I was reading the book Optimal Stopping and Free Boundary Problems by Peskir & Shiryaev and came across an interesting statement. The setting is as follows.
Let $(X_t)_{t\geq0}$ with state space $E=\mathbb{R}^d$ equipped with the $\sigma$-field $\mathcal{B} = \mathcal{B}(\mathbb{R}^d)$, be a strong Markov Process with right-continuous paths defined on the canonical pathspace space $(E^{[0,\infty)},\mathcal{B}^{[0,\infty)}, > P_x)$. Furthermore let $(\mathcal{F}_t)_{t\geq 0}$ be a right-continuous filtration and let $X_t$ be measurable w.r.t $\mathcal{F}_t$ for all $t\geq 0$. Let $G: E \to \mathbb{R}$ be measurable and upper-semi-continuous and $V(x) = \sup_{\tau} \mathbb{E}_x(G(X_{\tau}))$ (where $\tau $ is an almost surely finite stopping time) be lower-semi-continuous. Consider the closed set $D = \{x\in E \mid V(x) = G(x) \}$.
The author states that $\tau_{D} = \inf\{t\geq 0 \mid X_t \in D \}$ is a stopping time w.r.t. $(\mathcal{F}_t)_{t\geq 0 }$, since $(X_t)_{t\geq 0}$ has right-continuous paths and $(\mathcal{F}_t)_{t\geq 0}$ is right-continuous.
My question is why is that so ? I know that $\tau_{D}$ is a stopping time if, D is open or $X$ has continuous paths or $\mathcal{F_0}$ is $P_x$-complete for all $x\in E$. I would appreciate either reference or a solution for this statement.