Invariant Measures for Continuous Maps
4.1 Existence of Invariant Measures
The connection between ergodic theory and the dynamics of continuous maps on compact metric spaces begins with the next result, which shows that invariant measures can always be found.
Theorem 4.1.Let T : X → X be a continuous map of a compact metric space, and let(νn)be any sequence inM(X). Then any weak*-limit point of the sequence(μn)defined byμn =n1n−1
j=0 T∗jνn is a member ofMT(X).
An immediate consequence is the following important general statement, which shows that measure-preserving transformations are ubiquitous. It is known as the Kryloff–Bogoliouboff Theorem [214].
Corollary 4.2 (Kryloff–Bogoliouboff ). Under the hypotheses of Theo- rem4.1,MT(X)is non-empty.
Proof.SinceM(X) is weak*-compact, the sequence (μn) must have a limit
point.
Writef∞= sup{|f(x)| |x∈X}as usual.
Proof of Theorem4.1.Letμn(j)→μbe a convergent subsequence of (μn) and letf ∈C(X). Then, by applying the definition ofT∗μn, we get
f◦Tdμn(j)−
fdμn(j) = 1
n(j)
n(j)−1
i=0
f◦Ti+1−f◦Ti dνn(j)
= 1
n(j)
f◦Tn(j)+1−f
dνn(j)
2
n(j)f∞−→0 as j → ∞, for all f ∈C(X). It follows that
f ◦Tdμ =
fdμ, so μ is a
member ofMT(X) by Lemma B.12.
ThusMT(X) is a non-empty compact convex set, since convex combina- tions of elements ofMT(X) belong toMT(X). It follows thatMT(X) is an infinite set unless it comprises a single element. For many maps it is difficult to describe the space of invariant measures. The next example has very few ergodic invariant measures, and we shall see later many maps that have only one invariant measure.
4.1 Existence of Invariant Measures 99
Example 4.3 (North–South map). Define the stereographic projectionπfrom the circleX ={z ∈ C| |z−i|= 1} to the real axis by continuing the line from 2i through a unique point onX{2i} until it meets the line (z) = 0 (see Fig.4.1).
Fig. 4.1 The North-South map on the circle; forz= 2i,Tnz→0 asn→ ∞
The “North–South” mapT :X →X is defined by T(z) =
2i ifz= 2i;
π−1(π(z)/2) ifz= 2i
as shown in Fig.4.1. Using Poincar´e recurrence (Theorem 2.11) it is easy to show that MT(X) comprises the measures pδ2i+ (1−p)δ0, p∈ [0,1] that are supported on the two points 2i and 0. Only the measures corresponding top= 0 andp= 1 are ergodic.
It is in general difficult to identify measures with specific properties, but the ergodic measures are readily characterized in terms of the geometry of the space of invariant measures.
Theorem 4.4.Let X be a compact metric space and let T : X → X be a measurable map. The ergodic elements of MT(X) are exactly the extreme points ofMT(X).
That is, T is ergodic with respect to an invariant probability measure if and only if that measure cannot be expressed as a strict convex combination of two differentT-invariant probability measures. For any measurable setA, defineμ
Abyμ
A(C) =μ(A∩C).IfT is not assumed to be continuous, then we do not know thatMT(X)=∅, so without the assumption of continuity Theorem4.4may be true but vacuous (see Exercise4.1.1).
Proof of Theorem 4.4.Letμ∈MT(X) be a non-ergodic measure. Then there is a measurable setBwithμ(B)∈(0,1) and withT−1B=B. It follows that
1 μ(B)μ
B, 1 μ(XB)μ
XB∈MT(X), so
μ=μ(B) 1
μ(B)μ
B
+μ(XB) 1 μ(XB)μ
XB
expressesμas a strict convex combination of the invariant probability mea- sures
1 μ(B)μ
B
and 1
μ(XB)μ
XB,
which are different since they give different measures to the setB.
Conversely, letμbe an ergodic measure and assume that μ=sν1+ (1−s)ν2
expressesμas a strict convex combination of the invariant measuresν1andν2. Sinces >0,ν1μ, so there is a positive functionf ∈L1μ (f is the Radon–
Nikodym derivative dνdμ1; see Theorem A.15) with the property that ν1(A) =
A
fdμ (4.1)
for any measurable set A. The set B = {x∈ X | f(x) <1} is measurable sincef is measurable, and
B∩T−1B
fdμ+
BT−1B
fdμ=ν1(B)
=ν1(T−1B)
=
B∩T−1B
fdμ+
(T−1B)B
fdμ,
so
BT−1B
fdμ=
(T−1B)B
fdμ. (4.2)
By definition,f(x)<1 forx∈B(T−1B) whilef(x)1 for x∈T−1BB.
On the other hand,
μ((T−1B)B) =μ(T−1B)−μ((T−1B)∩B)
=μ(B)−μ((T−1B)∩B)
=μ(BT−1B)
4.1 Existence of Invariant Measures 101
so (4.2) implies that μ(BT−1B) = 0 and μ((T−1B)B) = 0. There- foreμ((T−1B)B) = 0, so by ergodicity ofμ we must haveμ(B) = 0 or 1.
Ifμ(B) = 1 then
ν1(X) =
X
fdμ < μ(B) = 1, which is impossible. Soμ(B) = 0.
A similar argument shows that μ({x∈X |f(x)>1}) = 0, so f(x) = 1 almost everywhere with respect toμ. By (4.1), this shows that
ν1=μ,
soμis an extreme point inMT(X).
WriteET(X) for the set of extreme points inMT(X)—by Theorem4.4, this is the set of ergodic measures forT.
Example 4.5.Let X = {1, . . . , r}Z and let T : X → X be the left shift map. In Example 2.9 we defined for any probability vectorp= (p1, . . . , pr) a T-invariant probability measure μ = μp on X, and by Proposition 2.15 all these measures are ergodic. Thus for this example the space ET(X) of ergodic invariant measures is uncountable. This collection of measures is an inconceivably tiny subset of the set of all ergodic measures—there is no hope of describing all of them.
Measures μ1 and μ2 are called mutually singular if there exist disjoint measurable setsAandB withA∪B =X for whichμ1(B) =μ2(A) = 0 (see Sect. A.4).
Lemma 4.6.If μ1, μ2 ∈ET(X) andμ1 =μ2 then μ1 and μ2 are mutually singular.
Proof.Let f ∈ C(X) be chosen with
fdμ1 =
fdμ2 (such a function exists by Theorem B.11). Then by the ergodic theorem (Theorem 2.30)
Afn(x)→
fdμ1 (4.3)
forμ1-almost everyx∈X, and Afn(x)→
fdμ2
forμ2-almost everyx∈X. It follows that the setA={x∈X |(4.3) holds} is measurable and hasμ1(A) = 1 butμ2(A) = 0.
Some of the problems for this section make use of the topological analog of Definition 2.7, which will be used later.
Definition 4.7.Let T : X → X and S : Y → Y be continuous maps of compact metric spaces (that is, topological dynamical systems). Then a homeomorphismθ:X→Y withθ◦T =S◦θis called atopological conjugacy, and if there such a conjugacy thenT and S are topologically conjugate. A continuous surjective mapφ:X →Y withφ◦T =S◦φis called atopological factor map, and in this caseS is said to be afactor ofT.
Exercises for Sect. 4.1
Exercise 4.1.1.LetX ={0,1n |n1}with the compact topology inherited from the reals. SinceX is countable, there is a bijectionθ : X → Z. Show that the map T : X → X defined by T(x) = θ−1(θ(x) + 1) is measurable with respect to the Borel σ-algebra on X but has no invariant probability measures.
Exercise 4.1.2.Show that a weak*-limit of ergodic measures need not be an ergodic measure by the following steps. Start with a pointxin the full 2- shift σ : X → X with the property that any finite block of symbols of length appears in x with asymptotic frequency 21 (such points certainly exist; indeed the ergodic theorem says that almost every point with respect to the (1/2,1/2) Bernoulli measure will do). Write (x1. . . xn0. . .0)∞for the pointy∈ {0,1}Z determined by the two conditions
y|[0,2n−1] =x1. . . xn0. . .0
andσ2n(y) =y. Now for eachnconstruct an ergodicσ-invariant measureμn
supported on the orbit of the periodic point (x1. . . xn0. . .0)∞in which there are n 0 symbols in every cycle of the periodic point under the shift. Show that μn converges to some limit ν and use Theorem 4.4to deduce that ν is not ergodic.
Exercise 4.1.3.For a continuous map T : X → X of a compact metric space (X,d), define the invertible extensionT:X →X as follows. Let
• X={x∈XZ|xk+1=T xk for allk∈Z};
• (T x) k=xk+1 for allk∈Zandx∈X;
with metric d(x, y) =
k∈Z2−|k|d(xk, yk). Write π : X → X for the map sendingxto x0. Prove the following.
(1)Tis a homeomorphism of a compact metric space, and π:X →X is a topological factor map.
(2) If (Y, S) is any homeomorphism of a compact metric space with the prop- erty that there is a topological factor map (Y, S)→(X, T), then (X, T) is a topological factor of (Y, S).