Harmonic measure as exit distribution

Remarks

Proof idea: by Kakutani's theorem, $E^x[\varphi(B_\tau)]$ is the solution of the Dirichlet problem with data $\varphi$, for each continuous $\varphi$. So the law of $B_\tau$ and $\omega_x$ give the same integral to each continuous function on the compact set $\partial D$, and the two Borel measures are equal. Example: for the ball $\abs{y} < 1$ in $\R^d$, the exit point from $x$ has density, with respect to normalized surface measure, equal to the Poisson kernel $(1 - \abs{x}^2)/\abs{x - \zeta}^d$. From the centre $x = 0$ this density is $1$, so the exit point is uniform on the sphere. Without regularity, the identity still holds when $\omega_x$ is the harmonic measure defined through Perron solutions. The set of irregular boundary points is then polar, and it gets no mass.

Going deeper

For a student

The probabilistic picture explains many properties of harmonic measure at once. It is a probability measure, it depends harmonically on the starting point, and it is monotone in the domain. In the plane, Brownian motion is conformally invariant up to a time change, which explains the conformal invariance of harmonic measure. Estimates such as Beurling's projection theorem have natural proofs and interpretations in this language.