Let $\{N(t); t > 0\}$ be a renewal counting process with inter-renewal rv s $\{X_n; n \geq 1\}$. Then, $lim_{t\rightarrow \infty} N(t) = \infty$ with probability 1.
Proof:
We only need to show $P\{\omega: lim_{t \rightarrow \infty} N(t, \omega) < \infty \} = 0$;
$P\{\omega: lim_{t \rightarrow \infty} N(t, \omega) < \infty \} = P\{\omega: \bigcup_{n \geq 1} \{lim_{t \rightarrow \infty} N(t, \omega) < n \}\} \leq \sum_{n\geq 1} P\{\omega:lim_{t \rightarrow \infty} N(t, \omega) < n \}.$
if we are able to prove $P\{\omega:lim_{t \rightarrow \infty} N(t, \omega) < n \} = 0$ for each $n\geq 1$, then we will get the desired result.
the following statements are form the textbook:
$\lim_{t\rightarrow \infty} P\{ N(t) < n\} = lim_{t\rightarrow \infty}P\{ S_{n} > t\} = 1 - \lim_{t\rightarrow \infty} P\{S_{n} \leq t\} = 1 - 1 = 0$,
here $S_{n} = X_1 + \dots +X_n,$ then we've proved the theorem.
My question is:
I think that $\lim_{t\rightarrow \infty} P\{ N(t) < n\}$ is not the same as $P\{\omega:lim_{t \rightarrow \infty} N(t, \omega) < n \}$, because the former is a limit of a function of t and the latter is the probability of the event$\{\omega:lim_{t \rightarrow \infty} N(t, \omega) < n \}$. how could they be equal to one and the other? maybe I ignore some condition so that the operation limit can be put inside the braces. can anyone tell me why they are equal? thanks~~~
Using Fatou's lemma, you have $P\{\omega:\lim_{t \rightarrow \infty} N(t, \omega) < n \} \le \lim_{t \rightarrow \infty} P\{\omega: N(t, \omega) < n \}$.