I want to show (if it is true) that if $\tau$ is a stopping time and $\mathbb{E}\tau < \infty$ then $\mathbb{E}N_\tau = \mathbb{E}\tau$ for $N_t$ - Poisson process with parameter $1$.
I started like this. I know that $N_t - t$ is a martingale. Then by Doob's optional sampling we have that $\mathbb{E}N_{\tau \wedge n} = \mathbb{E}(\tau \wedge n) \le \mathbb{E}\tau < \infty$. Now, by monotone convergence theorem $\mathbb{E}(\tau \wedge n) \to \mathbb{E}\tau$ and here i stopped. I have a problem with showing that $\mathbb{E}N_{\tau \wedge n} \to \mathbb{E}N_{\tau}$.
In the case of Wiener process, I would use maximal inequality for martingales to get the dominator, but here it won't work, I guess.
Since the Poisson process $(N_t)_{t \geq 0}$ has non-decreasing sample paths, we have $$N_{\tau} = \sup_{n \in \mathbb{N}} N_{\tau \wedge n}.$$ By the monotone convergence theorem, this implies that
$$\mathbb{E}(N_{\tau}) = \sup_{n \in \mathbb{N}} \mathbb{E}(N_{n \wedge \tau}).$$