Show that there exists an $\epsilon > 0$ such that $P(X_n > \epsilon, \text{ i.o.}) = 1$. 
Let $X_1, X_2, X_3, ...$ be a sequence of i.i.d. random variables. Suppose $P(X_n \ge 0) = 1$ and $P(X_n > 0) > 0$, for each $n \in \mathbb{N} := \mathbb{Z} \cap [1, \infty)$.
Show that there exists an $\epsilon > 0$ such that $P(X_n > \epsilon, \text{ i.o.}) = 1$.

My Attempt:
For each $n \in \mathbb{N}$, put $\epsilon_n = \inf\{X_n(\omega) : \omega \in \Omega\}$. Note that, since $P(X_n > 0) > 0$ for each $n$, there exists an $\alpha_n > 0$ such that $P(X_n > \alpha_n) > 0$; and so we are guaranteed that infinitely-many of the $\epsilon_n$ are positive (I think...?)
Now, put $\epsilon = \sup_{n}\epsilon_n$ and, for each $n$, define the event $E_n = \{X_n > \epsilon\}$. It is clear that $\epsilon$ is positive (since infinitely-many of the $\epsilon_n$ are positive). It would now be nice if I could argue (via Borel-Cantelli) that $\sum P(E_n) = \infty$, but I haven't had much luck doing so...
 A: There is problem. It can happen that $\varepsilon_n = 0$ for every $n \in \mathbb N$ (variables $X_n$ can take values $0$). But the answer is simple, note that since they are i.i.d, so every $X_n$ has the same distribution and $\mathbb P(X_n > 0 ) = \delta > 0$. It is enough to prove that there exists $\eta$ such that for any $n \in \mathbb N$ we have $\mathbb P(X_n > \eta) > 0$. Assume contrary that for every $\eta > 0$ we get $\mathbb P(X_n > \eta) = 0$. But then $\mathbb P(X_n > \frac{1}{m}) = 0$ for every $m \in \mathbb N$. Note that events $\{X_n > \frac{1}{m}\}$ are increasing (in $m$) and their sum is $\{X_n > 0\}$ so by continuity of $\mathbb P$ we get $\mathbb P(X_n > 0) = \lim_{m \to \infty} \mathbb P(X_n > \frac{1}{m}) = 0$ -contrary. So there exists $\eta>0$ such that for every $n \in \mathbb N$ we get $\mathbb P(X_n > \eta) = \alpha > 0$. Now all you need to do is apply Borel cantelli, since $\sum_n \mathbb P(X_n > \eta) = \sum_n \alpha =\infty$ and events $\{X_n > \eta\}$ are independent (due to independence of $\{X_n\}$), so borel cantelli tells us that $\mathbb P( \limsup \{X_n > \eta \}) = 1$, which by definition is $\mathbb P( \{X_n > \eta \}$ i.o$) = 1$
