Consider a homogeneous Poisson point process as function of time, $N(t)$ with the parameter $\lambda$, where $N(t)$ represents the total number of occurrences (e.g. phone calls) up to and including time $t$.
Then, $E(N(t))=\lambda t$ with $N(t)$ is Poisson $(\lambda t).$
For a nonhomogeneous Poisson point process with the intensity function, $\lambda(t)$, we have:
$E(N(k))=\int_o^k \lambda(s) \, ds.$ with $N(t+h)-N(t)$ is Poisson $(\int_t^{t+h} \lambda(s) \, ds)$ for a small $h>0$.
I want to understand why in the inhomogeneous case, we are looking at the difference between the total number of arrivals of now and a tiny bit in the future.
To homogeneous case makes sense to me, because if the arrival follows the Poisson distribution with the average number of arrivals per unit time, $\lambda$, then the expectation of that random variable standing at time $t$ should be simply the average times the time. I understand this as a special case of the Lebeague measure; $t$ can be spatial or finite-higher dimensional.
My questions are:
(1) In the non-homogeneous case, why is that we are looking at for a small $h>0$, the difference between $N(t+h)$ and $N(t)$ to characterize the this small difference follows a Poisson distribution?
(2) In the non-homogeneous case, is it problematic to make any characterization on $N(t)$?