Here is the example, which is said to be very easy but I have no idea about it.
$X_1...X_n$ are integrable and $$\lim_{t \to \infty}\limsup_{n \to \infty}E[|X_n|I\{|X_n|\ge t|\}]=0$$ Show that $X_n, n\ge 1$ are uniformly integrable.
Any help would be appreciated!
We want to show that for every $\varepsilon > 0$ there is a $K > 0$ such that $$\sup_n E[|X_n| \mathbb{1}\{|X_n| > K\}] \leq \varepsilon$$
The assumption $\lim_{t \to \infty}\limsup_{n \to \infty}E[|X_n|I\{|X_n|\ge t|\}]=0$ tells us that there exists $K_\infty > 0$ such that $\lim \sup_{n \to \infty} E[|X_n| \mathbb{1}\{|X_n| > K_\infty\}] \leq \frac{\varepsilon}{2}$ and hence there exists an $N$ such that for every $n \geq N$, $E[|X_n| \mathbb{1}\{|X_n| > K_\infty\}] \leq \varepsilon$.
This leaves us to deal with the finitely many $n < N$. Since each $X_i$ is integrable, there is a $K_i$ such that $E[|X_i| \mathbb{1}\{|X_n| > K_i\}] \leq \varepsilon$. So for $K \geq \max\{K_1, \dots, K_N, K_\infty\}$ $$E[|X_n| \mathbb{1}\{|X_n| > K\}] \leq \varepsilon$$ for every $n$ which is the desired result.