Proof:Let f ≥ 0 and f is measurable. Let A ∈ℱ. Then from 10.8.19,
∫ ∫ ∫
XA (ω )dμ = μ (A) = μ (T−1(A)) = X −1 (ω)dμ = XA (T (ω))dμ.
Ω Ω T (A) Ω
It follows that whenever s is a simple function,
∫ ∫
s(ω)dμ = s(Tω )dμ
If f ≥ 0 and measurable, Theorem 9.3.9 on Page 644, implies there exists an increasing
sequence of simple functions,
{sn}
converging pointwise to f. Then the result follows
from monotone convergence theorem. Splitting f ∈ L1 into real and imaginary parts we
apply this to the positive and negative parts of these and obtain 10.8.20 in this case also.
■
Definition 10.8.3A measurable function f, is said to be invariant if
f (Tω ) = f (ω).
A set, A ∈ℱ is said to be invariant if XAis an invariant function. Thus a set isinvariant if and only if T−1A = A. (XA
(Tω )
= XT−1
(A)
(ω)
so to say that XAisinvariant is to say that T−1A = A.)
The following theorem, the individual ergodic theorem, is the main result. Define
T0
(ω)
= ω. Let
n
Snf (ω) ≡ ∑ f (T k− 1ω ), S0f (ω) ≡ 0.
k=1
Also define the following maximal type function M∞f
(ω)
M ∞f (ω) ≡ sup{Skf (ω ) : 0 ≤ k} (10.8.21)
(10.8.21)
and let
Mnf (ω) ≡ sup{Skf (ω) : 0 ≤ k ≤ n} (10.8.22)
(10.8.22)
Then one can prove the following interesting lemma.
Lemma 10.8.4Let f ∈ L1
(μ)
where f has real values. Then∫
[M∞f>0]
fdμ ≥ 0.
Proof: First note that Mnf
(ω)
≥ 0 for all n and ω. This follows easily from the
observation that by definition, S0f
(ω )
= 0 and so Mnf
(ω)
is at least as large. There is
certainly something to show here because the integrand is not known to be nonnegative.
The integral involves f not M∞f.
Let T∗h ≡ h ∘ T. Thus T∗ is linear and maps measurable functions to measurable
functions by Lemma 10.8.1. It is also clear that if h ≥ 0, then T∗h ≥ 0 also. Therefore, for
large k ≤ n,
k k
S f (ω) ≡ ∑ f (T j− 1ω ) = f (ω )+ ∑ f (Tj−1ω)
k j=1 j=2
k−1
= f (ω)+ T ∗∑ f (T j− 1ω) (factored out T∗)
j=1
∗ ∗
= f (ω)+ T Sk−1f (ω) ≤ f (ω)+ T Mnf