Ergodicity
Denition: A measure µfor a dynamical system(X,T) isergodicif µ(A) =0 orµ(Ac) =0
for every measurableA⊂X such that T−1(A) =Amodµ.
I Here T−1(A) =Amodµmeans that the mass of the symmetric dierence µ(T−1(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 T−1(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.
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.
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:=ψ−1((−∞,a]) and Ac =ψ−1((a,∞))
both have positive measure. ByT-invariance,T−1A=A (mod µ), and we have a contradiction to ergodicity.
⇐Let Abe a set of positive measure such thatT−1A=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.
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.
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.
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.
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?
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.
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.
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.
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.
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?
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).
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.
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.
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.
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
ni−1
X
k=0
f ◦Tk(yi)6=
Z
X
f dµ.
Dene probability measuresνi := n1
i
Pni−1
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.