32 I. MEASURE PRESERVING AUTOMORPHISMS

Then λ → gλ is a continuous path in ([E],u) and g0 = 1,g1 = g. Moreover,

Xλ is gλ,S-invariant for each λ and gλ|

¯λ

X = id|

¯λ,

X where

¯λ

X = X \ Xλ.

We check that δu(S, gλSgλ

−1)

α + 2δ0. Fix A ∈ MALGµ. Then

gλSgλ

−1(A)∆S(A)

= [gλSgλ

−1(A

∩ Xλ) ∪ gλSgλ

−1(A

∩

¯λ)]∆

X

[S(A ∩ Xλ) ∪ S(A ∩

¯λ)]

X

= [gλSgλ

−1(A

∩ Xλ) ∪ S(A ∩

¯

X λ)]∆

[S(A ∩ Xλ) ∪ S(A ∩

¯

X λ)]

⊆ gλSgλ

−1(A

∩ Xλ)∆S(A ∩ Xλ)

=

[gSg−1(A)∆S(A)]

∩ Xλ,

so µ(gλSgλ

−1(A)∆S(A))

≤

µ(gSg−1(A)∆S(A))

≤

δu(gSg−1,S)

α + 2δ0

and we are done. ✷

Remark. I would like to thank Greg Hjorth for suggesting a modifi-

cation that simplified considerably my original calculation in the preceding

proof.

We next show that a variation of the proof of 5.1 allows one to show

the following stronger statement. Recall that an equivalence relation E is

E0-ergodic (or strongly ergodic) if for every Borel homomorphism π : E →

E0, (i.e., a Borel map π : X →

2N

with xEy ⇒ π(x)E0π(y)) the preimage of

some E0-class is conull. If E = EΓ

X

for a Borel action of a countable group

Γ on X and E is E0-ergodic, we will also call this action E0-ergodic. See

Jones-Schmidt [JS] and Hjorth-Kechris [HK3], Appendix A, for some basic

properties of this concept.

Theorem 5.2 (Kechris). Let E be an ergodic equivalence relation which

is not E0-ergodic. Then the conjugacy action of ([E],u) on (APER ∩ [E],u)

is turbulent.

Proof. Since E is not E0-ergodic, there is a Borel homomorphism π :

E → E0 such that the preimage of every E0-class is µ-null. Write E0 =

∞

n=1

En,E1 ⊆ E2 ⊆ . . . , En Borel with finite classes. We claim that the

set

n

[π−1(En)

∩ E] is dense in ([E],u). Indeed, if PER denotes the set of

periodic elements of Aut(X, µ), i.e.,

S ∈ PER ⇔ ∀x∃n(T

n(x)

= x),

then, by 3.3, PER∩[E] is dense in ([E],u). So fix a periodic S ∈ [E] and let

Xn = {x : π([x]S) is contained in a single En-class},

where [x]S is the S-orbit of x. Then X1 ⊆ X2 ⊆ . . . and X =

n

Xn.

Also each Xn is S-invariant and if Sn = S|Xn ∪ id|(X \ Xn), then Sn ∈

[π−1(En)

∩ E] and Sn → S uniformly.

As in the proof of 5.1, whose notation we keep below, fix T ∈ APER ∩

[E],U , and g ∈

[π−1(En)

∩ E] with

gTg−1

∈ U . Then choose δ0 small

enough so that α + 4δ0 and N ≥ n large enough so that for some

S ∈

[π−1(EN

) ∩ E],δu(S, T) δ0. It is again enough to find a continuous

path λ → gλ with g0 = 1,g1 = g and δu(S, gλSgλ

−1)

α + 2δ0.