5. TURBULENCE OF CONJUGACY 33

Fix a Borel transversal A for EN and let ν = π∗µ. Then ν(C) = 0 for

each EN -class C. Define the measure ρ on A by

ρ(B) = ν([B]EN ),B ⊆ A Borel.

Clearly it is non-atomic, so there is a continuous map λ → Aλ from [0, 1]

to MALGρ with A0 = ∅,A1 = A and 0 ≤ λ ≤ λ ≤ 1 ⇒ Aλ ⊆ Aλ and

ρ(Aλ \ Aλ) ≤ λ − λ. Let Xλ =

π−1([Aλ]EN

), so that µ(Xλ) = ρ(Aλ), thus

λ → Xλ is continuous from [0, 1] to MALGµ,X0 = ∅,X1 = X, 0 ≤ λ ≤ λ ≤

1 ⇒ Xλ ⊆ Xλ and µ(Xλ \Xλ) ≤ λ −λ. Moreover Xλ is

π−1(EN

)-invariant,

so g, S-invariant. Then define as before gλ = g|Xλ ∪ id|(X \ Xλ), so that Xλ

is also gλ,S-invariant. The proof then proceeds exactly as in 5.1. ✷

With essentially the same proof one can show the result of Foreman-

Weiss that the conjugacy action of (Aut(X, µ),w) on (ERG,w) is turbulent.

Actually one can prove a somewhat more precise version (replacing ERG by

APER) by using an additional fact (Lemma 5.4 below).

Theorem 5.3 (Foreman-Weiss [FW] for ERG). The conjugacy action

of the group (Aut(X, µ),w) on (APER,w) is turbulent.

Proof. We have already seen that every conjugacy class in APER is

dense and meager. We finally fix T ∈ APER in order to show that it

is turbulent for the conjugacy action. The following lemma was proved

together with Ben Miller.

Lemma 5.4. Every aperiodic, hyperfinite equivalence relation E is con-

tained in an ergodic, hyperfinite equivalence relation F .

Granting this, let S ∈ ERG be such that [T] ⊆ [S], where for T ∈

Aut(X, µ) we denote by [T] = [ET ] the full group of the equivalence relation

ET induced by T. Now [S] is dense in (Aut(X, µ),w). Then repeat the

argument in the proof of 5.1, starting with g ∈ [S].

Proof of the Lemma. Consider the ergodic decomposition of E. This

is given by a Borel E-invariant surjection π : X → E, where E is the space

of ergodic invariant measures for E. If Xe = {x ∈ X : π(x) = e}, then e is

the unique invariant ergodic measure for E|Xe and if ν = π∗µ, a probability

measure on E, then µ = edν(e), i.e., for A ⊆ X Borel,

µ(A) = e(A ∩ Xe)dν(e).

In particular, modulo µ-null sets, every E-invariant set A is of the form

π−1(B),

for some B ⊆ E.

Case 1. E is finite.

We will prove then the result by induction on card(E). If card(E) = 1,

there is nothing to prove. Assume now card(E) = 2, say E = {e1,e2}, with

µ(Xe1 ) ≥ µ(Xe2 ). Note that ei = (µ|Xei )/µ(Xei ). By Dye’s Theorem, we

can find X1 ⊆ Xe1 , with µ(X1) = µ(Xe2 ) and a Borel isomorphism ϕ from

E|X1 to E|Xe2 (modulo null sets) which preserves µ. Then define F to