If $E(X_n) \to E(X)$, does this imply convergence in $L^1$?

Question

My question is basically if there exists any counterexample where we have that $\mathbb{E}(X_n)\to \mathbb{E}(X)$ as $n\to \infty$, but $X_n$ does not converge to $X$ in $L^1$. If not, then it should be true that $\mathbb{E}(X_n)\to \mathbb{E}(X)$ as $n\to \infty$ implies that $X_n \overset{L^1}{\rightarrow}X$ but I cannot really get to prove that one.

Any tips on how to approach this problem? Or any counterexample you could maybe think of?

Answer 1

Let $X$ be 1 with probability 0.5 and 0 with probability 0.5. Then let $X_n$ be copies of $X$. Let $Y = 1-X$. Then $\mathbb E[X_n] = 0.5 = \mathbb E[Y]$ but $\lVert X_n - Y \rVert_{L^1} = \Vert X - Y \rVert_{L^1} = 1$ for all $n$, so we don't have $L^1$ convergence.

Answer 2

Suppose that all $X_n$ and $X$ are defined on different probability spaces and have standard normal distribution. In this case $E X_n = 0 \to 0 = EX$ and there's no convergence in $L_1$.