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Invariant Measures

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Ergodicity

Denition: A measure µfor a dynamical system(X,T) isergodicif µ(A) =0 orµ(Ac) =0

for every measurableA⊂X such that T1(A) =Amodµ.

I Here T1(A) =Amodµmeans that the mass of the symmetric dierence µ(T1(A)4A) =0.

I If µis a probability measure, the we can write µ(A) =0 or 1 for every measurable A⊂X such that T1(A) =Amodµ. As stated, the denition also applies to innite measures.

I Usually ergodicity is stated for invariant measures, but the denition wroks for non-invariant measures too.

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Ergodicity

Ergodicity means that the spaceX doesn't fall apart in two separate parts.

Example 1: T is the doubling map onX =S1 andµ= 12(Leb+δ0). This measure is not ergodic, becauseA={0}andAc =S1\ {0} are both invariant modµ, butµ(A) =µ(Ac) = 12.

Example 2: X = [0,1]and

A A

A AA

A A

A AA T(x) =





12 −2x ifx ∈[0,14];

2x−12 ifx ∈[14,34];

52 −2x ifx ∈[0,14].

Lebesgue measureT-invariant but not ergodic, becauseA= [0,12] andAc = (12,1]are both invariant modµ, butµ(A) =µ(Ac) = 12.

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Ergodicity

Proposition: Let µbe an invariant measure for(X,T). Thenµis ergodic if and only if the onlyT-invariant functionsψ∈L1(µ) i.e., ψ◦T =ψ µ-a.e., are constantµ-a.e.

Proof: ⇒ Let ψ:X →Rbe T-invariantµ-a.e., but not constant.

Thus there existsa∈Rsuch that

A:=ψ1((−∞,a]) and Ac1((a,∞))

both have positive measure. ByT-invariance,T1A=A (mod µ), and we have a contradiction to ergodicity.

⇐Let Abe a set of positive measure such thatT1A=A. Let ψ=1A be its indicator function; it is T-invariant because Ais T-invariant. By assumption,ψ is constantµ-a.e., but as ψ(x) =0 forx ∈Ac, it follows that µ(Ac) =0.

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Proving Ergodicity

Proving that a measure is ergodic is not always simple. We illustrate two dierent proofs, applicable to dierent systems.

Lemma 1: Lebesgue measure ergodic for the doubling map T :S1→S1,x 7→2xmod1.

Proof: Suppose by contradiction thatAandB are disjoint T-invariant sets, both of positive measure.

Lebesgue measure has the property that if Leb(A)>0, then Leb-a.e. x∈Ais a density point, which means that

ε→lim0 sup

x∈J,diam(J)<ε

Leb(A∩J Leb(J) =1.

That is, the closer tox ∈A, the more points belong to A, relatively.

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Invariant Measures

Takex andy density points of AandB respectively. LetJx 3x andJy 3y be dyadic intervals of length 2−nwhere n ∈Nis so large that

Leb(Jx∩A) Leb(Jx) > 2

3 and Leb(Jy ∩B) Leb(Jy) > 2

3. By linearity and T-invariance ofAandB also:

Leb(Tn(Jx∩A)) Leb(Tn(Jx))) > 2

3 and Leb((Tn(Jy∩B)) Leb(Tn(Jy))) > 2

3. But Tn(Jx) =Tn(Jy) =S1. Therefore

Leb(A)> 2

3 and Leb(B)> 2 3.

This contradicts thatAandB are disjoint. This concludes this proof by contradiction.

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Proving Ergodicity

Lemma 2: Let α∈Rbe irrational. Lebesgue measure ergodic for the doubling mapRα :S1→S1,x 7→x+αmod1.

Exercise 3.7: Show that Lebesgue measure is not ergodic if α∈Q.

Proof of Lemma 2: We show that everyT-invariant function ψ∈L2 must be constant. Indeed, write

ψ(x) =X

n∈Z

ane2πinx

as a Fourier series. TheT-invariance implies ψ◦T(x) =X

n∈Z

ane2πin(x+α) =X

n∈Z

ane2πinαe2πinx =ψ(x)

soane2πinα=an for all n∈Z. Sinceα /∈Q, we havean=0 for all n6=0, so ψ(x)≡a0 is indeed constant. Finally,L2 is dense inL1, so the same conclusion holds forψ∈L1.

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Circle Rotations

Another Lebesgue preserving map that we will frequently use as example is the circle rotation:

Rα :S1→S1, Rα(x) =x+αmod1.

I Lebesgue measure is always Rα-invariant, regardless whatα is.

I If α∈Q, then every orbit is periodic. Hence for every x ∈S1, there is an atomic Rα-measure such thatµ({x})>0.

I If α /∈Q, then every orbit is dense inS1.

Exercise 3.2: If α /∈Qandµ is anRα-invariant atomic measure, show thatµis an innite measure: µ(S1) =∞. Are all innite Rα-invariant measure atomic?

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More on Irrational Rotations

The rotationRα :S1→S1 is dened asRα(x) =x+α (mod 1).

Letα beirrational.

Theorem (Poincaré): Every orbit is dense inS1, and for every intervalJ and every x∈S1, the visit frequency

v(J) := lim

n→∞

1

n#{0≤i <n :Rαi(x)∈J}=|J|.

Figure:Henri Poicaré (1854-1912): pioneer of dynamical systems.

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More on Irrational Rotations

Proof: Asα /∈Q, thenx cannot be periodic, so its orbit is innite.

Letε >0. SinceS1 is compact, there must be m<n such that 0< δ:=d(Rαm(x),Rαn(x))< ε.

SinceRα is an isometry,

|Rαk(n−m)(x)−Rα(k+1)(n−m)(x)|=δ

for everyk ∈Z, and

{Rαk(n−m)(x) :k ∈Z}

is a collection of points such that every two neighbours are exactly δ apart. Since ε > δis arbitrary, this shows that orb(x) is dense.

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More on Irrational Rotations

LetJδ0 = [Rαm(x),Rαn(x)) andJδk =Rαk(n−m)(Jδ). Then for K =b1/δc,{Jδk}Kk=0 is a cover S1 of adjacent intervals, each of lengthδ, and Rαj(n−m) is an isometry fromJδi toJδi+j. Therefore the visit frequencies

vk = lim inf

n

1

n#{0≤i <n :Rαi(x)∈Jδk}

are all the same for 0≤k ≤K, and together they add up to at most 1+K1. This shows for example that

1

K +1 ≤vk ≤vk := lim sup

n

1

n#{0≤i <n :Rαi(x)∈Jδk} ≤ 1 K, and these inequalities areindependent of the point x.

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More on Irrational Rotations

Now an arbitrary intervalJ can be covered byb|J|/δc+2 such adjacentJδk, so

v(J)≤ |J|

δ +2

1

K ≤(|J|(K +1) +2) 1

K ≤ |J|+ 3 K.

A similar computation givesv(J)≥ |J| −K3.

Takingε→0 (henceδ →0 andK → ∞), we nd that the limit v(J) indeed exists, and is equal to|J|. This concludes the proof.

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More on Irrational Rotations

Example: Consider the rst digits of the powers of 2.

1 16 256 4096 2 32 512 8192 4 64 1024 16384 8 128 2048 32768

etc.

Exercise 3.3: Does 9 ever appear as rst digit?

Exercise 3.4: Does 2 appear innitely often?

Exercise 3.5: With which frequency does 1appear?

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More on Irrational Rotations

Hint for Solution: Dene h:R+→S1 ash(x) = log10x mod1.

Then

h(2x) =h(x) + log102. so the following diagram commutes:

R+ ×2 - R+

S?1 + log102 - S1 h

? h

and note also thatlog1026=Q.

The rst digit of 2n is b∈ {1, . . . ,9}if and only if nlog102∈[log10b,log10b+1).

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Unique ergodicity

Denition: A system (X,T) is calleduniquely ergodicif there is exactly oneT-invariant probability measure.

Irrational rotationsRα are uniquely ergodic, with Lebesgue as uniqueRα-invariant probability measure.

Exercise 3.6: Show that the unique invariant meaure of a uniquely ergodic system is necessarily ergodic.

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Unique ergodicity

Oxtoby's Theorem: Let X be a compact space and T :X →X continuous. A transformation(X,T) is uniquely ergodic if and only if, for every continuous functionψ andevery pointx ∈X, the Birkho averages

1 n

n−1

X

i=0

ψ◦Ti(x) convergeuniformly to a constant function.

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Unique ergodicity

Proof: Ifµ andν were two dierent ergodic measures, then we can nd a continuous functionf :X →Rsuch thatR

fdµ6=R fdν. Using Birkho's Ergodic Theorem for both measures (with their own typical pointsx andy), we see that

limn

1 n

n−1

X

k=0

f ◦Tk(x) = Z

fdµ6=

Z

fdν = lim

n

1 n

n−1

X

k=0

f ◦Tk(y), so there is not even convergence to a constant function.

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Unique ergodicity

Conversely, we know by the Ergodic Theorem that limnn1Pn−1

k=0f ◦Tk(x) =R

fdµ is constantµ-a.e. But if the convergence is not uniform, then there is a sequence(yi)⊂X and (ni)⊂N, such that

limi

1 ni

ni1

X

k=0

f ◦Tk(yi)6=

Z

X

f dµ.

Dene probability measuresνi := n1

i

Pni1

k=0 δTk(xi). This sequence (νi) has a weak accumulation pointsν which is shown to be T-invariant measures in the same way as in the proof of Krylov-Bogol'ubov Theorem. Butν 6=µ becauseR

f dν 6=R f dµ. Hence(X,T) cannot be uniquely ergodic.

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