Understanding $\mathcal F_\tau$ and showing that $X_\tau$ is $\mathcal F_\tau$-measurable where $\tau$ is a stopping time. Let $(\Omega ,\mathcal F,\{\mathcal F_t\},\mathbb P)$ be a filtered probability space. Let $\tau$ be a stopping time. We have that $$\mathcal F_\tau:=\{A\in \mathcal F\mid A\cap \{\tau\leq t\}\in \mathcal F_t\}.$$
Q1) What exactly is $\mathcal F_\tau$? I know it's a $\sigma $-algebra, but don't really see the motivation behind its definition. My teacher says that $A\in \mathcal F_\tau$ means that even if $A\notin \mathcal F$, you know that $A$ occurs or not whenever $\tau\leq t$. But to be honest, I don't really understand what it means. Can someone illustrate or explain it a bit more?
Q2) Let $(X_t)$ a stochastic process. Then $X_\tau$ is $\mathcal F_\tau$ -measurable. How can I prove that? I guess that I have to prove that $\{X_\tau\leq x\}\cap\{\tau\leq t\}\in \mathcal F_t$.
I know that $\{X_s\leq x\}\in \mathcal F_t$ for all $s\leq t$. I guess that $\{X_\tau\leq x\}\cap \{\tau\leq t\}$ if and only if there is $s\leq t$ s.t. $\{X_s\leq x\}$, i.e. $$\{X_\tau\leq x\}\cap \{\tau\leq t\}=\bigcup_{s\leq t}\{X_s\leq x\},$$
but I'm not so sure how to continue. Maybe if $(X_t)$ is continuous, then $$\bigcup_{s\leq t}\{X_s\leq x\}=\bigcup_{\substack{s\leq t\\ s\in \mathbb Q}}\{X_s\leq x\}\in \mathcal F_t,$$
but how can I conclude this whenever $(X_t)$ is not continuous?
 A: $\def\F{\mathscr{F}}\def\Ω{{\mit Ω}}\def\B{\mathscr{B}}\def\R{\mathbb{R}}$For Q1, since a stopping time are intuitively the (random) time when one stops observing the stochastic process according to some given rules with information available up to that time, then $\F_τ$ is the collection of events that can be checked to see whether they have happened or not by observation up to $τ$. Now for this remark:

$A\in \F_τ$ means that even if $A \notin \F$, you know that $A$ occur or not whenever $τ \leqslant t$.

The wording is incorrect because the definition of $\F_τ$ requires that $A \in \F$ for any $A \in \F_τ$, but what your professor said is more likely to be this:

$A\in \F_τ$ means that even if $A \notin \F_{\color{red}{t}}$, you know that $A$ occur or not whenever $τ \leqslant t$.

This interpretation is indeed correct since $A \cap \{τ \leqslant t\} \in \F_t$ exactly means that the knowledge on whether the event $A \cap \{τ \leqslant t\}$ has happened or not is available in the information up to time $t$.
For Q2, progressive measurability of $X$ is usually assumed to prove that $X_τ$ is $\F_τ$-measurable. Assuming this, the mapping$$
\begin{matrix}
([0, t] × \Ω, \B([0, t]) × \F) & \longrightarrow & (\R, \B(\R))\\
(s, ω) & \longmapsto & X(s, ω)
\end{matrix}
$$
is measurable for any $t \geqslant 0$. Note that$$
\begin{matrix}
(\Ω, \F) & \longrightarrow & ([0, t] × \Ω, \B([0, t]) × \F)\\
ω & \longmapsto & (τ(ω) ∧ t, ω)
\end{matrix}
$$
is also measurable since $τ ∧ t$ is also a stopping time, then the composition$$
\begin{matrix}
(\Ω, \F) & \longrightarrow & ([0, t] × \Ω, \B([0, t]) × \F) & \longrightarrow & (\R, \B(\R))\\
ω & \longmapsto & (τ(ω) ∧ t, ω) & \longmapsto & X(τ(ω) ∧ t, ω)
\end{matrix}
$$
is measurable, which implies that the stopped process $\{X_{τ ∧ t}\}$ is progressively measurable.
Now for any $t \geqslant 0$ and $B \in \B(\R)$, because $\{X_{τ ∧ t} \in B\} \in \F_t$, so$$
\{X_τ \in B\} \cap \{τ \leqslant t\} = \{X_{τ ∧ t} \in B\} \cap \{τ \leqslant t\} \in \F_t.
$$
Thus $X_τ$ is $\F_τ$-measurable.
A: I know that it's not really an answer, but I think that it's necessary to point out the following fact. So, I put it as an answer, and I hope that it will be fine with the community.

I'm not an expert in stopping time, but I don't think that it's true that $$\bigcup_{s<t}\{X_s\leq x\}=\bigcup_{\substack{s<t\\ s\in \mathbb Q}}\{X_s\leq x\}.$$
Take $X_s=|s-\sqrt 2|$. Then if $t>\sqrt 2$, $$\bigcup_{s\leq t}\{X_s\leq 0\}=\{X_{\sqrt 2}=0\}$$
but $$\bigcup_{\substack{s\leq t\\s\in \mathbb Q}}\{X_s\leq 0\}=\varnothing .$$
Nevertheless, this can be fix in considering $\{X_{\tau}<x\}$ instead of $\{X_\tau\leq x\}$ (since $\{(-\infty ,a)\mid a\in\mathbb R\}$ is a $\pi-$system generating Borel sets).
