I came across the following problem on Ergodic Theory with a view towards number theory textbook(Exercise 4.2.5.):
Let $T: X\to X$ be a continuous map on a compact metric space. Show that the measures in $\mathscr E^T(X)$ constrain all the ergodic averages in the following sense. For $f\in C(X)$, define
$$ m(f)=\inf_{u\in\mathscr E^T(X)}\left\{\int f d\mu\right\} $$ and $$ M(f)=\sup_{u\in\mathscr E^T(X)}\left\{\int f d\mu\right\}. $$ Prove that
$$ m(f)\le\liminf_{N\to\infty} A_N^f(x)\le\limsup_{N\to\infty} A_N^f(x)\le M(f) $$ for any $x\in X$.
We know that $ \limsup_{N\to\infty} A_N^f(x) $ converges to $\int f d\mu$ pointwise a.e. for any ergodic measure. But how to show that it is actually bounded for every point? I have no idea how to prove this.
Fix $f \in C(X)$. Fix $x \in X$. Say $\frac{1}{N_k}\sum_{n \le N_k} f(T^nx) \to \limsup_{N \to \infty} A_N^f(x)$. Let $\mu$ be a weak-$*$ limit point of the measures $\frac{1}{N_k}\sum_{n \le N_k} \delta_{T^n x}$. It's easy to see $\mu$ is $T$-invariant. Let $\lambda$, a probability measure on $\mathscr{M}^T(X)$, be the ergodic decomposition of $\mu$ (c.f. Theorem 4.8). Then, $$\limsup_{N \to \infty} A_N^f(x) = \int_X fd\mu = \int_{\mathscr{E}^T(X)} \left(\int_X fd\nu\right) d\lambda(\nu) \le \int_X f d\nu_0$$ for some $\nu_0 \in \mathscr{E}^T(X)$, as desired.