Consider a particle undergoing Brownian motion in $\mathbb{R}^n$, starting at the origin, and let $B(t)$ denote its position at time $t$. Let $X$ be an arbitrary subset of $\mathbb{R}^n$. I am trying to understand what properties $X$ needs to have so that the probability of the Brownian particle striking $X$ within time $t$ is non-zero, that is, $\mathbb{P}(B(s) \in X, s \leq t) \neq 0$. It seems to me that a sufficient condition on $X$ could be that it contains a subset $Y$ which is homeomorphic to $\mathbb{R}^{n - 1}$. Is this necessary too? Is there a more relaxed sufficient condition? Thanks in advance.



Brownian motion in $\mathbb{R}^n$, probability of hitting a set
It's not that simple. See about polar/nonpolar points/sets e.g. in <http://wiki.math.toronto.edu/TorontoMathWiki/index.php/Brownian_Motion_and_Harmonic_functions>


If I remember correctly, a set is not polar iff it has positive capacity (w.r.t. logarithmic potential in two dimensions, and Newton potential for $d\geq 3$).


