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?

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.
