Let $\{B_t : t \ge 0 \}$ be a standard Brownian Motion, and $T = \inf\{ n \ge 1 : B_{n} = n+1 \}$. Show that $P(T < \infty) < 1$.
I believe $P(T < n) = P\{ B_n = n+1 \text{ for at least one } k < n \}$. However, we know that Brownian motion is normally distributed $(B_n \sim N(0,n))$ and therefore continuous. In this case, the measure of a singleton is zero which implies that $P(B_n = n+1) = 0 \; \forall n$.
Therefore, I'm confused about how I can calculate $P(T < n)$ if each event seems to have probability zero. I'm not sure if I missed something?