This is motivated by Showing that Brownian motion will hit sphere with an exact integer distance.
The density of a set $A \subset \mathbb{R}^n$ is given by
$$d(A) = \limsup_{R \to \infty} \frac{\mu(A \cap B_R(0))}{\mu(B_R(0))}.$$
where $B_R(0)$ is the ball of radius $R$ centered at the origin.
Question: if $A$ is open and $d(A) > 0$, is it true that for the Brownian motion $B_t$ in $\mathbb{R}^n$, the hitting time $T_A$ to $A$ is finite almost surely?
Best Answer
It is true. By Fubini's Theorem, there exist spheres $S(0,R_j)$ with $R_j \to \infty$ such that the surface area of $A \cap S(0,R_j)$ exceeds $d(A)/2$ times the surface area of $S(0,R_j)$. Hence the probability that BM $B$ visits $A \setminus B(0,R)$ exceeds $d(A)/2$ for each $R$. The Blumenthal zero-one law (applied to the BM $\; t \mapsto tB(1/t)$) then implies that $P(T(A)<\infty)=1$.