12.8. THE ERGODIC THEOREM 305

Thus, T nN = N for all n ∈ N. For ω /∈ N, limn→∞1n Sn f (ω) exists. Now let

g(ω)≡{

0 if ω ∈ Nlimn→∞

1n Sn f (ω) if ω /∈ N

.

Then it is clear g satisfies the conditions of the theorem because if ω ∈ N, then T ω ∈ Nalso and so in this case, g(T ω) = g(ω)≡ 0. On the other hand, if ω /∈ N, then

g(T ω) = limn→∞

1n

Sn f (T ω) = limn→∞

1n

Sn f (ω) = g(ω) .

Which shows that g is invariant. Also, from Lemma 12.8.2,∫Ω

|g|dµ ≤ lim infn→∞

∫Ω

∣∣∣∣1nSn f∣∣∣∣dµ ≤ lim inf

n→∞

1n

n

∑k=1

∫Ω

∣∣∣ f (T k−1ω

)∣∣∣dµ

= lim infn→∞

1n

n

∑k=1

∫Ω

| f (ω)|dµ = ∥ f∥L1

so g ∈ L1 (Ω,µ).The last claim about convergence in L1 follows from the Vitali convergence theorem

if we verify the sequence,{ 1

n Sn f}∞

n=1 is uniformly integrable. To see this is the case, weknow f ∈ L1 (Ω) and so if ε > 0 is given, there exists δ > 0 such that whenever B ∈F andµ (B)≤ δ , then |

∫B f (ω)dµ|< ε . Taking µ (A)< δ , it follows∣∣∣∣∫A

1n

Sn f (ω)dµ

∣∣∣∣=∣∣∣∣∣1n n

∑k=1

∫A

f(

T k−1ω

)dµ

∣∣∣∣∣=∣∣∣∣∣1n n

∑k=1

∫Ω

XA (ω) f(

T k−1ω

)dµ

∣∣∣∣∣=

∣∣∣∣∣1n n

∑k=1

∫Ω

XA

(T k−1T−(k−1)

ω

)f(

T k−1ω

)dµ

∣∣∣∣∣=

∣∣∣∣∣1n n

∑k=1

∫Ω

XA

(T−(k−1)

ω

)f (ω)dµ

∣∣∣∣∣=

∣∣∣∣∣1n n

∑k=1

∫T k−1(A)

f (ω)dµ

∣∣∣∣∣≤ 1n

n

∑k=1

∣∣∣∣∫T k−1(A)f (ω)dµ

∣∣∣∣< 1n

n

∑k=1

ε = ε

because µ

(T (k−1)A

)= µ (A) by assumption. This proves the above sequence is uniformly

integrable and so, by the Vitali convergence theorem,

limn→∞

∫Ω

∣∣∣∣1nSn f −g∣∣∣∣dµ = 0.

This proves the theorem in the case the function has real values. In the case where f hascomplex values, apply the above result to the real and imaginary parts of f .