We will now apply the machinery already developed to a few concrete problems.

Proposition 1 Let {G} be a compact abelian group and {T} the rotation by {a \in G}. Then {T} is uniquely ergodic (with the Haar measure invariant) if {a^{\mathbb{Z}}} is dense in {G}.


The proof is straightforward. Suppose {\mu} is invariant with respect to rotations by {a}. Then for {f \in C(G)}, we have

\displaystyle \int f(a^m x ) d \mu = \int f(x) d \mu, \quad \forall m \in \mathbb{Z}

and hence

\displaystyle \int f(bx ) d \mu = \int f(x) d \mu, \quad \forall m \in \mathbb{Z},

for any {b \in G}, which means that {\mu} must be Haar measure (which is unique).

Corollary 2 An irrational rotation of the unit circle {S^1} is uniquely ergodic.


Application: Equidistribution


Theorem 3 Let {\xi \in \mathbb{R}} be irrational and let {f: \mathbb{R} \rightarrow \mathbb{C}} be continuous and {2 \pi }-periodic. Then\displaystyle \boxed{ \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{i=0}^{N-1} f( n \xi) = \int_0^1 f(x) dx .} (more…)

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. (more…)