3. FULL GROUPS OF EQUIVALENCE RELATIONS 15

Theorem 3.5. If E is ergodic (resp., aperiodic), then there is T ∈ [E] which

is ergodic (resp., aperiodic).

Proof. (i) We first give the proof for the aperiodic case, which works in

the pure Borel context (with no measures present). The proof below is due

to Dougherty.

Let E be a countable aperiodic Borel equivalence relation on a standard

Borel space X. Let be a Borel linear order on X and let a countable

group Γ act in a Borel way on X so that E = EΓ

X

. Put Γ = {γn}. Let

A = {x : [x]E has a largest element in }. Then E|A admits a Borel

transversal, so clearly there is an aperiodic Borel automorphism of A, say

TA, with TA(x)Ex, ∀x ∈ A. Let now B = X \ A. Define U : B → B

by U(x) = γn(x), where n is least with γn(x) x. Consider then the

equivalence relation on B:

xEU y ⇔ ∃m∃n(U

n(x)

= U

m(y)).

There is a Borel automorphism of B, say TB, that induces EU (see Dougherty-

Jackson-Kechris [DJK], 8.2). Clearly TB(x)Ex, ∀x ∈ B, and TB is aperiodic.

So T = TA ∪ TB works.

(ii) For the ergodic case, see Zimmer [Zi1], 9.3.2 or Kechris [Kec1], 5.66.

Alternatively, one can go over the proof of Dye’s Theorem, 3.13 below, given

in Kechris-Miller [KM], 7.13, and notice that if condition (4) is omitted in

that argument, one obtains a proof that every ergodic E contains an ergodic

subequivalence relation isomorphic to E0, defined before 3.8 below, which is

thus generated by an ergodic automorphism. ✷

As an immediate consequence, we obtain the next result.

Theorem 3.6. If E is ergodic, then ERG∩[E] is dense Gδ in (APER ∩

[E],u).

Proof. Since ERG is Gδ in w ⊆ u, it is clearly Gδ in u. Density follows

from the two preceding results. ✷

Similarly (using also 3.14 below) one can show the same fact for the set

of weak mixing T ∈ [E].

It will be useful, given a countable group Γ ≤ Aut(X, µ) and letting

E = EΓ

X

= {(x, y) : ∃γ ∈ Γ(γ(x) = y)}, to have a criterion for a countable

subgroup ∆ ≤ [E] to be dense in ([E],u). Note that if ∆ ≤ [E] is dense

in ([E],u), then E = E∆

X

. (However, if E = EΓ

X

, Γ may not be dense in

([E],u).)

Proposition 3.7. Suppose Γ ≤ Aut(X, µ) is a countable group and E =

EΓ

X

. Let also A ⊆ MALGµ be a countable Boolean algebra dense in MALGµ

and closed under Γ. Let ∆ ≤ [E] be a countable subgroup of [E]. Then ∆

is dense in ([E],u) provided that for every 0 and every A0,...,An−1 ∈

A,γ0,...,γn−1 ∈ Γ with

{Ai}i=01, n− {γi(An)}i=01 n−

each pairwise disjoint, there

is δ ∈ ∆ such that µ({x ∈ Ai : γi(x) = δ(x)}) , i = 0,...,n − 1.