It's known that, given $\Gamma \in \mathcal B (\mathbb R ^d)$ and $X = > (X_t)_{t\geq 0}$ with right-continuous path, the random time
$$T_{\Gamma} = \inf \{ t\geq 0 : X_t (\omega) \in \Gamma\} $$
- is an optional time if $\Gamma$ is open,
- it's a stopping time if $\Gamma$ is closed.
this result comes from I. Karatzas, S. E. Shreve, Brownian Motion and Stochastic Calculus.
My first question is whether this result holds also for left-continuous paths processes.
More generally, if the filtration $(\mathcal F_t)_{t\geq 0}$ is right-continuous, then for any adapted, optional process $X = (X_t)_{t\geq 0}$ and any borelien $\Gamma$, $T_{\Gamma}$ is a stopping time.
A result from P. A. Meyer, Dellacherie, Probabilités et Potentiel. (I wish hardly to see this proof so if anyone has a numerical version of this book I'd appreciate it very much).
An exercice which serves as an exemple is the following.
Let be $\alpha, \beta \in \mathbb R$ such that $\alpha < \beta $ and $x \in [\alpha, \beta ]$. Consider the random time
$$T_x = \inf \{ t\geq 0 : x+ B_t \notin [\alpha, \beta]\},$$ where $B=(B_t)_{t\geq 0}$ is a standard brownian motion in $\mathbb R$ starting from zero. Show that $T_x$ is a stopping time and $\mathbb P -\text{a.e.}$ finite.
The solution I've seen say so.
Consider $\tilde{ T_x} = \inf \{ t\geq 0 : x+ B_t \in (-\infty,\alpha[\cup ]\beta, +\infty) \},$ then $\{T_x \leq t\}=\{\tilde{T_x} \leq t\}$ without justifying. HERE the first question about this example.
Then we start to consider $\{T_x \leq t\}^c$
\begin{align} \{T_x > t\}&= \{\forall s \in [0,t], x+B_s \in [\alpha, \beta] \}\cup \{x+B_t =\alpha\}^c \cup \{x+B_t =\alpha\}^c \\ &= \underset{\{q\in [0,t],\\ q \in \mathbb Q\}}{\cap}\{x+B_q \in [\alpha, \beta]\}\end{align}
since $\{x+B_q \in [\alpha, \beta]\}\in \mathcal F_q \subset \mathcal F_t, \ \forall q \in \mathbb Q $, we conclude that $T_x$ is a stopping time.
In addition, in the solution one remarks that $\{x+B_t =\alpha\}^c ,\ \{x+B_t =\alpha\}^c \in \mathcal F_0$.
The point is I did not understand why the first line is right even less why the last remark is needed, obviously supposing that there is no mistake and no redundancy. I don't see why the current version wold be correct. More precisely, why would not be correct to write $\{T_x > t\}= \{\forall s \in [0,t], x+B_s \in [\alpha, \beta] \}$ simply.
I didn't get neither why must we index the intersection by rational numbers. Is it really necessary? The only answer I find to myself is that is just a technical issue related countable sets and indexation. But we could simply "index ate" by $s \in \mathbb R$, don't we ? Or should it be technically wrong by some reason I am unaware?
Edit: I made showing that $T_x < + \infty \ \mathbb P -\text{a.e.}$ subject of another question
I would appretiate if someone could help me whith all those questions.
Thanks in advance.