6 I. MEASURE PRESERVING AUTOMORPHISMS

j∈B
µ(T(Dj)) =
M
2N
+

j∈B
µ(Dj)
M
2N
+
2N
−M
2N
= 1, a contradiction.
Similarly reversing the roles of i, j.
In particular, 2.1 implies that (Aut(X, µ),w) is topologically locally finite,
i.e., has a locally finite countable dense subgroup.
We also have the following result which is a consequence of the Rokhlin
Lemma.
Theorem 2.2 (Uniform Approximation Theorem, Rokhlin, Halmos
[Ha]). If T Aut(X, µ) is aperiodic, then for each N 1, 0 there is a
periodic S Aut(X, µ) of period N such that δu(S, T)
1
N
+ .
Consider now the set APER of all aperiodic elements of Aut(X, µ). Then
as
T APER ∀n (δu(T
n,
1) = 1) ∀n∀m δu(T
n,
1) 1
1
m
,
APER is in (Aut(X, µ),w). (It is also clearly closed in (Aut(X, µ),u).)
In fact the following holds.
Proposition 2.3. APER is dense in (Aut(X, µ),w).
Proof. Take X =
2N
with the usual measure µ. Then the sets of the
form
k
i=1
{T : dµ(T(Di),T0(Di)) },
Di finite unions of basic nbhds, T0 Aut(X, µ), 0 form a nbhd basis
for T0 in w. Thus it is enough for each dyadic permutation π, of rank say
n, to find T APER such that for each s
2n,T(Ns)
= Tπ(Ns) = Nπ(s).
Let ϕ :
2N

2N
be any aperiodic element of Aut(X, µ), e.g., the odometer:
ϕ(1nˆ0ˆx)
=
0nˆ1ˆx, ϕ(1∞)
=
0∞.
Define then T(sˆx) = π(s)ˆϕ(x). Clearly
this works.
Moreover, using the Uniform Approximation Theorem, one obtains the
following result.
Theorem 2.4 (Conjugacy Lemma, Halmos [Ha]). Let T APER.
Then its conjugacy class
{STS−1
: S Aut(X, µ)} is uniformly dense in
APER. Therefore its conjugacy class is weakly dense in Aut(X, µ). Con-
versely, if the conjugacy class of T is weakly dense, T APER.
Proof. Observe that if U, V are periodic of period N, then they are
conjugate. Indeed let A X be a Borel set that meets (almost) every orbit
of U in exactly one point and let B X be defined similarly for V . Then
µ(A) = µ(B) =
1
N
. Thus there is S0 Aut(X, µ) with S0(A) = B. Define
now S Aut(X, µ) by
S(x) = S0(x), if x A,
S(U
n(x))
= V
n(S0(x)),
if x A, 0 n N.
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