Almost sure convergence of $\max(X_1, X_2,\ldots,X_n)$. $X_1, X_2,\ldots, X_n$ are independent with uniform distribution on $[0,a]$.
Prove that $\max(X_1, X_2,\ldots,X_n) \rightarrow a$ almost surely.
Ar first I look for the probability distribution i.e. $$F_\max(t)=P(\max(X_1, X_2,\ldots,X_n)<t)=P(X_1<t)^n=(F_{X_1}(t))^n=\left(\frac{t}{a}\right)^n.$$
So if $t<a$ it converges to $0$ and if $t\geq a$ it converges to $1$. 
How can I say that $\max(X_1, X_2,\ldots,X_n) \rightarrow a$ almost surely. Is there a better way to do it?
 A: The solution was mostly covered by the comments, but here is a more complete answer:
Your solution does not show a.s. convergence, as convergence in distribution does not imply a.s. convergence. What needs to be shown is that (denote $\max(X_1, \dots X_n)$ by $M_n$)
$$
P(\lim_{n \rightarrow \infty} M_n = a) = 1
$$
This happens to be the equivalent of (this is not directly evident but can be shown)
$$
P(\mid M_n -a \mid > \epsilon \text{  infinitely often as } n\rightarrow \infty) = 0
$$
By Borel Cantelli Lemma, this holds if 
$$
\sum^{\infty}_{n=1}P(\mid M_n -a \mid > \epsilon) <\infty
$$
As you showed yourself, $P(\mid M_n -a \mid > \epsilon) = \left(\frac{a-\epsilon}{a} \right)^n$ so you have
$$
\sum^{\infty}_{n=1}P(\mid M_n -a \mid > \epsilon)  = \frac{a}{\epsilon}<\infty
$$
and we are done.
A: Another argument is to note that $P(|M_n - a| < \epsilon) \to 1$ as $n \to \infty$ for $\epsilon > 0$ and so $M_n \to a$ in probability.  But since the events $\{ |M_n - a| < \epsilon \}$ are monotone increasing this implies $M_n \to a$ with probability one.
