Progressive measurability of a specific set related to Brownian motion 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.
 A: This follows from two quite standard results.


*

*All continuous and adapted processes are predictable.

*All predictable processes are progressively measurable.


So $W$, and hence $Z=W^{-1}(\mathbb{R}\setminus\{0\})$, is progressively measurable. However, as predictable processes are also optional, the set in question is optional. It, therefore, does not provide an example of a progressively measurable set which is not optional.
For an example of a set which is progressively measurable but not optional, take the set of left limit points of the connected components of $Z$. That is, take the set of points $(t,\omega)\in([0,\infty)\times\Omega)\setminus Z$ for which $(t,t+\epsilon)\times\{\omega\}\subset Z$ for some $\epsilon\gt0$ (depending on $t$ and $\omega$). It can be seen that this is progressively measurable but not optional from the optional projection theorem. In fact, the optional projection of its indicator function is $0$.
A: 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 $$[W=0] := \{(t,\omega) \in [0,\infty) \times \Omega; W(t,\omega)=0\}$$
