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)}