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)}
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