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\documentclass{rae}
\usepackage{amsmath,amsthm,amssymb}
%?% missing
%\coverauthor{A. B. Kharazishvili}
%\covertitle{Some Remarks on Quasiinvariant and Invariant Measures}
\received{September 16, 1997}
\MathReviews{Primary: 28A05, 28D05}
\keywords{Invariant measure, quasiinvariant measure,
nonmeasurable set, Haar\break measure}
\firstpagenumber{1}
%?% missing
\markboth{A. B. Kharazishvili}
{Some Remarks on Quasiinvariant and Invariant Measures}
%?% missing
\author{A. B. Kharazishvili, Institute of Applied
Mathematics, Tbilisi State University, University str.~2,
380043~~Tbilisi~~43, Republic of Georgia}
\title{SOME REMARKS ON QUASIINVARIANT AND INVARIANT MEASURES}
%%%%%%Put Author's Definitions Below Here %%%%%%%%%%%
\def\dom{\mbox{\rm dom\,}}
\def\card{\mbox{\rm card\,}}
\def\cl{\mbox{\rm cl\,}}
\def\pf{\noindent{\sc Proof.} }
\newcommand{\consecutive}
{\newtheorem{theorem}{Theorem}
\newtheorem{example}{Example}
\newtheorem{lemma}{Lemma}
\newtheorem{problem}{Problem}
}
\newcounter{ite}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\consecutive
\maketitle
\begin{abstract}
We discuss two natural questions concerning some
properties of\break quasiinvariant and invariant measures:
the existence of nonmeasurable sets with respect to such measures
and the existence of a nonzero $\sigma$-finite quasiinvariant measure
on a proper Borel subgroup of the real line.
\end{abstract}
\bigskip
Let $E$ be a nonempty set, let $G$ be a group of transformations
of $E$ and let
$\mu$ be a $\sigma$-finite measure defined on some $\sigma$-algebra
of subsets of $E$. Denote by $I(\mu)$ the family of all $\mu$-measure
zero subsets of $E$. We recall that
\begin{enumerate}
\item $\mu$ is $G$-quasiinvariant if $\dom(\mu)$ and $I(\mu)$ are
$G$-invariant classes of sets;
\item $\mu$ is $G$-invariant if $\dom(\mu)$ is a $G$-invariant class of
sets and, for each $X \in \dom(\mu)$ and for each $g \in G$, we have
$\mu(g(X)) = \mu(X)$.
\end{enumerate}
The most well-known example of an invariant measure is the left (right)
invariant Haar measure on a locally compact topological group
(see, for instance,~\cite2). An important special case of a Haar
measure is the standard Borel measure on
a finite-dimensional Euclidean space (in particular, on the real line).
A Haar measure $\mu$ on a $\sigma$-compact locally compact topological
group $(G,\cdot)$ has the so-called Steinhaus property: if $X \in
\dom(\mu)$
and $\mu(X) > 0$, then the set $X \cdot X^{-1}$ is a neighbourhood of
the
neutral element of~$G$. Note that the category analogue
of the Steinhaus property holds true for any topological group. Namely,
if
$(\Gamma,\cdot)$ is a topological group and $Y$ is a second category
subset
of $\Gamma$ possessing the Baire property, then the set $Y \cdot Y^{-1}$
is a neighbourhood of the neutral element of $\Gamma$.
Evidently, in order to obtain a quasiinvariant
measure, it is sufficient to take an arbitrary invariant measure $\mu$
and
to consider any measure equivalent to~$\mu$. However, there are
some quasiinvariant probability measures which cannot
be obtained in such a way.
This fact implies, in particular, that some natural questions
(e.g.\ concerning the existence of nonmeasurable sets or
the existence of a measure with some additional properties)
can relatively easily be solved for invariant measures, but turn out
to be rather difficult for quasiinvariant measures.
The main goal of this paper is to illustrate an essential difference
between invariant and quasiinvariant measures from this point of view.
In our further considerations, the symbol $\mathbb N$ denotes the
set of all natural numbers, $\mathbb Z$ -- the set of all integers,
$\mathbb Q$ -- the set of all rational numbers
and the symbol $\mathbb R$ denotes the real line.
We begin with one problem concerning the existence of nonmeasurable
sets with respect to nonzero $\sigma$-finite
quasiinvariant measures. First, let us recall that a group $G$
acts $I(\mu)$-freely in $E$ if, for any two distinct transformations
$g$ and $h$ from $G$, the set $\{x \in E~:~g(x) = h(x)\}$ is of
$\mu^*$-measure zero (where $\mu^*$ denotes, as usual, the outer
measure canonically associated with~$\mu$).
The following result was obtained in~\cite{1}, \cite{3} and~\cite{9} and can be
regarded as an abstract version of the classical Vitali theorem
on the existence of Lebesgue nonmeasurable subsets of the real line
(see~\cite{12}).
\begin{theorem} Let $(E,G)$ be a space with a transformation
group, let $\mu$ be a\break $\sigma$-finite $G$-quasiinvariant measure
on $E$ and let $X$ be a subset of $E$ with $\mu^{*}(X) > 0$.
Suppose also that $G$ is uncountable and acts $I(\mu)$-freely in~$E$.
Then there
exists a subset of $X$ nonmeasurable with respect to~$\mu$.
In particular, if $\mu$ is not identically zero, then $\dom(\mu)$
differs from the family of all subsets of $E$ (in other words,
$\mu$ is not an universal measure on~$E$).
\end{theorem}
For the proof of Theorem~1, see the works mentioned above.
Here we only wish to notice that the proof of this theorem
is based on the classical result of Ulam~\cite{11}, stating that
the first uncountable cardinal $\omega_1$ is not real-valued measurable,
and on certain properties of selectors associated with a subgroup
of $G$ of cardinality~$\omega_1$.
In connection with Theorem~1, let us also remark that it can be
essentially strengthened for $\sigma$-finite
invariant measures. Namely, the following statement was recently
obtained in~\cite{10}.
\begin{theorem} Let $(E,G)$ be a space with a transformation
group,
let $\mu$ be a\break $\sigma$-finite $G$-invariant measure on $E$ and let
$X$ be a $\mu$-measurable set with $\mu(X) > 0$.
Suppose, in addition, that $G$ is uncountable and acts
$I(\mu)$-freely in~$E$. Then there exists a subset $Y$ of $X$
nonmeasurable with respect to every $G$-invariant
measure on $E$ extending $\mu$ (in other words, $Y$ is absolutely
nonmeasurable with respect to the class of all $G$-invariant extensions
of~$\mu$).
\end{theorem}
\begin{example}\rm Let $E = \mathbb R \times \mathbb R$ denote
the Euclidean plane, let
$G$ be the group of all translations of $E$ and let $\nu$ be
the classical two-dimensional Lebesgue measure on $E$.
It is not difficult to show (by using the method of transfinite
induction)
that there exists a function $f:\mathbb R \rightarrow \mathbb R$
whose graph $Z = Z_{f}$ is a $\nu$-thick subset of~$E$. In particular,
we have
the equality $\nu^{*}(Z) = +\infty$. At the same time, by
applying the standard methods of extending invariant measures
(see, for example, \cite[pp.~276--287]{6} or~\cite{4}),
it can be proved that there exists a complete $G$-invariant
measure $\nu'$ on $E$ satisfying the following two conditions:
\begin{list}{\alph{ite})}{\itemsep0.15truecm\topsep0.15truecm\parsep0pt\usecounter{ite}}
\item $\nu'$ is an extension of $\nu$;
\item $Z \in \dom(\nu')$ and $\nu'(Z) = 0$.
\end{list}
We thus see that the set $Z$ (being of strictly
positive outer $\nu$-measure) contains no subset absolutely
nonmeasurable with respect to the class of all $G$-invariant extensions
of $\nu$. Therefore, we can conclude that the assumption $X \in
\dom(\mu)$
is rather essential in the formulation of Theorem~2. On the other
hand, it is obvious that Theorem~1 does not need such an assumption.
\end{example}
A detailed proof of Theorem~2 is given in~\cite{10}.
Unfortunately, the
argument used in~\cite{10} is heavily based on the assumption of the
invariance of $\mu$ and does not work for
$\sigma$-finite quasiinvariant measures. Consequently, the
following problem remains unsolved.
\begin{problem} Let $E$ be a set, let $G$ be an uncountable
group of transformations of $E$ and let $\mu$ be a $\sigma$-finite
$G$-quasiinvariant measure on $E$ such that $G$ acts $I(\mu)$-freely
in~$E$. Can one assert that, for any $\mu$-measurable set
$X$ with $\mu(X) > 0$, there exists a subset of $X$ absolutely
nonmeasurable with respect to the class of all $G$-quasiinvariant
extensions of~$\mu$?
\end{problem}
Moreover, this problem remains open even for
the classical situation where $E$ coincides with the real line
$\mathbb~R$ and $G$ is some uncountable group of translations of~$\mathbb R$.
\begin{example}\rm It is well known that any Vitali subset of the
real line $\mathbb R$ is not measurable in the Lebesgue sense and does not
possess the Baire property with respect to the standard topology
of $\mathbb R$. Furthermore, any Vitali subset
of is absolutely nonmeasurable
with respect to the class of all $\mathbb Q$-invariant
measures extending the
Lebesgue measure on~$\mathbb R$.
At the same time, it is not hard to demonstrate that
there exist a Vitali set $X$ and a nonzero $\sigma$-finite
${\mathbb R}$-invariant measure $\mu$ on ${\mathbb R}$, for which we have
$X \in \dom(\mu)$ (see~\cite[Chapter~4]{4}).
In other words, the set $X$ is not absolutely
nonmeasurable with respect to the class of all nonzero $\sigma$-finite
${\mathbb R}$-invariant measures on ${\mathbb R}$ and,
consequently, $X$ is not absolutely nonmeasurable with respect to
the class of all ${\mathbb R}$-quasiinvariant probability measures
on~${\mathbb R}$.
\end{example}
In connection with Example 2, the following problem may be posed.
\begin{problem} \rm Let $(E,G)$ be a space with a transformation group
and let $M$ ($M'$)
denote the class of all nonzero $\sigma$-finite $G$-invariant
($G$-quasiinvariant) measures on $E$. Give a characterization of all
those subsets of $E$ which are absolutely nonmeasurable with respect
to $M$~($M'$).
\end{problem}
Now, we wish to discuss one question concerning the existence
of invariant and quasiinvariant measures on some good (from the
set-theoretical point of view) subgroups of
the real line and the one-dimensional torus. First, let us
consider the following simple example.
\begin{example}\rm Let ${\bf S}^1$ denote, as usual, the unit circle
on the Euclidean plane, regarded as a compact topological Abelian
group, and let $G$ be an uncountable
Borel subgroup of ${\bf S}^1$.
Suppose also that there exists a
$G$-invariant Borel probability measure $\mu$ on~$G$.
Then one can assert that
$G = {\bf S}^{1}$.
Indeed, let us put $\mu'(X) = \mu(X \cap G)$ for any Borel subset $X$
of ${\bf S}^1$. Clearly, $\mu'$ is a $G$-invariant Borel probability
measure on ${\bf S}^1$. Since $G$ is uncountable, there exists
an element $g \in G$ of infinite order. Obviously, the set
$\{g^{n}~:~n \in {\mathbb N}\}$ is dense in ${\bf S}^1$. Starting with
this fact and taking account of the invariance of $\mu'$ with respect to
each $g^{n}~(n \in {\mathbb N})$, we can easily infer that $\mu'$ is also
invariant with respect to each element of ${\bf S}^1$. Consequently,
$\mu'$ must coincide with the standard
Borel probability measure on ${\bf S}^1$.
Suppose for a moment that $G \ne {\bf S}^{1}$. Then, using the
Steinhaus
property of the standard Borel probability measure,
we obtain that $\mu'(G) = 0$. But this
is impossible since the measure $\mu'$ is concentrated on $G$.
The contradiction yields us the required equality $G = {\bf S}^{1}$.
A similar argument works for ${\mathbb R}$, and we easily obtain that if
$G$ is an uncountable Borel subgroup of ${\mathbb R}$
for which there exists at least one $G$-invariant Borel measure
$\mu$ on $G$ satisfying the relation
$$
0 < \mu([0,1] \cap G) < +\infty,
$$
then $G$ coincides with ${\mathbb R}$.
\end{example}
If we want to establish
an analogous result for nonzero $\sigma$-finite quasiinvariant measures
given on proper Borel subgroups of~${\mathbb R}$ (${\bf S}^1$, respectively),
then we need a much more difficult technique.
In order to demonstrate this, let us recall one simple notion
from the general theory of topological groups (see, e.g.,
\cite[Chapter~2]{2}.
We say that a given topological group $G$ is compactly generated
if there exists a compact subset of $G$ algebraically generating~$G$.
\begin{lemma} Let $G$ be a locally compact topological group
and let $K$ be a compact subset of $G$. Then there
exists a clopen compactly generated subgroup of $G$ containing~$K$.
\end{lemma}
\pf Denote by $e = e_{G}$ the neutral element of $G$. Clearly,
the set $\{e\} \cup K$ is compact, too. Consequently, there exists an
open set $U \subset G$ containing $\{e\} \cup K$ and having the compact
closure. Let us put $V = \cl(U)$. Then it is easy to
check that the group generated by $V$ is the required
one.\qed
\medskip
We also need the following well-known statement
which describes the structure of compactly generated locally compact
Abelian groups.
\begin{lemma} Let $G$ be an arbitrary compactly generated
locally compact Abelian group. Then $G$ is topologically isomorphic to
the product group ${\mathbb R}^{n} \times {\mathbb Z}^{m} \times G_{0}$ where
$n$ and $m$ are some natural numbers and $G_0$ is a compact
Abelian group.
\end{lemma}
\pf The proof of this fundamental result (essentially due to
Pontryagin and van Kampen) is presented in Chapter~2
of~\cite{2}.\qed
\medskip
The next classical statement is due to Mackey (see~\cite5).
\begin{lemma} Suppose that $\Gamma$ is a standard topological
group (i.e.\ $\Gamma$ is a Borel subgroup of some Polish topological
group) and suppose that $\mu$ is a left\break $\Gamma$-quasiinvariant
Borel probability measure on~$\Gamma$.
Then there exists a locally compact Polish
topological group $G$ with the left $G$-invariant Haar measure $\nu$,
such that
\begin{enumerate}
\item there is an algebraic isomorphism $f:\Gamma \rightarrow G$
which simultaneously is a Borel isomorphism between the topological
spaces $\Gamma$ and~$G$;
\item the measure $f(\mu)$ is equivalent to the Haar measure~$\nu$.
\end{enumerate}
\end{lemma}
Actually, we need only relation~1 of Lemma~3. The fact that the
measures $f(\mu)$ and $\nu$ are equivalent does not play an essential
role in our further considerations.
Notice that Lemma~3 with some related results is discussed in the
well-known textbook by Parthasarathy~\cite7. However, the presentation
of this material in~\cite7 is not quite correct.
\begin{lemma} Let $\Gamma$,~$G$ and $f$ be as in the previous
lemma.
Then the mapping $f^{-1}$ is a continuous bijection from $G$ onto
$\Gamma$.
\end{lemma}
\pf Evidently, $f^{-1}$ is an algebraic isomorphism between
the groups $G$ and $\Gamma$. Also,
$f^{-1}$ is measurable with respect to the Haar
measure $\nu$ on~$G$. Hence, applying the Steinhaus
property of $\nu$, we
obtain that $f^{-1}$ is continuous. Actually, the same
argument yields the well-known fact that any measurable
(with respect to the completion of a Haar measure)
homomorphism
from a\break $\sigma$-compact
locally compact topological group into a separable
topological\break group is continuous. Moreover, the category analogue
of the Steinhaus property implies that an algebraic homomorphism
from a Baire topological group into a separable topological group,
having the Baire property, is necessarily continuous.\qed
\medskip
Now, we are ready to prove the following result (we do not assert that
it is new, but we could not find its precise proof in the literature).
\begin{theorem} Let $E$ be either ${\mathbb R}$ or ${\bf S}^1$
and let $\Gamma$ be a proper Borel subgroup of~$E$.
Suppose, in addition, that there exists a nonzero
$\sigma$-finite
$\Gamma$-\hskip0cmquasiinvariant Borel measure $\mu$ on $\Gamma$.
Then $\Gamma$ is at most countable.
\end{theorem}
\pf Without loss of generality we may assume that $\mu$ is
a probability measure. According to Lemma~3, there exists a
locally compact Polish topological Abelian group $G$ (equipped with
the Haar measure $\nu$) such that
\begin{enumerate}
\item there is an algebraic isomorphism $f:\Gamma \rightarrow G$
which simultaneously is a Borel isomorphism between the topological
spaces $\Gamma$ and~$G$;
\item the measure $f(\mu)$ is equivalent to the Haar measure~$\nu$.
\end{enumerate}
Suppose for a moment that our group $\Gamma$ is uncountable,
i.e. $\card(\Gamma) = {\bf c}$ where {\bf c} denotes the cardinality
of the continuum. Then we also have $\card(G) = {\bf c}$.
Let $K$ be an arbitrary uncountable compact subset of~$G$.
Applying Lemma~1, we can find a clopen compactly generated
subgroup $G'$ of $G$ containing $K$. In particular, for $G'$,
we have the equality $\card(G') = {\bf c}$.
Now, according to Lemma~2, the group $G'$ is topologically
isomorphic to the product group
${\mathbb R}^{n} \times {\bf Z}^{m} \times G_{0}$
where $n$ and $m$ are some natural
numbers and $G_0$ is a compact Abelian group.
Further, as we know (see Lemma~4), the mapping
$f^{-1}:G' \rightarrow E$ is an injective continuous
homomorphism from $G'$ onto a proper subgroup of $E$.
Consequently, $f^{-1}({\mathbb R}^{n})$ is a proper connected subgroup
of $E$ and $f^{-1}(G_{0})$ is a proper compact subgroup of~$E$.
>From these facts we can easily deduce that $n = 0$ and that the group
$G_0$ is finite. But this immediately implies that the group
$G'$ is at most countable, which is impossible.
The contradiction obtained finishes the proof of
Theorem~3.\qed
\medskip
Evidently, Theorem 3 remains true for any proper analytic (coanalytic)
subgroup $\Gamma$ of $E$ equipped with a nonzero $\sigma$-finite
$\Gamma$-quasiinvariant Borel measure $\mu$. Indeed, since $\Gamma$
is a Radon topological space, it
contains a subgroup $\Gamma'$ such that $\card(\Gamma') = \card(\Gamma)$
and $\Gamma'$ is an $F_{\sigma}$-subset
of $E$, and the restriction of $\mu$ to the Borel $\sigma$-algebra of
$\Gamma'$ is not identically equal
to zero. So we may directly apply Theorem~3 to~$\Gamma'$.
However, Theorem 3 cannot be extended to the class of all
projective subgroups of $E$ where
$E = {\mathbb R}$ or $E = {\bf S}^1$. Indeed, in the Constructible
Universe we have a projective subgroup $\Gamma$ of ${\mathbb R}$ nonmeasurable
in the Lebesgue sense. For such a group $\Gamma$,
the relations
$$
\Gamma \ne {\mathbb R},\quad\lambda^{*}(\Gamma) > 0,\quad\card(\Gamma) >
\omega
$$
are fulfilled,
where $\lambda$ denotes the standard Lebesgue measure on ${\mathbb R}$,
and it is clear that there exists a nonzero
$\sigma$-finite $\Gamma$-invariant Borel measure on~$\Gamma$.
\medskip
The following problem is of some interest from the
methodological point of view.
\begin{problem} Give a relatively elementary proof
of Theorem~{\rm3} by using only the methods of real analysis
and classical measure theory, which do not rely on the Mackey theorem
and on the deep results
concerning the structure of locally compact topological Abelian
groups.
\end{problem}
Let us remark that in~\cite8 the author
tries to present such a proof, but his argument is not quite
correct.
We conclude with an example
which shows that, for the Euclidean plane,
the situation essentially differs from the situation of the real line.
\begin{example}\rm Let $E = {\mathbb R} \times {\mathbb R}$ and let
$\Gamma = ({\mathbb R} \times \{0\}) + (\{0\} \times {\mathbb Q})$.
Obviously, $\Gamma$ is a proper uncountable
dense $F_{\sigma}$-subgroup
of~$E$. We equip $\Gamma$ with the induced topology. Then
it is not hard to define a nonzero $\sigma$-finite Borel measure
$\mu$ on $\Gamma$ satisfying these two conditions:
\begin{list}{\alph{ite})}{\itemsep0.15truecm\topsep0.15truecm\parsep0pt\usecounter{ite}}
\item
$\mu$ is $\Gamma$-invariant;
\item $\mu$ does not have the Steinhaus property.
\end{list}
More precisely, condition b) says that there exists a $\mu$-measurable
set $X$ with $\mu(X) > 0$ for which the difference set
$$
X - X = \{x - y : x \in X,~y \in X\}
$$
contains no neighbourhood of the neutral element of~$\Gamma$.
\end{example}
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\end{document}