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 from [0, 1]
to MALGρ with A0 = ∅,A1 = A and 0 λ λ 1 and
ρ(Aλ \ Aλ) λ λ. Let =
π−1([Aλ]EN
), so that µ(Xλ) = ρ(Aλ), thus
λ is continuous from [0, 1] to MALGµ,X0 = ∅,X1 = X, 0 λ λ
1 and µ(Xλ \Xλ) λ −λ. Moreover is
π−1(EN
)-invariant,
so g, S-invariant. Then define as before = g|Xλ id|(X \ Xλ), so that
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
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