Let $\{X_i\}_{i=1}^{\infty}$ be a sequence of iid random variables. If $P(X_i\geq0)=1$ and $P(X_i\neq0)>0$ for all $i$, then $\sum X_i=\infty$ a.s. I came across a proposition, which is stated without proof. It appears in a section on the 0-1 Laws.

Proposition. Let $\{X_i\}_{i=1}^{\infty}$ be a sequence of iid random variables on some probability space. If $P(X_i \geq 0) = 1$ and $P(X_i \neq 0) > 0$ for all $i$, then the sum of the $X_i$'s is $\infty$ a.s. (almost surely).

This doesn't quite seem trivial to me. So, I try to prove it:
Proof (so far): Let $S = \sum_{i=1}^{\infty} X_i$. If $P(S = \infty) = 1$, it follows that $S = \infty$ a.s. So, the problem boils down to showing that:

*

*$S \in \mathcal{T}$, where $\mathcal{T}$ is the tail $\sigma$-field, and $S$ is considered a tail event.

*Any $A \in \mathcal{T}$ has probability $P(A) \in \{0, 1\}$. We need to show that specifically $P(S = \infty) = 1$.

Since $\{X_i\}_{i=1}^{\infty}$ iid, by Kolmogorov's 0-1 Law, there exists a $\mathcal{T}$ defined as in (1). Not sure how to show that $S$ is a tail event, and that $P(S = \infty) = 1$? Can we use the fact that $P(X_i \neq 0) > 0$ and $P(X_i \neq 0) > 0$ in some way to conclude this? I don't see how.
 A: The fact that $(S=\infty)$ is tail event follows from the fact that for any $N$,  $S=\infty$ iff $ \sum\limits_{k=N}^{\infty} X_i=\infty$  and the event $\sum\limits_{k=N}^{\infty} X_i=\infty$ belongs to $\sigma (X_N,X_{N+1},...)$.
One way of showing that $S=\infty$ a.s is to show that $Ee^{-S}=0$.
Note that $Ee^{-S}=\prod Ee^{-X_i}=\lim_N (Ee^{-X_1})^{N}$. Can you use the hypothesis $P(X_i \neq 0 ) >0$ to check that $Ee^{-X_1}<1$ (which shows that $\lim_N (Ee^{-X_1})^{N}=0$)?
A: Maybe it's worth discussing what it means intuitively for $S = \infty$ to be a tail event. $S = \infty$ is the event that the series $\sum_{i = 1}^\infty X_i$ diverges to positive infinity.

*

*If I tell you the values of $X_1, X_2, X_3, X_4 \dots$, it's possible for you to determine whether $\sum_{i = 1}^\infty X_i$ diverges to positive infinity. (i.e. $S = \infty$ is in the sigma algebra $\sigma(X_1, X_2, X_3, X_4, \dots)$).

*If I only tell you the values of $X_2, X_3, X_4, \dots$, it's still possible for you to determine whether $\sum_{i = 1}^\infty X_i$ diverges to positive infinity. (i.e. $S = \infty$ is in the sigma algebra $\sigma(X_2, X_3, X_4, \dots)$).

*If I only tell you the values of $X_3, X_4, \dots$, it's still possible for you to determine whether $\sum_{i = 1}^\infty X_i$ diverges to positive infinity. (i.e. $S = \infty$ is in the sigma algebra $\sigma(X_3, X_4, \dots)$).

And so on. Since the event $S = \infty$ is in all of the sigma algebras of the form $\sigma(X_n , X_{n+ 1}, \dots)$, for all $n \in \mathbb N$, the event $S = \infty$ is in the tail sigma algebra.

As for showing that $P(S = \infty)$ cannot be zero, the method that sprang to my mind is more pedestrian than the one suggested by Kavi Rama Murthy.
First, observe that since $P(X_i > 0) > 0$, there exists a $c > 0$ and a $p > 0$ such that $P(X_i \geq c ) = p$, for each $i$. [If this wasn't true, then $P(X_i \geq \tfrac 1 n ) = 0$ for all $n \in \mathbb N$, hence $P(X_i > 0) = P(\cup_{n \in \mathbb N} \{X_i \geq \tfrac 1 n\}) = 0$, a contradiction.]
Therefore, the probability that all but finitely many of the $X_i$'s are less than $c$ is $0$, since there are countable many ways of choosing a finite subset of the $X_i$'s, and for each choice, the probability that all of the $X_i$'s outside of our finitely chosen subset are less than $c$ is $\lim_{k\to\infty} (1-p)^k = 0$. This implies that $S=\infty$ with probability $1$.
Although now that I have written this answer, I realise that I've shown that $P(S = \infty) = 1$ rather than that $P(S = \infty) > 0$, i.e. I've proven the statement directly without appealing to the Kolmogorov zero-one law. Never mind.
