# Equivalent definitions of uniform integrability

According to Wikipedia,

A class $$\mathcal{C}$$ of random variables is $$\textbf{uniformly integrable}$$ if given $$\epsilon > 0$$, there exists $$K \in [0. \infty)$$ such that $$\textbf{E}(|X|I_{|X| \geq K}) \leq \epsilon$$ for all $$X \in \mathcal{C}$$.

It says this is equivalent to saying that $$\lim_{K \to \infty} \sup_{x \in \mathcal{C}}\textbf{E}(|X|I_{|X| \geq K})=0$$. How to show this?

My attempt: It is clear that for $$K < M$$, $$|X|I_{|X| \geq M} \leq |X|I_{|X| \geq K}$$. Therefore, for this $$\epsilon$$, $$\textbf{E}(|X|I_{|X| \geq M}) \leq \epsilon$$ for all $$M \geq K$$. Hence, $$\lim_{K \to \infty} \sup_{x \in \mathcal{C}}\textbf{E}(|X|I_{|X| \geq K})=0$$. Is my reasoning correct? Also, is there an easier proof?

• @bitesizebo Yeah, you're right. Edited. Thanks for the clarification. Commented Sep 12, 2019 at 19:30

## 1 Answer

Your reasoning is correct but shows only one direction. For the other, assume that $$\lim_{R \to \infty} \sup_{x \in \mathcal{C}}\textbf{E}(|X|I_{|X| \geq R})=0$$. For a fixed $$\varepsilon>0$$, choose $$K$$ such that for $$R\geq K$$, $$\sup_{x \in \mathcal{C}}\textbf{E}(|X|I_{|X| \geq R})\leq\varepsilon$$.