8. COSTS AND THE OUTER AUTOMORPHISM GROUP 51
Then
{z An,m : Tk(z) = z} {z An,m : Tkhn,mTk
−1(z)
= hn,m(z)},
thus
δu(Tk, 1)
n,m
µ(An,m {z : Tkhn,mTk
−1(z)
= hn,m(z)}).
Fix 0 and find a finite set F
N2
such that
(n,m)∈F
µ(An,m)
and then K0 large enough so that for all (n, m) F ,
k K0 µ({z : Tkhn,mTk
−1(z)
= hn,m(z)})
card(F )
.
Then for such k,
δu(Tk, 1) + card(F ) ·
card(F )
= 2.
Thus δu(Tk, 1) 0, a contradiction.
Remark. Note that if δu(Tn, 1) 0, then there is a subsequence {Tni }
such that E
Tni
i
is not aperiodic (in fact it is equality on a set of posi-
tive measure). To this, find {ni} so that

i
δu(Tni , 1) ∞. Then if
Bni = supp(Tni ),
∑see
i
µ(Bni ) ∞, so, by Borel-Cantelli, µ({x : ∀i∃j i(x
Bnj )}) = 0, or (neglecting as usual null sets)
∀x∃i∀j i(Tnj (x) = x).
Then for some set B of positive measure and some i,
x B ∀j i(Tnj (x) = x),
i.e., E
Tnj
j≥i
|B = equality|B.
Thus we see that, given any sequence {Tn} E with Tn 1 in N[E],
the following are equivalent:
(i) 0∀n(δu(Tn, 1) ) (i.e., {Tn} is good),
(ii) For every subsequence {ni},E
Tni i
is aperiodic.
Remark. By definition, [E] not closed in N[E] means that there is a
sequence {Tn} [E] with δu(Tn, 1) 0 but Tn 1 in N[E]. We note here
that if E is actually E0-ergodic, then, if [E] is not closed in N[E], there is
such {Tn} with δu(Tn, 1) = 1, i.e., supp(Tn) = X.
Indeed we have some {Sn} [E] and 0 with δu(Sn, 1) and
Sn 1 in N[E]. Let An = {x : Sn(x) = x}, so that µ(X \ An) .
If δu(Sn, 1) 1, then we can assume that for some δ 0,δu(Sn, 1)
1 δ, so µ(An) δ. Thus µ(An)(1 µ(An)) 0.
Fix T [E]. If
Bn = {x : T
−1Sn(x)
= SnT
−1(x)},
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