Application of Dominate Convergence Theorem

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If $\{X_n\}$ is a sequence of identical distributed random variables with finite mean, then $$\underset{n\rightarrow\infty}{\lim}\frac{1}{n}\mathscr{E}\left(\underset{1\leq j\leq n}{\max}|X_j|\right)=0.$$

$\bullet$ Proof.

Let $Y_n=\underset{1\leq j\leq n}{\max}|X_j|$ and define $$g_n(t)=\frac{1}{n}\mathscr{P}\left\{\underset{1\leq j\leq n}{\max}|X_j|>t\right\}.$$We have $g_n(t)\rightarrow0$ as $n\rightarrow\infty$ for all $t$ and $$0\leq g_n(t)\leq\frac{1}{n}\sum_{j=1}^n\mathscr{P}\left\{|X_j|>t\right\}<\infty$$since $X_n$'s have finite mean (tail probability converges).  

Next, the expectation of a positive random variable with d.f. $F(x)$ can be transformed as $$\begin{array}{rl}\mathscr{E}(X)&=\int_0^\infty x\,dF(x)\\ &=\int_0^\infty\int_0^x 1\,dt\,dF(x)\\ & =\int_0^\infty\int_t^\infty 1\,dF(x)\,dt\quad\left(\because\,\mbox{Fubini}\right)\\ &=\int_0^\infty\mathscr{P}\{X>t\}\,dt. \end{array}$$Thus, we have by Dominate Convergence Theorem $$\underset{n\rightarrow\infty}{\lim}\frac{1}{n}\mathscr{E}(Y_n)=\underset{n\rightarrow\infty}{\lim}\int_0^\infty g_n(t)\,dt=\int_0^\infty\underset{n\rightarrow\infty}{\lim}g_n(t)\,dt=0.$$

$\Box$