Number Theory: An Introduction to Mathematics

3 Birkhoff’s Ergodic Theorem 467

Put

f ̄(x)= lim

n→∞

n−^1

n∑− 1

k= 0

f(Tkx), f(x)= lim

n→∞

n−^1

n∑− 1

k= 0

f(Tkx).

Thenf ̄andfareμ-measurable functions since, for any sequence(gn),

lim

n→∞

gn(x)=inf

m

(sup

n≥m

gn(x)), lim

n→∞

gn(x)=sup

m

(inf

n≥m

gn(x)).

Moreoverf ̄(x)=f ̄(Tx),f(x)=f(Tx)for everyx∈X,since

(n+ 1 )−^1

∑n

k= 0

f(Tkx)=(n+ 1 )−^1 f(x)+( 1 + 1 /n)−^1 n−^1

n∑− 1

k= 0

f(Tk+^1 x).

It is sufficient to show that
∫

X

f ̄dμ≤

∫

X

fdμ≤

∫

X

fdμ.

For then, sincef≤f ̄, it follows thatf ̄(x)= f(x)=f∗(x)forμ-almost allx∈X
and
∫

X

f∗dμ=

∫

X

fdμ.

Fix someM>0 and define the ‘cut-off’ functionf ̄Mby

f ̄M(x)=min{M,f ̄(x)}.

Thenf ̄Mis bounded andf ̄M(Tx)=f ̄M(x)for everyx∈X. Fix also anyε>0. By
the definition off ̄(x), for eachx∈Xthere exists a positive integernsuch that

f ̄M(x)≤n−^1

∑n−^1

k= 0

f(Tkx)+ε. (∗)

Thus ifFnis the set of allx ∈Xfor which (∗) holds and ifEn=

⋃n
k= 1 Fk,then
E 1 ⊆E 2 ⊆ ···andX=

⋃

n≥ 1 En. Since the setsEnareμ-measurable, we can
chooseNso large thatμ(EN)> 1 −ε/M.
Put

f ̃(x)=f(x) ifx∈EN,

=max{f(x),M} ifx∈/EN.

Also, letτ(x)be the least positive integern≤Nfor which (∗) holds ifx∈EN,and
letτ(x)=1ifx∈/EN.Sincef ̄MisT-invariant, (∗) implies

n∑− 1

k= 0

f ̄M(Tkx)≤

n∑− 1

k= 0

f(Tkx)+nε