Supremem of Uniformly Integrable Sequences I'm doing an exercise about uniformly integrable random variables:

Suppose $(\xi_n)$ is a sequence of uniformly integrable random variables, then
  \begin{equation}
\lim_{n\rightarrow\infty}\mathbb{E}\bigg[\frac{1}{n}\sup_{1\leq k\leq n}|\xi_k|\bigg]=0.
\end{equation}

I try to apply the definition and some other basic properties of uniformly integrable random variables, but it seems like I need more tricks. 
Any hint would be appreciated.
 A: Let $R$ be fixed and define $X_n:=\left\lvert \xi_n\right\rvert\mathbf 1\left\{\left\lvert \xi_n\right\rvert\leqslant R\right\}$ and $Y_n:=\left\lvert \xi_n\right\rvert\mathbf 1\left\{\left\lvert \xi_n\right\rvert\gt R\right\}$. Then 
$$
\mathbb{E}\bigg[\frac{1}{n}\sup_{1\leq k\leq n}|\xi_k|\bigg]\leqslant \frac Rn+\mathbb{E}\bigg[\frac{1}{n}\sup_{1\leq k\leq n}Y_k\bigg]
\leqslant \frac Rn+\mathbb{E}\bigg[\frac{1}{n}\sum_{1\leq k\leq n}Y_k\bigg]
$$
hence 
$$\mathbb{E}\bigg[\frac{1}{n}\sup_{1\leq k\leq n}|\xi_k|\bigg]\leqslant \frac Rn+\max_{1\leqslant k\leqslant n}\mathbb E\left[\left\lvert \xi_k\right\rvert\mathbf 1\left\{\left\lvert \xi_k\right\rvert\gt R\right\}\right].$$
Taking the $\limsup_{n\to\infty}$ yields 
$$\limsup_{n\to\infty}\mathbb{E}\bigg[\frac{1}{n}\sup_{1\leq k\leq n}|\xi_k|\bigg]\leqslant \sup_{ k\geqslant 1}\mathbb E\left[\left\lvert \xi_k\right\rvert\mathbf 1\left\{\left\lvert \xi_k\right\rvert\gt R\right\}\right].$$
Since $R$ is arbitrary, we can conclude using uniform integrability.
A: First note that if the random variables are uniformly integrable, then $$S:=\sup_{n\in \mathbb{N}} \mathbb{E}|\xi_n|<\infty.$$ 
Fix $\epsilon>0$. Since the random variables are uniformly integrable, there exists $\delta>0$ such that whenever $\mathbb{P}(A)<\delta$, $\sup_m \mathbb{E}1_A|\xi_m|<\epsilon$.   For each $n\in\mathbb{N}$, let $$X_n=\max_{1\leqslant m\leqslant n}|\xi_m|.$$   We want to show that $\lim_n \frac{1}{n}\mathbb{E}X_n=0$. 
For each $n\in\mathbb{N}$, let us define $A_{1,n}, \ldots, A_{n,n}$ by letting $$A_{1,n}=(|\xi_1|=X_n)$$ and $$A_{m+1,n}=(|\xi_{m+1}|=X_n)\setminus \cup_{i=1}^m A_{i,n}.$$  That is, $A_{k,n}$ is the set where the maximum $X_n=\max_{1\leqslant m\leqslant n}|\xi_m|$ is obtained for the first time at $m=k$. Then $A_{1,n}, \ldots, A_{n,n}$ is a partition of the probability space (let's call it $\Omega$).   Now let $m_0$ be a natural number greater than $1/\delta$. Then for any $n\in \mathbb{N}$, there are fewer than $m_0$ values of $m$ such that $\mathbb{P}(A_{m,n})\geqslant \delta$. This is because the sets $A_{1,n}, \ldots, A_{n,n}$ are disjoint, so if $J_n=\{m\leqslant n: \mathbb{P}(A_m)\geqslant \delta\}$, $$1=\mathbb{P}(\Omega)=\sum_{i=1}^n \mathbb{P}(A_{i,n}) \geqslant \delta |J_n|,$$ which would give a contradiction if $|J_n|\geqslant m_0$. 
Let us also note that $X_n=\sum_{i=1}^n 1_{A_{i,n}}|\xi_i|$.   Then if $i\in \{1, \ldots, n\}\setminus J_n$, since $\mathbb{P}(A_{i,n})<\delta$, $\mathbb{E}1_{A_{i,n}}|\xi_i|<\epsilon$.    As mentioned in the previous paragraph, $|J_n|\leqslant m_0$. Let $K_n=\{1, \ldots, n\}\setminus J_n$. Then  \begin{align*} \mathbb{E}X_n & = \sum_{i=1}^n \mathbb{E}1_{A_{i,n}}|\xi_i| = \sum_{i\in J_n}\mathbb{E}1_{A_{i,n}}|\xi_i| + \sum_{i\in K_n} \mathbb{E}1_{A_{i,n}}|\xi_n| \\ & \leqslant \sum_{i\in J_n}\mathbb{E}|\xi_i|+\sum_{i\in K_n}\epsilon \leqslant |J_n|S+n \epsilon \leqslant m_0 S+n\epsilon. \end{align*} 
Now dividing by $n$, we see that $$\mathbb{E}\frac{1}{n}X_n \leqslant \frac{m_0 S}{n}+\epsilon.$$  Since $m_0$ is fixed, $$\lim\sup_n \mathbb{E}\frac{1}{n}X_n \leqslant \lim\sup_n \frac{m_0 S}{n}+\epsilon=\epsilon.$$  Since $\epsilon>0$ was arbitrary, this shows that $\lim_n \mathbb{E}\frac{1}{n}X_n=0$. 
