Showing that Brownian motion will hit sphere with an exact integer distance

Question

Let $B=(B_t)_{t\geq0}$ be a three-dimensional Brownian motion starting from $x\in\mathbb{R}^3$. Show that, for all $r>0$ and all $x$, $$\mathbb{P}_x(|B_t+z|=r\text{ for some }t\geq0\text{ and some }z\in\mathbb{Z}^3)=1.$$ I really can't see how to get started with this - any advice would be greatly appreciated!

Answer 1

First, since the BM will eventually move arbitrarily far from the origin, we have $\limsup\lvert B_t + z \rvert = \infty$, so if $\lvert{B_0 + z \rvert} < r$ for any integral point $z$ then we are done by IVT.

Now suppose $\lvert{B_0 + z \rvert} \geq r$ for any integral point $z$. It suffices to prove that there exists an integer $n$ such that $\lvert B_n + z\rvert \leq r$. To show this, note that conditioned on $B_{n - 1} = x$, $B_n$ follows a normal distribution with mean $x$ and covariance matrix $I$. Under this distribution, the probability that $\lvert B_n + z\rvert \leq r$ for some integral $z$ is at least $\epsilon(r)$ for some constant $\epsilon(r) > 0$ (You can consider the sphere closest to $x$). Thus, for any $x$, $$\mathbb{P}(\forall z \in \mathbb{Z}^3, \lvert B_n + z\rvert \geq r \vert B_{n - 1} = x) \leq 1 - \epsilon(r).$$ Now taking the product with respect to $n$ gives the desired claim.