I have been checked by the following proposition when reading a textbook on stochastic calculus.
The problem is about a Poisson Process $N_t$ with parameter $\lambda$. In definition the authors emphasize that $N_t$ is cadlag and $N_t - N_{t-} \in \{0,1\}$ up to indistinguishability (I can't see where it is used in the sequel). Now suppose there are two stopping times $S \leq T$. The authors maintain that for $\delta > 0$, the following equality can be proved by virtue of Fatou's lemma:
$$ \limsup_{\delta \rightarrow 0} \frac{\mathbb{E}[(N_{T+\delta}-N_T-1)^{+} \mid \mathscr{F}_S]}{\lambda\delta} = 0\text{, a.s.} $$
I got lost here.. What I can see is merely the following inequalities:
$$ 0 \leq \limsup_{\delta \rightarrow 0} \frac{\mathbb{E}[(N_{T+\delta}-N_T-1)^{+} \mid \mathscr{F}_S]}{\lambda\delta} \leq \frac{\mathbb{E}[\limsup(N_{T+\delta}-N_T-1)^{+} \mid \mathscr{F}_S]}{\lambda\delta} $$
But how can we conclude that the numerator tends to $0$ more rapidly than the denominator $\lambda\delta$? Any hint will be greatly appreciated!
$$ (N_{T+\delta} -N_T -1)^+ \quad \begin{cases} =0 & \text{if } N_{T+\delta} = N_T, \\ \\ =0 & \text{if } N_{T+\delta} = N_T+1, \\ \\ >0 & \text{only if } N_{T+\delta} - N_T \ge 2. \end{cases} $$ So it's actually about the improbability of two or more arrivals in a short time, by comparison to the length of time. Suppose, for example, that $\lambda=4\text{ per hour}.$ The expected number of arrivals in one second, divided by one second, is $(4 \text{ per hour}).$ But the probability of more than one arrival in a mere one second is so small that the expected number of arrivals in one second is about the same as $1\times{}$the probability that the number of arrivals is $1.$