38 I. MEASURE PRESERVING AUTOMORPHISMS
If σ, τ are symmetric Rajchman (and thus non-atomic) probability mea-
sures supported by Λ
¯
Λ for the same independent set Λ, then Tσ,Tτ are
mixing and
σ τ Tσ,Tτ are unitarily equivalent.
As σ is clearly Borel, we have shown that restricted to symmetric
Rajchman probability measures supported by Λ∪
¯
Λ, for a fixed independent
set Λ, is Borel reducible to unitary equivalence on MIX. It only remains to
show that can be Borel reduced to restricted to such measures.
We will use here the following result of Rudin [Ru1] (see also Kahane-
Salem [KS], VIII, 3, Th´ eor` eme II): There is a closed set Λ T which is inde-
pendent and supports a Rajchman measure, i.e., Λ is in the class M0 of closed
sets of restricted multiplicity (see Kechris-Louveau [KL]). By a theorem of
Kaufman (see Kechris-Louveau [KL], VII. 1, Theorem 7), there is a Borel
function x σx from
2N
to the standard Borel space of probability measures
on T such that each σx is a Rajchman measure, supp(σx) supp(σy) = if
x = y (where supp(σ) = T \ {U open : σ(U) = 0}), and supp(σx) Λ.
Then, since Λ is independent, we have A∩B = ∅, for disjoint A, B Λ, so, by
replacing σx by
1
2
(σx + ¯x) σ (where ¯(A) σ = σ(
¯)),
A we have the following con-
clusion: There is a Borel map x σx from
2N
to symmetric Rajchman prob-
ability Borel measures with supp(σx) Λ
¯
Λ and supp(σx) supp(σy) =
if x = y.
Now consider an arbitrary probability Borel measure µ on
2N.
Put
σµ = σxdµ(x)
i.e., σµ(A) = σx(A)dµ(x), for each Borel set A T, or equivalently
fdσµ = (fdσx)dµ(x) for every continuous function f on T. It follows
that
ˆµ(n) σ = ˆx(n)dµ(x) σ
and since ˆx(n) σ = fn(x) 0 as |n| and |fn(x)| 1 we have, by
Lebesgue Dominated Convergence, that ˆµ(n) σ 0, i.e., σµ is a Rajchman
measure. Clearly σµ is symmetric and σµ is supported by Λ
¯
Λ. It only
remains to verify that
µ ν σµ σν.
Assume µ ν and let A T be Borel with σν(A) = 0, so that
σx(A)dν(x) = 0.
Then σx(A) = 0,ν−a.e.(x), thus σx(A) = 0,µ−a.e.(x), therefore σµ(A) = 0.
Thus σµ σν. Conversely, assume that σµ σν and let X
2N
be Borel
with ν(X) = 0. Let
X∗
=
x∈X
supp(σx), which is a Borel set in T. Then
σν(X∗)
=
σx(X∗)dν(x)
= ν(X) = 0. Thus
σµ(X∗)
=
σx(X∗)dµ(x)
= 0,
so
σx(X∗)
= 0,µ−a.e.(x), therefore x X, µ−a.e.(x), i.e., µ(X) = 0. So
µ ν.
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