               \input amsppt.sty
               \PSAMSFonts
	       \tenpoint
               %\font\eightmsb=msym8
               \font\eightmsb=msbm10
               \font\eighteufm=eufm10 at 8 pt
               \font\eighteurm=eurm8
               \font\eighteusm=eusm8
               \font\varseveneufm=eufm10 at 7 pt
               \scriptfont\eufmfam=\varseveneufm
   \def\.#1{{\accent"5F #1}}
   \def\=#1{{\accent"16 #1}}
   \gdef\nopagenumbers{\NoPageNumbers}
   \else\fi

\newcount\footmarkcount
\ifDHFamstexloaded
    \loadmsBm
    \def\eightBbb#1{{\fam\msBfam#1}}
    \loadmsFm
    \def\eightfrak#1{{\fam\msFfam#1}}
  \else
    \def\eightBbb#1{{\scriptstyle\Bbb#1}}
    \def\eightfrak#1{{\scriptstyle\frak#1}}
  \fi
\ifathome\athome\fi

%interword glue (\tenrm standard):
%  (space) 3.333 + 1.667 - 1.111
%  (,)     3.333 + 2.083 - 0.889
%  (.)     4.444 + 5.    - 0.370

%very odd fact:  \discrversionA{2.}{}  doesn't work,
%   \discrversionA{ 2.}{} does

%The false line feed introduced by my Fortran programs, leading to
%a surplus  ae  at the top of the next page, seems to be associated
%with  Fk(\032)3081  or something similar in the PostScript file.


\immediate\write0{Extract from MEASURE THEORY,
   by D.H.Fremlin, University of Essex, Colchester.
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   http://dsl.org/copyleft/dsl.txt.   This is a development version and the
   source files are not permanently archived.
   For further information contact david\@fremlin.org.}

%hyphenations
\hyphenation{Archi-me-dean}

%new style
\def\smallprintersize{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{468pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \nologo
     \def\copyrightdate##1{}
     \def\folio{\ifnum\pageno<0{}\else\number\pageno\fi}
     \def\mtfootline{}
     \stylenumber=10
     \pageno=-1000}

%With-proofs, results-only versions
\newif\ifwithproofs
\withproofstrue
\newif\ifresultsonly
\resultsonlyfalse
\def\noproofs{\withproofsfalse}
\def\resultsonly{\resultsonlytrue\withproofsfalse}
\long\def\proof#1{\ifwithproofs{\medskip

     \noindent{\bf proof} #1}\else\fi}
\long\def\prooflet#1{\ifwithproofs{#1}\else\fi}
%\long\def\Prooflet#1{\prooflet{\Prf\ #1 \Qed\ }}
\long\def\endnotes#1{\ifresultsonly\else{#1}\fi}
\long\def\exercises#1{\ifresultsonly\else{#1}\fi}
\long\def\references#1{\ifresultsonly\else{#1}\fi}
\long\def\dvro#1#2{\ifresultsonly{#1}\else{#2}\fi}

%Headline assignments

\def\topparagraph{}
\def\bottomparagraph{}
\def\newparagraph{}
\newif\ifpageturned

\newdimen\rhoffset
\newdimen\lhoffset
\lhoffset=0cm
\rhoffset=0cm

\newcount\volumeno
\volumeno=5
\newcount\luluvolumeno
\luluvolumeno=0

\let\plainpagebody=\pagebody
\def\pagebody{\ifodd\pageno\global\hoffset=\rhoffset
         \else\global\hoffset=\lhoffset\fi
     \plainpagebody
     \turnpage}
\def\turnpage{
     \iffirstpage\else{\baselineskip=24pt\line{\mtfootline}}\fi
%     \immediate\write0{turning page \folio,
%           top paragraph \topparagraph,   bottom \bottomparagraph,
%           new \newparagraph, pagetotal \the\pagetotal}
     \global\pageturnedtrue
     \global\firstpagefalse
     \global\let\topparagraph=\bottomparagraph
     \global\let\bottomparagraph=\newparagraph
     }

\newdimen\pagetotalplus
\newdimen\pagetotalminus

%wheader parameters:  #1 new paragraph name
 % #2#3#4  \vskip parameters if page break not forced (no. of points)
 % #5  spare height demanded if page break not to be forced
\long\def\wheader#1#2#3#4#5{\gdef\newparagraph{#1}
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by #5
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  %\showthe\pagetotal \showthe\pageshrink \showthe\pagegoal
  \global\pageturnedfalse
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal
%     \immediate\write0{forcing page break}
     \eject
     \else\vskip #2pt plus #3pt minus #4pt\fi
     \else\vskip #2pt plus #3pt minus #4pt\fi
  \ifpageturned\global\let\topparagraph=\newparagraph
%     \immediate\write0{paragraph change on turning page}
     \fi
%  \immediate\write0{top paragraph \topparagraph, bottom
%       \bottomparagraph, pagetotal \the\pagetotal}
  \ifdim\pagetotal>\pagegoal\else\gdef\bottomparagraph{#1}\fi
%   \immediate\write0{top paragraph \topparagraph, bottom
%       \bottomparagraph, pagetotal \the\pagetotal}
  }
\def\header#1{
     \wheader{#1}{10}{4}{4}{24pt}}
\def\leader#1#2{\immediate\write0{#1 #2}
     \wheader{#1}{14}{7}{5}{24pt}
     {\bf #1 #2}}
\def\vleader#1#2#3{\immediate\write0{#2 #3}
     \wheader{#2}{14}{7}{5}{#1}
     {\bf #2 #3}}
\def\Notesheader#1{\immediate\write0{#1 Notes and comments}
     \wheader{#1 Notes}{16}{8}{5}{48pt}
     \noindent{\bf #1 Notes and comments}}
\def\vNotesheader#1#2{\immediate\write0{#2 Notes and comments}
     \wheader{#2 Notes}{16}{8}{5}{#1}
     \noindent{\bf #2 Notes and comments}}
\def\spheader#1#2#3#4#5{\header{#1#2#3#4#5}{\bf (#5)}}
\def\spstheader#1#2#3#4#5{\header{*#1#2#3#4#5}{\bf (#5)}}
\def\vspheader#1#2#3#4#5#6{
     \wheader{#2#3#4#5#6}{10}{4}{4}{#1}{\bf (#6)}}
\def\sqheader#1#2#3#4#5{\header{#1#2#3#4#5}{$\pmb{>}${\bf (#5)}}}
\def\errheader#1#2#3{
     \wheader{#3}{6}{4}{2}{24pt}
     {\bf p #1 l #2}}
\long\def\woddheader#1#2#3#4#5{
   \ifodd\pageno\wheader{#1}{#2}{#3}{#4}{#5}
   \else\vskip #2pt plus #3pt minus #4pt\fi}

\def\indexmedskip{
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by 18 pt
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal
      \else\medskip\fi\else\medskip\fi}

\def\indexheader#1{\immediate\write0{#1}\vindexheader{#1}{24}}
\def\vindexheader#1#2{
  \gdef\newparagraph{#1}
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by #2 pt
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  %\showthe\pagetotal \showthe\pageshrink \showthe\pagegoal
  \global\pageturnedfalse
%  \ifdim\pagetotal>\pagegoal
%      \immediate\write0{warning: pagetotal greater than pagegoal}
%      \immediate\write0{...   last paragraph reached is #1}
%      \immediate\write0{...   check for underfull vbox here}
%      \else\fi
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal{\eject}
      \fi\fi
  \ifdim\pagetotal>\pagegoal\else\gdef\bottomparagraph{#1}\fi
  \ifpageturned{\global\let\topparagraph=\newparagraph
%     \immediate\write0{paragraph change on turning page}
      }\else\fi
%  \immediate\write0{top paragraph \topparagraph, bottom
%    \bottomparagraph}
  }

\def\leftheadline{\rlap{\folio}\hfill{\eightit\chaptername}\hfill
       \llap{\eightbf\topparagraph}}
\def\rightheadline{\rlap{\eightbf\bottomparagraph}\hfill
       {\eightit\headlinesectionname}\hfill\llap{\folio}}

%Footline assignments
\def\mtfootline{\ifodd\pageno\rightfootline\else\leftfootline\fi}
\def\rightfootline{\hfill\eightsmc D.H.Fremlin}
\def\leftfootline{\ifwithproofs{\eightsmc Measure Theory\hfill}
    \else{\eightsmc Measure Theory \eightit (abridged version)\hfill}
    \fi}

%Style assignments
\def\newchapter#1{\gdef\topparagraph{}
   \gdef\bottomparagraph{Chap.\ #1 {\it intro.}}
   \gdef\newparagraph{Chap.\ #1 {\it intro.}}
   \footmarkcount=0
   \gdef\sectionname{Introduction}
   \gdef\headlinesectionname{Introduction}
   \centerline{\bf Chapter #1}
   \medskip
   \centerline{\bf \chaptername}
   \medskip
   }
\def\newsection#1{\gdef\bottomparagraph{\S#1 {\it intro.}}
   \noindent{\bf #1 \sectionname}
   \smallskip
   \let\headlinesectionname=\sectionname}
\def\newvolume#1{\gdef\topparagraph{}
   \gdef\bottomparagraph{Vol.\ #1 {\it intro.}}
   \Loadfont{\twelvebf=cmbx12}
   \def\chaptername{Volume #1}
   \def\sectionname{\volumename}
   \def\headlinesectionname{\volumename}
   \centerline{\twelvebf Volume #1:  \volumename}
   \bigskip
   }

%font variations
\newif\ifvolwp
\volwpfalse
\def\tenrmstretch#1{\fontdimen3\tenrm=#1}
\font\elevenex=cmex10 at 11pt

%contents and introductions
\newif\iflargelogo
\def\vtmpb#1{}
\def\newvolume#1{}
\def\pagereference#1#2{\hfill\hbox{\ifresultsonly#2\qquad \hskip 1truein
                                                 \else#1\qquad \fi}}
\long\def\section#1#2#3#4#5#6{\ifresultsonly
    \qquad #1 #2\pagereference{#4}{#5}
  \else\vskip 2pt plus 1pt minus 0pt
    \vbox{\qquad #1 #2\pagereference{#4}{#5}\par
    \abstr{#6}
    }\vskip -1pt plus 1pt minus 0pt
  \fi}
\long\def\chapintrosection#1#2#3{\ifresultsonly
    \qquad Introduction\pagereference{#2}{#3}
  \else\vskip 1pt plus 1pt minus 0pt
    \qquad Introduction\pagereference{#2}{#3}
    \vskip 1pt plus 1pt minus 0pt
  \fi}
\def\abstr#1{\ifresultsonly{}
   \else{\advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \vskip1pt plus 0pt minus0pt
     \parindent=0pt
     \smallerfonts #1%
     \hbox{\vrule height0pt depth2pt width0pt}
     \vskip1pt plus 0pt minus0pt}\fi}
     
%language variations
\newif\ifUSEnglish
\USEnglishfalse
%\USEnglishtrue

%"Later editions only"
\newcount\latereditionpage
\newcount\latereditionmark
\latereditionpage=-1
\latereditionmark=0
\newcount\newpageno

%removing the \write0 instruction altogether disorganizes it!
\def\Latereditions{\newpageno=\pageno%
  \ifdim\pagetotal>\pagegoal\advance\newpageno by 1\fi%
     \divide\newpageno by 2%
  \immediate\write0{%
%     "Later editions only";
%      pagetotal \the\pagetotal, pagegoal \the\pagegoal,
%      pageno \number\pageno,\par newpageno \number\newpageno,
%      latereditionpage \number\latereditionpage%
  }\ifnum\latereditionpage=\newpageno%
     \ifDHFamspptloaded$^{\number\latereditionmark}$%
     \else\footnote{Later editions only.}\fi%
  \else\footnote{Later editions only.}%
  \global\latereditionpage=\newpageno%
  \global\latereditionmark=\footmarkcount\fi}%
%  \immediate\write0{newpageno \number\newpageno,
%      latereditionpage \number\latereditionpage}

   
%general definitions
\def\action{\mathchoice{\lower-.3ex\hbox{$\ssbullet$}}
                       {\lower-.3ex\hbox{$\ssbullet$}}
                       {\ssbullet}
                       {\ssbullet}}  %see also \varaction
\def\add{\mathop{\text{add}}\nolimits}
\long\def\allowmorestretch#1#2{
  \ifdim\pagewidth<#1pt\else
    \fontdimen3\tenrm=2pt\fontdimen4\tenrm=1.67pt\fi
  #2
    \fontdimen3\tenrm=1.67pt\fontdimen4\tenrm=1.11pt}
\def\am{atom\discretionary{-}{}{}lessly-measurable}
\def\AmuA{\Aut_{\bar\mu}\mskip-2mu\frak A}
\def\ASint{\mathchoice{\Bbb A\mskip-5mu}
                  {\Bbb A\mskip-3mu}
                  {\Bbb A\mskip-2mu}
                  {\Bbb A\mskip-2mu}\Sint}
\def\Asisfour{$(\frak A,\bar\mu,T,\family{t}{T}{\frak A_t})$}
\def\Asissix{$(\frak A,\bar\mu,T,
   \penalty-100\family{t}{T}{\frak A_t},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$}
\def\Asisfourt{$(\frak A,\bar\mu,\coint{0,\infty},
   \langle\frak A_t\rangle_{t\ge 0})$}
\def\Asissixt{$(\frak A,\bar\mu,\coint{0,\infty},
   \penalty-100\langle\frak A_t\rangle_{t\ge 0},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$}
\def\Atop#1#2{{\scriptstyle #1}\atop{\scriptstyle #2}}
\def\Aut{\mathop{\text{Aut}}}

\def\Bcap{\Booleanop{\cap}}
\def\Bcup{\Booleanop{\cup}}
\def\Booleanop#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\hskip.2em\scriptfont2=\eightsy$\scriptstyle#1$\hskip.2em}}
    {\hbox{\hskip.2em\scriptfont2=\eightsy$\scriptstyle#1$\hskip.2em}}
 %   {{\mskip2mu\scriptstyle#1\mskip2mu}}  %until 15.10.17
    {{\scriptscriptstyle#1}}  %from 15.10.17
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\Bsetminus{\Booleanop{\setminus}}
\def\Bsubseteq{\hskip.1em\Booleanop{\subseteq}\hskip.1em}
\def\Bsubset{\Booleanop{\subset}}
\def\Bsupseteq{\Booleanop{\supseteq}}
\def\Bsupset{\Booleanop{\supset}}
\def\Bsymmdiff{\Booleanop{\symmdiff}}
\def\Bcapshort{\Booleanopshort{\cap}}
\def\Bcupshort{\Booleanopshort{\cup}}
\def\Booleanopshort#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {\hskip-1pt\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {{\scriptstyle#1}}
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\Bsetminusshort{\Booleanopshort{\setminus}}
\def\BSint{\mathchoice{\Bbb B\mskip-5mu}
                  {\Bbb B\mskip-3mu}
                  {\Bbb B\mskip-2mu}
                  {\Bbb B\mskip-2mu}\Sint}
%\def\Bsubseteqshort{\hskip.1em\Booleanopshort{\subseteq}\hskip.1em}
\def\Bsubseteqshort{\hskip.1em\Booleanopshort{\subseteq}}
\def\Bsubsetshort{\Booleanopshort{\subset}}
\def\Bsupseteqshort{\Booleanopshort{\supseteq}}
\def\Bsymmdiffshort{\Booleanopshort{\symmdiff}}

\def\closeBooleanop#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {{\scriptstyle#1}}
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\closeBcup{\closeBooleanop{\cup}}
\def\closeBcap{\closeBooleanop{\cap}}
\def\closeBsetminus{\closeBooleanop{\setminus}}
\def\closeBsymmdiff{\closeBooleanop{\symmdiff}}

\def\biggerHeint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 2.6pt{\sevenrm H}\hss}
   \intop\nolimits}
\def\biggerPfint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}
   \intop\nolimits\hskip 1.2pt}
\def\biggerSint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 3.6pt{\sevenrm S}\hss}
   \intop\nolimits}
\def\bu{\mathop{\text{bu}}}
\def\Bvalue#1{\mathchoice
   {\hbox{$[\![#1]\!]$}}
   {\hbox{$[\![#1]\!]$}}
   {\hbox{$\scriptstyle[\![#1]\!]$}}
   {\hbox{$\scriptscriptstyle[\![#1]\!]$}}}
\def\cadlag{c\`adl\`ag}
\def\CalBa{\Cal{B}\eurm{a}}
   %note: neither  \cal  nor  \eusm  has lower case
\def\CalFr{\Cal F_{\text{Fr}}}
\def\callal{c\`all\`al}
\def\CalNwd{\Cal{N}\hskip-1.2pt\eurm{w}\hskip-.8pt\eurm{d}}

\def\CalRbg{\Cal{R}\mskip-1mu\eurm{b}\mskip-1mu\eurm{g}}
\def\CalSmz{\Cal{S}\mskip-2mu\eurm{mz}}
\def\CalUn{\Cal{U}\eurm{n}}
\def\capacity{\mathop{\text{cap}}\nolimits}
\def\Caratheodory{Cara\-th\'eo\-dory}
\def\Cdlg{C_{\text{dlg}}}
\def\cff{\cf\mskip1mu}
\def\checkpmb#1{\check{\pmb #1}}
\def\ci{\mathop{\text{ci}}}
\def\cl{\mathop{\text{cl}}}
\def\Clll{\hbox{$C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\clMsimp{\overline{M}_{\text{simp}}}
\def\closeuparrow{\mathchar"0222}
\def\closedownarrow{\mathchar"0223}
\def\clstar{\text{cl*}}
\def\cov{\mathop{\text{cov}}}
\def\covar#1#2{\mathchoice
  {[#1{\lower4pt\hbox{${\scriptstyle *}\atop{\vrule height 18pt depth -10pt}$}}#2]}
  {[#1{*\atop{\vrule height 8pt depth 0pt}}#2]}
  {[#1{*\atop{\vrule height 6pt depth 0pt}}#2]}
  {[#1{*\atop{\vrule height 6pt depth 0pt}}#2]}}
\def\covSh{\mathop{\text{cov}}\nolimits_{\text{Sh}}}
\def\cdownarrow{c\mskip2mu^{\downarrow}}
\def\cycle#1{(\overleftarrow{#1})}
\def\ddaggerloc{$(\ddagger)_{\text{loc}}$}
\def\ddownarrow{d\mskip2mu^{\downarrow}}
%\def\DiniD{\setover{
%  \hbox{\hskip.1em\vrule height-1pt width.5em depth1.4pt\hfil}}D}
%\def\Dinid{\hskip.1em
%  \hbox{\hskip.05em\vrule height-1.3pt width.5em depth1.7pt}\hskip-.65em D}
\def\Dinid{\hbox{$D$\hskip-8.3pt\vrule height-1pt width7pt depth1.85pt}
   \hskip1pt}
\def\DiniD{\hbox{$D$\hskip-8.3pt\vrule height9.2pt width7ptdepth-8.35pt}
   \hskip1pt}

\def\discrcenter#1#2{\ifdim\pagewidth>#1{#2}\else{

  \Centerline{#2}

  \noindent}\fi}
    %note:  in \discrcenter, make sure that #2 ends in a space
    % (if you want one to appear) and that the continuation
    % follows without
    % a space after the final } of #2, so that it will be
    % truly unindented
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
        \else\ifnum\stylenumber=3{\frnewpage}
        \else\ifnum\stylenumber=5{\frnewpage}
        \else\wheader{}{18}{14}{2}{72pt}\fi\fi\fi}
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\Displaycause#1{\cr\displaycause{#1}&}
\def\dotproduct{\hskip.1em\tbf{.}\hskip.1em}
\def\doubleheadrightarrow{\rightarrow\mskip-12mu\rightarrow}
\long\def\doubleinset#1{\inset{\inset{\parindent=-20pt #1}}}
\def\duparrow{d\mskip2mu^{\uparrow}}
\def\dvAformerly#1{\discrversionA{\footnote{Formerly #1.}}{}}
\def\dvAnew#1{\discrversionA{\footnote{New #1.}}{}}
\def\dvArevised#1{\discrversionA{\footnote{Revised #1.}}{}}
\def\dvrocolon#1{\dvro{#1: }{#1}}
\def\eightVVdash{\mskip5mu\vrule height 6pt depth 2pt width 0.4pt
  \mskip2mu\vrule height 6pt depth 2pt width 0.4pt
  \vrule height 2.25pt depth -1.8pt width 3.2pt\mskip2mu}
\def\embedsinto{\mskip5mu\raise1.5pt\hbox{$\subset$}\mskip-15mu
   \lower3.5pt\hbox{$\rightarrow$}\mskip5mu}
\def\energy{\mathop{\text{energy}}\nolimits}
%\eqalign:  see \S478 for breaks during \eqalign
\def\equivGT{\equiv_{\text{GT}}}
\def\equivT{\equiv_{\text{T}}}
\def\Eta{\text{H}}
\def\Expn{\Bbb E}
\long\def\Extract{\oldfootnote{}{
   \noindent{\eightit
       \ifresultsonly{Extract from {\eightsmc Measure Theory,}
          results-only version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else\ifwithproofs{Extract from {\eightsmc Measure Theory,}
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else{Extract from {\eightsmc Measure Theory,} abridged version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \fi\fi
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   {\eighttt http://dsl.org/copyleft/dsl.txt}.
   This is a development version and the
   source files are not permanently archived, but current versions are
normally
   accessible through
   {\eighttt https://www1.essex.ac.uk/maths/people/fremlin/mt.htm}.
   For further information contact {\eighttt david\@fremlin.org}.}
   }}
\def\Fn{\mathop{\text{Fn}}\nolimits}
\def\FN{\mathop{\text{FN}}\nolimits}
\def\formerly#1{\footnote{Formerly #1.}}
  %use only once in a chapter, never in same volume as reference
\def\fraction#1{\mathchoice{\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$\scriptstyle<$}#1\hbox{$\scriptstyle>$}}
     {\hbox{$\scriptscriptstyle<$}#1\hbox{$\scriptscriptstyle>$}}}
\def\frakmctbl{\frak m_{\text{countable}}}
\def\frmedcirc{{\raise 2.5pt\hbox{$_{\bigcirc}$}}}
\def\grad{\mathop{\text{grad}}\nolimits}
\def\hatpmb#1{\hat{\pmb{#1}}}
\def\hattilde#1{\skew2\hat{\tilde#1}}
\def\hattildepmb#1{\skew1\hat{\tildepmb#1}}
\def\hcov{\mathop{\text{hcov}}\nolimits}
\def\Heint{\mathchoice{\hbox to 0pt{\hskip 3.1pt H\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     \intop\nolimits}
\def\high#1{\lceil#1\rceil}
\def\highoverline#1{\overline{\vrule height6pt width0pt depth 0pt#1}}
\def\Hint#1{({\it Hint\/}: #1)}
\def\hL{\mathop{\text{hL}}}
\def\Hmp{\text{Hom}_{\bar\mu,\pi}}
\def\hp{\mathop{\text{hp}}\nolimits}
\def\hwtr{hereditarily weakly $\theta$-{\vthsp}refinable}
\def\iiv{ii_{\pmb{v}}}
\def\iivu{\iiv(\pmb{u})}
\def\imp{inverse-{\vthsp}measure-{\vthsp}preserving}
\def\innerprod#1#2{(#1|#2)}
\def\intstar{\text{int*}}
\def\CalIS{\Cal I(\Cal S)}
\def\jump{\mathop{\text{jump}}\nolimits}
\def\Jump{\mathop{\text{Jump}}\nolimits}
\def\Krein{Kre\v{\i}n}
\def\Lambdapv{\Lambda_{\text{pv}}}
\def\leRB{\le_{\text{RB}}}
\def\leRK{\le_{\text{RK}}}
\def\link{\mathord{\text{link}}}
\def\locmafp{\widehat{\otimes}_{\text{loc}}}
\def\Locmafp{\widehat{\bigotimes}^{\text{loc}}}
\def\low#1{\lfloor#1\rfloor}
\def\lti{left-\vthsp translation-\vthsp invariant}
\def\magnitude{\mathop{\text{mag}}}
\def\Mah{\mathop{\text{Mah}}}
\def\Mahcr{\Mah_{\text{cr}}}
\def\MahcrR{\Mah_{\text{crR}}}
\def\MahqR{\Mah_{\text{qR}}}
\def\MahR{\Mah_{\text{R}}}
\def\MahspcrR{\Mah_{\text{spcrR}}}
\def\Malm{M_{\text{alm}}}
\def\Marik{Ma\v r\'\i k} 
\def\Mbv{M_{\text{bv}}}
\def\Mcdlg{M_{\text{cdlg}}}
\def\Mdm{M_{\text{D-M}}}
\def\med{\mathop{\text{med}}\nolimits}
\def\Mfa{M_{\text{fa}}}
\def\Mhsr{\mathop{\text{Mhsr}}\nolimits}
\def\Mjf{\mathop{M_{\text{j-f}}}\nolimits}
\def\Mlbv{M_{\text{lbv}}}
\def\Mlob{M_{\text{lo-b}}}
\def\Mlns{M_{\text{ln-s}}}
\def\Mmo{M_{\text{mo}}}
\def\Mlmo{M_{\text{lmo}}}
\def\Mns{M_{\text{n-s}}}
\def\Mnsu{M_{\text{n-s}}^{\uparrow}}
%\def\Mnsu{M_{\text{n-s}\uparrow}}
\def\MMah{M_{\text{Mah}}}
\def\Mob{M_{\text{o-b}}}
\def\Mpob{M_{\text{po-b}}}
\def\Mpv{M_{\text{pv}}}
\def\MSi{M_{\text{S-i}}}
\def\Msimp{M_{\text{simp}}}
\def\Mth{Ma\-ha\-ram-{\vthsp}type-{\vthsp}homogeneous}
\def\Mvlm{M_{\text{vlm}}}
\def\negquad{\hskip-1em\relax}
\def\NN{\BbbN^{\Bbb N}}
\def\non{\mathop{\text{non}}}
\def\notBooleanop#1{
   \ifamstexsupported{\mathchoice
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
    {{\not\scriptstyle#1}}
    {{\not\scriptscriptstyle#1}}}
   \else{\mathchoice{{\not\scriptstyle#1}}{{\not\scriptstyle#1}}
    {{\not\scriptstyle#1}}{{\not\scriptscriptstyle#1}}}
    \fi}
\def\notBsubseteq{\notBooleanop{\subseteq}}
\def\notBsupseteq{\notBooleanop{\supseteq}}
\def\notVVdash{\not\mskip4mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip-2mu}
\def\nuprime{\nu^{\mskip0.7mu\prime}} %better  \nu'  in superscripts
\def\nw{\mathop{\text{nw}}}
\def\Osclln{\mathop{\text{Osclln}}\nolimits}
\def\otp{\mathop{\text{otp}}}
\def\partstar{\mathchoice{\partial\hbox{*}}
     {\partial\hbox{*}}
     {\partial\hbox{\sevenrm *}}
     {\partial\hbox{\fiverm *}}}
\def\per{\mathop{\text{per}}}
\def\Pfint{\mathchoice{\hbox to 0pt{\hskip 3.1pt Pf\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     \intop\nolimits
     \mathchoice{\hskip 2pt}{\hskip 1.2pt}{\hskip 1pt}{\hskip 1pt}}
\def\Pou{\mathop{\text{Pou}}\nolimits}
\def\prforc{\preccurlyeq_{\text{forcing}}}
\def\prGT{\preccurlyeq_{\text{GT}}}
\def\prT{\preccurlyeq_{\text{T}}}
\def\pssqa{power set $\sigma$-quotient algebra}
\def\qm{quasi-{\vthsp}measurable}
\def\Reverse#1{\setover
   {\lower0.4ex\hbox{$\scriptscriptstyle\leftrightarrow$}}{#1}}
\def\rhoKR{\rho_{\text{KR}}} 
\def\rhotv{\rho_{\text{tv}}} 
\def\Rint{\mathchoice{\hbox to 0pt{\hskip 3.1pt R\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     \intop\nolimits}
\def\RO{\text{RO}}
\def\rti{right-{\vthsp}translation-{\vthsp}invariant}
\def\rvm{real-{\vthsp}valued-{\vthsp}measurable}
\def\sat{\mathop{\text{sat}}}
\def\sfamilysS#1{\family{\sigma}{\Cal S}{#1_{\sigma}}}
\def\sfcc{$\sigma$-finite-cc}
\def\sgn{\mathop{\text{sgn}}}
\def\shr{\mathop{\text{shr}}\nolimits}
\def\Sii{\text{S}ii}
\def\Siiv{\Sii_{\pmb v}}
\def\Sint{\mathchoice{\hbox to 0pt{\hskip 4.1pt\ninerm S\hss}} 
     {\raise 0.6pt\hbox to 0pt{\hskip 3.0pt{\sevenrm S}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     \intop\nolimits}
\def\Sinteightpoint{\raise 0.6pt\hbox to 0pt{\hskip 2.3pt{\fiverm S}\hss}
     \intop\nolimits}
\def\SL{\text{SL}}
\def\sm{semi-martingale} %hyphen to remind me that it's the "virtual" kind
\def\spread{\mathop{\text{spr}}}
\def\ssplrarrow{^{\scriptscriptstyle\leftrightarrow}}
\def\ssptilde{{\raise1.2ex\hbox{$\scriptscriptstyle\sim$}}}
%\def\sspvec{{\raise1.2ex\hbox{$\scriptscriptstyle\rightarrow$}}}
\def\sspvec{\mskip3mu\vec{\null}\mskip5mu}
\def\Sti{\mathop{\text{Sti}}\nolimits}
\def\strbar{^{\scriptscriptstyle\pmb{-}}}
\def\strplus{^{\scriptscriptstyle\pmb{+}}}
\def\strprime{^{\mathchoice{\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1mu\pmb{\prime}}}}
%\def\strprime{\,\pmb{\prime}}
\def\supp{\mathop{\text{supp}}}
\def\tausm{\tau_{\sigma\text{-m}}}
\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_<$}
   \phantom{\tbf 1^{(#1)}}}
%\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_-$}
%   \phantom{\tbf 1^{(#1)}}}
\def\tensorhat{\widehat{\otimes}}
\def\Tensorhat{\widehat{\bigotimes}}
\def\tensorhatsigma{\widehat{\otimes}_{\sigma}}
\def\tildeClll{\hbox{$\tilde C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\tildehat#1{\skew2\tilde{\hat#1}}
\def\tildehatpmb#1{\tilde{\hatpmb#1}}
\def\tildepmb#1{\tilde{\pmb{#1}}}
\def\Tr{\mathop{\text{Tr}}}
\def\trs{^{\top}\hskip-1pt}
\def\triplepc#1#2#3{(#1,#2,\hbox{$<$}#3)}
\def\TSi{\frak T_{\text{S-i}}}
\def\undphi{\underline{\phi}{}}
\def\undpsi{\underline{\psi}{}}
\def\undtheta{\underline{\theta}{}}
\def\updown{\mathchoice
  {\hbox{\hskip-1.5pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-3.3pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %OK for subscript, fails for superscript
    \hskip-1.7pt\vrule height6.5pt width.7pt depth0pt
    \hskip.9pt\vrule height6.5pt width.7pt depth0pt
    \hskip-2.5pt\lower5.4pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %not tested
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}}
\def\updownarrows{\uparrow\mskip-4mu\downarrow}
\def\updowncgt{$\updown$-convergent}
\def\upr{\mathop{\text{upr}}\nolimits}
\def\ureg{\mathop{\text{ureg}}}

\def\Var{\mathop{\text{Var}}\nolimits}

\def\varaction{\mathchoice{\lower-.2ex\hbox{$\smallcirc$}}
                       {\lower-.2ex\hbox{$\smallcirc$}}
                       {\ssbullet}
                       {\ssbullet}}
\def\varcheck#1{{\setover
       {\lower 0.4ex\hbox{$\scriptscriptstyle\vee$}}{#1}}}
\def\varcheckf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\vee$}}{f}}}
\def\varchecki{\mskip-3mu\text{\it{\accent"14{\hbox{\hskip-3.5pt\i}}}}\mskip2mu}
\def\varspcheck{^{\scriptscriptstyle\vee}}
%\def\varspcheck{{\setover{\scriptscriptstyle\vee}{\phantom{g}}}}
%\def\varspcheckf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\vee$}}{\phantom{f}}}}
\def\varhat#1{{\setover{\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{#1}}}
\def\varhatf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{f}}}
\def\varinnerprod#1#2{#1\dotproduct#2}
\def\varsphat{^{\scriptscriptstyle\wedge}}
% or use  \hbox{\^{\null}}
%\def\varsphat{{\setover{\scriptscriptstyle\wedge}{\phantom{g}}}}
%\def\varsphatf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\wedge$}}{\phantom{f}}}}
\def\varBspcheck{\mathchoice
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}}
\def\varCspcheck{\mathchoice
  {\hskip-1pt\check{\phantom x}}   %not tested
  {\hskip-1pt\check{\phantom x}}
  {\hskip-1.2pt\check{\phantom x}}
  {\hskip-1.5pt\check{\phantom x}}}  %not tested
\def\varCsphat{\mathchoice
  {\hskip-1pt\hat{\phantom x}}   %not tested
  {\hskip-1pt\hat{\phantom x}}
  {\hskip-1.2pt\hat{\phantom x}}  %not tested
  {\hskip-1.5pt\hat{\phantom x}}}  %not tested
\def\VeqL{$V\hbox{$=$}L$}
\def\VVdash{\mskip5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip2mu}
\def\VVdP{\VVdash_{\Bbb P}}
\def\VVdPk{\VVdash_{\Bbb P_{\kappa}}}
\def\wdistr{\mathop{\text{wdistr}}}
\def\weight{\mathop{\text{wt}}}
\def\w1lim{\mathop{\text{w1lim}}}
\def\wsid{weakly $(\sigma,\infty)$-{\vthsp}distributive}
\def\1lim{\mathop{1\mskip-1.5mu\text{lim}}}
\def\2vm{two-{\vthsp}valued-{\vthsp}measurable}

%Cara\-th\'e\-o\-dory

\def\chaptername{}
\def\sectionname{}
\def\headlinesectionname{}

\loadeusm\loadeurm
\Loadfourteens
\Loadfont{\ninerm=cmr9}

%note:  $..\cmmnt{=..}..$  is better as  \cmmnt{\mskip5mu
 
  
%----------------------------------------------------------%  
  
\smallprint  
%\USsmallprint  
%\noproofs  
%\resultsonly  
%\atUEssex  
\usegraphicx
  
\Extract  
  
%----------------------------------------------------------%  

\frfilename{mt44.tex}
\versiondate{18.4.08}
\copyrightdate{1998}

\def\chaptername{Topological groups}
\def\sectionname{Introduction}

\newchapter{44}

Measure theory begins on the real line, which is of course a group;  and
one of the most fundamental properties of Lebesgue measure is its
translation-invariance\cmmnt{ (134A)}.   Later we come to the standard
measure on the unit circle\cmmnt{ (255M)}, and counting measure on the
integers is also translation-invariant, if we care to notice;  moreover,
Fourier series and transforms clearly depend utterly
on the fact that shift operators don't disturb the measure-theoretic
structures we are building.   Yet another example appears in the usual
measure on $\{0,1\}^I$, which is translation-invariant if we identify
$\{0,1\}^I$ with the group $\Bbb Z_2^I$\cmmnt{ (345Ab)}.   Each of
these examples is special in many other ways.   But it turns out that
a particular combination of properties which they share,
all being locally
compact Hausdorff spaces with group operations for which multiplication
and inversion are continuous, is the basis of an extraordinarily
powerful theory of invariant measures.

As usual, I have no choice but to move rather briskly through a wealth
of ideas.   The first step is to set out a suitably general existence
theorem, assuring us that every locally compact Hausdorff topological
group has non-trivial invariant Radon measures, that is, `Haar
measures'\cmmnt{ (441E)}.   As remarkable as the existence of
Haar measures is their (essential) uniqueness\cmmnt{ (442B)};  the
algebra, topology and measure theory of a topological group are linked
in so many ways that they form a peculiarly solid structure.
I investigate a miscellany of facts about this structure in \S443,
including the basic theory of the modular functions linking left-invariant
measures with right-invariant measures.

I have already mentioned that Fourier analysis depends on the
translation-invariance of Lebesgue measure.   It turns out that
substantial parts of the abstract theory of Fourier series and
transforms can be generalized to arbitrary locally compact groups.   In
particular, convolutions\cmmnt{ (\S255)} appear again, even in
non-abelian groups\cmmnt{ (\S444)}.   But for the central part of the
theory, a transform relating functions on a group $X$ to functions on
its `dual' group $\Cal X$, we do need the group to be abelian.
Actually I give only the foundation of this theory:  if $X$ is an
abelian locally compact Hausdorff group, it is the dual of its
dual\cmmnt{ (445U)}.   (In `ordinary' Fourier theory, where we are
dealing with the cases $X=\Cal X=\Bbb R$ and $X=S^1$, $\Cal X=\Bbb Z$,
this duality is so straightforward that one hardly notices it.)   But on
the way to the duality theorem we necessarily see many of the themes of
Chapter 28 in more abstract guises.

A further remarkable
fact is that any Haar measure has a translation-invariant
lifting\cmmnt{ (447J)}.   The proof demands a union between the ideas
of the ordinary Lifting Theorem\cmmnt{ (\S341)} and some of the
elaborate structure theory which has been developed for locally compact
groups\cmmnt{ (\S446)}.

For the last two sections of the chapter, I look at groups which are not
locally compact, and their actions on appropriate spaces.
For a particularly important class of group actions, Borel measurable
actions of Polish groups on Polish spaces,
we have a natural necessary and
sufficient condition for the existence of an invariant
measure\cmmnt{ (448P)}, complementing the result for locally compact
spaces in 441C.
In a slightly different direction, we can look at those groups, the
`amenable' groups, for which all actions (on compact Hausdorff spaces)
have invariant measures.   This again leads to some very remarkable
ideas, which I sketch in \S449.

\discrpage


\frfilename{mt441.tex}
\versiondate{27.6.16}
\copyrightdate{1997}

\def\high#1{\lceil#1\rceil}
\def\rti{right-{\vthsp}translation-{\vthsp}invariant}
\def\tr{\mathop{\text{tr}}\nolimits}

\def\chaptername{Topological groups}
\def\sectionname{Invariant measures on locally compact spaces}

\newsection{441}

I begin this chapter with the most important theorem on
the existence of invariant measures:  every locally compact Hausdorff
group has left and right Haar measures (441E).   I derive this as a
corollary of a general result concerning invariant measures on locally
compact spaces (441C), which has other interesting consequences (441H).

\leader{441A}{Group actions} I repeat the definitions\cmmnt{ from \S4A5} on
which this chapter is based.

\medskip

{\bf (a)} If $G$ is a group and $X$ is a set, an {\bf action} of $G$ on
$X$ is a function $(a,x)\mapsto a\action x:G\times X\to X$ such that

\Centerline{$(ab)\action x=a\action(b\action x)$ for all $a$, $b\in G$,
$x\in X$,}

\Centerline{$e\action x=x$ for every $x\in X$}

\noindent where $e$ is the identity of $G$\cmmnt{ (4A5Ba)}.   In this
context I write

\Centerline{$a\action A=\{a\action x:x\in A\}$}

\noindent for $a\in G$, $A\subseteq X$\cmmnt{ (4A5Bc)}.
If $f$ is a function defined on a
subset of $X$, then $(a\action f)(x)=f(a^{-1}\action x)$ whenever
$a\in G$ and $x\in X$ and $a^{-1}\action x\in\dom f$\cmmnt{ (4A5C(c-i))}.

\spheader 441Ab If a group $G$ acts on a set $X$, a measure $\mu$ on $X$
is {\bf $G$-invariant} if $\mu(a^{-1}\action E)$ is defined and equal to
$\mu E$ whenever $a\in G$ and $\mu$ measures $E$.
%former 4{}41Ac-4{}41Ad now in 4A5C

\cmmnt{(Of course this is the same thing as saying that
$\mu(a\action E)=\mu E$ for every $a\in G$ and measurable set $E$;  I
use the formula
with $a^{-1}$ so as to match my standard practice when $a$ is actually a
function from $X$ to $X$.)}

\leader{441B}{}\cmmnt{ It will be useful later to be able to quote the
following elementary results.

\medskip

\noindent}{\bf Lemma} Let $X$ be a topological space, $G$ a group, and
$\action$ an action of $G$ on $X$ such that $x\mapsto a\action x$ is
continuous for every $a\in G$.

(a) If $\mu$ is a quasi-Radon measure on $X$ such that
$\mu(a\action U)\le\mu U$ for every open set $U\subseteq X$ and every
$a\in G$, then $\mu$ is $G$-invariant.

(b) If $\mu$ is a Radon measure on $X$ such that
$\mu(a\action K)\le\mu K$ for every compact set $K\subseteq X$ and every
$a\in G$, then $\mu$ is $G$-invariant.

\proof{ Note first that the maps $x\mapsto a\action x$ are actually
homeomorphisms (4A5Bd), so that $a\action U$ and $a\action K$ will be
open, or compact, as $U$ and $K$ are.   Next, the inequality $\le$ in
the hypotheses is an insignificant refinement;  since we must also have

\Centerline{$\mu U=\mu(a^{-1}\action a\action U)\le\mu(a\action U)$}

\noindent in (a),

\Centerline{$\mu K=\mu(a^{-1}\action a\action K)\le\mu(a\action K)$}

\noindent in (b), we always have equality here.

Now fix $a\in G$, and set $T_a(x)=a\action x$ for $x\in X$.   Then
$T_a$ is a homeomorphism, so the image measure $\mu T_a^{-1}$ will be
quasi-Radon, or Radon, if $\mu$ is.   In (a), we are told that
$\mu T_a^{-1}$ agrees with $\mu$ on the open sets, while in (b) we are
told that they agree on the compact sets;  so in both cases we have
$\mu=\mu T_a^{-1}$, by 415H(iii) or 416E(b-ii).   Consequently we have
$\mu T_a^{-1}[E]=\mu E$ whenever $\mu$ measures $E$.   As $a$ is
arbitrary, $\mu$ is $G$-invariant.
}%end of proof of 441B

\leader{441C}{Theorem}\cmmnt{ ({\smc Steinlage 75})} Let $X$ be a
non-empty locally compact Hausdorff
space and $G$ a group acting on $X$.   Suppose that

\inset{(i) $x\mapsto a\action x$ is continuous for every $a\in G$;

(ii) every orbit $\{a\action x:a\in G\}$ is dense;

(iii) whenever $K$ and $L$ are disjoint compact subsets of $X$ there is
a non-empty open subset $U$ of $X$ such that, for every $a\in G$, at
most one of $K$, $L$ meets $a\action U$.}

\noindent Then there is a non-zero $G$-invariant Radon measure $\mu$ on
$X$.

\proof{{\bf (a)} $\bigcup_{a\in G}a\action U=X$
for every non-empty open $U\subseteq X$.   \Prf\ If $x\in X$, then the
orbit of $x$ must meet $U$, so there is a $a\in G$ such that
$a\action x\in U$;  but this means that $x\in a^{-1}\action U$.\ \QeD

Fix some point $z_0$ of $X$ and write $\Cal V$ for the set of open sets
containing $z_0$.   Then if $K$, $L$ are disjoint compact subsets of $X$
there is a $U\in\Cal V$ such that, for every $a\in G$, at most one of
$K$, $L$ meets $a\action U$.   \Prf\ By hypothesis, there is a non-empty
open set $V$ such that, for every $a\in G$, at most one of $K$, $L$
meets $a\action V$.   Now there is an $b\in G$ such that
$b\action z_0\in V$;  set
$U=b^{-1}\action V$.   Because $b^{-1}$ acts on $X$ as a homeomorphism,
$U\in\Cal V$;  and if $a\in G$, then
$a\action U=(ab^{-1})\action V$ can meet at most one of $K$ and
$L$.\ \Qed

\medskip

{\bf (b)} If $U\in\Cal V$ and $A\subseteq X$ is any relatively compact
set, then $\{a\action U:a\in G\}$ is an open cover of $X$, so there is a
finite set $I\subseteq G$ such that
$\overline{A}\subseteq\bigcup_{a\in I}a\action U$.   Write $\high{A:U}$
for $\min\{\#(I):I\subseteq G,\,A\subseteq\bigcup_{a\in I}a\action U\}$.

\medskip

{\bf (c)} The following facts are now elementary.

\medskip

\quad{\bf (i)} If $U\in\Cal V$ and $A$, $B\subseteq X$ are relatively
compact, then

\Centerline{$0\le\high{A:U}\le\high{A\cup B:U}
\le\high{A:U}+\high{B:U}$,}

\noindent and $\high{A:U}=0$ iff $A=\emptyset$.

\medskip

\quad{\bf (ii)} If $U$, $V\in\Cal V$ and $V$ is relatively compact, and
$A\subseteq X$ also is relatively compact, then

\Centerline{$\high{A:U}\le\high{A:V}\,\high{V:U}$.}

\noindent\Prf\ If $A\subseteq\bigcup_{a\in I}a\action V$ and
$V\subseteq\bigcup_{b\in J}b\action U$, then
$A\subseteq\bigcup_{a\in I,b\in J}(ab)\action U$.\ \Qed

\medskip

\quad{\bf (iii)} If $U\in\Cal V$, $A\subseteq X$ is relatively compact
and $b\in G$, then $\high{b\action A:U}=\high{A:U}$.   \Prf\ If
$I\subseteq G$ and $A\subseteq\bigcup_{a\in I}a\action U$, then
$b\action A\subseteq\bigcup_{a\in I}(ba)\action U$, so
$\high{b\action A:U}\le\#(I)$;  as $I$ is arbitrary,
$\high{b\action A:U}\le\high{A:U}$.
On the other hand, the same argument shows that

\Centerline{$\high{A:U}=\high{b^{-1}\action b\action A:U}
\le\high{b\action A:U}$,}

\noindent so we must have equality.\ \Qed

\medskip

{\bf (d)} Fix a relatively compact $V_0\in\Cal V$.   (This is the first
place where we use the hypothesis that $X$ is locally compact.)   For
every $U\in\Cal V$ and every relatively compact set $A\subseteq X$ write

$$\lambda_UA=\bover{\high{A:U}}{\high{V_0:U}}.$$

\noindent Then (c) tells us immediately that

\quad(i) if $A$, $B\subseteq X$ are relatively compact,

\Centerline{$0\le\lambda_UA\le\lambda_U(A\cup B)
\le\lambda_UA+\lambda_UB$;}

\quad(ii) $\lambda_UA\le\high{A:V_0}$ for every relatively compact
$A\subseteq X$;

\quad(iii) $\lambda_U(b\action A)=\lambda_UA$ for every relatively
compact $A\subseteq X$ and every $b\in G$;

\quad(iv) $\lambda_UV_0=1$.

\medskip

{\bf (e)} Now for the point of the hypothesis (iii) of the theorem.   If
$K$, $L$ are disjoint compact subsets of $X$, there is a $V\in\Cal V$
such that $\lambda_U(K\cup L)=\lambda_UK+\lambda_UL$ whenever
$U\in\Cal V$ and $U\subseteq V$.   \Prf\ By (a), there is a $V\in\Cal V$
such that
any translate $a\action V$ can meet at most one of $K$ and $L$.   Take
any $U\in\Cal V$ included in $V$.   Let $I\subseteq G$ be such that
$\bigcup_{a\in I}a\action U\supseteq K\cup L$ and
$\#(I)=\high{K\cup L:U}$.   Then

\Centerline{$I'=\{a:a\in I,\,K\cap a\action U\ne\emptyset\}$,
\quad$I''=\{a:a\in I,\,L\cap a\action U\ne\emptyset\}$}

\noindent are disjoint.  $K\subseteq\bigcup_{a\in I'}a\action U$, so
$\high{K:U}\le\#(I')$, and similarly $\high{L:U}\le\#(I'')$.   But this
means that

\Centerline{$\high{K:U}+\high{L:U}\le\#(I')+\#(I'')\le\#(I)
=\high{K\cup L:U}$,}

\Centerline{$\lambda_UK+\lambda_UL\le\lambda_U(K\cup L)$.}

\noindent As we already know that
$\lambda_U(K\cup L)\le\lambda_UK+\lambda_UL$, we must have equality, as
claimed.\ \Qed

\medskip

{\bf (f)} Let $\Cal F$ be an ultrafilter on $\Cal V$ containing all
sets of the form $\{U:U\in\Cal V,\,U\subseteq V\}$ for $V\in\Cal V$.
If $A\subseteq X$ is relatively compact, $0\le\lambda_UA\le\high{A:V_0}$
for every $U\in\Cal V$, so $\lambda A=\lim_{U\to\Cal F}\lambda_UA$ is
defined in $[0,\high{A:V_0}]$.   From (d-i) and (d-iii) we see that

\Centerline{$0\le\lambda(b\action A)=\lambda A
\le\lambda(A\cup B)\le\lambda A+\lambda B$}

\noindent for all relatively compact $A$, $B\subseteq X$ and $b\in G$.
From (d-iv) we see that $\lambda V_0=1$.   Moreover, from (e) we see
that if $K$, $L\subseteq X$ are disjoint compact sets,

\Centerline{$\{U:U\in\Cal V,\,\lambda_U(K\cup L)
=\lambda_UK+\lambda_UL\}\in\Cal F$,}

\noindent so $\lambda(K\cup L)=\lambda K+\lambda L$.

\medskip

{\bf (g)} By 416M, there is a Radon measure $\mu$ on $X$ such that

\Centerline{$\mu K=\inf\{\lambda L:L\subseteq X$ is compact,
$K\subseteq\interior L\}$}

\noindent for every compact set $K\subseteq X$.   Now $\mu$ is
$G$-invariant.   \Prf\ Take $b\in G$.   If $K$, $L\subseteq X$ are
compact and $K\subseteq\interior L$, then
$b\action K\subseteq\interior b\action L$, because $x\mapsto b\action x$
is a homeomorphism;  so

\Centerline{$\mu(b\action K)\le\lambda(b\action L)
=\lambda L$.}

\noindent As $L$ is arbitrary, $\mu(b\action K)\le\mu K$.   As $b$ and
$K$ are arbitrary, $\mu$ is $G$-invariant, by 441Bb.\ \Qed

\medskip

{\bf (h)} Finally, $\mu V_0\ge\lambda V_0\ge 1$, so $\mu$ is non-zero.
}%end of proof of 441C

\leader{441D}{}\cmmnt{ The hypotheses of 441C are deliberately drawn
as widely as possible.   The principal application is the one for which
the chapter is named.

\medskip

\noindent}{\bf Definition} If $G$ is a topological group, a {\bf left
Haar measure} on $G$ is a non-zero quasi-Radon measure $\mu$ on $G$
which is invariant for the left action of $G$ on itself\cmmnt{, that
is, $\mu(aE)=\mu E$ whenever $\mu$ measures $E$ and $a\in G$}.

Similarly, a {\bf right Haar measure} is a non-zero quasi-Radon measure
$\mu$ such that $\mu(Ea)=\mu E$ whenever $E\in\dom\mu$ and $a\in G$.

\cmmnt{(My reasons for requiring `quasi-Radon' here will appear in
\S\S442 and 443.)}

\leader{441E}{Theorem} A locally compact Hausdorff topological group has
left and right Haar measures, which are both Radon measures.

\proof{ Both the left and right actions of $G$ on itself satisfy the
conditions of 441C.   \Prf\ In both cases, condition (i) is just the
(separate) continuity of multiplication, and (ii) is trivial, as every
orbit is the whole of $G$.   As for (iii), let us take the left action
first.  Given disjoint compact subsets $K$, $L$ of $G$, then
$M=\{y^{-1}z:y\in K,\,z\in L\}$ is a compact subset of $G$ not
containing the identity $e$.   Because the topology is Hausdorff, $M$ is
closed and $X\setminus M$ is a neighbourhood of $e$.   Because
multiplication and inversion are continuous, there are open
neighbourhoods $V$, $V'$ of $e$ such that $u^{-1}v\in G\setminus M$
whenever $u\in V$ and $v\in V'$.   Set $U=V\cap V'$;  then $U$ is a
non-empty open set in $G$.

\Quer\ Suppose, if possible, that there is a $a\in G$ such that $a\action U=aU$
meets both $K$ and $L$.   Take $y\in K\cap aU$ and $z\in L\cap aU$.
Then $a^{-1}y\in U\subseteq V$ and $a^{-1}z\in U\subseteq V'$, so

\Centerline{$y^{-1}z=(a^{-1}y)^{-1}a^{-1}z\in G\setminus M$;}

\noindent but also $y^{-1}z\in M$.\ \Bang

Thus $aU$ meets at most one of $K$, $L$ for any $a\in G$.   As $K$ and
$L$ are arbitrary, condition (iii) of 441C is satisfied.

For the right action, we use the same ideas, but vary the formulae.
Set $M=\{yz^{-1}:y\in K,\,z\in L\}$, and choose $V$ and $V'$ such that
$uv^{-1}\in G\setminus M$ for $u\in V$, $v\in V'$.   Then if $a\in G$,
$y\in K$ and $z\in L$, $za(ya)^{-1}\in M$ and one of $za$, $ya$ does not
belong to $U=V\cap V'$, that is, one of $z$, $y$ does not belong to
$Ua^{-1}=a\action U$.\ \Qed

Then 441C provides us with non-zero left and right Haar measures on $G$,
and also tells us that they are Radon measures.
}%end of proof of 441E

\leader{441F}{}\cmmnt{ A different type of example is provided by
locally compact metric spaces.

\medskip

\noindent}{\bf Definition} If $(X,\rho)$ is any metric space, its {\bf
isometry group} is the set of permutations $g:X\to X$ which are {\bf
isometries}, that is, $\rho(g(x),g(y))=\rho(x,y)$ for all $x$, $y\in X$.

\leader{441G}{The topology of an isometry group} Let $(X,\rho)$ be a
metric space and $G$ the isometry group of $X$.

\spheader 441Ga Give $G$ the topology
of pointwise convergence inherited from the product topology of $X^X$.
Then $G$ is a Hausdorff topological group and the action of $G$ on $X$
is continuous.    \prooflet{\Prf\ If
$x\in X$, $g_0$, $h_0\in G$ and $\epsilon>0$, then
$V=\{g:\rho(gh_0(x),g_0h_0(x))\le\bover12\epsilon\}$ is a neighbourhood
of $g_0$ and $V'=\{h:\rho(h(x),h_0(x))\le\bover12\epsilon\}$ is a
neighbourhood of $h_0$.   If $g\in V$ and $h\in V'$ then

$$\eqalign{\rho(gh(x),g_0h_0(x))&
\le\rho(gh(x),gh_0(x))+\rho(gh_0(x),g_0h_0(x))\cr&
\le\rho(h(x),h_0(x))+\Bover12\epsilon
\le\epsilon.\cr}$$

\noindent As $g_0$, $h_0$ and $\epsilon$ are arbitrary, the function
$(g,h)\mapsto gh(x)$ is continuous;  as $x$ is arbitrary, multiplication
on $G$ is continuous.    As for inversion, suppose that $g_0\in G$,
$\epsilon>0$ and $x\in X$.   Then
$V=\{g:\rho(gg_0^{-1}(x),x)\le\epsilon\}$ is a neighbourhood of $g_0$,
and if $g\in V$ then

\Centerline{$\rho(g^{-1}(x),g_0^{-1}(x))
=\rho(x,gg_0^{-1}(x))\le\epsilon$.}

\noindent Because $g_0$, $\epsilon$ and $x$ are arbitrary, inversion on
$G$ is continuous, and $G$ is a topological group.   Because $X$ is
Hausdorff, so is $G$.

To see that the action is continuous, take $g_0\in G$, $x_0\in X$ and
$\epsilon>0$.   Then $V=\{g:g\in G,\,\rho(g(x_0),g_0(x_0))\penalty-100
\le\bover12\epsilon\}$ is a neighbourhood of
$g_0$.   If $g\in V$ and $x\in U(x_0,\bover12\epsilon)$, then

\Centerline{$\rho(g(x),g_0(x_0))
\le\rho(g(x),g(x_0))+\rho(g(x_0),g_0(x_0))
\le\rho(x,x_0)+\Bover12\epsilon
\le\epsilon$.}

\noindent As $g_0$, $x_0$ and $\epsilon$ are arbitrary,
$(g,x)\mapsto g(x):G\times X\to X$ is continuous.\ \Qed}

\spheader 441Gb If $X$ is compact, so is $G$.   \prooflet{\Prf\
By Tychonoff's theorem (3A3J), $X^X$ is compact.   Suppose that
$g\in X^X$ belongs to the closure of $G$ in $X^X$.   For any $x$,
$y\in X$, the set $\{f:f\in X^X$, $\rho(f(x),f(y))=\rho(x,y)\}$ is closed
and includes $G$, so contains $g$;  thus $g$ is an isometry.   \Quer\ If
$g[X]\ne X$, take $x\in X\setminus g[X]$ and set $x_n=g^n(x)$ for every
$n\in\Bbb N$.   Because $g$ is continuous and $X$ is compact, $g[X]$ is
closed and there is some $\delta>0$ such that
$U(x,\delta)\cap g[X]=\emptyset$.   But this means that

\Centerline{$\rho(x_m,x_n)=\rho(g^m(x),g^m(x_{n-m}))
=\rho(x,x_{n-m})\ge\delta$}

\noindent whenever $m<n$, so that $\sequencen{x_n}$ can have no cluster
point in $X$;  which is impossible, because $X$ is supposed to be
compact.\ \BanG\   This shows that $g$ is surjective and belongs to $G$.
As $g$ is arbitrary, $G$ is closed in $X^X$, therefore compact.\ \Qed}

\leader{441H}{Theorem} If $(X,\rho)$ is a non-empty locally compact
metric space with isometry group $G$, then there is a non-zero
$G$-invariant Radon measure on $X$.

\proof{{\bf (a)} Fix any $x_0\in X$, and set
$Z=\overline{\{g(x_0):g\in G\}}$;  then $Z$ is a closed subset of $X$,
so is in itself locally
compact.   Let $H$ be the isometry group of $Z$.

\medskip

{\bf (b)} We need to know that $g\restr Z\in H$ for every $g\in G$.
\Prf\ Because $g:X\to X$ is a homeomorphism,

\Centerline{$g[Z]=\overline{\{gg'(x_0):g'\in G\}}=Z$,}

\noindent so $g\restr Z$ is a permutation of $Z$, and of
course it is an isometry, that is, belongs to $H$.\ \Qed

\medskip

{\bf (c)} Now $Z$ and $H$ satisfy the conditions of 441C.

\medskip

\Prf{\bf (i)} is true just because all isometries are continuous.

\medskip

\quad{\bf (ii)} Take $z\in Z$ and let $U$ be a
non-empty relatively open subset of $Z$.   Then $U=Z\cap V$ for some
open set $V\subseteq X$;  as $Z\cap V\ne\emptyset$, there must be a
$g_0\in G$ such that $g_0(x_0)\in V$.   At the same time, there is a
sequence $\sequencen{h_n}$ in $G$ such that
$z=\lim_{n\to\infty}h_n(x_0)$.   Now

\Centerline{$\rho(g_0(x_0),g_0h_n^{-1}(z))
=\rho(h_n(x_0),z)\to 0$}

\noindent as $n\to\infty$,
so there is some $n$ such that $g_0h_n^{-1}(z)\in V$;  of
course $g_0h_n^{-1}(z)$ also belongs to $U$, while $g_0h_n^{-1}\restr Z$
belongs to $H$, by (b) above.   As $U$ is arbitrary, the $H$-orbit of
$z$ is dense in $Z$;  as $z$ is arbitrary, $H$ satisfies condition (ii)
of 441C.

\medskip

\quad{\bf (iii)} Given that $K$ and $L$ are disjoint compact subsets of
$Z$, there must be a $\delta>0$ such that $\rho(y,z)\ge\delta$ for every
$y\in K$, $z\in L$.   Let $U$ be the relatively open ball
$\{z:z\in Z,\,\rho(z,x_0)<\bover12\delta\}$.   Then for any $h\in H$,
$\rho(y,z)<\delta$ for any $y$, $z\in h[U]$, so $h[U]$ cannot
meet both $K$ and $L$.\ \Qed

\medskip

{\bf (d)} 441C therefore provides us with a non-zero $H$-invariant Radon
measure $\nu$ on $Z$.   Setting $\mu E=\nu(E\cap Z)$ whenever
$E\subseteq X$ and $E\cap Z\in\dom\nu$, it is easy to check that $\mu$
is $G$-invariant (using (b) again) and is a Radon measure on $X$.
}%end of proof of 441H

\cmmnt{
\leader{441I}{Remarks (a)} Evidently there is a degree of overlap
between the cases above.   In an abelian group, for instance, the left
and right group actions necessarily give rise to the same invariant
measures.   If we take $X=\BbbR^2$, it has a group structure (addition)
for which we have invariant measures (e.g., Lebesgue measure);  these
are just the translation-invariant measures.   But 441H tells us that we
also have measures which are invariant under all isometries (rotations
and reflections as well as translations);  from where we now stand,
there is no surprise remaining in the fact that Lebesgue measure is
invariant under this much larger group.   (Though if you look back at
Chapter 26, you will see that a bare-handed proof of this takes a
certain amount of effort.)   If we turn next to the unit sphere
$\{x:\|x\|=1\}$ in $\BbbR^3$, we find that there is no useful group
structure, but it is a compact metric space, so carries invariant
measures, e.g., two-dimensional Hausdorff measure.

\spheader 441Ib
The arguments of 441C leave open the question of how far the invariant
measures constructed there are unique.   Of course any scalar multiple
of an invariant measure will again be invariant.   It is natural to give
a special place to invariant probability measures, and call them
`normalized';  whenever we have a non-zero totally finite invariant
measure we
shall have an invariant probability measure.   Counting measure on any
set will be invariant under any action of any group, and it is natural
to say that these measures also are `normalized';  when faced with a
finite set with two or more elements, we have to choose which
normalization seems most reasonable in the context.

\spheader 441Ic We shall see in 442B that Haar measures (with a given
handedness) are necessarily scalar multiples of each other.   In
442Ya, 443Ud and 443Xy we have further situations
in which invariant measures are essentially unique.   If, in
441C, there are non-trivial $G$-invariant subsets of $X$, we do not
expect such a result.   But there are interesting cases in which the
question seems to be open.
}%end of comment

\leader{441J}{}\cmmnt{ Of course we shall be much concerned with
integration with respect to invariant measures.   The results we need
are elementary corollaries of theorems already dealt with at length, but
it will be useful to have them spelt out.

\medskip

\noindent}{\bf Proposition} Let $X$ be a set, $G$ a group acting on $X$,
and $\mu$ a $G$-invariant measure on $X$.   If $f$ is a real-valued
function defined on a subset of $X$, and $a\in G$, then
$\int f(x)\mu(dx)=\int f(a\action x)\mu(dx)$ if either integral is
defined in $[-\infty,\infty]$.

\proof{ Apply 235G to the \imp\ functions $x\mapsto a\action x$ and
$x\mapsto a^{-1}\action x$.
}%end of proof of 441J

\vleader{72pt}{441K}{Theorem} Let $X$ be a set, $G$ a group acting on $X$, and
$\mu$ a $G$-invariant measure on $X$ with measure algebra $\frak A$.

(a) We have an action of $G$ on $\frak A$ defined by setting
$a\action E^{\ssbullet}=(a\action E)^{\ssbullet}$ whenever $a\in G$ and
$\mu$ measures $E$.

(b) We have an action of $G$ on $L^0=L^0(\mu)$ defined by setting
$a\action f^{\ssbullet}=(a\action f)^{\ssbullet}$ for every $a\in G$,
$f\in\eusm L^0(\mu)$.

(c) For $1\le p\le\infty$ the formula of (b) defines actions of $G$ on
$L^p=L^p(\mu)$, and $\|a\action u\|_p=\|u\|_p$ whenever $u\in L^p$ and
$a\in G$.

\proof{{\bf (a)} If $E$, $F\in\dom\mu$ and
$E^{\ssbullet}=F^{\ssbullet}$, then (because $x\mapsto a^{-1}\action x$
is \imp) $(a\action E)^{\ssbullet}=(a\action F)^{\ssbullet}$ for every $a\in G$.
So the
given formula does define a function from $G\times\frak A$ to $\frak A$.
It is now easy to check that it is an action.

\medskip

{\bf (b)} Take $f\in\eusm L^0=\eusm L^0(\mu)$ and $a\in G$.   Set
$\phi_a(x)=a^{-1}\action x$ for $x\in X$, so that $\phi_a:X\to X$ is
\imp.   Then $a\action f=f\phi_a$ belongs to $\eusm L^0$.   If $f$,
$g\in\eusm L^0$ and $f\eae g$, then $f\phi_a\eae g\phi_a$, so
$(a\action f)^{\ssbullet}=(a\action g)^{\ssbullet}$.   This shows that
the given formula defines a function from $G\times L^0$ to $L^0$, and
again it is easy to see that it is an action.

\medskip

{\bf (c)} If $f\in\eusm L^p(\mu)$ then

\Centerline{$\int|a\action f|^pd\mu=\int|f(a^{-1}\action x)|^pd\mu
=\int|f|^pd\mu$}

\noindent by 441J.   So $a\action f\in\eusm L^p(\mu)$ and
$\|a\action f^{\ssbullet}\|_p=\|f^{\ssbullet}\|_p$.   Thus we have a
function from $G\times L^p$ to $L^p$, and once more it must be an
action.
}%end of proof of 441K

\leader{441L}{Proposition} Let $X$ be a locally compact Hausdorff space
and $G$ a group acting on $X$ in such a way that $x\mapsto a\action x$
is continuous for every $a\in G$.   If $\mu$ is a Radon measure on $X$,
then $\mu$ is $G$-invariant iff
$\int f(x)\mu(dx)=\int f(a\action x)\mu(dx)$ for every $a\in G$ and
every continuous function $f:X\to\Bbb R$ with compact support.

\proof{ For $a\in G$, set $T_a(x)=a\action x$ for every $x\in X$.
Then $\nu_a=\mu T_a^{-1}$ is a Radon measure on $X$.   If $f\in C_k(X)$,
then

\Centerline{$\int fd\nu_a=\int f T_ad\mu$}

\noindent by 235G.   Now

$$\eqalignno{\mu\text{ is }G\text{-invariant}
&\iff\nu_a=\mu\text{ for every }a\in G\cr
&\iff\int fd\nu_a=\int fd\mu\text{ for every }a\in G,\,f\in C_k(X)\cr
\displaycause{416E(b-v)}
&\iff\int fT_ad\mu=\int fd\mu\text{ for every }a\in G,\,f\in
C_k(X)\cr}$$

\noindent as claimed.
}%end of proof of 441L

\exercises{\leader{441X}{Basic exercises $\pmb{>}$(a)}
%\spheader 441Xa
Let $X$ be a set.   (i) Show that there is a one-to-one correspondence
between actions $\action$ of the group $\Bbb Z$ on $X$ and permutations
$f:X\to X$ defined by the formula
$n\action x=f^n(x)$.   (ii) Show that if $f:X\to X$ is a permutation, a
measure $\mu$ on $X$ is $\Bbb Z$-invariant for the corresponding action
iff $f$ and $f^{-1}$ are both
\imp.   (iii) Show that if $X$ is a compact Hausdorff space and
$\action$ is a continuous action of $\Bbb Z$ on $X$, then there is a
$\Bbb Z$-invariant Radon probability
measure on $X$.   \Hint{437T.}
%441A

\spheader 441Xb
Let $(X,\Tau,\nu)$ be a measure space and $G$ a group acting on $X$.
Set $\Sigma=\{E:E\subseteq X,\,g\action E\in\Tau$ for every $g\in G\}$,
and for $E\in\Sigma$ set

$$\eqalign{\mu E=\sup\{\sum_{i=0}^n\nu(g_i\action F_i):n\in\Bbb N,
  &\,F_0,\ldots,F_n\text{ are disjoint subsets of }E\cr
&\text{belonging to }\Sigma,\,
  g_i\in G\text{ for each }i\le n\}\cr}$$

\noindent  (cf.\ 112Yd, 234Xl).
Show that $\mu$ is a $G$-invariant measure on $X$.
%441A

\spheader 441Xc Let $r\ge 1$ be an integer, and
$X=\coint{0,1}^r$.
Let $G$ be the set of $r\times r$ matrices with integer
coefficients and determinant $\pm 1$, and for $A\in G$, $x\in X$ say that
$A\action x=\Matrix{\fraction{\eta_1}\\ \ldots\\ \fraction{\eta_r}}$ where
$\Matrix{\eta_1\\ \ldots\\ \eta_r}=Ax$ and $\fraction{\alpha}$ is the
fractional part of $\alpha$ for each $\alpha\in\Bbb R$.
(i) Show that $\action$ is an action of $G$ on $X$, and that
Lebesgue measure on $X$ is $G$-invariant.
(ii) Show that if
$X$ is given the compact Hausdorff topology corresponding to the bijection
$\alpha\mapsto(\cos 2\pi\alpha,\sin 2\pi\alpha)$ from $X$ to the unit
circle in $\Bbb R^2$, and $G$ is given its discrete topology, the
action is continuous.
%441A

\spheader 441Xd
Let $X$ be a topological space and $G$ a group acting on
$X$ such that ($\alpha$) all the maps $x\mapsto a\action x$ are
continuous ($\beta$) all the orbits of $G$ are dense.   Show that any
non-zero $G$-invariant quasi-Radon measure on $X$ is strictly positive.
%441C

\sqheader 441Xe Let $G$ be a compact Hausdorff topological group.  (i)
Show that its conjugacy classes are closed.   (ii) Show that if $K$,
$L\subseteq G$ are disjoint compact sets then
$\{ac^{-1}da^{-1}:a\in G,\,c\in K,\,d\in L\}$ is a compact set not
containing $e$, so that
there is a neighbourhood $U$ of $e$ such that whenever $c^{-1}d\in U$
and $a\in G$ then either $aca^{-1}\notin K$ or $ada^{-1}\notin L$.
(iii) Show that every conjugacy class of $G$ carries a Radon probability
measure which is invariant under the conjugacy action of $G$.
%441C

\spheader 441Xf Let $(G,\cdot)$ be a topological group.   (i) On $G$
define a binary operation
$\diamond$ by saying that $x\diamond y=y\cdot x$ for all $x$,
$y\in G$.   Show that $(G,\diamond)$ is a topological group isomorphic
to $(G,\cdot)$, and that any element of $G$ has the same inverse for
either group operation.   (ii) Suppose that $\mu$
is a left Haar measure on $(G,\cdot)$.   Show that $\mu$ is a right Haar
measure on $(G,\diamond)$.   (iii) Set $\phi(a)=a^{-1}$ for
$a\in G$.   Show that if $\mu$ is a left Haar measure on $(G,\cdot)$ then
the image measure
$\mu\phi^{-1}$ is a right Haar measure on $(G,\cdot)$.   (iv)
Show that $(G,\cdot)$ has a left Haar measure iff it has a right Haar
measure.
(v) Show that $(G,\cdot)$ has a left Haar probability measure iff it has
a totally finite left Haar measure iff it has a right Haar probability
measure.   (iv) Show that $(G,\cdot)$ has a $\sigma$-finite left Haar
measure iff it has a $\sigma$-finite right Haar measure.
%441D

\sqheader 441Xg(i) For Lebesgue measurable
$E\subseteq\Bbb R\setminus\{0\}$, set $\nu E=\int_E\bover1{|x|}dx$.
Show that $\nu$ is
a (two-sided) Haar measure if $\Bbb R\setminus\{0\}$ is given the group
operation of multiplication.   (ii) For Lebesgue measurable
$E\subseteq\Bbb C\setminus\{0\}$, identified with
$\Bbb R^2\setminus\{0\}$, set $\nu E=\int_E\bover1{|z|^2}\mu(dz)$, where
$\mu$ is two-dimensional Lebesgue measure.   Show that $\nu$ is a
(two-sided)
Haar measure on $\Bbb C\setminus\{0\}$ if we take complex multiplication
for the group operation.   \Hint{263D.}
%441D

\sqheader 441Xh(i) Show that Lebesgue measure on $\BbbR^r$ is a (two-sided) Haar
measure, for any $r\ge 1$, if we take addition for the
group operation.   (ii) Show that the usual measure on $\{0,1\}^I$ is a
two-sided Haar measure, for any set $I$, if we give
$\{0,1\}^I$ the group operation corresponding to its identification with
$\Bbb Z_2^I$.   (iii) Describe the corresponding Haar measure on
$\Cal PI$ when $\Cal PI$ is given the group operation $\symmdiff$.
%441D

\spheader 441Xi Let $G$ be a locally compact Hausdorff topological
group.   (i) Show that any (left) Haar measure on $G$ must be strictly
positive.   (ii) Show that $G$ has a totally finite (left) Haar measure
iff it is compact.   (iii) Show that $G$ has a $\sigma$-finite (left)
Haar measure iff it is $\sigma$-compact.
%441D

\sqheader 441Xj(i) Let $G$ and $H$ be topological groups with left Haar
measures $\mu$ and $\nu$.   Show that the quasi-Radon product measure on
$G\times H$ (417N) is a left Haar measure on $G\times H$.   (ii) Let
$\familyiI{G_i}$ be a family of topological groups, and suppose that
each $G_i$ has a left Haar probability measure (as happens, for
instance, if each $G_i$ is compact).   Show that the quasi-Radon product
measure on $\prod_{i\in I}G_i$ (417O) is a left Haar measure on
$\prod_{i\in I}G_i$.
%441D

\spheader 441Xk(i) Show that any (left) Haar measure on a topological
group, as defined in 441D, must be locally finite.   (ii) Show that any
(left) Haar measure on a locally compact Hausdorff group must be a Radon
measure.
%441D

\spheader 441Xl Let $r\ge 1$ be an integer, and set
$X=\{x:x\in\Bbb R^r,\,\|x\|=1\}$.   Let $G$ be the group of orthogonal
$r\times r$ real
matrices, so that $G$ acts transitively on $X$.   Show that (when given
its natural topology as a subset of $\BbbR^{r^2}$) $G$ is a compact
Hausdorff topological group.   Let $\mu$ be a left Haar measure on $G$,
and $x$ any point of $X$;  set $\phi_x(T)=Tx$ for $T\in G$.   Show that
the image measure
$\mu\phi_x^{-1}$ is a $G$-invariant measure on $X$, independent of the
choice of $x$.
%441E

\spheader 441Xm Let $X$ be a non-abelian Hausdorff
topological group with a left Haar probability measure $\mu$.   Let
$\lambda$ be the quasi-Radon product measure on $X^2$.   Show that
$\lambda\{(x,y):xy=yx\}\le\bover58$.   \Hint{if $Z$ is the centre of $X$,
$X/Z$ is not cyclic, so $\mu Z\le\bover14$.}
%441E

\spheader 441Xn Let $X$ be a compact metric space, and $g:X\to X$ any
isometry.   Show that $g$ is surjective.   \Hint{if $x\in X$, then
$\rho(g^mx,g^nx)\ge\rho(x,g[X])$ for any $m<n$.}
%441F

\spheader 441Xo Let $(X,\rho)$ be a locally compact metric space,
and $\Cal C$ the set of closed subsets of $X$ with its Fell
topology (4A2T).   Show that if $G$ is the isometry group of $X$ with
its topology of pointwise convergence, then $(g,F)\mapsto g[F]$
is a continuous action of $G$ on $\Cal C$.
%441F

\spheader 441Xp Let $(X,\rho)$ be a metric space and $\Cal K$ the family
of compact subsets of $X$ with the topology induced by
the Vietoris topology on the space of closed subsets of $X$ (4A2T).
Show that if $G$ is the isometry group of $X$ with its topology of
pointwise convergence, then $(g,K)\mapsto g[K]$ is a continuous
action of $G$ on $\Cal K$.
%441F

\sqheader 441Xq Let $(X,\rho)$ be a metric space, and $G$ its isometry
group with the topology $\frak T$ of pointwise convergence.   (i) Show
that if $X$ is compact, $\frak T$ can
be defined by the metric $(g,h)\mapsto\max_{x\in X}\rho(g(x),h(x))$.
(ii) Show that if
$\{y:\rho(y,x)\le\gamma\}$ is compact for every $x\in X$ and $\gamma>0$,
then $G$ is locally compact.   (iii) Show that if $X$ is separable then
$G$ is metrizable.   (iv) Show that if $(X,\rho)$ is complete then
$G$ is complete under its bilateral uniformity.
(v) Show that if $X$ is separable and $(X,\rho)$
is complete then $G$ is Polish.
%441G

\sqheader 441Xr Give $\Bbb N$ the zero-one metric $\rho$.
Let $G$ be the isometry group of $\Bbb N$ (that
is, the group of all permutations of $\Bbb N$) with its
topology of pointwise convergence.
(i) Show that $G$ is a G$_{\delta}$ subset of $\NN$, so is a Polish
group.   (ii) Show that if we set $\Delta(g,h)=\min\{n:n\in\Bbb N$,
$g(n)\ne h(n)\}$
and $\sigma(g,h)=1/(1+\Delta(g^{-1},h^{-1}))$ for distinct $g$,
$h\in G$, then $\sigma$ is a \rti\ metric on $G$ inducing its topology.
(iii) Show that there
is no complete \rti\ metric on $G$ inducing its topology.   \Hint{any
such metric must have the same Cauchy sequences as $\sigma$.}
(iv) Show that $G$ is not locally compact.
%441G

\sqheader 441Xs Let $r\ge 1$ be an integer, and $S_{r-1}$ the sphere
$\{x:x\in\BbbR^r,\,\|x\|=1\}$.   (i) Show that every isometry $\phi$
from $S_{r-1}$ to itself corresponds to an orthogonal $r\times r$ matrix
$T$.   \Hint{$T=\langle\phi(e_i)\dotproduct e_j\rangle_{i,j<r}$.}
(ii) Show that the topology of pointwise convergence on the isometry
group of $S_{r-1}$ corresponds to the topology on the set of $r\times r$
matrices regarded as a subset of $\BbbR^{r^2}$.
%441G

\spheader 441Xt Let $X$ be a locally compact metric space and $G$ a
subgroup of the isometry group of $X$.   Show that for every $x\in X$
there is a non-zero $G$-invariant Radon measure on
$\overline{\{g(x):g\in G\}}$.
%441H

\spheader 441Xu\dvAnew{2018} Let $X$ be a set with its zero-one metric
and $G$ the group of permutations
of $X$ with its topology of pointwise convergence.   Let
$\Cal W\subseteq\Cal P(X^2)$ be the set of total orderings of $X$.
Show that $\Cal W$ is compact for the usual topology of
$\Cal P(X^2)$.   For $g\in G$ and $W\in\Cal W$ write
$g\action W=\{(g(x),g(y)):(x,y)\in W\}$;  show
that $\action$ is a continuous action of $G$ on $\Cal W$.
Show that there is a unique
$G$-invariant Radon probability measure $\mu$ on $\Cal W$ such that
$\mu\{W:(x_i,x_{i+1})\in W$ for every $i<n\}=\Bover1{(n+1)!}$ whenever
$x_0,\ldots,x_n\in X$ are distinct.
%441H

\leader{441Y}{Further exercises (a)}
%\spheader 441Ya
Let $(X,\rho)$ be a metric space, and $\Cal C$ the family of
non-empty closed
subsets of $X$, with its Hausdorff metric $\tilde\rho$ (4A2T).   Show
that if $G$ is the isometry group of $X$, $(g,F)\mapsto g[F]$ is an
action of $G$ on $\Cal C$.
%441A query ?continuous

\spheader 441Yb (M.Elekes \& T.Keleti) Let $X$ be a set, $G$ a group
acting on $X$,  $\Sigma$ a
$\sigma$-algebra of subsets of $X$ such that $g\action E\in\Sigma$
whenever $E\in\Sigma$ and $g\in G$, and $H$ an element of $\Sigma$.
Suppose that
$\mu$ is a measure, with domain the subspace $\sigma$-algebra
$\Sigma_H$, such that $\mu(g\action E)=\mu E$ whenever $E\in\Sigma_H$
and $g\in G$ are such that $g\action E\subseteq H$.   (i) Show that
$\sum_{n=0}^{\infty}\mu E_n=\sum_{n=0}^{\infty}\mu E'_n$ whenever
$\sequencen{E_n}$ and $\sequencen{E'_n}$ are sequences in $\Sigma_H$ for
which there are sequences $\sequencen{g_n}$, $\sequencen{g'_n}$ in $G$
such that $\sequencen{g_n\action E_n}$ and
$\sequencen{g'_n\action E'_n}$ are partitions of the same subset of $X$.
(ii) Show that there is a $G$-invariant measure with domain $\Sigma$
which extends $\mu$.
%441A

\spheader 441Yc Take $1\le s\le r\in\Bbb N$.   Let $\Cal C$ be the set of
closed subsets of $\BbbR^r$ with its Fell topology.   Let
$\Cal C_s\subseteq\Cal C$ be the set of $s$-dimensional linear subspaces
of $\BbbR^r$.    Show that $\Cal C_s$ is a closed subset of
$\Cal C$, therefore
a compact metrizable space in its own right, and that the group $G$ of
orthogonal $r\times r$ matrices acts transitively on $\Cal C_s$,
so that there is a $G$-invariant
Radon probability measure on $\Cal C_s$.
%441Ya, 441C

\spheader 441Yd For $w$, $z\in\Bbb C\setminus\{0\}$ set
$\rho(w,z)=|\Ln(\bover{w}{z})|$, where $\Ln v=\ln|v|+i\arg v$ for
non-zero complex numbers $v$.   (i) Show that $\rho$ is a metric on
$\Bbb C\setminus\{0\}$.   (ii) Show that the 2-dimensional Hausdorff
measure $\mu^{(\rho)}_{H2}$ derived from $\rho$ (264K, 471A) is a Haar measure
for the multiplicative group $\Bbb C\setminus\{0\}$.   (iii) Show that
$\mu^{(\rho)}_{H2}=\bover4{\pi}\nu$, where $\nu$ is the measure of
441Xg(ii).
%441Xg  441D  264I  263D

\spheader 441Ye Let $X$ be the group of real $r\times r$ orthogonal
matrices, where $r\ge 2$ is an integer.   Give $X$ the Euclidean metric,
regarding it as a subset of $\BbbR^{r^2}$.   (i) Show that the left and
right actions of $X$ on itself are distance-preserving.   (ii) Show that
$\bover{r(r-1)}2$-dimensional Hausdorff measure on $X$ is a two-sided
Haar measure.
%441D

\spheader 441Yf Let $X=\text{SO}(3)$ be the set of real $3\times 3$
orthogonal matrices with determinant $1$.   Give $X$ the metric
corresponding to its embedding in $9$-dimensional Euclidean space.
(i) Show that $X$ can be parametrized as the set of matrices

$$\phi\Matrix{z\\ \alpha\\ \theta}
=\Matrix{z&-\cos\theta\sqrt{1-z^2}&\sin\theta\sqrt{1-z^2}\\
\cos\alpha\sqrt{1-z^2}&z\cos\alpha\cos\theta-\sin\alpha\sin\theta
  &-z\cos\alpha\sin\theta-\sin\alpha\cos\theta\\
\sin\alpha\sqrt{1-z^2}&z\sin\alpha\cos\theta+\cos\alpha\sin\theta
  &-z\sin\alpha\sin\theta+\cos\alpha\cos\theta}$$

\noindent with $z\in[-1,1]$, $\alpha\in[-\pi,\pi]$ and
$\theta\in[-\pi,\pi]$.   (ii) Show that if $T$ is the $9\times 3$ matrix
which is the derivative of $\phi$ at $\Matrix{z\\ \alpha\\ \theta}$,
then $T\trs T=\Matrix{\Bover2{1-z^2}&0&0\\0&2&2z\\0&2z&2}$ has
constant determinant,
so that if $\mu$ is Lebesgue measure on $[-1,1]\times[-\pi,\pi]^2$
then the image measure $\mu\phi^{-1}$ is a Haar measure on $X$.
\Hint{441Ye, 265E.}   (iii)
Show that if $A\in X$ corresponds to a rotation through an angle
$\gamma(A)\in[0,\pi]$
then its trace $\tr(A)$ (that is, the sum of its diagonal entries) is
$1+2\cos\gamma(A)$.   \Hint{$\tr(AB)=\tr(BA)$ for any square matrices $A$ and
$B$ of the same size.}   (iv) Show that if $X$ is given its Haar
probability measure then $\cos\gamma(A)$ has expectation $-\bover12$.
%441Ye 441D

\spheader 441Yg\dvAnew{2016}
Let $\Bbb H$ be the division ring of the {\bf quaternions}, that is,
$\BbbR^4$ with its usual addition and with multiplication defined by the rule

$$\eqalign{(\xi_0,\xi_1,\xi_2,\xi_3)\times(\eta_0,\eta_1,\eta_2,\eta_3)
&=(\xi_0\eta_0-\xi_1\eta_1-\xi_2\eta_2-\xi_3\eta_3,
\xi_0\eta_1+\xi_1\eta_0+\xi_2\eta_3-\xi_3\eta_2,\cr&
\mskip50mu\xi_0\eta_2-\xi_1\eta_3+\xi_2\eta_0+\xi_4\eta_1,
\xi_0\eta_3+\xi_1\eta_2-\xi_2\eta_1+\xi_3\eta_0).\cr}$$

\noindent For Lebesgue measurable $E\subseteq\Bbb H\setminus\{0\}$,
set $\nu E=\int_E\Bover{1}{\|x\|^4}dx$.   Show that
(i) $\|x\times y\|=\|x\|\|y\|$ for all $x$, $y\in\Bbb H$
(ii) $\Bbb H\setminus\{0\}$ is a group
(iii) $\nu$ is a (two-sided)
Haar measure on $\Bbb H\setminus\{0\}$.
%441D 441Xg

\spheader 441Yh In 441Xl, show that $\mu\phi_x^{-1}$ is a scalar
multiple of Hausdorff $(r-1)$-dimensional measure on $X$.
%441Xl, 441E

\spheader 441Yi For any topological spaces $X$ and $Y$, and any set $G$
of functions from $X$ to $Y$, the {\bf compact-open} topology on $G$ is
the topology generated by sets of the form
$\{g:g\in G,\,g[K]\subseteq H\}$, where $K\subseteq X$ is compact and
$H\subseteq Y$ is open.   Show that if $(X,\rho)$ is a metric space and
$G$ is the isometry group of $X$, then the topology of pointwise
convergence on $G$ is its compact-open topology.
%441G

\spheader 441Yj Let $X$ be a compact Hausdorff space and $G$ the group
of all homeomorphisms from $X$ to itself.   (i) Let $\Rho$ be the family
of all continuous pseudometrics on $X$ (see 4A2Jg).   For $\rho\in\Rho$ and
$g$, $h\in G$, set $\tau_{\rho}(g,h)=\max_{x\in X}\rho(g(x),h(x))$.
Show that every $\tau_{\rho}$ is a \rti\ pseudometric on $G$, and that
$G$ with the topology generated by $\{\tau_{\rho}:\rho\in\Rho\}$ is a
topological group.
(ii) Show that this is the compact-open topology as defined in 441Yi.
(iii) Show that if $X$ is metrizable then $G$ is Polish.
%441G 441Yi

\leaveitout{\spheader 441Y? Let $U$ be a Hilbert space, and write $S$
for the unit sphere $\{u:\|u\|=1\}$.   Show that any isometry of $S$
corresponds to a bounded linear operator from $U$ to itself.   Show that
the topology of pointwise convergence on the isometry group of $S$
corresponds to the strong operator topology on $\eurm B(U;U)$.
%441Xs
}%end of leaveitout

\spheader 441Yk Let $(X,\Cal W)$ be a locally compact Hausdorff uniform
space, and suppose that $G$ is a group acting on $X$ `uniformly
equicontinuously';  that is, for every $W\in\Cal W$ there is a
$V\in\Cal W$ such that $(a\action x,a\action y)\in W$ whenever
$(x,y)\in V$ and
$a\in G$.   Show that there is a non-zero $G$-invariant Radon measure on
$X$.
%441H

\spheader 441Yl Let $X$ be a compact metric space and $G$ the isometry
group of $X$.   Show that every $G$-orbit in $X$ is closed.
%441H

\spheader 441Ym Let $T$ be any set, and $\rho$ the $\{0,1\}$-valued
metric on $T$.   Let $X$ be the set of partial orders on $T$, regarded
as a subset of $\Cal P(T\times T)$.   Show that $X$ is compact (cf.\
418Xy).   Let $G$ be the group of isometries of $T$ with its topology of
pointwise convergence.   Set
$\pi\action x=\{(t,u):(\pi^{-1}(t),\pi^{-1}(u))\in x\}$ for
$\pi\in G$ and $x\in X$.   Show that $\action$ is a continuous action of
$G$ on $X$.   Show that there is a strictly positive $G$-invariant Radon
probability measure $\mu$ on $X$.
%441I  what are the extreme points of $X$?

\spheader 441Yn Let $X$ be a set, $G$ a group acting on $X$, and $\mu$ a
$G$-invariant measure on $X$ with measure algebra $\frak A$.   Show that
if $\tau$ is any rearrangement-invariant extended Fatou norm on
$L^0(\frak A)$ then the formula of 441Kb defines a norm-preserving
action of $G$ on the Banach space $L^{\tau}$.
%441K

\spheader 441Yo Let $\action_X$ be an action of a group $G$ on a set $X$,
$\mu$ a $G$-invariant measure on $X$, $(\frak A,\bar\mu)$ its measure
algebra and $\action_{\frak A}$ the induced action on $\frak A$.
Set $Z=X^G$;  define $\phi:X\to Z$ by
setting $\phi(x)=\family{g}{G}{g^{-1}\action x}$ for $x\in X$;
let $\nu$ be the image
measure $\mu\phi^{-1}$, and $(\frak B,\bar\nu)$ its measure algebra.
Let $\action_Z$ be the left shift action of $G$ on $Z$;  show
that $\nu$ is $\action_Z$-invariant, so that there is an induced action
$\action_{\frak B}$ on $\frak B$.   Show that
$(\frak A,\bar\mu,\action_{\frak A})$ and
$(\frak B,\bar\nu,\action_{\frak B})$ are isomorphic.
%441K

\spheader 441Yp
Let $X$ be a topological space, $G$ a topological group and $\action$ a
continuous action of $G$ on $X$.   Let $M_{\text{qR}}^+$ be
the set of totally
finite quasi-Radon measures on $X$.   (i) Show that we have
an action $\action$ of $G$ on $M_{\text{qR}}^+$ defined by saying that
$(a\action\nu)(E)=\nu(a^{-1}\action E)$ whenever $a\in G$,
$\nu\in M_{\text{qR}}^+$ and $E\subseteq X$ are such that $\nu$ measures
$a^{-1}\action E$.
(ii) Show that this action is continuous if we give $M_{\text{qR}}^+$
its narrow topology.   (iii) Show that if $\nu\in M_{\text{qR}}^+$,
$f\in\eusm L^1(\nu)$ is non-negative
and $f\nu$ is the corresponding indefinite-integral measure, then
$a\action(f\nu)$ is the
indefinite-integral measure $(a\action f)(a\action\nu)$ for every $a\in G$.

\spheader 441Yq
Let $X$ be a set, $G$ a group acting on $X$
and $\mu$ a totally finite $G$-invariant measure on
$X$ with domain $\Sigma$.   Suppose there is a probability measure $\nu$ on
$G$, with domain $\Tau$, such that
$(a,x)\mapsto a^{-1}\action x:G\times X\to X$ is
$(\Tau\tensorhat\Sigma,\Sigma)$-measurable and $\nu$ is invariant under the
left action of $G$ on itself.   Let $u\in L^0(\mu)$ be such that
$a\action u=u$ for every $a\in G$.  Show that there is an
$f\in\eusm L^0(\mu)$ such that $f^{\ssbullet}=u$ and $a\action f=f$ for
every $a\in G$.   \Hint{if $u=g^{\ssbullet}$ where $g:X\to\Bbb R$ is
$\Sigma$-measurable and $\mu$-integrable, try
$f(x)=\int g(a^{-1}\action x)\nu(da)$ when this is defined.}
%441K

\spheader 441Yr
Let $X$ be a topological space, $G$ a compact Hausdorff
group, $\action$ a continuous action of $G$ on $X$,
and $\mu$ a $G$-invariant quasi-Radon measure on $X$.
Let $u\in L^0(\mu)$ be such that
$a\action u=u$ for every $a\in G$.  Show that there is an
$f\in\eusm L^0(\mu)$ such that $f^{\ssbullet}=u$ and $a\action f=f$ for
every $a\in G$.
%441K
}%end of exercises

\endnotes{
\Notesheader{441}
The proof I give of 441C is essentially the same as the proof of 441E in
{\smc Halmos 50}, \S58.

In part (f) of the proof of 441C I use an ultrafilter, relying on a
fairly strong consequence of the Axiom of Choice.   In this volume, as
in the last, I generally employ the axiom of choice without stopping to
consider whether it is really needed.   But Haar measure, at least, is
so important that I point out now that it can be built with much weaker
principles.   For a construction of Haar measure not dependent on
choice, see 561G in Volume 5.   I
ought to remark that the argument there leads us to a translation-invariant
linear functional rather than a measure, and that while there is still a
version of the Riesz representation theorem (564I), we
may get something less
than a proper countably additive measure if we do not have countable
choice.   Moreover, in the absence of
the full axiom of choice, we may find that we have fewer locally compact
topological groups than we expect.

While Haar measure is surely the pre-eminent application of the theory
here, I think that some of the other consequences of 441C (441H, 441Xe,
441Xl, 441Yc, 441Yk) are sufficiently striking to justify the trouble
involved in the extra generality.   I ought to remark that there are
important examples of invariant measures which have nothing to do with
441C.   Some of these
will appear in \S449;  for the moment I note only 441Xa.

{\smc Federer 69}, \S2.7, develops a general theory of `covariant'
measures $\mu$ (`relatively invariant' in {\smc Halmos 50}) such that
$\mu(a\action E)=\psi(a)\mu E$ for appropriate
sets $E\subseteq X$ and $a\in G$, where $\psi:G\to\ooint{0,\infty}$ is a
homomorphism;  for instance, taking $\mu$ to be Lebesgue measure on
$\BbbR^r$, we have $\mu T[E]=|\det T|\mu E$ for every linear space
isomorphism $T:\BbbR^r\to\BbbR^r$ and every measurable set $E$ (263A).
The theory I have described here can deal only with the subgroup of
isometric linear isomorphisms (that is, the orthogonal group).
Covariant measures arise naturally in many other contexts, such as 443T
below.

Hausdorff measures, being defined in terms of metrics, are necessarily
invariant under isometries, so appear naturally in this context,
starting with 264I.   There are interesting challenges both in finding
suitable metrics and in establishing exact constants, as in
441Yd-441Ye.

It is worth pausing over the topology of an isometry group, as described
in 441G.   It is quite surprising that such an elementary idea should
give us a topological group at all.   I offer some exercises
(441Xq-441Xs, %441Xq 441Xr 441Xs
441Yi) to help you relate the construction to material which may
be more familiar.
It is of course a `weak' topology, except when the underlying space is
compact (441Xq(i)).   See also 441Yj.   These groups are rarely locally
compact, and you may find them pushed out of your mind by the
extraordinary theory which you
will see developed in the next hundred pages; %104 or so
but in the last two sections of this chapter, and in \S493, they will
become leading examples.
}%end of notes

\discrpage

\frfilename{mt442.tex}
\versiondate{21.3.07}
\copyrightdate{1998}

\def\chaptername{Topological groups}
\def\sectionname{Uniqueness of Haar measures}

\newsection{442}

Haar measure has an extraordinary wealth of special properties, and it
will be impossible for me to cover them all properly in this chapter.
But surely the second thing to take on board, after the existence of
Haar measures on locally compact Hausdorff groups\cmmnt{ (441E)}, is
the fact that they are, up to scalar multiples, unique.   This is the
content of 442B.   We find also that while left and right Haar measures
can be different\cmmnt{ (442Xf)}, they are not only direct mirror
images of each other (442C)
-- as is, I suppose, to be expected --  but even more closely related
(442F, 442H, 442L).   Investigating this relation, we are led naturally
to the `modular function' of a group (442I).

\leader{442A}{Lemma} Let $X$ be a topological group and $\mu$ a left
Haar measure on $X$.

(a) $\mu$ is strictly positive and locally finite.

(b) If $G\subseteq X$ is open and $\gamma<\mu G$, there are an open set
$H$ and an open neighbourhood $U$ of the identity such that
$HU\subseteq G$\cmmnt{ (writing $HU$ for $\{xy:x\in H,\,y\in U\}$)} and $\mu H\ge\gamma$.

(c) If $X$ is locally compact and Hausdorff, $\mu$ is a Radon measure.

\proof{{\bf (a)(i)} \Quer\ If $G\subseteq X$ were a non-empty open set
such that $\mu G=0$, then we should have $\mu(xG)=0$ for every $x\in X$,
so that $X$ would be covered by negligible open sets;  but as $\mu$ is
supposed to be $\tau$-additive, $\mu X=0$.\ \Bang

\medskip

\quad{\bf (ii)} Because $\mu$ is effectively locally finite, there is
some non-empty open set $G$ such that $\mu G<\infty$;  but now
$\{xG:x\in X\}$ is a cover of $X$ by open sets of finite measure.

\medskip

{\bf (b)} Let $\Cal U$ be the family
of open sets containing the identity $e$, and $\Cal H$ the family of
open sets $H$ such that $HU\subseteq G$ for some $U\in\Cal U$.   Because
$\Cal U$ is downwards-directed,
$\Cal H$ is upwards-directed;  because $e\in U$ for every $U\in\Cal U$,
$\bigcup\Cal H\subseteq G$.   If $x\in G$, then $x^{-1}G\in\Cal U$, and
there is a $U\in\Cal U$ such that $UU\subseteq x^{-1}G$;  but now
$xUU\subseteq G$, so $xU\in\Cal H$.   Thus $\bigcup\Cal H=G$.   Because
$\mu$ is $\tau$-additive, there is an $H\in\Cal H$ such that $\mu
H\ge\gamma$.

\medskip

{\bf (c)} Use (a) and 416G.
}%end of proof of 442A

\leader{442B}{Theorem} Let $X$ be a topological group.   If $\mu$ and
$\nu$ are left Haar measures on $X$, they are multiples of each other.

\proof{{\bf (a)} Let $\Cal G$ be the family of non-empty open sets $G$
such that $\mu G$ and $\nu G$ are both finite;  because $\mu$ and $\nu$
are locally finite (442Aa), $\Cal G$ is a base for
the topology of $X$.   Note that $G\cup H\in\Cal G$ for all $G$,
$H\in\Cal G$.   Set $\Cal U=\{U:U\in\Cal G,\,U=U^{-1},\,e\in U\}$,
where $e$ is the identity of $G$;
then $\Cal U$ is a base of neighbourhoods of $e$ (4A5Ec).   Let $\Cal F$
be the filter on $\Cal U$ generated by the sets
$\{U:U\in\Cal U,\,U\subseteq V\}$ as $V$ runs over $\Cal U$.

\medskip

{\bf (b)} (The key.)  If $G\in\Cal G$ and $0<\epsilon<1$, there is a
$V_1\in\Cal U$ such that

\Centerline{$(1-\epsilon)\Bover{\mu G}{\nu G}\le\Bover{\mu U}{\nu U}$}

\noindent whenever $U\in\Cal U$ and $U\subseteq V_1$.   \Prf\ By 442Aa,
$\mu G$ and $\nu G$ are both non-zero.   By 442Ab, there are an open set
$H$ and a neighbourhood $V_1$ of $e$ such that $HV_1\subseteq G$ and
$\mu H\ge(1-\epsilon)\mu G$;  shrinking $V_1$ if need be, we may suppose
that $V_1\in\Cal U$.   Take any $U\in\Cal U$ such that $U\subseteq V_1$,
so that $HU\subseteq G$.   Consider the product quasi-Radon measure
$\lambda$ of $\mu$ and $\nu$ on $X\times X$ (417R), and the set
$W=\{(x,y):x,\,y\in G,\,x^{-1}y\in U\}$.   Because the function
$(x,y)\mapsto x^{-1}y$ is continuous (4A5Eb), $W$ is open.
Consequently

\Centerline{$\int\mu\{x:(x,y)\in W\}\nu(dy)
=\lambda W=\int\nu\{y:(x,y)\in W\}\mu(dx)$}

\noindent (417C(b-v-$\beta$)).   But we see that if $x\in H$ then
$xU\subseteq G$, so $(x,y)\in W$ whenever $y\in xU$, and

\Centerline{$\int\nu\{y:(x,y)\in W\}\mu(dx)
\ge\int_H\nu(xU)\mu(dx)=\mu H\cdot\nu U$.}

\noindent On the other hand,

$$\eqalign{\int\mu\{x:(x,y)\in W\}\nu(dy)
&\le\int_G\mu\{x:x^{-1}y\in U\}\nu(dy)
=\int_G\mu\{x:y^{-1}x\in U\}\nu(dy)\cr
&=\int_G\mu(yU)\nu(dy)
=\mu U\cdot\nu G,\cr}$$

\noindent so that

\Centerline{$(1-\epsilon)\mu G\cdot\nu U
\le\mu H\cdot\nu U
\le\lambda W
\le\mu U\cdot\nu G$.}

\noindent  Dividing both sides by $\nu U\cdot\nu G$, we have the
result.\ \Qed

\medskip

{\bf (c)} In the same way, there is a $V_2\in\Cal V$ such that

\Centerline{$(1-\epsilon)\Bover{\nu G}{\mu G}\le\Bover{\nu U}{\mu U}$}

\noindent whenever $U\in\Cal U$ and $U\subseteq V_2$.   So if
$U\in\Cal U$ and $U\subseteq V_1\cap V_2$, we have

\Centerline{$(1-\epsilon)\Bover{\mu G}{\nu G}\le\Bover{\mu U}{\nu U}
\le\Bover{1}{1-\epsilon}\Bover{\mu G}{\nu G}$.}

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\lim_{U\to\Cal F}\Bover{\mu U}{\nu U}
=\Bover{\mu G}{\nu G}$.}

\noindent And this is true for every $G\in\Cal G$.

\medskip

{\bf (d)} So if we set $\alpha=\lim_{U\to\Cal F}\bover{\mu U}{\nu U}$,
we shall have $\mu G=\alpha\nu G$ for every $G\in\Cal G$.   Now $\mu$
and $\alpha\nu$ are quasi-Radon measures agreeing on the
base $\Cal G$, which is closed under finite unions, so are identical, by
415H(iv).


}%end of proof of 442B

\leader{442C}{Proposition} Let $X$ be a topological group and $\mu$ a
left Haar measure on $X$.
Setting $\nu E=\mu(E^{-1})$ whenever $E\subseteq X$ is such that
$E^{-1}=\{x^{-1}:x\in E\}$ is measured by $\mu$, $\nu$ is a right Haar
measure on $X$.

\proof{ Set $\phi(x)=x^{-1}$ for $x\in X$.   Then $\phi$ is a
homeomorphism, so the image measure $\nu=\mu\phi^{-1}$ is a quasi-Radon
measure.   It is
non-zero because $\nu X=\mu X$.   If $E\in\dom\nu$ and $x\in X$, then

\Centerline{$\nu(Ex)=\mu(x^{-1}E^{-1})=\mu E^{-1}=\nu E$.}

\noindent So $\nu$ is a right Haar measure.
}%end of proof of 442C

\leader{442D}{Remark} Clearly all the arguments of 442A-442C must be
applicable to right Haar
measures\cmmnt{;  that is, any right Haar measure must be locally
finite and strictly
positive;  two right Haar measures on the same group must be multiples
of each other;  and if $X$ carries a right Haar measure $\nu$ then
$E\mapsto\nu E^{-1}$ will be a left Haar measure on $X$}.
\cmmnt{(If you are unhappy with such a bold appeal to the symmetry
between `left' and `right' in topological groups, write the reflected
version of 442C out in full, and use it to reflect 442A-442B.)

}Thus we may say that a topological group {\bf carries Haar measures} if
it has either a left or a right Haar measure.   \cmmnt{These can, of
course, be the same;  in fact it takes a certain amount of exploration
to find a group in which they are different (e.g., 442Xf).}

\leader{442E}{Lemma} Let $X$ be a topological group, $\mu$ a left Haar
measure on $X$ and $\nu$ a right Haar measure on $X$.
If $G$, $H\subseteq X$ are open, then

\Centerline{$\mu G\cdot\nu H
=\int_H\nu(xG^{-1})\mu(dx)$.}

\proof{ Let $\lambda$ be the quasi-Radon product measure of
$\mu$ and $\nu$ on
$X\times X$.   The sets $W_1=\{(x,y):y^{-1}x\in G,\,x\in H\}$ and
$W_2=\{(x,y):x\in G,\,yx\in H\}$ are both open, so 417C(b-v-$\beta$) tells
us that

$$\eqalign{\int_H\nu(xG^{-1})\mu(dx)
&=\int\nu\{y:x\in H,\,y^{-1}x\in G\}\mu(dx)
=\lambda W_1\cr
&=\int\mu\{x:x\in H,\,y^{-1}x\in G\}\nu(dy)
=\int\mu(H\cap yG)\nu(dy)\cr
&=\int\mu(y^{-1}H\cap G)\nu(dy)
=\int\mu\{x:x\in G,\,yx\in H\}\nu(dy)\cr
&=\lambda W_2
=\int_G\nu\{y:yx\in H\}\mu(dx)\cr
&=\int_G\nu(Hx^{-1})\mu(dx)
=\mu G\cdot\nu H.\cr}$$
}%end of proof of 442E

\leader{442F}{Domains of Haar \dvrocolon{measures}}\cmmnt{ 442B tells
us, in part, that any two left Haar measures on a topological
group must have the same domain and the same negligible sets;
similarly, any two right Haar measures have the same domain and the same
negligible sets.   In fact left and right Haar measures agree on both.

\medskip

\noindent}{\bf Proposition} Let $X$ be a topological group which carries
Haar measures.   If $\mu$ is a left Haar measure and $\nu$ is a right
Haar measure on $X$, then they have the same domains and the same null ideals.

\proof{{\bf (a)} Suppose that $F\subseteq X$ is a closed set such that
$\mu F=0$.   Then $\nu F=0$.   \Prf\Quer\ Otherwise, there is an open
set $H$ such that $\nu H<\infty$ and $\nu(F\cap H)>0$.   Let $G$ be any
open set such that $0<\mu G<\infty$.   By 442E,

\Centerline{$\mu G\cdot\nu H
=\int_H\nu(xG^{-1})\mu(dx)$.}

\noindent But also

\Centerline{$\mu G\cdot\nu(H\setminus F)
=\int_{H\setminus F}\nu(xG^{-1})\mu(dx)
=\int_H\nu(xG^{-1})\mu(dx)=\mu G\cdot\nu H$,}

\noindent so $\mu G\cdot\nu(H\cap F)=0$.\ \Bang\Qed

\medskip

{\bf (b)} It follows that

\Centerline{$\nu E=\sup_{F\subseteq E\text{ is closed}}\nu F=0$}

\noindent whenever $E$ is a Borel set such that $\mu E=0$.   Now take
any $E\in\dom\mu$.   Set

\Centerline{$\Cal G=\{G:G\subseteq X$ is open, $\mu G<\infty$,
$\nu G<\infty\}$.}

\noindent Because both $\mu$ and $\nu$ are locally finite, $\Cal G$
covers $X$.   If $G\in\Cal G$, there are Borel sets $E'$, $E''$ such
that $E'\subseteq E\cap G\subseteq E''$ and $\mu(E''\setminus E')=0$.
In this case $\nu(E''\setminus E')=0$ so $E\cap G\in\dom\nu$.   Because
$\nu$ is complete, locally determined and
$\tau$-additive, $E\in\dom\nu$ (414I).   If $\mu E=0$, it follows that

\Centerline{$\nu E=\sup_{F\subseteq E\text{ is closed}}\nu F=0$}

\noindent just as above.

\medskip

{\bf (c)} Thus $\nu$ measures $E$ whenever $\mu$ measures $E$, and $E$
is $\nu$-negligible whenever it is $\mu$-negligible.   I am sure you
will have no difficulty in believing that all the arguments above, in
particular that of 442E, can be re-cast to show that
$\dom\nu\subseteq\dom\mu$;  alternatively, apply the result in the form
just demonstrated to the left Haar measure $\nuprime$ and the right Haar
measure $\mu'$, where

\Centerline{$\nuprime E=\nu E^{-1}$,
\quad$\mu'E=\mu E^{-1}$}

\noindent as in 442C.
}%end of proof of 442F

\vleader{60pt}{442G}{Corollary}
Let $X$ be a topological group and $\mu$ a left
Haar measure on $X$ with domain $\Sigma$.   Then, for $E\subseteq X$ and
$a\in X$,

\Centerline{$E\in\Sigma\iff E^{-1}\in\Sigma\iff Ea\in\Sigma$,}

\Centerline{$\mu E=0\iff\mu E^{-1}=0\iff\mu(Ea)=0$.}

\proof{ Apply 442F with $\nu E=\mu E^{-1}$.}

\leader{442H}{Remark} \dvro{If}{From 442F-442G we see that if} $X$ is
any topological group which
carries Haar measures, there is a distinguished $\sigma$-algebra
$\Sigma$ of subsets of $X$, which we may call the algebra of {\bf Haar
measurable sets}, which is the domain of any Haar measure on $X$.
Similarly, there is a $\sigma$-ideal $\Cal N$ of $\Cal PX$, the ideal of
{\bf Haar negligible sets}\footnote{{\bf Warning!} do not confuse with the
`Haar null' sets described in 444Ye below.}, which is the null ideal for
any Haar measure on $X$.   Both $\Sigma$ and $\Cal N$ are
translation-invariant and also invariant under the inversion operation
$x\mapsto x^{-1}$.

If we form the quotient $\frak A=\Sigma/\Cal N$,
then we have a fixed Dedekind complete Boolean algebra which is the
{\bf Haar measure algebra} of the group $X$ in the sense that any Haar
measure on $X$, whether left or right, has measure algebra based on
$\frak A$.   If $a\in X$, the maps $x\mapsto ax$, $x\mapsto xa$,
$x\mapsto x^{-1}$ give rise to Boolean automorphisms of $\frak A$.

\cmmnt{For a member of $\Sigma$, we have a notion of `$\sigma$-finite'
which is symmetric between left and right (442Xd).   We do not in
general have a corresponding two-sided notion of `finite measure'
(442Xg(i));  but of course we can if we wish speak of a set as having
`finite left Haar measure' or `finite right Haar measure' without
declaring which Haar measure we are thinking of.
It is the case, however, that if the group $X$ itself has finite left
Haar measure, it also has finite right Haar measure;  see
442Ic-d below.}

\leader{442I}{The modular function} Let $X$ be a topological group which
carries Haar measures.

\spheader 442Ia There is a group homomorphism
$\Delta:X\to\ooint{0,\infty}$ defined by the formula

\Centerline{$\mu(Ex)=\Delta(x)\mu E$ whenever $\mu$ is a left Haar
measure on $X$ and $E\in\dom\mu$.}

\noindent\prooflet{\Prf\ Fix on a left Haar measure $\tilde\mu$ on $X$.
For $x\in X$, let $\mu_x$ be the function defined by saying

\Centerline{$\mu_xE=\tilde\mu(Ex)$ whenever $E\subseteq X$,
$Ex\in\dom\mu$,}

\noindent that is, for every Haar measurable set $E\subseteq X$.
Because the function $\phi_x:X\to X$ defined by setting
$\phi_x(y)=yx^{-1}$ is a homeomorphism, $\mu_x=\tilde\mu\phi_x^{-1}$ is
a quasi-Radon measure on $X$;  and

\Centerline{$\mu_x(yE)=\tilde\mu(yEx)=\tilde\mu(Ex)=\mu_xE$}

\noindent whenever $\mu_x$ measures $E$, so $\mu_x$ is a left Haar
measure on $X$.   By 442B, there is a $\Delta(x)\in\ooint{0,\infty}$
such that $\mu_x=\Delta(x)\tilde\mu$;  because $\tilde\mu$ surely takes
at least one value in $\ooint{0,\infty}$, $\Delta(x)$ is uniquely
defined.

If $\mu$ is any other left Haar measure on $X$, then
$\mu=\alpha\tilde\mu$ for some $\alpha>0$, so that

\Centerline{$\mu(Ex)=\alpha\tilde\mu(Ex)
=\alpha\Delta(x)\tilde\mu E=\Delta(x)\mu E$.}

\noindent Thus $\Delta:X\to\ooint{0,\infty}$ has the property asserted
in the formula offered.

To see that $\Delta$ is a group homomorphism, take any $x$, $y\in X$ and
a Haar measurable set $E$ such that $0<\tilde\mu E<\infty$, and observe
that

\Centerline{$\Delta(xy)\tilde\mu E
=\tilde\mu(Exy)
=\Delta(y)\tilde\mu(Ex)
=\Delta(y)\Delta(x)\tilde\mu E$,}

\noindent so that
$\Delta(xy)=\Delta(y)\Delta(x)=\Delta(x)\Delta(y)$.\ \Qed
}%end of prooflet

$\Delta$ is called the {\bf left modular function} of $X$.

\spheader 442Ib We find now that $\nu(xE)=\Delta(x^{-1})\nu E$ whenever
$\nu$ is a right Haar measure on $X$, $x\in X$ and $E\subseteq X$ is
Haar measurable.
\prooflet{\Prf\ Let $\mu$ be the left Haar measure derived from $\nu$,
so that $\mu E=\nu E^{-1}$ whenever $E$ is Haar measurable.   If $x\in
X$ and $E\in\dom\nu$, then

\Centerline{$\nu(xE)=\mu(E^{-1}x^{-1})
=\Delta(x^{-1})\mu E^{-1}=\Delta(x^{-1})\nu E$.  \Qed}
}%end of prooflet

Thus we may call $x\mapsto\Delta(x^{-1})=\Bover1{\Delta(x)}$ the {\bf
right modular function} of $X$.

\spheader 442Ic If $X$ is abelian, then\cmmnt{ obviously}
$\Delta(x)=1$ for
every $x\in X$\cmmnt{, because $\mu(Ex)=\mu(xE)=\mu E$ whenever
$x\in X$, $\mu$ is a left Haar measure on $X$ and $E$ is Haar measurable}.
Equally, if any\cmmnt{ (therefore every)} left (or right) Haar measure
$\mu$ on $X$ is totally
finite, then\cmmnt{ $\mu(Xx)=\mu(xX)=\mu X$, so again} $\Delta(x)=1$
for every $x\in X$.   This will be the case\cmmnt{, in particular,}
for any compact
Hausdorff topological group\cmmnt{ (recall that by 441E any such group
carries Haar measures), because its Haar measures are locally finite,
therefore totally finite}.

Groups in which $\Delta(x)=1$ for every $x$ are called {\bf unimodular}.

\spheader 442Id \dvro{A}{From the definition of $\Delta$, we see
that a} topological group carrying Haar measures is
unimodular iff every left Haar measure is a right Haar measure.

\spheader 442Ie In particular, if a group has any totally finite (left
or right) Haar measure, its left and right Haar measures are the same,
and it has a unique Haar probability measure, which we may call its {\bf
normalized Haar measure}.

In the other direction, any group with its discrete topology is unimodular, since counting measure is a two-sided Haar measure.

\leader{442J}{Proposition} For any topological group carrying Haar
measures, its left modular function is continuous.

\proof{ Let $X$ be a topological group carrying Haar measures, with left
modular function $\Delta$.

{\bf (a)} If $\epsilon>0$, there is an open set $U_{\epsilon}$
containing the identity $e$ of $X$ such that $\Delta(x)\le 1+\epsilon$
for every $x\in U$.   \Prf\ Take any left Haar measure $\mu$ on $X$, and
an open set $G$ such that $0<\mu G<\infty$.   By 442Ab, there are an
open set $H$ and a neighbourhood $U_{\epsilon}$ of the identity such
that $HU_{\epsilon}\subseteq G$ and $\mu G\le(1+\epsilon)\mu H$.   If
$x\in U_{\epsilon}$, then $Hx\subseteq G$, so
$\Delta(x)=\Bover{\mu(Hx)}{\mu H}\le 1+\epsilon$.\ \Qed

\medskip

{\bf (b)} Now, given $x_0\in X$ and $\epsilon>0$, $V=\{x:x^{-1}x_0\in
U_{\epsilon},\,x_0^{-1}x\in U_{\epsilon}\}$ is an open set containing
$x_0$.   If $x\in V$, then

\Centerline{$\Delta(x)
=\Delta(x_0)\Delta(x_0^{-1}x)\le(1+\epsilon)\Delta(x_0)$,}

\Centerline{$\Delta(x_0)
=\Delta(x)\Delta(x^{-1}x_0)\le(1+\epsilon)\Delta(x)$,}

\noindent so

\Centerline{$\Bover1{1+\epsilon}\Delta(x_0)\le\Delta(x)
\le(1+\epsilon)\Delta(x)$.}

\noindent As $\epsilon$ is arbitrary, $\Delta$ is continuous at $x_0$;
as $x_0$ is arbitrary, $\Delta$ is continuous.
}%end of proof of 442J

\leader{442K}{Theorem} Let $X$ be a topological group and $\mu$ a
left Haar measure on $X$.   Let $\Delta$ be the left modular function of
$X$.

(a) $\mu(E^{-1})=\int_E\Delta(x^{-1})\mu(dx)$ for every $E\in\dom\mu$.

(b)(i) $\int f(x^{-1})\mu(dx)=\int\Delta(x^{-1})f(x)\mu(dx)$
whenever $f$ is a real-valued function such that either integral is
defined in $[-\infty,\infty]$;

\quad(ii) $\int f(x)\mu(dx)=\int\Delta(x^{-1})f(x^{-1})\mu(dx)$
whenever $f$ is a real-valued function such that either integral is
defined in $[-\infty,\infty]$.

(c) $\int f(xy)\mu(dx)=\Delta(y^{-1})\int f(x)\mu(dx)$ whenever $y\in X$
and $f$ is a real-valued function such that either integral is defined
in $[-\infty,\infty]$.

\proof{{\bf (a)(i)} Setting $\nu_1 E=\mu E^{-1}$ for Haar measurable
sets $E\subseteq X$, we
know that $\nu_1$ is a right Haar measure, so 442E tells us that

\Centerline{$\mu G\cdot\nu_1 H
=\int_H\nu_1(xG^{-1})\mu(dx)
=\int_H\mu(Gx^{-1})\mu(dx)
=\mu G\int_H\Delta(x^{-1})\mu(dx)$}

\noindent for all open sets $G$, $H\subseteq X$.   Since there is an
open set $G$ such that $0<\mu G<\infty$,
$\mu H^{-1}=\int_H\Delta(x^{-1})\mu(dx)$ for every open set
$H\subseteq X$.

\medskip

\quad{\bf (ii)} Now let $\nu_2$ be the indefinite-integral measure
defined by setting
\discrcenter{390pt}{$\nu_2 E=\int\Delta(x^{-1})\chi E(x)\mu(dx)$ }whenever
this is defined in $[0,\infty]$ (234J\formerly{2{}34B}).
Then $\nu_2$ is effectively
locally finite.   \Prf\ If $\nu_2 E>0$, then $\mu E>0$, so there is an
$n\in\Bbb N$ such that $\mu(E\cap H)>0$, where
$H$ is the open set $\{x:\Delta(x^{-1})<n\}$.   Now there is an open set
$G\subseteq H$
such that $\mu G<\infty$ and $\mu(E\cap G)>0$, in which case
$\nu_2 G\le n\mu G<\infty$ and $\nu_2(E\cap G)>0$.\ \Qed

Accordingly $\nu_2$ is a quasi-Radon measure (415Ob).   Since it agrees
with the quasi-Radon measure $\nu_1$ on open sets, by (i), the two are
equal;  that is, $\mu E^{-1}=\int_E\Delta(x^{-1})\mu(dx)$ for every
$E\in\dom\mu$.

\medskip

{\bf (b)(i)} Apply 235E with $X=Y$, $\Sigma=\Tau=\dom\mu$, $\mu=\nu$ and
$\phi(x)=x^{-1}$, $J(x)=\Delta(x^{-1})$, $g(x)=\Delta(x^{-1})f(x)$ for
$x\in X$.   From (a) we have

\Centerline{$\int J\times\chi(\phi^{-1}[F])d\mu
=\int_{F^{-1}}\Delta(x^{-1})\mu(dx)=\mu F=\nu F$}

\noindent for every $F\in\Tau$ (using 442G to see that
$F^{-1}\in\Sigma$).   So we get

$$\eqalign{\int f(x^{-1})\mu(dx)
&=\int\Delta(x^{-1})g(x^{-1})\mu(dx)
=\int J\times g\phi\,d\mu\cr
&=\int g\,d\nu
=\int\Delta(x^{-1})f(x)\mu(dx)\cr}$$

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

\medskip

\quad{\bf (ii)} Set $\Reverse{f}(x)=f(x^{-1})$ whenever this is defined
(4A5C(c-ii)), and apply (i) to $\Reverse{f}$.

\wheader{442K}{4}{2}{2}{30pt}

{\bf (c)} Similarly, apply 235E with $\mu=\nu$, $\phi(x)=xy$,
$J(x)=\Delta(y)$ for every $x\in X$;  then

\Centerline{$\int J\times\chi(\phi^{-1}[F])d\mu
=\Delta(y)\mu(Fy^{-1})=\mu F$}

\noindent for every $F\in\dom\mu$, so

\Centerline{$\int f(xy)\mu(dx)
=\Delta(y^{-1})\int J\times f\phi\,d\mu
=\Delta(y^{-1})\int f(x)\mu(dx)$.}
}%end of proof of 442K

\leader{442L}{Corollary} Let $X$ be a group carrying Haar measures.   If
$\mu$ is a left Haar measure on $X$ and $\nu$ is a right Haar measure,
then each is an indefinite-integral measure over the other.

\proof{ Let $\Reverse\mu$ be the right Haar measure defined by setting
$\Reverse{\mu} E=\mu E^{-1}$ for every Haar measurable $E\subseteq X$.
Then $\Reverse{\mu} E=\int_E\Delta(x^{-1})\mu(dx)$ for every
$E\in\dom\mu=\dom\Reverse{\mu}$, so $\Reverse{\mu}$ is an
indefinite-integral
measure over $\mu$;  because $\nu$ is a multiple of $\Reverse{\mu}$,
it also
is an indefinite-integral measure over $\mu$.   Similarly, or because
$\Delta$ is strictly positive, $\mu$ is an indefinite-integral measure
over $\nu$.
}%end of proof of 442L

\exercises{\leader{442X}{Basic exercises $\pmb{>}$(a)}
%\spheader 442Xa
Let $X$ and $Y$ be topological groups with (left) Haar probability
measures $\mu$ and $\nu$, and $\phi:X\to Y$ a continuous surjective
group homomorphism.   Show that $\phi$ is \imp\ for $\mu$ and $\nu$.
%442B

\spheader 442Xb
(i) Let $X$ and $Y$ be two topological groups carrying Haar measures.
Show that the product topological group $X\times Y$ (4A5G) carries Haar
measures.
(ii) Let $\familyiI{X_i}$ be any family of topological groups carrying
totally finite Haar measures.   Show that the product group
$\prod_{i\in I}X_i$ carries a totally finite Haar measure.
\Hint{417O.}
%442D

\spheader 442Xc Let $X$ be a subgroup of the group $(\Bbb R,+)$.   Show
that $X$ carries Haar measures iff it is either discrete (so that
counting measure is a Haar measure on $X$) or of full outer Lebesgue
measure (so that the subspace measure on $X$ is a Haar measure).
\Hint{if $G$ has a Haar measure $\nu$ and is not discrete, then
$\nu(G\cap[\alpha,\beta])=(\beta-\alpha)\nu(G\cap[0,1])$ whenever
$\alpha\le\beta$.}   In particular, $\Bbb Q$ does not carry Haar
measures.
%442D

\spheader 442Xd Let $X$ be a topological group carrying Haar measures;
let $\Sigma$ be the algebra of Haar measurable subsets of $X$.   Let
$\mu$ and $\nu$ be any Haar measures on $X$ (either left or right).
Show that a set $E\in\Sigma$ can be covered by a sequence of sets of
finite measure for $\mu$ iff it can be covered by a sequence of sets of
finite measure for $\nu$.
%442H

\spheader 442Xe Let $X$ be a topological group carrying Haar measures
and $\frak A$ its Haar measure algebra (in the sense of 442H).   Show
that we have left, right and conjugacy actions of $X$ on $\frak A$ given
by the formulae $z\action E^{\ssbullet}=(zE)^{\ssbullet}$,
$z\action E^{\ssbullet}=(Ez^{-1})^{\ssbullet}$ and
$z\action E^{\ssbullet}=(zEz^{-1})^{\ssbullet}$ for every Haar
measurable $E\subseteq X$ and every $z\in X$.
%442H

\sqheader 442Xf On $\BbbR^2$ define a binary operation $*$ by setting
$(\xi_1,\xi_2)*(\eta_1,\eta_2)=(\xi_1+\eta_1,\xi_2+e^{\xi_1}\eta_2)$.
(i) Show that $*$ is a group operation under which $\BbbR^2$ is a
locally compact topological group.   (ii) Show that Lebesgue measure
$\mu$ is a right Haar measure for $*$.   (iii) Let $\nu$
be the indefinite-integral measure on $\BbbR^2$ defined by setting $\nu
E=\int_Ee^{-\xi_1}d\xi_1d\xi_2$ for Lebesgue measurable sets
$E\subseteq\BbbR^2$.   Show that $\nu$ is a left Haar measure for $*$.
\Hint{263D.}
(iv) Thus $(\BbbR^2,*)$ is not unimodular.   (v) Show that the left
modular function of $(\BbbR^2,*)$ is $(\xi_1,\xi_2)\mapsto e^{-\xi_1}$.

\sqheader 442Xg Let $X$ be any topological group which is not
unimodular.   (i) Show that there is an open subset of $X$ which is of
finite measure for all left Haar measures on $X$ and of infinite measure
for all right Haar measures.   \Hint{the modular function is unbounded.}
(ii) Let $\mu$ be a left Haar measure on $X$ and $\nu$ a right Haar
measure.   Show that $L^0(\mu)=L^0(\nu)$ and
$L^{\infty}(\mu)=L^{\infty}(\mu)$, but that $L^p(\mu)\ne L^p(\nu)$ for
any $p\in\coint{1,\infty}$.
%442I

\spheader 442Xh Let $X$ and $Y$ be topological groups carrying Haar
measures, with left modular functions $\Delta_X$ and $\Delta_Y$
respectively.   Show that the left modular function of $X\times Y$ is
$(x,y)\mapsto\Delta_X(x)\Delta_Y(y)$.
%442I

\spheader 442Xi Let $X$ be any topological group and
$\Delta:X\to\ooint{0,\infty}$ a group homomorphism such that
$\{x:\Delta(x)\le 1+\epsilon\}$ is a neighbourhood of the identity in
$X$ for every $\epsilon>0$.    Show that $\Delta$ is continuous.
%442J

\spheader 442Xj Let $X$ be a topological group with a right Haar measure
$\nu$ and left modular function $\Delta$.   Show that
$\nu E^{-1}=\int_E\Delta(x)\nu(dx)$ for every Haar measurable set
$E\subseteq X$.
%442K

\leader{442Y}{Further exercises (a)}
%\spheader 442Ya
In 441Yc, show that the only $G$-invariant Radon measures on $\Cal C_s$
are multiples of Hausdorff $s(r-s)$-dimensional measure on $\Cal C_s$.
\Hint{$G$ itself is $\bover{r(r-1)}2$-dimensional (cf.\ 441Ye), and for
any $C\in\Cal C_s$ the stabilizer of $C$ is
$\bover{s(s-1)}2+\bover{(r-s)(r-s-1)}2$-dimensional.
See {\smc Federer 69}, 3.2.28.}
%442B

\spheader 442Yb Let $r\ge 1$, and let $X$ be the group of
non-singular $r\times r$ real matrices.   Regarding $X$ as an open
subset of
$\BbbR^{r^2}$, show that a two-sided Haar measure $\mu$ can be defined
on $X$ by setting $\mu E=\int_E\Bover1{|\det A|^r}\mu_L(dA)$, where
$\mu_L$ is Lebesgue measure on $\BbbR^{r^2}$;  so that $X$ is
unimodular.
%442I

\spheader 442Yc Show that there is a set $A\subseteq[0,1]$, of Lebesgue
outer measure $1$, such that no countable set of translates of $A$
covers any set of Lebesgue measure greater than $0$.   \Hint{let
$\langle F_{\xi}\rangle_{\xi<\frak c}$ run over the uncountable closed
subsets of $[0,1]$ with cofinal repetitions (4A3Fa), and enumerate the
countable subsets of $\Bbb R$ as $\langle I_{\xi}\rangle_{\xi<\frak c}$.
Choose inductively $x_{\xi}$, $x'_{\xi}\in F_{\xi}$ such that
$x_{\xi}\notin\bigcup_{\eta,\zeta<\xi}x'_{\eta}-I_{\zeta}$,
$x'_{\xi}\notin\bigcup_{\eta,\zeta\le\xi}x_{\eta}+I_{\zeta}$;  set
$A=\{x_{\xi}:\xi<\frak c\}$.}   Show that we can extend Lebesgue measure
on $\Bbb R$ to a translation-invariant measure for which $A$ is
negligible.   \Hint{417A.}
%+

\spheader 442Yd\dvAformerly{4{}71Yl}
Let $(X,\rho)$ be a metric space, and $\mu$,
$\nu$ two non-zero quasi-Radon measures on $X$
such that $\mu B(x,\delta)=\mu B(y,\delta)$
and $\nu B(x,\delta)=\nu B(y,\delta)$ for all $\delta>0$ and $x$, $y\in X$.
Show that $\mu$ is a multiple of $\nu$.
%442B out of order query mt42bits
}%end of exercises

\leader{442Z}{Problem} Let $X$ be a compact Hausdorff space, and $G$ the
group of autohomeomorphisms of $X$.   Suppose that $G$ acts transitively
on $X$.   Does it follow that there is at most one $G$-invariant Radon
probability measure on $X$?

\endnotes{
\Notesheader{442} Haar measure dominates the theory of locally compact
topological groups for two reasons:  it is ubiquitous (the existence
theorem, 441E) and essentially defined by the group structure (the
uniqueness theorem, 442B).   I have tried to show that these are rather
different results by setting the theorems out with different hypotheses.
I presented the existence theorem as a special case of 441C, which
demands a locally compact space and a group, but allows them to be
different.
In the uniqueness theorem (roughly following {\smc Halmos 50}) I demand
a group with an invariant quasi-Radon measure, but do not (at this
point) ask for any hypothesis of compactness.   In fact it will become
apparent in the next section that this is a somewhat spurious
generality;  442B and 442I here can be deduced from the traditional
forms in which the group is assumed to be locally compact and Hausdorff.
From the point of view of the topological measure theory to which this
volume is devoted, however, I think the small extra labour involved in
tracing through the arguments without relying on the Riesz
Representation Theorem is instructive.   For instance, it emphasizes
interesting features of the domains and null ideals of Haar
measures (442H).

There is however a more serious question concerning the uniqueness
theorem.   I do not know whether it really belongs to the theory of
topological groups, as described here, or to the theory of group actions
along with 441C.   The trouble is that I know of no example of a
Hausdorff space $X$ and a transitive group $G$ of homeomorphisms of $X$
such that $X$ carries $G$-invariant Radon measures which are not
multiples of each other (see 442Z).   443U and 443Xy below eliminate the
simplest possibilities.
We do need to put some restriction on the measures;  for instance,
counting measure on $\Bbb R$ is translation-invariant, but has nothing
to do with Lebesgue measure.   There are also proper
translation-invariant extensions of Lebesgue measure (442Yc);  for
far-reaching elaborations of this idea see {\smc Hewitt \& Ross 63},
\S16.
}%end of notes

\discrpage

\frfilename{mt443.tex}
\versiondate{14.1.13}
\copyrightdate{2001}

\def\lti{left-\vthsp translation-\vthsp invariant}

\def\chaptername{Topological groups}
\def\sectionname{Further properties of Haar measure}

\newsection{443}

I devote a section to filling in some details of the general theory of
Haar measures before turning to the special topics dealt with in the
rest of the chapter.   The first question concerns the left and right
shift operators acting on sets, on elements of the measure algebra, on
measurable functions and on function spaces.   All these operations can
be regarded as group actions, and, if appropriate topologies are
assigned, they
are continuous actions (443C, 443G).   As an immediate consequence
of this I give an important result about product sets
$\{ab:a\in A,\,b\in B\}$ in a
topological group carrying Haar measures (443D).

The second part of the section revolves around a basic structure
theorem:  all the Haar measures considered here can be reduced to Haar
measures on locally compact Hausdorff groups (443L).   The argument
involves two steps:  the reduction to the Hausdorff case, which is
elementary, and the completion of a Hausdorff topological group.   Since
a group carries more than one natural uniform structure we must take
care to use the correct one, which in this context is the `bilateral'
uniformity (443H-443I, 443K).   On the way I pick up an essential fact
about the approximation of Haar measurable sets by Borel sets (443J).
Finally, I give Halmos' theorem that Haar measures are completion
regular (443M) and a note on the complementary nature of the meager and
null ideals for atomless Haar measure (443O).

In the third part of the section I turn to the special properties of
quotient groups of locally compact groups and the corresponding actions,
following A.Weil.   If $X$ is a locally compact Hausdorff group and $Y$
is a closed subgroup of $X$, then $Y$ is again a locally compact
Hausdorff group, so has Haar measures and a modular function;  at the
same time, we have a natural action of $X$ on the set of left cosets of
$Y$.   It turns out that there is an invariant Radon measure for this
action if and only if the modular function of $Y$ matches that of $X$
(443R).   In this case we can express a left Haar measure of $X$ as an
integral of measures supported by the cosets of $Y$ (443Q).   When $Y$
is a normal subgroup, so that $X/Y$ is itself a locally compact
Hausdorff group, we can relate the modular functions of $X$ and $X/Y$
(443T).   We can apply these results whenever we
have a continuous transitive action of a compact group on a compact
space (443U).


\leader{443A}{Haar measurability}\cmmnt{ I recall and expand on some
facts already noted in 442H.}  Let $X$ be a topological group carrying
Haar measures.

\spheader 443Aa All Haar measures on $X$, whether left or right, have
the same domain
$\Sigma$,\cmmnt{ which I call} the algebra of `Haar measurable' sets,
and the same null ideal $\Cal N$,\cmmnt{ which I call} the ideal
of `Haar negligible' sets.   The corresponding quotient algebra
$\frak A=\Sigma/\Cal N$, the `Haar measure algebra', is the Boolean
algebra underlying
the measure algebra of any Haar measure.   \cmmnt{Because Haar
measures are (by the definition in 441D) quasi-Radon, therefore complete
and strictly localizable
(415A), }$\Sigma_G$ is closed under Souslin's operation\cmmnt{ (431A)} and
$\frak A$ is Dedekind complete\cmmnt{ (322Be)}.   Recall that
any semi-finite measure on $\frak A$\cmmnt{, and in particular any
Haar measure on $X$,} gives rise to the same
measure-algebra topology and uniformity on
$\frak A$\cmmnt{ (324H)}, so we may speak of `the' topology and
uniformity of $\frak A$.

\cmmnt{Because $\Sigma$ is the domain of a left Haar measure,}
$xE\in\Sigma$ whenever $E\in\Sigma$ and $x\in X$;  \cmmnt{because
$\Sigma$ is the domain of a right Haar measure,} $Ex\in\Sigma$ whenever
$E\in\Sigma$ and $x\in X$.
\cmmnt{Similarly,} $xE$ and $Ex$ are Haar negligible whenever $E$ is
Haar negligible and $x\in X$.
\cmmnt{Moreover,} $E^{-1}\cmmnt{\mskip5mu =\{x^{-1}:x\in E\}}$ is
Haar measurable or Haar negligible whenever $E$ is.

Note that $\Sigma$ and $\Cal N$ are invariant in the strong sense that if
$\phi:X\to X$ is any group automorphism which
is also a homeomorphism, then $\Sigma=\{\phi[E]:E\in\Sigma\}$ and
$\Cal N=\{\phi[E]:E\in\Cal N\}$.   \prooflet{\Prf\ If $\mu$ is a left
Haar measure on $X$, let $\nu$ be the image measure $\mu\phi^{-1}$.
Because $\phi$ is a homeomorphism, $\nu$ is a non-zero quasi-Radon measure.
If $\nu$ measures $E$ and $a\in X$, then

\Centerline{$\nu(aE)=\mu\phi^{-1}[aE]=\mu((\phi^{-1}a)(\phi^{-1}[E]))
=\mu\phi^{-1}[E]=\nu E$.}

\noindent So $\nu$ is again a left Haar measure, and has domain $\Sigma$
and null ideal $\Cal N$.   But
$\dom\nu=\{E:\phi^{-1}[E]\in\Sigma\}=\{\phi[E]:E\in\Sigma\}$ and
$\nu^{-1}[\{0\}]=\{E:\phi^{-1}[E]\in\Cal N\}=\{\phi[E]:E\in\Cal N$.\ \Qed}

\spheader 443Ab We\cmmnt{ even} have a symmetric notion of `measurable
envelope' in $X$:  for any $A\subseteq X$, there is a Haar measurable
set $E\supseteq A$ such that $\mu(E\cap F)=\mu^*(A\cap F)$ for any Haar
measurable $F\subseteq X$ and any Haar measure $\mu$ on $X$.
\prooflet{\Prf\ Start with a fixed Haar measure $\mu_0$.   Then $A$ has
a measurable envelope $E$ for $\mu_0$, by 213J and 213L.
Now to say that `$E$
is a measurable envelope for $A$' means just that (i)
$A\subseteq E\in\Sigma$ (ii) if $F\in\Sigma$ and $F\subseteq E\setminus A$
then
$F\in\Cal N$, so $E$ is also a measurable envelope for $A$ for any other
Haar measure on $X$.\ \Qed
}%end of prooflet

In this context I will call $E$ a {\bf Haar measurable envelope} of $A$.

\spheader 443Ac Similarly, we have a notion of {\bf full outer Haar
measure}:  a subset $A$ of $X$ is of full outer Haar measure if $X$ is a
Haar measurable envelope of $A$\cmmnt{, that is, $A\cap E\ne\emptyset$
whenever $E$ is a Haar measurable set which is not Haar negligible, that
is, $A$ is of full outer measure for any (left or right) Haar measure on
$X$}.

\spheader 443Ad For any Haar measure $\mu$ on $X$, we can identify
$L^{\infty}(\mu)$ with $L^{\infty}(\frak A)$\cmmnt{ (363I)} and
$L^0(\mu)$ with $L^0(\frak A)$\cmmnt{ (364Ic)}.
Thus these constructions are independent of $\mu$.   The topology of
convergence in measure of $L^0$ is\cmmnt{ determined by its Riesz
space structure (367T);  so that this also is} independent of the
particular Haar measure we may select.     Of course the same is true of
the norm of $L^{\infty}$.   \cmmnt{Note however that the spaces $L^p$,
for $1\le p<\infty$, are different for left and right Haar measures on
any group which is not unimodular (442Xg), and that even for left
Haar measures $\mu$ the norm on $L^p(\mu)$ changes if $\mu$ is
re-normalized.}

\spheader 443Ae \cmmnt{When it seems appropriate,} I will use the
phrases {\bf
Haar measurable function}, meaning a function measurable with respect to
the $\sigma$-algebra of Haar measurable sets, and {\bf Haar almost
everywhere},
meaning `on the complement of a Haar negligible set'.   Note that we can
identify $L^0(\frak A)$ with the set of equivalence classes in the space
$\eusm L^0$, where $\eusm L^0$ is the space of Haar measurable
real-valued functions defined Haar-a.e.\ in $X$, and $f\sim g$ if $f=g$
Haar-a.e.   \cmmnt{In the language of \S241,
$\eusm L^0=\eusm L^0(\mu)$ for any Haar measure $\mu$ on $X$.}

\spheader 443Af \dvro{We}{Because $E^{-1}\in\Sigma$ for every
$E\in\Sigma$, and $E^{-1}\in\Cal N$ whenever $E\in\Cal N$, we} have a
canonical automorphism $a\mapsto\Reverse{a}:\frak A\to\frak A$ defined by
writing $(E^{\ssbullet})\ssplrarrow=(E^{-1})^{\ssbullet}$ for every
$E\in\Sigma$.   Being an automorphism, this must be a homeomorphism for
the measure-algebra topology of $\frak A$\cmmnt{ (324G)}.
\dvro{If}{In the same way, if} $f\in\eusm L^0$ then
$\Reverse{f}\in\eusm L^0$, where $\Reverse{f}(x)=f(x^{-1})$ whenever this
is defined\cmmnt{ (4A5C(c-ii)), since
$\{x:x\in\dom\Reverse{f},\,\Reverse{f}(x)>\alpha\}
=\{x:x\in\dom f,\,f(x)>\alpha\}^{-1}$ belongs to $\Sigma$ for every
$\alpha$};  and we can define an $f$-algebra automorphism
$u\mapsto\Reverse{u}:L^0\to L^0$ by saying that
$(f^{\ssbullet})\ssplrarrow=(\Reverse{f})^{\ssbullet}$ for $f\in\eusm L^0$.   If
we identify $L^0$ with $L^0(\frak A)$\cmmnt{ rather than with a set of
equivalence classes in $\eusm L^0$}, then we can define the map
$u\mapsto\Reverse{u}$ as the Riesz homomorphism associated with the Boolean
homomorphism $a\mapsto\Reverse{a}:\frak A\to\frak A$\cmmnt{, as in
364P}.
\cmmnt{Note that $\|\Reverse{u}\|_{\infty}=\|u\|_{\infty}$ for every
$u\in L^{\infty}$, but that (unless $X$ is unimodular) other $L^p$
spaces are not invariant under the involution $\ssplrarrow$ (442Xg again).}

\spheader 443Ag If $X$ carries any totally finite\cmmnt{ (left or
right)} Haar measure, it is unimodular, and has a
unique, two-sided, Haar probability measure\cmmnt{ (442Ie)}.
\cmmnt{(In particular, this is
the case whenever $X$ is compact.)}   For such
groups we have $L^p$-spaces, for $1\le p\le\infty$, defined by the group
structure, with canonical norms.

\leader{443B}{Lemma} Let $X$ be a topological group and $\mu$ a left
Haar measure on $X$.   If $E\subseteq X$ is measurable and
$\mu E<\infty$, then for any $\epsilon>0$ there is a neighbourhood $U$
of the
identity $e$ such that $\mu(E\symmdiff xEy)\le\epsilon$ whenever $x$,
$y\in U$.

\proof{ Set $\delta=\Bover{\min(1,\epsilon)}{10+3\mu E}>0$.
Write $\Cal U$ for the
family of open neighbourhoods of $e$.   Because $\mu$ is effectively
locally finite, there is an open set $G_0$ of finite measure such that
$\mu(E\setminus G_0)\le\delta$.   Let $F\subseteq G_0\setminus E$ be a
closed set such that $\mu F\ge\mu(G_0\setminus E)-\delta$, and set
$G=G_0\setminus F$, so that $\mu(G\setminus E)\le\delta$
and $\mu(E\setminus G)\le\delta$.   For
$U\in\Cal U$ set $H_U=\interior\{x:UxU\subseteq G\}$.   Then $\Cal
H=\{H_U:U\in\Cal U\}$ is upwards-directed, and has union $G$, because if
$x\in G$ there is a $U\in\Cal U$ such that $UxUU\subseteq G$, so that
$x\in H_U$.   So there is a $V\in\Cal U$ such that
$\mu(G\setminus H_V)\le\delta$.
Recall that the left modular function $\Delta$ of $X$ is continuous
(442J).   So there is a $U\in\Cal U$ such that $U\subseteq V$ and
$|\Delta(y)-1|\le\delta$ for every $y\in U$.

Now suppose that $x$, $y\in U$.   Set $E_1=E\cap H_V$.   Then
$xE_1y\subseteq G$, so

\Centerline{$\mu(E_1\cup xE_1y)\le\mu G\le\mu E+\delta$.}

\noindent   On the other hand,

\Centerline{$\mu E_1\ge\mu E-\mu(E\setminus G)-\mu(G\setminus H_V)
\ge\mu E-2\delta$,}

\Centerline{$\mu(xE_1y)=\Delta(y)\mu E_1
\ge(1-\delta)(\mu E-2\delta)\ge\mu E-(2+\mu E)\delta$.}

\noindent So

\Centerline{$\mu(E\cap xEy)\ge\mu(E_1\cap xE_1y)
=\mu E_1+\mu(xE_1y)-\mu(E_1\cup xE_1y)
\ge\mu E-(5+\mu E)\delta$.}

\noindent At the same time,

\Centerline{$\mu(xEy)=\Delta(y)\mu E\le(1+\delta)\mu E$.}

\noindent So

\Centerline{$\mu(E\symmdiff xEy)=\mu E+\mu(xEy)-2\mu(E\cap xEy)
\le(10+3\mu E)\delta\le\epsilon$,}

\noindent as required.
}%end of proof of 443B

\leader{443C}{Theorem}  Let $X$ be a topological group carrying Haar
measures, and $\frak A$ its Haar measure algebra.   Then we have
continuous actions of $X$ on $\frak A$ defined by writing

\Centerline{$x\action_lE^{\ssbullet}=(xE)^{\ssbullet}$,
\quad$x\action_rE^{\ssbullet}=(Ex^{-1})^{\ssbullet}$,
\quad$x\action_cE^{\ssbullet}=(xEx^{-1})^{\ssbullet}$}

\noindent for Haar measurable sets $E\subseteq X$ and $x\in X$.

\proof{{\bf (a)} The functions $\action_l$, $\action_r$ and $\action_c$
are all well defined because the maps $E\mapsto xE$, $E\mapsto Ex^{-1}$ and
$E\mapsto xEx^{-1}$ are all Boolean automorphisms of the algebra
$\Sigma$ of Haar measurable sets preserving the ideal of Haar negligible
sets (442G).   It is elementary to check that they are actions of
$X$ on $\frak A$.

Fix a left Haar measure $\mu$ on $X$ and let $\bar\mu$ be the
corresponding measure on $\frak A$.   Then the topology of $\frak A$ is
defined by the pseudometrics $\rho_a$, for $\bar\mu a<\infty$, where
$\rho_a(b,c)=\bar\mu(a\Bcap(b\Bsymmdiff c))$.

\medskip

{\bf (b)} Now suppose that $x_0\in X$, $b_0\in\frak A$, $\bar\mu
a<\infty$ and $\epsilon>0$.   Let $E$, $F_0\in\Sigma$ be such that
$E^{\ssbullet}=a$ and $F_0^{\ssbullet}=b_0$;  set
$\delta=\bover14\epsilon>0$.   Note that

\Centerline{$\mu(x_0^{-1}E\cap F_0)\le\mu(x_0^{-1}E)=\mu E<\infty$.}

\noindent Let $U$ be a neighbourhood of the identity $e$ such that

\Centerline{$\mu(E\symmdiff yE)\le\delta$,
\quad$\mu((x_0^{-1}E\cap F_0)\symmdiff y(x_0^{-1}E\cap F_0))\le\delta$}

\noindent whenever $y\in U$ (443B).   Set

\Centerline{$a'=x_0^{-1}\action_la=(x_0^{-1}E)^{\ssbullet}$.}

\noindent Now suppose that
$x\in Ux_0\cap x_0U^{-1}$ and that $\rho_{a'}(b,b_0)\le\delta$.   Then
$\rho_a(x\action_lb,x_0\action_lb_0)\le\epsilon$.   \Prf\ Let
$F\in\Sigma$ be such that $F^{\ssbullet}=b$.   Then $xx_0^{-1}$ and
$x^{-1}x_0$ both belong to $U$, so

$$\eqalignno{\rho_a(x\action_lb,x_0\action_lb_0)
&=\mu(E\cap(xF\symmdiff x_0F_0))\cr
&=\mu(E\cap xF)
    +\mu(E\cap x_0F_0)
    -2\mu(E\cap xF\cap x_0F_0)\cr
&=\mu(x^{-1}E\cap F)
    +\mu(E\cap x_0F_0)
    -2\mu(x^{-1}x_0(x_0^{-1}E\cap F_0)\cap F)\cr
&\le\mu(x_0^{-1}E\cap F)
    +\mu(x^{-1}E\symmdiff x_0^{-1}E)
    +\mu(x_0^{-1}E\cap F_0)\cr
  &\qquad -2\mu(x_0^{-1}E\cap F_0\cap F)
    +2\mu(x^{-1}x_0(x_0^{-1}E\cap F_0)\symmdiff(x_0^{-1}E\cap F_0))\cr
&\le\mu(x_0^{-1}E\cap(F\symmdiff F_0))
    +\mu(E\symmdiff xx_0^{-1}E)
    +2\delta\cr
&\le\delta+\delta+2\delta
=4\delta=\epsilon. \text{ \Qed}\cr}$$

As $x_0$, $b_0$, $a$ and $\epsilon$ are arbitrary, $\action_l$ is
continuous.

\medskip

{\bf (c)} The same arguments, using a right Haar measure to provide
pseudometrics defining the topology of $\frak A$, show that $\action_r$
is continuous.   (Or use the method of 443X(d-ii).)

\medskip

{\bf (d)} Accordingly the map $(x,y,a)\mapsto x\action_l(y\action_ra)$
is continuous.   So

\Centerline{$(x,a)\mapsto x\action_l(x\action_ra)=x\action_ca$}

\noindent is continuous.
}%end of proof of 443C

\leader{443D}{Proposition} Let $X$ be a topological group carrying Haar
measures.
If $E\subseteq X$ is Haar measurable but not Haar negligible, and
$A\subseteq X$ is not Haar negligible, then

(a) there are $x$, $y\in X$ such that $A\cap xE$, $A\cap Ey$ are not
Haar negligible;

(b) $EA$ and $AE$ both have non-empty interior;

(c) $E^{-1}E$ and $EE^{-1}$ are neighbourhoods of the identity.

\proof{{\bf (a)(i)} Let $\mu$ be any left Haar measure on $X$, and for
Borel sets $F\subseteq X$ set

\Centerline{$\nu F=\sup\{\mu(F\cap IE):I\subseteq X$ is finite$\}$.}

\noindent It is easy to check that $\nu$ is an effectively locally
finite $\tau$-additive Borel measure, inner regular with respect to the
closed sets, because $\{IE:I\in[X]^{<\omega}\}$ is upwards-directed and
$\mu$ is quasi-Radon.   Moreover,

\Centerline{$\nu(xF)=\sup_{I\in[X]^{<\omega}}\mu(xF\cap IE)
=\sup_{I\in[X]^{<\omega}}\mu(F\cap x^{-1}IE)
=\nu F$}

\noindent for every Borel set $F\subseteq X$ and every $x\in X$.
Accordingly the c.l.d.\ version $\tilde\nu$ of $\nu$ is a
left-translation-invariant quasi-Radon measure on $X$ (415Cb);  since
$\nu E>0$, $\tilde\nu$ is non-zero and is itself a left Haar measure.
Consequently $A$ is not
$\tilde\nu$-negligible.   Let $H$ be a measurable envelope of $A$ for
Haar measure (443Ab).   Then $H$ is not Haar negligible, so there is a
closed set $F\subseteq H$ which is not Haar negligible, and
$\nu F=\tilde\nu F>0$.   Thus there is an $x\in X$ such that

\Centerline{$0<\mu(F\cap xE)\le\mu(H\cap xE)=\mu^*(A\cap xE)$,}

\noindent and $A\cap xE$ is not Haar negligible.

\medskip

\quad{\bf (ii)} Applying the same arguments, but starting with a right
Haar measure $\mu$, we see that there is a $y\in X$ such that $A\cap Ey$
is not Haar negligible.

\medskip

{\bf (b)} Let $\mu$ be a left Haar measure on $X$, and $F$ a Haar
measurable envelope of $A$.   The function
$x\mapsto(xE^{-1})^{\ssbullet}:X\to\frak A$ is continuous, where
$\frak A$ is the Haar measure algebra of $X$ (443C), so

\Centerline{$H=\{x:\mu^*(A\cap xE^{-1})>0\}
=\{x:F^{\ssbullet}\cap(xE^{-1})^{\ssbullet}\ne 0\}$}

\noindent is open.   Now

\Centerline{$H\subseteq\{x:A\cap xE^{-1}\ne\emptyset\}=AE$,}

\noindent so $H\subseteq\interior AE$;  and $E^{-1}$ is Haar measurable
and not Haar negligible, so $H\ne\emptyset$, by (a).   Thus $\interior
AE\ne\emptyset$.

Similarly, using a right Haar measure (or observing that
$EA=(A^{-1}E^{-1})^{-1}$), we see that $EA$ has non-empty interior.

\medskip

{\bf (c)} Again taking a left Haar measure $\mu$, $\mu$ is semi-finite, so
there is an $F\subseteq E$ such that $0<\mu F<\infty$.   By 443B, there is
a neighbourhood $U$ of the identity such that $\mu(F\symmdiff xFy)<\mu F$
for all $x$, $y\in U$.   In particular, if $x\in U$,
$\mu(F\setminus xF)<\mu F$, so $F\cap xF\ne\emptyset$ and
$x\in FF^{-1}\subseteq EE^{-1}$;  at the same time,
$\mu(F\setminus Fx)<\mu F$, $F\cap Fx\ne\emptyset$ and
$x\in F^{-1}F\subseteq E^{-1}E$.   So $E^{-1}E$ and $EE^{-1}$ both include
$U$ and are neighbourhoods of the identity.
}%end of proof of 443D

%look:  really need to know that if $E$ has non-zero inner measure,
%$D$ is dense then $D+E$ is conegligible

\vleader{72pt}{443E}{Corollary} Let $X$ be a Hausdorff topological group
carrying Haar measures.   Then the following are equiveridical:

(i) $X$ is locally compact;

(ii) every Haar measure on $X$ is a Radon measure;

(iii) there is some compact subset of $X$ which is not Haar negligible.

\proof{{\bf (i)$\Rightarrow$(ii)} Haar measures are locally finite
quasi-Radon measures (441D, 442Aa), so on locally compact Hausdorff
spaces must be Radon measures (416G).

\medskip

{\bf (ii)$\Rightarrow$(iii)} is obvious, just because
Haar measures are non-zero and any Radon measure
is tight (that is, inner regular with respect to the
closed compact sets).

\medskip

{\bf (iii)$\Rightarrow$(i)} If $K\subseteq X$ is a compact set which is
not Haar negligible, then $KK$ is a compact set with non-empty interior,
so $X$ is locally compact (4A5Eg).
}%end of proof of 443E

\leader{443F}{}\cmmnt{ Later in the chapter we shall need the
following straightforward fact.

\medskip

\noindent}{\bf Lemma} Let $X$ be a topological group carrying Haar
measures, and $Y$ an open subgroup of $X$.   If $\mu$ is a left Haar
measure on $X$, then the subspace measure $\mu_Y$ is a left Haar measure
on $Y$.   Consequently a subset of $Y$ is Haar measurable or Haar
negligible, when regarded as a subset of the topological group $Y$, iff
it is Haar measurable or Haar negligible when regarded as a subset of
the topological group $X$.

\proof{ By 415B, $\mu_Y$ is a quasi-Radon measure;  because $\mu$ is
strictly positive, $\mu_Y$ is non-zero, and it is easy to check that
it is \lti.   So
it is a Haar measure on $Y$.   The rest follows at once from 442H/443A.
}%end of proof of 443F

\leader{443G}{}\cmmnt{ We can repeat the ideas of 443C in terms of
function spaces, as follows.

\medskip

\noindent}{\bf Theorem} Let $X$ be a topological group with a left Haar
measure $\mu$.   Let $\Sigma$ be the domain of $\mu$,
$\eusm L^0=\eusm L^0(\mu)$ the space of $\Sigma$-measurable real-valued
functions defined
almost everywhere in $X$, and $L^0=L^0(\mu)$ the corresponding space of
equivalence classes\cmmnt{ (\S241)}.

(a) $a\action_lf$, $a\action_rf$ and
$a\action_cf$\cmmnt{ (definitions:  4A5C(c-ii))} belong to $\eusm L^0$ for
every $f\in\eusm L^0$ and $a\in X$.

(b) If $a\in X$, then
$\esssup|a\action_lf|=\esssup|a\action_rf|=\esssup|f|$ for every
$f\in\eusm L^{\infty}=\eusm L^{\infty}(\mu)$\cmmnt{, where
$\esssup|f|$ is the essential supremum of $|f|$ (243Da)}.   For
$1\le p<\infty$, $\|a\action_lf\|_p=\|f\|_p$ and
$\|a\action_rf\|_p=\Delta(a)^{-1/p}\|f\|_p$ for every
$f\in\eusm L^p=\eusm L^p(\mu)$, where $\Delta$ is the left modular
function of $X$.

(c) We have shift actions of $X$ on $L^0$ defined by setting

\Centerline{$a\action_lf^{\ssbullet}=(a\action_lf)^{\ssbullet}$,
\quad$a\action_rf^{\ssbullet}=(a\action_rf)^{\ssbullet}$,
\quad$a\action_cf^{\ssbullet}=(a\action_cf)^{\ssbullet}$}

\noindent for $a\in X$ and $f\in\eusm L^0$.   If
$\Reverse{\phantom{x}}$ is
the reversal operator on $L^0$ defined in 443Af, we have

\Centerline{$a\action_l\Reverse{u}=(a\action_ru)\ssplrarrow$,
\quad$a\action_c\Reverse{u}=(a\action_cu)\ssplrarrow$}

\noindent for every $a\in X$ and $u\in L^0$.

(d)\dvAnew{2010} If we give $L^0$ its
topology of convergence in measure these three actions, and also the
reversal operator $\ssplrarrow$, are continuous.

(e) For $1\le p\le\infty$ the formulae of (c) define actions of $X$ on
$L^p=L^p(\mu)$, and $\|a\action_lu\|_p=\|u\|_p$ for every $u\in L^p$,
$a\in X$;  interpreting $\Delta(a)^{-1/\infty}$ as $1$ if necessary,
$\|a\action_ru\|_p=\Delta(a)^{-1/p}\|u\|_p$ whenever $u\in L^p$ and
$a\in X$.

(f)\dvAformerly{4{}43Ge}
For $1\le p<\infty$ these actions are continuous.

\proof{{\bf (a)} Let $f\in\eusm L^0$.   Then $F=\dom f$ is conegligible,
so $aF=\dom a\action_lf$ and $Fa^{-1}=\dom a\action_rf$ are conegligible
(442G).
For any $\alpha\in\Bbb R$, set $E_{\alpha}=\{x:x\in F,\,f(x)<\alpha\}$;
then $\{x:(a\action_lf)(x)<\alpha\}=aE_{\alpha}$ and
$\{x:(a\action_rf)(x)<\alpha\}=E_{\alpha}a^{-1}$ are measurable, so
$a\action_lf$ and
$a\action_rf$ are measurable.   Thus $a\action_lf$ and $a\action_rf$
belong to $\eusm L^0$.   It follows at once that
$a\action_cf=a\action_l(a\action_rf)$ belongs to $\eusm L^0$.

\medskip

{\bf (b)(i)} For $\alpha\ge 0$,

$$\eqalign{\esssup|f|\le\alpha
&\iff|f(x)|\le\alpha\text{ for almost all }x\cr
&\iff|(a\action_lf)(x)|\le\alpha\text{ for almost all }x\cr
&\iff|(a\action_rf)(x)|\le\alpha\text{ for almost all }x\cr}$$

\noindent because the null ideal of $\mu$ is invariant under both
left and right translations.   So
$\esssup|f|=\esssup|a\action_lf|=\esssup|a\action_rf|$.

\medskip

\quad{\bf (ii)} For $1\le p<\infty$,

$$\eqalignno{\|a\action_lf\|_p^p
&=\int|(a\action_lf)(x)|^p\mu(dx)
=\int|f(a^{-1}x)|^p\mu(dx)
=\int|f(x)|^p\mu(dx)\cr
\noalign{\noindent (441J)}
&=\|f\|_p^p,\cr
\|a\action_rf\|_p^p
&=\int|(a\action_rf)(x)|^p\mu(dx)
=\int|f(xa)|^p\mu(dx)
=\Delta(a^{-1})\int|f(x)|^p\mu(dx)\cr
\noalign{\noindent (442Kc)}
&=(\Delta(a)^{-1/p}\|f\|_p)^p.\cr}$$

\medskip

{\bf (c)(i)} I have already checked that $a\action_lf$, $a\action_rf$
and $a\action_cf$ belong to
$\eusm L^0$ whenever $f\in\eusm L^0$ and $a\in X$.   If $f$,
$g\in\eusm L^0$ and $f\eae g$, let $E$ be the conegligible set
$\{x:x\in\dom f\cap\dom g,\,f(x)=g(x)\}$;  then $aE$ and $Ea^{-1}$ and
$aEa^{-1}$ are conegligible and

\Centerline{$(a\action_lf)(x)=(a\action_lg)(x)$ for every $x\in aE$,
\quad$(a\action_rf)(x)=(a\action_rg)(x)$ for every $x\in Ea^{-1}$,}

\Centerline{$(a\action_cf)(x)=(a\action_cg)(x)$ for every
$x\in aEa^{-1}$,}

\noindent so $a\action_lf\eae a\action_lg$,
$a\action_rf\eae a\action_rg$ and
$a\action_cf\eae a\action_cg$.   Accordingly the formulae given
define functions $\action_l$, $\action_r$ and $\action_c$ from
$X\times L^0$ to $L^0$.   They are actions just because the original
$\action_l$, $\action_r$ and $\action_c$ are actions of $X$ on
$\eusm L^0$ (4A5Cc-4A5Cd).

\medskip

\quad{\bf (ii)} If $f\in\eusm L^0$, then

\Centerline{$(a\action_l\Reverse{f})(x)
=\Reverse{f}(a^{-1}x)
=f(x^{-1}a)
=(a\action_rf)(x^{-1})
=(a\action_rf)\ssplrarrow(x)$}

\noindent when any of these is defined, which is almost everywhere, so
$a\action_lu=(a\action_ru)\ssplrarrow$ for every $u\in L^0$.
Similarly,

\Centerline{$(a\action_c\Reverse{f})(x)
=\Reverse{f}(a^{-1}xa)
=f(a^{-1}x^{-1}a)
=(a\action_cf)(x^{-1})
=(a\action_cf)\ssplrarrow(x)$}

\noindent and $a\action_c\Reverse{u}=(a\action_cu)\ssplrarrow$.

\medskip

{\bf (d)(i)} In 367T there is a description of convergence in measure on
$L^0$ in terms of its Riesz space structure.   As $\ssplrarrow$ is a Riesz
space automorphism of $L^0$, it must also be a homeomorphism for the
topology of convergence in measure.

\medskip

\quad{\bf (ii)} To see that $\action_l$ is continuous,
it will be convenient to work
with the space $\eusm L^0_{\Sigma}$ of Haar measurable
real-valued functions
defined on the whole of $X$.   I will use a characterization of
convergence in measure from 245F:  a subset $W$ of $L^0$ is open iff
whenever $f_0^{\ssbullet}\in W$ there are a set $E$ of finite measure
and an $\epsilon>0$ such that $f^{\ssbullet}\in W$ whenever
$\mu\{x:x\in E,\,|f(x)-f_0(x)|>\epsilon\}\le\epsilon$.   Now if $E$ is a
measurable
set of finite measure, $f\in\eusm L^0_{\Sigma}$ and $\epsilon>0$,
there is a neighbourhood $U$ of the identity $e$ of $X$ such that
$\mu\{x:x\in E,\,|f(ax)-f(x)|\ge\epsilon\}\le\epsilon$ for every $a\in U$.
\Prf\ Let $m\ge 1$ be such that
$\mu\{x:x\in E,\,|f(x)|\ge m\epsilon\}\le\bover12\epsilon$.   For
$-m\le k<m$, set
$E_k=\{x:x\in E,\,k\epsilon\le f(x)<(k+1)\epsilon\}$.   By 443B, there
is a neighbourhood $U$ of $e$ such that

\Centerline{$\mu(E_k\symmdiff a^{-1}E_k)\le\Bover{\epsilon}{4m}$}

\noindent whenever $a\in U$ and $-m\le k<m$.   Now, for $a\in U$,

\Centerline{$\{x:x\in E,\,|f(ax)-f(x)|\ge\epsilon\}
\subseteq\{x:x\in E,\,|f(x)|\ge m\epsilon\}
  \cup\bigcup_{k=-m}^{m-1}(E_k\symmdiff a^{-1}E_k)$}

\noindent has measure at most

\Centerline{$\Bover{\epsilon}2+2m\Bover{\epsilon}{4m}=\epsilon$.  \Qed}

\medskip

\quad{\bf (iii)} Let $E$ be a measurable set of finite measure, $a_0\in X$,
$f_0\in\eusm L^0_{\Sigma}$ and $\epsilon>0$.
Set $\delta=\epsilon/3>0$.   Note that $\mu(a_0^{-1}E)=\mu E$ is finite.
Let $U$ be a neighbourhood of $e$ such that

\Centerline{$\mu\{x:x\in a_0^{-1}E,\,
|f_0(yx)-f_0(x)|\ge\delta\}\le\delta$,
\quad$\mu(yE\symmdiff E)\le\delta$}

\noindent whenever $y\in U$.

Now suppose that $a\in Ua_0\cap a_0U^{-1}$
and that $f\in\eusm L^0_{\Sigma}$ is such that
$\mu\{x:x\in a_0^{-1}E,\,|f(x)-f_0(x)|\ge\delta\}\le\delta$.
In this case,

$$\eqalign{\{x:x&\in E,\,|f(a^{-1}x)-f_0(a_0^{-1}x)|\ge\epsilon\}\cr
&\subseteq\{x:x\in E,\,|f(a^{-1}x)-f_0(a^{-1}x)|\ge\delta\}\cr
  &\qquad\qquad  \cup\{x:x\in E,\,
     |f_0(a^{-1}x)-f_0(a_0^{-1}x)|\ge\delta\}\cr
&\subseteq(E\symmdiff aa_0^{-1}E)
  \cup\{x:x\in aa_0^{-1}E,\,
     |f(a^{-1}x)-f_0(a^{-1}x)|\ge\delta\}\cr
  &\qquad\qquad  \cup a_0\{w:w\in a_0^{-1}E,\,
     |f_0(a^{-1}a_0w)-f_0(w)|\ge\delta\}\cr
&\subseteq(E\symmdiff aa_0^{-1}E)
  \cup a\{w:w\in a_0^{-1}E,\,\,
     |f(w)-f_0(w)|\ge\delta\}\cr
  &\qquad\qquad  \cup a_0\{w:w\in a_0^{-1}E,\,
     |f_0(a^{-1}a_0w)-f_0(w)|\ge\delta\}\cr}$$

\noindent has measure at most $3\delta=\epsilon$
because $aa_0^{-1}$, $e$ and $a^{-1}a_0$ all belong to $U$.
Because $E$ and $\epsilon$ are arbitrary, the function
$(a,u)\mapsto a\action_lu$ is continuous at
$(a_0,f_0^{\ssbullet})$;
as $a_0$ and $f_0$ are arbitrary, $\action_l$ is continuous.

\medskip

\quad{\bf (iv)} Now

\Centerline{$(a,u)\mapsto a\action_ru
=(a\action_l\Reverse{u})\ssplrarrow$}

\noindent must also be continuous.   It follows at once that $\action_c$
is continuous, since
$a\action_cu=a\action_l(a\action_ru)$.

\medskip

{\bf (e)} follows at once from (b) and (c).

\medskip

{\bf (f)} Fix $p\in\coint{1,\infty}$.

\medskip

\quad{\bf (i)} If $u\in L^p$ and $\epsilon>0$, there is a neighbourhood
$U$ of $e$ such that
$\|u-y\action_l(z\action_ru)\|_p\le\epsilon$ whenever $y$, $z\in U$.
\Prf\ When $u$ is of the form $(\chi E)^{\ssbullet}$, where
$\mu E<\infty$, we have

\Centerline{$y\action_l(z\action_ru)=\chi(yEz^{-1})^{\ssbullet}$,
\quad$\|u-y\action_l(z\action_ru)\|_p=\mu(E\symmdiff yEz^{-1})^{1/p}$,}

\noindent so the result is immediate from 443B.   If
$u=\sum_{i=0}^n\alpha_i(\chi E_i)^{\ssbullet}$, where every $E_i$ has
finite measure, then, setting $u_i=(\chi E_i)^{\ssbullet}$ for each $i$,

\Centerline{$\|u-y\action_l(z\action_ru)\|_p
\le\sum_{i=0}^n|\alpha_i|\|u_i-y\action_l(z\action_ru_i)\|_p
\le\epsilon$}

\noindent whenever $y$ and $z$ are close enough to $e$.
In general, there is a
$v$ of this form such that $\|u-v\|_p\le\bover14\epsilon$.   If we take
a neighbourhood $U$ of $e$ such that
$\|v-y\action_l(z\action_rv)\|_p\le\bover14\epsilon$ and
$\Delta(z)^{-1/p}\le 2$ whenever $y$, $z\in U$, then

\Centerline{$\|y\action_l(z\action_ru)-y\action_l(z\action_rv)\|_p
=\Delta(z)^{-1/p}\|u-v\|_p\le \Bover12\epsilon$}

\noindent whenever $z\in U$, so

\Centerline{$\|u-y\action_l(z\action_ru)\|_p
\le\|u-v\|_p+\|v-y\action_l(z\action_r v)\|_p
  +\|y\action_l(z\action_rv)-y\action_l(z\action_ru)\|_p
\le\epsilon$}

\noindent whenever $y$, $z\in U$.\ \Qed

\medskip

\quad{\bf (ii)} Now suppose that $u_0\in L^p$, $a_0$, $b_0\in X$ and
$\epsilon>0$.   Set $v_0=a_0\action_l(b_0\action_ru_0)$ and
$\delta=\epsilon/(1+2\Delta(b_0)^{-1/p})>0$.   Let $U$ be a
neighbourhood of $e$ such that
$\Delta(y)^{-1/p}\le 2$ and
$\|v_0-y\action_l(z\action_rv_0)\|_p\le\delta$ whenever $y$, $z\in U$.
If $a\in Ua_0$, $b\in Ub_0$ and $\|u-u_0\|_p\le\delta$, then

$$\eqalign{\|a\action_l(b\action_ru)-v_0\|_p
&\le\|a\action_l(b\action_ru)-a\action_l(b\action_ru_0)\|_p
 +\|a\action_l(b\action_ru_0)-v_0\|_p\cr
&=\Delta(b)^{-1/p}\|u-u_0\|_p
  +\|aa_0^{-1}\action_l(bb_0^{-1}\action_rv_0)
       -v_0\|_p\cr
&\le\Delta(bb^{-1}_0)^{-1/p}\Delta(b_0)^{-1/p}\delta
  +\delta
\le\delta(1+2\Delta(b_0)^{-1/p})
=\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$(a,b,u)\mapsto a\action_l(b\action_ru)$ is continuous at
$(a_0,b_0,u_0)$.

As in (c), this is enough to show that $\action_l$, $\action_r$ and
$\action_c$ are all continuous actions.
}%end of proof of 443G

\cmmnt{\medskip

\noindent{\bf Remark} I have written this out for a left Haar measure
$\mu$, since the spaces $L^p(\mu)$ depend on this;  if $\nu$ is a right
Haar measure, and $X$ is not unimodular, then $L^p(\mu)\ne L^p(\nu)$ for
$1\le p<\infty$.   But recall that the topology of convergence in
measure on $L^0$ is the same for all Haar measures (443Ad),
so (c) above, and the case $p=\infty$ of (b) and (d), are two-sided;
they belong to the theory of the Haar measure algebra.
}%end of comment

\leader{443H}{Theorem} Let $X$ be a topological group carrying Haar
measures.   Then there is a neighbourhood of the identity which is
totally bounded for the bilateral uniformity on $X$.

\proof{ Let $\mu$ be a left Haar measure on $X$.   Let $V_0$ be a
neighbourhood of the identity $e$ such that $\mu V_0<\infty$ (442Aa).
Let $V$ be a neighbourhood of $e$ such that $VV\subseteq V_0$ and
$V^{-1}=V$.

\Quer\ Suppose, if possible, that $V$ is not totally bounded for the
bilateral uniformity on $X$.   By 4A5Oa, one of the following must
occur:

\medskip

{\bf case 1} There is an open neighbourhood $U$ of $e$ such that
$V\not\subseteq IU$ for any finite set $I\subseteq X$.   In this case,
we may choose a sequence $\sequencen{x_n}$ in $V$ inductively such that
$x_n\notin x_iU$ whenever $i<n$.   Let $U_1$ be an open neighbourhood of
$e$ such that $U_1\subseteq V$ and $U_1U_1^{-1}\subseteq U$;  then
$\sequencen{x_nU_1}$ is disjoint.   Since $\mu(x_nU_1)=\mu U_1>0$ for
every $n$ (by the other clause in 442Aa),
$\mu(\bigcup_{n\in\Bbb N}x_nU_1)=\infty$;  but
$x_nU_1\subseteq V_0$ for every $n$, so this is impossible.

\medskip

{\bf case 2} There is an open neighbourhood $U$ of $e$ such that
$V\not\subseteq UI$ for any finite set $I\subseteq X$.   So we may
choose a sequence $\sequencen{x_n}$ in $V$ inductively such that
$x_n\notin Ux_i$ whenever $i<n$.   Let $U_1$ be an open neighbourhood of
$e$ such that $U_1\subseteq V$ and $U_1U_1^{-1}\subseteq U$;  then
$\sequencen{U_1^{-1}x_n}$ is disjoint, so $\sequencen{x_n^{-1}U_1}$ is
also disjoint.   Since $\mu(x_n^{-1}U_1)=\mu U_1>0$ for every $n$,
$\mu(\bigcup_{n\in\Bbb N}x_n^{-1}U_1)=\infty$;  but
$x_n^{-1}U_1\subseteq V_0$ for every $n$, so this also is
impossible.\ \Bang

Thus $V$ is totally bounded for the bilateral uniformity on $X$, and we
have the required totally bounded neighbourhood of $e$.
}%end of proof of 443H

\leader{443I}{Corollary}
Let $X$ be a topological group.   If $A\subseteq X$
is totally bounded for the bilateral
uniformity of $X$, it has finite outer measure for any (left or right)
Haar measure on $X$.

\proof{ If $\mu$ is a Haar measure on $X$, let $U$ be an open
neighbourhood of the identity $e$ of finite measure.   There is a finite
set $I$ such that $A\subseteq IU\cap UI$ (4A5Oa again), so that
$\mu^*A\le\#(I)\mu U$ is finite.
}%end of proof of 443I

\vleader{48pt}{443J}{Proposition}
Let $X$ be a topological group carrying Haar measures, and $\frak A$ its
Haar measure algebra.

(a) There is an open-and-closed subgroup $Y$ of $X$ such that, for any
Haar measure $\mu$ on $X$, $Y$ can be covered by countably many open
sets of finite measure.

(b)(i) If $E\subseteq X$ is any Haar measurable set, there are an
F$_{\sigma}$ set $E'\subseteq E$ and a G$_{\delta}$ set $E''\supseteq E$
such that $E''\setminus E'$ is Haar negligible.

\quad(ii) Every Haar negligible set is included in a Haar negligible
Borel set, and for every Haar measurable set $E$ there is a Borel set
$F$ such that $E\symmdiff F$ is Haar negligible.

\quad(iii) The Haar measure algebra $\frak A$ of $X$ may be identified
with $\Cal B/\Cal I$, where $\Cal B$ is the Borel $\sigma$-algebra of
$X$ and
$\Cal I$ is the ideal of Haar negligible Borel sets.

\quad(iv) Every member of $L^0(\frak A)$ can be identified with the
equivalence class of some Borel measurable function from $X$ to
$\Bbb R$.
Every member of $L^{\infty}(\frak A)$ can be identified with the
equivalence class of a bounded Borel measurable function from $X$ to
$\Bbb R$.

\proof{{\bf (a)} Let $V$ be an open neighbourhood of the identity which
is totally bounded for the bilateral uniformity of $X$ (443H);  we may
suppose that $V^{-1}=V$.   Set $Y=V\cup VV\cup VVV\cup VVVV\cup\ldots$.
Then $Y$ is an open subgroup of $X$, therefore also closed (4A5Ek).
By 4A5Ob, every power of $V$ is totally bounded, so $Y$ is a countable
union of totally bounded sets.   If $\mu$
is any left Haar measure on $X$, then any totally bounded set has finite
outer measure for $\mu$ (443I).   Thus $Y$ is a countable union of
sets of finite measure for $\mu$.   The same argument applies to right
Haar measures, so $Y$ is a subgroup of the required form.

\medskip

{\bf (b)(i)} Let $E\subseteq X$ be a Haar measurable set, and fix a left
Haar measure $\mu$ on $X$.   Take the open subgroup $Y$ of (a), and
index the set of its left cosets as $\familyiI{Y_i}$;  because any
translate of a totally bounded set is totally bounded (4A5Ob again), each
$Y_i$ is an open set expressible as $\bigcup_{n\in\Bbb N}H_{in}$, where
every $H_{in}$ is a totally bounded open set, so that $\mu H_{in}$ is
finite.

For $i\in I$ and $m$, $n\in\Bbb N$ there is a closed set
$F_{imn}\subseteq E\cap H_{im}$ such that
$\mu F_{imn}\ge\mu(E\cap H_{im})-2^{-n}$.   Set
$F_{mn}=\bigcup_{i\in I}F_{imn}$ for each $m$, $n\in\Bbb N$;   then
$F_{mn}$ is closed (4A2Bb).   So $E'=\bigcup_{m,n\in\Bbb N}F_{mn}$ is
F$_{\sigma}$.   For each $i\in I$,

\Centerline{$(E\setminus E')\cap Y_i
\subseteq\bigcup_{m\in\Bbb N}
\bigl(E\cap H_{im}\setminus\bigcup_{n\in\Bbb N}F_{imn}\bigr)$}

\noindent is negligible;  thus
$\{G:G\subseteq X$ is open, $\mu(G\cap E\setminus E')=0\}$ covers $X$
and $E\setminus E'$ must be negligible (414Ea).

In the same way, there is an F$_{\sigma}$ set $F^*\subseteq X\setminus
E$ such that $(X\setminus E)\setminus F^*$ is negligible;  now
$E''=X\setminus F^*$ is G$_{\delta}$ and $E''\setminus E$ is negligible,
so $E''\setminus E'$ also is.   (I am speaking here as if `negligible'
meant `$\mu$-negligible'.   But of course this is the same thing as the
`Haar negligible' of the statement of the proposition.)

\medskip

\quad{\bf (ii), (iii), (iv)} follow at once.
}%end of proof of 443J

\vleader{48pt}{443K}{Theorem}
Let $X$ be a Hausdorff topological group carrying
Haar measures.   Then the completion $\widehat{X}$ of $X$ under its
bilateral uniformity is a locally compact Hausdorff group, and $X$ is of
full outer Haar measure in $\widehat{X}$.   Any\cmmnt{ (left or
right)} Haar
measure on $X$ is the subspace measure corresponding to a Haar measure
\cmmnt{(of the same chirality)} on $\widehat{X}$.

\proof{{\bf (a)} By 443H and 4A5N,
$\widehat{X}$ is a locally compact Hausdorff
group in which $X$ is embedded as a dense subgroup.

\medskip

{\bf (b)} Let $\mu$ be a left Haar measure on $X$.   Then there is a
Radon measure $\lambda$ on $\widehat{X}$ such that $\mu$ is the
subspace measure $\lambda_X$.   \Prf\ For Borel sets
$E\subseteq\widehat{X}$, set $\nu E=\mu(X\cap E)$.   Then $\nu$ is a
Borel measure, and it is $\tau$-additive because $\mu$ is.
Any point of $\widehat{X}$
has a compact neighbourhood $V$ in $\widehat{X}$;  now $V$ must be
totally bounded
for the bilateral uniformity of $\widehat{X}$ (4A2Je), so $V\cap X$ is
totally bounded for the bilateral uniformity of $X$ (4A5Ma), and
$\nu V=\mu(V\cap X)$ is
finite (443I).   Thus $\nu$ is locally finite.   If $\nu E>0$, there is
an open set $H\subseteq X$ such that $\mu H<\infty$ and
$\mu(H\cap X\cap E)>0$, because $\mu$ is effectively locally finite;
now there is an
open set $G\subseteq\widehat{X}$ such that $H=X\cap G$, so that
$\nu G<\infty$ and $\nu(G\cap E)>0$.   Thus $\nu$ is effectively locally
finite.

By 416H, the c.l.d.\ version $\lambda$ of $\nu$ is a Radon measure on
$\widehat{X}$.   Since $\lambda K=\nu K=0$ whenever
$K\subseteq\widehat{X}\setminus X$ is compact, $X$ is of full outer
measure for $\lambda$.   Accordingly

\Centerline{$\lambda_X(G\cap X)=\lambda G=\nu G=\mu(G\cap X)$}

\noindent for every open set $G\subseteq\widehat{X}$, and
$\lambda_X=\mu$, because they are quasi-Radon measures agreeing on the
open sets (415B, 415H(iii)).\ \Qed

\medskip

{\bf (c)} Continuing the argument of (b), $\lambda$ is a left Haar
measure on $\widehat{X}$.   \Prf\ Let $G\subseteq\widehat{X}$ be open,
and $z\in\widehat{X}$.   If $K\subseteq zG$ is compact, then
$z^{-1}K\subseteq G$, and $\{w:w\in\widehat{X},\,wK\subseteq G\}$ is a
non-empty open set (4A5Ei), so meets $X$.   Take $x\in X$ such that
$xK\subseteq G$;  then

\Centerline{$\lambda G=\mu(X\cap G)\ge\mu(X\cap xK)
=\mu(x(X\cap K))=\mu(X\cap K)=\lambda K$.}

\noindent As $K$ is arbitrary, $\lambda(zG)\le\lambda G$.   By
441Ba, $\lambda$ is invariant under the left action of $\widehat{X}$
on itself, that is, is a left Haar measure.\ \Qed

We know that $X$ is of full outer measure for $\lambda$, so this shows
that it has full outer Haar measure in $\widehat{X}$.

\medskip

{\bf (d)} The same arguments, looking at $Gz$ and $Kz^{-1}$ in (c), show
that if $\mu$ is a right Haar measure on $X$ it is the subspace measure
$\lambda_X$ for a right Haar measure $\lambda$ on $\widehat{X}$.
}%end of proof of 443K

\leader{443L}{Corollary}\discrversionA{\footnote{Expanded 2009.}}{}
Let $X$ be any topological group with a Haar
measure $\mu$.   Then we can find $Z$, $\lambda$ and $\phi$ such that

(i) $Z$ is a locally compact Hausdorff topological group;

(ii) $\lambda$ is a Haar measure on $Z$;

(iii) $\phi:X\to Z$ is a continuous homomorphism,
\imp\ for $\mu$ and $\lambda$;

(iv) $\mu$ is inner regular with respect to
$\{\phi^{-1}[K]:K\subseteq Z$ is compact$\}$;

(v) if $E\subseteq X$ is Haar measurable, we can find a Haar measurable set
$F\subseteq Z$ such that $\phi^{-1}[F]\subseteq E$ and
$E\setminus\phi^{-1}[F]$ is Haar negligible;

(vi) a set $G\subseteq X$ is an open set in $X$ iff it is of the
form $\phi^{-1}[H]$ for some open set $H\subseteq Z$;

(vii) a set $G\subseteq X$ is a regular open set in $X$ iff it is of the
form $\phi^{-1}[H]$ for some regular open set $H\subseteq Z$;

(viii) a set $A\subseteq X$ is nowhere dense in $X$ iff $\phi[A]$ is
nowhere dense in $Z$.

%\query How far are (Z,\lambda,\phi)  unique?
%indeed:  if  G  is the subgroup of  \Aut\frak A  generated by the
%action of  X , how far does  (\Aut\frak A,G)  determine a
%locally compact Hausdorff topological group  X ?

\proof{{\bf (a)} Let $Y\subseteq X$ be the closure of $\{e\}$, where $e$
is the identity of $X$.   Then $Y$ is a closed normal subgroup of $X$,
and if $\phi:X\to X/Y$ is the quotient map, every open (or closed)
subset of $X$ is of the form $\phi^{-1}[H]$ for some open (or closed)
set $H\subseteq Y$ (4A5Kb).

Consider the image measure $\nu=\mu\phi^{-1}$ on $X/Y$.   This is
quasi-Radon.   \Prf\ Because $\mu$ is a complete $\tau$-additive
topological measure, so is $\nu$.   If $F\in\dom\nu$ and $\nu F>0$,
there is an open set $G\subseteq X$ such that $\mu G<\infty$ and
$\mu(G\cap\phi^{-1}[F])>0$;  now $G=\phi^{-1}[H]$ for some open set
$H\subseteq X/Y$, and $\nu H=\mu G$ is finite, while $\nu(H\cap
F)=\mu(G\cap\phi^{-1}[F])>0$.   Thus $\nu$ is effectively locally finite
(therefore semi-finite).   Again, if $F\in\dom\nu$ and $\nu F>\gamma$,
there is a closed set $E\subseteq\phi^{-1}[F]$ such that $\mu
E\ge\gamma$;  now $E$ is expressible as $\phi^{-1}[H]$ for some closed
set $H\subseteq X/Y$;  because $\phi$ is surjective, $H\subseteq F$, and
$\nu H=\mu E\ge\gamma$.   Thus $\nu$ is inner regular with respect to
the closed sets.   Finally, suppose that $F\subseteq X/Y$ is such that
$F\cap F'\in\dom\nu$ whenever $\nu F'<\infty$.  If $E\subseteq X$ is a
closed set of finite measure, it is of the form $\phi^{-1}[F']$ where
$\nu F'=\mu E<\infty$, so $F'\cap F\in\dom\nu$ and
$E\cap\phi^{-1}[F]\in\dom\mu$;  by 412Ja, we can conclude that
$\phi^{-1}[F]\in\dom\mu$ and $F\in\dom\nu$.   Thus $\nu$ is locally
determined and is a quasi-Radon measure.\ \Qed

We find also that $\nu$ is a left Haar measure.   \Prf\ If $z\in X/Y$
and $F\in\dom\nu$, express $z$ as $\phi(x)$ where $x\in X$;  then
$\phi^{-1}[zF]=x\phi^{-1}[F]$, so

\Centerline{$\nu(zF)=\mu(x\phi^{-1}[F])=\mu\phi^{-1}[F]=\nu F$. \Qed}

\medskip

{\bf (b)} Thus $X/Y$ is a topological group with a left Haar measure
$\nu$.   Because $Y$ is closed, $X/Y$ is Hausdorff (4A5J(b-ii-$\alpha$)).
We can
therefore form its completion $Z=\widehat{X/Y}$, a locally compact
Hausdorff group, and find a left Haar measure $\lambda$ on $Z$ such
that $\nu$ is the corresponding subspace measure on $X/Y$, which is of
full outer measure for $\lambda$ (443K).
The embedding $X/Y\embedsinto Z$ is
therefore \imp\ for $\nu$ and $\lambda$, so that $\phi$, regarded as a
map from $X$ to $Z$, is \imp\ for $\mu$ and $\lambda$.   Also, of
course, $\phi:X\to Z$ is a continuous homomorphism.

If $E\subseteq X$ and $\mu E>\gamma$, there is a closed set
$E'\subseteq E$ such that $\mu E'>\gamma$.
Now $E'$ is of the form $\phi^{-1}[F]$
where $F\subseteq X/Y$ is closed and $\nu F=\mu E'>\gamma$.   Next, $F$
is of the form $(X/Y)\cap F'$ where $F'\subseteq Z$ is closed and
$\lambda F'=\nu F>\gamma$.   So there is a compact set $K\subseteq F'$
such that $\lambda K\ge\gamma$, and we have

\Centerline{$\phi^{-1}[K]\subseteq\phi^{-1}[F']
=\phi^{-1}[F]\subseteq E$,
\quad$\mu\phi^{-1}[K]=\nu(K\cap(X/Y))=\lambda K\ge\gamma$.}

\noindent As $E$ and $\gamma$ are arbitrary, $\mu$ is inner regular with
respect to $\{\phi^{-1}[K]:K\subseteq Z$ is compact$\}$.
So (i)-(iv) are true.

\medskip

{\bf (c)} If $E\subseteq X$ is Haar measurable, then by 443J(b-i) there
is an F$_{\sigma}$ set $E'\subseteq E$ such that $E\setminus E'$ is Haar
negligible.   Now there are an F$_{\sigma}$ set $H\subseteq X/Y$ such
that $E'=\phi^{-1}[H]$ and an F$_{\sigma}$ set $F\subseteq Z$ such that
$H=(X/Y)\cap F$, in which case $E'=\phi^{-1}[F]$, and
$E\setminus\phi^{-1}[F]$ is Haar negligible.   This deals with (v).

\medskip

{\bf (d)} Concerning (vi)-(viii), we just have to put 4A2B and 4A5Kb
together.   If $G$, $A\subseteq X$, then

$$\eqalignno{G\text{ is open in }X
&\iff\text{ there is an open }V\subseteq X/Y
  \text{ such that }G=\phi^{-1}[V]\cr
&\iff\text{ there are a }V\subseteq X/Y\text{ and an open }H\subseteq Z\cr
&\mskip100mu
  \text{ such that }G=\phi^{-1}[V]\text{ and }V=(X/Y)\cap H\cr
&\iff\text{ there is an open }H\subseteq Z
  \text{ such that }G=\phi^{-1}[H];}$$

$$\eqalignno{G\text{ is a regular open set in }X
&\iff\phi[G]\text{ is a regular open subset of }X/Y\cr
\displaycause{4A5K(b-iii), 4A2B(f-iii)}
&\iff\text{ there is a regular open }H\subseteq Z\cr
&\mskip100mu
  \text{ such that }\phi[G]=(X/Y)\cap H\cr
\displaycause{4A2B(j-ii)}
&\iff\text{ there is a regular open }H\subseteq Z\cr
&\mskip100mu
  \text{ such that }G=\phi^{-1}[H];}$$

$$\eqalignno{A\text{ is nowhere dense in }X
&\iff\phi[A]\text{ is nowhere dense in }X/Y\cr
\displaycause{4A5K(iv)-(v)}
&\iff\phi[A]\text{ is nowhere dense in }Z\cr}$$

\noindent(4A2B(j-i)).
}%end of proof of 443L

\leader{443M}{Theorem}\cmmnt{ ({\smc Halmos 50})} Let $X$ be a
topological group and $\mu$ a Haar measure on $X$.   Then $\mu$ is
completion regular.

\proof{{\bf (a)} Suppose first that $\mu$ is a left Haar measure and
that $X$ is locally compact and Hausdorff.   In this case any
self-supporting compact set $K\subseteq X$ is a zero set.   \Prf\ For
each $n\in\Bbb N$, there is an open neighbourhood $U_n$ of the identity
such that $\mu(K\symmdiff xK)\le 2^{-n}$ for every $x\in U_n$ (443B);
we may suppose that $\overline{U}_{\mskip-3mu n+1}\subseteq U_n$
for each $n$.
Each set $U_nK$ is open (4A5Ed), so $\bigcap_{n\in\Bbb N}U_nK$ is a
G$_{\delta}$
set.   \Quer\ If $K\ne\bigcap_{n\in\Bbb N}U_nK$, there is an
$x\in\bigcap_{n\in\Bbb N}U_nK\setminus K$.   For each $n\in\Bbb N$,
there are $y_n\in U_n$, $z_n\in K$ such that $x=y_nz_n$.   Let $\Cal F$
be any non-principal ultrafilter on $\Bbb N$.   Then
$z=\lim_{n\to\Cal F}z_n$ is defined in $K$, so

\Centerline{$xz^{-1}=\lim_{n\to\Cal F}xz_n^{-1}=\lim_{n\to\Cal F}y_n$}

\noindent is defined in $X$;  because $y_n\in\overline{U}_{\mskip-3mu i}$
for every $i\le n$,

\Centerline{$xz^{-1}
\in\bigcap_{i\in\Bbb N}\overline{U}_{\mskip-3mu i}
=\bigcap_{i\in\Bbb N}U_i$.}

\noindent Consequently $\mu(xz^{-1}K\symmdiff K)=0$;  because $\mu$ is
left-translation-invariant, $\mu(K\setminus zx^{-1}K)=0$.   But as
$x\notin K$, $z\in K\setminus zx^{-1}K$ and $K\cap(X\setminus zx^{-1}K)$
is non-empty.   And $zx^{-1}K$ is closed, so $X\setminus zx^{-1}K$ is
open and $K$ is not self-supporting, contrary to hypothesis.\ \Bang

Thus $K=\bigcap_{n\in\Bbb N}U_nK$ is a G$_{\delta}$ set.   Being a
compact G$_{\delta}$ set in a completely regular Hausdorff space, it is
a zero set (4A2F(h-v)).\ \Qed

Since $\mu$ is surely inner regular with respect to the compact
self-supporting sets (414F), it is inner regular with respect to the
zero sets, and is completion regular.

\medskip

{\bf (b)} Now suppose that $\mu$ is a left Haar measure on an arbitrary
topological group $X$.   By 443L, we can find a locally compact
Hausdorff topological group $Z$, a continuous homomorphism $\phi:X\to Z$
and a left Haar measure $\lambda$ on $Z$ such that $\phi$ is \imp\ for
$\mu$ and $\lambda$ and $\mu$ is inner regular with respect to
$\{\phi^{-1}[K]:K\subseteq Z$ is compact$\}$.   Now if $E\in\dom\mu$ and
$\gamma<\mu E$, there is a compact set $K\subseteq Z$ such that
$\phi^{-1}[K]\subseteq E$ and $\nu K>\gamma$.   Next, there is a zero
set $L\subseteq K$ such that $\nu L\ge\gamma$;  in which case
$\phi^{-1}[L]\subseteq E$ is a zero set and $\mu\phi^{-1}[L]\ge\gamma$.
Thus $\mu$ is inner regular with respect to the zero sets and is
completion regular.

\medskip

{\bf (c)} Finally, if $\mu$ is a right Haar measure on a topological
group $X$, let $\Reverse{\mu}$ be the corresponding left Haar measure, setting
$\Reverse{\mu} E=\mu E^{-1}$ for Haar measurable sets $E$.   Then
$\Reverse{\mu}$ is
inner regular with respect to the zero sets;  because
$x\mapsto x^{-1}:X\to X$ is a homeomorphism, so is $\mu$.
}%end of proof of 443M

\leader{443N}{}\cmmnt{ I give a simple result showing how the
measure-theoretic properties of groups carrying Haar measures have
topological consequences which might not be expected.

\medskip

\noindent}{\bf Proposition} Let $X$ be a topological group carrying
Haar measures\cmmnt{ (for instance, $X$ might be any locally
compact Hausdorff group)}.

(i) Let $G$ be a regular open subset of $X$.   Then $G$ is a cozero set.

(ii) Let $F$ be a nowhere dense subset of $X$.   Then $F$ is included in
a nowhere dense zero set.

\proof{{\bf (a)} Suppose to begin with that $X$ is locally compact,
$\sigma$-compact and Hausdorff.   Let $\mu$ be a left
Haar measure on $X$;  then $\mu$ is $\sigma$-finite, because $X$ is
covered by a sequence of compact sets, which must all have finite
measure.

\medskip

\quad{\bf (i)} Write $\Cal U$ for the
family of open neighbourhoods of
the identity $e$ of $X$.   For each $U\in\Cal U$, set
$H_U=\interior\{x:xU\subseteq G\}$;   then $\{H_U:U\in\Cal U\}$ is an
upwards-directed family of open sets with union $G$, as in the proofs of
442Ab and 442B, so $G^{\ssbullet}=\sup_{U\in\Cal U}H_U^{\ssbullet}$ in
the measure algebra $\frak A$ of $\mu$.   Because $\mu$ is
$\sigma$-finite, $\frak A$ is ccc (322G) and there is a sequence
$\sequencen{U_n}$ in $\Cal U$ such that
$G^{\ssbullet}=\sup_{n\in\Bbb N}H_{U_n}^{\ssbullet}$ (316E).   In this
case, $G\setminus\bigcup_{n\in\Bbb N}H_{U_n}$ is negligible, so must
have empty interior.

By 4A5S, there is a closed normal subgroup $Y$ of $X$, included in
$\bigcap_{n\in\Bbb N}U_n$, such that $X/Y$ is metrizable.   Let
$\pi:X\to X/Y$ be the canonical map.

For each $n\in\Bbb N$, $Q_n=\pi[H_{U_n}]$ is open (4A5J(a-i)), and

\Centerline{$H_{U_n}\subseteq\pi^{-1}[Q_n]=H_{U_n}Y
\subseteq H_{U_n}U_n\subseteq G$.}

\noindent So

\Centerline{$\overline{G}
=\overline{\bigcupop_{n\in\Bbb N}H_{U_n}}
=\overline{\bigcupop_{n\in\Bbb N}\pi^{-1}[Q_n]}$.}

\noindent Setting $Q=\interior\overline{\bigcup_{n\in\Bbb N}Q_n}$, and
using 4A5J(a-i) and 4A2B(f-ii), we see that

\Centerline{$\pi^{-1}[Q]
=\interior\overline{\bigcupop_{n\in\Bbb N}\pi^{-1}[Q_n]}
=\interior\overline{G}=G$}

\noindent (this is where I use the hypothesis that $G$ is a regular open
set).   But $Q$, being an open set in a metrizable space, is a cozero
set (4A2Lc), so $G=\pi^{-1}[Q]$ is a cozero set (4A2C(b-iv)), as
required by (i).

\medskip

\quad{\bf (ii)} Now consider the nowhere dense set $F\subseteq X$.
This
time, let $\Cal G$ be a maximal disjoint family of cozero subsets of
$X\setminus F$.   Then $\Cal G$ is countable, again because $\mu$ is
$\sigma$-finite, and $\bigcup\Cal G$ is dense, because the topology of
$X$ is completely regular.   So $X\setminus\bigcup\Cal G$ is a nowhere
dense zero set including $F$.

\medskip

{\bf (b)} Next, suppose that $X$ is any locally compact Hausdorff
topological group.   Then $X$ has a $\sigma$-compact open subgroup $X_0$
(4A5El).   By (a), any
regular open set in $X_0$ is a cozero set in $X_0$.   The same applies
to all the (left) cosets of $X_0$, because these are homeomorphic to
$X_0$.

If $C$ is any coset
of $X_0$, then $G\cap C$ is a regular open set in $C$, so is a cozero
set in $C$.   But as the left cosets of $X_0$ form a partition of $X$
into open sets, $G$ is also a cozero set in $X$ (4A2C(b-vii)).

Similarly, $F\cap C$ is nowhere dense in $C$ for every left coset $C$ of
$X_0$, so is included in a nowhere dense zero set in $C$, and the union
of these will be a nowhere dense zero set in $X$ including $F$.

\medskip

{\bf (c)} Now suppose that $X$ is any group carrying Haar
measures.   Let $Z$, $\lambda$ and $\phi:X\to Z$ be as in 443L.
Then $G$ is expressible as $\phi^{-1}[H]$ for some regular open set
$H\subseteq Z$ (443L(vii));  by (b), $H$ is a cozero set, so $G$ also is
(4A2C(b-iv) again).   As for $F$, $\phi[F]$ is nowhere dense
in $Z$, by 443L(viii).   Let $F'\supseteq\phi[F]$ be a nowhere dense zero
set;  then $\phi^{-1}[F']\supseteq F$ is a zero set, and
$\phi[\phi^{-1}[F']]\subseteq F'$ is nowhere dense, so $\phi^{-1}[F']$ is
nowhere dense.
}%end of proof of 443N

\leader{443O}{}\discrversionA{\footnote{New 2009.}}
{}\cmmnt{ An expected result, well known for Lebesgue measure,
but which seems to need a little attention for the non-metrizable case, is
the following.

\medskip

\noindent}{\bf Proposition} Let $X$ be a topological group and $\mu$ a left
Haar measure on $X$.   Then the following are equiveridical:

(i) $\mu$ is not purely atomic;

(ii) $\mu$ is atomless;

(iii) there is a non-negligible nowhere dense subset of $X$;

(iv) $\mu$ is inner regular with respect to the nowhere dense sets;

(v) there is a conegligible meager subset of $X$;

(vi) there is a negligible comeager subset of $X$.

\noindent If $X$ is Hausdorff, we can add

(vii) the topology of $X$ is not discrete.

\proof{ Write $\Sigma$ for the domain of $\mu$.

\medskip

{\bf (a)(i)$\Rightarrow$(ii)} If $\mu$ is not purely atomic, let
$E\in\Sigma$ be a non-negligible set not including any atom.   If
$F\in\Sigma$ is any other non-negligible set, then there is an $x\in X$
such that $F\cap xE$ is not negligible, by 443Da.
Now the subspace measures on
$F\cap xE$ and $x^{-1}F\cap E$ are isomorphic, and the latter is atomless,
so $F\cap xE$ is not an atom and $F$ is not an atom.   As $F$ is arbitrary,
$\mu$ is atomless.

\medskip

{\bf (b)(iii)$\Rightarrow$(iv)} The argument is similar.   Suppose that $A$
is a non-negligible nowhere dense subset of $X$;  then $E=\overline{A}$ is
a non-negligible closed nowhere dense set.   If $F\in\Sigma$ is
non-negligible, there is an $x\in X$ such that $F\cap xE$ is
non-negligible;  as $y\mapsto xy:X\to X$ is a homeomorphism, $xE$ and
$F\cap xE$ are nowhere dense.   Thus every non-negligible measurable set
includes a nowhere dense non-negligible measurable set;  as the family of
nowhere dense sets is an ideal, $\mu$ is inner regular with respect to the
nowhere dense sets (412Aa).

\medskip

{\bf (c)(iv)$\Rightarrow$(i)} Suppose that $\mu$ is inner regular with
respect to the nowhere dense sets.
Let $U$ be an open neighbourhood of the identity $e$ of $X$ with
finite measure (442Aa once more),
and $V$ an open neighbourhood of $e$ such that
$VV^{-1}V\subseteq U$.   Then $\mu V>0$, by the other half of 442Aa.

\Quer\ If there is a $\mu$-atom $E\subseteq V$, let $F_0\subseteq E$ be a
non-negligible measurable nowhere dense set, $F_1\subseteq F_0$ a
non-negligible closed set and $F\subseteq F_1$ a non-empty self-supporting
closed set (414F).   Because $y\mapsto xy$ is a measure-preserving
automorphism, $xF$ is a self-supporting closed set, and an atom for $\mu$,
for every
$x\in X$.   So if $x$, $y\in X$ and $xF\cap yF$ is non-negligible,
then $xF\setminus yF$ is negligible and $xF\subseteq yF$;  similarly,
$yF\subseteq xF$, so $xF=yF$.   Now no finite number of translates of
the nowhere dense set $F$
can cover the non-empty open set $V$, while
$V\subseteq\bigcup_{x\in X}xF$, so we must have a sequence
$\sequencen{x_n}$ in $X$ such that
$x_nF\cap V\setminus\bigcup_{i<n}x_iF\ne\emptyset$ for every $n\in\Bbb N$.
In this case, $x_iF\cap x_nF$ is negligible whenever $i<n$, so

\Centerline{$\mu(\bigcup_{n\in\Bbb N}x_nF)
=\sum_{n\in\Bbb N}\mu(x_nF)=\infty$.}

\noindent However $V\cap x_nF\ne\emptyset$, so
$x_n\in VF^{-1}\subseteq VV^{-1}$ and $x_nF\subseteq VV^{-1}V\subseteq U$
for every $n\in\Bbb N$;  and $U$ is supposed to have finite measure.\ \Bang

Accordingly $V$ does not include any atom and $\mu$ cannot be purely
atomic.

\medskip

{\bf (d)(iv)$\Rightarrow$(v)} Suppose that $\mu$ is inner regular with
respect to the nowhere dense sets.   Let $\Cal G$ be a maximal disjoint
family of open sets of non-zero finite measure.   Then
$\interior(X\setminus\bigcup\Cal G)$ is negligible, because $\mu$ is
effectively locally finite, so must be empty, and
$X\setminus\bigcup\Cal G$ is nowhere dense.   For each $G\in\Cal G$, let
$\sequencen{F_{Gn}}$ be a sequence of nowhere dense measurable
subsets of $G$ such that $\mu G=\lim_{n\to\infty}\mu F_{Gn}$.   For
$n\in\Bbb N$, set $A_n=\bigcup_{G\in\Cal G}F_{Gn}$.   Then $A_n$ is nowhere
dense.   \Prf\ If $H\subseteq X$ is open and not empty, either
$H\cap A_n=\emptyset$ or there is a $G\in\Cal G$ such that
$H\cap G\ne\emptyset$, in which case
$H\cap G\setminus\overline{F}_{\mskip-3mu Gn}\subseteq H\setminus A_n$
is open and non-empty.\ \QeD\   So
$D=(X\setminus\bigcup\Cal G)
  \cup\bigcup_{n\in\Bbb N}\overline{A}_{\mskip-3mu n}$ is
meager.   \Quer\ If $D$ is not conegligible, let $H$ be an open set of
finite measure such that $H\setminus D$ is non-negligible.   As
$H\setminus D\subseteq\bigcup\Cal G$, there is a $G\in\Cal G$ such that
$\mu(H\cap G\setminus D)>0$;  but
$H\cap G\setminus D\subseteq G\setminus\bigcup_{n\in\Bbb N}F_{Gn}$.\ \Bang

Thus $D$ is a conegligible meager set.

\medskip

{\bf (e)(v)$\Leftrightarrow$(vi)} The complement of a witness for (v)
witnesses (vi), and conversely.

\medskip

{\bf (f)(v)$\Rightarrow$(iii)} is elementary, since $\mu$ is non-zero.

\medskip

{\bf (g)(ii)$\Rightarrow$(iii)} Suppose that $\mu$ is atomless.
I proceed through an expanding series of special cases,
as in 443N.

\medskip

\quad\grheada\ Suppose to begin with that
$X$ is locally compact, Hausdorff and
$\sigma$-compact.   In this case $X$ has a closed negligible normal
subgroup $Y$ such that $X/Y$ is separable and metrizable.   \Prf\ Since
$\mu$ is atomless, $\{e\}$ must be negligible.   Since $\mu$ is locally
finite and inner regular with respect to the closed sets,
there must be a sequence $\sequencen{U_n}$ of open neighbourhoods of $e$
such that $\inf_{n\in\Bbb N}\mu U_n=0$.   By 4A5S again,
there is a closed normal
subgroup $Y\subseteq\bigcap_{n\in\Bbb N}U_n$ such that $X/Y$ is metrizable,
and of course $Y$ is negligible.   Since the canonical map from $X$ onto
$X/Y$ is continuous (4A5J(a-i) again),
$X/Y$ is $\sigma$-compact, therefore separable (4A2Hd, 4A2Pd).\ \Qed

Write $\pi:X\to X/Y$ for the canonical map.
Let $D\subseteq X/Y$ be a countable dense set, and consider $\pi^{-1}[D]$.
This is a countable union of
translates of $Y$, so is negligible;  let
$F\subseteq X\setminus\pi^{-1}[D]$
be a closed non-negligible set.   Then $\pi[F]$ does not meet $D$.
Because $\pi$ is an open mapping (4A5Ja once more),
$\interior F=\emptyset$ and $F$ is nowhere dense.

Thus in this case we have a non-negligible nowhere dense set, as required.

\medskip

\quad\grheadb\ Now suppose just that $X$ is locally compact and Hausdorff.   In
this case it has an open $\sigma$-compact subgroup $X_0$ say.
The subspace measure $\mu_{X_0}$
on $X_0$ is a left Haar measure on $X_0$ (443F), and
is atomless;  by ($\alpha$),
there is a nowhere dense set $F\subseteq X_0$ such
that $0<\mu_{X_0}F=\mu F$.   So in this case too we have a non-negligible
nowhere dense set.

\medskip

\quad\grheadc\
For the general case, let $Z$, $\lambda$ and $\phi:X\to Z$ be as
in 443L.   Then $\lambda$ is atomless.   \Prf\ 443L(v) implies that
the measure-preserving Boolean
homomorphism from the measure algebra of $\lambda$ to the measure
algebra of $\mu$ induced by $\phi$ is surjective, therefore an isomorphism;
so both measure algebras are atomless and $\lambda$ is atomless (322Bg).\
\Qed

By ($\beta$), there is a nowhere dense non-negligible subset $H$ of $Z$;
replacing $H$ by its closure, if necessary, we may suppose that $H$ is
closed.   Set $F=\phi^{-1}[H]$;  then $F\subseteq X$ is closed and
non-negligible because $\phi$ is continuous and \imp.   Since
$\phi[F]\subseteq H$ is nowhere dense, so is $F$ (443L(viii)).
Thus we have the required non-negligible nowhere dense set in the general
case also.

\medskip

{\bf (h)} Now suppose that $X$ is Hausdorff.   If $\mu$ is atomless, then
$\mu\{x\}=0$ for every $x$, so $\{x\}$ is never open and the topology is
not discrete.   If $\mu$ has an atom $E$, let $F\subseteq E$ be a closed
self-supporting set of non-zero measure;  then $F$ also is an atom, so
cannot have two disjoint non-empty relatively open sets, and must be a
singleton.   Thus we have an $x_0$ such that $\mu\{x_0\}>0$;  as $\mu$ is
\lti, $\mu\{x\}=\mu\{x_0\}$ for every $x\in X$.
We know also that there is an open set $G$ of non-zero finite measure,
which must be finite;  so every singleton subset of $G$ is open.   It
follows that every singleton subset of $X$ is open, and $X$ has its
discrete topology.   Thus (ii)$\Leftrightarrow$(vii) when $X$ is Hausdorff.
}%end of proof of 443O


\leader{443P}{Quotient
\dvrocolon{spaces}}\discrversionA{\footnote{Formerly 4{}43O.}}
{}\cmmnt{ I come now to the
relationship between the modular functions of \S442, normal subgroups
and quotient spaces.

\medskip

\noindent}{\bf Lemma} Let $X$ be a locally compact Hausdorff topological
group and $Y$ a closed subgroup of $X$.   Let $Z=X/Y$ be the set of left
cosets of $Y$ in $X$ with the quotient topology and $\pi:X\mapsto Z$ the
canonical map, so that $Z$ is a
locally compact Hausdorff space and we have a continuous action of $X$
on $Z$ defined by writing $a\action\pi x=\pi(ax)$ for $a$,
$x\in X$\cmmnt{ (4A5J(b-iii))}.   Let $\nu$ be a left Haar measure on $Y$
and write $C_k(X)$, $C_k(Z)$ for the spaces of continuous
real-valued functions with compact supports on $X$, $Z$ respectively.

(a) We have a positive linear operator $T:C_k(X)\to C_k(Z)$
defined by writing

\Centerline{$(Tf)(\pi x)=\int_Yf(xy)\nu(dy)$}

\noindent for every $f\in C_k(X)$ and $x\in X$.   If $f>0$ in $C_k(X)$
then $Tf>0$ in $C_k(Z)$.
If $h\ge 0$ in $C_k(Z)$ then there is an $f\ge 0$ in $C_k(X)$ such
that $Tf=h$.

(b) If $a\in X$ and $f\in C_k(X)$, then
$T(a\action_lf)(z)=(Tf)(a^{-1}\action z)$ for every $z\in Z$.

(c) Now suppose that $a$ belongs to the normalizer of
$Y$\cmmnt{ (that is, $aYa^{-1}=Y$)}.   In this case, we can define
$\psi(a)\in\ooint{0,\infty}$ by the formula

\Centerline{$\nu(aFa^{-1})=\psi(a)\nu F$ for every $F\in\dom\nu$,}

\noindent and

\Centerline{$T(a\action_rf)(\pi x)=\psi(a)\cdot(Tf)(\pi(xa))$}

\noindent for every $x\in X$ and $f\in C_k(X)$.

\proof{ I should begin by remarking that because $Y$ is a closed
subgroup of a locally compact Hausdorff group, it is itself a locally
compact Hausdorff group, so does have a left Haar measure, which is a
Radon measure.

\medskip

{\bf (a)(i)} The first thing to check is that if $f\in C_k(X)$ then $Tf$
is well-defined as a member of $\BbbR^{Z}$.   \Prf\ ($\alpha$) If $x\in
X$, then $y\mapsto f(xy):Y\to\Bbb R$ is a continuous function with
compact support, so $\int f(xy)\nu(dy)$ is defined in $\Bbb R$.
($\beta$) If
$x_1$, $x_2\in X$ and $\pi x_1=\pi x_2$, then $x_1^{-1}x_2\in Y$, and

\Centerline{$\int f(x_2y)\nu(dy)
=\int f(x_1(x_1^{-1}x_2y))\nu(dy)
=\int f(x_1y)\nu(dy)$,}

\noindent applying 441J to the function $y\mapsto f(x_1y)$ and
the left action of $Y$ on itself.   Thus we can safely write
$(Tf)(\pi x)=\int f(xy)\nu(dy)$ for every $x\in X$, and $Tf$ will be a
real-valued function on $Z$.\ \Qed

\medskip

\quad{\bf (ii)} Now $Tf$ is continuous for every $f\in C_k(X)$.   \Prf\
Given $z_0\in Z$, take $x_0\in X$ such that $z=\pi x_0$.   We have an
$h\in C_k(X)^+$ such that for every $\epsilon>0$ there is an open set
$U_{\epsilon}$ containing $x_0$ such that
$|f(x_0y)-f(xy)|\le\epsilon h(y)$ whenever $x\in U_{\epsilon}$ and
$y\in X$ (4A5Pb).   In this case,

\Centerline{$|(Tf)(\pi x)-(Tf)(\pi x_0)|
=|\int f(xy)-f(x_0y)\nu(dy)|
\le\epsilon\int_Yh(y)\nu(dy)$}

\noindent for every $x\in U_{\epsilon}$, so that
$|(Tf)(z)-(Tf)(z_0)|\le\epsilon\int_Y h\,d\nu$ for every
$z\in\pi[U_{\epsilon}]$.   Since each $\pi[U_{\epsilon}]$ is an open
neighbourhood of $z_0$ (4A5J(a-i), as always),
$Tf$ is continuous at $z_0$;  as $z_0$
is arbitrary, $Tf$ is continuous.\ \Qed

\medskip

\quad{\bf (iii)} Since

$$\eqalign{\{z:(Tf)(z)\ne 0\}
&=\{\pi x:\int f(xy)\nu(dy)\ne 0\}
\subseteq\{\pi x:f(xy)\ne 0\text{ for some }y\in Y\}\cr
&=\{\pi x:f(x)\ne 0\}
\subseteq\pi[\overline{\{x:f(x)\ne 0\}}]\cr}$$

\noindent is relatively compact, $Tf\in C_k(Z)$ for every
$f\in C_k(X)$.

\medskip

\quad{\bf (iv)} The formula for $Tf$ makes it plain that
$T:C_k(X)\to C_k(Z)$ is a positive linear operator.

\medskip

\quad{\bf (v)} If $f\in C_k(X)^+$ and $x\in X$ are such that $f(x)>0$, then
$\{y:y\in Y,\,f(xy)>0\}$ is a non-empty open set in $Y$;  because $\nu$
is strictly positive,

\Centerline{$(Tf)(\pi x)=\int f(xy)\nu(dy)>0$.}

\noindent In particular, $Tf>0$ if $f>0$.   Moreover, if $z\in Z$
there is an $f\in C_k(X)^+$ such that $(Tf)(z)>0$.   Now

\Centerline{$\{\{z:(Tf)(z)>0\}:f\in C_k(X)^+\}$}

\noindent is an upwards-directed family of open subsets of $Z$, so if
$L\subseteq Z$ is any compact set there is an $f\in C_k(X)^+$ such
that $(Tf)(z)>0$ for every $z\in L$.

\medskip

\quad{\bf (vi)} Now suppose that $h\in C_k(Z)^+$.   By (v), there is
an $f_0\in C_k(X)^+$ such that $(Tf_0)(z)>0$ whenever
$z\in\overline{\{w:h(w)\ne 0\}}$.   Setting $h'(z)=h(z)/(Tf_0)(z)$ when
$h(z)\ne 0$, $0$ for other $z\in Z$, we have $h'\in C_k(Z)$ and
$h=h'\times Tf_0$.   Set

\Centerline{$f(x)=f_0(x)h'(\pi x)\ge 0$}

\noindent for every $x\in X$.   Because $h'$ and $\pi$ are continuous,
$f\in C_k(X)$.   For any $x\in X$,

$$\eqalign{(Tf)(\pi x)
&=\int f_0(xy)h'(\pi(xy))\nu(dy)
=h'(\pi x)\int f_0(xy)\nu(dy)\cr
&=h'(\pi x)(Tf_0)(\pi x)
=h(\pi x).\cr}$$

\noindent Thus $Tf=h$.

\medskip

{\bf (b)} If $z=\pi x$, then $a^{-1}\action z=\pi(a^{-1}x)$, so

\Centerline{$(Tf)(a^{-1}\action z)
=\int f(a^{-1}xy)\nu(dy)
=\int (a\action_lf)(xy)\nu(dy)
=T(a\action_lf)(z)$.}

\medskip

{\bf (c)} Define $\phi:Y\to Y$ by writing $\phi(y)=a^{-1}ya$ for
$y\in Y$.   Because $\phi$ is a homeomorphism, the image measure
$\nu\phi^{-1}$ is a Radon
measure on $Y$;  because $\phi$ is a group automorphism, $\nu\phi^{-1}$
is a left Haar measure.   (If $F\in\dom\nu\phi^{-1}$ and $y\in Y$, then

\Centerline{$\nu\phi^{-1}[yF]=\nu(ayFa^{-1})=\nu(Fa^{-1})
=\nu(aFa^{-1})=\nu\phi^{-1}[F]$.)}

\noindent $\nu\phi^{-1}$ must therefore be a multiple of $\nu$;  say
$\nu\phi^{-1}=\psi(a)\nu$.

If $g\in C_k(Y)$, then

\Centerline{$\int g(a^{-1}ya)\nu(dy)
=\int g\phi\,d\nu
=\int g\,d(\nu\phi^{-1})
=\psi(a)\int g\,d\nu$.}

\noindent Now take $f\in C_k(X)$.   Then

$$\eqalignno{T(a\action_rf)(\pi x)
&=\int(a\action_rf)(xy)\nu(dy)
=\int f(xya)\nu(dy)\cr
&=\int f(xa(a^{-1}ya))\nu(dy)
=\psi(a)\int f(xay)\nu(dy)\cr
\noalign{\noindent (using the remark above with $g(y)=f(xay)$)}
&=\psi(a)(Tf)(\pi(xa))\cr}$$

\noindent for every $x\in X$, as claimed.
}%end of proof of 443P

\leader{443Q}{Theorem}\discrversionA{\footnote{Formerly 4{}43P.}}
{} Let $X$ be a locally compact Hausdorff
topological group and $Y$ a closed subgroup of $X$.   Let $Z=X/Y$ be the
set of left cosets of $Y$ in $X$ with the quotient topology, and
$\pi:X\to Z$ the canonical map, so that $Z$ is
a locally compact Hausdorff space and we have a continuous action of $X$
on $Z$ defined by writing $a\action\pi x=\pi(ax)$ for $a$,
$x\in X$.   Let $\nu$ be a left Haar measure on $Y$.   Suppose that
$\lambda$ is a non-zero $X$-invariant Radon measure on $Z$.

(a) For each $z\in Z$, we have a Radon measure $\nu_z$ on $X$ defined
by the formula

\Centerline{$\nu_zE=\nu(Y\cap x^{-1}E)$}

\noindent whenever $\pi x=z$ and the right-hand side is defined.   In
this case, for a real-valued function $f$ defined on a subset of $X$,

\Centerline{$\int f\,d\nu_z=\int f(xy)\nu(dy)$}

\noindent whenever either side is defined in $[-\infty,\infty]$.

(b) We have a left Haar measure $\mu$ on $X$ defined by the formulae

\Centerline{$\int f\,d\mu=\biggeriint f\,d\nu_z\lambda(dz)$}

\noindent for every $f\in C_k(X)$, and

\Centerline{$\mu G=\int\nu_zG\,\lambda(dz)$}

\noindent for every open set $G\subseteq X$.

(c) If $D\subseteq Z$, then $D\in\dom\lambda$ iff
$\pi^{-1}[D]\subseteq X$ is
Haar measurable, and $\lambda D=0$ iff $\pi^{-1}[D]$ is Haar negligible.

(d) If $\nu Y=1$, then $\lambda$ is the image measure $\mu\pi^{-1}$.

(e) Suppose now that $X$ is $\sigma$-compact.   Then
$\mu E=\int\nu_zE\,\lambda(dz)$ for every Haar measurable set
$E\subseteq X$.   If $f\in\eusm L^1(\mu)$, then
$\int f\,d\mu=\iint f\,d\nu_z\lambda(dz)$.

(f) Still supposing that $X$ is $\sigma$-compact, take
$f\in\eusm L^1(\mu)$, and for $a\in X$ set $f_a(y)=f(ay)$ whenever
$y\in Y$ and
$ay\in\dom f$.  Then $Q_f=\{a:a\in X,\,f_a\in\eusm L^1(\nu)\}$ is
$\mu$-conegligible, and the function
$a\mapsto f_a^{\ssbullet}:Q_f\to L^1(\nu)$ is almost continuous.

\proof{{\bf (a)} First, we do have a function $\nu_z$ depending only on
$z$, because if $z=\pi x_1=\pi x_2$ then
$x_2^{-1}x_1\in Y$, so

\Centerline{$\nu(Y\cap x_1^{-1}E)
=\nu(x_2^{-1}x_1(Y\cap x_1^{-1}E))
=\nu(Y\cap x_2^{-1}E)$}

\noindent whenever either side is defined.   Of course $\nu_z$, being
the image of the Radon measure $\nu$ under
the continuous map $y\mapsto xy:Y\to X$ whenever $\pi x=z$, is always
a Radon measure on $X$ (418I).   We also have

\Centerline{$\int_Yf(xy)\nu(dy)=\int_Xf\,d\nu_{\pi x}$}

\noindent whenever $x\in X$ and $f$ is a real-valued function such that
either side is defined in $[-\infty,\infty]$, by 235J.

I remark here that if $z\in Z$ then the coset $C=\pi^{-1}[\{z\}]$ is
$\nu_z$-conegligible, because if $\pi x=z$ then $Y=Y\cap x^{-1}C$.

\medskip

{\bf (b)(i)} Let $T:C_k(X)\to C_k(Z)$ be the positive linear operator
of 443P;  that is,

\Centerline{$(Tf)(z)=\int f(xy)\nu(dy)=\int f\,d\nu_z$}

\noindent whenever $f\in C_k(X)$, $x\in X$ and $z=\pi x$.   Then we have
a positive linear functional $\theta:C_k(X)\to\Bbb R$ defined by setting
$\theta(f)=\int Tf\,d\lambda$ for every $f\in C_k(X)$.   By the Riesz
Representation Theorem (436J), there is a Radon measure $\mu$ on $X$
defined by saying that $\int f\,d\mu=\theta(f)$ for every $f\in C_k(X)$.
Note that $\mu$ is non-zero.   \Prf\ Because $\lambda$ is non-zero,
there is some $h\in C_k(Z)^+$ such that $\int h\,d\lambda\ne 0$;  now
there is some $f\in C_k(X)$ such that $Tf=h$, by 443Pa, and
$\int f\,d\mu\ne 0$.\ \Qed

\medskip

\quad{\bf (ii)} $\mu$ is a left Haar measure.   \Prf\ If $f\in C_k(X)$
and $a\in X$, then we have
$T(a\action_lf)(z)=(Tf)(a^{-1}\action z)$ for every $z\in Z$, by 443Pb.
So

$$\eqalignno{\int a\action_lfd\mu
&=\int T(a\action_lf)d\lambda
=\int Tf(a^{-1}\action z)\lambda(dz)
=\int Tf(z)\lambda(dz)\cr
\noalign{\noindent (by 441J or 441L, because $\lambda$ is
$X$-invariant)}
&=\int f\,d\mu.\cr}$$

\noindent By 441L in the other direction, $\mu$ is invariant under the
left action of $X$ on itself, that is, is a left Haar measure.\ \Qed

\medskip

\quad{\bf (iii)} If $G\subseteq X$ is open then
$\mu G=\int\nu_zG\,\lambda(dz)$.   \Prf\ Set
$A=\{f:f\in C_k(X),\,0\le f\le\chi G\}$.   Then
$\mu G=\sup_{f\in A}\int f\,d\mu$ and

\Centerline{$\nu_zG=\sup_{f\in A}\int f\,d\nu_z
=\sup_{f\in A}(Tf)(z)$}

\noindent for every $z\in Z$, by 414Ba, because $\nu_z$ is
$\tau$-additive.   But as $Tf$ is continuous
for every $f\in A$, and $\lambda$ is $\tau$-additive, we also have

\Centerline{$\int\nu_zG\,\lambda(dz)
=\sup_{f\in A}\int Tf\,d\lambda
=\sup_{f\in A}\int f\,d\mu
=\mu G$.  \Qed}

\medskip

{\bf (c)(i)} Let $\Cal A$ be the family of those sets $A\subseteq X$
such that $\mu A$ and $\int\nu_z(A)\lambda(dz)$ are defined in
$[0,\infty]$ and equal.   Then $\bigcup_{n\in\Bbb N}A_n$ belongs to
$\Cal A$ whenever $\sequencen{A_n}$ is a non-decreasing sequence in
$\Cal A$, and $A\setminus B\in\Cal A$ whenever $A$, $B\in\Cal A$,
$B\subseteq A$ and $\mu A<\infty$.   Moreover, if $A\in\Cal A$ and
$\mu A=0$, then every subset of $A$ belongs to $\Cal A$, since $A$ must
be $\nu_z$-negligible for $\lambda$-almost every $z$.   We also know
from (b) that every open set belongs to $\Cal A$.

Applying the Monotone Class Theorem (136B) to
$\{A:A\in\Cal A,\,A\subseteq G\}$, we see that if $E\subseteq X$ is a
Borel set
included in an open set $G$ of finite measure, then $E\in\Cal A$.   So
if $E$ is a relatively compact Haar measurable set, $E\in\Cal A$ (using
443J(b-i), or otherwise).

\medskip

\quad{\bf (ii)} If $D\in\dom\lambda$ then $\pi^{-1}[D]\in\dom\mu$.
\Prf\ Let $K\subseteq X$ be compact.   Then $\pi[K]\subseteq Z$ is
compact, so there are Borel sets $F_1$, $F_2\subseteq Z$ such that
$F_1\subseteq D\cap\pi[K]\subseteq F_2$ and $F_2\setminus F_1$ is
$\lambda$-negligible.   Now
$\nu_z(K\cap\pi^{-1}[F_2\setminus F_1])=0$ whenever
$z\notin F_2\setminus F_1$, by the remark added to the proof of (a)
above, so

\Centerline{$\mu(K\cap\pi^{-1}[F_2\setminus F_1])
=\int\nu_z(K\cap\pi^{-1}[F_2\setminus F_1])\lambda(dz)
=0$.}

\noindent Since

\Centerline{$K\cap\pi^{-1}[F_1]\subseteq K\cap\pi^{-1}[D]
\subseteq K\cap\pi^{-1}[F_2]$,}

\noindent $K\cap\pi^{-1}[D]\in\dom\mu$.   As $K$ is arbitrary,
$\pi^{-1}[D]$ is measured by $\mu$, so is Haar measurable.\ \Qed

If $\lambda D=0$ then the same arguments show that
$\mu(K\cap\pi^{-1}[D])=0$ for every compact $K\subseteq X$, so that
$\mu\pi^{-1}[D]=0$ and $\pi^{-1}[D]$ is Haar negligible.

\medskip

\quad{\bf (iii)} Now suppose that $D\subseteq Z$ is such that
$\pi^{-1}[D]\in\dom\mu$.   Let $L\subseteq Z$ be compact.   Then there
is a relatively compact open set $G\subseteq X$ such that
$\pi[G]\supseteq L$ (because $\{\pi[G]:G\subseteq X$ is open and
relatively compact$\}$ is an upwards-directed family of open sets
covering $Z$).   In this case,

\Centerline{$\int\nu_z(G\cap\pi^{-1}[D\cap L])\lambda(dz)
=\mu(G\cap\pi^{-1}[D]\cap\pi^{-1}[L])$}

\noindent is well-defined, by (i).   But  if $z=\pi x$ then

$$\eqalign{\nu_z(G\cap\pi^{-1}[D\cap L])
&=0\text{ if }z\notin D\cap L,\cr
&=\nu_zG=\nu(Y\cap x^{-1}G)>0\text{ if }z\in D\cap L,\cr}$$

\noindent because if $z\in L$ then $\pi x\in\pi[G]$ and
$Y\cap x^{-1}G\ne\emptyset$.   So

\Centerline{$D\cap L
=\{z:\nu_z(G\cap\pi^{-1}[D\cap L])>0\}$}

\noindent is measured by $\lambda$.   As $L$ is arbitrary, and $\lambda$
is a Radon measure, $D\in\dom\lambda$.

\medskip

\quad{\bf (iv)} If $\pi^{-1}[D]$ is Haar negligible, then, in (iii) above,
we shall have $\int\nu_z(G\cap\pi^{-1}[D\cap L])\lambda(dz)=0$, so
that $\lambda(D\cap L)=0$;  as $L$ is arbitrary, $\lambda D=0$,
by 412Ib or 412Jc.

\medskip

{\bf (d)} If $\nu Y=1$, then, for any open set $H\subseteq Z$,
$\nu_z\pi^{-1}[H]=1$ if $z\in H$, $0$ otherwise.   So

\Centerline{$\mu\pi^{-1}[H]
=\int\nu_z(\pi^{-1}[H])\lambda(dz)
=\lambda H$.}

\noindent Thus $\lambda$ and the image measure $\mu\pi^{-1}$ agree on
the open sets and, being Radon measures (418I again), must be equal
(416E(b-iii)).

\medskip

{\bf (e)} If $X$ is actually $\sigma$-compact, then (c)(i) of this proof
tells us that
$\mu E=\int\nu_zE\,\lambda(dz)$ for every Haar measurable set
$E\subseteq X$, since $E$ is the union of an increasing sequence of
relatively compact measurable sets.   Consequently
$\int f\,d\mu=\iint f\,d\nu_z\lambda(dz)$ for every $\mu$-simple
function $f$.
Now suppose that $f$ is a non-negative
$\mu$-integrable function.   Then
there is a non-decreasing sequence $\sequencen{f_n}$ of non-negative
$\mu$-simple functions converging to $f$ everywhere in $\dom f$.   If we
set $A=\{z:\nu_z^*(X\setminus\dom f)>0\}$, then

\Centerline{$\int\nu_z(X\setminus\dom f)\lambda(dz)
=\mu(X\setminus\dom f)=0$,}

\noindent so $\lambda A=0$.  Since
$\int f\,d\nu_z=\lim_{n\to\infty}\int f_nd\nu_z$ for every
$z\in Z\setminus A$,

$$\eqalign{\int f\,d\mu
&=\lim_{n\to\infty}\int f_n\,d\mu\cr
&=\lim_{n\to\infty}\iint f_nd\nu_z\lambda(dz)
=\iint f\,d\nu_z\lambda(dz).\cr}$$

\noindent Applying this to the positive and negative parts of $f$, we
see that the same formula is valid for any $\mu$-integrable function
$f$.

\medskip

{\bf (f)(i)} If $f\in\eusm L^1(\mu)$ and $a\in X$, then

\Centerline{$\int f_ad\nu=\int f(ay)\nu(dy)
=\int f\,d\nu_{\pi a}$}

\noindent if any of these are defined.   So if $f$,
$g\in\eusm L^1(\mu)$, $\|f_a-g_a\|_1=\int|f-g|d\nu_{\pi a}$ if either is
defined.

\medskip

\quad{\bf (ii)} Let $\Phi$ be the set of all almost continuous functions
from $\mu$-conegligible subsets of $X$ to $L^1(\nu)$, where $L^1(\nu)$
is given its norm topology.   (In terms of the
definition in 411M, a member $\phi$ of $\Phi$ is to be almost continuous
with respect to the subspace measure on $\dom\phi$.)   If $\phi\in\Phi$
and $\psi$ is a function from a conegligible subset of $X$ to $L^1(\nu)$
which is equal almost everywhere to $\phi$, then $\psi\in\Phi$.   If
$\sequencen{\phi_n}$ is a sequence in $\Phi$ converging $\mu$-almost
everywhere to $\psi$, then $\psi\in\Phi$ (418F).

\medskip

\quad{\bf (iii)} For $f\in\eusm L^1(\mu)$, set
$\phi_f(a)=f_a^{\ssbullet}$ whenever this is defined in $L^1(\nu)$.
Set $M=\{f:f\in\eusm L^1(\mu),\,\phi_f\in\Phi\}$.   If
$\sequencen{f^{(n)}}$ is a sequence in $M$, $f\in\eusm L^1(\mu)$ and
$\|f^{(n)}-f\|_1\le 4^{-n}$ for every $n$, then $f\in M$.   \Prf\ Set
$g=\sum_{n=0}^{\infty}2^n|f^{(n)}-f|$, defined on

\Centerline{$\{x:x\in\dom f\cap\bigcap_{n\in\Bbb N}\dom f^{(n)},
\,\sum_{n=0}^{\infty}2^n|f^{(n)}(x)-f(x)|<\infty\}$;}

\noindent then $g\in\eusm L^1(\mu)$.   Now

\Centerline{$D=\{z:z\in Z,\,g$ is $\nu_z$-integrable$\}$}

\noindent is $\lambda$-conegligible, by (e), and
$E=\{a:a\in X,\,\pi a\in D\}$ is $\mu$-conegligible, by (c).

If $a\in E$, then

\Centerline{$|f^{(n)}_a-f_a|\le 2^{-n}g\,\,\nu_{\pi a}$-a.e.}

\noindent for every $n\ge 1$.   So

$$\eqalign{\|\phi_f(a)-\phi_{f^{(n)}}(a)\|_1
&=\int_Y|f_a-f_a^{(n)}|d\nu
=\int_X|f-f^{(n)}|d\nu_{\pi a}\cr
&\le 2^{-n}\int g\,d\nu_{\pi a}
\to 0\cr}$$

\noindent as $n\to\infty$.   Thus
$\phi_f=\lim_{n\to\infty}\phi_{f^{(n)}}$ almost everywhere, and
$\phi\in\Phi$, by (ii).\ \Qed

\medskip

\quad{\bf (iv)} $C_k(X)\subseteq M$.   \Prf\ If $f\in C_k(X)$ and
$a_0\in X$, then there is an $h\in C_k(X)^+$ such that for every
$\epsilon>0$ there is an open set $G_{\epsilon}$ containing $a_0$ such
that $|f(a_0y)-f(ay)|\le\epsilon h(y)$ whenever
$a\in G_{\epsilon}$ and $y\in X$ (4A5Pb again).   In this case,

\Centerline{$\|\phi_f(a_0)-\phi_f(a)\|_1
=\int|f(a_0y)-f(ay)|\nu(dy)
\le\epsilon\int_Yh\,d\nu$}

\noindent whenever $a\in G_{\epsilon}$.   As $\epsilon$ is arbitrary,
$\phi_f$ is continuous at $a_0$;  as $a_0$ is arbitrary, $\phi_f$ is
continuous, and $f\in M$.\ \Qed

\medskip

\quad{\bf (v)} Now take any $f\in\eusm L^1(\mu)$.   Then for each
$n\in\Bbb N$ we can find $f^{(n)}\in C_k(X)$ such that
$\|f^{(n)}-f\|_1\le 4^{-n}$ (416I), so $f\in M$, by (iii).
This completes the proof.
}%end of proof of 443Q

\vleader{72pt}{443R}{Theorem}\dvAformerly{4{}43Q}
Let $X$ be a locally compact Hausdorff
topological group and $Y$ a closed subgroup of $X$.   Let $Z=X/Y$ be the
set of left cosets of $Y$ in $X$ with the quotient topology, so that
$Z$ is a locally compact Hausdorff space and we have a continuous
action of $X$ on $Z$ defined by writing $a\action(xY)=axY$ for $a$,
$x\in X$.   Let $\Delta_X$ be the left modular function
of $X$ and $\Delta_Y$ the left modular function of $Y$.   Then the
following are equiveridical:

(i) there is a non-zero $X$-invariant Radon measure $\lambda$ on $Z$;

(ii) $\Delta_Y$ is the restriction of $\Delta_X$ to $Y$.

\proof{ Fix a left Haar measure $\nu$ on $Y$, and let
$T:C_k(X)\to C_k(Z)$ be the corresponding linear operator as defined in
443Pa.

\medskip

{\bf (a)(i)$\Rightarrow$(ii)} Suppose that $\lambda$ is a
non-zero $X$-invariant Radon measure on $Z$.   Construct a left Haar
measure $\mu$ on $X$ as in 443Q.
In the notation of part (b-i) of the proof of 443Q, we have

\Centerline{$\int fd\mu=\iint fd\nu_z\lambda(dz)=\int Tfd\lambda$}

\noindent for every $f\in C_k(X)$.

Suppose that $a\in Y$.   In this case, $a$ surely belongs to the
normalizer of $Y$, and, in the language of 443Pc, we have
$\nu(aFa^{-1})=\psi(a)\nu F$ for every $F\in\dom\nu$.   But as

\Centerline{$\nu(aFa^{-1})=\nu(Fa^{-1})=\Delta_Y(a^{-1})\nu F$,}

\noindent we must have $\psi(a)=\Delta_Y(a^{-1})$.

Fix some $f>0$ in $C_k(X)$.   We have

$$\eqalignno{T(a\action_rf)(\pi x)
&=\psi(a)\cdot(Tf)(\pi(xa))
=\psi(a)\int f(xay)\nu(dy)
=\psi(a)\int f(xy)\nu(dy)\cr
\displaycause{because $a\in Y$}
&=\psi(a)\cdot(Tf)(\pi x)\cr}$$

\noindent for every $x$, so that (using 442Kc) we have

$$\eqalign{\Delta_X(a^{-1})\int f\,d\mu
&=\int a\action_rfd\mu
=\int T(a\action_rf)d\lambda\cr
&=\psi(a)\int Tf\,d\lambda
=\psi(a)\int f\,d\mu
=\Delta_Y(a^{-1})\int f\,d\mu.\cr}$$

\noindent As $\int f\,d\mu>0$, we must have
$\Delta_X(a^{-1})=\Delta_Y(a^{-1})$;  as $a$ is arbitrary,
$\Delta_Y=\Delta_X\restrp Y$, as required by (ii).

\medskip

{\bf (b)(ii)$\Rightarrow$(i)} Now suppose that
$\Delta_Y=\Delta_X\restrp Y$.   This time, start
with a left Haar measure $\mu$ on $X$.

\medskip

\quad\grheada\ (The key.)  If $f\in C_k(X)$ is such that $Tf\ge 0$ in
$C_k(Z)$, then $\int f\,d\mu\ge 0$.   \Prf\ There is an
$h\in C_k(Z)^+$ such that $h(z)=1$ whenever $z\in Z$ and $(Tf)(z)\ne 0$;
by 443Pa, we can
find a $g\in C_k(X)^+$ such that $Tg=h$.   Now observe that
$x\mapsto(Tf)(\pi x)$ is a
non-negative continuous real-valued function on $X$, so

$$\eqalignno{0
&\le\int_Xg(x)(Tf)(\pi x)\mu(dx)\cr
&=\int_Xg(x)\int_Yf(xy)\nu(dy)\mu(dx)
=\int_Y\int_Xg(x)f(xy)\mu(dx)\nu(dy)\cr
\noalign{\noindent (by 417Ga or 417Gb, because
$(x,y)\mapsto g(x)f(xy):X\times Y\to\Bbb R$ is a continuous function
with compact support)}
&=\int_Y\Delta_X(y^{-1})\int_Xg(xy^{-1})f(x)\mu(dx)\nu(dy)\cr
\noalign{\noindent (applying 442Kc to the function
$x\mapsto g(xy^{-1})f(x)$)}
&=\int_Xf(x)\int_Y\Delta_X(y^{-1})g(xy^{-1})\nu(dy)\mu(dx)\cr
\noalign{\noindent (because $(x,y)\mapsto\Delta_X(y^{-1})g(xy^{-1})f(x)$
is continuous and has compact support)}
&=\int_Xf(x)\int_Y\Delta_Y(y^{-1})g(xy^{-1})\nu(dy)\mu(dx)\cr
\noalign{\noindent (because $\Delta_X\restrp Y=\Delta_Y$, by
hypothesis)}
&=\int_Xf(x)\int_Yg(xy)\nu(dy)\mu(dx)\cr
\noalign{\noindent (applying 442K(b-ii) to the function
$y\mapsto g(xy)$)}
&=\int_Xf(x)(Tg)(\pi x)\mu(dx)
=\int_Xf(x)\mu(dx)\cr}$$

\noindent because $(Tg)(\pi x)=1$ whenever $f(x)\ne 0$.\ \Qed

\medskip

\quad\grheadb\ Applying this to $f$ and $-f$, we see that
$\int f\,d\mu=0$ whenever $Tf=0$, so that $\int f\,d\mu=\int g\,d\mu$
whenever $f$, $g\in C_k(X)$ and $Tf=Tg$.   Accordingly (because $T$ is
surjective)
we have a functional $\theta:C_k(Z)\to\Bbb R$ defined by saying that
$\theta(Tf)=\int f\,d\mu$ whenever $f\in C_k(X)$, and $\theta$ is
positive and linear.   By the Riesz Representation Theorem again, there
is a Radon measure $\lambda$ on $Z$ such that
$\theta(h)=\int h\,d\lambda$ for every $h\in C_k(Z)$.

If $a\in X$ and $h\in C_k(Z)$ take $f\in C_k(X)$ such that $Tf=h$.
Then, for any $x\in X$,

\Centerline{$(Tf)(a\action\pi x)=(Tf)(\pi(ax))
=\int f(axy)\nu(dy)=\int (a^{-1}\action_lf)(xy)\nu(dy)
=T(a^{-1}\action_lf)(\pi x)$.}

\noindent So

$$\eqalign{\int h(a\action z)\lambda(dz)
&=\int (Tf)(a\action z)\lambda(dz)
=\int T(a^{-1}\action_lf)(z)\lambda(dz)\cr
&=\int a^{-1}\action_lfd\mu
=\int f\,d\mu
=\int h(z)\lambda(dz).\cr}$$

\noindent By 441L again, $\lambda$ is $X$-invariant.   Also $\lambda$ is
non-zero because there is surely some $f$ such that $\int f\,d\mu\ne 0$.
So we have the required non-zero $X$-invariant Radon measure on $Z$.
}%end of proof of 443R

\leader{443S}{Applications}\discrversionA{\footnote{Formerly 4{}43R.}}
{}\cmmnt{ This theorem applies in a variety
of cases.}   Let $X$ be a locally compact Hausdorff
topological group and $Y$ a closed subgroup of $X$.

\spheader 443Sa If $Y$ is a normal subgroup of $X$, then
$\Delta_Y=\Delta_X\restrp Y$.
\prooflet{\Prf\ $X/Y$ has a group structure under which it is a locally
compact Hausdorff group (4A5J(b-ii)).   It therefore has a left Haar
measure, which is surely $X$-invariant in the sense of 443R.\ \Qed
}%end of prooflet

Note that in this context any of the invariant measures $\lambda$ of 443Q
must be left Haar measures on the quotient group.

\spheader 443Sb If $Y$ is compact, then $\Delta_Y=\Delta_X\restrp Y$.
\prooflet{\Prf\ Both $\Delta_Y$ and $\Delta_X\restrp Y$ are
continuous homomorphisms from $Y$ to $\ooint{0,\infty}$;  since the only
compact subgroup of $\ooint{0,\infty}$ is $\{1\}$, they are both
constant with value $1$.\ \QeD}   So\cmmnt{ 443R tells us that} we
have an $X$-invariant Radon measure $\lambda$ on $X/Y$.   Since $Y$ has
a Haar probability measure, $\lambda$ will be the image of
a left Haar measure under the canonical map\cmmnt{ (443Qd)}.

\spheader 443Sc If, in (b), $Y$ is a normal subgroup,
then\cmmnt{ we find that, for $W\in\dom\lambda$ and $x\in X$,

\Centerline{$\lambda(W\cdot\pi x)=\mu(\pi^{-1}[W]\cdot x)
=\Delta_X(x)\lambda W$,}

\noindent so that} $\Delta_{X/Y}\pi=\Delta_X$, writing
$\pi:X\to X/Y$ for the canonical map.   \cmmnt{This is
a special case of 443T below, because (in the terminology there)
$\psi(a)=\nu Y/\nu Y=1$ for every $a\in X$.}

\spheader 443Sd If $Y$ is open, $\Delta_Y=\Delta_X\restrp Y$.
\prooflet{\Prf\ If $\mu$ is a left Haar measure on $X$, then the
subspace measure $\mu_Y$ is a left Haar measure on $Y$ (443F).   There
is an open set $G\subseteq Y$ such that $0<\mu G<\infty$, and now
$\mu(Gy)=\Delta_X(y)\mu G=\Delta_Y(y)\mu G$ for every
$y\in Y$.\ \QeD}   \cmmnt{This time, $X/Y$ is discrete, so counting
measure is an $X$-invariant Radon measure on $X/Y$.}

\leader{443T}{Theorem}\discrversionA{\footnote{Formerly 4{}43S.}}
{} Let $X$ be a locally compact Hausdorff
topological group and
$Y$ a closed normal subgroup of $X$;  let $Z=X/Y$ be the quotient group,
and $\pi:X\to Z$ the canonical map.   Write $\Delta_X$,
$\Delta_{Z}$ for the left modular functions of $X$, $Z$
respectively.   Define $\psi:X\to\ooint{0,\infty}$ by the formula

\Centerline{$\nu(aFa^{-1})=\psi(a)\nu F$ whenever $F\in\dom\nu$ and
$a\in X$,}

\noindent where $\nu$ is a left Haar measure on
$Y$\cmmnt{ (cf.\ 443Pc)}.   Then

\Centerline{$\Delta_{Z}(\pi a)=\psi(a)\Delta_X(a)$}

\noindent for every $a\in X$.

\proof{ Let $T:C_k(X)\to C_k(Z)$ be the map defined in 443P, and
$\lambda$ a
left Haar measure on $Z$;  then, as in 443Qb, we have a left Haar
measure $\mu$ on $X$ defined by the formula
$\int f\,d\mu=\int Tf\,d\lambda$ for every $f\in C_k(X)$.   Fix on some
$f>0$ in $C_k(X)$ and $a\in X$, and set $w=\pi a$.   By 443Pc, we have

\Centerline{$T(a\action_rf)(z)=\psi(a)(Tf)(\pi(xa))
=\psi(a)(Tf)(zw)$}

\noindent \noindent whenever $\pi x=z$.   So

$$\eqalignno{\Delta_X(a^{-1})\int f\,d\mu
&=\int a\action_rfd\mu\cr
\noalign{\noindent (442Kc once more)}
&=\int T(a\action_rf)d\lambda=\psi(a)\int(Tf)(zw)\lambda(dz)\cr
&=\psi(a)\Delta_{Z}(w^{-1})\int Tf(z)\lambda(dz)
=\psi(a)\Delta_{Z}(w^{-1})\int f\,d\mu.\cr}$$

\noindent Thus $\Delta_X(a^{-1})=\psi(a)\Delta_{Z}(w^{-1})$;  because
both $\Delta_X$ and $\Delta_{Z}$ are multiplicative,

\Centerline{$\Delta_{Z}(\pi a)=\Delta_{Z}(w)=\psi(a)\Delta_X(a)$.}
}%end of proof of 443T

\leader{443U}{Transitive
\dvrocolon{actions}}\discrversionA{\footnote{Formerly 4{}43T.} }
{}\cmmnt{ All the results
from 443P onwards have been expressed in terms of groups acting on
quotient groups.   But the same structures can appear if we start from a
group action.   To simplify the hypotheses, I give the following result
for compact groups only.

\medskip

\noindent}{\bf Theorem} Let $X$ be a compact Hausdorff topological
group, $Z$ a non-empty compact Hausdorff space, and $\action$ a
transitive continuous action of $X$ on $Z$.
Write $\pi_z(x)=x\action z$ for $z\in Z$ and $x\in X$.

(a) For every $z\in Z$, $Y_z=\{x:x\in X,\,x\action z=z\}$ is a compact
subgroup of $X$.   If we give the set $X/Y_z$ of left cosets of $Y_z$ in
$X$ its quotient topology, we have a homeomorphism $\phi_z:X/Y_z\to Z$
defined by the formula $\phi_z(xY_z)=x\action z$ for every $x\in X$.

(b) Let $\mu$ be a Haar probability measure on $X$.   Then the image
measure $\mu\pi^{-1}_z$ is an $X$-invariant Radon probability measure on
$Z$, and $\mu\pi^{-1}_w=\mu\pi^{-1}_z$ for all $w$, $z\in Z$.

(c) Every non-zero $X$-invariant Radon measure on $Z$ is of the form
$\mu\pi^{-1}_z$ for a Haar measure $\mu$ on $X$ and some (therefore any)
$z\in Z$.

(d) There is a strictly positive $X$-invariant Radon probability measure
on $Z$, and any two non-zero $X$-invariant Radon measures on $Z$ are
scalar multiples of each other.

(e) Take any $z\in Z$, and let $\nu$ be the Haar probability measure of
$Y_z$.   If $\mu$ is a Haar measure on $X$, then

\Centerline{$\mu E=\int\nu(Y_z\cap x^{-1}E)\mu(dx)$}

\noindent whenever $E\subseteq X$ is Haar measurable.

\proof{{\bf (a)} Because $\action$ is an action of $X$ on $Z$,
$Y_z$ is always a subgroup;  because $\action$ is continuous, $Y_z$ is
closed, therefore compact.
Given $z\in Z$, then for $x$, $y\in X$ we have

\Centerline{$x\action z=y\action z
\iff x^{-1}y\in Y_z\iff xY_z=yY_z$.}

\noindent So the formula given for $\phi_z$ defines an injection from
$Z/Y_z$ to $Z$, which is surjective because $\action$ is transitive.
To see that $\phi_z$ is continuous, take any open set
$H\subseteq Z$.   Then

\Centerline{$\{x:xY_z\in\phi_z^{-1}[H]\}=\{x:x\action z\in H\}
=\pi_z^{-1}[H]$}

\noindent is open in $X$ (because $\action$ is continuous), so
$\phi_z^{-1}[H]$ is open in $X/Y_z$.
Because $X/Y_z$ is compact and $\phi_z$ is a bijection, $\phi_z$ is a
homeomorphism (3A3Dd).

\medskip

{\bf (b)} Because $X$ is compact, therefore unimodular (442Ic), we can
speak of `Haar measures' on $X$ without specifying `left' or `right'.
If $\mu$ is the Haar probability measure on $X$, then the image measure
$\mu\pi_z^{-1}$ is a Radon probability measure on $Z$ (418I once more).
To see
that the measures $\mu\pi_z^{-1}$ are $X$-invariant, take any Borel set
$H\subseteq Z$ and $y\in X$, and consider

$$\eqalign{(\mu\pi_z^{-1})(y^{-1}\action H)
&=\mu\{x:x\action z\in y^{-1}\action H\}
=\mu\{x:yx\action z\in H\}
=\mu\{x:yx\in\pi_z^{-1}[H]\}\cr
&=\mu(y^{-1}\pi_z^{-1}[H])
=\mu(\pi_z^{-1}[H])
=(\mu\pi_z^{-1})(H).\cr}$$

\noindent By 441B, this is enough to ensure that $\mu\pi_z^{-1}$ is
invariant.

If $w$, $z\in Z$ and $H\subseteq Z$ is a Borel set, then there is a
$y\in X$ such that $y\action w=z$, and now

\Centerline{$\pi_w^{-1}[H]=\{x:x\action w\in H\}
=\{x:(xy^{-1})\action z\in H\}=\{x:xy^{-1}\in\pi_z^{-1}[H]\}
=(\pi_z^{-1}[H])y$.}

\noindent But $\mu$ is a two-sided Haar measure, so

\Centerline{$(\mu\pi_w^{-1})(H)=\mu(\pi_w^{-1}[H])
=\mu((\pi_z^{-1}[H])y)=\mu(\pi_z^{-1}[H])=(\mu\pi_z^{-1})(H)$.}

\noindent Thus $\mu\pi_w^{-1}$ and $\mu\pi_z^{-1}$ agree on the Borel
sets and must be equal (416Eb).

\medskip

{\bf (c)} Now we come to the interesting bit.   Suppose that $\lambda$
is a non-zero $X$-invariant Radon measure on $Z$.   Take any $z\in Z$
and consider the Radon measure $\lambda'$ on $X/Y_z$ got by setting
$\lambda'H=\lambda\phi_z[H]$ whenever $H\subseteq X/Y_z$ and $\phi_z[H]$
is measured by $\lambda$.   In this case, if $x\in X$ and
$H\subseteq X/Y_z$ is measured by $\lambda'$,

$$\eqalignno{\lambda'(x\action H)
&=\lambda\{\phi_z(x\action w):w\in H\}
=\lambda\{\phi_z(x\action yY_z):y\in X,\,yY_z\in H\}\cr
&=\lambda\{\phi_z(xyY_z):y\in X,\,yY_z\in H\}
=\lambda\{xy\action z:y\in X,\,yY_z\in H\}\cr
&=\lambda\{x\action(y\action z):y\in X,\,yY_z\in H\}
=\lambda(x\action\phi_z[H])
=\lambda\phi_z[H]\cr
\displaycause{because $\lambda$ is $X$-invariant}
&=\lambda'H.\cr}$$

\noindent So $\lambda'$ is $X$-invariant.

Now let $\nu$ be the Haar probability measure on the compact Hausdorff
group $Y_z$.   By 443Qb, we have a (left) Haar measure $\mu$ on $X$
defined by the formula
$\mu G=\int\nu_wG\,\lambda'(dw)$ for every open $G\subseteq X$, where
$\nu_{xY_z}G=\nu(Y_z\cap x^{-1}G)$ for every $y\in X$ and every open
$G\subseteq X$.   Let $H\subseteq Z$ be an open set.   Then for any
$x\in X$,

$$\eqalign{\nu_{xY_z}(\pi_z^{-1}[H])
&=\nu\{y:y\in Y_z,\,xy\in\pi_z^{-1}[H]\}
=\nu\{y:y\in Y_z,\,xy\action z\in H\}\cr
&=\nu\{y:y\in Y_z,\,x\action(y\action z)\in H\}
=\nu\{y:y\in Y_z,\,x\action z\in H\}\cr
&=\nu Y_z
=1\text{ if }x\action z\in H,\cr
&=\nu\emptyset=0\text{ otherwise}.\cr}$$

\noindent So

\Centerline{$\mu(\pi_z^{-1}[H])
=\lambda'\{xY_z:x\in X,\,x\action z\in H\}
=\lambda'\phi_z^{-1}[H]
=\lambda H$.}

\noindent As $H$ is arbitrary, the image measure $\mu\pi_z^{-1}$ agrees
with $\lambda$ on the open subsets of $Z$;  as they are both Radon
measures, $\mu\pi_z^{-1}=\lambda$, as required.

\medskip

{\bf (d)} This is now easy.   $X$ carries a non-zero Haar measure, so by
(b) there is an $X$-invariant Radon probability measure on $Z$.
If $\lambda_1$ and $\lambda_2$ are non-zero $X$-invariant Radon measures
on $Z$, then they are of the form $\mu_1\pi_w^{-1}$ and
$\mu_2\pi_z^{-1}$ where $\mu_1$ and $\mu_2$ are Haar measures on $X$ and
$w$, $z\in Z$.
By (b) again, $\mu_1\pi_w^{-1}=\mu_1\pi_z^{-1}$, and since $\mu_1$ and
$\mu_2$ are multiples of each other (442B), so are $\lambda_1$ and
$\lambda_2$.

To see that the invariant probability measure $\lambda$ on $X$ is
strictly positive, take any non-empty open set $H\subseteq Z$.   Then
$Z$ is
covered by the open sets $x\action H$, as $x$ runs over $X$.
Because $Z$ is compact, it is covered by finitely many of these, so at
least one of them has non-zero measure.   But they all have the same
measure as $H$, so $\lambda H>0$.

\medskip

{\bf (e)} Write $\theta:X\to X/Y_z$ for the canonical map.   For
$w\in X/Y_z$ we have a Radon measure $\nu_w$ on $X$ defined by setting
$\nu_wE=\nu(Y_z\cap x^{-1}E)$ whenever $\theta x=w$ and the right-hand
side is defined (443Qa).   By (a)-(b) above, or otherwise, there is a
non-zero $X$-invariant Radon measure $\lambda$ on $X/Y_z$;  re-scaling
if necessary, we may suppose that $\lambda(X/Y_z)=\mu X$.   By 443Qe, we
have a Haar measure $\mu'$ on $X$ defined by setting
$\mu'E=\int\nu_wE\,\lambda(dw)$ for every Haar measurable $E$;  since

\Centerline{$\mu'X=\int\nu_wX\,\lambda(dw)
=\int\nu Y_z\,\lambda(dw)=\lambda(X/Y_z)=\mu X$,}

\noindent $\mu'=\mu$.   Moreover, $\lambda=\mu'\theta^{-1}$ (443Qd).
So

$$\eqalign{\mu E
&=\mu'E
=\int\nu_wE\,\lambda(dw)\cr
&=\int\nu_{\theta x}E\,\mu'(dx)
=\int\nu(Y_z\cap x^{-1}E)\mu(dx)\cr}$$

\noindent for every Haar measurable $E\subseteq X$.
}%end of proof of 443U

\exercises{\leader{443X}{Basic exercises $\pmb{>}$(a)}%
%\sqheader 443Xa
\dvAformerly{4{}43Xb} Let $X$ be a topological group, $\mu$ a left
Haar measure on $X$ and $\lambda$ the corresponding quasi-Radon product
measure on $X\times X$.   (i) Show that the maps
$(x,y)\mapsto(y,x)$,
$(x,y)\mapsto(x,xy)$, $(x,y)\mapsto(y^{-1}x,y)$ are automorphisms of the
measure space $(X\times X,\lambda)$.   \Hint{use 417C(b-v-$\beta$) to show that
they preserve the measures of open sets.}   (ii) Show that the maps
$(x,y)\mapsto(y^{-1},xy)$, $(x,y)\mapsto(yx,x^{-1})$,
$(x,y)\mapsto(y^3x,x^{-1}y^{-2})$ are automorphisms
of $(X\times X,\lambda)$.   \Hint{express them as compositions of maps
of the forms in (ii).}
%- %

\spheader 443Xb
Let $X$ be a topological group carrying Haar measures
and $A\subseteq X$.   (i) Show that
$A$ is self-supporting (definition:  411Na) for one Haar
measure on $X$ iff it is self-supporting for every Haar measure on $X$.
(ii) Show that $A$ has non-zero inner measure for one
Haar measure on $X$ iff it has non-zero inner measure for every Haar
measure on $X$.
%443A

\spheader 443Xc Let $X$ be a topological group, $\mu$ a Haar measure on
$X$, and $E$, $F$ measurable subsets of $X$.   Show that
$(x,y,w,z)\mapsto\mu(xEy\cap wFz):X^4\to[0,\infty]$ is lower
semi-continuous.
%443B

\spheader 443Xd Let $X$ be a topological group carrying Haar measures
and $\frak A$ its Haar measure algebra.
(i) Show that we have a continuous
action of $X\times X$ on $\frak A$ defined by the formula
$(x,y)\action E^{\ssbullet}=(xEy^{-1})^{\ssbullet}$ for $x$, $y\in X$
and Haar measurable sets $E\subseteq X$.
(ii) Show that if $x\in X$ and
$a\in\frak A$ then
$x\action_ra=(x\action_l\Reverse{a})\ssplrarrow$, where $\Reverse{a}$ is as
defined in 443Af.
%443C

\spheader 443Xe Let $(\frak A,\bar\mu)$ be a measure algebra.   Show that
it is isomorphic to the measure algebra of a topological group with a Haar
measure iff it is localizable and quasi-homogeneous in the sense of
374G-374H.
%443D

\spheader 443Xf Let $X$ be a topological group with a left Haar measure
$\mu$.   (i) Show that if $Y$ is a subgroup of $X$ such that $\mu_*Y>0$,
then $Y$ is open.   In particular, any non-negligible closed subgroup of
$X$ is open.
(ii) Let $Y$ be any subgroup of $X$ which is not Haar negligible.   Show
that the subspace measure $\mu_Y$ is a left Haar measure on $Y$.   Show
that $\overline{Y}$ is a Haar measurable envelope of $Y$.   \Hint{apply
443Db inside the topological group $\overline{Y}$.}
%443D %443F

\spheader 443Xg Write out a version of 443G for right Haar measures.
%443G

\spheader 443Xh Let $X$ be a topological group carrying Haar measures,
and $L^0$ the space of equivalence classes of Haar measurable functions,
as in 443A;  let $u\mapsto\Reverse{u}:L^0\to L^0$ be the operator of 443Af.
Show that if
$\mu$ is a left Haar measure on $X$ and $\nu$ is a right Haar measure,
$p\in[1,\infty]$ and $u\in L^0$, then $\Reverse{u}\in L^p(\nu)$ iff
$u\in L^p(\mu)$.
%443G

\spheader 443Xi Let $X$ be a topological group carrying Haar measures,
and $\frak A$ its Haar measure algebra.   Show that, in the language of
443C and 443G, $\chi(x\action_la)=x\action_l\chi a$ and
$\chi(x\action_ra)=x\action_r\chi a$ for every $x\in X$ and
$a\in\frak A$.
%443G

\spheader 443Xj Let $X$ be a topological group carrying Haar measures.
Show that $X$ is totally bounded for its bilateral uniformity
iff $X$ is totally bounded for its right uniformity\cmmnt{ (definition:
4A5Ha)} iff its Haar measures are totally finite.
%443H

\spheader 443Xk Let $X$ be a topological group, $\mu$ a left Haar
measure on $X$, and $A\subseteq X$ a set which is self-supporting for
$\mu$.   Show that the following are equiveridical:  (i) for every
neighbourhood $U$ of the identity $e$, there is a countable set
$I\subseteq X$ such that $A\subseteq UI$;  (ii) for every neighbourhood
$U$ of $e$, there is a countable set $I\subseteq X$ such that
$A\subseteq IU$;
(iii) for every neighbourhood $U$ of $e$, there is a countable set
$I\subseteq X$ such that $A\subseteq IUI$;   (iv) $A$ can be covered by
countably many sets of finite measure for $\mu$;  (v) $A$ can be covered
by countably many open sets of finite measure for $\mu$;  (vi) $A$ can
be covered by countably many sets which are totally bounded for the
bilateral uniformity on $X$.
%443I

\sqheader 443Xl Let $X$ be a topological group carrying Haar measures.
(i) Show that the following are equiveridical:  ($\alpha$) $X$ is ccc;
($\beta$) $X$ has a $\sigma$-finite Haar measure;  ($\gamma$) every Haar
measure on $X$ is $\sigma$-finite.
(ii) Show that if $X$ is locally compact and Hausdorff, we can add
($\delta$) $X$ is $\sigma$-compact.
%443Xk 443I

\spheader 443Xm Let $X$ be a topological group carrying Haar measures.
Show that every subset of $X$ has a Haar measurable envelope which is a
Borel set.
%443J

\spheader 443Xn In 443L, show that (i) $\phi[A]$ is Haar negligible in $Z$
whenever $A$ is Haar negligible in $X$ (ii) $\Delta_X=\Delta_Z\phi$, where
$\Delta_X$, $\Delta_Z$ are the left modular functions of $X$, $Z$
respectively (iii) $\phi[X]$ is dense in $Z$ (iv) $Z$ is unimodular iff $X$
is unimodular.
%443L

\sqheader 443Xo Let $X$ and $Y$ be topological groups with left Haar
measures $\mu$ and $\nu$.   Show that the c.l.d.\ and quasi-Radon
product measures of $\mu$ and $\nu$ on $X\times Y$ coincide.
\Hint{start with locally compact Hausdorff spaces, and show that a
compact G$_{\delta}$ set in $X\times Y$ belongs to
$\Cal B(X)\tensorhat\Cal B(Y)$, where $\Cal B(X)$ and $\Cal B(Y)$ are
the Borel $\sigma$-algebras of $X$ and $Y$;  now use 441Xj and 443L.}
%443M

\spheader 443Xp Let $\familyiI{X_i}$ be a family of topological groups
and $X=\prod_{i\in I}X_i$ their product.   Suppose that each $X_i$ has a
Haar probability measure $\mu_i$.   Show that the ordinary and
quasi-Radon product measures on $X$ coincide.
%443M

\spheader 443Xq Let $X$ be a locally compact Hausdorff group and $Y$ a
closed subgroup of $X$;  write $X/Y$ for the space of left cosets of $Y$
in $X$, with its quotient topology.   Show that if $\lambda_1$ and
$\lambda_2$ are non-zero $X$-invariant Radon measures on $X/Y$, then
each is a multiple of the other.   \Hint{look at the Haar measures they
define on $X$.}
%443Q

\sqheader 443Xr Write $S^1$ for the circle group
$\{s:s\in\Bbb C,\,|s|=1\}$, and set $X=S^1\times S^1$, where the first copy of $S^1$
is given its usual topology and the second copy is given its discrete
topology, so that $X$ is an abelian locally compact Hausdorff group.
Set $E=\{(s,s):s\in S^1\}$.
(i) Show that $E$ is a closed Haar negligible subset of $X$.
(ii) Set $Y=\{(1,s):s\in S^1\}$;  check that $Y$ is a closed
normal subgroup of $X$, and that the quotient group $X/Y$ can be
identified with $S^1$ with its usual topology;  let $\lambda$ be the
Haar probability measure of $X/Y$.   Let $\nu$ be counting measure on
$Y$.   Show that, in the language of 443Q, $\nu_zE=1$ for every
$z\in X/Y$, so that $\mu E\ne\int\nu_zE\,\lambda(dz)$.
(iii) Setting $f=\chi E$,
show that the map $a\mapsto f_a^{\ssbullet}$ described in 443Qf is not
almost continuous.
%443Q

\spheader 443Xs(i) In 443P, suppose that $G\subseteq X$ is an open set
such that $GY=X$.   Show that for every $h\in C_k(Z)^+$ there is an
$f\in C_k(X)^+$ such that $Tf=h$
and $\overline{\{x:f(x)>0\}}\subseteq G$.
(ii) In 443Pc, show that $\psi$ is multiplicative.
(iii) In 443R, suppose that
there is an open set $G\subseteq X$
such that $GY$ has finite measure for the left Haar measures
of $X$.   Show that $Z$ has an $X$-invariant Radon probability measure.
\Hint{$Y$ is totally bounded for its right uniformity.}
%443R

\spheader 443Xt Let $X$ be a locally compact Hausdorff group.   Show
that it has a closed normal subgroup $Y$ such that $Y$ and $X/Y$ are
both unimodular.   \Hint{take $Y=\{x:\Delta(x)=1\}$.}
%443Sa mt44bits

\sqheader 443Xu Let $X=\BbbR^2$ be the example of 442Xf.   (i) Let
$Y_1$ be the subgroup $\{(\xi,0):\xi\in\Bbb R\}$.   Describe the left
cosets of $Y_1$ in $X$.   Show that there is no non-trivial
$X$-invariant Radon measure on the set $X/Y_1$ of these left cosets.
Find a base $\Cal U$ for the topology of $X/Y_1$ such that you can
identify the sets $x\action U$, where $x\in X$ and $U\in\Cal U$, with
sufficient precision to explain why the hypothesis (iii) of 441C is not
satisfied.  (ii) Let $Y_2$ be the normal subgroup
$\{(0,\xi):\xi\in\Bbb R\}$.   Find
the associated function $\psi:X\to\ooint{0,\infty}$ as described in
443Pc and 443T.
%443T, 442Xf

\spheader 443Xv Let $X$ be a locally compact Hausdorff group and $Y$ a
compact normal subgroup of $X$.   Show that $X$ is unimodular iff the
quotient group $X/Y$ is unimodular.   \Hint{the function $\psi$ of 443T
must be constant.}
%443T

\spheader 443Xw Take any integer $r\ge 1$, and let $G$ be the isometry
group of $\BbbR^r$ with its topology of pointwise convergence (441G).
(i) Show that $G$ is metrizable and locally compact.   \Hint{441Xq.}
(ii) Let $H\subseteq G$ be the set of translations.   Show that
$H$ is an abelian closed normal subgroup of $G$, and that Lebesgue
measure on $\BbbR^r$ can be regarded as a Haar measure on $H$.   (iii)
Show that the quotient group $G/H$ is compact.   (iv) Show that $G$ is
unimodular.   \Hint{the function $\psi$ of 443T is constant.}
%443T

\sqheader 443Xx Set $X=\BbbR^3$ with the operation

\Centerline{$(\xi_1,\xi_2,\xi_3)*(\eta_1,\eta_2,\eta_3)
=(\xi_1+\eta_1,\xi_2+e^{\xi_1}\eta_2,\xi_3+e^{-\xi_1}\eta_3)$.}

\noindent (i) Show that (with the usual topology of $\BbbR^3$) $X$ is a
topological group.   (ii) Show that it is unimodular.   \Hint{Lebesgue
measure is a two-sided Haar measure.}
(iii) Show that $X$ has both a closed subgroup and a
Hausdorff quotient group which are not unimodular.
%443T, 442Xf

\sqheader 443Xy Let $(X,\rho)$ be a non-empty compact metric space such
that the group $G$ of isometries of $X$ is transitive.   Show that any
two non-zero
$G$-invariant Radon measures on $X$ must be multiples of each other.
\Hint{441Gb, 443U.}
%443U

\sqheader 443Xz Show that 443G is equally valid if we take functions to be
complex-valued rather than real-valued, and work with $L^p_{\Bbb C}$
rather than $L^p$.
%used in \S445

\leader{443Y}{Further exercises (a)}
%\spheader 443Ya
Let $X$ be a topological group carrying Haar measures
and $\frak A$ its Haar measure algebra.   Show that two principal ideals
of $\frak A$ are isomorphic (as Boolean algebras) iff they have the same
cellularity.
%443D

\spheader 443Yb Let $X$ be a topological group carrying Haar measures,
and $E\subseteq X$ a Haar measurable set such that $E\cap U$ is not Haar
negligible for any neighbourhood $U$ of the identity.   Show that for
any $A\subseteq X$ the set
$A'=\{x:x\in A,\,A\cap xE$ is Haar negligible$\}$ is Haar negligible.
\leaveitout{otherwise,
take a closed self-supporting non-negligible set $F$ of finite measure,
which is a Haar measurable envelope of $F\cap A'$,
and show that $F\cap xE$ is
negligible for every $x\in F$.   Consider the set
$\{(x,y):x\in F,\,y\in E,\,xy\in F\}$;  show that its vertical
sections are negligible but some of its horizontal sections are not.}
%443D

\spheader 443Yc Let $X$ be a locally compact Hausdorff group.   Show
that we have continuous shift
actions $\action_l$, $\action_r$ and $\action_c$
of $X$ on the Banach space $C_0(X)$ defined by formulae corresponding to
those of 443G.
%443G

\spheader 443Yd Let $X$ be a compact Hausdorff topological group and
$\frak A$ its Haar measure algebra.   Let $Y$ be a subgroup of $X$;
for $y\in Y$, define $\hat y\in\Aut\frak A$ by setting
$\hat y(a)=y\action_la$ for $a\in\frak A$.
Show that $\{\hat y:y\in Y\}$ is ergodic (definition:
395Ge) iff $Y$ is dense in $X$.
%443G

\spheader 443Ye Let $\frak A$ be a Boolean algebra, $G$ a group, and
$\action$ an action of $G$ on $\frak A$ such that $a\mapsto g\action a$ is
a Boolean automorphism for every $g\in G$.   (i) Show that we have a
corresponding action of $G$ on $L^{\infty}=L^{\infty}(\frak A)$
defined by saying that, for every $g\in G$,
$g\action\chi a=\chi(g\action a)$ for $a\in\frak A$ and
$u\mapsto g\action u$ is a positive linear operator on
$L^{\infty}$.   (ii) Show that if $\frak A$ is
Dedekind $\sigma$-complete,
this action on $L^{\infty}$ extends to an action on
$L^0=L^0(\frak A)$ defined by saying that
$\Bvalue{g\action u>\alpha}=g\action\Bvalue{u>\alpha}$ for $g\in G$,
$u\in L^0$ and $\alpha\in\Bbb R$.
%443G

\spheader 443Yf Let $(\frak A,\bar\mu)$ be a measure algebra,
$G$ a topological group, and
$\action$ a continuous action of $G$ on $\frak A$ (when $\frak A$ is
given its measure-algebra topology) such that $a\mapsto g\action a$ is
a measure-preserving Boolean automorphism for every $g\in G$.
(i) Show that the corresponding action of $G$ on $L^0=L^0(\frak A)$, as
defined in
443Ye, is continuous when $L^0$ is given the topology of convergence in
measure, and induces continuous actions of $G$ on
$L^p=L^p(\frak A,\bar\mu)$ for $1\le p<\infty$.
(ii) Show that if we give the
unit ball $B$ of $L^{\infty}=L^{\infty}(\frak A)$ the topology induced by
$\frak T_s(L^{\infty},L^1)$, then the action of $G$ on $L^0$ induces a
continuous action of $G$ on $B$.
%443Ye 443G

\spheader 443Yg Let $X$ be a topological group with a left Haar measure
$\mu$, and $A\subseteq X$.   Show that the following are equiveridical:
(i) $A$ is totally bounded for the bilateral uniformity of $X$ (ii)
there are non-empty open sets $G$, $H\subseteq X$ such that $\mu(AG)$,
$\mu(A^{-1}H)$ are both finite.
%443I

\spheader 443Yh
Give an example of a locally compact Hausdorff group, with left Haar
measure $\mu$, such that no open normal subgroup can be covered by a
sequence of sets of finite measure for $\mu$.
%443J mt43bits

\spheader 443Yi Let $X$ be a topological group.   Let $\Sigma$ be the
family of subsets of $X$ expressible in the form $\phi^{-1}[F]$ for some
Borel subset $F$ of a separable metrizable topological group $Y$ and
some continuous homomorphism $\phi:X\to Y$.   Show that $\Sigma$ is a
$\sigma$-algebra of subsets of $X$ and that multiplication, regarded as
a function from $X\times X$ to $X$, is
$(\Sigma\tensorhat\Sigma,\Sigma)$-measurable.   Show that any compact
G$_{\delta}$ set belongs to $\Sigma$.   Show that if $X$ is
$\sigma$-compact, then $\Sigma$ is the Baire $\sigma$-algebra of $X$.
%443L

\spheader 443Yj Let $X$ be any Hausdorff topological group with cardinal
greater than $\frak c$.   Let $\Cal B$ be the Borel $\sigma$-algebra of
$X$.   Show that $(x,y)\mapsto xy$ is not
$(\Cal PX\tensorhat\Cal PX,\Cal B)$-measurable.
\leaveitout{show that for
any sequence $\sequencen{A_n}$ in $\Cal PX$ there are distinct $x$,
$y\in X$ such that $\{n:x\in A_n\}=\{n:y\in A_n\}$;  consequently
$\{(x,y):xy^{-1}=e\}\notin\Cal PX\tensorhat\Cal PX$.}
%443Yi (443L)

\spheader 443Yk Let $X$ be a topological group and $\mu$ a left Haar
measure on $X$.   Show that $\mu$ is inner regular with respect to the
family of closed sets $F\subseteq X$ such that
$F=\bigcap_{n\in\Bbb N}FU_n$ for some sequence $\sequencen{U_n}$ of
neighbourhoods of the identity.   \leaveitout{use 443M
for the locally compact
case, and 443L to extend the result to general Haar measures.}
%443L, 443M

\spheader 443Yl
Let $X$ be a topological group carrying Haar measures.   Let
$E\subseteq X$ be a Haar
measurable set such that $E\cap U$ is not Haar negligible for any
neighbourhood $U$ of the identity.   Show that there is a sequence
$\sequencen{x_n}$ in $E$ such that
$x_{i_0}x_{i_1}\ldots x_{i_n}\in E$ whenever $n\in\Bbb N$ and
$i_0<i_1<\ldots<i_n$ in $\Bbb N$.   (See {\smc Plewik \& Voigt 91}.)
%443C, 443Yb, 443L, 443M, 443Yk mt43bits

\spheader 443Ym Let $X$ be a topological group and $\mu$ a Haar
measure on $X$.   Show that any closed self-supporting subset of $X$ is
a zero set.   \leaveitout{start with the case in which $X$ is
locally compact and $\sigma$-compact.}
%443M

\spheader 443Yn Find a compact Hausdorff space $X$ with a strictly
positive Radon measure such that there is a regular open set
$G\subseteq X$ which is not a cozero set.
%443N %mt44bits 419A

\spheader 443Yo Let $X$ be a locally compact Hausdorff topological
group which is not discrete (as topological space).   (i) Show that there
is a Haar negligible zero set containing the identity of $X$.   (ii) Show
that if $X$ is $\sigma$-compact, it
has a Haar negligible compact normal subgroup $Y$ which is a zero
set in $X$, so that $X/Y$ is metrizable.   (iii) Show that there is a Haar
negligible set $A\subseteq X$ such that $AA$ is not Haar measurable.
%419Xe, 443O(v) for (iii)

\spheader 443Yp Find a
non-discrete locally compact Hausdorff topological group $X$ such
that if $Y$ is a normal subgroup of $X$ which is a zero set in $X$ then $Y$
is open.
%443Yo 443O

\spheader 443Yq Show that there is a subgroup $X$ of the additive group
$\BbbR^2$ such that $X$ has full outer Lebesgue measure but
$\{\xi:(\xi,0)\in X\}=\Bbb Q$.   \leaveitout{enumerate the family
$\Cal K$ of
compact subsets of $\BbbR^2$ of positive Lebesgue measure as
$\langle K_{\xi}\rangle_{\xi<\frak c}$ (cf.\ 4A3Fa).   Show that
$\#(\pi_2[K])=\frak c$ for every $K\in\Cal K$, where
$\pi_2(\alpha_1,\alpha_2)=\alpha_2$ (Fubini's
theorem and 419H).   Choose $x_{\xi}\in K_{\xi}$, for $\xi<\frak c$,
such that $\pi_2x_{\xi}$ is never a rational linear combination of
$\{\pi_2x_{\eta}:\eta<\xi\}$.   Let $X$ be the group generated by
$\{x_{\xi}:\xi<\frak c\}\cup\{(q,0):q\in\Bbb Q\}$.}
Show that $X$ carries Haar measures, but that its closed subgroup
$X\cap(\Bbb R\times\{0\})$ does not.
%443P

\spheader 443Yr Show that if $X$ is a locally compact Hausdorff group,
$Y$ a compact subgroup of $X$, $Z=X/Y$ the set of left cosets of $Y$ and
$\pi:X\to Z$ the canonical map, and $Z$ is given its quotient
topology, then $R=\{(\pi x,x):x\in X\}$ is an usco-compact relation in
$Z\times X$ and $R[L]=\pi^{-1}[L]$ is compact
for every compact $L\subseteq Z$.
%443S

\spheader 443Ys Let $G$ be the isometry group of $\BbbR^r$, as in
443Xw.   (i) Show that
if we set $\rho(g,h)=\sup_{\|x\|\le 1}\|g(x)-h(x)\|$, then $\rho$ is a
metric on $G$ defining its topology.   (ii) Describe Haar measures
on $G$ ($\alpha$) in terms of Hausdorff measure of an appropriate
dimension for the metric $\rho$
($\beta$) in terms of a parametrization of $G$ and
Lebesgue measure on a suitable Euclidean space.
%443Xw, 443T

\spheader 443Yt Let $r\ge 1$ be an integer, and $G$ the group of
invertible affine transformations of $\BbbR^r$, with the topology of
pointwise convergence inherited from $(\BbbR^r)^{\BbbR^r}$.   (i) Show
that $G$ is a locally compact Hausdorff topological group.
(ii) Show that $G$ is
not unimodular, and find its modular functions.
%443Xw (443T) 443Ys 442Yb 443Xu

\spheader 443Yu Let $X$ be a topological group with a left Haar measure
$\mu$ and left modular function $\Delta$;  suppose that $X$ is not
unimodular.   Show that
$\mu\{x:\alpha<\Delta(x)<\beta\}=\infty$ whenever $\alpha<\beta$
and $\{x:\alpha<\Delta(x)<\beta\}$ is non-empty.
%443T %443L
}%end of exercises

\endnotes{
\Notesheader{443} Most of us, by the time we come to study measures on
general topological groups, have come to trust our intuition concerning
the behaviour of Lebesgue measure on $\Bbb R$, and the principal
discipline imposed by the subject is the search for the true path
between hopelessness and over-confidence when extending this intuition
to the general setting.   The biggest step is the loss of commutativity,
especially as the non-abelian groups of elementary courses in group
theory are mostly finite, and are therefore untrustworthy guides to the
concerns of this chapter.   Accordingly we find ourselves going rather
slowly and carefully through the calculations in such results as 443C.
Note, for instance, that in an abelian group the actions I call
$\action_l$ and $\action_r$ are still different;  $x\action_lE$
corresponds to $x+E=E+x$, but $x\action_rE$ corresponds to $E-x$.   (The
action $\action_c$ becomes trivial, of course.)   When we come to
translate the formulae of Fourier analysis in the next section,
manoeuvres of this kind will often be necessary.   In the present
section, I have done my best to give results in `symmetric' forms;  you
may therefore take it that when the words `left' or `right' appear in
the statement of a proposition, there is some real need to break the
symmetry.   Subject to these remarks, such results as 443C and 443G are
just a matter of careful conventional analysis.   I see that I have used
slightly different techniques in the two cases.   Of course 443C can be
thought of as a special case of 443Gc-443Gd, if we remember that
$\chi:\frak A\to L^0(\frak A)$ embeds $\frak A$ topologically as a
subspace of $L^0$ (367R).

443B, 443D and 443E belong to a different family;  they deal with actual
sets rather than with members of a measure algebra or a function space.
I suppose it is 443D which will most often be quoted.
%check when possible!  443D (2) 443B (0) 443E (0) 28.10.01
Its corollary 443E deals with the obvious question of when Haar measures
are tight.

Readers who have previously encountered the theory of Haar measures on
locally compact groups will have been struck by how closely the more
general theory here
is able to follow it.   The explanation lies in 443K-443L;  my general
Haar measures are really just subspace measures on subgroups of full
Haar measure in locally compact groups.   Knowing this, it is easy to
derive all the results above from the locally compact theory.   I use
this method only in 443M-443O, %443M 443N 443O
because (following {\smc Halmos 50}) I feel
that from the point of view of pure measure theory the methods show
themselves more clearly if we do not use ideas depending on compactness,
but instead rely directly on $\tau$-additivity.   But I note that 443Xo
and 443Yi also go faster with the aid of 443L.

{\smc Halmos 50} goes a little farther, with a theory of groups carrying
translation-invariant measures for which the operation $(x,y)\mapsto xy$
is $(\Sigma\tensorhat\Sigma,\Sigma)$-measurable, where $\Sigma$ is the
domain of the measure.   The essence of the theory here, and my reason
for insisting on $\tau$-additive measures, is that for these we
have a suitable theory of product measures.   If we start with
quasi-Radon measures and use the quasi-Radon product measure, then
multiplication is measurable just because it is continuous.   In order
to achieve similar results without either assuming metrizability or
using the theory of $\tau$-additive product measures, we have to
restrict the measure to something smaller than the Borel
$\sigma$-algebra, as in 443Yg.   (See 443Yj.)
443J and 443Xo show that certain disconcerting features of general
Radon measures (419C, 419E) cannot arise in the case of Haar measures.

When I come to look at subgroups and quotient groups, I do specialize to
the locally compact case.   One obstacle to generalization is the fact
that a closed subgroup
of a group carrying Haar measures need not itself carry Haar measures
(443Yq).   443P and 443R are taken from {\smc Federer 69}, who goes
farther, with many other applications.   I give a fairly detailed
analysis of the relationship between a Haar measure on a group $X$ and a
corresponding $X$-invariant measure on a family of left cosets (443Q)
for the sake of applications in \S447.   One of the challenges here is
to distinguish clearly those results which apply to all locally compact
groups from those which rely on $\sigma$-compactness or some such
limitation.   There is a standard example (443Xr) which provides a
useful test in such questions.   You may recognise this as a version of
a fundamental example related to Fubini's theorem, given in 252K.

Most of the results of this section begin with a topological group $X$.
But starting from \S441 it is equally natural to start with a group
action.   If we have a topological group $X$ acting continuously on a
topological space $Z$, and $X$ carries Haar measures, then we have a
good chance of finding an invariant measure on $Z$.   In the simplest
case, in which $X$ and $Z$ are both compact and the action is
transitive, there is a unique invariant Radon probability measure on $Z$
(443U).
}%end of notes

\discrpage



\frfilename{mt444.tex}
\versiondate{23.7.07/4.6.13}
\copyrightdate{1999}

\def\esssup{\mathop{\text{ess sup}}}
\def\ssplrarrow{^{\scriptscriptstyle\leftrightarrow}}

\def\chaptername{Topological groups}
\def\sectionname{Convolutions}

\newsection{444}

In this section, I look again at the ideas of \S\S255 and 257, seeking
the appropriate generalizations to topological groups other than
$\Bbb R$.   Following {\smc Hewitt \& Ross 63}, I begin with
convolutions of measures (444A-444E) %444A 444B 444C 444D 444E
before proceeding to convolutions of functions
(444O-444V); %444O 444P 444Q 444R 444T 444U 444S 444V
in between, I mention the convolution of a function and a measure
(444G-444M) %444G 444H 444I 444J 444K 444L 444M
and a general result concerning continuous group actions on
quasi-Radon measure spaces (444F).

While I continue to give the results in terms of
real-valued functions, the applications
of the ideas here in the next section will be to complex-valued functions;
so you may wish to keep the complex case in mind\cmmnt{ (444Xx)}.

\leader{444A}{Convolution of measures:  Proposition} If $X$ is a
topological group and $\lambda$ and $\nu$ are two totally finite
quasi-Radon measures on $X$, we have a quasi-Radon measure $\lambda*\nu$
on $X$ defined by saying that

$$\eqalign{(\lambda*\nu)(E)
&=(\lambda\times\nu)\{(x,y):xy\in E\}\cr
&=\int\nu(x^{-1}E)\lambda(dx)
=\int\lambda(Ey^{-1})\nu(dy)\cr}$$

\noindent for every $E\in\dom(\lambda*\nu)$, where $\lambda\times\nu$ is
the quasi-Radon product measure on $X\times X$.

\proof{ Set $\phi(x,y)=xy$ for $x$, $y\in X$.   Then $\phi$ is continuous,
while $X$, being a topological group, is regular (4A5Ha);  so
so 418Hb tells us that there is a unique quasi-Radon measure $\lambda*\nu$
on $X$ such that $\phi$ is \imp\ for $\lambda\times\nu$ and $\lambda*\nu$,
that is, $(\lambda*\nu)(E)=(\lambda\times\nu)\{(x,y):xy\in E\}$ whenever
$E$ is measured by $\lambda*\nu$.

As for the other formulae, Fubini's theorem (417Ga) tells us that

$$\eqalign{(\lambda*\nu)(E)
&=(\lambda\times\nu)\phi^{-1}[E]\cr
&=\int\nu(\phi^{-1}[E])[\{x\}]\lambda(dx)
   =\int\nu(x^{-1}E)\lambda(dx)\cr
&=\int\lambda(\phi^{-1}[E])^{-1}[\{y\}]\nu(dy)
   =\int\lambda(Ey^{-1})\nu(dy)\cr}$$

\noindent for any $E\in\dom(\lambda*\nu)$.
}%end of proof of 444A

\leader{444B}{Proposition} If $X$ is a topological group,
$\lambda_1*(\lambda_2*\lambda_3)=(\lambda_1*\lambda_2)*\lambda_3$ for
all totally finite quasi-Radon measures $\lambda_1$, $\lambda_2$ and
$\lambda_3$ on $X$.

\proof{ If $E\subseteq X$ is Borel, then

$$\eqalign{(\lambda_1*(\lambda_2*\lambda_3))(E)
&=\int(\lambda_2*\lambda_3)(x^{-1}E)\lambda_1(dx)\cr
&=\int(\lambda_2\times\lambda_3)\{(y,z):xyz\in E\}\lambda_1(dx)\cr
&=(\lambda_1\times(\lambda_2\times\lambda_3))\{(x,(y,z)):xyz\in E\}\cr
&=((\lambda_1\times\lambda_2)\times\lambda_3)\{((x,y),z):xyz\in E\}\cr
&=\int(\lambda_1\times\lambda_2)\{(x,y):xyz\in E\}\lambda_3(dz)\cr
&=((\lambda_1*\lambda_2)*\lambda_3)(E).\cr}$$

\noindent (For the central identification between
$\lambda_1\times(\lambda_2\times\lambda_3)$ and
$(\lambda_1\times\lambda_2)\times\lambda_3$, observe that as both are
quasi-Radon measures it is enough to check that they agree on sets of
the type $G_1\times(G_2\times G_3)\cong (G_1\times G_2)\times G_3$ where
$G_1$, $G_2$ and $G_3$ are open, as in 417J\query.)

So $\lambda_1*(\lambda_2*\lambda_3)$ and
$(\lambda_1*\lambda_2)*\lambda_3$ agree on the Borel sets and must be
identical.
}%end of proof of 444B

\leader{444C}{Theorem} Let $X$ be a topological group and $\lambda$,
$\nu$ two totally finite quasi-Radon measures on $X$.
Then

\Centerline{$\int fd(\lambda*\nu)
=\int f(xy)(\lambda\times\nu)d(x,y)
=\iint f(xy)\lambda(dx)\nu(dy)
=\iint f(xy)\nu(dy)\lambda(dx)$}

\noindent for any $(\lambda*\nu)$-integrable real-valued function $f$.
In particular, $(\lambda*\nu)(X)=\lambda X\cdot\nu X$.

%what about [-\infty,\infty]-valued functions?  see 447

\proof{ If $f$ is of the form $\chi E$, so that

\Centerline{$\int fd(\lambda*\nu)=(\lambda*\nu)(E)$,
\quad$\int f(xy)(\lambda\times\nu)d(x,y)
=(\lambda\times\nu)\{(x,y):xy\in E\}$,}

\Centerline{$\iint f(xy)\lambda(dx)\nu(dy)
=\int\lambda(Ey^{-1})\nu(dy)$,
\quad$\iint f(xy)\nu(dy)\lambda(dx)
=\int\nu(x^{-1}E)\lambda(dx)$}

\noindent this is covered by the result in 444A.   Now it is easy to run
through the standard progression to the cases of (i) simple functions
(ii) non-negative Borel measurable functions defined everywhere (iii)
functions defined, and zero, almost everywhere (iv) non-negative
integrable functions and (v) arbitrary integrable functions.
}%end of proof of 444C

\leader{444D}{Proposition} Let $X$ be an abelian topological group.
Then $\lambda*\nu=\nu*\lambda$ for all totally finite quasi-Radon
measures $\lambda$, $\mu$ on $X$.

\proof{ For any Borel set $E\subseteq X$,

$$\eqalign{(\lambda*\nu)(E)
&=(\lambda\times\nu)\{(x,y):xy\in E\}
=(\nu\times\lambda)\{(y,x):xy\in E\}\cr
&=(\nu\times\lambda)\{(y,x):yx\in E\}
=(\nu*\lambda)(E).\cr}$$
}%end of proof of 444D

\leader{444E}{The Banach algebra of $\tau$-additive
measures (a)} Let $X$ be a topological group.   \cmmnt{Recall from
437Ab that we have a band $C_b(X)^{\sim}_{\tau}$ in the $L$-space
$C_b(X)^{\sim}$
consisting of those order-bounded linear functionals $f:C_b(X)\to\Bbb R$
such that $|f|$
is smooth (equivalently, $f^+$ and $f^-$ are both smooth);  that is,
such that $|f|$, $f^+$ and $f^-$ can be represented by totally finite
quasi-Radon measures on $X$.}
\cmmnt{Because $X$ is completely regular (4A5Ha again),}
$C_b(X)^{\sim}_{\tau}$ can be identified with
the band $M_{\tau}$ of signed $\tau$-additive Borel measures on $X$,
that is, the set of those countably additive functionals $\nu$
defined on the Borel $\sigma$-algebra of $X$ such that $|\nu|$
is $\tau$-additive\cmmnt{ (437G)}.

\spheader 444Eb For any $\tau$-additive totally finite Borel measures
$\lambda$, $\nu$ on $X$ we can define their convolution $\lambda*\nu$ by
\cmmnt{the formulae of 444A, that is,}

\Centerline{$(\lambda*\nu)(E)
=\int\nu(x^{-1}E)\lambda(dx)
=\int\lambda(Ey^{-1})\nu(dy)$}

\noindent for any Borel set $E\subseteq X$\cmmnt{, if we note that the
completions $\hat\lambda$, $\hat\nu$ of $\lambda$ and $\nu$ are
quasi-Radon measures (415Cb), so that $\lambda*\nu$, as defined by these
formulae, is just the restriction of $\hat\lambda*\hat\nu$, as defined
in 444A, to the Borel $\sigma$-algebra}.   Now\cmmnt{ the formulae
make it
obvious that} the map $*$ is bilinear in the sense that

\Centerline{$(\lambda_1+\lambda_2)*\nu=\lambda_1*\nu+\lambda_2*\nu$,}

\Centerline{$\lambda*(\nu_1+\nu_2)=\lambda*\nu_1+\lambda*\nu_2$,}

\Centerline{$(\alpha\lambda)*\nu=\lambda*(\alpha\nu)
=\alpha(\lambda*\nu)$}

\noindent for all totally finite $\tau$-additive Borel measures
$\lambda$, $\lambda_1$, $\lambda_2$, $\nu$, $\nu_1$, $\nu_2$ and all
$\alpha\ge 0$.   Consequently\cmmnt{, regarding elements of $M_{\tau}$
as functionals on the Borel $\sigma$-algebra,} we have a bilinear operator
$*:M_{\tau}\times M_{\tau}\to M_{\tau}$ defined by saying that

\Centerline{$(\lambda_1-\lambda_2)*(\nu_1-\nu_2)
=\lambda_1*\nu_1-\lambda_1*\nu_2-\lambda_2*\nu_1+\lambda_2*\nu_2$}

\noindent for all $\lambda_1$, $\lambda_2$, $\nu_1$, $\nu_2\in
M_{\tau}^+$.

\spheader 444Ec
\cmmnt{We see from 444B that} $*$ is associative\cmmnt{ on
$M_{\tau}^+$, so it will be associative on the whole of $M_{\tau}$}.
\cmmnt{Observe
that $*$ is positive in the sense that $\lambda*\nu\ge 0$ if $\lambda$,
$\nu\ge 0$;  so that}
\dvro{$|\lambda*\nu|
\le|\lambda|*|\nu|$}
{
$$\eqalign{|\lambda*\nu|
&=|\lambda^+*\nu^+-\lambda^+*\nu^--\lambda^-*\nu^+
    +\lambda^-*\nu^-|\cr
&\le\lambda^+*\nu^++\lambda^+*\nu^-+\lambda^-*\nu^++\lambda^-*\nu^-\cr
&=|\lambda|*|\nu|\cr}$$

\noindent} for any $\lambda$, $\nu\in M_{\tau}$.

\spheader 444Ed
If $\lambda$, $\nu\in M_{\tau}^+$ then\cmmnt{ $\|\lambda\|=\lambda X$
and $\|\nu\|=\nu X$ (362Ba), so}

\dvro{\Centerline{$\|\lambda*\nu\|
=\|\lambda\|\|\nu\|$.}}
{\Centerline{$\|\lambda*\nu\|
=(\lambda*\nu)(X)
=(\lambda\times\nu)(X\times X)
=\lambda X\cdot\nu X
=\|\lambda\|\|\nu\|$.}}

\noindent Generally, for any $\lambda$, $\nu\in M_{\tau}$,

\Centerline{$\|\lambda*\nu\|
\cmmnt{\mskip5mu =\||\lambda*\nu|\|}
\le\cmmnt{\||\lambda|*|\nu|\|
=\||\lambda|\|\||\nu|\|
=}\|\lambda\|\|\nu\|$.}

\noindent\cmmnt{Thus }$M_{\tau}$ is a Banach algebra under the operation
$*$\cmmnt{, as well as being an $L$-space}.   If $X$ is abelian
then $M_{\tau}$ will be a commutative algebra\cmmnt{, by 444D}.

\leader{444F}{}\cmmnt{ In preparation for the next construction I give
a general result extending ideas already touched on in 443C and 443G.

\medskip

\noindent}{\bf Theorem} Let $X$ be a topological space, $G$ a
topological group and $\action$ a continuous action of $G$ on $X$.   For
$A\subseteq X$, $a\in G$ write $a\action A=\{a\action x:x\in A\}$.   Let
$\nu$ be a   measure on $X$.

(a) If $f:X\to[0,\infty]$ is lower semi-continuous, then
$a\mapsto\int a\action f\,d\nu:G\to[0,\infty]$ is lower semi-continuous.
\cmmnt{ (See 4A5Cc for the definition of $a\action f$.)}   In
particular, if $V\subseteq X$ is open, then
$a\mapsto\nu(a\action V):G\to[0,\infty]$ is lower semi-continuous.

(b) If $f:X\to\Bbb R$ is continuous, then
$a\mapsto(a\action f)^{\ssbullet}:G\to L^0$ is continuous, if
$L^0=L^0(\nu)$ is given the topology of convergence in measure.

(c) If $\nu$ is $\sigma$-finite and $E\subseteq X$ is a Borel set, then
$a\mapsto(a\action E)^{\ssbullet}:G\to\frak A$ is Borel
measurable, if
the measure algebra $\frak A$ of $\nu$ is given its measure-algebra
topology.

(d) If $\nu$ is $\sigma$-finite and $f:X\to\Bbb R$ is Borel measurable,
then $a\mapsto (a\action f)^{\ssbullet}:G\to L^0$ is Borel measurable.

(e) If $\nu$ is $\sigma$-finite, then

\quad(i)
$a\mapsto\nu(a\action E):G\to[0,\infty]$ is Borel measurable for any
Borel set $E\subseteq X$;

\quad (ii)
if $f:X\to\Bbb R$ is Borel measurable, then $Q=\{a:\int a\action f\,d\nu$
is defined in $[-\infty,\infty]\}$ is a Borel set, and
$a\mapsto\int a\action f\,d\nu:Q\to[-\infty,\infty]$ is Borel measurable.

\proof{{\bf (a)(i)} Note first that if $f:X\to[0,\infty]$ is Borel
measurable, then, for each $a\in G$, $a\action f$ is the composition of
$f$ with the continuous function $x\mapsto a^{-1}\action x$, so is Borel
measurable, and if $f$ is finite-valued then $(a\action f)^{\ssbullet}$
is defined in $L^0=L^0(\nu)$.

\medskip

\quad{\bf (ii)} Suppose that $f:X\to[0,\infty]$ is lower
semi-continuous, $\gamma\in\coint{0,\infty}$, $a\in G$ and
$\int a\action f>\gamma$.   Let
$\Cal U$ be the set of open neighbourhoods of $a$ in $G$.   For
$U\in\Cal U$, $x\in X$ set

\Centerline{$\phi_U(x)=\sup\{\inf_{b\in U,y\in V}(b\action f)(y):V$ is
an open neighbourhood of $x$ in $X\}$.}

\noindent Then $\phi_U$ is lower semi-continuous.   \Prf\ If
$\phi_U(x)>\alpha$, there is an open neighbourhood $V$ of $x$ such that
$\inf_{b\in U,y\in V}(b\action f)(y)\penalty-100>\alpha$;
now $\phi_U(y)>\alpha$ for every $y\in V$.\ \QeD\   If $U'\subseteq U$ then
$\phi_{U'}\ge\phi_U$,
so $\{\phi_U:U\in\Cal U\}$ is upwards-directed.   Also
$\sup_{U\in\Cal U}\phi_U=a\action f$ in $[0,\infty]^X$.   \Prf\ Of
course $\phi_U(x)\le (a\action f)(x)$ for every $x$.   If $x\in X$ and
$(a\action f)(x)>\alpha$, then
$\{y:f(y)>\alpha\}$ is an open set containing $a^{-1}\action x$, so
(because $\action$ is continuous) there are a $U\in\Cal U$ and an open
neighbourhood $V$ of $x$ such that $f(b^{-1}\action y)>\alpha$ whenever
$b\in U$ and $y\in V$;  in which case $\phi_U(x)\ge\alpha$.   As $\alpha$
is arbitrary, $\sup_{U\in\Cal U}\phi_U(x)=(a\action f)(x)$.\ \Qed

By 414Ba, $\int a\action f\,d\nu=\sup_{U\in\Cal U}\int\phi_Ud\nu$, and
there is a $U\in\Cal U$ such that $\int\phi_Ud\nu>\gamma$.   Now suppose
that $b\in U$;  then $\phi_U(x)\le (b\action f)(x)$ for every $x$, so
$\int b\action f\,d\nu>\gamma$.
This shows that $\{a:\int a\action f\,d\nu>\gamma\}$ is an open set in
$G$, so that $a\mapsto\int a\action f\,d\nu$ is lower semi-continuous.

\medskip

\quad{\bf (iii)} If $V\subseteq X$ is open, then $\chi V$ is lower
semi-continuous, and $\chi(a\action V)=a\action(\chi V)$ for every
$a\in G$.   So $a\mapsto\nu(a\action V)=\int a\action(\chi V)d\nu$ is
lower semi-continuous.

Thus (a) is true.

\medskip

{\bf (b)} Take any $a\in G$, $E\in\dom\nu$ such that $\nu E<\infty$ and
$\epsilon>0$.   Let $\Cal U$ be the family of open neighbourhoods of $a$
in $G$, and for $U\in\Cal U$ set

\Centerline{$H_U
=\interior\{x:|(b\action f)(x)-(a\action f)(x)|\le\epsilon$ whenever
$b\in U\}$.}

\noindent Then $\{H_U:U\in\Cal U\}$ is upwards-directed.   Also,
it has union $X$.   \Prf\ If $x\in X$ then, because
$(b,y)\mapsto f(b^{-1}\action y)$ is continuous, there are a
$U\in\Cal U$ and an open neighbourhood $V$ of $x$ such that
$|(b\action f)(y)-(a\action f)(x)|\le\bover12\epsilon$ whenever
$b\in U$ and $y\in V$.
But now $|(b\action f)(y)-(a\action f)(y)|\le\epsilon$ whenever
$b\in U$ and
$y\in V$, so that $H_U$ includes $V$, which contains $x$.\ \Qed

So there is a $U\in\Cal U$ such that
$\nu(E\setminus H_U)\le\epsilon$ (414Ea).   In this case, for any
$b\in U$, we must have

\Centerline{$\int_E\min(1,|b\action f-a\action f|)d\nu
\le\epsilon(1+\nu E)$.}

\noindent As $E$ and $\epsilon$ are arbitrary,
$b\mapsto (b\action f)^{\ssbullet}$ is continuous at $a$;  as $a$ is
arbitrary, it is continuous everywhere.   Thus (b) is true.

\medskip

{\bf (c)(i)} Let us start by supposing that $E$ is an open set and that
$\nu$ is totally finite.   In this case the function
$a\mapsto\nu(a\action E)$
is lower semi-continuous, by (a) above, therefore Borel measurable.
Now let $W\subseteq\frak A$ be an open set, and write
$H=\{a:a\in G,\,(a\action E)^{\ssbullet}\in W\}$.   For $m$,
$k\in\Bbb N$ set
$H_{mk}=\{a:2^{-m}k\le\nu(a\action E)<2^{-m}(k+1)\}$, so that $H_{mk}$
is Borel, and
$U_{mk}=G\setminus\overline{H_{mk}\setminus H}$, so that $U_{mk}$ is
open.   Let $H'$ be $\bigcup_{m,k\in\Bbb N}H_{mk}\cap U_{mk}$;  then
$H'$ is Borel and $H'\subseteq H$.   In fact $H'=H$.   \Prf\ If
$a\in H$, then $W$ is an open
set containing $(a\action E)^{\ssbullet}$.   Let $\delta>0$ be such that
$a\in W$ whenever
$\bar\nu(a\Bsymmdiff(a\action E)^{\ssbullet})\le\delta$,
where $\bar\nu$ is the measure on $\frak A$;
let $m$, $k\in\Bbb N$ be such that $2^{-m}\le\bover14\delta$ and
$2^{-m}k\le\nu(a\action E)<2^{-m}(k+1)$.   If we take $\nuprime$ to be
the indefinite-integral measure over $\nu$ defined by $\chi(a\action E)$,
then $\nuprime$ is a quasi-Radon measure (415Ob), so (by (a) again)
$U=\{b:\nuprime(b\action E)>2^{-m}(k-1)\}$ is an open set, and of course
it
contains $a$.   If $b\in U\cap H_{mk}$, then

$$\eqalign{\nu((b\action E)\symmdiff(a\action E))
&=\nu(b\action E)+\nu(a\action E)-2\nu((b\action E)\cap(a\action E))\cr
&\le 2^{-m}(k+1)+2^{-m}(k+1)-2\cdot 2^{-m}(k-1)
\le 4\cdot 2^{-m}
\le\delta,\cr}$$

\noindent so $(b\action E)^{\ssbullet}\in W$ and $b\in H$.   This shows
that $U\cap(H_{mk}\setminus H)=\emptyset$ and $U\subseteq U_{mk}$ and
$a\in U\cap H_{mk}\subseteq H'$.   As $a$ is arbitrary, $H=H'$.\ \QeD\
Thus $H$ is a
Borel subset of $G$.   As $W$ is arbitrary, the map
$a\mapsto(a\action E)^{\ssbullet}$ is Borel measurable.

\medskip

\quad{\bf (ii)} To extend this to a general $\sigma$-finite quasi-Radon
measure $\nu$, still
supposing that $E$ is open, let $h:X\to\Bbb R$ be a strictly positive
integrable function (215B(viii)) and $\nuprime$ the corresponding
indefinite-integral measure.   As in (i), this $\nuprime$ also is a
quasi-Radon measure.   Since $\nu$
and $\nuprime$ have the same domains and the same null ideals,
the Boolean algebra $\frak A$ is still the underlying algebra of the
measure algebra of $\nuprime$;  by 324H, the topologies on $\frak A$
induced by the measures $\bar\nu$, $\bar\nuprime$ are the same.   So we
can apply (i) to the measure $\nuprime$ to see that
$a\mapsto(a\action E)^{\ssbullet}:G\to\frak A$ is still Borel
measurable.

\medskip

\quad{\bf (iii)} Next, suppose that $E$ is expressible as
$\bigcup_{i\le n}V_{2i}\setminus V_{2i+1}$ where each $V_i$ is open.
Then $a\mapsto(a\action E)^{\ssbullet}$ is Borel measurable.   \Prf\ Set
$X'=X\times\{0,\ldots,2n+1\}$, with the product topology (giving
$\{0,\ldots,2n+1\}$ its discrete topology) and define a measure
$\nuprime$ on $X$ and an action of $G$ on $X'$ by setting

\Centerline{$\nuprime F=\sum_{i=0}^{2n+1}\nu\{x:(x,i)\in F\}$}

\noindent whenever $F\subseteq X'$ is such that
$\{x:(x,i)\in F\}\in\dom\nu$ for every $i\le 2n+1$,

\Centerline{$a\action(x,i)=(a\action x,i)$}

\noindent whenever $a\in G$, $x\in X$ and $i\le 2n+1$.   Then
$V=\{(x,i):i\le 2n+1,\,x\in V_i\}$ is an open set in $X'$, while
$\nuprime$
is a $\sigma$-finite quasi-Radon measure, as is easily checked;  so, by
(ii), the map
$a\mapsto(a\action V)^{\ssbullet}:G\to\frak A'$ is Borel measurable,
where $\frak A'$ is the measure algebra of $\nuprime$.   On the other
hand, we can identify $\frak A'$ with the simple power $\frak A^{2n+2}$
(322Lb), and the map

\Centerline{$\langle c_i\rangle_{i\le 2n+1}
\mapsto\sup_{i\le n}c_{2i}\Bsetminus c_{2i+1}:\frak A^{2n+2}\to\frak A$}

\noindent is continuous, by 323B.   So the map

\Centerline{$a\mapsto(a\action E)^{\ssbullet}
=\sup_{i\le n}(a\action V_{2i})^{\ssbullet}
  \Bsetminus(a\action V_{2i+1})^{\ssbullet}$}

\noindent is the composition of a Borel measurable function with a
continuous function, and is Borel measurable.\ \Qed

\medskip

\quad{\bf (iv)} Now the family $\Cal E$ of all those Borel sets
$E\subseteq X$ such that $a\mapsto(a\action E)^{\ssbullet}$ is Borel
measurable is closed under unions and intersections of monotonic
sequences.   \Prf\ ($\alpha$) If $\sequencen{E_n}$ is a non-decreasing
sequence in $\Cal E$ with union $E$, then

\Centerline{$(a\action E)^{\ssbullet}
=\sup_{n\in\Bbb N}(a\action E_n)^{\ssbullet}
=\lim_{n\to\infty}(a\action E_n)^{\ssbullet}$}

\noindent (323Ea) for every $a\in G$.   So
$a\mapsto(a\action E)^{\ssbullet}$ is
the pointwise limit of a sequence of Borel measurable functions into a
metrizable space (323Gb, because $(\frak A,\bar\nu)$ is
$\sigma$-finite), and is Borel measurable, by 418Ba.   Thus
$E\in\Cal E$.   ($\beta$) If $\sequencen{E_n}$ is a non-increasing
sequence in $\Cal E$ the same argument applies, since this time

\Centerline{$(a\action E)^{\ssbullet}
=\inf_{n\in\Bbb N}(a\action E_n)^{\ssbullet}
=\lim_{n\to\infty}(a\action E_n)^{\ssbullet}$}

\noindent (323Eb) for every $a\in G$.\ \Qed

Since $\Cal E$ contains all sets of the form
$\bigcup_{i\le n}V_i\cap F_i$ where every $V_i$ is open and every $F_i$
is closed, by (iii), $\Cal E$ must be the whole Borel $\sigma$-algebra,
by 4A3C(g-ii).

This completes the proof of (c).

\medskip

{\bf (d)(i)} We need the following extension of (c):  if
$\sequencen{E_n}$ is any sequence of Borel sets in $X$, then
$a\mapsto\sequencen{(a\action E_n)^{\ssbullet}}:
  G\mapsto\frak A^{\Bbb N}$ is Borel measurable.   \Prf\ I repeat the
idea of (c-iii) above.
On $X'=X\times\Bbb N$ define a measure $\nuprime$ by setting

\Centerline{$\nuprime F=\sum_{n=0}^{\infty}\nu\{x:(x,n)\in F\}$}

\noindent whenever $F\subseteq X'$ is such that
$\{x:(x,n)\in F\}\in\dom\nu$ for every $n\in\Bbb N$.   As before, it is
easy to check
that $\nuprime$ is a $\sigma$-finite quasi-Radon measure, if we give
$\Bbb N$ its discrete
topology and $X'$ the product topology.   As before, set
$a\action(x,n)=(a\action x,n)$ for $a\in G$, $x\in X$ and $n\in\Bbb N$,
to obtain a continuous action of $G$ on $X'$.   Applying (c) to this
action, the map $a\mapsto(a\action E)^{\ssbullet}:G\to\frak A'$ is
Borel measurable, where $\frak A'$ is the measure algebra of $\nuprime$
and $E=\{(x,n):n\in\Bbb N,\,x\in E_n\}$.   But we can identify $\frak A'$
(as Boolean algebra) with $\frak A^{\Bbb N}$, by 322L, as before;  so
that if we re-interpret $a\mapsto(a\action E)^{\ssbullet}:G\to\frak A'$
as $a\mapsto\sequencen{(a\action E_n)^{\ssbullet}}:G\to\frak A^{\Bbb N}$
it is still Borel measurable.   (As in (c-ii), this time using
323L, the measure-algebra topology of $\frak A'$ matches the product
topology on $\frak A^{\Bbb N}$.)\ \Qed

\medskip

\quad{\bf (ii)} Now suppose that $f:X\to[0,1]$ is Borel measurable.
Define $\sequencen{E_n}$ inductively by the formula

\Centerline{$E_n=\{x:x\in X,\,
(f-\sum_{i<n}2^{-i-1}\chi E_i)(x)\ge 2^{-n-1}\}$.}

\noindent Then every $E_n$ is a Borel set and
$f=\sum_{n=0}^{\infty}2^{-n-1}\chi E_n$.   Next, observe that the
function

\Centerline{$\sequencen{c_n}\mapsto\sum_{n=0}^{\infty}2^{-n-1}\chi c_n:
\frak A^{\Bbb N}\to L^0$}

\noindent is continuous, because each of the maps
$c\mapsto 2^{-n-1}\chi c$ is continuous (367R), addition is continuous
(245Da, 367Ma) and the
series is uniformly summable.   Accordingly we may think of the map
$a\mapsto (a\action f)^{\ssbullet}$ as the composition of the continuous
function
$\sequencen{c_n}\mapsto\sum_{n=0}^{\infty}2^{-n-1}\chi c_i$ with the
Borel measurable function
$a\mapsto\sequencen{(a\action E_n)^{\ssbullet}}$, and it is Borel
measurable.

\wheader{444F}{4}{2}{2}{30pt}

\quad{\bf (iii)} For general Borel measurable $f:X\to\Bbb R$, set
$q(t)=\bover12(1+\bover{t}{|t|+1})$, so that $q:\Bbb R\to\ooint{0,1}$ is
a homeomorphism, and set $p=q^{-1}:\ooint{0,1}\to\Bbb R$.   Then the
function $\bar p:P\to L^0$ is continuous, where
$P=\{u:u\in L^0,\,\Bvalue{u\in\ooint{0,1}\,}=1\}$ (367S).   But now

\Centerline{$a\action f=p\frsmallcirc q\frsmallcirc(a\action f)
=p\frsmallcirc(a\action(q\frsmallcirc f))$}

\noindent for every $a$, so that $a\mapsto (a\action f)^{\ssbullet}$ is
the composition of the Borel map
$a\mapsto(a\action(q\frsmallcirc f))^{\ssbullet}$ with the
continuous map $\bar p$, and is Borel measurable.

\medskip

{\bf (e)(i)} We need only recall that $\bar\nu:\frak A\to\Bbb R$ is
lower semi-continuous (323Cb), so that (applying (c) above)

\Centerline{$a\mapsto\nu(a\action E)=\bar\nu(a\action E)^{\ssbullet}$}

\noindent is a composition of Borel measurable functions and is Borel
measurable.   (Of course there are
much more direct arguments, using fragments of the proof above.)

\medskip

\quad{\bf (ii)} The point is that the maps $u\mapsto u^+$,
$u\mapsto u^-:L^0\to(L^0)^+$ are continuous (245Db, 367M), while
$u\mapsto\int u:(L^0)^+\to[0,\infty]$ is lower semi-continuous
(369Mb), therefore Borel measurable.   Accordingly
$a\mapsto\int (a\action f)^+=\int((a\action f)^{\ssbullet})^+$ and
$a\mapsto\int (a\action f)^-$ are Borel measurable functions from $G$ to
$[0,\infty]$, so that

\Centerline{$Q
=\{a:\min(\int (a\action f)^+,\int (a\action f)^-)<\infty\}$}

\noindent is a Borel set, and

\Centerline{$a\mapsto\int a\action f
=\int (a\action f)^+-\int (a\action f)^-:
Q\to[-\infty,\infty]$}

\noindent is Borel measurable.
}%end of proof of 444F

\vleader{48pt}{444G}{Corollary} Let $X$ be a topological group and $\nu$ a
$\sigma$-finite quasi-Radon measure on $X$.

(a) If $f:X\to\Bbb R$ is a Borel measurable function, then
$\{x:\int f(y^{-1}x)\nu(dy)$ is defined in $[-\infty,\infty]\}$ is a
Borel set in
$X$ and $x\mapsto\int f(y^{-1}x)\nu(dy)$ is Borel measurable.

(b) If $f$, $g:X\to\Bbb R$ are Borel measurable functions, then
$\{x:\int f(xy^{-1})g(y)\nu(dy)$ is defined in $[-\infty,\infty]\}$ is a
Borel set and $x\mapsto\int f(xy^{-1})g(y)\nu(dy)$ is Borel measurable.

(c) If $\nu$ is totally finite and $f:X\to\Bbb R$ is a bounded
continuous function, then $x\mapsto\int f(y^{-1}x)\nu(dy):X\to\Bbb R$ is
continuous.

\proof{{\bf (a)} Set $\Reverse{f}(x)=f(x^{-1})$ for $x\in X$ (4A5C(c-ii));
then $\Reverse{f}$ is Borel
measurable.   Let $\action_l$ be the left action of $X$ on itself.
Then, in the language of 444Fe,
$Q=\{x:\int x\action_l\Reverse{f}\,d\nu$ is defined in $[-\infty,\infty]\}$
is a Borel set, and $x\mapsto\int x\action_l\Reverse{f}\,d\nu$ is Borel
measurable.   But

\Centerline{$(x\action_l\Reverse{f})(y)
=\Reverse{f}(x^{-1}y)=f(y^{-1}x)$}

\noindent for all $x$, $y$, so
$\int x\action_l\Reverse{f}\,d\nu=\int f(y^{-1}x)\nu(dy)$ if either
integral is defined.

\medskip

{\bf (b)(i)} Set $\Reverse{\nu}E=\nu E^{-1}$ when this is defined, writing
$E^{-1}=\{x^{-1}:x\in E\}$ for $E\subseteq X$;  that is, $\Reverse{\nu}$ is
the image measure $\nu\phi^{-1}$, where $\phi(x)=x^{-1}$ for $x\in X$.
Because $\phi$ is a homeomorphism, $\Reverse{\nu}$ is a quasi-Radon measure.
By 235J,
$\int h\,d\Reverse{\nu}=\int h(x^{-1})\nu(dx)$ for any real-valued function
$h$ for which either integral is defined in $[-\infty,\infty]$.

\medskip

\quad{\bf (ii)} Now suppose that $f$ and $g$ are non-negative Borel
measurable functions from $X$ to $\Bbb R$.
Then $\Reverse{g}$ also is a non-negative Borel
measurable function.   We know from 444Fd that
$x\mapsto (x^{-1}\action_lf)^{\ssbullet}:X\to L^0(\Reverse{\nu})$
is a Borel measurable function;  now multiplication in $L^0$ is
continuous (367Mb), so the map
$x\mapsto((x^{-1}\action_lf)\times\Reverse{g})^{\ssbullet}$
is Borel measurable;  since integration is lower semi-continuous on
$(L^0)^+$,
$x\mapsto\int(x^{-1}\action_lf)\times\Reverse{g}\,d\Reverse{\nu}:
X\to[0,\infty]$ is Borel measurable.   But

\Centerline{$\int(x^{-1}\action_lf)\times\Reverse{g}\,d\Reverse{\nu}
=\int f(xy)g(y^{-1})\Reverse{\nu}(dy)
=\int f(xy^{-1})g(y)\nu(dy)$}

\noindent whenever any of the integrals is defined in
$[-\infty,\infty]$, so this is the function we needed to know about.

\medskip

\quad{\bf (iii)} For general Borel measurable functions $f$ and $g$, we
have

$$\eqalign{\int f(xy^{-1})g(y)\nu(dy)
&=\bigl(\int f^+(xy^{-1})g^+(y)\nu(dy)
+\int f^-(xy^{-1})g^-(y)\nu(dy)\bigr)\cr
&\qquad\qquad-\bigl(\int f^+(xy^{-1})g^-(y)\nu(dy)
+\int f^-(xy^{-1})g^+(y)\nu(dy)\bigr)\cr}$$

\noindent exactly when the subtraction can be done in
$[-\infty,\infty]$, so that $x\mapsto\int f(xy^{-1})g(y)\nu(dy)$ is a
difference of Borel measurable functions and is Borel measurable, with
Borel measurable domain.

\medskip

{\bf (c)} Continuing the argument of (a), if $f$ is bounded and
continuous and $\nu$ is
totally finite then all the functions $x\action_l\Reverse{f}$ are bounded
and continuous, so $\int x\action_l\Reverse{f}\,d\nu$ is defined and finite
for every $x$.   Next, the function
$x\mapsto(x\action_l\Reverse{f})^{\ssbullet}:X\to L^0(\nu)$ is
continuous for the topology of convergence in measure (444Fb), which
agrees with the norm topology of $L^1(\nu)$ on $\|\,\|_{\infty}$-bounded
sets (246Jb).   It follows that

\Centerline{$x\mapsto\int (x\action_l\Reverse{f})^{\ssbullet}
=\int f(y^{-1}x)\nu(dy)$}

\noindent is continuous.
}%end of proof of 444G

\leader{444H}{Convolutions of measures and functions}\cmmnt{ I
introduce some notation which I shall use for the rest of the section.}
Let $X$ be a topological group.   If $f$ is a real-valued function
defined on a subset of $X$, and $\nu$ is a measure on $X$, set

\Centerline{$(\nu*f)(x)=\int f(y^{-1}x)\nu(dy)$}

\noindent whenever the integral is defined in $\Bbb R$.

\leader{444I}{Proposition} Let $X$ be a topological group and $\lambda$,
$\nu$ two totally finite quasi-Radon measures on $X$.

(a) For any Borel measurable function $f:X\to\Bbb R$, $\nu*f$ is a Borel
measurable function with a Borel domain.

(b) $\nu*f\in C_b(X)$ for every $f\in C_b(X)$.

(c) For any real-valued function $f$ defined on a subset of $X$,
$(\lambda*(\nu*f))(x)=((\lambda*\nu)*f)(x)$ whenever the right-hand side
is defined.

\proof{{\bf (a)} This follows at once from 444Ga.

\medskip

{\bf (b)} This is just a restatement of 444Gc.

\medskip

{\bf (c)} If $((\lambda*\nu)*f)(x)$ is defined, then

$$\eqalignno{((\lambda*\nu)*f)(x)
&=\int f(t^{-1}x)(\lambda*\nu)(dt)
=\iint f((yz)^{-1}x)\nu(dz)\lambda(dy)\cr
\displaycause{444C}
&=\iint f(z^{-1}y^{-1}x)\nu(dz)\lambda(dy)\cr
&=\int(\nu*f)(y^{-1}x)\lambda(dy)
=(\lambda*(\nu*f))(x).\cr}$$
}%end of proof of 444I

\leader{444J}{Convolutions of functions and measures}
Let $X$ be a topological group carrying Haar measures;  let $\Delta$ be
its left modular function\cmmnt{ (442I)}.   If $f$ is a real-valued
function defined on a subset of $X$, and $\nu$ is a measure on $X$, set

\Centerline{$(f*\nu)(x)=\int f(xy^{-1})\Delta(y^{-1})\nu(dy)$}

\noindent whenever the integral is defined in $\Bbb R$.
\cmmnt{From 444Gb we see that if $\nu$ is a $\sigma$-finite
quasi-Radon measure and
$f$ is Borel measurable, then $f*\nu$ is a Borel measurable function
with a Borel domain.}   If $f$ is
non-negative and $\nu$-integrable, write $f\nu$ for the corresponding
indefinite-integral measure over $\nu$\cmmnt{ (234J\formerly{2{}34B})}.

\leader{444K}{Proposition} Let $X$ be a topological group with a left
Haar measure $\mu$.   Let $\nu$ be a totally finite
quasi-Radon measure on $X$.   Then for any non-negative $\mu$-integrable
real-valued function $f$, $f\mu$ is a quasi-Radon measure;  moreover,
$\nu*f$ and $f*\nu$ are $\mu$-integrable, and we have

\Centerline{$(\nu*f)\mu=\nu*f\mu$,
\quad$(f*\nu)\mu=f\mu*\nu$.}

\noindent In particular,
$\int\nu*f\,d\mu=\int f*\nu\,d\mu=\nu X\cdot\int fd\mu$.

\proof{{\bf (a)} $f\mu$ is a quasi-Radon measure by 415Ob.

\medskip

{\bf (b)} Suppose first that $f$ is Borel measurable and
defined everywhere on $X$, as well as being non-negative and
$\mu$-integrable.

\medskip

\quad{\bf (i)} Let $E\subseteq X$ be a Borel set such that $\mu E<\infty$.
The function
$(x,y)\mapsto f(x^{-1}y)\chi E(y):X\times X\to\coint{0,\infty}$ is Borel
measurable, so

$$\eqalignno{(\nu*f\mu)(E)
&=\int(f\mu)(x^{-1}E)\nu(dx)\cr
\noalign{\noindent (444A)}
&=\iint f(y)\chi(x^{-1}E)(y)\mu(dy)\nu(dx)\cr
\noalign{\noindent (by the definition of $f\mu$, 234I\formerly{2{}34A})}
&=\iint f(x^{-1}y)\chi(x^{-1}E)(x^{-1}y)\mu(dy)\nu(dx)\cr
\noalign{\noindent (441J)}
&=\iint f(x^{-1}y)\chi E(y)\mu(dy)\nu(dx)
=\iint f(x^{-1}y)\chi E(y)\nu(dx)\mu(dy)\cr}$$

\noindent (by Fubini's theorem, 417Ga, because
$(x,y)\mapsto f(x^{-1}y)\chi E(y)$ is non-negative and Borel measurable
and zero outside $X\times E$).   So
$\int f(x^{-1}y)\nu(dx)$ must be finite for $\mu$-almost every $y\in E$.
Because $\mu$ is complete and locally determined and inner regular with
respect to the Borel sets of finite measure,
$(\nu*f)(y)=\int f(x^{-1}y)\nu(dx)$ is defined in $\Bbb R$ for $\mu$-almost
every $y\in X$.   So we have an indefinite-integral measure $(\nu*f)\mu$.
Next, we have

$$\eqalign{\infty
&>\nu X\int fd\mu
=(\nu*f\mu)(X)
\ge(\nu*f\mu)(E)\cr
&=\iint f(x^{-1}y)\chi E(y)\nu(dx)\mu(dy)
=\iint(\nu*f)(y)\chi E(y)\mu(dy)
=((\nu*f)\mu)(E)\cr}$$

\noindent for every Borel set $E$ such that $\mu E$ is finite.   Again
because $\mu$ is inner regular with respect to the Borel sets of finite
measure, $\nu*f$ is $\mu$-integrable and $(\nu*f)\mu$ is totally finite.
Since $\nu*f\mu$ and $(\nu*f)(\mu)$ are totally finite quasi-Radon measures
agreeing on open sets of finite measure for $\mu$, and $\mu$ is locally
finite (442Aa), 415H(iv) assures us that $\nu*f\mu=(\nu*f)(\mu)$.

\medskip

\quad{\bf (ii)} Now consider $f*\nu$.   This time, if
$E\subseteq X$ is Borel and $\mu E<\infty$,

$$\eqalignno{(f\mu*\nu)(E)
&=\int(f\mu)(Ey^{-1})\nu(dy)
=\iint_{Ey^{-1}}f(x)\mu(dx)\nu(dy)\cr
&=\int\Delta(y^{-1})\int_Ef(xy^{-1})\mu(dx)\nu(dy)\cr
\noalign{\noindent (by 442Kc)}
&=\int_E\int\Delta(y^{-1})f(xy^{-1})\nu(dy)\mu(dx).\cr}$$

\noindent Once again, we see that $\int\Delta(y^{-1})f(xy^{-1})\nu(dy)$ is
defined for $\mu$-almost every $x\in E$;  as $E$ is arbitrary,
$(f*\nu)(x)$ is defined in $\Bbb R$ for $\mu$-almost every $x\in X$;
and 444Gb tells us that $f*\nu$ is Borel measurable.   As before,

\Centerline{$\int_Ef*\nu\,d\mu\le(f\mu*\nu)(X)
=\int fd\mu\cdot\nu X<\infty$}

\noindent for Borel sets $E$ with $\mu E<\infty$, so that $f*\nu$ is
$\mu$-integrable;  as before, the quasi-Radon measures $(f*\nu)\mu$ and
$f\mu*\nu$ agree on open sets of finite $\mu$-measure, so coincide.

\medskip

{\bf (c)} Next, suppose that $f$ is defined and zero $\mu$-a.e.   In
this case there is a Borel set $E$ such that $\mu E=0$ and $f(x)$ is
defined and equal to zero for every $x\in X\setminus E$ (443J(b-ii)).
Set $g=\chi E$.   Then $g\mu$ is the zero measure, so
$(\nu*g)\mu=\nu*g\mu$, $(g*\nu)\mu=g\mu*\nu$ are all zero;  that is,
there is some $\mu$-conegligible set $F$ such that

\Centerline{$0=(\nu*g)(x)=\int\chi E(y^{-1}x)\nu(dy)
=\nu(xE^{-1})$,}

\Centerline{$0=(g*\nu)(x)=\int\chi E(xy^{-1})\Delta(y^{-1})\nu(dy)
=\int_{E^{-1}x}\Delta(y^{-1})\nu(dy)$}

\noindent for every $x\in F$.   But now, for $x\in F$, we must have
$\nu(xE^{-1})=\nu(E^{-1}x)=0$ (since $\Delta$ is strictly positive), so
that

\Centerline{$(\nu*f)(x)=\int f(y^{-1}x)\nu(dy)
=\int_{xE^{-1}}f(y^{-1}x)\nu(dy)=0$,}

\noindent because if $y\notin xE^{-1}$ then $y^{-1}x\notin E$ and
$f(y^{-1}x)=0$.   Similarly,

\Centerline{$(f*\nu)(x)=\int f(xy^{-1})\Delta(y^{-1})\nu(dy)
=\int_{E^{-1}x}f(xy^{-1})\Delta(y^{-1})\nu(dy)=0$.}

\noindent Thus $\nu*f$ and $f*\nu$ are defined, and zero, $\mu$-almost
everywhere.

\medskip

{\bf (d)} For an arbitrary non-negative $\mu$-integrable function $f$,
we can express it in the form $g+h$ where $g$ is a non-negative
$\mu$-integrable Borel measurable function defined everywhere, and $h$
is zero almost everywhere.   In this case, $\nu*h^+$, $\nu*h^-$,
$h^+*\nu$ and $h^-*\nu$ are defined, and zero, $\mu$-a.e., so
$\nu*f\eae\nu*g$ and $f*\nu\eae g*\nu$.   We therefore have

\Centerline{$(\nu*f)\mu=(\nu*g)\mu=\nu*g\mu=\nu*f\mu$,
\quad$(f*\nu)\mu=(g*\nu)\mu=g\mu*\nu=f\mu*\nu$,}

\noindent as required.

\medskip

{\bf (e)} Finally, we have

\Centerline{$\int\nu*fd\mu=((\nu*f)\mu)(X)=(\nu*f\mu)(X)
=\nu X\cdot(f\mu)(X)=\nu X\cdot\int fd\mu$,}

\Centerline{$\int f*\nu\,d\mu=((f*\nu)\mu)(X)=(f\mu*\nu)(X)
=(f\mu)(X)\cdot\nu X=\nu X\cdot\int fd\mu$.}
}%end of proof of 444K

\leader{444L}{Corollary} Let $X$ be a topological group carrying Haar
measures.   Suppose that $\nu$ is a non-zero quasi-Radon measure on $X$
and $E\subseteq X$ is a Haar measurable set such that $\nu(xE)=0$ for every
$x\in X$.   Then $E$ is Haar negligible.

\proof{ Let $\mu$ be a left Haar measure on $X$.   There is a non-zero
totally finite quasi-Radon measure $\nuprime$ on $X$ such that
$\nuprime(xE)=0$
for every $x\in X$.   \Prf\ Take any $F$ such that $0<\nu F<\infty$, and
set $\nuprime H=\nu(H\cap F)$ whenever this is defined.\ \QeD\   Let $G$
be
any Borel set such that $\mu G<\infty$, and set $f=\chi(G\cap E^{-1})$.
Then $f$ is $\mu$-integrable, and

\Centerline{$(\nuprime*f)(x)
=\int\chi(G\cap E^{-1})(y^{-1}x)\nuprime(dy)
=\nuprime(xG^{-1}\cap xE)=0$}

\noindent for every $x\in X$.   By 444K,
$\nuprime*f\mu=(\nuprime*f)\mu$ is the zero
measure, and $(f\mu)(X)=0$, that is, $\mu(G\cap E^{-1})=0$.   As $G$ is
arbitrary, $\mu E^{-1}=0$ and $E$ is Haar negligible (442H).
}%end of proof of 444L

\leader{444M}{Proposition} Let $X$ be a topological group and $\mu$ a
left Haar measure on $X$.   Let $\nu$ be a quasi-Radon
measure on $X$ and $p\in[1,\infty]$.

(a) Suppose that $\nu X<\infty$.   Then
we have a bounded positive linear operator
$u\mapsto\nu*u:L^p(\mu)\to L^p(\mu)$, of norm at most $\nu X$, defined
by saying that $\nu*f^{\ssbullet}=(\nu*f)^{\ssbullet}$ for every
$f\in\eusm L^p(\mu)$.

(b) Set $\gamma=\int\Delta(y)^{(1-p)/p}\nu(dy)$ if $p<\infty$,
$\int\Delta(y)^{-1}\nu(dy)$ if $p=\infty$, where $\Delta$ is the left
modular function of $X$.   Suppose that $\gamma<\infty$.   Then we have
a bounded positive linear operator
$u\mapsto u*\nu:L^p(\mu)\to L^p(\mu)$, of norm at most $\gamma$, defined
by saying that $f^{\ssbullet}*\nu=(f*\nu)^{\ssbullet}$ for every
$f\in\eusm L^p(\mu)$.

\proof{ I will write $\eusm L^p$, $L^p$ for $\eusm L^p(\mu)$, $L^p(\mu)$.
Note that if $f_1$, $f_2\in\eusm L^0(\mu)$ and $f_1=f_2\,\,\mu$-a.e., then
444L tells us that $\nu*|f_1-f_2|$ and $|f_1-f_2|*\nu$ are both zero
$\mu$-almost everywhere, so that $\nu*f_1\eae\nu*f_2$ and
$f_1*\nu\eae f_2*\nu$, in the sense that there is a $\mu$-conegligible set
$F$ such that $(\nu*f_1)\restr F=(\nu*f_2)\restr F$ and
$(f_1*\nu)\restr F=(f_2*\nu)\restr F$.   In particular, if we are told that
$\nu*f_1$ belongs to $\eusm L^p$, and that
$f_1^{\ssbullet}=f_2^{\ssbullet}$ in $L^0(\mu)$, then we can be sure that
$\nu*f_2\in\eusm L^p$ and $(\nu*f_2)^{\ssbullet}=(\nu*f_1)^{\ssbullet}$;
and similarly for $f_1*\nu$, $f_2*\nu$.

\wheader{444M}{6}{2}{2}{48pt}

{\bf (a)} If $\nu X=0$ the result is trivial.   Multiplying $\nu$ by a
positive scalar does not affect the inequalities we need, so we may suppose
that $\nu X=1$.   If $f\ge 0$ is $\mu$-integrable, then 444K tells us that
$\nu*f$ is $\mu$-integrable and that

$$\eqalign{\|\nu*f\|_1&=\int_X(\nu*f)(x)\mu(dx)=((\nu*f)\mu)(X)\cr
&=(\nu*f\mu)(X)=\nu X\cdot(f\mu)(X)=\|f\|_1,\cr}$$

\noindent using 444C or 444A for the penultimate equality.   Since
evidently $\nu*(f+g)=\nu*f+\nu*g$, $\nu*(\alpha f)=\alpha\nu*f$ at any
point where
the right-hand sides of the equations are defined in $\Bbb R$,
we have a positive
linear operator $T_1:L^1\to L^1$ defined by saying that
$T_1g^{\ssbullet}=(\nu*g)^{\ssbullet}$ for every $\mu$-integrable Borel
measurable function $g$, with $\|T_1\|=1$.

Similarly, if $h:X\to\Bbb R$ is a bounded Borel measurable function,
then $\nu*h$ also is a Borel
measurable function, by 444Ga.   Of course it is bounded, since

\Centerline{$|(\nu*h)(x)|=|\int h(y^{-1}x)\nu(dy)|
\le\sup_{y\in X}|h(y)|$}

\noindent for every $x$.   So we have a positive linear operator
$T_{\infty}:L^{\infty}\to L^{\infty}$ defined by saying that
$T_{\infty}h^{\ssbullet}=(\nu*h)^{\ssbullet}$ for every bounded Borel
measurable function $h$.   Moreover, if $u\in L^{\infty}$, there is a
Borel measurable $h$ such that $h^{\ssbullet}=u$ and
$\sup_{y\in X}|h(y)|=\|u\|_{\infty}$, so that

\Centerline{$\|T_{\infty}u\|_{\infty}
\le\sup_{x\in X}|(\nu*h)(x)|\le\|u\|_{\infty}$;}

\noindent thus $\|T_{\infty}\|\le 1$.

Since $T_1$ and $T_{\infty}$ agree on $L^1\cap L^{\infty}$, they have a
common extension to a linear operator
$T:L^1+L^{\infty}\to L^1+L^{\infty}$.   By 371Gd, $\|Tu\|_p\le\|u\|_p$
whenever $p\in[1,\infty]$ and $u\in L^p$.   (Strictly speaking, I am
relying on the standard identifications of $L^1$, $L^{\infty}$ and $L^p$
with the corresponding subspaces of $L^0(\frak A)$, where
$(\frak A,\bar\mu)$ is the measure algebra of $\mu$.   Of course the
argument for 371Gd applies equally well in $L^0(\mu)$.)   Now suppose
that $f\in\eusm L^p$.   Then it is expressible as $g+h$ where
$g\in\eusm L^1$ and $h:X\to\Bbb R$ is a bounded Borel measurable
function, so we shall have

\Centerline{$\nu*f=\nu*g+\nu*h$ wherever the right-hand side is
defined;}

\noindent accordingly $\nu*f$ is defined $\mu$-a.e.\ and is measurable, and

$$\eqalign{\|\nu*f\|_p
&=\|(\nu*g)^{\ssbullet}+(\nu*h)^{\ssbullet}\|_p
=\|T_1g^{\ssbullet}+T_{\infty}h^{\ssbullet}\|_p\cr
&=\|Tf^{\ssbullet}\|_p
\le\|f^{\ssbullet}\|_p
=\|f\|_p,\cr}$$

\noindent as required.

\medskip

{\bf (b)(i)} As in (a), the case $\nu X=0$ is trivial.   Otherwise, because
$\Delta$ is strictly positive, $\gamma>0$;  again considering a scalar
multiple of $\nu$ if necessary, we may suppose that $\gamma=1$.   Note that
$\nu$ is surely $\sigma$-finite.

\medskip

\quad{\bf (ii)} If $p=1$, then $\nu X=\gamma=1$.   If $f\in\eusm L^1$ is
non-negative, then, by 444K, as in (a) above,

\Centerline{$\|f*\nu\|_1=((f*\nu)\mu)(X)
=(f\mu*\nu)(X)=(f\mu)(X)\cdot\nu X=\|f\|_1$.}

\noindent For general $\mu$-integrable $f$,

\Centerline{$\|f*\nu\|_1\le\|f^+*\nu\|_1+\|f^-*\nu\|_1
=\|f^+\|_1+\|f^-\|_1=\|f\|_1$.}

\medskip

\quad{\bf (iii)} If $p=\infty$, then directly from the formula
$(f*\nu)(x)=\int f(xy^{-1})\Delta(y^{-1})\nu(dy)$ we see that if
$f:X\to\Bbb R$ is a bounded Borel measurable function then

\Centerline{$|(f*\nu)(x)|
\le\int\Delta(y)^{-1}\nu(dy)\cdot\sup_{y\in X}|f(y)|
=\sup_{y\in X}|f(y)|$}

\noindent for every $x$.   Since changing $f$ on a $\mu$-negligible set
changes $f*\nu$ on a $\mu$-negligible set, we can argue as in (a) above
to see that $f^{\ssbullet}\mapsto(f*\nu)^{\ssbullet}$ defines a linear
operator from $L^{\infty}$ to itself of norm at most $1$.

\medskip

\quad{\bf (iv)} Now suppose that $1<p<\infty$.   Set
$q=\bover{p}{p-1}$, so that $\int\Delta(y^{-1})^{1/q}\nu(dy)=\gamma=1$.
Suppose for the moment that $f\in\eusm L^p$ is a non-negative Borel
measurable function, and let $h:X\to\Bbb R$
be another non-negative Borel measurable function such that
$\int h^qd\mu\le 1$.   In this case

$$\eqalignno{\int(f*\nu)\times h\,d\mu
&=\iint h(x)f(xy^{-1})\Delta(y^{-1})\nu(dy)\mu(dx)\cr
&=\iint h(x)f(xy^{-1})\Delta(y^{-1})\mu(dx)\nu(dy)\cr
\noalign{\noindent (by 417Ga, because
$(x,y)\mapsto h(x)f(xy^{-1})\Delta(y^{-1})$ is Borel measurable and
$\{x:h(x)\ne 0\}$
is a countable union of sets of finite measure for $\mu$, while $\nu$ is
$\sigma$-finite)}
&=\iint h(xy)f(x)\mu(dx)\nu(dy)\cr}$$

\noindent by 442Kc, as usual, at least if the last integral is finite.
But, for any $y\in X$,

$$\eqalign{\int h(xy)f(x)\mu(dx)
&\le\|f\|_p\bigl(\int|h(xy)|^q\mu(dx)\bigr)^{1/q}\cr
&=\|f\|_p\bigl(\Delta(y^{-1})\int|h(x)|^q\mu(dx)\bigr)^{1/q}
\le\|f\|_p\Delta(y^{-1})^{1/q}.\cr}$$

\noindent So

\Centerline{$\int(f*\nu)\times h\,d\mu
=\iint h(xy)f(x)\mu(dx)\nu(dy)
\le\int\|f\|_p\Delta(y^{-1})^{1/q}\nu(dy)
=\|f\|_p$.}

\noindent Because $\mu$ (being a quasi-Radon measure)
is semi-finite, this means that
$f*\nu\in\eusm L^p$ and that $\|f*\nu\|_p\le\|f\|_p$ (366D-366E,
or 244Xe and 244Fa).
(Once again, we need to know that every member of $L^q$ can be
represented by a Borel measurable function;  this is a consequence of
443J or 412Xe.)

For general Borel measurable $f:X\to\Bbb R$ such that
$\int|f|^pd\mu<\infty$, we know that from 444G that $f*\nu$ is Borel
measurable, while
$|f*\nu|\le|f|*\nu$ (and $f*\nu$ is defined wherever $|f|*\nu$ is
finite), so that

\Centerline{$\|f*\nu\|_p\le\||f|*\nu\|_p\le\||f|\|_p
=\|f\|_p$.}

\noindent Finally, if $f\in\eusm L^p$ is arbitrary, then there is a
Borel measurable $g:X\to\Bbb R$ such that $f\eae g$, so that
$f*\nu\eae g*\nu$ and

\Centerline{$\|f*\nu\|_p=\|g*\nu\|_p\le\|g\|_p=\|f\|_p$.}

\noindent It follows at once that we have a bounded linear operator
$f^{\ssbullet}\mapsto(f*\nu)^{\ssbullet}:L^p\to L^p$, of norm at most
$1=\gamma$.
}%end of proof of 444M

\leader{444N}{}\cmmnt{ The following lemma on exchanging the order of
repeated integrals will be fundamental to the formulae in the rest of
the section.

\medskip

\noindent}{\bf Lemma} Let $X$ be a topological group and $\mu$ a left
Haar measure on $X$.   Suppose that $f$, $g$,
$h\in\eusm L^0(\mu)$\cmmnt{ (the space of measurable real-valued
functions defined $\mu$-a.e.\ in $X$)} are non-negative.   Then, writing
$\int\ldots d(x,y)$ to denote integration with respect to the
quasi-Radon product measure $\mu\times\mu$,

\Centerline{$\iint f(x)g(y)h(xy)dxdy=\iint f(x)g(y)h(xy)dydx
=\int f(x)g(y)h(xy)d(x,y)$}

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

\proof{ Following the standard pattern in results of this type, I deal
with successively more complicated functions $f$, $g$ and $h$.
Evidently the situation is symmetric, so that it will be enough if I can
show that $\iint f(x)g(y)h(xy)dxdy=\int f(x)g(y)h(xy)d(x,y)$.

\medskip

{\bf (a)} Suppose first that $f=\chi F$, $g=\chi G$ and $h=\chi H$,
where $F$, $G$, $H$ are Borel subsets of $X$.   In this case

\Centerline{$\iint f(x)g(y)h(xy)dxdy
=\sup_{U,V\in\Sigma^f}\int_V\int_Uf(x)g(y)h(xy)dxdy$,}

\noindent where $\Sigma^f$ is the ideal of measurable sets of finite
measure for $\mu$.   \Prf\ For $y\in X$, $n\in\Bbb N$ and $U\in\Sigma^f$
write

\Centerline{$q(y)=\int f(x)h(xy)dx=\mu(F\cap Hy^{-1})$,
\quad$q_U(y)=\int_Uf(x)h(xy)dx=\mu(U\cap F\cap Hy^{-1})$,}

\Centerline{$q^{(n)}(y)=\min(n,q(y))$,
\quad$q^{(n)}_U(y)=\min(n,q_U(y))$.}

\noindent Then every $q_U$ is continuous, by 443C (with a little help
from 323Cc), while
$\sup_{U\in\Sigma^f}q_U(y)=q(y)$ for every $y$, because $\mu$ is
semi-finite.   Because $\mu$ is $\tau$-additive and effectively locally
finite,
$(q^{(n)})^{\ssbullet}=\sup_{U\in\Sigma^f}(q_U^{(n)})^{\ssbullet}$ in
$L^0(\mu)$ for every $n$ (414Ab);  because $\Sigma^f$ is
upwards-directed,

$$\eqalign{\int q(y)g(y)dy
&=\sup_{n\in\Bbb N}\int q^{(n)}(y)g(y)dy\cr
&=\sup_{n\in\Bbb N,U\in\Sigma^f}\int q^{(n)}_U(y)g(y)dy
=\sup_{U\in\Sigma^f}\int q_U(y)g(y)dy,\cr}$$

\noindent that is,

\Centerline{$\iint f(x)g(y)h(xy)dxdy
=\sup_{U\in\Sigma_f}\int\int_Uf(x)g(y)h(xy)dxdy$.}

\noindent On the other hand, for any $U\in\Sigma^f$, we surely have

\Centerline{$\int\int_Uf(x)g(y)h(xy)dxdy
=\sup_{V\in\Sigma^f}\int_V\int_Uf(x)g(y)h(xy)dxdy$,}

\noindent again because $\mu$ is semi-finite.   Putting these together,
we have the result.\ \Qed

Looking at the other side of the equation,
$\int f(x)g(y)h(xy)d(x,y)=(\mu\times\mu)W$, where
$W=(F\times G)\cap\{(x,y):xy\in H\}$ is a Borel set;  so that

$$\eqalign{\iint f(x)g(y)h(xy)dxdy
&=\sup_{U,V\in\Sigma^f}(\mu\times\mu)((U\times V)\cap W)\cr
&=\sup_{U,V\in\Sigma^f}\int_{U\times V}f(x)g(y)h(xy)d(x,y)\cr}$$

\noindent (417C(b-iii)).   But now we can apply 417Ga to see that, for any
$U$, $V\in\Sigma^f$,

\Centerline{$\int_{U\times V}f(x)g(y)h(xy)d(x,y)
=\int_V\int_Uf(x)g(y)h(xy)dxdy$.}

\noindent Taking the supremum over $U$ and $V$, we get

\Centerline{$\int f(x)g(y)h(xy)d(x,y)
=\iint f(x)g(y)h(xy)dxdy$.}

\medskip

{\bf (b)} Clearly both sides of our equation

\Centerline{$\int f(x)g(y)h(xy)d(x,y)
=\iint f(x)g(y)h(xy)dxdy$}

\noindent are additive in $f$, $g$ and $h$ separately (subtraction, of
course, will be another matter, as I am allowing $\infty$ to appear
without restriction);  and also behave identically if $f$ or $g$ or $h$
is multiplied by a non-negative scalar.   So the identity will be valid
if $f$, $g$ and $h$ are all finite sums of non-negative multiples of
indicator functions of Borel sets.   Moreover, by repeated use of
B.Levi's theorem, we see that if $\sequencen{f_n}$, $\sequencen{g_n}$
and $\sequencen{h_n}$ are non-decreasing sequences of such functions
with suprema $f$, $g$ and $h$, then

$$\eqalign{\int f(x)g(y)h(xy)d(x,y)
&=\sup_{n\in\Bbb N}\int f_n(x)g_n(y)h_n(xy)d(x,y)\cr
&=\sup_{n\in\Bbb N}\iint f_n(x)g_n(y)h_n(xy)dxdy
=\iint f(x)g(y)h(xy)dxdy.\cr}$$

\noindent So the identity is valid for all non-negative Borel
measurable functions $f$, $g$ and $h$.

\medskip

{\bf (c)} Finally, suppose only that $f$, $g$ and $h$ are non-negative,
measurable and defined almost everywhere.   In this case, by 443J(b-iv),
there are Borel measurable functions $f_0$, $g_0$ and $h_0$,
non-negative, defined everywhere on $X$ and equal almost everywhere to
$f$, $g$ and $h$ respectively.   Let $E$ be the conegligible set

\Centerline{$\{x:x\in\dom f\cap\dom h\cap\dom g,\,
f(x)=f_0(x),\,g(x)=g_0(x),\,h(x)=h_0(x)\}$.}

We find that $\int f(x)h(xy)dx=\int f_0(x)h_0(xy)dx$ for every $y\in X$.
\Prf\ $E\cap Ey^{-1}$ is conegligible (see 443A), and
$f(x)h(xy)=f_0(x)h_0(xy)$ for every $x\in E\cap Ey^{-1}$.\ \QeD\
Consequently

\Centerline{$\iint f(x)g(y)h(xy)dxdy
=\iint f_0(x)g_0(y)h_0(xy)dxdy$.}

Secondly, $f(x)g(y)h(xy)=f_0(x)g_0(y)h_0(xy)\,\,(\mu\times\mu)$-a.e.
\Prf\ Set $W=\{(x,y):x\in E,\,y\in E,\,xy\in E\}$.   Fubini's theorem,
applied to $(U\times V)\setminus W$ where $U$, $V\in\Sigma^f$, shows
that $W$ is conegligible;  but of course
$f(x)g(y)h(xy)=f_0(x)g_0(y)h_0(xy)$ whenever $(x,y)\in W$.\ \QeD\
Accordingly

\Centerline{$\int f(x)g(y)h(xy)d(x,y)
=\int f_0(x)g_0(y)h_0(xy)d(x,y)$.}

Combining this with the result of (b), applied to $f_0$, $g_0$ and
$h_0$, we see that once again

\Centerline{$\int f(x)g(y)h(xy)d(x,y)
=\iint f(x)g(y)h(xy)dxdy$,}

\noindent as required.
}%end of proof of 444N

\leader{444O}{Convolutions of functions:  Theorem} Let $X$ be a
topological group and $\mu$ a left Haar measure on $X$.   For $f$,
$g\in\eusm L^0=\eusm L^0(\mu)$,  write
$(f*g)(x)=\int f(y)g(y^{-1}x)dy$ whenever this is defined in
$\Bbb R$, taking the integral with respect to $\mu$.

(a) Writing $\Delta$ for the left modular function of $X$,

$$\eqalign{(f*g)(x)
&=\int f(y)g(y^{-1}x)dy
=\int f(xy)g(y^{-1})dy\cr
&=\int\Delta(y^{-1})f(y^{-1})g(yx)dy
=\int\Delta(y^{-1})f(xy^{-1})g(y)dy\cr}$$

\noindent whenever any of these integrals is defined in $\Bbb R$.

(b) If $f\eae f_1$ and $g\eae g_1$, then $f*g=f_1*g_1$.

(c)(i) $|(f*g)(x)|\le(|f|*|g|)(x)$ whenever either is defined in $\Bbb R$.

\quad(ii)

\Centerline{$((f_1+f_2)*g)(x)=(f_1*g)(x)+(f_2*g)(x)$,}

\Centerline{$(f*(g_1+g_2))(x)=(f*g_1)(x)+(f*g_2)(x)$,}

\Centerline{$((\alpha f)*g)(x)=(f*(\alpha g))(x)=\alpha(f*g)(x)$}

\noindent whenever the right-hand expressions are defined in $\Bbb R$.

(d) If $f$, $g$ and $h$ belong to $\eusm L^0$ and any of

\Centerline{$\int(|f|*|g|)(x)|h|(x)dx$,
\quad$\iint|f(x)g(y)h(xy)|dxdy$,}

\Centerline{$\iint|f(x)g(y)h(xy)|dydx$,
\quad$\int|f(x)g(y)h(xy)|d(x,y)$}

\noindent is defined in $\coint{0,\infty}$ (writing $\int\ldots d(x,y)$ for
integration with respect to the quasi-Radon product measure
$\mu\times\mu$ on $X\times X$), then

\Centerline{$\int(f*g)(x)h(x)dx$,
\quad$\iint f(x)g(y)h(xy)dxdy$,}

\Centerline{$\iint f(x)g(y)h(xy)dydx$,
\quad$\int f(x)g(y)h(xy)d(x,y)$}

\noindent are all defined, finite and equal, provided that in the
expression $(f*g)(x)h(x)$ we interpret the product as $0$ when $h(x)=0$
and $(f*g)(x)$ is undefined.

(e) If $f$, $g$ and $h$ belong to $\eusm L^0$, $f*g$ and $g*h$ are
defined a.e.\ and $x\in X$ is such that either $(|f|*(|g|*|h|))(x)$ or
$((|f|*|g|)*|h|)(x)$ is defined in $\Bbb R$, then $(f*(g*h))(x)$ and
$((f*g)*h)(x)$ are defined and equal.

(f) If $a\in X$ and $f$, $g\in\eusm L^0$,

\Centerline{$a\action_l(f*g)=(a\action_lf)*g$,
\quad$a\action_r(f*g)=f*(a\action_rg)$,}

\Centerline{$(a\action_rf)*g=\Delta(a^{-1})f*(a^{-1}\action_lg)$,}

\Centerline{$\Reverse{f}*\Reverse{g}=(g*f)\ssplrarrow$.}

(g) If $X$ is abelian then $f*g=g*f$ for all $f$ and $g$.

\proof{{\bf (a)} Use 441J and 442Kb to see that the
four formulae for $f*g$ coincide.

\medskip

{\bf (b)} Setting

\Centerline{$E=\{y:y\in\dom f\cap\dom f_1\cap\dom g\cap\dom g_1$,
$f(y)=f_1(y)$, $g(y)=g_1(y)\}$,}

\noindent $E$ is conegligible.   If $x\in X$, then
$f(y)g(y^{-1}x)=f_1(y)g_1(y^{-1}x)$ for every $y\in E\cap xE^{-1}$,
which is also conegligible, by 443A;  so $(f*g)(x)=(f_1*g_1)(x)$ if
either is defined.

\medskip

{\bf (c)} These are all elementary.

\medskip

{\bf (d)} First consider non-negative $f$, $g$ and $h$.   The point is
that, if any of the integrals is defined and finite,

$$\eqalignno{\int(f*g)(x)h(x)dx
&=\iint\Delta(x^{-1})f(x^{-1})g(xy)h(y)dxdy\cr
&=\iint\Delta(x^{-1})f(x^{-1})g(xy)h(y)dydx\cr
\noalign{\noindent (by 444N, recalling that
$x\mapsto\Delta(x^{-1})f(x^{-1})$ will belong to $\eusm L^0$ if $f$
does, by 442J and 442H)}
&=\iint f(x)g(x^{-1}y)h(y)dydx
=\iint f(x)g(y)h(xy)dydx\cr}$$

\noindent (substituting $xy$ for $y$ in the inner integral, as permitted
by 441J).   (The `and finite' at the beginning of the last sentence is
there because I have changed the rules since the last paragraph, and
$f*g$ is not allowed to take the
value $\infty$.   So we have to take care that

\Centerline{$\{y:h(y)>0,\,\int f(x)g(x^{-1}y)dx=\infty\}$}

\noindent is negligible.)   Now applying 444N again, we get

$$\eqalign{\int(f*g)(x)h(x)dx
&=\iint f(x)g(y)h(xy)dxdy\cr
&=\iint f(x)g(y)h(xy)dydx
=\int f(x)g(y)h(xy)d(x,y)\cr}$$

\noindent if any of these integrals is finite.

For the general case, the hypothesis on $|f|$, $|g|$ and $|h|$ is
sufficient to ensure that the four expressions are equal for any
combination of $f^{\pm}$, $g^{\pm}$ and $h^{\pm}$;  adding and
subtracting these combinations appropriately, we get the result.

\medskip

{\bf (e)} The point is that, for non-negative $f$, $g$ and $h$,

$$\eqalignno{((f*g)*h)(x)
&=\int(f*g)(z)h(z^{-1}x)dz
=\int(f*g)(z)h'(z)dz\cr
\noalign{\noindent (setting $h'(z)=h(z^{-1}x)$)}
&=\iint f(y)g(z)h'(yz)dzdy\cr
\noalign{\noindent (using (d);  to see that $h'$ is measurable,
refer to 443A as usual)}
&=\iint f(y)g(z)h(z^{-1}y^{-1}x)dzdy\cr
&=\int f(y)(g*h)(y^{-1}x)dy
=(f*(g*h))(x)\cr}$$

\noindent at least as long as one of the expressions here is finite.
(Note that, as in 255J, we need to suppose that $f*g$ and $g*h$ are
defined a.e.\ when moving from
$\int(f*g)(z)h(z^{-1}x)dz$ to $\iint f(y)g(z)h(z^{-1}y^{-1})dydz$ and
from $\iint f(y)g(z)h(z^{-1}y^{-1}x)dzdy$ to
$\int f(y)(g*h)(y^{-1}x)dy$, since in part (d) I am more tolerant
of infinities in the repeated integrals than I was in the definition of
$f*g$.)   Once again, subject to the inner integrals implicit in the
formulae $f*(g*h)$ and $(f*g)*h$ being adequately defined, we can
use addition and subtraction to obtain the result for general $f$, $g$
and $h$.

\medskip

{\bf (f)} These are immediate from the formulae in (a), using 442K if
necessary.

\medskip

{\bf (g)} If $X$ is abelian, then $\Delta(y)=1$ for every $y$, so

\Centerline{$(g*f)(x)
=\int g(y)f(y^{-1}x)dy
=\int g(y)\Delta(y^{-1})f(xy^{-1})dy
=(f*g)(x)$}

\noindent if either $(f*g)(x)$ or $(g*f)(x)$ is defined in $\Bbb R$.
}%end of proof of 444O

\leader{444P}{Proposition} Let $X$ be a topological group and $\mu$ a
left Haar measure on $X$.

(a) If $f\in\eusm L^1(\mu)^+$ and $g\in\eusm L^0(\mu)$ then
$f*g$\cmmnt{, as
defined in 444O,} is equal to $(f\mu)*g$\cmmnt{ as defined in 444H}.

(b) If $f\in\eusm L^0(\mu)$ and $g\in\eusm L^1(\mu)^+$ then
$f*g=f*(g\mu)$.

\proof{ Again, these are immediate from the formulae above:

\Centerline{$(f*g)(x)
=\int g(y^{-1}x)f(y)\mu(dy)
=\int g(y^{-1}x)(f\mu)(dy)
=(f\mu*g)(x)$,}

\Centerline{$(f*g)(x)
=\int f(xy^{-1})\Delta(y^{-1})g(y)\mu(dy)
=\int f(xy^{-1})\Delta(y^{-1})(g\mu)(dy)
=(f*g\mu)(x)$}

\noindent whenever these are defined, using 235K, as usual, to calculate
$\int\ldots d(f\mu)$, $\int\ldots d(g\mu)$.   (Note that as we assume throughout
that $f$ and $g$ are defined $\mu$-almost everywhere, all the functions
$y\mapsto g(y^{-1}x)$, $y\mapsto f(xy^{-1})$ are also defined
$\mu$-a.e., by the results set out in 443A.)
}%end of proof of 444P

\leader{444Q}{Proposition} Let $X$ be a topological group and $\mu$ a
left Haar measure on $X$.

(a) Let $f$, $g$ be non-negative
$\mu$-integrable functions.   Then\cmmnt{, defining $f*g$ as in 444O,
we have}
$f*g\in\eusm L^1\cmmnt{=\eusm L^1(\mu)}$ and

\Centerline{$(f\mu)*(g\mu)=(f*g)\mu$.}

(b) For any $f$, $g\in\eusm L^1$, $f*g\in\eusm L^1$ and

\Centerline{$\int f*g\,d\mu=\int fd\mu\int g\,d\mu$,
\quad$\|f*g\|_1\le\|f\|_1\|g\|_1$.}

\proof{{\bf (a)} Putting 444K and 444P together,
$f\mu*g\mu=(f\mu*g)\mu$, so that $f*g=f\mu*g$ is $\mu$-integrable, and

\Centerline{$(f*g)\mu=(f\mu*g)\mu=f\mu*g\mu$.}

\medskip

{\bf (b)} Taking $h=\chi X$ in 444Od, we get
$\int f*g\,d\mu=\int fd\mu\int g\,d\mu$.   Now

\Centerline{$\|f*g\|_1=\int|f*g|\le\int|f|*|g|=\int|f|
\int|g|
=\|f\|_1\|g\|_1$.}

}%end of proof of 444Q

\leader{444R}{Proposition} Let $X$ be a topological group and $\mu$ a
left Haar measure on $X$.   Take any $p\in[1,\infty]$.

(a) If $f\in\eusm L^1(\mu)$ and $g\in\eusm L^p(\mu)$, then
$f*g\in\eusm L^p(\mu)$ and $\|f*g\|_p\le\|f\|_1\|g\|_p$.

(b) $f*\Reverse{g}=(g*\Reverse{f})\ssplrarrow$ for all $f$,
$g\in\eusm L^0$.   If $X$
is unimodular then $\|\Reverse{f}\|_p=\|f\|_p$ for every $f\in\eusm L^0$.

(c) Set $q=\infty$ if $p=1$, $p/(p-1)$ if $1<p<\infty$, $1$ if
$p=\infty$.   If $f\in\eusm L^p(\mu)$ and $g\in\eusm L^q(\mu)$, then
$f*\Reverse{g}$ is defined everywhere in $X$ and is continuous, and
$\|f*\Reverse{g}\|_{\infty}\le\|f\|_p\|g\|_q$.   If $X$ is unimodular, then
$f*g\in C_b(X)$ and $\|f*g\|_{\infty}\le\|f\|_p\|g\|_q$ for every
$f\in\eusm L^p(\mu)$, $g\in\eusm L^q(\mu)$.

\medskip

\noindent{\bf Remark} In the formulae above, interpret $\|g\|_{\infty}$
as $\|g^{\ssbullet}\|_{\infty}=\esssup|g|$ for
$g\in\eusm L^{\infty}=\eusm L^{\infty}(\mu)$,
and as $\infty$ for $g\in\eusm L^0\setminus\eusm L^{\infty}$.
\cmmnt{Because $\mu$ is strictly positive, this agrees with the usual
definition $\|g\|_{\infty}=\sup_{x\in X}|g(x)|$ when $g$ is continuous
and defined everywhere on $X$.}

\proof{{\bf (a)} If $f\ge 0$, then $f*g=(f\mu)*g$ belongs to
$\eusm L^p(\mu)$, and

\Centerline{$\|f*g\|_p=\|f\mu*g\|_p\le(f\mu)(X)\|g\|_p=\|f\|_1\|g\|_p$,}

\noindent by 444Pa and 444Ma.   Generally, $f*g\eae f^+*g-f^-*g$ belongs to
$\eusm L^p(\mu)$ and

\Centerline{$\|f*g\|_p\le\|f^+*g\|_p+\|f^-*g\|_p
\le(\|f^+\|_1+\|f^-\|_1)\|g\|_p=\|f\|_1\|g\|_p$.}

\medskip

{\bf (b)(i)} By 443A once more, $\Reverse{f}\in\eusm L^0$ whenever
$f\in\eusm L^0$.   For $x\in X$,

\Centerline{$(f*\Reverse{g})(x)
=\int f(y)\Reverse{g}(y^{-1}x)dy
=\int\Reverse{f}(y^{-1})g(x^{-1}y)dy
=(g*\Reverse{f})(x^{-1})
=(g*\Reverse{f})\ssplrarrow(x)$}

\noindent if any of these are defined.

\medskip

\quad{\bf (ii)} If $X$ is unimodular, then, for any $f\in\eusm L^0$,

\Centerline{$\|\Reverse{f}\|_p^p
=\int|f(x^{-1})|^pdx
=\int\Delta(x^{-1})|f(x)|^pdx
=\|f\|_p^p$;}

\noindent while $\esssup|\Reverse{f}|=\esssup|f|$ because $E^{-1}$ is
conegligible whenever $E\subseteq X$ is conegligible.

\medskip

{\bf (c)(i)} For any $x\in X$,

\Centerline{$(f*\Reverse{g})(x)
=\int f(y)g(x^{-1}y)dy
=\int f\times(x\action_lg)$}

\noindent in the language of 443G and 444O.   By 443Gb,
$x\action_lg\in\eusm L^q$, so
$(f*\Reverse{g})(x)=\int f\times(x\action_lg)$ is defined.

\medskip

\quad{\bf (ii)} If $p>1$, so that $q<\infty$, then
$x\mapsto(x\action_lg)^{\ssbullet}:X\to L^q$ is continuous (443Gf), so

\Centerline{$x\mapsto(f*\Reverse{g})(x)
=\int f^{\ssbullet}\times(x\action_lg)^{\ssbullet}$}

\noindent is continuous, because $f^{\ssbullet}\in L^p\cong(L^q)^*$.
If $p=1$, then

\Centerline{$(f*\Reverse{g})(x)=\int f(xy)g(y)dy
=\int(x^{-1}\action_lf)\times g$}

\noindent for every $x$;  since
$x\mapsto(x^{-1}\action_lf)^{\ssbullet}:X\to L^1$ is
continuous, so is $f*\Reverse{g}$.

\medskip

\quad{\bf (iii)} If $X$ is unimodular then $f*g=f*\Tilde{\Tilde g}$ is
continuous, because $\Reverse{g}\in\eusm L^q$ by (b), and
$\|f*g\|_{\infty}\le\|f\|_p\|\Reverse{g}\|_q=\|f\|_p\|g\|_q$.
}%end of proof of 444R

\leader{444S}{Remarks} Let $X$ be a topological group and $\mu$ a left Haar
measure on $X$.

\spheader 444Sa From 444Ob and 444Ra we see that we have a
bilinear operator $(u,v)\mapsto u*v:L^1(\mu)\times L^p(\mu)\to L^p(\mu)$
defined by saying that $f^{\ssbullet}*g^{\ssbullet}=(f*g)^{\ssbullet}$ for
every $f\in\eusm L^1(\mu)$ and $g\in\eusm L^p(\mu)$.
\cmmnt{Indeed, 444Ob tells us that} $*$ can\cmmnt{ actually} be
regarded as a function from $L^1\times L^p$ to $\eusm L^p$.
Putting 443Ge together with 444Oe and 444Of, we have

\Centerline{$u*(v*w)=(u*v)*w$,}

\Centerline{$a\action_l(u*w)=(a\action_lu)*w$,
\quad$a\action_r(u*w)=u*(a\action_rw)$,}

\Centerline{$(a\action_ru)*w=\Delta(a^{-1})u*(a^{-1}\action_lw)$}

\noindent whenever $u$, $v\in L^1$, $w\in L^p$ and $a\in X$.

Similarly, if the group is unimodular, and $\bover1p+\bover1q=1$, the
map $*:\eusm L^p\times\eusm L^q\to C_b(X)$\cmmnt{ (444Rc)}
factors through a map from $L^p\times L^q$ to $C_b(X)$.

\spheader 444Sb In particular, $*:L^1\times L^1\to L^1$ is
associative;  evidently it is bilinear;  and $\|u*v\|_1\le\|u\|_1\|v\|_1$
for all $u$, $v\in L^1$.   So $L^1$ is a\cmmnt{ normed algebra;
since $L^1$ is $\|\,\|_1$-complete, it is a} Banach
algebra.   By 444Qb, $\int u*v=\int u\int v$ for all $u$, $v\in L^1$.
$L^1$ is commutative if $X$ is abelian\cmmnt{ (444Og)}.

\spheader 444Sc
Let $\Cal B$ be the Borel $\sigma$-algebra of $X$ and
$M_{\tau}$ the Banach algebra of signed $\tau$-additive Borel
measures on $X$, as in 444E.   If, for $f\in\eusm L^1=\eusm L^1(\mu)$
and $E\in\Cal B$, we
write $(f\mu\restr\Cal B)(E)=\int_Efd\mu$, then
$f\mu\restr\Cal B\in M_{\tau}$\cmmnt{;  for $f\ge 0$, this is because the
indefinite-integral measure $f\mu$ is a quasi-Radon measure, and in general
it is because $f\mu\restr\Cal B=f^+\mu\restr\Cal B-f^-\mu\restr\Cal B$}.
For $f$, $g\in\eusm L^1$, we have

\Centerline{$f^{\ssbullet}=g^{\ssbullet}\text{ in }L^1
\Longrightarrow f\eae g
\Longrightarrow f\mu\restr\Cal B=g\mu\restr\Cal B$,}

\noindent so we have an operator $T:L^1\to M_{\tau}$ defined by setting
$T(f^{\ssbullet})=f\mu\restr\Cal B$ for $f\in\eusm L^1$.\cmmnt{   It
is easy to check that

\Centerline{$(f+g)\mu\restr\Cal B=f\mu\restr\Cal B+g\mu\restr\Cal B$,
\quad$(\alpha f)\mu\restr\Cal B=\alpha(f\mu\restr\Cal B)$,}

\Centerline{$(|f|\mu\restr\Cal B)(E)=\int_E|f|
=\sup_{F\in\Cal B,F\subseteq E}\int_Ff-\int_{E\setminus F}f
=|f\mu\restr\Cal B|(E)$}

\noindent for $f$, $g\in\eusm L^1$, $\alpha\in\Bbb R$ and $E\in\Cal B$, so
that} $T$ is a Riesz homomorphism;  moreover,\cmmnt{

\Centerline{$\|f\mu\restr\Cal B\|
=|f\mu\restr\Cal B|(X)
=(|f|\mu\restr\Cal B)(X)
=\int|f|d\mu
=\|f^{\ssbullet}\|_1$}

\noindent for $f\in\eusm L^1$, so that} $T$ is
norm-preserving.\cmmnt{   Finally, for non-negative $f$, $g\in\eusm L^1$,
we have

$$\eqalignno{Tf^{\ssbullet}*Tg^{\ssbullet}
&=(f\mu\restr\Cal B)*(g\mu\restr\Cal B)
=(f\mu*g\mu)\restr\Cal B\cr
\displaycause{444Eb, since the completions of $f\mu\restr\Cal B$,
$g\mu\restr\Cal B$ are the quasi-Radon indefinite-integral measures $f\mu$,
$g\mu$}
&=(f*g)\mu\restr\Cal B\cr
\displaycause{444Qa}
&=T(f*g)^{\ssbullet}
=T(f^{\ssbullet}*g^{\ssbullet}).\cr}$$

\noindent Thus $Tu*Tv=T(u*v)$ for non-negative $u$, $v\in L^1$;  by
linearity,} $Tu*Tv=T(u*v)$ for all $u$, $v\in L^1$, and $T$ is an embedding
of $L^1$ as a subalgebra of $M_{\tau}$.

\leader{444T}{Proposition} Let $X$ be a topological group and $\mu$ a
left Haar measure on $X$.   Then for any $p\in\coint{1,\infty}$,
$f\in\eusm L^p(\mu)$ and $\epsilon>0$ there is a neighbourhood $U$ of
the identity $e$ in $X$ such that
$\|\nu*f-f\|_p\le\epsilon$ and $\|f*\nu-f\|_p\le\epsilon$ whenever $\nu$
is a quasi-Radon measure on $X$ such that $\nu U=\nu X=1$.

\proof{{\bf (a)} To begin with, suppose that $f$ is non-negative,
continuous and bounded, and that $G=\{x:f(x)>0\}$ has finite measure;
set $M=\sup_{x\in X}f(x)$.    Write $\Cal U$ for the family of
neighbourhoods of $e$.   Take $\delta>0$, $\eta\in\ocint{0,1}$ such that

\Centerline{$(2\delta+(1+\delta)^p-1)^{1/p}\|f\|_p\le\epsilon$,}

\Centerline{$(1-\eta)^p\int((f-\eta\chi X)^+)^pd\mu-M^p\eta
\ge(1-\delta)\int f^pd\mu$,
\quad$(1-\eta)^{(1-p)/p}\le 1+\delta$.}

For each $U\in\Cal U$, set

\Centerline{$H_U=\interior\{x:f(y)\ge f(x)-\eta$ for every
$y\in xU^{-1}\cup U^{-1}x\}$.}

\noindent Then $H_U$ is open and for every $x\in X$ there is a $U\in\Cal
U$ such that $|f(y)-f(x)|\le\bover12\eta$ whenever
$y\in xUU^{-1}\cup U^{-1}xU$, so that $x\in\interior xU\subseteq H_U$.
Thus $\{H_U:U\in\Cal U\}$
is an upwards-directed family of open sets with union $X$, and there is
a $U\in\Cal U$ such that $\mu(G\setminus H_U)\le\eta$;  moreover,
because $\Delta$ is continuous, we can suppose that
$\Delta(y^{-1})\ge 1-\eta$ for every $y\in U$.

Now suppose that $\nu$ is a quasi-Radon measure on $X$ such that
$\nu U=\nu X=1$.   Then, for any $x\in H_U$,

\Centerline{$(\nu*f)(x)
=\int f(y^{-1}x)\nu(dy)
=\int_U f(y^{-1}x)\nu(dy)
\ge f(x)-\eta$}

\noindent because $x\in H_U$ and $y^{-1}x\in U^{-1}x$ whenever $y\in U$.
Similarly,

\Centerline{$(f*\nu)(x)
=\int_U f(xy^{-1})\Delta(y^{-1})\nu(dy)
\ge (f(x)-\eta)(1-\eta)$}

\noindent for every $x\in H_U$.   Now this means that, setting
$h_1=\nu*f$, $h_2=f*\nu$ we have
$(f\wedge h_i)(x)\ge (f(x)-\eta)(1-\eta)$ for every $x\in H_U$ and
both $i$.   Accordingly

$$\eqalign{\int(f\wedge h_i)^pd\mu
&\ge(1-\eta)^p\int_{G\cap H_U}((f-\eta\chi X)^+)^pd\mu\cr
&\ge(1-\eta)^p\int((f-\eta\chi X)^+)^pd\mu-\int_{G\setminus H_U}f^p\cr
&\ge(1-\eta)^p\int((f-\eta\chi X)^+)^pd\mu-M^p\eta
\ge (1-\delta)\int f^pd\mu.\cr}$$

Now, just because $f$ and $h_i$ are non-negative, and $p\ge 1$,

\Centerline{$|f-h_i|^p+2(f\wedge h_i)^p\le f^p+h_i^p$.}

\noindent Also, writing

\Centerline{$\gamma=\int\Delta(y)^{(1-p)/p}\nu(dy)
=\int_U\Delta(y)^{(1-p)/p}\nu(dy)
\le(1-\eta)^{(1-p)/p}
\le 1+\delta$,}

\noindent we have

\Centerline{$\|h_1\|_p=\|\nu*f\|_p\le\|f\|_p$,
\quad $\|h_2\|_p=\|f*\nu\|_p\le\gamma\|f\|_p$}

\noindent (444M), so that $\int h_i^pd\nu\le(1+\delta)^p\int f^pd\mu$
for both $i$, and

\Centerline{$\int|f-h_i|^pd\mu
\le\int f^pd\mu+\int h_i^pd\mu-2\int(f\wedge
h_i)^pd\mu
\le(2\delta+(1+\delta)^p-1)\int f^pd\mu$}

\noindent for both $i$.   But this means that

\Centerline{$\max(\|f-f*\nu\|_p,\|f-\nu*f\|_p)
\le(2\delta+(1+\delta)^p-1)^{1/p}\|f\|_p\le\epsilon$.}

\noindent As $\nu$ is arbitrary, we have found a suitable $U$.

\medskip

{\bf (b)} For any continuous bounded function $f$ such that
$\mu\{x:f(x)\ne 0\}<\infty$, we can find neighbourhoods $U_1$, $U_2$ of
$e$ such that

\Centerline{$\|f^+-\nu*f^+\|_p\le\bover12\epsilon$,
\quad$\|f^+-f^+*\nu\|_p\le\bover12\epsilon$}

\noindent whenever $\nu U_1=\nu X=1$,

\Centerline{$\|f^--\nu*f^-\|_p\le\bover12\epsilon$,
\quad$\|f^--f^-*\nu\|_p\le\bover12\epsilon$}

\noindent whenever $\nu U_2=\nu X=1$.   So we shall have

\Centerline{$\|f-\nu*f\|_p\le\epsilon$,
\quad$\|f-f*\nu\|_p\le\epsilon$}

\noindent whenever $\nu(U_1\cap U_2)=\nu X=1$.

\medskip

{\bf (c)} For general $f\in\eusm L^p(\mu)$, there is a bounded
continuous function $g:X\to\Bbb R$ such that $\mu\{x:g(x)\ne 0\}<\infty$
and $\|f-g\|_p\le\bover14\epsilon$ (415Pa).   Now there is a
neighbourhood $U_1$ of $e$ such that

\Centerline{$\|g-\nu*g\|_p\le\bover14\epsilon$,
\quad$\|g-g*\nu\|_p\le\bover14\epsilon$}

\noindent whenever $\nu U_1=\nu X=1$.   There is also a neighbourhood
$U_2$ of $e$ such that $\Delta(y^{-1})^{(1-p)/p}\le 2$ for every
$y\in U_2$, so that

\Centerline{$\|g*\nu-f*\nu\|_p\le 2\|g-f\|_p\le\bover12\epsilon$}

\noindent whenever $\nu U_2=\nu X=1$.   Since we have

\Centerline{$\|\nu*g-\nu*f\|_p\le\|g-f\|_p\le\bover14\epsilon$}

\noindent whenever $\nu X=1$, we get $\|f-\nu*f\|_p\le\epsilon$,
$\|f=f*\nu\|_p\le\epsilon$ whenever $\nu(U_1\cap U_2)=\nu X=1$.

This completes the proof.
}%end of proof of 444T

\vleader{60pt}{444U}{Corollary} Let $X$ be a topological group and $\mu$ a left
Haar measure on $X$.   For any Haar measurable $E\subseteq X$ such that
$0<\mu E<\infty$, and any $f\in\bigcup_{1\le p<\infty}\eusm L^p(\mu)$,
write

\Centerline{$f_E(x)=\Bover1{\mu E}\int_{xE}fd\mu$,
\quad$f'_E(x)=\Bover1{\mu(Ex)}\int_{Ex}fd\mu$}

\noindent for $x\in X$.  Then, for any $p\in\coint{1,\infty}$,
$f\in\eusm L^p$ and
$\epsilon>0$, there is a neighbourhood $U$ of the identity in $X$ such
that $\|f_E-f\|_p\le\epsilon$ and $\|f'_E-f\|_p\le\epsilon$ whenever
$E\subseteq U$ is a non-negligible Haar measurable set.

\proof{ Take $\delta\in\ooint{0,1}$ such that
$\delta(1-\delta)^{(1-p)/p}\|f\|_p\le\bover12\epsilon$.
By 444T, there is a neighbourhood $U$ of the identity such that
$\|f-f*\nu\|_p\le\bover12\epsilon$, $\|f-\nu*f\|_p\le\epsilon$ whenever
$\nu$ is a quasi-Radon measure on $X$ such that $\nu U=\nu X=1$.
Shrinking $U$ if necessary, we may suppose also that $U=U^{-1}$, that
$\mu U<\infty$
and that $|\Delta(x)-1|\le\delta$ for every $x\in U$, where $\Delta$ is
the left modular function of $X$.   If $E\subseteq U$ and $\mu E>0$,
consider the quasi-Radon measures $\nu$, $\nuprime$, $\Reverse{\nu}$ and
$\Reverse{\nu}'$ on $X$ defined by setting

\Centerline{$\nu F
 =\Bover1{\mu E^{-1}}\int_{E\cap F}\Delta(x^{-1})\mu(dx)$,
\quad$\nuprime F=\Bover{\mu(E\cap F)}{\mu E}$,
\quad$\Reverse{\nu} F=\nu F^{-1}$,
\quad$\Reverse{\nu}' F=\nuprime F^{-1}$}

\noindent whenever these are defined.   (They are quasi-Radon measures
because $\nu$ and $\nuprime$ are totally finite indefinite-integral
measures over $\mu$ and
the map $x\mapsto x^{-1}$ is a homeomorphism.)   Because
$E\subseteq U=U^{-1}$, we have

\Centerline{$\Reverse{\nu} U=\nu U^{-1}
=\Bover1{\mu E^{-1}}\int_E\Delta(x^{-1})\mu(dx)=1=\Reverse{\nu} X$}

\noindent by 442Ka, while

\Centerline{$\Reverse{\nu}' X=\Reverse{\nu}' U=\nuprime U^{-1}=1$.}

\noindent Now consider
$f*\Reverse{\nu}$ and $\Reverse{\nu}'*f$.   For any $x\in X$,

$$\eqalignno{(f*\Reverse{\nu})(x)
&=\int f(xy^{-1})\Delta(y^{-1})\Reverse{\nu}(dy)
=\int f(xy)\Delta(y)\nu(dy)\cr
\noalign{\noindent (because $\Reverse{\nu}$ is the image of $\nu$ under the
map $y\mapsto y^{-1}$)}
&=\Bover1{\mu E^{-1}}\int\chi E(y)\Delta(y^{-1})f(xy)\Delta(y)\mu(dy)\cr
\noalign{\noindent (noting that $\nu$ is an indefinite-integral measure
over $\mu$, and using 235K)}
&=\Bover1{\mu E^{-1}}\int\chi E(x^{-1}y)f(y)\mu(dy)
=\Bover1{\mu E^{-1}}\int_{xE}f(y)\mu(dy)
=\Bover{\mu E}{\mu E^{-1}}f_E(x),\cr
(\Reverse{\nu}'*f)(x)
&=\int f(y^{-1}x)\Reverse{\nu}'(dy)
=\int f(yx)\nuprime(dy)\cr
\noalign{\noindent (because $\Reverse{\nu}'$ is the image of $\nuprime$
under
the map $y\mapsto y^{-1}$)}
&=\Bover1{\mu E}\int\chi E(y)f(yx)\mu(dy)
=\Bover{\Delta(x^{-1})}{\mu E}\int\chi E(yx^{-1})f(y)\mu(dy)\cr
\noalign{\noindent (by 442Kc)}
&=\Bover1{\mu(Ex)}\int_{Ex}f(y)\mu(dy)
=f'_E(x).\cr}$$

So

\Centerline{$\|f-f_E\|_p
\le\|f-f*\Reverse{\nu}\|_p
 +\bigl|\Bover{\mu E^{-1}}{\mu E}-1\bigr|\|f*\Reverse{\nu}\|_p$.}

\noindent Now $\mu E^{-1}=\int_E\Delta(y^{-1})\mu(dy)$, so that

\Centerline{$|\mu E-\mu E^{-1}|\le\int_E|\Delta(y^{-1})-1|\mu(dy)
\le\delta\mu E$,
\quad$|\Bover{\mu E^{-1}}{\mu E}-1|\le\delta$;}

\noindent also

\Centerline{$\|f*\Reverse{\nu}\|_p
\le\|f\|_p\int\Delta(y)^{(1-p)/p}\nu(dy)
\le\|f\|_p(1-\delta)^{(1-p)/p}$}

\noindent by 444Mb, so

\Centerline{$\bigl|\Bover{\mu E^{-1}}{\mu E}-1\bigr|\|f*\Reverse{\nu}\|_p
\le\delta(1-\delta)^{(1-p)/p}\|f\|_p\le\Bover12\epsilon$,}

\noindent and $\|f-f_E\|_p\le\epsilon$.

On the other hand,

\Centerline{$\|f-f'_E\|_p=\|f-\Reverse{\nu}'*f\|_p\le\epsilon$;}

\noindent as $E$ is arbitrary, we have found a suitable $U$.
}%end of proof of 444U

 \leader{444V}{}\cmmnt{ So far I have not emphasized the special
properties of compact groups.   But of course they are the centre of the
subject, and for the sake of a fundamental theorem in \S446 I give the
following result.

\wheader{444V}{4}{2}{2}{72pt}

\noindent}{\bf Theorem} Let $X$ be a compact topological group and $\mu$
a left Haar measure on $X$.

(a) For any $u$, $v\in L^2=L^2(\mu)$ we can interpret their convolution
$u*v$ either as a member of the space $C(X)$ of continuous real-valued
functions on $X$, or as a member of the space $L^2$.

(b) If $w\in L^2$, then $u\mapsto u*w$ is a compact linear operator
whether regarded as a map from $L^2$ to $C(X)$ or as a map from $L^2$ to
itself.

(c) If $w\in L^2$ and $w=\Reverse{w}$\cmmnt{ (as defined in 443Af)}, then
$u\mapsto u*w:L^2\to L^2$ is a self-adjoint operator.

\proof{{\bf (a)} Being compact, $X$ is unimodular (442Ic).   As noted in
444Sa, $*$ can be regarded as a bilinear operator from $L^2\times L^2$ to
$C_b(X)=C(X)$.   Because $\mu X$ must be finite, we now have a natural
map $f\mapsto f^{\ssbullet}$ from $C(X)$ to $L^2$, so that we can think
of $u*v$ as a member of $L^2$ for $u$, $v\in L^2$.

\medskip

{\bf (b)(i)} Evidently $u\mapsto u*w:L^2\to C(X)$ is linear,
for any $w\in L^2$.

\medskip

\quad{\bf (ii)} Let $B$ be the unit ball of $L^2$, and give it the
topology induced by the weak topology $\frak T_s(L^2,L^2)$, so that $B$
is compact (4A4Ka).   Let $\action_l$ be the left action of $X$ on $L^2$
as in 443G and 444S.

If $f$, $g\in\eusm L^2$ and $a\in X$, then

\Centerline{$(f*g)(a)
=\int f(x)g(x^{-1}a)dx
=\int f(x)\Reverse{g}(a^{-1}x)dx
=\int f\times a\action_l\Reverse{g}$.}

\noindent (Note that because $X$ is unimodular, $\Reverse{g}$ and
$a\action_l\Reverse{g}$ are square-integrable whenever $g$ is.)   So
if $u$, $w\in L^2$ and $a\in X$,
$(u*w)(a)=\innerprod{u}{a\action_l\Reverse{w}}$.
It follows that, for any $w\in L^2$, the function
$(a,u)\mapsto(u*w)(a):X\times B\to\Bbb R$ is continuous.   \Prf\ We know
that $\action_l:X\times L^2\to L^2$ is continuous when $L^2$ is given
its norm topology (443Gf).   Now $(u,v)\mapsto\innerprod{u}{v}$ is
continuous, so $(a,u)\mapsto(u*w)(a)=\innerprod{u}{a\action_l\Reverse{w}}$
must be continuous.\ \Qed

Because $X$ is compact, this means that $u\mapsto u*w:B\to C(X)$ is
continuous when $C(X)$ is given its norm topology and $B$ is given the
weak topology (4A2G(g-ii)).   Because $B$ is compact in the weak
topology, $\{u*w:u\in B\}$ is compact in $C(X)$.   But this implies that
$u\mapsto u*w$ is a compact linear operator (definition:  3A5La).

\medskip

\quad{\bf (iii)} Again
because $X$ is compact, $\mu$ is totally finite, so,
for $f\in C(X)$, $\|f\|_2\le\|f\|_{\infty}\sqrt{\mu X}$, and the natural
map $f\mapsto f^{\ssbullet}:C(X)\to L^2$ is a bounded linear operator.
Consequently the map $u\mapsto(u*w)^{\ssbullet}:L^2\to L^2$ is a compact
operator, by 4A4La.

\medskip

{\bf (c)} Now suppose that $w=\Reverse{w}$.   In this case
$\innerprod{u*w}{v}=\innerprod{u}{v*w}$ for all $u$, $v\in L^2$.   \Prf\
Express $u$, $v$ and $w$ as $f^{\ssbullet}$, $g^{\ssbullet}$ and
$h^{\ssbullet}$ where $f$, $g$ and $h$ are square-integrable Borel
measurable functions defined everywhere on $X$.   We have

$$\eqalignno{\innerprod{u*w}{v}
&=\int(f*h)(x)g(x)dx
=\iint f(y)h(y^{-1}x)g(x)dydx\cr
&=\iint f(y)h(y^{-1}x)g(x)dxdy\cr
\noalign{\noindent (because $(x,y)\mapsto f(y)h(y^{-1}x)g(x)$ is Borel
measurable, $\mu$ is totally finite and
\discrversionA{\break}{}$\iint|f(y)h(y^{-1}x)g(x)|dydx
=\innerprod{|u|*|w|}{|v|}$ is finite)}
&=\iint f(y)\Reverse{h}(x^{-1}y)g(x)dxdy
=\int f(y)(g*\Reverse{h})(y)dy\cr
&=\innerprod{u}{v*\Reverse{w}}
=\innerprod{u}{v*w}.\text{ \Qed}\cr}$$

\noindent As $u$ and $v$ are arbitrary, this shows that
$u\mapsto u*w:L^2\to L^2$ is self-adjoint.
}%end of proof of 444V

\exercises{\leader{444X}{Basic exercises $\pmb{>}$(a)}
%\spheader 444Xa
Let $X$ be a Hausdorff topological group.    Show that if $\lambda$ and
$\nu$ are totally finite Radon measures on $X$ then $\lambda*\nu$ is the
image measure $(\lambda\times\nu)\phi^{-1}$, where $\phi(x,y)=xy$ for
$x$, $y\in X$, and in particular is a Radon measure.
%444A

\sqheader 444Xb Let $X$ be a topological group and $\lambda$, $\nu$ two
totally finite quasi-Radon measures on $X$.   Writing $\supp\lambda$ for
the support of $\lambda$, show that
$\supp(\lambda*\nu)=\overline{(\supp\lambda)(\supp\nu)}$.
%444A

\spheader 444Xc Let $X$ be a topological group and
$M^+_{\text{qR}}$ the family of totally finite quasi-Radon measures on
$X$.   Show that $(\lambda,\nu)\mapsto\lambda*\nu:
M^+_{\text{qR}}\times M^+_{\text{qR}}\to M^+_{\text{qR}}$ is continuous
for the narrow topology on $M^+_{\text{qR}}$.   \Hint{437Ma, 437N.}
%444A

\spheader 444Xd Let $X$ be a Hausdorff topological group.   Show that
$X$ is abelian iff its Banach algebra of signed $\tau$-additive Borel
measures is commutative.
%444E

\spheader 444Xe Let $X$ be a topological group, and $M_{\tau}$ its
Banach algebra of signed $\tau$-additive Borel measures.   (i) Show that
we have actions $\action_l$, $\action_r$ of $X$ on $M_{\tau}$ defined by
writing $(a\action_l\nu)(E)=\nu(aE)$, $(a\action_r\nu)(E)=\nu(Ea^{-1})$.
(ii) Show that $(a\action_l\lambda)*\nu=a\action_l(\lambda*\nu)$,
$\lambda*(a\action_r\nu)=a\action_r(\lambda*\nu)$ for all $a\in X$ and
$\lambda$, $\nu\in M_{\tau}$.
%444E

\spheader 444Xf Let $X$ be a compact Hausdorff topological group, and $B$ a
norm-bounded subset of the Banach algebra $M_{\tau}$ of signed
$\tau$-additive Borel measures on $X$.   Show that
$(\lambda,\nu)\mapsto\lambda*\nu:B\times B\to M_{\tau}$ is continuous
for the vague topology on $M_{\tau}$.   \Hint{437Md.}
%444E 444Xc

\spheader 444Xg Let $X$ be a topological group, and $\nu$ a totally
finite quasi-Radon measure on $X$.   Show that for any
Borel sets $E$, $F\subseteq X$, the function
$(g,h)\mapsto\nu(gE\cap Fh)$ is Borel measurable.   \Hint{for Borel sets
$W\subseteq X\times X$,
set $\nuprime W=\nu\{x:(x,x)\in W\}$.   Consider the action of
$X\times X$ on
itself defined by writing $(g,h)\action(x,y)=(gx,yh^{-1})$.}
%444F

\spheader 444Xh Let $X$ be a topological group and $f$ a real-valued
function defined on a subset of $X$.   (i) Show that
$a\action_r(\nu*f)=\nu*(a\action_rf)$ (definition:  4A5Cc) whenever
$a\in X$ and $\nu$ is a
measure on $X$.   (ii) Show that if $X$ carries Haar measures, then
$a\action_l(f*\nu)=(a\action_lf)*\nu$ whenever $a\in X$ and $\nu$ is a
measure on $X$.
%444J

\spheader 444Xi Let $X$ be a topological group carrying Haar measures,
$f:X\to\Bbb R$ a bounded continuous function and $\nu$ a totally finite
quasi-Radon measure on $X$.   Show that $f*\nu$ is continuous.
%444J 444Fe 444Ib

\spheader 444Xj Let $X$ be a topological group carrying Haar measures,
$f$ a real-valued function defined on a subset of $X$, and $\lambda$,
$\nu$ totally finite quasi-Radon measures on $X$.   Show that
$((f*\nu)*\lambda)(x)=(f*(\nu*\lambda))(x)$ whenever the right-hand side
is defined.   (See also 444Yj.)
%444Ic 444J

\spheader 444Xk Let $X$ be an abelian topological group carrying Haar
measures.   Show that $f*\nu=\nu*f$ for every measure $\nu$ on $X$ and
every real-valued function $f$ defined on a subset of $X$.
%444J

\sqheader 444Xl Let $X$ be a topological group and $\mu$ a left Haar
measure on $X$.   (i) Let $\nu$ be a totally finite quasi-Radon measure
on $X$ such that $x\mapsto\nu(xF)$ is continuous for every closed set
$F\subseteq X$.   Show that $\nu$ is truly continuous with respect to
$\mu$.   \Hint{if $\mu F=0$, apply 444K to $\nu*\chi F^{-1}$ to see that
$\nu(xF)=0$ for $\mu$-almost every $x$.}
(ii) Let $\nu$ be a totally finite Radon measure on $X$ such that
$x\mapsto\nu(xK)$ is continuous for every compact set $K\subseteq X$.
Show that $\nu$ is truly continuous with respect to $\mu$.
%444K

\spheader 444Xm Let $X$ be a topological group carrying Haar measures,
$E\subseteq X$ a Haar negligible set and $\nu$ a $\sigma$-finite
quasi-Radon measure on $X$.   Show that $\nu(xE)=\nu(Ex)=0$ for
Haar-a.e.\ $x\in X$.
%444K

\spheader 444Xn Let $X$ be a topological group carrying Haar measures,
and $\nu$ a non-zero totally finite quasi-Radon measure on $X$ such that
$\nu(xE)=0$
whenever $x\in X$ and $\nu E=0$.   (i) Show that $\nu$ is strictly
positive, so that $X$ is ccc.
(ii) Show that a subset of $X$ is $\nu$-negligible iff it is Haar
negligible.
%444L

\spheader 444Xo Use the method of part (b) of the proof of 444M to prove
part (a) there.
%444M

\sqheader 444Xp Let $X$ be the group $S^1\times S^1$, with the topology
defined by giving the first coordinate the usual topology of $S^1$ and
the second coordinate its discrete topology, so that $X$ is a locally
compact abelian group.   Let $\mu$ be a Haar measure on $X$.   (i) Find
a Borel measurable function $f:X\times X\to\{0,1\}$ such that
$\iint f(x,y)\mu(dx)\mu(dy)\ne\iint f(x,y)\mu(dy)\mu(dx)$.   (ii) Let
$\nu$ be the Radon measure on $X$ defined by setting
$\nu E=\#(\{s:(s,s^{-1})\in E\})$ if this is finite, $\infty$ otherwise.
Define $g:X\to\{0,1\}$ by setting $g(s,t)=1$ if $s=t$, $0$ otherwise.
Show that $\iint g(xy)\nu(dy)\mu(dx)=\infty$,
$\iint g(xy)\mu(dx)\nu(dy)=0$.   (iii) Find a closed set $F\subseteq X$
such that $x\mapsto\nu(xF)$ is not Haar measurable.
%444N  mt44bits

\sqheader 444Xq Let $X$ be a Hausdorff topological group and for
$a\in X$ write $\delta_a$ for the Dirac measure on $X$ concentrated at $a$.
(i) Show that $\delta_a*\delta_b=\delta_{ab}$ for all
$a$, $b\in X$.   (ii) Show that, in the notation of 444Xe,
$\delta_a*\hat\nu$ is the completion of $a^{-1}\action_l\nu$ and
$\hat\nu*\delta_a$ is the completion of $a\action_r\nu$ for every
$a\in X$ and every totally finite $\tau$-additive Borel measure $\nu$ on
$X$ with completion $\hat\nu$.   (iii) Show that
$\delta_a*f=a\action_lf$ for every $a\in X$ and every real-valued
function $f$ defined on a subset of $X$.   (iv) Show that if $X$ carries
Haar measures, and has left modular function $\Delta$,
$f*\delta_a=\Delta(a^{-1})a^{-1}\action_rf$ for every $a\in X$ and
every real-valued function $f$ defined on a subset of $X$.   (v) Use
these formulae to relate 444Of to 444B.
%444O

\spheader 444Xr Let $X$ be a topological group and $\mu$ a left Haar
measure on $X$.   Show that if $f$, $g\in\eusm L^0(X)$ then
$(f*g)\ssplrarrow=\Reverse{g}*\Reverse{f}$.
%444O

\spheader 444Xs Let $X$ be a locally compact Hausdorff topological group
and $\mu$ a left Haar measure on $X$.   Show that if $f$, $g:X\to\Bbb R$
are continuous functions with compact support, then $f*g$ is a
continuous function with compact support.
%444R 444Xb

\spheader 444Xt In 444Rc, show that $f*\Reverse{g}$ is uniformly continuous
for the bilateral uniformity.   \Hint{in 443Gf, $x\mapsto x\action_lu$
is uniformly continuous.}
%444R

\spheader 444Xu Let $X$ be a topological group with a totally finite
Haar measure $\mu$.   Show that (i)
$\innerprod{u*w}{v}=\innerprod{u}{v*\Reverse{w}}$ for any $u$, $v$,
$w\in L^2=L^2(\mu)$, where $\Reverse{w}$ and $u*v$ are defined as in 443Af
and 444V (ii) the map $u\mapsto u*w:L^2\to L^2$ is a compact linear
operator for any $w\in L^2$.   \Hint{for (ii), use 443L.}
%444V

\spheader 444Xv Let $X$ be a topological group with a Haar probability
measure $\mu$.   Show that $L^2(\mu)$ with convolution is a Banach
algebra.
%444V

\sqheader 444Xw(i) Let $X_1$, $X_2$ be topological groups with totally
finite quasi-Radon measures $\lambda_i$, $\nu_i$ on $X_i$ for each $i$.
Let $\lambda=\lambda_1\times\lambda_2$, $\nu=\nu_1\times\nu_2$ be the
quasi-Radon product measure on the topological group
$X=X_1\times X_2$.   Show that
$\lambda*\nu=(\lambda_1*\nu_1)\times(\lambda_2*\nu_2)$.
(ii) Let $\familyiI{X_i}$ be a family of topological groups, and
$\lambda_i$, $\nu_i$ quasi-Radon probability measures
on $X_i$ for each $i$.   Let $\lambda=\prod_{i\in I}\lambda_i$,
$\nu=\prod_{i\in I}\nu_i$ be the
quasi-Radon product measures on the topological group
$\prod_{i\in I}X_i$.   Show that
$\lambda*\nu=\prod_{i\in I}\lambda_i*\nu_i$.
%444A out of order query

\sqheader 444Xx Show that 444C, 444O, 444P, 444Qb and
444R-444U %444R 444S 444T 444U
remain valid if we work with complex-valued, rather than real-valued,
functions, and with $\eusm L^p_{\Bbb C}$ and $L^p_{\Bbb C}$
rather than $\eusm L^p$ and $L^p$.

\spheader 444Xy\dvAnew{2010} Let $X$ be a topological group with a left
Haar measure $\mu$ and left modular function $\Delta$.   Write
$\pmb{\Delta}\in L^0=L^0(\mu)$ for the equivalence class of the function
$\Delta$.   For $u\in L^0$ write $u^*$ for
$\Reverse{u}\times\Reverse{\pmb{\Delta}}$.   Show that (i) $(u^*)^*=u$ for
every $u\in L^0$ (ii) $u\mapsto u^*:L^0\to L^0$ is a Riesz space
automorphism (iii) $u^*\in L^1$ for every
$u\in L^1=L^1(\mu)$ (iv) $u\mapsto u^*:L^1\to L^1$ is an $L$-space
automorphism (v) $u^**v^*=(v*u)^*$ for all $u$, $v\in L^1$ (v) defining
$T:L^1\to M_{\tau}$ as in 444Sc, show that $Tu^*=\Reverse{Tu}$
(that is, $(Tu^*)(E)=(Tu)(E^{-1})$ for Borel sets $E$) for every
$u\in L^1$.

\leader{444Y}{Further exercises (a)}
%\spheader 444Ya
Find a subgroup $X$ of $\{0,1\}^{\Bbb N}$ and
quasi-Radon probability measures $\lambda$, $\nu$ on $X$ and a set
$A\subseteq X$ such that $(\lambda*\nu)^*(A)=1$ but
$(\lambda\times\nu)\{(x,y):x,\,y\in X,\,x+y\in A\}=0$.
%444A  %mt44bits

\spheader 444Yb Let $X$ be a {\bf topological semigroup}, that is, a
semigroup with a topology such that multiplication is continuous.
(i) For totally finite
$\tau$-additive Borel measures $\lambda$, $\nu$ on $X$, show that there
is a $\tau$-additive Borel measure $\lambda*\nu$ defined by saying that
$(\lambda*\nu)(E)=(\lambda\times\nu)\{(x,y):xy\in E\}$ for every Borel
set $E\subseteq X$.   (ii) Show that in this context
$(\lambda_1*\lambda_2)*\lambda_3=\lambda_1*(\lambda_2*\lambda_3)$.
(iii) Show that $\int fd(\lambda*\nu)=\int f(xy)\lambda(dx)\nu(dy)$
whenever $f$ is $(\lambda*\nu)$-integrable.   (iv) Show that if the
topology is Hausdorff and $\lambda$ and $\nu$ are tight (that is, inner
regular with respect to the compact sets) so is $\lambda*\nu$.   (v)
Show that we have
a Banach algebra of signed $\tau$-additive Borel measures on $X$, as in
444E.
%444E

\spheader 444Yc Let $X$ be a topological group, and write
$M_{\tau}^{(\Bbb C)}$ for the complexification of the $L$-space
$M_{\tau}$ of 444E, as described in 354Yl.   Show that
$M_{\tau}^{(\Bbb C)}$, with the natural extension of the convolution
operator of 444E, is a complex Banach algebra, and that we still have
$|\lambda*\nu|\le|\lambda|*|\nu|$ for $\lambda$,
$\nu\in M_{\tau}^{(\Bbb C)}$.
%444E

\spheader 444Yd Find a locally compact Hausdorff topological group $X$,
a Radon probability measure $\nu$ on $X$ and an open set $G\subseteq X$
such that $\{(xGx^{-1})^{\ssbullet}:x\in X\}$ is not a separable subset
of the measure algebra of $\nu$.
%444G  %mt44bits

\spheader 444Ye Let $X$ be a metrizable group.
We say that a subset $A$ of $X$ is {\bf Haar null}
%or {\bf shy} watch it:  maybe "shy" demands a shy Borel superset
if there are a universally Radon-measurable set $E\supseteq A$ and a
non-zero Radon measure
$\nu$ on $X$ such that $\nu(xEy)=0$ for every $x$, $y\in X$.   (i) Show
that the family of Haar null sets is a translation-invariant
$\sigma$-ideal of subsets of $X$.
\Hint{if $\sequencen{E_n}$ is a sequence of universally
Radon-measurable Haar null sets, we can find Radon
probability measures $\nu_n$ concentrated on compact sets
near the identity such that
$\nu_n(xE_ny)=0$ for every $x$, $y$ and $n$;  now construct an infinite
convolution product $\nu=\nu_0*\nu_1*\ldots$ from the
probability product of the $\nu_n$ and show that
$\nu(xE_ny)=0$ for every $x$, $y$ and $n$.}   (ii) Show that if $X$ and $Y$
are Polish groups, $\phi:X\to Y$ is a surjective continuous
homomorphism and
$B\subseteq Y$ is Haar null, then $\phi^{-1}[B]$ is Haar null in $X$.
(iii) Show that if $X$ is a
locally compact Polish group then a subset of $X$ is Haar null iff it is
Haar negligible in the sense of 442H.   (See {\smc Solecki 01}.)
%444L

\spheader 444Yf Suppose that the continuum hypothesis is true.   Let $\nu$
be Cantor measure on $\Bbb R$ (256Hc).   Show that there is a set
$A\subseteq\Bbb R$ such that $\nu(x+A)=0$ for every $x\in\Bbb R$, but
$A$ is not Haar negligible.
%444L

\spheader 444Yg Let $X$ be a topological group and $\mu$ a left Haar
measure on $X$.   Let $\tau$ be a $\Cal T$-invariant extended Fatou norm
on $L^0(\mu)$ (\S374).   Show that if $\nu$ is any totally finite
quasi-Radon measure on $X$, then we have a linear operator
$f^{\ssbullet}\mapsto(\nu*f)^{\ssbullet}$ from $L^{\tau}$ to itself, of
norm at most $\nu X$.
%444M

\spheader 444Yh Let $X$ be a topological group with a left Haar measure
$\mu$, $M_{\tau}$ the Banach algebra of signed $\tau$-additive Borel
measures on $X$, and $p\in[1,\infty]$.   (i) Show that we have a
multiplicative linear operator $T$ from $M_{\tau}$ to the Banach algebra
$\eurm B(L^p(\mu);L^p(\mu))$ defined by writing
$(T\nu)(f^{\ssbullet})=(\hat\nu*f)^{\ssbullet}$ whenever $\nu$ is a
totally finite $\tau$-additive Borel measure on $X$ with completion
$\hat\nu$ and $f\in\eusm L^p(\mu)$.   \Hint{Use 444K and 444B to show
that $(\lambda*\nu)*f\eae\lambda*(\nu*f)$ for enough $\lambda$, $\nu$
and $f$.   See also 444Yj.}   (ii) Show that $\|T\nu\|\le\|\nu\|$ for
every $\nu\in M_{\tau}^+$.
(iii) Give an example in which $\|T\nu\|<\|\nu\|$.
%444M

\spheader 444Yi Let $X$ be a unimodular topological group with left Haar
measure $\mu$.   Suppose that $p$, $q$, $r\in[1,\infty]$ are such that
$\bover1p+\bover1q=1+\bover1r$, interpreting $\bover1{\infty}$ as $0$.
Show that if $f\in\eusm L^p(\mu)$ and $g\in\eusm L^q(\mu)$ then
$f*g\in\eusm L^r(\mu)$ and $\|f*g\|_r\le\|f\|_p\|g\|_q$.   \Hint{255Yl.
Take care to justify any changes in order of integration.}
%444O, 444Xt

\spheader 444Yj Let $X$ be a topological group carrying Haar measures.
Investigate conditions under which the associative laws

\Centerline{$\lambda*(\nu*f)=(\lambda*\nu)*f$,
\quad$\lambda*(f*\nu)=(\lambda*f)*\nu$,
\quad$f*(\lambda*\nu)=(f*\lambda)*\nu$,}

\Centerline{$f*(g*\nu)=(f*g)*\nu$,
\quad$f*(\nu*g)=(f*\nu)*g$,
\quad$\nu*(f*g)=(\nu*f)*g$}

\noindent will be valid, where $\lambda$ and $\nu$ are quasi-Radon
measures on $X$ and $f$, $g$ are real-valued functions.   Relate your
results to 444Xq.
%444O

\spheader 444Yk Let $X$ be a topological group with a left Haar measure
$\mu$ and left modular function $\Delta$.   (i) Suppose that
$f\in\eusm L^0(\mu)$.   Show that the following are equiveridical:
($\alpha$) $f(yx)=\Delta(y)f(xy)$ for $(\mu\times\mu)$-almost every
$x$, $y\in X$;
($\beta$) $(a\action_cf)^{\ssbullet}=\Delta(a^{-1})f^{\ssbullet}$
for every $a\in X$.
(ii) Show that in this case $f(x)=0$ for almost every $x$ such that
$\Delta(x)\ne 1$.
(iii) Suppose that $f\in\eusm L^1(\mu)$.   Show that the following are
equiveridical:
($\alpha$) $f(yx)=\Delta(y)f(xy)$ for $(\mu\times\mu)$-almost every
$x$, $y\in X$;
($\beta$) $(f*g)^{\ssbullet}=(g*f)^{\ssbullet}$ for every
$g\in\eusm L^1(\mu)$.
%444O

\spheader 444Yl Let $X$ be a topological group and $\mu$ a left Haar
measure on $X$.   Let $\tau$ be a $\Cal T$-invariant extended Fatou
norm on $L^0(\mu)$ such that $\tau\restr L^{\tau}$ is an
order-continuous norm.   For a totally finite quasi-Radon measure $\nu$
on $X$, let $T_{\nu}:L^{\tau}\to L^{\tau}$ be the corresponding linear
operator (444Yg).   Show that for any $u\in L^{\tau}$ and $\epsilon>0$
there is a neighbourhood $U$ of the identity in $X$ such that
$\tau(T_{\nu}u-u)\le\epsilon$ whenever $\nu U=\nu X=1$.
%444T

\spheader 444Ym Let $X$ be a topological group with a left Haar measure
$\mu$.   For $u\in L^2=L^2(\mu)$, set $A_u=\{a\action_lu:a\in X\}$ (443G)
in $L^2$, and $D=\{v*u:v\in L^1(\mu),\,v\ge 0,\,\int v=1\}$.   (i) Show
that the closed convex hull of $A_u$ in $L^2$ is the closure of $D$.
\Hint{($\alpha$) use 444Od to show that if $w\in L^2$ and
$\innerprod{w'}{w}\ge\gamma$ for every $w'\in A_u$, then
$\innerprod{w'}{w}\ge\gamma$ for every $w'\in D$ ($\beta$) use 444U to
show that $A_u\subseteq\overline{D}$.}   (ii) Show that the closed linear
subspace $W_u$ generated by $A_u$ is the closure of $\{v*u:v\in L^1\}$.
(iii) Show that if $w\in L^2$ and $w\in A_u^{\perp}$, that is,
$\innerprod{u'}{w}=0$ for every $u'\in A$, then
$W_w\subseteq W_u^{\perp}$.
(iv) Show that if $X$ is $\sigma$-compact, then $W_u$ is
separable.   \Hint{$A_u$ is $\sigma$-compact, by 443Gf.}   (v) Set
$C=\{f^{\ssbullet}:f\in\eusm L^2\cap C(X)\}$.   Show that $C\cap W_u$ is
dense in $W_u$.   \Hint{$v*u\in C$ for many $v$, by 444Rc.}   (vi) Show
that if $X$ is $\sigma$-compact, then $W_u$ has an orthonormal basis in
$C$.   (vii) Show that $L^2$ has an orthonormal basis in $C$.   \Hint{if
$X$ is $\sigma$-compact, take a maximal orthogonal family of subspaces
$W_u$, find a suitable orthonormal basis of each, and use (iii) to see
that these assemble to form a basis of $L^2$.   For a general locally
compact Hausdorff group, start with a $\sigma$-compact open subgroup,
and then deal with its cosets.   For a general topological group with a
Haar measure, use 443L.}   (Compare 416Yh.)
%444S

\spheader 444Yn Let $X$ be a topological group with a left Haar measure
$\mu$.   Let $\lambda$ be the quasi-Radon product measure on
$X\times X$.   Let $\eusm U$ be the set of those
$h\in\eusm L^1(\lambda)$ such
that $(X\times X)\setminus\{(x,y):(x,y)\in\dom h,\,h(x,y)=0\}$ can be
covered by a sequence of open sets of finite measure.
(i) Show that if $h\in\eusm U$, then $(x,y)\mapsto h(y,y^{-1}x)$ belongs
to $\eusm U$.   \Hint{443Xa.}
(ii) Show that if
$h\in\eusm U$, then $(Th)(x)=\int h(y,y^{-1}x)\mu(dy)$ is defined for
almost every $x\in X$ and $Th$
is $\mu$-integrable, with $\|Th\|_1\le\|h\|_1$.   \Hint{255Xj.}
(iii) Show that if $h_1$, $h_2\in\eusm U$ are equal
$\lambda$-a.e.\ then $Th_1=Th_2\,\,\mu$-a.e.
(iv) Show that every
member of $L^1(\lambda)$ can be represented by a member of $\eusm U$.
\Hint{443Xk.}
(v) Show that if $f$, $g\in\eusm L^1(\mu)$ and both are
zero outside some countable union of open sets of finite measure, then
$T(f\otimes g)=f*g$, where $(f\otimes g)(x,y)=f(x)g(y)$.
(vi) Show that
if we set $\tilde T(h^{\ssbullet})=(Th)^{\ssbullet}$ for $h\in\eusm U$,
then $\tilde T:L^1(\lambda)\to L^1(\mu)$ is the unique continuous linear
operator such that $\tilde T(u\otimes v)=u*v$ for all $u$,
$v\in L^1(\mu)$, where $u*v$ is defined in 444S and
$\otimes:L^1(\mu)\times L^1(\mu)\to L^1(\lambda)$ is the canonical
bilinear operator (253E).
%444S

\spheader 444Yo In 444Yn, suppose that $\mu X=1$.   (i) Show that the
map $\tilde T$ belongs to the
class $\Cal T_{\bar\lambda,\bar\mu}$ of \S373.   (ii) Show that if
$p\in[1,\infty]$ then $\|Th\|_p\le\|h\|_p$ whenever
$h\in\Cal U\cap\eusm L^p(\lambda)$.
%444Yn, 444S

\spheader 444Yp Rewrite this section in terms of right Haar measures
instead of left Haar measures.
%444+

\spheader 444Yq\dvAnew{2010}
Let $X$ be a topological group and $M_{\text{qR}}^+$ the set of totally
finite quasi-Radon measures on $X$.   For $\nu\in M_{\text{qR}}^+$,
define $\Reverse{\nu}\in M_{\text{qR}}^+$ by
saying that $\Reverse{\nu}(E)=\nu E^{-1}$ whenever $E\subseteq X$ and
$\nu$ measures $E^{-1}$.   (i) Show that if $\lambda$,
$\nu\in M_{\text{qR}}^+$ then
$\Reverse{\lambda}*\Reverse{\nu}=(\nu*\lambda)\ssplrarrow$.
(ii) Taking $\action_l$, $\action_r$ to be the left and right actions of
$X$ on itself, and defining corresponding actions of $X$ on
$M_{\text{qR}}^+$ as in 441Yp, show that
$a\action_l(\lambda*\nu)=(a\action_l\lambda)*\nu$ and
$a\action_r(\lambda*\nu)=\lambda*(a\action_r\nu)$ for $\lambda$,
$\nu\in M_{\text{qR}}^+$ and $a\in X$.
}%end of exercises

\endnotes{
\Notesheader{444} The aim of this section and the next is to work
through ideas from the second half of Chapter 25, and Chapter 28, in
forms natural in the context of general topological groups.   (It is of
course possible to go farther;  see 444Yb.   It is the glory and
confusion of twentieth-century mathematics that it has no firm stopping
points.)   The move from $\Bbb R$ to an arbitrary topological group is a
large one, and I think it is worth examining the various aspects of this
leap as they affect the theorems here.   The most conspicuous change,
and the one which most greatly affects the forms of the results, is the
loss of commutativity.   We are forced to re-examine every formula to
determine exactly which manipulations can still be justified.
Multiplications must be written the right way round, and inversions
especially must be watched.   But while there are undoubtedly some
surprises, we find that in fact (provided we take care over the
definitions) the most important results survive.   Of course I wrote the
earlier results out with a view to what I expected to do here, but no
dramatic manoeuvers are needed to turn the fundamental results 255G,
255H, 255J, 257B, 257E, 257F into the new versions 444Od, 444Qb, 444Oe,
444C, 444B, 444Qa.   (The changed order of presentation is an indication
of the high
connectivity of the web here, not of any new pattern.)   In fact what
makes the biggest difference is not commutativity, as such, but
unimodularity.   In groups which are not unimodular we do have new
phenomena, as in 444Mb and 444Of,
and these lead to complications in the proofs of such results as 444U,
even though the result there is exactly what one would expect.

In this section I ignore right Haar measures entirely.   I do not even
put them in the exercises.   If you wish to take this theory farther,
you may some day have to work out the formulae appropriate to
right Haar measures.   (You can check your results in
{\smc Hewitt \& Ross 63}, 20.32.)   But for the moment, I think that
they are likely
to be just a source of confusion.   There is one point which you may
have noticed.   The theory of groups is essentially symmetric.   In the
definition of `group' there is no distinction between left and right.
In the formulae defining group actions, we do have such a distinction,
because they must reflect the fact that we write $g\action x$ rather
than $x\action g$.
With $\action_l$ and $\action_r$, for instance (444Of), if we want them
to be actions in the standard sense we have to put
an \hbox{$^{-1}$} into the definition of $\action_l$ but not into the
definition of $\action_r$.   But we still expect that, for instance,
$\lambda*\nu$ and $\nu*\lambda$ will be related in some transparent way.
However there is an exception to this rule in the definitions of $\nu*f$
and $f*\nu$ (444H, 444J).   The modular function appears in the latter,
so in fact the definition applies only in a more restricted class of
groups.   In abelian groups we assume that $f*\nu$ and $\nu*f$ will be
equal, and they are (444Xk), but strictly speaking, on the definitions
here, we can write
$f*\nu=\nu*f$ only for abelian topological groups carrying Haar
measures.

From the point of view of the proofs in this section, the principal
change is that the Haar measures here are no longer assumed to be
$\sigma$-finite.   I am well aware that non-$\sigma$-finite measures are
a minority interest, especially in harmonic analysis, but I do think it
interesting that $\sigma$-finiteness is not relevant to the main
results, and the techniques required to demonstrate this are very much
in the spirit of this treatise (see, in particular, the proof of 444N,
and the repeated applications of 443Jb).
The basic difficulty is that we can no longer exchange repeated
integrals, even of non-negative Borel measurable functions, quite
automatically.   Let me emphasize that the result in 444N is really
rather special.   If we try to generalize it to other measures or other
types of function we encounter the usual obstacles (444Xp).

A difficulty of a different kind arises in the proof of 444Fc.   Here I
wish to show that the function
$g\mapsto(g\action E)^{\ssbullet}:G\to\frak A$ is Borel measurable for
every Borel
measurable set $E$.   The first step is to deal with open sets $E$, and
it would be nice if we could then apply the Monotone Class Theorem.
But the difficulty is that even though the map
$(a,b)\mapsto a\Bsetminus b:\frak A\times\frak A\to\frak A$ is
continuous, it does not quite follow that the map

\Centerline{$g\mapsto(g\action(E\setminus F))^{\ssbullet}
=(g\action E)^{\ssbullet}\Bsetminus(g\action F)^{\ssbullet}$}

\noindent is Borel measurable whenever
$g\mapsto(g\action E)^{\ssbullet}$ and
$g\mapsto(g\action F)^{\ssbullet}$ are, because the map
$g\mapsto((g\action E)^{\ssbullet},(g\action F)^{\ssbullet}):
G\to\frak A\times\frak A$ might conceivably fail to be Borel measurable,
if the metric space $\frak A$ is not separable, that is, if the Maharam
type of the measure $\nu$ is uncountable.   Of course the difficulty is
easily resolved by an extra twist in the argument.

I use different techniques for the two parts of 444M as an excuse to
recall the ideas of \S371;  in fact part (a) is slightly easier than
part (b) when proved by the method of the latter (444Xo).

444U is a kind of density theorem.   Compared with the density theorems
in \S\S223 and 261, it is a `mean' rather than `pointwise' density
theorem;  if $E$ is concentrated near the identity, then
$f_E^{\ssbullet}$ approximates $f^{\ssbullet}$ in $L^p$, but there is no
suggestion that we can be sure that $f_E(x)\bumpeq f(x)$ for any
particular $x$s unless we know much more about the set $E$.   In fact
this is to be expected from the form of the results concerning Lebesgue
measure.   The sets $E$ considered in Volume 2 are generally intervals or
balls, and even in such a general form as 223Ya we need a notion of
scalar multiplication separate from the group operation.
}%end of notes

\discrpage

\frfilename{mt445.tex}
\versiondate{20.3.08}
\copyrightdate{1998}

\def\varhatbarf{{\setover{\,\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{\bar f}}}

\def\chaptername{Topological groups}
\def\sectionname{The duality theorem}

\newsection{445}

In this section I present a proof of the Pontryagin-van Kampen duality
theorem (445U).   As in Chapter 28, and for the same reasons, we need to
use complex-valued functions;  the relevant formulae in \S\S443 and 444
apply
unchanged, and I shall not repeat them here, but you may wish to re-read
parts of those sections taking functions to be complex- rather than
real-valued.   (It {\it is} possible to avoid complex-valued measures,
which I relegate to the exercises.)
The duality theorem itself applies only to abelian locally compact
Hausdorff groups, and it would be reasonable, on first reading, to take
it for granted that all groups here are of this type, which simplifies
some of the proofs a little.

My exposition is based on that of {\smc Rudin 67}.   I start with the
definition of `dual group', including a description of a topology on the
dual (445A), and the simplest examples (445B), with a mention of
Fourier-Stieltjes transforms of measures (445C-445D).   The elementary
special properties of dual groups of groups carrying Haar measures are
in 445E-445G;  in particular, in these cases, the bidual of a group
begins to make sense, and we can start talking about Fourier transforms
of functions.

Serious harmonic analysis begins with the identification of the dual
group with the maximal ideal space of $L^1$ (445H-445K).   The next idea
is that of `positive definite' function (445L-445M).   Putting these
together, we get the first result here which asserts that the
dual group of an abelian group $X$ carrying Haar measures is
sufficiently large to effectively describe functions on $X$ (Bochner's
theorem, 445N).   It is now easy to establish that $X$ can be faithfully
embedded in its bidual (445O).   We also have most of the machinery
necessary to describe the correctly normalized Haar measure of the dual
group, with a first step towards identifying functions whose Fourier
transforms will have inverse Fourier transforms (the Inversion Theorem,
445P).   This leads directly to the Plancherel Theorem, identifying the
$L^2$ spaces of $X$ and its dual (445R).   At this point it is clear
that the bidual $\frak X$ cannot be substantially larger than $X$, since
they must have essentially the same $L^2$ spaces.   A little
manipulation of
shifts and convolutions in $L^2$ (445S-445T) shows that $X$ must be
dense in $\frak X$, and a final appeal to local compactness
shows that $X$ is closed in $\frak X$.

\leader{445A}{Dual groups} Let $X$ be any topological group.

\spheader 445Aa A {\bf character} on $X$ is a continuous group
homomorphism from $X$ to $S^1=\{z:z\in\Bbb C,\,|z|=1\}$.   It is easy to
see that the set $\Cal X$ of all characters on $X$ is a subgroup of the
group $(S^1)^X$\ifwithproofs{, just because $S^1$ is an abelian
topological
group.   (If $\chi$, $\theta\in\Cal X$, then $x\mapsto\chi(x)\theta(x)$
is continuous, and

\Centerline{$(\chi\theta)(xy)
=\chi(xy)\theta(xy)
=\chi(x)\chi(y)\theta(x)\theta(y)
=\chi(x)\theta(x)\chi(y)\theta(y)
=(\chi\theta)(x)(\chi\theta)(y)$.)}

\noindent}\else{.  }\fi   So $\Cal X$ itself is an abelian group.

\spheader 445Ab Give $\Cal X$ the topology of uniform convergence on
subsets of $X$ which are totally bounded for the bilateral uniformity on
$X$\cmmnt{ (4A5Hb, 4A5O)}.   \cmmnt{(If $\Cal E$ is the set of
totally bounded subsets of $X$, then the topology of $\Cal X$ is
generated by the pseudometrics $\rho_E$, where
$\rho_E(\chi,\theta)=\sup_{x\in E}|\chi(x)-\theta(x)|$ for $E\in\Cal E$
and $\chi$, $\theta\in\Cal X$.
It will be useful, in this formula, to interpret $\sup\emptyset=0$, so
that $\rho_{\emptyset}$ is the zero pseudometric.   Note that $\Cal E$
is closed under finite unions, so $\{\rho_E:E\in\Cal E\}$ is
upwards-directed, as in 2A3Fe.)}   Then $\Cal X$ is a Hausdorff
topological group.   \prooflet{(If $x\in E\in\Cal E$ and $\chi$,
$\chi_0$, $\theta$, $\theta_0\in\Cal X$,

$$\eqalign{|(\chi\theta)(x)-(\chi_0\theta_0)(x)|
&=|\chi(x)(\theta(x)-\theta_0(x))+\theta_0(x)(\chi(x)-\chi_0(x))|\cr
&\le|\theta(x)-\theta_0(x)|+|\chi(x)-\chi_0(x)|,\cr}$$

\Centerline{$|\chi^{-1}(x)-\chi_0^{-1}(x)|
=|\overline{\chi(x)}-\overline{\chi_0(x)}|=|\chi(x)-\chi_0(x)|$,}

\noindent so

\Centerline{$\rho_E(\chi\theta,\chi_0\theta_0)
\le\rho_E(\chi,\chi_0)+\rho_E(\theta,\theta_0)$,
\quad$\rho_E(\chi^{-1},\chi_0^{-1})=\rho_E(\chi,\chi_0)$.}

\noindent If $\chi\ne\theta$ then there is an $x\in X$ such that
$\chi(x)\ne\theta(x)$, and now $\{x\}\in\Cal E$ and
$\rho_{\{x\}}(\chi,\theta)>0$.)
}%end of prooflet

\spheader 445Ac Note that if $X$ is locally compact, then\cmmnt{ its
totally bounded sets are just its relatively compact sets (4A5Oe),
so} the topology of $\Cal X$ is the topology of uniform convergence on
compact subsets of $X$.

\spheader 445Ad If $X$ is compact, then $\Cal X$ is discrete.
\prooflet{\Prf\ $X$ itself is totally bounded, so
$U=\{\chi:|\chi(x)-1|\le 1$ for every $x\in X\}$ is a neighbourhood of
the identity $\iota$ in $\Cal X$.   But if $\chi\in U$ and $x\in X$ then
$|\chi(x)^n-1|\le 1$ for every $n\in\Bbb N$, so $\chi(x)=1$.   Thus
$U=\{\iota\}$ and $\iota$ is an isolated point of $\Cal X$;  it follows
that every point of $\Cal X$ is isolated.\ \Qed}

\spheader 445Ae If $X$ is discrete then $\Cal X$ is compact.
\prooflet{\Prf\ The only totally bounded sets in $X$ are the finite
sets, so the topology of $\Cal X$ is just that induced by its embedding
in $(S^1)^X$.   On the other hand, every homomorphism from $X$ to $S^1$ is
continuous, so $\Cal X$ is a closed set in $(S^1)^X$, which is compact
by Tychonoff's theorem.\ \Qed}

\cmmnt{\spheader 445Af I ought to remark that to most group theorists
the word `character' means
something rather different.   For a finite abelian
group $X$ with its discrete topology, the `characters' on $X$,
as defined in (a) above, are just the group homomorphisms from
$X$ to $\Bbb C\setminus\{0\}$, which in this context
can be identified with the characters of the irreducible
complex representations of $X$.}

\leader{445B}{Examples (a)} If $X=\Bbb R$ with addition, then $\Cal X$
can also be identified with the additive group $\Bbb R$, if we write
$\chi_y(x)=e^{iyx}$ for $x$, $y\in\Bbb R$.

\prooflet{\Prf\ It is easy to check that every $\chi_y$, so defined, is
a character on $\Bbb R$, and that $y\mapsto\chi_y:\Bbb R\to\Cal X$ is a
homomorphism.   On the other hand, if $\chi$ is a character, then
(because it is continuous) there is a $\delta\ge 0$ such that
$|\chi(x)-1|\le 1$ whenever $|x|\le\delta$.   $\chi(\delta)$ is uniquely
expressible as $e^{i\alpha}$ where $|\alpha|\le\bover{\pi}2$.   Set
$y=\alpha/\delta$, so that $\chi(\delta)=\chi_y(\delta)$.   Now
$\chi(\bover12\delta)$ must be one of the square roots of
$\chi(\delta)$, so is $\pm\chi_y(\bover12\delta)$;  but as
$|\chi(\bover12\delta)-1|\le 1$, it must be $+\chi_y(\bover12\delta)$.
Inducing on $n$, we see that $\chi(2^{-n}\delta)=\chi_y(2^{-n}\delta)$
for every $n\in\Bbb N$, so that
$\chi(2^{-n}k\delta)=\chi_y(2^{-n}k\delta)$ for every $k\in\Bbb Z$,
$n\in\Bbb N$;  as $\chi$ and $\chi_y$ are continuous, $\chi=\chi_y$.
Thus the map $y\mapsto\chi_y$ is surjective and is a group isomorphism
between $\Bbb R$ and $\Cal X$.

As for the topology of $\Cal X$, $\Bbb R$ is a locally compact
topological group, so
the totally bounded sets are just the relatively compact sets
(4A5Oe again),
that is, the bounded sets in the usual sense (2A2F).   Now a
straightforward calculation shows that for any $\alpha\ge 0$ in
$\Bbb R$ and $\epsilon\in\ooint{0,2}$,

\Centerline{$\rho_{[-\alpha,\alpha]}(\chi_y,\chi_z)\le\epsilon
\iff\alpha|y-z|\le 2\arcsin\Bover{\epsilon}2$,}

\noindent so that the topology of $\Cal X$ agrees with that of
$\Bbb R$.\ \Qed}%end of prooflet

\spheader 445Bb Let $X$ be the group $\Bbb Z$ with its discrete
topology.   Then we
may identify its dual group $\Cal X$ with $S^1$ itself, writing
$\chi_{\zeta}(n)=\zeta^n$ for $\zeta\in S^1$, $n\in\Bbb Z$.
\prooflet{\Prf\ Once again, it is elementary to check that every
$\chi_{\zeta}$ is a character, and that $\zeta\mapsto\chi_{\zeta}$ is an
injective group homomorphism from $S^1$ to $\Cal X$.   If $\chi\in\Cal
X$,
set $\zeta=\chi(1)$;  then $\chi=\chi_{\zeta}$.   So $\Cal X\cong S^1$.
And because the only totally bounded sets in $X$ are finite,
$\chi\mapsto\chi_{\zeta}$ is continuous, therefore a
homeomorphism.\ \Qed}%end of prooflet

\spheader 445Bc On the other hand, if $X=S^1$ with its usual topology,
then we may identify its dual group $\Cal X$ with $\Bbb Z$, writing
$\chi_n(\zeta)=\zeta^n$ for $n\in\Bbb Z$, $\zeta\in S^1$.
\prooflet{\Prf\ The verification follows the same lines as in (a) and
(b).   As usual, the key step is to show that the map
$n\mapsto\chi_n:\Bbb Z\to\Cal X$ is surjective.   We can do this by applying (a).   If
$\chi\in\Cal X$, then $x\mapsto\chi(e^{ix})$ is a character of $\Bbb R$,
so there is a $y\in\Bbb R$ such that $\chi(e^{ix})=e^{iyx}$ for every
$x\in\Bbb R$.   In particular, $e^{2iy\pi}=\chi(1)=1$, so $y\in\Bbb Z$,
and $\chi=\chi_y$.   Concerning the topology of $\Cal X$, we know from
445Ad that it must be discrete, so that also matches the usual topology
of $\Bbb Z$.\ \Qed}

\spheader 445Bd Let $\family{j}{J}{X_j}$ be any family of topological
groups, and $X$ their product\cmmnt{ (4A5G)}.   For each $j\in J$ let
$\Cal X_j$ be the dual group of $X_j$.   Then the dual group of $X$ can
be identified with the subgroup $\Cal X$ of $\prod_{j\in I}\Cal X_j$
consisting
of those $\chi\in\prod_{j\in J}\Cal X_j$ such that $\{j:\chi(j)$ is not
the identity$\}$ is finite;  the action of $\Cal X$ on $X$ is defined by
the formula

\Centerline{$\chi\action x=\prod_{j\in J}\chi(j)(x(j))$.}

\noindent\cmmnt{(This is well-defined because only finitely many
terms in the product are not equal to $1$.) }If $I$ is finite, so that
$\Cal X=\prod_{j\in I}\Cal X_j$, the topology of $\Cal X$ is the product
topology.

\prooflet{\Prf\ As usual, it is easy to check that $\action$, as defined
above, defines an injective homomorphism from $\Cal X$ to the dual group
of $X$.   If $\theta$ is any character on $X$, then for each $j\in I$ we
have a continuous group homomorphism $\varepsilon_j:X_j\to X$ defined by
setting $\varepsilon_j(\xi)(j)=\xi$, $\varepsilon_j(\xi)(k)=e_k$, the
identity of $X_k$, for every $k\ne j$.   Setting
$\chi(j)=\theta\varepsilon_j$ for each $j$, we obtain
$\chi\in\prod_{j\in I}\Cal X_j$.   Now there is a neighbourhood $U$ of
the
identity of $X$ such that $|\theta(x)-1|\le 1$ for every $x\in U$, and
we may suppose that $U$ is of the form $\{x:x(j)\in G_j$ for every $j\in
J\}$, where $J\subseteq I$ is finite and $G_j$ is a neighbourhood of
$e_j$ for every $j\in J$.   If $k\in I\setminus J$,
$\varepsilon_k(\xi)\in U$ for every $\xi\in X_k$, so that
$|\chi(k)(\xi)-1|\le 1$ for every $\xi$, and $\chi(k)$ must be the
identity character on $X_k$;  this shows that $\chi\in\Cal X$.
If $x\in X$ and $x(j)=e_j$ for $j\in J$, then again $|\theta(x^n)-1|\le
1$ for every $n\in\Bbb N$, so $\theta(x)=\chi\action x=1$.   For any
$x\in X$, we can express it as a finite product $y\prod_{j\in
J}\varepsilon_j(x(j))$ where $y(j)=e_j$ for every $j\in J$, so that

\Centerline{$\theta(x)=\theta(y)\prod_{j\in J}\theta\varepsilon_j(x(j))
=\prod_{j\in J}\chi(j)(x(j))
=\chi\action x$.}

\noindent Thus $\action$ defines an isomorphism between $\Cal X$ and the
dual group of $X$.

As for the topology of $\Cal X$, a subset of $X$ is totally bounded iff
it is included in a product of totally bounded sets (4A5Od).
If $E=\prod_{j\in I}E_j$ is such a product (and not empty), then for
$\chi$, $\theta\in\Cal X$

\Centerline{$\sup_{j\in I}\rho_{E_j}(\chi(j),\theta(j))
\le\rho_E(\chi,\theta)
\le\sum_{j\in I}\rho_{E_j}(\chi(j),\theta(j))$,}

\noindent so (if $I$ is finite) the topology on $\Cal X$ is just the
product topology.\ \Qed
}%end of prooflet

\leader{445C}{Fourier-Stieltjes transforms} Let $X$ be a topological
group, and $\Cal X$ its dual group.
For any totally finite topological measure $\nu$ on $X$, we can form its
`characteristic function' or {\bf Fourier-Stieltjes transform}
$\varhat{\nu}:\Cal X\to\Bbb C$ by writing
$\varhat{\nu}(\chi)=\int\chi(x)\nu(dx)$.   \cmmnt{ (This generalizes the
`characteristic functions' of \S285.)}

\leader{445D}{Theorem} Let $X$ be a topological group, and $\Cal X$ its
dual group.   If $\lambda$ and $\nu$ are totally finite quasi-Radon
measures on $X$, then
$(\lambda*\nu)\varsphat=\varhat{\lambda}\times\varhat{\nu}$.

\proof{ If $\chi\in\Cal X$, then, by 444C,

$$\eqalign{(\lambda*\nu)\varsphat(\chi)
&=\int\chi\,d(\lambda*\nu)
=\iint\chi(xy)\lambda(dx)\nu(dy)\cr
&=\iint\chi(x)\chi(y)\lambda(dx)\nu(dy)
=\int\chi(x)\lambda(dx)\cdot\int\chi(y)\nu(dy)
=\varhat{\lambda}(\chi)\varhat{\nu}(\chi).\cr}$$

}%end of proof of 445D


\leader{445E}{}\cmmnt{ Let us turn now to groups carrying Haar
measures.   I start with three welcome properties.

\medskip

\noindent}{\bf Proposition} Let $X$ be a topological group with a
neighbourhood of the identity which is totally bounded for the bilateral
uniformity on $X$, and $\Cal X$ its dual group\cmmnt{, with its dual
group topology}.

(a) The map $(\chi,x)\mapsto \chi(x):\Cal X\times X\to S^1$ is
continuous.

(b) Let $\frak X$ be the dual group of $\Cal X$\cmmnt{, again with its
dual group topology, the topology of uniform convergence on totally
bounded subsets of $\Cal X$}.   Then we have a continuous homomorphism
$x\mapsto\hat x:X\to\frak X$ defined by setting $\hat x(\chi)=\chi(x)$
for $x\in X$ and $\chi\in\Cal X$.

(c) For any totally finite quasi-Radon measure $\nu$ on $X$, its
Fourier-Stieltjes
transform $\varhat{\nu}:\Cal X\to\Bbb C$ is uniformly continuous.

\cmmnt{\medskip

\noindent{\bf Remark} Note that the condition here is satisfied by any
topological group $X$ carrying Haar measures (443H).}

\proof{ Fix an open totally bounded set $U_0$ containing the identity.

\medskip

{\bf (a)} Let $\chi_0\in\Cal X$, $x_0\in X$ and $\epsilon>0$.   Then
$x_0U_0$ is totally bounded, so

\Centerline{$V=\{\chi:|\chi(y)-\chi_0(y)|\le\Bover12\epsilon$ for every
$y\in x_0U_0\}$}

\noindent is a neighbourhood of $\chi_0$.   Also

\Centerline{$U=\{x:x\in x_0U_0$,
$|\chi_0(x)-\chi_0(x_0)|\le\Bover12\epsilon\}$}

\noindent is a neighbourhood of $x_0$.   And if $\chi\in V$, $x\in U$ we
have

\Centerline{$|\chi(x)-\chi_0(x_0)|
\le|\chi(x)-\chi_0(x)|+|\chi_0(x)-\chi_0(x_0)|\le\epsilon$.}

\noindent As $\chi_0$, $x_0$ and $\epsilon$ are arbitrary,
$(\chi,x)\mapsto\chi(x)$ is continuous.

\medskip

{\bf (b)(i)} It is easy to check that $\hat x$, as defined above, is
always a homomorphism from $\Cal X$ to $S^1$, and that $x\mapsto\hat
x:X\to(S^1)^{\Cal X}$ is a homomorphism.
Because $\rho_{\{x\}}$ is always one of the defining pseudometrics for
the topology of $\Cal X$ (445Ab), $\hat x$ is always continuous, so
belongs to $\frak X$.

\medskip

\quad{\bf (ii)} To see that \hbox{$\sphat\,$} is continuous, I argue as
follows.   Take an open set $H\subseteq\frak X$ and $x_0\in X$ such that
$\hat x_0\in H$.   Then there are a totally bounded set
$F\subseteq\Cal X$ and
an $\epsilon>0$ such that $\frak x\in H$ whenever $\frak x\in\frak X$
and $\rho_F(\frak x,\hat x_0)\le\epsilon$.   Now $x_0U_0$ is a totally
bounded neighbourhood of $x_0$, so

\Centerline{$V=\{\theta:\theta\in\Cal X$,
$|\theta(y)-1|\le\Bover12\epsilon$ for every $y\in x_0U_0\}$}

\noindent is a neighbourhood of the identity in $\Cal X$.   There are
therefore $\chi_0,\ldots,\chi_n\in\Cal X$ such that
$F\subseteq\bigcup_{k\le n}\chi_kV$.   Set

\Centerline{$U=\{x:x\in x_0U_0$,
$|\chi_k(x)-\chi_k(x_0)|<\Bover12\epsilon$ for every $k\le n\}$.}

\noindent Then $U$ is an open neighbourhood of $x_0$ in $X$.

If $x\in U$ and $\chi\in F$ then there is a $k\le n$ such that
$\theta=\chi_k^{-1}\chi\in V$, so that

$$\eqalign{|\chi(x)-\chi(x_0)|
&=|\chi_k(x)\theta(x)-\chi_k(x_0)\theta(x_0)|\cr
&\le|\chi_k(x)-\chi_k(x_0)|+|\theta(x)-\theta(x_0)|
\le\Bover12\epsilon+\Bover12\epsilon
=\epsilon.\cr}$$

\noindent But this shows that $\rho_F(\hat x,\hat x_0)\le\epsilon$, so
$\hat x\in H$.

So we have $x_0\in U\subseteq\{x:\hat x\in H\}$.   As $x_0$ is
arbitrary, $\{x:\hat x\in H\}$ is open;  as $H$ is arbitrary,
$x\mapsto\hat x$ is continuous.

\medskip

{\bf (c)} Let $\epsilon>0$.   Because $\nu$ is $\tau$-additive, there
are $x_0,\ldots,x_n\in X$ such that $\nu(X\setminus\bigcup_{k\le
n}x_kU_0)\le\bover13\epsilon$.   Set $E=\bigcup_{k\le n}x_kU_0$;  then
$E$ is totally bounded.   So

\Centerline{$V=\{\theta:|\theta(x)-1|\le\Bover{\epsilon}{1+3\nu X}$
for every $x\in E\}$}

\noindent is a neighbourhood of the identity in $\Cal X$.   If $\chi$,
$\chi'\in\Cal X$ are such that $\theta=\chi^{-1}\chi'$ belongs to $V$,
then

$$\eqalign{|\varhat{\nu}(\chi)-\varhat{\nu}(\chi')|
&\le\int|\chi(x)-\chi'(x)|\nu(dx)
=\int|1-\theta(x)|\nu(dx)\cr
&\le 2\nu(X\setminus E)+\Bover{\epsilon\nu E}{1+3\nu X}
\le\epsilon.\cr}$$

\noindent Since $\epsilon$ is arbitrary (and $\Cal X$ is abelian), this
is enough to show that $\varhat{\nu}$ is uniformly continuous.
}%end of proof of 445E

\leader{445F}{Fourier transforms of functions} Let $X$ be a topological
group with a left Haar measure $\mu$.   For any
$\mu$-integrable complex-valued function $f$, define its {\bf Fourier
transform} $\varhatf:\Cal X\to\Bbb C$ by setting
$\varhatf(\chi)=\int f(x)\chi(x)\mu(dx)$ for every character $\chi$ of
$X$.  \cmmnt{(Compare 283A.   If $f$ is real and non-negative, then
$\varhatf=(f\mu)\varsphat$ as defined in 445C, where $f\mu$ is the
indefinite-integral measure, as in 444K.)}
\cmmnt{Note that} $\varhatf=\varhat{g}$ whenever $f\eae g$, so we
can\cmmnt{ equally
well} write $\varhat{u}(\chi)=\varhatf(\chi)$ whenever $u=f^{\ssbullet}$
in $L_{\Bbb C}^1(\mu)$.

\leader{445G}{Proposition} Let $X$ be a topological group with a left
Haar measure $\mu$.   Then for any $\mu$-integrable complex-valued
functions $f$ and $g$,
$(f*g)\varsphat=\varhatf\times\varhat{g}$;  \cmmnt{that
is,} $(u*v)\varsphat=\varhat{u}\times\varhat{v}$ for all $u$, $v\in
L_{\Bbb C}^1(\mu)$.

\proof{ For any character $\chi$ on $X$,

$$\eqalign{\int\chi(x)(f*g)(x)dx
&=\iint\chi(xy)f(x)g(y)dxdy\cr
&=\iint\chi(x)\chi(y)f(x)g(y)dxdy
=\varhatf(\chi)\varhat{g}(\chi)\cr}$$
%too long for Centerline in full size

\noindent (using 444Od).
}%end of proof of 445G

\leader{445H}{Theorem} Let $X$ be a topological group with a left Haar
measure $\mu$;  let $\Cal X$ be its dual group and let $\Phi$ be the set
of non-zero multiplicative linear functionals on the complex Banach
algebra $\cmmnt{L_{\Bbb C}^1=}L_{\Bbb C}^1(\mu)$\cmmnt{ (444Sb)}.
Then there is a one-to-one
correspondence between $\Cal X$ and $\Phi$, defined by the formulae

\Centerline{$\phi(f^{\ssbullet})
=\int f\times\chi\,d\mu=\varhatf(\chi)$
for every $f\in\eusm L_{\Bbb C}^1=\eusm L_{\Bbb C}^1(\mu)$,}

\Centerline{$\phi(a\action_lu)=\chi(a)\phi(u)$ for every
$u\in L_{\Bbb C}^1$, $a\in X$,}

\noindent for $\chi\in\Cal X$ and $\phi\in\Phi$.

\cmmnt{\medskip

\noindent{\bf Remark} I follow 443G in writing
$a\action_lf^{\ssbullet}=(a\action_lf)^{\ssbullet}$, where
$(a\action_lf)(x)=f(a^{-1}x)$ for $f\in\eusm L_{\Bbb C}^1$ and $a$,
$x\in X$, as in 4A5Cc.
}%end of comment

\proof{{\bf (a)} If $\chi\in\Cal X$ then we can think of its equivalence
class $\chi^{\ssbullet}$ as a member of
$L_{\Bbb C}^{\infty}=L_{\Bbb C}^{\infty}(\mu)$, so
that we can define $\phi_{\chi}\in(L_{\Bbb C}^1)^*$ by writing
$\phi_{\chi}(u)=\int\chi^{\ssbullet}\times u$ for every
$u\in L_{\Bbb C}^1$;
that is, $\phi_{\chi}(f^{\ssbullet})=\int f\times\chi=\varhatf(\chi)$
for every $f\in\eusm L_{\Bbb C}^1$.
445G tells us that $\phi_{\chi}$ is multiplicative.   To see that it is
non-zero, recall that $\mu$ is strictly positive (442Aa) and that $\chi$
is continuous.   Let $G$ be an open set containing the identity $e$ of
$X$ such that $|\chi(x)-1|\le\bover12$ for every $x\in G$;  then
$\Real(\chi(x))\ge\bover12$ for every $x\in G$, so

\Centerline{$|\int_G\chi(x)dx|\ge\Real\int_G\chi(x)dx
=\int_G\Real(\chi(x))dx\ge\bover12\mu G>0$.}

\noindent Accordingly $\phi_{\chi}(\chi G)^{\ssbullet}\ne 0$ and
$\phi_{\chi}\ne 0$.   (I hope that no confusion will arise if I continue
occasionally to write $\chi E$ for the indicator function of a set
$E$, even if the symbol $\chi$ is already active in the sentence.)

\medskip

{\bf (b)} Now suppose that $\phi$ is a non-zero multiplicative linear
functional on $L_{\Bbb C}^1$.   Fix on some $g_0\in\eusm L_{\Bbb C}^1$
such that $\phi(g_0^{\ssbullet})=1$.
Let $\Delta$ be the left modular function of $X$.   (If you are reading
this proof on the assumption that $X$ is abelian, then
$a\action_rf=a^{-1}\action_lf$
and $\Delta\equiv 1$, so the argument below simplifies usefully.)

\medskip

\quad{\bf (i)} For any $u\in L_{\Bbb C}^1$ and $a\in X$,
$\phi(a^{-1}\action_lu)
=\Delta(a)\phi(a\action_ru)$.
\Prf\ Let $f\in\eusm L^1_{\Bbb C}$ be such that $f^{\ssbullet}=u$.
Take any $\epsilon>0$.   Then for any sufficiently small open
neighbourhood $U$
of the identity, if we set $h=\Bover1{\mu U}\chi U$, we shall have

\Centerline{$\|(a\action_rf)*h-a\action_rf\|_1\le\epsilon$,
\quad$\|h*f-f\|_1\le\epsilon$}

\noindent (444T, with 444P;
see 444U).   Setting $w=h^{\ssbullet}$, we have

\Centerline{$\|(a\action_ru)*w-a\action_ru\|_1\le\epsilon$,
\quad$\|w*u-u\|_1\le\epsilon$,}

$$\eqalignno{
|\phi(a^{-1}\action_lu)-\Delta(a)\phi(a\action_ru)|
&\le|\phi(a^{-1}\action_lu)-\phi(u*(a^{-1}\action_lw))|\cr
\displaycause{444Sa}
&\mskip100mu +|\Delta(a)\phi((a\action_ru)*w)
     -\Delta(a)\phi(a\action_ru)|\cr
&=|\phi(a^{-1}\action_lu)
  -\phi(u)\phi(a^{-1}\action_lw)|\cr
&\mskip100mu
  +\Delta(a)|\phi((a\action_ru)*w)-\phi(a\action_ru)|\cr
&\le|\phi(a^{-1}\action_lu)
  -\phi(a^{-1}\action_lw)\phi(u)|\cr
&\mskip100mu  +\Delta(a)\|(a\action_ru)*w-a\action_ru\|_1\cr
\displaycause{because $\|\phi\|\le 1$ in $(L_{\Bbb C}^1)^*$, by 4A6F}
&\le|\phi(a^{-1}\action_lu)
  -\phi((a^{-1}\action_lw)*u)|+\epsilon\Delta(a)\cr
&\le\|a^{-1}\action_lu-(a^{-1}\action_lw)*u\|_1
  +\epsilon\Delta(a)\cr
&=\|a^{-1}\action_l(u-w*u)\|_1
  +\epsilon\Delta(a)\cr
\displaycause{by another of the formulae in 444Sa}
&=\|u-w*u\|_1
  +\epsilon\Delta(a)\cr
\displaycause{443Ge}
&\le(1+\Delta(a))\epsilon.\cr}$$

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

\medskip

\quad{\bf (ii)} For any $u$, $v\in L_{\Bbb C}^1$ and $a\in X$,
$\phi(a\action_lu)\phi(v)
=\phi(u)\phi(a\action_lv)$.   \Prf\

$$\eqalign{\phi(a\action_lu)\phi(v)
&=\phi((a\action_lu)*v)
=\phi(a\action_l(u*v))
=\Delta(a^{-1})\phi(a^{-1}\action_r(u*v))\cr
&=\Delta(a^{-1})\phi(u*a^{-1}\action_rv)
=\phi(u)\Delta(a^{-1})\phi(a^{-1}\action_rv)
=\phi(u)\phi(a\action_lv),\cr}$$

\noindent using (i) for the third and sixth equalities, and 444Sa for
the second and fourth.\ \Qed

\medskip

\quad{\bf (iii)} Let $v_0$ be $g_0^{\ssbullet}$, so that $\phi(v_0)=1$,
and set $\chi(a)=\phi(a\action_lv_0)$ for
every $a\in X$.   Then if $a$, $b\in X$,

$$\eqalignno{\chi(ab)
&=\phi(ab\action_lv_0)
=\phi(a\action_l(b\action_lv_0))\cr
\noalign{\noindent (because $\action_l$ is an action of $X$ on
$L_{\Bbb C}^1$, as noted in 443Ge for $L^1$)}
&=\phi(a\action_l(b\action_lv_0))\phi(v_0)
=\phi(b\action_lv_0)\phi(a\action_lv_0)\cr
\noalign{\noindent (by (ii) above)}
&=\chi(a)\chi(b).\cr}$$

\noindent So $\chi:X\to\Bbb C$ is a group homomorphism.   Moreover,
because $\action_l$ is continuous (443Gf), and $\phi$ also is continuous
(indeed, of norm at most $1$), $\chi$ is continuous.   Finally,

\Centerline{$|\chi(a)|\le\|a\action_lv_0\|_1=\|v_0\|_1$}

\noindent for every $a\in X$, by 443Gb;  it follows at once that
$\{\chi(a)^n:n\in\Bbb Z\}$ is bounded, so that $|\chi(a)|=1$, for every
$a\in X$.   Thus $\chi\in\Cal X$.   Moreover, for any $u\in L_{\Bbb C}^1$,

\Centerline{$\phi(a\action_lu)=\phi(a\action_lu)\phi(v_0)
=\phi(u)\phi(a\action_lv_0)=\chi(a)\phi(u)$.}

\medskip

\quad{\bf (iv)} Now $\phi=\phi_{\chi}$.   \Prf\  Because
$\phi\in(L_{\Bbb C}^1)^*$, there is some
$h\in\eusm L_{\Bbb C}^{\infty}(\mu)$ such that
$\phi(f^{\ssbullet})=\int h(x)f(x)dx$ for every $f\in\eusm L_{\Bbb C}^1$
(243Gb/243K;
recall that by the rules of 441D, $\mu$ is suppose to be a quasi-Radon
measure, therefore strictly localizable, by 415A).   In this case, for
any $f\in\eusm L_{\Bbb C}^1$,

$$\eqalignno{\phi(f^{\ssbullet})
&=\phi(f*g_0)^{\ssbullet}
=\int h(x)(f*g_0)(x)dx
=\iint h(xy)f(x)g_0(y)dydx\cr
\noalign{\noindent (444Od)}
&=\iint h(y)f(x)g_0(x^{-1}y)dydx
=\int\phi(x\action_lg_0)^{\ssbullet}f(x)dx\cr
&=\int\chi(x)f(x)dx
=\phi_{\chi}(f^{\ssbullet}).\text{ \Qed}\cr}$$

\medskip

{\bf (c)} Thus we see that the formulae announced do define a surjection
from $\Cal X$ onto $\Phi$.   We have still to confirm that it is
injective.   But if $\chi$, $\theta$ are distinct members of $\Cal X$,
then $\{x:\chi(x)\ne\theta(x)\}$ is a non-empty open set, so has
positive measure, because $\mu$ is strictly positive;  because
$\mu$ is semi-finite, they represent different linear functionals on
$L_{\Bbb C}^1$, and $\phi_{\chi}\ne\phi_{\theta}$.

This completes the proof.
}%end of proof of 445H

\leader{445I}{The topology of the dual group:  Proposition} Let $X$ be a
topological group with a left Haar measure $\mu$, and $\Cal X$ its dual
group.   For $\chi\in\Cal X$, let $\chi^{\ssbullet}$ be its equivalence
class in $L_{\Bbb C}^0=L_{\Bbb C}^0(\mu)$, and $\phi_{\chi}\in(L_{\Bbb
C}^1)^*=(L_{\Bbb C}^1(\mu))^*$ the
multiplicative linear functional corresponding to $\chi$\cmmnt{, as in
445H}.   Then the
maps $\chi\mapsto\chi^{\ssbullet}$ and $\chi\mapsto\phi_{\chi}$ are
homeomorphisms between $\Cal X$ and its images in $L_{\Bbb C}^0$ and
$(L_{\Bbb C}^1)^*$,
if we give $L_{\Bbb C}^0$ the topology of convergence in
measure\cmmnt{ (245A/245M)}
and $(L_{\Bbb C}^1)^*$ the weak* topology\cmmnt{ (2A5Ig)}.

\proof{{\bf (a)} Note that $\chi\mapsto\chi^{\ssbullet}$ is injective
because $\mu$ is strictly positive, so that if $\chi$, $\theta$ are
distinct members of $\Cal X$ then the non-empty open set
$\{x:\chi(x)\ne\theta(x)\}$ has non-zero measure;  and that
$\chi\mapsto\phi_{\chi}$ is injective by 445H.   So we have one-to-one
correspondences between $\Cal X$ and its images in $L_{\Bbb C}^0$ and
$(L_{\Bbb C}^1)^*$.

Write $\frak T$ for the topology of $\Cal X$ as defined in 445Ab, $\frak
T_{m}$ for the topology induced by its identification with its image in
$L_{\Bbb C}^0$, and $\frak T_{w}$ for the topology induced by its
identification
with its image in $(L_{\Bbb C}^1)^*$.   Let $\Cal E$ be the family of
non-empty
totally bounded subsets of $X$ and $\Sigma^f$ the set of measurable sets
of finite measure;  for $E\in\Cal E$, $F\in\Sigma^f$ and $f\in\eusm
L_{\Bbb C}^1=\eusm L_{\Bbb C}^1(\mu)$ set

\Centerline{$\rho_E(\chi,\theta)=\sup_{x\in E}|\chi(x)-\theta(x)|$,}

\Centerline{$\rho'_F(\chi,\theta)
=\int_F\min(1,|\chi(x)-\theta(x)|)\mu(dx)$,}

\Centerline{$\rho''_f(\chi,\theta)
=|\int f(x)\chi(x)\mu(dx)-\int f(x)\theta(x)\mu(dx)|$}

\noindent for $\chi$, $\theta\in\Cal X$.   Then $\frak T$ is generated
by the pseudometrics $\{\rho_E:E\in\Cal E\}$, $\frak T_m$ is generated by
$\{\rho'_F:F\in\Sigma^f\}$ and $\frak T_w$ is generated by
$\{\rho''_f:f\in\eusm L_{\Bbb C}^1\}$.

\medskip

{\bf (b)} $\frak T_m\subseteq\frak T$.   \Prf\ Suppose that
$F\subseteq X$ is a measurable set of finite measure, and $\epsilon>0$.
There is a
non-empty totally bounded open set $U\subseteq X$ (443H).   Since
$\{xU:x\in X\}$ is an open cover of $X$ and $\mu$ is
$\tau$-additive, there are $y_0,\ldots,y_n\in X$ such that
$\mu(F\setminus\bigcup_{j\le n}y_jU)\le\bover13\epsilon$;  set
$E=\bigcup_{j\le n}y_jU$.   Then $E$ is totally bounded, and
$\rho'_F(\chi,\theta)\le\epsilon$ whenever
$\rho_E(\chi,\theta)\le\Bover{\epsilon}{1+3\mu E}$.   As $F$ and
$\epsilon$ are
arbitrary, the identity map $(\Cal X,\frak T)\to(\Cal X,\frak T_m)$ is
continuous (2A3H), that is, $\frak T_m\subseteq\frak T$.\ \Qed

\medskip

{\bf (c)} $\frak T_w\subseteq\frak T_m$.   \Prf\
If $f\in\eusm L_{\Bbb C}^1$ and $\epsilon>0$ let $F\in\Sigma^f$, $M>0$
be such that
$\int(|f|-M\chi F)^+d\mu\le\bover14\epsilon$.   If $\chi$,
$\theta\in\Cal X$ and $\rho'_F(\chi,\theta)\le\Bover{\epsilon}{4M}$,
then

$$\eqalign{\rho''_f(\chi,\theta)
&=\bigl|\int(\chi-\theta)\times f\bigr|
\le\int|\chi-\theta|\times|f|\cr
&\le 2\int(|f|-M\chi F)^++M\int_F|\chi-\theta|\cr
&\le\Bover12\epsilon+2M\int_F\min(1,|\chi-\theta|)
=\Bover12\epsilon+2M\rho'_F(\chi,\theta)
\le\epsilon.\cr}$$

\noindent As $f$ and $\epsilon$ are arbitrary, this shows that
$\frak T_w\subseteq\frak T_m$.\ \Qed

\medskip

{\bf (d)} Finally, $\frak T\subseteq\frak T_w$.   \Prf\ Fix
$\chi\in\Cal X$, $E\in\Cal E$ and $\epsilon>0$.   Let
$u\in L_{\Bbb C}^1$ be such
that $\phi_{\chi}(u)=1$, and represent $u$ as $f^{\ssbullet}$ where
$f\in\eusm L_{\Bbb C}^1(\mu)$.   Set
$U=\{a:a\in X,\,\|a\action_lu-u\|_1<\bover14\epsilon\}$;  then $U$ is an
open neighbourhood of the identity $e$ of $X$, because
$a\mapsto a\action_lu$ is continuous (443Gf).   Because $E$ is totally
bounded, there are $y_0,\ldots,y_n\in X$ such that
$E\subseteq\bigcup_{k\le n}y_kU$.
Set $f_k=y_k\action_lf$, so that $f_k^{\ssbullet}=y_k\action_lu$ for each $k\le n$.

Now suppose that $\theta\in\Cal X$ is such that

\Centerline{$\rho''_f(\theta,\chi)\le\Bover{\epsilon}4$,
\quad$\rho''_{f_k}(\theta,\chi)\le\Bover{\epsilon}4$ for every
$k\le n$.}

\noindent Take any $x\in E$.   Then there is a $k\le n$ such that
$x\in y_kU$, so that $y_k^{-1}x\in U$ and

\Centerline{$\|x\action_lu-y_k\action_lu\|_1
=\|y_k\action_l(y_k^{-1}x\action_lu-u)\|_1
=\|y_k^{-1}x\action_lu-u\|_1
\le\Bover{\epsilon}4$}

\noindent (using 443Ge for the second equality).   Now
$\phi_{\chi}(x\action_lu)=\chi(x)$ (445H), so

$$\eqalign{|\phi_{\theta}(x\action_lu)-\chi(x)|
&\le|\phi_{\theta}(x\action_lu-y_k\action_lu)|
   +|\phi_{\theta}(y_k\action_lu)-\phi_{\chi}(y_k\action_lu)|
   +|\phi_{\chi}(y_k\action_lu-x\action_lu)|\cr
&\le 2\|x\action_lu-y_k\action_lu\|_1+\rho''_{f_k}(\theta,\chi)
\le \Bover34\epsilon.\cr}$$

\noindent On the other hand,

\Centerline{$|\theta(x)-\phi_{\theta}(x\action_lu)|
=|\theta(x)||1-\phi_{\theta}(u)|
=\rho''_f(\theta,\chi)\le\Bover{\epsilon}4$.}

\noindent So $|\theta(x)-\chi(x)|\le\epsilon$.   As $x$ is arbitrary,
$\rho_E(\theta,\chi)\le\epsilon$.

As $\chi$, $E$ and $\epsilon$ are arbitrary, this shows that
$\frak T\subseteq\frak T_w$.\ \Qed
}%end of proof of 445I

\leader{445J}{Corollary} For any topological group $X$ carrying Haar
measures, its dual group $\Cal X$ is locally compact and Hausdorff.

\proof{ Let $\Phi$ be the set of non-zero multiplicative linear
functionals on $L_{\Bbb C}^1=L_{\Bbb C}^1(\mu)$, for some left Haar
measure $\mu$ on $X$,
and give $\Phi$ its weak* topology.   Then
$\Phi\cup\{0\}\subseteq(L_{\Bbb C}^1)^*$ is the set of all
multiplicative linear
functionals on $L_{\Bbb C}^1$, and is closed for the weak* topology,
because

\Centerline{$\{\phi:\phi\in(L_{\Bbb C}^1)^*,\,
\phi(u*v)=\phi(u)\phi(v)\}$}

\noindent is closed for all $u$, $v\in L_{\Bbb C}^1$.   Because the unit
ball of $(L_{\Bbb C}^1)^*$ includes $\Phi$ (4A6F again), and
is a compact Hausdorff space for the weak* topology (3A5F),
so is $\Phi\cup\{0\}$.   So $\Phi$ itself is an open subset of a compact
Hausdorff space and is a locally compact Hausdorff space (3A3Bg).
Since the topology on $\Cal X$ can be identified with the weak*
topology on $\Phi$ (445I), $\Cal X$ also is locally compact and
Hausdorff.
}%end of proof of 445J

\leader{445K}{Proposition} Let $X$ be a topological group and $\mu$ a
left Haar measure on $X$.   Let $\Cal X$ be the dual group of $X$, and
write $C_0=C_0(\Cal X;\Bbb C)$ for the Banach algebra of continuous
functions $h:\Cal X\to\Bbb C$ such that $\{\chi:|h(\chi)|\ge\epsilon\}$
is compact for every $\epsilon>0$.

(a) For any $u\in L_{\Bbb C}^1=L_{\Bbb C}^1(\mu)$, its Fourier transform
$\varhat{u}$ belongs to $C_0$.

(b) The map $u\mapsto\varhat{u}:L_{\Bbb C}^1\to C_0$ is a multiplicative
linear
operator, of norm at most $1$.

(c) Suppose that $X$ is abelian.   For
$f\in\eusm L_{\Bbb C}^1=\eusm L_{\Bbb C}^1(\mu)$,
set $\tilde f(x)=\overline{f(x^{-1})}$ whenever this is defined.   Then
$\tilde f\in\eusm L_{\Bbb C}^1$ and $\|\tilde f\|_1=\|f\|_1$.   For
$u\in L_{\Bbb C}^1$,
we may define $\tilde u\in L_{\Bbb C}^1$ by setting $\tilde u=\tilde
f^{\ssbullet}$ whenever $u=f^{\ssbullet}$.   Now $\varhat{\tilde u}$ is
the complex conjugate of $\varhat{u}$, so
$(u*\tilde u)\varsphat=|\varhat{u}|^2$.

(d) Still supposing that $X$ is abelian,
$\{\varhat{u}:u\in L_{\Bbb C}^1\}$ is a
norm-dense subalgebra of $C_0$, and $\|\varhat{u}\|_{\infty}=r(u)$, the
spectral radius of $u$\cmmnt{ (4A6G)}, for every $u\in L_{\Bbb C}^1$.

\proof{{\bf (a)} As in 445H and 445J, let $\Phi$ be the set of non-zero
multiplicative linear functionals on $L_{\Bbb C}^1$, so that
$\Phi\cup\{0\}$ is
compact for the weak* topology of $(L_{\Bbb C}^1)^*$, and
$\varhat{u}(\chi)=\phi_{\chi}(u)$
for every $\chi\in\Cal X$.   By the definition of the weak* topology,
$\phi\mapsto\phi(u)$ is continuous;  since we can identify the weak*
topology on $\Phi$ with the dual group topology of $\Cal X$ (445I),
$\varhat{u}$ is continuous.   Also, for any $\epsilon>0$,

\Centerline{$\{\chi:\chi\in\Cal X,\,|\varhat{u}(\chi)|\ge\epsilon\}
\cong\{\phi:\phi\in\Phi,\,|\phi(u)|\ge\epsilon\}$,}

\noindent which is a closed subset of $\Phi\cup\{0\}$, therefore
compact.

\medskip

{\bf (b)} It is immediate from the definition of $\varsphat$ that it is
a linear operator from $L_{\Bbb C}^1$ to $\Bbb C^{\Cal X}$, and therefore
from $L_{\Bbb C}^1$ to $C_0$;   it is multiplicative
by 445G, and of norm at most $1$ because all the multiplicative linear
functionals $u\mapsto\varhat{u}(\chi)$ must be of norm at most $1$.

\woddheader{445K}{4}{2}{2}{30pt}

{\bf (c)} Now suppose that $X$ is abelian.   If
$f\in\eusm L_{\Bbb C}^1$, then

\Centerline{$\int\tilde f(x)dx
=\overline{\intop f(x^{-1})dx}
=\overline{\intop f(x)dx}$}

\noindent by 442Kb, so $\tilde f\in\eusm L_{\Bbb C}^1$;  the same
formulae tell us that $\|\tilde f\|_1=\|f\|_1$.   If $f\eae g$ then
$\tilde f\eae\tilde g$ (442G, or otherwise), so $\tilde u$ is
well-defined.   If $\chi\in\Cal X$, and $u=f^{\ssbullet}$, then

$$\eqalign{\varhat{\tilde u}(\chi)
&=\int\tilde f(x)\chi(x)dx
=\int\overline{f(x^{-1})}\chi(x)dx
=\int\overline{f(x)}\chi(x^{-1})dx\cr
&=\int\overline{f(x)\chi(x)}dx
=\overline{\int f(x)\chi(x)dx}
=\overline{\varhat{u}(\chi)},\cr}$$

\noindent so $\varhat{\tilde u}$ is the complex conjugate of
$\varhat{u}$, and

\Centerline{$(u*\tilde u)\varsphat
=\varhat{u}\times\varhat{\tilde u}=|\varhat{u}|^2$.}

\medskip

{\bf (d)} To see that $A=\{\varhat{u}:u\in L_{\Bbb C}^1\}$ is dense in
$C_0$, we
can use the Stone-Weierstrass theorem in the form 4A6B.   $A$ is a
subalgebra of $C_0$;  it separates the points (because the canonical map
from $\Cal X$ to $(L_{\Bbb C}^1)^*$ is injective);  if $\chi\in\Cal X$,
there is
an $h\in A$ such that $h(\chi)\ne 0$ (because elements of $\Cal X$ act
on $L_{\Bbb C}^1$ as non-zero functionals);  and the complex conjugate
of any function in $A$ belongs to $A$, by (c) above.

Accordingly $A$ is dense in $C_0$, by 4A6B.

The calculation of $\|\varhat{u}\|_{\infty}$ is an immediate consequence
of the characterization of $r(u)$ as
$\max\{|\phi(u)|:\phi\in\Phi\}$ (4A6K) and the identification of $\Phi$
with $\Cal X$.
}%end of proof of 445K

\cmmnt{\medskip

\noindent{\bf Remark} This is the first point in this section where we
really need to know whether or not our group is abelian.
}%end of comment

\leader{445L}{Positive definite functions} Let $X$ be a group.

\spheader 445La A function $h:X\to\Bbb C$ is called {\bf positive
definite} if

\Centerline{$\sum_{j,k=0}^n\zeta_j\bar\zeta_k h(x_k^{-1}x_j)\ge 0$}

\noindent for all $\zeta_0,\ldots,\zeta_n\in\Bbb C$ and
$x_0,\ldots,x_n\in X$.

\spheader 445Lb Suppose that $h:X\to\Bbb C$ is positive definite.
Then, writing $e$ for the identity of $X$,

\quad(i) $|h(x)|\le h(e)$ for every $x\in X$;

\quad(ii) $h(x^{-1})=\overline{h(x)}$ for every $x\in X$.

\noindent\prooflet{\Prf\ If $\zeta\in\Bbb C$ and $x\in X$, take $n=1$,
$x_0=e$, $x_1=x$, $\zeta_0=1$ and $\zeta_1=\zeta$ in the definition in
(a) above, and observe that

\Centerline{$(1+|\zeta|^2)h(e)+\zeta h(x)+\bar\zeta h(x^{-1})
=h(e^{-1}e)+\zeta h(e^{-1}x)+\bar\zeta h(x^{-1}e)
  +\zeta\bar\zeta h(x^{-1}x)
\ge 0$.}

\noindent Taking $\zeta=0$, $x=e$ we get $h(e)\ge 0$.   Taking $\zeta=1$
we see that $h(x)+h(x^{-1})$ is real, and taking $\zeta=i$, we see that
$h(x)-h(x^{-1})$ is purely imaginary;   that is,
$h(x^{-1})=\overline{h(x)}$, for any $x$.   Taking $\zeta$ such that
$|\zeta|=1$, $\zeta h(x)=-|h(x)|$ we get $2h(e)-2|h(x)|\ge 0$, that is,
$|h(x)|\le h(e)$ for every $x\in X$.\ \Qed}

\spheader 445Lc If $h:X\to\Bbb C$ is positive definite and
$\chi:X\to S^1$ is a homomorphism, then $h\times\chi$ is positive definite.   \prooflet{\Prf\
If $\zeta_0,\ldots,\zeta_n\in\Bbb C$ and $x_0,\ldots,x_n\in X$ then

\Centerline{$\sum_{j,k=0}^n\zeta_j\bar\zeta_k(h\times\chi)(x_k^{-1}x_j)
=\sum_{j,k=0}^n\zeta_j\chi(x_j)\overline{\zeta_k\chi(x_k)}
  h(x_k^{-1}x_j)\ge 0$.  \Qed}
}%end of prooflet

\spheader 445Ld If $X$ is an abelian topological group and $\mu$ a Haar
measure on $X$, then for any $f\in\eusm L_{\Bbb C}^2(\mu)$ the
convolution
$f*\tilde f:X\to\Bbb C$ is continuous and positive definite, where
$\tilde f(x)=\overline{f(x^{-1})}$ whenever this is defined.
\prooflet{\Prf\ As in 444Rc, $f*\tilde f$ is defined everywhere on
$X$ and is continuous.   (The definition of $\ssptilde$ has shifted since
\S444, but the argument there applies unchanged to the present
situation.)   Now, if $x_0,\ldots,x_n\in X$ and
$\zeta_0,\ldots,\zeta_n\in\Bbb C$,

$$\eqalign{\sum_{j,k=0}^n\zeta_j\bar\zeta_k(f*\tilde f)(x_k^{-1}x_j)
&=\sum_{j,k=0}^n\zeta_j\bar\zeta_k\int
  f(y)\tilde f(y^{-1}x_k^{-1}x_j)dy\cr
&=\int\sum_{j,k=0}^n\zeta_j\bar\zeta_k
  f(y)\overline{f(x_j^{-1}x_ky)}dy\cr
&=\int\sum_{j,k=0}^n\zeta_j\bar\zeta_k
  f(x_jy)\overline{f(x_j^{-1}x_kx_jy)}dy\cr
&=\int\sum_{j,k=0}^n\zeta_j\bar\zeta_k
  f(x_jy)\overline{f(x_ky)}dy\cr
&=\int|\sum_{j=0}^n\zeta_j f(x_jy)|^2dy
\ge 0.\cr}$$

\noindent So $f*\tilde f$ is positive definite.\ \Qed
}%end of prooflet

\leader{445M}{Proposition} Let $X$ be a topological group and $\nu$ a
quasi-Radon measure on $X$.   If $h:X\to\Bbb C$ is a continuous positive
definite function, then
$\iint h(y^{-1}x)f(x)\overline{f(y)}\nu(dx)\nu(dy)$\vthsp$\ge 0$ for
every $\nu$-integrable function $f$.

\proof{{\bf (a)} Extend $f$, if necessary, to the whole of $X$;  since
the hypothesis implies that $\dom f$ is conegligible, this does not
affect the integrals.  Let $\lambda$ be the product quasi-Radon measure
on $X\times X$;  because $h$ is continuous (by hypothesis) and bounded
(by 445L(b-i)), the function $(x,y)\mapsto
h(y^{-1}x)f(x)\overline{f(y)}$ is
$\lambda$-integrable, and (because $\{x:f(x)\ne 0\}$ can be covered by a
sequence of sets of finite measure)

\Centerline{$I=\iint h(y^{-1}x)f(x)\overline{f(y)}\nu(dx)\nu(dy)
=\int h(y^{-1}x)f(x)\overline{f(y)}\lambda(d(x,y))$}

\noindent (417G).

\medskip

{\bf (b)} Let $\epsilon>0$.   Set $\gamma=\sup_{x\in X}|h(x)|=h(e)$
(445L(b-i)).   Let
$F\subseteq X$ be a non-empty measurable set of finite measure for $\nu$
such that $\gamma\int_{(X\times X)\setminus(F\times
F)}|f(x)\overline{f(y)}|\lambda(d(x,y))\le\bover12\epsilon$ and $f$ is
bounded on $F$;  say $|f(x)|\le M$ for every $x\in F$.  Let $\delta>0$
be such that

\Centerline{$\delta(M^2+2M\gamma)(\nu F)^2+2M^2\gamma\delta
\le\bover12\epsilon$.}

Let $\Cal G$ be the set

$$\eqalign{\{G\times H:G,\,H\subseteq X\text{ are open},\,
&|h(y^{-1}x)-h(y_1^{-1}x_1)|\le\delta\cr
&\text{ whenever }x,\,x_1\in G,\,y,\,y_1\in H\}.\cr}$$

\noindent Because $h$ is continuous, $\Cal G$ is a cover of $X\times X$.
Because $\lambda$ is $\tau$-additive, there is a finite set $\Cal
G_0\subseteq\Cal G$ such that
$\lambda((F\times F)\setminus\bigcup\Cal G_0)\le\delta$;  we may suppose
that $\Cal G_0$ is non-empty.   Set
$W=(F\times F)\cap\bigcup\Cal G_0$.   Enumerate $\Cal G_0$ as
$\langle G_i\times H_i\rangle_{i\le n}$.

Let $\Cal F$ be a finite partition of $F$ into measurable sets such that
$|f(x)-f(x')|\le\delta$ whenever $x$, $x'$ belong to the same member of
$\Cal F$.   Let $\Cal E$ be the partition of $F$ generated by
$\Cal F\cup\{F\cap G_j:j\le n\}\cup\{F\cap H_j:j\le n\}$.   Enumerate
$\Cal E$ as $\langle E_k\rangle_{k\le m}$;  for each $k\le m$ choose
$x_k\in E_k$.   Set
$J=\{(j,k):j\le m,\,k\le m,\,E_j\times E_k\subseteq W\}$;  then
$W=\bigcup_{(j,k)\in J}E_j\times E_k$.

\medskip

{\bf (c)} If $(j,k)\in J$, $x\in E_j$ and $y\in E_k$ then

\Centerline{$|h(y^{-1}x)f(x)\overline{f(y)}
   -h(x_k^{-1}x_j)f(x_j)\overline{f(x_k)}|
\le\delta(M^2+2M\gamma)$,}

\noindent because there must be some $r\le n$ such that
$E_j\times E_k\subseteq G_r\times H_r$, so that
$|h(y^{-1}x)-h(x_k^{-1}x_j)|\le\delta$, while there are members of
$\Cal F$ including $E_j$ and $E_k$, so that $|f(x)-f(x_j)|\le\delta$ and
$|\overline{f(y)}-\overline{f(x_k)}|\le\delta$;  at the same time,

\Centerline{$|f(x)\overline{f(y)}|\le M^2$,
\quad$|h(x_k^{-1}x_j)||\overline{f(y)}|\le M\gamma$,
\quad$|h(x_k^{-1}x_j)||f(x_j)|\le M\gamma$}

\noindent because $x$, $y$, $x_j$ and $x_k$ all belong to $F$.

\woddheader{445M}{4}{2}{2}{72pt}

{\bf (d)} Set $\zeta_j=f(x_j)\nu E_j$ for $j\le m$, so that
$\bar\zeta_j=\overline{f(x_j)}\nu E_j$.   Now consider

$$\eqalign{\bigl|\int_W h(&y^{-1}x)f(x)\overline{f(y)}\lambda(d(x,y))
  -\sum_{(j,k)\in J}\zeta_j\bar\zeta_kh(x_k^{-1}x_j)\bigr|\cr
&=\bigl|\sum_{(j,k)\in J}\int_{E_j\times E_k}
      h(y^{-1}x)f(x)\overline{f(y)}
  -h(x_k^{-1}x_j)f(x_j)\overline{f(x_k)}\lambda(d(x,y))\bigr|\cr
&\le\sum_{(j,k)\in J}\int_{E_j\times E_k}|h(y^{-1}x)f(x)\overline{f(y)}
  -h(x_k^{-1}x_j)f(x_j)\overline{f(x_k)}|\lambda(d(x,y))\cr
&\le\sum_{(j,k)\in J}\delta(M^2+2M\gamma)\nu E_j\nu E_k\cr
&\le\delta(M^2+2M\gamma)\lambda W
\le\delta(M^2+2M\gamma)(\nu F)^2.\cr}$$

\noindent On the other hand,

$$\eqalign{\bigl|\int_{(X\times X)\setminus W}
   h(y^{-1}x)f(x)\overline{f(y)}\lambda(d(x,y))\bigr|
&\le\gamma\int_{(X\times X)\setminus(F\times
F)}|f(x)\overline{f(y)}|\lambda(d(x,y))\cr
&\qquad\qquad+\gamma\int_{(F\times F)\setminus W}|f(x)
     \overline{f(y)}|\lambda(d(x,y))\cr
&\le\Bover12\epsilon+\gamma\delta M^2,\cr}$$

\noindent and

$$\eqalign{\bigl|\sum_{j\le m,k\le m,(j,k)\notin J}
    h(x_k^{-1}x_j)f(x_j)\tilde f(x_k)\bigr|
&\le\gamma M^2\sum_{j\le m,k\le m,(j,k)\notin J}\nu E_j\nu E_k\cr
&=\gamma M^2\lambda((F\times F)\setminus W)
\le\gamma M^2\delta.\cr}$$

\noindent Putting these together,

$$\eqalign{|I-\sum_{j,k=0}^m\zeta_j\bar\zeta_kh(x_k^{-1}x_j)|
&\le\delta(M^2+2M\gamma)(\nu F)^2
+\Bover12\epsilon+\gamma\delta M^2
+\gamma M^2\delta
\le\epsilon.\cr}$$

\noindent But $\sum_{j,k=0}^m\zeta_j\bar\zeta_kh(x_k^{-1}x_j)\ge 0$,
because $h$ is positive definite.   As $\epsilon$ is arbitrary,
$I\ge 0$, as required.
}%end of proof of 445M

\leader{445N}{Bochner's theorem}\cmmnt{ ({\smc Herglotz 1911},
{\smc Bochner 1933}, {\smc Weil 1940})} Let $X$ be an abelian topological group
with a Haar measure $\mu$, and $\Cal X$ its dual group.   Then for
any continuous positive definite function $h:X\to\Bbb C$ there is a
unique totally finite Radon measure $\nu$ on $\Cal X$ such that

\Centerline{$\int h\times f\,d\mu=\int\varhatf\,d\nu$
for every $f\in\eusm L_{\Bbb C}^1=\eusm L_{\Bbb C}^1(\mu)$,}

\Centerline{$h(x)=\int\chi(x)\nu(d\chi)$ for every $x\in X$.}

\proof{{\bf (a)} If $h(e)=0$, where $e$ is the identity in $X$, then
$h=0$, by 445L(b-i), and we can take $\nu$ to be the zero measure.
Otherwise, since
multiplying $h$ by a positive scalar leaves $h$ positive definite and
does not affect the result, we may suppose that $h(e)=1$.   For $f$,
$g\in\eusm L_{\Bbb C}^1=\eusm L_{\Bbb C}^1(\mu)$ set

\Centerline{$(f|g)
=\biggeriint f(x)\overline{g(y)}h(y^{-1}x)dxdy
=\iint f(x)\overline{g(y^{-1})}h(yx)dxdy$}

\noindent (by 442Kb, since $X$ is unimodular).   Then, by 445M,
$(f|f)\ge 0$ for every $f\in\eusm L_{\Bbb C}^1$.   Also
$(f_1+f_2|g)=(f_1|g)+(f_2|g)$, $(\zeta f|g)=\zeta(f|g)$ and
$(g|f)=\overline{(f|g)}$
for all $f$, $g$, $f_1$, $f_2\in\eusm L_{\Bbb C}^1$ and
$\zeta\in\Bbb C$.  \Prf\
Only the last is anything but trivial, and for this we have

$$\eqalignno{(g|f)
&=\iint g(x)\overline{f(y)}h(y^{-1}x)dxdy\cr
&=\iint g(x)\overline{f(y)}h(y^{-1}x)dydx\cr
\noalign{\noindent (by 417Ga, because
$(x,y)\mapsto g(x)\overline{f(y)}h(y^{-1}x)$ is integrable for the
product measure and
zero off the square of a countable union of sets of finite measure)}
&=\iint g(x)\overline{f(y)}\overline{h(x^{-1}y)}dydx\cr
\noalign{\noindent (using 445L(b-ii))}
&=\overline{\iint f(y)\overline{g(x)}
    h(x^{-1}y)dydx}\cr
&=\overline{(f|g)}.\text{ \Qed}\cr}$$

\medskip

{\bf (b)} If $f$, $g\in\eusm L_{\Bbb C}^1$, $|(f|g)|^2\le(f|f)(g|g)$.
\Prf\
(Really this is just Cauchy's inequality.)   For
any $\alpha$, $\beta\in\Bbb C$,

\Centerline{$|\alpha|^2(f|f)+2\Real(\alpha\bar\beta(f|g))+|\beta|^2(g|g)
=(\alpha f+\beta g|\alpha f+\beta g)
\ge 0$.}

\noindent If $(f|f)=0$ we have $2\Real(\alpha(f|g))+(g|g)\ge 0$ for
every $\alpha\in\Bbb C$ so in this case $(f|g)=0$;  similarly
$(f|g)=0$ if $(g|g)=0$;  otherwise we can find non-zero $\alpha$,
$\beta$ such that $|\alpha|^2=(g|g)$, $|\beta|^2=(f|f)$ and
$\alpha\bar\beta(f|g)=-|\alpha\beta(f|g)|$, in which case the inequality
simplifies
to $|(f|g)|\le|\alpha\beta|$ and $|(f|g)|^2\le(f|f)(g|g)$, as
required.\ \Qed

\medskip

{\bf (c)} Now consider the functional
$\psi\in(L_{\Bbb C}^1)^*=(L_{\Bbb C}^1(\mu))^*$
corresponding to $h$, so that $\psi(f^{\ssbullet})=\int h\times f\,d\mu$
for every $f\in\eusm L_{\Bbb C}^1$.   Then
$|\psi(f^{\ssbullet})|^2\le(f|f)$ for
every $f\in\eusm L_{\Bbb C}^1$.   \Prf\ Let $\epsilon>0$.   Then there
is an open neighbourhood $U$ of $e$ such that $U=U^{-1}$ and

\Centerline{$|h(y^{-1}x)-h(e)|\le\epsilon$ whenever $x$, $y\in U$,}

\Centerline{$\|a\action_lf-f\|_1\le\epsilon$ for every $a\in U$}

\noindent where $(a\action_lf)(x)=f(a^{-1}x)$ whenever this is defined, as
usual (443Gf).   Shrinking $U$ if need be, we may suppose that
$\mu U<\infty$, and of course $\mu U>0$.   Set
$g=\Bover1{\mu U}\chi U\in\eusm L_{\Bbb C}^1$.   Then

$$\eqalign{|(f|g)-\psi(f^{\ssbullet})|
&=\Bover1{\mu U}\bigl|\int_U\int_Xf(x)h(y^{-1}x)dxdy
  -\int_U\int_Xf(x)h(x)dxdy\bigr|\cr
&=\Bover1{\mu U}\bigl|\int_U\int_Xf(yx)h(x)dxdy
  -\int_U\int_Xf(x)h(x)dxdy\bigr|\cr
&=\Bover1{\mu U}\bigl|\int_U\int_X
  ((y^{-1}\action_lf)(x)-f(x))h(x)dxdy\bigr|\cr
&\le\Bover1{\mu U}\int_U\int_X
  |(y^{-1}\action_lf)(x)-f(x)||h(x)|dxdy\cr
&\le\Bover1{\mu U}\int_U\|y^{-1}\action_lf-f\|_1\mu(dy)
\le\epsilon.\cr}$$

\noindent Also

$$\eqalign{|(g|g)-1|
&=\bigl|\Bover1{\mu U^2}\int_U\int_U
   (h(y^{-1}x)-1)dxdy\bigr|\cr
&\le\Bover1{\mu U^2}\int_U\int_U|h(y^{-1}x)-1|dxdy
\le\epsilon.\cr}$$

\noindent So

\Centerline{$\max(0,|\psi(f^{\ssbullet})|-\epsilon)^2
\le|(f|g)|^2
\le(f|f)(g|g)
\le(1+\epsilon)(f|f)$.}

\noindent Letting $\epsilon\downarrow 0$ we have the result.\ \Qed

\woddheader{445N}{4}{2}{2}{40pt}

{\bf (d)} If we look at $(f|f)$, however, and apply 444Od, we see that

$$\eqalign{(f|f)
&=\iint f(x)\overline{f(y)}h(y^{-1}x)dxdy\cr
&=\iint f(x)\overline{f(y^{-1})}h(yx)dxdy
=\int h(x)(f*\tilde f)(x)dx,\cr}$$

\noindent where $\tilde f(x)=\overline{f(x^{-1})}$ whenever this is
defined;  that is,

\Centerline{$(f|f)=\psi(f*\tilde f)^{\ssbullet}$.}

\noindent (Note that $\tilde f\in\eusm L_{\Bbb C}^1$ because $X$ is
unimodular,
as in part (c) of the proof of 445K.)    So (c) tells us that

\Centerline{$|\psi(f^{\ssbullet})|^2\le\psi(f*\tilde f)^{\ssbullet}$}

\noindent for every $f\in\eusm L_{\Bbb C}^1$, that is,
$|\psi(u)|^2\le\psi(u*\tilde u)$ for every $u\in L_{\Bbb C}^1$, defining
$\tilde
u$ as in 445Kc.

\medskip

{\bf (e)} In fact

\Centerline{$|\psi(u)|\le\|\varhat{u}\|_{\infty}$}

\noindent for every $u\in L_{\Bbb C}^1$.  \Prf\ Set $u_0=u$ and
$u_{k+1}=u_k*\tilde u_k$ for every $k\in\Bbb N$.   We need to know that
$u_k=\tilde u_k$ for $k\ge 1$.   To see this, represent $u_{k-1}$ as
$f^{\ssbullet}$ where $f\in\eusm L_{\Bbb C}^1$, so that $u_k=(f*\tilde
f)^{\ssbullet}$.   Now

$$\eqalign{(f*\tilde f)\ssptilde(x)
&=\overline{(f*\tilde f)(x^{-1})}
=\overline{\int f(x^{-1}y)\tilde f(y^{-1})dy}\cr
&=\int\overline{f(x^{-1}y)}f(y)dy
=\int\tilde f(y^{-1}x)f(y)dy
=(f*\tilde f)(x)\cr}$$

\noindent for every $x$, so $(f*\tilde f)\ssptilde=f*\tilde f$ and
$\tilde u_k=u_k$.   Accordingly $u_{k+1}=u_k*u_k$ for $k\ge 1$ and we
have $u_k=(u_1)^{2^{k-1}}$ for every $k\ge 1$.

At the same time, we have $|\psi(u_k)|^2\le\psi(u_{k+1})$ for every $k$,
by (d), so that, for $k\ge 1$,

\Centerline{$|\psi(u)|^{2^k}\le\psi(u_k)\le\|u_k\|_1
=\|u_1^{2^{k-1}}\|_1$,}

\Centerline{$|\psi(u)|\le\|u_1^{2^{k-1}}\|_1^{1/2^k}$.}

\noindent Letting $k\to\infty$, $|\psi(u)|\le\sqrt{r(u_1)}$, where
$r(u_1)$ is the spectral radius of $u_1$.

At this point, recall that $r(u_1)=\|\varhat{u}_1\|_{\infty}$ (445Kd),
while $|\varhat{u}|^2=\varhat{u}_1$ (445Kc), so
$r(u_1)=\|\varhat{u}\|_{\infty}^2$ and
$|\psi(u)|\le\|\varhat{u}\|_{\infty}$.\ \Qed

\medskip

{\bf (f)} Now consider $\varsphat$ as a linear operator from
$L_{\Bbb C}^1$ to
$C_0=C_0(\Cal X;\Bbb C)$, as in 445K.   If $\varhat{u}=0$ then
$\psi(u)=0$, by (e), so setting $A=\{\varhat{u}:u\in L_{\Bbb C}^1\}$ we
have a
linear functional $\psi_0:A\to\Bbb C$ defined by saying that
$\psi_0(\varhat{u})=\psi(u)$ for every $u\in L_{\Bbb C}^1$.   By (e),
$\|\psi_0\|\le 1$.   Now
$\psi_0$ has an extension to a bounded linear operator $\psi_1$,
still of norm at most $1$, from $C_0$ to $\Bbb C$ (3A5Ab).

\medskip

{\bf (g)} Suppose that $q\in C_0$ and $0\le q\le\tbf{1}$,
writing $\tbf{1}$
for the constant function with value $1$;  set $\alpha=\psi_1(q)$.
Then for any $\zeta\in\Bbb C$ and $\gamma\ge 0$ we have
$|\zeta-\gamma\alpha|\le\max(|\zeta|,|\gamma|,|\zeta-\gamma|)$.   \Prf\
Let $\epsilon>0$.   Set $V=\{x:|1-h(x)|<\epsilon\}$;  then $V$ is an
open neighbourhood of $e$;  set $f=\bover1{\mu V}\chi V$ and
$u=f^{\ssbullet}$, so that

\Centerline{$\|\varhat{u}\|_{\infty}=r(u)\le\|u\|_1=1$,}

\Centerline{$|1-\psi(u)|=|\Bover1{\mu V}\int_V(1-h(x))dx|
\le\epsilon$.}

\noindent Set $v=u*\tilde u$;  then

\Centerline{$\psi_1(\varhat{v})=\psi(v)
\ge|\psi(u)|^2\ge(1-\epsilon)^2$,}

\noindent using part (d) for the central inequality.
But $\varhat{v}=|\varhat{u}|^2$, so that $0\le\varhat{v}\le\tbf{1}$ and
$\psi_1(\varhat{v})\le 1$.

Now consider $\|\zeta\varhat{v}-\gamma q\|_{\infty}$.   If $\chi\in\Cal
X$, then $\zeta\varhat{v}(\chi)$ and $\gamma q(\chi)$ both lie in the
triangle with vertices $0$, $\zeta$ and $\gamma$, because
$0\le\varhat{v}\le\tbf{1}$ and $0\le q\le\tbf{1}$.   So

\Centerline{$|\zeta\varhat{v}(\chi)-\gamma q(\chi)|
\le\max(|\gamma|,|\zeta|,|\gamma-\zeta|)$.}

\noindent As $\chi$ is arbitrary,

\Centerline{$\|\zeta\varhat{v}-\gamma q\|_{\infty}
\le\max(|\gamma|,|\zeta|,|\gamma-\zeta|)$.}

\noindent Accordingly

$$\eqalign{|\zeta-\gamma\alpha|
&\le|\zeta-\zeta\psi_1(\varhat{v})|
  +|\psi_1(\zeta\varhat{v}-\gamma q)|\cr
&\le|\zeta|(1-(1-\epsilon)^2)+\|\zeta\varhat{v}-\gamma q\|_{\infty}
\le 2\epsilon|\zeta|+\max(|\zeta|,|\gamma|,|\zeta-\gamma|).\cr}$$

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

Taking $\zeta=\gamma=1$ we see that
$|1-\alpha|\le 1$, so that $\Real\alpha\ge 0$.   Taking $\zeta=\pm i$,
we see that $|i\pm\gamma\alpha|\le\sqrt{1+\gamma^2}$ for every
$\gamma\ge 0$, so that $\Imag\alpha=0$.   Thus $\psi_1(q)\ge 0$;  and
this is true whenever $0\le q\le\tbf{1}$ in $C_0$.

\medskip

{\bf (h)} It follows at once that $\psi_1(q)\ge 0$ whenever $q\ge 0$
in $C_0$.   Applying the Riesz Representation Theorem, in the form
436K, to the restriction of $\psi_1$ to $C_0(\Cal X;\Bbb R)$, we
see that there is a totally finite Radon measure $\nu$ on
$\Cal X$ such that $\psi_1(q)=\int q\,d\nu$ for every real-valued
$q\in C_0$;
of course it follows that $\psi_1(q)=\int q\,d\nu$ for every $q\in C_0$.
Unwrapping the definition of $\psi_1$, we see that

\Centerline{$\int h(x)f(x)dx
=\psi(f^{\ssbullet})=\psi_1(\varhatf)=\int\varhatf\,d\nu$}

\noindent for every $f\in\eusm L_{\Bbb C}^1(\mu)$.

\medskip

{\bf (i)} For the second formula, argue as follows.   Given
$f\in\eusm L_{\Bbb C}^1(\mu)$, consider the function
$(x,\chi)\mapsto f(x)\chi(x):X\times\Cal X\to\Bbb C$.   Because
$(\chi,x)\mapsto\chi(x)$ is
continuous (445Ea), this is $\Lambda$-measurable, where
$\Lambda$ is the domain of the product quasi-Radon
measure $\mu\times\nu$ on $X\times\Cal X$.   It is integrable because
$\nu\Cal X<\infty$ and $|\chi(x)|=1$ for every $\chi$, $x$;  moreover,
it is zero off the set $\{x:f(x)\ne 0\}\times\Cal X$, which is a
countable
union of products of sets of finite measure.   Note also that because
$\chi\mapsto\chi(x)$ is continuous and bounded for every $x\in X$,
$h_1(x)=\int\chi(x)\nu(d\chi)$ is defined, and $|h_1(x)|\le\nu X$,  for
every $x\in X$.   What is more, $h_1$ is continuous.   \Prf\ Let
$\frak X$ be the dual group of $\Cal X$, and for $x\in X$ let $\hat x$
be the corresponding member of $\frak X$.   Then, in the language of
445C, applied to the topological group $\Cal X$,

\Centerline{$h_1(x)=\int\hat x\,d\nu=\varhat{\nu}(\hat x)$}

\noindent for every $x\in X$.   But $\varhat{\nu}:\frak X\to\Bbb C$ is
continuous, by 445Ec, and $x\mapsto\hat x:X\to\frak X$ is continuous, by
445Eb;  so $h_1$ also is continuous.\ \Qed\

We may therefore apply Fubini's theorem (417G) to see that

$$\eqalign{\int f(x)h_1(x)\mu(dx)
&=\iint f(x)\chi(x)\nu(d\chi)\mu(dx)
=\iint f(x)\chi(x)\mu(dx)\nu(d\chi)\cr
&=\int\varhatf(\chi)\nu(d\chi)
=\int f(x)h(x)\mu(dx).\cr}$$

\noindent Since this is true for every $f\in\eusm L_{\Bbb C}^1$,
$h_1\eae h$;   since both are continuous, $h_1=h$, as required.

\medskip

{\bf (j)} Finally, to see that $\nu$ is uniquely defined, note that
$\{\varhat{f}:f\in\eusm L^1_{\Bbb C}\}$ is $\|\,\|_{\infty}$-dense in $C_0$
(445Kd), so 436K tells us that there can be at most one totally finite
Radon measure $\nu$ on $\Cal X$ such that
$\int h\times f\,d\mu=\int\varhatf\,d\nu$ for every
$f\in\eusm L^1_{\Bbb C}$.
}%end of proof of 445N

\leader{445O}{Proposition} Let $X$ be a Hausdorff abelian topological
group carrying Haar measures.   Then the map $x\mapsto\hat x$ from $X$
to its bidual group $\frak X$ is a homeomorphism between $X$ and its
image in $\frak X$.   In particular, the dual group $\Cal X$ of $X$
separates the points of $X$.

\proof{ We already know that \hbox{$\sphat\,$} is continuous (445Eb) and
that $\Cal X$ is locally compact and Hausdorff (445J).
Now let $U$ be any neighbourhood of the identity $e$ of $X$.   Let
$V\subseteq U$ be an open neighbourhood of $e$ such that
$VV^{-1}\subseteq U$ and
$\mu V<\infty$.   Then $f=\chi V\in\eusm L_{\Bbb C}^2(\mu)$, so
$f*\tilde f$ is positive definite and continuous (445Ld) and there is a
totally finite Radon measure
$\nu$ on $\Cal X$ such that $(f*\tilde f)(x)=\int\chi(x)\nu(d\chi)$ for
every $x\in X$ (445N).   Note that, writing $\frak e$ for the identity
of $\frak X$ and $\varhat{\nu}:\frak X\to\Bbb C$ for the Fourier-Stieltjes
transform of $\nu$,

$$\eqalign{\varhat{\nu}(\frak e)
&=\int\frak e(\chi)\nu(d\chi)
=\int\chi(e)\nu(d\chi)
=(f*\tilde f)(e)\cr
&=\int f(y)\tilde f(y^{-1})\mu(dy)
=\int|f(y)|^2\mu(dy)
\ne 0.\cr}$$

\noindent Now $\varhat{\nu}$ is continuous (445Ec), so
$W=\{\frak x:\varhat{\nu}(\frak x)\ne 0\}$ is a neighbourhood of
$\frak e$.   If $x\in X$ and $\hat x\in W$, then

\Centerline{$(f*\tilde f)(x)=\int\chi(x)\nu(d\chi)
=\int\hat x(\chi)\nu(d\chi)=\varhat{\nu}(\hat x)\ne 0$,}

\noindent so there is some $y\in X$ such that
$f(y)\tilde f(y^{-1}x)\ne 0$, that is, $f(y)\ne 0$ and
$f(x^{-1}y)\ne 0$, that is, $y$ and
$x^{-1}y$ both belong to $V$;  in which case $x\in VV^{-1}\subseteq U$.

Thus $U\supseteq\{x:\hat x\in W\}$.   This means that, writing $\frak S$
for $\{\{x:\hat x\in H\}:H\subseteq\frak X$ is open$\}$, every
neighbourhood of $e$ for the original topology $\frak T$ of $X$ is a
neighbourhood of $e$ for $\frak S$.   But (it is easy to check)
$(X,\frak S)$ is a topological group because $\frak X$ is a topological
group and \hbox{$\sphat\,$} is a homomorphism.   So
$\frak T\subseteq\frak S$ (4A5Fb).    As we know already that
$\frak S\subseteq\frak T$, the two topologies are equal.

It follows at once that if $\frak T$ is Hausdorff, then (because
$\frak S$ is Hausdorff) the map \hbox{$\sphat\,$} is an injection and is a
homeomorphism between $X$ and its image in $\frak X$.
}%end of proof of 445O

\vleader{60pt}{445P}{The Inversion Theorem}
Let $X$ be an abelian topological
group and $\mu$ a Haar measure on $X$.   Then there is a unique Haar
measure $\lambda$ on the dual group $\Cal X$ of $X$ such that whenever
$f:X\to\Bbb C$ is continuous, $\mu$-integrable and positive definite,
then $\varhatf:\Cal X\to\Bbb C$ is $\lambda$-integrable and

\Centerline{$f(x)
=\int\varhatf(\chi)\overline{\chi(x)}\lambda(d\chi)$}

\noindent for every $x\in X$.

\proof{{\bf (a)} Write $P$ for the set of $\mu$-integrable positive
definite continuous functions $h:X\to\Bbb C$.   For $h\in P$, let
$\nu_h$ be the corresponding totally finite Radon measure on $\Cal X$
defined in 445N, so that

\Centerline{$\int f\times h\,d\mu=\int\varhatf\,d\nu_h$}

\noindent for every $f\in\eusm L_{\Bbb C}^1=\eusm L_{\Bbb C}^1(\mu)$.

\medskip

{\bf (b)} The basis of the argument is the following fact.   If
$f\in\eusm L_{\Bbb C}^1$ and $h_1$, $h_2\in P$, then

\Centerline{$\int\varhatf\times\varhat{\bar h}_2d\nu_{h_1}
=\int\varhatf\times\varhat{\bar h}_1d\nu_{h_2}$.}


$$\eqalignno{\text{\Prf }\int\varhatf\times\varhat{\bar h}_1d\nu_{h_2}
&=\int(f*\bar h_1)\varsphat d\nu_{h_2}\cr
\noalign{\noindent (445G)}
&=\int h_2(x)(f*\bar h_1)(x)\mu(dx)
=\int \bar h_2(x^{-1})(f*\bar h_1)(x)\mu(dx)\cr
\noalign{\noindent (by 445Lb)}
&=((f*\bar h_1)*\bar h_2)(e)
=((f*\bar h_2)*\bar h_1)(e)\cr
\noalign{\noindent (because $*$ is associative and commutative, by 444Oe
and 444Og)}
&=\int\varhatf\times\varhat{\bar h}_2d\nu_{h_1}.\text{ \Qed}\cr}$$

Now because $\varhat{\bar h}_1$ and $\varhat{\bar h}_2$ are both bounded
(by $\int|h_1|d\mu$ and $\int|h_2|d\mu$ respectively), and $\nu_{h_1}$
and $\nu_{h_2}$ are both totally finite measures, and
$\{\varhatf:f\in\eusm L_{\Bbb C}^1(\mu)\}$ is $\|\,\,\|_{\infty}$-dense
in
$C_0=C_0(\Cal X;\Bbb C)$ (445Kd), we must have

\Centerline{$\int p\times\varhat{\bar h}_2d\nu_{h_1}=\int
p\times\varhat{\bar h}_1d\nu_{h_2}$}

\noindent for every $p\in C_0$.

\wheader{445P}{6}{2}{2}{48pt}

{\bf (c)} Let $\Cal K$ be the family of compact subsets of $\Cal X$.
For $K\in\Cal K$ set

\Centerline{$P_K=\{h:h\in P,\,\varhat{\bar h}(\chi)>0$ for every
$\chi\in K\}$.}

\noindent Then $P_K$ is non-empty.   \Prf\ Set

\Centerline{$U=\{x:x\in X$, $|1-\chi(x)|\le\bover12$ for every $\chi\in
K\}$.}

\noindent Then $U$ is a neighbourhood of the identity $e$ of $X$, by
445Eb.   Let $V$ be an open neighbourhood of $e$, of finite measure,
such that $VV^{-1}\subseteq U$, set $g=\bover1{\mu V}\chi V$, and try
$h=g*\tilde g$.   Then $h$ is continuous and positive definite (445Ld),
real-valued and non-negative (because $g$ and $\tilde g$ are),
zero outside $U$ (because
$VV^{-1}\subseteq U$, as in the proof of 445O),  and

\Centerline{$\int h\,d\mu=\int g\,d\mu\cdot\int\tilde g\,d\mu=1$}

\noindent (444Qb).   Next,

\Centerline{$\varhat{\bar h}
=\varhat{h}=|\varhat{g}|^2$}

\noindent (445Kc) is non-negative, and if $\chi\in K$ then

\Centerline{$|1-\varhat{h}(\chi)|
=|\int h(x)-h(x)\chi(x)\mu(dx)|
\le\int h(x)|1-\chi(x)|\mu(dx)
\le\bover12$}

\noindent because $|1-\chi(x)|\le\bover12$ if $x\in U$ and $h(x)=0$ if
$x\in X\setminus U$.   So

\Centerline{$\varhat{\bar h}(\chi)=\varhat{h}(\chi)\ge\bover12$}

\noindent for every $\chi\in K$, and $h\in P_K$.\ \Qed

\medskip

{\bf (d)} Because $\Cal K$ is upwards-directed, $\{P_K:K\in\Cal K\}$ is
downwards-directed and generates a filter $\Cal F$ on $P$.   Let
$C_k=C_k(\Cal X;\Bbb C)$ be the space of continuous complex-valued
functions on $\Cal X$ with compact support.   If $q\in C_k$, then

\Centerline{$\phi(q)
=\lim_{h\to\Cal F}\int\bover{q}{\varhat{\bar h}}\,d\nu_h$}

\noindent is defined in $\Bbb C$, where in the division
$q/\varhat{\bar h}$ we interpret $0/0$ as $0$ if necessary.   \Prf\
Setting
$K=\overline{\{\chi:q(\chi)\ne 0\}}$, we see in fact that for any $h_1$,
$h_2\in P_K$ we may define a function $p\in C_k$ by setting

$$\eqalign{p(\chi)
&=\bover{q(\chi)}{\varhat{\bar h}_1(\chi)\varhat{\bar h}_2(\chi)}
   \text{ if }\chi\in K,\cr
&=0\text{ if }q(\chi)=0,\cr}$$

\noindent so that

$$\eqalignno{\int\bover{q}{\varhat{\bar h}_1}\,d\nu_{h_1}
&=\int p\times\varhat{\bar h}_2\,d\nu_{h_1}
=\int p\times\varhat{\bar h}_1\,d\nu_{h_2}\cr
\noalign{\noindent (by (b) above)}
&=\int\bover{q}{\varhat{\bar h}_2}\,d\nu_{h_2}.\cr}$$

\noindent So this common value must be $\phi(q)$.\ \Qed

If $q$, $q'\in C_k$ and $\alpha\in\Bbb C$, then

$$\int\bover{q+q'}{\varhat{\bar h}}\,d\nu_h
=\int\bover{q}{\varhat{\bar h}}\,d\nu_h
  +\int\bover{q'}{\varhat{\bar h}}\,d\nu_h,$$

$$\int\bover{\alpha q}{\varhat{\bar h}}\,d\nu_h
=\alpha\int\bover{q}{\varhat{\bar h}}\,d\nu_h$$

\noindent whenever $h\in P_K$, where
$K=\overline{\{\chi:|q(\chi)|+|q'(\chi)|>0\}}$;  so
$\phi(q+q')=\phi(q)+\phi(q')$ and $\phi(\alpha q)=\alpha\phi(q)$.
Moreover, if $q\ge 0$, then
$q/\varhat{\bar h}\ge 0$ for every $h\in P_K$, so $\phi(q)\ge 0$.

\medskip

{\bf (e)} By the Riesz Representation Theorem (in the form 436J) there
is a Radon measure $\lambda$ on $\Cal X$ such that
$\int q\,d\lambda=\phi(q)$ for any continuous function $q$ of compact
support.   (As in part (h) of the proof of 445N, the shift from
real-valued $q$ to complex-valued $q$ is elementary.)

\medskip

{\bf (f)} Now $\lambda$ is translation-invariant.   \Prf\ Take
$\theta\in\Cal X$ and $q\in C_k$.   Set
$K=\overline{\{\chi:q(\chi)\ne 0\}}$ and $L=\theta^{-1}K$, and
take any $h\in P_K$.  Set $h_1(x)=h\times\theta^{-1}$.   Then
$h_1$ is positive definite (445Lc);  of course it is continuous and
$\mu$-integrable;  and for any $\chi\in L$,

\Centerline{$\varhat{\bar h}_1(\chi)
=\int\overline{h(x)}\theta(x)\chi(x)\mu(dx)
=\varhat{\bar h}(\theta\chi)
>0$.}

\noindent So $h_1\in P_L$.

To relate $\nu_{h_1}$ to $\nu_h$, observe that if
$f\in\eusm L_{\Bbb C}^1$ then

\Centerline{$\varhatf(\theta\chi)
=\int f(x)\theta(x)\chi(x)\mu(dx)
=(f\times\theta)\varsphat(\chi)$,}

\noindent so

\Centerline{$\int\varhatf(\theta\chi)\nu_{h_1}(d\chi)
=\int (f\times\theta)(x)h_1(x)\mu(dx)
=\int f(x)h(x)\mu(dx)
=\int\varhatf(\chi)\nu_h(d\chi)$.}

\noindent So we see that the equation

\Centerline{$\int p(\theta\chi)\nu_{h_1}(d\chi)
=\int p(\chi)\nu_h(d\chi)$}

\noindent is valid whenever $p$ is of the form $\varhatf$, for some
$f\in\eusm L_{\Bbb C}^1$, and therefore for every $p\in C_0$.

Set $q_1(\chi)=q(\theta\chi)$ for every $\chi\in\Cal X$, so that $q_1\in
C_k$ and $L=\overline{\{\chi:q_1(\chi)\ne 0\}}$.   Accordingly

$$\eqalign{\phi(q_1)
&=\int\bover{q_1(\chi)}{\varhat{\bar h}_1(\chi)}\nu_{h_1}(d\chi)
=\int\bover{q(\theta\chi)}{\varhat{\bar h}(\theta\chi)}
  \nu_{h_1}(d\chi)\cr
&=\int\bover{q(\chi)}{\varhat{\bar h}(\chi)}
  \nu_h(d\chi)
=\phi(q).\cr}$$

\noindent So

\Centerline{$\int q(\theta\chi)\lambda(d\chi)
 =\int q(\chi)\lambda(d\chi)$.}

\noindent As $q$ and $\theta$ are arbitrary, $\lambda$ is
translation-invariant (441L).\ \Qed

\medskip

{\bf (g)} Thus $\lambda$ is either zero or a Haar measure on $\Cal X$.
I have still to confirm that

\Centerline{$f(x)
=\int\varhatf(\chi)\overline{\chi(x)}\lambda(d\chi)$}

\noindent whenever $f$ is continuous, positive definite and
$\mu$-integrable, and $x\in X$.   But recall the formula from (b) above.
If $q\in C_k$, $K=\overline{\{\chi:q(\chi)\ne 0\}}$ and $h\in P_K$, then
we must have

$$\int q\times\varhatbarf d\lambda
=\int\bover{q\times\varhatbarf}{\varhat{\bar h}}d\nu_h
=\int\bover{q\times\varhat{\bar h}}{\varhat{\bar h}}d\nu_f
=\int q\,d\nu_f.$$

\noindent In particular, $\int q\times\varhatbarf d\lambda\ge 0$ whenever
$q\ge 0$;  since $\varhatbarf$ is continuous (445Ka), and $\lambda$,
being a Radon measure, is strictly positive, this shows that
$\varhatbarf\ge 0$.   Also

\Centerline{$\int\varhatbarf d\lambda
=\sup\{\int q\times\varhatbarf d\lambda:q\in C_k,\,0\le q\le\tbf{1}\}
=\nu_f(\Cal X)<\infty$,}

\noindent so we have an indefinite-integral measure
$\varhatbarf\lambda$;  since this is a Radon measure (416Sa), and acts on
$C_k$ in the same way as $\nu_f$, it is actually equal to $\nu_f$
(by the uniqueness guaranteed in 436J).   In particular, for any $x\in X$,

\Centerline{$f(x)=\int\chi(x)\nu_f(d\chi)
=\int\chi(x)\varhatbarf(\chi)\lambda(d\chi)$}

\noindent by the second formula in 445N, and 235K.   But

$$\eqalignno{\varhatbarf(\chi)
&=\int\overline{f(x)}\chi(x)\mu(dx)
=\int f(x^{-1})\chi(x)\mu(dx)\cr
\noalign{\noindent (by 445Lb)}
&=\int f(x)\chi(x^{-1})\mu(dx)\cr
\noalign{\noindent (because $X$ is abelian, therefore unimodular)}
&=\varhatf(\chi^{-1})\cr}$$

\noindent for every $\chi\in\Cal X$.   So

$$\eqalignno{f(x)
&=\int\chi(x)\varhatf(\chi^{-1})\lambda(d\chi)
=\int\chi^{-1}(x)\varhatf(\chi)\lambda(d\chi)\cr
\noalign{\noindent (because $\Cal X$ is abelian)}
&=\int\varhatf(\chi)\overline{\chi(x)}\lambda(d\chi).\cr}$$

\medskip

{\bf (h)} We should check that $\lambda$ is non-zero
and unique.   But the construction in part (c) of the proof shows that
there are many $f\in P$ such that $f(e)\ne 0$, and for any such $f$ we have
$f(e)=\int\varhatf\,d\lambda$.   This shows simultaneously that $\lambda$
is non-zero, therefore a Haar measure;  and as all the Haar measures on
$\Cal X$ are scalar multiples of each other, there is at most one suitable
$\lambda$.
}%end of proof of 445P

\leader{445Q}{Remark}\cmmnt{ We can extract the following useful fact
from part (g) of the proof above.}   If $h:X\to\Bbb C$ is
$\mu$-integrable, continuous and positive definite, then
$\varhat{\bar h}$ is non-negative and $\lambda$-integrable, and the
Radon measure $\nu_h$ of 445N is just the indefinite-integral measure
$\varhat{\bar h}\lambda$.

\cmmnt{Note also that} $\lambda$ is\cmmnt{ actually} a Radon
measure\cmmnt{;  of
course it has to be, because $\Cal X$ is locally compact and Hausdorff
(445J)}.

\leader{445R}{The Plancherel Theorem} Let $X$ be an abelian topological
group with a
Haar measure $\mu$, and $\Cal X$ its dual group.   Let $\lambda$ be the
Haar measure on $\Cal X$ corresponding to $\mu$\cmmnt{ (445P)}.
Then there is a normed space isomorphism
$T:L_{\Bbb C}^2(\mu)\to L_{\Bbb C}^2(\lambda)$
defined by setting $T(f^{\ssbullet})=\varhatf\vthsp^{\ssbullet}$
whenever
$f\in\eusm L_{\Bbb C}^1(\mu)\cap\eusm L_{\Bbb C}^2(\mu)$.

\proof{{\bf (a)} Since $\varhatf=\varhat{g}$ whenever $f\eae g$, the
formula certainly defines an operator $T$ from
$L_{\Bbb C}^1(\mu)\cap L_{\Bbb C}^2(\mu)$
to $L_{\Bbb C}^0(\lambda)$, and of course it is linear.

If $f\in\eusm L_{\Bbb C}^1(\mu)\cap\eusm L_{\Bbb C}^2(\mu)$,
$h=f*\tilde f$ is continuous and positive definite (445Ld) and integrable,
and $\varhat{h}=|\varhatf|^2$ (445Kc).   Now

$$\eqalignno{\int|\varhatf|^2\,d\lambda
&=\int\varhat{h}\,d\lambda
=h(e)\cr
\noalign{\noindent (445P)}
&=\int f(x)\tilde f(x^{-1})\mu(dx)
=\int|f|^2d\mu.\cr}$$

\allowmorestretch{468}{
\noindent Thus $\|Tu\|_2=\|u\|_2$ whenever
$u\in L_{\Bbb C}^1(\mu)\cap L_{\Bbb C}^2(\mu)$;
since $L_{\Bbb C}^1(\mu)\cap L_{\Bbb C}^2(\mu)$ is $\|\,\,\|_2$-dense in
$L_{\Bbb C}^2(\mu)$
(244Ha/244Pb, or otherwise), we have a unique isometry
$T:L_{\Bbb C}^2(\mu)\to L_{\Bbb C}^2(\lambda)$ agreeing with the given
formula on $L_{\Bbb C}^1(\mu)\cap L_{\Bbb C}^2(\mu)$ (2A4I).
}%end of allowmorestretch

\medskip

{\bf (b)} The meat of the theorem is of course the proof that $T$ is
surjective.   \Prf\Quer\ Suppose, if possible, that
$W=T[L_{\Bbb C}^2(\mu)]$ is not equal to $L_{\Bbb C}^2(\lambda)$.
Because $T$ is linear, $W$ is a linear
subspace of $L_{\Bbb C}^2(\lambda)$;  because $T$ is an isometry, $W$ is
isometric to $L_{\Bbb C}^2(\mu)$, and in particular is complete,
therefore closed
in $L_{\Bbb C}^2(\lambda)$ (3A4Fd).   There is therefore a non-zero
continuous
linear functional on $L_{\Bbb C}^2(\lambda)$ which is zero on $W$
(3A5Ad), and this is of the form $u\mapsto\int u\times v$ for some
$v\in L_{\Bbb C}^2(\lambda)$ (244J/244Pc).   What this means is that
there is
a $g\in\eusm L_{\Bbb C}^2(\lambda)$ such that $g^{\ssbullet}\ne 0$ in
$L_{\Bbb C}^2(\lambda)$ but $\int\varhatf\times g\,d\lambda=0$ for every
$f\in\eusm L_{\Bbb C}^1(\mu)\cap\eusm L_{\Bbb C}^2(\mu)$.

Suppose that $f\in\eusm L_{\Bbb C}^1(\mu)$ and that $h$ is any
$\mu$-integrable
continuous positive definite function on $X$.   Then $h$ is bounded
(445Lb), so $|h|^2$ also is $\mu$-integrable, and
$f*\bar h\in\eusm L_{\Bbb C}^1(\mu)\cap\eusm L_{\Bbb C}^2(\mu)$ (444Ra).
Accordingly

\Centerline{$\int g\times\varhatf\times\varhat{\bar h}\,d\lambda
=\int g\times(f*\bar h)\varsphat d\lambda=0$.}

\noindent Thus $\int g\times\varhatf\,d\nu_h=0$, where
$\nu_h=\varhat{\bar h}\lambda$ is the
Radon measure on $\Cal X$ corresponding to $h$ constructed in 445N (see
445Q).   And this is true for every $f\in\eusm L_{\Bbb C}^1(\mu)$.   But
as $\{\varhatf:f\in\eusm L_{\Bbb C}^1(\mu)\}$ is
$\|\,\,\|_{\infty}$-dense in
$C_0(\Cal X;\Bbb C)$ (445Kd), and $g$ is $\nu_h$-integrable (because
$\varhat{\bar h}\in\eusm L_{\Bbb C}^2(\lambda)$, by (a), so
$\int|g\times\varhat{\bar h}|d\lambda<\infty$), $g$ must be zero
$\nu_h$-a.e., that is,
$g\times\varhat{\bar h}=0\,\,\lambda$-a.e.   Now recall (from part (c)
of the proof of 445P, for instance) that for every compact set
$K\subseteq\Cal X$ there is a $\mu$-integrable continuous positive
definite $h$ such that $\varhat{\bar h}(\chi)\ne 0$ for every
$\chi\in K$, so $g=0$ a.e.\ on $K$.   Since $\lambda$ is tight,
$g=0$ a.e.\ (412Jc), which is impossible.\ \Bang\Qed

Thus $T$ is surjective and we have the result.
}%end of proof of 445R

\leader{445S}{}\cmmnt{ While we do not have a direct definition of
$\varhatf$ for $f\in\eusm L_{\Bbb C}^2\setminus\eusm L_{\Bbb C}^1$, the
map $T:L_{\Bbb C}^2(\mu)\to L_{\Bbb C}^2(\lambda)$ does correspond to
the map $f\mapsto\varhatf$ in many
ways.   In particular, we have the following useful properties.

\medskip

\noindent}{\bf Proposition}   Let $X$ be an abelian topological group
with a Haar measure $\mu$, $\Cal X$ its dual group, $\lambda$ the
associated Haar measure on $\Cal X$ and
$T:L_{\Bbb C}^2(\mu)\to L_{\Bbb C}^2(\lambda)$ the
standard isometry\cmmnt{ described in 445R}.   Suppose that $f_0$,
$f_1\in\eusm L_{\Bbb C}^2(\mu)$ and $g_0$,
$g_1\in\eusm L_{\Bbb C}^2(\nu)$
are such that $Tf_0^{\ssbullet}=g_0^{\ssbullet}$ and
$Tf_1^{\ssbullet}=g_1^{\ssbullet}$, and take any $\theta\in\Cal X$.
Then

(a) setting $f_2=\bar f_0$, $g_2(\chi)=\overline{g_0(\chi^{-1})}$
whenever this is defined, $Tf_2^{\ssbullet}=g_2^{\ssbullet}$;

(b) setting $f_3=f_1\times\theta$, $g_3(\chi)=g_1(\theta\chi)$ whenever
this is defined, $Tf_3^{\ssbullet}=g_3^{\ssbullet}$;

(c) setting $f_4=f_0\times f_1\in\eusm L_{\Bbb C}^1(\mu)$,
$\varhatf_4(\theta)=(g_0*g_1)(\theta)$.

\proof{{\bf (a)} We have isometries $R_1:L_{\Bbb C}^2(\mu)\to L_{\Bbb
C}^2(\mu)$ and
$R_2:L_{\Bbb C}^2(\lambda)\to L_{\Bbb C}^2(\lambda)$ defined by setting
$R_1f^{\ssbullet}=(\bar f)^{\ssbullet}$ for every $f\in\eusm L_{\Bbb
C}^2(\mu)$
and $R_2g^{\ssbullet}=(\tilde g)^{\ssbullet}$ for every $g\in\eusm
L_{\Bbb C}^2(\lambda)$, where $\tilde g(\chi)=\overline{g(\chi^{-1})}$
whenever
this is defined.   Now if $f\in\eusm L_{\Bbb C}^1(\mu)$, then

\Centerline{$\varhatbarf(\chi)
=\int\overline{f(x)}\chi(x)\mu(dx)
=\overline{\intop f(x)\chi^{-1}(x)\mu(dx)}
=\tilde{\varhatf}(\chi)$}

\noindent for every $\chi\in\Cal X$.   So $TR_1u=R_2Tu$ for every
$u\in L_{\Bbb C}^1(\mu)\cap L_{\Bbb C}^2(\mu)$;  as
$L_{\Bbb C}^1\cap L_{\Bbb C}^2$ is dense in $L_{\Bbb C}^2$,
$TR_1=R_2T$, which is what we need to know.

\medskip

{\bf (b)} This time, set $R_1f^{\ssbullet}=(f\times\theta)^{\ssbullet}$
for every $f\in\eusm L_{\Bbb C}^2(\mu)$ and
$R_2g^{\ssbullet}=(\theta^{-1}\action_lg)^{\ssbullet}$ for
$g\in\eusm L_{\Bbb C}^2(\lambda)$, where
$(\theta\action_lg)(\chi)=g(\theta^{-1}\chi)$ whenever this
is defined.   Once again, $R_1:L_{\Bbb C}^2(\mu)\to L_{\Bbb C}^2(\mu)$
and $R_2:L_{\Bbb C}^2(\lambda)\to L_{\Bbb C}^2(\lambda)$ are isometries.
If $f\in\eusm L_{\Bbb C}^1(\mu)$, then

\Centerline{$(f\times\theta)\varsphat(\chi)
=\int f(x)\theta(x)\chi(x)\mu(dx)
=\varhatf(\theta\chi)$}

\noindent for every $\chi$, so $TR_1f^{\ssbullet}=R_2Tf^{\ssbullet}$;
once again, this is enough to prove that $TR_1=R_2T$, as required.

\medskip

{\bf (c)} We have

$$\eqalignno{(g_0*g_1)(\theta)
&=\int g_0(\chi^{-1})g_1(\theta\chi)\lambda(d\chi)
=\int\overline{g_2(\chi)}g_3(\chi)\lambda(d\chi)
=(g_3^{\ssbullet}|g_2^{\ssbullet})\cr
\noalign{\noindent (where $(\,\,|\,\,)$ is the standard inner product of
$L_{\Bbb C}^2(\lambda)$)}
&=(Tf_3^{\ssbullet}|Tf_2^{\ssbullet})\cr
\noalign{\noindent (using (a) and (b))}
&=(f_3^{\ssbullet}|f_2^{\ssbullet})\cr
\noalign{\noindent (because linear isometries of Hilbert space preserve
inner products, see 4A4Jc)}
&=\int f_3(x)\overline{f_2(x)}\mu(dx)
=\int f_1(x)\theta(x)f_0(x)\mu(dx)
=(f_0\times f_1)\varsphat(\theta).\cr}$$

}%end of proof of 445S

\leader{445T}{Corollary} Let $X$ be an abelian topological group with a
Haar measure
$\mu$, and $\lambda$ the corresponding Haar measure on the dual group
$\Cal X$ of $X$\cmmnt{ (445P)}.   Then for any non-empty open set
$H\subseteq\Cal X$, there is an $f\in\eusm L_{\Bbb C}^1(\mu)$ such that
$\varhatf\ne 0$ and $\varhatf(\chi)=0$ for $\chi\in\Cal X\setminus H$.

\proof{ Let $V_1$ and $V_2$ be non-empty open sets of finite measure
such that $V_1V_2\subseteq H$, and let $g_1$, $g_2$ be their
indicator functions.   Then there are $f_1$, $f_2\in\eusm L_{\Bbb
C}^2(\mu)$
such that $Tf_j^{\ssbullet}=g_j^{\ssbullet}$ for both $j$, where
$T:L_{\Bbb C}^2(\mu)\to L_{\Bbb C}^2(\lambda)$ is the isometry of 445R.
In this case, by
445Sc, $(f_1\times f_2)\varsphat=g_1*g_2$.   But it is easy to check
that $g_1*g_2$ is non-zero and zero outside $H$.
}%end of proof of 445T

\leader{445U}{The Duality Theorem}\cmmnt{ ({\smc Pontryagin 1934},
{\smc Kampen 1935})} Let $X$ be a locally compact Hausdorff abelian
topological group.   Then the canonical map $x\mapsto\hat x$ from $X$ to
its bidual $\frak X$\cmmnt{ (445E)} is an isomorphism between $X$ and
$\frak X$ as topological groups.

\proof{ By 445O, \hbox{$\sphat\,$} is a homeomorphism between $X$ and its
image $\hat X\subseteq\frak X$.   Accordingly $\hat X$ is itself, with
its subspace topology, a locally compact topological group, and is
closed in $\frak X$ (4A5Mc).

\Quer\ Suppose, if possible, that $\hat X\ne\frak X$.   Let $\mu$ be a
Haar measure on $X$ (441E) and $\lambda$ the associated Haar measure on
the dual $\Cal X$ of $X$ (445P).   By 445T, there is a
$g\in\eusm L_{\Bbb C}^1(\lambda)$ such that $\varhat{g}$ is zero on
$\hat X$ but not zero
everywhere, so that $g$ is not zero a.e.   We may suppose that $g$ is
defined everywhere on $\Cal X$.   In this case we have

\Centerline{$0=\varhat{g}(\hat x)
=\int g(\chi)\hat x(\chi)\lambda(d\chi)
=\int g(\chi)\chi(x)\lambda(d\chi)$}

\noindent for every $x\in X$.

By 418J, $g:\Cal X\to\Bbb C$ is almost continuous.   Let
$K\subseteq\{\chi:\chi\in\Cal X$, $g(\chi)\ne 0\}$ be a compact set such
that $\int_K|g|d\lambda\ge\bover34\int_{\Cal X}|g|d\lambda$ and
$g\restr K$ is continuous.   Set $q(\chi)=\overline{g(\chi)}/|g(\chi)|$
for $\chi\in K$.   Now consider the linear span $A$ of
$\{\hat x:x\in X\}$ as a linear space of complex-valued functions on
$\Cal X$.   Since
$\widehat{xy}=\hat x\times\hat y$ for all $x$, $y\in X$, $A$ is a
subalgebra
of $C_b=C_b(\Cal X;\Bbb C)$;  since $\widehat{x^{-1}}=\bar{\hat x}$ for
every $x\in X$, $\bar h\in A$ for every $h\in A$;  the constant function
$\hat e$ belongs to $A$;  and $A$ separates the points of
$\Cal X$.   By the Stone-Weierstrass theorem, in the form 281G, there is
an $h\in A$ such that $|h(\chi)-q(\chi)|\le\bover12$ for every
$\chi\in K$ and $|h(\chi)|\le 1$ for every $\chi\in\Cal X$.   Of course
$\int g\times h\,d\lambda=0$ for every $h\in A$ because
$\int g\times\hat x\,d\lambda=0$ for every $x\in X$.


Now, however,

$$\eqalign{\int_K|g|d\lambda
&=\int_Kg\times q\,d\lambda
\le\bigl|\int_Kg\times h\,d\lambda\bigr|+\int_K|g|\times|h-q|d\lambda\cr
&\le\bigl|\int_{\Cal X\setminus K}g\times h\,d\lambda\bigr|
  +\Bover12\int_K|g|d\lambda
\le\int_{\Cal X\setminus K}|g|d\lambda+\Bover12\int_K|g|d\lambda
<\int_K|g|d\lambda,\cr}$$

\noindent which is impossible.\ \Bang

Thus $\hat X=\frak X$ and the proof is complete.
}%end of proof of 445U

\exercises{\leader{445X}{Basic exercises (a)}
%\spheader 445Xa
Consider the additive group $\Bbb Q$ with its usual topology.   Show
that its dual group can be identified with the additive group $\Bbb R$
with its usual topology.
%445B

\spheader 445Xb Let $X$ be any topological group, and $\Cal X$ its dual
group.   Show that if $\nu$ is a totally finite Radon measure on $X$,
then its Fourier-Stieltjes
transform $\varhat{\nu}:\Cal X\to\Bbb C$ is continuous.
%445E

\spheader 445Xc Let $X$ be a topological group carrying Haar measures
and $\Cal X$ its dual group.   For a totally finite quasi-Radon measure
$\nu$ on $\Cal X$ set $\varhat{\nu}(x)=\int\chi(x)\nu(d\chi)$ for every
$x\in X$.   (i) Show that $\varhat{\nu}:X\to\Bbb C$ is continuous.
(ii) Show that
for any totally finite quasi-Radon measure $\mu$ on $X$ with
Fourier-Stieltjes
transform $\hat\mu$, $\int\varhat{\mu}\,d\nu=\int\varhat{\nu}\,d\mu$.
%445E

\sqheader 445Xd Let $X$ be a group and $h_1$, $h_2:X\to\Bbb C$ positive
definite functions.   Show that $h_1+h_2$, $\alpha h_1$ and $\bar h_1$
are also positive definite for any $\alpha\ge 0$.
%445L

\spheader 445Xe(i) Let $X$ be a group, $Y$ a subgroup of $X$ and
$h:Y\to\Bbb C$ a positive definite function.   Set $h_1(x)=h(x)$ if
$x\in Y$, $h_1(x)=0$ if $x\in X\setminus Y$.   Show that $h_1$ is
positive definite.   (ii) Let $X$ and $Y$ be groups, $\phi:X\to Y$ a
group homomorphism and $h:Y\to\Bbb C$ a positive definite function.
Show that $h\phi:X\to\Bbb C$ is positive definite.

\spheader 445Xf Let $X$ be a topological group and $\Cal X$ its dual
group.   (i) Let $\nu$ be any totally finite topological measure on
$\Cal X$ and set $h(x)=\int\chi(x)\nu(d\chi)$ for $x\in X$.   Show that
$h:X\to\Bbb C$ is positive definite.   (ii) Let $\nu$ be any totally
finite topological measure on $X$.   Show that its Fourier transform
$\varhat{\nu}:\Cal X\to\Bbb C$ is positive definite.
%445L

\sqheader 445Xg Let $X$ be a topological group with a left Haar measure,
and $h:X\to\Bbb C$ a bounded continuous function.   Show that $h$ is
positive definite iff $\int h(x^{-1}y)f(y)\overline{f(x)}dxdy\ge 0$ for
every integrable function $f$.
%445M

\sqheader 445Xh Let $\phi:\BbbR^r\to\Bbb C$ be a function.   Show that
it is the characteristic function of a probability distribution on
$\Bbb R^r$ iff it is continuous and positive definite and $\phi(0)=1$.
%445N

\sqheader 445Xi Let $X$ be a compact Hausdorff abelian topological
group, and $\mu$ the Haar probability measure on $X$.   Show that the
corresponding Haar measure on the dual group $\Cal X$ of $X$ is just
counting measure on $\Cal X$.
%445P

\sqheader 445Xj Let $X$ be an abelian group with its discrete topology,
and $\mu$ counting measure on $X$.   Let $\Cal X$ be the dual group and
$\lambda$ the corresponding Haar measure on $\Cal X$.   Show that
$\lambda\Cal X=1$.
%445P

\spheader 445Xk Let $X$ be the topological group $\Bbb R$, and
$\mu=\Bover1{\sqrt{2\pi}}\mu_L$, where $\mu_L$ is Lebesgue measure.
(i) Show that if we identify the dual group $\Cal X$ of $X$ with
$\Bbb R$, writing $\chi(x)=e^{-i\chi x}$ for $x$, $\chi\in\Bbb R$, then the
Haar measure on $\Cal X$ corresponding to the Haar measure $\mu$ on $X$
is $\mu$ itself.   (ii) Show that if we change the action of $\Bbb R$ on
itself by setting $\chi(x)=e^{-2\pi i\chi x}$, then the Haar measure on
$\Cal X$ corresponding to $\mu_L$ is $\mu_L$.
%445P

\spheader 445Xl Let $X_0,\ldots,X_n$ be abelian topological groups with
Haar measures $\mu_0,\ldots,\mu_n$, and let $X=X_0\times\ldots\times
X_n$ be the product group with its Haar measure
$\mu=\mu_0\times\ldots\times\mu_n$.   For each $k\le n$ let $\Cal X_k$
be the dual group of $X_k$ and $\lambda_k$ the Haar measure on $\Cal
X_k$ corresponding to $\mu_k$.   Show that if we identify
$\Cal X=\Cal X_0\times\ldots\times\Cal X_n$ with the dual group of $X$,
then the Haar measure on $\Cal X$ corresponding to $\mu$ is just the
product measure $\lambda_0\times\ldots\times\lambda_n$.
%445P

\sqheader 445Xm Let $X$ be a compact Hausdorff abelian topological
group, with dual group $\Cal X$, and $\mu$ the Haar probability measure
on $X$.   (i) Show that $\int\chi\,d\mu=0$ for every $\chi\in\Cal X$
except the identity.   (ii) Show that $\{\chi^{\ssbullet}:\chi\in\Cal
X\}$ is an orthonormal basis of the Hilbert space $L_{\Bbb C}^2(\mu)$.
\Hint{$(u|\chi^{\ssbullet})=(Tu|\varhat{\chi}\vthsp^{\ssbullet})$
where $T$ is the operator of 445R.}
%445R

\spheader 445Xn Let $X$ be an abelian topological group with a Haar
measure $\mu$, $\lambda$ the associated Haar measure on the dual group
$\Cal X$ and $T:L_{\Bbb C}^2(\mu)\to L_{\Bbb C}^2(\lambda)$ the standard
isometry.
Suppose that $u$, $v\in L_{\Bbb C}^2(\mu)$.   Suppose that $f_0$,
$f_1\in\eusm L_{\Bbb C}^2(\mu)$, $g_0$, $g_1\in\eusm L_{\Bbb C}^2(\nu)$
are such that
$Tf_0^{\ssbullet}=g_0^{\ssbullet}$ and
$Tf_1^{\ssbullet}=g_1^{\ssbullet}$.   Show that
$(f_0*f_1)(x)=\int g_0(\chi)g_1(\chi)\overline{\chi(x)}\lambda(d\chi)$
for any $x\in X$.
%445S

\spheader 445Xo Let $X$ be a locally compact Hausdorff abelian
topological group and $\Cal X$ its dual group.   Show that a function
$h:\Cal X\to\Bbb C$ is the Fourier-Stieltjes transform of a totally
finite Radon measure on $X$ iff it is continuous and positive definite.
%445U, 445N

\def\pltwadic{+_{\text{2adic}}}

\spheader 445Xp(i) Show that we can define a binary operation
$\pltwadic$ on $X=\{0,1\}^{\Bbb N}$ by setting $x\pltwadic y=z$ whenever
$x$, $y$, $z\in X$ and $\sum_{i=0}^k2^i(x(i)+y(i)-z(i))$ is divisible by
$2^{k+1}$ for every $k$.   (ii) Show that if we give $X$ its usual
topology then $(X,\pltwadic)$ is an abelian topological group.  (iii)
Show that the usual measure on $X$ is the Haar probability measure for
this group operation.   (iv) Show that
$G=\{\zeta:\zeta\in\Bbb C,\,\Exists n\in\Bbb N,\,\zeta^{2^n}=1\}$ is a
subgroup of $\Bbb C$.   (v) Show that the dual of $(X,\pltwadic)$ is
$\{\chi_{\zeta}:\zeta\in G\}$ where
$\chi_{\zeta}(x)=\prod_{i=0}^{\infty}\zeta^{2^ix(i)}$ for $\zeta\in G$
and $x\in X$.   (vi) Show that the functions $f$, $g:X\to X$ described
in 388E are of the form $x\mapsto x\pm_{\text{2adic}}x_0$ for a certain
$x_0\in X$.
% +

\spheader 445Xq Let $X$ be a locally compact Hausdorff abelian topological
group.   Show that if two totally finite Radon measures on
$X$ have the same Fourier-Stieltjes transform, they are equal.
\Hint{281G.}
%445O out of order query

\leader{445Y}{Further exercises (a)}
%\spheader 445Ya
Let $X$ be any Hausdorff topological group.   Let $\widehat{X}$ be its
completion under its bilateral uniformity.   Show that the dual
groups of $X$ and $\widehat{X}$ can be identified as groups.   Show that
they can be identified as topological groups if {\it either} $X$ is
metrizable {\it or} $X$ has a totally bounded neighbourhood of the
identity.
%445B

\spheader 445Yb Let $\family{j}{I}{X_j}$ be a countable family of
topological groups, with product $X$;  let $\Cal X_j$ be the dual group
of each $X_j$, and $\Cal X$ the dual group of $X$.   Show that the
topology of $\Cal X$ is generated by sets of the form $\Cal
X\cap\prod_{j\in I}H_j$
where $H_j\subseteq\Cal X_j$ is open for each $j$.
%445B

\spheader 445Yc Let $X$ be a real linear topological space, with
addition as its group operation.   Show that its dual group is just the
set of functionals $x\mapsto e^{if(x)}$ where $f:X\to\Bbb R$ is a
continuous linear functional.   Hence show that there are abelian groups
with trivial duals.
%445B

\spheader 445Yd Let $X$ be the group of rotations of $\BbbR^3$, that
is, the group of orthogonal real $3\times 3$ matrices with determinant
$1$, and give $X$ its usual topology, corresponding to its embedding as
a subspace of $\BbbR^9$.   Show that the only character on $X$ is the
constant function $1$.   \Hint{(i) show that two rotations through the
same angle
are conjugate in $X$;  (ii) show that if $0<\theta\le\bover{\pi}2$ then
the product of two rotations through the angle $\theta$ about
orthogonal axes is not a rotation through an angle $2\theta$.}
%445B

\spheader 445Ye Let $X$ be a finite abelian group, endowed with its
discrete topology.   Show that its dual is isomorphic to $X$.
\Hint{$X$ is a product of cyclic groups.}
%445B

\spheader 445Yf Show that if $I$ is any uncountable set, then there is a
quasi-Radon probability measure $\nu$ on the topological group
$\Bbb R^I$ such that its Fourier-Stieltjes transform
$\varhat{\nu}$ is not continuous.   \Hint{take $\nu$ to be a power of a
suitably widely spread probability distribution on $\Bbb R$.}
%445E

\spheader 445Yg Let $X$ be an abelian topological group with a Haar
measure $\mu$, and $\lambda$ the associated Haar measure on the dual
group $\Cal X$ of $X$.   Let
$T:L_{\Bbb C}^2(\mu)\to L_{\Bbb C}^2(\lambda)$ be the
standard isomorphism.   Suppose that $f\in\eusm L_{\Bbb C}^2(\mu)$ and
$g\in\eusm L_{\Bbb C}^1(\lambda)\cap\eusm L_{\Bbb C}^2(\lambda)$ are such
that $Tf^{\ssbullet}=g^{\ssbullet}$.   Show that
$f(x)=\int g(\chi)\overline{\chi(x)}\lambda(d\chi)$ for almost every
$x\in X$.   \Hint{look first at locally compact Hausdorff $X$.}
%445S

\spheader 445Yh Let $X$ be a locally compact Hausdorff abelian
topological group
with dual group $\Cal X$, $P_{\text{R}}$ the set of Radon probability
measures on $X$, $\sequencen{\nu_n}$ a sequence in $P_{\text{R}}$ and
$\nu$ a member of $P_{\text{R}}$.
Show that the following are equiveridical:  (i)
$\sequencen{\nu_n}\to\nu$ for the narrow topology on $P_{\text{R}}$;
(ii) $\lim_{n\to\infty}\varhat{\nu}_n(\chi)=\varhat{\nu}(\chi)$ for every
$\chi\in\Cal X$.   \Hint{compare 285L.   For the critical step, showing
that $\{\nu_n:n\in\Bbb N\}$ is uniformly tight, use the formulae in
445N to show that there is an integrable $f:\Cal X\to\Bbb C$ such that
$0\le\varcheckf\le\tbf{1}$ and
$\int\varcheckf(x)\nu(dx)\ge 1-\bover12\epsilon$.}
%445U

\spheader 445Yi Let $X$ be a topological group carrying Haar measures
and $\Cal X$ its dual group.   Let $M_{\tau}$ be the complex Banach
space of signed totally finite $\tau$-additive Borel measures on
$\Cal X$ (put the ideas of 437F and 437Yb together).   Show that $X$
separates
the points of $M_{\tau}$ in the sense that if $\nu\in M_{\tau}$ is
non-zero, there is an $x\in X$ such that $\int\chi(x)\nu(d\chi)\ne 0$,
if the integral is appropriately interpreted.
\Hint{use the method in the proof of 445U.}
%445U

\spheader 445Yj Let $X$ be an abelian topological group carrying Haar
measures.   Let $M_{\tau}$ be the complex Banach space of signed totally
finite $\tau$-additive Borel measures on $X$.   Show that the dual
$\Cal X$ of $X$ separates the points of $M_{\tau}$ in the sense that if
$\nu\in M_{\tau}$ is non-zero, there is a $\chi\in\Cal X$ such that
$\int\chi(x)\nu(dx)\ne 0$.   \Hint{use 443L to reduce to the case in
which $X$ is locally compact and Hausdorff;  now use 445U and 445Yi.}
%445U, 445Yi

\spheader 445Yk Let $X$ be an abelian topological group and $\mu$ a Haar
measure on $X$.   Show that the spectral radius of any non-zero element
of $L_{\Bbb C}^1(\mu)$ is non-zero.   \Hint{445Yj, 445Kd.}
%445U, 445Yj

\spheader 445Yl Show that for any integer $p\ge 2$ there is an operation
$+_{p\text{adic}}$ on $\{0,\ldots,p-1\}^{\Bbb N}$ with properties
similar to those of the operation $\pltwadic$ of 445Xp.
%445Xp

\spheader 445Ym Let $\mu$ be Lebesgue measure on $\coint{0,\infty}$.
(i) For $f$, $g\in\eusm L^1(\mu)$ define
$(f*g)(x)=\int_0^xf(y)g(x-y)\mu(dy)$
whenever the integral is defined.   Show that $f*g\in\eusm L^1(\mu)$.
(ii) Show that we can define a bilinear operator $*$ on $L^1(\mu)$ by
setting $f^{\ssbullet}*g^{\ssbullet}=(f*g)^{\ssbullet}$ for $f$,
$g\in\eusm L^1(\mu)$, and that under this multiplication $L^1(\mu)$ is a
Banach algebra.   (iii) Show that if $\phi:L^1(\mu)\to\Bbb R$ is a
multiplicative linear operator then there is some $s\ge 0$ such that
$\phi(f^{\ssbullet})=\int_0^{\infty}f(x)e^{-sx}\mu(dx)$ for every
$f\in\eusm L^1(\mu)$.
%445H out of order
}%end of exercises

\endnotes{
\Notesheader{445} I repeat that this section is intended to be a more or
less direct attack on the duality theorem.   At every point the clause
`let $X$ be a locally compact Hausdorff abelian topological group' is
present in
spirit.   The actual statement of each theorem involves some subset of
these properties, purely in accordance with the principle of omission
of irrelevant hypotheses, not because I expect to employ the results in
any more general setting.

In 445Ab I describe a topology on the dual group in a context so
abstract that we have rather a lot of choice.   For groups carrying Haar
measures, the alternative descriptions of the topology on the dual
(445I) make it plain that this must be the first topology to study.   By
445E it is already becoming fairly convincing.
But it is not clear that there is any such pre-eminent topology in the
general case.

Fourier-Stieltjes transforms hardly enter into the arguments of this
section;  I mention them mostly because they form the obvious
generalization of the ideas in \S285.   But I note that the principal
theorem of \S285 (that sequential convergence of characteristic
functions corresponds to sequential convergence of distributions, 285L)
generalizes directly to the context here (445Yh).

I have tried to lay this treatise out in such a way that we periodically
return to themes from past chapters at a higher level of sophistication.
There seem to be four really important differences between this section
and the previous treatment in Chapter 28.   (i) The first is the obvious
one;  we are dealing with general locally compact Hausdorff abelian
groups, rather than with $\Bbb R$ and $S^1$ and $\Bbb Z$.   Of course
this puts much heavier demands on our technique, and, to begin with,
leaves our imaginations unfocused.   (ii) The second concerns
differentiation, or rather its absence;  since we no longer have any
differential structure on our groups, a substantial part of the theory
evaporates, and we are forced to employ new tactics in the rest.   (iii)
The third concerns the normalization of the measure on the dual group.
As soon as we know that $\Cal X$ is a locally compact group (445J) we
know that it carries Haar measures.   The problem is to describe the
particular one we need in appropriate terms.   In the case of the dual
pairs $(\Bbb R,\Bbb R)$ or $(S^1,\Bbb Z)$, we have measures already
presented (counting measure on $\Bbb Z$, Lebesgue measure on
$\ocint{-\pi,\pi}$ and $\Bbb R$).   (They are not in fact dual in the
sense we need here, at least not if we use the simplest formulae for the
duality, and have to be corrected in each case by a factor of $2\pi$.
See 445Xk.)   But since we do have dual pairs already to hand, we can
simultaneously develop theories of Fourier transforms and inverse
Fourier transforms (for the pair $(S^1,\Bbb Z)$ the inverse Fourier
transform is just summation of trigonometric series), and the problem is
to successfully match operations which have independent existences.
(iv) The final change concerns an interesting feature of $\Bbb Z$ and
$\Bbb R$.   Repeatedly, in \S\S282-283, the formulae invoked symmetric
limits $\lim_{n\to\infty}\sum_{k=-n}^n$ or $\lim_{a\to\infty}\int_{-a}^a$
to approach some conditionally convergent sum or integral.   Elsewhere
one sometimes deals with singularities by examining `Cauchy principal
values';  if $\int_{-1}^1f$ is undefined, try
$\lim_{a\downarrow 0}(\int_{-1}^{-a}f+\int_a^1f)$.   This particular
method seems to disappear in the general context.   But the general
challenge of the subject remains the same:  to develop a theory of the
transform $u\mapsto\varhat{u}$ which will apply to the largest possible
family of objects $u$ and will enable us to justify, in the widest
possible contexts, the manipulations listed in the notes to \S284.   The
calculations in 445S and 445Xn, treating `shift' and `convolution' in
$L^2$, are typical.

In terms of the actual proofs of the results here, `test functions'
(284A) have gone, and in their place we take a lot more trouble over the
Banach algebra $L^1$.   This algebra is the key to one of the magic
bits, which turns up in rather undignified corners in 445Kd and part (e)
of the proof of 445N.   Down to 445O, the dominating problem is that we
do not know that the dual group $\Cal X$ of a group $X$ is large enough
to tell us anything interesting about $X$.   (After that, the problem
reverses;  we have to show that $\Cal X$ is not too big.)   We find that
(under rather specially arranged circumstances) we are able to say
something useful about the spectral radius of a member of $L^1$, and we
use this to guarantee that it has a non-trivial Fourier transform.   If
we identify $\Cal X$ with the maximal ideal space of $L^1$ (445H), then
the Fourier transform on $L^1$ becomes the `Gelfand map', a general
construction of great power in the theory of commutative Banach
algebras.

There is one similarity between the methods of this section and those of
\S284.   In both cases we have isomorphisms between $L_{\Bbb C}^2(\mu)$
and $L_{\Bbb C}^2(\lambda)$ (the Plancherel Theorem), but cannot define
the Fourier
transform of a function in $\eusm L_{\Bbb C}^2$ in any direct way;
indeed, while
the Fourier transform of a function in $\eusm L_{\Bbb C}^1$, or even of
a (totally finite) measure, can really be thought of as a (continuous)
function, the transform of a function in $L^2$ is at best a member of
$L^2$, not a function at all.   We manoeuvre around this difficulty by
establishing that our (genuine) Fourier transforms match dense subspaces
isometrically.   In \S284 I used test functions, and in the present
section I use $\eusm L^1\cap\eusm L^2$.   Test functions are easier
partly because the Fourier transform of a test function is again a test
function, and all the formulae we need are easy to establish for such
functions.

Searching for classes of functions which will be readily manageable in
general locally compact abelian groups, we come to the `positive
definite' functions.   The phrase is unsettling, since the functions
themselves are in no obvious sense positive (nor even, as a rule,
real-valued).   Also their natural analogues in the theory of bilinear
forms are commonly called `positive semi-definite'.   However, their
Fourier transforms, whether regarded as
measures (445N) or as functions (445Q), are positive, and, as a bonus,
we get a characterization of the Fourier transforms of measures (445Xf,
445Xh), answering a question left hanging in 285Xu.
}%end of notes

\discrpage

\frfilename{mt446.tex}
\versiondate{8.10.13}
\copyrightdate{1998}

\def\chaptername{Topological groups}
\def\sectionname{The structure of locally compact groups}

\def\ker{\mathop{\text{ker}}}

\newsection{446}

I develop those fragments of the structure theory of locally compact
Hausdorff topological groups which are needed for the main theorem of
the next section.   Theorem 446B here is of independent interest, being
both itself important and with a proof which uses the measure theory
of this chapter in an interesting way;  but the rest of the section,
from 446D on, is starred.   Note that in this section, unlike the last,
groups are not expected to be abelian.

\leader{446A}{Finite-dimensional representations (a) Definitions (i)}
For any $r\in\Bbb N$, write $M_r=M_r(\Bbb R)$ for the space of
$r\times r$ real matrices.   If we identify it with the space
$\eurm B(\BbbR^r;\BbbR^r)$, where $\BbbR^r$ is given its Euclidean norm,
then $M_r$ becomes a unital Banach algebra\cmmnt{ (4A6C)}, with
identity $I$, the $r\times r$ identity matrix.   Write $GL(r,\Bbb R)$
for the group of invertible elements of $M_r$.

\medskip

\quad{\bf (ii)} Let $X$ be a topological group.   A
{\bf finite-dimensional representation} of $X$ is a continuous
homomorphism from $X$ to a group of the form $GL(r,\Bbb R)$ for some
$r\in\Bbb N$.   If the homomorphism is injective the representation is
called {\bf faithful}\cmmnt{ (cf.\ 4A5Be)}.

\medskip

{\bf (b)} Observe that if $X$ is any topological group and $\phi$ is a
finite-dimensional representation with kernel $Y$, then $X/Y$ has a
faithful finite-dimensional representation $\psi$ defined by writing
$\psi(x^{\ssbullet})=\phi(x)$ for every $x\in X$\cmmnt{ (4A5La)}.

\leader{446B}{Theorem} Let $X$ be a compact Hausdorff topological group.
Then for any $a\in X$, other than the identity, there is a
finite-dimensional representation $\phi$ of $X$ such that
$\phi(a)\ne I$;  and we can arrange that $\phi(x)$ is an orthogonal matrix
for every $x\in X$.

\proof{{\bf (a)} Let $U$ be a symmetric neighbourhood of the identity
$e$ in $X$ such that $a\notin UU$.   Because $X$ is completely regular
(3A3Bb), there is a non-zero continuous function
$h:X\to\coint{0,\infty}$ such that $h(x)=0$ for every
$x\in X\setminus U$;  replacing $h$ by $x\mapsto h(x)+h(x^{-1})$ if
necessary, we may suppose that
$h(x)=h(x^{-1})$ for every $x$.   Let $\mu$ be a (left) Haar measure on
$X$ (441E), and set $w=h^{\ssbullet}$ in $L^0(\mu)$.

\medskip

{\bf (b)} Define an operator $T$ from $L^2=L^2(\mu)$ to itself by
setting

\Centerline{$Tu=u*w$ for every $u\in L^2$,}

\noindent where $*$ is convolution;  that is,
$Tf^{\ssbullet}=(f*h)^{\ssbullet}$ for every
$f\in\eusm L^2=\eusm L^2(\mu)$.  Then $T$ is a compact self-adjoint
operator on the real Hilbert space $L^2$ (444V).

\medskip

{\bf (c)} For any $z\in X$, define $S_z:L^2\to L^2$ by setting
$S_zu=z\action_lu$ for $u\in L^2$,
where $\action_l$ is the left shift action, so that $S_z$ is a
norm-preserving linear operator (443Ge).
Also $S_z$ commutes with $T$.   \Prf\  (443Ge).  ThenIf $f\in\eusm L^2$, then

$$\eqalignno{S_zTf^{\ssbullet}
&=(z\action_l(f*h))^{\ssbullet}
=((z\action_lf)*h)^{\ssbullet}\cr
\noalign{\noindent (444Of)}
&=TS_zf^{\ssbullet}.\text{ \Qed}\cr}$$

\medskip

{\bf (d)} Now $S_aTw\ne Tw$.   \Prf\ Set $g=h*h$, so that
$Tw=g^{\ssbullet}$ and $g$, $a\action_lg$ are both continuous functions
(444Rc).   Then

\Centerline{$(a\action_lg)(e)=g(a^{-1})
=\biggerint h(y)h(y^{-1}a^{-1})dy=0$}

\noindent because if $y\in U$ then $y^{-1}a^{-1}\notin U$, while

\Centerline{$g(e)=\biggerint h(y)h(y^{-1})dy=\int h(y)^2dy>0$.}

\noindent So the open set $\{x:g(x)\ne (a\action_lg)(x)\}$ is non-empty;
because $\mu$ is strictly positive (442Aa), it is non-negligible, and

\Centerline{$S_aTw=(a\action_lg)^{\ssbullet}\ne g^{\ssbullet}=Tw$.  \Qed}

\medskip

{\bf (e)} The closed linear subspace $\{u:S_au=u\}$ therefore does not
include $T[L^2]$.   But the linear span of
$\{Tv:v$ is an eigenvector of $T\}$ is dense in $T[L^2]$ (4A4M),
so there is an eigenvector
$v^*$ of $T$ such that $S_aTv^*\ne Tv^*$.   Let $\gamma\in\Bbb R$ be
such that $Tv^*=\gamma v^*$;  since $Tv^*\ne 0$, $\gamma\ne 0$, and
$V=\{u:u\in L^2$, $Tu=\gamma u\}$ is finite-dimensional (4A4Lb).

\medskip

{\bf (f)} $S_z[V]\subseteq V$ for every $z\in X$.   \Prf\

\Centerline{$TS_zu=S_zTu=S_z(\gamma u)=\gamma S_zu$}

\noindent for every $u\in V$.\ \Qed

We therefore have a map $z\mapsto S_z\restrp V:X\to\eurm B(V;V)$.
As observed in 443Gc, this is actually a semigroup
homomorphism, and of course $S_e\restrp V$ is the identity of
$\eurm B(V;V)$, so $S_z\restrp V$ is always invertible, and we have a
group homomorphism from $X$ to the group of invertible elements of
$\eurm B(V;V)$.   Taking an orthonormal basis $(v_1,\ldots,v_r)$ of $V$,
we have a homomorphism $\phi$ from $X$ to $GL(r,\Bbb R)$, defined by
setting $\phi(z)=\langle\innerprod{S_zv_i}{v_j}\rangle_{1\le i,j\le r}$ for
every $z\in X$.   Moreover, $\phi$ is continuous.   \Prf\ For any
$u\in L^2$,
$z\mapsto S_zu:X\to L^2$ is continuous, by 443Gf.   But this means that
all the maps $z\mapsto\innerprod{S_zv_i}{v_j}$ are continuous;
since the topology
of $GL(r,\Bbb R)$ can be defined in terms of these functionals (see the
formulae in 262H), $\phi$ is continuous.\ \Qed

Thus $\phi$ is a finite-dimensional representation of $X$.   But $V$ was
chosen to contain $v^*$;  of course $Tv^*\in V$, while
$S_aTv^*\ne Tv^*$;  so that $\phi(a)$ is not the identity.

\medskip

{\bf (g)} Finally, $\phi(z)$ is an orthogonal matrix for every $z\in X$.
\Prf\ As observed in (c), $S_z$ is norm-preserving, so $S_z\restrp V$ is
again norm-preserving.   By 4A4Jb,
$\innerprod{S_zv_i}{S_zv_j}=\innerprod{v_i}{v_j}$ for $1\le i,j\le r$, that
is, $\phi(z)$ is orthogonal.\ \Qed
}%end of proof of 446B

\leader{446C}{Corollary} Let $X$ be a compact Hausdorff topological
group.   Then for any neighbourhood $U$ of the identity of $X$ there is
a finite-dimensional representation of $X$ with kernel included in $U$.

\proof{ Let $\Phi$ be the set of finite-dimensional representations of
$X$.   By 446B, $\bigcap_{\phi\in\Phi}\ker(\phi)=\{e\}$, where $e$ is
the identity of $X$.   Because $X\setminus\interior U$ is compact and
disjoint from $\bigcap_{\phi\in\Phi}\ker(\phi)$ (and $\ker(\phi)$ is
closed for every $\phi$), there must be $\phi_0,\ldots,\phi_n\in\Phi$
such that $\bigcap_{i\le n}\ker(\phi_i)\subseteq U$.   For each $i\le
n$, let $r_i\in\Bbb N$ be the integer such that $\phi_i$ is a continuous
homomorphism from $X$ to $GL(r_i,\Bbb R)$.   Set $r=\sum_{i=0}^nr_i$.
Then we have a map $\phi:X\to GL(r,\Bbb R)$
given by the formula

$$\phi(x)=\Matrix{\phi_0(x)&\tbf{0}&\ldots&\tbf{0}\\
   \tbf{0}&\phi_1(x)&\ldots&\tbf{0}\\
   \ldots&\ldots&\ldots&\ldots\\
   \tbf{0}&\tbf{0}&\ldots&\phi_n(x)}$$

\noindent for every $x\in X$.   It is easy to check that $\phi$ is a
continuous homomorphism, and
$\ker(\phi)=\bigcap_{i\le n}\ker(\phi_i)\subseteq U$.
So we have an appropriate representation of $X$.
}%end of proof of 446C


\leader{*446D}{Notation (a)}\cmmnt{ It will help to be clear on an
elementary point of notation.}   If $X$ is a group and $A$ is a subset
of $X$ I will write
$A^0=\{e\}$\cmmnt{, where $e$ is the identity of $X$,}
%observe that  \emptyset^0=\{e\}  cf  0^0=1
and $A^{n+1}=AA^n$ for $n\in\Bbb N$\cmmnt{,
so that $A^3=\{x_1x_2x_3:x_1,\,x_2,\,x_3\in A\}$, etc}.
\cmmnt{Now we find that $A^{m+n}=A^mA^n$ and $A^{mn}=(A^m)^n$ for all
$m$, $n\in\Bbb N$.   Writing $A^{-1}=\{x^{-1}:x\in A\}$ as usual, we
have $(A^r)^{-1}=(A^{-1})^r$.   But note that if we also continue to
write $A^{-1}=\{x^{-1}:x\in A\}$, then $AA^{-1}$ is not in general equal
to $A^0$;  and that there is no simple relation between $A^r$, $B^r$ and
$(AB)^r$, unless $X$ is abelian.}

\wheader{446D}{0}{0}{0}{72pt}

\spheader 446Db\cmmnt{ In the rest of this section, I shall make
extensive use
of the following device.}   If $X$ is a group with identity $e$, $e\in
A\subseteq X$ and
$n\in\Bbb N$, write $D_n(A)=\{x:x\in X,\,x^i\in A$ for every $i\le n\}$.

\inset{(i) $D_0(A)=X$.}

\inset{(ii) $D_1(A)=A$.}

\inset{(iii) $D_n(A)\subseteq D_m(A)$ whenever $m\le n$.}

\inset{(iv) $D_{mn}(A)\subseteq D_m(D_n(A))$ for all $m$, $n\in\Bbb N$.}

\cmmnt{\noindent \Prf\ If $x\in D_{mn}(A)$ then $(x^i)^j\in A$
whenever $j\le n$ and $i\le m$.\ \Qed}

\inset{(v) If $r\in\Bbb N$ and $A^r\subseteq B$, then $D_n(A)\subseteq
D_{nr}(B)$ for every $n\in\Bbb N$;  in particular, $A\subseteq D_r(B)$.}

\prooflet{\noindent\Prf\ For $r=0$ this is trivial.   Otherwise, take
$x\in D_n(A)$ and $i\le nr$.   Then
we can express $i$ as $i_1+\ldots+i_r$ where $i_j\le n$ for each $j$, so
that

\Centerline{$x^i=\prod_{j=1}^rx^{i_j}\in A^r\subseteq B$.  \Qed}
}%end of prooflet

\inset{(vi) If $A=A^{-1}$ then $D_n(A)=D_n(A)^{-1}$ for every $n\in\Bbb
N$.}

\inset{(vii) If $D_m(A)\subseteq B$ where $m\in\Bbb N$, then
$D_{mn}(A)\subseteq D_n(B)$ for every $n\in\Bbb N$.}

\prooflet{\noindent\Prf\ If $x\in D_{mn}(A)$ and $i\le n$, then
$x^{ij}\in A$ for
every $j\le m$, so $x^i\in D_m(A)\subseteq B$.\ \Qed}

\medskip

\spheader 446Dc In (b), if $X$ is a topological group and $A$ is closed,
then every $D_n(A)$ is closed;  if moreover $A$ is compact, then
$D_n(A)$ is compact for every $n\ge 1$.   If $A$ is a neighbourhood of
$e$, then so is every $D_n(A)$\prooflet{, because the map $x\mapsto x^i$
is continuous for every $i\le n$}.

\leader{*446E}{Lemma} Let $X$ be a group, and $U\subseteq X$.   Let
$f:X\to\coint{0,\infty}$ be a bounded function such that $f(x)=0$ for
$x\in X\setminus U$;  set $\alpha=\sup_{x\in X}f(x)$.   Let
$A\subseteq X$ be a symmetric set containing $e$, and $K$ a set
including $A^k$, where $k\ge 1$.   Define $g:X\to\coint{0,\infty}$ by
setting

\Centerline{$g(x)=\Bover1k\sum_{i=0}^{k-1}\sup\{f(yx):y\in A^i\}$}

\noindent for $x\in X$.   Then

(a) $f(x)\le g(x)\le\alpha$ for every $x\in X$, and $g(x)=0$ if
$x\notin K^{-1}U$.

(b) $|g(ax)-g(x)|\le\Bover{j\alpha}k$ if $j\in\Bbb N$, $a\in A^j$ and
$x\in X$.

(c) For any $x$, $z\in X$,
$|g(x)-g(z)|\le\sup_{y\in K}|f(yx)-f(yz)|$.

\proof{{\bf (a)} Of course

\Centerline{$f(x)=\Bover1k\sum_{i=0}^{k-1}f(ex)
\le g(x)\le\Bover1k\sum_{i=0}^{k-1}\alpha=\alpha$}

\noindent for every $x$.   Suppose that $g(x)\ne 0$.   Then there must
be an $i<k$ and a
$y\in A^i$ such that $f(yx)\ne 0$, so that $yx\in U$.   But also,
because $e\in A$, $y\in A^k\subseteq K$, so
$x\in y^{-1}U\subseteq K^{-1}U$.

\medskip

{\bf (b)} Suppose first that $j=1$, so that $a\in A$.   If $\epsilon>0$
there are $y_i\in A^i$, for $i<k$, such that
$g(ax)\le\Bover1k\sum_{i=0}^{k-1}f(y_iax)+\epsilon$.   Now $y_ia\in
A^{i+1}$ for each $i$, so

\Centerline{$g(x)\ge\Bover1k\sum_{i=0}^{k-2}f(y_iax)
\ge g(ax)-\epsilon-\Bover{\alpha}k$.}

\noindent As $\epsilon$ is arbitrary, $g(x)\ge g(ax)-\bover{\alpha}k$.
Similarly, as $a^{-1}\in A$ (because $A$ is symmetric), $g(ax)\ge
g(x)-\bover{\alpha}k$ and
$|g(ax)-g(x)|\le\bover{\alpha}k$.

For the general case, induce on $j$.   (If $j=0$, then $a=e$ and the
result is trivial.)

\medskip

{\bf (c)} Set $\gamma=\sup_{y\in K}|f(yx)-f(yz)|$.   If $\epsilon>0$,
there are $y_i\in A^i$, for $i<k$, such that
$g(x)\le\Bover1k\sum_{i=0}^{k-1}f(y_ix)+\epsilon$.   Now every $y_i$
belongs to $K$, so

\Centerline{$g(z)\ge\Bover1k\sum_{i=0}^{k-1}f(y_iz)
\ge\Bover1k\sum_{i=0}^{k-1}(f(y_ix)-\gamma)
\ge g(x)-\epsilon-\gamma$.}

\noindent As $\epsilon$ is arbitrary, $g(z)\ge g(x)-\gamma$;
similarly, $g(x)\ge g(z)-\gamma$.
}%end of proof of 446E

\leader{*446F}{Lemma} Let $X$ be a locally compact Hausdorff
topological group and $\sequencen{A_n}$ a sequence of
closed symmetric subsets
of $X$ all containing the identity $e$ of $X$.   Suppose that for every
neighbourhood $W$ of $e$ there is
an $n_0\in\Bbb N$ such that $A_n\subseteq W$ for every $n\ge n_0$.   Let
$U$ be a compact neighbourhood of $e$ and suppose that for each
$n\in\Bbb N$ we have $k(n)\in\Bbb N$ such that $A_n^{k(n)}\subseteq U$
and $A_n^{k(n)+1}\not\subseteq U$.   Let $\Cal F$ be a non-principal
ultrafilter on $\Bbb N$ and write $Q$ for the limit
$\lim_{n\to\Cal F}A_n^{k(n)}$ in the space $\Cal C$
of closed subsets of $X$ with the Fell topology.

(i) If $Q^2=Q$ then $Q$ is a compact subgroup of $X$ included in $U$ and
meeting the boundary of $U$.

(ii) If $Q^2\ne Q$ then there are a neighbourhood $W$ of $e$ and an
infinite set $I\subseteq\Bbb N$ such that for every $n\in I$ there are
an $x\in A_n$ and an $i\le k(n)$ such that $x^i\notin W$.

\proof{{\bf (a)} I ought to begin by explaining why the limit
$\lim_{n\to\Cal F}A_n^{k(n)}$ is defined;  this is just because the
Fell topology is always compact (4A2T(b-iii)) and when based on
a locally compact Hausdorff space is Hausdorff (4A2T(e-ii)).

Because $U$ is closed, $\{F:F\in\Cal C$, $F\subseteq U\}$ is closed,
by the definition of the Fell topology (4A2T(a-ii));
because every $A_n^{k(n)}$ is included in
$U$, so is $Q$, and $Q$ is compact.   Because
$x\mapsto x^{-1}$ is a homeomorphism of $X$,
$F\mapsto F^{-1}$ is a homeomorphism of $\Cal C$, and

\Centerline{$Q^{-1}=\lim_{n\to\Cal F}(A_n^{k(n)})^{-1}
=\lim_{n\to\Cal F}(A_n^{-1})^{k(n)}
=\lim_{n\to\Cal F}A_n^{k(n)}=Q$.}

\noindent And of course $e\in Q$ because $e\in A_n^{k(n)}$ for every
$n$ and $\{(x,F):x\in F\in\Cal C\}$ is closed in $X\times\Cal C$
(4A2T(e-i)).

For each $n\in\Bbb N$, we have an $a_n\in A_n^{k(n)}$ and an
$x_n\in A_n$ such that $a_nx_n\notin U$.   Now $a=\lim_{n\to\Cal F}a_n$
is defined (because $U$ is compact), and belongs to $Q$.   Also
$\lim_{n\to\infty}x_n=e$, because every neighbourhood of $e$ includes
all but finitely many of the $A_n$, so
$a=\lim_{n\to\Cal F}a_nx_n\notin\interior U$, and $a$ belongs to the
boundary of $U$.   Thus $Q$ meets the boundary of $U$.

\medskip

{\bf (b)} From (a) we see that if $Q^2=Q$ then $Q$ is a compact subgroup
of $X$, included in $U$ and meeting the boundary of $U$.   So henceforth
let us suppose that $Q^2\ne Q$ and seek to prove (ii).

Let $w\in Q^2\setminus Q$.   Let $W_0\subseteq U$ be an open
neighbourhood of $e$ such that $W_0wW_0^2\cap QW_0^2=\emptyset$ (4A5Ee).

\medskip

{\bf (c)} Fix a left Haar measure $\mu$ on $X$.   Let
$f:X\to\coint{0,\infty}$ be a continuous function such that
$\{x:f(x)>0\}\subseteq W_0$ and $\int f(x)dx=1$.   Set
$\alpha=\sup_{x\in X}f(x)$ and $\beta=\int f(x)^2dx$, so that $\alpha$ is
finite and $\beta>0$.   $W_0U^2W_0\subseteq U^4$ is open and
relatively compact, so has
finite measure, and there is an $\eta>0$ such that

\Centerline{$2\eta(1+\alpha\mu(W_0U^2W_0))<\beta$.}

\noindent Let $W$ be a neighbourhood of $e$ such that $W\subseteq W_0$
and $|f(yax)-f(ybx)|\le\eta$ whenever $y\in(U^{-1})^2\cup U$, $x\in X$
and $ab^{-1}\in W$ (4A5Pa).

\medskip

{\bf (d)} Express $w$ as $w'w''$ where $w'$, $w''\in Q$.   Then

\Centerline{$\{n:A_n^{k(n)}\cap W_0w'\ne\emptyset\}$,
\quad$\{n:A_n^{k(n)}\cap w''W_0\ne\emptyset\}$,}

\Centerline{$\{n:A_n^{k(n)}\subseteq QW_0\}
=\{n:A_n^{k(n)}\cap(U\setminus QW_0)=\emptyset\}$}

\noindent all belong to $\Cal F$, by the definition of the Fell topology.
Also

\Centerline{$\{n:k(n)\ge 1\}$}

\noindent is cofinite in $\Bbb N$, because $A_n\subseteq U$ for all $n$
large enough.   Let $I$ be the intersection of these four sets, so that
$I$ belongs to $\Cal F$ and must be infinite.

\medskip

{\bf (e)} Let $n\in I$.   \Quer\ Suppose, if possible, that $x^i\in W$
for every $x\in A_n$ and $i\le k(n)$.   (The rest of the proof will be a
search for a contradiction.)   Note that $k(n)\ge 1$.

Choose $x_j\in A_n$, for $j<2k(n)$, such that the products
$x_{2k(n)-1}x_{2k(n)-2}\ldots x_{k(n)}$,
$x_{k(n)-1}\ldots x_{0}$ belong to $W_0w'$, $w''W_0$ respectively;  set
$\tilde w=x_{2k(n)-1}\ldots x_0$, so that

\Centerline{$\tilde w\in A_n^{2k(n)}\cap W_0w'w''W_0\subseteq W_0wW_0$.}

\noindent Since $A_n^{k(n)}\subseteq QW_0$ and $W_0wW_0^2\cap QW_0^2$ is
empty, $\tilde wW_0$ does not meet $A_n^{k(n)}W_0$.

\medskip

{\bf (f)} Define $g:X\to\coint{0,\infty}$ by setting

\Centerline{$g(x)
=\Bover1{k(n)}\sum_{i=0}^{k(n)-1}\sup\{f(yx):y\in A_n^i\}$.}

\noindent Then $g(\tilde wx)f(x)=0$ for every $x\in X$.   \Prf\ If
$f(x)\ne 0$, then $x\in W_0$, so $\tilde wx\notin A_n^{k(n)}W_0$, and
$g(\tilde wx)=0$, by 446Ea.\ \QeD\  Accordingly

\Centerline{$\biggerint(g(x)-g(\tilde wx))f(x)dx
=\int g(x)f(x)dx\ge\beta$}

\noindent since $g\ge f$ (446Ea).

Set $y_0=e$ and $y_{i+1}=x_{i}y_i$ for $i<2k(n)$, so that
$y_{2k(n)}=\tilde w$ and

\Centerline{$y_i\in A_n^i\subseteq A_n^{2k(n)}
=(A_n^{k(n)})^2\subseteq U^2$}

\noindent for every $i\le 2k(n)$.   Then

\Centerline{$g(x)-g(\tilde wx)
=\sum_{i=0}^{2k(n)-1}g(y_ix)-g(y_{i+1}x)$}

\noindent for every $x\in X$.   Let $i<2k(n)$ be such that

\Centerline{$\biggerint(g(y_ix)-g(y_{i+1}x))f(x)dx
\ge\Bover1{2k(n)}\int(g(x)-g(\tilde wx))f(x)dx
\ge\Bover{\beta}{2k(n)}$.}

\noindent Set $u=x_i$ and $v=y_i$, so that $u\in A_n$, $v\in U^2$ and

$$\eqalign{\int(g(x)-g(ux))f(v^{-1}x)dx
&=\int(g(vx)-g(uvx))f(x)dx\cr
&=\int(g(y_ix)-g(y_{i+1}x))f(x)dx
\ge\Bover{\beta}{2k(n)}.\cr}$$

\medskip

{\bf (g)} We have

$$\eqalign{\int(g(x)-g&(u^{k(n)}x))f(v^{-1}x)dx\cr
&=\sum_{j=0}^{k(n)-1}\int(g(u^jx)-g(u^{j+1}x))f(v^{-1}x)dx\cr
&=\sum_{j=0}^{k(n)-1}\int(g(x)-g(ux))f(v^{-1}u^{-j}x)dx\cr
&=k(n)\int(g(x)-g(ux))f(v^{-1}x)dx\cr
  &\qquad\qquad +\sum_{j=0}^{k(n)-1}\int(g(x)-g(ux))
                    (f(v^{-1}u^{-j}x)-f(v^{-1}x))dx,\cr}$$

\noindent that is,

$$\eqalign{k(n)\int&(g(x)-g(ux))f(v^{-1}x)dx\cr
&=\int(g(x)-g(u^{k(n)}x))f(v^{-1}x)dx\cr
 &\qquad\qquad -\sum_{j=0}^{k(n)-1}\int(g(x)-g(ux))
        (f(v^{-1}u^{-j}x)-f(v^{-1}x))dx.\cr}$$

\noindent Set

\Centerline{$\beta_1=\sum_{j=0}^{k(n)-1}$$\biggerint(g(x)-g(ux))
        (f(v^{-1}u^{-j}x)-f(v^{-1}x))dx$,}

\Centerline{$\beta_2=\biggerint(g(x)-g(u^{k(n)}x))f(v^{-1}x)dx$;}

\noindent then

\Centerline{$\beta_2-\beta_1
=k(n)\biggerint(g(x)-g(ux))f(v^{-1}x)dx
\ge\bover12\beta$.}

\medskip

{\bf (h)}\grheada\ Examine $\beta_1$.   We know that, because $u\in A_n$,
$|g(x)-g(ux)|\le\Bover{\alpha}{k(n)}$ for every $x$ (see 446Eb).   On
the other hand, we are supposing that $x^j\in W$ for every $j\le k(n)$
and every $x\in A_n$, so, in particular, $u^j\in W\subseteq W_0$ for
every $j\le k(n)$.   Also, as noted in (f), $v\in U^2$.  So for any
$j<k(n)$ we must have
$|f(v^{-1}u^{-j}x)-f(v^{-1}x)|\le\eta$ for every $x\in X$, by the choice
of $W$, while $f(v^{-1}u^{-j}x)-f(v^{-1}x)=0$ unless $x\in W_0U^2W_0$.
So

$$\eqalign{|\beta_1|
&\le\sum_{j=0}^{k(n)-1}\int|g(x)-g(ux)|
        |f(v^{-1}u^{-j}x)-f(v^{-1}x)|dx\cr
&\le\sum_{j=0}^{k(n)-1}\Bover{\alpha}{k(n)}\eta\mu(W_0U^2W_0)
=\alpha\eta\mu(W_0U^2W_0).\cr}$$

\medskip

\quad\grheadb\ Now consider $\beta_2$.   As $u^{k(n)}\in W$,
$|f(zu^{k(n)}x)-f(zx)|\le\eta$ for every $z\in U$ and $x\in X$,
by the choice of $W$, so (because $A_n^{k(n)}\subseteq U$)
$|g(u^{k(n)}x)-g(x)|\le\eta$ for every $x$ (446Ec).   Accordingly

\Centerline{$|\beta_2|\le\eta\biggerint f(v^{-1}x)dx
=\eta\int f(x)dx=\eta$.}

\medskip

{\bf (i)} But this means that

\Centerline{$\beta\le 2(|\beta_1|+|\beta_2|)
\le 2\eta(1+\alpha\mu(W_0U^2W_0))<\beta$,}

\noindent which is absurd.\ \Bang

\medskip

{\bf (j)} Thus for every $n\in I$ there are an $x\in A_n$ and an $i\le
k(n)$ such that $x^i\notin W$, and (ii) is true.   This completes the
proof.
}%end of proof of 446F

\leader{*446G}{`Groups with no small subgroups' (a) Definition} Let
$X$ be a topological group.   We say that $X$ {\bf has no small
subgroups} if there is a neighbourhood $U$ of the identity $e$ of $X$
such that the only subgroup of $X$ included in $U$ is $\{e\}$.

\spheader 446Gb If $X$ is a Hausdorff topological group and $U$ is a
compact symmetric neighbourhood of the identity $e$ such that the only
subgroup of $X$ included in $U$ is $\{e\}$, then $\{D_n(U):n\in\Bbb N\}$
is a base of neighbourhoods of
$e$\cmmnt{, where $D_n(U)=\{x:x\in X,\,x^i\in U$
for every $i\le n\}$}.   \prooflet{\Prf\ By 446Dc,
$\langle D_n(U)\rangle_{n\ge 1}$ is a non-increasing
sequence of compact neighbourhoods of $e$, and if
$x\in\bigcap_{n\in\Bbb N}D_n(U)$ then $x^i\in U$ for every $i\in\Bbb N$;
as $U^{-1}=U$, $U$
includes the subgroup $\{x^i:i\in\Bbb Z\}$, so $x=e$.   Thus
$\bigcap_{n\in\Bbb N}D_n(U)=\{e\}$ and $\{D_n(U):n\in\Bbb N\}$ is a base
of neighbourhoods of $e$ (4A2Gd).\ \Qed}

\spheader 446Gc In particular, a locally compact Hausdorff topological
group with no small subgroups is metrizable\cmmnt{ (4A5Q)}.

\leader{*446H}{Lemma} Let $X$ be a locally compact Hausdorff
topological group.   \cmmnt{For $A\subseteq X$, $n\in\Bbb N$ set
$D_n(A)=\{x:x^i\in A$ for every $i\le n\}$.}
If $U\subseteq X$ is a compact symmetric
neighbourhood of the identity which does not include any subgroup of $X$
other than $\{e\}$, then
there is an $r\ge 1$ such that $D_{rn}(U)^n\subseteq U$ for every
$n\in\Bbb N$.

% and $D_n(U)^n\subseteq U^r$ for every $n\ge 1$, not needed I think

\proof{\Quer\ Suppose, if possible, that for every $r\ge 1$
there is an $n_r\in\Bbb N$ such that $D_{rn_r}(U)^{n_r}\not\subseteq U$.
Of course $n_r\ge 1$.   Set $A_0=U$ and
$A_r=D_{rn_r}(U)$ for $r\ge 1$.   Note that $D_{n_1}(U)\subseteq A_0$ but
$D_{n_1}(U)^{n_1}\not\subseteq U$, so $A_0^{n_1}\not\subseteq U$.
We therefore have, for every $r\in\Bbb N$, a $k_r$ such that
$A_r^{k(r)}\subseteq U$ but $A_r^{k(r)+1}\not\subseteq U$.   Also, by
446Gb, every neighbourhood of $e$ includes all but finitely many of the
$A_r$.   We can therefore apply 446F to the sequence
$\sequence{r}{A_r}$.   Of course $k(r)\ge 1$ for every $r$, while
$k(r)<n_r$ for $r\ge 1$.   Since $U$ includes no
non-trivial subgroup, (i) of 446F is impossible, and we are left with
(ii).   Let $W$, $I$ be as declared there, so that
$A_r\not\subseteq D_{k(r)}(W)$ for every $r\in I$.   There must be some
$m\ge 1$ such that $D_m(U)\subseteq W$ (446Gb).
Take $r\in I$ such that $r\ge m$;  then $rn_r\ge mk(r)$, so

$$\eqalignno{A_r
&=D_{rn_r}(U)
\subseteq D_{mk(r)}(U)
\subseteq D_{k(r)}(D_m(U))\cr
\noalign{\noindent (446D(b-iv))}
&\subseteq D_{k(r)}(W),\cr}$$

\noindent which is impossible.\ \Bang
}%end of proof of 446H

\leaveitout{(proof of other bit)
{\bf (b)} Thus we have some $l\ge 1$ such that
$D_{ln}(U)^n\subseteq U$ for every $n\in\Bbb N$.   Set $r=2l$;  we surely have
$D_{rn}(U)^n\subseteq U$ for every $n\ge 1$.   If $1\le n\le l$, then
$D_n(U)^n\subseteq U^n\subseteq U^r$.   If $n\ge l$, then $n$ is
expressible as $il+j$ where
$i\ge 1$, $j<l$.   So

\Centerline{$D_n(U)^n
\subseteq D_{il}(U)^{(i+1)l}
\subseteq D_{il}(U)^{2il}
=(D_{il}(U)^i)^{r}
\subseteq U^r$.}

\noindent So we have the result.}

\leader{*446I}{Lemma} Let $X$ be a locally compact Hausdorff
topological
group and $U$ a compact symmetric neighbourhood of the identity in $X$
such that $U$ does not include any subgroup of $X$ other than $\{e\}$.
\dvro{Let}{For $n\in\Bbb N$,
set $D_n(U)=\{x:x^i\in U$ for every $i\le n\}$, and
let} $\Cal F$ be any non-principal ultrafilter on $\Bbb N$.
Suppose that $\sequencen{x_n}$ is a sequence in $X$ such that
$x_n\in D_n(U)$ for every $n\in\Bbb N$.   Then we have a continuous
homomorphism
$q:\Bbb R\to X$ defined by setting $q(t)=\lim_{n\to\Cal F}x_n^{i(n)}$
whenever $\sequencen{i(n)}$ is a sequence in $\Bbb Z$ such that
$\lim_{n\to\Cal F}\Bover{i(n)}n=t$ in $\Bbb R$.

\proof{{\bf (a)} If $\sequencen{i(n)}$ is any sequence in $\Bbb N$ such
that $\lim_{n\to\Cal F}\Bover{i(n)}n$ is defined in $\Bbb R$, then
$\lim_{n\to\Cal F}x_n^{i(n)}$ is defined in $X$.   \Prf\ There is some
$m\in\Bbb N$ such that $m>\lim_{n\to\Cal F}\Bover{i(n)}n$, so that
$J=\{n:i(n)\le mn\}\in\Cal F$;  but if $n\in J$, then

\Centerline{$x_n\in D_n(U)\subseteq D_{mn}(U^m)\subseteq
D_{i(n)}(U^m)$,}

\noindent by 446D(b-v), and $x_n^{i(n)}\in U^m$.   But this means that
$\Cal F$ contains $\{n:x_n^{i(n)}\in U^m\}$;  as $U^m$ is compact,
$\lim_{n\to\Cal F}x_n^{i(n)}$ is defined in $X$.\ \Qed

More generally, if $\sequencen{i(n)}$ is any sequence in $\Bbb Z$ such
that $\lim_{n\to\Cal F}\Bover{i(n)}n$ is defined in $\Bbb R$, then
$\lim_{n\to\Cal F}x_n^{i(n)}$ is defined in $X$.   \Prf\ At least one of
$\{n:i(n)\ge 0\}$, $\{n:i(n)\le 0\}$ belongs to $\Cal F$.   In the
former case,
$\lim_{n\to\Cal F}x_n^{i(n)}=\lim_{n\to\Cal F}x_n^{\max(0,i(n))}$ is
defined;  in the latter case,

\Centerline{$\lim_{n\to\Cal F}x_n^{i(n)}
=\lim_{n\to\Cal F}(x_n^{\max(0,-i(n))})^{-1}
=(\lim_{n\to\Cal F}x_n^{\max(0,-i(n))})^{-1}$}

\noindent is defined.\ \Qed

\medskip

{\bf (b)} If $V$ is any neighbourhood of $e$, there is a $\delta>0$ such
that $\lim_{n\to\Cal F}x_n^{i(n)}\in V$ whenever $\sequencen{i(n)}$ is a
sequence in $\Bbb Z$ such that $\lim_{n\to\Cal F}|\Bover{i(n)}n|\le\delta$.
\Prf\ By 446Gb, there is an $m\ge 1$ such that $D_m(U)\subseteq V$.
Take $\delta<\bover1m$.   If $\lim_{n\to\Cal F}|\Bover{i(n)}n|\le\delta$, then
$J=\{n:m|i(n)|\le n\}\in\Cal F$.   But if $n\in J$, then

\Centerline{$x_n\in D_n(U)\subseteq D_{m|i(n)|}(U)
\subseteq D_{|i(n)|}(D_m(U))$}

\noindent and $x_n^{|i(n)|}\in D_m(U)$;  since $D_m(U)$, like $U$, is
symmetric (446D(b-vi)), $x_n^{i(n)}\in D_m(U)$.   This is true for every
$n\in J$, so $\lim_{n\to\Cal F}x_n^{i(n)}\in D_m(U)\subseteq V$, as
required.\ \Qed

\medskip

{\bf (c)} It follows at once that if $\sequencen{i(n)}$ is a
sequence in $\Bbb Z$ such that $\lim_{n\to\Cal F}\Bover{i(n)}n=0$, then
$\lim_{n\to\Cal F}x_n^{i(n)}=e$.   Consequently,
if $\sequencen{i(n)}$, $\sequencen{j(n)}$ are sequences
in $\Bbb Z$ such that $\lim_{n\to\Cal F}\Bover{i(n)}n$ and
$\lim_{n\to\Cal F}\Bover{j(n)}n$ both exist in $\Bbb R$ and are equal, then
$\lim_{n\to\Cal F}x_n^{i(n)}=\lim_{n\to\Cal F}x_n^{j(n)}$.   \Prf\ Set
$k(n)=i(n)-j(n)$.   Then $\lim_{n\to\Cal F}x_n^{k(n)}=e$ because
$\lim_{n\to\Cal F}\Bover{k(n)}n=0$.   But now

\Centerline{$\lim_{n\to\Cal F}x_n^{i(n)}
=\lim_{n\to\Cal F}x_n^{j(n)}x_n^{k(n)}
=\lim_{n\to\Cal F}x_n^{j(n)}$.\ \Qed}

\medskip

{\bf (d)} We do therefore have a function $q:\Bbb R\to X$ defined by the
given formula.   Now $q(s+t)=q(s)q(t)$ for all $s$, $t\in\Bbb R$.
\Prf\ Take sequences $\sequencen{i(n)}$, $\sequencen{j(n)}$ such that
$s=\lim_{n\to\infty}\Bover{i(n)}n$, $t=\lim_{n\to\infty}\Bover{j(n)}n$;  then
$s+t=\lim_{n\to\infty}\Bover{i(n)+j(n)}n$, so

\Centerline{$q(s+t)
=\lim_{n\to\Cal F}x_n^{i(n)+j(n)}
=\lim_{n\to\Cal F}x_n^{i(n)}x_n^{j(n)}
=q(s)q(t)$.  \Qed}

\medskip

{\bf (e)} Thus $q$ is a homomorphism.   Finally, (b) shows that it is
continuous at $0$, so it must be continuous (4A5Fa).
}%end of proof of 446I

\leader{*446J}{Lemma} Let $X$ be a locally compact Hausdorff
topological
group with no small subgroups.   Then there is a neighbourhood $V$ of
the identity $e$ such that $x=y$ whenever $x$, $y\in V$ and $x^2=y^2$.

\proof{ Let $U$ be a symmetric compact neighbourhood of $e$ not
including any subgroup of $X$ except $\{e\}$.   Let $\sequence{n}{U_n}$
be a non-increasing sequence of neighbourhoods of $e$ comprising a base
of neighbourhoods of $e$ (446Gc), and with $U_0=U$.

\Quer\ Suppose, if possible, that for each $n\in\Bbb N$ there are
distinct $x_n$, $y_n\in U_n$ such that $x_n^2=y_n^2$.   Set
$a_n=x_n^{-1}y_n$;  then

\Centerline{$x_n^{-1}a_nx_n=x_n^{-2}y_nx_n=y_n^{-1}x_n=a_n^{-1}$.}

\noindent Accordingly

\Centerline{$x_n^{-1}a_n^mx_n=(x_n^{-1}a_nx_n)^m=a_n^{-m}$}

\noindent for every $m\in\Bbb N$.

Since $U$ includes no non-trivial subgroup, and $a_n\ne e$, there is a
$k(n)\in\Bbb N$ such that $a_n^i\in U$ for $i\le k(n)$ and
$a_n^{k(n)+1}\notin U$.   Let $\Cal F$ be any non-principal ultrafilter
on $\Bbb N$;  then $a=\lim_{n\to\Cal F}a_n^{k(n)}$ is defined and
belongs to $U$.   Also, because $\sequencen{x_n}$ and $\sequencen{y_n}$
both converge to $e$, so does $\sequencen{a_n}$, and

\Centerline{$a=\lim_{n\to\Cal F}a_n^{k(n)+1}\in X\setminus\interior U$;}

\noindent thus $a$ cannot be $e$.

However, $x_n^{-1}a_n^{k(n)}x_n=a_n^{-k(n)}$ for each $n$, so

\Centerline{$e^{-1}ae=\lim_{n\to\Cal F}x_n^{-1}a_n^{k(n)}x_n
=\lim_{n\to\Cal F}(a_n^{k(n)})^{-1}
=a^{-1}$.}

\noindent So $a=a^{-1}$ and $\{e,a\}$ is a non-trivial subgroup of $X$
included in $U$, which is supposed to be impossible.\ \Bang

Thus some $U_n$ serves for $V$.
}%end of proof of 446J

\leader{*446K}{Lemma} Let $X$ be a locally compact Hausdorff
topological
group with no small subgroups.   \cmmnt{For $A\subseteq X$ set
$D_n(A)=\{x:x^i\in A$ for every $i\le n\}$.}   Then there is a compact
symmetric neighbourhood $U$ of the identity $e$ such that whenever $V$
is a neighbourhood of $e$ there are an $n_0\in\Bbb N$ and a
neighbourhood $W$ of $e$ such that whenever $n\ge n_0$, $x\in D_n(U)$,
$y\in D_n(U)$ and $x^ny^n\in W$, then $xy\in D_n(V)$.

\proof{{\bf (a)} Let $U_0$ be a compact symmetric neighbourhood of $e$
such that ($\alpha$) $U_0^3$ includes no subgroup of $X$ other than
$\{e\}$ ($\beta$) whenever $x$, $y\in U_0$ and $x^2=y^2$, then $x=y$;
such a neighbourhood exists by 446J.
Let $r\ge 1$ be such that $D_{rn}(U_0)^n\subseteq U_0$ for every
$n\in\Bbb N$ (446H).   Let $U$ be a compact symmetric neighbourhood of
$e$ such that $U^r\subseteq U_0$.   In this case
$D_n(U)\subseteq D_{rn}(U_0)$ for
every $n$, by 446D(b-v).   So $D_n(U)^n\subseteq U_0$ for every $n$.

\medskip

{\bf (b)} Fix a left Haar measure $\mu$ on $X$.   Let
$f:X\to\coint{0,\infty}$ be a continuous function such that
$\int f(x)dx=1$ and $f(x)=0$ for $x\in X\setminus U_0$.   Set
$\alpha=\sup_{x\in X}|f(x)|$,
$\beta=\int f(x)^2dx$, so that $\alpha$ is finite and $\beta>0$.   For
$n\ge 1$, set

\Centerline{$f_n(x)
=\Bover1n\sum_{i=0}^{n-1}\sup\{f(yx):y\in D_n(U)^i\}$}

\noindent for $x\in X$.   Because $D_n(U)^n\subseteq U_0$, we can
apply 446E to see that, for each $n$,

\inset{(i) $f_n\ge f$,}

\inset{(ii) $f_n(x)=0$ if $x\notin U_0^2$,}

\inset{(iii) $|f_n(ax)-f_n(x)|\le\Bover{j\alpha}n$ if $j\in\Bbb N$,
$a\in D_n(U)^j$ and $x\in X$,}

\inset{(iv) for any $x$, $z\in X$,
$|f_n(x)-f_n(z)|\le\sup_{y\in U_0}|f(yx)-f(yz)|$.}

\noindent It follows that

\inset{(v) for any $\epsilon>0$ there is a
neighbourhood $W$ of $e$ such that $|f_n(ax)-f_n(bx)|\le\epsilon$
whenever $a$, $b$, $x\in X$, $n\in\Bbb N$ and $ab^{-1}\in W$}

\noindent (4A5Pa again).

\medskip

{\bf (c)} It will help to have the following fact available.   Suppose
we are given sequences $\sequencen{x_n}$, $\sequencen{y_n}$ such that
$x_n$ and $y_n$ belong to $D_n(U)$ for every $n\in\Bbb N$ and
$\lim_{n\to\infty}x_n^ny_n^n=e$.   Write

\Centerline{$\gamma_n
=\sup\{|f_n(y_n^jx)-f_n(x_n^{-j}x)|:j\le n,\,x\in X\}$}

\noindent for each $n\in\Bbb N$.   Then $\lim_{n\to\infty}\gamma_n=0$.
\Prf\Quer\ Otherwise, there is an $\eta>0$ such that
$J=\{n:\gamma_n>\eta\}$ is infinite.   Let $W$ be a neighbourhood of $e$
such that $|f_n(ax)-f_n(bx)|\le\eta$ whenever $n\in\Bbb N$, $x$, $a$,
$b\in X$ and
$ab^{-1}\in W$ ((b-v) above);  let $W'$ be a neighbourhood of $e$ such
that $ab\in W$ whenever $a$, $b\in U$ and $ba\in W'$ (4A5Ej).
Then for each $n\in J$ there must be a
$j(n)\le n$ such that $y_n^{j(n)}x_n^{j(n)}\notin W$, while $x_n^{j(n)}$
and $y_n^{j(n)}$ both belong to $U$, so that
$x^{j(n)}y^{j(n)}\notin W'$.   Let $\Cal F$ be
a non-principal ultrafilter on $\Bbb N$ containing $J$.   By 446I, there
are continuous homomorphisms $q$, $\tilde q$ from $\Bbb R$ to $X$ such
that $q(t)=\lim_{n\to\Cal F}x_n^{-i(n)}$,
$\tilde q(t)=\lim_{n\to\Cal F}y_n^{i(n)}$ whenever $\sequencen{i(n)}$
is a sequence in $\Bbb Z$ such
that $\lim_{n\to\Cal F}\Bover{i(n)}n=t$ in $\Bbb R$.   (Of course $x_n^{-1}\in
D_n(U)$ for every $n$, by 446D(b-vi).)   Setting
$t_0=\lim_{n\to\Cal F}\Bover{j(n)}n\in[0,1]$,

\Centerline{$q(-t_0)\tilde q(t_0)
=\lim_{n\to\Cal F}x_n^{j(n)}y_n^{j(n)}
\notin\interior W'$,}

\noindent so $q(t_0)\ne \tilde q(t_0)$ and $q\ne\tilde q$.   But

\Centerline{$q(-1)\tilde q(1)=\lim_{n\to\Cal F}x_n^ny_n^n=e$,}

\noindent so $q(1)=\tilde q(1)$.   Now if $0\le i(n)\le n$, then
$x_n^{-i(n)}\in D_n(U)^n\subseteq U_0$;  so if $0\le t\le 1$, $q(t)\in
U_0$.   Similarly, $\tilde q(t)\in U_0$ whenever $t\in[0,1]$.   But
recall that $U_0$ was chosen so that if $x$, $y\in U_0$ and $x^2=y^2$
then $x=y$.   In particular, since $q(\bover12)$ and $\tilde
q(\bover12)$ both belong to $U_0$, and their squares $q(1)$, $\tilde
q(1)$ are equal, $q(\bover12)=\tilde q(\bover12)$.   Repeating this
argument, we see that $q(2^{-k})=\tilde q(2^{-k})$ for every $k\in\Bbb
N$, so that $q(2^{-k}i)=\tilde q(2^{-k}i)$ for every $k\in\Bbb N$,
$i\in\Bbb Z$;  since $q$ and $\tilde q$ are supposed to be continuous,
they must be equal;  but $q(t_0)\ne\tilde q(t_0)$.\ \Bang\Qed

\medskip

{\bf (d)} Now let $V$ be any neighbourhood of $e$.

\Quer\ Suppose, if possible, that for every neighbourhood $W$ of $e$ and
$n_0\in\Bbb N$ there are $n\ge n_0$ and $x$, $y\in D_n(U)$ such that
$x^ny^n\in W$ but $xy\notin D_n(V)$.   For $k\in\Bbb N$ choose
$n_k\in\Bbb N$ and $\tilde x_k$, $\tilde y_k\in D_{n_k}(U)$ such that
$\tilde x_k^{n_k}\tilde y_k^{n_k}\in D_k(U)$ but
$\tilde x_k\tilde y_k\notin D_{n_k}(V)$, and $n_k>n_{k-1}$ if $k\ge 1$.
Now we know that
$\sequence{k}{D_k(U)}$ is a non-increasing sequence constituting a base
of neighbourhoods of $e$ (446Gb), so
$\lim_{k\to\infty}\tilde x_k^{n_k}\tilde y_k^{n_k}=e$.    Set
$J=\{n_k:k\in\Bbb N\}$.   For
$n=n_k$, set $x_n=\tilde x_k$, $y_n=\tilde y_k$;  for
$n\in\Bbb N\setminus J$, set $x_n=y_n=e$.
Then $x_n$, $y_n\in D_n(U)$ for every
$n\in\Bbb N$ and $\sequencen{x_n^ny_n^n}\to e$ as $n\to\infty$, while
$x_ny_n\notin D_n(V)$ for $n\in J$.

We know from (c) that

\Centerline{$\gamma_n
=\sup\{|f_n(y_n^jx)-f_n(x_n^{-j}x)|:j\le n,\,x\in X\}\to 0$}

\noindent as $n\to\infty$;  for future reference, take a sequence
$\langle j(n)\rangle_{n\ge 1}$ such that $1\le j(n)\le n$ for every
$n\ge 1$, $\lim_{n\to\infty}\Bover{j(n)}n=0$ and
$\lim_{n\to\infty}\Bover{n\gamma_n}{j(n)}=0$.

\medskip

{\bf (e)} Let $l\ge 1$ be such that $D_l(U_0^3)\subseteq V$ (446Gb
again), and set
$K=U_0^{2l+1}$.   Then $D_{nl}(U_0^3)\subseteq D_n(V)$
for every $n$ (446D(b-vii)).   So $x_ny_n\notin D_{nl}(U_0^3)$ for
$n\in J$.   For each $n\in J$ choose $m(n)\le nl$ such that
$(x_ny_n)^{m(n)}\notin U_0^3$;  for $n\in\Bbb N\setminus J$, set
$m(n)=1$.
Then $(x_ny_n)^{-m(n)}x\notin U_0^2$ for any $x\in U_0$ and $n\in J$, and
$f_n((x_ny_n)^{-m(n)}x)f(x)=0$ for any $x\in X$ and $n\in J$.   So

\Centerline{$|\biggerint(f_n((x_ny_n)^{-m(n)}x)-f_n(x))f(x)dx|
=\int f_n(x)f(x)dx\ge\beta$}

\noindent for $n\in J$, because $f_n\ge f$ ((b-i) above).

\medskip

{\bf (f)} Of course $m(n)>0$ for every $n\in J$, therefore for every
$n$.   So we can set

\Centerline{$g_n(x)=\Bover1{m(n)}\sum_{i=0}^{m(n)-1}f((x_ny_n)^ix)$}

\noindent for $x\in X$ and $n\in\Bbb N$.   Note that $g_n$, like $f$, is
non-negative, and also that $\int g_n(x)dx=\int f(x)dx=1$.   We need to
know that $g_n(x)=0$ if $x\notin K$;  this is because
$(x_ny_n)^i\in D_n(U)^{2nl}\subseteq U_0^{2l}$ for every $i\le m(n)$,
while $f(x)=0$ if
$x\notin U_0$, and $U_0$ is symmetric.

We also have

$$\eqalign{|g_n(ax)-g_n(x)|
&\le\sup_{i<m(n)}|f((x_ny_n)^iax)-f((x_ny_n)^ix)|\cr
&\le\sup\{|f(wax)-f(wx)|:w\in U_0^{2l}\}\cr}$$

\noindent for every $n\in\Bbb N$ and $a$, $x\in X$ (cf.\ 446Ec), so for
every $\eta>0$ there must be a neighbourhood $W$ of $e$ such that
$|g_n(ax)-g(x)|\le\eta$ whenever $n\in\Bbb N$, $a\in W$ and $x\in X$, by
4A5Pa once more.   Since also $g_n(ax)=g_n(x)=0$ if
$a\in U_0$ and $x\notin
U_0^{-1}K$, and $U_0^{-1}K$ has finite measure, we see that for every
$\eta>0$ there is a neighbourhood $W$ of $e$ such that
$\int|g_n(ax)-g(x)|dx\le\eta$ for every $a\in W$, $n\in\Bbb N$.

\medskip

{\bf (g)} Returning to the formula in (e), we see that, for any
$n\in J$,

$$\eqalignno{\beta
&\le\bigl|\int\bigl(f_n((x_ny_n)^{-m(n)}x)-f_n(x)\bigr)
                                                  f(x)dx\bigr|\cr
&=\bigl|\sum_{i=0}^{m(n)-1}
  \int\bigl(f_n((x_ny_n)^{-i-1}x)-f_n((x_ny_n)^{-i}x)\bigr)
                                                 f(x)dx\bigr|\cr
&=\bigl|\sum_{i=0}^{m(n)-1}
  \int\bigl(f_n((x_ny_n)^{-1}x)-f_n(x)\bigr)f((x_ny_n)^ix)dx\bigr|\cr
&=m(n)\bigl|\int\bigl(f_n((x_ny_n)^{-1}x)-f_n(x)\bigr)g_n(x)dx\bigr|\cr
&\le ln\bigl|\int\bigl(f_n(y_n^{-1}x_n^{-1}x)-f_n(x)\bigr)
                                               g_n(x)dx\bigr|\cr
&\le ln\bigl|\int\bigl(f_n(y_n^{-1}x)-f_n(x)-f_n(y_n^{-1}x_n^{-1}x)
   +f_n(x_n^{-1}x)\bigr)g_n(x)dx\bigr|\cr
&\qquad\qquad\qquad
   +ln\bigl|\int\bigl(2f_n(x)-f_n(x_n^{-1}x)-f_n(y_n^{-1}x)\bigr)
                                                g_n(x)dx\bigr|.\cr}$$

\noindent Next,

$$\eqalign{j(n)\bigl(2&f_n(x)-f_n(x_n^{-1}x)-f_n(y_n^{-1}x)\bigr)\cr
&=j(n)\bigl(f_n(x)-f_n(x_n^{-1}x)\bigr)
  -f_n(x)+f_n(x_n^{-j(n)}x)\cr
&\qquad\qquad +j(n)\bigl(f_n(x)-f_n(y_n^{-1}x)\bigr)
  -f_n(x)+f_n(y_n^{-j(n)}x)\cr
&\qquad\qquad +2f_n(x)-f_n(x_n^{-j(n)}x)
  -f_n(y_n^{-j(n)}x)\cr}$$

\noindent for every $x$, and finally

$$\eqalign{\int\bigl(2f_n(x)&-f_n(x_n^{-j(n)}x)-f_n(y_n^{-j(n)}x)\bigr)
                                                 g_n(x)dx\cr
&=\int\bigl(f_n(x)-f_n(y_n^{-j(n)}x)\bigr)g_n(x)dx
  -\int\bigl(f_n(y_n^{j(n)}x)-f_n(x)\bigr)g_n(x)dx\cr
&\qquad\qquad\qquad
  +\int\bigl(f_n(y_n^{j(n)}x)-f_n(x_n^{-j(n)}x)\bigr)g_n(x)dx\cr
&=\int\bigl(f_n(y_n^{j(n)}x)-f_n(x)\bigr)g_n(y_n^{j(n)}x)dx
   -\int\bigl(f_n(y_n^{j(n)}x)-f_n(x)\bigr)g_n(x)dx\cr
&\qquad\qquad\qquad
  +\int\bigl(f_n(y_n^{j(n)}x)-f_n(x_n^{-j(n)}x)\bigr)g_n(x)dx\cr
&=\int\bigl(f_n(y_n^{j(n)}x)-f_n(x)\bigr)
                    \bigl(g_n(y_n^{j(n)}x)-g_n(x)\bigr)dx\cr
&\qquad\qquad\qquad
  +\int\bigl(f_n(y_n^{j(n)}x)-f_n(x_n^{-j(n)}x)\bigr)g_n(x)dx.\cr}$$

So if we write

\Centerline{$\beta_{1n}
  =ln\biggerint\bigl(f_n(y_n^{-1}x)-f_n(x)-f_n(y_n^{-1}x_n^{-1}x)
     +f_n(x_n^{-1}x)\bigr)g_n(x)dx$,}

\Centerline{$\beta_{2n}
  =\Bover{ln}{j(n)}\biggerint\bigl(j(n)(f_n(x)-f_n(x_n^{-1}x))
    -f_n(x)+f_n(x_n^{-j(n)}x)\bigr)g_n(x)dx$,}

\Centerline{$\beta_{3n}
  =\Bover{ln}{j(n)}\biggerint\bigl(j(n)(f_n(x)-f_n(y_n^{-1}x))
    -f_n(x)+f_n(y_n^{-j(n)}x)\bigr)g_n(x)dx$,}

\Centerline{$\beta_{4n}
  =\Bover{ln}{j(n)}\biggerint\bigl(f_n(y_n^{j(n)}x)-f_n(x)\bigr)
                        \bigl(g_n(y_n^{j(n)}x)-g_n(x)\bigr)dx$,}

\Centerline{$\beta_{5n}=\Bover{ln}{j(n)}
  \biggerint\bigl(f_n(y_n^{j(n)}x)-f_n(x_n^{-j(n)}x)\bigr)g_n(x)dx$}

\noindent for $n\ge 1$, we have

\Centerline{$|\beta_{1n}|+|\beta_{2n}|+|\beta_{3n}|
+|\beta_{4n}|+|\beta_{5n}|\ge\beta$}

\noindent for every $n\in J$.

\medskip

{\bf (h)} Now $\beta_{1n}\to 0$ as $n\to\infty$.   \Prf\

$$\eqalign{\bigl|\int\bigl(f_n&(y_n^{-1}x)-f_n(x)-f_n(y_n^{-1}x_n^{-1}x)
     +f_n(x_n^{-1}x)\bigr)g_n(x)dx\bigr|\cr
&=\bigl|\int\bigl(f_n(y_n^{-1}x)-f_n(x)\bigr)g_n(x)dx
   -\int\bigl(f_n(y_n^{-1}x_n^{-1}x)-f_n(x_n^{-1}x)\bigr)
                                                  g_n(x)dx\bigr|\cr
&=\bigl|\int\bigl(f_n(y_n^{-1}x)-f_n(x)\bigr)g_n(x)dx
   -\int\bigl(f_n(y_n^{-1}x)-f_n(x)\bigr)g_n(x_nx)dx\bigr|\cr
&=\bigl|\int\bigl(f_n(y_n^{-1}x)-f_n(x)\bigr)
                        \bigl(g_n(x)-g_n(x_nx)\bigr)dx\bigr|\cr
&\le\Bover{\alpha}n\int|g_n(x)-g_n(x_nx)|dx\cr}$$

\noindent by (b-iii).   So

\Centerline{$|\beta_{1n}|
\le\alpha l\biggerint|g_n(x)-g_n(x_nx)|dx\to 0$}

\noindent as $n\to\infty$, by (f), since surely
$\sequencen{x_n}\to e$.\ \Qed

\medskip

{\bf (i)} Now look at $\beta_{2n}$.   If we set

\Centerline{$\gamma'_n
=\sup\{\biggerint|g_n(ax)-g(x)|dx:a\in D_n(U)^{j(n)}\}$,}

\noindent then $\gamma'_n\to 0$ as $n\to\infty$.   \Prf\ Given
$\epsilon>0$, there is a neighbourhood $W$ of $e$ such that
$\int|g_n(ax)-g_n(x)|dx\le\epsilon$ whenever $n\ge 1$ and $a\in W$,
as noted at the end of (f).   Let $p$ be such that $D_p(U_0)\subseteq
W$.   Then, for all $n$ large enough, $pj(n)\le n$, so that
$D_n(U)^{pj(n)}\subseteq U_0$ and $D_n(U)^{j(n)}\subseteq
D_p(U_0)\subseteq W$ (446D(b-v)) and
$\int|g_n(ax)-g(x)|dx\le\epsilon$
for every $a\in D_n(U)^{j(n)}$.\ \Qed

We have

$$\eqalignno{\bigl|\int\bigl(&j(n)(f_n(x)-f_n(x_n^{-1}x))-f_n(x)
  +f_n(x_n^{-j(n)}x)\bigr)g_n(x)dx\bigr|\cr
&=\bigl|\sum_{i=0}^{j(n)-1}\int\bigl(f_n(x)-f_n(x_n^{-1}x)
  -f_n(x_n^{-i}x)+f_n(x_n^{-i-1}x)\bigr)g_n(x)dx\bigr|\cr
&=\bigl|\sum_{i=0}^{j(n)-1}\int\bigl(f_n(x)-f_n(x_n^{-1}x)\bigr)g_n(x)dx
  -\int\bigl(f_n(x_n^{-i}x)-f_n(x_n^{-i-1}x)\bigr)g_n(x)dx\bigr|\cr
&=\bigl|\sum_{i=0}^{j(n)-1}\int\bigl(f_n(x)-f_n(x_n^{-1}x)\bigr)g_n(x)dx
  -\int\bigl(f_n(x)-f_n(x_n^{-1}x)\bigr)g_n(x_n^ix)dx\bigr|\cr
&=\bigl|\sum_{i=0}^{j(n)-1}\int\bigl(f_n(x)-f_n(x_n^{-1}x)\bigr)
                                 \bigl(g_n(x)-g_n(x_n^ix)\bigr)dx|\cr
&\le\sum_{i=0}^{j(n)-1}\int|f_n(x)-f_n(x_n^{-1}x)|
  |g_n(x)-g_n(x_n^ix)|dx
\le j(n)\Bover{\alpha}n\gamma'_n.\cr}$$

\noindent So

\Centerline{$|\beta_{2n}|\le l\alpha\gamma'_n\to 0$}

\noindent as $n\to\infty$.   Similarly, $\langle\beta_{3n}\rangle_{n\ge
1}\to 0$.

\woddheader{446K}{4}{2}{2}{30pt}

{\bf (j)} As for $\beta_{4n}$, we have

\Centerline{$\biggerint|f_n(y_n^{j(n)}x)-f_n(x)|
  |g_n(y_n^{j(n)}x)-g_n(x)|dx
\le\Bover{\alpha j(n)}n\gamma'_n$,}

\noindent putting (b-iii) and the definition of
$\gamma'_n$ together.   So

\Centerline{$|\beta_{4n}|\le l\alpha\gamma'_n\to 0$}

\noindent as $n\to\infty$.

\medskip

{\bf (k)} We come at last to $\beta_{5n}$.   Here, for every $n\ge 1$,

$$\eqalign{\bigl|\int\bigl(f_n(y_n^{j(n)}x)-f_n(x_n^{-j(n)}x)\bigr)
                                             g_n(x)dx\bigr|
&\le\int\bigl|f_n(y_n^{j(n)}x)-f_n(x_n^{-j(n)}x)\bigr|g_n(x)dx\cr
&\le\gamma_n\int g_n(x)dx
=\gamma_n\cr}$$

\noindent by the definition of $\gamma_n$ in (c) above.   So

\Centerline{$|\beta_{5n}|\le l\Bover{n}{j(n)}\gamma_n\to 0$}

\noindent by the choice of the $j(n)$.

\medskip

{\bf (l)} Thus $\beta_{in}\to 0$ as $n\to\infty$ for every $i$.   But
this is impossible, because $0<\beta\le\sum_{i=1}^5|\beta_{in}|$ for
every $n\in J$.\ \Bang

This contradiction shows that we must be able to find a neighbourhood
$W$ of $e$ and an $n_0\in\Bbb N$ such that $xy\in D_n(V)$ whenever $n\ge
n_0$, $x$, $y\in D_n(U)$ and $x^ny^n\in W$;  as $V$ is arbitrary, $U$
has the property required.
}%end of proof of 446K

\leader{*446L}{Definition} Let $X$ be a topological group.   A
{\bf $B$-sequence} in $X$ is a non-increasing sequence $\sequencen{V_n}$
of closed neighbourhoods of the identity, constituting a base of
neighbourhoods of the identity, such that there is some $M$ such that
for every $n\in\Bbb N$ the set $V_nV_n^{-1}$ can be covered by at most
$M$ left translates of $V_n$.

\leader{*446M}{Proposition} Let $X$ be a locally compact Hausdorff
topological group with no small subgroups.   Then it has a $B$-sequence.

\proof{{\bf (a)} For $A\subseteq X$ set $D_n(A)=\{x:x^i\in A$ for every
$i\le n\}$.   We know from 446K that there is a compact symmetric
neighbourhood $U$ of the identity $e$ such that whenever $V$ is a
neighbourhood of $e$ there are an $n_0\in\Bbb N$ and a neighbourhood $W$
of $e$ such that whenever $n\ge n_0$, $x\in D_n(U)$, $y\in D_n(U)$ and
$x^ny^n\in W$, then $xy\in D_n(V)$.   Shrinking $U$ if necessary, we may
suppose that $U$ includes no subgroup of $X$ other than $\{e\}$, so that
there is an $r\ge 1$ such that $D_{rn}(U)^n\subseteq U$ for every
$n\in\Bbb N$ (446H).

Let $V$ be a closed symmetric neighbourhood of $e$ such that
$V^{2r}\subseteq U$.   Then $D_n(V)^2\subseteq D_n(U)$ for every
$n\in\Bbb N$.   \Prf\ $D_n(V)\subseteq D_{2rn}(U)$, by 446D(b-v), so

\Centerline{$(D_n(V)^2)^n\subseteq D_{2rn}(U)^{2n}\subseteq U$,}

\noindent and $D_n(V)^2\subseteq D_n(U)$ (446D(b-v) again).\ \QeD\
Take $n_0\in\Bbb N$ and a neighbourhood $W$ of $e$ such that
whenever $n\ge n_0$, $x$, $y\in D_n(U)$ and $x^ny^n\in W^{-1}W$,
then $xy\in D_n(V)$.

\medskip

{\bf (b)} Let $M$ be so large that
$U$ can be covered by $M$ left translates of $W$.   Then for any
$n\ge n_0$, $D_n(V)D_n(V)^{-1}=D_n(V)^2$ can be covered by $M$ left
translates of $D_n(V)$.

\Prf\ Let $z_0,\ldots,z_{M-1}$ be such that
$U\subseteq\bigcup_{i\le M}z_iW$.   For each $i<M$, set
$A_i=\{x:x\in D_n(U)$, $x^n\in z_iW\}$;  if $A_i\ne\emptyset$ choose
$x_i\in A_i$;  otherwise, set $x_i=e$.

For any $y\in D_n(V)^2$, $y\in D_n(U)$, so $y^n\in U$ and there is some
$i<M$ such that $y\in A_i$.   In this
case $x_i$ also belongs to $A_i$.   Now $z_i^{-1}y^n$ and
$z_i^{-1}x_i^n$ both belong to $W$, so $x_i^{-n}y^n$ belongs to
$W^{-1}W$, and $x_i^{-1}y\in D_n(V)$, by the choice of $W$ and $n_0$.
But this means that $y\in x_iD_n(V)$.   As $y$ is arbitrary,
$D_n(V)^2\subseteq\bigcup_{i<M}x_iD_n(V)$ is covered by $M$ left
translates of $D_n(V)$.\ \Qed

\medskip

{\bf (c)} But this means that $\sequencen{D_{n+n_0}(V)}$ is a
$B$-sequence in $X$.   (It constitutes a base of neighbourhoods of $e$
by 446Gb, as usual.)
}%end of proof of 446M

\leader{*446N}{Proposition} Let $X$ be a locally compact Hausdorff
topological group with a faithful finite-dimensional representation.
Then it has a $B$-sequence.

\proof{{\bf (a)} Let $\phi:X\to GL(r,\Bbb R)$ be a faithful
finite-dimensional representation.   Identifying $M_r$ with the Banach
algebra $\eurm B=\eurm B(\BbbR^r;\BbbR^r)$, where
$\BbbR^r$ is given the Euclidean norm, we see that $GL(r,\Bbb R)$
is an open subset of $\eurm B$
(4A6H).   Note also that the operator norm $\|\,\|$
of $\eurm B$ is equivalent
%4{}A4Ia
to its `Euclidean' norm corresponding to an
identification with $\BbbR^{r^2}$, that is, writing
$\|T\|_{HS}=\sqrt{\sum_{i=1}^r\sum_{j=1}^r\tau_{ij}^2}$ if
$T=\langle\tau_{ij}\rangle_{1\le i,j\le r}$, $\|\,\,\|_{HS}$ is
equivalent to $\|\,\,\|$.    (See the inequalities in 262H.)   In
particular, all the balls $B(T,\delta)=\{S:\|S-T\|\le\delta\}$ are
closed for the Euclidean norm (4A2Lj).
If we write $\mu_L$ for Lebesgue measure on
$\eurm B$, identified with $\BbbR^{r^2}$, and set
$\gamma=\mu_LB(\tbf{0},1)$, then $0<\gamma<\infty$ (because
$B(\tbf{0},1)$ includes, and is included in, non-trivial Euclidean
balls) and $\mu_LB(T,\delta)=\delta^{r^2}\gamma$ for every
$T\in\eurm B$ and $\delta\ge 0$ (using 263A, or otherwise).

\medskip

{\bf (b)} We need to recall a basic inequality concerning
inversion in Banach algebras.   If $T\in\eurm B$ and
$\|T-I\|\le\bover12$, then $T$ is invertible and

\Centerline{$\|T^{-1}-I\|\le\Bover{\|T-I\|}{1-\|T-I\|}\le 1$}

\noindent (4A6H), so $\|T^{-1}\|\le 2$.

\medskip

{\bf (c)} Now let $V$ be a compact neighbourhood of the identity
$e$ of $X$.   Let $V_1$ be a neighbourhood of $e$ such that
$(V_1V_1^{-1})^{-1}V_1V_1^{-1}\subseteq V$.   For $\delta>0$, set
$U_{\delta}=\{x:x\in V$, $\|\phi(x)-I\|\le\delta\}$.   Then each
$U_{\delta}$ is a compact neighbourhood of $e$, because $\phi$ is
continuous.   Also

\Centerline{$\bigcap_{\delta>0}U_{\delta}
=\{x:x\in V,\,\phi(x)=I\}=\{e\}$.}

\noindent So
$\{U_{\delta}:\delta>0\}$ is a base of neighbourhoods of $e$ (4A2Gd),
and there is a $\delta_1>0$ such that $U_{\delta_1}\subseteq V_1$;  of
course we may suppose that $\delta_1\le\bover18$.

\medskip

{\bf (d)} If $\delta\le\bover12$ and
$x\in U_{\delta}U_{\delta}^{-1}$, then $\|\phi(x)-I\|\le 4\delta$.
\Prf\ Express $x$ as $yz^{-1}$ where $y$,
$z\in U_{\delta}$.   Then $\|\phi(z)-I\|\le\bover12$, so
$\|\phi(z^{-1})\|\le 2$ and

\Centerline{$\|\phi(x)-I\|
=\|(\phi(y)-\phi(z))\phi(z^{-1})\|
\le 2\|\phi(y)-\phi(z)\|\le 4\delta$.  \Qed}

\medskip

{\bf (e)} Now if $\delta\le\delta_1$,
$U_{\delta}U_{\delta}^{-1}$ can be covered by at most $m=17^{r^2}$ left
translates of $U_{\delta}$.   \Prf\Quer\ Suppose, if possible,
otherwise.   Then we can choose
$x_0,\ldots,x_m\in U_{\delta}U_{\delta}^{-1}$ such that
$x_j\notin x_iU_{\delta}$ whenever
$i<j\le m$.   If $i<j\le m$, then

\Centerline{$x_i^{-1}x_j\in
  (U_{\delta}U_{\delta}^{-1})^{-1}U_{\delta}U_{\delta}^{-1}
\subseteq (V_1V_1^{-1})^{-1}V_1V_1^{-1}
\subseteq V$,}

\noindent and $x_i^{-1}x_j\notin U_{\delta}$, so
$\|\phi(x_i^{-1}x_j)-I\|>\delta$.
Set $T_i=\phi(x_i)$ for each $i\le m$;  then

\Centerline{$\|T_i-I\|\le 4\delta\le\bover12$}

\noindent for each $i$, by (d), while

\Centerline{$\delta<\|\phi(x_i^{-1}x_j)-I\|=\|T_i^{-1}T_j-I\|
\le\|T_i^{-1}\|\|T_j-T_i\|\le 2\|T_j-T_i\|$}

\noindent whenever $i<j\le m$.   Write
$B_i=B(T_i,\bover{\delta}4)$ for each $i$;  then all the $B_i$ are
disjoint.   But also they are all included in $B(I,\bover{17\delta}4)$,
so we have

\Centerline{$(17^{r^2}+1)\bigl(\Bover{\delta}4\bigr)^{r^2}\gamma
\le\bigl(\Bover{17\delta}4\bigr)^{r^2}\gamma$,}

\noindent which is impossible.\ \Bang\Qed

\medskip

{\bf (f)} Accordingly, setting $W_n=U_{2^{-n}\delta_1}$,
$\sequencen{W_n}$ is a $B$-sequence in $X$.
}%end of proof of 446N


\leader{*446O}{Theorem} Let $X$ be a locally compact Hausdorff
topological group.   Then it has an open subgroup $Y$ which has a
compact normal subgroup $Z$ such that $Y/Z$ has no small subgroups.

\proof{{\bf (a)} Let $U$ be a compact neighbourhood of the identity $e$
of $X$.   Then there are a subgroup $Y_0$ of $X$, included in $U$, and a
neighbourhood $W_0$ of $e$ such that every subgroup of $X$ included in
$W_0$ is also included in $Y_0$.

\medskip\Prf {\bf (i)} To begin with (down to the end of (iii)) let us
suppose that $X$ is metrizable.   Let $\sequencen{V_n}$ be a
non-increasing sequence of closed
symmetric neighbourhoods of $e$ running over a base of
neighbourhoods of $e$, and such that $V_1^2\subseteq V_0\subseteq U$.
For each $n\in\Bbb N$, set
$A_n=\{x:x^i\in V_n$ for every $i\in\Bbb N\}$.

\medskip

\quad{\bf (ii)}\Quer\ Suppose, if possible, that
$\bigcup_{k\in\Bbb N}A_n^k\not\subseteq U$ for every $n\in\Bbb N$.   For
each $n\in\Bbb N$ let $k(n)\in\Bbb N$ be such that
$A_n^{k(n)}\subseteq U$,
$A_n^{k(n)+1}\not\subseteq U$.   Then $\sequencen{A_n}$ and $U$ satisfy
the conditions of 446F (because $A_m\subseteq V_n$ whenever $m\ge n$,
and $\{V_n:n\in\Bbb N\}$ is a base of neighbourhoods of $e$).   Let
$\Cal F$ be a non-principal ultrafilter on $\Bbb N$ and set
$Q=\lim_{n\to\Cal F}A_n^{k(n)}$ in the space $\Cal C$
of closed subsets of $X$ with the Fell topology.

If $W$ is any neighbourhood of $e$, there is an $n\in\Bbb N$ such that
$V_n\subseteq W$, so that $x^i\in W$ whenever $x\in\bigcup_{m\ge n}A_m$ and
$i\in\Bbb N$.   Thus (ii) of 446F is not true, and $Q$ must be a closed
subgroup of $X$ included in $U$ and meeting the boundary of $U$.

By 446C, there are an $r\in\Bbb N$ and a continuous homomorphism
$\phi:Q\to GL(r,\Bbb R)$ such that the kernel $Z$ of $\phi$ is included
in $\interior V_1$.   Let $G\subseteq X$ be an open set including $Z$
and with closure disjoint from
$(X\setminus\interior V_1)\cup\{x:x\in Q,\,\|\phi(x)-I\|\ge\bover16\}$;
such can be found because $Z$ is compact and $X$ is regular (4A2F(h-ii)).
Then $Z\subseteq G$, and any subgroup $Z'$ of $Q$ included in
$\overline{G}$ has
$\|\phi(x)^i-I\|\le\bover16$ for every $x\in Z'$ and $i\in\Bbb N$,
so that $Z'\subseteq Z$, by 4A6N.   Set $V=\overline{G}$.

Since $V\subseteq U$ and $A_n^{k(n)+1}\not\subseteq U$ for every $n$, we
can find $j(n)\le k(n)$ such that $A_n^{j(n)}\subseteq V$ and
$A_n^{j(n)+1}\not\subseteq V$ for every $n$.   Set
$Q'=\lim_{n\to\Cal F}A_n^{j(n)}$.   As before, (ii) of 446F cannot be
true of $Q'$, and $Q'$ must be a closed subgroup of $X$
meeting the boundary of $V$.   Because $e\in A_n$,
$A_n^{j(n)}\subseteq A_n^{k(n)}$ for every $n$, and $Q'\subseteq Q$,
because $\{(E,F):E\subseteq F\}$ is closed in $\Cal C$ (4A2T(e-i));  also
$Q'\subseteq V$, so $Q'\subseteq Z$.   But $Z$ does not meet the
boundary of $V$.\ \Bang

\medskip

\quad{\bf (iii)} So there is some $n\in\Bbb N$ such that
$A_n^k\subseteq U$ for every $k\in\Bbb N$.   Because $A_n^{-1}=A_n$,
$Y_0=\bigcup_{k\in\Bbb N}A_n^k$ is a subgroup of $X$.
Any subgroup of $X$ included in $V_n$ is a subset of $A_n$ so is
included in $Y_0$.   Thus we have a pair $Y_0$, $W_0=V_n$ of the kind
required, at least when $X$ is metrizable.

\medskip

\quad{\bf (iv)} Now suppose that $X$ is $\sigma$-compact.   Let $U_1$ be
a neighbourhood of $e$ such that $U_1^2\subseteq U$.   Then there is a
closed normal subgroup $X_0$ of $X$ such that $X_0\subseteq U_1$ and
$X'=X/X_0$ is metrizable (4A5S).   By (i)-(iii), there are a subgroup
$Y'_0$ of $X'$, included in the image of $U_1$ in $X'$, and a
neighbourhood $W'_0$ of the identity in $X'$ such that any subgroup of
$X'$ included in $W'_0$ must also be included in $Y'_0$.   Write
$\pi:X\to X'$ for the canonical homomorphism and consider
$Y_0=\pi^{-1}[Y'_0]$, $W_0=\pi^{-1}[W'_0]$.   Then $W_0$ is a
neighbourhood of $e$ and $Y_0$ is a subgroup of $X$ included in

\Centerline{$\pi^{-1}[\pi[U_1]]=U_1X_0\subseteq U_1^2\subseteq U$.}

\noindent And if $Z$ is any subgroup of $X$ included in $W_0$, then
$\pi[Z]\subseteq W'_0$ so $\pi[Z]\subseteq Y'_0$ and $Z\subseteq Y_0$.
Thus in this case also we have the result.

\medskip

\quad{\bf (v)} Finally, for the general case, observe that $X$ has a
$\sigma$-compact open subgroup $X_1$ (4A5El).   So we can find a
subgroup
$Y_0$ of $X_1$, included in $U\cap X_1$, and a neighbourhood $W_0$ of
the identity in $X_1$ such that any subgroup of $X_1$ included in $W_0$
is also included in $Y_0$.   But of course $Y_0$ and $W_0$ also serve
for $X$ and $U$.

This completes the proof of (a).\ \Qed

\medskip

{\bf (b)} Of course $\overline{Y_0}$ is a subgroup of $X$ (4A5Em);
being included in $U$, it is compact.
By 446C, there is a finite-dimensional representation
$\phi:\overline{Y_0}\to GL(r,\Bbb R)$, for some $r\in\Bbb N$, such that the kernel
$Z$ of $\phi$ is included in $\interior W_0$.   Let $W_1$ be a
neighbourhood of $e$ in $X$ such that $\|\phi(x)-I\|\le\bover16$ for
every $x\in W_1\cap \overline{Y_0}$, and set $W=W_1Z\cap W_0$.   Note that if
$x\in W\cap\overline{Y_0}$, there is a $z\in Z$ such that $xz\in W_1\cap \overline{Y_0}$, so that
$\|\phi(x)-I\|=\|\phi(xz)-I\|\le\bover16$.   Of course
$Z\subseteq\interior W_1Z$, so $Z\subseteq\interior W$.

If $Y'$ is a subgroup of $\overline{Y_0}$ included in $W$, then
$\|\phi(x)^i-I\|\le\bover16$ for every $i\in\Bbb N$ and $x\in Y'$, so
$Y'\subseteq Z$.   Consequently any subgroup of $X$ included in $W$ is a
subgroup of $Z$, since by the choice of $\overline{Y_0}$ and $W_0$ it is a subgroup
of $\overline{Y_0}$.

Now let $Y$ be the normalizer of $Z$ in $X$.   $Z$ is compact, so
$G=\{x:xZx^{-1}\subseteq\interior W\}$ is open (4A5Ei), and contains $e$;  but
also $G\subseteq Y$, because if $x\in G$ then $xZx^{-1}$ is a subgroup
of $X$ included in $W$, and must be included in $Z$.
Accordingly $Y=GY$ is open.

Since any subgroup of $Y$ included in $W$ is a subgroup of $Z$, we see
that any subgroup of $Y/Z$ included in the image of $W$ is the trivial
subgroup, and $Y/Z$ has no small subgroups, as required.
}%end of proof of 446O

\vleader{72pt}{*446P}{Corollary} Let $X$ be a locally compact Hausdorff
topological group.   Then it has a chain
$\langle X_{\xi}\rangle_{\xi\le\kappa}$ of closed subgroups,
where $\kappa$ is an infinite cardinal, such that

\quad(i) $X_0$ is open,

\quad (ii) $X_{\xi+1}$ is a normal subgroup of $X_{\xi}$ for every
$\xi<\kappa$,

\quad (iii) $X_{\xi}$ is compact for $\xi\ge 1$,

\quad (iv) $X_{\xi}=\bigcap_{\eta<\xi}X_{\eta}$ for non-zero limit
ordinals $\xi\le\kappa$,

\quad (v) $X_{\xi}/X_{\xi+1}$ has a $B$-sequence for every $\xi<\kappa$,

\quad (vi) $X_{\kappa}=\{e\}$, where $e$ is the identity of $X$.

\proof{{\bf (a)} By 446O, $X$ has an open subgroup $X_0$ with a compact
normal subgroup $X_1$ such that $X_0/X_1$ has no small subgroups.   By
446M, $X_0/X_1$ has a $B$-sequence.

\medskip

{\bf (b)} Let $\Phi$ be the set of finite-dimensional representations of
$X_1$;   if we distinguish the trivial homomorphisms from $X_1$ to each
$GL(r,\Bbb R)$, $\Phi$ is infinite.   Set $\kappa=\#(\Phi)$ and let
$\langle\phi_{\xi}\rangle_{1\le\xi<\kappa}$ run over $\Phi$.
For $1\le\xi\le\kappa$, set

\Centerline{$X_{\xi}=\{x:x\in X_1,\,\phi_{\eta}(x)=I$ for
$1\le\eta<\xi\}$.}

\noindent   Then $\langle X_{\xi}\rangle_{\xi\le\kappa}$ satisfies
conditions (i)-(iv).
As for (v), I have already checked the case $\xi=0$, and
if $1\le\xi<\kappa$, then $\phi_{\xi}\restr X_{\xi}$ is a
finite-dimensional representation of $X_{\xi}$ with kernel $X_{\xi+1}$,
so $X_{\xi}/X_{\xi+1}$ has a faithful finite-dimensional representation
(446Ab), and therefore has a $B$-sequence (446N).

Finally, $X_{\kappa}=\{e\}$ by 446B;  if $x\in X_1$ and $x\ne e$, there
is a $\phi\in\Phi$ such that $\phi(x)\ne\phi(e)$, so that $x\notin
X_{\kappa}$.
}%end of proof of 446P

\exercises{\leader{446X}{Basic exercises (a)}
%\spheader 446Xa
Let $X$ be a locally compact Hausdorff abelian topological group.
Show that for every element $a$ of $X$, other than the identity, there
is a two-dimensional representation $\phi$ of $X$ such that
$\phi(a)\ne I$.   \Hint{445O.}
%446B

\spheader 446Xb Let $\sequencen{X_n}$ be any sequence of groups with 
their discrete topologies, and $X$ the product topological group 
$\prod_{n\in\Bbb N}X_n$.   Show that $X$ has a $B$-sequence.   
\Hint{set $V_n=\{x:x(i)=e(i)$ for $i<n\}$.}
%446L

\leader{446Y}{Further exercises (a)}
%\spheader 446Ya
Let $X$ be the countable group of all permutations of $\Bbb N$
which are
products of an even number of transpositions.   Give $X$ its discrete
topology, so that it is a locally compact topological group.   Show that
any finite-dimensional representation of $X$ is trivial.
\Hint{$X$ is simple and has many commuting involutions.}
%446Xa 446B a nontrivial representation must be faithful.
% If  T\in\GL(\Bbb R,r)  is an involution, then  
% \BbbR^r=N_{T-I}\oplus N_{T+I}  and  T  has lots of eigenvectors.
% If  T_0,\ldots,T_n  are commuting involutions then 
% \BbbR^r  is the linear span of  \bigcap_{i\le n}\{x:T_ix=\pm x\} .
% So there cannot be a set of more than  2^r  commuting involutions.

\spheader 446Yb Let $X$ be a compact Hausdorff topological group,
$f\in C(X)$ and $\epsilon>0$.
Show that there are a finite-dimensional representation
$\phi:X\to GL(r,\Bbb R)$ and $a$, $b\in\BbbR^r$ such that
$|f(x)-\innerprod{\phi(x)(a)}{b}|\le\epsilon$ for every $x\in X$.
%446B  Stone-Weierstrass:  need to know about tensor product of
% Hilbert spaces,
%\innerprod{u\otimes u'}{v\otimes v'}
%=\innerprod{u}{v}\cdot\innerprod{u'}{v'}

\spheader 446Yc Let $\kappa$ be an infinite cardinal,
and $(\frak B_{\kappa},\bar\nu_{\kappa})$ the measure
algebra of the usual measure on $\{0,1\}^{\kappa}$.   Give the group
$\Aut_{\bar\nu_{\kappa}}(\frak B_{\kappa})$ of measure-preserving
automorphisms of $\frak B_{\kappa}$ its topology of pointwise convergence.
Let $X$ be a compact Hausdorff topological group of weight
at most $\kappa$.   Show that there is a continuous injective homomorphism
from $X$ to $\Aut_{\bar\nu_{\kappa}}(\frak B_{\kappa})$.
%446B we have lots of homomorphisms into orthogonal groups which
% have Haar probability measures

%\spheader 446Yb Improve the constant $17^{r^2}$ in part (e) of the
%proof of 446N.  \Hint{\S263.}
%%446N
}%end of exercises

\endnotes{
\Notesheader{446} The ideas above are extracted from the structure
theory for locally compact groups, as described in
{\smc Montgomery \& Zippin 55}.
(A brisker and sometimes neater, but less complete,
exposition can be found in {\smc Kaplansky 71}.)   The full theory goes
very much deeper into the analysis of groups with no small subgroups.
One of the most important ideas, hidden away in 446I and part (c) of the
proof of 446K, is that of `one-parameter subgroup';  if $X$ is a group
with no small subgroups, there are enough continuous homomorphisms from
$\Bbb R$ to $X$ not only to provide a great deal of information on the
topological group structure of $X$, but even to set up a differential
structure ({\smc Kaplansky 71}, \S II.3).
For our purposes here, however, all we need to know is that groups with
no small subgroups have `$B$-sequences' (446L-446M), which can form the
basis of a theory corresponding to Vitali's theorem and Lebesgue's
Density Theorem in $\BbbR^r$ (447C-447D below).

There are four essential elements in the argument here.   Working from
the outside, the first step is 446O:  starting from a locally compact
Hausdorff group $X$, we can find an open subgroup $X_0$ of $X$ and a
compact normal subgroup $X_1$ of $X_0$ such that $X_0/X_1$ has no small
subgroups.   This depends on a subtle argument based on the first key
lemma, the dichotomy in 446F, which in turn uses the `smoothing'
construction in 446E and a careful analysis of inequalities involving
integrals.   (Naturally enough, the translation-invariance of the Haar
integral is a leitmotiv of this investigation.)   Note the remarkable
transition in 446H.   The sets $D_n(U)$ are defined solely in terms of
powers, while the sets $D_n(U)^n$ involve products.   We are able to
obtain information about products $x_1\ldots x_n$ from information about
the powers $x_j^i$ for $i$, $j\le n$.

Next, we need to find a chain of closed subnormal subgroups of $X_1$,
decreasing to $\{e\}$, such that the quotients all have faithful
finite-dimensional representations (in this context, this means that
they are isomorphic to compact subgroups of $GL(r,\Bbb R)$).   This step
depends on the older ideas in 446B-446C, where we use the theory of
compact operators on Hilbert spaces to show that a compact group has
many representations as actions on finite-dimensional subspaces of its
$L^2$ space.   (Observe that in this section I revert to real-valued,
rather than complex-valued, functions.)   This can be thought of as a
development of the result of 445O.   If $X$
is a locally compact abelian group, its characters separate its points
(cf.\ 446Xa);  if $X$ is compact but not necessarily abelian, its
finite-dimensional representations separate its points.   (But if $X$ is
neither compact nor abelian, there are further difficulties;  see
446Ya.)

The other two necessary facts are that both groups with no small
subgroups, and groups with faithful finite-dimensional representations,
have $B$-sequences.   The latter is reasonably straightforward (446N);
any complications are due entirely to the fact that the natural measure
on $GL(r,\Bbb R)$, inherited from $\BbbR^{r^2}$, is not quite invariant
under multiplication, so we have to manipulate some inequalities.   For
groups with no small subgroups (446M) we have much more to do.   The
proof I give here depends on a second key lemma, 446K, refining the
methods of 446F;  a slightly stronger version of this result is the
basis of the analysis of one-parameter subgroups in the general theory
(compare {\smc Montgomery \& Zippin 55}, \S3.8).

}%end of notes

\discrpage

\frfilename{mt447.tex}
\versiondate{7.1.10}
\copyrightdate{1998}

\def\undphi{\underline{\phi}\vthsp}
\def\undpsi{\underline{\psi}\vthsp}
\def\undtheta{\underline{\theta}}
\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}}}

\def\chaptername{Topological groups}
\def\sectionname{Translation-invariant liftings}

\newsection{447}

I devote a section to the main theorem of
{\smc Ionescu Tulcea \& Ionescu Tulcea 67}:
a group carrying Haar measures has a
translation-invariant lifting (447J).   The argument uses an inductive
construction of the same type as that used in \S341 for the ordinary
Lifting Theorem.   It depends on the structure theory for locally
compact groups described in \S446.   On the way I describe a Vitali
theorem for certain metrizable groups (447C), with a corresponding
density theorem (447D).


\leader{447A}{Liftings and lower densities} Let $X$ be a group carrying
Haar measures, $\Sigma$ its algebra of Haar measurable sets and $\frak A$
its Haar measure algebra\cmmnt{ (442H, 443A)}.

\spheader 447Aa Recall that a {\bf lifting} of $\frak A$ is either a
Boolean homomorphism $\theta:\frak A\to\Sigma$ such that
$(\theta a)^{\ssbullet}=a$ for every $a\in\frak A$,
or a Boolean homomorphism
$\phi:\Sigma\to\Sigma$ such that $E\symmdiff\phi E$ is Haar negligible
for every $E\in\Sigma$ and $\phi E=\emptyset$ whenever $E$ is Haar
negligible\cmmnt{ (341A)}.   Such a lifting $\theta$ or $\phi$ is {\bf
\lti} if $\theta((xE)^{\ssbullet})=x(\theta E^{\ssbullet})$ or
$\phi(xE)=x(\phi E)$ for every $E\in\Sigma$ and $x\in X$.
\cmmnt{(In the notation of 443C, a lifting $\theta:\frak A\to\Sigma$
is \lti\ if $\theta(x\action_la)=x\theta a$ for every $x\in X$,
$a\in\frak A$.)}

\cmmnt{The language of 341A demanded a named measure;  I spoke there
of a lifting for a measure space $(X,\Sigma,\mu)$ or a measure $\mu$.
But (as noted in 341Lh) what the concept really depends on is a triple
$(X,\Sigma,\Cal I)$, where $\Sigma$ is an algebra of subsets of $X$ and
$\Cal I$ is an ideal of $\Sigma$.   Variations in the measure which do
not affect the algebra of measurable sets or the null ideal are
irrelevant.   So, in the present context, we can speak of a
`lifting for Haar measure' without declaring which Haar measure we are
using, nor even whether it is a left or right Haar measure.}

\spheader 447Ab Now suppose that $\Sigma_0$ is a $\sigma$-subalgebra of
$\Sigma$.   In this case, a {\bf partial lower density} on $\Sigma_0$ is
a function $\undphi:\Sigma_0\to\Sigma$ such that $\undphi E=\undphi F$
whenever $E$, $F\in\Sigma_0$ and $E\symmdiff F$ is negligible,
$E\symmdiff\undphi E$ is negligible for every $E\in\Sigma_0$,
$\undphi\emptyset=\emptyset$ and
$\undphi(E\cap F)=\undphi E\cap\undphi F$ for all $E$, $F\in\Sigma_0$.
\cmmnt{(See 341C-341D.)}  As in (a), such
a function is {\bf \lti} if $xE\in\Sigma_0$ and
$\undphi(xE)=x(\undphi E)$ for every $x\in X$ and $E\in\Sigma_0$.

\leader{447B}{Lemma} Let $X$ be a group carrying Haar measures and $Y$ a
subgroup of $X$.   Write $\Sigma_Y$ for the algebra of Haar
measurable subsets $E$ of $X$ such that $EY=E$, and suppose that
$\undphi:\Sigma_Y\to\Sigma_Y$ is a \lti\ partial lower density.   Then
$G\subseteq\undphi(GY)$ for every open set $G\subseteq X$.

\proof{ Of course $GY$ is open (4A5Ed), so belongs to $\Sigma_Y$.   Let
$a\in G$ and let $U$ be an open neighbourhood of the identity in $X$
such that $U^{-1}Ua\subseteq G$.
Then $UaY$ is a non-empty open set, therefore not negligible (442Aa),
and there is an $x\in UaY\cap\undphi(UaY)$.   Express $x$ as $uay$ where
$u\in U$ and $y\in Y$;  then $ua=xy^{-1}\in Ua\cap\undphi(UaY)$, because
$\undphi(UaY)\in\Sigma_Y$.   So

\Centerline{$a=u^{-1}ua\in u^{-1}\undphi(UaY)
=\undphi(u^{-1}UaY)\subseteq\undphi(GY)$.}

\noindent As $a$ is arbitrary, $G\subseteq\undphi(GY)$.
}%end of proof of 447B

\leader{447C}{Vitali's theorem} Let $X$ be a topological group with a
left Haar measure $\mu$, and $\sequencen{V_n}$ a $B$-sequence in
$X$\cmmnt{ (definition:  446L)}.
If $A\subseteq X$ is any set and $K_x$ is an infinite subset of $\Bbb N$
for every $x\in A$, then there is a disjoint family $\Cal V$ of sets
such that $A\setminus\bigcup\Cal V$ is negligible and every member of
$\Cal V$ is of the form $xV_n$ for some $x\in A$ and $n\in K_x$.

\proof{{\bf (a)} There is surely some $r$ such that $V_r$ is totally
bounded for the bilateral uniformity on $X$ (443H);
replacing $V_i$ by $V_r$ for $i<r$ and $K_x$ by $K_x\setminus r$
for each $x$, we may suppose that $V_0$ is totally bounded.

\medskip

{\bf (b)} Choose $\sequencen{I_n}$ inductively, as follows.   Given
$I_j\subseteq X$ for $j<n$, choose a set $I_n\subseteq A$ which is
maximal subject to the conditions

\Centerline{$n\in K_x$ for every $x\in I_n$,}

\Centerline{$xV_n\cap yV_j=\emptyset$ whenever $x\in I_n$, $j<n$ and
$y\in I_j$,}

\Centerline{$xV_n\cap yV_n=\emptyset$ whenever $x$, $y\in I_n$ are
distinct.}

\noindent On completing the induction, set
$\Cal V=\{xV_n:n\in\Bbb N,\,x\in I_n\}$;  this is a disjoint family.

\medskip

{\bf (c)} \Quer\ Suppose, if possible, that $A\setminus\bigcup\Cal V$ is
not negligible.   By 415B, the subspace measure on
$A\setminus\bigcup\Cal V$ is $\tau$-additive and has a non-empty support.   
Take any $a$ belonging to this support
and set $G=\interior(aV_0)$;  then $G$ is totally bounded and
$\delta=\mu^*(G\cap A\setminus\bigcup\Cal V)$ is non-zero.   Let $M$ be
such that every $V_nV_n^{-1}$ can be covered by $M$ left translates of
$V_n$, so that $\mu(V_nV_n^{-1})\le M\mu V_n$ for every $n$.   Set

\Centerline{$J_n=I_n\cap GV_0^{-1}$,
\quad$E_n=J_nV_n$,
\quad$\tilde E_n=E_nV_n^{-1}$}

\noindent for each $n$.

If $n\in\Bbb N$ and $x\in J_n$, then $xV_n\subseteq GV_0^{-1}V_0$.
Accordingly (because $\family{x}{J_n}{xV_n}$ is disjoint)

\Centerline{$\#(J_n)\mu V_n=\sum_{x\in J_n}\mu(xV_n)
\le\mu(GV_0^{-1}V_0)<\infty$}

\noindent because $GV_0^{-1}V_0$ is totally bounded (4A5Ob).   So $J_n$ is
finite and $E_n$ is closed.
Note that if $x\in I_n$ and $G\cap xV_n\ne\emptyset$, then
$x\in GV_n^{-1}\subseteq GV_0^{-1}$ and $x\in J_n$;  so $G\cap\bigcup\Cal V=G\cap\bigcup_{n\in\Bbb N}E_n$.
Also, $\sequencen{E_n}$ is a disjoint sequence of subsets of
$GV_0^{-1}V_0$;  accordingly $\sum_{n=0}^{\infty}\mu E_n$ is finite,
and there is an $m\in\Bbb N$ such that
$M\sum_{n=m}^{\infty}\mu E_n<\delta$.

Observe next that, for any $x\in X$ and $n\in\Bbb N$,

\Centerline{$\mu(xV_nV_n^{-1})=\mu(V_nV_n^{-1})\le M\mu
V_n=M\mu(xV_n)$.}

\noindent So

\Centerline{$\mu\tilde E_n\le\sum_{x\in J_n}\mu(xV_nV_n^{-1})
\le M\sum_{x\in J_n}\mu(xV_n)=M\mu E_n$}

\noindent for each $n$, and $\mu(\bigcup_{n\ge m})\tilde E_n<\delta$.

This means that $\bigcup_{n\ge m}\tilde E_n$ cannot include $A\cap
G\setminus\bigcup\Cal V$, and there is a $z$ belonging to

$$\eqalign{A\cap G\setminus(\bigcup\Cal V\cup\bigcup_{n\ge m}\tilde E_n)
&=A\cap G\setminus(\bigcup_{n\in\Bbb N}E_n\cup\bigcup_{n\ge m}\tilde
E_n)\cr
&=A\cap G\setminus(\bigcup_{n<m}E_n\cup\bigcup_{n\ge m}\tilde
E_n).\cr}$$

\noindent Now there must be a first $k\ge m$ such that $k\in K_z$ and
$zV_k\subseteq G\setminus\bigcup_{n<m}E_n$.   (This is where we use the
hypothesis that $\{V_n:n\in\Bbb N\}$ is a base of neighbourhoods of the
identity.)   Since $z\in G\setminus J_k$, $z\notin I_k$, and there are
$j\le k$, $x\in I_j$ such that $zV_k\cap xV_j\ne\emptyset$.   In this
case, $x\in GV_0^{-1}$, so $x\in J_j$.   Accordingly

\Centerline{$z\in xV_jV_k^{-1}\subseteq xV_jV_j^{-1}
\subseteq\tilde E_j$}

\noindent and $j<m$;  but this means that
$zV_k\cap xV_j\subseteq zV_k\cap E_j$ must be empty, which is
impossible.\ \Bang

\medskip

{\bf (d)} Thus $\mu(A\setminus\bigcup\Cal V)=0$, and $\Cal V$ is an
appropriate family.
}%end of proof of 447C

\leader{447D}{Theorem} Let $X$ be a topological group with a left Haar
measure $\mu$, and $\sequencen{V_n}$ a $B$-sequence in $X$.
Then for any Haar measurable set $E\subseteq X$,

$$\lim_{n\to\infty}\bover{\mu(E\cap xV_n)}{\mu V_n}=\chi E(x)$$

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

\proof{{\bf (a)} Let $\alpha<1$, and set

$$A=\{x:x\in E,\,\liminf_{n\to\infty}\bover{\mu(E\cap xV_n)}{\mu
V_n}<\alpha\}.$$

\noindent\Quer\ Suppose, if possible, that $A$ is not negligible.   Then
there is an open set $G$ of finite measure such that $\mu^*(G\cap
A)=\gamma>0$.   Let $\delta>0$ be such that
$\gamma>\alpha(\gamma+\delta)+\delta$.
Take a Borel set $F$ which is a measurable envelope of $G\cap A$ and a
closed set $F_1\subseteq G\setminus F$ such that $\mu
F_1\ge\mu(G\setminus F)-\delta$.
Writing $H=G\setminus F_1$, we see that $H\cap A=G\cap A$ and

\Centerline{$\mu H\le\mu^*(H\cap A)+\delta=\gamma+\delta$.}

For each $x\in H\cap A$, set

\Centerline{$K_x=\{n:xV_n\subseteq H,\,\mu(E\cap xV_n)\le\alpha\mu
V_n\}$.}

\noindent Then $K_x$ is infinite.   By Vitali's theorem in the form
447C, there is a disjoint family $\Cal V\subseteq\{xV_n:x\in H\cap
A,\,n\in K_x\}$ such that $(H\cap A)\setminus\bigcup\Cal V$ is
negligible.   Since every member of $\Cal V$ has non-zero measure, while
$\mu H$ is finite, $\Cal V$ is countable.   Now $\mu(\bigcup\Cal
V)\ge\mu^*(H\cap A)$, so $\mu(H\setminus\bigcup\Cal V)\le\delta$;  also,
because $\mu(E\cap V)\le\alpha\mu V$ for every $V\in\Cal V$, and $\Cal
V$ is disjoint,

\Centerline{$\mu(E\cap\bigcup\Cal V)
\le\alpha\mu(\bigcup\Cal V)\le\alpha\mu H$}

\noindent and

\Centerline{$\gamma=\mu^*(A\cap H)\le\mu(E\cap H)
\le\mu(E\cap\bigcup\Cal V)+\delta\le\alpha\mu H+\delta
\le\alpha(\gamma+\delta)+\delta$,}

\noindent which is impossible, by the choice of $\delta$.\ \Bang

\medskip

{\bf (b)} As $\alpha$ is arbitrary,

$$\liminf_{n\to\infty}\bover{\mu(E\cap xV_n)}{\mu V_n}=1$$

\noindent for almost every $x\in E$.   Similarly,

$$\liminf_{n\to\infty}\bover{\mu(xV_n\setminus E)}{\mu V_n}=1,
\quad\limsup_{n\to\infty}\bover{\mu(E\cap xV_n)}{\mu V_n}=0
$$

\noindent for almost every $x\in X\setminus E$, so

$$\lim_{n\to\infty}\bover{\mu(E\cap xV_n)}{\mu V_n}=\chi E(x)$$

\noindent for almost every $x\in X$.
}%end of proof of 447D

\leader{447E}{}\cmmnt{ We need to recall some results from
443P-443R.}   %443P 443Q 443R
Let $X$ be a locally compact Hausdorff topological group,
and $Y$ a closed subgroup of $X$ such that the modular function of $Y$
is the restriction to $Y$ of the modular function of $X$.   Let $\mu$ be
a left Haar measure on $X$ and $\mu_Y$ a left Haar measure on $Y$.

\spheader 447Ea Writing $C_k(X)$ for the space of continuous real-valued
functions on $X$ with compact support, and $X/Y$ for the set of left
cosets of $Y$ in $X$ with the quotient topology, we have a linear
operator $T:C_k(X)\to C_k(X/Y)$ defined by writing
$(Tf)(x^{\ssbullet})=\int_Yf(xy)\mu_Y(dy)$ whenever $x\in X$ and
$f\in C_k(X)$\cmmnt{ (443P)};  moreover,
$T[C_k(X)^+]=C_k(X/Y)^+$\cmmnt{ (443Pa)}, and we have an invariant
Radon measure
$\lambda$ on $X/Y$ such that $\int Tfd\lambda=\int fd\mu$ for every
$f\in C_k(X)$\cmmnt{ (see part (b) of the proof of 443R)}.
\cmmnt{Turning this
structure round, we see from 443Qb that} $\mu$,
$\mu_Y$ and $\lambda$ here are related
in exactly the same way as $\mu$, $\nu$ and $\lambda$ in 443Q.
If $Y$ is a normal subgroup of $X$, so that $X/Y$ is the quotient group,
$\lambda$ is a left Haar measure.   If $Y$ is compact and $\mu_Y$ is the
Haar probability measure on $Y$, then $\lambda$ is the image measure
$\mu\pi^{-1}$, where $\pi(x)=x^{\ssbullet}=xY$ for every
$x\in X$\cmmnt{ (443Qd)}.

\spheader 447Eb If $E\subseteq X$ and $EY=Y$, then $E$ is Haar
measurable iff $\tilde E=\{x^{\ssbullet}:x\in E\}$ belongs to the domain
of $\lambda$, and $E$ is Haar negligible iff $\tilde E$ is
$\lambda$-negligible\cmmnt{ (443Qc)}.
%If $E\subseteq X$ is Haar measurable, and $\{y:xy\in E\}$ is
%$\mu_Y$-negligible for $\mu$-almost every $x\in X$, then $E$ is
%$\mu$-negligible (??).

\spheader 447Ec Now suppose that $X$ is $\sigma$-compact.   Then for any
Haar measurable $E\subseteq X$, $\mu E=\int g\,d\lambda$ in
$[0,\infty]$, where $g(x^{\ssbullet})=\mu_Y(Y\cap x^{-1}E)$ is defined for
almost every $x\in X$\cmmnt{ (443Qe)}.   \cmmnt{In particular, }$E$
is Haar negligible
iff $\mu_Y(Y\cap x^{-1}E)=0$ for almost every $x\in X$.

\spheader 447Ed Again suppose that $X$ is $\sigma$-compact.   Then we
can extend the operator $T$ of part (a) to an operator from
$\eusm L^1(\mu)$ to $\eusm L^1(\lambda)$ by writing
$(Tf)(x^{\ssbullet})=\int f(xy)\mu_Y(dy)$ whenever $f\in\eusm L^1(\mu)$,
$x\in X$ and the integral is defined, and $\int Tfd\lambda=\int fd\mu$ for every
$f\in\eusm L^1(\mu)$\cmmnt{ (443Qe)}.   If $f\in\eusm L^1(\mu)$, and
we set $f_x(y)=f(xy)$
for all those $x\in X$, $y\in Y$ for which $xy\in\dom f$, then
$Q=\{x:f_x\in\eusm L^1(\mu_Y)\}$ is $\mu$-conegligible, and
$x\mapsto f_x^{\ssbullet}:Q\to L^1(\mu_Y)$ is almost
continuous\cmmnt{ (443Qf)}.

\spheader 447Ee If $X$ is $\sigma$-compact, $Y$ is compact and $\mu_Y$
is the Haar probability measure on $Y$\cmmnt{, so that $\lambda$ is
the image
measure $\mu\pi^{-1}$}, then\cmmnt{ we can apply 235G to
the formula in (d) to see that}

\Centerline{$\iint f(xy)\mu_Y(dy)\mu(dx)=\int(Tf)(x^{\ssbullet})\mu(dx)
=\int Tf\,d\lambda=\int fd\mu$}

\noindent for every\cmmnt{ $\mu$-integrable function $f$, and
therefore (because $\mu$ is $\sigma$-finite) for every} function $f$
such that
$\int fd\mu$ is defined in $[-\infty,\infty]$.   In particular,
$\mu E=\int\nu(Y\cap x^{-1}E)\mu(dx)$ for every Haar measurable set $E\subseteq X$.

\leader{447F}{Lemma} Let $X$ be a $\sigma$-compact locally compact
Hausdorff topological group and $Y$ a closed subgroup of $X$ such that
the modular function of $Y$ is the restriction to $Y$ of the modular
function of $X$.   Let $Z$ be a compact normal subgroup of $Y$ such that
the quotient group $Y/Z$ has a $B$-sequence.   Let $\Sigma_Y$ be the
$\sigma$-algebra of those Haar measurable subsets $E$ of $X$ such that
$EY=E$, and $\Sigma_Z$ the algebra of Haar measurable sets $E\subseteq X$
such that $EZ=E$.   Let $\undphi:\Sigma_Y\to\Sigma_Y$ be a \lti\
partial lower density.   Then there is a \lti\ partial lower density
$\undpsi:\Sigma_Z\to\Sigma_Z$ extending $\undphi$.

\proof{{\bf (a)} Let $\mu$ be a left Haar measure on $X$, $\mu_Y$ a left
Haar measure on $Y$ and $\mu_Z$ the Haar probability measure on $Z$;  then there
is a left Haar measure $\nu$ on $Y/Z$ such that
$\int g(y)\mu_Y(dy)=\int (Tg)(u)\nu(du)$ for every $g\in C_k(Y)$, where
$(Tg)(y^{\ssbullet})=\int g(yz)\mu_Z(dz)$ for every $y\in Y$ (447Ea).
We are supposing that
$Y/Z$ has a $B$-sequence $\sequencen{V_n}$.   It follows that there is
a sequence $\sequencen{h_n}$ in $C_k(Y)^+$ such that
(i) $\int h_n(y)\mu_Y(dy)=1$ for every $n$
(ii) whenever $F\subseteq Y$ is Haar
measurable (regarded as a subset of $Y$, that is), and $FZ=Z$, then

\Centerline{$\lim_{n\to\infty}\int\chi F(by)h_n(y)\mu_Y(dy)
=\chi F(b)$}

\noindent for $\mu_Y$-almost every $b\in Y$.

\Prf\ Since any subsequence of $\sequencen{V_n}$ is a $B$-sequence, and $Y/Z$ is locally compact, we
may suppose that every $V_n$ is compact.   For each
$n\in\Bbb N$, choose a non-negative $h'_n\in C_k(Y/Z)$ such that

\Centerline{$\int h'_nd\nu=1$,
\quad$\int|h'_n-\Bover1{\nu V_n}\chi V_n|d\nu\le 2^{-n}$.}

\noindent (This is possible by 416I, or otherwise.)   Let
$h_n\in C_k(Y)^+$ be such that $Th_n=h'_n$ (447Ea again);  then
$\int h_nd\mu_Y=\int h'_nd\nu=1$.   Now
if $F\subseteq Y$ is Haar measurable and $FZ=Z$, there is a Haar
measurable $\tilde F\subseteq Y/Z$ such that
$F=\{y:y^{\ssbullet}\in\tilde F\}$ (447Eb).   Take $b\in Y$ and
$n\in\Bbb N$.   Because

\Centerline{$\int\chi F(byz)h_n(yz)\mu_Z(dz)
=\int\chi\tilde F(b^{\ssbullet}y^{\ssbullet})h_n(yz)\mu_Z(dz)
=\chi\tilde F(b^{\ssbullet}y^{\ssbullet})h'_n(y^{\ssbullet})$}

\noindent for every $y\in Y$,

$$\eqalignno{\int\chi F(by)h_n(y)\mu_Y(dy)
&=\iint\chi F(byz)h_n(yz)\mu_Z(dz)\mu_Y(dy)\cr
\displaycause{447Ee}
&=\int\chi\tilde F(b^{\ssbullet}y^{\ssbullet})h'_n(y^{\ssbullet})\mu_Y(dy)
=\int\chi\tilde F(b^{\ssbullet}u)h'_n(u)\nu(du)\cr}$$

\noindent because $y\mapsto y^{\ssbullet}$ is \imp\ for $\mu_Y$ and $\nu$ (447Ee).   So

$$\eqalign{|\chi F(b)-\int\chi F(by)h_n(y)\mu_Y(dy)|
&=|\chi\tilde F(b^{\ssbullet})-\int\chi\tilde
F(b^{\ssbullet}u)h'_n(u)\nu(du)|\cr
&\le|\chi\tilde F(b^{\ssbullet})-\Bover1{\nu V_n}\int\chi\tilde
F(b^{\ssbullet}u)\chi V_n(u)\nu(du)|\cr
&\qquad\qquad\qquad\qquad
  +\int|h'_n(u)-\Bover1{\nu V_n}\chi V_n(u)|\nu(du)\cr
&\le|\chi\tilde F(b^{\ssbullet})
  -\Bover{\nu(\tilde F\cap b^{\ssbullet}V_n)}{\nu V_n}|
  +2^{-n}.\cr}$$

\noindent Since

\Centerline{$\{v:v\in Y/Z,\,\lim_{n\to\infty}\Bover{\nu(\tilde F\cap vV_n)}{\nu V_n}\ne\chi\tilde F(v)\}$}

\noindent is $\nu$-negligible (447D), its inverse image in $Y$ is
$\mu_Y$-negligible, so
$\chi F(b)=
\discrversionA{\break}{}
\lim_{n\to\infty}\int\chi F(by)h_n(y)\mu_Y(dy)$
for almost every $b$, as claimed.\ \Qed

\medskip

{\bf (b)} We find now that if $E\in\Sigma_Z$, then
$\lim_{n\to\infty}\int\chi E(xy)h_n(y)\mu_Y(dy)=\chi E(x)$ for
$\mu$-almost every $x\in X$.   \Prf\ Set $E_x=Y\cap x^{-1}E$ for
$x\in X$.   Because $X$ is $\sigma$-compact, we can express $E$ as the union of
a non-decreasing sequence $\sequence{k}{F^{(k)}}$ where each $F^{(k)}$
is Haar measurable and relatively compact;  set
$F_x^{(k)}=Y\cap x^{-1}F^{(k)}$ for each $x$.   In this case, for any $k\in\Bbb N$,
$Q_k=\{x:F^{(k)}_x\in\dom(\mu_Y)\}$ is conegligible, and
$x\mapsto(\chi F^{(k)}_x)^{\ssbullet}:Q_k\to L^1(\mu_Y)$ is almost
continuous (447Ed), so that
$x\mapsto\int\chi F^{(k)}_x(y)h_n(y)\mu_Y(dy):Q_k\to[0,1]$ is
measurable, for each
$n$.   But this means that, setting $Q=\bigcap_{k\in\Bbb N}Q_k$,

\Centerline{$x\mapsto\int\chi E_x(y)h_n(y)\mu_Y(dy)
=\lim_{k\to\infty}\int\chi F^{(k)}_x(y)h_n(y)\mu_Y(dy):Q\mapsto[0,1]$}

\noindent is measurable, for every $n\in\Bbb N$.   Note that if
$x\in X$ and $y\in Y$
then $F^{(k)}_{xy}=y^{-1}F^{(k)}_x$, so that $Q_kY=Q_k$ for every
$k\in\Bbb N$, and $QY=Q$.

Now consider $\tilde Q=\{x:x\in Q,\,
\lim_{n\to\infty}\int\chi E(xy)h_n(y)\mu_Y(dy)=\chi E(x)\}$.   This is a Haar measurable subset of $X$.   If $a\in Q$, then

\Centerline{$Y\cap a^{-1}\tilde Q
=\{y:y\in Y$, $\lim_{n\to\infty}\chi E(ays)h_n(s)\mu_Y(ds)=\chi E(ay)\}$}

\noindent is $\mu_Y$-conegligible, by the choice of the $h_n$ in (a)
above.   Because $Q$ is $\mu$-conegligible, $Q\setminus\tilde Q$ is $\mu$-negligible
(447Ec) and $\tilde Q$ is conegligible, as required.\ \Qed

\medskip

{\bf (c)} We are now ready for the formulae at the centre of this proof.
For any Haar measurable set $E\subseteq X$, $n\in\Bbb N$ and $\gamma<1$,
set

$$\eqalign{\psi_{n\gamma}(E)
&=\bigcup\{G\cap\undphi F:G\subseteq X\text{ is open, }F\in\Sigma_Y,\cr
&\qquad\qquad\int\chi E(xy)h_n(y)\mu_Y(dy)
  \text{ is defined and at least }\gamma
  \text{ for every }x\in G\cap F\},\cr}$$

\Centerline{$\undpsi E=\bigcap_{\gamma<1}\bigcup_{n\in\Bbb N}
  \bigcap_{m\ge n}\psi_{m\gamma}(E)$.}

\noindent The rest of the proof is devoted to checking that
$\undpsi\restr\Sigma_Z$ is a \lti\ partial lower density extending
$\undphi$.

\medskip

{\bf (d)} I had better make one remark straight away.   If
$G\subseteq X$ is open, then $G\subseteq\undphi(GY)$ (447B).
It follows that if $G\subseteq X$ is open, $F\in\Sigma_Y$ and
$G\cap\undphi F\ne\emptyset$, then $G\cap F\ne\emptyset$.   \Prf\ If
$a\in G\cap\undphi F$, then $a\in\undphi(GY)\cap\undphi F=\undphi(GY\cap
F)$, so $GY\cap F\ne\emptyset$, that is, $G\cap F=G\cap FY^{-1}$ is
non-empty.\ \QeD\  I mention this now because we need to know that the
condition

\Centerline{$\int\chi E(xy)h_n(y)\mu_Y(dy)$ is defined and at least $\gamma$ for every $x\in G\cap F$}

\noindent in the definition of $\psi_{n\gamma}(E)$ is never vacuously
satisfied if $G\cap\undphi F\ne\emptyset$.   In particular,
$\psi_{n\gamma}(\emptyset)=\emptyset$ whenever $n\in\Bbb N$ and
$0<\gamma<1$, so that $\undpsi\emptyset=\emptyset$.

\medskip

{\bf (e)} If $E\subseteq X$ is Haar measurable, $n\in\Bbb N$ and
$\epsilon>0$, then for almost every $a\in X$ there are an open set
$G\subseteq X$ and an $F\in\Sigma_Y$ such that $a\in G\cap\undphi F$ and
$\int|\chi E(xy)-\chi E(ay)|h_n(y)\mu_Y(dy)\le\epsilon$ whenever
$x\in G\cap F$.   \Prf\ Let
$\tilde\lambda$ be the invariant Radon measure on $X/Y$ derived from $\mu$
and $\mu_Y$ as in 447Ea.
Take $\delta>0$ such that
$\delta(1+2\|h_n\|_{\infty})\le\epsilon$.   Because $X$ is
$\sigma$-compact, locally compact and Hausdorff, therefore Lindel\"of
and regular (4A2Hd, 3A3Bb), there is a sequence
$\sequence{r}{f_r}$ of continuous functions from $X$ to $[0,1]$ such
that $\int|\chi E(x)-f_r(x)|\mu(dx)\le 2^{-r}$ for every $r$
(415Pb).   Set $g_r=|\chi E-f_r|$ for each $r$.
For $f\in\eusm L^1(\mu)$, define $\tilde Tf\in\eusm L^1(\tilde\lambda)$ by writing $(\tilde Tf)(x^{\ssbullet})=\int f(xy)\mu_Y(dy)$ whenever this is defined (447Ed);
we have $\int\tilde Tfd\tilde\lambda=\int fd\mu$.

Set $Q=\{x:Y\cap x^{-1}E\in\dom\mu_Y\}$, so that
$Q\in\Sigma_Y$ is conegligible (447Ec).   For each $r$, set
$\tilde F_r=\{u:u\in X/Y$, $\tilde Tg_r(u)$ is defined and at least
$\delta\}$;   then

\Centerline{$\tilde\lambda\tilde F_r
\le\Bover1{\delta}\int Tg_rd\tilde\lambda
=\Bover1{\delta}\int g_rd\mu\le 2^{-r}/\delta$.}

\noindent So $\bigcap_{r\in\Bbb N}\tilde F_r$ is $\tilde\lambda$-negligible.
Set

$$\eqalign{F_r
&=\{x:x\in X,\,x^{\ssbullet}\in\tilde F_r\}
=\{x:\tilde Tg_r(x^{\ssbullet})\ge\delta\}\cr
&=\{x:\int g_r(xy)\mu_Y(dy)\ge\delta\}
=\{x:\int|\chi E(xy)-f_r(xy)|\mu_Y(dy)\ge\delta\}\cr}$$

\noindent for each $r$;  then $F_rY=F_r$, $F_r$ is Haar measurable (447Eb) and
$\bigcap_{r\in\Bbb N}F_r$ is $\mu$-negligible (also by 447Eb).
Since $(X\setminus F_r)\symmdiff\undphi(X\setminus F_r)$ is negligible for each $r$,
$Q_1=Q\cap\bigcup_{r\in\Bbb N}\undphi(X\setminus F_r)\setminus F_r$ is
conegligible.   Note that $Q_1Y=Q_1$.

Suppose that $a\in Q_1$.   Then there is an $r\in\Bbb N$ such that

\Centerline{$a\in Q_1\cap\undphi(X\setminus F_r)\setminus F_r
=(Q_1\setminus F_r)\cap\undphi(Q_1\setminus F_r)$.}

\noindent Set $F=Q_1\setminus F_r\in\Sigma_Y$.    Consider the function
$x\mapsto\int f_r(xy)h_n(y)\mu_Y(dy)$.   We chose $h_n$ with compact
support $L\subseteq Y$ say.  If $V$ is a compact neighbourhood of $a$ in
$X$, then $f_r$ is uniformly continuous on $VL$ for the right
uniformity on
$X$ (4A5Ha, 4A2Jf).   There is therefore an open neighbourhood $U$ of
the identity of
$X$ such that $|f_r(x')-f_r(x)|\le\delta$ whenever $x$, $x'\in VL$ and
$x'x^{-1}\in U$;  of course we may suppose that $G=Ua$ is a subset of
$V$.

Take any $x\in G\cap F$.   Then if $y\in L$, we have $ay$, $xy$ both in
$VL$, while $xy(ay)^{-1}\in U$, so that $|f_r(xy)-f_r(ay)|\le\delta$.
Accordingly $|f_r(xy)-f_r(ay)|h_n(y)\le\delta h_n(y)$ for every $y\in
Y$, and

\Centerline{$\int|f_r(ay)-f_r(xy)|h_n(y)\mu_Y(dy)
\le\delta$.}

\noindent At the same time, because both $x$ and $a$ belong to
$F=Q_1\setminus F_r$,

\Centerline{$\int|\chi E(ay)-f_r(ay)|h_n(y)\mu_Y(dy)
\le\delta\|h_n\|_{\infty}$,}

\Centerline{$\int|\chi E(xy)-f_r(xy)|h_n(y)\mu_Y(dy)
\le\delta\|h_n\|_{\infty}$.}

\noindent Putting these together,

\Centerline{$\int|\chi E(ay)-\chi E(xy)|h_n(y)\mu_Y(dy)
\le\delta(1+2\|h_n\|_{\infty})\le\epsilon$.}

\noindent Thus $G$ and $F$ witness that $a$ has the property required;
as $a$ is any member of the conegligible set $Q_1$, we have the
result.\ \Qed

\medskip

{\bf (f)} If $E\in\Sigma_Z$ then
$E\symmdiff\undpsi E$ is negligible.   \Prf\ By (e), applied in turn to
every $n$ and every $\epsilon$ of the form $2^{-i}$, there is a
conegligible set $Q_1\subseteq X$ such that whenever $a\in Q_1$,
$n\in\Bbb N$ and $\epsilon>0$ there are an open set $G$ containing $a$
and an $F\in\Sigma_Y$ such that $a\in\undphi F$ and
$\int|\chi E(xy)-\chi E(ay)|h_n(y)\mu_Y(dy)\le\epsilon$ for every
$x\in G\cap F$.   By (b), there is a conegligible set $Q_2\subseteq X$
such that $\lim_{n\to\infty}\int\chi E(ay)h_n(y)\mu_Y(dy)=\chi E(a)$ for
every $a\in Q_2$.

Suppose that $a\in Q_1\cap Q_2\cap E$.   Let $\gamma<1$;  set
$\epsilon=\bover12(1-\gamma)$.   Because $a\in Q_2$, there is an
$n\in\Bbb N$ such that
$\int\chi E(ay)h_m(y)\mu_Y(dy)\ge 1-\epsilon$ for every $m\ge n$.   Take
any $m\ge n$.   Because $a\in Q_1$, there are an open set $G$ and an
$F\in\Sigma_Y$ such that $a\in G\cap\undphi F$ and
$\int|\chi E(ay)-\chi E(xy)|h_m(y)\mu_Y(dy)\le\epsilon$ whenever
$x\in G\cap F$.   But now

\Centerline{$\int\chi E(xy)h_m(y)\mu_Y(dy)\ge 1-2\epsilon=\gamma$}

\noindent for every $x\in G\cap F$, so $a\in\psi_{m\gamma}(E)$.   This
is true for every $m\ge n$;  as $\gamma$ is arbitrary, $a\in\undpsi E$.
As $a$ is arbitrary, $Q_1\cap Q_2\cap E\subseteq\undpsi E$.

Now suppose that $a\in Q_1\cap Q_2\cap\undpsi E$.   Then there is an
$n\in\Bbb N$ such that $a\in\psi_{m,3/4}(E)$ for every $m\ge n$.   There
is an $m\ge n$ such that $|\int\chi E(ay)h_m(y)\mu_Y(dy)-\chi
E(a)|\le\bover14$.   There are an open set $G_1$ and an $F_1\in\Sigma_Y$
such that $a\in G_1\cap\undphi F_1$ and
$\int\chi E(xy)h_m(y)\mu_Y(dy)\ge\bover34$ for every $x\in G_1\cap F_1$.   There are also an open set $G_2$ and an $F_2\in\Sigma_Y$ such that
$a\in G_2\cap\undphi F_2$ and
$\int|\chi E(ay)-\chi E(xy)|h_m(y)\mu_Y(dy)\le\bover14$ for every
$x\in G_2\cap F_2$.   Set
$G=G_1\cap G_2$, $F=F_1\cap F_2$;  then $a\in G\cap\undphi F$, so $G\cap
F$ is not empty ((d) above).   Take $x\in G\cap F$.   Then

\Centerline{$\int\chi E(xy)h_m(y)\mu_Y(dy)\ge\Bover34$,}

\Centerline{$\int|\chi E(ay)-\chi E(xy)|h_m(y)\mu_Y(dy)
\le\Bover14$,}

\Centerline{$|\int\chi E(ay)h_m(y)\mu_Y(dy)-\chi E(a)|
\le\Bover14$,}

\noindent so $\chi E(a)\ge\bover14$ and $a\in E$.   This shows that
$Q_1\cap Q_2\cap\undpsi E\subseteq E$.

Accordingly $E\symmdiff\undpsi E\subseteq X\setminus(Q_1\cap Q_2)$ is
negligible, as required.\ \QeD\

In particular, $\undpsi E$ is Haar measurable for every $E\in\Sigma_Z$.

\medskip

{\bf (g)} If $E\in\Sigma_Z$ then $\undpsi E\in\Sigma_Z$.   \Prf\ We have just seen that $\undpsi E$ is Haar measurable.   Take
$z\in Z$, $n\in\Bbb N$, $\gamma<1$ and $a\in\psi_{n\gamma}(E)$.   Then
there are an open $G\subseteq X$ and an $F\in\Sigma_Y$ such that $a\in
G\cap\undphi F$ and $\int\chi E(xy)h_n(y)\mu_Y(dy)\ge\gamma$ for
every $x\in G\cap F$.   Because $F$ and $\undphi F$ belong to
$\Sigma_Y$, $az\in\undphi F$.   Of course $Gz$ is an open set containing
$az$.   If $x\in Gz\cap F$, then $xz^{-1}\in G\cap F$ and
$\int\chi E(xz^{-1}y)h_n(y)\mu_Y(dy)\ge\gamma$.   But

\Centerline{$\chi E(xz^{-1}y)=\chi E(xy\cdot y^{-1}z^{-1}y)=\chi E(xy)$}

\noindent for every $y\in Y$, because $Z\normalsubgroup Y$ (so
$z'=y^{-1}z^{-1}y\in Z$) and we are supposing that $E\in\Sigma_Z$ (so
$xyz'\in E$ iff $xy\in E$).   So

\Centerline{$\int\chi E(xy)h_n(y)\mu_Y(dy)
=\int\chi E(xz^{-1}y)h_n(y)\mu_Y(dy)\ge\gamma$.}

\noindent As $x$ is arbitrary, $Gz$ and $F$ witness that
$az\in\psi_{n\gamma}E$.

This shows that, for any $n$ and $\gamma$, $az\in\psi_{n\gamma}(E)$
whenever $a\in\psi_{n\gamma}(E)$ and $z\in Z$.   It follows at once that
$az\in\undpsi E$ whenever $a\in\undpsi E$ and $z\in Z$, as claimed.\
\Qed

\medskip

{\bf (h)} For any Haar measurable $E\subseteq X$ and $c\in X$,
$\undpsi(cE)=c\undpsi E$.   \Prf\ Suppose that $n\in\Bbb N$, $\gamma<1$
and $a\in\psi_{n\gamma}(E)$.   Then there are an open set $G\subseteq X$
and an $F\in\Sigma_Y$ such that $a\in G\cap\undphi F$ and
$\int\chi E(xy)h_n(y)\mu_Y(dy)\ge\gamma$ for every $x\in G\cap F$.   Now
$cG$ is an open set containing $ca$, $cF\in\Sigma_Y$ and
$\undphi(cF)=c\undphi F$ contains $ca$, and if $x\in cG\cap cF$ we have

\Centerline{$\int\chi(cE)(xy)h_n(xy)\mu_Y(dy)
=\int\chi E(c^{-1}xy)h_n(xy)\mu_Y(dy)\ge\gamma$}

\noindent because $c^{-1}x\in G\cap F$.   But this means that $cG$, $cF$
witness that $ca\in\psi_{n\gamma}(cE)$.   Since $a$ is arbitrary,
$c\psi_{n\gamma}(E)\subseteq\psi_{n\gamma}(cE)$;  as $n$ and $\gamma$
are arbitrary, $c\undpsi E\subseteq\undpsi(cE)$.   Similarly, of course,
$c^{-1}\undpsi(cE)\subseteq\undpsi E$, so in fact
$\undpsi(cE)=c\undpsi E$, as claimed.\ \Qed

\medskip

{\bf (i)} If $E_1$, $E_2\subseteq X$ are Haar measurable and
$E_1\setminus E_2$ is $\mu$-negligible,
$\undpsi E_1\subseteq\undpsi E_2$.   \Prf\ Take $n\in\Bbb N$, $\gamma<1$
and $a\in\psi_{n\gamma}(E_1)$.   Then there are an open set
$G\subseteq X$
and an $F\in\Sigma_Y$ such that $a\in G\cap\undphi F$ and
$\int\chi E_1(xy)h_n(y)\mu_Y(dy)\ge\gamma$ for every $x\in G\cap F$.
Let $Q$ be the set of those $x\in X$ such that
$\mu_Y$ measures $Y\cap x^{-1}(E_1\setminus E_2)$ and
$\int\chi(E_1\setminus E_2)(xy)\mu_Y(dy)=0$;  then $QY=Q$ is
conegligible (447Ec).   Now $Q\cap F\in\Sigma_Y$, and
$\undphi(Q\cap F)=\undphi F$ contains $a$.   But if $x\in G\cap Q\cap F$, $\chi E_1(xy)\le\chi E_2(xy)$ for $\mu_Y$-almost every $y$.   So

\Centerline{$\int\chi E_2(xy)h_n(y)\mu_Y(dy)
\ge\int\chi E_1(xy)h_n(y)\mu_Y(dy)\ge\gamma$.}

\noindent Thus $G$ and $Q\cap F$ witness that $a\in\psi_{n\gamma}(E_2)$.
As $a$ is arbitrary, $\psi_{n\gamma}(E_1)\subseteq\psi_{n\gamma}(E_2)$;
as $n$ and $\gamma$ are arbitrary, $\undpsi E_1\subseteq\undpsi E_2$.\
\Qed

In particular, (i) $\undpsi E_1=\undpsi E_2$ whenever $E_1\symmdiff E_2$
is negligible (ii) $\undpsi E_1\subseteq\undpsi E_2$ whenever
$E_1\subseteq E_2$.

\medskip

{\bf (j)} If $E_1$, $E_2\subseteq X$ are Haar measurable,
$\undpsi(E_1\cap E_2)=\undpsi E_1\cap\undpsi E_2$.   \Prf\ By (i),
$\undpsi(E_1\cap E_2)\subseteq\undpsi E_1\cap\undpsi E_2$.   So take
$a\in\undpsi E_1\cap\undpsi E_2$.   Let $\gamma<1$.   Set
$\delta=\bover12(1+\gamma)<1$.   Then there are $n_1$, $n_2\in\Bbb N$
such that $a\in\psi_{m\delta}(E_1)$ for every $m\ge n_1$ and
$a\in\psi_{m\delta}(E_2)$ for every $m\ge n_2$.   Set $n=\max(n_1,n_2)$
and take any $m\ge n$.   Then there are open sets $G_1$,
$G_2\subseteq X$ and $F_1$, $F_2\in\Sigma_Y$ such that
$a\in G_1\cap G_2\cap\undphi F_1\cap\undphi F_2$,
$\int\chi E_1(xy)h_m(y)\mu_Y(dy)\ge\delta$ for
every $x\in G_1\cap F_1$ and $\int\chi E_2(xy)h_m(y)\mu_Y(dy)\ge\delta$ for
every $x\in G_2\cap F_2$.   Let $Q\in\Sigma_Y$ be the conegligible set
of those $x\in X$ such that $\int\chi(E_1\cap E_2)(xy)\mu_Y(dy)$
is defined.   Set $G=G_1\cap G_2$, $F=F_1\cap F_2\cap Q$;  then $G$ is
open, $F\in\Sigma_Y$ and

\Centerline{$\undphi F=\undphi(F_1\cap F_2)=\undphi F_1\cap\undphi
F_2$,}

\noindent so that $a\in G\cap\undphi F$.   Now take any $x\in G\cap F$.
We have

$$\eqalign{\int(1-\chi&(E_1\cap E_2))(xy)h_m(y)\mu_Y(dy)\cr
&\le \int(1-\chi E_1(xy))h_m(y)\mu(dy)
  +\int(1-\chi E_2(xy))h_m(y)\mu(dy)\cr
&\le 2(1-\delta)=1-\gamma,\cr}$$

\noindent and $\int\chi(E_1\cap E_2)(xy)h_m(y)\mu_Y(dy)\ge\gamma$,
because $\int h_m(y)\mu_Y(dy)=1$.

As $x$ is arbitrary, $G$ and $F$ witness that
$a\in\psi_{m\gamma}(E_1\cap E_2)$.   And this is true for every
$m\ge n$.   As $\gamma$ is arbitrary, $a\in\undpsi(E_1\cap E_2)$.   As
$a$ is arbitrary,
$\undpsi E_1\cap\undpsi E_2\subseteq\undpsi(E_1\cap E_2)$ and
the two are equal.\ \Qed

\medskip

{\bf (k)} If $E\in\Sigma_Y$, $\undpsi E=\undphi E$.   \Prf\ (i) Suppose
$a\in\undphi E$, $n\in\Bbb N$ and $\gamma<1$.   Set $F=E\cap\undphi E\in\Sigma_Y$, $G=X$.   Then $G\cap\undphi F=\undphi(E\cap\undphi E)=\undphi E$ contains $a$.
Take any $x\in F$.   Then

\Centerline{$\int\chi E(xy)h_n(y)\mu_Y(dy)
=\int h_n(y)\mu_Y(dy)=1$;}

\noindent as $x$ is arbitrary, $a\in\psi_{n\gamma}(E)$.   As $n$ and
$\gamma$ are arbitrary, $a\in\undpsi E$;  as $a$ is arbitrary,
$\undphi E\subseteq\undpsi E$.   (ii) Suppose $a\in\undpsi E$.   Then
there must
be some open $G\subseteq X$ and $F\in\Sigma_Y$ and $n\in\Bbb N$ such
that $a\in G\cap\undphi F$ and $\int\chi E(xy)h_n(y)\mu_Y(dy)>0$ for
every $x\in G\cap F$.   This surely implies that
$G\cap F\subseteq EY=E$, so that $GY\cap F\subseteq E$.   But
$a\in G\subseteq\undphi(GY)$, by 447B, so

\Centerline{$a\in\undphi(GY)\cap\undphi F=\undphi(GY\cap
F)\subseteq\undphi E$.}

\noindent This shows that $\undpsi E\subseteq\undphi E$.\ \Qed

\medskip

{\bf (l)} Thus we have assembled all the facts required to establish
that $\undpsi\restr\Sigma_Z$ is a
\lti\  partial lower density extending $\undphi$.
}%end of proof of 447F

\leader{447G}{Lemma} Let $X$ be a $\sigma$-compact locally compact
Hausdorff topological group, and $\sequencen{Y_n}$ a non-increasing
sequence of compact subgroups of $X$ with intersection $Y$.   Let
$\Sigma$ be the algebra of Haar measurable subsets of $X$;  set
$\Sigma_{Y_n}=\{E:E\in\Sigma,\,EY_n=E\}$ for each $n$, and
$\Sigma_Y=\{E:E\in\Sigma,\,EY=E\}$.   Suppose that for each $n\in\Bbb N$
we are given a \lti\ partial lower density
$\undphi_n:\Sigma_{Y_n}\to\Sigma_{Y_n}$, and that $\undphi_{n+1}$
extends $\undphi_n$ for every $n$.   Then there is a \lti\ partial lower
density $\undphi:\Sigma_Y\to\Sigma_Y$ extending every $\undphi_n$.

\proof{{\bf (a)} Fix a left Haar measure $\mu$ on $X$, and for each
$n\in\Bbb N$ let $\nu_n$ be the Haar probability measure on $Y_n$
(442Ie).   As noted in 443Sb, the modular function of $X$ must be equal
to $1$, and equal to the modular function of $Y_n$, everywhere in every
$Y_n$.

\medskip

{\bf (b)} We need to know that for any $E\in\Sigma_Y$ there is an $F$ in
the $\sigma$-algebra $\Lambda$ generated by
$\bigcup_{n\in\Bbb N}\Sigma_{Y_n}$ such that $E\symmdiff F$ is
negligible.   \Prf\ Because
$X$ is $\sigma$-compact and $\mu$ is a Radon measure (442Ac), there is a
sequence $\sequence{i}{K_i}$ of compact sets such
that $K_i\subseteq E$ for every $i$ and
$E\setminus\bigcup_{i\in\Bbb N}K_i$ is negligible.   For each
$i\in\Bbb N$, $\bigcap_{n\in\Bbb N}K_iY_n=K_iY$ (4A5Eh), so is included
in $E$.   Set $F=\bigcup_{i\in\Bbb N}\bigcap_{n\in\Bbb N}K_iY_n$;  then
$F$ belongs to the $\sigma$-algebra generated by $\bigcup_{n\in\Bbb N}\Sigma_{Y_n}$, and $K_i\subseteq F\subseteq E$ for every $i$, so
$E\symmdiff F$ is negligible.\ \Qed

\medskip

{\bf (c)} For each $E\in\Sigma_Y$, $n\in\Bbb N$ set

\Centerline{$g_{En}(x)=\nu_n(Y_n\cap x^{-1}E)$ whenever this is
defined.}

\noindent By 447Ee, $g_{En}$ is defined $\mu$-almost everywhere and is
$\Sigma$-measurable.   In fact $g_{En}$ is $\Sigma_{Y_n}$-measurable,
because
$g_{En}(xy)=g_{En}(x)$ whenever $x\in X$, $y\in Y_n$ and either is defined.
If $F\in\Sigma_{Y_n}$ then
$g_{En}(x)\chi F(x)=\nu_n(Y_n\cap x^{-1}(E\cap F))$ whenever this is
defined, which is almost everywhere;  so
$\int_Fg_{En}d\mu=\mu(E\cap F)$, by 447Ee.   If $E$,
$E'\in\Sigma_Y$ and $E\symmdiff E'$ is negligible, then
$g_{En}\eae g_{E'n}$, because $g_{E\symmdiff E',n}=0$ a.e.

It follows that $\sequencen{g_{En}}\to\chi E\,\,\mu$-a.e.\ for every
$E\in\Sigma_Y$.   \Prf\ Let $G\subseteq X$ be any non-empty relatively
compact open set, and set $U=GY_0$, so that $U$ also is a non-empty
relatively compact open set, $UY_n=U$ for every $n$ and $UY=U$.   Set
$\mu_U(F)=\Bover{\mu F}{\mu U}$ whenever $F\in\Sigma$ and $F\subseteq U$,
so that $\mu_U$ is a probability measure on $U$.

Writing $\Sigma_{Y_n}^{(U)}$, $\Sigma_Y^{(U)}$ and $\Lambda^{(U)}$ for
the subspace $\sigma$-algebras on $U$ generated by $\Sigma_{Y_n}$,
$\Sigma_Y$ and $\Lambda$, we see that if $F\in\Sigma_{Y_n}^{(U)}$ then

\Centerline{$\int_Fg_{En}d\mu_U
=\bover1{\mu U}\int_Fg_{En}d\mu
=\mu_U(E\cap F)$.}

\noindent So $g_{En}\restr U$ is a conditional expectation of
$\chi(E\cap U)$ on $\Sigma^{(U)}_{Y_n}$.   By L\'evy's martingale
theorem (275I), $\sequencen{g_{En}}$ converges almost everywhere in $U$
to a conditional expectation $g$ of $\chi(E\cap U)$ on $\Lambda^{(U)}$,
because of course $\Lambda^{(U)}$ is the $\sigma$-algebra of subsets of
$U$ generated by $\bigcup_{n\in\Bbb N}\Sigma^{(U)}_{Y_n}$.   But as
there is an $F\in\Lambda$ such that $E\symmdiff F$ is negligible, by (b)
above, $g$ must be equal to $\chi E$ almost everywhere in $U$.

Thus $g_{En}\to\chi E$ almost everywhere in $U$ and therefore almost
everywhere in $G$.   As $G$ is arbitrary, $g_{En}\to\chi E$ almost
everywhere in $X$, by 412Jb (applied to the family $\Cal K$ of subsets
of relatively compact open sets).\ \Qed

\medskip

{\bf (d)} Now we can use the method of 341G, as follows.   For
$E\in\Sigma_Y$, $k\ge 1$ and $n\in\Bbb N$ set

\Centerline{$H_{kn}(E)=\{x:x\in\dom(g_{En}),\,g_{En}(x)\ge 1-2^{-k}\}
\in\Sigma_{Y_n}$,}

\Centerline{$\tilde H_{kn}(E)=\undphi_n(H_{kn}(E))$,
\quad$\undphi E
=\bigcap_{k\ge 1}\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}\tilde H_{km}(E)$.}

\noindent By the arguments of parts (e)-(i) of the proof of 341G,
$\undphi$ is a lower density on $\Sigma_Y$ extending every $\undphi_n$.

\medskip

{\bf (e)} To see that $\undphi$ is \lti, we may argue as follows.   Let
$E\in\Sigma_Y$ and $a\in X$.   Then, for any $n$,

\Centerline{$g_{aE,n}(x)=\nu_n(Y_n\cap x^{-1}aE)=g_{En}(a^{-1}x)$}

\noindent for almost every $x$, so
$aH_{kn}(E)\symmdiff H_{kn}(aE)$ is negligible, and

\Centerline{$\tilde H_{kn}(aE)
=\undphi_n(H_{kn}(aE))=\undphi_n(aH_{kn}(E))=a\undphi_n(H_{kn}(E))
=a\tilde H_{kn}(E)$,}

\noindent for every $k$.   Accordingly

%$$\eqalign{\undphi(aE)
%&=\bigcap_{k\ge 1}\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}
%  \tilde H_{km}(aE)\cr
%&=a(\bigcap_{k\ge 1}\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}
%  \tilde H_{km}(E))
%=a\undphi(E),\cr}$$

\Centerline{$\undphi(aE)
=\bigcap_{k\ge 1}\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}
  \tilde H_{km}(aE)
=a(\bigcap_{k\ge 1}\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}
  \tilde H_{km}(E))
=a\undphi(E)$,}

\noindent as required.
}%end of proof of 447G

\leader{447H}{Lemma} Let $X$ be a locally compact Hausdorff topological
group, and $\Sigma$ the algebra of Haar measurable sets in $X$.   Then
there is a \lti\ lower density $\undphi:\Sigma\to\Sigma$.

\proof{{\bf (a)} To begin with (down to the end of (c) below) let us
suppose that $X$ is $\sigma$-compact.   By 446P, there is a family
$\langle X_{\xi}\rangle_{\xi\le\kappa}$ of closed subgroups of $X$,
where $\kappa$ is an infinite cardinal, such that

\inset{$X_0$ is an open subgroup of $X$,

for every $\xi<\kappa$, $X_{\xi+1}$ is a normal subgroup of
$X_{\xi}$ and $X_{\xi}/X_{\xi+1}$ has a $B$-sequence,

for every non-zero limit ordinal $\xi\le\kappa$,
$X_{\xi}=\bigcap_{\eta<\xi}X_{\eta}$,

$X_1$ is compact,

$X_{\kappa}=\{e\}$, where $e$ is the identity of $X$.}

\noindent Note that for every $\xi\le\kappa$, the modular function
$\Delta_{\xi}$ of $X_{\xi}$ is just the restriction to $X_{\xi}$ of the
modular function $\Delta$ of $X$.   \Prf\ For $\xi=0$ this is because
$X_0$ is an open subgroup of $X$ (443Sd).   For $\xi\ge 1$,
$X_{\xi}$ is compact, so $\Delta_{\xi}$ and $\Delta\restr X_{\xi}$
are both constant with value $1$, as noted in 443Sb.\ \Qed

\medskip

{\bf (b)} For each $\xi\le\kappa$, write $\Sigma_{\xi}$ for the
$\sigma$-algebra $\{E:E\in\Sigma$, $EX_{\xi}=E\}$.    I seek to choose
inductively a family $\langle\undphi_{\xi}\rangle_{\xi\le\kappa}$ such
that each $\undphi_{\xi}:\Sigma_{\xi}\to\Sigma_{\xi}$ is a \lti\ partial
lower density, and $\undphi_{\xi}$ extends $\undphi_{\eta}$ whenever
$\eta<\xi$.

\medskip

\quad{\bf (i)} {\it Start\/} Since $X_0$ is an open subgroup of $X$,
every
member of $\Sigma_0$ is open, and we can start the induction by setting
$\undphi_0E=E$ for every $E\in\Sigma_0$.

\medskip

\quad{\bf (ii)} {\it Inductive step to a successor ordinal\/} If we have
defined $\langle\undphi_{\eta}\rangle_{\eta\le\xi}$, where $\xi<\kappa$,
then $X_{\xi+1}\normalsubgroup X_{\xi}$ is compact and $X_{\xi}/X_{\xi+1}$ has a
$B$-sequence.   So the conditions of 447F are satisfied and
$\undphi_{\xi}$ has an extension to a \lti\ partial lower density
$\undphi_{\xi+1}:\Sigma_{\xi+1}\to\Sigma_{\xi+1}$.   Of course
$\undphi_{\xi+1}$ extends $\undphi_{\eta}$ for every $\eta\le\xi$
because $\undphi_{\xi}$ does.

\medskip

\quad{\bf (iii)} {\it Inductive step to a limit ordinal of countable
cofinality\/} If we have defined
$\langle\undphi_{\eta}\rangle_{\eta<\xi}$, where $\xi\le\kappa$ is a
non-zero limit ordinal of countable cofinality, let $\sequencen{\xi_n}$
be a strictly increasing sequence with supremum $\xi$;  we may suppose
that $\xi_0\ge 1$, so that every $X_{\xi_n}$ is compact.   Then
$X_{\xi}=\bigcap_{n\in\Bbb N}X_{\xi_n}$, so by 447G there is a \lti\
partial lower density $\undphi:\Sigma_{\xi}\to\Sigma_{\xi}$ extending
every $\undphi_{\xi_n}$, and therefore extending $\undphi_{\eta}$
whenever $\eta<\xi$.

\medskip

\quad{\bf (iv)} {\it Inductive step to a limit ordinal of uncountable
cofinality\/} Suppose we have defined
$\langle\undphi_{\eta}\rangle_{\eta<\xi}$, where $\xi\le\kappa$ is a
limit ordinal of uncountable cofinality.   Then for every
$E\in\Sigma_{\xi}$ there are an $\eta<\xi$ and an $F\in\Sigma_{\eta}$
such that $E\symmdiff F$ is negligible.   \Prf\ (Cf.\ part (b) of the
proof of 447G.)   Because $X$ is $\sigma$-compact, there are
non-decreasing sequences $\sequence{i}{K_i}$, $\sequence{i}{L_i}$ of
compact subsets of $E$, $X\setminus E$ respectively such that
$E\setminus\bigcup_{i\in I}K_i$ and
$(X\setminus E)\setminus\bigcup_{i\in\Bbb N}L_i$ are negligible.   For
each $i\in\Bbb N$, $K_iX_{\xi}\cap L_i\subseteq EX_{\xi}\setminus E$ is
empty;   by 4A5Eh again, there is an $\eta_i<\xi$ such that
$K_iX_{\eta_i}\cap L_i$ is empty.
Set $\eta=\sup_{i\in\Bbb N}\eta_i$, $F=\bigcup_{i\in\Bbb N}K_iX_{\eta}$;
this works.\ \Qed

Accordingly we have a function
$\undphi_{\xi}:\Sigma_{\xi}\to\Sigma_{\xi}$ defined by writing
$\undphi_{\xi}(E)=\undphi_{\eta}(F)$ whenever $E\in\Sigma_{\xi}$,
$\eta<\xi$, $F\in\Sigma_{\eta}$ and $E\symmdiff F$ is negligible.
\Prf\ If $\eta\le\eta'<\xi$ and $F\in\Sigma_{\eta}$,
$F'\in\Sigma_{\eta'}$ are such that $E\symmdiff F$ and $E\symmdiff F'$
are both negligible, then $F\symmdiff F'$ is negligible so
$\undphi_{\eta}(F)=\undphi_{\eta'}(F)=\undphi_{\eta'}(F')$.\ \QeD\   It
is easy to check that $\undphi_{\xi}$ is a \lti\ partial lower density
(cf.\ part (A-d) of the proof of 341H), and of course it extends
$\undphi_{\eta}$ for every $\eta<\xi$.

\medskip

{\bf (c)} On completing the induction, we see that
$\Sigma_{\kappa}=\Sigma$, so that $\undphi_{\kappa}:\Sigma\to\Sigma$ is
a \lti\ lower density.

\medskip

{\bf (d)} For the general case, recall that $X$ certainly has an open
$\sigma$-compact subgroup $Y$ say (4A5El).   If $\Sigma$ is the algebra
of Haar measurable subsets of $X$, and $\Tau$ is the algebra of Haar
measurable subsets of $Y$, then $\Tau$ is just
$\Sigma\cap\Cal PY=\{E\cap Y:E\in\Sigma\}$, and the Haar negligible
subsets of $Y$ are just sets of the form $E\cap Y$ where $E$ is a Haar
negligible subset of $X$ (443F).

Let $\undpsi:\Tau\to\Tau$ be a \lti\ lower density.   For
$E\in\Sigma$ set

\Centerline{$\undphi E=\{x:x\in X,\,e\in\undpsi(Y\cap x^{-1}E)\}$,}

\noindent where $e$ is the identity of $X$.   It is easy to check that

\Centerline{$\undphi\emptyset=\emptyset$,}

\Centerline{$\undphi E=\undphi F$ if $E\symmdiff F$ is negligible,}

\Centerline{$\undphi(E\cap F)=\undphi E\cap\undphi F$ for all $E$,
$F\in\Sigma$,}

\noindent directly from the corresponding properties of $\undpsi$.   If
$E\in\Sigma$ and $a\in X$, then

\Centerline{$x\in\undphi E
\iff e\in\undpsi(Y\cap x^{-1}E)
\iff e\in\undpsi(Y\cap(ax)^{-1}aE)
\iff ax\in\undphi(aE)$,}

\noindent so $\undphi(aE)=a\undphi E$.

I have not yet checked that $E\symmdiff\undphi E$ is always negligible.
But if $E\in\Sigma$, then

\Centerline{$E\symmdiff\undphi E
=\{x:e\in\undpsi(Y\cap x^{-1}E)\symmdiff(Y\cap x^{-1}E)\}$,}

\noindent so

$$\eqalign{(E\symmdiff\undphi E)\cap Y
&=\{x:x\in Y,\,e\in\undpsi(x^{-1}(E\cap Y))\}
  \symmdiff(E\cap Y)\cr
&=\{x:x\in Y,\,e\in x^{-1}\undpsi(E\cap Y)\}\symmdiff(E\cap Y)
=\undpsi(E\cap Y)\symmdiff(E\cap Y)\cr}$$

\noindent is negligible.   Moreover, for any $a\in X$,

\Centerline{$(E\symmdiff\undphi E)\cap aY
=a((a^{-1}E\symmdiff\undphi(a^{-1}E))\cap Y)$}

\noindent because $\undphi$ is translation-invariant, so
$(E\symmdiff\undphi E)\cap aY$ is negligible.   Since $\{aY:a\in X\}$ is
an open cover of $X$, $E\symmdiff\undphi E$ is negligible (412Jb again).
In particular, $\undphi E\in\Sigma$.   So $\undphi:\Sigma\to\Sigma$ is a
\lti\ lower density, as required.
}%end of proof of 447H

\leader{447I}{Theorem}\cmmnt{ ({\smc Ionescu Tulcea \&
Ionescu Tulcea 67})}
Let $X$ be a locally compact Hausdorff topological group.   Then
it has a \lti\ lifting for its Haar measures.

\proof{ (Cf. 345B-345C.)   Write $\Sigma$ for the algebra of Haar
measurable subsets of $X$, and let $\undphi:\Sigma\to\Sigma$ be a \lti\
lower density (447H).   Let $\phi_0:\Sigma\to\Sigma$ be any lifting such
that $\phi_0E\supseteq\undphi E$ for every $E\in\Sigma$ (341Jb).   For
$E\in\Sigma$, set

\Centerline{$\phi E=\{x:e\in\phi_0(x^{-1}E)\}$,}

\noindent where $e$ is the identity of $X$.   It is easy to check that
$\phi:\Sigma\to\Cal PX$ is a Boolean homomorphism.   Also

\Centerline{$x\in\undphi E
\Longrightarrow e\in\undphi(x^{-1}E)
\Longrightarrow e\in\phi_0(x^{-1}E)
\Longrightarrow x\in\phi E$.}

\noindent So $\phi$ is a lifting (341Ib).   Finally, $\phi$ is \lti\ by
the argument used in (d) of the proof of 447H (and also in (e) of the
proof of 345B).
}%end of proof of 447I

\leader{447J}{Corollary} Let $X$ be any topological group carrying Haar
measures.   Then it has a \lti\ lifting for its left Haar measures.

\proof{ Let $\mu$ be a left Haar measure on $X$.   By 443L, we have a
locally compact Hausdorff topological group $Z$ and a continuous
homomorphism $f:X\to Z$, \imp\ for $\mu$ and an appropriate left Haar
measure $\nu$ on $Z$, such that for every $E$ in the domain $\Sigma$ of
$\mu$ there is an $F$ in the domain $\Tau$ of $\nu$ such that
$f^{-1}[F]\subseteq E$ and $E\setminus f^{-1}[F]$ is negligible.   Let
$\psi$ be a \lti\ lifting for $\nu$.
Since $F^{\ssbullet}\mapsto f^{-1}[F]^{\ssbullet}$ is an isomorphism
between the measure algebras of $\mu$ and $\nu$, we have a lifting
$\phi:\Sigma\to\Sigma$ given by saying that
$\phi E=f^{-1}[\psi F]$ whenever
$F\in\Tau$ and $E\symmdiff f^{-1}[F]$ is negligible (346D).
Now $\phi$ is \lti\ because $f$ is a group homomorphism and $\psi$ is \lti.
}%end of proof of 447J

\exercises{\leader{447X}{Basic exercises $\pmb{>}$(a)}
%\spheader 447Xa
Let $X=\Bbb R\times\{-1,1\}$, with its usual topology,
and define a multiplication on $X$ by setting
$(x,\delta)(y,\epsilon)=(x+\delta y,\delta\epsilon)$.   Show that $X$ is
a locally compact topological group.   Show that there is no lifting for
the Haar measure algebra of $X$ which is both left- and
right-translation-invariant.   \Hint{345Xc.}
%447A  %mt44bits

\spheader 447Xb Let $X$ be a topological group carrying Haar measures
which has a $B$-sequence.   Show that it has a $B$-sequence
$\sequencen{V_n}$ such that
$\sup_{n\in\Bbb N}\mu(V_nV_n^{-1})/\mu V_n$ is finite for any Haar
measure $\mu$ on $X$, whether left or right.
%447C

\spheader 447Xc Let $X$ be a topological group with a left Haar measure
$\mu$, and $\sequencen{V_n}$ a $B$-sequence for $X$.   Show that if
$f\in\eusm L^0(\mu)$ is locally integrable, then
$f(x)=\lim_{n\to\infty}\Bover1{\mu V_n}\biggerint_{xV_n}fd\mu$ for almost
every $x$.
%447C 261C

\leader{447Y}{Further exercises (a)}
%\spheader 447Ya
Describe a compact Hausdorff topological group such that
its Haar measure has no lifting which is both left- and
right-translation-invariant.
%447Xa, 447A   S^1\times\{-1,1\}

\spheader 447Yb Let $(X,\Sigma,\mu)$ be a measure space, with measure
algebra $\frak A$, and $\undtheta:\frak A\to\Sigma$ a lower density.
Show that we have a function
$q:L^{\infty}(\frak A)^+\to L^{\infty}(\Sigma)^+$ such that
$\{x:q(u)(x)>\alpha\}=\bigcup_{\beta>\alpha}\undtheta\Bvalue{u>\beta}$
for every $\alpha\ge 0$ and $u\in L^{\infty}(\frak A)^+$.   Show that
$q(u)^{\ssbullet}=u$, $q(\alpha u)=\alpha q(u)$, $q(u\wedge v)=q(u)\wedge q(v)$ and $q(\chi a)=\chi(\undtheta a)$ for every $u$,
$v\in L^{\infty}(\frak A)^+$, $\alpha\ge 0$ and $a\in\frak A$.
%447I

}%end of exercises

\endnotes{
\Notesheader{447} The structure of the proof of 447I is exactly that of
the proof of the ordinary Lifting Theorem in \S341;  the lifting is
built from a lower density which is constructed inductively on a family
of sub-$\sigma$-algebras.   To get a translation-invariant lifting it is
natural to look for a translation-invariant lower density, and a simple
trick (already used in \S345) ensures that this is indeed enough.   The
refinements we need here are dramatic but natural.   To make the
final lower density $\undphi$ (in 447H) translation-invariant, it is
clearly sensible (if we can do it) to keep all the partial lower
densities $\undphi_{\xi}$ translation-invariant.   This means that their
domains $\Sigma_{\xi}$ should be translation-invariant.   It does not
quite follow that they have to be of the form
$\Sigma_{\xi}=\{E:EX_{\xi}=X_{\xi}\}$ for closed subgroups $X_{\xi}$,
but if we look at the leading example of $\{0,1\}^I$ (345C) this also is
a reasonable thing to try first.   So now we have to consider what extra
hypotheses will be needed to make the induction work.   The inductive
step to limit ordinals of uncountable cardinality remains elementary, at
least if the $X_{\xi}$ are compact (part (b-iv) of the proof of 447H).
The inductive step to limit ordinals of countable cofinality (447G) is
harder, but can be managed with ideas already presented.   Indeed,
compared with the version in 341G, we have the advantage of a formula
for the auxiliary functions $g_{En}$, which is very helpful when we come
to translation-invariance.   We have to do something
about the fact that we are no longer working with a probability
space -- that is the point of the $\mu_U$ in part (c)
of the proof of 447G.
(Another expression of the manoeuvre here is in 369Xq.)

Where we do need a new idea is in the inductive step to a successor
ordinal.   If $\Sigma_{\xi+1}$ is to be translation-invariant, it must
be much bigger than the $\sigma$-algebra generated by
$\Sigma_{\xi}\cup\{E\}$, as discussed in 341F.   To make the step a
small one (and therefore presumably easier), we want $X_{\xi+1}$ to be a
large subgroup of $X_{\xi}$ in some sense;  as it turns out, a helpful
approach is to ask for $X_{\xi+1}$ to be a normal subgroup of $X_{\xi}$
and for $X_{\xi}/X_{\xi+1}$ to be small.   At this point we have to know
something of the structure theory of locally compact topological groups.
The right place to start is surely the theory of compact
Hausdorff groups.   Such a group $X$ actually has a continuous
decreasing chain $\langle X_{\xi}\rangle_{\xi\le\kappa}$ of closed
normal subgroups, from $X_0=X$ to $X_{\kappa}=\{e\}$, such that all the
quotients $X_{\xi}/X_{\xi+1}$ are Lie groups.   I do not define `Lie
group' here, because for our purposes it is enough to know that the
quotients have faithful finite-dimensional representations, and
therefore have `$B$-sequences' in the sense of 446L.   Having
identified this as a relevant property, it is not hard to repeat
arguments from \S221 and \S261 to prove versions of Vitali's theorem
and Lebesgue's Density Theorem in such groups (447C-447D).   This will
evidently provide translation-invariant lower densities for groups of
this special type, just as Lebesgue lower density is a
translation-invariant lower density on $\BbbR^r$ (345B).

Of course we still have to find a way of combining this construction
with a translation-invariant lower density on $\Sigma_{\xi}$ to produce
a translation-invariant lower density on $\Sigma_{\xi+1}$, and this is
what I do in 447F.   The argument I offer is essentially that of
{\smc Ionescu Tulcea \& Ionescu Tulcea 67}, \S7, and is the deepest part of
this section.

For compact groups, these ideas are all we need, and indeed the step to
a limit ordinal of countable cofinality is a little easier, since we
have a Haar probability measure on the whole group.   The next step, to
general $\sigma$-compact locally compact groups, demands much deeper
ideas from the structure theory, but from the point of view of the
present section the modifications are minor.   The subgroups $X_{\xi}$
are now not always normal subgroups of $X$, which means that we have to
be more careful in the description of the quotient spaces $X/X_{\xi}$
(they must consist of {\it left} cosets), and we have to watch the
modular functions of the $X_{\xi}$ in order to be sure that there are
invariant measures on the quotients.   An extra obstacle at the
beginning is that we may have to start the chain with a proper subgroup
$X_0$ of $X$, but since $X_0$ can be taken to be open, it is pretty
clear that this will not be serious, and in fact it gives no trouble
(part (b-i) of the proof of 447H).   For $\xi\ge 1$, the $X_{\xi}$ are
compact, so the
inductive steps to limit ordinals are nearly the same.

The step to a general locally compact Hausdorff topological group (part
(d) of the proof of 447H) is essentially elementary.   And finally I
note that the whole thing applies to general topological groups with
Haar measures (447J), for the usual reasons.   There is an implicit
challenge here:  find expressions of the arguments used in this section
which will be valid in the more general context.   The measure-theoretic
part of such a programme might be achievable, but I do not see any hope
of a workable structure theory to match that of \S446 which does not use
443L or something like it.
}%end of notes

\discrpage



\frfilename{mt448.tex}
\versiondate{12.4.13}
\copyrightdate{1997}

\def\Gse{$G$-\vthsp$\sigma$-\vthsp equidecomposable}
\def\IMPLY#1#2{{\bf (#1)$\Rightarrow$(#2)}}
\def\psG{\preccurlyeq^{\sigma}_G}
\def\psH{\preccurlyeq^{\sigma}_H}
\def\pstildeG{\preccurlyeq^{\sigma}_{\tilde G}}
\def\ptG{\preccurlyeq^{\tau}_G}
\def\low#1{\lfloor#1\rfloor}
\def\high#1{\lceil#1\rceil}

\def\chaptername{Topological groups}
\def\sectionname{Polish group actions}

\newsection{448}

I devote this section to two quite separate theorems.   The first is
an interesting result about measures on
Polish spaces which are invariant under actions of Polish groups.   In
contrast to \S441, we no longer have a strong general
existence theorem for such measures, but instead have a natural
necessary and sufficient condition in terms of countable dissections:
there is an invariant probability measure on $X$ if and only if there is no
countable dissection of $X$ into Borel sets which can be rearranged, by
the action of the group, into two copies of $X$ (448P).

\cmmnt{The principal ideas needed here have already been set out in
\S395, and in many of the proofs I allow myself to direct you to the
corresponding arguments there rather than write the formulae out again.
I do not think you need read through \S395 before embarking on this
section;  I will try to give sufficiently detailed references so that
you can take them one paragraph at a time, and many of the arguments
referred to are in any case elementary.   But unless you are already
familiar with this topic, you will need a copy of \S395 to hand to fully
follow the proofs below.
}%end of comment

The second theorem concerns the representation of group actions on measure
algebras in terms of group actions on measure spaces.   If we have a
locally compact Polish group $G$ (so that we do have Haar measures), and
a Borel measurable action of $G$ on the measure algebra of a Radon measure
$\mu$ on a Polish space $X$, then it can be represented by a Borel
measurable action of $G$ on $X$ (448S).   The proof is mostly descriptive
set theory based on \S\S423-424, but it also uses some interesting facts
about $L^0$ spaces (448Q-448R).

\leader{448A}{Definitions}\cmmnt{ (Compare 395A.)}   Let $\frak A$ be a
Dedekind $\sigma$-complete Boolean algebra, and $G$ a subgroup of
$\Aut\frak A$.   For $a$, $b\in\frak A$ I will say that an isomorphism
$\phi:\frak A_a\to\frak A_b$ between the corresponding principal ideals
belongs to the {\bf countably full local semigroup generated by $G$} if
there are a countable partition of unity $\langle a_i\rangle_{i\in I}$
in $\frak A_a$ and a family
$\langle\pi_i\rangle_{i\in I}$ in $G$ such that $\phi c=\pi_ic$
whenever $i\in I$ and $c\Bsubseteq a_i$.   If such an isomorphism exists
I will say that $a$ and $b$ are {\bf \Gse}.

I write $a\psG b$ to mean that there is a $b'\Bsubseteq b$ such
that $a$ and $b'$ are \Gse.

\cmmnt{As in \S395, }I will say that a function $f$ with domain
$\frak A$ is {\bf $G$-invariant} if $f(\pi a)=f(a)$ whenever
$a\in\frak A$ and $\pi\in G$.

\cmmnt{I have expressed these definitions, and most of the work below,
in terms of abstract Dedekind $\sigma$-complete Boolean algebras.   The
applications I have in mind for this section are to
$\sigma$-algebras of sets.   If you have already worked through \S395,
the version here should come very easily;  but even if you have not, I
think that the extra abstraction clarifies some of the ideas.
}

\leader{448B}{}\cmmnt{ I begin with results corresponding to
395B-395D;  there is hardly any difference, except that we must now
occasionally pause to check that a partition of unity is countable.

\medskip

\noindent}{\bf Lemma} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra and $G$ a subgroup of $\Aut\frak A$.   Write
$G^*_{\sigma}$ for the countably full local semigroup generated by $G$.

(a) If $a$, $b\in\frak A$ and $\phi:\frak A_a\to\frak A_b$ belongs to
$G^*_{\sigma}$, then $\phi^{-1}:\frak A_b\to\frak A_a$ also belongs to
$G^*_{\sigma}$.

(b) Suppose that $a$, $b$, $a'$, $b'\in\frak A$ and that
$\phi:\frak A_a\to\frak A_{a'}$, $\psi:\frak A_{b}\to\frak A_{b'}$
belong to $G^*_{\sigma}$.   Then $\psi\phi\in G^*_{\sigma}$;
its domain is $\frak A_c$ where $c=\phi^{-1}(b\Bcap a')$, and its
set of values is $\frak A_{c'}$ where $c'=\psi(b\Bcap a')$.

(c) If $a$, $b\in\frak A$ and $\phi:\frak A_a\to\frak A_b$ belongs to
$G^*_{\sigma}$, then $\phi\restrp\frak A_c\in G^*_{\sigma}$ for any
$c\Bsubseteq a$.

(d) Suppose that $a$, $b\in\frak A$ and that
$\psi:\frak A_a\to\frak A_b$ is an isomorphism such that there are a
countable partition of unity $\langle a_i\rangle_{i\in I}$ in
$\frak A_a$ and a family
$\langle\phi_i\rangle_{i\in I}$ in $G^*_{\sigma}$ such that
$\psi c=\phi_ic$ whenever $i\in I$ and $c\Bsubseteq a_i$.   Then
$\psi\in G^*_{\sigma}$.

\proof{{\bf (a)} As 395Bb.

\medskip

{\bf (b)} As 395Bc.

\medskip

{\bf (c)} As 395Bd.

\medskip

{\bf (d)} For each $i\in I$, let $\family{j}{J(i)}{a_{ij}}$,
$\family{j}{J(i)}{\pi_{ij}}$ witness that $\phi_i\in G^*_{\sigma}$;
then $\langle a_i\Bcap a_{ij}\rangle_{i\in I,j\in J(i)}$ and $\langle
\pi_{ij}\rangle_{i\in I,j\in J(i)}$ witness that $\psi\in G^*_{\sigma}$.
}%end of proof of 448B

\leader{448C}{Lemma} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra and $G$ a subgroup of $\Aut\frak A$.   Write
$G^*_{\sigma}$ for the countably full local semigroup generated by $G$.

(a) For $a$, $b\in\frak A$, $a\psG b$ iff there is a
$\phi\in G^*_{\sigma}$ such that $a\in\dom\phi$ and
$\phi a\Bsubseteq b$.

(b)(i) $\psG$ is transitive and reflexive;

\quad(ii) if $a\psG b$ and $b\psG a$ then $a$ and $b$ are \Gse.

(c) $G$-$\sigma$-equidecomposability is an equivalence relation on
$\frak A$.

(d) If $\langle a_i\rangle_{i\in I}$ and $\langle b_i\rangle_{i\in I}$
are countable families in $\frak A$, of which
$\langle b_i\rangle_{i\in I}$ is disjoint, and $a_i\psG b_i$ for every
$i\in I$, then $\sup_{i\in I}a_i\psG\sup_{i\in I}b_i$.

\proof{ The arguments of 395C apply unchanged, calling on 448B in place
of 395B.
}%end of proof of 448C

\vleader{72pt}{448D}{Theorem} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra and $G$ a subgroup of $\Aut\frak A$.   Then the
following are equiveridical:

(i) there is an $a\ne 1$ such that $a$ is \Gse\
with $1$;

(ii) there is a disjoint sequence $\sequencen{a_n}$ of non-zero elements
of $\frak A$ which are all \Gse;

(iii) there are non-zero \Gse\ $a$, $b$, $c\in\frak A$ such that
$a\Bcap b=0$ and $a\Bcup b\Bsubseteq c$;

(iv) there are \Gse\ $a$, $b\in\frak A$ such
that $a\Bsubset b$.

\proof{ As 395D.
}%end of proof of 448D

\leader{448E}{Definition}\cmmnt{ (Compare 395E.)}   Let $\frak A$ be a
Dedekind $\sigma$-complete Boolean algebra and $G$ a subgroup of
$\Aut\frak A$.   I will say that  $G$ is {\bf countably non-paradoxical}
if\cmmnt{ the statements of 448D are false;  that
is, if} one of the following equiveridical statements is true:

\inset{(i)
if $a$ is \Gse\ with $1$ then $a=1$;}

\inset{(ii) there is no disjoint sequence $\sequencen{a_n}$ of
non-zero elements of $\frak A$ which are all \Gse;}

\inset{(iii) there are no non-zero \Gse\ $a$, $b$, $c\in\frak A$ such
that $a\Bcap b=0$ and $a\Bcup b\Bsubseteq c$;}

\inset{(iv) if $a$, $b\in\frak A$ are \Gse\ and
$a\Bsubseteq b$ then $a=b$.}

%add: there are $a\in\frak A$, $\phi$ in the
%countably full subgroup generated
%by $\pi$ which is not recurrent on $a$.


\leader{448F}{}\cmmnt{ We now come to one of the points where we need
to find a new path because we are looking at algebras which need not be
Dedekind complete.   Provided the original group $G$ is {\it countable},
we can still follow the general line of \S395, as follows.

\medskip

\noindent}{\bf Lemma}\cmmnt{ (Compare 395G.)}   Let $\frak A$ be a
Dedekind $\sigma$-complete Boolean algebra and $G$ a countable subgroup
of $\Aut\frak A$.   Let $\frak C$ be the fixed-point subalgebra of $G$.

(a) For any $a\in\frak A$, $\upr(a,\frak C)$\cmmnt{ (313S)} is
defined, and is given by the formula

\Centerline{$\upr(a,\frak C)=\sup\{\pi a:\pi\in G\}$.}

(b) If $G^*_{\sigma}$ is the countably full local semigroup generated by
$G$, then $\phi(c\Bcap a)=c\Bcap\phi a$ whenever
$\phi\in G^*_{\sigma}$, $a\in\dom\phi$ and $c\in\frak C$.

(c) $\upr(\phi a,\frak C)=\upr(a,\frak C)$ whenever
$\phi\in G^*_{\sigma}$ and $a\in\dom\phi$;\cmmnt{  consequently,}
$\upr(a,\frak C)\Bsubseteq\upr(b,\frak C)$ whenever
$a\psG b$.

(d) If $a\psG b$ and $c\in\frak C$ then $a\Bcap c\psG b\Bcap c$.
So $a\Bcap c$ and $b\Bcap c$ are \Gse\ whenever $a$ and $b$ are
\Gse\ and $c\in\frak C$.

\proof{{\bf (a)} As remarked in 395Ga, $\frak C$ is order-closed.
Because $G$ is countable and $\frak A$ is Dedekind
$\sigma$-complete, $c^*=\sup\{\pi a:\pi\in G\}$ is defined in $\frak A$
If $\phi\in G$, then

\Centerline{$\phi c^*=\sup\{\phi\pi a:\pi\in G\}\Bsubseteq c^*$}

\noindent because $\phi$ is order-continuous and $\phi\pi\in G$ for every
$\pi\in G$.   Similarly $\phi^{-1}c^*\Bsubseteq c^*$ and
$c^*\Bsubseteq\phi c^*$.   Thus $\phi c^*=c^*$;  as $\phi$ is arbitrary,
$c^*\in\frak C$.

If $c\in\frak C$, then

$$\eqalign{a\Bsubseteq c
&\iff\pi a\Bsubseteq\pi c\text{ for every }\pi\in G\cr
&\iff\pi a\Bsubseteq c\text{ for every }\pi\in G\cr
&\iff c^*\Bsubseteq c,\cr}$$

\noindent so $c^*=\inf\{c:a\Bsubseteq c\in\frak C\}$, taking the infimum
in $\frak C$, as required in the definition of $\upr(a,\frak C)$.

\medskip

{\bf (b)} Suppose that $\familyiI{a_i}$, $\familyiI{\pi_i}$ witness that
$\phi\in G^*_{\sigma}$.   Then

\Centerline{$\phi(a\Bcap c)=\sup_{i\in I}\pi_i(a_i\Bcap a\Bcap c)
=\sup_{i\in I}\pi_i(a_i\Bcap a)\Bcap c=c\Bcap\phi a$.}

\medskip

{\bf (c)} For $c\in\frak C$,

\Centerline{$a\Bsubseteq c
\iff a\Bcap c=a
\iff\phi(a\Bcap c)=\phi a
\iff c\Bcap\phi a=\phi a
\iff\phi a\Bsubseteq c$.}

\medskip

{\bf (d)} There is a $\phi\in G^*_{\sigma}$ such that
$\phi a\Bsubseteq b$;  now

\Centerline{$a\Bcap c\psG\phi(a\Bcap c)=c\Bcap\phi a
\Bsubseteq b\Bcap c$.}
}%end of proof of 448F

\leader{448G}{}\cmmnt{ With this support, we can now continue with the
ideas of 395H-395L, adding at each step the hypothesis `$G$ is
countable' to compensate for the weakening of the hypotheses `$\frak A$
is Dedekind complete, $G$ is fully non-paradoxical' to `$\frak A$ is
Dedekind $\sigma$-complete, $G$ is countably non-paradoxical'.

\medskip

\noindent}{\bf Lemma}\cmmnt{ (Compare 395H.)}   Let $\frak A$ be a
Dedekind $\sigma$-complete Boolean algebra and $G$ a countable countably
non-paradoxical subgroup of $\Aut\frak A$.   Write $\frak C$ for the
fixed-point subalgebra of $G$.   Take any $a$, $b\in\frak A$.   Then
$c_0=\sup\{c:c\in\frak C,\,a\Bcap c\psG b\}$ is defined in $\frak A$ and
belongs to $\frak C$;  $a\Bcap c_0\psG b$ and $b\Bsetminus c_0\psG a$.

\proof{ Let $\sequencen{\pi_n}$ be a sequence running over $G$.   Define
$\sequencen{a_n}$, $\sequencen{b_n}$ inductively,  setting

\Centerline{$a_n=(a\Bsetminus\sup_{i<n}a_i)
   \Bcap\pi_n^{-1}(b\Bsetminus\sup_{i<n}b_i)$,
\quad$b_n=\pi_na_n$.}

\noindent Then $\sequencen{a_n}$ is a disjoint sequence in $\frak A_a$
and $\sequencen{b_n}$ is a
disjoint sequence in $\frak A_b$, and $\sup_{n\in\Bbb N}a_n$ is \Gse\
with $\sup_{n\in\Bbb N}b_n$.
Set

\Centerline{$a'=a\Bsetminus\sup_{n\in\Bbb N}a_n$,
\quad$b'=b\Bsetminus\sup_{n\in\Bbb N}b_n$,
\quad$c_0=1\Bsetminus\upr(a',\frak C)\Bsubseteq\sup_{n\in\Bbb N}a_n$.}

\noindent Then

\Centerline{$a\Bcap c_0\Bsubseteq\sup_{n\in\Bbb N}a_n\psG b$.}

Now $b'\Bsubseteq c_0$.   \Prf\Quer\ Otherwise, because
$c_0=1\Bsetminus\sup_{n\in\Bbb N}\pi_na'$, there must be an
$n\in\Bbb N$ such that $b'\Bcap\pi_na'\ne 0$.   But in this case
$d=a'\Bcap\pi_n^{-1}b'\ne 0$, and we have

\Centerline{$d\Bsubseteq(a\Bsetminus\sup_{i<n}a_i)
  \Bcap\pi_n^{-1}(b\Bsetminus\sup_{i<n}b_i)$,}

\noindent so that $d\Bsubseteq a_n$, which is absurd.\ \Bang\QeD\
Consequently

\Centerline{$b\Bsetminus c_0
\Bsubseteq b\Bsetminus b'=\sup_{n\in\Bbb N}b_n\psG a$.}

Now take any $c\in\frak C$ such that $a\Bcap c\psG b$, and consider
$c'=c\Bsetminus c_0$.   Then $b'\Bcap c'=0$, that is, $b\Bcap
c'=\sup_{n\in\Bbb N}b_n\Bcap c'$, which is \Gse\ with
$\sup_{n\in\Bbb N}a_n\Bcap c'=(a\Bsetminus a')\Bcap c'$.   But now

\Centerline{$a\Bcap c'=a\Bcap c_0\Bcap c'
\psG b\Bcap c'\psG(a\Bcap c')\Bsetminus(a'\Bcap c')$;}

\noindent because $G$ is countably non-paradoxical, $a'\Bcap c'$ must be
$0$, that is, $c'\Bsubseteq c_0$ and $c\Bsubseteq c_0$.
So $c_0$ has the required properties.
}%end of proof of 448G

\leader{448H}{Lemma}\cmmnt{ (Compare 395I.)}   Let $\frak A$ be a
Dedekind $\sigma$-complete Boolean
algebra, not $\{0\}$, and $G$ a countable countably non-paradoxical
subgroup of $\Aut\frak A$.   Let $\frak C$ be the fixed-point subalgebra
of $G$.   Suppose that $a$, $b\in\frak A$ and that $\upr(a,\frak C)=1$.
Then there are non-negative $u$, $v\in L^0(\frak C)$ such that

$$\eqalign{\Bvalue{u\ge n}
&=\max\{c:c\in\frak C,
\text{ there is a disjoint family }\langle d_i\rangle_{i<n}\cr
&\qquad\qquad\qquad\qquad
\text{such that }a\Bcap c\psG d_i\Bsubseteq b\text{ for every }i<n\},\cr}$$

$$\eqalign{\Bvalue{v\le n}
&=\max\{c:c\in\frak C,
\text{ there is a family }\langle d_i\rangle_{i<n}\cr
&\qquad\qquad\qquad\qquad\text{ such that }d_i\psG a
\text{ for every }i<n\text{ and }b\Bcap c\Bsubseteq\sup_{i<n}d_i\}\cr}$$

\noindent for every $n\in\Bbb N$.   Moreover, we have

\quad(i) $\Bvalue{u\in\Bbb N}=\Bvalue{v\in\Bbb N}=1$,

\quad(ii) $\Bvalue{v>0}=\upr(b,\frak C)$,

\quad(iii) $u\le v\le u+\chi 1$.

\proof{ The argument of 395I applies unchanged, except that every $\ptG$
must be replaced with a $\psG$, and we use 448F and 448G in place of
395G and 395H.   $\frak C$ is Dedekind $\sigma$-complete because it is
order-closed in the Dedekind $\sigma$-complete algebra $\frak A$
(314Eb).
}%end of proof of 448H

\leader{448I}{Notation}\cmmnt{ (Compare 395J.)}   In the context of
448H, I will write $\low{b:a}$ for $u$, $\high{b:a}$ for $v$.

\leader{448J}{Lemma}\cmmnt{ (Compare 395K-395L.)}   Let $\frak A$ be a
Dedekind $\sigma$-complete Boolean algebra, not $\{0\}$, and $G$ a
countable countably non-paradoxical subgroup of
$\Aut\frak A$ with fixed-point subalgebra $\frak C$.   Suppose that $a$,
$a_1$, $a_2$, $b$, $b_1$, $b_2\in\frak A$ and that

\Centerline{$\upr(a,\frak C)=\upr(a_1,\frak C)=\upr(a_2,\frak C)=1$.}

\noindent Then

(a) $\low{0:a}=\high{0:a}=0$ and $\low{1:a}\ge\chi 1$.

(b) If $b_1\psG b_2$ then $\low{b_1:a}\le\low{b_2:a}$ and
$\high{b_1:a}\le\high{b_2:a}$.

(c) $\high{b_1\Bcup b_2:a}\le\high{b_1:a}+\high{b_2:a}$.

(d) If $b_1\Bcap b_2=0$, then
$\low{b_1:a}+\low{b_2:a}\le\low{b_1\Bcup b_2:a}$.

(e) If $c\in\frak C$ is such that $a\Bcap c$ is a
relative atom over $\frak C$, then
$c\Bsubseteq\Bvalue{\high{b:a}-\low{b:a}=0}$.

(f) $\low{b:a_2}\ge\low{b:a_1}\times\low{a_1:a_2}$,
$\high{b:a_2}\le\high{b:a_1}\times\high{a_1:a_2}$.

\proof{ As in 395K-395L.
}%end of proof of 448J.

\leader{448K}{}\cmmnt{ For the result corresponding to 395Mb, we again
need to find a new approach;  I deal with it by adding a further
hypothesis to the list which has already accreted.

\medskip

\noindent}{\bf Definition}  Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra and $G$ a
countable subgroup of $\Aut\frak A$
with fixed-point subalgebra $\frak C$.   I will
say that $G$ has the {\bf $\sigma$-refinement property} if for every
$a\in\frak A$ there is a $d\Bsubseteq a$ such that $d\psG a\Bsetminus d$
and $a'=a\Bsetminus\upr(d,\frak C)$ is a relative atom over
$\frak C$\cmmnt{, that is, every $b\Bsubseteq a'$ is expressible as
$a'\Bcap c$ for some $c\in\frak C$}.

\cmmnt{(If we replace $\psG$ with $\ptG$, as used in \S395, we see
that 395Ma could be read as `if $\frak A$ is a Dedekind complete Boolean
algebra, then any subgroup of $\Aut\frak A$ has the $\tau$-refinement
property'.)}

\leader{448L}{}\cmmnt{ I give the principal case in which the
`$\sigma$-refinement property' just defined arises.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra with countable Maharam
type\cmmnt{ (definition:  331F)}.   Then any countable
subgroup of $\Aut\frak A$ has the $\sigma$-refinement property.

\proof{{\bf (a)} Let $E$ be a countable subset of $\frak A$ which
$\tau$-generates $\frak A$, and $\frak E$ the subalgebra of $\frak A$
generated by $E$;  then $\frak E$ is countable (331Gc), and the smallest
order-closed subset of $\frak A$ including $\frak E$ is a subalgebra of
$\frak A$ (313Fc), so must be $\frak A$ itself.

\medskip

{\bf (b)} Suppose that $b\in\frak A\setminus\{0\}$ and
$\pi\in\Aut\frak A$
are such that $b\Bcap\pi b=0$.   Then there is an $e\in\frak E$ such
that $b\Bcap e\Bsetminus\pi e\ne 0$.   \Prf\Quer\ Otherwise, set

\Centerline{$D=\{d:d\in\frak A,\,b\Bcap d\Bsetminus\pi d=0\}$.}

\noindent Then $\frak E\subseteq D$, but $b\notin D$.   So $D$ cannot be
order-closed.   {\bf case 1} If $D_0\subseteq D$ is a non-empty
upwards-directed set with supremum $d_0\notin D$, then $b\Bcap
d_0\Bsetminus\pi d_0\ne 0$, so there is a $d\in D_0$ such that $b\Bcap
d\Bsetminus\pi d_0\ne 0$;  but now $d\notin D$, which is impossible.
{\bf case 2} If $D_0\subseteq D$ is a non-empty
downwards-directed subset of $D$ with infimum $d_0\notin D$, then
$b\Bcap d_0\Bsetminus\pi d_0\ne 0$.   But $\pi$ is
order-continuous, so there is a $d\in D_0$ such that $b\Bcap
d_0\Bsetminus\pi d\ne 0$;  and now $d\notin D$, which is impossible.
Thus in either case we have a contradiction.\ \Bang\Qed

\medskip

{\bf (c)} Now let $G$ be a countable subgroup of $\Aut\frak A$, with
fixed-point subalgebra $\frak C$, and let $\sequencen{(\pi_n,e_n)}$ be a
sequence running over $G\times\frak E$.  Take any $a\in\frak A$.   For
$k\in\Bbb N$ set

\Centerline{$a_k=a\Bcap e_k\Bcap\pi_k^{-1}(a\Bsetminus e_k)$,}

\Centerline{$a'_k=a_k\Bsetminus\sup_{j<k}\upr(a'_j,\frak C)$.}

\noindent Then

\Centerline{$a'_k\Bcap\pi_ka'_k\Bsubseteq a_k\Bcap\pi_ka_k=0$}

\noindent for every $k\in\Bbb N$, and whenever $j<k$ in $\Bbb N$ we have

\Centerline{$a'_j\Bcap a'_k=0$,
\quad$\pi_ja'_j\Bcap a'_k=0$,}

\Centerline{$a'_j\Bcap\pi_ka'_k
=\pi_k(\pi_k^{-1}a'_j\Bcap a'_k)=0$,
\quad$\pi_ja'_j\Bcap\pi_ka'_k
=\pi_k(a'_k\Bcap\pi_k^{-1}\pi_ja'_j)=0$.}

\noindent So, setting $d=\sup_{k\in\Bbb N}a'_k$ and $d'=\sup_{k\in\Bbb
N}\pi_ka'_k$, $d$ and $d'$ are disjoint and \Gse\ and included in $a$,
and $d\psG a\Bsetminus d$.

Consider $a'=a\Bsetminus\upr(d,\frak C)$.   Since
$a'_k=a_k\Bsetminus\sup_{j<k}\upr(a'_j,\frak C)$ for each $k$,

\Centerline{$\upr(d,\frak C)
=\sup_{k\in\Bbb N}\upr(a'_k,\frak C)=\sup_{k\in\Bbb N}\upr(a_k,\frak C)$.}

\noindent\Quer\ Suppose, if possible, that $a'$ is not a relative atom
over $\frak C$;  that is, that there is a $b\Bsubseteq a'$ such that
$b\ne a'\Bcap c$ for any $c\in\frak C$.   Then, in particular,
$b\ne a'\Bcap\upr(b,\frak C)$, and there is a $\pi\in G$ such that
$b'=a'\Bcap\pi b\Bsetminus b\ne 0$.   Then
$b'\Bcup\pi^{-1}b'\Bsubseteq a$, while $b'\Bcap\pi^{-1}b'=0$,
so $\pi b'\Bcap b'=0$.   By (b), there
is an $e\in\frak E$ such that $b''=b'\Bcap e\Bsetminus\pi e\ne 0$.   Let
$k$ be such that $\pi^{-1}=\pi_k$ and $e=e_k$, so that

\Centerline{$b''=b'\Bcap e_k\Bsetminus\pi_k^{-1}e_k
\Bsubseteq a\Bcap e_k\Bcap\pi_k^{-1}(a\Bsetminus e_k)=a_k$.}

\noindent(Because $\pi^{-1}b'\Bsubseteq a$, $b'\Bsubseteq\pi_k^{-1}a$.)
Since also

\Centerline{$b''\Bcap\upr(a'_j,\frak C)
\Bsubseteq a'\Bcap\upr(d,\frak C)=0$}

\noindent for every $j$, $b''\Bsubseteq a'_k\Bsubseteq d$, which is
impossible, because $b''\Bsubseteq a'$.\ \Bang

Thus $a'$ is a relative atom over $\frak C$, as required.
}%end of proof of 448L


\leader{448M}{Lemma} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra, not $\{0\}$, and $G$ a countable countably
non-paradoxical subgroup of $\Aut\frak A$ with fixed-point subalgebra
$\frak C$.   If $G$ has the $\sigma$-refinement property, then for any
$\epsilon>0$ there is an $a^*\in\frak A$ such that $\upr(a^*,\frak C)=1$
and $\high{b:a^*}\le\low{b:a^*}+\epsilon\low{1:a^*}$ for every
$b\in\frak A$.

\proof{ As part (b) of the proof of 395M.
}%end of proof of 448M

\leader{448N}{Theorem}\cmmnt{ (Compare 395N.)}  Let $\frak A$ be a
Dedekind $\sigma$-complete Boolean algebra and $G$ a countable countably
non-paradoxical subgroup of $\Aut\frak A$ with
fixed-point subalgebra $\frak C$.   Suppose that $G$ has the
$\sigma$-refinement
property\cmmnt{ of 448K}.   Then there is a function
$\theta:\frak A\to L^{\infty}(\frak C)$ such that

(i) $\theta$ is additive, non-negative and sequentially
order-continuous;

(ii) $\theta a=0$ iff $a=0$, $\theta 1=\chi 1$;

(iii) $\theta(a\Bcap c)=\theta a\times\chi c$ for every $a\in\frak A$,
$c\in\frak C$;  in particular, $\theta c=\chi c$ for every
$c\in\frak C$;

(iv) if $a$, $b\in\frak A$ are \Gse, then $\theta a=\theta b$;  in
particular, $\theta$ is $G$-invariant.

\proof{ The arguments of the proof of 395N apply here also, though we
have to take things in a slightly different order.   As in 395N, set

$$\theta_a(b)=\bover{\high{b:a}}{\low{1:a}}\in L^0(\frak C)$$

\noindent whenever $\upr(a,\frak C)=1$ and $b\in\frak A$.   This time,
turn immediately to part (c) of the proof to see that if $e_n$ is chosen
(using 448M) such that $\upr(e_n,\frak C)=1$ and
$\high{b:e_n}\le\low{b:e_n}+2^{-n}\low{1:e_n}$ for every $b\in\frak A$,
then $\theta_{e_n}b\le\theta_ab+2^{-n}\high{1:a}$ whenever
$\upr(a,\frak C)=1$ and $b\in\frak A$.   So we can write

\Centerline{$\theta b=\inf_{n\in\Bbb N}\theta_{e_n}b
=\inf_{\upr(a,\frak C)=1}\theta_ab$}

\noindent for every $b\in\frak A$, and we have a function
$\theta:\frak A\to L^0$ as before.   The rest of the proof is unchanged,
except that
we have a simplification in (h), since we need consider only the case
$\kappa=\omega$.
}%end of proof of 448N

\leader{448O}{}\cmmnt{ This concludes the adaptations we need from
\S395.   I now return to the specific problem addressed in the present
section.   The first step is a variation on 448N.

\medskip

\noindent}{\bf Theorem} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra, not $\{0\}$, and $G$ a countable subgroup of
$\Aut\frak A$ with the $\sigma$-refinement property.   Let $\frak C$ be
the fixed-point subalgebra of $G$.   Then the following are equiveridical:

(i) there are a Dedekind $\sigma$-complete Boolean algebra $\frak D$,
not $\{0\}$, and a $G$-invariant sequentially order-continuous
non-negative additive function $\theta:\frak A\to L^{\infty}(\frak D)$
such that $\theta 1=\chi 1$;

(ii) if $a\in\frak A$ and $1\psG a$, then
$\upr(1\Bsetminus a,\frak C)\ne 1$;

(iii) if $a\in\frak A$ and $1\psG a$, then $1\not\psG 1\Bsetminus a$.

\proof{{\bf (a)(i)$\Rightarrow$(iii)} Take
$\theta:\frak A\to L^{\infty}(\frak D)$ as in (i).   If $1\psG a$, then
there is a
$b\Bsubseteq a$ which is \Gse\ with $1$, so that $\theta b=\chi 1$, just
as in 395N(v)/448N(iv).   But this means that $\theta a=\chi 1$;  so
that $\theta(1\Bsetminus a)\ne\chi 1$ and $1\not\psG 1\setminus a$.

\medskip

{\bf (b)not-(ii)$\Rightarrow$not-(iii)} Suppose that (ii) is false;
that there is an $a\in\frak A$ such that $1\psG a$ and
$\upr(1\Bsetminus a,\frak C)=1$.   Let $G^*_{\sigma}$ be the countably
full local
semigroup generated by $G$;  then there is a $\psi\in G^*_{\sigma}$ such
that $\psi 1\Bsubseteq a$.   Set $b_0=1\Bsetminus\psi 1$ and
$b_n=\psi^nb_0$ for every $n\ge 1$;  then

\Centerline{$b_0\Bcap b_n\Bsubseteq b_0\Bcap\psi 1=0$}

\noindent for every $n\ge 1$, so

\Centerline{$b_m\Bcap b_n=\psi^m(b_0\Bcap b_{n-m})=0$}

\noindent whenever $m<n$.

Let $\sequence{i}{\pi_i}$ be a sequence running over $G$;  then

\Centerline{$\sup_{i\in\Bbb N}\pi_ib_0=\upr(b_0,\frak C)
\Bsupseteq\upr(1\Bsetminus a,\frak C)=1$.}

\noindent Set

\Centerline{$a_j=\pi_jb_0\Bsetminus\sup_{i<j}\pi_ib_0$}

\noindent for every $j\in\Bbb N$, so that $\sequence{j}{a_j}$ is a
partition of unity in $\frak A$.   Define $\psi_1$,
$\psi_2:\frak A\to\frak A$ by setting

\Centerline{$\psi_1d=\sup_{i\in\Bbb N}\psi^{2i}\pi_i^{-1}(d\Bcap a_i)$,
\quad$\psi_2d=\sup_{i\in\Bbb N}\psi^{2i+1}\pi_i^{-1}(d\Bcap a_i)$}

\noindent for every $d\in\frak A$.   Because
$\psi^{2i}\pi_i^{-1}a_i\Bsubseteq b_{2i}$ for every $i$,
$\sequence{i}{\psi^{2i}\pi_i^{-1}a_i}$ is disjoint, so $\psi_1\in
G^*_{\sigma}$ (448Bd);  similarly, $\psi_2\in G^*_{\sigma}$.   Thus

\Centerline{$1\psG\psi_11\Bsubseteq\sup_{i\in\Bbb N}b_{2i}$,}

\Centerline{$1\psG\psi_21\Bsubseteq\sup_{i\in\Bbb N}b_{2i+1}
\Bsubseteq 1\Bsetminus\psi_11$}

\noindent and (iii) is false.

\medskip

{\bf (c)} For the rest of this proof I will suppose that (ii) is
true and seek to prove (i).

Let $\Cal I$ be the $\sigma$-ideal of $\frak A$ generated by
$\{1\Bsetminus a:a\in\frak A,\,1\psG a\}$.   Then $1\notin\Cal I$.
\Prf\Quer\   Otherwise, there is a sequence $\sequencen{a_n}$ such that
$1\psG a_n$ for every $n$ and $\sup_{n\in\Bbb N}1\Bsetminus a_n=1$.
Choose $\psi_n\in G^*_{\sigma}$ such that $\psi_n1\Bsubseteq a_n$, and
set $c_n=\upr(1\Bsetminus\psi_n1,\frak C)$ for each $n$, so that
$\sup_{n\in\Bbb N}c_n=1$.   Set $c'_n=c_n\Bsetminus\sup_{i<n}c_i$ for
each $n$, and write

\Centerline{$\psi d=\sup_{n\in\Bbb N}\psi_n(d\Bcap c'_n)$}

\noindent for each $d\in\frak A$.
Because every $c'_n$ belongs to $\frak C$,
$\sequencen{\psi_nc'_n}=\sequencen{c'_n}$ is disjoint, and
$\psi\in G^*_{\sigma}$.

By (ii), $c=\upr(1\Bsetminus\psi 1,\frak C)$ is not $1$;  let $n$
be such that $c'=c'_n\Bsetminus c\ne 0$.   Because $c'\subseteq c'_n$,
$c'\Bsetminus\psi_n1=c'\Bsetminus\psi 1$;  because $c'\in\frak C$,

$$\eqalign{0
&\ne c'\Bcap c'_n
\Bsubseteq c'\Bcap\upr(1\Bsetminus\psi_n1,\frak C)
=\upr(c'\Bsetminus\psi_n1,\frak C)\cr
&=\upr(c'\Bsetminus\psi 1,\frak C)
=c'\Bcap\upr(1\Bsetminus\psi 1,\frak C)
=0\cr}$$

\noindent which is absurd.\ \Bang\Qed

Let $\frak B$ be the quotient Boolean algebra $\frak A/\Cal I$;  then
$\frak B$ is Dedekind $\sigma$-complete and the canonical homomorphism
$a\mapsto a^{\ssbullet}:\frak A\to\frak B$ is sequentially
order-continuous (314C, 313Qb).

\medskip

{\bf (d)} Next, $\pi b\in\Cal I$ whenever $b\in\Cal I$ and $\pi\in G$.
\Prf\ The sets $\{a:1\psG a\}$ and $\{1\Bsetminus a:1\psG a\}$ are both
invariant under the action of $G$, so $\Cal I$ also must be
invariant.\ \QeD\   We
can therefore define, for each $\pi\in G$, a Boolean automorphism
$\tilde\pi:\frak B\to\frak B$, setting $\tilde\pi a^{\ssbullet}=(\pi
a)^{\ssbullet}$ for every $a\in\frak A$.    Because
$(\pi\phi)\ssptilde=\tilde\pi\tilde\phi$ for all $\pi$, $\phi\in G$,
$\tilde G=\{\tilde\pi:\pi\in G\}$ is a subgroup of $\Aut\frak B$;  of
course it is countable.   Let $\frak D$ be the fixed-point subalgebra of
$\tilde G$ in $\frak B$.   Because $\frak B$ is not $\{0\}$, nor is
$\frak D$.

\medskip

{\bf (e)} $\tilde G$ is countably non-paradoxical.   \Prf\ Suppose that
$b$ is $\tilde G$-$\sigma$-equidecomposable with $1$ in $\frak B$.   Let
$\sequencen{b_n}$ be a partition of unity in $\frak B$ and
$\sequencen{\pi_n}$ a sequence in $G$ such that
$\sequencen{\tilde\pi_nb_n}$ is disjoint and has supremum $b$.
For each $n\in\Bbb N$, let
$a_n\in\frak A$ be such that $a_n^{\ssbullet}=b_n$.   We have

\Centerline{$(a_m\Bcap a_n)^{\ssbullet}=(\pi_m a_m\Bcap\pi_n
a_n)^{\ssbullet}=0$}

\noindent whenever $m\ne n$, so

\Centerline{$d=\sup_{m\ne n}(a_m\Bcap a_n)
  \Bcup\sup_{m\ne n}(a_m\Bcap\pi_m^{-1}\pi_na_n)$}

\noindent belongs to $\Cal I$, while $\sequencen{a_n\Bsetminus d}$,
$\sequencen{\pi_n(a_n\Bsetminus d)}$ are disjoint.

Because $\sup_{n\in\Bbb N}b_n=1$ in $\frak B$,
$d'=1\Bsetminus\sup_{n\in\Bbb N}a_n\in\Cal I$.   Because $\Cal I$ is
a $\sigma$-ideal and $G$ is countable,

\Centerline{$c^*=\upr(d\Bcup d',\frak C)=\sup_{\pi\in G}\pi(d\Bcup d')$}

\noindent belongs to $\Cal I$, while
$\{c^*\}\cup\{a_n\Bsetminus c^*:n\in\Bbb N\}$ is a partition of unity in
$\frak A$.

Define $\psi\in G^*_{\sigma}$ by setting

\Centerline{$\psi a
=\sup_{n\in\Bbb N}\pi_n(a\Bcap a_n\Bsetminus c^*)\Bcup(a\Bcap c^*)$}

\noindent for every $a\in\frak A$.   Then $1\Bsetminus\psi 1\in\Cal I$,
by the definition of $\Cal I$, so $(\psi 1)^{\ssbullet}=1$ in $\frak B$.   But

$$\eqalign{(\psi 1)^{\ssbullet}
&=\sup_{n\in\Bbb N}(\pi_n(a_n\Bsetminus c^*))^{\ssbullet}
=\sup_{n\in\Bbb N}(\pi_na_n)^{\ssbullet}\cr
&=\sup_{n\in\Bbb N}\tilde \pi_nb_n
=b.\cr}$$

\noindent So $b=1$.   As $b$ is arbitrary, $\tilde G$ is countably
non-paradoxical.\ \Qed

\medskip

{\bf (f)} $\tilde G$ has the $\sigma$-refinement property.   \Prf\ Let
$b\in\frak
B$.   Then there is an $a\in\frak A$ such that $a^{\ssbullet}=b$.
Because $G$ is supposed to have the $\sigma$-refinement property, there
is a
$d\Bsubseteq a$ such that $d\psG a\Bsetminus d$ and
$a\Bsetminus\upr(d,\frak C)$ is a relative atom over $\frak C$.
Set $e=d^{\ssbullet}\Bsubseteq b$.

We know that there are a partition of unity $\sequencen{d_n}$ in $\frak
A_d$ and a sequence $\sequencen{\pi_n}$ in $G$ such that
$\pi_nd_n\Bsubseteq a\Bsetminus d$ for every $n$ and
$\sequencen{\pi_nd_n}$ is disjoint.   Now $\sequencen{d_n^{\ssbullet}}$
is a partition of unity in $\frak B_e$,
$\tilde\pi_nd_n^{\ssbullet}\Bsubseteq b\setminus e$ for every $n$, and
$\sequencen{\tilde\pi_nd_n^{\ssbullet}}$ is disjoint;  so $e\pstildeG
b\Bsetminus e$.

Suppose that $b_0\Bsubseteq b\Bsetminus\upr(e,\frak D)$.   Then it is
expressible as $a_0^{\ssbullet}$ where $a_0\Bsubseteq a$ and

\Centerline{$(a_0\Bcap\pi d)^{\ssbullet}=b_0\Bcap\tilde\pi e=0$}

\noindent for every $\pi\in G$.   So if we set
$a_1=a_0\Bsetminus\sup_{\pi\in G}\pi d$, we shall have $a_1\Bsubseteq
a\Bsetminus\upr(d,\frak C)$ and $a_1^{\ssbullet}=b_0$.   Now
$a\Bsetminus\upr(d,\frak C)$ is supposed to be a relative atom over
$\frak C$, so $a_1=a\Bcap c$ for some $c\in\frak C$.   In this case,

\Centerline{$\tilde\pi c^{\ssbullet}=(\pi c)^{\ssbullet}=c^{\ssbullet}$}

\noindent for every $\pi\in G$, so $c^{\ssbullet}\in\frak D$, while
$b_0=b\Bcap c^{\ssbullet}$.   As $b_0$ is arbitrary,
$b\Bsetminus\upr(e,\frak D)$ is a relative atom over $\frak D$.

Thus $e$ has both the properties required by the definition 448K.   As
$b$ is arbitrary, $\tilde G$ has the $\sigma$-refinement property.\ \Qed

\medskip

{\bf (g)} 448N now tells us that there is a sequentially
order-continuous non-negative additive functional $\theta_0:\frak B\to
L^{\infty}(\frak D)$ such that $\theta_01=\chi 1$ and
$\theta_0(\tilde\pi b)=\theta_0b$ whenever $b\in\frak B$ and $\pi\in G$.
If we set $\theta a=\theta_0a^{\ssbullet}$ for $a\in\frak A$,
it is easy to see that
$\theta$ has all the properties required by (i) of this theorem.   Thus
(ii)$\Rightarrow$(i), and the proof is complete.
}%end of proof of 448O

\leader{448P}{}\cmmnt{ At last we come to Polish spaces.

\medskip

\noindent}{\bf Theorem}\cmmnt{ ({\smc Nadkarni 90},
{\smc Becker \& Kechris 96})}  Let $G$ be a Polish group acting on a non-empty Polish
space $(X,\frak T)$ with a Borel measurable action $\action$.   For
Borel sets $E$, $F\subseteq X$ say that $E\psG F$ if there are a
countable partition $\familyiI{E_i}$ of $E$ into Borel sets, and a
family $\familyiI{g_i}$ in $G$, such that $g_i\action E_i\subseteq F$
for every $i$ and $\familyiI{g_i\action E_i}$ is disjoint.
Then the following are equiveridical:

(i) there is a $G$-invariant Radon probability measure $\mu$ on $X$;

(ii) if $F\subseteq X$ is a Borel set such that $X\psG F$, then
$\bigcap_{n\in\Bbb N}g_n\action F\ne\emptyset$ for any sequence
$\sequencen{g_n}$ in $G$;

(iii) there are no disjoint Borel sets $E$, $F\subseteq X$ such that
$X\psG E$ and $X\psG F$.

\proof{{\bf (a)} Let us start with the easy parts.

\medskip

\quad{\bf (i)$\Rightarrow$(ii)} Let $\mu$ be a $G$-invariant Radon
probability measure on $X$, and suppose that $X\psG F$.   Let
$\familyiI{E_i}$ be a countable partition of $X$ into Borel sets and
$\familyiI{h_i}$ a family in $G$ such that $\familyiI{h_i\action E_i}$
is disjoint and $h_i\action E_i\subseteq F$ for every $i$.   Then

\Centerline{$\mu F
\ge\sum_{i\in I}\mu(h_i\action E_i)
=\sum_{i\in I}\mu E_i
=\mu X$,}

\noindent so $F$ is conegligible.   Consequently
$\bigcap_{n\in\Bbb N}g_n\action F$ must be conegligible and cannot be
empty, for any sequence $\sequencen{g_n}$ in $G$.   As $F$ and
$\sequencen{g_n}$ are arbitrary, (ii) is true.

\medskip

\quad{\bf (ii)$\Rightarrow$(iii)} Assume (ii).   \Quer\ If there are
disjoint $E$, $F$ such that $X\psG E$ and $X\psG F$, then we have a
countable partition $\familyiI{E_i}$ of $X$ into Borel sets and a family
$\familyiI{g_i}$ in $G$ such that $g_i\action E_i\subseteq E$ for every
$i\in I$.   But there is an $x\in\bigcap_{i\in I}g_i^{-1}\action F$, by
(ii).   In this case there is a $j\in I$ such that $x\in E_j$ and
$g_j\action x\in E\cap F$, which is impossible.\ \BanG\  So (iii) must
be true.

\medskip

{\bf (b)} For the rest of the proof, therefore, I shall assume (iii) and
seek to prove (i).

Let $\frak T$ be the topology of $X$, and $\Cal B=\Cal B(X)$ its Borel
$\sigma$-algebra.
For $g\in G$ define $\pi_g:\Cal B\to\Cal B$ by writing $\pi_gE=g\action
E$ for every $E\in\Cal B$.   Then $\pi_{gh}=\pi_g\pi_h$ for all $g$,
$h\in G$, so $\tilde G=\{\pi_g:g\in G\}$ is a subgroup of $\Aut\Cal B$.
Observe that for $E$, $F\in\Cal B$, $E\psG F$, in the sense here, iff
$E\pstildeG F$ in the sense of 448A.

By the Becker-Kechris theorem (424H), there is a Polish topology
$\frak T_1$ on $X$, giving rise to the same Borel $\sigma$-algebra
$\Cal B$ as
the original topology, for which the action of $G$ is continuous.   Let
$\Cal U$ be a countable base for $\frak T_1$ closed under finite unions.
(We are going to have three Polish topologies on $X$ in this
proof, so watch carefully.)

\medskip

{\bf (c)} For the time being (down to the end of (f) below) let us
suppose that $G$, and therefore $\tilde G$, are countable.   In this
case, because $\Cal B$ is countably generated,
$\tilde G$ has the $\sigma$-refinement property, by 448L.
We can therefore apply 448O to see that (iii) implies that

\inset{(i)$'$ there are a Dedekind $\sigma$-complete Boolean algebra
$\frak D$, not $\{0\}$, and a $\tilde G$-invariant sequentially
order-continuous non-negative additive functional
$\theta:\Cal B\to L^{\infty}(\frak D)$ such that $\theta X=\chi 1$.}

Express
$\frak D$ as $\Sigma/\Cal J$ where $\Sigma$ is a $\sigma$-algebra of
subsets of
a set $Z$ and $\Cal J$ is a $\sigma$-ideal of $\Sigma$ (314N).  Then we
can identify $L^{\infty}(\frak D)$ with the quotient $\eusm L^{\infty}/W$,
where $\eusm L^{\infty}$ is the space of bounded $\Sigma$-measurable
real-valued functions on $Z$ and $W$ is the set
$\{f:f\in \eusm L^{\infty},\,\{z:f(z)\ne 0\}\in\Cal J\}$ (363Hb).   For
each $E\in\Cal B$, let $f_E\in\eusm L^{\infty}$ be a representative of
$\theta E\in L^{\infty}(\frak D)$;  because $\theta(\pi E)=\theta E$
whenever $E\in\Cal B$ and $\pi\in\tilde G$, we may suppose that
$f_{\pi E}=f_E$ whenever $E\in\Cal B$ and $\pi\in\tilde G$.

Let $\frak B$ be the subalgebra of $\Cal B$ generated by
$\{\pi U:U\in\Cal U,\,\pi\in\tilde G\}$.   Then $\frak B$ is countable
and $\pi E\in\frak B$ for every $E\in\frak B$, $\pi\in \tilde G$.   By
4A3I, there is yet
another Polish topology $\frak S$ on $X$ which is zero-dimensional and
such that every member of $\frak B$ is open-and-closed for $\frak S$.
Of course $\frak S\supseteq\Cal U$, so $\Cal B$ is still the algebra of
$\frak S$-Borel sets (423Fb).   Let $\Cal W$ be a countable base
for $\frak S$ consisting of sets which are open-and-closed for
$\frak S$, and let $\frak B_1$ be the subalgebra of $\Cal B$ generated
by $\Cal W\cup\frak B$;  then $\frak B_1$ is countable and consists of
open-and-closed sets for $\frak S$.   Let $\sequencen{W_n}$ be a
sequence running over $\Cal W$.   Let $\rho$ be a complete metric on $X$
defining the topology $\frak S$, and for $m$, $n\in\Bbb N$ set

\Centerline{$W_{mn}=\bigcup\{W_i:i\le n,\,
   \diam_{\rho}(W_i)\le 2^{-m}\}$;}

\noindent then for each $m\in\Bbb N$, $\sequencen{W_{mn}}$ is a
non-decreasing sequence in $\frak B_1$ with union $X$.

\wheader{448P}{4}{2}{2}{72pt}

{\bf (d)} Consider the subsets of $Z$ of the following types:

\Centerline{$P_E=\{z:f_E(z)<0\}$, where $E\in\frak B_1$,}

\Centerline{$Q_{EF}=\{z:f_{E\cup F}(z)\ne f_E(z)+f_F(z)\}$, where $E$,
$F\in\frak B_1$ and $E\cap F=\emptyset$,}

\Centerline{$R=\{z:f_X(z)\ne 1\}$,}

\Centerline{$S_m=\{z:\sup_{n\in\Bbb N}f_{W_{mn}}(z)\ne 1\}$,
 where $m\in\Bbb N$.}

\noindent Because

\Centerline{$f_E^{\ssbullet}=\theta E\ge 0$ for every $E\in\Cal B$,}

\Centerline{$f_{E\cup F}^{\ssbullet}=\theta(E\cup F)=\theta E+\theta F
=f_E^{\ssbullet}+f_F^{\ssbullet}$ whenever $E\cap F=\emptyset$,}

\Centerline{$f_X^{\ssbullet}=\theta X=\chi 1$,}

\Centerline{$\sup_{n\in\Bbb N}f_{W_{mn}}^{\ssbullet}
=\sup_{n\in\Bbb N}\theta W_{mn}
=\theta(\bigcup_{n\in\Bbb N}W_{mn})
=\theta X=\chi 1$}

\noindent for every $m\in\Bbb N$, all the sets $P_E$, $Q_{EF}$, $R$ and
$S_m$ belong to $\Cal J$.   Since $\frak D\ne\{0\}$, $Z\notin\Cal J$;
so there is a $z_0\in Z$ not belonging to $R$ or $P_E$ or $Q_{EF}$ or
$S_m$ whenever $m\in\Bbb N$ and $E$, $F\in\frak B_1$ are disjoint.

Set $\nu E=f_E(z_0)$ for every $E\in\frak B_1$.   If $E$,
$F\in\frak B_1$ are disjoint, then $\nu(E\cup F)=\nu E+\nu F$ because
$z_0\notin Q_{EF}$;  thus $\nu:\frak B_1\to\Bbb R$ is additive.   If
$E\in\frak B_1$ then $\nu E\ge 0$ because $z_0\notin P_E$, so $\nu$ is
non-negative.   $\nu X=1$ because $z_0\notin R$.   For each
$m\in\Bbb N$, $\sup_{n\in\Bbb N}\nu W_{mn}=1$ because $z_0\notin S_m$.

\medskip

{\bf (e)} For any $\epsilon>0$ there is an $\frak S$-compact set
$K\subseteq X$ such that $\nu E\ge 1-\epsilon$ whenever $E\in\frak B_1$
and $E\supseteq K$.   \Prf\ For each $m\in\Bbb N$ we have a
$k(m)\in\Bbb N$ such that $\nu W_{m,k(m)}\ge 1-2^{-m-1}\epsilon$.   Set
$K=\bigcap_{m\in\Bbb N}W_{m,k(m)}$.   Because every $W_{m,k(m)}$ is
$\frak S$-closed, $K$ is $\frak S$-closed, therefore $\rho$-complete;
because every $W_{m,k(m)}$ is a finite union of sets of diameter at most
$2^{-m}$, $K$ is $\rho$-totally bounded, therefore $\frak S$-compact
(4A2Je).   \Quer\ Suppose, if possible, that $E\in\frak B_1$ is
such that $K\subseteq E$ and $\nu E<1-\epsilon$.   For every
$m\in\Bbb N$,

\Centerline{$\nu(\bigcap_{i\le m}W_{i,k(i)})
\ge 1-\sum_{i=0}^m\nu(X\setminus W_{i,k(i)})
\ge 1-\sum_{i=0}^m2^{-i-1}\epsilon
>1-\epsilon
>\nu E$}

\noindent because $\nu$ is non-negative and finitely additive.   So
$\bigcap_{i\le m}W_{i,k(i)}\setminus E$ must be non-empty.   There is
therefore an ultrafilter $\Cal F$ on $X$ containing
$W_{i,k(i)}\setminus E$ for every $i\in\Bbb N$.   Now for each $i$ there
must be a $j\le k(i)$ such that $\diam W_j\le 2^{-i}$ and
$W_j\in\Cal F$, so $\Cal F$ is
a $\rho$-Cauchy filter, and $\frak S$-converges to $x$ say.   Because
every $W_{i,k(i)}$ is $\frak S$-closed,
$x\in\bigcap_{i\in\Bbb N}W_{i,k(i)}=K$;  because $E\in\frak S$,
$x\notin E$;  but $K$ is
supposed to be included in $E$.\ \Bang

Thus $\inf\{\nu E:K\subseteq E\in\frak B_1\}\ge 1-\epsilon$.   As
$\epsilon$ is arbitrary, we have the result.\ \Qed

\medskip

{\bf (f)} By 416O, there is an $\frak S$-Radon measure $\mu$ on
$X$ extending $\nu$.   Because $\mu$ is just the completion of its
restriction to $\Cal B$, it is also $\frak T$-Radon and
$\frak T_1$-Radon (433Cb).

Now $\mu$ is $G$-invariant.   \Prf\ Take any $g\in G$.   Set
$\mu_gE=\mu(g\action E)$ whenever $E\subseteq X$ and $g\action
E\in\dom\mu$.   The map $x\mapsto g\action x$ is a homeomorphism for
$\frak T_1$, so $\mu_g$ also is a $\frak T_1$-Radon measure.   (Setting
$\phi(x)=g^{-1}\action x$, $\mu_g$ is the image measure $\mu\phi^{-1}$.)
Again because $\frak T$ and $\frak T_1$ have the same Borel
$\sigma$-algebras,
$\mu_g$ is $\frak T$-Radon.   If $E\in\frak B$, then $E$ and
$g\action E$ belong to $\frak B\subseteq\frak B_1$, so

$$\eqalignno{\mu_gE
&=\mu(g\action E)
=\nu(g\action E)
=\nu(\pi_gE)
=f_{\pi_gE}(z_0)
=f_E(z_0)\cr
\noalign{\noindent (because $f_{\pi_gE}=f_E$, as declared in (c) above)}
&=\nu E
=\mu E.\cr}$$

\noindent In particular, $\mu_gE=\mu E$ for every $E$ in the algebra
generated by $\Cal U$.   But $\mu_g$ and $\mu$ are both $\frak T$-Radon
measures, and $\Cal U$ is a base for $\frak T_1$ closed under finite
unions, so $\mu_g=\mu$ (415H(iv)).
As $g$ is arbitrary, $\mu$ is $G$-invariant.\ \Qed

Thus we have found a $G$-invariant Radon probability measure, and (i) is
true.

\medskip

{\bf (g)} Thus (iii)$\Rightarrow$(i) if $G$ is countable.   Now let us
consider the general case.   Because $G$ is a Polish group, it has a
countable dense subgroup $H$.   (Take $H$ to be the subgroup generated
by any countable dense subset of $G$.)   Of course there can be no
disjoint $E$, $F\in\Cal B$ such that $X\psH E$ and $X\psH F$, so there
must be an $H$-invariant Radon probability
measure $\mu$ on $X$, by the arguments of
(b)-(f).   ($H$ need not be a Polish group in its subspace topology.
But if we give it its discrete topology, then $x\mapsto h\action x$ is
still a $\frak T_1$-homeomorphism for every $h\in H$, so the action of
$H$ on $X$ is still continuous if $H$ is given its discrete topology and
$X$ is given $\frak T_1$.)

Now $\mu$ is $G$-invariant.   \Prf\ For any $g\in G$, let $\mu_g$ be the
Radon probability measure defined by setting $\mu_gE=\mu(g\action E)$
whenever this is defined.   (As in (f) above, this formula does define a
probability measure which is Radon for either $\frak T$ or $\frak T_1$.)
Let $f:X\to\Bbb R$ be any bounded $\frak T_1$-continuous function.
Then

\Centerline{$\int f d\mu_g=\int f(g^{-1}\action x)\mu(dx)$}

\noindent (applying 235G with
$\phi(x)=g^{-1}\action x$).   Now there is
a sequence $\sequencen{h_n}$ in $H$ converging to $g$.   In this case,
because $G$ is a topological group, $g^{-1}=\lim_{n\to\infty}h_n^{-1}$.
Because the action of $G$ on $X$ is $\frak T_1$-continuous,
$g^{-1}\action x=\lim_{n\to\infty}h^{-1}_n\action x$, for $\frak T_1$,
for every $x\in X$.   Because $f$ is $\frak T_1$-continuous,
$f(g^{-1}\action x)=\lim_{n\to\infty}f(h_n^{-1}\action x)$ in $\Bbb R$
for every $x\in X$.   By Lebesgue's Dominated Convergence Theorem,

\Centerline{$\int fd\mu_g
=\int f(g^{-1}\action x)\mu(dx)
=\lim_{n\to\infty}f(h_n^{-1}\action x)\mu(dx)
=\lim_{n\to\infty}\int fd\mu_{h_n}
=\int fd\mu$}

\noindent because $\mu$ is $H$-invariant, so $\mu_{h_n}=\mu$ for every
$n$.   As $f$ is arbitrary, $\mu_g=\mu$, by 415I.   As $g$ is arbitrary,
$\mu$ is $G$-invariant.\ \Qed

Thus (iii)$\Rightarrow$(i) in all cases, and the proof is complete.
}%end of proof of 448P

\leader{448Q}{}\dvAnew{2011}\cmmnt{ I turn now to Mackey's theorem.
I pave the
way with a couple of lemmas which are of independent interest.

\medskip

\noindent}{\bf Lemma}
Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure
space with countable Maharam type.   Write $L^0(\Sigma)$ for the
set of $\Sigma$-measurable functions from $X$ to $\Bbb R$.
Then there is a function $T:L^0(\mu)\to L^0(\Sigma)$ such that

($\alpha$) $u=(Tu)^{\ssbullet}$ for every $u\in L^0$,

($\beta$) $(u,x)\mapsto(Tu)(x):L^0\times X\to\Bbb R$ is
$(\Cal B\tensorhat\Sigma)$-measurable,

\noindent where $\Cal B=\Cal B(L^0)$ is the Borel $\sigma$-algebra of
$L^0$ with its topology of convergence in measure.

\proof{{\bf (a)} Consider first the case in which $\mu$ is a probability
measure.

\medskip

\quad{\bf (i)} Let $\sequencen{E_n}$ be a sequence in $\Sigma$ such that
the measure algebra $\frak A$ of $\mu$ is $\tau$-generated by
$\{E_n^{\ssbullet}:n\in\Bbb N\}$.
For $n\in\Bbb N$ let $\Sigma_n$ be
the finite subalgebra of $\Sigma$ generated by $\{E_i:i\le n\}$, and
for $n\in\Bbb N$, $u\in L^{\infty}=L^{\infty}(\mu)$ and $x\in X$ set

$$\eqalign{(S_nu)(x)
&=\Bover1{\mu E}\int_Eu\text{ if }E\text{ is the atom of
}\Sigma_n\text{ containing }x\text{ and }\mu E>0,\cr
&=0\text{ if the atom of }\Sigma_n\text{ containing }x\text{ is
negligible}.\cr}$$

\noindent Then $(u,x)\mapsto(S_nu)(x)$ is
$(\Cal B\tensorhat\Sigma)$-measurable, because
$u\mapsto\int_Eu:L^{\infty}\to\Bbb R$ is
continuous (for the topology of convergence in measure) for every
$E\in\Sigma$.   So if we set
$Su=\limsup_{n\to\infty}S_nu$ for $u\in L^{\infty}$,
$(u,x)\mapsto(Su)(x)$ will be $(\Cal B\tensorhat\Sigma)$-measurable.

On the other hand, if $f\in\eusm L^{\infty}$, $S_nf^{\ssbullet}$ is a
conditional expectation of $f$ on $\Sigma_n$ for each $n$.
So L\'evy's martingale theorem (275I) tells us that
if $f\in\eusm L^{\infty}$ then $\sequencen{S_nf^{\ssbullet}}$ converges
a.e.\ to a conditional expectation $g$ of $f$ on the $\sigma$-algebra
$\Sigma_{\infty}$ generated by $\bigcup_{n\in\Bbb N}\Sigma_n$.   But we
chose $\sequencen{E_n}$ to generate $\frak A$, so
$\frak A=\{E^{\ssbullet}:E\in\Sigma_{\infty}\}$.   If now $E\in\Sigma$,
there is an $F\in\Sigma_{\infty}$ such that $E\symmdiff F$ is negligible,
so

\Centerline{$\int_Eg=\int_Fg=\int_Ff=\int_Ef$.}

\noindent As $E$ is arbitrary,

\Centerline{$f\eae g\eae\limsup_{n\to\infty}S_nf^{\ssbullet}
=Sf^{\ssbullet}$.}

\noindent Turning this round, $(Su)^{\ssbullet}=u$ for every
$u\in L^{\infty}$.

\medskip

\quad{\bf (ii)} Now define $R:L^0\to L^{\infty}$ by setting

\Centerline{$Rf^{\ssbullet}=(\arctan f)^{\ssbullet}$}

\noindent for $f\in\eusm L^0$ (see 241I).   Then $R$ is continuous for
the topology of convergence in measure (245Dd), so
$(u,x)\mapsto(SRu)(x):L^0\times X\to\Bbb R$ is
$(\Cal B\tensorhat\Sigma)$-measurable.   Note that if $u\in L^0$, then
$-\bover{\pi}2<(S_nRu)(x)<\bover{\pi}2$ for every $x$ and $n$, so
$-\bover{\pi}2\le(SRu)(x)\le\bover{\pi}2$ for every $x$;  also, if
$u=f^{\ssbullet}$, then $-\bover{\pi}2<\arctan f(x)<\bover{\pi}2$
whenever $f(x)$ is defined,
so $-\bover{\pi}2<(SRu)(x)<\bover{\pi}2$ for almost every $x$.
If now we set

$$\eqalign{\tan_0t
&=\tan t\text{ for }-\Bover{\pi}2<t<\Bover{\pi}2,\cr
&=0\text{ for }t=\pm\Bover{\pi}2,\cr}$$

\noindent and $Tu=\tan_0SRu$, we shall have $Tu\in L^0(\Sigma)$ and
$(Tu)^{\ssbullet}=u$ for every
$u\in L^0$, while $(u,x)\mapsto(Tu)(x)$ is
$(\Cal B\tensorhat\Sigma)$-measurable.

\medskip

{\bf (b)} For the general case, if $\mu X=0$ the result is trivial,
as we can just set
$(Tu)(x)=0$ for all $u$ and $x$.   So suppose otherwise.   Let
$\nu$ be a probability measure with the same domain and the same negligible
sets as $\mu$ (215B(vii)).   Then the measure algebra of $\nu$, regarded as a
Boolean algebra, is the same as that of $\mu$, so
$\nu$ also has countable Maharam type;  similarly,
$L^0=L^0(\nu)$.   Moreover, the topology of convergence in measure on $L^0$
is the same, whichever measure we take to define it (245Xm, 367T).
So we can apply (a) to $(X,\Sigma,\nu)$.
}%end of proof of 448Q

\leader{448R}{Lemma}\dvAnew{2011}
Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure
space with countable Maharam type.

(a) $L^0=L^0(\mu)$, with its topology of convergence in measure,
is a Polish space.

(b) Let $\frak A$ be the measure algebra
of $\mu$, and $\frak A^f$ the set $\{a:a\in\frak A$, $\bar\mu a<\infty\}$.
Then the Borel $\sigma$-algebra $\Cal B=\Cal B(L^0)$
is the $\sigma$-algebra
of subsets of $L^0$ generated by sets of the form
$\{u:\bar\mu(a\Bcap\Bvalue{u\in F})>\alpha\}$,
where $a\in\frak A^f$, $F\subseteq\Bbb R$ is Borel, and
$\alpha\in\Bbb R$.

\proof{{\bf (a)} By 245Eb, $L^0$ is metrizable, and complete when
regarded as a linear topological space;  so by 4A4Bj there is a metric on
$L^0$, defining its topology, under which $L^0$ is complete.
By 367Rb, $L^0$ is separable, so it is a Polish space.

\medskip

{\bf (b)} Write $\Upsilon$ for the $\sigma$-algebra
of subsets of $L^0$ generated by sets of the form
$\{u:\bar\mu(a\Bcap\Bvalue{u\in F})>\alpha\}$,
where $a\in\frak A^f$, $F\subseteq\Bbb R$ is Borel, and
$\alpha\in\Bbb R$.

\medskip

\quad{\bf (i)}
If $a\in\frak A^f$, $\alpha\in\Bbb R$ and $H\subseteq\Bbb R$ is open, then
$U=\{u:\bar\mu(a\Bcap\Bvalue{u\in H})>\alpha\}$ is open in $L^0$.
\Prf\ If $u\in U$, there are a compact set
$K\subseteq H$ and a $\delta>0$ such
that $\bar\mu(a\Bcap\Bvalue{u\in K})>\alpha+\delta$.   Now there is an
$\eta\in\ocint{0,1}$
such that $|\alpha-\beta|>\eta$ whenever $\alpha\in K$ and
$\beta\in\Bbb R\setminus H$.   In this case,
$V=\{v:v\in L^0$, $\bar\mu(a\Bcap\Bvalue{|u-v|>\eta})\le\delta\}$
is a neighbourhood of $u$ in $L^0$ (367L).   If $v\in V$, then

\Centerline{$\Bvalue{v\in H}
\Bsupseteq\Bvalue{u\in K}\Bcap\Bvalue{|u-v|\le\eta}$,}

$$\eqalign{\bar\mu(a\Bcap\Bvalue{v\in H})
&\ge\bar\mu(a\Bcap\Bvalue{u\in K})-\bar\mu(a\Bcap\Bvalue{|u-v|>\eta})\cr
&\ge\bar\mu(a\Bcap\Bvalue{u\in K})-\delta
>\alpha.\cr}$$

\noindent Thus $V\subseteq U$ and $U$ is a neighbourhood of $u$;  as $u$ is
arbitrary, $U$ is open.\ \Qed

\medskip

\quad{\bf (ii)} Thus $u\mapsto\bar\mu(a\Bcap\Bvalue{u\in H})$ is
$\Cal B$-measurable for every $a\in\frak A^f$ and open $H\subseteq\Bbb R$.
Now the set

\Centerline{$\{F:F\subseteq\Bbb R$ is Borel,
$u\mapsto\bar\mu(a\Bcap\Bvalue{u\in F})$ is $\Cal B$-measurable
for every $a\in\frak A^f\}$}

\noindent is a Dynkin class containing all open sets, so is the Borel
$\sigma$-algebra of $\Bbb R$ (136B), and
$u\mapsto\bar\mu(a\Bcap\Bvalue{u\in F})$ is
$\Cal B$-measurable for every $a\in\frak A^f$ and Borel
$F\subseteq\Bbb R$.   Thus $\Upsilon\subseteq\Cal B$.

\medskip

\quad{\bf (iii)} In the other direction, we know that $\frak A$ is
separable, by 331O;  let $\sequence{k}{c_k}$ run over a dense
subset of $\frak A$.   We also know that there is a sequence
$\sequencen{a_n}$ in $\frak A^f$ with supremum $1$.   Set

\Centerline{$E_{nkqq'}=\{u:u\in L^0$,
$\bar\mu(a_n\Bcap c_k\Bcap\Bvalue{u>q})>q'\}\in\Upsilon$}

\noindent for $n$, $k\in\Bbb N$ and $q$, $q'\in\Bbb Q$.   If $u$,
$v\in L^0$ are different, there are $n$, $k$, $q$ and $q'$ such that
$E_{nkqq'}$ contains one of $u$, $v$ and not the other.   \Prf\ Choose
$q\in\Bbb Q$ such that
$\Bvalue{u>q}\ne\Bvalue{v>q}$.   Suppose for the moment that
$c=\Bvalue{u>q}\Bsetminus\Bvalue{v>q}\ne 0$.   Let $n\in\Bbb N$ be such
that $\bar\mu(a_n\Bcap c)>0$.   Let $k\in\Bbb N$ be such that
$\bar\mu(a_n\Bcap(c\Bsymmdiff c_k))<\bar\mu(a_n\Bcap c)$.   Then

$$\eqalign{\bar\mu(a_n\Bcap c_k\Bcap\Bvalue{v>q})
&\le\bar\mu(a_n\Bcap c_k\Bsetminus c)\cr
&<\bar\mu(a_n\Bcap c)-\bar\mu(a_n\Bcap c\Bsetminus c_k)
\le\bar\mu(a_n\Bcap c_k\Bcap\Bvalue{u>q}),\cr}$$

\noindent so there is a $q'\in\Bbb Q$ such that $u\in E_{nkqq'}$ and
$v\notin E_{nkqq'}$.   Similarly, if
$\Bvalue{v>q}\notBsubseteq\Bvalue{u>q}$ there are $n$, $k\in\Bbb N$ and
$q'\in\Bbb Q$ such that $v\in E_{nkqq'}$ and
$u\notin E_{nkqq'}$.\ \Qed

By 423J, the $\sigma$-algebra
generated by $\{E_{nkqq'}:n$, $k\in\Bbb N$, $q$, $q'\in\Bbb Q\}$
is the whole of
$\Cal B$, and $\Upsilon$ must be equal to $\Cal B$, as claimed.
}%end of proof of 448R

\leader{448S}{Mackey's theorem}\dvAnew{2011} ({\smc Mackey 62})
Let $G$ be a locally compact Polish group,
$(X,\Sigma)$ a standard Borel space
and $\mu$ a $\sigma$-finite measure with domain $\Sigma$.
Let $(\frak A,\bar\mu)$ be the measure algebra of $\mu$
with its measure-algebra topology.   Let $\varaction$
be a Borel measurable action of $G$ on $\frak A$ such that
$a\mapsto g\varaction a$ is a Boolean automorphism for every $g\in G$.
Then we have a $(\Cal B(G)\tensorhat\Sigma,\Sigma)$-measurable action
$\action$ of $G$ on $X$ such that

\Centerline{$g\varaction E^{\ssbullet}=(g\action E)^{\ssbullet}$}

\noindent for every $g\in G$ and $E\in\Sigma$,
writing $g\action E$ for $\{g\action x:x\in E\}$ as usual.

\proof{{\bf (a)} To begin with (down to the end of (j) below) suppose
that $X=\Bbb R$,
with $\Sigma=\Cal B(\Bbb R)$ its Borel $\sigma$-algebra, and that $\mu$ is
totally finite.   The first thing to note is
that for every $g\in G$ the automorphism $a\mapsto g\varaction a$ can be
represented by a Borel automorphism $f_g:\Bbb R\to\Bbb R$ such that
$g\varaction E^{\ssbullet}=f_g^{-1}[E]^{\ssbullet}$ for every
$E\in\Cal B(\Bbb R)$ (425Ac).   Of course $f_g$ belongs to the
space $L^0(\Sigma)$ of $\Sigma$-measurable
functions from $\Bbb R$ to itself, so we can speak of its
equivalence class
$f_g^{\ssbullet}\in L^0(\mu)$.   If we give $L^0(\mu)$ its topology of
convergence in measure, it is a Polish space (448Ra).

The function $g\mapsto f_g^{\ssbullet}:G\to L^0(\mu)$ is Borel measurable.
\Prf\ If $E\in\Cal B(\Bbb R)$, $a\in\frak A$
and $\alpha\in\Bbb R$, then, setting $b=E^{\ssbullet}$,

\Centerline{$\Bvalue{f_g^{\ssbullet}\in E}=f_g^{-1}[E]^{\ssbullet}
=g\varaction b$}

\noindent for every $g\in G$, so

\Centerline{$\{g:\bar\mu(a\Bcap\Bvalue{f_g^{\ssbullet}\in E})>\alpha\}
=\{g:\bar\mu(a\Bcap(g\varaction b))>\alpha\}$}

\noindent is a Borel set in $G$, because $\varaction$ is Borel
measurable and $\{c:\bar\mu(a\Bcap c)>\alpha\}$ is open in $\frak A$.
Thus $\Cal B(G)$ contains the inverse images of the sets generating
$\Cal B(L^0(\mu))$ described in 448Rb,
and therefore the inverse image of every set in $\Cal B(L^0(\mu))$,
as required.\ \Qed

\medskip

{\bf (b)} By 448Q, there is a function
$T:L^0(\mu)\to L^0(\Sigma)$
such that $(Tu)^{\ssbullet}=u$ for every $u\in L^0(\mu)$ and
$(u,x)\mapsto (Tu)(x)$ is
$(\Cal B(L^0(\mu))\tensorhat\Cal B(\Bbb R))$-measurable.
Define $\phi:G\times\Bbb R\to\Bbb R$ by setting

\Centerline{$\phi(g,x)=(Tf_g^{\ssbullet})(x)$}

\noindent for $g\in G$ and $x\in\Bbb R$.   Then $\phi$ is a composition of
the Borel measurable functions $(g,x)\mapsto(f_g^{\ssbullet},x)$ and
$(u,x)\mapsto(Tu)(x)$, so is Borel measurable;  and if $g\in G$ then
$\phi(g,x)=f_g(x)$ for
$\mu$-almost every $x$, because $f_g\eae Tf_g^{\ssbullet}$.   Now

\Centerline{$g\varaction E^{\ssbullet}
=(f_g^{-1}[E])^{\ssbullet}
=\{x:\phi(g,x)\in E\}^{\ssbullet}$}

\noindent for every $g\in G$ and $E\in\Cal B(\Bbb R)$.

\medskip

{\bf (c)} Let $\lambda$ be a Haar measure on $G$.   Because $G$ is a Polish
space and $\lambda$ is a Radon measure on $G$, $\lambda$ is
$\sigma$-finite (411Ge) and $L^0(\lambda)$, with its
topology of convergence in measure, is a Polish space (448Ra again).
For $x\in\Bbb R$, set
$\phi_x(g)=\phi(g,x)$ for $g\in G$;  then $\phi_x:G\to\Bbb R$ is Borel
measurable.    Set $\theta(x)=\phi_x^{\ssbullet}$ in $L^0(\lambda)$.
Then $\theta:\Bbb R\to L^0(\lambda)$ is Borel measurable.
\Prf\ Again I use the
characterization of the Borel $\sigma$-algebra of $L^0(\lambda)$ in
448Rb.   Let $(\frak C,\bar\lambda)$ be the measure algebra of
$\lambda$.
If $c\in\frak C$, $\bar\lambda c<\infty$, $E\subseteq\Bbb R$ is Borel, and
$\alpha\in\Bbb R$, take a Borel set $F\subseteq G$ such that
$c=F^{\ssbullet}$;  then

$$\eqalign{\{x:\bar\lambda(c\Bcap\Bvalue{\theta(x)\in E})>\alpha\}
&=\{x:\lambda(F\cap\phi_x^{-1}[E])>\alpha\}\cr
&=\{x:\lambda\{g:g\in F,\,\phi_x(g)\in E\}>\alpha\}\cr
&=\{x:\lambda\{g:g\in F,\,\phi(g,x)\in E\}>\alpha\}\cr
&=\{x:\lambda W^{-1}[\{x\}]>\alpha\}\cr}$$

\noindent where $W=\{(g,x):g\in F$, $x\in\Bbb R$, $\phi(g,x)\in E\}$ is a
Borel subset of $G\times\Bbb R$.   But this means that
$W\in\Cal B(G)\tensorhat\Cal B(\Bbb R)$ (4A3Ga) and
$x\mapsto\lambda W^{-1}[\{x\}]$ is Borel measurable (252P),
so $\{x:\bar\lambda(c\Bcap\Bvalue{\theta(x)\in E})>\alpha\}$ is a Borel
subset of $\Bbb R$.   Thus the inverse image of every set in the generating
family for the Borel $\sigma$-algebra of $L^0(\lambda)$
is a Borel set, and we have a Borel measurable function.\ \Qed

Let $\nu$ be the totally finite Borel measure on $L^0(\lambda)$ defined
by setting
$\nu F=\mu\theta^{-1}[F]$ for every Borel set $F\subseteq L^0(\lambda)$.

\medskip

{\bf (d)} If $E\subseteq\Bbb R$ is Borel, there is a set
$A\subseteq L^0(\lambda)$ such that $E\symmdiff\theta^{-1}[A]$ is
$\mu$-negligible.   \Prf\ Let $\hat\mu$,
$\hat\nu$ be the completions of $\mu$ and $\nu$, so that $\theta$ is \imp\
for $\hat\mu$ and $\hat\nu$ (234Ba).
Because $E$ is a Borel subset of $\Bbb R$,
$\theta[E]$ is an analytic subset
of $L^0(\lambda)$ (423Gb), therefore Souslin-F (423Eb);
accordingly $\hat\nu$ measures $\theta[E]$
(431B).   Let $\Tau_0$ be the
$\sigma$-algebra of subsets of $L^0(\lambda)$ generated by the Souslin-F
subsets of $L^0(\lambda)$.   By 423Q, there is a $\Tau_0$-measurable
function $\theta':\theta[E]\to E$ such that $\theta\theta'$ is the identity
on $\theta[E]$.   Now $\Tau_0$ is included in the domain $\hat\Tau$ of
$\hat\nu$, so $\theta'$ is $\hat\Tau$-measurable, and there is a
Borel set $F_0\subseteq\theta[E]$ such that
$\theta'\restr F_0$ is Borel measurable and
$\theta[E]\setminus F_0$ is $\hat\nu$-negligible (212Fa).
Since $\theta'$ is
surely injective, $E_0=\theta'[F_0]$ is a Borel subset of $E$ (423Ib)
and $\theta\restr E_0$ is a bijection from $E_0$ to $F_0$ with inverse
$\theta'$.   Note that

\Centerline{$\mu(E\setminus\theta^{-1}[F_0])
\le\hat\mu(\theta^{-1}[\theta[E]]\setminus\theta^{-1}[F_0])
=\hat\nu(\theta[E]\setminus F_0)
=0$.}

Define $\psi:\Bbb R\to\Bbb R$ by setting

$$\eqalign{\psi(x)
&=\theta'\theta(x)\text{ if }x\in\theta^{-1}[F_0],\cr
&=x\text{ otherwise}.\cr}$$

\noindent Then $\psi$ is a Borel measurable
function and $\theta\psi=\theta$, that is,
$\phi_x^{\ssbullet}=\phi_{\psi(x)}^{\ssbullet}$ for every $x\in\Bbb R$.
Consequently

\Centerline{$\{(g,x):g\in G$, $x\in\Bbb R$,
$\phi(g,x)\ne\phi(g,\psi(x))\}$}

\noindent has $\lambda$-negligible horizontal sections.
Since it is a Borel set, it must have many $\mu$-negligible vertical
sections;  let $g_0\in G$ be such that
$\{x:\phi(g_0,x)\ne\phi(g_0,\psi(x))\}$ is $\mu$-negligible.
By (b), we also have $\phi(g_0,x)=f_{g_0}(x)$ for $\mu$-almost every $x$.
So the Borel set
$H=\{x:f_{g_0}(x)=\phi(g_0,x)=\phi(g_0,\psi(x))\}$
is $\mu$-conegligible.

Set $A=\theta[E\cap\theta^{-1}[F_0]\cap H]$.
Of course $A\subseteq F_0$.   If $x\in\theta^{-1}[A]\setminus E$,
then there is a
$y\in E\cap\theta^{-1}[F_0]\cap H$ such that $\theta(y)=\theta(x)$;  now
$\psi(y)=\psi(x)$ and $x\ne y$, so

\Centerline{$\phi(g_0,\psi(x))=\phi(g_0,\psi(y))=f_{g_0}(y)\ne f_{g_0}(x)$}

\noindent and $x\notin H$.   Thus
$\theta^{-1}[A]\setminus E\subseteq\Bbb R\setminus H$ is $\mu$-negligible.
On the other hand,
$E\setminus\theta^{-1}[A]\subseteq E\setminus(\theta^{-1}[F_0]\cap H)$
is also $\mu$-negligible.   So
$E\symmdiff\theta^{-1}[A]$ is $\mu$-negligible, as required.\ \Qed

\medskip

{\bf (e)} There is a $\mu$-conegligible Borel set
$H\subseteq\Bbb R$ such that
$\theta\restr H$ is injective.   \Prf\ Let $\sequencen{E_n}$ be a sequence
of Borel sets in $\Bbb R$ such that whenever $x$, $y\in\Bbb R$ are
distinct there is an $n$ such that $x\in E_n$ and $y\notin E_n$.
For each $n\in\Bbb N$
let $A_n\subseteq L^0(\lambda)$ be such that
$E_n\symmdiff\theta^{-1}[A_n]$ is
$\mu$-negligible;  let $H$ be a $\mu$-conegligible Borel set disjoint from
$\bigcup_{n\in\Bbb N}(E_n\symmdiff\theta^{-1}[A_n])$.   If $x$, $y\in H$
are distinct, there is an $n\in\Bbb N$ such that $x\in E_n$ and
$y\notin E_n$;  now $x\in\theta^{-1}[A_n]$ and $y\notin\theta^{-1}[A_n]$,
so $\theta(x)\ne\theta(y)$.\ \Qed

\medskip

{\bf (f)} Of course $\theta[H]$ is now a Borel subset of $L^0(\lambda)$,
and must be
$\hat\nu$-conegligible.   Let $\frak B$ be the measure algebra of $\nu$,
and $\pi:\frak B\to\frak A$ the measure-preserving homomorphism defined by
setting $\pi F^{\ssbullet}=\theta^{-1}[F]^{\ssbullet}$ for every Borel set
$F$.   If $E\subseteq\Bbb R$ is Borel, then
$E^{\ssbullet}=\pi(\theta[E\cap H])^{\ssbullet}$ belongs to $\pi[\frak B]$,
so $\pi$ is surjective and is an isomorphism.

\medskip

{\bf (g)} Recall that
we have a continuous action $\action_l$ of $G$ on $L^0(\lambda)$
defined as in 443G.   If $g\in G$, then

\Centerline{$g\action_l\theta(x)=\theta(\phi(g^{-1},x))
=\theta(f_{g^{-1}}(x))$}

\noindent for $\mu$-almost
every $x\in\Bbb R$.   \Prf\ Consider the set
$\{(h,x):\phi(g^{-1}h,x)=\phi(h,\phi(g^{-1},x))\}
\subseteq G\times\Bbb R$.
Because $h\mapsto g^{-1}h$ is continuous, it is Borel measurable,
so $(h,x)\mapsto\phi(g^{-1}h,x)$ is Borel measurable;
the same is true of $(h,x)\mapsto\phi(h,\phi(g^{-1},x))$, so
$\{(h,x):\phi(g^{-1}h,x)=\phi(h,\phi(g^{-1},x))\}$ is a Borel set.
For given $h\in G$ and $E\in\Cal B(\Bbb R)$, set
$F=\{x:\phi(h,x)\in E\}$;   then

$$\eqalign{\{x:\phi(g^{-1}h,x)\in E\}^{\ssbullet}
&=(g^{-1}h)\varaction E^{\ssbullet}
=g^{-1}\varaction(h\varaction E^{\ssbullet})
=g^{-1}\varaction\{x:\phi(h,x)\in E\}^{\ssbullet}\cr
&=g^{-1}\varaction F^{\ssbullet}
=\{x:\phi(g^{-1},x)\in F\}^{\ssbullet}
=\{x:\phi(h,\phi(g^{-1},x))\in E\}^{\ssbullet}\cr}$$

\noindent so
$\{x:\phi(g^{-1}h,x)\in E\}\symmdiff\{x:\phi(h,\phi(g^{-1},x))\in E\}$
is $\mu$-negligible.   As $E$ is arbitrary,
$\phi(g^{-1}h,x)=\phi(h,\phi(g^{-1},x))$ for $\mu$-almost
every $x$.

This is true for every $h\in G$.   So there is a
$\mu$-conegligible Borel set $H'\subseteq\Bbb R$ such that
if $x\in H'$ then
$\phi(g^{-1}h,x)=\phi(h,\phi(g^{-1},x))$ for $\lambda$-almost every $h$.
But this means that if $x\in H'$ then

\Centerline{$(g\action_l\phi_x)(h)
=\phi_x(g^{-1}h)
=\phi(g^{-1}h,x)
=\phi(h,\phi(g^{-1},x))
=\phi_{\phi(g^{-1},x)}(h)$}

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

\Centerline{$g\action_l\theta(x)
=g\action_l\phi_x^{\ssbullet}
=(g\action_l\phi_x)^{\ssbullet}
=\phi_{\phi(g^{-1},x)}^{\ssbullet}
=\theta(\phi(g^{-1},x))$.}

\noindent Thus $g\action_l\theta(x)=\theta(\phi(g^{-1},x))$ for almost
every $x$.   And of course we already know from (b) that
$\phi(g^{-1},x)=f_{g^{-1}}(x)$ for almost every $x$.\ \Qed

\medskip

{\bf (h)} We have a function $\varaction_l:G\times\frak B\to\frak B$
defined by setting

\Centerline{$g\smallcirc_lF^{\ssbullet}=(g\action_lF)^{\ssbullet}$}

\noindent for every Borel set $F\subseteq L^0(\lambda)$ and $g\in G$,
writing
$g\action_lF=\{g\action_lu:u\in F\}$ as in 4A5Bc.   \Prf\ Take any
$g\in G$.   By (g) just above, applied to $g^{-1}$,
$g^{-1}\action_l\theta(x)=\theta(f_g(x))$
for $\mu$-almost every $x$.   Because the shift
operator $u\mapsto g\action_lu:L^0(\lambda)\to L^0(\lambda)$ is a
homeomorphism, it is a
Borel automorphism, and $g\action_lF$ is a Borel set for every Borel set
$F\subseteq L^0(\lambda)$.   If $\nu F=0$, then

$$\eqalign{\nu(g\action_lF)
&=\mu\{x:\theta(x)\in g\action_lF\}
=\mu\{x:g^{-1}\action_l\theta(x)\in F\}\cr
&=\mu\{x:\theta(f_g(x))\in F\}
=\mu(f_g^{-1}[\theta^{-1}[F]])
=0\cr}$$

\noindent because $\theta^{-1}[F]$ is $\mu$-negligible and $f_g$
represents an automorphism of the measure algebra $\frak A$ of $\mu$.
It follows that
$(g\action_lF_0)\symmdiff(g\action_lF_1)=g\action_l(F_0\symmdiff F_1)$ is
$\nu$-negligible and
$(g\action_lF_0)^{\ssbullet}=(g\action_lF_1)^{\ssbullet}$ whenever
$F_0^{\ssbullet}=F_1^{\ssbullet}$, which is what we need to know.\ \Qed

\medskip

{\bf (i)} For any $b\in\frak B$ and $g\in G$,
$g\varaction\pi b=\pi(g\varaction_lb)$.   \Prf\ Let
$F\subseteq L^0(\lambda)$ be
a Borel set such that $b=F^{\ssbullet}$, and set $E=\theta^{-1}[F]$,
$a=E^{\ssbullet}=\pi b$.   Then
$g\varaction_lb=(g\action_lF)^{\ssbullet}$, so

$$\eqalign{\pi(g\varaction_lb)
&=\{x:\theta(x)\in g\action_lF\}^{\ssbullet}
=\{x:g^{-1}\action_l\theta(x)\in F\}^{\ssbullet}
=\{x:\theta(f_g(x))\in F\}^{\ssbullet}\cr
&=\{x:f_g(x)\in E\}^{\ssbullet}
=(f_g^{-1}[E])^{\ssbullet}
=g\varaction a
=g\varaction\pi b,\cr}$$

\noindent as required.\ \Qed

\medskip

{\bf (j)} Now observe that because $\lambda$ is a Haar measure,
$\lambda G>0$, so $L^0(\lambda)\ne\{0\}$, $L^0(\lambda)$ is uncountable and
$\#(L^0(\lambda))=\frak c=\#(\Bbb R)$ (423L).   By
425Ad, there is a Borel isomorphism
$\tilde\theta:\Bbb R\to L^0(\lambda)$ which represents $\pi$.   Set

\Centerline{$g\action x=\tilde\theta^{-1}(g\action_l\tilde\theta(x))$}

\noindent for $g\in G$ and $x\in\Bbb R$.   Then
$\action:G\times\Bbb R\to\Bbb R$ is a composition of the Borel measurable
functions $(g,x)\mapsto(g,\tilde\theta(x))$, $(g,u)\mapsto g\action_lu$ and
$u\mapsto\tilde\theta^{-1}(u)$, so is Borel measurable.
Because $\Sigma=\Cal B(\Bbb R)$ and
$\Cal B(G\times\Bbb R)=\Cal B(G)\tensorhat\Cal B(\Bbb R)$ (4A3Ga again),
$\action$ is $(\Cal B(G)\tensorhat\Sigma,\Sigma)$-measurable.   If $g$,
$h\in G$ and $x\in\Bbb R$,

\Centerline{$gh\action x
=\tilde\theta^{-1}(gh\action_l\tilde\theta(x))
=\tilde\theta^{-1}(g\action_l(h\action_l\tilde\theta(x)))
=\tilde\theta^{-1}(g\action_l\tilde\theta(h\action x))
=g\action(h\action x)$,}

\noindent and if $e$ is the identity of $G$ and $x\in\Bbb R$,

\Centerline{$e\action x
=\tilde\theta^{-1}(e\action_l\tilde\theta(x))
=\tilde\theta^{-1}\tilde\theta(x)
=x$.}

\noindent Thus $\action$ is an action of $G$ on $\Bbb R$.   If
$g\in G$ and $E\in\Cal B(\Bbb R)$, set $F=\tilde\theta[E]$.   Then
$F^{\ssbullet}=\pi^{-1}[E^{\ssbullet}]$, so

\Centerline{$(\tilde\theta^{-1}[g\action_lF])^{\ssbullet}
=\pi((g\action_lF)^{\ssbullet})
=\pi(g\varaction_lF^{\ssbullet})
=g\varaction\pi F^{\ssbullet}
=g\varaction E^{\ssbullet}$.}

\noindent As

\Centerline{$g\action E
=\{g\action x:x\in E\}
=\{\tilde\theta^{-1}(g\action_l\tilde\theta(x)):x\in E\}
=\{\tilde\theta^{-1}(g\action_l u):u\in F\}
=\tilde\theta^{-1}[g\action_l F]$,}

\noindent we see that

\Centerline{$g\varaction E^{\ssbullet}=(g\action E)^{\ssbullet}$}

\noindent as required in the statement of this theorem.

\medskip

{\bf (k)} Thus the result is true if
$(X,\Sigma)=(\Bbb R,\Cal B(\Bbb R))$ and $\mu$ is totally finite.   As for
non-totally-finite $\mu$, there will always be a totally finite measure
$\mu_1$ with the same domain and the same null ideal (215B again), in which
case the measure algebra of $\mu_1$ will have the same Boolean algebra
$\frak A$, though with a different measure $\bar\mu_1$.   However the
measure-algebra topology of $\frak A$ is unchanged (324H), so
$\varaction$ is still Borel measurable, and we can use the Borel
measurable action of $G$ on $\Bbb R$ found by the method of (a)-(j) above.
Since we are assuming that $(X,\Sigma)$ is a standard Borel space,
this covers all the cases in which $X$ is uncountable, by 424C-424D.

\medskip

{\bf (l)} We are left with the case of countable $X$.   This is of course
essentially trivial.
$\Sigma=\Cal PX$ and $\mu$ is a point-supported measure.   Let $Y$ be the
set of atoms of $\mu$, that is, the set $\{x:\mu\{x\}>0\}$.   Then we can
identify the measure algebra $\frak A=\Cal PX/\Cal P(X\setminus Y)$
with $\Cal PY$, in which case the equivalence class $E^{\ssbullet}$ of any
$E\subseteq X$ becomes identified with $E\cap Y$.
As in part (a) of the proof above, we can represent each automorphism
$a\mapsto g\varaction a:\frak A\to\frak A$ by a
permutation $f_g:X\to X$, and we must have
$f_g^{-1}[Y]=Y$.   Try

$$\eqalign{g\action x
&=f_g^{-1}(x)\text{ if }x\in Y,\cr
&=x\text{ if }x\in X\setminus Y\cr}$$

\noindent for every $g\in G$.   If $g$, $h\in G$ and $x\in Y$,

$$\eqalign{\{gh\action x\}
&=\{f_{gh}^{-1}(x)\}
=gh\varaction\{x\}
=g\varaction(h\varaction\{x\})
=g\varaction f_h^{-1}[\{x\}]\cr
&=f_g^{-1}[f_h^{-1}[\{x\}]]
=f_g^{-1}[\{f_h^{-1}(x)\}]
=f_g^{-1}\{h\action x\}
=\{g\action(h\action x)\}\cr}$$

\noindent and $gh\action x=g\action(h\action x)$;  if $x\in X\setminus Y$,
then $gh\action x=x=g\action(h\action x)$.
Of course $e\action x=x$ for every $x\in X$.
So $\action$ is an action of
$G$ on $X$.   To see that it is
$(\Cal B(G)\tensorhat\Sigma,\Sigma)$-measurable, note that the
measure-algebra topology of $\frak A\cong\Cal PY$ is the discrete topology.
If $y\in Y$, then $\{g:g\action y=z\}=\{g:g\varaction\{y\}=\{z\}\}$ is a
Borel set for every $z\in X$;  if $x\in X\setminus Y$, then
$\{g:g\action x=z\}$ is either $G$ or $\emptyset$ for every $z\in X$.
So

\Centerline{$\{(g,x):g\action x\in W\}
=\bigcup_{z\in W,x\in X}\{g:g\action x=z\}\times\{x\}$}

\noindent belongs to $\Cal B(G)\tensorhat\Sigma$
for every subset $W$ of $X$, and
$\action$ is $(\Cal B(G)\tensorhat\Sigma,\Sigma)$-measurable.
Finally, if $g\in G$,

\Centerline{$(g\action E)^{\ssbullet}
=(g\action E)\cap Y
=Y\cap f_g^{-1}[E]
=(f_g^{-1}[E])^{\ssbullet}
=g\varaction E^{\ssbullet}$}

\noindent for every $E\subseteq X$.   So again we have a suitable action of
$G$ on $X$.

This completes the proof.
}%end of proof of 448S

\leader{448T}{Corollary}\dvAnew{2011} Let $G$ be a $\sigma$-compact
locally compact Hausdorff group, $X$ a Polish space,
$\mu$ a $\sigma$-finite Borel measure on $X$,
and $(\frak A,\bar\mu)$ the measure algebra of $\mu$,
with its measure-algebra topology.   Let $\varaction$
be a continuous action of $G$ on $\frak A$ such that
$a\mapsto g\varaction a$ is a Boolean automorphism for every $g\in G$.
Then we have a Borel measurable action
$\action$ of $G$ on $X$ such that

\Centerline{$g\varaction E^{\ssbullet}=(g\action E)^{\ssbullet}$}

\noindent for every $g\in G$ and $E\in\Cal B(X)$.

\proof{ We know that $\frak A$ is separable (331O again)
and metrizable (323Gb);   let $\sequencen{a_n}$ run over a
topologically dense subset of $\frak A$ and $\sequencen{U_n}$ over a base
for its topology.   For each $(m,n)$ such that $a_m\in U_n$,
$V_{mn}=\{g:g\varaction a_m\in U_n\}$ is a neighbourhood of the identity
$e$ of $G$.   By 4A5S, there is a compact normal subgroup $H$ of $G$ such
that $H\subseteq\bigcap\{V_{mn}:m$, $n\in\Bbb N$, $a_m\in U_n\}$
and $G/H$ is Polish.
Now we have a continuous action $\bar{\varaction}$ of $G/H$ on $\frak A$
such that $g^{\ssbullet}\bar{\varaction} a=g\varaction a$ for every
$g\in G$ and $a\in\frak A$.   \Prf\ If $g$, $h\in G$ are such that
$g^{\ssbullet}=h^{\ssbullet}$, $m\in\Bbb N$, and $a_m\in U_n$, then
$g^{-1}h\in V_{mn}$ so $g^{-1}h\varaction a_m\in U_n$.
As $n$ is arbitrary, $g^{-1}h\varaction a_m=a_m$;  as
$a\mapsto g^{-1}h\varaction a$ is continuous, and $m$ is arbitrary,
$g^{-1}h\varaction a=a$ and $h\varaction a=g\varaction a$ for every
$a\in\frak A$.   This shows that the given formula defines a function
$\bar{\varaction}$ from $(G/H)\times\frak A$ to $\frak A$.   It is easy to
check that $\bar{\varaction}$ is an action of $G/H$ on $\frak A$.

Now suppose that
$v\in G/H$, $a\in\frak A$ and $U$ is a neighbourhood of
$v\bar{\varaction}a$
in $\frak A$.   Let $g\in G$ be such that $g^{\ssbullet}=v$;  then
$g\varaction a=v\bar{\varaction} a$, so there are
open sets $V\subseteq G$ and $U'\subseteq\frak A$ such that $g\in V$,
$a\in U'$ and $h\varaction b\in U$ whenever $h\in V$ and $b\in U'$.
By 4A5Ja, $W=\{h^{\ssbullet}:h\in V\}$ is open in $G/H$;  now
$v\in W$ and $w\bar{\varaction} b\in U$ whenever $w\in W$ and
$b\in U'$.   As $v$, $a$ and $U$ are arbitrary, $\bar{\varaction}$ is
continuous.\ \Qed

There is therefore a Borel measurable action
$\bar{\action}:(G/H)\times X\to X$ such that
$v\bar{\varaction} E^{\ssbullet}=(v\bar{\action}E)^{\ssbullet}$ whenever
$v\in G/H$ and $E\in\Cal B(X)$ (448S).
Set $g\action x=g^{\ssbullet}\bar{\action}x$ for $g\in G$ and $x\in X$.
It is elementary to check that $\action$ is an action of $G$ on $X$.
Also it is Borel measurable, because $(g,x)\mapsto(g^{\ssbullet},x)$ is
continuous, therefore Borel measurable, and
$(g^{\ssbullet},x)\mapsto g^{\ssbullet}\bar{\action}x$ is Borel measurable.
If $g\in G$ and $E\in\Cal B(X)$, then

\Centerline{$g\varaction E^{\ssbullet}
=g^{\ssbullet}\bar{\varaction} E^{\ssbullet}
=(g^{\ssbullet}\bar{\action}E)^{\ssbullet}
=(g\action E)^{\ssbullet}$,}

\noindent so $\action$ is an action of the kind we seek.
}%end of proof of 448T

\exercises{
\leader{448X}{Basic exercises (a)}
%\spheader 448Xa
Show that the results in 448Fb and 448Fd remain true if $G$ is not
assumed to be countable.
%448F

\spheader 448Xb
In part (c) of the proof of 448O, show that $\Cal I$ is just the set of
those $d\in\frak A$ such that $d\Bsubseteq\upr(1\Bsetminus a,\frak C)$
for some $a$ such that $1\psG a$.
%448O

\spheader 448Xc   Show that, in part (c) of the proof of 448P, we can
if we wish take $Z=X$ and $\Sigma=\Cal B$.
%448P

\sqheader 448Xd Let $(X,\Sigma)$ be a standard Borel space and
$\Sigma_0$ a countable subalgebra of $\Sigma$.   Show that there is a
sequence $\sequencen{\sequence{i}{E_{ni}}}$ of partitions of unity in
$\Sigma$ such that whenever $\nu:\Sigma\to\Bbb R$ is a finitely additive
functional and $\nu X=\sum_{i=0}^{\infty}\nu E_{ni}$ for every $n\in\Bbb
N$, then $\nu\restr\Sigma_0$ is countably additive.
%448P

\sqheader 448Xe  Set $X=[0,1]\setminus\Bbb Q$, $G=\Bbb Q$ and define
$\action:G\times X\to X$ by requiring that $g\action x-g-x\in\Bbb Z$ for
$g\in G$ and $x\in X$.   Show that this is a Borel measurable action and that Lebesgue
measure on $X$ is $G$-invariant.   Find a metric on $X$, inducing its
topology, for which all the maps $x\mapsto g\action x$ are isometries.
%448P

\sqheader 448Xf Show that a Polish group carries Haar measures iff it is
locally compact.   \Hint{443E.}

\spheader 448Xg Give $\Bbb Z^{\Bbb N}$ its usual (product) topology and
abelian group structure.   Show that it is a Polish group,
and has no Haar measure.
%448P

\sqheader 448Xh Let $(X,\rho)$ be a metric space, $G$ a group and
$\action$ an action of $G$ on $X$ such that $x\mapsto g\action x$ is an
isometry for every $g\in G$.   (i) Show that if $\mu$ is a $G$-invariant
quasi-Radon probability measure on $X$ then $\{g\action x:g\in G\}$ is
totally bounded for every $x$ in the support of $\mu$.   (ii) Show that
if the action is transitive and there is a non-zero $G$-invariant
quasi-Radon measure on $X$, then $X$ is covered by totally bounded open
sets.   (iii) Suppose that $X$ has measure-free weight (see \S438;  for
instance, $X$ could be separable).   Show that if the action is
transitive and there is a $G$-invariant topological
probability measure on $X$ then $X$ is totally bounded.
%448P

\spheader 448Xi Let $(\frak A,\bar\mu)$ be a measure algebra, with the
operation $\Bsymmdiff$ and the measure-algebra topology.   (i) Show that
$\frak A$ is a topological group.   (ii) Show that if $\bar\mu$ is
$\sigma$-finite and $\frak A$ has countable Maharam type, it is a Polish
group.   (iii) Show that if $(\frak A,\bar\mu)$ is semi-finite and not
purely atomic, then $\frak A$ has no Haar measure.
%448Xh 448P

\spheader 448Xj Let $(\frak A,\bar\mu)$ be the measure algebra of
Lebesgue measure on $[0,1]$, with its measure metric, and
$G=\Aut_{\bar\mu}\frak A$ the group of measure-preserving automorphisms
on $\frak A$.   (i) Show that if we give $G$ the topology induced by the
topology of pointwise convergence on
the isometry group of $\frak A$, then it is a
Polish group.   \Hint{441Xq.}  (ii) Show that if $\nu$ is a
$G$-invariant topological probability measure on $\frak A$, then
$\nu\{0,1\}=1$.
%448Xh 448P

\leader{448Y}{Further exercises (a)}
%\spheader 448Ya
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean algebra, and $G$ a
subgroup of $\Aut\frak A$;  let $G^*_{\sigma}$ be the countably full
local semigroup generated by $G$, and write $H$ for the union of all the
full local semigroups generated by countable subgroups of $G$ (following
the definition in 395A as written, without troubling about whether
$\frak A$ is Dedekind complete).   (i) Show that
$G^*_{\sigma}\subseteq H$.   (ii) Find an example in which
$H\ne G^*_{\sigma}$.   (iii) Show
that if $\frak A$ is Dedekind complete then $G^*_{\sigma}=H$.   (iv)
Show that if $\frak A$ is ccc then $G^*_{\sigma}=H$ is the full local
semigroup generated by $G$.
%448A %mt44bits

\spheader 448Yb In 448N, show that $\theta$ is uniquely defined.
%448N

\spheader 448Yc Let $(X,\Sigma)$ be a standard Borel space, $Y$ any
set, $\Tau$ a $\sigma$-algebra of subsets of $Y$ and $\Cal J$ a
$\sigma$-ideal of subsets of $\Tau$.
Let $\theta:\Sigma\to L^{\infty}(\Tau/\Cal J)$ be a non-negative,
sequentially
order-continuous additive function.   Show that there is a
non-negative, sequentially order-continuous additive function
$\phi:\Sigma\to L^{\infty}(\Tau)$ such that (identifying
$L^{\infty}(\Tau/\Cal J)$ with a quotient space of $L^{\infty}(\Tau)$)
$\theta E=(\phi E)^{\ssbullet}$ for every $E\in\Sigma$.
%448P
}%end of exercises

%\query:  Mackey's theorem for semigroup actions?

\cmmnt{\Notesheader{448} The keys to the first part of the section
are in 448F, 448G and 448L.   Even though we no longer have
a Dedekind complete algebra, the fact that we are working with countable
groups means that the suprema we actually need are defined.   The final
step, however, uses yet another idea.   In a standard Borel space, given
a finitely additive functional on the $\sigma$-algebra, we can sometimes
confirm an adequate approximation to countable additivity by looking at
only countably many sequences
(448Xd).   This enables us to pass from a $G$-invariant map
$\theta:\frak A\to L^{\infty}(\frak D)$ to a $G$-invariant Radon measure
(parts (d)-(f) of the proof of 448P), without needing to know
anything about the algebra $\frak D$ except that it is Dedekind
$\sigma$-complete.   In particular (and in contrast to the corresponding
step in 395P) we do not need to suppose that $\frak D$ is a measurable
algebra.   I do not know whether there is a useful ergodicity condition
which could be added to the hypotheses of 448O to ensure that $\frak D$
there becomes $\{0,1\}$.

448P was proved in the case $G=\Bbb Z$ by {\smc Nadkarni 90};  the
extension to general Borel actions by Polish groups is due to
{\smc Becker \& Kechris 96}.   (See {\smc Nadkarni 90} for notes on the
history of the problem, and {\smc Kechris 95} for the basic general
theory of Polish groups and Borel actions.)
It is a remarkable result, but its application is limited by the
difficulty of determining whether either condition (ii) or condition
(iii) is satisfied.   Much commoner situations are those like
448Xe-448Xj, %448Xe 448Xf 448Xg 448Xh 448Xi 448Xj
where either there is no invariant measure or we can find one easily.

The second main theorem makes no reference to the first.   But it has
something in common.   It is an example of the power of descriptive set
theory to dramatically extend a result on group actions, which is
comparatively straightforward when the group in question is $\Bbb Z$, to
arbitrary Polish groups.   Nadkarni's theorem is not obvious, but it is a
lot easier than the general result here.   Mackey's theorem for countable
groups also requires a little care, but is essentially covered (in usefully
greater generality) by 344C.   The descriptive set theory
the theorem here relies on does
not go as deep as the Becker-Kechris theorem, but in exchange it calls on a
kind of lifting theorem quite different from those in Chapter 34.
Looked at from the standpoint of Chapter 34, 448Q is a
rank impossibility (see 341Xg);
but the point is that we have abandoned the ordinary
algebraic requirements on a lifting and replaced them by a strong
measurability property.

Of course a lifting was used in 344C as well, but in a quite different way.
There the hypotheses were adjusted to give a slightly more general context
in which we could be sure that individual homomorphisms from the measure
algebra to itself were representable by functions from the measure space to
itself;  and I relied indirectly on the lifting theorem 341K
to set up the functions.   For the context of the present section,
this step was done in 425A with no mention of
liftings, but using the classification of standard Borel spaces in 424D.
In view of 424Yf, it is plain that we do not get much extra generality by
using the argument through 341K.   The real difference in 344B-344C is that
we can deal with semigroups of homomorphisms as well as groups of
automorphisms.

The proof of Mackey's theorem is based on there being a Haar measure on
$G$, so that we can use Fubini's theorem
(three times, in parts (c), (d) and (g) of the
proof).   There are non-locally-compact groups $G$ for which a
corresponding result is true ({\smc Kwiatowska \& Solecki 11});
it remains quite unclear when to expect this.
}%end of notes

\discrpage


\frfilename{mt449.tex}
\versiondate{13.6.13}
\copyrightdate{2003}

\def\chaptername{Topological groups}
\def\sectionname{Amenable groups}

\def\length{\mathop{\text{length}}\nolimits}
\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}}}

\newsection{449}

I end this chapter with a brief introduction to `amenable' topological
groups.   I start with the
definition (449A) and straightforward results assuring us that there
are many amenable groups (449C).   At a slightly deeper level we have a
condition for a group to be amenable in terms of a universal object
constructible from the group, not invoking `all compact Hausdorff
spaces' (449E).   I give some notes on amenable locally compact
groups, concentrating on a long list of properties equivalent to
amenability (449J), and a version of Tarski's theorem characterizing
amenable discrete groups (449M).   I end with Banach's theorem on extending
Lebesgue measure in one and two dimensions.

\leader{449A}{Definition} A topological group $G$ is {\bf amenable} if
whenever $X$ is a non-empty compact Hausdorff space and $\action$ is a
continuous action of $G$ on $X$, then there is a $G$-invariant
Radon probability measure on $X$.

\cmmnt{
\medskip

\noindent{\bf Warning}:  other definitions have been used, commonly
based on conditions equivalent to amenability for locally compact
Hausdorff groups, such as those listed in 449J(ii)-449J(xiv).   In
addition, many authors use the phrase `amenable group' to
mean a group which is amenable in its discrete topology.   The danger
of this to the non-specialist is that many theorems concerning amenable
discrete groups do not generalize in the ways one might expect.}

\leader{449B}{Lemma} Let $G$ be a topological group, $X$ a locally
compact Hausdorff space, and $\action$ a continuous action of $G$ on
$X$.

(a) Writing $C_0$ for the Banach space of continuous real-valued
functions on $X$ vanishing at $\infty$\cmmnt{ (436I)}, the map
$a\mapsto a^{-1}\action f:C_0\to C_0$\cmmnt{ (definition:  4A5Cc)}
is uniformly continuous for the right uniformity on $G$ and the norm
uniformity of $C_0$, for any $f\in C_0$.

(b) If $\mu$ is a $G$-invariant Radon measure on $X$ and
$1\le p<\infty$, then
$a\mapsto a^{-1}\action u:G\to L^p$\cmmnt{ (definition: 441Kc)} is
uniformly
continuous for the right uniformity on $G$ and the norm uniformity of
$L^p=L^p(\mu)$, for any $u\in L^p$.

\proof{{\bf (a)(i)} Note first that if $a\in G$ and $f\in C_0$, then
$x\mapsto a\action x:X\to X$ is a homeomorphism (4A5Bd), so
$x\mapsto f(a\action x)$ belongs to $C_0$;  but this is just the
function $a^{-1}\action f$.

\medskip

\quad{\bf (ii)} For any $\epsilon>0$ and $f\in C_0$ there is a
neighbourhood $V$ of the identity $e$ of $G$ such that
$\|f-a^{-1}\action_lf\|_{\infty}\le\epsilon$ for every $a\in V$.
\Prf\Quer\
Suppose, if possible, otherwise.   For each symmetric neighbourhood
$V$ of $e$ set

\Centerline{$Q_V
=\{(a,x):a\in V,\,x\in X,\,|f(x)|\ge\Bover{\epsilon}2,\,
 |f(x)-f(a\action x)|\ge\epsilon\}$.}

\noindent We are supposing that there are $a\in V$ and $x\in X$ such
that $|f(x)-f(a\action x)|\ge\epsilon$.   If
$|f(x)|\ge\bover12\epsilon$
then $(a,x)\in Q_V$.   Otherwise, $|f(a\action
x)|\ge\bover12\epsilon$,
$a^{-1}\in V$ and $|f(a\action x)-f(a^{-1}a\action x)|\ge\epsilon$, so
$(a^{-1},a\action x)\in Q_V$.   Thus $Q_V$ is never empty.
Since $Q_V\subseteq Q_{V'}$
whenever $V\subseteq V'$, there is an ultrafilter $\Cal F$ on
$G\times X$ such that $Q_V\in\Cal F$ for every neighbourhood $V$ of
$e$.
Setting $\pi_1(a,x)=a$ and $\pi_2(a,x)=x$ for $(a,x)\in G\times X$, we
see that the image filter $\pi_1[[\Cal F]]$ contains every
neighbourhood
of $e$, so converges to $e$, while $\pi_2[[\Cal F]]$ contains the
compact set $\{x:|f(x)|\ge\bover12\epsilon\}$, so must have a limit
$x_0$ in $X$.   So $\Cal F\to(e,x_0)$ in
$G\times X$.   Next, because the action is continuous,
$\action[[\Cal F]]\to e\action x_0=x_0$, and there must be an
$F\in\Cal F$ such that
$|f(x_0)-f(a\action x)|\le\bover13\epsilon$ for
every $(a,x)\in F$.   Also, of course, there is an $F'\in\Cal F$ such
that $|f(x_0)-f(x)|\le\bover13\epsilon$ whenever $(a,x)\in F'$.   But
now there is an $(a,x)\in Q_G\cap F\cap F'$, and we have

\Centerline{$|f(x)-f(a\action x)|\ge\epsilon$,
\quad$|f(x_0)-f(a\action x)|\le\bover13\epsilon$,
\quad$|f(x_0)-f(x)|\le\bover13\epsilon$}

\noindent simultaneously, which is impossible.\ \Bang\Qed

Now we find that if $a$, $b\in G$, $ab^{-1}\in V$ and $x\in X$, then

\Centerline{$|(a^{-1}\action f)(x)-(b^{-1}\action f)(x)|
=|f(a\action x)-f(b\action x)|
=|f(ab^{-1}\action(b\action x))-f(b\action x)|
\le\epsilon$.}

\noindent As $x$ is arbitrary,
$\|a^{-1}\action f-b^{-1}\action f\|_{\infty}\le\epsilon$;  as
$\epsilon$ is arbitrary, $a\mapsto a^{-1}\action f$ is uniformly
continuous for the right uniformity.

\medskip

{\bf (b)(i)} Suppose that $f:X\to\Bbb R$ is continuous and has compact
support $K=\overline{\{x:f(x)\ne 0\}}$.
Let $H\supseteq K$ be an open set of finite measure.
Then $V_0=\{a:a\in G,\,a\action x\in H$ for every $x\in K\}$ is a
neighbourhood of $e$.   \Prf\ If we take a continuous function $f_0$
with compact support such that $\chi K\le f_0\le\chi H$
(4A2G(e-i)), then
$V_0\supseteq\{a:\|f_0-a^{-1}\action f_0\|_{\infty}<1\}$, which
is a neighbourhood of $e$ by (a).\ \QeD\  Let $\epsilon>0$.   By (a)
again, there is a symmetric neighbourhood $V_1$ of $e$ such that
$(\|f-a^{-1}\action f\|_{\infty})^p\mu H\le\epsilon^p$ for every
$a\in V_1$;  we
may suppose that $V_1\subseteq V_0$.   If $a\in V_1$,
$f(x)=f(a\action x)=0$ for every $x\in X\setminus H$, so that

\Centerline{$\|f-a^{-1}\action f\|_p^p
=\biggerint_H|f-a^{-1}\action f|^pd\mu
\le(\|f-a^{-1}\action f\|_{\infty})^p\mu H\le\epsilon^p$.}

Now suppose that $a$, $b\in G$ and that $ab^{-1}\in V_1$.   Then

$$\eqalignno{\|a^{-1}\action f-b^{-1}\action f\|_p^p
&=\int|f(a\action x)-f(b\action x)|^p\mu(dx)\cr
&=\int|f(a\action(b^{-1}\action x))
  -f(b\action(b^{-1}\action x))|^p\mu(dx)\cr
\displaycause{because $\mu$ is $G$-invariant, see 441L}
&=\int|ba^{-1}\action f-f)|^pd\mu
\le\epsilon^p,\cr}$$

\noindent and $\|a^{-1}\action f-b^{-1}\action f\|_p\le\epsilon$.   As
$\epsilon$ is
arbitrary, $a\mapsto(a^{-1}\action f)^{\ssbullet}$ is uniformly
continuous for the right uniformity.

\medskip

\quad{\bf (ii)} In general, given $u\in L^p(\mu)$ and $\epsilon>0$,
there is an $f\in C_k(X)$ such that
$\|u-f^{\ssbullet}\|_p\le\epsilon$
(416I).   Let $V$ be a neighbourhood of $e$ such that
$\|a^{-1}\action f-a^{-1}\action f\|_{\infty}\le\epsilon$ whenever
$ab^{-1}\in V$;  then
$\|a^{-1}\action u-(a^{-1}\action f)^{\ssbullet}\|_p
=\|u-f^{\ssbullet}\|_p$ (because $\mu$ is $G$-invariant), so

\Centerline{$\|a^{-1}\action u-b^{-1}\action u\|_p
\le\|a^{-1}\action u-a^{-1}\action f^{\ssbullet}\|_p
+\|a^{-1}\action f-b^{-1}\action f\|_p
+\|b^{-1}\action f^{\ssbullet}-b^{-1}\action u\|_p
\le 3\epsilon$}

\noindent whenever $ab^{-1}\in V$ (using 441Kc).   As $\epsilon$ is
arbitrary,
$a\mapsto a^{-1}\action u$ is uniformly continuous for the right
uniformity.   This completes the proof.
}%end of proof of 449B

\leader{449C}{Theorem} (a) Let $G$ and $H$ be topological groups such
that there is a continuous surjective homomorphism from $G$ onto $H$.
If $G$ is amenable, so is $H$.

(b) Let $G$ be a topological group and suppose that there is a dense
subset $A$ of $G$ such that every finite subset of $A$ is included in
an amenable subgroup of $G$.    Then $G$ is amenable.

(c) Let $G$ be a topological group and $H$ a normal subgroup of $G$.
If $H$ and $G/H$ are both amenable, so is $G$.

(d) Let $G$ be a topological group with two amenable subgroups $H_0$
and
$H_1$ such that $H_0$ is normal and $H_0H_1=G$.   Then $G$ is
amenable.

(e) The product of any family of amenable topological groups is
amenable.

(f) Any abelian topological group is amenable.

(g) Any compact Hausdorff topological group is amenable.

\proof{{\bf (a)} Let $\phi:G\to H$ be a continuous surjective
homomorphism.   Let $X$ be a non-empty compact Hausdorff space and
$\action:H\times X\to X$ a continuous action.   For $a\in G$ and
$x\in X$, set $a\action_1x=\phi(a)\action x$.   Then $\action_1$ is a
continuous action of $G$ on $X$, so there is a $G$-invariant Radon
probability measure $\mu$ on $X$.   Because $\phi[G]=H$, $\mu$ is also
$H$-invariant;  as $X$ and $\action$ are arbitrary, $H$ is amenable.

\medskip

{\bf (b)(i)} Let $X$ be a non-empty compact Hausdorff space and
$\action$ a continuous action of $G$ on $X$.   Let $P=P_{\text{R}}(X)$
be the set of
Radon probability measures on $X$ with the narrow topology (437Jd),
so that $P$
is a compact Hausdorff space (437R(f-ii));  recall that in this context the
vague and narrow topologies coincide (437Kc).   For $a\in G$ and
$x\in X$, set
$T_a(x)=a\action x$, so that $T_a:X\to X$ is a homeomorphism.
For $a\in G$ and $\mu\in P$ write  $a\action\mu$ for the image measure
$\mu T_a^{-1}$, so that $a\action\mu\in P$ (418I);  it is easy to
check that $(a,\mu)\mapsto a\action\mu:G\times P\to P$ is an action.
The point is that it is continuous.   \Prf\ Let $f\in C(X)$, $a_0\in
G$,
$\mu_0\in P$ and $\epsilon>0$.   By 449Ba, there is a neighbourhood
$V$
of $a_0$ in $G$ such that
$\|a^{-1}\action f-a_0^{-1}\action f\|_{\infty}\le\bover12\epsilon$
for
every $a\in V$.   Next, there is a neighbourhood $W$ of $\mu_0$ in $P$
such that
$|\int a_0^{-1}\action fd\mu-\int a_0^{-1}\action fd\mu_0|
\le\bover12\epsilon$
for every $\mu\in W$.   But now, if $a\in V$ and $\mu\in W$,

$$\eqalignno{|\int fd(a\action\mu)-\int fd(a_0\action\mu_0)|
&=|\int fT_ad\mu-\int fT_{a_0}d\mu_0|\cr
\displaycause{235J}
&=|\int a^{-1}\action fd\mu-\int a_0^{-1}\action fd\mu_0|\cr
&\le|\int a^{-1}\action fd\mu-\int a_0^{-1}\action fd\mu|
  +|\int a_0^{-1}\action fd\mu-\int a_0^{-1}\action fd\mu_0|\cr
&\le\|a^{-1}\action f-a_0^{-1}\action f\|_{\infty}+\Bover12\epsilon
\le\epsilon.\cr}$$

\noindent As $\epsilon$, $a_0$ and $\mu_0$ are arbitrary,
$(a,\mu)\mapsto\int fd(a\action\mu)$ is continuous;  as $f$ is
arbitrary, $\action$ is continuous.\ \Qed

\medskip

\quad{\bf (ii)} Because the topology of $P$ is Hausdorff, it follows
that $Q_a=\{\mu:\mu\in P,\,a\action\mu=\mu\}$ is closed in $P$ for any
$a\in G$, and that  $G_{\mu}=\{a:a\in G,\,a\action\mu=\mu\}$ is closed
in $G$ for any $\mu\in P$.
Now for any finite subset $I$ of $A$ there is an amenable subgroup
$H_I$ of $G$ including $I$.   The restriction of the action to
$H_I\times X$ is a continuous action of $H_I$ on $X$, so has an
$H_I$-invariant Radon
probability measure, and
$\bigcap_{a\in I}Q_a\supseteq\bigcap_{a\in H_I}Q_a$ is non-empty.
Because $P$ is compact, there is a $\mu\in\bigcap_{a\in A}Q_a$.
Since $G_{\mu}$ includes the dense set $A$, it is the whole
of $G$, and $\mu$ is $G$-invariant.   As $X$ and $\action$ are
arbitrary, $G$ is amenable.

\medskip

{\bf (c)} Let $X$ be a compact Hausdorff space and
$\action$ a continuous action of $G$ on $X$.   Let $P$ be the space of
Radon probability measures on $X$ with its narrow topology.   Define
$a\action\mu$, for $a\in G$ and $\mu\in P$, as in (b-i) above, so that
this is a continuous action of $G$ on $P$.   Set
$Q=\{\mu:\mu\in P,\,a\action\mu=\mu$ for every $a\in H\}$;  then $Q$
is a closed subset of $P$ and, because $H$ is amenable, is non-empty,
since it is the set of $H$-invariant Radon probability measures on
$X$.   Next, $b\action\mu\in Q$ for every $\mu\in Q$ and $b\in G$.
\Prf\ If $a\in H$, then

\Centerline{$a\action(b\action\mu)=(ab)\action\mu
=(bb^{-1}ab)\action\mu=b\action((b^{-1}ab)\action\mu)=b\action\mu$,}

\noindent because $H$ is normal, so $b^{-1}ab\in H$.  As $a$ is
arbitrary, $b\action\mu\in Q$.\ \QeD\    Accordingly we have a
continuous action of $G$ on the compact Hausdorff space $Q$.

If $a\in H$ and $b\in G$, then $b\action\mu=(ba)\action\mu$ for every
$\mu\in Q$.   We therefore have a map $\varaction:G/H\times Q\to Q$
defined
by setting $b^{\ssbullet}\varaction\mu=b\action\mu$ whenever $b\in G$
and $\mu\in Q$.   It is easy to check that this is an action.
Moreover, it is continuous, because if
$W\subseteq Q$ is relatively open then $\{(b,\mu):b\action\mu\in W\}$
is open in $G\times Q$, so its image
$\{(b^{\ssbullet},\mu):b^{\ssbullet}\varaction\mu\in W\}$ is open in
$(G/H)\times Q$ (using 4A2B(f-iv)).
Because $G/H$ is amenable, there is a
$(G/H)$-invariant Radon probability measure $\lambda$ on $Q$.

Now consider the formula
$p(f)=\int_Q(\int_X f(x)\mu(dx))\lambda(d\mu)$.   If $f\in C(X)$, then
$\mu\mapsto\int_Xf(x)\mu(dx)$ is continuous for the vague topology on
$Q$, so $p(f)$ is well-defined.   Clearly $p$ is a linear functional,
$p(f)\ge 0$ if $f\ge 0$, and $p(\chi X)=1$;  so there is a Radon
probability measure $\nu$ on $X$ such that $p(f)=\int fd\nu$ for every
$f\in C(X)$ (436J/436K).   If $b\in G$, then, in the language of (b) above,

$$\eqalignno{\int fd(b\action\nu)
&=\int fT_bd\nu
=p(fT_b)
=\int_Q\bigl(\int_X fT_bd\mu\bigr)\lambda(d\mu)\cr
&=\int_Q\int_X fd(b\action\mu)\lambda(d\mu)
=\int_Q\int_X fd(b^{\ssbullet}\varaction\mu)\lambda(d\mu)
=\int_Q\int_X fd\mu\,\lambda(d\mu)\cr
\displaycause{because $\lambda$ is $G/H$-invariant}
&=\int fd\nu\cr}$$

\noindent for every $f\in C(X)$, so that $b\action\nu=\nu$.   Thus
$\nu$
is $G$-invariant.   As $X$ and $\action$ are arbitrary, $G$ is
amenable.

\medskip

{\bf (d)} The canonical map from $H_1$ to $G/H_0$ is a continuous
surjective homomorphism.   By (a), $G/H_0$ is amenable;  by (c), $G$
is
amenable.

\medskip

{\bf (e)} By (c) or (d), the product of two amenable topological
groups
is amenable, since each can be regarded as a normal subgroup of the
product.   It follows that the product of finitely
many amenable topological groups is amenable.   Now let
$\familyiI{G_i}$
be any family of amenable topological groups with product $G$.   For
finite $J\subseteq I$ let $H_J$ be the set of those $a\in G$ such that
$a(i)$ is the identity in $G_i$ for every $i\in I\setminus J$.   Then
$H_J$ is isomorphic (as topological group) to $\prod_{i\in J}G_i$, so
is
amenable.   Since $\{H_J:J\in[I]^{<\omega}\}$ is an upwards-directed
family of subgroups of $G$ with dense union, (b) tells us that $G$ is
amenable.

\medskip

{\bf (f)(i)} The first step is to observe that the group $\Bbb Z$,
with
its discrete topology, is amenable.   \Prf\ Let $X$ be a compact
Hausdorff space and $\action$ a continuous action of $\Bbb Z$ on $X$.
Set $\phi(x)=1\action x$ for $x\in X$.   Then $\phi:X\to X$ is
continuous, so by
437T there is a Radon probability measure $\mu$ on $X$ such that
$\mu$ is equal to the image measure $\mu\phi^{-1}$.   Because $\phi$
is
bijective, we see that, for $E\subseteq X$,
$E\in\dom\mu$ iff $\phi[E]\in\dom(\mu\phi^{-1})=\dom\mu$, and in this
case $\mu\phi[E]=\mu E$;  that is, $\phi^{-1}$, like $\phi$, is \imp.
Now we can induce on $n$ to see
that $\mu(\phi^n)^{-1}$ and $\mu(\phi^{-n})^{-1}$ are equal to $\mu$
for
every $n$.   Since $n\action x=\phi^n(x)$ for every $x\in X$ and
$n\in\Bbb Z$, $\mu$ is $\Bbb Z$-invariant.
As $X$ and $\action$ are arbitrary, $\Bbb Z$ is amenable.\ \Qed

\medskip

\quad{\bf (ii)} Now let $G$ be any abelian topological group.   For
each
finite set $I\subseteq G$ let $\phi_I:\BbbZ^I\to G$ be the continuous
homomorphism defined by setting $\phi_I(z)=\prod_{a\in I}a^{z(a)}$ for
$z\in\BbbZ^I$.   By (i) just above and (e), we know that $\BbbZ^I$
(with
its discrete topology) is amenable, so (a) tells us that the subgroup
$G_I=\phi_I[\BbbZ^I]$ is amenable.   But now
$\{G_I:I\in[G]^{<\omega}\}$
is an upwards-directed family of amenable subgroups of $G$ with union
$G$, so from (b) we see that $G$ is amenable.

\medskip

{\bf (g)} This is immediate from 443Ub.   (See also 449Xe.)
}%end of proof of 449C

\leader{449D}{Theorem} Let $G$ be a topological group.

(a) Write $U$ for the set of bounded real-valued functions on
$G$ which are uniformly continuous for the right uniformity of $G$.
Then $U$ is an $M$-space, and we have an action $\action_l$ of $G$ on
$U$ defined by
the formula $(a\action_lf)(y)=f(a^{-1}y)$ for $a$, $y\in G$ and
$f\in U$.

(b) Let $Z\subseteq\BbbR^U$ be the set of Riesz homomorphisms
$z:U\to\Bbb R$ such that $z(\chi G)=1$.   Then $Z$ is a compact
Hausdorff space, and we have a continuous action of $G$ on $Z$ defined
by the formula $(a\action z)(f)=z(a^{-1}\action_lf)$ for
$a\in G$, $z\in Z$ and $f\in U$.

(c) Setting $\hat a(f)=f(a)$ for $a\in G$ and $f\in U$, the map
$a\mapsto\hat a:G\to Z$ is a continuous function from $G$ onto a dense
subset of $Z$.   If $a$, $b\in G$ then
$a\action\hat b=\widehat{ab}$.

(d) Now suppose that $X$ is a compact Hausdorff space,
$(a,x)\mapsto a\action x$ is a continuous action of $G$ on $X$, and
$x_0\in X$.    Then there is a unique
continuous function $\phi:Z\to X$ such
that $\phi(\hat e)=x_0$ and $\phi(a\action z)=a\action\phi(z)$ for
every $a\in G$ and $z\in Z$.

(e) If $G$ is Hausdorff then the action of $G$ on $Z$ is faithful and
the map $a\mapsto\hat a$ is a homeomorphism between $G$ and its image
in
$Z$.

\proof{{\bf (a)} Because $U$ is a norm-closed Riesz
subspace of $C_b(G)$ containing the constant functions (4A2Jh), it is
an $M$-space.   To see that the given
formula defines an action, we need to check that $a\action_lf$
belongs to $U$ whenever $a\in G$ and $f\in U$.   Of course
$a\action_lf$
is continuous and $\|a\action_lf\|_{\infty}=\|f\|_{\infty}$ is finite.
If $\epsilon>0$ there is a neighbourhood $V$ of the identity $e$ in
$G$
such that
$|f(b)-f(c)|\le\epsilon$ whenever $b$, $c\in G$ and
$bc^{-1}\in V$;  now $a^{-1}Va$ is a neighbourhood of $e$, and if
$bc^{-1}\in aVa^{-1}$ then $(a^{-1}b)(a^{-1}c)^{-1}\in V$, so
$|(a\action_lf)(b)-(a\action_lf)(c)|
=|f(a^{-1}b)-f(a^{-1}c)|\le\epsilon$.   As $\epsilon$ is arbitrary,
$a\action_lf$ is uniformly continuous with respect to the right
uniformity.   Now $\action_l$ is an action, just as in 4A5Cc.

\medskip

{\bf (b)(i)} Because $U$ is an $M$-space with standard order unit
$\chi G$, $Z$ is a compact Hausdorff space and $U$ can be identified,
as normed Riesz space, with $C(Z)$ (354L).
For any $a\in G$, the map $f\mapsto a\action_lf:U\to U$ is a Riesz
homomorphism leaving the constant functions fixed.   So we can define
$a\action z$, for $z\in Z$, by saying that
$(a\action z)(f)=z(a^{-1}\action_lf)$ for any $f\in U$, and
$a\action z$
will belong to $Z$ for any $a\in G$ and $z\in Z$.   As usual, it is
easy to check that this formula defines an action of $G$ on $Z$.

\medskip

\quad{\bf (ii)} $(a,z)\mapsto a\action z$ is continuous.   \Prf\ Take
$a_0\in G$, $z_0\in Z$ and any neighbourhood $W$ of $a_0\action z_0$
in $Z$.   Because $U$ corresponds to the whole of $C(Z)$, and $Z$ is
completely regular, there is an $f\in U^+$ such that
$(a_0\action z_0)(f)=1$ and $z(f)=0$ for every $z\in Z\setminus W$.
Set $W_0=\{z:z(a_0^{-1}\action_lf)>\bover12\}$.
Observe that $z_0(a_0^{-1}\action_lf)=1$, so
$W_0$ is an open subset of $Z$ containing $z_0$.   Next, set
$V_0=\{a:a\in G,\,\|a^{-1}\action_lf-a_0^{-1}\action_lf\|_{\infty}
\le\bover12\}$.
There is a neighbourhood $V$ of $e$ such that
$|f(b)-f(c)|\le\bover12$ whenever $b$, $c\in G$ and
$bc^{-1}\in V$.   If $a\in Va_0$ then
$ab(a_0b)^{-1}=aa_0^{-1}\in V$ so

\Centerline{$|(a^{-1}\action_lf)(b)-(a_0^{-1}\action_lf)(b)|
=|f(ab)-f(a_0b)|\le\Bover12$}

\noindent for every $b\in G$, and $a\in V_0$.   Thus $V_0\supseteq Va_0$
is a neighbourhood of $a_0$.

Now if $a\in V_0$ and $z\in W_0$ we shall have

\Centerline{$(a\action z)(f)=z(a^{-1}\action_lf)
\ge z(a_0^{-1}\action_lf)-\Bover12>0$}

\noindent and $a\action z\in W$.
As $a_0$, $z_0$ and $W$ are arbitrary, the action of $G$ on $Z$ is
continuous.\ \Qed

\medskip

{\bf (c)} Of course $\hat a$, as defined, is a Riesz homomorphism
taking
the correct value at $\chi G$, so belongs to $Z$.   Because
$U\subseteq C(X)$, the map
$a\mapsto\hat a$ is continuous.   \Quer\ If
$\{\hat a:a\in G\}$ is not dense in $Z$, there is a non-zero $h\in
C(Z)$
such that $h(\hat a)=0$ for every $a\in G$;  but as $U$ is identified
with $C(Z)$, there is an $f\in U$ such that $z(f)=h(z)$ for every
$z\in Z$.   In this case, $f$ cannot be the zero function, but
$f(a)=\hat a(f)=h(\hat a)=0$ for every $a\in G$.\ \BanG\  Thus the
image of $G$ is dense, as claimed.

If $a$, $b\in G$ and $f\in U$ then

\Centerline{$(a\action\hat b)(f)
=\hat b(a^{-1}\action_lf)
=(a^{-1}\action_lf)(b)
=f(ab)
=\widehat{ab}(f)$,}

\noindent so $a\action\hat b=\widehat{ab}$.

\medskip

{\bf (d)} We have a Riesz homomorphism $T:C(X)\to\BbbR^G$ defined by
setting $(Tg)(a)=g(a\action x_0)$ for every $g\in C(X)$ and $a\in G$.
Now $Tg\in U$ for every $g\in C(X)$.   \Prf\
$Tg(a)=(a^{-1}\action g)(x_0)$;    since the map
$a\mapsto a^{-1}\action g$ is
uniformly continuous (449Ba), so is $Tg$.
$\|Tg\|_{\infty}\le\|g\|_{\infty}$ is finite, so $Tg\in U$.\ \Qed

Of course $T(\chi X)=\chi G$.   So if $z\in Z$, $zT:C(X)\to\Bbb R$ is
a Riesz homomorphism such that $(zT)(\chi X)=1$.   There is therefore a
unique $\phi(z)\in X$ such that $(zT)(g)=g(\phi(z))$ for every
$g\in C(X)$ (354L again).   Since the function $z\mapsto g(\phi(z))=z(Tg)$
is continuous for every $g\in C(X)$, $\phi$ is continuous.

Now suppose that $a\in G$.   Then $\phi(\hat a)=a\action x_0$.   \Prf\
If $g\in C(X)$, then

\Centerline{$g(\phi(\hat a))=\hat a(Tg)=(Tg)(a)
=g(a\action x_0)$.  \Qed}

\noindent So if $a$, $b\in G$, then

\Centerline{$\phi(a\action\hat b)
=\phi(\widehat{ab})
=(ab)\action x_0
=a\action(b\action x_0)=a\action\phi(b)$.}

\noindent Since $\{\hat b:b\in G\}$ is dense in $Z$, and all the
functions here are continuous, $\phi(a\action z)=a\action\phi(z)$ for
all $a\in G$ and $z\in Z$.

To see that $\phi$ is unique, observe that if $a\in G$ then
$\phi(\hat a)=\phi(\widehat{ae})=\phi(a\action\hat{e})$ must be
$a\action\phi(\hat e)=a\action x_0$;  since $\{\hat a:a\in G\}$ is dense in
$Z$, $X$ is Hausdorff and $\phi$ is declared to be continuous, $\phi$ is
uniquely defined.

\medskip

{\bf (e)} Now suppose that the topology of $G$ is Hausdorff.   Then it
is defined by the bounded uniformly continuous functions (4A2Ja);
the map $a\mapsto\hat a$ is therefore injective and is a homeomorphism
between $G$ and its image in $Z$.   If $a$, $b\in G$ are distinct,
then
$a\action\hat e=\hat a\ne\hat b=b\action\hat e$, so the action is
faithful.
}%end of proof of 449D

\medskip

\noindent{\bf Definition} \dvro{The}{Following {\smc Brook 70}, the} space
$Z$, together with the canonical action of $G$ on it and the map
$a\mapsto\hat a:G\to Z$, is called the {\bf greatest ambit} of the
topological group $G$.

\leader{449E}{Corollary} Let $G$ be a topological group.   Then the
following are equiveridical:

(i) $G$ is amenable;

(ii) there is a $G$-invariant Radon probability measure on the
greatest ambit of $G$;

(iii) writing $U$ for the space of bounded real-valued functions on
$G$ which are uniformly continuous for the right uniformity, there is
a
positive linear functional $p:U\to\Bbb R$ such that $p(\chi G)=1$ and
$p(a\action_lf)=p(f)$ for every $f\in U$ and $a\in G$.

\proof{ Let $Z$ be the greatest ambit of $G$.

\medskip

{\bf (i)$\Rightarrow$(ii)} As soon as we know that $Z$ is a
compact Hausdorff space and the action of $G$ on $Z$ is continuous
(449Db), this becomes a special case of the definition of `amenable
topological group'.

\medskip

{\bf (ii)$\Rightarrow$(i)} Let $\mu$ be a $G$-invariant Radon
probability measure on $Z$.   Given any continuous action of $G$ on a
non-empty compact
Hausdorff space $X$, fix $x_0\in X$ and let $\phi:Z\to X$ be
a continuous function such that
$\phi(a\action z)=a\action\phi(z)$ for every $a\in G$ and $z\in Z$,
as in 449Dd.   Let $\nu$ be the image measure $\mu\phi^{-1}$.   Then
$\nu$ is a Radon probability measure on $X$ (418I again).
If $F\in\dom\nu$ and $a\in G$, then

\Centerline{$\nu(a\action F)=\mu\phi^{-1}[a\action F]
=\mu(a\action\phi^{-1}[F])=\mu\phi^{-1}[F]=\nu F$.}

\noindent As $a$ and $F$ are arbitrary, $\nu$ is $G$-invariant;  as
$X$
and $\action$ are arbitrary, $G$ is amenable.

\medskip

{\bf (ii)$\Leftrightarrow$(iii)} The identification of $U$ with
$C(Z)$ (see (b-i)
of the proof of 449D) means that we have a one-to-one correspondence
between Radon probability
measures $\mu$ on $Z$ and positive linear functionals $p$ on $U$ such
that $p(\chi G)=1$, given by the formula $p(f)=\int z(f)\mu(dz)$ for
$f\in U$ (436J/436K again).   Now

$$\eqalignno{\mu\text{ is }&G\text{-invariant}\cr
&\iff\int (a\action z)(f)\mu(dz)=\int z(f)\mu(dz)
  \text{ for every }f\in U,\,a\in G\cr
\displaycause{441L}
&\iff\int z(a^{-1}\action_lf)\mu(dz)=\int z(f)\mu(dz)
  \text{ for every }f\in U,\,a\in G\cr
&\iff\int z(a\action_lf)\mu(dz)=\int z(f)\mu(dz)
  \text{ for every }f\in U,\,a\in G\cr
&\iff p(a\action_lf)=p(f)
  \text{ for every }f\in U,\,a\in G.\cr}$$

\noindent So there is a $G$-invariant $\mu$, as required by (ii), iff
there is a $G$-invariant $p$ as required by (iii).
}%end of proof of 449E

\leader{449F}{Corollary} Let $G$ be a topological group.

(a) If $G$ is amenable, then

\quad(i) every open subgroup of $G$ is amenable;

\quad(ii)\dvAformerly{4{}49Xg} every dense subgroup of $G$ is amenable.

(b)\dvAnew{2010}
Suppose that for every sequence $\sequencen{V_n}$ of neighbourhoods of
the identity $e$ of $G$ there is a normal subgroup $H$ of $G$ such that
$H\subseteq\bigcap_{n\in\Bbb N}V_n$ and $G/H$ is amenable.   Then $G$ is
amenable.

\proof{ Write $U_G$ for
the set of bounded real-valued functions on $G$ which are uniformly
continuous for the right uniformity of $G$;  if $H$ is a subgroup of $G$,
let $U_H$ the set of
bounded real-valued functions on $H$ which are uniformly continuous
for the right uniformity of $H$;  and if
$H\normalsubgroup G$, let $U_{G/H}$ the set of
bounded real-valued functions on the quotient
$G/H$ which are uniformly continuous
for the right uniformity of $G/H$.

\medskip

{\bf (a)(i)}\grheada\ Let $H$ be an open subgroup of $G$.
Take a set $A\subseteq G$ meeting
each right coset of $H$ in just one point, so that each member of $G$ is
uniquely expressible as $ya$ where $y\in H$ and $a\in A$.   Define
$T:U_H\to\BbbR^G$ by setting $(Tf)(ya)=f(y)$ whenever $f\in U_H$, $y\in H$
and $a\in A$.   Then $T$ is a positive linear operator.   Also
$T[U_H]\subseteq U_G$.   \Prf\ Let
$f\in U_H$.   Of course $Tf$ is bounded.   If $\epsilon>0$, there is a
neighbourhood $W$ of the identity in $H$ such that
$|f(x)-f(y)|\le\epsilon$ whenever $x$, $y\in H$ and $xy^{-1}\in W$.
Because $H$ is open, $W$ is also a neighbourhood of the identity in
$G$.   Now suppose that $x$, $y\in G$ and $xy^{-1}\in W$.   Express $x$ as
$x_0a$ and $y$ as $y_0b$ where $x_0$, $y_0\in H$ and $a$, $b\in A$.
Then

\Centerline{$x_0ab^{-1}y_0^{-1}\in W\subseteq H$,}

\noindent so $ab^{-1}\in H$ and
$a\in Hb$ and $a=b$ and $x_0y_0^{-1}\in W$ and

\Centerline{$|(Tf)(x)-(Tf)(y)|=|f(x_0)-f(y_0)|\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $Tf$ is uniformly continuous and
belongs to $U_G$.\ \Qed

\medskip

\qquad\grheadb\ Next, $b\action_l(Tf)=T(b\action_lf)$ whenever $f\in U_H$
and $b\in H$.   \Prf\ If $x\in G$, express it as $ya$ where $y\in H$ and
$a\in A$.   Then

\Centerline{$(b\action_lTf)(x)=(Tf)(b^{-1}x)=(Tf)(b^{-1}ya)
=f(b^{-1}y)=(b\action_lf)(y)=T(b\action_lf)(x)$.  \Qed}

\medskip

\qquad\grheadc\ By 449E, there is a positive linear functional
$p:U_G\to\Bbb R$
such that $p(\chi G)=1$ and $p(a\action_lf)=p(f)$ whenever $f\in U_G$
and $a\in G$.   Set $q(f)=p(Tf)$ for $f\in U_H$;  then $q$ is a positive
linear operator, $q(\chi H)=1$ and $q$ is $H$-invariant, by (ii).   So
by 449E in the other direction, $H$ is amenable.

\medskip

\quad{\bf (ii)} Now suppose that $H$ is a dense subgroup of $G$.
It is easy to see that the right
uniformity of $H$ is the subspace uniformity induced by the right
uniformity of $G$ (3A4D), so that $f\restr H\in U_H$ for every $f\in U_G$.
In the other direction, if $g\in U_H$,
then $g$ extends uniquely to a member
of $U_G$, by 3A4G;  write $Tg$ for the extension.   In this case,
$b\action_lTg=T(b\action_lg)$ for every $g\in U_H$ and $b\in H$.   \Prf\
$b\action_lTg$ and $T(b\action_lg)$ are continuous, and for $a\in H$,

\Centerline{$(b\action_lTg)(a)=Tg(b^{-1}a)=g(b^{-1}a)
=(b\action_lg)(a)=T(b\action_lg)(a)$;}

\noindent as $H$ is dense in $G$, $b\action_lTg=T(b\action_lg)$.\ \Qed

Now we can use the same argument as in (i-$\gamma$) above to see that $H$
is amenable.

\woddheader{449F}{4}{2}{2}{48pt}

{\bf (b)(i)} Let $\Cal H$ be the family of normal subgroups $H$ of $G$ such
that $G/H$ is amenable.

\medskip

\qquad\grheada\ For $H\in\Cal H$,
let $\pi_H:G\to G/H$ be the canonical homomorphism and
$p_H:U_{G/H}\to\Bbb R$ a
positive linear functional such that $p_H(\chi(G/H))=1$ and
$p_H(c\action_lg)=p_H(g)$ whenever $g\in U_{G/H}$ and $c\in G/H$.
Let $U'_H$ be
$\{f:f\in U_G$, $f(x)=f(y)$ whenever $x$, $y\in G$ and $xy^{-1}\in H\}$.
Then $U'_H$ is a linear subspace of $U_G$ containing $\chi G$.

If $f\in U'_H$
then there is a unique $g\in U_{G/H}$ such that $f=g\pi_H$.   \Prf\
Because $f(x)=f(y)$ whenever $\pi_Hx=\pi_Hy$, and $\pi_H$ is surjective,
there is a unique function $g:G/H\to\Bbb R$ such that $f=g\pi_H$;  because
$f$ is bounded, so is $g$.   Given $\epsilon>0$, there is an open
neighbourhood
$W$ of $e$ such that $|f(x)-f(y)|\le\epsilon$ whenever $xy^{-1}\in W$.
In this case, $\pi_H[W]$ is a neighbourhood of the identity in $G/H$
(4A5J(a-i)).   Suppose that $c_0$, $c_1\in G/H$ are such that
$c_0c_1^{-1}\in\pi_H[W]$.   Then there are $x_0$, $x_1\in G$ and $x\in W$
such that $\pi_Hx_0=c_0$, $\pi_Hx_1=c_1$ and $\pi_Hx=c_0c_1^{-1}$.
As $\pi_H(x_0x_1^{-1})=\pi_Hx$, there is a $y\in H$ such that
$yx_0x_1^{-1}=x$ belongs to $W$;  so that $\pi_H(yx_0)=c_0$ and

\Centerline{$|g_H(c_0)-g_H(c_1)|
=|f(yx_0)-f(x_1)|\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $g\in U_{G/H}$.\ \Qed

We therefore have a functional $p'_H:U'_H\to\Bbb R$ defined by setting
$p'_H(g\pi_H)=p_H(g)$ whenever $g\in U_{G/H}$.   Of course $g\ge 0$
whenever $g\pi_H\ge 0$, so $p'_H$ is a positive linear functional, and
$p'_H(\chi G)=1$.

\medskip

\qquad\grheadb\
If $f\in U'_H$ and $a\in G$ then $a\action_lf\in U'_H$ and
$p'_H(a\action_lf)=p'_H(f)$.   \Prf\ Let $g\in U_{G/H}$ be such that
$f=g\pi_H$.   Then

\Centerline{$(a\action_lf)(x)
=f(a^{-1}x)
=g\pi_H(a^{-1}x)
=g(\pi_H(a)^{-1}\pi_H(x))
=(\pi_H(a)\action_lg)(\pi_H(x))$}

\noindent for every $x\in G$;  so $a\action_lf=(\pi_H(a)\action_lg)\pi_H$
belongs to $U'_H$, and

\Centerline{$p'_H(a\action_lf)=p_H(\pi_H(a)\action_lg)
=p_H(g)=p'_H(f)$.  \Qed}

\medskip

\quad{\bf (ii)} For any family $\Cal V$ of neighbourhoods of $e$,
set $\Cal H_{\Cal V}=\{H:H\in\Cal H$, $H\subseteq\bigcap\Cal V\}$.
Now for any $f\in U_G$ there is a countable family $\Cal V$ of
neighbourhoods of $e$ such that $f\in U'_H$ for every
$H\in\Cal H_{\Cal V}$.
\Prf\ For each $n\in\Bbb N$ choose a
neighbourhood $V_n$ of $e$ such that $|f(x)-f(y)|\le 2^{-n}$ whenever
$xy^{-1}\in V_n$, and set $\Cal V=\{V_n:n\in\Bbb N\}$.\ \Qed

\medskip

\quad{\bf (iii)} We are supposing that $\Cal H_{\Cal V}$ is non-empty for
every countable family $\Cal V$ of neighbourhoods of $e$.   There is
therefore an ultrafilter $\Cal F$ on $\Cal H$ containing $\Cal H_{\Cal V}$
for every countable $\Cal V$.   Now we see
from (ii) that for any $f\in U_G$ the set
$\{H:H\in\Cal H$, $f\in U'_H\}$ belongs to $\Cal F$, while
$|p'_H(f)|\le\|f\|_{\infty}$ for any $H$ such that $f\in U'_H$;  so we can
set $p(f)=\lim_{H\to\Cal F}p'_H(f)$ for every $f\in U_G$.    In this case,
of course, $p$ is a positive linear functional and $p(\chi G)=1$.
Also, given $f\in U_G$ and $a\in G$, then
$p'_H(f)=p'_H(a\action_lf)$ whenever $f\in U'_H$, by (i-$\beta$), so
$p(f)=p(a\action_lf)$.   Thus $p$ satisfies (iii) of 449E and $G$ is
amenable.
}%end of proof of 449F

\leader{449G}{Example} Let $F_2$ be the free group on two generators,
with its discrete topology.   Then $F_2$ is a $\sigma$-compact unimodular
locally compact
Polish group.   But it is not amenable.   \prooflet{\Prf\ Let $a$
and $b$ be the generators of $F_2$.   Then every element of $F_2$ is
uniquely expressible as a word (possibly empty) in the letters $a$,
$b$,
$a^{-1}$, $b^{-1}$ in which the letters $a$, $a^{-1}$ are never
adjacent
and the letters $b$, $b^{-1}$ are never adjacent.   Write $A$ for the
set of elements of $F_2$ for which the canonical word does not begin
with either $b$ or $b^{-1}$, and $B$ for the set of elements of $F_2$
for which
the canonical word does not begin with either $a$ or $a^{-1}$.   Then
$A\cup B=F_2$ and $A\cap B=\{e\}$.   \Quer\ Suppose, if possible, that
$F_2$ is amenable.   Every member of $\ell^{\infty}(F_2)$ is uniformly
continuous with
respect to the right uniformity.   So there is an $F_2$-invariant
positive linear functional $p:\ell^{\infty}(F_2)\to\Bbb R$ such that
$p(\chi F_2)=1$.   Let $\nu$ be the corresponding non-negative
additive
functional on $\Cal PF_2$, so that $\nu C=p(\chi C)$ for every
$C\subseteq F_2$.   For $c\in F_2$ and $C\subseteq F_2$,
$c\action_l\chi C=\chi(cC)$, so $\nu(cC)=\nu C$ for every
$C\subseteq F_2$
and $c\in F_2$.   In particular, $\nu(b^nA)=\nu A$ for every
$n\in\Bbb Z$;
but as all the $b^nA$, for $n\in\Bbb Z$, are disjoint, $\nu A=0$.
Similarly $\nu B=0$ and

\Centerline{$0=\nu(A\cup B)=\nu F_2=p(\chi F_2)=1$,}

\noindent which is absurd.\ \BanG\
Thus $F_2$ is not amenable, as claimed.\ \Qed}

\leader{449H}{}\cmmnt{ In this section so far, I have taken care to
avoid assuming that groups are locally compact.   Some of the most
interesting amenable groups are very far from being locally compact
(e.g., 449Xh).
But of course a great deal of work has been done on amenable locally
compact groups.   In particular, there is a remarkable list of equivalent
properties, some of which I will present in the next theorem.
It will be useful to have the following facts to hand.

\medskip

\noindent}{\bf Lemma} Let $G$ be a locally compact Hausdorff topological
group, and $U$ the space of bounded real-valued functions on $G$
which are uniformly continuous for the right uniformity\cmmnt{, as
in 449D-449E}.   Let $\mu$ be a left Haar measure on $G$, and $*$ the
corresponding convolution on $\eusm L^0(\mu)$\cmmnt{ (444O)}.

(a) If $h\in\eusm L^1(\mu)$ and $f\in\eusm L^{\infty}(\mu)$ then
$h*f\in U$.

(b) Let $p:U\to\Bbb R$ be a positive linear functional such that
$p(a\action_lf)=p(f)$ whenever $f\in U$ and $a\in G$.   Then
$p(h*f)=p(f)\int h\,d\mu$ for every $h\in\eusm L^1(\mu)$ and
$f\in U$.

%I suppose this works for groups carrying Haar measures?

\proof{{\bf (a)} Recall that we know from 444Rc that $h*f$ is defined
everywhere in $G$ and is continuous.   For any $x\in G$,

\Centerline{$(h*f)(x)=\int h(xy)f(y^{-1})\mu(dy)
=\int (x^{-1}\action_lh)\times\Reverse{f}$,}

\noindent where $\Reverse{f}(y)=f(y^{-1})$ whenever $y^{-1}\in\dom f$
(4A5C(c-ii)).   By 449Bb, applied to the left action of $G$ on itself,
$x\mapsto(x^{-1}\action_lh)^{\ssbullet}:G\to L^1(\mu)$ is uniformly
continuous for
the right uniformity of $G$ and the norm uniformity of $L^1(\mu)$.
Since $u\mapsto\int u\times v:L^1(\mu)\to\Bbb R$ is uniformly
continuous for every $v\in L^{\infty}(\mu)$, $x\mapsto(h*f)(x)
=\int(x^{-1}\action_lh)^{\ssbullet}\times\Reverse{f}^{\ssbullet}$ is
uniformly continuous for the right uniformity (3A4Cb).
Of course $\sup_{x\in G}|(h*f)(x)|\le\|h\|_1\|f\|_{\infty}$ is finite,
so $h*f\in U$.

\medskip

{\bf (b)} Let $\epsilon>0$.   Then there are a compact set
$K\subseteq G$ such that $\int_{G\setminus K}|h|d\mu\le\epsilon$
(412Je) and a symmetric open neighbourhood $V_0$ of $e$ such that
$|f(x)-f(y)|\le\epsilon$ whenever $xy^{-1}\in V_0$.   Let
$a_0,\ldots,a_n\in G$ be such that
$K\subseteq\bigcup_{i\le n}a_iV_0$, and set
$E_i=a_iV_0\setminus\bigcup_{j<i}a_jV_0$, $\alpha_i=\int_{E_i}h\,d\mu$
for each $i\le n$ and $F=G\setminus\bigcup_{j\le n}a_jV_0$.
If $x\in G$ and $y\in E_i$, then
$y^{-1}x(a_i^{-1}x)^{-1}=(a_i^{-1}y)^{-1}$ belongs to $V_0$, so
$|f(y^{-1}x)-f(a_i^{-1}x)|\le\epsilon$.   So, for any $x\in G$,

$$\eqalignno{\bigl|(h*f)(x)&-\sum_{i=0}^n\alpha_if(a_i^{-1}x)\bigr|\cr
&=\bigl|\int h(y)f(y^{-1}x)\mu(dy)
    -\sum_{i=0}^n\alpha_if(a_i^{-1}x)\bigr|\cr
&=\bigl|\int_Fh(y)f(y^{-1}x)\mu(dy)
  +\sum_{i=0}^n(\int_{E_i}h(y)f(y^{-1}x)\mu(dy)
     -\alpha_if(a_i^{-1}x))\bigr|\cr
&\le\|f\|_{\infty}\int_F|h|d\mu
  +\sum_{i=0}^n\bigl|\int_{E_i}h(y)
    (f(y^{-1}x)-f(a_i^{-1}x))\mu(dy)\bigr|\cr
&\le\|f\|_{\infty}\int_{X\setminus K}|h|d\mu
  +\epsilon\sum_{i=0}^n\int_{E_i}|h|d\mu
\le\epsilon(\|f\|_{\infty}+\|h\|_1).\cr}$$

\noindent Thus

\Centerline{$\|h*f-\sum_{i=0}^n\alpha_ia_i\action_lf\|_{\infty}
\le\epsilon(\|f\|_{\infty}+\|h\|_1)$.}

\noindent Since $p(a_i\action_lf)=p(f)$ for every $i$, it follows that

$$\eqalign{|p(h*f)-p(f)\int h\,d\mu|
&\le\epsilon(\|f\|_{\infty}+\|h\|_1)p(\chi G)
   +|\sum_{i=0}^n\alpha_i-\int h\,d\mu||p(f)|\cr
&\le\epsilon(\|f\|_{\infty}+\|h\|_1)p(\chi G)
   +|\int_Fh\,d\mu|\|f\|_{\infty}p(\chi G)\cr
&\le\epsilon(2\|f\|_{\infty}+\|h\|_1)p(\chi G).\cr}$$

\noindent As $\epsilon$ is arbitrary, $p(h*f)=p(f)\int h\,d\mu$, as
claimed.
}%end of proof of 449H

\leader{449I}{Notation}\dvAnew{2010}
It will save repeated explanations if I say now
that for the next two results,
given a locally compact Hausdorff group $G$,
$\Sigma_G$ will be the algebra of Haar measurable subsets of $G$ and
$\Cal N_G$ the ideal of Haar negligible subsets of $G$\cmmnt{ (443A)},
while $\Cal B_G$ will be
the Borel $\sigma$-algebra of $G$.   Recall that all three
are left- and right-translation-invariant and inversion-invariant, and
indeed autohomeomorphism-invariant, in that if $\gamma:G\to G$ is a
function of any of the types

\Centerline{$x\mapsto ax$,\quad$x\mapsto xa$,
\quad$x\mapsto x^{-1}$}

\noindent or is a group automorphism which is also a homeomorphism, and
$E\subseteq G$, then
$\gamma[E]$ belongs to $\Sigma_G$, $\Cal N_G$ or $\Cal B_G$ iff $E$
does\cmmnt{ (443Aa)}.

\woddheader{449J}{0}{0}{0}{108pt}

\leader{449J}{Theorem} Let $G$ be a locally compact Hausdorff group;
fix a left Haar measure $\mu$ on $G$.   Write $\eusm L^1$ for
$\eusm L^1(\mu)$ and $L^{\infty}$ for $L^{\infty}(\mu)$, etc.
Let $C_{k1}^+$ be the set of continuous
functions $h:G\to\coint{0,\infty}$ with compact supports such that
$\int h\,d\mu=1$, and suppose that $q\in\coint{1,\infty}$.
Then the following are equiveridical:

(i) $G$ is amenable;

(ii) there is a positive linear functional $p:C_b(G)\to\Bbb R$ such
that $p(\chi G)=1$ and $p(a\action_lf)=p(f)$ for every $f\in C_b(G)$ and
every $a\in G$;

(iii)\dvAnew{2010} there is a finitely additive functional
$\phi:\Cal B_G\to[0,1]$ such that
$\phi G=1$, $\phi(aE)=\phi E$ for every $E\in\Cal B_G$ and $a\in G$, and
$\phi E=0$ for every Haar negligible $E\in\Cal B_G$;

(iv)\dvAnew{2010}
there is a finitely additive functional $\phi:\Sigma_G\to[0,1]$ such that
$\phi G=1$, $\phi(aE)=\phi(Ea)=\phi(E^{-1})=\phi E$ for every
$E\in\Sigma_G$ and $a\in G$, and $\phi E=0$ for every $E\in\Cal N_G$;

(v)\dvAnew{2010} there is a positive
linear functional $\tilde p:L^{\infty}\to\Bbb R$ such that
$\tilde p(\chi G^{\ssbullet})=1$ and
$\tilde p(a\action_lu)=\tilde p(a\action_ru)=\tilde p(a\action_cu)
=\tilde p(\Reverse{u})=\tilde p(u)$ for
every $u\in L^{\infty}$ and every $a\in G$\cmmnt{, where $\action_l$,
$\action_r$, $\action_c$ and $\ssplrarrow$ are defined as in
443Af and 443Gc};

(vi) there is a positive
linear functional $\tilde p:L^{\infty}\to\Bbb R$ such that
$\tilde p(\chi G^{\ssbullet})=1$ and
$\tilde p(a\action_lu)=\tilde p(u)$ for
every $u\in L^{\infty}$ and every $a\in G$;

(vii) there is a positive
linear functional $\tilde p:L^{\infty}\to\Bbb R$ such that
$\tilde p(\chi G^{\ssbullet})=1$ and
$\tilde p(\nu*u)=\nu G\cdot\tilde p(u)$ for
every $u\in L^{\infty}$ and every totally finite Radon measure
$\nu$ on $G$\cmmnt{, where $\nu*u$ is defined as in 444Ma};

(viii) there is a positive linear functional
$\tilde p:L^{\infty}\to\Bbb R$ such that
$\tilde p(\chi G^{\ssbullet})=1$ and
$\tilde p(v*u)=\tilde p(u)\int v$ for
every $v\in L^1$ and $u\in L^{\infty}$;

(ix) for every finite set $J\subseteq\eusm L^1$ and
$\epsilon>0$, there is an $h\in C_{k1}^+$ such that
$\|g*h-(\int g\,d\mu)h\|_1\le\epsilon$ for every $g\in J$;

(x) for every compact set $K\subseteq G$ and $\epsilon>0$, there is
an $h\in C_{k1}^+$ such that $\|a\action_lh-h\|_1\le\epsilon$ for
every $a\in K$;

(xi)\dvAnew{2010}
for any finite set $I\subseteq G$ and $\epsilon>0$, there is a
$u\in L^q$ such that $\|u\|_q=1$ and $\|u-a\action_lu\|_q\le\epsilon$
for every $a\in I$;

(xii)\dvAnew{2010}
for any finite set $I\subseteq G$ and $\epsilon>0$, there is a
compact set $L\subseteq G$ with non-zero measure such that
$\mu(L\symmdiff aL)\le\epsilon\mu L$ for every $a\in I$;

(xiii)\dvArevised{2010}
for every compact set $K\subseteq G$ and $\epsilon>0$, there is a
symmetric compact neighbourhood $L$ of the identity $e$ in $G$
such that $\mu(L\symmdiff aL)\le\epsilon\mu L$ for every $a\in K$;

(xiv)\cmmnt{ ({\smc Emerson \& Greenleaf 67})} for every compact set
$K\subseteq G$ and $\epsilon>0$, there is a
compact set $L\subseteq G$ with non-zero measure such that
$\mu(KL)\le(1+\epsilon)\mu L$.

\proof{{\bf (a)(i)$\Rightarrow$(vii)} Write $U$ for the space
of bounded real-valued functions on $G$ which are uniformly continuous
for the right uniformity.   Then we have a positive linear functional
$p:U\to\Bbb R$ such that $p(\chi G)=1$ and $p(a\action_lf)=p(f)$ for
every $f\in U$ and $a\in G$ (449E).   Now if
$f\in\eusm L^{\infty}$, $h_1$,
$h_2\in\eusm L^1$ and $\int h_1d\mu=\int h_2d\mu$, then
$p(h_1*f)=p(h_2*f)$.   \Prf\ By 449Ha, both $h_1*f$ and $h_2*f$ belong
to $U$.   Set $h=h_1-h_2$.   By 444T, there is a neighbourhood $V$ of
$e$ such that $\|h*\nu-h\|_1\le\epsilon$ whenever
$\nu$ is a quasi-Radon measure on $G$ such that $\nu V=\nu G=1$,
defining $h*\nu$ as in 444J.   In
particular, taking $\nu$ to be the indefinite-integral measure over
$\mu$ defined from
$g=\Bover1{\mu V}\chi V$, $\|h*g-h\|_1\le\epsilon$ (using 444Pb).   Now

$$\eqalignno{|p(h_1*f)-p(h_2*f)|
&=|p(h*f)|
\le |p((h*g)*f)|+|p((h*g-h)*f)|\cr
&\le |p(h*(g*f))|+\|(h*g-h)*f\|_{\infty}\cr
\displaycause{because $*$ is associative, 444Oe}
&\le|p(g*f)\int h\,d\mu|+\|h*g-h\|_1\|f\|_{\infty}\cr
\displaycause{449Hb}
&\le\epsilon\|f\|_{\infty}.\cr}$$

\noindent As $\epsilon$ is arbitrary, $p(h_1*f)=p(h_2*f)$,
as claimed.\ \Qed

Of course $p(h*f)=0$ whenever $h\in\eusm L^1$, $f\in\eusm L^{\infty}$
and $f=0$ a.e.\ (444Ob).   We can therefore define a functional
$\tilde p:L^{\infty}\to\Bbb R$ by saying that
$\tilde p(f^{\ssbullet})=p(h*f)$ whenever $f\in\eusm L^{\infty}$,
$h\in\eusm L^1$ and $\int h\,d\mu=1$.   $\tilde p$ is positive and
linear because $p$ is.   It follows that
$p(h*f)=\tilde p(f^{\ssbullet})\int h\,d\mu$ whenever $h\in\eusm L^1$
and $f\in\eusm L^{\infty}$.   Also
$\tilde p(\chi G^{\ssbullet})=p(\chi G)=1$ because
$h*\chi G=(\int h\,d\mu)\chi G$ for every $h\in\eusm L^1$.

If $u\in L^{\infty}$ and $\nu$ is a totally finite Radon measure on
$G$,
express $u$ as $f^{\ssbullet}$ where
$f\in\eusm L^{\infty}$, so that $\nu*u=(\nu*f)^{\ssbullet}$ (444Ma).
Taking any non-negative
$h\in\eusm L^1$ such that $\int h\,d\mu=1$, we have

$$\eqalignno{h*(\nu*f)
&=h\mu*(\nu*f)\cr
\displaycause{444Pa;  here $h\mu$ is the indefinite-integral measure,
as in 444J}
&=(h\mu*\nu)*f\cr
\displaycause{444Ic}
&=(h*\nu)\mu*f\cr
\displaycause{444K}
&=(h*\nu)*f}$$

\noindent (444Pa again).   So

$$\eqalignno{\tilde p(\nu*u)
&=\tilde p((\nu*f)^{\ssbullet})
=p(h*(\nu*f))\cr
&=p((h*\nu)*f)
=\int h*\nu\,d\mu\cdot\tilde p(u)
=\nu G\cdot\tilde p(u)\cr}$$

\noindent (444K).  As $\nu$ and $u$ are arbitrary, $\tilde p$ has the
required properties.

\medskip

{\bf (b)(vii)$\Rightarrow$(vi)} Take $\tilde p$ from (vii).
If $a\in G$ and $u\in L^{\infty}$, consider the Dirac measure $\delta_a$
on $G$ concentrated at $a$.   Then $\delta_a*u=a\action_lu$.
\Prf\ Take $f\in\eusm L^{\infty}$ such that $f^{\ssbullet}=u$.
Then

\Centerline{$(\delta_a*f)(x)
=\int f(y^{-1}x)\delta_a(dy)=f(a^{-1}x)=(a\action_lf)(x)$}

\noindent whenever $a^{-1}x\in\dom f$, so $\delta_a*f=a\action_lf$ and

\Centerline{$\delta_a*u=\delta_a*f^{\ssbullet}=(\delta_a*f)^{\ssbullet}
=(a\action_lf)^{\ssbullet}=a\action_lu$.  \Qed}

\noindent Accordingly, using (vii),

\Centerline{$\tilde p(a\action_lu)
=\tilde p(\delta_a*u)=\delta_a(G)\tilde p(u)=\tilde p(u)$,}

\noindent as required by (vi).

\medskip

{\bf (c)(vi)$\Rightarrow$(v)}\grheada\ The first step is to note that
since there is a left-invariant mean there must also be a right-invariant
mean, that is, a positive linear functional
$\tilde q:L^{\infty}\to\Bbb R$ such that
$\tilde q(\chi G^{\ssbullet})=1$ and $\tilde q(a\action_ru)=\tilde q(u)$
for every $u\in L^{\infty}$ and every $a\in G$.   \Prf\ Set
$\tilde q(u)=\tilde p(\Reverse{u})$ for $u\in L^{\infty}$.
Evidently $\tilde q$ is a positive linear functional and
$\tilde q(\chi G^{\ssbullet})=1$.   By 443Gc,

\Centerline{$\tilde q(a\action_ru)
=\tilde p((a\action_ru)\ssplrarrow)
=\tilde p(a\action_l\Reverse{u})
=\tilde p(\Reverse{u})
=\tilde q(u)$}

\noindent whenever $u\in L^{\infty}$ and $a\in G$.\ \Qed

\medskip

\qquad\grheadb\ At this point, recall that $L^1$ is a Banach algebra
under convolution (444Sb), and that $L^{\infty}$ can be identified
with its normed space dual, because $\mu$ is a quasi-Radon measure,
therefore localizable (415A), and we can use 243Gb.
We therefore have an Arens
multiplication on $(L^{\infty})^*\cong(L^1)^{**}$
defined by the formulae of 4A6O.
Of course $\tilde p$ and $\tilde q$ both belong to $(L^{\infty})^*$;
write $\tilde r_0=\tilde p\frsmallcirc\tilde q$ for their Arens product.
To see that $\tilde r_0(\chi G^{\ssbullet})=1$,
note that if $u$, $v\in L^1$ then, defining
$\chi G^{\ssbullet}\frsmallcirc u$ and
$\tilde q\frsmallcirc\chi G^{\ssbullet}$ as in 4A6O, we have

\Centerline{$\int(\chi G^{\ssbullet}\frsmallcirc u)\times v
=\int\chi G^{\ssbullet}\times(u*v)
=\int u\int v$}

\noindent as noted in 444Sb;  consequently
$\chi G^{\ssbullet}\frsmallcirc u=(\int u)\chi G^{\ssbullet}$,

\Centerline{$\int(\tilde q\frsmallcirc\chi G^{\ssbullet})\times u
=\tilde q(\chi G^{\ssbullet}\frsmallcirc u)
=\tilde q((\int u)\chi G^{\ssbullet})
=\int u$}

\noindent and
$\tilde q\frsmallcirc\chi G^{\ssbullet}=\chi G^{\ssbullet}$.   Now, of
course,

\Centerline{$\tilde r_0(\chi G^{\ssbullet})
=\tilde p(\tilde q\frsmallcirc\chi G^{\ssbullet})
=\tilde p(\chi G^{\ssbullet})
=1$.}

\noindent As noted in
4A6O, $\|\tilde r_0\|\le\|\tilde p\|\|\tilde q\|=1$, so
$\tilde r_0$ must be a positive linear functional.

\medskip

\qquad\grheadc\ We find next that
$\tilde r_0(a\action_lu)
=\tilde r_0(u)$ whenever $u\in L^{\infty}$ and
$a\in G$.   \Prf\ By 443Ge, we have a bounded linear operator
$S:L^1\to L^1$ defined by setting $Sv=a^{-1}\action_lv$ for every
$v\in L^1$.    By 444Sa, $S(u*v)=(Su)*v$ for all $u$, $v\in L^1$.
Identifying $L^{\infty}$ with
$(L^1)^*$, we have the adjoint operator
$S':L^{\infty}\to L^{\infty}$ given by saying that

$$\eqalign{\int S'u\times v
&=\int u\times Sv
=\int u\times(a^{-1}\action_lv)\cr
&=\int a\action_l(u\times a^{-1}\action_lv)
=\int(a\action_lu)\times v\cr}$$

\noindent whenever $u\in L^{\infty}$ and $v\in L^1$, so that
$S'u=a\action_lu$ for every $u\in L^{\infty}$.   But this means that

\Centerline{$(S''\tilde p)(u)=\tilde p(a\action_lu)=\tilde p(u)$}

\noindent for every $u$, so that $S''\tilde p=\tilde p$.   By 4A6O(b-i),

\Centerline{$S''\tilde r_0
=S''(\tilde p\frsmallcirc\tilde q)
=(S''\tilde p)\frsmallcirc\tilde q
=\tilde p\frsmallcirc\tilde q=\tilde r_0$,}

\noindent that is, $\tilde r_0(a\action_lu)=\tilde r_0(u)$ for every
$u\in L^{\infty}$.\ \Qed

\medskip

\qquad\grheadd\ In the same way,
$\tilde r_0(a\action_ru)
=\tilde r_0(u)$ whenever $u\in L^{\infty}$ and
$a\in G$.   \Prf\ This time, define $T:L^1\to L^1$ by
setting $Tv=\Delta(a^{-1})a^{-1}\action_rv$ for every
$v\in L^1$, where $\Delta$ is the left modular function of $G$;
444Sa tells us that $T(u*v)=u*Tv$ for all $u$, $v\in L^1$.
Since $\int fd\mu=\Delta(a)\int a\action_rfd\mu$
for every $f\in\eusm L^1$ (442Kc),
$\int v=\Delta(a)\int a\action_rv$ for every $v\in L^1$, and

$$\eqalign{\int T'u\times v
&=\int u\times Tv
=\Delta(a^{-1})\int u\times(a^{-1}\action_rv)\cr
&=\int a\action_r(u\times a^{-1}\action_rv)
=\int(a\action_ru)\times v\cr}$$

\noindent whenever $u\in L^{\infty}$ and $v\in L^1$.   Thus
$T'u=a\action_ru$ for every $u\in L^{\infty}$.   But now we have

\Centerline{$(T''\tilde q)(u)=\tilde q(a\action_ru)=\tilde q(u)$}

\noindent for every $u$, so that $T''\tilde q=\tilde q$.   By 4A6O(b-ii),
$T''\tilde r_0=\tilde r_0$, that is,
$\tilde r_0(a\action_ru)=\tilde r_0(u)$ for every
$u\in L^{\infty}$.\ \Qed

\medskip

\qquad\grheade\ Thus $\tilde r_0$ is both left- and
right-invariant.   To get reversal-invariance, set

\Centerline{$\tilde r(u)=\Bover12(\tilde r_0(u)+\tilde r_0(\Reverse{u}))$}

\noindent for $u\in L^{\infty}$.   Then $\tilde r_0$ is a
positive linear functional and $\tilde r_0(\chi G^{\ssbullet})=1$.
Because

\Centerline{$a\action_l\Reverse{u}=(a\action_ru)\ssplrarrow$,
\quad$a\action_r\Reverse{u}=(a\action_lu)\ssplrarrow$,}

\noindent $u\mapsto\tilde r_0(\Reverse{u})$ and $\tilde r$ are also
both left- and right-invariant, and of course
$\tilde r(\Reverse{u}) =\tilde r(u)$ for every $u$.
Finally,

\Centerline{$\tilde r(a\action_cu)=\tilde r(a\action_l(a\action_ru))
=\tilde r(u)$}

\noindent for every $u\in L^{\infty}$ and $a\in G$, so
$\tilde r$ has all the properties required by (v).

\medskip

{\bf (d)(v)$\Rightarrow$(iv)}
Take $\tilde p$ from (v), and set
$\phi E=\tilde p(\chi E^{\ssbullet})$ for every $E\in\Sigma_G$.
Then $\phi:\Sigma_G\to[0,1]$ is additive and $\phi G=1$;  also, if
$E\in\Cal N_G$, $\chi E^{\ssbullet}=0$ in $L^{\infty}$ and
$\phi E=0$.   If $E\in\Sigma_G$ and $a\in G$,
then  $\chi(aE)=a\action_l(\chi E)$ (4A5C(c-ii)) and

\Centerline{$\phi(aE)
=\tilde p(\chi(aE)^{\ssbullet})
=\tilde p((a\action_l\chi E)^{\ssbullet})
=\tilde p(a\action_l(\chi E^{\ssbullet}))
=\tilde p(\chi E^{\ssbullet})
=\phi E$.}

\noindent Next, $\chi E^{-1}=(\chi E)\ssplrarrow$ and
$(\chi E^{-1})^{\ssbullet}=(\chi E^{\ssbullet})\ssplrarrow$, so

\Centerline{$\phi(E^{-1})
=\tilde p((\chi E^{\ssbullet})\ssplrarrow)
=\tilde p(\chi E^{\ssbullet})
=\phi E$.}

\noindent Consequently, for $E\in\Sigma_G$ and $a\in G$,

\Centerline{$\phi(Ea)=\phi((Ea)^{-1})
=\phi(a^{-1}E^{-1})=\phi(E^{-1})=\phi E$.}

\noindent Thus $\phi$ satisfies the requirements of (iv).

\medskip

{\bf (e)(iv)$\Rightarrow$(iii)} This is trivial;  we have only to take
$\phi:\Sigma_G\to[0,1]$ as in (iv)
and consider $\phi\restr\Cal B_G$.

\medskip

{\bf (f)(iii)$\Rightarrow$(ii)} Given $\phi:\Cal B_G\to[0,1]$ as in (iii),
set $p(f)=\dashint fd\phi$ for $f\in C_b(G)$, where $\dashint\,d\phi$
is as defined in 363L, that is, the unique $\|\,\|_{\infty}$-continuous
linear functional on the space
$L^{\infty}(\Cal B_G)$ of bounded Borel measurable functions from
$G$ to $\Bbb R$ such that $\dashint\chi E\,d\phi=\phi E$ for every
$E\in\Cal B_G$.   $p$ is positive because $\phi$ is non-negative (363Lc),
and $p(\chi G)=\phi G=1$.   If $a\in G$, then

\Centerline{$\dashint a\action_l\chi E\,d\phi=\dashint\chi(aE)d\phi
=\phi(aE)=\phi E=\dashint\chi E\,d\phi$}

\noindent for every $E\in\Cal B_G$;  because
$f\mapsto\dashint fd\phi$ and $f\mapsto\dashint a\action_lfd\phi$ are both
linear and $\|\,\|_{\infty}$-continuous, they agree on
$L^{\infty}(\Cal B_G)\supseteq C_b(G)$, and

\Centerline{$p(a\action_lf)=\dashint a\action_lfd\phi
=\dashint fd\phi=p(f)$}

\noindent for every $f\in C_b(G)$, as required.

\medskip

{\bf (g)(ii)$\Rightarrow$(i)} Given $p$ as in (ii), its restriction to
the space of bounded right-uniformly-continuous functions  is
positive, linear and $G$-invariant, so $G$ is amenable, by 449E.

\medskip

{\bf (h)(vii)$\Rightarrow$(viii)} Take $\tilde p$ as in (vii).   If
$g\in\eusm L^1$, $f\in\eusm L^{\infty}$ and $g\ge 0$, then

$$\eqalign{\tilde p(g*f)^{\ssbullet}
&=\tilde p(g\mu*f)^{\ssbullet}
=\tilde p(g\mu*f^{\ssbullet})\cr
&=(g\mu)(G)\tilde p(f^{\ssbullet})
=\int g\,d\mu\cdot\tilde p(f^{\ssbullet});\cr}$$

\noindent translating into terms of $L^1$ and $L^{\infty}$ as in 444Sa,
we get $\tilde p(v*u)=\int v\cdot\tilde p(u)$ for all
$u\in L^{\infty}$ and $v\in(L^1)^+$.   By linearity, the same is
true for all $v\in L^1$, as required by (viii).

\woddheader{449J}{4}{2}{2}{60pt}

{\bf (i)(viii)$\Rightarrow$(ix)} Suppose that (viii) is true.

\medskip

\quad\grheada\ Note first that if
$J\subseteq\eusm L^{\infty}$ is finite and $\epsilon>0$, then

\Centerline{$A(J,\epsilon)
=\{h:h\in C_{k1}^+,\,|\int f\times h\,d\mu-\tilde p(f^{\ssbullet})|
  \le\epsilon$ for every $f\in J\}$}

\noindent is non-empty.   \Prf\ It is enough to consider the case in which
$\chi G\in J$.   Let $\eta\in\ooint{0,\bover12}$ be such
that $\eta+5\eta\sup_{f\in J}\|f^{\ssbullet}\|_{\infty}\le\epsilon$.
Because
$\tilde p\in(L^{\infty})^*\cong(L^1)^{**}$, there is a $u_0\in L^1$ such
that $\|u_0\|_1\le 1$ and
$|\tilde p(f^{\ssbullet})-\int f^{\ssbullet}\times u_0|\le\eta$ for
every $f\in J$ (4A4If).   In particular,
$\int u_0\ge\tilde p(\chi G^{\ssbullet})-\eta=1-\eta$.   By 416I
again, there is a continuous $h_0:G\to\Bbb R$ with compact support such that
$\|u_0-h_0^{\ssbullet}\|_1\le\eta$.   Now $\int h_0d\mu\ge 1-2\eta$
and $\int|h_0|d\mu\le 1+\eta$.   So if we set $h_0^+=h_0\vee 0$,
$\gamma=\int h_0^+d\mu$ and $h=\Bover1{\gamma}h_0^+$, we shall have

\Centerline{$\gamma\le 1+\eta$,
\quad$\|h_0^+-h_0\|_1=\Bover12\int|h_0|-h_0\le 2\eta$,
\quad$\|h-h_0^+\|=|\gamma-1|\le 2\eta$,}

\noindent so
$\|u_0-h^{\ssbullet}\|_1\le 5\eta$, while $h\in C_{k1}^+$.   This will mean
that

\Centerline{$|\tilde p(f^{\ssbullet})-\int f\times h\,d\mu|
\le\eta+5\eta\|f^{\ssbullet}\|_{\infty}\le\epsilon$}

\noindent for every $f\in J$.\ \QeD\

We therefore have a filter $\Cal F$ on $C_{k1}^+$
containing every $A(J,\epsilon)$, and
$\tilde p(f^{\ssbullet})=\lim_{h\to\Cal F}\int f\times h\,d\mu$ for
every $f\in\eusm L^{\infty}$.

\medskip

\quad\grheadb\ Now
$0=\lim_{h\to\Cal F}(g*h)^{\ssbullet}-(\int g\,d\mu)h^{\ssbullet}$ for
the weak topology of $L^1$, for every $g\in\eusm L^1$.   \Prf\ Set
$\gamma=\int g\,d\mu$.   Let $f\in\eusm L^{\infty}$.   Define $g'$ by
setting $g'(x)=\Delta(x^{-1})g(x^{-1})$ whenever this is defined,
where $\Delta$ is the left modular function of $G$, as before;
then $g'\in\eusm L^1$
and $\int g'd\mu=\gamma$ (442K(b-ii)).
Set $v=(g')^{\ssbullet}\in L^1$.   If $h\in C_{k1}^+$, then

$$\eqalignno{\int f\times(g*h)d\mu
&=\iint f(xy)g(x)h(y)\mu(dx)\mu(dy)\cr
\displaycause{444Od}
&=\int\bigl(\int\Delta(x^{-1})g'(x^{-1})f(xy)\mu(dx)\bigr)h(y)\mu(dy)\cr
&=\int (g'*f)(y)h(y)\mu(dy)\cr
\displaycause{444Oa}
&=\int (g'*f)\times h\,d\mu.\cr}$$

\noindent By (viii), we have

\Centerline{$\tilde p(g'*f)^{\ssbullet}
=\tilde p(v*f^{\ssbullet})
=\tilde p(f^{\ssbullet})\int v
=\gamma\tilde p(f^{\ssbullet})$,}

\noindent so we get

$$\eqalign{\lim_{h\to\Cal F}\int f\times(g*h-\gamma h)d\mu
&=\lim_{h\to\Cal F}\int (g'*f-\gamma f)\times h\,d\mu\cr
&=\tilde p(g'*f)^{\ssbullet}-\gamma\tilde p(f^{\ssbullet})
=0.\cr}$$

\noindent As $f$ is arbitrary, and $(L^1)^*$ can be identified with
$L^{\infty}$, this is all we need.\ \Qed

\medskip

\quad\grheadc\ Now take any finite set $J\subseteq\eusm L^1$ and
$\epsilon>0$.   On $(L^1)^J$ let $\frak T$ be the locally convex linear
space topology which is the product topology when each copy of $L^1$ is
given the norm topology, and $\frak S$ the corresponding weak topology.
Define $T:C_k(G)\to(L^1)^J$ by setting

\Centerline{$Th
=\bigl\langle\hbox{$(g*h)^{\ssbullet}
  -(\int g\,d\mu)h^{\ssbullet}$}\bigr\rangle_{g\in J}$}

\noindent for $h\in C_k(G)$, where $C_k(G)$ is the linear space of
continuous real-valued functions on $G$ with compact support.    Then
$T$ is linear.   Moreover, by ($\beta$), $\lim_{h\to\Cal F}Th=0$ in
$(L^1)^J$ for the product topology, if each copy of $L^1$ is given its
weak topology.   By 4A4Be, this is just $\frak S$.   In particular,
$0$ belongs to the $\frak S$-closure of $T[C_{k1}^+]$.   But $C_{k1}^+$
is convex and
$T$ is linear, so $T[C_{k1}^+]$ is convex;  by 4A4Ed, $0$ belongs to the
$\frak T$-closure of $T[C_{k1}^+]$.   There is therefore an
$h\in C_{k1}^+$ such that
$\|(g*h)^{\ssbullet}-(\int g\,d\mu)h^{\ssbullet}\|_1\le\epsilon$ for
every $g\in J$.   As $J$ and $\epsilon$ are arbitrary, (ix) is true.

\medskip

{\bf (j)(ix)$\Rightarrow$(x)} Suppose that (ix) is true and that
we are given a compact set $K\subseteq G$ and $\epsilon>0$.   Set
$\eta=\bover13\epsilon$.
Fix any $h_0\in C_{k1}^+$.   Let $V$ be a neighbourhood of $e$ such
that $\|c\action_lh_0-h_0\|_1\le\eta$ whenever $c\in V$ (443Gf).   Let
$I\subseteq G$ be a finite set such that $K\subseteq IV$.   By (ix),
there is an $h_1\in C_{k1}^+$ such that

\Centerline{$\|b\action_lh_0*h_1-h_1\|_1\le\eta$ for every $b\in I$,
\quad$\|h_0*h_1-h_1\|_1\le\eta$.}

\noindent (I omit brackets because
$(b\action_lh_0)*h_1=b\action_l(h_0*h_1)$, see 444Of.)
Set $h=h_0*h_1$.   Then $h\in C_{k1}^+$.   \Prf\ $h$ is
continuous, by
444Rc (or otherwise).   If we write $M_i$ for the support $\supp(h_i)$ of
$h_i$ for both $i$, then $h(x)=0$ for every $x\in G\setminus M_0M_1$,
so $h$ has compact support.   Of course $h\ge 0$, and
$\int h\,d\mu=\int h_0d\mu\int h_1d\mu=1$ (444Qb), so $h\in C_{k1}^+$.\
\Qed

If $a\in K$, there are $b\in I$, $c\in V$ such that $a=bc$, so that

$$\eqalignno{\|a\action_lh-h\|_1
&=\|b\action_l(c\action_lh_0*h_1)-h_0*h_1\|_1\cr
&\le\|b\action_l(c\action_lh_0-h_0)*h_1\|_1
  +\|b\action_lh_0*h_1-h_1\|_1+\|h_1-h_0*h_1\|_1\cr
&\le\|c\action_lh_0-h_0\|_1+\eta+\eta
\le 3\eta=\epsilon.\cr}$$

\noindent So this $h$ will serve.

\medskip

{\bf (k)(x)$\Rightarrow$(xiii)} Suppose that (x) is true.

\medskip

\quad\grheada\ I show first that for any compact set $K\subseteq G$ and
$\epsilon>0$ there is an $h\in C_{k1}^+$ such that

\Centerline{$\Reverse{h}=h$,
\quad$h(e)=\|h\|_{\infty}$,}

\Centerline{$\|a\action_lh-h\|_1\le\epsilon$ for every $a\in K$.}

\noindent \Prf\ By (x), there is an $h_0\in C_{k1}^+$ such that
$\|a\action_lh_0-h_0\|_1\le\epsilon$ for every $a\in K$.   Set
$\gamma=\int\Reverse{h}_0d\mu$;  because
$\Reverse{h}_0\in C_k(X)^+\setminus\{0\}$, $\gamma$ is finite and not $0$.
Try $h=h_0*\bover1{\gamma}\Reverse{h}_0$, so that
$h(x)=\bover1{\gamma}\int h_0(y)h_0(x^{-1}y)\mu(dy)$ for every $x\in G$.
By 444Rc, $h\in C_b(G)$ and

\Centerline{$\|h\|_{\infty}\le\|h_0\|_2\|\Bover1{\gamma}h_0\|_2
=\Bover1{\gamma}\int h_0^2d\mu=h(e)$.}

\noindent Because $h_0\ge 0$, $h\ge 0$;   by 444Qb, $\int h\,d\mu=1$;  and
(as in (j) above) $\supp(h)$ is included in the compact set
$\supp(h_0)\supp(\Reverse{h}_0)$, so
$h\in C_{k1}^+$.   By 444Rb, or otherwise, $h=\Reverse{h}$.

Finally, if $a\in K$, then

$$\eqalignno{\|a\action_lh-h\|_1
&=\Bover1{\gamma}\|a\action_l(h_0*\Reverse{h}_0)-h_0*\Reverse{h}_0\|_1
=\Bover1{\gamma}\|(a\action_lh_0-h_0)*\Reverse{h}_0\|_1\cr
\displaycause{444Of once more}
&\le\Bover1{\gamma}\|a\action_lh_0-h_0\|_1\|\Reverse{h}_0\|_1\cr
\displaycause{444Qb again}
&=\|a\action_lh_0-h_0\|_1
\le\epsilon,\cr}$$

\noindent as required.\ \Qed

\medskip

\quad\grheadb\ Next, for any $\epsilon$, $\delta>0$ and
any compact set $K\subseteq G$ there are an open
symmetric neighbourhood $V$ of $e$ and a closed set $F$ such
that $\mu V<\infty$, $\mu F\le\delta$ and
$\mu(aV\symmdiff V)\le\epsilon\mu V$ whenever $a\in K\setminus F$.
\Prf\ Of course it is enough to deal with the case in which $\mu K>0$.
Set $\eta=\epsilon\delta/\mu K$.   By ($\alpha$), there is an
$h\in C_{k1}^+$ such that
$h(e)=\|h\|_{\infty}$, $h=\Reverse{h}$ and
$\|a\action_lh-h\|_1<\eta$ for every $a\in K$.

Set $K_0=\supp(h)$ and $K^*=K_0\cup KK_0$, so that
$K^*\subseteq G$ is compact.   Set

$$\eqalign{Q&=\{(a,x,t):a\in K,\,x\in G,\,t\in\Bbb R,\cr
&\mskip150mu\text{either }h(x)\le t<h(a^{-1}x)
  \text{ or }h(a^{-1}x)\le t<h(x)\}.\cr}$$

\noindent Then $Q$ is a Borel subset of $G\times G\times\Bbb R$ included
in the compact set $K\times K^*\times[0,h(e)]$.   Let $\mu_L$
be Lebesgue measure on $\Bbb R$, and let $\mu\times\mu\times\mu_L$ be
the $\tau$-additive product measure on $G\times G\times\Bbb R$ (417D).
(Of course this is actually a Radon measure.)   For $t\in\Bbb R$ let
$V_t$ be the open set $\{x:h(x)>t\}$.   Now 417G tells us that

$$\eqalignno{(\mu\times\mu\times\mu_L)(Q)
&=\int_{G\times G}\mu_L\{t:(a,x,t)\in Q\}(\mu\times\mu)(d(a,x))\cr
\displaycause{where $\mu\times\mu$ is the $\tau$-additive product
measure on $G\times G$, so that we can identify
$\mu\times\mu\times\mu_L$ with $(\mu\times\mu)\times\mu_L)$, as in
417Db}
&=\int_K\int_G|h(a^{-1}x)-h(x)|\mu(dx)\mu(da)\cr
\displaycause{we can use 417G again because
$\{x:h(a^{-1}x)\ne h(x)\}\subseteq K^*$ if $a\in K$, and $\mu K^*$ is
finite}
&=\int_K\|a\action_lh-h\|_1\mu(da)
<\eta\mu K\cr
\displaycause{by the choice of $h$}
&=\eta\mu K\int h\,d\mu
=\eta\mu K(\mu\times\mu_L)\{(x,t):0\le t<h(x)\}\cr
&=\eta\mu K\int_0^{h(e)}\mu V_t\,\mu_L(dt)\cr}$$

\noindent as in 252N.   (The c.l.d.\ and $\tau$-additive product
measures on $G\times\Bbb R$ coincide, by 417T.)   On the other hand,

$$\eqalignno{(\mu\times\mu\times\mu_L)(Q)
&=\int_{K\times\Bbb R}\mu\{x:(a,x,t)\in Q\}(\mu\times\mu_L)(d(a,t))\cr
\displaycause{again, we can use 417G because $x\in K^*$ whenever
$(a,x,t)\in Q$}
&=\int_{K\times\Bbb R}\mu(V_t\symmdiff aV_t)(\mu\times\mu_L)(d(a,t))\cr
&=\int_0^{\infty}\int_K
  \mu(V_t\symmdiff aV_t)\mu(da)\mu_L(dt)\cr}$$

\noindent because $V_t=aV_t=G$ whenever $t<0$ and $a\in G$.   So there
must be some $t\in\ooint{0,h(e)}$ such that

\Centerline{$\int_K\mu(V_t\symmdiff aV_t)\mu(da)
<\eta\mu K\mu V_t=\epsilon\delta\mu V_t$.}

\noindent Set $V=V_t$ and
$F=\{a:a\in K,\,\mu(V_t\symmdiff aV_t)\ge\epsilon\mu V_t\}$;  then
$V$ is open, $F$ is closed (443C) and $\mu(V\symmdiff aV)\le\epsilon\mu V$
for every $a\in K\setminus F$;  also $0<\mu V<\infty$,
$V$ is symmetric (because $h=\Reverse{h}$) and $e\in V$
(because $t<h(e)$).\ \Qed

\medskip

\quad\grheadc\ Now let $K\subseteq G$ be a compact set and
$\epsilon>0$,
as in the statement of (xiii);  enlarging $K$ and lowering $\epsilon$ if
necessary, we may suppose that $\mu K>0$ and $\epsilon\le 1$.   Set
$K_1=K\cup KK$, so that $K_1$ is still compact.   By ($\beta$), we
have a symmetric open neighbourhood $V$ of $e$, of finite measure,
such that
$W=\{a:a\in K_1,\,\mu(aV\symmdiff V)>\bover13\epsilon\mu V\}$ has
measure less than $\bover12\mu K$.   If $a\in K$, then $W\cup a^{-1}W$
cannot cover $K$, so there is a $b\in K\setminus W$ such that
$ab\notin W$;  thus $b$ and $ab$ both belong to $K_1\setminus W$, and

\Centerline{$\mu(aV\symmdiff V)
\le\mu(aV\symmdiff abV)+\mu(abV\symmdiff V)
\le\mu(V\symmdiff bV)+\Bover13\epsilon\mu V
\le\Bover23\epsilon\mu V$.}

We can now find a compact symmetric neighbourhood $L$ of $e$, included in
$V$, with $8\mu(V\setminus L)\le\epsilon\mu L$.   In this case,
we shall have $\mu L>0$ and

$$\eqalign{\mu(aL\symmdiff L)
&\le\mu(aV\setminus aL)+\mu(aV\symmdiff V)+\mu(V\setminus L)\cr
&\le 2\mu(V\setminus L)+\Bover23\epsilon(\mu L+\mu(V\setminus L))
\le\epsilon\mu L\cr}$$

\noindent for every $a\in K$, as required.

\medskip

{\bf (l)(xiii)$\Rightarrow$(xiv)} Suppose that (xiii) is true, and that
$K\subseteq G$ is compact and $\epsilon>0$.   Enlarging $K$ if
necessary, we may
suppose that it includes a neighbourhood of $e$.   Of course we may
also suppose that $\epsilon\le 1$.

\medskip

\quad\grheada\ The first thing to note is that there is a set
$I\subseteq G$ such that $KI=G$ and
$m=\sup_{y\in G}\#(\{x:x\in I,\,y\in Kx\})$ is finite.   \Prf\ Let $V$
be an open neighbourhood of $e$ such that $VV^{-1}\subseteq K$.   Let
$I\subseteq G$ be maximal subject to $x^{-1}V\cap y^{-1}V=\emptyset$
for all distinct $x$, $y\in I$.   If $x\in G$, there must be a $y\in I$
such that $x^{-1}V\cap y^{-1}V\ne\emptyset$, so that
$x^{-1}\in y^{-1}VV^{-1}$ and $x\in VV^{-1}y\subseteq Ky\subseteq KI$;
as $x$ is arbitrary, $G\subseteq KI$.   If $y\in G$ and
$I_y=\{x:x\in I,\,y\in Kx\}$, then $I_y\subseteq K^{-1}y$, so that
$\{x^{-1}V:x\in I_y\}$ is a disjoint family of subsets of $y^{-1}KV$.
But this means that $\#(I_y)\mu V\le\mu(y^{-1}KV)=\mu(KV)$.
Accordingly $\sup_{y\in G}\#(I_y)\le\Bover{\mu(KV)}{\mu V}$ is
finite.\
\Qed

\medskip

\quad\grheadb\ Set $\gamma=\sup_{a\in K}\Delta(a)$, $K^*=KKK^{-1}$.
Let $\delta>0$ be such that

\Centerline{$\delta\gamma m<\epsilon(\mu K-\delta\gamma)$,}

\noindent and let $\eta>0$ be such that

\Centerline{$1+\Bover{\delta\gamma m}{\mu K-\delta\gamma}
\le(1+\epsilon)(1-\Bover{\eta}{\delta}\mu K^*)$.}

\noindent By (xiii), there is a non-negligible
compact set $L^*\subseteq G$ such that
$\mu(aL^*\symmdiff L^*)\le\eta\mu L^*$ for every $a\in K^*$.
Set $L=\{x:x\in L^*,\,\mu(K^*\setminus L^*x^{-1})\le\delta\}$;  note
that $L$ is closed (because $x\mapsto\mu(K^*\cap L^*x^{-1})$ is
continuous, see 443C), therefore compact.

\medskip

\quad\grheadc\ $\mu L\ge(1-\Bover{\eta}{\delta}\mu K^*)\mu L^*$.
\Prf\
Set $W=\{(x,y):x\in K^*,\,y\in L^*\setminus L,\,xy\notin L^*\}$.
Then $W$ is a relatively compact Borel subset of $G\times G$, so we may
apply Fubini's theorem (in the form 417G, as usual) to see that

$$\eqalign{\delta\mu(L^*\setminus L)
&\le\int_{L^*\setminus L}\mu(K^*\setminus L^*y^{-1})\mu(dy)
=\int\mu W^{-1}[\{y\}]\mu(dy)\cr
&=\int\mu W[\{x\}]\mu(dx)
=\int_{K^*}\mu((L^*\setminus L)\setminus x^{-1}L^*)\mu(dx)\cr
&\le\int_{K^*}\mu(L^*\setminus x^{-1}L^*)\mu(dx)
=\int_{K^*}\mu(xL^*\setminus L^*)\mu(dx)
\le\eta\mu K^*\mu L^*\cr}$$

\noindent by the choice of $L^*$.   Accordingly

\Centerline{$\mu L=\mu L^*-\mu(L^*\setminus L)
\ge(1-\Bover{\eta}{\delta}\mu K^*)\mu L^*$.  \Qed}

\noindent In particular, $\mu L>0$.

\medskip

\quad\grheadd\ Set $J=\{x:x\in I,\,L\cap Kx\ne\emptyset\}$.
For each $x\in J$, choose $z_x\in Kx\cap L$.   Then
$\Delta(z_x)\le\gamma\Delta(x)$ and
$\Delta(x)(\mu K-\gamma\delta)\le\mu(Kx\cap L^*)$ for every $x\in J$.
\Prf\ $\Delta(z_x)=\Delta(z_xx^{-1})\Delta(x)\le\gamma\Delta(x)$ because
$z_xx^{-1}\in K$.   Next, because $z_x\in L$,

\Centerline{$\mu(K^*z_x\setminus L^*)
=\Delta(z_x)\mu(K^*\setminus L^*z_x^{-1})\le\delta\Delta(z_x)$.}

\noindent Since $x\in K^{-1}z_x$, $Kx\subseteq KK^{-1}z_x\subseteq K^*z_x$,

\Centerline{$\mu(Kx\setminus L^*)
\le\mu(K^*z_x\setminus L^*)\le\delta\Delta(z_x)\le\delta\gamma\Delta(x)$}

\noindent and

\Centerline{$\mu(Kx\cap L^*)=\mu(Kx)-\mu(Kx\setminus L^*)
\ge\Delta(x)\mu K-\delta\gamma\Delta(x)
=\Delta(x)(\mu K-\delta\gamma)$.  \Qed}

\wheader{449J}{6}{2}{2}{48pt}

\quad\grheade\ Now recall that $\sum_{x\in I}\chi(Kx)\le m\chi G$, by
the definition of $m$, so that

\Centerline{$(\mu K-\delta\gamma)\sum_{x\in J}\Delta(x)
\le\sum_{x\in J}\mu(Kx\cap L^*)
\le m\mu L^*$.}

\noindent
Since $KI=G$, $L\subseteq KJ$ and

$$\eqalign{KL&\subseteq\bigcup_{x\in J}KKx
\subseteq\bigcup_{x\in J}KKK^{-1}z_x=\bigcup_{x\in J}K^*z_x,\cr
\cr
\mu(KL\setminus L^*)
&\le\sum_{x\in J}\mu(K^*z_x\setminus L^*)
=\sum_{x\in J}\Delta(z_x)\mu(K^*\setminus L^*z_x^{-1})\cr
&\le\delta\sum_{x\in J}\Delta(z_x)
\le\delta\gamma\sum_{x\in J}\Delta(x)
\le\Bover{\delta\gamma m}{\mu K-\delta\gamma}\mu L^*.\cr}$$

\noindent Accordingly

$$\eqalignno{\mu(KL)
&\le\mu L^*(1+\Bover{\delta\gamma m}{\mu K-\delta\gamma})
\le\mu L\cdot\bover{1+\bover{\delta\gamma m}
  {\mu K-\delta\gamma}}{1-\bover{\eta}{\delta}\mu K^*}\cr
\displaycause{by ($\gamma$) above}
&\le(1+\epsilon)\mu L,\cr}$$

\noindent by the choice of $\delta$ and $\eta$.   Thus we have found
an
appropriate set $L$.

\medskip

{\bf (m)(xiv)$\Rightarrow$(xii)} If $I\subseteq G$ is finite and
$\epsilon>0$,
$I\cup\{e\}$ is compact, so there is a compact set $L\subseteq G$, of
non-zero measure, such that
$\mu(IL\cup L)\le(1+\bover12\epsilon)\mu L$.   Consequently

\Centerline{$\mu(L\symmdiff aL)
=2\mu(aL\setminus L)
\le 2\mu(IL\setminus L)
\le\epsilon\mu L$}

\noindent for every $a\in I$, as required by (xii).

\medskip

{\bf (n)(xii)$\Rightarrow$(ii)} Write $\Cal L$ for the family of all
compact subsets of $G$ with non-zero measure.   For $L\in\Cal L$,
define $p_L:C_b(G)\to\Bbb R$ by setting $p_L(f)=\Bover1{\mu L}\int_Lfd\mu$
for $f\in C_b(G)$.   Of course $|p_L(f)|\le\|f\|_{\infty}$.
For finite $I\subseteq G$, $\epsilon>0$ set

\Centerline{$\Cal A(I,\epsilon)
=\{L:L\in\Cal L,\,\mu(aL\symmdiff L)\le\epsilon\mu L$ for
every $a\in I\}$.}

\noindent By (xii), no $\Cal A(I,\epsilon)$ is empty.   So we have an
ultrafilter $\Cal F$ on $\Cal L$ containing every $\Cal A(I,\epsilon)$.
Set $p(f)=\lim_{L\to\Cal F}p_L(f)$ for $f\in C_b(G)$;  then
$p:C_b(G)\to\Bbb R$ is a positive linear functional and $p(\chi G)=1$.

If $a\in G$, $f\in C_b(G)$ and $L\in\Cal A(\{a\},\epsilon)$, then

$$\eqalign{|p_L(a\action_lf)-p_L(f)|
&=\Bover1{\mu L}\bigl|\int_Lf(a^{-1}x)\mu(dx)-\int_Lf(x)\mu(dx)\bigr|\cr
&=\Bover1{\mu L}\bigl|\int_{aL}fd\mu-\int_Lfd\mu\bigr|
\le\Bover1{\mu L}\|f\|_{\infty}\mu(aL\symmdiff L)
\le\epsilon\|f\|_{\infty}.\cr}$$

\noindent
Since $\Cal F$ contains $\Cal A(\{a\},\epsilon)$ for every
$\epsilon>0$,

\Centerline{$|p(a\action_lf)-p(f)|
=\lim_{L\to\Cal F}|p_L(a\action_lf)-p_L(f)|=0$.}

\noindent As $f$ and $a$ are arbitrary, $p$ witnesses that (ii) is true.

\medskip

{\bf (o)(xii)$\Rightarrow$(xi)} Given a finite set $I\subseteq G$ and
$\epsilon>0$, (xii) tells us that there is a compact set $L\subseteq G$
of non-zero measure such
that $\mu(aL\symmdiff L)\le\epsilon^q\mu L$ for every $a\in I$.   Try
$u=\Bover1{(\mu L)^{1/q}}(\chi L)^{\ssbullet}$.   Then $\|u\|_q=1$.
If $a\in I$, then $a\action_lu=\Bover1{(\mu L)^{1/q}}\chi(aL)^{\ssbullet}$,
so

\Centerline{$\int|u-a\action_lu|^q
=\int(\Bover1{(\mu L)^{1/q}}\chi(aL\symmdiff L)^{\ssbullet})^q
=\Bover{\mu(aL\symmdiff L)}{\mu L}
\le\epsilon^q$}

\noindent and $\|u-a\action_lu\|_q\le\epsilon$, as required by (xi).

\medskip

{\bf (p)(xi)$\Rightarrow$(xii)} Now assume that (xi) is true.   Let
$I\subseteq G$ be a finite set, and $\epsilon>0$.   Set
$\delta=\Bover{\epsilon}{4+\epsilon}$, and let
$\eta>0$ be such that that $(1+\eta\#(I))^q\le 1+\delta$.

Take $u\in L^q$ such that $\|u\|_q=1$ and
$\|u-a\action_lu\|_q\le\eta$ for every $a\in I$.   Setting $v=|u|$, we
see that $a\action_lv=|a\action_lu|$, so
$|v-a\action_lv|\le|u-a\action_lu|$ and $\|v-a\action_lv\|_q\le\eta$ for
every $a\in I$, while $\|v\|_q=1$.   Let $f:G\to\coint{0,\infty}$ be
a function such
that $f^{\ssbullet}=v$ in $L^q$;  then $\|f\|_q=1$ while
$\|f-a\action_lf\|_q\le\eta$ for every $a\in I$.   Set
$g=\sup_{a\in I\cup\{e\}}a\action_lf$;  then

\Centerline{$f\le g\le f+\sum_{a\in I}(a\action_lf-f)^+$,}

\noindent so

\Centerline{$\|g\|_q\le 1+\sum_{a\in I}\|a\action_lf-f\|_q
\le 1+\eta\#(I)$,
\quad$\int g^qd\mu\le(1+\eta\#(I))^q\le 1+\delta$.}

\noindent For $t>0$, set $E_t=\{x:f(x)^q\ge t\}$, $F_t=\{t:g(x)^q\ge t\}$.
Then

\Centerline{$\int_0^{\infty}\mu E_tdt=\int f^qd\mu=1$,
\quad$\int_0^{\infty}\mu F_tdt=\int g^qd\mu\le 1+\delta$,}

\noindent where the integrals here are with respect to Lebesgue measure
(252O).   There must therefore be a $t>0$ such that $\mu E_t>0$ and
$\mu F_t\le(1+\delta)\mu E_t$.

If $a\in I$,
then $a\action_lf\le g$, so $aE_t=\{x:(a\action_lf)(x)\ge t\}$
is included in $F_t$;  also, of course, $E_t\subseteq F_t$.
We therefore have

\Centerline{$\mu(E_t\symmdiff aE_t)=2\mu(aE_t\setminus E_t)
\le 2\mu(F_t\setminus E_t)\le 2\delta\mu E_t$.}

\noindent There is no reason why $E_t$ should be compact, so it may
not itself be the $L$ we seek.   However, $\mu E_t$ is certainly finite, so
there must be a compact $L\subseteq E$ such that
$\mu L\ge(1-\delta)\mu E_t$.   In this case, $\mu L>0$ and

$$\eqalign{\mu(aL\symmdiff L)
&\le\mu(aL\symmdiff aE_t)+\mu(aE_t\symmdiff E_t)+\mu(E_t\symmdiff L)\cr
&\le 2\mu(E_t\setminus L)+2\delta\mu E_t
\le 4\delta\mu E_t
\le\Bover{4\delta}{1-\delta}\mu L
=\epsilon\mu L\cr}$$

\noindent for every $a\in I$.   So this $L$ will serve.
}%end of proof of 449J

\cmmnt{\medskip

\noindent{\bf Remark} Of course there are many variations possible in the
conditions listed above, some of which are in
449Xk-449Xm.  %449Xk 449Xl 449Xm
}

\leader{449K}{Proposition} Let $G$ be an amenable locally compact
Hausdorff group, and $H$ a subgroup of $G$.   Then $H$ is amenable.

\proof{\cmmnt{({\smc Paterson 88}, 1.12)}
{\bf (a)} For most of the proof (down to the end of (g) below), suppose
that $H$ is closed.   Let $V$ be a compact neighbourhood of the identity in
$G$.   Let $I\subseteq G$ be a maximal set such that
$VzH\cap Vz'H=\emptyset$ for all distinct $z$, $z'\in I$.
Then $V^{-1}VIH=G$.   \Prf\ If $x\in G$, there is a $z\in I$ such that
$VxH\cap VzH\ne\emptyset$, that is,

\Centerline{$x\in V^{-1}VzHH^{-1}=V^{-1}VzH\subseteq V^{-1}VIH$.  \Qed}

\medskip

{\bf (b)} If $x\in G$ then

\Centerline{$I\cap V^{-1}xH
=\{z:z\in I$, $z\in V^{-1}xH\}
=\{z:z\in I$, $x\in VzH^{-1}=VzH\}$}

\noindent has at most one element.   If $K\subseteq G$ is compact, then
there is a finite set $J\subseteq G$ such that $K\subseteq V^{-1}J$, and
now

\Centerline{$I\cap KH\subseteq\bigcup_{x\in J}I\cap V^{-1}xH$}

\noindent is finite.

\medskip

{\bf (c)} Let $h\in C_k(X)^+$ be such that $h\ge\chi(V^{-1}V)$;  write
$W$ for the support of $h$.   Set $g(x)=\sum_{z\in I}h(xz^{-1})$ for
$x\in G$.

If $K\subseteq G$ is compact, then

$$\eqalign{&\{z:z\in I,\,h(xyz^{-1})\ne 0
   \text{ for some }x\in K\text{ and }y\in H\}\cr
&\mskip100mu
\subseteq I\cap\{z:KHz^{-1}\cap W\ne\emptyset\}\cr
&\mskip100mu
=I\cap\{z:zH^{-1}K^{-1}\cap W^{-1}\ne\emptyset\}
=I\cap W^{-1}KH\cr}$$

\noindent is finite.   In particular, $\{z:z\in I$, $h(xz^{-1})\ne 0\}$
and $g(x)$ are finite for every $x\in G$.   Next, if $x_0\in G$,
then $J=\{z:h(xz^{-1})\ne 0$ for some $x\in x_0V\}$ is finite, and
$g(x)=\sum_{z\in J}h(xz^{-1})$ for $x\in x_0V$, so $g$ is continuous at
$x_0$;  as $x_0$ is arbitrary, $g\in C(G)$.   Of course $g\ge 0$ because
$h\ge 0$.

\medskip

{\bf (d)(i)} For $x\in G$, set $g_x(y)=g(x^{-1}y)$ for $y\in H$.   Then
$g_x\in C_k(H)$.   \Prf\ $g_x$ is continuous because $g$ is.   Now
$J=\{z:z\in I,\,h(x^{-1}yz^{-1})\ne 0$ for some $y\in H\}$ is finite, and

$$\eqalign{\{y:g_x(y)\ne 0\}
&\subseteq\{y:\text{ there is a }z\in I
   \text{ such that }h(x^{-1}yz^{-1})\ne 0\}
   \cr
&\subseteq\{y:\text{ there is a }z\in J
   \text{ such that }x^{-1}yz^{-1}\in W\}
=xWJ\cr}$$

\noindent which is compact.\ \Qed

\medskip

\quad{\bf (ii)} Moreover, for any $x_0\in G$ there is a compact set
$L\subseteq H$ such that
$\{x:|g_x-g_{x_0}|\le\epsilon\chi L\}$ is a neighbourhood of $x_0$ for
every $\epsilon>0$.   \Prf\

\Centerline{$J=\{z:z\in I,\,h(x^{-1}yz^{-1})\ne 0
   \text{ for some }x\in x_0V\text{ and }y\in H\}$}

\noindent is finite, by (c) in its full strength.   Let $L$ be the compact
set $H\cap x_0VWJ$.

Take any $\epsilon>0$.   If $x\in x_0V$, then
$g_x(y)=\sum_{z\in J}h(x^{-1}yz^{-1})$ for every $y\in H$, and
$g_x(y)=0$ for $y\in H\setminus L$.   Moreover, setting
$g'_x(y)=\sum_{z\in J}h(x^{-1}yz^{-1})$ for $x\in G$ and $y\in H$,
$(x,y)\mapsto g'_x(y)$ is continuous, so $x\mapsto g'_x:G\to C(H)$ is
continuous if we give $C(H)$ the topology of uniform convergence on compact
subsets of $H$ (4A2G(g-i)).   In particular, $x\mapsto g'_x\restr L$ is
continuous for the norm topology of $C(L)$, and
$U
=\{x:x\in x_0V$, $\|g'_x\restr L-g'_{x_0}\restr L\|_{\infty}\le\epsilon\}$
is a neighbourhood of $x_0$.   But if $x\in U$, then

\Centerline{$g_x(y)=g_{x_0}(y)=0$ for $y\in H\setminus L$,}

\Centerline{$|g_x(y)-g_{x_0}(y)|=|g'_x(y)-g'_{x_0}(y)|\le\epsilon$
for $y\in L$,}

\noindent so $|g_x-g_{x_0}|\le\epsilon\chi L$.\ \Qed

\medskip

{\bf (e)} Now take a left Haar measure $\nu$ on $H$.   (This is where it
really matters whether $H$ is closed.)   Define $T:C_b(H)\to\BbbR^G$ by
setting

\Centerline{$(Tf)(x)=\int g_x\times fd\nu
=\int_Hg(x^{-1}y)f(y)\nu(dy)$}

\noindent for $f\in C_b(H)$ and $x\in G$.   Then $Tf\in C(G)$ for every
$f\in C_b(H)$.   \Prf\ Given $x_0\in G$ and $\epsilon>0$, let
$L\subseteq H$ be a compact set as in (d-ii).   Let $\delta>0$ be such that
$\delta\int_L|f|d\nu\le\epsilon$.   Then

\Centerline{$\{x:|(Tf)(x)-(Tf)(x_0)|\le\epsilon\}
\supseteq\{x:|g_x-g_{x_0}|\le\delta\chi L\}$}

\noindent is a neighbourhood of $x_0$.   As $x_0$ and $\epsilon$ are
arbitrary, $Tf$ is continuous.\ \Qed

Clearly, $T:C_b(H)\to C(G)$ is a positive linear operator.   Next, if
$f\in C_b(H)$ and $b\in H$, $T(b\action_lf)=b\action_l(Tf)$.    \Prf\

$$\eqalign{T(b\action_lf)(x)
&=\int_Hg(x^{-1}y)(b\action_lf)(y)\nu(dy)
=\int_Hg(x^{-1}y)f(b^{-1}y)\nu(dy)\cr
&=\int_Hg(x^{-1}by)f(y)\nu(dy)
=(Tf)(b^{-1}x)
=(b\action_lTf)(x)\cr}$$

\noindent for every $x\in G$.\ \Qed

We need to know that $T(\chi H)(x)>0$ for every $x\in G$.   \Prf\
There is a $z\in I$ such that $x^{-1}\in V^{-1}VzH$, as remarked in (a).
Now $x^{-1}Hz^{-1}$ meets $V^{-1}V$, so there is a $y\in H$ such that

\Centerline{$1\le h(x^{-1}yz^{-1})\le g(x^{-1}y)=g_x(y)$}

\noindent and $T(\chi H)(x)=\int_Hg_xd\nu>0$
because $g_x$ is continuous and non-negative and $\nu$ is
strictly positive.\ \Qed

\medskip

{\bf (f)} We therefore have a positive linear operator
$S:C_b(H)\to C(G)$ defined by setting $Sf=\Bover{Tf}{T(\chi H)}$ for
$f\in C_b(H)$.   Since $S(\chi H)=\chi G$,
$S[C_b(H)]\subseteq C_b(G)$;  moreover, for $f\in C_b(H)$ and
$b\in H$,

$$\eqalign{S(b\action_lf)
&=\Bover{T(b\action_lf)}{T(\chi H)}
=\Bover{T(b\action_lf)}{T(b\action_l\chi H)}\cr
&=\Bover{b\action_l(Tf)}{b\action_l(T\chi H)}
=b\action_l\Bover{Tf}{T\chi H}
=b\action_lSf.\cr}$$

\medskip

{\bf (g)} At this point, recall that by 449J(ii) there is a positive linear
functional $p:C_b(G)\to\Bbb R$ such that $p(\chi G)=1$ and
$p(a\action_lf)=p(f)$ whenever $f\in C_b(G)$ and $a\in G$.
Set $q=pS:C_b(H)\to\Bbb R$;  then
$q$ is a positive linear functional, $q(\chi H)=1$ and
$q(b\action_lf)=q(f)$ whenever $f\in C_b(H)$ and $b\in H$.   So $q$
witnesses that 449J(ii) is true for $H$, and $H$ is amenable.

\medskip

{\bf (h)} All this has been on the assumption that $H$ is closed.   But in
general $\overline{H}$ is a closed subgroup of $G$, therefore amenable by
(a)-(g) here, and $H$ is dense in $\overline{H}$, therefore amenable by
449F(a-ii).
}%end of proof of 449K

%can't we improve the argument in (b), or at least move much of it to 4A5,
%by showing that  \phi  is continuous in a more general case?

%Very likely the method of Paterson 88, 1.12, would be better still,
%if I knew what a Bruhat function was.

\leader{449L}{}\cmmnt{ If we make a further step back towards the
origin of this topic, and suppose that our group is discrete, then we
have a striking further condition to add to the lists above.   I give
this as a corollary of a general result on group actions recalling the
main theorems of \S\S395 and 448.

\medskip

\noindent}{\bf Tarski's theorem} Let $G$ be a group acting on a
non-empty set $X$.   Then the following are equiveridical:

(i) there is an additive functional
$\nu:\Cal PX\to[0,1]$ such that $\nu X=1$ and $\nu(a\action A)=\nu A$
whenever $A\subseteq X$ and $a\in G$;

(ii) there are no
$A_0,\ldots,A_n$, $a_0,\ldots,a_n$, $b_0,\ldots,b_n$ such that
$A_0,\ldots,A_n$ are subsets of $X$ covering $X$,
$a_0,\ldots,a_n,b_0,\penalty-100\ldots,b_n$ belong to $G$, and
$a_0\action A_0$, $b_0\action A_0$, $a_1\action A_1$,
$b_1\action A_1,\ldots,b_n\action A_n$ are all disjoint.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} This is elementary, for if
$\nu:\Cal PX\to[0,1]$ is a non-zero additive functional and
$A_0,\ldots,A_n$ cover $X$ and $a_0,\ldots,b_n\in G$, then

\Centerline{$\sum_{i=0}^n\nu(a_i\action A_i)
  +\sum_{i=0}^n\nu(b_i\action A_i)
=2\sum_{i=0}^n\nu A_i\ge 2\nu X>\nu X$,}

\noindent and $a_0\action A_0,\ldots,b_n\action A_n$ cannot all be
disjoint.

\medskip

{\bf (b)(ii)$\Rightarrow$(i)} Now suppose that (ii) is true.

\medskip

\quad\grheada\ Suppose that $c_0,\ldots,c_n\in G$.   Then there is a
finite set $I\subseteq X$ such that
$\#(\{c_i\action x:i\le n,\,x\in I\})<2\#(I)$.   \Prf\Quer\ Otherwise,
by the Marriage Lemma in the form 4A1H, applied to the set

\Centerline{$R=\{((x,j),c_i\action x):x\in X,\,j\in\{0,1\},\,i\le n\}
\subseteq(X\times\{0,1\})\times X$,}

\noindent there is an injective function $\phi:X\times\{0,1\}\to X$
such
that $\phi(x,j)\in\{c_i\action x:i\le n\}$ for every $x\in X$ and
$j\in\{0,1\}$.   Now set
$B_{ij}=\{x:\phi(x,0)=c_i\action x,\,\phi(x,1)=c_j\action x\}$ for $i$,
$j\le n$, so that $X=\bigcup_{i,j\le n}B_{ij}$.   Let
$A_{ij}\subseteq B_{ij}$ be such that $\langle A_{ij}\rangle_{i,j\le n}$
is a partition of $X$, and set $a_{ij}=c_i$, $b_{ij}=c_j$ for $i$,
$j\le n$;  then $a_{ij}\action A_{ij}\subseteq\phi[A_{ij}\times\{0\}]$,
$b_{ij}\action A_{ij}\subseteq\phi[A_{ij}\times\{1\}]$ are all
disjoint, which is supposed to be impossible.\ \Bang\Qed

\medskip

\quad\grheadb\ Suppose that $J\subseteq G$ is finite and $\epsilon>0$.
Then there is a non-empty finite set $I\subseteq X$ such that
$\#(I\symmdiff c\action I)\le\epsilon\#(I)$ for every $c\in J$.
\Prf\ It is enough to consider the case in which the identity $e$ of
$G$ belongs to $J$.   \Quer\ Suppose, if possible, that there is no such
set $I$.    Let $m\ge 1$ be such that $(1+\bover12\epsilon)^m\ge 2$.   Set
$K=J^m=\{c_1c_2\ldots c_m:c_1,\ldots,c_m\in J\}$.   By ($\alpha$),
there
is a finite set $I_0\subseteq X$ such that $\#(I_0^*)<2\#(I_0)$, where
$I_0^*=\{a\action x:a\in K,\,x\in I_0\}$.   Choose $c_1,\ldots,c_m$
and $I_1,\ldots,I_m$ inductively such that

\inset{\noindent given that $I_k$ is a non-empty finite subset of $X$,
where $0\le k<m$, take $c_{k+1}\in J$ such that
$\#(I_k\symmdiff c_{k+1}\action I_k)>\epsilon\#(I_k)$ and set
$I_{k+1}=I_k\cup c_{k+1}\action I_k$.}

\noindent Then $I_k\subseteq\{a\action x:a\in J^k,\,x\in I_0\}$ for
each $k\le m$, and in particular $I_m\subseteq I_0^*$.   But also

$$\eqalign{\#(I_{k+1})
&=\#(I_k)+\#((c_{k+1}\action I_k)\setminus I_k)\cr
&=\#(I_k)+\Bover12\#((c_{k+1}\action I_k)\symmdiff I_k)
\ge(1+\Bover12\epsilon)\#(I_k)\cr}$$

\noindent for every $k<m$, so

\Centerline{$\#(I^*_0)\ge\#(I_m)\ge(1+\Bover12\epsilon)^m\#(I_0)
\ge 2\#(I_0)$,}

\noindent contrary to the choice of $I_0$.\ \Bang\Qed

\medskip

\quad\grheadc\ There is therefore an ultrafilter $\Cal F$ on
$[X]^{<\omega}\setminus\{\emptyset\}$ such that

\Centerline{$\Cal A_{c\epsilon}
=\{I:I\in[X]^{<\omega}\setminus\{\emptyset\},\,
\#(I\symmdiff c\action I)\le\epsilon\#(I)\}$}

\noindent belongs to $\Cal F$ for every $c\in G$ and $\epsilon>0$.
For
$I\in[X]^{<\omega}\setminus\{\emptyset\}$ and $A\subseteq X$ set
$\nu_I(A)=\#(A\cap I)/\#(I)$, and set $\nu A=\lim_{I\to\Cal F}\nu_IA$
for every $A\subseteq X$, so that $\nu:\Cal PX\to[0,1]$ is an additive
functional and $\nu X=1$.

Now $\nu$ is $G$-invariant.   \Prf\ If $A\subseteq X$ and $c\in G$ and
$\epsilon>0$, then $\Cal A_{c^{-1},\epsilon}\in\Cal F$.   If
$I\in\Cal A_{c^{-1},\epsilon}$, then

$$\eqalign{|\nu_I(c\action A)-\nu_I(A)|
&=\Bover1{\#(I)}|\#(I\cap(c\action A))-\#(I\cap A)|\cr
&=\Bover1{\#(I)}|\#((c^{-1}\action I)\cap A)-\#(I\cap A)|\cr
&\le\Bover1{\#(I)}\#((c^{-1}\action I)\symmdiff I)
\le\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\lim_{I\to\Cal F}\nu_I(c\action A)-\nu_I(A)=0$, and
$\nu(c\action A)=\nu A$.   As $A$ and $c$ are arbitrary, $\nu$ is
$G$-invariant.\ \Qed

So $\nu$ witnesses that (i) is true, and the proof is complete.
}%end of proof of 449L

\leader{449M}{Corollary} Let $G$ be a group with its discrete
topology.   Then the following are equiveridical:

(i) $G$ is amenable;

(ii) there are no
$A_0,\ldots,A_n$, $a_0,\ldots,a_n$, $b_0,\ldots,b_n$ such that
$G=\bigcup_{i\le n}A_i$
$a_0,\ldots,a_n,b_0,\ldots,b_n$ belong to $G$, and $a_0A_0$,
$b_0A_0$, $a_1A_1$,
$b_1A_1,\ldots,b_nA_n$ are disjoint.

\proof{ All we have to observe is that ($\alpha$) every function from
$G$ to $\Bbb R$ is uniformly continuous for the right uniformity of
$G$,
so that $G$ is amenable iff there is an invariant positive linear
functional $p:\ell^{\infty}(G)\to\Bbb R$ such that $p(\chi G)=1$
($\beta$) that a positive linear functional on $\ell^{\infty}(G)$ is
$G$-invariant iff the corresponding additive functional on $\Cal PG$
is
$G$-invariant.   So (i) of 449L is equivalent to amenability of $G$ as
defined in 449A.
}%end of proof of 449M

\leader{449N}{Theorem} Let $G$ be a group which is amenable in its
discrete topology, $X$ a set, and $\action$ an action of $G$ on $X$.
Let $\Cal E$ be a subring of $\Cal PX$ and
$\nu:\Cal E\to\coint{0,\infty}$ a finitely additive functional which
is $G$-invariant in the sense that $g\action E\in\Cal E$ and
$\nu(g\action E)=\nu E$ whenever $E\in\Cal E$ and $g\in G$.   Then
there is an extension of $\nu$ to a $G$-invariant non-negative finitely
additive functional $\tilde\nu$ defined on the ideal $\Cal I$ of
subsets of $X$ generated by $\Cal E$.

\proof{{\bf (a)} There is a non-negative finitely additive functional
$\theta:\Cal I\to\Bbb R$ extending $\nu$.   \Prf\ Let $V$ be the
linear subspace of $\ell^{\infty}(X)$ generated by $\{\chi E:E\in\Cal E\}$,
so that $V$ can be identified with the Riesz space $S(\Cal E)$ (361L).
Let $U$ be the solid linear subspace of $\ell^{\infty}(X)$ generated
by $V$.   For $u\in U$ set
$q(u)=\inf\{\dashint v\,d\nu:v\in V$, $|u|\le v\}$, where
$\dashint\,\,d\nu:S(\Cal E)\to\Bbb R$ is the positive linear functional
corresponding to $\nu:\Cal E\to\coint{0,\infty}$ as in 361F-361G.
Then $q$ is a seminorm, and $|\dashint v\,d\nu|\le q(v)$ for every
$v\in V$.   So there is a linear functional $f:U\to\Bbb R$ such that
$f(v)=\dashint v\,d\nu$ for every $v\in V$ and $|f(u)|\le q(u)$ for every
$u\in U$ (4A4D(a-i)).   Set $\theta A=f(\chi A)$ for $A\in\Cal I$.
Then $\theta:\Cal I\to\Bbb R$ is additive and extends $\nu$.   If
$A\in\Cal I$, there is an $E\in\Cal E$ including $A$.   Now we have

\Centerline{$\theta(E\setminus A)=f(\chi E-\chi A)
\le q(\chi E-\chi A)\le\dashint\chi E\,d\nu=\nu E=\theta E$,}

\noindent so $\theta A\ge 0$.   So $\theta$ is non-negative.\ \Qed

\medskip

{\bf (b)} As in the proof of 449M, we have a positive $G$-invariant
linear functional $p:\ell^{\infty}(G)\to\Bbb R$ such that $p(\chi G)=1$.
For $A\in\Cal I$, set $f_A(a)=\theta(a^{-1}\action A)$ for $a\in G$, and
$\tilde\nu A=p(f_A)$.   Then $\tilde\nu:\Cal I\to\coint{0,\infty}$ is
additive.   If $E\in\Cal E$ then $f_A(a)=\nu E$ for every $a$, so
$\tilde\nu$ extends $\nu$.   If $A\in\Cal I$ and $a$, $b\in G$, then
$f_{aA}(b)=\theta(b^{-1}aA)=f_A(a^{-1}b)$, so $f_{aA}=a\action_lf_A$
and

\Centerline{$\tilde\nu(aA)=p(f_{aA})=p(f_A)=\tilde\nu A$.}

\noindent Thus $\tilde\nu$ is $G$-invariant, as required.
}%end of proof of 449N

\leader{449O}{Corollary}\cmmnt{ ({\smc Banach 1923})} If $r=1$ or $r=2$,
there is a functional $\theta:\Cal P\BbbR^r\to[0,\infty]$ such that
(i) $\theta(A\cup B)=\theta A+\theta B$ whenever $A$,
$B\subseteq\BbbR^r$ are disjoint (ii) $\theta E$ is the Lebesgue
measure of $E$ whenever $E\subseteq\BbbR^r$ is Lebesgue measurable
(iii) $\theta(g[A])=\theta A$ whenever $A\subseteq\BbbR^r$ and
$g:\BbbR^r\to\BbbR^r$ is an isometry.

\proof{{\bf (a)} The point is that the group $G$ of all isometries of
$\BbbR^r$, with its discrete topology, is amenable.   \Prf\ Let
$G_0\subseteq G$ be the subgroup consisting of rotations about
$\tbf{0}$;  because $r\le 2$, this is abelian, therefore amenable
(449Cf).   Let $G_1\subseteq G$ be the subgroup consisting of
isometries
keeping $\tbf{0}$ fixed;  then $G_0$ is a normal subgroup of $G_1$,
and $G_1/G_0$ is abelian, so $G_1$ is amenable (449Cc).   Let
$G_2\subseteq G$ be the normal subgroup consisting of translations;
then $G_2$ is abelian, therefore amenable.   Now $G=G_1G_2$, so $G$ is
amenable (449Cd).\ \Qed

\medskip

{\bf (b)} Let $\Cal E$ be the ring of subsets of $\BbbR^r$ with finite
Lebesgue measure, and let $\nu$ be the restriction of Lebesgue measure
to $\Cal E$.   Then $\nu$ is $G$-invariant.   By 449N, there is a
$G$-invariant extension $\tilde\nu$ of $\nu$ to the ideal $\Cal I$
generated by
$\Cal E$.   Setting $\theta A=\tilde\nu A$ for $A\in\Cal I$,
$\infty$
for $A\in\Cal P\BbbR^r\setminus\Cal I$, we have a suitable functional
$\theta$.
}%end of proof of 449O

\exercises{\leader{449X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 449Xa
Let $G$ be a topological group.   On $G$ define a binary operation
$\diamond$ by saying that $x\diamond y=yx$ for all $x$,
$y\in G$.   Show that $(G,\diamond)$ is a topological group isomorphic
to $G$, so is amenable iff $G$ is.
%449A

\spheader 449Xb
Show that any finite topological group is amenable.
%449C

\sqheader 449Xc Show that, for any $r\in\Bbb N$, the isometry group of
$\BbbR^r$, with the topology of pointwise convergence, is amenable.
\Hint{443Xw, 449Cd.}
%441G isometry group, 449C

\spheader 449Xd Find a locally compact Polish
group which is amenable but not unimodular.
\Hint{442Xf, 449Cd.}
%449C

\spheader 449Xe\dvAnew{2013} Prove 449Cg directly from 441C, without
mentioning Haar measure.
%449C

\spheader 449Xf Let $G$ be a topological group and $U$ the space of
bounded real-valued functions on $G$ which are uniformly continuous
for the right uniformity of $G$.   Show that $\action_r$, as defined in
4A5Cc, gives an action of $G$ on $U$.
%449D

%find example in which it is not continuous for $\|\,\|_{\infty}$ on
%$U$?

\spheader 449Xg Let $\action$ be an action of a group $G$ on a set
$X$, and $U$ a Riesz subspace of $\ell^{\infty}(X)$, containing the
constant functions, such that $a\action f\in U$ whenever $f\in U$
and $a\in G$.
Show that the following are equiveridical:  (i) there is a
$G$-invariant
positive linear functional $p:U\to\Bbb R$ such that $p(\chi X)=1$;
(ii)
$\sup_{x\in X}\sum_{i=0}^nf_i(x)-f_i(a_i\action x)\ge 0$ whenever
$f_0,\ldots,f_n\in U$ and $a_0,\ldots,a_n\in G$.   \Hint{if (ii) is
true, let $V$ be the linear subspace generated by
$\{f-a\action f:f\in U$, $a\in G\}$ and show that
$\inf_{g\in V}\|g-\chi X\|_{\infty}=1$.}
%449E

\sqheader 449Xh Let $X$ be a set and $G$ the group of all
permutations of $X$.   (i) Give $X$ the zero-one metric,
so that $G$ is the
isometry group of $X$.   Show that $G$, with the topology of
pointwise convergence (441G), is amenable.   \Hint{for any
$I\in[X]^{<\omega}$,
$\{a:a\in G,\,a(x)=x$ for every $x\in X\setminus I\}$ is amenable.}
(ii) Show that if $X$ is infinite then $G$, with its discrete
topology, is not amenable.   \Hint{the left action of $F_2$ on itself
can be regarded as an embedding of $F_2$ in $G$.}
%449C 449Fa 449G

\spheader 449Xi Let $G$ be a Hausdorff topological group, and $\hat G$
its completion with respect to its bilateral uniformity (definition:
4A5Hb).   Show that $G$ is amenable iff $\hat G$ is.
%449Cb 449Fa

\spheader 449Xj(i) Let $G$ be the group with generators $a$, $b$ and
relations $a^2=b^3=e$ (that is, the quotient of the free group on two
generators $a$ and $b$ by the normal subgroup generated by
$\{a^2,b^3\}$).   Show that $G$, with its discrete topology, is not
amenable.
(ii) Let $G$ be the group with generators $a$, $b$ and
relations $a^2=b^2=e$.   Show that $G$, with its discrete topology, is
amenable.   \Hint{we have a function $\length:G\to\Bbb N$ such that
$\length(ab)\le\length(a)+\length(b)$ for all $a$, $b\in G$ and
$\limsup_{n\to\infty}\Bover1n\#(\{a:\length(a)\le n\})$ is finite.
See also 449Yf.}
%449Xj 449G

\spheader 449Xk Let $G$ be a locally compact Hausdorff group, and $\mu$ a
left Haar measure on $G$.   Show that $G$ is amenable iff
for every finite set $I\subseteq G$, finite set
$J\subseteq\eusm L^{\infty}(\mu)$ and $\epsilon>0$, there is an
$h\in C_{k1}(G)^+$ (definition:  449J) such that
$|\int f(ax)h(x)\mu(dx)-\int f(x)h(x)\mu(dx)|\le\epsilon$
for every $a\in I$ and $f\in J$.   \Hint{the image of the unit ball in
$L^1$ is weak* dense in the unit ball of $(L^{\infty})^*$.}
%449J

\spheader 449Xl Let $G$ be a locally compact Hausdorff group,
and $\mu$ a left Haar measure on $G$.   Show that the following are
equiveridical:
(i) $G$ is amenable;
(ii) there is a positive linear functional
$p^{\#}:L^{\infty}(\mu)\to\Bbb R$ such that
$p^{\#}(\chi G^{\ssbullet})=1$ and
$p^{\#}(a\action_ru)=p^{\#}(u)$ for every $u\in L^{\infty}(\mu)$ and
every $a\in G$;
(iii) for every finite set $I\subseteq G$, finite set
$J\subseteq\eusm L^{\infty}(\mu)$ and $\epsilon>0$, there is an
$h\in C_{k1}^+$
such that $|\int f(xa)h(x)\mu(dx)-\int f(x)h(x)\mu(dx)|\le\epsilon$
for every $a\in I$ and $f\in J$.
%449J

\spheader 449Xm\dvAnew{2010} Let $G$ be a locally compact Hausdorff
group and
$\CalBa_G$ its Baire $\sigma$-algebra.    Show that
$G$ is amenable iff there is a non-zero finitely additive
$\phi:\CalBa_G\to[0,1]$
such that $\phi(aE)=\nu E$ for every $a\in G$ and $E\in\CalBa_G$.
%449J

\spheader 449Xn A {\bf symmetric F{\o}lner sequence} in a group $G$ is a
sequence $\sequencen{L_n}$ of non-empty finite
symmetric subsets of $G$ such that
$\lim_{n\to\infty}\Bover{\#(L_n\symmdiff aL_n)}{\#(L_n)}=0$ for every
$a\in G$.   Show that a group $G$ has a symmetric F{\o}lner sequence iff
it is countable and amenable when given its discrete topology.
%449J

\sqheader 449Xo Let $G$ be a group which is amenable when given its
discrete topology.   Let $\phi:\Cal PG\to[0,1]$ be an additive
functional such that $\phi G=1$ and $\phi(aE)=\phi E$ whenever
$E\subseteq G$ and $a\in G$.   For $E\subseteq G$ set
$\psi E=\dashint\phi(Ex)\phi(dx)$.   Show that
$\psi:\Cal PG\to[0,1]$ is an additive functional, that
$\psi G=1$ and that $\psi(aE)=\psi(Ea)=\psi E$ for
every $E\subseteq G$ and $a\in G$.
%449M

\spheader 449Xp Let $X$ be a non-empty set, $G$ a group and $\action$
an action of $G$ on $X$.   Suppose that $G$ is an amenable group when
given its discrete topology.   Show that there is an additive functional
$\nu:\Cal PX\to[0,1]$ such that $\nu X=1$ and
$\nu(a\action A)=\nu A$ for every $A\subseteq X$ and every $a\in G$.
%449M

\spheader 449Xq
Let $G$ be a locally compact Hausdorff group and
$\mu$ a left Haar measure on $G$.   Show that if $G$, with its discrete
topology, is amenable, then there is a functional
$\phi:\Cal PG\to[0,\infty]$, extending $\mu$, such that
$\phi(A\cup B)=\phi A+\phi B$ whenever $A$, $B\subseteq G$
are disjoint and $\phi(xA)=\phi A$ for every $x\in G$ and
$A\subseteq G$.
%449N

\spheader 449Xr Let $X$ be a compact metrizable space, $\phi:X\to X$ a
continuous function and $\mu$ a Radon probability measure on $X$ such
that $\mu\phi^{-1}=\mu$.   Show that for $\mu$-almost every $x\in X$,
$\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(\phi^i(x))$ is defined for
every $f\in C(X)$.   \Hint{4A2Pe, 372J.}
%+  372J Ergodic Theorem %4A2Pe  $C(X)$ is separable.

\leader{449Y}{Further exercises (a)}
%\spheader 449Ya
If $S$ is a semigroup with identity $e$ and $X$ is a set, an
{\bf action} of $S$ on $X$ is a map
$(s,x)\mapsto s\action x:S\times X\to X$
such that $s\action(t\action x)=(st)\action x$ and $e\action x=x$ for
every $s$, $t\in S$ and $x\in X$.   A topological semigroup $S$ with
identity is {\bf amenable} if for every non-empty compact Hausdorff
space $X$ and every continuous action of $S$ on $X$ there is a Radon
probability measure $\mu$ on $X$ such that
$\int f(s\action x)\mu(dx)=\int f(x)\mu(dx)$ for every $s\in S$ and
$f\in C(X)$.   Show that

\quad(i) $(\Bbb N,+)$, with its discrete topology, is amenable;

\quad(ii) if $S$ is a topological semigroup and $\Cal S$ is an
upwards-directed family of amenable sub-semigroups of $S$ with dense
union in $S$, then $S$ is amenable;

\quad(iii) if $\familyiI{S_i}$ is a family of amenable topological
semigroups with product $S$ then $S$ is amenable;

\quad(iv) if $S$ is an amenable topological semigroup, $S'$ is a
topological
semigroup, and there is a continuous multiplicative surjection from
$S$ onto $S'$, then $S'$ is amenable;

\quad(v) if $S$ is an abelian topological semigroup, then it is
amenable.
%449C

\spheader 449Yb Give an example of a topological semigroup  $S$ with
identity such that $S$ is amenable
in the sense of 449Ya but $(S,\diamond)$ is not, where $a\diamond b=ba$
for $a$, $b\in S$.
%449Ya 449C 44bits

\spheader 449Yc Let $G$ be a topological group and $U$ the space of
bounded real-valued functions on $G$ which are uniformly
continuous for the right uniformity.   Let $M_{\text{qR}}^+$ be the
space of totally finite quasi-Radon measures on $G$.   (i) Show that if
$\nu\in M_{\text{qR}}^+$ then $\nu*f$ (definition:
444H) belongs to $U$ for every $f\in U$.
(ii) Show that $(\nu,f)\mapsto\nu*f:M_{\text{qR}}^+\times U\to U$ is
continuous if $M_{\text{qR}}^+$ is given its narrow topology and $U$ is
given its norm topology.   (iii) Show that if
$p:U\to\Bbb R$ is a continuous linear functional such that
$p(a\action_lf)=p(f)$ for every $f\in U$ and $a\in G$, then
$p(\nu*f)=\nu G\cdot p(f)$ for every $f\in U$ and every totally finite
quasi-Radon measure $\nu$ on $G$.
%449H

\spheader 449Yd Re-work 449J for general groups carrying Haar
measures.
%449J

\spheader 449Ye Let $G$ be a group with a symmetric F{\o}lner sequence
$\sequencen{L_n}$ (449Xn), and $\action$ an action of $G$ on a reflexive
Banach space $U$ such that $u\mapsto a\action u$ is a linear operator of
norm at most $1$ for every $a\in G$.   For $n\in\Bbb N$ set
$T_nu=\Bover1{\#(L_n)}\sum_{a\in L_n}a\action u$ for $u\in U$.
Show that
for every $u\in U$ the sequence $\sequencen{T_nu}$ is norm-convergent
to a $v\in U$ such that $a\action v=v$ for every $a\in G$.   \Hint{372A.
See also 461Yg below.}
%449Xn 372A 449J

\spheader 449Yf
Let $G$ be a locally compact Hausdorff group and
$\mu$ a left Haar measure on $G$.   Suppose that $G$ is {\bf exponentially
bounded}, that is,
$\limsup_{n\to\infty}(\mu(K^n))^{1/n}\le 1$ for every compact set
$K\subseteq G$.    Show that $G$ is amenable.
%449J(xiv)

\spheader 449Yg\dvAnew{2010} Let
$G$ be a group and $\action$ an action of $G$ on a set
$X$.   Let $\Tau$ be an algebra of subsets of $X$ such that
$g\action E\in\Tau$ for every $E\in\Tau$ and $g\in G$, and $H$ a member
of $\Tau$;   write $\Tau_H$ for $\{E:E\in\Tau$, $E\subseteq H\}$.
Let $\nu:\Tau_H\to[0,\infty]$ be
a functional which is additive in the sense that $\nu\emptyset=0$ and
$\nu(E\cup F)=\nu E+\nu F$ whenever $E$, $F\in\Tau_H$ are disjoint,
and locally $G$-invariant in the sense that $g\action E\in\Tau$ and
$\nu(g\action E)=\nu E$ whenever $E\in\Tau_H$, $g\in G$ and
$g\action E\subseteq H$.   Show that
there is an extension of $\nu$ to a $G$-invariant additive functional
$\tilde\nu:\Tau\to[0,\infty]$.
%for 449Yh

\spheader 449Yh
Let $X$ be a set, $A$ a subset of $X$, and $\action$ an
action of a group $G$ on $X$.
Show that the following are equiveridical:  (i)
there is a functional $\theta:\Cal PX\to[0,\infty]$ such that
$\theta A=1$, $\theta(B\cup C)=\theta B+\theta C$ and
$\theta(a\action B)=\theta B$ for all disjoint
$B$, $C\subseteq X$ and $a\in G$;   (ii) there are no
$A_0,\ldots,A_n$, $a_0,\ldots,a_n$, $b_0,\ldots,b_n$ such that
$A_0,\ldots,A_n$ are subsets of $G$ covering $A$,
$a_0,\ldots,b_n$ belong to $G$, and $a_0\action A_0$,
$b_0\action A_0$,
$a_1\action A_1$, $b_1\action A_1,\ldots,b_n\action A_n$ are
disjoint subsets of $A$.
%449Yg 449L mt44bits

\spheader 449Yi ({\smc \'Swierczkowski 58}) Let $G$ be the group of
orthogonal $3\times 3$ matrices.   Set
$S=\Matrix{\bover35&\bover45&0\\-\bover45&\bover35&0\\0&0&1}$ and
$T=\Matrix{1&0&0\\0&\bover35&\bover45\\0&-\bover45&\bover35}$.   Show
that $S$ and $T$ are free in $G$ (that is, no non-trivial product
of the form $S^{n_0}T^{n_1}S^{n_2}T^{n_3}\ldots
S^{n_{2k}}$ can be the identity), so that
$G$ is not amenable in its discrete topology.   \Hint{let $R$ be the
ring of $3\times 3$ matrices over the field $\Bbb Z_5$.   In $R$ set
$\sigma=\Matrix{3&4&0\\1&3&0\\0&0&0}$,
$\tau=\Matrix{0&0&0\\0&3&4\\0&1&3}$.   Show that
$\sigma^2=\sigma$.   Now suppose $\rho\in R$ is defined as a
non-trivial
product of the elements $\sigma$, $\tau$ and their transposes $\sigma\trs$,
$\tau\trs$ in which $\sigma$ and $\sigma\trs$ are never adjacent,
$\tau$ and $\tau\trs$ are never adjacent, and the last term is
$\sigma$.   Set $\Matrix{a\\b\\c}=\rho\Matrix{1\\0\\0}$.   Show that if
the first term in the product is $\sigma$ or $\sigma\trs$, then $c=0$
and $b\ne 0$, and otherwise $a=0$ and $b\ne 0$.}
%449O

\spheader 449Yj
Let $F_2$ be the free group on two generators $a$,
$b$.   (i) Show that there is a partition $(A,B,C,D)$ of
$F_2$ such that $aA=A\cup B\cup C$ and $bB=A\cup B\cup D$.
(ii) Let $S_2$ be the unit
sphere in $\BbbR^3$.   Show that if $S$, $T$ are the matrices of
449Yi, there is a partition $(A,B,C,D,E)$ of $S^2$ such that
$E$ is countable, $S[A]=A\cup B\cup C$ and $T[B]=A\cup B\cup D$.
(This is a version of the {\bf Hausdorff paradox}.)
(iii) Show that there is no non-zero rotation-invariant additive
functional from $\Cal PS_2$ to $[0,1]$.   (iv) Show that there is no
rotation-invariant additive
extension of Lebesgue measure to all subsets of the
unit ball $B(\tbf{0},1)$.   (See {\smc Wagon 85}.)
%449Yi 449O Wagon
}%end of exercises

\endnotes{
\Notesheader{449}
The general theory of amenable groups is outside the scope of this book.
Here I have tried only to indicate some of the specifically
measure-theoretic arguments which are used in the theory.   Primarily we
have the Riesz representation theorem, enabling us to move between
linear functionals and measures.   Since the invariant measures
considered in the definition of `amenable group' are all Radon measures
on compact Hausdorff spaces, they can equally well be thought of as
linear functionals on spaces of continuous functions.   What is striking
is that the definition in terms of continuous actions on arbitrary
compact Hausdorff spaces can be reduced to a question concerning an
invariant mean on a single space easily constructed from the group
(449E).

The first part of this section deals with general topological groups.
It is a remarkable fact that some of the most important
non-locally-compact topological groups are amenable.   For most of
these we shall have to wait until we have done `concentration of measure'
(\S\S476, 492) and can approach `extremely amenable' groups (\S493).
But there is an easy example in 449Xh which already indicates one of
the basic methods.

For a much fuller account of the theory of amenable locally compact groups,
see {\smc Paterson 88}.
Theorem 449J here is mostly taken from {\smc Greenleaf 69}, where you
will find many references to its development.   Historically the subject
was dominated by the case of discrete groups, in which combinatorial
rather than measure-theoretic formulations seem more appropriate.   In
449J, conditions (ii)-(viii) relate to invariant means of one kind or
another, strengthening that of 449E.
Note that the means of
449E(iii) and 449J(ii)-(viii) are normalized by conditions
$p(\chi G)=\tilde p(\chi G^{\ssbullet})=\phi G=1$,
while the left Haar
measure $\mu$ of 449J has a degree of freedom;  so that when they come
together in 449J(viii) the two sides of the equation
$\tilde p(g*f)^{\ssbullet}=\tilde p(f^{\ssbullet})\int g\,d\mu$ must
move together if we change $\mu$ by a scalar factor.   Of course this
happens through the hidden dependence of the convolution operation on
$\mu$.   (The convolutions in 449J(vii) do {\it not} involve $\mu$.)
Between 449J(i) and 449J(viii) there is a double step.   First
we note that a convolution $g*f$ is a kind of weighted average of left
translates of $f$, so that if we have a mean which is invariant under
translations we can hope that it will be invariant under convolutions
(449H, 449Yc).   What is more remarkable is that an invariant mean on
the space $U$ of bounded uniformly continuous functions should in the first
place give rise to a left-invariant mean on the space
$L^{\infty}$ of (equivalence classes of) bounded Haar measurable
functions (449J(vi)), and then even a two-sided-invariant mean (449J(v)).

Condition (xii) in 449J looks at a different aspect of the
phenomenon.   In effect, it amounts to saying that not only is there
an invariant mean, but that there is an invariant mean defined by the
formula $p(f)=\lim_{L\to\Cal F}\Bover1{\mu L}\int_Lf$ for a suitable
filter $\Cal F$ on the family of sets of non-zero finite measure.
This may be called a `F{\o}lner condition', following
{\smc F{\o}lner 55}.   (449Xk looks for an invariant mean of
the form $p(f)=\lim_{h\to\Cal F}\int f\times h\,d\mu$, where $\Cal F$
is a suitable filter on $\eusm L^1(\mu)$.)   In 449J(xi), the case $q=1$
is just a weaker version of condition (x), but the case $q=2$ tells us
something new.

The techniques developed in \S444 to handle Haar measures on groups
which are not locally compact can also be used in 449H-449J, using
`totally bounded for the bilateral uniformity' in place of `compact'
when appropriate (449Yd).   443L provides another route to the same
generalization.

In 449K, it is natural to ask whether the hypothesis `locally compact' is
necessary.   It certainly cannot be dropped completely;  there is an
important
amenable Polish group with a closed subgroup which is not amenable
(493Xf).

The words `right' and `left' appear repeatedly in this section, and it
is not perhaps immediately clear which of the ordinary symmetries we can
expect to find.   The fact that the operation $x\mapsto x^{-1}:G\to G$
always gives us an isomorphism between a group and the same set with the
multiplication reversed (449Xa) means that we do not have to distinguish
between `left amenable' and `right amenable' groups, at least if we
start from the definition in 449A.   In 449C also there is nothing to
break the symmetry between left and right.   In 449B and 449D-449E,
however, we must commit ourselves to the {\it left} action of the group
on the space of functions which are uniformly continuous with respect to
the {\it right} uniformity.   If we wish to change one, we must also
expect to change the other.   In the list of conditions
in 449J, some can be reflected straightforwardly (see 449Xl),
but in groups which are not unimodular there seem to be difficulties.
For semigroups we do have a difference
between `left' and `right' amenability (449Yb).

It is not surprising that in the search for invariant means we should
repeatedly use averaging and limiting processes.   The
`finitely additive integrals' $\dashint\,d\phi$, $\dashint\,d\nu$
in part (f) the proof of 449J and part (a) of the proof of 449N
are an effective way of using one
invariant additive functional $\phi$ or $\nu$ to build another.
Similarly, because we are looking only for finite additivity,
we can be optimistic about taking cluster
points of families of almost-invariant functionals, as in the proofs of
449F, 449J and 449L.

In the case of discrete groups, in which all considerations of
measurability and continuity evaporate, we have a completely different
technique available, as in 449L.   Here we can go directly from a
non-paradoxicality condition, a weaker version of conditions already
introduced in 395E and 448E,
to a F{\o}lner condition (($\beta$) in part
(b) of the proof of 449L) which easily implies amenability.   I remind
you that I still do not know how far these ideas can be applied to
other algebras than $\Cal PX$ (395Z).   The difficulty is that the
unscrupulous use of the axiom of choice in the infinitary Marriage
Lemma seems to give us no control over the nature of the sets
$A_{ij}$ described in (b-$\alpha$) of the proof of 449L;  moreover, the
structure
of the proof depends on having a suitable invariant measure (counting
measure on $X$) to begin with.   For more on amenable discrete groups
and their connexions with measure theory see {\smc Laczkovich 02}.
}%end of notes

\discrpage

\ifresultsonly\else
\frfilename{mt44conc.tex}
\versiondate{31.7.09}
\copyrightdate{2002}
     
\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
     
\def\chaptername{Topological groups}
\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{}41A}{Shift actions} The definition of shift actions in 4{}41Ac
of the 2003 and 2006 editions, called on in the 2008 edition of Volume 5,
has been moved to 4A5Cc.
     
\leader{4{}44Xn}{Orthonormal bases} The sketch of a construction of an
orthonormal basis in $L^2$ consisting of equivalence classes of continuous
functions,
referred to in {\smc Bogachev 07}, is now 
444Ym.

\leader{\S4{}45}{Convolutions} The material mentioned in
the notes to \S257 in the May 2001 edition of Volume 2 has been moved to
\S444.   The particular result referred to as `4{}45K' is now 444R.

\leader{4{}45Xq}{} The exercise 4{}45Xq, referred to in the
May 2002 edition of Volume 3, %in 3{}87H 
has been moved to 445Xp.
     
\leader{4{}49I, 4{}49J}{Tarski's theorem} The proof of Tarski's theorem,
referred to in the 2002 and 2004 editions of Volume 3, is now in 449L.     

\bigskip

\noindent{\bf References}

\medskip

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

\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{[491Yt.]}

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.}
Springer, 1969 (reprinted 1996).
\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}
F{\o}lner E.\ [55] `On groups with full Banach mean value',
Math.\ Scand.\ 3 (1955) 243-254.
\cmmnt{[\S449 {\it notes\/}.]}

\indexheader{Ford}
Ford L.R.\ \& Fulkerson D.R.\ [56] `Maximal flow through a network',
Canadian J.\ Math.\ 8 (1956) 399-404.
\cmmnt{[4A4N.]}

Frankl P.\ \& R\"odl V.\ [02] `Extremal problems on set systems',
Random Structures Algorithms 20 (2002) 131-164.
\cmmnt{[497L.]}

\indexheader{Fremlin}
Fremlin D.H.\ [74] {\it Topological Riesz Spaces and
Measure Theory.}   Cambridge U.P., 1974.
\cmmnt{[\S462 {\it notes\/}.]}

Fremlin D.H.\ [75a] `Pointwise compact sets of measurable functions',
Manuscripta Math.\ 15 (1975) 219-242.
\cmmnt{[463K.]}

Fremlin D.H.\ [75b] `Topological measure spaces:  two counter-examples',
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
functions', Manuscripta Math.\ 33 (1981) 387-405.
\cmmnt{[451T.]}

Fremlin D.H.\ [84] {\it Consequences of Martin's Axiom.}
Cambridge U.P., 1984.
\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
Symposium), Heldermann, 1988.

Furstenberg H.\ [77]
`Ergodic behavior of diagonal measures and a theorem of Szemer\'edi
on arithmetic progressions', J. d'Analyse Math.\ 31 (1977), 204-256.
\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.]}

Glasner S.\ [98] `On minimal actions of Polish groups', Topology and its Applications 85 (1998) 119-125.
\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\/}.]}

Greenleaf F.P.\ [69] {\it Invariant Means on Topological Groups and Their
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\cmmnt{[\S449 {\it notes.}]}

Grothendieck A. [92] {\it Topological Vector Spaces.}  Gordon \& Breach,
1992.
\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.\
Sci.\ (Math.) 29 (1981) 1-5.
\cmmnt{[439F, \S439 {\it notes\/}.]}

\medskip%H

\indexheader{Halmos}
Halmos P.R.\ [1944] `In general a measure-preserving transformation is
mixing', Annals of Math.\ 45 (1944) 786-792.
\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.]}

Hart J.E.\ \& Kunen K.\ [99] `Orthogonal continuous functions',
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.]}

Hewitt E.\ \& Ross K.A.\ [63] {\it Abstract Harmonic Analysis I.}
Springer, 1963 and 1979.
\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}
Ionescu Tulcea A.\ [73] `On pointwise convergence, compactness and
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\cmmnt{[463C.]}

Ionescu Tulcea A.\ [74] `On pointwise convergence, compactness, and
equicontinuity II', Advances in Math.\ 12 (1974) 171-177.
\cmmnt{[463G.]}

Ionescu Tulcea A.\ \& Ionescu Tulcea C.\ [67] `On the existence of a
lifting commuting with the left translations of an arbitrary locally
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\cmmnt{[\S447 {\it intro.}, 447I, \S447 {\it notes\/}.]}

Ionescu Tulcea A.\ \& Ionescu Tulcea C.\ [69] {\it Topics in the Theory
of Lifting.}  Springer, 1969.
\cmmnt{[\S453 {\it intro.}]}

\indexheader{Ismail}
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(1980) 281-292.
\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.

Jameson G.J.O.\ [74] {\it Topology and Normed Spaces.}  Chapman \& Hall,
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\cmmnt{[\S4A4.]}

\indexheader{Jayne}
Jayne J.E.\ [76] `Structure of analytic Hausdorff spaces', Mathematika
23 (1976) 208-211.
\cmmnt{[422Yc.]}

Jayne J.E.\ \& Rogers C.A.\ [95] `Radon measures on Banach spaces with
their weak topologies', Serdica Math.\ J.\ 21 (1995) 283-334.
\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.]}

Johnson R.A.\ [82] `Products of two Borel measures', Trans.\ Amer.\
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\cmmnt{[434R.]}

Judah H.\ [93] (ed.) {\it Proceedings of the Bar-Ilan Conference on Set
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Juh\'asz I., Kunen K.\ \& Rudin M.E.\ [76] `Two more hereditarily
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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}
Kaplansky I.\ [71] {\it Lie Algebras and Locally Compact Groups.}
Univ.\ of Chicago Press, 1971.
\cmmnt{[\S446 {\it notes.}]}

Kechris A.S.\ [95] {\it Classical Descriptive Set Theory.}  Springer,
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\cmmnt{[Chap.\ 42 {\it intro.}, 423T, \S423 {\it notes\/},
\S448 {\it notes\/}, \S4A2, 4A3S, 4A3T.]}

%Kechris A.S.\ [p09] {\it Global Aspects of Ergodic Group Actions.}
%Amer.\ Math.\ Soc.\

Kellerer H.G.\ [84] `Duality theorems for marginal problems', Zeitschrift
f\"ur Wahrscheinlichkeitstheorie verw.\ Gebiete 67 (1984) 399-432.
\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',
Math.\ Annalen 190 (1971) 233-241.
\cmmnt{[413Xr.]}

Kittrell J.\ \& Tsankov T.\ [09] `Topological properties of full groups',
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.]}

Koumoullis G.\ \& Prikry K.\ [83] `The Ramsey property and measurable
selectors', J.\ London Math.\ Soc.\ (2) 28 (1983) 203-210.
\cmmnt{[451T.]}

Kuipers L.\ \& Niederreiter H.\ [74] {\it Uniform Distribution of
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

Laczkovich M.\ [02] `Paradoxes in measure theory', pp.\ 83-123 in
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\cmmnt{[\S449 {\it notes\/}.]}

LeCam L.M.\ \& Neyman J.\ [67] (eds.) {\it Proc.\ Fifth Berkeley
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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}}
Luke\v{s} J., Mal\'y J.\ \& Zaj\'\i\v{c}ek L.\ [86]
{\it Fine topology methods
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\cmmnt{[414Ye.]}

\medskip%M

\indexheader{Mackey}
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\cmmnt{[448S.]}

Marczewski E.\ [53] `On compact measures', Fundamenta Math.\ 40 (1953)
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\cmmnt{[413T, Chap.\ 45 {\it intro.}, \S451 {\it intro.}, 451J.]}

Marczewski E.\ \& Ryll-Nardzewski C.\ [53] `Remarks on the compactness
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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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\cmmnt{[Chap.\ 47 {\it intro.}, \S471 {\it intro.}, \S472 {\it notes\/},
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%QA311.M2 9520 6456 1

Mauldin R.D. \& Stone A.H.\ [81] `Realization of maps', Michigan Math.\
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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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Monthly 80 (1973) 349-359.
\cmmnt{[481M.]}

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\cmmnt{[\S495 {\it notes\/}.]}

Milman V.D.\ \& Schechtman G.\ [86]
{\it Asymptotic Theory of Finite Dimensional Normed Spaces.}
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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
Notes in Mathematics 1951).
\cmmnt{[\S467 {\it notes.}]}

Montgomery D.\ \& Zippin L.\ [55] {\it Topological Transformation
Groups.}   Interscience, 1955.
\cmmnt{[\S446 {\it notes.}]}

\indexheader{Moran}
Moran W.\ [68] `The additivity of measures on completely regular
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\cmmnt{[439P.]}

Moran W.\ [69] `Measures and mappings on topological spaces', Proc.\
London Math.\ Soc.\ (3) 19 (1969) 493-508.
\cmmnt{[435Xk.]}

Moran W.\ [70] `Measures on metacompact spaces', Proc.\ London Math.\
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\cmmnt{[438J.]}

\indexheader{Moschovakis}
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\cmmnt{[Chap.\ 42 {\it intro.}, \S423 {\it notes\/}.]}

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\cmmnt{[481P, \S482 {\it notes\/}.]}

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%ISBN 5-900975-01-0

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\cmmnt{[451U.]}

\medskip%N

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%Naimark M.A. [70] {\it Normed Rings.}  Wolters-Noordhoff, 1970.


Neyman J.\ [56] (ed.) {\it Proceedings of the Third Berkeley Symposium
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\medskip%O
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\cmmnt{[434Yq.]}

\medskip%P

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\cmmnt{[452I, 452S, 452Ye, \S452 {\it notes\/}.]}

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\cmmnt{[452R.]}

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\cmmnt{[449K, \S449 {\it notes\/}.]}
% QA 403.P2 store A

Perron O.\ [1914] `\"Uber den Integralbegriff', Sitzungsberichte der
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\cmmnt{[\S493 {\it notes\/}.]}

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\cmmnt{[484Xb.]}

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\cmmnt{[484B.]}

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\cmmnt{[\S461 {\it notes\/}.]}

\indexheader{Plewik}
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\indexheader{Pontryagin}
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\cmmnt{[445U.]}

\indexheader{Preiss}
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Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 27 (1973) 109-116.
\cmmnt{[439S.]}
%ILL 31.7.15

Preiss D.\ \& Ti\v{s}er J.\ [91]
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\cmmnt{[\S466 {\it notes\/}.]}

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\cmmnt{[462B, 462C.]}

\medskip%Q\medskip%R

Radon J.\ [1913] `Theorie und Anwendungen der absolut additivien
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\cmmnt{[\S416 {\it notes\/}.]}

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\cmmnt{[\S451 {\it notes\/}.]}

\indexheader{Rao B.V.}
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\cmmnt{[419F.]}

Rao B.V. \& Srivastava S.M.\ [94] `Borel isomorphism theorem',
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\cmmnt{[424C.]}

\indexheader{Rao M.}
Rao M.\ [77] {\it Brownian motion and classical potential theory.}
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{\tt http://www1.essex.ac.uk/maths/people{\bsp}fremlin{\bsp}rao.htm}.
%used for \S477

\indexheader{Ressel}
Ressel P.\ [77] `Some continuity and measurability results on spaces of
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\cmmnt{[417C, 437M.]}

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\cmmnt{[\S494 {\it notes\/}.]}

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\cmmnt{[\S471 {\it notes\/}.]}

Rogers C.A.\ [80] (ed.) {\it Analytic Sets.}   Academic, 1980.
\cmmnt{[Chap.\ 42 {\it intro.}, \S422 {\it notes\/}, \S423 {\it notes\/}.]}

Rogers L.C.J.\ \& Williams D.\ [94] {\it Diffusions, Markov Processes
and Martingales,} vol.\ I.   Wiley, 1994.
\cmmnt{[\S455 {\it notes\/}.]}
%HA33.4.W5

Rokhlin V.A.\ [1948] `A ``general'' measure-preserving transformation is not
mixing', Doklady Akademii Nauk SSSR 60 (1948) 349-351.
\cmmnt{[494E.]}

Rosendal C.\ [09]
`The generic isometry and measure preserving homeomorphism are conjugate
to their powers',  Fundamenta Math.\ 205 (2009) 1-27.
%({\tt http://www.math.uic.edu{\bsp}$\sptilde$rosendal/}).
\cmmnt{[494Ye.]}

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fixed points on metric compacta', Israel J.\ Math.\ 162 (2007) 349-371.
\cmmnt{[494Z, \S494 {\it notes}.]}

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normal', Fund.\ Math.\ 73 (1971) 179-186.
\cmmnt{[439O.]}

Rudin M.E.\ [84] `Dowker spaces', pp.\ 761-780 in
{\smc Kunen \& Vaughan 84}.
\cmmnt{[439O.]}

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\cmmnt{[\S445 {\it intro\/}.]}

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\medskip%S

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\indexheader{Sierpi\'nski}
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\indexheader{Szemer\'edi}
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\indexheader{Talagrand} %? before \medskip to act correctly

\medskip%T

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%465Xm

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%used in chap 41

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\medskip%U

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\medskip%V

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% `Ta{sh}ken{t}skogo instituta' taken from ref in Fund Math

\medskip%W

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\cmmnt{[\S434 {\it notes\/}.]}

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\indexheader{Wheeler}
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  % 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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