Convergence in distribution and moments convergence

Question

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 $(X_n)_n$?

Answer 1

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 $$ \limsup_{n\to +\infty}\int_{|X_n|>k}|X_n|dP= E[|X|]-E[\min(|X|,k)]. $$