So, let’s fix a compact metric space and a transformation which is continuous. We defined the space of probability Borel measures which are -invariant, showed it was nonempty, and proved that the extreme points correspond to ergodic measures (i.e. measures with respect to which is ergodic). We are interested in knowing what looks like, based solely on the **topological** properties of . Here are some techniques we can use:

1) If has no fixed points, then cannot have any atoms (i.e. ). Otherwise would have infinite measure.

2) The set of recurrent points in (i.e. such that there exists a sequence with ) has -measure one. We proved this earlier.

3) The set of non-wandering points has measure one. We define this notion now. Say that is **wandering** if there is a neighborhood of such that . In other words, the family of sets is disjoint. If not, say that 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 can be identified with the one-point compactification of the real line , with corresponding to the point at infinity and to (e.g. by stereographic projection). So, define the map by sending . The associated map is called the **north-south map**.

It is clear that the iterates of send any point other than the north pole 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 must be a convex combination of the point masses at each pole, and one can check easily that any such convex combination is in . The extreme points (i.e. ergodic measures) are precisely these point masses.

**Unique ergodicity **

We now consider a special extreme case. Suppose consists of one point . Equivalently, by the Krein-Milman theorem, there is precisely one which is ergodic. It then follows that for every continuous ,

for almost all (relative to !).

In many cases, will be highly singular. If not, we can get minimality.

Proposition 1Suppose for all nonempty open . Then is minimal.

If is not minimal, there is a proper closed with . Take and let extend the measure to all of . This measure is -invariant and supported in , so it is not , and is not uniquely ergodic, contradiction.

**Everywhere convergence **

One of the nice consequences of minimality is that the -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 2Suppose is a continuous map of a compact metric space , and is uniquely ergodic with measure . Then for all and continuous ,

Suppose the contrary, and there existed a continuous function and a subsequence that converged to something other than . Consider the measures

where is the point mass at . 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 is -invariant, because

Let such a weak* limit point be . Then but

because of the choice of the sequence .

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

April 1, 2010 at 8:11 am

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