Let $E$ be a finite dimensional Euclidean space. We take $X:=(X_t)_{t\geq 0}$ to be an $\Bbb{F}:=(\mathcal{F}_t)_{t\geq 0}$ adapted process with values in $E$. Let $G\subset E$ be open and assume that $X$ has either right- or left-continuous trajectories. Now define $T_{X,G}:=\inf\{t>0: X_t(\omega)\in G\}$. I want to show that this is an $\Bbb{F}$-optional time.
By definition $T$ is an $\Bbb{F}$-optional time if for all $k>0$, $\{T<k\}\in \mathcal{F}_k$. As I understood the part
assume that $X$ has either right- or left-continuous trajectories
they mean that for a fixed $\omega\in \Omega$, $t\mapsto X_t(\omega)$ is right- or left-continuous.
But now if I fix an arbitrary $k\geq 0$ and consider $$\{T_{X,G}<k\}=\{\inf\{t>0: X_t(\omega)\in G\}<k\}$$ then I don't know how to continue. So I mean if we would work in the discrete case then $\{\inf\{t>0: X_t(\omega)\in G\}<k\}=\bigcup_{1\leq n\leq k-1}\{\inf\{t>0: X_t(\omega)\in G\}=k\}$ but in our case we are not in the discrete one. I think therefore we need this assumption of right- or left-continuous trajectories, but I don't see how this is useful in the proof.
Can maybe someone help me?
$$\inf\{t>0\mid X_t\in G\}<k\iff\exists t\in(0,k),\ X_t\in G,$$ and because $G$ is open and $X_t$ is either left- or right-continuous, the latter is equivalent to $$ \exists q\in(0,k)\cap\mathbb Q,\ X_q\in G. $$
Hence $$ \{T_{X,G}<k\}=\bigcup_{q\in(0,k)\cap\mathbb Q}\{X_q\in G\}\in\mathcal F_k. $$