If $\mathbb{E}(\sup_n |X_n|)< \infty$ then $(X_n)_n$ is uniformly integrable

Question

Let $\{X_n: n\ge 1\}$ be a sequence of random variables satisfying $E [ \sup | X_n|] < \infty $. Show that $\{X_n\}$ is uniformly integrable.

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Answer 1

Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of random variables. If there exists an integrable random variable $Y$ such that $|X_n| \leq Y$ for all $n \in \mathbb{N}$, then $(X_n)_{n \in \mathbb{N}}$ is uniformly integrable.

Proof: Since $|X_n| \leq Y$, we have $$\{|X_n|>R\} \subseteq \{Y>R\}.$$ Thus, $$\int_{|X_n|>R} |X_n| \, d\mathbb{P} \leq \int_{|X_n|>R} Y \, d\mathbb{P} \leq \int_{Y>R} |Y| \, d\mathbb{P}$$

implying

$$\sup_{n \in \mathbb{N}} \int_{|X_n|>R} |X_n| \, d\mathbb{P} \leq \int_{Y>R} |Y| \, d\mathbb{P}.$$

Since $Y \in L^1$ it follows from the dominated convergence theorem that the rigt-hand side converges to $0$ as $R \to \infty$, and this finishes the proof.