Infinity as supremum for a sequence of numbers Let $X_1, X_2, · · ·$ be sequence of independent nonnegative random variables having a common distribution function $F$ such that $P(X_1=0)<1, E[X_1]=a<\infty$ and $\mathrm{Var}[X_1]=\sigma^2<\infty$. Let's denote $$S_0 = 0, S_n = X_1 + \cdots + X_n$$ for $n = 1, 2, \ldots $ and $$N_t =\sup\left\{n:S_n ≤t\right\}$$ for any $t≥0$.
I want to show that $N_t\rightarrow\infty$ as $t\rightarrow\infty$.
By common sense, this is obvious but I do not know how to write it down in a mathematical way. Any help would be appreciated!
 A: First observe that the condition $P(X_1=0)<1$ implies that there exists $k_0$ such that $P(0\leqslant X_1\leqslant 1/k_0)<1$. Indeed, let $A_k$ be the event $\{0\leqslant X_1\leqslant 1/k\}$; the sequence $(A_k)$ is non-increasing and its intersection has probability smaller than $1$, hence so should have one of the $A_k$.
Consider now the event $B_j:=\{  X_j\gt 1/k_0\}$. These events are independent and $\sum_{j\geqslant 1}P(B_j)$ diverges, hence the Borel-Cantelli lemma show that $B_j$ happens infinetely often, which means that for (almost every) $\omega$, the set of $j$ such that $\omega$ belongs to $B_j$ is infinite. This shows that $S_n\to \infty$ almost surely as $n$ goes to infinity.
Now suppose that we do not have $N_t\to\infty$ almost surely. This means that there is a set $E$ of positive probability such that for each $\omega\in E$, $N_t(\omega)$ doesn't go to infinity as $t$ goes to infinity. Since $t\mapsto N_t(\omega)$ is non-decreasing, this means that $N_t(\omega)$ is bounded for each $\omega\in E$, i.e. $N_t(\omega)\leqslant C(\omega)$ for each $t\geqslant 0$. Therefore, by definition of $N_t$, we would have for each $t$ that $S_{C(\omega)+1}\gt t$, which is a contradiction.
A: We want to prove that $P(lim_{t\rightarrow\infty}N_t=\infty)=1$. $lim_{t\rightarrow\infty}N_t<\infty$ can happen only when $X_i=\infty$ for some $i$.
$$\{lim_{t\rightarrow\infty}N_t<\infty\}\subseteq\cup_{i=1}^\infty\{X_i=\infty\}$$
On the other hand we have $P(X_i=\infty)=0$, therefore:
$$P(lim_{t\rightarrow\infty}N_t<\infty)\leq P(\cup_{i=1}^\infty\{X_i=\infty\})\leq \Sigma_{i=1}^\infty P(X_i=\infty)=0$$
