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For three-dimensional Brownian motion, every radius $r>0$ and every initial point satisfyIf the initial point is on one of the spheres, it is already a hit at time zero. If it is inside a lattice-centered open ball, its eventual exit hits that sphere. Otherwise choose at each integer time a nearest lattice center. The current displacement from it has norm at most $\sqrt{3}/2$, so the multivariate normal distribution of the next unit increment gives a uniform positive chance of entering its radius-$r/2$ open ball. The geometric tail bound from a uniform escape probability makes eventual entry certain, and continuity forces a boundary crossing. This periodic-family recurrence does not contradict the Brownian sphere-hitting probability in dimension three for one fixed sphere.

For $m\ge 1$, the increments have exponential momentThus $Z_n=2^{S_n}$ is an exponential martingale of a random walk relative to the natural filtration. We justify its stopping limit. From any surviving state in $(-m,m)$, a block of $2m$ successive increments all equal to $-1$ forces an exit. The block has probability $q=(6/7{)}^{2m}>0$, independently of the preceding increments. The geometric tail bound from a uniform escape probability givesso $T$ is finite almost surely and has finite expectation.

At the lower exit $S_T=-m$ exactly; at the upper exit $S_T$ is either $m$ or $m+1$. Before exit the same overall bound $-m\le S_{n\wedge T}\le m+1$ holds. Hence $Z_{n\wedge T}$ is bounded by $2^{m+1}$, uniformly in $n$. The bounded optional stopping theorem gives $\mathbb{E}{Z}_{n\wedge T}=Z_0=1$, and the dominated convergence theorem now yieldsAllowing the upper overshoot by one is essential; the terminal state need not equal $m$ on upper exit.

Let ${\tau }_r=\inf \{t\ge 0:|B_t|=r\}$, ${\tau }_R=\inf \{t\ge 0:|B_t|=R\}$, and $\tau ={\tau }_r\wedge {\tau }_R$. The function $u(y)=|y{|}^{-1}$ is harmonic away from the origin: its radial Laplacian in dimension three isThe Itô formula therefore makes $u(B_{t\wedge \tau })$ a local martingale, bounded between $1/R$ and $1/r$, hence a true martingale. The annulus exit time is finite almost surely. For instance, exit from the containing open ball is finite because from every point in the open ball a sufficiently large Gaussian vector increment has a uniform positive probability of leaving it in unit time; iterate the geometric tail bound from a uniform escape probability.

The optional stopping theorem at $t\wedge \tau$, followed by the dominated convergence theorem, gives $\mathbb{E}u(B_{\tau })=u(x)$. By continuity, there is no radial overshoot and the two exit spheres cannot be hit simultaneously. Writing $p={\mathbb{P}}_x({\tau }_r<{\tau }_R)$ givesAll harmonicity and boundedness statements apply only in the annulus, which stays away from the singularity at zero.