Topics In Analysis

10.8 The Ergodic Theorem

I am putting this theorem here because it seems to fit in well with the material of this chapter.

In this section

will be a finite measure space. This means that μ

< ∞. The mapping, T : Ω → Ω will satisfy the following condition.

T (A),T- 1(A ) \in ℱ whenever A \in ℱ, T is one to one. (10.8.18)

(10.8.18)

For example, you could have T a homeomorphism on some topological space X and the σ algebra could be the Borel sets.

Lemma 10.8.1 If T satisfies 10.8.18, then f ∘ T is measurable whenever f is measurable.

Proof: Let U be an open set. Then

(f \circT)-1 (U ) = T- 1(f-1(U)) \in ℱ

by 10.8.18. ■

Now suppose that in addition to 10.8.18, T also satisfies

( ) μ T -1A = μ(A), (10.8.19)

(10.8.19)

for all A ∈ℱ. In words, T⁻¹ is measure preserving. Note that also

( -1 ) μ(TA ) = μ T TA = μ(A )

so also T is measure preserving. Then for T satisfying 10.8.18 and 10.8.19, we have the following simple lemma.

Lemma 10.8.2 If T satisfies 10.8.18 and 10.8.19 then whenever f is nonnegative and measurable,

\int \int f (ω)dμ = f (T ω)dμ. (10.8.20) Ω Ω

(10.8.20)

Also 10.8.20 holds whenever f ∈ L¹

Proof: Let f ≥ 0 and f is measurable. Let A ∈ℱ. Then from 10.8.19,

\int \int \int 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,

\int \int 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,

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 10.8.3 A measurable function f, is said to be invariant if

f (Tω ) = f (ω).

A set, A ∈ℱ is said to be invariant if X_A is an invariant function. Thus a set is invariant if and only if T⁻¹A = A. (X_A

= X_T⁻¹(A)

so to say that X_A is invariant is to say that T⁻¹A = A.)

The following theorem, the individual ergodic theorem, is the main result. Define T⁰

= ω. Let

n Snf (ω) \equiv \sum f (T k- 1ω ), S0f (ω) \equiv 0. k=1

Also define the following maximal type function M_∞f

M \inftyf (ω) \equiv sup{Skf (ω ) : 0 \leq k} (10.8.21)

(10.8.21)

and let

Mnf (ω) \equiv sup{Skf (ω) : 0 \leq k \leq n} (10.8.22)

(10.8.22)

Then one can prove the following interesting lemma.

Lemma 10.8.4 Let f ∈ L¹

where f has real values. Then ∫ _[M∞f>0]fdμ ≥ 0.

Proof: First note that M_nf

≥ 0 for all n and ω. This follows easily from the observation that by definition, S₀f

= 0 and so M_nf

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) \equiv \sum f (T j- 1ω ) = f (ω )+ \sum f (Tj-1ω) k j=1 j=2 k-1 = f (ω)+ T *\sum f (T j- 1ω) (factored out T*) j=1 * * = f (ω)+ T Sk-1f (ω) \leq f (ω)+ T Mnf

and so, taking the supremum for k ≤ n,

* Mnf (ω) \leq f (ω)+ T Mnf (ω).

Now since M_nf ≥ 0,

\int \int M f (ω)dμ = M f (ω)dμ Ω n [Mnf >0] n

\int \int \leq f (ω)dμ+ T*Mnf (ω) dμ [Mnf >0] Ω

\int \int = f (ω)dμ + Mnf (ω)dμ [Mnf >0] Ω

by Lemma 10.8.2. It follows that

\int f (ω)dμ \geq 0 [Mnf>0]

for each n. Also, since M_nf

→ M_∞f

, the following pointwise convergence holds.

X [Mnf>0](ω)f (ω ) \to X [M \inftyf>0](ω )f (ω)

Since f is in L¹, the dominated convergence theorem implies

\int \int f (ω)dμ = lni\tom\infty f (ω)dμ \geq 0.■ [M \inftyf>0] [Mnf>0]

Theorem 10.8.5 Let

be a probability space and let T : Ω → Ω satisfy 10.8.18 and 10.8.19, T⁻¹ is measure preserving and T⁻¹ maps ℱ to ℱ and T is one to one. Then if f ∈ L¹

having real or complex values and

\sumn Snf (ω ) \equiv f (T k-1ω),S0f (ω) \equiv 0, (10.8.23) k=1

(10.8.23)

it follows there exists a set of measure zero N, and an invariant function g such that for all ω

1- nl\toim\infty nSnf (ω) = g (ω ). (10.8.24)

(10.8.24)

and also

1 1 ln\toim\infty nSnf = g in L (Ω)

Proof: To begin with, we assume f has real values. Now if A is an invariant set, X_A

= X_A

and so

\sumn ( k- 1 ) ( k-1 ) n\sum ( k-1 ) Sn (XAf)(ω) \equiv f T ω XA T ω = f T ω XA (ω ) k=1 k=1

\sumn ( k-1 ) = XA (ω) f T ω = XA (ω )Snf (ω). k=1

Therefore, for such an invariant set,

Mn (XAf )(ω ) = XA (ω)Mnf (ω),M \infty (XAf)(ω) = XA (ω )M \inftyf (ω ). (10.8.25)

(10.8.25)

Let −∞ < a < b < ∞ and define

[ ] N \equiv - \infty < lim inf -1S f (ω) < a < b < lim sup 1-S f (ω) < \infty (10.8.26) ab n\to\infty n n n\to\infty n n

(10.8.26)

Observe that from the definition,

1- 1- lim nin\tof\infty n Snf (ω) = lim inn\tof\infty nSnf (Tω )

and

1 1 lim sup -Snf (ω) = lim sup -Snf (Tω). n\to \infty n n\to \infty n

Thus if ω ∈ N_ab, it follows that Tω ∈ N_ab and if Tω ∈ N_ab, then so is ω. Thus N_ab is an invariant set. Also, if ω ∈ N_ab, then

( ) a- lim inf 1-Snf (ω) = lim sup a - 1Snf (ω) > 0 n\to\infty n n\to\infty n

and

( 1 ) lim sup nSnf (ω)- b > 0 n\to\infty

It follows that

Nab \subseteq [M \infty (f - b) > 0]\cap [M \infty (a- f) > 0].

Consequently, since N_ab is invariant, argued above,

XNabM \infty (f - b) = M \infty (XNab (f - b))

and so from Lemma 10.8.4

\int \int (f (ω)- b)dμ = X (ω)(f (ω)- b)dμ Nab [XNabM\infty (f-b)>0] Nab

\int = XNab (ω)(f (ω)- b)dμ \geq 0 (10.8.27) [M\infty (XNab(f-b))>0]

(10.8.27)

and

\int \int N (a - f (ω))dμ = X M (a-f)>0 XNab (ω)(a- f (ω ))dμ ab [Nab \infty ]

\int = XNab (ω)(a- f (ω ))dμ \geq 0 (10.8.28) [M\infty(XNab(a-f))>0]

(10.8.28)

It follows that

\int aμ(Nab) \geq fdμ \geq bμ(Nab). (10.8.29) Nab

(10.8.29)

Since a < b, it follows that μ

= 0.

Now let

N \equiv \cup {Nab : a < b,a,b \in ℚ }.

It follows that μ

= 0. Now TN_a,b = N_a,b and so

T (N ) = \cupa,bT (Na,b) = \cupa,bNa,b = N.

Thus, TⁿN = N for all n ∈ ℕ. For ω

N,lim_n→∞

S_nf

exists. Now let

{ g (ω ) \equiv 0 if ω \in N . limn \to\infty n1Snf (ω) if ω ∕\in N

Then it is clear g satisfies the conditions of the theorem because if ω ∈ N, then Tω ∈ N also and so in this case, g

= g

≡ 0. On the other hand, if ω

N, then

1- 1- g(Tω) = ln\toim\infty nSnf (Tω) = lnim\to\infty n Snf (ω) = g(ω).

Which shows that g is invariant. Also, from Lemma 10.8.2,

\int \int ||1 || 1\sumn \int | ( )| |g|dμ \leq lim ni\tonf\infty ||nSnf ||dμ \leq lim ni\tonf\infty n |f Tk-1ω |dμ Ω Ω k=1 Ω 1\sumn \int = lim ni\tonf\infty n Ω|f (ω)|dμ = ∥f∥L1 k=1

so g ∈ L¹

The last claim about convergence in L¹ follows from the Vitali convergence theorem if we verify the sequence,

_n=1^∞ is uniformly integrable. To see this is the case, we know f ∈ L¹

and so if ε > 0 is given, there exists δ > 0 such that whenever B ∈ℱ and μ

≤ δ, then

< ε. Taking μ

< δ, it follows

||\int 1 || ||1 \sumn \int ( ) || ||1\sumn \int ( ) || || -Snf (ω)dμ|| = ||- f Tk-1ω dμ ||= ||-- XA (ω)f T k-1ω dμ|| A n |n k=1 A | |nk=1 Ω |

| | ||1\sumn \int ( k- 1 -(k- 1) ) ( k-1 ) || = ||n- XA T T ω f T ω dμ|| | k=1 Ω | ||1\sumn \int ( - (k-1) ) || = ||n ΩXA T ω f (ω)dμ|| k=1