Brownian motion in $\mathbb{R}^n$, probability of hitting a set 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.
 A: 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 $d\geq 3$).
A: Serguei Popov has the "right" answer.  Just to give a quick counterexample, consider the set $X = (\mathbb{R} \setminus \mathbb{Q})^n$ of points with all coordinates irrational.  For every $s$ we have $\mathbb{P}(B(s) \in X) = 1$, so we strike $X$ almost immediately, almost surely.  Yet $X$ is totally disconnected so it doesn't even contain a homeomorphic copy of $\mathbb{R}^1$.  So your sufficient condition is certainly not necessary.
Also, I wanted to comment on one technical point.  You say $X$ is an "arbitrary" subset, but for the question to make sense, you really need $X$ to be Borel.  Even then it is not trivial to see that your "event" $A = \{\exists s \le t : B(s) \in X\}$ is measurable, since the $\exists$ makes it an uncountable union.  But you can show it as follows: let our sample space $\Omega$ be the path space $C([0,\infty), \mathbb{R}^n)$ which is standard Borel.  Then the set $C = \{(\omega, t) : \omega(t) \in X\} \subset \Omega \times [0,\infty)$ is Borel as soon as $X$ is.  And $A$ is the projection of $C$ onto the first coordinate, so $A$ is analytic, and analytic sets are universally measurable.
You might also like to look up the debut theorem, which states that $\tau_X = \inf\{t : B(t) \in X\}$ is a stopping time for any Borel set $X$.  Again, because of measurability considerations, this is a nontrivial theorem.
