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)},