I am trying to prove that the first hitting time of a closed set H by a Brownian motion is a stopping time. I have found a proof that states:
$$\{\mathcal{T}\leqslant t\} = \bigcap_{n=1}^{\infty}\bigcup_{s\in\mathbb{Q}\cap(0,t)}\bigcup_{x\in\mathbb{Q}^d\cap H}\{B(s)\in Ball(x,1/n)\} \in \mathcal{F}(t) $$
After looking at this expression for a while, I think I see logic in it, namely why it takes countable unions of all "points" that give us Brownian motion inside this set at times before t.
However, I don't understand why we cannot simply write:
$$\{\mathcal{T}\leqslant t\} = \bigcup_{ s\in\mathbb{Q}\cap(0,t)} \{B(s)\in H \} \in \mathcal{F}(t)$$
My question is, why is it wrong?
So far what I think the reason is that $s\in\mathbb{Q}\cap(0,t)$ and not $(0,t]$ , so we have a limit of sets from the left and it has something to do with left-continuity of the filtration $\mathcal{F}(t)$.
Thank you.
Just consider $H:=\{1\}$ and suppose that a sample path $t \mapsto B_t(\omega)$ of the Brownian motion (which starts at time $0$ at $0$) satisfies
$$B_T(\omega)=1$$
for a certain $T>0$ and
$$B_t(\omega)<1$$
for all $t \in [0,T)$, i.e. the sample path hits the set $H=\{1\}$ at time $T$ for the first time. Then $\omega \in \{\tau \leq T\}$, but $\omega$ is not contained in the union
$$\bigcup_{s \in \mathbb{Q} \cap (0,T)} \{B_s \in H\}.$$
This means that the identity
$$\{\tau \leq t\} = \bigcup_{s \in \mathbb{Q} \cap (0,t)} \{B_s \in H\}$$
does not hold true.
Remark: It is possible to show that
$$\{\tau \leq t\} = \left\{\omega; \inf_{s \in \mathbb{Q} \cap (0,t)} d(B_s(\omega),H)=0 \right\}$$
where
$$d(x,H) := \inf_{y \in H} |x-y|.$$
This is another possibility to prove that $\tau$ is a stopping time (see e.g. Brownian Motion - An Introduction to Stochastic Processes by René Schilling & Lothar Partzsch, Chapter 5, for details).