I have learned the definition of uniformly integrable and its sufficient and necessary condition:
We call $\{X_n, n\in \mathbb{N}\}$ uniformly integrable, if
$$
\lim_{x \to \infty} \sup_{n\in \mathbb{N}} E(|X_n| 1_{\{|X_n| \geq x\}})=0
$$
And it has a sufficient and necessary condition:
$\{X_n, n\in \mathbb{N}\}$ is uniformly integrable if and only if $\forall \epsilon > 0$, there exists $\delta > 0$ such that $\forall A$ satisfying $P(A) < \delta$, we have $\sup_{n\in \mathbb{N}} E(|X_n| 1_{A}) < \epsilon$ and $\sup_{n\in \mathbb{N}} E(|X_n|) < \infty$.
I need prove:
If there exists $Y \in L_1$ such that $\forall x > 0$, we have
$$
\sup_{n\in \mathbb{N}} P(|X_n|>x) \leq P(Y>x)
$$
then $\{X_n, n\in \mathbb{N}\}$ is uniformly integrable.
I want to first prove $\sup_{n\in \mathbb{N}} E(|X_n|) < \infty$. I tried this:
\begin{align*}
E|X_n| &= E(|X_n|1_{\{ X_n>x \}}) + E(|X_n|1_{\{ X_n \leq x \}})\\
&\leq E(|X_n|1_{\{ X_n>x \}}) + xP(X_n \leq x)\\
&\leq E(|X_n|1_{\{ X_n>x \}}) + x
\end{align*}
I have difficulty to move on. Could anyone give me some advice? Thank you!
Best Answer
It is more straightforward to work with the definition than the equivalent characterisation you give in this case. We have, for $x > 0$, \begin{align*} \mathbb{E}[|X_n| 1_{\{|X_n| \geq x\}}] &= \int_x^\infty \mathbb{P}(|X_n| > y) dy \\& \leq \int_x^\infty \mathbb{P}(Y>y) dy \\& = \mathbb{E}[Y 1_{\{Y \geq x\}}] \to 0 \end{align*} as $x \to \infty$ since $Y \in L^1$. Since this bound is true independent of $n$, this gives the desired result.