Number Theory: An Introduction to Mathematics

3 Birkhoff’s Ergodic Theorem 471

Suppose now that (i) holds and let f∈L(X,B,μ). Then the functionf∗in the
statement of Theorem 17 must be constant a.e. Moreover, ifγis its constant value, we
must have

γ=

∫

X

f∗dμ=

∫

X

fdμ.

Thus (i) implies (ii).
Suppose next that (ii) holds and letA,B∈B. Then, for almost allx∈X,

lim

n→∞

n−^1

n∑− 1

k= 0

χA(Tkx)=

∫

X

χAdμ=μ(A).

Hence, for almost allx∈X,

lim

n→∞

n−^1

∑n−^1

k= 0

χA(Tkx)χB(x)=μ(A)χB(x)

and so, by the dominated convergence theorem,

μ(A)μ(B)=

∫

X

lim

n→∞

n−^1

n∑− 1

k= 0

χA(Tkx)χB(x)dμ(x)

= lim

n→∞

n−^1

n∑− 1

k= 0

∫

X

χA(Tkx)χB(x)dμ(x)

= lim

n→∞

n−^1

n∑− 1

k= 0

μ(T−kA∩B).

Thus (ii) implies (iii).

⋃Suppose now that (iii) holds and chooseC ∈ Bwithμ(C)>0. PutA =
n≥ 0 T

−nCandB=(⋃

n≥ 1 T

−nC)c. Then, for everyk≥1,T−kA⊆⋃

n≥ 1 T

−nC

and henceμ(T−kA∩B)=0. Thus

n−^1

n∑− 1

k= 0

μ(T−kA∩B)=μ(A∩B)/n→0asn→∞.

Sinceμ(A)≥μ(C)>0, it follows from (iii) thatμ(B)=0. Thus (iii) implies (iv).
Next choose anyA,B∈Bsuch thatμ(A)>0,μ(B)>0. If (iv) holds, then
μ(

⋃

n≥ 1 T

−nA)=1 and hence

μ(B)=μ

(

B∩∪

n≥ 1

T−nA

)

=μ

(

∪

n≥ 1

(B∩T−nA)

)

.

Sinceμ(B)>0, it follows thatμ(B∩T−nA)>0forsomen>0. Thus (iv) implies
(v).
Finally chooseA∈BwithT−^1 A=Aand putB=Ac. Then, for everyn≥1,
we haveμ(T−nA∩B)=μ(A∩B)=0. If (v) holds, it follows that eitherμ(A)= 0
orμ(B)=0. Hence (v) implies thatTis ergodic.