Harmonic measure

Remarks

The measure exists by the Riesz representation theorem. Indeed, by the maximum principle, $\varphi \mapsto u_\varphi(z)$ is a positive linear functional on the continuous functions on $\partial\Omega$, and it sends $1$ to $1$. For fixed $E$, the function $z \mapsto \omega(z, E, \Omega)$ is harmonic on $\Omega$ with values in $[0, 1]$. Example: for the unit disc $\mathbb{D}$, the Poisson kernel gives $d\omega_z(e^{i\theta}) = \frac{1 - \abs{z}^2}{2\pi \abs{e^{i\theta} - z}^2} \, d\theta$. At $z = 0$ this is $d\theta / 2\pi$, so the harmonic measure of an arc is its length divided by $2\pi$. Non-example: the punctured disc $\mathbb{D} \setminus \set{0}$ does not satisfy the hypothesis. The data $1$ at $0$ and $0$ on the unit circle have no harmonic solution that is continuous up to the boundary. With the more general Perron solution, the point $0$ gets harmonic measure $0$. If $f \colon \Omega \to \Omega'$ is a conformal map that extends to a homeomorphism of the closures, then $\omega(f(z), f(E), \Omega') = \omega(z, E, \Omega)$. Probabilistically, $\omega(z, E, \Omega)$ is the probability that a Brownian motion started at $z$ first leaves $\Omega$ through $E$.