This is from a second undergraduate course in probability.
Show if $X_n \overset{L^2}{\rightarrow} X$, $f$ a uniformly continuous and bounded function, then $\mathbb{E}f(X_n) \rightarrow \mathbb{E}f(X)$.
I suppose having $f$ bounded means the expected values are finite. Uniform continuity is used to prove that if $X_n \overset{d}{\rightarrow} X$, then $g(X_n) \overset{d}{\rightarrow} g(X)$. And characteristic functions are uniformly continuous. But that material from the course doesn't seem helpful.
How do you start?
$X_n\to^P X$ is sufficient. Let $\varepsilon >0$. There exists $\delta>0$ s.t. for all $x,y$ we have $|x-y|<\delta \implies |f(x)-f(y)|<\varepsilon$ so $$\begin{aligned} |E[f(X_n)]-E[f(X)]|&\stackrel{\textrm{Jensen}}\leq E[|f(X_n)-f(X)|]\\ &=E[\mathbf{1}_{\{|X_n-X|<\delta\}}|f(X_n)-f(X)|]+E[\mathbf{1}_{\{|X_n-X|\geq \delta\}}|f(X_n)-f(X)|]\\ &\leq \varepsilon P(|X_n-X|<\delta)+2\|f\|_\infty P(|X_n-X|\geq \delta)\end{aligned}$$ Since $X_n\to ^P X$ we have $\limsup_{n\to \infty}|E[f(X_n)]-E[f(X)]|\leq \varepsilon$.