3. FULL GROUPS OF EQUIVALENCE RELATIONS 17
Otherwise, find (S0,T0) (APER
[E])2,
0 such that for every
(S, T)
(APER∩[E])2
with δu(S, S0) , δu(T, T0) , we have w(S, T) =
1, i.e., w(S, T)(x) = x, µ−a.e. (x). Let w1,w2,...,wn = w be the co-initial
(non-∅) subwords of w (e.g., if w =
xy2x−1,w1
=
x−1,w2
=
yx−1,w3
=
y2x−1,w4
=
xy2u−1
= w). Clearly n 2.
Lemma 3.10. If P1,...,Pk Aut(X, µ), and on a Borel set of positive
measure A and every x A we have that P1(x),P2P1(x),...,Pk · · · P1(x)
are distinct, then there is B A of positive measure such that the sets
P1(B),P2P1(B),...,Pk · · · P1(B) are pairwise disjoint.
Proof. We can assume that X is a Polish space, A X is clopen
and P1,...,Pk are homeomorphisms. Fix a countable basis {Un} for X.
Then every point x A is contained in a basic open set Un(x) such that
P1(Un(x)),...,Pk · · · P1(Un(x)) are pairwise disjoint. As A {Un(x) : x
A}, for some x A, B = Un(x) A has positive measure.
For each T Aut(X, µ), let ET be the equivalence relation induced by
T,
xET y ∃n Z(T
n(x)
= y)
and let
[T] = [ET ].
Lemma 3.11. Given aperiodic T Aut(X, µ) and a Borel set with µ(A)
0, there is a Borel set A A with µ(A ) 0 and aperiodic T [T] such
that T(x) = T (x),µ-a.e. x A , T(x) = T (x),µ-a.e. x A .
Proof. We can assume that A intersects every orbit of T and, by
Poincar´ e recurrence, that for (almost) every x A there are k, 0 with
T
k(x),T
(x) A. We can also assume that if B = X \ A, then µ(B) 0
and ∀x B∃k, 0(T
k(x)
B, T

(x) B) (since if µ(B) = 0 we can
take A = A = X, T (x) = T
2(x)).
Then we can find A A [B]T ,
where [B]T denotes the T-saturation of B, such that [A ]T = [B]T , so that
µ(A ) 0, ∀x A ∃k 0∃ 0(T
k(x),T
(x) A ) and finally x A
T(x) A . Then we put T (x) = T(x) if x A , and T (x) = T(T
k(x)),
where k 0 is least with T k(x) A , if x A .
For convenience, in the rest of the argument, we will write, for any
reduced word v(x, y),
vS,T
instead of v(S, T). Continuing the proof, let
0 k0 n be least such that on a set of positive measure A0 we have
x A0 x, w1
S0,T0
(x),...,wk0−1
S0,T0
(x0) are distinct and
wk0
S0,T0
(x) {x,...,wk0−1
S0,T0
(x)}.
Clearly k0 2 and actually
wk0
S0,T0
(x) {x,...,wk0−2
S0,T0
(x)}
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