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.