So, let’s fix a compact metric space {X} and a transformation {T: X \rightarrow X} which is continuous. We defined the space {M(X,T)} of probability Borel measures which are {T}-invariant, showed it was nonempty, and proved that the extreme points correspond to ergodic measures (i.e. measures with respect to which {T} is ergodic). We are interested in knowing what {M(X,T)} looks like, based solely on the topological properties of {T}. Here are some techniques we can use:

1) If {T} has no fixed points, then {\mu \in M(X,T)} cannot have any atoms (i.e. {\mu(\{x\})=0, x \in X}). Otherwise {\{x, Tx , T^2x, \dots \}} would have infinite measure.

2) The set of recurrent points in {X} (i.e. {x \in X} such that there exists a sequence {n_i \rightarrow \infty} with {T^{n_i}x \rightarrow x}) has {\mu}-measure one. We proved this earlier.

3) The set of non-wandering points has measure one. We define this notion now. Say that {x \in X} is wandering if there is a neighborhood {U} of {X} such that {T^{-n}(U) \cap U = \emptyset, \forall n \in \mathbb{N}}. In other words, the family of sets {T^{i}(U), i \in \mathbb{Z}_{\geq 0}} is disjoint. If not, say that {x} is non-wandering. Any recurrent point, for instance, is non-wandering, which implies that the set of non-wandering points has measure one.

Here is an example.

The North-South map

Recall that the unit circle {S^1} can be identified with the one-point compactification of the real line {\mathbb{R}}, with {(0,1)} corresponding to the point at infinity and {0} to {(0,-1)} (e.g. by stereographic projection). So, define the map {\mathbb{R} \cup \{\infty\} \rightarrow \mathbb{R} \cup \{\infty\}} by sending {x \rightarrow x/2, \infty \rightarrow \infty}. The associated map {T: S^1 \rightarrow S^1} is called the north-south map.

It is clear that the iterates of {T} send any point other than the north pole {(0,1)} to the south pole in the limit. The only recurrent points are thus the north and south poles. So it follows that the elements of {M(X,T)} must be a convex combination of the point masses at each pole, and one can check easily that any such convex combination is in {M(X,T)}. The extreme points (i.e. ergodic measures) are precisely these point masses.

Unique ergodicity

We now consider a special extreme case. Suppose {M(X,T)} consists of one point {\mu}. Equivalently, by the Krein-Milman theorem, there is precisely one {\mu \in M(X,T)} which is ergodic. It then follows that for every continuous {f: X \rightarrow \mathbb{C}},

\displaystyle \frac{1}{N} \sum_{i=0}^{N-1} f(T^i(x)) \rightarrow \int f d \mu,

for almost all {x} (relative to {\mu}!).

In many cases, {\mu} will be highly singular. If not, we can get minimality.

Proposition 1 Suppose {\mu(U)>0} for all nonempty open {U}. Then {T} is minimal.


If {T} is not minimal, there is a proper closed {F \subset X} with {T^{-1}F = F}. Take {\mu_1 \in M(F,T)} and let {\mu'(E) := \mu_1(F \cap E)} extend the measure to all of {X}. This measure is {T}-invariant and supported in {F}, so it is not {\mu}, and {T} is not uniquely ergodic, contradiction.

Everywhere convergence

One of the nice consequences of minimality is that the {T}-averages of a continuous function actually converge everywhere to a constant. This is a strengthening of the usual conclusion, and is another reason we want ergodicity.


Theorem 2

Suppose {T: X \rightarrow X} is a continuous map of a compact metric space {X}, and {T} is uniquely ergodic with measure {\mu}. Then for all {x \in X} and continuous {f: X \rightarrow \mathbb{C}},\displaystyle A_N(f,x) := \frac{1}{N} \sum_{i=0}^{N-1} f(T^i(x)) \rightarrow \int f d \mu. 

Suppose the contrary, and there existed a continuous function {f} and a subsequence {A_{N_j}(f,x)} that converged to something other than {\int f d \mu}. Consider the measures

\displaystyle \mu_j = \frac{1}{N_j} \sum_{i=0}^{N_j-1} T^{-i}\delta_x = \frac{1}{N_j} \sum_{i=0}^{N_j-1} \delta_{T^jx}

where {\delta_x} is the point mass at {x}. It is easy to see, by the same kind of argument in the proof of the fixed point lemma, that any weak* limit point of {\mu_j} is {T}-invariant, because

\displaystyle ||T^{-1} \mu_j - \mu_j ||_{M(X)} \leq \frac{2}{N_j}.

Let such a weak* limit point be {\mu'}. Then {\mu, \mu' \in M(X,T)} but

\displaystyle \int f d \mu \neq \int f d \mu' = \lim_j A_{N_j}(f,x)

because of the choice of the sequence {N_j}

Next time, we’ll see some applications of this.