Harmonic measure of the two annulus boundary circles

The annulus $A$ is connected: for $w_k={\rho }_k{e}^{i{\theta }_k}\in A$, $k=1,2$, the path $w(t)=((1-t){\rho }_1+t{\rho }_2)\exp \left(i\right((1-t){\theta }_1+t{\theta }_2\left)\right)$, $t\in [0,1]$, has modulus between ${\rho }_1$ and ${\rho }_2$ and hence values in $A$, and it joins $w_1$ to $w_2$. Together with openness and boundedness from [F1] this makes $A$ a bounded complex domain.

Boundary points of the outer circle are regular. Fix $\xi \in C_R$ and set ${\zeta }_{\ast }:=2\xi$, so that $\mathrm{\mid }{\zeta }_{\ast }\mathrm{\mid }=2R$ and $\mathrm{\mid }{\zeta }_{\ast }-\xi \mathrm{\mid }=R$. Define $b(w):=\log R-\log \mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }$ for $w\in A$. Since ${\zeta }_{\ast }\mathrm{\notin }\bar{A}$, the map $w\mapsto w-{\zeta }_{\ast }$ is a holomorphic map with values in $\mathbb{C}\setminus \{0\}$ on the open set containing $\bar{A}$, so $w\mapsto \log \mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }$ is harmonic on $A$ by [F4] and $b$ is harmonic, hence subharmonic, on $A$ by [F4]. For $w\in A$ the reverse triangle inequality gives $\mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }\ge \mathrm{\mid }{\zeta }_{\ast }\mathrm{\mid }-\mathrm{\mid }w\mathrm{\mid }>2R-R=R$, so $b(w)<0$; and $b(\xi )=\log R-\log R=0$ with $b$ continuous at $\xi$, so $b(w)\to 0$ as $w\to \xi$. Finally $b\le 0$ on $\mathrm{\partial }A$ with equality only at $\xi$: on $C_R$ the identity $\mathrm{\mid }\eta -{\zeta }_{\ast }\mathrm{\mid }=R$ holds exactly when $\eta =\xi$, and on $C_r$ one has $\mathrm{\mid }\eta -{\zeta }_{\ast }\mathrm{\mid }\ge 2R-r>R$. Hence for every neighbourhood $V$ of $\xi$ the continuous function $b$ is strictly negative on the compact set $\mathrm{\partial }A\setminus V$ (if that set is empty there is nothing to bound), so its maximum there is some $c_V<0$; thus $b$ is a barrier at $\xi$, and $\xi$ is regular by [F3].

Boundary points of the inner circle are regular. Fix $\xi \in C_r$ and set ${\zeta }_{\ast }:=\xi \mathrm{/}2$, so that $\mathrm{\mid }{\zeta }_{\ast }\mathrm{\mid }=r\mathrm{/}2$ and $\mathrm{\mid }\xi -{\zeta }_{\ast }\mathrm{\mid }=r\mathrm{/}2$. Define $b(w):=\log (r\mathrm{/}2)-\log \mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }$ for $w\in A$. Here ${\zeta }_{\ast }\mathrm{\notin }\bar{A}$, so $b$ is harmonic on $A$ by [F4] and hence subharmonic. For $w\in A$ the reverse triangle inequality gives $\mathrm{\mid }w-{\zeta }_{\ast }\mathrm{\mid }\ge \mathrm{\mid }w\mathrm{\mid }-\mathrm{\mid }{\zeta }_{\ast }\mathrm{\mid }>r-r\mathrm{/}2=r\mathrm{/}2$, so $b(w)<0$; and $b(\xi )=0$ with $b$ continuous at $\xi$, so $b(w)\to 0$ as $w\to \xi$. On $\mathrm{\partial }A$ one has $b\le 0$ with equality only at $\xi$: the equality $\mathrm{\mid }\eta -{\zeta }_{\ast }\mathrm{\mid }=r\mathrm{/}2$ forces $\mathrm{\mid }\eta \mathrm{\mid }=r$ and $\eta$ on the ray through ${\zeta }_{\ast }$, that is $\eta =\xi$, while for $\mathrm{\mid }\eta \mathrm{\mid }=R$ one has $\mathrm{\mid }\eta -{\zeta }_{\ast }\mathrm{\mid }\ge R-r\mathrm{/}2>r\mathrm{/}2$. So for every neighbourhood $V$ of $\xi$ the maximum of $b$ over the compact set $\mathrm{\partial }A\setminus V$ is a $c_V<0$, and $b$ is a barrier at $\xi$; by [F3] the point $\xi$ is regular.

Define $s(w):=(\log R-\log \mathrm{\mid }w\mathrm{\mid })\mathrm{/}\log (R\mathrm{/}r)$ for $w\in \bar{A}$. On the annulus $A$ the function $s$ is a real linear combination of the harmonic function $\log \mathrm{\mid }\cdot \mathrm{\mid }$ restricted to $A$ and the constant $\log R$, divided by the nonzero constant $\log (R\mathrm{/}r)>0$, so $s$ is harmonic on $A$ by [F4]. On $\mathrm{\partial }A$ it takes the values $s(\eta )=(\log R-\log r)\mathrm{/}\log (R\mathrm{/}r)=1$ for $\eta \in C_r$ and $s(\eta )=0$ for $\eta \in C_R$; since $\mathrm{\mid }\cdot \mathrm{\mid }$ is continuous and positive on the compact annulus $\bar{A}$, the formula extends $s$ continuously to $\bar{A}$. Thus $s$ is a continuous harmonic extension to $\bar{A}$ of the continuous boundary function $\phi$ of [F5], so $H_{\phi }=s$ on $A$ by the uniqueness in [F2].

Steps 1.2 and 1.3 cover every point of $\mathrm{\partial }A=C_r\cup C_R$, so every boundary point of the bounded complex domain $A$ is regular. By [F2] there is Dependent Choice used here to obtain the harmonic measure ${\omega }_A^z$, the unique Radon probability measure on $\mathrm{\partial }A$ with $H_{\psi }(z)=\int \psi d{\omega }_A^z$ for every continuous $\psi$, and the Perron envelope of a continuous datum is its unique continuous harmonic extension.

Evaluating the representing identity of [F2] for $\phi$ at the point $z$, and using $\phi ={\mathbf{1}}_{C_r}$ on $\mathrm{\partial }A$ by [F5], gives ${\omega }_A^z(C_r)={\int }_{\mathrm{\partial }A}{\mathbf{1}}_{C_r}d{\omega }_A^z={\int }_{\mathrm{\partial }A}\phi d{\omega }_A^z=H_{\phi }(z)=s(z)=\frac{\log R-\log \mathrm{\mid }z\mathrm{\mid }}{\log R-\log r}\mathrm{.}$

The complementary indicator ${\mathbf{1}}_{C_R}=1-\phi$ is likewise continuous on $\mathrm{\partial }A$, and ${\omega }_A^z$ is a probability measure, so ${\omega }_A^z(C_R)={\int }_{\mathrm{\partial }A}(1-\phi )d{\omega }_A^z=1-{\omega }_A^z(C_r)=1-\frac{\log R-\log \mathrm{\mid }z\mathrm{\mid }}{\log R-\log r}=\frac{\log \mathrm{\mid }z\mathrm{\mid }-\log r}{\log R-\log r}\mathrm{.}$

Since $r<\mathrm{\mid }z\mathrm{\mid }<R$, both numerator quantities $\log R-\log \mathrm{\mid }z\mathrm{\mid }$ and $\log \mathrm{\mid }z\mathrm{\mid }-\log r$ are positive and their sum is $\log R-\log r>0$, so the two masses lie in $(0,1)$ and sum to $1$, as claimed. Dependent Choice was used only in step 2.1 through the existence and uniqueness theorem [F2]; the explicit barriers, the logarithmic function $s$ and all identities above are choice-free.