Show that $\lim\limits_{n\to\infty} \frac{1}{n} \mathbb{E}(\max\limits_{k\leq n} |X_{k}|) = 0.$ 
Let $\{X_{n}\}_{n\in \mathbb{N}}$ be a sequence of i.i.d. random
  variables in $L^{1}(\Omega)$. Show that
$$\lim_{n\to\infty} \frac{1}{n} \mathbb{E}\left(\max_{k\leq n}
 |X_{k}|\right) = 0.$$

I am really not so sure how to approach this problem. I am given the hint to use the following result for any non-negative random variable $X$.
$$\mathbb{E}X = \int_{0}^{\infty} \operatorname P(X > x) \mathop{dx}. $$
So following this approach I get
$$\lim_{n\to\infty} \frac{1}{n} \int_{0}^{\infty} \operatorname P\left(\max_{k\leq n} |X_{k}| > x\right) \mathop{dx} $$
And also I think this integral 
$$\int_0^{\infty} (\operatorname P(|X_{1}| \leq x)^{k} \mathop{dx} $$
will be useful but I'm not sure how to continue and will really appreciate your help.
I guess we may need law of large numbers. Not sure though.
 A: Since $|\sigma(|X_k|)|<|\sigma(X_k)|$, given that $X_k$ are i.i.d. we can infer that $|X_k|$ are i.i.d. Thus $$\mathbb{E}\left[\max_{k\leq n}
 |X_{k}|\right]=\int_0^\infty\operatorname P\left(\max_{k\leq n}
 |X_{k}|>x\right)\,dx=\int_0^\infty\prod_{k\le n}\left(1-\operatorname P(|X_k|\le x)\right)\,dx.$$ Since $\mu:=\Bbb E[|X_i|]<\infty$ for $i\le k$, splitting $[0,\infty)$ into $[0,\mu)\cup[\mu,\infty)$ yields \begin{align}\mathbb{E}\left[\max_{k\leq n}
 |X_{k}|\right]&=\int_0^\infty\left(1-F_{|X_i|}(x)\right)^n\,dx\\&\le\int_0^{\mu}\left(1-F_{|X_i|}(x)\right)^n\,dx+\int_{\mu}^\infty\left(\frac{\mu}x\right)^n\,dx\end{align} by Markov's inequality. The function $\left(1-F_{|X_i|}(x)\right)^n$ is decreasing, so \begin{align}\mathbb{E}\left[\max_{k\leq n}
 |X_{k}|\right]&\le \mu\left(1-F_{|X_i|}(0)\right)^n+\mu^n\int_{\mu}^\infty\frac1{x^n}\,dx\le\mu+\frac{\mu^{1-n}}{n-1}\cdot\mu^n.\end{align} Hence $$\lim_{n\to\infty} \frac{1}{n} \mathbb{E}\left(\max_{k\leq n}
 |X_{k}|\right)=\lim_{n\to\infty}\frac\mu{n-1}=0.$$
