Unique ergodicity

Remarks

By the Krylov–Bogolyubov theorem the set of invariant Borel probability measures is never empty, so unique ergodicity asks that it has a single element.

Example: let $X = \R/\Z$, let $\alpha$ be irrational and let $T(x) = x + \alpha \bmod 1$. If $\mu$ is invariant, its Fourier coefficients $\hat\mu(h) = \int e^{-2\pi i h x} \, d\mu(x)$ satisfy $\hat\mu(h) = e^{-2\pi i h \alpha} \hat\mu(h)$. For $h \ne 0$ the factor $e^{-2\pi i h \alpha}$ is not $1$, so $\hat\mu(h) = 0$. A measure on the circle is determined by its Fourier coefficients, so $\mu$ is Lebesgue measure, and $T$ is uniquely ergodic.

Non-example: the doubling map $T(x) = 2x \bmod 1$ on $\R/\Z$ is continuous and fixes $0$. Both the point mass at $0$ and Lebesgue measure are invariant, so $T$ is not uniquely ergodic.

Equivalent form: $T$ is uniquely ergodic if and only if, for every continuous $f \colon X \to \R$, the averages $\frac{1}{N} \sum_{n=0}^{N-1} f(T^n x)$ converge uniformly in $x$ to a constant. The constant is then $\int f \, d\mu$ for the unique invariant measure $\mu$.