(2)measure preserving dynamical system

In this lecture we prove Birkho's Ergodic Theorem (also called the Pointwise Ergodic Theorem):

Theorem: Let µbe a probability measure andψ∈L¹(µ). Then the ergodic average

ψ^∗(x) := lim

n→∞

1 n

n−1

X

i=0

ψ◦Tⁱ(x)

existsµ-a.e., andψ^∗ is T -invariant, i.e.,ψ^∗◦T =ψ^∗ µ-a.e.

If in additionµis ergodicthenψ^∗ =R

X ψdµ forµ-a.e. x Or in short:

Space Average = Time Average (for typical points).

measure preserving dynamical system. Take M_N =max{S_n :0≤n≤N}. Then

Z

AN

f dµ≥0 for A_N ={x ∈X :M_N(x)>0}.

TheKoopman operatorU_T :L¹(µ)→L¹(µ) is dened as

U_Tf =f ◦T.

Clearly U_T is linear andpositive, i.e., f ≥0 implies U_Tf ≥0.

We write theergodic sumas

S_n=S_nf =

n−1

X

k=0

f ◦T^k and S₀ ≡0.

all 0≤n ≤N and by positivity of the Koopman operator, also U_TM_N ≥U_TS_n. Add f : U_TM_N+f ≥U_TS_n+f =S_n+1. For x ∈A_N, this means

U_TM_N(x) +f(x) ≥ max

1≤n≤NS_n(x)

≥_x∈AN max

0≤n≤NS_n(x) =M_N(x).

Therefore f ≥M_N −U_TM_N on A_N, and (since M_N ≥S₀=0) Z

AN

f dµ ≥ Z

AN

M_N dµ− Z

AN

U_TM_N dµ

= Z

X M_N dµ− Z

AN

U_TM_N dµ

= Z

X M_N dµ− Z

X U_TM_N dµ=0.

This completes the proof.

Lemma: Let(X,T,B, µ)be a probability measure preserving dynamical system, and E ⊂X a T -invariant subset. Let

Bα:={x ∈X :sup

n

1

nS_ng(x)> α}.

Then Z

Bα∩Eg dµ≥αµ(Bα∩E).

For a measurable set E ⊂X , we dene a probability measure µ_E(B) = 1

µ(E)µ(B∩E) If E is T -invariant thenµ_E is T -invariant too.

Proof: Ifµ(E) =0 then there is nothing to prove. So assume that µ(E)>0. Take f =g−α, so

Bα=∪_NA_N for A_N ={x ∈X :M_N(x)>0}.

Note also that A_N ⊂A_N+1 for all N. Therefore for each ε >0 there exists N ∈Nsuch that

Z

Bα

f dµ_E ≥ Z

AN

f dµ_E ≥ −ε.

Sinceεis arbitrary, R

Bαf dµ_E ≥0. Addingα again we have Z

Bα

g dµ_E = Z

Bα

f +α dµ_E ≥αµ_E(Bα∩E).

Multiply everything byµ(E) to get the lemma.

Proof of Birkho's Ergodic Theorem:

Recallψ∈L¹(µ). Dene ψ=lim sup

n→∞

1

nS_nψ and ψ=lim inf

n→∞

1 nS_nψ.

Since

|n+1 n

1

n+1Sn+1ψ−1

nSnψ◦T|= 1

n|ψ(x)| →0 as n→ ∞, we have ψ◦T =ψand similarly ψ◦T =ψ. We want to show thatψ=ψ µ-a.e.

Let

Eα,β ={x ∈X :ψ(x)< β, α < ψ(x)}

Then Eα,β is T -invariant, and

{x ∈X :ψ(x)< ψ(x)}= [

α,β∈Q,β<α

Eα,β.

This is a countable union, and therefore it suces to show that µ(Eα,β) =0 for every pair of rationals β < α.

WriteBα :={x ∈X :sup_n¹_nS_nψ(x)> α} as in our Lemma. Since Eα,β =Eα,β∩Bα, this Lemma gives

Z

Eα,β

ψ dµ= Z

Eα,β∩B^αψdµ≥αµ(Eα,β∩Bα) =αµ(Eα,β).

From the previous slide:

Z

Eα,β

ψ dµ≥αµ(Eα,β).

We repeat this argument replacingψ, α, β by −ψ,−α,−β. Note that−ψ=−ψ and−ψ=−ψ. This gives

Z

Eα,β

ψ dµ≤βµ(Eα,β).

Sinceβ < α, this can only be true if µ(Eα,β) =0.

Thereforeψ=ψ=ψ^∗, i.e., the lim sup and lim inf are actually limitsµ-a.e.

The next step is to show thatψ^∗ ∈L¹(µ). From Measure Theory we have:

Fatou's LemmaIf(g_n)_n∈N are non-negative L¹(µ)-functions, then lim inf

n gn∈L¹(µ) and Z

X lim inf

n gn dµ≤lim inf

n

Z

X gn dµ.

Here we apply this to g_n=|¹_nS_nψ|, which belong to L¹(µ) because (by T -invariance)

Z

X

|1

nSnψ|dµ≤ 1 n

n−1

X

k=0

Z

X

|ψ| ◦T^kdµ= Z

|ψ|dµ <∞.

Hence in the limit: R

X|ψ|dµ≤lim inf_nR

X |ψ|dµ <∞.

Next, we need to show that _X ψ dµ= _X ψdµ(so without absolute value signs). Take

D_k,n={x ∈X : k

n ≤ψ^∗(x)< k+1 n }.

Then D_k,n is T -invariant, and ∪_k∈ZD_k,n =X modµ. Also Bk

n−ε∩D_k,n=D_k,n forε >0. Therefore our Lemma gives Z

Dk,n

ψ dµ = Z

Bk n−ε∩Dk,n

ψ dµ

≥ (k

n −ε)µ(Bk

n−ε∩D_k,n)

= (k

n −ε)µ(D_k,n).

n Dk,n

Z

Dk,n

ψ^∗ dµ≤ k+1

n µ(D_k,n)≤ 1

nµ(D_k,n) + Z

Dk,n

ψdµ.

Summing over all k ∈Z, we ndR

Xψ^∗ dµ≤ ¹_n+R

X ψdµ. Since n∈Nis arbitrary, also

Z

X ψ^∗ dµ≤ Z

X ψ dµ, By the same argument for−ψ, we ndR

X ψ^∗ dµ≥R

Xψ dµ.

Hence Z

Xψ^∗= Z

Xψ dµ.

Finally, ifµis ergodic, the T -invariant functionψ^∗ has to be constantµ-a.e., soψ^∗ =R

ψdµ. This completes the proof of Birkho's Ergodic Theorem.

The L^p Ergodic Theorem is a generalisation of Von Neumann's L² version of the Ergodic Theorem, which predates¹ Birkho's Theorem, but nowadays, it is usually proved as a corollary of the pointwise ergodic theorem.

Theorem: Let (X,T,B, µ) be a probability measure preserving dynamical system. Ifµis ergodic, andψ∈L^p(µ)for some 1≤p <∞ then there existsψ^∗ ∈L^p(µ) with ψ^∗◦T =ψ^∗ µ-a.e.

such that

k1

nSnψ−ψ^∗k_p→0 as n→ ∞.

1John von Neumann was earlier in proving his L¹-version, but Birkho delayed its publication until after the appearance of his own paper.

First assume thatψis bounded (and hence in L^p(µ)). By Birkho's Ergodic Theorem there isψ^∗ such that _n¹Snψ(x)→ψ^∗(x) µ-a.e., andψ^∗ is bounded (and hence in L^p(µ) too). In particular,

|1

nS_nψ(x)−ψ^∗(x)|^p→0 µ-a.e.

By the Bounded Convergence Theorem, we can swap the limit and the integral:

nlim→∞k1

nSnψ−ψ^∗k_p

= lim

n→∞

Z

X

|1

nS_nψ(x)−ψ^∗(x)|^p dµ ₁/p

= Z

X lim

n→∞|1

nS_nψ(x)−ψ^∗(x)|^p dµ ₁/p

=0.

Convergent sequences are Cauchy, so for everyε >0 there is N =N(ε, ψ) such that

k1

mSmψ− 1

nSnψk_p< ε

2 (*)

for all m,n≥N.

Now ifφ∈L^p(µ) is unbounded, we want to show that ¹_nS_nφis a Cauchy sequence ink k_p. Letε >0 be arbitrary, and takeψ bounded such thatkφ−ψk_p< ε/4. By T -invariance,

k1

nS_nφ−1

nS_nψk_p≤ kφ−ψk_p for all n≥1. (**)

k1

mS_mψ−1

nS_nψk_p < ε

2 for all m,n ≥N(ε, ψ) (*) and

k1

nS_nφ−1

nS_nψk_p ≤ kφ−ψk_p < ε

4 for all n≥1. (**) By the triangle inequality,

k1

mS_mφ−1

nS_nφk_p ≤ k1

mS_mφ− 1

mS_mψk_p +k1

mS_mψ− 1 nS_nψk_p +k1

nS_nφ−1 nS_nψk_p

< ε 4 +ε

2 +ε 4 =ε for all m,n≥N(ε, ψ).

Hence ¹_nS_nφis Cauchy in k k_p and thus converges to some φ^∗ ∈L^p(µ). We have

n+1 n

1

n+1Sn+1φ(x)− 1

nSnφ◦T(x) =|1

nφ(x)|

for all x. Taking the limit n→ ∞ givesφ^∗=φ^∗◦T µ-a.e.

This concludes the proof of the L^p Ergodic Theorem.