In $d\geq 3$, we have that BM is transient a.s. i.e. $lim_{t\to \infty}|B_{t}|=\infty$.
But does this imply $P_{x}(T_{A}<\infty)=0$ for some type of Borel sets $A\subset \mathbb{R}^{d}$ with $vol_{d}(A)>0$.
For example, if $B_{t}$ starts at origin, it will hit shells centered at origin a.s.. Also, for balls centered at origin $P_{x}(T_{B_{r}(0)}<\infty)=\frac{r^{d}}{|x|^{d}}>0$ where $|x|>r$.
Question: So if it is positive for balls centered origin, shouldn't $P_{x}(T_{A}<\infty)$ be positive for other types of sets with positive volume containing the origin because I can just fit a ball inside them. And since there is nothing special about the origin, we can move the set around.
Thanks
For any $A$ of positive measure, and any $\epsilon > 0$, we have $P(T_A \leq \epsilon)>0$
Because $P(T_A \leq \epsilon) \geq P(B_{\epsilon} \in A) >0$ since $B_\epsilon$ has a strictly positive density in $R^d$.
In fact $\lim_{t\to \infty}|B_t| =\infty$ a.s. doesn't not imply $P(T_A < +\infty) = 0$ for some $A$ of positive measure. You can only say that almost surely we can find some $A(\omega)$ of positive measure such that the Brownian motion does not pass through $A(\omega)$, but $A$ depends on $\omega$