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 = EΓ

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.