Let $(X_n)_n$ be a sequence of $L^1.$ Prove that $(X_n)_n$ is uniformly integrable if and only if $$\lim_{p \to +\infty}\limsup_n\int_{|X_n|>p}|X_n|dP=0.$$
the implication is easy since $$\limsup_n\int_{|X_n|>p}|X_n|dP\leq \sup_n\int_{|X_n|>p}|X_n|dP$$
The converse is more difficult. I wrote $$\sup_{k \in \mathbb{N}}\int_{|X_k|>p}|X_k|dP\leq \max_{1 \leq k \leq n} \int_{|X_k|>p}|X_k|dP+\sup_{k \geq n} \int_{|X_k|>p}|X_k|dP,$$ if we took the limit $p \to +\infty,$ we will have $$\max_{1 \leq k \leq n} \int_{|X_k|>p}|X_k|dP \to0$$ using the dominated convergence theorem $\max_{1 \leq k \leq n}X_n \in L^1,$ but I can't see how to prove that $\sup_{k \geq n} \int_{|X_k|>p}|X_k|dP \to 0$ ($n$ is fix), and also if we took the limit in $n,$ we will have a problem with $\max_{1 \leq k \leq n} \int_{|X_k|>p}|X_k|dP$.
Fix $\epsilon>0$, then there is some $p >0$ such that
$$\limsup_{n \to \infty} \int_{|X_n|>p} |X_n| \, d\mathbb{P} \leq \epsilon.$$
Consequently, we can find $N=N(p) \in \mathbb{N}$ such that
$$\sup_{n \geq N} \int_{|X_n|>p} |X_n| \, d\mathbb{P} \leq 2 \epsilon.\tag{1}$$
Since the random variables $X_1,\ldots,X_{N-1}$ are integrable, it follows from the dominated convergence theorem that there is some $r \geq p$ such that
$$\int_{|X_n|>r} |X_n| \, d\mathbb{P} \leq 2 \epsilon$$
for all $n=1,\ldots,N-1$. By $(1)$ and the monotoncity of the integral, we also have
$$\int_{|X_n| >r} |X_n| \,d\mathbb{P} \leq \int_{|X_n|>p} |X_n| \, d\mathbb{P} \leq 2 \epsilon$$
for all $n \geq N$. Thus,
$$\sup_{n \in \mathbb{N}} \int_{|X_n|>r} |X_n| \, d\mathbb{P} \leq 2 \epsilon.$$
This implies (again by the monotonicity of the integral)
$$\sup_{n \in \mathbb{N}} \int_{|X_n|>R} |X_n| \, d\mathbb{P} \leq 2 \epsilon$$
for all $R \geq r$, which proves that
$$\lim_{R \to \infty} \sup_{n \in \mathbb{N}} \int_{|X_n|>R} |X_n| \, d\mathbb{P}=0.$$