 Solution

Let $D$ be the unit disc and normalize its positive Poisson kernel at the boundary point $1$ byIt is harmonic in $D$. Let $p_D(t,z,w)$ be the killed Brownian transition density, for planar Brownian motion with generator $\frac{1}{2}\Delta$. The Doob h-transform has sub-Markov transition densityIts missing mass represents paths already absorbed at the distinguished boundary point. Equivalently, for a stopping time $\sigma$ before exit from a compact subdomain, its law is weighted relative to ordinary killed Brownian motion by $h(B_{\sigma })/h(B_0)$. These stopped laws are consistent and define a diffusion up to its lifetime. Its generator isThis is a rigorous Brownian motion conditioned to exit at a boundary point, not conditioning on an event of positive probability. Exhaustion of $D$ shows that its terminal boundary limit is $1$; its lifetime is finite. For example its expected lifetime from $0$ is ${\int }_D{G}_D^{\mathrm{BM}}(0,w)h(w)dw$, which is finite: the Green function vanishes linearly near the smooth boundary and cancels the Poisson-kernel singularity there, while its logarithmic singularity at $0$ is integrable.

One can also condition ordinary Brownian motion on exiting through an arc $I_{ϵ}$ about $1$, and then let its length decrease to zero. The conditional harmonic functions ${\omega }_D(z,I_{ϵ})/{\omega }_D(0,I_{ϵ})$ converge locally uniformly to $h(z)$. Thus their stopped conditional laws converge to this same diffusion. Both constructions specify the conditioning unambiguously.