Let $X_1,X_2,\cdots$ be i.i.d $p^{\text{th}}$ ($p\in (0,\infty)$) integrable random variables. Show that $\cfrac{1}{n^{1/p}} \sup_{k\le n} |X_k| \to 0$ almost surely.
I "think" I got the problem right, but I'm really unsure of the last few bits.
Proof Let $\varepsilon>0$ and $A_n=\{|X_n| \ge \varepsilon n^{1/p}\}$. We can prove that, given $X_n$ are $p$-th integrable, that
$$\sum_{n=1}^{\infty} P(A_n) <\infty$$
(The proof of this I'm quite sure so I'm not going to recite it). So applying Borel Cantelli we have
$$ P\left(\limsup_{n\to\infty} A_n\right)=0 $$
Therefore, there are at most finitely many $n$ such that $|X_n|\ge \varepsilon n^{1/p}$. Then there exists $\alpha$ such that $\sup_{k\le n} |X_k|=|X_{\alpha}|$ for sufficiently large $n$. But then there exists $N$ such that
$$ |X_{\alpha}| < \varepsilon N^{1/p} $$
So that
$$ P\left(\left\{\frac{1}{n^{1/p}}\sup_{k\le n} |X_k| \to 0\right\}\right)=1 $$
But I'm not particularly sure about the last bit of the proof, mainly because I've never used the fact that r.v's are independent. It seems to suggest me to apply another version of Borel-Cantelli, but I don't see the way I could.
It'd be nice if someone could point out wrong bits/proof read
Lemma: Let $\{a_n\}_{n\ge 1}$ be a sequence of non-negative real numbers such that $\frac{a_n}{n}\to 0$ as $n\to\infty$. Then $$\frac{\sup_{k\le n}a_k}{n}\to 0$$.
Proof: Given $\epsilon >0$, $\exists N$ such that $\forall n\ge N$, $\frac{a_n}{n}<\epsilon$. Now, let $b_n=\sup_{k\le n} a_k$. Now, for $n\ge N$, $b_n\le\max\{b_N,n\epsilon\}$. Hence, $\frac{b_n}{n}<\max\{b_N/n,\epsilon\}$. So, choose $N'$ such that $\forall n>N'$, $\frac{b_N}{n}<\epsilon$. Then, $\forall n>N'$, $\frac{b_n}{n}\epsilon$. So, $\frac{b_n}{n}\to 0$.
Now, for this problem you already have $\frac{|X_n|^p}{n}\to 0$ a.e. So, by previous lemma, $\frac{\sup{k\le n}|X_k|^p}{n}\to 0$ a.e .