We're given the following assumption:
Let $X = [0, 1]$ and $T: X \rightarrow X$ be a continuous map. Suppose that there exists a measure $\mu$ which is ergodic with respect to $T$ such that for every $x \in X$ there exists constant $c > 0$ such that for all $f \in C(X)$ with $f \geq 0$ $$\limsup_{N \rightarrow \infty} \frac{1}{n} \sum_{i=0}^{n-1}f(T^ix) \leq c \int_X f d\mu$$
Now I want to show that $\mu$ is the unique ergodic measure.
I found a helpful answer here: https://math.stackexchange.com/a/3346509/715494
However, the proof is not quite clear to me. Firstly, I think there was a typo when they wrote "so in fact for $v$-a.e". They meant "$\nu$-a.e." here, correct? If so, how do we conclude from the fact that $C(X, \mathbb{R})$ is separable, that the statement actually holds $\nu$-a.e. and not only $\mu$-a.e.?
Then we reach $\int f\nu \leq C \int f d\mu$ and they say that this contradicts the lemma. Can someone explain how this contradicts the lemma?
$C(X,\mathbb{R})$ being separable means that there is a countable dense subset of it, say $(f_n)_{n\in \mathbb{N}}$. Let $$A_n=\{x\in X: \frac{1}{M} \sum_{m=1}^M f_n(T^mx) \xrightarrow{M\to \infty} \int f_n\ d\nu \}\ ,$$ for $n\in \mathbb{N}$. Then $\nu(A_n)=1$, for any $n\in \mathbb{N}$, hence $\nu(\bigcap_{n\in \mathbb{N}} A_n)=1$. Now notice that for $x\in \bigcap_{n\in \mathbb{N}} A_n $ (i.e. for $\nu$-a.e. $x$ we have that $$\frac{1}{M} \sum_{m=1}^M f_n(T^mx) \xrightarrow{M\to \infty} \int f_n\ d\nu\ ,\ \text{simultaneously for all} \ n\in \mathbb{N}.$$ By the density of $(f_n)_{n\in \mathbb{N}}$ w.r.t. the $|| \cdot ||_{\infty}$-norm and a typical approximation argument we can recover the first claim (you should be able to work this out).
For the second claim, observe first of all that two distinct ergodic measures $\mu , \nu$ must be mutually singular (to see this argue by contradiction and use the ergodic theorem which implies that for any measurable $A \subset X$ we have that the limit $\lim_{N\to \infty} \frac{1}{N} \sum_{n=1}^N \mathbb{1}_A(T^nx)$ is a constant for both $\mu$ and $\nu$-almost all $x\in X$).
Then, for any $f \in C(X,\mathbb{R}^{\geq 0})$ and $f\not\equiv 0$ we have shown that $\int f\ d\nu \leq C \int f\ d\mu \implies \frac{\int f\ d\nu}{\int f\ d\mu} \leq C$, contradicting the lemma.