My class notes and wikipedia both say that $X_n \xrightarrow{L^1} X$ $\Leftrightarrow \; X_n \xrightarrow{P} X$ and $X_n$ are uniformly integrable.
I am trying to work through the proof. I am able to show the $\Leftarrow$ direction and I am able to show that $X_n \xrightarrow{L^1} X$ $\Rightarrow \; X_n \xrightarrow{P} X$. However, I am not able to show that $X_n \xrightarrow{L^1} X$ $\Rightarrow X_n$ are uniformly integrable.
Definitions I'm using:
$X_n \xrightarrow{L^1} X$ means $\mathbb{E}[|X_n-X|] \rightarrow 0$
$X_n$ uniformly integrable means $\lim\limits_{b\rightarrow \infty} \sup\limits_{n\geq 1} \mathbb{E}[|X_n|I_{|X_n|>b}] = 0$.
Wikipedia Link: http://en.wikipedia.org/wiki/Convergence_of_random_variables#Properties_4l ; last bullet point/
Best Answer
Here are some steps:
If $\{Y_n,n\geqslant 1\}$ is a uniformly integrable (UI) family and $Y$ an integrable random variable, then the family $\{Y_n+Y,n\geqslant 1\}$ is UI. Hence we can assume that $X=0$ and the $X_n$'s are non-negative.
Fix $\varepsilon\gt 0$; there is $N$ such that if $n\geqslant N$, then $\mathbb E[X_n]\lt \varepsilon$. We have $$\sup_n\mathbb E\left[X_n\chi_{\{X_n\gt b\}}\right]\leqslant \varepsilon+\max_{1\leqslant n\leqslant N}\mathbb E\left[X_n\chi_{\{X_n\gt b\}}\right].$$
Since $N$ is fixed (it depends on $\varepsilon$ and $(X_k)_{k\geqslant 1}$ but not on $b$, we have
$$\limsup_{b\to \infty}\sup_n\mathbb E\left[X_n\chi_{\{X_n\gt b\}}\right]\leqslant \varepsilon.$$