Theorem: Vague Convergence of Uniformly Bounded Sequence of Finite Borel Measures implies Convergence of Integrals of Bounded Continuous Functions Vanishing at Infinity

Let $M > 0$.
Let $\mu$ be a finite Borel measure on $\R$ with $\map \mu \R \le M$. 
Let $\sequence {\mu_n}_{n \mathop \in \N}$ be a sequence of finite Borel measures converging vaguely to $\mu$ such that:
:$\map {\mu_n} \R \le M$ for each $n \in \N$.
Let $f : \R \to \R$ be continuous such that:
:$\ds \lim_{\size x \mathop \to \infty} \map f x = 0$
Then:
:$\ds \int f \rd \mu_n \to \int f \rd \mu$

Proof:
Let $\epsilon > 0$.
Since:
:$\ds \lim_{\size x \mathop \to \infty} \map f x = 0$
there exists $K > 0$ such that:
:$\ds \size {\map f x} < \frac \epsilon 2$ for $\size x \ge K$.
Define $f_1 : \R \to \R$ by:
:$\ds \map {f_1} x = \begin {cases} \map f x & : x \in \closedint {-K} K \\ \paren {x - K + 1} \map f K & : x \in \closedint K {K + 1} \\ \paren {x + K + 1} \map f {-K} & : x \in \closedint {-K - 1} {-K} \\ 0 & : \text {otherwise} \end {cases}$
Note that since $\map {f_1} x = 0$ for $x \in \R \setminus \closedint {-K - 1} {K + 1}$, $f_1$ has compact support.
Further, we can see that $f_1$ is continuous. 
Hence, from the definition of vague convergence, we have:
:$\ds \int f_1 \rd \mu_n \to \int f_1 \rd \mu$
as $n \to \infty$.  
Define $f_2 : \R \to \R$ by $\map {f_2} x = \map f x - \map {f_1} x$ for each $x \in \R$.
We show that $\size {\map {f_2} x} < \epsilon$ for all $x \in \R$. 
For $x \in \closedint {-K} K$, we have:
:$\map {f_1} x = \map f x$
and hence:
:$\map {f_2} x = 0$
For $x \in \R \setminus \closedint {-K - 1} {K + 1}$, we have:
:$\map {f_1} x = 0$
and hence:
:$\map {f_2} x = \map f x$
Since $\size x \ge K$ in this case, we have:
:$\size {\map {f_2} x} < \dfrac \epsilon 2 < \epsilon$
Now, for $x \in \closedint K {K + 1}$, we have:
{{begin-eqn}}
{{eqn	| l = \size {\map {f_2} x}
		| o = \le
		| r = \size {x - K + 1} \size {\map f K} + \size {\map f x}
		| c = Triangle Inequality
}}
{{eqn	| o = <
		| r = \frac \epsilon 2 + \frac \epsilon 2
		| c = since $\size {x - K + 1} \le 1$ for $x \in \closedint K {K + 1}$ and $\ds \size {\map f x} < \frac \epsilon 2$ for $\size x \ge K$
}}
{{eqn	| r = \epsilon 
}}
{{end-eqn}}
Finally, for $x \in \closedint {-K - 1} {-K}$, we have:
{{begin-eqn}}
{{eqn	| l = \size {\map {f_2} x}
		| o = \le
		| r = \size {x + K + 1} \size {\map f {-K} } + \size {\map f x}
		| c = Triangle Inequality
}}
{{eqn	| o = <
		| r = \frac \epsilon 2 + \frac \epsilon 2
		| c = since $\size {x + K + 1} \le 1$ for $x \in \closedint {-K - 1} {-K}$ and $\ds \size {\map f x} < \frac \epsilon 2$ for $\size x \ge K$
}}
{{eqn	| r = \epsilon 
}}
{{end-eqn}}
We now have:
{{begin-eqn}}
{{eqn	| l = \size {\int f \rd \mu - \int f \rd \mu_n}
		| r = \size {\int f_1 \rd \mu + \int f_2 \rd \mu - \int f_1 \rd \mu_n - \int f_2 \rd \mu_n}
		| c = Integral of Integrable Function is Additive
}}
{{eqn	| o = \le
		| r = \size {\int f_1 \rd \mu - \int f_1 \rd \mu_n} + \size {\int f_2 \rd \mu - \int f_2 \rd \mu_n}
}}
{{end-eqn}}
Since:
:$\ds \int f_1 \rd \mu_n \to \int f_1 \rd \mu$ as $n \to \infty$
there exists $N \in \N$ such that:
:$\ds \size {\int f_1 \rd \mu - \int f_1 \rd \mu_n} < \epsilon$
for $n \ge N$. 
From Triangle Inequality and Triangle Inequality for Integrals, we have:
:$\ds \size {\int f_2 \rd \mu - \int f_2 \rd \mu_n} \le \int \size {f_2} \rd \mu + \int \size {f_2} \rd \mu_n$
From Integral of Integrable Function is Monotone and Integral of Characteristic Function, we have:
:$\ds \int \size {f_2} \rd \mu + \int \size {f_2} \rd \mu_n \le \epsilon \map \mu \R + \epsilon \map {\mu_n} \R \le 2 M \epsilon$ 
Hence we have:
:$\ds \size {\int f \rd \mu - \int f \rd \mu_n} \le \paren {2 M + 1} \epsilon$
for $n \ge N$.
Since $\paren {2 M + 1} \epsilon \to 0$ as $\epsilon \to 0^+$, and $\epsilon > 0$ was arbitrary, we have:
:$\ds \size {\int f \rd \mu - \int f \rd \mu_n} \to 0$
as desired.
{{qed}}