42 I. MEASURE PRESERVING AUTOMORPHISMS

∑

iM

µ(Ai) . Then for some N and all n ≥ N,

µ(Tn(Ai)∆Ai)

(M + 1)

, ∀i ≤ M,

TnγiTn

−1(x)

= γi(x), ∀i ≤ M,

for all x ∈ An,

i

where µ(An)

i

(M+1)

, i ≤ M. Thus off a set of measure

2 , we have that if i ≤ M and x ∈ Ai, then Tn

−1(x)

∈ Ai and

TTn

−1(x)

= γi(Tn

−1(x))

= Tn

−1(γi(x))

= Tn

−1T

(x).

Thus µ({x : TTn

−1(x)

= Tn

−1T

(x)}) 3 , if n ≥ N, so δu(TTn

−1,Tn −1T

) =

δu(TnTTn

−1,T

) → 0. ✷

Thus the following is also a complete metric for N[E]:

¯

δ N[E](T1,T2) =

¯w(T1,T2)

δ +

n

2−nδu(T1γnT1 −1,T2γnT2 −1),

where Γ = {γn}.

Remark. It is clear that in the preceding one can equivalently use δu

instead of δu.

(B) There is also another way to understand the topology of N[E].

Consider the measure M on E defined by

M(A) = card(Ax)dµ(x),

where Ax = {y : (x, y) ∈ A}, for any Borel set A ⊆ E. This is a σ-finite

Borel measure on E.

Given now T ∈ N[E] consider the map T × T on E defined by

T × T(x, y) = (T(x),T(y)).

Then it is easy to see that T ×T preserves the measure M, therefore induces

a unitary operator on the Hilbert space

L2(E,

M)

denoted by UT

×T

:

UT

×T

(f)(x, y) = f(T

−1(x),T −1(y)).

The map T → UT

×T

is a group isomorphism between N[E] and a sub-

group of

U(L2(E,

M)). (To see that it is 1-1, assume that UT

×T

= 1, i.e.,

f(T

−1(x),T −1(y))

= f(x, y), M-a.e., for all f ∈

L2(E,

M). Then for any

g ∈

L2(X,

µ), if

f(x, y) =

g(x), if x = y,

0, otherwise,

then f ∈

L2(E,

M) and g(T

−1(x))

= f(T

−1(x),T −1(x))

= f(x, x) = g(x), µ-

a.e., so T = 1.)

Now notice that UTn×Tn → 1 in

U(L2(E,

M)) ⇒ Tn → 1 in N[E]. This

is because for any T ∈ N[E],S ∈ [E], if f is the characteristic function of