# Weak and weak-* convergence of probability measures

Assume $$\mu_n,\mu$$ are Borel probability measures on $$X=\mathbb{R}$$ (or more generally on a Polish space $$X$$). Assume also that $$\mu_n\stackrel{*}{\rightharpoonup}\mu$$, i.e., $$\int f\ d\mu_n\to\int f\ d\mu$$ for all continuous functions $$f\in C_0(X)$$ vanishing at infinity. Are there necessary conditions, either on $$\mu$$ or $$X$$, implying that $$\mu_n\rightharpoonup\mu$$, i.e., $$\int f\ d\mu_n\to\int f\ d\mu$$ for all continuous bounded functions $$f\in C_b(X)$$? I know that if $$X$$ is compact then these two notions of convergence are equivalent, so I am looking for weaker conditions than compactness of $$X$$ or $$\text{supp}(\mu)$$. In light of Prokhorov's theorem, we can also ask: under what conditions on $$X$$ or $$\mu$$ is a weak-* convergent sequence tight? Perhaps if $$\mu$$ satisfies some integrability condition, e.g., finiteness of the first moment?

Since you already know that $$\mu$$ is a probability, it means that no measure is "leaking" to $$\infty$$. For example, take $$\mu_n(A) = \begin{cases}1,&n \in A \\ 0,& n \not \in A\end{cases}.$$ In this case, $$\mu_n \xrightarrow{*} 0$$.
Now, for any $$\varepsilon > 0$$, you can choose a compact set $$K \subset \mathbb{R}$$ such that $$\mu(K) > 1 - \varepsilon$$. And you can pick up some $$g \in C_0$$ such that $$g|_K = 1$$, and $$0 \leq g \leq 1$$. Notice that if $$n$$ is big enough, $$\int g \,\mathrm{d}\mu_n \geq \int g \,\mathrm{d}\mu - \varepsilon \geq \mu(K) > 1 - \varepsilon.$$ Therefore, for any big enough $$n$$, $$\int (1-g) \,\mathrm{d}\mu_n < \varepsilon.$$
So, for any bounded continuous $$f$$, for any big enought $$n$$, $$\left| \int f \,\mathrm{d}\mu_n - \int gf \,\mathrm{d}\mu_n \right| = \left| \int (1-g)f \,\mathrm{d}\mu_n \right| \leq \sup |f| \varepsilon.$$ Consider $$\left| \int f \,\mathrm{d}\mu_n - \int f \,\mathrm{d}\mu \right| \leq \left| \int f \,\mathrm{d}\mu_n - \int gf \,\mathrm{d}\mu_n \right| + \left| \int gf \,\mathrm{d}\mu_n - \int gf \,\mathrm{d}\mu \right| + \left| \int gf \,\mathrm{d}\mu - \int f \,\mathrm{d}\mu \right|.$$ We have already shown that the first term of the right side can be made less then $$\sup |f| \varepsilon$$ for $$n$$ big enough. The second term can be made small, for $$\mu_n \xrightarrow{*} \mu$$, and $$gf \in C_0$$. And the last term is also less then $$\sup |f| \varepsilon$$, by the choice of $$K$$. Since $$\varepsilon$$ was arbitrary, $$\int f \,\mathrm{d}\mu_n \rightarrow \int f \,\mathrm{d}\mu$$.
The key point is that $$\mu$$ is a probability measure. The difference between the two kinds of convergence is that one might converge to a non probability while the other diverges.