# Convergence of measures on a compact metric space

In the paper "Ergodic optimization" by Oliver Jenkinson, Proposition 2.1 says (among other things):

Let $$T:X\to X$$ be a continuous map on a compact metric space. If $$f:X\to\mathbb{R}$$ is either continuous, or the characteristic function of a closed subset, then

$$\sup_{\mu\in\mathcal{M}_{T}}\int fd\mu = \limsup_{n\to\infty}\frac{1}{n}\sup_{x\in X}S_{n}f(x).$$

Here $$\mathcal{M}_{T}$$ is the collection of Borel probability measures invariant under $$T$$ in $$X$$, and $$S_{n}f := \sum_{i=0}^{n-1} f\circ T^{i}$$.

I want to prove that

$$\sup_{\mu\in\mathcal{M}_{T}}\int fd\mu \geq \limsup_{n\to\infty}\frac{1}{n}\sup_{x\in X}S_{n}f(x)$$

to understand the proof. My doubt is:

Why do you have an accumulation point $$\mu$$ with respect to the weak-$$\ast$$ topology? Is it some usual property of compact metric spaces? If you could give me some reference to read about it I would appreciate it very much?

The proof in the paper: Compactness of $$X$$ means that the set $$\mathcal{M}$$ of Borel probability measures on $$X$$ is compact with respect to the weak-$$\ast$$ topology. If

$$\mu_{n} := \frac{1}{n}\sum_{i=0}^{n-1}\delta_{T^{i}x_{n}},$$

where $$x_{n}$$ is such that

$$\max_{x\in X}\frac{1}{n}S_{n}f(x) = \frac{1}{n}S_{n}f(x_{n}) = \int fd\mu_{n},$$

then the sequence $$(\mu_{n})$$ has a weak-$$\ast$$ accumulation point $$\mu$$. It is easy to see that in fact $$\mu\in\mathcal{M}_{T}$$.

Without loss of generality we shall suppose that $$\mu_{n}\to\mu$$ in the weak-$$\ast$$ topology. If $$f$$ is continuous, this means that

$$\lim_{n\to\infty}\int fd\mu_{n} = \int fd\mu \leq \sup_{\mu\in\mathcal{M}}\int fd\mu.$$

• Are you sure you have defined $\mathcal M_T$ properly? Is this the space of probability measures or the space of real/complex measures? Commented May 17, 2022 at 23:16
• This is a special case of Prokhorov's theorem. When $X$ is compact then the tightness hypothesis becomes trivial. Commented May 18, 2022 at 7:16

For a metric space $$X$$, $$X$$ is compact iff the set $$\operatorname{Prob}(X)$$ of Borel probability measures endowed with vague topology (= weakstar topology) is a compact metric space, whence compactness (w/r/t vague topology) implies sequential compactness (In a metric space, compact implies sequentially compact), that is, any sequence of Borel probability measures will be convergent to some Borel probability measure up to taking a subsequence.