32 I. MEASURE PRESERVING AUTOMORPHISMS
Then λ is a continuous path in ([E],u) and g0 = 1,g1 = g. Moreover,
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 =
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 with g0 = 1,g1 = g and δu(S, gλSgλ
−1)
α + 2δ0.
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