14 I. MEASURE PRESERVING AUTOMORPHISMS
Proposition 3.2. In the uniform topology, [E] is separable, thus ([E],u) is
Polish.
Proof. Since E is a countable equivalence relation, there is a countable
group Γ Aut(X, µ) such that E =
X
= the equivalence relation induced
by Γ. Thus T [E] iff there is a Borel partition {An}n=1

of X and {γn}n=1


Γ such that T|An = γn|An. Fix also a countable Boolean algebra A of Borel
sets closed under the Γ-action and dense in the measure algebra.
Given C1,...,CN A and γ1,...,γN Γ such that C1,...,CN are
pairwise disjoint and γ1(C1),...,γN (CN ) are pairwise disjoint, let C0 =
X \
i≤N
Ci,D0 = X \
i≤N
γi(Ci), so that µ(C0) = D0, and pick TC0,D0
Aut(X, µ) with T(C0) = D0. Put finally SC,
γ
= TC0,D0 |C0
i≤N
γi|Ci.
Clearly there are only countably many SC,
γ
. We will show that for any
T [E] and 0 there are C, γ such that δu(T, SC,
γ
) . It follows then
that [E] is separable (being contained in the uniform closure of the SC,
γ
).
So fix T [E], 0, a Borel partition {An}n=1

of X and {γn}n=1

Γ
such that T|An = γn|An, ∀n 1. For each δ1,δ2 0, choose N = N(δ1)
large enough so that

iN
µ(Ai) δ1, and then B1,...,BN A with
µ(Ai∆Bi) δ2. Let for 1 i N,
Ci = {x Bi : x
ji
Bj & γi(x)
ji
γj(Cj)},
so that {Ci}i≤N A are pairwise disjoint and so are {γi(Ci)}i≤N A.
Then if δ1,δ2 are chosen small enough, δu(T, SC,
γ
) .
Since the identity is a continuous map from the Polish group ([E],u) onto
the Borel subgroup [E] of (Aut(X, µ),w), it follows that [E] is a Polishable
subgroup of (Aut(X, µ),w) with the unique corresponding Polish topology
being the uniform topology.
(B) Some of the basic facts we mentioned earlier for Aut(X, µ) “localize”
to [E].
First the proof of Rokhlin’s Lemma actually shows that the Uniform
Approximation Theorem is true in each [E].
Theorem 3.3 (Uniform Approximation Theorem for [E]). If T [E]
is aperiodic, then for each N 1, 0 there is a periodic S [E] of period
N such that δu(S, T)
1
N
+ .
From this we also have the following, in case E is also ergodic.
Theorem 3.4 (Conjugacy Lemma for [E]). If E is ergodic, and T [E]
is aperiodic,
{STS−1
: S [E]} is uniformly dense in APER [E] (which
is a closed subset of ([E],u)).
Proof. In the proof of 2.4 note that, since E is ergodic, there is S0 [E]
with S0(A) = B.
We also note another well-known basic fact.
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