Does convergence of $\lim_{n \rightarrow \infty} \int h d\mu_n$ for all continuous, bounded $h$ imply weak convergence of $(\mu_n)$? Let $(\mu_n)$ be a sequence of positive finite Borel measures on $\mathbb{R}$. Suppose $\lim_{n \rightarrow \infty} \int h d\mu_n$ converges for every bounded, continuous function $h$ on $\mathbb{R}$. Must there exist a finite Borel measure $\mu$ on $\mathbb{R}$ such that 
$$
\lim_{n \rightarrow \infty} \int h d\mu_n = \int h d\mu
$$
for every bounded, continuous function $h$ on $\mathbb{R}$?
My initial thought was that the Lévy–Prokhorov metric for finite Borel measures might be useful. But then I realized that Cauchy-ness of the sequence $\lim_{n \rightarrow \infty} \int h d\mu_n$ may not be the same thing as Cauchy-ness of the sequence $(\mu_n)$ in the Lévy–Prokhorov metric.
 A: Let $X\sim N(0,1)$ and $\mu$ its distribution. Then $\mu$ is a probability measure in $\mathbb{R}$.
Define $\mu_n(A) = \mu(A \cap [n,n+1])$ and note that
$$ \int f d\mu_n = \int f\chi_{[n,n+1]}d\mu$$
but $f\chi_{[n,n+1]} \rightarrow 0$ a.s. and $f\chi_{[n,n+1]}$ is bounded once $f$ itself is bounded. Then by the Theorem of Dominated convergence ($\mathbb{R}$ has finite measure through $\mu$) we have that $$ \lim_n \int fd\mu_n = 0 $$ for all limited functions.
But if $\mu_n \rightarrow \eta$ weakly, then by Portmanteau, we should have $$ 0=\liminf_n \mu_n((-k,k))\ge\eta((-k,k))$$ for all $k$. And this implies $\eta \equiv 0 $
I know that $0$ is trivially a measure, but this shows that if you require a non-trivial measure the statement is false.
If you allow this kind of situation, I still don't know. Note that the hypothesis implies existence of limit for the characteristic functions. If you prove the sequence of measures is rigid so your statement is true (it is a Lévy's continuity for rigid sequences)
Sorry if I don't give you a answer, but I don't have the privilege to comment the questions.
A: We prove that the assertion is true. This has already been proved in the literature, such as in [1, Corollary 8.6.3], and the following proof is also partially motivated by the proof therein.
1. Define $T : C_b(\mathbb{R}) \to \mathbb{R}$ by
$$ T(f) = \lim_{n\to\infty} \int_{\mathbb{R}} f \, \mathrm{d}\mu_n, \qquad \forall f \in C_b(\mathbb{R}). $$
This gives rise to a well-defined functional on $C_b(\mathbb{R})$ by the assumption. It is also clear that $T$ is a positive linear functional. Moreover, if $\| f \| := \sup \left| f \right| $ denotes the supremum norm, then
$$ \left| T (f) \right| \leq T(\left| f \right|) \leq \|f\| T(1), $$
and so, $T$ is continuous. Now by the Riesz-Markov-Kakutani representation theorem, there exists a unique Borel measure $\mu$ on $\mathbb{R}$ such that
$$ T(f) = \int_{\mathbb{R}} f \, \mathrm{d}\mu, \qquad \forall f \in C_c(\mathbb{R}). $$
In other words, we know that $\mu_n \to \mu$ in a vague topology. Moreover, it is clear that $\mu(\mathbb{R}) \leq T(1)$. Indeed, if $f \in C_c(\mathbb{R})$ and $0 \leq f \leq 1$, then
$$ \int_{\mathbb{R}} f \, \mathrm{d}\mu = T(f) \leq T(1), $$
and letting $ f \uparrow 1$ along a subsequence together with the monotone convergence theorem shows the desired inequality. This $\mu$ is certainly the very candidate for the weak limit of $\mu_n$.
2. To establish this convergence, however, we need the uniform tightness of $(\mu_n)_{n\geq 1}$. Considering that the tightness condition intuitively means "no mass escapes towards the infinity", this essentially amounts to showing that
$$\mu(\mathbb{R}) = T(1) = \lim_{n\to\infty} \mu_n(\mathbb{R}). $$
In order to show this, assume otherwise that $\mu(\mathbb{R}) < T(1)$. We will prove that this leads to a contradiction. Intuitively, we will construct a sequence of "bumps" $(\rho_j)_{j\geq 1}$ that moves along with the position of the mass escaping towards the infinity. This will then be used to derive a contradiction.
To this end, write $\epsilon = \frac{T(1) - \mu(\mathbb{R})}{2020} > 0$. Then we will recursively construct a seqeunce $(\rho_j)_{j\geq 1}$ in $C_c(\mathbb{R})$ and a subsequence $\mu_{n_j}$ as follows:

*

*We will conveniently set $\chi_0 = 0$ and $n_0 = 0$.


*Suppose that $j \in \{1, 2, \dots \}$, and $\chi_{j-1} \in C_c(\mathbb{R})$ with $0 \leq \chi_{j-1} \leq 1$ and $n_{j-1}$ has been chosen. Since
$$ \lim_{n\to\infty} \int_{\mathbb{R}} \chi_{j-1} \, \mathrm{d}\mu_n
= \int_{\mathbb{R}} \chi_{j-1} \, \mathrm{d}\mu
\leq \mu(\mathbb{R})
\qquad\text{and}\qquad
\lim_{n\to\infty} \mu_n(\mathbb{R}) = T(1), $$
we can find $n_j > n_{j-1}$ such that
$$ \int_{\mathbb{R}} \chi_{j-1} \, \mathrm{d}\mu_{n_j} < \mu(\mathbb{R}) + \epsilon \qquad \text{and} \qquad \mu_{n_j}(\mathbb{R}) > T(1) - \epsilon. $$
Also, choose $r > 0$ such that $\chi_{j-1}(x) = 0$ whenever $\left| x \right| > r$. Then we can find $\chi_j \in C_c(\mathbb{R})$ such that
$$ \mathbf{1}_{[-r-1,r+1]} \leq \chi_j \leq 1 \qquad \text{and} \qquad \int_{\mathbb{R}} \chi_{j} \, \mathrm{d}\mu_{n_j} > T(1) - 2\epsilon. $$
Finally, set $\rho_j = \chi_j - \chi_{j-1}$ for each $j \in \{ 1, 2, \dots \}$. Then
$$ \int_{\mathbb{R}} \rho_{j} \, \mathrm{d}\mu_{n_j} > T(1) - \mu(\mathbb{R}) - 3\epsilon, \qquad \forall j \geq 1. \tag{*} $$
Also, the construction ensures that $0 \leq \rho_j \leq 1$ for all $j $ and that for each given $x \in \mathbb{R}$ there is at most $2$ values of $j$ for which $\rho_j(x) > 0$. Altogether, for any bounded sequence $a = (a_j)_{j\geq 1} \in \ell^{\infty}$, the function
$$ f_a = \sum_{k \geq 1} a_k \rho_k $$
defines a bounded continuous function on $\mathbb{R}$. Then the assumption tells that
$$ \int_{\mathbb{R}} f_a \, \mathrm{d}\mu_{n_j}
= \sum_{k \geq 1} a_k \int_{\mathbb{R}} \rho_k \, \mathrm{d}\mu_{n_j} $$
converges as $j\to\infty$ for any $a \in \ell^{\infty}$. This shows that the sequence $c^{(j)} = (c^{(j)}_k)_{k\geq 1} \in \ell^1$ defined by $c^{(j)}_k = \int_{\mathbb{R}} \rho_k \, \mathrm{d}\mu_{n_j} $ converges weakly as $j\to\infty$. Now it is well-known that this implies that $c^{(j)}$ converges in $\ell^1$. (See this posting, for instance.) Moreover, since
$$ \lim_{j\to\infty} c^{(j)}_k = \int_{\mathbb{R}} \rho_k \, \mathrm{d}\mu, $$
the $\ell^1$-limit of $c^{(j)}$ is the sequence $c = (c_k)_{k\geq 1}$ defined by $c_k = \int_{\mathbb{R}} \rho_k \, \mathrm{d}\mu$. So we must have
$$ \lim_{j\to\infty} \left| c^{(j)}_j - c_j \right|
\leq \lim_{j\to\infty} \| c^{(j)} - c \|_{\ell^1}
= 0. $$
On the other hand, $\lim_{j\to\infty} c_j = 0$ and $\text{(*)}$ shows that
$$ \lim_{j\to\infty} \left| c^{(j)}_j - c_j \right|
\geq T(1) - \mu(\mathbb{R}) - 3\epsilon
> 0 $$
which is a a contradiction.
3. Now we are ready to prove the claim. Fix a sequence $\chi \in C_c(\mathbb{R})$ such that $0 \leq \chi(x) \leq 1$ for all $x \in \mathbb{R}$. Then for any $ f \in C_b(\mathbb{R})$ and $n$,
\begin{align*}
\left| \int f \, \mathrm{d}\mu_n - \int f \, \mathrm{d}\mu \right|
&\leq \left| \int f \chi \, \mathrm{d}\mu_n - \int f  \chi \, \mathrm{d}\mu \right| \\
&\quad + \left| \int f(1 - \chi) \, \mathrm{d}\mu_n \right| + \left| \int f(1 - \chi) \, \mathrm{d}\mu \right| \\
&\leq \left| \int f \chi \, \mathrm{d}\mu_n - \int f  \chi \, \mathrm{d}\mu \right| \\
&\quad + \|f\| \left( \mu_n(\mathbb{R}) - \int \chi \, \mathrm{d}\mu_n \right) + \|f\| \left( \mu(\mathbb{R}) - \int \chi \, \mathrm{d}\mu \right)
\end{align*}
Letting $n\to\infty$ and using the fact that $\mu(\mathbb{R}) = \lim_{n\to\infty} \mu_n(\mathbb{R})$,
\begin{align*}
\limsup_{n\to\infty} \left| \int f \, \mathrm{d}\mu_n - \int f \, \mathrm{d}\mu \right|
&\leq 2 \|f\| \left( \mu(\mathbb{R}) - \int \chi \, \mathrm{d}\mu \right)
\end{align*}
Since the left-hand side is independent of $\chi$, letting $\chi \uparrow 1$ along a subsequence proves the desired claim.

[1] Bogachev, Vladimir Igorevich. Measure Theory. Springer, 2007.
