5. TURBULENCE OF CONJUGACY 31
Clearly D APER [E] is conjugacy invariant in [E] and, being non-
empty, it is dense in (APER [E],u) by 3.4. Finally as w u, C is in
(Aut(X, µ),u) and so D is in (APER [E],u). As D is disjoint from the
conjugacy class of T in [E], we are done.
We finally verify that if T APER [E], then T is turbulent (for the
conjugacy action of ([E],u) on (APER [E],u)).
Fix 0 and let
U = {S APER [E] : δu(S, T) }.
Fix also a nbhd V of 1 in ([E],u). It is enough to show that the local orbit
O(T, U , V )
is dense in U .
Since
{gTg−1
: g [E]} is dense in (APER [E],u), it is clearly dense
in (U , u). So fix any g [E] with
gTg−1
U . It is enough to show that
gTg−1
O(T, U , V )
u
.
Let E =

n=1
En, where E1 E2 . . . are Borel with card([x]En )
n, ∀x. Then
n
[En] is dense in ([E],u). Indeed, given S [E], if Xn = {x :
S(x) [x]En }, then X1 X2 . . . and
n
Xn = X. So for any ρ 0,
choose n large enough so that µ(Xn) 1 ρ. Then, as x Xn S(x)
[x]En , we can find S [En] such that S|Xn = S |Xn and so δu(S, S ) ρ.
So we can clearly assume that g [En], for some large enough n.
Notice that it is enough to find a continuous path λ in ([E],u),λ
[0, 1], such that g0 = 1,g1 = g and gλTgλ
−1
U , ∀λ. Because then we
can find λ0 = 0 λ1 · · · λk = 1 with gλi+1
gλi1
V, ∀i k. If
T1 = gλ1 , T2 = gλ2
gλ11,...,Tk
= gλk
gλk1
−1
, then Ti V and
TiTi−1 . . . T1TT1
−1
. . .
Ti−1
−1
Ti−1
= gλi
Tgλi1
U , ∀i k.
Choose δ0 small enough so that if α =
δu(gTg−1,T),
then α + 4δ0 .
Find then N n large enough, so that there is S [EN ] with δu(S, T) δ0.
Then δu(S,
gSg−1)
δ0 + δu(T,
gTg−1)
+ δ0 = α + 2δ0.
Claim. There is a continuous path λ in ([E],u) with g0 = 1,g1 = g
and δu(S, gλSgλ
−1)
α + 2δ0.
Then δu(T, gλTgλ
−1)
δ0 + α + 2δ0 + δ0 = α + 4δ0 , i.e., gλTgλ
−1

U , ∀λ.
Proof of the claim. Let Z be a Borel transversal for EN and let
λ be continuous from [0, 1] into MALGµ, so that Z0 =
Z1 = Z, µ(Zν \ Zλ) λ), for 0 λ ν 1. Let = [Zλ]EN , so that
µ(Xν \ Xλ) N · λ). Finally, define
=
g on Xλ,
id on X \ Xλ.
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