This is primarily in reference to this question on MO. Serguei Popov's answer gives an explicit formula for the probability of a Brownian particle starting at the origin in $\mathbb{R}^n$ hitting the sphere $S^{n - 1}_r = \{x \in \mathbb{R}^n : \Vert x\Vert = r\}$ within time $t$, which is the same as $\mathbb{P}(\sup _{s \leq t}\Vert B(s)\Vert \geq r)$.

Now, it seems to me that $\mathbb{P}(\sup _{s \leq t}\Vert B(s)\Vert > r)$ should be strictly greater than $\mathbb{P}(\sup _{s \leq t}\Vert B(s)\Vert \geq r)$, because of the following intuitive reasoning: when the particle strikes the sphere, it can come back inside, or travel along the sphere, that is the boundary of the ball $B(0, r)$.

But, using the formula referenced in the answer for two distinct radii $r_1 < r_2$, and letting $r_2 \to r_1$, it seems that $\mathbb{P}(\sup _{s \leq t}\Vert B(s)\Vert > r) = \mathbb{P}(\sup _{s \leq t}\Vert B(s)\Vert \geq r)$. What I am missing here?