Let $\displaystyle (X,d)$ a metric space, $\displaystyle \mathcal{B}(X)$ the algebra of its Borel sets, $\displaystyle (\mathbb{P}_n)_n$ a sequence of probability measures on $\displaystyle (X,\mathcal{B}(X))$ and $\displaystyle C_b(X)$ the space of bounded continuous functions on $\displaystyle X$.

We say the sequence $\displaystyle \mathbb{P}_n$ converges weakly (e.g. in distribution) to the probability measure $\displaystyle \mathbb{P}$ if, for each$\displaystyle f\in C_b(X)$:

$\displaystyle \lim_{n\rightarrow\infty}\int_Xf(x)d\mathbb{P}_n(x )=\int_Xf(x)d\mathbb{P}(x)$

Now suppose that, given a sequence of measures $\displaystyle (\mathbb{P}_n)_n$ on $\displaystyle (\mathbb{R},\mathcal{B}(\mathbb{R}))$, we have:

$\displaystyle \lim_{n\rightarrow\infty}\int_{\mathbb{R}}f(x)d\ma thbb{P}_n(x)=\int_{\mathbb{R}}f(x)d\mathbb{P}(x)$ for every infinitely differentiable function $\displaystyle f$ with compact support.

How to you prove this implies that $\displaystyle (\mathbb{P}_n)_n$ converges weakly to $\displaystyle \mathbb{P}$?

The Weierstrass theorem only applies to continuous functions on a compact, so it can't be used here. I wouldn't be surprised if this is a classic, but I would greatly appreciate some help.