Convergence of measures — revisited In this thread, I asked a question about the convergence of measures. The conjecture I posed there, which turned out to be false, was supposed to be a lemma that I wanted to use to prove a proposition, outlined as follows.
Consider the measurable space $(\mathbb R,\mathscr B_{\mathbb R})$. Suppose that $\{\mu_n\}_{n=1}^{\infty}$ is a sequence of finite Radon measures and $\mu$ is a finite Radon measure on $\mathbb R$. Suppose also that $$\lim_{n\to\infty}\mu_n(\mathbb R)=\mu(\mathbb R)\tag{1}$$
and that $\{\mu_n\}_{n=1}^{\infty}$ converges to $\mu$ vaguely, in the sense that $$\lim_{n\to\infty}\int f\,\mathrm d\mu_n=\int f\,\mathrm d\mu$$
for any $f\in C_c$, where $C_c$ is the space of compactly supported, continuous, real-to-real functions. For any $x\in\mathbb R$ and $n\in\mathbb N$, define
\begin{align*}F_n(x)\equiv&\,\mu_n((-\infty,x]),\\F(x)\equiv&\,\mu((-\infty,x]).\end{align*}
Proposition:$\quad$ For any $a\in\mathbb R$ and $\varepsilon>0$, $$\limsup_{n\to\infty}F_n(a)\leq F(a+\varepsilon).$$
I figured out a proof that I think is correct, which I now share in order to seek feedback. Your thoughts are much appreciated.
Proof:$\quad$ Let $a\in\mathbb R$ and $\varepsilon>0$ be given. Also, fix $\delta>0$. Since $\mu$ is a finite Radon measure, there exists a compact set $C\subset\mathbb R$ such that $$\mu(\mathbb R)\geq\mu(C)>\mu(\mathbb R)-\frac{\delta}{5}.$$ Since compact sets are bounded, one can choose $K>0$ so large that $$-K<a<a+\varepsilon<K,$$ $C\subseteq[-K,K]$, and $$\mu(\mathbb R)\geq\mu([-K,K])>\mu(\mathbb R)-\frac{\delta}{5}.\tag{2}$$
Pick any $L>K$. Clearly, $[-K,K]\subseteq[-L,L]$. Construct a continuous function $h$ such that $h$ is 1 on $[-K,K]$, it vanishes outside $[-L,L]$, and it is linear on $[-L,-K]$ and on $[K,L]$. Then, $h\in C_c$. By vague convergence, there exists some $N_1\in\mathbb N$ such that $$\int h\,\mathrm d\mu<\int h\,\mathrm d\mu_n+\frac{2\delta}{5}\quad\forall n\geq N_1.\tag{3}$$
Also, by (1), there exists some $N_2\in\mathbb N$ such that $$\mu(\mathbb R)+\frac{\delta}{5}>\mu_n(\mathbb R)>\mu(\mathbb R)-\frac{\delta}{5}\quad\forall n\geq N_2\tag{4}.$$
Now, if $n\geq\max\{N_1,N_2\}$, then, on the one hand,
\begin{align*}\mu([-K,K])=&\,\int_{[-K,K]}h\,\mathrm d\mu\leq\int h\,\mathrm d\mu<\int h\,\mathrm d\mu_n+\frac{2\delta}5\\=&\,\int_{[-L,L]}h\,\mathrm d\mu_n+\frac{2\delta}5\leq\mu_n([-L,L])+\frac{2\delta}5;\end{align*}
and, on the other hand,
$$\mu_n([-L,L])\leq\mu_n(\mathbb R)<\mu(\mathbb R)+\frac{\delta}{5}<\mu([-K,K])+\frac{2\delta}{5},$$
by (2) and (4). Therefore,
$$\mu_n([-L,L])-\frac{2\delta}{5}<\mu([-K,K])<\mu_n([-L,L])+\frac{2\delta}{5}\quad\forall n\geq\max\{N_1,N_2\}\tag{5}.$$
Let $M>L$ be arbitrary. Construct a continuous function $g$ such that $g$ is 1 on $[-L,a]$, it vanishes outside $[-M,a+\varepsilon]$, and it is linear on $[-M,-L]$ and on $[a,a+\varepsilon]$. Then, $g\in C_c$. By vague convergence, there exists some $N_3\in\mathbb N$ such that $$\int g\,\mathrm d\mu_n<\int g\,\mathrm d\mu+\frac{\delta}{5}\quad\forall n\geq N_3.\tag{6}$$
Let $n\geq N^*\equiv\max\{N_1,N_2,N_3\}$. Then,
\begin{align*}
F_n(a)=&\,\mu_n((-\infty,a])=\mu_n((-\infty,-L))+\mu_n([-L,a])\\
=&\,\mu_n(\mathbb R)-\mu_n([-L,L])-\mu_n((L,\infty))+\mu_n([-L,a])\\
\leq&\,\mu_n(\mathbb R)-\mu_n([-L,L])+\mu_n([-L,a])\\
\underbrace{<}_{(5)}&\,\mu_n(\mathbb R)-\mu([-K,K])+\frac{2\delta}5+\mu_n([-L,a])\\
\underbrace{<}_{(4)}&\,\mu(\mathbb R)-\mu([-K,K])+\frac{3\delta}5+\mu_n([-L,a])\\
\underbrace{<}_{(2)}&\,\frac{4\delta}5+\mu_n([-L,a])=\int_{[-L,a]}g\,\mathrm d\mu_n+\frac{4\delta}{5}\\
\leq&\,\int g\,\mathrm d\mu_n+\frac{4\delta}{5}\underbrace{<}_{(6)}\int g\,\mathrm d\mu+\delta=\int_{[-M,a+\varepsilon]}g\,\mathrm d\mu+\delta\\
\leq&\,\mu([-M,a+\varepsilon])+\delta\leq\mu((-\infty,a+\varepsilon])+\delta=F(a+\varepsilon)+\delta.
\end{align*}
Since this is true for all $n\geq N^*$, it follows that
$$\limsup_{n\to\infty} F_n(a)=\inf_{k\in\mathbb N}\sup_{n\geq k}F_n(a)\leq \sup_{n\geq N^*}F_n(a)\leq F(a+\varepsilon)+\delta.$$
Given that $\delta>0$ can be made arbitrarily small, the claim follows. $\blacksquare$
Note: this proposition is needed to prove Proposition 7.19(b) in Folland (1999). The earlier printings report the result without the assumption (1), in which case the proposition I presented can be shown to be false. The errata sheet points out that the claim is true if (1) is assumed, but gives no proof. That's why I attempted to construct one. Any comments are welcome.
 A: This turned out to be too long to comment so I'm writing it as an answer even though it is a different question. In this case, it is not the escape of mass but the "placement" of mass. Usually we try to simplify weak convergence results by restricting to a nice subset. To do this we really require $\mu_n(X) = 1$ for all sufficiently large $n$.
To see this, take $X = \mathbb{Q} \cap [0,1]$ and define $\mu_n$ on $[0,1]$ such that $\mu_n(X) = 0$, $\mu_n([0,1]) = 1$ and take $\mu(X) = \mu([0,1]) = 1$. 
(For instance, one can take $\{r_i\}_{i \geq 1}$ to be an enumeration of the rationals, set $\mu = \sum_{i=1}^\infty \delta_{r_i}2^{-i}$, $\mu_n = \sum_{i=1}^\infty \delta_{r_i + \pi/n}2^{-i}$.)
Notice that for any $f \in C_b([0,1])$, 
$$\delta_{r_i + \pi/n} f = f(r_i + \pi/n) \rightarrow f(r_i) = \delta_{r_i}f,$$
and since we can bound the tail of $\mu$ and $\mu_n$, $\mu_n \Rightarrow \mu$. From the mapping theorem, one expects $\mu_n h^{-1} \Rightarrow \mu h^{-1}$ for $h$ such that $\mu(D_h) = 0$, where $D_h$ is the set of discontinuities of $h$. 
Consider
$$h(x) = \begin{cases} 0 & x \in \mathbb{Q} \\ 1 & x \in [0,1] \setminus\mathbb{Q}\end{cases}.$$
Notice that $h|_X$ is continuous in the sense that for any $x_n \rightarrow x$ for $x_n, x \in X$, we have $h(x_n) \rightarrow h(x)$. However, 
$$1 = \mu_n([0,1]\setminus \mathbb{Q}) = \mu_n h^{-1}(\{1\}) \geq \mu h^{-1}(\{1\}) = 0.$$
Since $\{1\}$ is a closed set, we do not have
$$\limsup_{n\rightarrow\infty} \mu_n h^{-1}(F) \leq \mu h^{-1}(F)$$
for all closed $F$, so $\mu_n h^{-1} \not\Rightarrow \mu h^{-1}.$ 
