Let $\{W_t: t \in R_+\} $ be a standard Brownian motion process on a given probability space. I am interested in assessing the progressive measurability of the following set: $Z(\omega) := \{t: W_t(\omega) \not= 0\}.$
I can see that the set $Z(\omega) $ is open and that I can therefore write it as a countable union of open sets (which would be basically the excursion time intervals between the times the process hits 0). However, I cannot seem to produce a rigorous argument to prove that the set above is progressive.
I think this example is due to Meyer or Dellacherie and Meyer and it is meant to supply an example of a process that is progressive and not optional. I have known it for some time, but now I would like to fill all the steps.
Thank you.
Let $T>0$ and define $$Y(t,\omega) := 1_{Z(\omega)}(t) = \begin{cases} 1 & W_t(\omega) \neq 0 \\ 0 & W_t(\omega) = 0 \end{cases}$$ for $t \in [0,T]$, $\omega \in \Omega$. Then
$$[Y=0] = \{(t,\omega); t \leq T, W_t(\omega) = 0\} = [W=0] \cap ([0,T] \times \Omega)$$
Since the Brownian motion $W$ is progressively measurable, we conclude that
$$[Y=0] \in \mathcal{B}[0,T] \otimes \mathcal{F}_T$$
By definition, $Y$ does only attain the values $0$ and $1$, thus
$$Y:([0,T] \times \Omega,\mathcal{B}[0,T] \otimes \mathcal{F}_T) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$ is measurable.
Notation $$[W=0] := \{(t,\omega) \in [0,\infty) \times \Omega; W(t,\omega)=0\}$$