Show that every measure is the vague limit of measures with finite support Let $\mu: \mathbb{B}(\mathbb{R}) \to [0, \infty[$ be a finite Borel-measure. Show that there exists a sequence of Borel-measures $\{\mu_n\}_{n=1}^\infty$ with finite supports such that 
$$\mu_n \stackrel{v}{\to} \mu, \quad\sup_{n=1}^\infty \mu_n(\mathbb{R}) < \infty$$
Here $\mu_n \stackrel{v}{\to} \mu$ means vague convergence. This means that for every $a,b \in \mathbb{R}$ with $\mu\{a\}= \mu\{b\}=0$ we have
$$\mu_n(]a,b])\stackrel{n \to \infty}\to \mu(]a,b])$$
Equivalently, 
$$\int fd\mu_n \to \int fd \mu$$
for all continuous compactly supported functions $f: \mathbb{R} \to \mathbb{R}$.
Attempt:
I am hinted to use the sequence
$$\mu_n = \sum_{j=-n2^n +1}^{n2^n}\mu\left(\left]\frac{j-1}{2^n}, \frac{j}{2^n}\right]\right) \delta_{\{j/2^n\}}$$
where $\delta_{a}$ is the Dirac measure at $a \in \mathbb{R}$.
I'm a bit unsure how to prove this. Using the definition seems a little bit tedious. Maybe I can associate the distribution functions 
$$F_n(x) = \mu_n(]-\infty, x]), \quad F(x) = \mu(]-\infty, x]),  x \in \mathbb{R}$$
and show that for $x \notin D(F)$ (the points where $F$ is not continuous, or equivalently where the $\mu$-measure of the singelton is non-zero) we have
$$F_n(x) \to F(x)$$
Any help into the right direction is appreciated!
 A: I will use the characterization in terms of compactly supported functions that you mention.
Let $\epsilon > 0$.
Since $f$ is uniformly continuous, there is $\delta > 0$ such that $|f(x) - f(y)| \leq \epsilon / (1 + \mu(\Bbb{R}))$ for $|x-y| \leq \delta$.
Now, choose $n_0 = n_0(f, \delta)$ so large that $2^{-n_0} \leq \delta$ and $\mathrm{supp} f \subset (-n_0,n_0]$, and let $n \geq n_0$ be arbitrary.
If you define $F := \sum_{j = -n 2^n + 1}^{n 2^n} 1_{((j-1)/2^n, j/2^n]} f(j/2^n)$, then $|F(x) - f(x)| \leq \epsilon$ for all $x$. Indeed, if $x \notin (-n,n] \supset (-n_0,n_0]$, then $F(x) = f(x) = 0$. Otherwise, we have $x \in ((j-1)/2^n, j/2^n]$ for a unique $j \in \{-n2^n + 1,\dots,n2^n\}$, and hence $|F(x) - f(x)| = |f(j/2^n) - f(x)| \leq \epsilon / (1 + \mu(\Bbb{R}))$, since $|x - j/2^n| \leq 2^{-n} \leq \delta$.
Finally, note that $\int F \, d \mu = \int f \, d \mu_n$ (why?!).
Therefore
$$
\bigg|
 \int f d \mu_n - \int f d \mu
\bigg|
= \bigg|
    \int F d \mu - \int f d \mu
  \bigg|
\leq \int |F - f| d \mu
\leq \frac{\epsilon}{1 + \mu(\Bbb{R})} \cdot \int 1 d \mu
\leq \epsilon,
$$
for all $n \geq n_0$.
Since this holds for any compactly supported, continuous $f$, we are done.
