50 I. MEASURE PRESERVING AUTOMORPHISMS

so (see Kechris-Miller [KM], 27.7) Em = E

S1,...,Sk

for some S1,...,Sk ∈

[Em], which by 8.5 we can assume are aperiodic. But [Em] is not closed

in N[Em] since {Tn} is good for Em as well. So, by 8.4, Cµ(Em) = 1. As

m

Em = E, it follows (see Kechris-Miller [KM], 23.5) that Cµ(E) = 1.

So it only remains to give the proof of 8.2.

Proof of Lemma 8.2. I will give an argument due to Ben Miller which

is simpler than my original one.

Assume, towards a contradiction, that there is a set of positive measure

on which F = E

Tn

n

is periodic. It follows that there is a partition X =

A∪B into two F -invariant Borel sets of positive measure with F |A periodic.

By ergodicity of E, we can assume that A, B are complete sections for E.

Let {gn} be a sequence of Borel involutions such that E =

n

graph(gn).

Also let {An} be a sequence of partial Borel F -transversals (i.e., each An

meets every F -class in at most one point) such that

n

An = A. Let finally

Gn,Hn be pairwise disjoint such that Gn∪Hn = supp(gn) and gn(Gn) = Hn.

Put

Gn,m = {x ∈ Gn ∩ Am : gn(x) ∈ B},Hn,m = Gn,m ∪ gn(Gn,m)

and let

hn,m = gn|Hn,m ∪ id|(X \ Hn,m).

Then {hn,m} is a family of involutions and

xEy & x ∈ A & y ∈ B ⇔ ∃n, m(hn,m(x) = y & x = y).

Also hn,m ∈ [E] and

n,m

supp(hn,m) = X.

Claim. If z ∈ supp(hn,m) and Tkhn,mTk

−1(z)

= hn,m(z), then Tk(z) = z.

Proof of the claim. Assume z ∈ supp(hn,m) and Tkhn,mTk

−1(z)

=

hn,m(z) but Tk(z) = z (and thus Tk

−1(z)

= z as well).

Case (1). z ∈ A. Either Tk

−1(z)

∈ supp(hn,m), so Tkhn,mTk

−1(z)

=

z = hn,m(z), which is impossible or Tk

−1(z)

∈ supp(hn,m), so z = Tk

−1(z)

are F -equivalent and in supp(hn,m) ∩ A ⊆ Am, which is an F -transversal, a

contradiction.

Case (2). z ∈ B. Let hn,m(z) = y, so y ∈ A and hn,m(y) = z. As

in Case (1), Tk

−1(z)

∈ supp(hn,m). Put hn,m(Tk

−1(z))

= w. Then wFy and

w = y. But also w, y ∈ supp(hn,m) ∩ A ⊆ Am, a contradiction.

So we see that

supp(hn,m) ∩ {z : Tk(z) = z} ⊆ {z : Tkhn,mTk

−1(z)

= hn,m(z)}.

Let An,m ⊆ supp(hn,m) be pairwise disjoint with

n,m

An,m =

n,m

supp(hn,m) = X.