# About stopping time in Brownian Motion

I am reading the book Brownian Motion by Yuval peres and Peter Morters. I am having trouble understanding the following remark. In perticular, How the set on the right equals the set on the left. The remark is as follows,

Suppose $$H$$ is a closed set, for example a singleton. Then the first hitting time $$T=\inf \{t \geqslant 0: B(t) \in H\}$$ of the set $$H$$ is a stopping time with respect to $$\left(\mathcal{F}^{0}(t): t \geqslant 0\right)$$. Indeed, we note that $$\{T \leqslant t\}=\bigcap_{n=1}^{\infty} \bigcup_{s \in \mathbb{Q} \cap(0, t)} \bigcup_{x \in \mathbb{Q}^{d} \cap H}\left\{B(s) \in \mathcal{B}\left(x, \frac{1}{n}\right)\right\} \in \mathcal{F}^{0}(t) .$$

Here, $$\mathcal{F}^{0}(t) = \sigma\left(B(s): 0 \leq s \leq t \right)$$

I don't get why it is true because, if H a closed set which does not intersect with $$\mathbb{Q}^d$$ then the intersection on RHS would be just empty.

• You are right. The identity is not correct. Commented Jun 7, 2022 at 7:17

As stated in the comments, the identity is not correct as written. The argument should just be that there exists a countable subset $$A \subseteq H$$ which is dense in $$H$$, and then $$\{T \leqslant t\}=\bigcap_{n=1}^{\infty} \bigcup_{s \in \mathbb{Q} \cap(0, t)} \bigcup_{x \in A}\left\{B(s) \in \mathcal{B}\left(x, \frac{1}{n}\right)\right\} \in \mathcal{F}^{0}(t) .$$
• Ah, this makes sense, Thanks! So intuitively we are just checking if for some s whether the Brownian motion gets arbitrarily close to an element of $A$ (and consequently and equivalently to the set $H$). And conclude with continuity of $B$ and closedness $H$. Commented Jun 7, 2022 at 16:50
• @Mathaddict Yes, exactly! The continuity of $B$ allows us to only check rational $s$, and closedness of $H$ lets us only check that we get arbitrarily close to $H$. Commented Jun 7, 2022 at 19:22