In my study of stochastic processes I find this property that gives the "form" that can take a stopping time.
Consider $A\subset\mathbb{R}$ a closed set and $ X = (X_t)_{t\geq 0}$ an adapted continuous process. Then, the hitting time of $A$ by $X$ defined as $T_{A}(X) := \inf\left\{t\geq 0 : X_t\in A\right\}$ is a stopping time.
The result is intuitive if we keep the intuition of a stopping time as an event for which we are able to know if it occured or not by looking at the information available so far. However, the proof is a little bit more involved.
The idea of the proof is to put $\left\{T_{A}(X)\leq t\right\} = \bigcap_{k=1}^{\infty}\bigcup_{s\in[0,t]\cap\mathbb{Q}}\left\{d(X_s, A)\leq\frac{1}{k}\right\}$
using the continuity of $X$ and the fact that $A$ is closed, but really I cannot see how it is used here ?
Then, they conclude that $\left\{T_{A}(X)\leq t\right\}\in\mathcal{F}_t$ using the fact that $X_t$ is adapted and the stability of the countable union and intersection which is a point that I can understand I think.
Thank you a lot !
Fix $\omega \in \Omega $. Then observe that if $T_A(X(\omega ))\in[0, t]$, when $s\mapsto X_s(\omega )$ is continuous, then there are rational numbers $q\in\mathbb{Q}\cap [0,t]$ such that $d(X_q(\omega ),A)$ is arbitrarily small. This is the case because $X$ being continuous means that $\{X_h(\omega ):h\in[0,t]\}=[a,b]$ for some $a,b\in \mathbb{R}$, so every point in $[a,b]$ is a limit point and $Q:=\{X_q(\omega ):q\in \mathbb{Q}\cap [0,t]\}$ is dense in $[a,b]$.
Finally, if $A$ is closed it means that $d(X_s(\omega ),A)=0$ if and only if $X_s(\omega )\in A$, so the "if... then..." statement of the first paragraph becomes an "if and only if" statement. To be more clear, we have that
$$ \begin{align*} \bigcap_{k=1}^{\infty}\bigcup_{s\in[0,t]\cap\mathbb{Q}}\left\{\omega \in \Omega :d(X_s(\omega ), A)\leq\frac{1}{k}\right\}&=\{\omega \in \Omega :d(X_s(\omega ),A)=0,s\in[0,t]\}\\ &=\{\omega \in \Omega : X_s(\omega )\in A, s\in[0,t]\}\\ &=\{\omega \in \Omega : T_A(\omega )\in[0,t]\} \end{align*} $$ where the first equality holds because $X$ is continuous, the second because $A$ is closed, and the third due to the definition of $T_A$.