Convergence of expecations implies convergence of positive and negative parts?

Question

If we have $E|X_n| \rightarrow E|X|$ does that imply \begin{equation} \lim_{n\rightarrow\infty} E X_n^\pm = X^\pm \end{equation} How about if we only have $EX_n \rightarrow EX$? Is this true in general for real numbers?

Edit: I think the answer is probably no, but how about if we have an additional condition, say $\liminf_{n\rightarrow\infty} EX_n^+ = EX^+$ and $\liminf_{n\rightarrow\infty}EX_n^- = EX^-$, does that imply the limsups are also equal and so we have the result?

Answer 1

1. $\mathbb{E}|X_n| \to \mathbb{E}|X|$: No, this does not imply the statement. Simply consider $X_n := 1$ for $n$ odd and $X_n := -1$ for $n$ even.
2. $\mathbb{E}X_n \to \mathbb{E}X$: Again, no. Consider $$X_n \sim \frac{1}{2} \delta_n + \frac{1}{2} \delta_{-n}.$$
3. With the additional condition: Consider the sequence $(X_1,\ldots,X_k,0,X_{k+1},\ldots,X_{2k},0,\ldots)$ for some $k \geq 1$ where $(X_n)_n$ is as in 2.

Remark. A sufficient condition is e.g. that $X_n \to X$ in $L^1$.