Assume $f_n, ~n\in \mathbb{N}$ and $f$ are continuous functions where $f_n$ converges uniformly to $f$ and $P_n, ~n\in\mathbb{N}$ and are probability measures on $(\mathbb{R},\mathcal{B})$ where $P_n$ converges weakly to $P$.
Edit: $f$ is also bounded.
I want to prove that $$\lim\limits_{n\rightarrow \infty} \int f_n ~dP_n = \int f ~dP.$$
Solution:
Uniform convergence of $f_n$ means $$\forall \varepsilon > 0 ~\exists n_1(\varepsilon) \in \mathbb{N} ~\forall n>n_1(\varepsilon): \Vert f_n - f \Vert_\infty < \frac{\varepsilon}{2}.$$ Weakly convegence of $P_n$ means $$\forall \varepsilon > 0 ~\exists n_2(\varepsilon) \in \mathbb{N} ~\forall n>n_2(\varepsilon): \vert \int g ~dP_n - \int g ~dP \vert < \frac{\varepsilon}{2}$$ for any bounded continous function $g$.
Now I fix $\varepsilon > 0$ and $k:=\max\{n_1(\varepsilon),n_2(\varepsilon)\}$.
\begin{align} \forall n > k:~~&\vert \int f_n ~dP_n - \int f ~dP\vert \\ &= \vert \int (f_n-f) ~dP_n + \int f ~dP_n -\int f ~dP\vert \\ &\leq \int \vert f_n-f \vert ~dP_n + \vert\int f ~dP_n -\int f ~dP\vert \\ &\leq \Vert f_n - f \Vert_\infty P_n(\mathbb{R}) + \vert \int f ~dP_n -\int f ~dP\vert \\ &\leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{align}
Since, $\varepsilon$ was arbitrary the claim follows.