Showing $(X-X_n)^+$ being uniformly integrable and $X_n\rightarrow X$ in $L^1$

Question

Suppose $X_n\rightarrow X$ in probability and all $X_n$ are non-negative. If $EX_n<\infty$, $EX<\infty$, and $\lim_{n\rightarrow \infty}EX_n=EX$, why can we say $(X-X_n)^+$ is uniformly integrable?

Can we further show that $X_n\rightarrow X$ in $L^1$?

Thanks!

Answer 1

The $X_n$'s are non-negative hence $X - X_n \le X$ and so $|(X- X_n)^+| = (X- X_n)^+ \le X^+$ and $X^+ \in L^1$ by assumption hence $$((X - X_n)^+)_{n\in\Bbb N}$$ is uniformly integrable.