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 =
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 in (APER
[E],u).
Proof. Since ERG is in w u, it is clearly 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 =
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 =
X
, Γ may not be dense in
([E],u).)
Proposition 3.7. Suppose Γ Aut(X, µ) is a countable group and E =

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.
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