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…)