Convergence of means and convergence in mean Assume $(X_n)$ for $n \geq 0$ uniformly integrable and  $X_n \to X$ in probability. I want to show that $X_n \to X$ in $L^1$. I will probably have to achieve this with some bounding of $E[|X-X_n|]$. Convergence in probability implies convergence in distribution, and together with Skorokhod's theorem, this implies $E[X_n]\to E[X]$, but I don't think this can be used to bound $E[|X-X_n|]$. Any idea how to proceed?
 A: First, observe that $X$ is integrable, by applying Fatou's lemma to $\lvert X_{n_k}\rvert$, where $(X_{n_k})$ is such that $X_{n_k}\to X$ almost surely. Hence the sequence $(Y_n)_{n\geqslant 1}$ defined by $Y_n=\lvert X_n-X\rvert$ is uniformly integrable and converges in probability to $0$. In order to show the convergence in $\mathbb L^1$, recall that $\mathbb EY_n=\int_0^\infty\mathbb P(Y_n>t)dt$. Let $f_n(t):= \mathbb P(Y_n>t)$. Since $Y_n\to 0$ in probability, $f_n(t)\to 0$ for each positive $t$ which is a good start, but we cannot conclude yet (which is normal, as we did not use uniform integrability yet). Write for a fixed $R$
$$
\mathbb EY_n=\int_0^R\mathbb P(Y_n>t)dt+ \int_R^\infty\mathbb P(Y_n\mathbf{1}_{\{Y_n>R\}}>t)dt.
$$
The second term of the right hand side is smaller than $\mathbb E\left[Y_n\mathbf{1}_{\{Y_n>R\}}\right]$ hence
$$\tag{*}
\mathbb EY_n\leqslant \int_0^R\mathbb P(Y_n>t)dt+\sup_{k\geqslant 1}\mathbb E\left[Y_k\mathbf{1}_{\{Y_k>R\}}\right].
$$
As $n$ goes to infinity (for fixed $R$), $\int_0^R\mathbb P(Y_n>t)dt\to 0$ because $\int_0^R\mathbb P(Y_n>t)dt\leqslant R\mathbb P(Y_n>\varepsilon)+\varepsilon$
hence taking the $\limsup_{n\to\infty}$ in (*) and using uniform integrability finishes the proof.
