# Prove that $T$ is uniquely ergodic

Let $$T:X\rightarrow X$$ be a continuous map on a compact metric space $$(X,d)$$. Suppose that $$\mu$$ is ergodic with respect to $$T$$ and for every $$x\in X$$ there exists a constant $$C=C(x)$$ such that for every $$f \in C(X), f \geq 0$$, \begin{align*} \limsup_{N \rightarrow \infty} \frac{1}{N} \sum_{n=0}^{N-1} f (T^nx) \leq C \int f d\mu. \end{align*}

Show that $$T$$ is uniquely ergodic.

I knew the following theorem

$$\textbf{Theorem}$$ the following properties are equivalent.

(i) $$T$$ is uniquely ergodic

(ii) For every $$f\in C(X)$$, \begin{align*} A_N^f:=\frac{1}{N} \sum_{n=0}^{N-1} f(T^nx) \rightarrow C_f, \end{align*}
where $$C_f$$ is a constant independent of $$x$$.

I don't know how to induce the convergence of $$A_N^f$$ from the assumption in the problem.

Any help is appreciated...

Thank you!!

• please see my answer below – mathworker21 Nov 3 at 23:06

Lemma: Let $$\nu,\mu$$ be mutually singular probability measures on a metric space $$X$$. Then $$\sup_{\substack{f \in C(X,\mathbb{R}^{\ge 0}) \\ f \not \equiv 0}} \frac{\int f d\nu}{\int f d\mu} = +\infty.$$
Proof: Take disjoint $$A,B$$ with $$\nu(A) = 1, \mu(B) = 1$$. Take $$\epsilon > 0$$. Since finite Borel measures on a metric space are inner regular, there are closed $$A' \subseteq A$$ and $$B' \subseteq B$$ with $$\nu(A') \ge 1-\epsilon$$ and $$\mu(B') \ge 1-\epsilon$$. By Urysohn's Lemma, there is $$f \in C(X,[0,1])$$ with $$f \equiv 1$$ on $$A'$$ and $$f \equiv 0$$ on $$B'$$. For this $$f$$, we have $$\frac{\int fd\nu}{\int fd\mu} \ge \frac{1-\epsilon}{\epsilon}$$.
Suppose $$T$$ is not uniquely ergodic. Let $$\nu$$ be another ergodic measure w.r.t. $$T$$. By the ergodic theorem, for any $$f \in C(X,\mathbb{R})$$, $$\frac{1}{N}\sum_{n \le N} f(T^nx) \to \int fd\nu$$ for $$\nu$$-a.e. $$x \in X$$. Since $$X$$ is compact, $$C(X,\mathbb{R})$$ is separable (w.r.t. $$||\cdot||_\infty$$ topology), so in fact for $$v$$-a.e. $$x \in X$$, all $$f \in C(X,\mathbb{R})$$ satisfy $$\frac{1}{N}\sum_{n \le N} f(T^nx) \to \int fd\nu$$. Take one such $$x$$. Then, by hypothesis, there is some $$C$$ with $$\int fd\nu \le C \int fd\mu$$ for every $$f \in C(X,\mathbb{R}^{\ge 0})$$. This contradicts the Lemma.