almost sure convergence and convergence of expectation implies convergence in expectation

Question

IF $X_n \xrightarrow{\text{a.s.}}X$ and $\mathbf{E}[X_n] \rightarrow \mathbf{E}[X] $ does this imply $\mathbf{E}[|X_n - X|] \rightarrow 0$?

Answer 1

As Will's answer above shows, the answer is no in general, but if you also assume that all of the $X_n$ are non-negative, then the answer is yes. Proof: apply Fatou's lemma to $X_n + X - |X_n - X|$.

Answer 2

No.

For instance let $U$ be uniformly distributed on $(0,1)$ and for all $n\in\mathbb N^*$ let $X_n$ be defined by $$ X_n=-n²1_{\{U<\frac1n\}}+n^21_{\{U>1-\frac1n\}}. $$

Then $X_n\underset{n\to+\infty}{\overset{\textrm{a.s.}}{\longrightarrow}}0$ and $\mathbb E[X_n]=0\underset{n\to+\infty}{\longrightarrow}\mathbb E[0]$ but $\mathbb E[\vert X_n-0\vert]=2n\underset{n\to+\infty}{\longrightarrow}+\infty$.