A sequence $X_k$ of real random variables on $(\Omega,F,\mathbb P)$ is uniformly integrable iff all $X_k$ are integrable and $$\lim_{n \to \infty}\limsup_{k\to\infty}\mathbb E(\vert X_k \vert \mathbb 1_{\vert X_k \vert \ge n})=0$$
$\textbf{Definition}:$ A familiy of real valued random variables $X_t, t\in I$ ($I$ arbitrary index set) is called uniformly integrable, if $$\lim_{n \to \infty}\sup_{t \in I}\int_{\{\vert X_t \vert \ge n\}}\vert X_t \vert \, d\mathbb P=0$$
Therefore '$\Leftarrow$' is clear. Since $\limsup_{k \to \infty}$ can be 'substitued' by $\sup_{t \in I}$. The problem is the other direction. Help is much appreciated!
Fix $\varepsilon>0$. By assumption, there exists $n_0$ such that $\limsup_{k\to\infty}\mathbb E(|X_k|1_{|X_k|\ge {n_0}})<\varepsilon$. (Notice that this is still true if one replaces $n_0$ by any $n>n_0$, since this sequence is non-increasing.) Thus, there exists $k_0$ such that $\mathbb E(|X_k|1_{|X_k|\ge n_0})<\varepsilon$ for all $k\ge k_0$. Since $X_1,\ldots,X_{k_0-1}$ are all integrable, there exist $n_1,\ldots,n_{k_0-1}$ such that $\mathbb E(|X_k|1_{|X_k|\ge n_k})<\varepsilon$ for $k=1,\ldots,k_0-1$. Put $N:=\max_{0\le k\le n}n_k$; if $n\ge N$, then $\mathbb E(|X_k|1_{|X_k|\ge n})<\varepsilon$ for all $k$. This completes the proof.