Let $B_t$ be a brownian motion started at 0. I am trying to prove that $\tau$, defined as: $$ \tau = \inf\{t > 0 \mbox{ }|\mbox{ } \left|B_t\right| \geq \frac{1}{1+t} \} $$ is a stopping time with respect to the filtration $(\mathscr{F}_{t}^B)_{t\geq 0}$.
I have some confusion concerning what if exactly measurable and what is not. I have done the following: $$ {\{\tau \leq t\}} = \bigcup_{s\in[0,t]\cap\mathbb Q } \{\left|B_s\right| \geq \frac{1}{1+s}\} = \bigcup_{s\in[0,t]\cap\mathbb Q } \left(\Omega\setminus\{\left|B_s\right| < \frac{1}{1+s}\}\right) $$
I can conclude by saying that $\{\left|B_s\right| < \frac{1}{1+s}\} \subseteq\mathscr{F}_{\left(\frac{1}{1+s}\right)} \subseteq \mathscr{F}_t$, but because I have "changed" the problem to $\Omega \setminus \{\left|B_s\right| < \frac{1}{1+s}\}$... haven't I modified the problem to the case a hitting time to an open set?
P.S: I know the answer is that I have not changed anything at all, I just want to make sure why.