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\title{The Furstenberg lemma characterizes amenability\footnote{Key
words and phrases: Amenability, orbit equivalence, mean,
homogeneous structure. 2000 Mathematics Subject Classification:
Primary 03E15, 37A20. Secondary 28A60, 03C15}}
\author{Greg Hjorth\footnote{NSF grant DMS 0140503}}
\date{\today}
\maketitle
\abstract{We characterize amenability in terms of the existence
of equivariant assignments of measures for cocycles into the homeomorphism
group of a {\it single} compact {\it metric} space.}
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\bigskip
\section{Introduction}
\bigskip
It is well known that every cocycle
from an amenable equivalence relation to the homeomorphism
group of a compact metric space or from an amenable
action of a group to the homeomorphism group of a
compact metric space admits a suitably equivariant
assignment of measures. This theorem in some form dates
back to \cite{Furstenberg}, though an explicit discussion can
be found in chapter 4 of \cite{zimmer}.
In this note we obtain a converse using ideas suggested by the
concept of {\it ultrahomogeneous structure} from
\cite{hodges}.
It is important to note here that the metric
space is fixed and {\it metrizable}.
If we have the cocycle into the homeomorphism groups
of different compact non-metrizable spaces, then the existence of
an equivariant assignment of measures
becomes closer to the definition of
$\S$4.3 \cite{zimmer} and the result is already
known; see \cite{bates}.
\begin{theorem}
Let $E$ be a countable, measurable equivalence relation on a
standard Borel probability space $(S, \mu)$ and suppose
that for every measurable cocycle
\[\alpha: E\rightarrow {\rm Hom}(K)\]
from $E$ to the homeomorphism group of a compact
metric space we have a measurable assignment
\[x\mapsto \mu_x,\]
\[S\rightarrow M(K),\]
from $S$ to the probability measures on $K$, which
is equivariant in the sense that
\[\mu_z=\alpha(z, x)\cdot \mu_x.\]
Then $E$ is amenable.
\end{theorem}
Recall that a {\it cocycle} from $E$ to a group $\Gamma$
consists in a function $\alpha: E\rightarrow \Gamma$
with
\[\alpha(x_3, x_2)\alpha(x_2, x_1)=\alpha(x_3, x_1).\]
In the case that a group $\Lambda$ acts on the space $X$ a
{\it cocycle} for the action is a function
\[\alpha: X\times \Lambda\rightarrow\Gamma\]
such that
\[\alpha(g, h\cdot x)\alpha(h, x)=\alpha(gh, x).\]
(In the below we generally do not distinguish between
something happening everywhere and almost everywere; the
distinction will be irrelevant.)
We take the Connes-Feldman-Weiss definition of amenability for
equivalence relations.
\begin{definition} An equivalence relation $E$ on $(S, \mu)$ is
{\it amenable} if there is a linear function
\[P:L^\infty(E)\rightarrow L^\infty(S)\] which is positive,
sends the constant function 1 on $E$ to the constant function
1 on $S$, and commutes with all projections and all morphisms
included in the equivalence relation.
Following \cite{zimmer}, we say that
the measurable action of a lcsc group $\Gamma$ on a space $X$ is
{\it amenable} if the induced orbit equivalence relation is amenable
and almost every point $x$ has its stabilizer,
\[\Gamma_x=\{g\in \Gamma: g\cdot x=x\}\]
amenable.
\end{definition}
\begin{notation} For $\psi\in L^\infty(E)$ and $f:A\rightarrow B$
a measurable bijection included in the graph of $E$, we
let $\psi^f$ be defined by $\psi^f(f(x), y)=\psi(x, y)$ for $(f(x), y)\in
E\cap B\times S$ and
$\psi^f(z, y)=0$ for $z$ not in $B$. For $\psi\in L^\infty(S)$ we
would let $\psi^f(f(x))=\psi(x)$ and $\psi^f(z)=0$ for $z$ outside
$B$.
\end{notation}
With this notation we can rephrase the commutativity requirement
for $P$: Given any such $f:A\rightarrow B$ a measurable bijection
with $f(x) E x$ all $x\in A$, and given an $\psi\in L^\infty(E)$,
\[ P(\psi^f)=(P(\psi))^f.\]
The Connes-Feldman-Weiss definition of amenability is equivalent
to the equivalence relation being measurably hyperfinite.
\begin{theorem}
Let $\Gamma$ be a countable group acting measurably on a
standard Borel probability space $(S, \mu)$ and suppose
that for every measurable cocycle
\[\alpha: S\times \Gamma\rightarrow {\rm Hom}(K)\]
we have a measurable assignment
\[x\mapsto \mu_x,\]
\[S\rightarrow M(K),\]
from $S$ to the probability measures on $K$, which
is equivariant in the sense that
\[\mu_{g\cdot x}=\alpha(x, g)\cdot \mu_x.\]
Then the action is amenable.
\end{theorem}
As a working definition, let us simply say that an action
is {\it amenable} if the resulting equivalence relation is amenable
and the stabilizers of a.e. point are amenable.
\bigskip
\bigskip
\section{Proof}
\bigskip
We need a number of specific definitions to support the
construction ahead.
\begin{definition} A {\it partition} of an infinite set $X$ consists
in a set $\{P_1, P_2, ..., P_\ell\}$ such that $P_i$ is disjoint from
$P_j$ for $i\neq j$ and
\[\bigcup_{i\leq \ell} P_i = X.\]
Given a finite sequence $\vec f = (f_1, f_2, ...f_n)$ of functions
from $X$ to $\{0, 1\}$, we say that
$\p= \{P_1, P_2, ..., P_\ell\}$ is the {\it partition generated by $\vec f$}
if each $P_i$ equals
\[\{a\in X: f_1(a) = k^i_1, f_2(a)=k^i_2, ... f_n(a)=k^i_n\}\]
for some choice of $k^i_1,...,k^i_n\in \{0, 1\}$.
\end{definition}
Remark: The trivial partition $\p=\{X\}$ is generated by the
empty sequence.
\begin{definition} Let $X$ be a countably infinite set.
A collection
\[\f\subset \{0, 1\}^X\] is said to be {\it random} if whenever
\leftskip 0.4in
\noindent (i) $\p=\{P_1, P_2, ..., P_\ell\}$ is a partition generated by
some finite sequence $\vec f$ from $\f$, and
\noindent (ii) we have
cardinals $\kappa_i, \lambda_i\in \{0, 1, 2, ... \aleph_0\}$,
each $i\leq n$, with $\kappa_i+\lambda_i$ always equal to $|P_i|$, and
\noindent (iii) at each $i$ we have finite subsets $\vec u_i \subset P_i$,
$\vec v_i\subset P_i$, each of cardinality at most $\kappa_i$ and
$\lambda_i$ respectively,
\leftskip 0in
\noindent there is some $g\in \f$, not appearing on the $\vec f$ sequence,
with
\[|\{n\in P_i : g(n)=1\}|=\kappa_i,\]
\[|\{n\in P_i : g(n)=0\}|=\lambda_i, \]
and $g$ assuming the value $1$ on each element of $\vec u_i$ and the
value 0 on each element of $\vec v_i$.
\end{definition}
{\bf Remarks:} (i) Here $|Y|$ denotes the cardinality of $Y$.
(ii) If one of $\kappa_i, \lambda_i$ equals $\aleph_0$ then
$\kappa_i+\lambda_i=\aleph_0$. ($\aleph_0+1=\aleph_0+2=...
=\aleph_0+\aleph_0=\aleph_0$).
(ii) The idea of the definition of a random collection $\f$ is this:
Given any finite sequence $\vec f$ from $\f$ and finite
$\vec a$ from $X$, we can find a new
function $g\in \f$ which behaves relative to $\vec f$ and
$\vec a$ in any previously
prescribed manner. Anything which can happen does.
\begin{lemma}
If $X$ is a countably infinite set, and $\f_0\subset\{0, 1\}^X$
is a countable collection of functions, then we can find a countable
random collection $\f\subset \{0, 1\}^X$ which includes $\f_0$.
\end{lemma}
\begin{proof}
We build $\f$ in stages. We first take some countable
$\f_1\supset \f_0$ which includes enough functions to satisfy
the definition of random for any choice of parameters from
$\f_0$. We then repeat, replacing $\f_0$ by $\f_1$ and obtain
$\f_2\supset \f_1$. In this way we obtain an ascending chain
\[\f_0\subset \f_1\subset\f_2\subset...\subset\f_n\subset\f_{n+1}...\]
and finish with
\[\f=\bigcup_{n\in\N}\f_n.\]
\end{proof}
\begin{lemma}
\label{homog}
If $X_1, X_2$ are countably infinite sets, $\f_1\subset\{0, 1\}^{X_1},
\f_2\subset \{0, 1\}^{X_2}$ both random, then there is a bijection
\[\pi: X_1\rightarrow X_2\]
such that the induced function
\[\varphi: \{0, 1\}^{X_1}\rightarrow \{0, 1\}^{X_2}\]
defined by
\[ \varphi(f)(a) =f(\pi^{-1}(a))\]
provides a bijection between $\f_1$ and $\f_2$.
\end{lemma}
\begin{proof} We build $\pi$ in a sequence of stages, with only a finite amount
of $\pi$ established at each stage. We simultaneosly build $\varphi$ finite
step by finite step. We guarantee an isomorphism at the very end stage by a
``back and forth argument", in the sense of $\S 3.2$ \cite{hodges}.
Let $(f_i)_{i\in \N}$, $(g_i)_{i\in \N}$ enumerate $\f_1$, $\f_2$ respectively,
and let $(a_i)_{i\in \N},$ $(b_i)_{i\in \N}$ enumerate $X_1$, $X_2$
respectively. Using that $X_2$ is random and considering the trivial partition
$\{X_2\}$ we can find some $g\in \f_2$ such that
\[ |\{b\in X_2: g(b)=1\}| = |\{a\in X_1: f_1(a)=1\}|,\]
\[ |\{b\in X_2: g(b)=0\}| = |\{a\in X_1: f_1(a)=0\}|.\]
Let $\varphi(f_1)=g$, choose $b'\in X_2$ with $g(b')=f_1(a_1)$ and let
$\pi(a_1)=b'$.
Now consider $g_1$. We want to find suitable $f\in \f_1$ with
$\varphi(f)=g_1$. First let us assume that $g_1\neq g$. Then by
considering the randomness of $\f_1$ we can find some $f\in \f_1$
with
\[f(a_1)=g_1(b')\]
and
\[ |\{b\in X_2: g(b)=\ell _1, g_1(b)=\ell _2\}|
= |\{a\in X_1: f_1(a)=\ell_1, f(a)=\ell_2\}|,\]
any choice of $\ell_1, \ell_2\in \{0, 1\}$. We then let $\varphi(f)=g_1$,
and go on to consider how to extend $\pi$ by looking at $b_1$. If
$b_1=b'$ then $\pi(a_1)=b_1$; otherwise we choose some $a'\neq a_1$
with
\[f(a')=g_1(b_1),\]
\[f_1(a')=g(b_1),\]
and let $\pi(a')=b_1$. (Note: We can find such an $a'$ since
$|\{a: f(a)=g_1(b_1), f_1(a)=g(b_1)\}|$ equals
$|\{a: g_1(a)=g_1(b_1), g(a)=g(b_1)\}|$.)
Having done all this under the assumption $g_1$ does not
equal $g$ we need to consider also the case
$g_1=g$. But here there is really nothing to do, since
we have already decided that $\varphi(f)=g$, and we simply go on to
find some suitable choice of $a'$ with $\pi(a')=b_1$ as above.
Now we go back to the $X_1$ side and try to find a suitable
value for $\varphi(f_2)$.
Again there is a split in cases. If $f_2\in \{f_1, f\}$ then
the
value of $\varphi(f_2)$ is
already determined and the whole process
is kind of trivial.
So suppose instead $f_2\neq f, f_1$. Then we can choose
$g'\in \f_2$, $g'\neq \{g_1, g\}$ so that
at each choice of $\ell_1, \ell_2, \ell_3$ we have
\[ |\{b\in X_2: g(b)=\ell _1, g_1(b)=\ell _2, g'(b)=\ell_3\}|
= |\{a\in X_1: f_1(a)=\ell_1, f(a)=\ell_2, f_2(a)=\ell_3\}|,\]
and
\[g'(b_1)=f_2(a'),\]
\[g'(b')=f_2(a_1).\]
We again choose a suitable value of
$\pi(a_2)$ so that
\[g(\pi(a_2))=f_1(a_2),\]
\[g_1(\pi(a_2))=f(a_2),\]
\[g'(\pi(a_2))=f_2(a_2).\]
With only futher notational complications, we keep going in this way for
infinitely many steps, back and forth between the two sides, and
thereby enforce an isomorphism.
\end{proof}
\begin{definition} A measurable equivalence relation $E$ on a standard
Borel probability space $(S, \mu)$ is said to have the
{\it Furstenberg property}
if for every measurable cocycle
\[\alpha: E\rightarrow {\rm Hom}(K)\]
from $E$ to the homeomorphism group of a compact
metric space we have a measurable assignment
\[x\mapsto \mu_x,\]
\[S\rightarrow M(K),\]
from $S$ to the probability measures on $K$, which
is equivariant in the sense that
\[\mu_z=\alpha(z, x)\cdot \mu_x.\]
A measurable action of a countable group
$\Gamma$ on standard Borel probability space $(S, \mu)$ is
said to have the {\it Furstenberg property}
if for every measurable cocycle
\[\alpha: S\times \Gamma\rightarrow {\rm Hom}(K)\]
we have a measurable assignment
\[x\mapsto \mu_x,\]
\[S\rightarrow M(K),\]
from $S$ to the probability measures on $K$, which
is equivariant in the sense that
\[\mu_{g\cdot x}=\alpha(x, g)\cdot \mu_x.\]
\end{definition}
Assume from now on that $E$ on $(S, \mu)$ is an
equivalence relation with
\leftskip 0.4in
\noindent (i) $(S, \mu)$ a standard Borel
probability space;
\noindent (ii) every $E$ equivalence class is
countably infinite;
\noindent (iii) $E$ is measurable (e.g. in the sense
of being a measurable subset of $(S^2, \mu^2)$);
\noindent (iv) $E$ has the Furstenberg property.
\leftskip 0in
We will work towards showing that $E$ is amenable. After all
that is said and finished, we will indicate how the argument
modifies for the case of group actions.
Note that the collection of means (i.e. positive, linear, norm
one, unit respecting)
\[P: L^\infty(E)\rightarrow L^\infty(S)\]
is compact in the topology given by the functions
\[P\mapsto \int_A P(\psi)d\mu\]
($\psi\in L^\infty(E), A\subset X$ measurable).
(The point here is that we
can represent $P$ by looking at its restriction to
the unit ball
$(L^\infty(E))_1$; then each
$P(\psi)$ lands in $(L^\infty(S))_1$ which is
compact in the weak-$*$ topology.)
Inside this collection, for any given
$\psi\in L^\infty(E)$, $f\subset E$, the ones which appropriately commute,
in the sense $P(\psi^f)=(P(\psi))^f$
form a closed subset.
For purely notational convenience, let us fix $G$ a group of
measurable
bijections
of $S$ whose orbit equivalence relation equals $E$. Appealing
to the compactness of the space of means, it suffices to show
that given a countable collection $\f_0\subset L^\infty(E)$
consisting of functions assuming only the values $0$ and
$1$ and closed under the $G$ action, we may find
\[P: \f_0\rightarrow L^\infty(S)\]
such that:
\leftskip 0.4in
\noindent (i) $P(\psi^f)=(P(\psi))^f$ all $\psi\in \f_0$ and
$f\in G$;
\noindent (ii) if $A\subset S$ measurable and $\psi\in \f_0$
has $\psi(x, y)=0$ all $x\in A$, then likewise
$P(\psi)(x)=0$ all $x\in A$;
\noindent (iii) if $\psi\leq \psi'$ both in $\f_0$,
then $P(\psi)\leq P(\psi')$;
\noindent (iv) if $\psi+ \psi'=\psi''$ all in $\f_0$, then
$P(\psi'')=P(\psi')+P(\psi)$;
\noindent (v) $P(1)=1$.
\leftskip 0in
\bigskip
For each $\psi\in \f_0$ and $x\in S$ we define $\psi_x\in
\ell^\infty([x]_E)$ by $\psi_x(y)=\psi(x, y)$. After possibly
expanding the set $\f_0$ we may assume that at each $x\in S$, the
collection
\[\{\psi_x: \psi\in \f_0\}\]
is random (in our earlier sense) for $\{0, 1\}^{[x]_E}$.
Now fix some arbitrary countably infinite set $X$ and let
$\f$ be a collection of random functions on $X$. Let
$K$ be the space of all means {\it just on} $\f$; that is to
say, all
\[m: \f\rightarrow [0, 1]\]
such that
\leftskip 0.4in
\noindent (a) if $\psi\leq \psi'$ both in $\f$,
then $m(\psi)\leq m(\psi')$;
\noindent (b) if $\psi+ \psi'=\psi''$ all in $\f$, then
$m(\psi'')=m(\psi')+m(\psi)$;
\noindent (c) $m(1)=1$.
\leftskip 0in
Since $\f$ is countable, $K$ is compact, metric.
At each $x\in S$ we may in a measurable manner assign a
bijection
\[\pi_x: [x]_E\rightarrow X\]
such that the induced map
\[\rho_x: \ell^\infty([x]_E)\rightarrow \ell^\infty(X)\]
defined by
\[(\rho_x(\psi))(a)=\psi(\pi_x^{-1}(a))\]
provides a bijection of $\{\psi_x:\psi\in \f_0\}$ with
$\f$. (The existence of some such assignment, without worrying
about measurability, follows by \ref{homog}. Then to see it
can be found measurably, we can either appeal to the Jankov
von Neumann uniformization theorem for measurable selectors,
or go back into the proof of \ref{homog} and see that it is
so uniform as to actually allow a {\it Borel} choice of
$x\mapsto \rho_x$.)
We can then define a measurable cocyle
\[\alpha: E\rightarrow {\rm Hom}(K)\]
by
\[(\alpha(x, z))(m)(\psi)=m(\rho_z(\rho_x^{-1}(\psi))).\]
We apply the assumption of Furstenberg's property, and
get a measurable assignment
\[S\rightarrow M(K)\]
\[x\mapsto \mu_x\]
with
\[\mu_x=\alpha(x, z)\cdot \mu_z.\]
We then let
\[(P(\psi))(x)=\int_K m(\rho_x(\psi_x))d\mu_x(m).\]
We want to check that the induced
\[P: L^\infty(E)\rightarrow L^\infty(S)\]
satisfies (i)-(v) above, and then we will be done.
(ii)-(v) follow at once from the definition, so we
only need to deal with (i).
So fix some $f:S\rightarrow S$ in $\Gamma$ and
$\psi\in \f_0$.
Then
\[(P(\psi^f))(x)=\int_K m \rho_x((\psi^f)_x)d\mu_x(m) \]
\[= \int_K m \rho_x(\psi_{f^{-1}(x)})d\mu_x(m), \]
since
\[(\rho_x((\psi^f)_x)))\pi_x(y)=(\psi^f)_x(y)=\psi^f(x, y)\]
\[=\psi(f^{-1}(x), y)=\psi_{f^{-1}(x)}(y)=
\rho_x(\psi_{f^{-1}(x)})\pi_x(y).\]
On the other hand,
\[(P(\psi))^f(x)=(P(\psi))(f^{-1}(x))\]
\[= \int_K m \rho_{f^{-1}(x)}((\psi)_{f^{-1}(x)})d\mu_{f^{-1}(x)}(m),\]
which in turn, by our assumption of $\alpha$-equivariance, equals
\[\int_K m \rho_{f^{-1}(x)}((\psi)_{f^{-1}(x)})
d\alpha(f^{-1}(x), x)\cdot\mu_{x}(m),\]
which by the definition of the cocycle unwinds as
\[ \int_K(\rho_x(\rho_{f^{-1}(x)}^{-1}(\rho_{f^{-1}(x)}(\psi_{f^{-1}(x)}))))
d\mu_x(m) = \int_K(\rho_x(\psi_{f^{-1}(x)})
d\mu_x(m),\]
as above and as required.
\bigskip
Thus we have established the amenability of $E$ assuming the
existence of equivariant cocycles for all the appropriate
cocycles. It remains now to indicate the modifications will
enable the same result to be proved for group actions.
From now on assume that $\Gamma$ is a countable group
acting measurably on a standard Borel $(S, \mu)$ and the
action of $\Gamma$ has the Furstenberg property. Given the
earlier result, we know that the orbit equivalence relation
$E_\Gamma$ is amenable, and it suffices to show that a.e. the stabilizers
are amenable. For this purpose it suffices to show that if
$\Gamma_0$ is a finitely generated subgroup of $\Gamma$ and there
is a non-null set of $x\in S$ whose stabilizer includes $\Gamma_0$
then $\Gamma_0$ is non-null.
So fix such a $\Gamma_0$. Let $\{f_i:i\in \N\}$ be a countable
collection of functions
\[\Gamma_0\rightarrow \{0, 1\}\]
closed under $\Gamma_0$-translation.
By compactness again, it suffices to show that there is
a $\Gamma_0$-invariant mean defined just on
$\{f_i: i\in \N\}$.
Let $(\Gamma_0\gamma_i)_{i\in \Lambda}$ enumerate
without repititions the left cosets of $\Gamma_0$. At each
$j\in \N$ define
\[\hat{f}_j: \Gamma\rightarrow \{0, 1\}\]
by
\[\hat{f}_j(\gamma\gamma_i)=f_j(\gamma)\]
for $\gamma\in \Gamma_0$; in other words, we
multiply out $f_j$ across the various cosets using
our representatives $(\gamma_i)_{i\in \Lambda}$.
Now take a countable collection
\[\f\subset L^\infty(S\times \Gamma)\]
such that
\leftskip 0.4in
\noindent (i) each $h\in \f$ assumes only the values 0 and 1;
\noindent (ii) if $x\in S$, $j\in \N$, then there is $h\in \f$
with
\[h_x=\hat{f}_j\]
(where as usual we define $h_x$ by $h_x(\gamma)=h(x, \gamma)$);
\noindent (iii) at each $x$, $\{h_x: h\in \f\}$ is a random
collection of functions from $\Gamma$ to $\{0, 1\}$;
\noindent (iv) if $\gamma\in \Gamma$, $h\in \f$, then
\[\gamma\cdot h\in L^\infty(S\times \Gamma)\]
(where as usual $\gamma\cdot h$ is defined by
$(\gamma\cdot h)(y, \gamma_1)=h(\gamma^{-1}\cdot y,
\gamma^{-1}\gamma_1)$).
\leftskip 0in
\bigskip
We again let $X$ be some countably infinite space and let
$\f$ be a random collection of functions from $X$ to
$\{0, 1\}$. At each $x\in S$ we again measurably assign a
corresponding bijection
\[\pi_x: \Gamma\rightarrow X\]
so that the induced
\[\rho_x: \ell^\infty(\Gamma)\rightarrow \ell^\infty(X)\]
provides a bijection. We again let $K$ be the space of
means on $\f$.
Given $x\in S, \gamma\in \Gamma$, we obtain a continuous
\[\alpha(x, \gamma): K\rightarrow K\]
defined by
\[(\alpha(x, \gamma))(m)(\psi)
= m(\rho_x(\gamma^{-1}\cdot (\rho_{\gamma\cdot x}^{-1}
(\psi)))), \]
where $\Gamma$ acts on $\ell^\infty(\Gamma)$ by left translation in
the usual way -- that is to say, $(\gamma\cdot \varphi)(\gamma')
=\varphi(\gamma^{-1}\gamma')$. It is routinely seen that
this provides a measurable cocycle
\[\alpha: X\times \Gamma\rightarrow {\rm Hom}(K).\]
We apply the assumption of the Furstenberg property and
get
\[S\rightarrow M(K)\]
\[x\mapsto \mu_x\]
with
\[\mu_{g\cdot x}=\alpha(x, g)\cdot \mu_x.\]
At each $x$ we can pull back and obtain a mean
$\nu_x$ defined on $\{\varphi_x: \varphi\in \f\}$
by
\[\nu_x(\varphi)=\int_K m(\rho_x(\varphi))d\mu_x(m).\]
It is then immediate that if $\Gamma_0$ is included in the
stabilizer of $x$ then $\nu_x$ is $\Gamma_0$-invariant.
From this we obtain a $\Gamma_0$-invariant mean
on $\{f_i: i\in \N\}$ with
\[\nu(f_j)=\nu_x(\hat{f}_j).\]
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\bibitem{bates} T. Bates, MSc, Thesis, University of Ottawa,
1994.
\bibitem{cofewe} A. Connes, J. Feldman, B. Weiss,
{\it An amenable equivalence relation is generated
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\bibitem{Furstenberg} H. Furstenberg,
{\it Boundary theory and stochastic processes on homogeneous spaces
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\bibitem{zimmer} R. Zimmer, {\bf Ergodic theory and semi-simple groups},
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\end{thebibliography}
6363 MSB
Mathematics
UCLA
405 Hilgard Avenue
Los Angeles
CA90095-1555
greg@math.ucla.edu
www.math.ucla.edu/\~{}greg
\end{document}