Measurability of point processes

Question

I am reading the "Construction of Diffusion processes with Wentzell's Boundary conditions by means of poisson point processes of Browninan excursions" of S. Watanabe and he says:

I understand that to say that $N_p((0,t]\times U)$ is $\mathcal{F}_t$ adapted means $$\forall a >0\qquad\{\omega\mid N_{p(\omega)}((0,t]\times U))>a\} \in \mathcal{F}_t $$ But what does it mean to say that $\{N_p((0,t]\times U)\}_{U \in \mathbb{B}(X)}$ is $\mathcal{F}_t$ adapted?

muaddib · Answer 1 · 2015-07-24T19:12:20.273

1

The only way I make sense of this is that for all $U \in \mathcal{B}(X)$ and $t$ it is required to be true that the random variables $f_{U, t}$ given by $$f_{U, t}(\omega) = N_{p(\omega)}((0, t] \times U)$$ are $\mathcal{F}_t$ measurable. I have placed $U$ and $t$ in the subscript of $f$ to highlight the dependence of its definition on those two parameters.

edited Jul 24 '15 at 19:12

answered Jul 23 '15 at 21:53

muaddib

8,267

It is the same as the first sentence I wrote ? – Conrado Costa Jul 23 '15 at 21:55
In a way. If you instead define $$f_t(\omega) = N_{p(\omega)}((0,t]\times U)$$ then $f_t$ can be considered an adapted process. – muaddib Jul 23 '15 at 21:59
@ConradoCosta - I realized given your questions I had misread the original question. I've modified my answer. – muaddib Jul 24 '15 at 19:12
I see. Anyway, it is strange to use such a notation, One often writes ${B_t}t$ to mean the brownian process. So ${N_p((0,t]\times U)}{U \in \mathbb{B}(X)}$ is a strange object... – Conrado Costa Jul 24 '15 at 19:16
@ConradoCosta Agreed. – muaddib Jul 24 '15 at 19:17

Measurability of point processes

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