Convergence in distribution and weak convergence implies convergence of expectations

Question

I'm preparing myself for the final exam of my graduate Probability Theory course and was stuck with one of the exercises our professor gave us.

Let $x_n, n=1,2,\cdots$, and $X$ be random variables satisfying \begin{equation}\lim_{n\rightarrow \infty}X_n = X\text{ in distribution.}\end{equation} Let $f_n,n=1,2\cdots,$ and $f$ be bounded and continuous functions on $(-\infty,\infty).$ Suppose that $f_n \rightarrow f$ uniformly.

Prove or disprove $$\lim_{n\rightarrow \infty}\mathbb{E}[f_n(X_n)] = \mathbb{E}[f(X)].$$

Since $f_n$ is bounded and continuous, I can easily show that $\mathbb{E}[f_n(X_k)] \rightarrow \mathbb{E}[f_n(X)]$ (by applying the continuous mapping theorem), but I am not sure how to include the uniform convergence property to finish the proof.

I've looked at other questions, but have not found anything resembling my question except Uniform convergence in distribution, which seems to leave out the part of $f_n \rightarrow f$.

Answer 1

$$|\mathbb{E}f_n(X_n)-\mathbb{E}f(X)|\le \mathbb{E}|f_n(X_n)-f(X)|\le \mathbb{E}[|f_n(X_n)-f(X_n)|+|f(X_n)-f(X)|]\le \sup_{x\in\mathbb{R}}|f_n(x)-f(x)|+\mathbb{E}|f(X_n)-f(X)|\rightarrow 0 \text{ as } n\rightarrow \infty$$

because $f_n$ converges uniformly. For the second term use a.s. representation and the dominated convergence theorem (because $f$ is bounded).