I'm trying to prove that if $(X_n)_{n\geq 1}$ is uniformly integrable, then $X_n$ almost surely converging to $X$ implies $X_n$ converges to $X$ in $L^{1}$.
How is this done?
Generally speaking:
Let $\{X_{n}\}_{n\in\mathbb N}$ be a sequence of random variables, in $L^{p}$, $p\geq 1$, which converges to $X\in L^{0}$ in probability. Then, the following statements are equivalent:
1.) The sequence $\{|X_{n}|^{p}\}_{n\in \mathbb N}$ is uniformly integrable.
2.) $X_{n}\xrightarrow{L^{p}}X$.
Proof: Since there exists a subsequence $\{X_{n_{k}}\}_{k\in\mathbb N}$ such that $X_{n_{k}}\xrightarrow{a.s}X$, Fatou's lemma implies that
$$\mathbb E[|X|^{p}]=\mathbb E[\liminf_{k}|X_{n_{k}}|^{p}]\leq\liminf_{k}\mathbb E[|X_{n_{k}}|^{p}]\leq\sup_{X\in\chi}\mathbb E[|X|^p]<\infty,$$
where the last inequality follows from the fact that uniformly integrable families are bounded in $L^{1}$. Now that we know that $X\in L^{p}$, uniform integrability of $\{|X_{n}|^{p}\}_{n\in\mathbb N}$ implies that the family $\{|X_{n}-X|^{p}\}_{n\in\mathbb N}$ is UI. Since $X_{n}\xrightarrow{\mathbb P}X$ if and only if $X_{n}-X\xrightarrow{\mathbb P}0$, we can assume without loss of generality that $X=0$ a.s., and, consequently, we need to show that $\mathbb E[|X_{n}|^{p}]\rightarrow 0$. We fix an $\epsilon>0$, and start by the following estimate:
$$\mathbb E[|X_{n}|^{p}]=\mathbb E[|X_{n}|^{p}1_{\{|X_{n}|^{p}\leq\frac{\epsilon}{2}\}}]+\mathbb E[|X_{n}|^{p}1_{\{|X_{n}|^{p}>\frac{\epsilon}{2}\}}]$$
$$\leq \frac{\epsilon}{2}+\mathbb E[|X_{n}|^{p}1_{\{|X_{n}|^{p}>\frac{\epsilon}{2}\}}]$$
By uniform integrability there exists $\rho>0$ such that
$$\sup_{n\in\mathbb N}\mathbb E[|X_{n}|^{p}1_{A}]<\frac{\epsilon}{2}$$
whenever $\mathbb P(A)\leq \rho$. Convergence in probability now implies that there exists $n_{0}\in\mathbb N$ such that for $n\geq n_{0}$, we have
$$\mathbb P[|X_{n}|^{p}>\frac{\epsilon}{2}]\leq \rho.$$
It follows directly from the estimate that for $n\geq n_{0}$, we have
$$\mathbb E[|X_{n}|^{p}]\leq \epsilon.$$