$\max\{X_1,X_2,\cdots ,X_n\}\xrightarrow{a.s.}x^*_F$ [closed]

I have the following question at hand:

Let $\{X_n\}_{n\ge 1}$ be an iid sequence of random variables with common distribution function $F$ ; such that $x^*_F:=\sup\{x\ : F(x)<1\}<\infty.$ Show that $\max\{X_1,X_2,\cdots ,X_n\}$ converges almost surely to $x^*_F.$

How to attack this problem?

• This appears to be a textbook-style problem. Please read the self-study tag-wiki, and edit to either more clearly indicate your thoughts on the problem and where you run into difficulties -- and add the tag. – Glen_b Oct 10 '16 at 22:06

The maximum $M_n:=\max(X_1,\cdots,X_n)$ has distribution $F^n(x)$. Fix $\epsilon>0$, so: \begin{align*} P(|M_n-x_F^*|>\epsilon)&=P(M_n-x_F^*>\epsilon)+P(M_n-x_F< -\epsilon)\\ &\leq 1-F^n(x_F^*+\epsilon)+F^n(x_F^*-\epsilon)\\ &=1-1+z^n, \end{align*}
where $z<1$. Thus:
$$\sum_nP(|M_n-x_F^*|>\epsilon)\leq \sum_n z^n=\frac{1}{1-z}<\infty.$$
Now use the fact that when the Borel Cantelli Lemma holds for all $\epsilon>0$, it follows that $M_n\rightarrow x_F^*$ almost surely.
• I know that $\sum_1^\infty P(A_n)\lt \infty\implies P(lim sup (A_n)=0$. How to derive Almost sure convergence? – Qwerty Oct 11 '16 at 4:59
• So Is the following correct? $$\forall\epsilon>0 \ ,\ \sum\limits_1^\infty P(|M_n-x^*_F|>\epsilon)<\infty\\\implies \forall\epsilon>0 \ ,\ P(\limsup_{n\to\infty}|M_n-x^*_F|>\epsilon)=0\\\implies \forall\epsilon>0 \ ,\ P(|M_n-x^*_F|>\epsilon \text{ for infinitely many } n )=0\\\implies P(M_n\to x^*_F \text{ as } n\to\infty)=1\\\implies M_n\xrightarrow{a.s.} x^*_F$$ – Qwerty Oct 12 '16 at 6:12