Uniqueness of vague limits. Let $\mu_n, n \geq 1, \mu$ be finite measures on $\mathbb{B}(\mathbb{R})$. We say that 
$$\mu_n \stackrel{v}{\to} \mu $$
i.e. $\mu_n\to \mu$ vaguely if $\mu_n]a,b] \to \mu]a,b]$ for all $a < b$ with $\mu\{a\}= \mu\{b\} = 0$. Equivalently, 
$$\int_\mathbb{R} fd  \mu_n \to \int_\mathbb{R} f d \mu$$
for all continuous compactly supported functions.
Is this limit measure unique? 
I know that if $\mu_n \to \mu$ weakly, then the limit is unique but this is stronger. Can I maybe use this result by approximating continuous functions by functions with compact support or something like that?
 A: Yes, it is unique. Suppose that $\mu_n \to \mu$ and $\mu_n \to \nu$ for finite measures $\mu$ and $\nu$. By the definition of vague convergence, we have
$$\int f \, d\mu = \int f \, d\nu$$
for all compactly supported continuous functions $f$. Now take a compact set $K \subseteq \mathbb{R}$ and define
$$U_k := K+B(0,1/k) \quad \text{and} \quad f_k(x) := \frac{d(x,U_k^c)}{d(x,U_k^c)+d(x,K)}$$
where
$$d(x,A) := \inf_{y \in A} |x-y|.$$
Then $f_k \downarrow 1_K$ as $n \to \infty$, and therefore it follows from the monotone convergence theorem that
$$\mu(K) = \inf_{k \in \mathbb{N}} \int f_k \, d\mu = \inf_{k \in \mathbb{N}} \int f_k \, d\nu = \nu(K).$$
This shows that $\mu(K)=\nu(K)$ for every compact set $K$. Since the compact sets are a $\cap$-stable generator of the Borel $\sigma$-algebra, it follows from the uniqueness of measure theorem (see below) that $\mu(B)=\nu(B)$ for all Borel sets $B$.

Theorem: Let $\mu,\nu$ be finite measures on a measurable space $(X,\mathcal{A})$. Let $\mathcal{G} \subset \mathcal{A}$ be such that $\mathcal{A}=\sigma(\mathcal{G})$ and $\mathcal{G}$ is finite under intersections (i.e. $G,H \in \mathcal{G} \implies G \cap H \in \mathcal{G}$). If $$\forall G \in \mathcal{G} \::\: \mu(G)=\nu(G),$$ then $$\forall A \in \mathcal{A}\::\: \mu(A)=\nu(A).$$

