Download 21. Birkhoff`s Ergodic Theorem

MATH41112/61112 Ergodic Theory
Lecture 21
21. Birkhoff ’s Ergodic Theorem
§21.1
Introduction
An ergodic theorem is a result that describes the limiting behaviour of the
sequence
n−1
1X
f ◦ Tj
(21.1)
n
j=0
as n → ∞. The precise formulation of an ergodic theorem depends on the
class of function f (for example, one could assume that f is integrable,
L2 , or continuous), and the notion of convergence that we use (for example, we could study pointwise convergence, L 2 convergence, or uniform
convergence). The result that we are interested here—Birkhoff’s Ergodic
Theorem—deals with pointwise convergence of (21.1) for an integrable function f .
§21.2
Conditional expectation
We will need the concepts of Radon-Nikodym derivates and conditional expectation.
Definition. Let µ be a measure on (X, B). We say that a measure ν
is absolutely continuous with respect to µ and write ν µ if ν(B) = 0
whenever µ(B) = 0, B ∈ B.
Remark. Thus ν is absolutely continuous with respect to µ if sets of µmeasure zero also have ν-measure zero (but there may be more sets of νmeasure zero).
For example, let f ∈ L1 (X, B, µ) be non-negative and define a measure ν by
Z
ν(B) =
f dµ.
B
Then ν µ.
The following theorem says that, essentially, all absolutely continuous
measures occur in this way.
Theorem 21.1 (Radon-Nikodym)
Let (X, B, µ) be a probability space. Let ν be a measure defined on B and
suppose that ν µ. Then there is a non-negative measurable function f
such that
Z
f dµ, for all B ∈ B.
ν(B) =
B
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MATH41112/61112 Ergodic Theory
Lecture 21
Moreover, f is unique in the sense that if g is a measurable function with
the same property then f = g µ-a.e.
Exercise 21.1
If ν µ then it is customary to write dν/dµ for the function given by the
Radon-Nikodym theorem, that is
Z
dν
ν(B) =
dµ.
B dµ
Prove the following relations:
(i) If ν µ and f is a µ-integrable function then
Z
Z
dν
dµ.
f dν = f
dµ
(ii) If ν1 , ν2 µ then
d(ν1 + ν2 )
dν1 dν2
=
+
.
dµ
dµ
dµ
(iii) If λ ν µ then
dλ dν
dλ
=
.
dµ
dν dµ
Let A ⊂ B be a sub-σ-algebra. Note that µ defines a measure on A by
restriction. Let f ∈ L1 (X, B, µ). Then we can define a measure ν on A by
setting
Z
f dµ.
ν(A) =
A
Note that ν µ|A . Hence by the Radon-Nikodym theorem, there is a
unique A-measurable function E(f | A) such that
Z
ν(A) = E(f | A) dµ.
We call E(f | A) the conditional expectation of f with respect to the σalgebra A.
So far, we have only defined E(f | A) for non-negative f . To define
E(f | A) for an arbitrary f , we split f into positive and negative parts
f = f+ − f− where f+ , f− ≥ 0 and define
E(f | A) = E(f+ | A) − E(f− | A).
Thus we can view conditional expectation as an operator
E(· | A) : L1 (X, B, µ) → L1 (X, A, µ).
Note that E(f | A) is uniquely determined by the two requirements that
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MATH41112/61112 Ergodic Theory
Lecture 21
(i) E(f | A) is A-measurable, and
R
R
(ii) A f dµ = A E(f | A) dµ for all A ∈ A.
Intuitively, one can think of E(f | A) as the best approximation to f in the
smaller space of all A-measurable functions.
Exercise 21.2
(i) Prove that f 7→ E(f | A) is linear.
(ii) Suppose that g is A-measurable and |g| < ∞ µ-a.e. Show that E(f g |
A) = gE(f | A).
(iii) Suppose that T is a measure-preserving transformation. Show that
E(f | A) ◦ T = E(f ◦ T | T −1 A).
(iv) Show that E(f | B) = f .
(v) Let N denote the trivial σ-algebra
consisting of all sets of measure 0
R
and 1. Show that E(f | N ) = f dµ.
To state Birkhoff’s Ergodic Theorem precisely, we will need the sub-σalgebra I of T -invariant subsets, namely:
I = {B ∈ B | T −1 B = B a.e.}.
Exercise 21.3
Prove that I is a σ-algebra.
§21.3
Birkhoff ’s Pointwise Ergodic Theorem
Birkhoff’s Ergodic Theorem deals with the behaviour of
µ-a.e. x ∈ X, and for f ∈ L1 (X, B, µ).
1
n
Pn−1
j=0
f (T j x) for
Theorem 21.2 (Birkhoff ’s Ergodic Theorem)
Let (X, B, µ) be a probability space and let T : X → X be a measurepreserving transformation. Let I denote the σ-algebra of T -invariant sets.
Then for every f ∈ L1 (X, B, µ), we have
n−1
1X
f (T j x) → E(f | I)
n
j=0
for µ-a.e. x ∈ X.
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MATH41112/61112 Ergodic Theory
Lecture 21
Corollary 21.3
Let (X, B, µ) be a probability space and let T : X → X be an ergodic
measure-preserving transformation. Let f ∈ L 1 (X, B, µ). Then
n−1
1X
f (T j x) →
n
j=0
Z
f dµ,
as n → ∞,
for µ-a.e. x ∈ X.
Proof. If T is ergodic then I is the trivial σ-algebra NR consisting of sets
of measure 0 and 1. If f ∈ L1 (X, B, µ) then E(f | N ) = f dµ. The result
follows from the general version of Birkhoff’s ergodic theorem.
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§21.4
Appendix: The proof of Birkhoff ’s Ergodic Theorem
The proof is something of a tour de force of hard analysis. It is based on
the following inequality.
Theorem 21.4 (Maximal Inequality)
Let (X, B, µ) be a probability space, let T : X → X be a measure-preserving
transformation and let f ∈ L1 (X, B, µ). Define f0 = 0 and, for n ≥ 1,
fn = f + f ◦ T + · · · + f ◦ T n−1 .
For n ≥ 1, set
Fn = max fj
0≤j≤n
(so that Fn ≥ 0). Then
Z
f dµ ≥ 0.
{x|Fn (x)>0}
Proof. Clearly Fn ∈ L1 (X, B, µ). For 0 ≤ j ≤ n, we have Fn ≥ fj , so
Fn ◦ T ≥ fj ◦ T . Hence
Fn ◦ T + f ≥ fj ◦ T + f = fj+1
and therefore
Fn ◦ T (x) + f (x) ≥ max fj (x).
1≤j≤n
If Fn (x) > 0 then
max fj (x) = max fj (x) = Fn (x),
1≤j≤n
0≤j≤n
so we obtain that
f ≥ F n − Fn ◦ T
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MATH41112/61112 Ergodic Theory
Lecture 21
on the set A = {x | Fn (x) > 0}.
Hence
Z
Z
Z
Fn ◦ T dµ
Fn dµ −
f dµ ≥
A
ZA
ZA
=
Fn dµ −
Fn ◦ T dµ
X
A
Z
Z
Fn ◦ T dµ
Fn dµ −
≥
X
X
= 0
where we have used
(i) Fn = 0 on X \ A
(ii) Fn ◦ T ≥ 0
(iii) µ is T -invariant.
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Corollary 21.5
If g ∈ L1 (X, B, µ) and if


n−1


X
1
Bα = x ∈ X | sup
g(T j x) > α


n≥1 n
j=0
then for all A ∈ B with T −1 A = A we have that
Z
g dµ ≥ αµ(Bα ∩ A).
Bα ∩A
Proof. Suppose first that A = X. Let f = g − α, then


n−1
∞ 

X
[
g(T j x) > nα
x|
Bα =


=
=
n=1
j=0
∞
[
{x | fn (x) > 0}
n=1
∞
[
{x | Fn (x) > 0}
n=1
(since fn (x) > 0 ⇒ Fn (x) > 0 and Fn (x) > 0 ⇒ fj (x) > 0 for some 1 ≤ j ≤
n). Write Cn = {x | Fn (x) > 0} and observe that Cn ⊂ Cn+1 . Thus χCn
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MATH41112/61112 Ergodic Theory
Lecture 21
converges to χBα and so f χCn converges to f χBα , as n → ∞. Furthermore,
|f χCn | ≤ |f |. Hence, by the Dominated Convergence Theorem,
Z
Z
Z
Z
f dµ =
f χCn dµ →
f χBα dµ =
f dµ, as n → ∞.
Cn
X
X
Bα
Applying the maximal inequality, we have, for all n ≥ 1,
Z
f dµ ≥ 0.
Cn
Therefore
Z
f dµ ≥ 0,
Bα
i.e.,
Z
g dµ ≥ αµ(Bα ).
Bα
For the general case, we work with the restriction of T to A, T : A → A,
and apply the maximal inequality on this subset to get
Z
g dµ ≥ αµ(Bα ∩ A),
Bα ∩A
as required.
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Proof of Birkhoff ’s Ergodic Theorem. Let
n−1
1X
f (T j x)
f (x) = lim sup
n
n→∞
∗
j=0
and
n−1
1X
f (T j x).
f∗ (x) = lim inf
n→∞ n
j=0
Writing
n−1
an (x) =
1X
f (T j x),
n
j=0
observe that
n+1
1
an+1 (x) = an (T x) + f (x).
n
n
Taking the lim sup and lim inf as n → ∞ gives us that f ∗ ◦ T = f ∗ and
f∗ ◦ T = f ∗ .
We have to show
(i) f ∗ = f∗ µ-a.e
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MATH41112/61112 Ergodic Theory
Lecture 21
(ii) f ∗ ∈ L1 (X, B, µ)
R
R
(iii) f ∗ dµ = f dµ.
We prove (i). For α, β ∈ R, define
Eα,β = {x ∈ X | f∗ (x) < β and f ∗ (x) > α}.
Note that
{x ∈ X | f∗ (x) < f ∗ (x)} =
[
Eα,β
β<α, α,β∈Q
f∗
(a countable union). Thus, to show that
= f∗ µ-a.e., it suffices to show
that µ(Eα,β ) = 0 whenever β < α. Since f∗ ◦ T = f∗ and f ∗ ◦ T = f ∗ , we
see that T −1 Eα,β = Eα,β . If we write


n−1


X
1
f (T j x) > α
Bα = x ∈ X | sup


n≥1 n
j=0
then Eα,β ∩ Bα = Eα,β .
Applying Corollary 21.5 we have that
Z
Z
f dµ =
f dµ
Eα,β ∩Bα
Eα,β
≥ αµ(Eα,β ∩ Bα ) = αµ(Eα,β ).
Replacing f , α and β by −f , −β and −α and using the fact that (−f ) ∗ =
−f∗ and (−f )∗ = −f ∗ , we also get
Z
f dµ ≤ βµ(Eα,β ).
Eα,β
Therefore
αµ(Eα,β ) ≤ βµ(Eα,β )
and since β < α this shows that µ(Eα,β ) = 0. Thus f ∗ = f∗ µ-a.e. and
n−1
1X
f (T j x) = f ∗ (x)
lim
n→∞ n
j=0
We prove (ii). Let
Then gn ≥ 0 and
n−1
1 X
j
gn = f ◦ T .
n j=0
Z
gn dµ ≤
7
Z
|f | dµ
µ-a.e.
MATH41112/61112 Ergodic Theory
Lecture 21
so we can apply Fatou’s Lemma to conclude that lim n→∞ gn = |f ∗ | is integrable, i.e., that f ∗ ∈ L1 (X, B, µ).
We prove (iii). For n ∈ N and k ∈ Z, define
k+1
k
∗
n
Dk = x ∈ X | ≤ f (x) <
.
n
n
For every ε > 0, we have that
Dkn ∩ B k −ε = Dkn .
n
Since T −1 Dkn = Dkn , we can apply Corollary 22.4 again to obtain
Z
k
− ε µ(Dkn ).
f dµ ≥
n
n
Dk
Since ε > 0 is arbitrary, we have
Z
k
f dµ ≥ µ(Dkn ).
n
n
Dk
Thus
Z
f ∗ dµ ≤
Dkn
≤
k+1
µ(Dkn )
n
Z
1
n
µ(Dk ) +
f dµ
n
Dkn
(where the first inequality follows from the definition of D kn ). Since
[
X=
Dkn
k∈Z
(a disjoint union), summing over k ∈ Z gives
Z
Z
1
∗
f dµ
µ(X) +
f dµ ≤
n
X
X
Z
1
=
+
f dµ.
n
X
Since this holds for all n ≥ 1, we obtain
Z
Z
f dµ.
f ∗ dµ ≤
X
X
Applying the same argument to −f gives
Z
Z
∗
(−f ) dµ ≤ −f dµ
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MATH41112/61112 Ergodic Theory
so that
Therefore
Z
∗
f dµ =
Z
Z
Lecture 21
f∗ dµ ≥
∗
f dµ =
Z
Z
f dµ.
f dµ,
as required.
Finally, we prove that f ∗ = E(f | I). First note that as f ∗ is T -invariant,
it is measurable with respect to I. Moreover, if I is any T -invariant set then
Z
Z
f, dµ = f ∗ dµ.
I
I
Hence f ∗ = E(f | I).
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