Conditions for uniform integrability of sequence of integrable random variables converging to integrable random variable?

Question

Let $X_n:\Omega\to \mathbb{R}$ be an integrable random variable converging to integrable $X$ almost surely. I know that this does not mean $\{X_n:n\in\mathbb{N}\}$ is uniformly integrable.

But is there are any conditions that we can impose on $X_n$ or $X$ to guarantee uniform integrability (except for being dominated by an integrable non-negative random variable $Y$ i.e $|X_n| < Y$)?

Answer 1

Let $\left(X_n\right)_{n\geqslant 1}$ be a sequence of random variables which converges to an integrable random variable $X$ almost surely. Then the following conditions are equivalent:

• (i) the sequence $\left(X_n\right)_{n\geqslant 1}$ is uniformly integrable;
• (ii) the sequence $\left(\mathbb E\left\lvert X_n\right\rvert\right)_{n\geqslant 1}$ converges to $\mathbb E\left\lvert X\right\rvert$;
• (iii) the sequence $\left(\mathbb E\left\lvert X_n-X\right\rvert\right)_{n\geqslant 1}$ converges to $0$.

Ideas of proof: (i) implies (iii). Fix $\varepsilon \gt 0$. By uniform integrability, there exists a $\delta$ such that if $\mathbb P(A)\leqslant \delta$, then $\mathbb E \left\lvert X\mathbf 1_A\right\rvert\lt\varepsilon$ and for all $n$, $\mathbb E \left\lvert X_n\mathbf 1_A\right\rvert\lt\varepsilon$. By Egoroff's theorem, there exists a set $S$ such that $\sup_{\omega\in S}\left\lvert X_n(\omega)-X(\omega)\right\rvert\to 0$ and the measure of $\Omega\setminus S$ is smaller than $\delta$. Then $$ \mathbb E\left\lvert X_n-X\right\rvert\leqslant \sup_{\omega\in S}\left\lvert X_n(\omega)-X(\omega)\right\rvert+\mathbb E \left\lvert X_n\mathbf 1_{\Omega\setminus S}\right\rvert+\mathbb E \left\lvert X\mathbf 1_{\Omega\setminus S}\right\rvert\leqslant \sup_{\omega\in S}\left\lvert X_n(\omega)-X(\omega)\right\rvert+2\varepsilon. $$

(iii) implies (i) does not require uniform integrability.

Implication $(ii) \Rightarrow (iii)$ can be established using Fatou's lemma with $Y_n:=\left\lvert X\right\rvert+\left\lvert X_n\right\rvert-\left\lvert X_n-X\right\rvert$. The converse is not hard.

Answer 2

The $X_n$ could all obey a uniform moment bound, such as $\exists a>1, B<\infty$ such that $|EX_n|^a\le B$ for all $n$.