Unique ergodicity is equivalent to uniform ergodic averages

Assume unique ergodicity with invariant probability $\mu$. If uniform convergence to $\int fd\mu$ failed for some continuous $f$, countable choice would give $\epsilon >0$, strictly increasing $n_j$, and $x_j\in K$ such that $\mid A_{n_j}f(x_j)-\int fd\mu \mid \ge \epsilon \mathrm{.}$ Define the empirical Borel probabilities ${\eta }_j=n_j^{-1}{\sum }_{k<n_j}{\delta }_{T^k{x}_j}$; finite additivity and countable additivity of each point mass make this a probability.

By [F1], pass to a subsequence, not relabelled, and a Borel probability $\eta$ such that $\int gd{\eta }_j\to \int gd\eta$ for every $g\in C(K,\mathbb{R})$. For such $g$, $\int g\circ Td{\eta }_j-\int gd{\eta }_j=\frac{g(T^{n_j}{x}_j)-g(x_j)}{n_j}⟶0,$ because $g$ is bounded by [F4]. Taking limits shows that $\eta$ and its pullback probability $E\mapsto \eta (T^{-1}E)$ have identical continuous test integrals; [F2] makes them equal. Hence $\eta$ is invariant.

Unique ergodicity gives $\eta =\mu$, but $\int fd{\eta }_j=A_{n_j}f(x_j)$ and the closed inequality in step 1.1 passes to the limit, contradicting $\int fd\eta =\int fd\mu$. Thus $A_{n}f\to \int fd\mu$ uniformly for every $f$.

Conversely, assume all continuous averages converge uniformly to constants $c(f)$. Fix $x_0\in K$ and form ${\eta }_n=n^{-1}{\sum }_{k<n}{\delta }_{T^k{x}_0}$. By [F1], a subsequence has a continuous-test limit probability $\eta$. The telescoping calculation of step 2.1 makes $\eta$ invariant, and uniform convergence gives $\int fd\eta ={\lim }_r{A}_{n_r}f(x_0)=c(f)\mathrm{.}$

If $\rho$ is any invariant Borel probability, then [F3] gives $\int A_{n}fd\rho =\int fd\rho$. Moreover [F5] gives $\mid \int A_{n}fd\rho -c(f)\mid \le \parallel A_{n}f-c(f){\parallel }_{\mathrm{\infty }}⟶0.$ Hence $\int fd\rho =c(f)=\int fd\eta$ for every continuous real $f$. By [F2], $\rho =\eta$. Thus an invariant probability exists and is unique, and its integral is the asserted constant.

Steps 1.1–3.1 prove the forward implication and steps 3.2–4.1 prove the converse. Countable choice is used precisely through [F1] and to select the failure witnesses in step 1.1; no stronger choice principle is invoked.