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 1 Suppose
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 2
Supposeis 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
[…] Climbing Mount Bourbaki Thoughts on mathematics « Unique ergodicity […]