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 → Tσ 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.