I am reading Jean Francois Le Gall's Brownian Motion, martingales and stochastic calculus. There is a Theorem here that states for every $a>0, $ $T_a = \inf\{t\ge 0: B_t = a\}$ has the same distribution as $\frac{a^2}{B_1^2}$ and has density $f(t)=\frac{a}{\sqrt{2\pi t^3}} \exp(-\frac{a^2}{2t})1_{\{t>0\}}$.
The proof is given by $$P(T_a\le t)=P(S_t\ge a)=P(|B_t|\ge a)=P(B_t^2\ge a^2)=P(tB_1^2\ge a^2)=P(\frac{a^2}{B_1^2}\le t).$$
However, I am not sure why the final equality holds as in how we can divide both sides by $B_1^2$ since $B_1$ can take the value $0$. Is this because $P(tB_1^2\ge a^2)=P(tB_1^2\ge a^2, B_1\neq 0)=P(t\ge \frac{a^2}{B_1^2}, B_1\neq 0)=P(t\ge \frac{a^2}{B_1^2})$ since $P(B_1 \neq 0)=1$?
It is simpler than that, in the event that:
$$tB_1^2 \geq a^2 $$
As $a > 0$ , $B_1$ is nonzero hence the probability of these events are exactly the same.