Denition: A measure µfor a dynamical system(X,T)is T-invariantif
µ(A) =µ(T−1(A)) for every measurable setA⊂X.
Lemma 1: Let T :X →X be a continuous map on a compact spaceX. Thenµis T-invariant if and only if
Z
X
f dµ= Z
X
f ◦T dµ
for every f ∈C(X). (Here C(X) is the space of all continuous functions onX, equipped with the normk k∞.)
Proof of Lemma 1:
Assume thatµis T-invariant andA∈ B, the σ-algebra of µ-measurable sets. Then
Z
X
1A◦T dµ=µ(T−1A) =µ(A) = Z
X
1A dµ.
A similar expression works for linear combinations of indicator sets.
These linear combinations of indicator sets lie dense inC(X), so the result carries over toC(X).
(NB: We used here that continuous function of a compact space are bounded).
For the other direction: for every closed setAandε >0, we can nd a functionf ∈C(X)such that f ≡1 onAand
R|f −1A|dµ < εas well as R
|f −1A| ◦T dµ < ε. Then
|µ(T−1A)−µ(A)| = | Z
1A◦T dµ− Z
1A dµ|
= | Z
(f −1A)◦T dµ− Z
(f −1A) dµ|
≤ 2ε.
Sinceεis arbitrary, µ(T−1A) =µ(A). The closed sets generate the Borelσ-algebra, so also µ(T−1A) =µ(A) for all A∈ B.
We have seen examples that Lebesgue measure isT-invariant for the doubling map and also for rotations
Rα(x) =x+αmod1 onS1. It is reasonable to ask:
Does every dynamical system have an invariant measure?
Theorem of Krylov-Bogol'ubov: IfT :X →X is a continuous map on a nonempty compact metric spaceX, then the set of invariant probability measuresM(X,T)6=∅.
T : (0,1)→(0,1), T(x) =x2.
0,1∈/ (0,1), so the Dirac measuresδ0 andδ1 are not allowed.
Suppose thatµ isT-invariant, and assume that A⊂(0,1) is a compact set withµ(A)>0. Then there isn1 ∈Nsuch that A1 :=T−n1(A)is disjoint from A. Hence µ(A1) =µ(A)>0.
Proceed by induction: ni ∈Nis such thatAi :=T−ni(Ai−1) is disjoint fromAi−1. Thenµ(A) =µ(Ai)>0 for alli ∈Nand all Ai are pairwise disjoint. Hence
µ((0,1))≥X
i
µ(Ai) =∞,
soµ is not a probability measure.
Exercise 2.1: Show that continuity is an essential assumption in the Theorem of Krylov-Bogol'ubov.
Proof of the Theorem of Krylov-Bogol'ubov:
Letν be any probability measure (invariant or not) and dene Cesaro means:
νn(A) = 1 n
n−1
X
j=0
ν(T−jA).
These are all probability measures.
The collection of probability measures on a compact metric space is known to be compact in the weak∗ topology.
This means that there is limit probability measureµand a subsequence(ni)i∈N such that for every continuous function ψ:X →R:
Z
X
ψdνni → Z
ψ dµasi → ∞.
continuous functionψA :X →[0,1]such that ψA(x) =1 if x ∈A and
µ(A)≤ Z
X
ψAdµ≤µ(A) +ε and
µ(T−1A)≤ Z
X
ψA◦T dµ≤µ(T−1A) +ε.
Here we use outer regularity of the measureµ:
µ(A) = inf{µ(G) :G ⊃A is open}.
We takeG ⊃A so small thatµ(G)−µ(A)< ε and make sure that ψA =0 for allx ∈/G. Note that it is important thatA is closed, because if there existsa∈∂A\A, then the above property fails for µ=δa.
Now by Lemma 1 and the denition ofµ
|µ(T−1(A)) − µ(A)| ≤ Z
ψA◦T dµ− Z
ψA dµ
+ε
= lim
i→∞
Z
ψA◦T dνni − Z
ψA dνni
+ε
= lim
i→∞
1 ni
ni−1
X
j=0
Z
ψA◦Tj+1 dν− Z
ψA◦Tj dν
+ε
≤ lim
i→∞
1 ni
Z
ψA◦Tni dν− Z
ψA dν
+ε
≤ lim
i→∞
2
nikψAk∞+ε=ε.
Sinceε >0 is arbitrary, µ(T−1(A)) =µ(A). The closed sets generate Borel sets, soµ(T−1(A)) =µ(A)for all A∈ B.
