Convergence in distribution and moments convergence Let $(X_n)_n$ be a sequence of random variables of $L^1$ converging in distribution to X. 
Prove that $(X_n)_n$ is uniformly integrable if and only if 
$$X \in L^1 \quad \text{and} \quad \lim_n E[|X_n|]=E[|X|].$$
If $(X_n)_n$ is uniformly integrable then by Fatou's lemma we have 
$$E[|X|] \leq \liminf_n E[|X_n|] \leq \sup_n E[|X_n|]<+\infty$$
and $$\forall n,k \in \mathbb{N}, \big|E[|X_n|]-E[|X|]\big| \leq \sup_jE\left[|X_j|1_{|X_j|>k}\right]+\big|E\left[\min(|X_n|,k)\right]-E[|X|]\big|$$
By taking the limit we have $\limsup_n \big|E[|X_n|]-E[|X|]\big|=0$
for the other side, we have $\forall n,k$
$$\int_{|X_n|>k}|X_n|dP=E[|X_n|]-E[\min(|X_n|,k)]$$
So how to obtain the uniform integrability of $(X_n)_n$?

For all fixed $k$, take the $\limsup_{n\to +\infty}$ in the equality 
$$
\int_{|X_n|>k}|X_n|dP=E[|X_n|]-E[\min(|X_n|,k)].
$$
By assumption, $E[|X_n|]\to E[|X|]$ and using the definition of the convergence in distribution gives $E[\min(|X_n|,k)]\to E[\min(|X|,k)]$ hence 
$$
\limsup_{n\to +\infty}\int_{|X_n|>k}|X_n|dP= E[|X|]-E[\min(|X|,k)].
$$
