TITLE: Convergence in probability of renormalized running maximum of i.i.d. square integrable random variables
QUESTION [5 upvotes]: Let $(X_n)_{n \ge1}$ be a sequence of i.i.d. real-valued random variables. Define
  $$M_n = \max_{1\le k \leq n}X_k$$
  Let $F$ be the distribution function of $X_1$. $F(x) < 1$ for all $x \in \mathbb{R}$. Assume that $EX_1^2$ is finite. Show that 
  $$\frac{M_n}{\sqrt{n}} \overset{P}{\to} 0$$

Attempt
Let $\varepsilon > 0$ be given and note that 
$$
P\left( \frac{|M_n|}{\sqrt{n}} \geq \varepsilon \right) = P(M_n^2 \geq n \varepsilon^2) = P\left(\bigcup_{k=1}^n (X_k^2 \geq n \varepsilon^2)\right)\\ \leq \sum_{k=1}^n P(X_k^2 \geq n \varepsilon^2) =  nP(X_1^2 \geq n \varepsilon^2)
$$
I'm not quite sure what to do from here. It seems like this final thing should converge to zero but how do I prove this?

REPLY [2 votes]: Indeed, if one manages to show that $$nP(X_1^2\geqslant n\epsilon^2)$$ converges to $0$ as soon as $E(X_1^2)$ is finite, your proof is complete. To do so, consider the random variables $$Y_n=n\mathbf 1_{X_1^2\geqslant n\epsilon^2}$$
and note that, for every $n$, $$Y_n\leqslant \epsilon^{-2}X_1^2$$
and that, when $n\to\infty$, $Y_n\to0$ almost surely. Hence, Lebesgue dominated convergence theorem shows that $$nP(X_1^2\geqslant n\epsilon^2)=E(Y_n)\to0$$