Let $X$ be a stochastic process with continuous sample paths and the canonical filtration. $\Gamma\in {\cal B} (\mathbb{R}^n)$ be a closed set and let $T \triangleq \inf\{t\ge 0:X_t \in \Gamma\}$. Show that $T$ is a stopping time.
I understand the proof using $\Gamma_n = \{x:$dist$(x,\Gamma)<\frac{1}{n}\}$ and the fact $T_n = \inf\{t\ge 0: X_t \in \Gamma_n\}$ is an optional time. My question is why is the following method wrong or not obvious?
Using continuity, if $T(\omega) \le t$, then given any $\epsilon > 0$, $A =\{X_s(\omega):s \in [T(\omega), T(\omega) +\epsilon)\} \cap \Gamma\neq \emptyset$. Then there exists $\{s_n\}\subset [T(\omega), T(\omega)+\epsilon)$ non-increasing (not necessarily distinct) such that $s_n \rightarrow T(\omega)$ with $X_{s_n}(\omega) \in \Gamma$, and since $\Gamma$ is closed, we know $\lim_{n\rightarrow \infty}X_{s_n}(\omega) = X_{T(\omega)}(\omega)\in \Gamma$. Therefore, $\{T \le t\} = \bigcup_{0 \le s \le t}\{X_s \in \Gamma\}\in \bigcup_{0\le s \le t}{\cal F}_s^X = {\cal F}_t^X$.
The reason I am asking is that I think this method is more straight forward than the one I mentioned before, and I don't see this method anywhere online for this problem, so there must be some problem with it.
So, please let me know if this is correct. Thanks in advance.
Best Answer
The issue with your argument is that $$\bigcup_{0 \le s \le t}\{X_s \in \Gamma\}$$ is an uncountable union and a priori may not be in $\mathcal{F}_t$. If you tried replacing the union with one over a countable subset of $[0,t]$ you may find that $X$ hits your set only at a single point $s$ that is not in your countable set, so the argument cannot be fixed this way either. Indeed, consider $X(t) = t$ deterministic and $\Gamma = \{x\}$ to be a single point.