How to prove the $\lim \inf$ and $\lim \sup$ for this expression involving a Poisson Process? How do I prove the two equalities in the following picture?
Let $N$ be a Poisson Process, then 
Property (ii) alluded in the picture is $\forall_t(N_t - N_{t-}) \in \{0,1\}$
For the first inequality, I tried using
$\lim \inf (N_{T+\delta}-N_T - 1)^+/ \lambda\delta = 0$, since  $\exists \epsilon \forall \delta \in ]0,\epsilon[$, we have $N_{T+\delta}=N_T$ ($N$ is cadlag and in the interval there's no jump), so by Fatou $0\leq \lim \inf E(...) \leq E(\lim \inf ...) = 0$
For the second equality, I thought of using $(N_{T+\delta}-N_T - 1)^+=(N_{T+\delta}-N_T - 1)I_{\{N_{T+\delta}-N_T > 1\}}$.
But then I obtain $P(N_{T+\delta}-N_T > 1\mid \mathcal{F}_S)$ which is not what we would want...
 A: Your application of Fatou's lemma doesn't seem right (the inequality goes the other way), and I think the question is a little misleading, sorry! (This is from Lemma 5.5.20 of my book with Robert Elliott, "Stochastic Calculus and Applications".)
To show the first inequality,  assume $T=S=0$ (for notational simplicity) so we want to compute
$E[(N_\delta -1)^+]$. Let $\tau_1$ be the time of the first jump of $N$, and $\tau = \min(\delta, \tau_1)$, which is a stopping time. Then we can write
$0\leq E[(N_\delta -1)^+] = E[(N_\delta -N_{\tau})N_{\tau}].$
By applying optional stopping at time $\tau$, (and remembering $|N_\tau|\leq 1$, so we don't need to worry about integrability problems) we get
$E[(N_\delta -N_{\tau})N_{\tau}] = E[ E[N_\delta - N_\tau|\mathcal{F}_\tau] N_\tau] = E[\lambda(\delta-\tau)N_\tau] \leq \lambda \delta E[N_\tau] \leq \lambda \delta E[N_\delta] = (\lambda\delta)^2.$
Therefore,
$\lim_{\delta\to 0} \frac{E[(N_\delta-1)^+]}{\lambda\delta} = 0.$
The second equality then comes from substituting this limit into the calculation on the previous page of the book.
Sorry for the error - I'll add this correction to the errata sheet on my webpage.
