If $\mathbb{E}\left[\left|X_n −X\right|\right] \to 0$, as $n \to\infty $ then, The family $\{X_n\}_{n\geq 1}$ is uniformly integrable.

Question

Let $X_n$ $n \ge 1$ be integrable random variables and $X\in L^1$. Show that :

If $\mathbb{E}|Xn −X| \to 0$, as $n \to\infty $ then, The family $X_n$, $n\geq 1 $ is uniformly integrable.

Answer 1

It's probably easiest to use the definition of uniform integrability that goes: For every $\varepsilon>0,$ there exists a $\delta>0$ such that $\mathbb{P}(A)<\delta$ implies $\mathbb{E} 1_A |X_n|\leq \varepsilon$.

Note that any finite sub-family $\mathcal{X}_N:=\{X,X_1,X_2,...,X_N\}$ is uniformly integrable. Now, let $\varepsilon>0$ and pick $N$ large enough that $\|X-X_n\|_{L^1}\leq \frac{\varepsilon}{2}$ for all $n\geq N$. Pick, furthermore, by uniform integrability of $\mathcal{X}_N,$ $\delta>0$ such that $\mathbb{P}(A)<\delta$ implies $\mathbb{E} 1_A |Y|\leq \frac{\varepsilon}{2}$ for all $Y\in\mathcal{X}_N$.

Now, fix any such $A$ and any $n$. If $n\leq N$, then $\mathbb{E} 1_A|X_n|\leq\frac{\varepsilon}{2}$ by our choice of $\delta$. If $n>N,$ we have that $$ \mathbb{E} 1_A|X_n|\leq \mathbb{E}1_A|X|+\mathbb{E}1_A|X-X_n|\leq \mathbb{E} 1_A|X|+\|X-X_n\|_{L^1}\leq \varepsilon, $$ which establishes the desired.