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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?

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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, $\mu_n \xrightarrow{*} 0$.

This does not happen in your case.

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.

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    $\begingroup$ Thanks for your detailed answer! This helped a lot. $\endgroup$ – njirons Aug 29 at 1:16

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