I am trying to go through the problems in Shreve & Oksendal, this is the problem 2.6
Prove that the first hitting time $\tau$ of $A$ an open subset of $\mathcal{B}(\mathbb{R}^d)$ is an optional time.
Here is the beginning of my intuitive argument: Since $A$ is open it can be rewritten as a countable intersection of closed borel sets i.e $A=\cap_n\{A_n\}$. Then for $\tau$ to be less than $t$ , it means that at some point before $t$ we were in all of the $\{A_n\}$. I translate my "at some point" by an union and the "in all of the" by an intersection; which gives $$\{\tau<t\} = \cup_{r\in\mathbb{Q}\cap\left[0,t\right)}\cap_n\{X_r\in A_n\}$$ and every sets on the r.h.s $\in\mathcal{F}_r\subset\mathcal{F}_t$
I am quite certain that my arguments is full of holes, I would appreciate any critics, thanks !
Edit: I feel my argument is missing the countable limit process to approach $X_t$, say $\{r_i\}_{i\geq 0}\in\mathbb{Q}\cap\left[0,t\right)$ with $\{r_i\}\downarrow t$ $$\{\tau<t\} = \cup_{i\geq 0}\cap_{j\geq i}\cap_n\{X_{r_i}\in A_n\}$$