Does $\max_{i\le n} \big|Z_i\big|=\frac 1 n \sum_{i\le n} \big|Z_i\big|$ as $n \to \infty$? (Convergence of R.V.s sequence $Z_1,Z_2,Z_3,...$) 
$Z_1,Z_2,Z_3,...$ are integrable R.V.s that are independent and identically distributed.
Show that the following expression converges to zero in probability: $\frac 1 n \max_\limits{i\le n} \left|Z_i\right|$.

What does $\max_\limits{i\le n} \big|Z_i\big|$ mean? Is it the biggest $Z_i$ for $i\le n$? If so, the law of large numbers state that $\frac 1 n \sum_\limits{i\le n} \big|Z_i\big| \to Z$ and I think it is true that $\max_\limits{i\le n} \big|Z_i\big|$ is equal to $\frac 1 n \sum_\limits{i\le n} \big|Z_i\big|$?
Then we have  $\frac 1 n\frac 1 n \sum_\limits{i\le n} \big|Z_i\big|$ and we have $\frac 1 n Z$ as $n \to \infty$. So since $Z$ has finite values (right?), this has to go to zero? If I am right, how can I write this in a more sophiscated way? Thanks.
 A: For $\epsilon>0$ we have:$$\begin{aligned}P\left(\frac{1}{n}\max\left(\left|Z_{1}\right|,\dots,\left|Z_{1}\right|\right)\geq\epsilon\right) & =P\left(\max\left(\left|Z_{1}\right|,\dots,\left|Z_{1}\right|\right)\geq n\epsilon\right)\\
 & =1-P\left(\max\left(\left|Z_{1}\right|,\dots,\left|Z_{1}\right|\right)<n\epsilon\right)\\
 & =1-P\left(\left|Z_{1}\right|< n\epsilon\wedge\cdots\wedge\left|Z_{n}\right|<n\epsilon\right)\\
 & =1-P\left(\left|Z_{1}\right|<n\epsilon\right)^{n}\\
 & =1-\left(1-P\left(\left|Z_{1}\right|\geq n\epsilon\right)\right)^{n}\\
 & =1-\left(1-\frac{nP\left(\left|Z_{1}\right|\geq n\epsilon\right)}{n}\right)^{n}\\
 & \approx1-e^{-nP\left(\left|Z_{1}\right|\geq n\epsilon\right)}
\end{aligned}
$$
It is our aim to prove that the expression tends to $0$ by increasing $n$.
For that it is enough now to prove that $\lim_{n\to\infty}nP\left(\left|Z_{1}\right|\geq n\epsilon\right)=0$
and this can be done on base of the fact that $Z_{1}$ is integrable. 
First give that a try yourself.

edit:
$$nP\left(\left|Z_{1}\right|\geq n\epsilon\right)=\epsilon^{-1}n\epsilon\mathbb{E}1_{\left[n\epsilon,\infty\right)}\left(\left|Z_{1}\right|\right)\leq\epsilon^{-1}\mathbb{E}\left[\left|Z_{1}\right|1_{\left[n\epsilon,\infty\right)}\left(\left|Z_{1}\right|\right)\right]=$$$$\epsilon^{-1}\left[\mathbb{E}\left|Z_{1}\right|-\mathbb{E}\left[\left|Z_{1}\right|1_{\left[0,n\epsilon\right)}\left(\left|Z_{1}\right|\right)\right]\right]\tag1$$
The monotone convergence theorem tells us that: $$\lim_{n\to\infty}\mathbb{E}\left|Z_{1}\right|1_{\left[0,n\epsilon\right)}\left(\left|Z_{1}\right|\right)=\mathbb{E}\left|Z_{1}\right|$$
Consequently the RHS of $(1)$ will tend to $0$ by increasing $n$.
