Uniform ball sampling by Brownian stopping ($W_{{\tau }_R}\sim \mathrm{Unif}(B(0,r))$)

For standard Brownian motion started at zero, choose an independent radius $R$ with $\mathbb{P}(R\le s)=(s/r{)}^n$, $0\le s\le r$, and stop on first reaching that radius. Rotational invariance of Brownian motion makes the conditional exit position uniform on each sphere. Mixing with radius density $n{s}^{n-1}/r^n$ gives the uniform probability distribution on the ball. In dimension one, the conditional sign has probabilities $1/2,1/2$. The stopped point samples volume measure even though each conditional exit law is supported on a boundary.

Let $g_D(x)={\mathbb{E}}_x{T}_D$, possibly infinite, and suppose the closed ball $\overline{B(x,r)}$ lies in the open set $D$. A Strong Markov property decomposition at an independent random-radius exit time, using uniform ball sampling by Brownian stopping, gives the displayed identity in the nonnegative extended reals. Indeed the expected exit time from a radius-$s$ ball started at its centre is $s^2/n$, and averaging this against $n{s}^{n-1}/r^n$ gives $r^2/(n+2)$. The identity gives local integrability whenever $g_D(x)$ is finite.

For a connected open set $D$, finiteness of the Brownian exit time expectation at one point implies finiteness everywhere. The Brownian exit-time ball averaging identity gives integrability on a ball centred at any finite point. Every point of that ball has a smaller centred ball contained in it; the same identity then gives a finite value there. The set of finite points is open. It is also relatively closed: near a limit point in $D$, choose a fixed ball radius smaller than a quarter of its distance to $D^c$ and centre it at a nearby finite point. The limit point lies in that ball and hence is finite. Connectedness completes the proof.