Harmonic measure of an arc of the unit circle

Let $A_E:=\{t\in [0,2\pi ):e^{it}\in E\}$. Choose the integer $m$ for which ${\alpha }^{\mathrm{\prime }}:=\alpha -2\pi m\in [0,2\pi )$, and put ${\beta }^{\mathrm{\prime }}:=\beta -2\pi m$, so ${\alpha }^{\mathrm{\prime }}<{\beta }^{\mathrm{\prime }}\le {\alpha }^{\mathrm{\prime }}+2\pi$. If ${\beta }^{\mathrm{\prime }}\le 2\pi$, then $A_E$ agrees with $[{\alpha }^{\mathrm{\prime }},{\beta }^{\mathrm{\prime }}]\cap [0,2\pi )$ up to the possible duplicate endpoint at $0$. If ${\beta }^{\mathrm{\prime }}>2\pi$, it agrees with $[{\alpha }^{\mathrm{\prime }},2\pi )\cup [0,{\beta }^{\mathrm{\prime }}-2\pi ]$ up to endpoints. In the full-circle case ${\beta }^{\mathrm{\prime }}-{\alpha }^{\mathrm{\prime }}=2\pi$, one has $A_E=[0,2\pi )$. In each case, splitting the integral at $2\pi$ if necessary and translating one part by $2\pi$, the periodicity in [F2] gives ${\int }_{A_E}{k}_z(t)dt={\int }_{\alpha }^{\beta }{k}_z(t)dt$. Endpoints have zero angular measure.

The kernel $k_z(t):=(1-\mathrm{\mid }z{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }{e}^{it}-z{\mathrm{\mid }}^2$ is continuous and positive on $\mathbb{R}$ by [F2], and $1-\mathrm{\mid }z{\mathrm{\mid }}^2>0$ for $z\in \mathbb{D}$, so $k_z$ is a nonnegative measurable function on the interval $[\alpha ,\beta ]$ of finite length $\beta -\alpha \le 2\pi$.

Combining [F1] with step 1.1 expresses the harmonic measure of the arc as an integral of $k_z$ over $[\alpha ,\beta ]$: ${\omega }_{\mathbb{D}}^z(E)={\int }_{A_E}{k}_z(t)\frac{dt}{2\pi }=\frac{1}{2\pi }{\int }_{\alpha }^{\beta }{k}_z(t)dt\mathrm{.}$

The right-hand integral is an ordinary integral of a continuous function on a compact interval: by [F2] $k_z$ is Riemann integrable on $[\alpha ,\beta ]$ and its Riemann and Lebesgue integrals over that interval coincide, so the value in step 2.1 is well defined and equals the displayed Riemann integral.

At $z=0$ the kernel is identically one, because $\mathrm{\mid }{e}^{it}-0\mathrm{\mid }=1$ and $1-\mathrm{\mid }0{\mathrm{\mid }}^2=1$; the identity of step 2.1 therefore gives ${\omega }_{\mathbb{D}}^0(E)=\frac{1}{2\pi }{\int }_{\alpha }^{\beta }1dt=\frac{\beta -\alpha }{2\pi },$ a number in $[0,1]$, equal to $1$ exactly when the arc is the full circle $\beta =\alpha +2\pi$ and equal to the normalized angular length otherwise.

The calculation used only the explicitly given Poisson density and the elementary integration of a continuous periodic kernel; Dependent Choice was used only through [F1]. In particular no harmonicity of $z\mapsto {\omega }_{\mathbb{D}}^z(E)$, and no regularity of an indicator of $E$ as a boundary datum, was used or asserted here.