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.
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