24 I. MEASURE PRESERVING AUTOMORPHISMS

Clearly Φ is a bijection of MALGµ with itself with inverse

Φ−1(B)

= supp(f

−1(T

)),

where T is an involution with support B. Also if supp(T) = A, supp(T ) =

B,

A ⊆ B ⇔ supp(T) ⊆ supp(T )

⇔ ⊥(T ) ⊆ ⊥(T)

⇔ ⊥(f(T )) ⊆ ⊥(f(T))

⇔ supp(f(T)) ⊆ supp(f(T ))

⇔ Φ(A) ⊆ Φ(B).

Thus Φ is an automorphism of the Boolean algebra MALGµ (not necessarily

the measure algebra MALGµ), so it is given by a (unique modulo null sets)

Borel automorphism ϕ : X → X which is non-singular, i.e., preserves null

sets, via

Φ(A) = ϕ(A).

We now want to verify that actually ϕ is measure preserving, i.e., ϕ ∈

Aut(X, µ), and moreover that f(S) =

ϕSϕ−1,

∀S ∈ [E].

Lemma 4.9. For each S ∈ [E],f(S) =

ϕSϕ−1.

Granting this we verify that ϕ ∈ Aut(X, µ). Since ϕ is non-singular,

there is measurable r : X →

R+

with µ(ϕ(A)) =

A

rdµ, ∀A ∈ MALGµ. We

will show that r is actually constant, r = k. Then µ(ϕ(A)) = kµ(A), so,

putting A = X, k = 1, thus ϕ ∈ Aut(X, µ).

If r is not constant, there are two disjoint sets of positive measure A, B

with 0 ≤ r(x) λ r(y), for x ∈ A, y ∈ B. Let C ⊆ A, D ⊆ B be such

that µ(C) = µ(D) 0 and let K = ϕ(C). Let also U ∈ [E] be such that

U(C) = D. Then µ(ϕ(D)) =

D

rdµ λµ(D) = λµ(C),µ(K) = µ(ϕ(C)) =

C

rdµ λµ(C) µ(ϕ(D)). So µ(K) µ(ϕ(D)) and ϕ(D) =

ϕUϕ−1(K),

so µ(K)

µ(ϕUϕ−1(K)),

thus

ϕUϕ−1

∈ Aut(X, µ), contradicting that

ϕUϕ−1

= f(U) ∈ [F ] ⊆ Aut(X, µ).

Proof of Lemma 4.9. As [E] is generated by involutions, it is enough

to prove this for S = T, an involution. Notice that if T ∈ [E] is an involution,

then

supp(ϕTϕ−1)

= ϕ(supp(T)) = Φ(supp(T)) = supp(f(T)),

i.e.,

ϕTϕ−1

and f(T) have the same support A. Assume, towards a contra-

diction, that

B = {x ∈ A :

ϕTϕ−1(x)

= f(T)(x)}