4 I. MEASURE PRESERVING AUTOMORPHISMS
so µ(Ai \ S(Ai)) µ(Ai) δ, ∀i n. Then δu(S, 1)
∑n
i=1
µ(Ai \ S(Ai))

n
i=1
µ(Ai) .
It follows that both δu and δu are lower-semicontinuous in the space
(Aut(X,
µ),w)2.
Let us finally observe that similar calculations show that the map
T supp(T)
Aut(X, µ) MALGµ
is continuous in (Aut(X, µ),u) and Baire class 1 in (Aut(X, µ),w). This
is clear for u. For the weak topology, we need to show that for each A
MALGµ, 0, the set {T : µ(supp(T)∆A) } = F is Fσ. Note that
T F ∃α, β[α, β Q and α, β 0 and α + β and
µ({x A : T(x) = x}) α and
µ({x A : T(x) = x}) β].
Now the previous argument also shows that for each γ 0,B MALGµ,
{T : µ({x B : T(x) = x}) γ}
is open in w. It follows that
{T : µ({x A : T(x) = x}) α}
is closed in w and since µ({x A : T(x) = x}) = µ(A) µ({x A : T(x) =
x}), the set
{T : µ({x A : T(x) = x}) β}
is open in w, thus F is in w.
Remark. One could also consider the operator norm topology on the
group Aut(X, µ) induced by the identification T UT . However, it is
easy to see that this is the discrete topology. We will show that T = 1
UT I 1 (where I is the identity operator on
L2(X,
µ)). Since T = 1
there is a set of positive measure on which T(x) = x, so there is a set A
of positive measure, say , such that T(A) A = ∅. Let f
L2(X,
µ) be
equal to
1

on A and 0 otherwise. Thus f = 1. Also UT (f) f
2
=
|f(T
−1(x))

f(x)|2dµ(x)

x∈A
|f(T
−1(x))

f(x)|2dµ(x)
= 1.
Convention. When we consider Aut(X, µ) as a topological group with-
out explicitly indicating which topology we use, it will be assumed that it is
equipped with the weak topology.
Comments. The basic facts about the topologies w, u can be found
in Halmos [Ha]. Note however that Halmos uses δu for our δu (and vice
versa). Since we more often use (our) δu we decided to leave it unprimed.
For the characterizations of the unitary operators UT , T Aut(X, µ), see,
e.g., Fleming-Jamison [FJ], p. 73, Glasner [Gl2], p. 366, and Walters [Wa],
Ch. 2.
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