Let $ \tau(\omega) =\inf \{ t \in (0, \infty): B_t(\omega) = 10\}$. Does it follow $P( \tau = s) = 0?$

Question

Consider the first time the Brownian motion reaches $10$, i.e. $$\tau(\omega) =\inf \{ t \in (0, \infty): B_t(\omega) = 10\}$$ Clearly, $P(B_t = 10) =0, \forall t \in \mathbb{R}^+$since the Brownian motion is normally distributed. But does it follow that $$P( \tau = s) =0, \forall s \in \mathbb{R}^+$$ and does considering the interval $[0, \infty)$ make a difference?

Answer 1

Note that $P(\tau = s) \leq P(B_s=10)=0$ since the event $\{\tau=s\}$ is contained in the event $\{B_s=10\}$, and since $B_s$ has a continuous distribution.

Note that this only shows that the law of $\tau$ is atomless, which doesn't necessarily imply that a density exists with respect to Lebesgue measure (think Cantor-type or local-time measures). If you want an explicit density for $\tau$ see theorem 2.32 on page 56 of this book.