locally uniformly in $\Omega$, by Harnack's theorem; as the limit is necessarily a non-negative harmonic minorant of $g_{\Omega}(\cdot, w)$, it must be identically zero.

### A.3. Green Potentials.
A Green potential in $\Omega$ is a non-negative superharmonic function whose largest harmonic minorant is identically zero. Given $\sigma$, a Radon measure in $\Omega$, let us put

$$ (A.3) \qquad g(\sigma, \Omega; z) := \int g_{\Omega}(z, w) d\sigma(w). $$

This is a superharmonic function of $z \in \Omega$ which is either identically $+\infty$, or locally integrable with distributional Laplacian $-\sigma$ by Fubini's theorem. If $g(\sigma, \Omega; \cdot)$ is not identically $+\infty$, using the monotone convergence and Fubini's theorem, the proof of [29, Lemma 4.3.6] carries over to integrals instead of sums to show that the largest harmonic minorant of $g(\sigma, \Omega; \cdot)$ is the integral against $d\sigma(w)$ of the largest harmonic minors of the $g_{\Omega}(\cdot, w)$, namely zero. Thus, $g(\sigma, \Omega; \cdot)$ is a Green potential. Conversely, it follows from the Riesz representation theorem stated below that every Green potential has the form (A.3). Notice that if $\sigma(\Omega) < \infty$, then $g(\sigma, \Omega; \cdot) \not\equiv +\infty$, for if $V \subset \Omega$ is as in (A.2) and $W$ is a nonempty open set with $\bar{W} \subset V$, then

$$ \int_W g(\sigma, \Omega; z)^* \mathbf{1}(z) < C\sigma(V) + C'\sigma(\Omega \setminus V) \int_W^* \mathbf{1} $$

by Fubini's theorem, where $C'$ is an upper bound for $g_{\Omega}(z, w)$ on $(\Omega \setminus V) \times \bar{W}$.

**Riesz Representation Theorem.** Let $u \not\equiv +\infty$ be a superharmonic function on $\Omega$ that has a harmonic minorant. Then $u = g(\sigma, \Omega; \cdot) + h$, where $h$ is the largest harmonic minorant of $u$ and $\sigma := -\Delta u$.

*Proof.* Assume first that $\sigma(\Omega) < \infty$. Then, $g(\sigma, \Omega; \cdot) \not\equiv +\infty$ and therefore $h := u - g(\sigma, \Omega; \cdot)$ is harmonic on $\Omega$. Clearly, $h$ is a minorant of $u$. Since the largest harmonic minorant of $g(\sigma, \Omega; \cdot)$ is zero, $h$ is the largest harmonic minorant of $u$. If $\sigma(\Omega) = \infty$, pick $\Omega_m$ to be an increasing exhaustion of $\Omega$ by relatively compact open sets. Put $\sigma_m := \sigma_{\lfloor \Omega_m}$, which are finite measures on $\Omega_m$ because $\sigma$ is a Radon measure. By what precedes, $u_{\lfloor \Omega_m} = g(\sigma_m, \Omega_m; \cdot) + h_m$, where $h_m$ is the largest harmonic minorant of $u_{\lfloor \Omega_m}$. As the functions $g_{\Omega_m}(\cdot, w)$ increase locally uniformly to $g_{\Omega}(\cdot, w)$ while $h_m$ decrease and are bounded below by any harmonic minorant of $u$, we get by monotone convergence and Harnack's theorem that $u = g(\sigma, \Omega; \cdot) + h$, where $h$ is harmonic and necessarily $g(\sigma, \Omega; \cdot) \not\equiv +\infty$. We now conclude the proof as in the first case. $\square$

This version of the Riesz representation theorem featuring the weak-Laplacian may be compared to the more abstract formulation for Green spaces (of which $\Omega$ is a special case) in [8, Section VI.7], that does not refer to the Laplacian; see also the planar statement of [53, Theorem 4.5.4].

The previous considerations allow us to simplify in our case the notion of admissibility of a measure given in [19, Section 1] and [20, Section I.3]. According to that definition, a measure $\sigma$ is *admissible* if it is integrable against continuous Green potentials with compactly supported Laplacian. Fubini's theorem immediately implies that $g(\sigma, \Omega; \cdot) \not\equiv +\infty$ if $\sigma$ is admissible. In the present Greenian context the condition $g(\sigma, \Omega; \cdot) \not\equiv +\infty$ is also sufficient for (and therefore equivalent to) admissibility of $\sigma$. Indeed, if $v$ is compactly supported in $\Omega$ with continuous potential, let $V$ be an open set such that $\text{supp } v \subset V \subset \bar{V} \subset \Omega$. By continuity, there exists $C$ such that $g(v, \Omega; z) \le C, z \in \Omega$. Since $g(\sigma, \Omega; \cdot) \not\equiv +\infty$, the same is true for the potential of $\sigma_{\lfloor \Omega \setminus V}$, As this potential is harmonic in $V$ and is not equal identically to $+\infty$ there, it is finite and locally bounded in $V$. Hence, $g(\sigma_{\lfloor \Omega \setminus V}, \Omega; z) \le C', z \in \text{supp } v$. Therefore, it follows from Fubini's theorem that

$$ \int g(v, \Omega; z) d\sigma(z) \le C\sigma(V) + C'v(\Omega). $$
locally uniformly in $\Omega$, by Harnack's theorem; as the limit is necessarily a non-negative harmonic minorant of $g_{\Omega}(\cdot, w)$, it must be identically zero.

**A.3. Green Potentials.** A Green potential in $\Omega$ is a non-negative superharmonic function whose largest harmonic minorant is identically zero. Given $\sigma$, a Radon measure in $\Omega$, let us put

$$ (A.3) \qquad g(\sigma, \Omega; z) := \int g_{\Omega}(z, w) d\sigma(w). $$

This is a superharmonic function of $z \in \Omega$ which is either identically $+\infty$, or locally integrable with distributional Laplacian $-\sigma$ by Fubini's theorem. If $g(\sigma, \Omega; \cdot)$ is not identically $+\infty$, using the monotone convergence and Fubini's theorem, the proof of [29, Lemma 4.3.6] carries over to integrals instead of sums to show that the largest harmonic minorant of $g(\sigma, \Omega; \cdot)$ is the integral against $d\sigma(w)$ of the largest harmonic minors of the $g_{\Omega}(\cdot, w)$, namely zero. Thus, $g(\sigma, \Omega; \cdot)$ is a Green potential. Conversely, it follows from the Riesz representation theorem stated below that every Green potential has the form (A.3). Notice that if $\sigma(\Omega) < \infty$, then $g(\sigma, \Omega; \cdot) \not\equiv +\infty$, for if $V \subset \Omega$ is as in (A.2) and $W$ is a nonempty open set with $\bar{W} \subset V$, then

$$ \int_W g(\sigma, \Omega; z)^* \mathbf{1}(z) < C\sigma(V) + C'\sigma(\Omega \setminus V) \int_W^* \mathbf{1} $$

by Fubini's theorem, where $C'$ is an upper bound for $g_{\Omega}(z, w)$ on $(\Omega \setminus V) \times \bar{W}$.

**Riesz Representation Theorem.** Let $u \not\equiv +\infty$ be a superharmonic function on $\Omega$ that has a harmonic minorant. Then $u = g(\sigma, \Omega; \cdot) + h$, where $h$ is the largest harmonic minorant of $u$ and $\sigma := -\Delta u$.

*Proof.* Assume first that $\sigma(\Omega) < \infty$. Then, $g(\sigma, \Omega; \cdot) \not\equiv +\infty$ and therefore $h := u - g(\sigma, \Omega; \cdot)$ is harmonic on $\Omega$. Clearly, $h$ is a minorant of $u$. Since the largest harmonic minorant of $g(\sigma, \Omega; \cdot)$ is zero, $h$ is the largest harmonic minorant of $u$. If $\sigma(\Omega) = \infty$, pick $\Omega_m$ to be an increasing exhaustion of $\Omega$ by relatively compact open sets. Put $\sigma_m := \sigma_{\lfloor \Omega_m}$, which are finite measures on $\Omega_m$ because $\sigma$ is a Radon measure. By what precedes, $u_{\lfloor \Omega_m} = g(\sigma_m, \Omega_m; \cdot) + h_m$, where $h_m$ is the largest harmonic minorant of $u_{\lfloor \Omega_m}$. As the functions $g_{\Omega_m}(\cdot, w)$ increase locally uniformly to $g_{\Omega}(\cdot, w)$ while $h_m$ decrease and are bounded below by any harmonic minorant of $u$, we get by monotone convergence and Harnack's theorem that $u = g(\sigma, \Omega; \cdot) + h$, where $h$ is harmonic and necessarily $g(\sigma, \Omega; \cdot) \not\equiv +\infty$. We now conclude the proof as in the first case. $\square$

This version of the Riesz representation theorem featuring the weak-Laplacian may be compared to the more abstract formulation for Green spaces (of which $\Omega$ is a special case) in [8, Section VI.7], that does not refer to the Laplacian; see also the planar statement of [53, Theorem 4.5.4].

The previous considerations allow us to simplify in our case the notion of admissibility of a measure given in [19, Section 1] and [20, Section I.3]. According to that definition, a measure $\sigma$ is *admissible* if it is integrable against continuous Green potentials with compactly supported Laplacian. Fubini's theorem immediately implies that $g(\sigma, \Omega; \cdot) \not\equiv +\infty$ if $\sigma$ is admissible. In the present Greenian context the condition $g(\sigma, \Omega; \cdot) \not\equiv +\infty$ is also sufficient for (and therefore equivalent to) admissibility of $\sigma$. Indeed, if $v$ is compactly supported in $\Omega$ with continuous potential, let $V$ be an open set such that $\text{supp } v \subset V \subset \bar{V} \subset \Omega$. By continuity, there exists $C$ such that $g(v, \Omega; z) \le C, z \in \Omega$. Since $g(\sigma, \Omega; \cdot) \not\equiv +\infty$, the same is true for the potential of $\sigma_{\lfloor \Omega \setminus V}$, As this potential is harmonic in $V$ and is not equal identically to $+\infty$ there, it is finite and locally bounded in $V$. Hence, $g(\sigma_{\lfloor \Omega \setminus V}, \Omega; z) \le C', z \in \text{supp } v$. Therefore, it follows from Fubini's theorem that

$$ \int g(v, \Omega; z) d\sigma(z) \le C\sigma(V) + C'v(\Omega). $$
A.4. Capacities. Given two Radon measures $\sigma_1$ and $\sigma_2$ on $\Omega$, we put

$$ (\sigma_1, \sigma_2)_\Omega := \int g(\sigma_1, \Omega; z) d\sigma_2(z) = \int g(\sigma_2, \Omega; z) d\sigma_1(z), $$

which is either a non-negative number or $+\infty$. The Green energy of $\sigma$ is defined as $I_\Omega(\sigma) := (\sigma, \sigma)_\Omega$. The Greenian capacity relative to $\Omega$ of a compact set $K \subset \Omega$ is a non-negative number

$$ (A.4) \qquad \mathrm{cap}_{\Omega}(K) := \frac{1}{\inf_{\mu \in \mathcal{P}(K)} I_{\Omega}(\mu)}, $$

where $\mathcal{P}(K)$ is the set of Borel probability measures on $K$. The Greenian capacity of a Borel set $B$ is given by

$$ (A.5) \qquad \mathrm{cap}_{\Omega}(B) := \sup_{K \subset B} \mathrm{cap}_{\Omega}(K) = \inf_{U \supset B} \mathrm{cap}_{\Omega}(U), $$

where the supremum is taken over all compact subsets of $B$, the infimum is taken over all open sets containing $B$, and the equality is due to a theorem by Choquet [8, Section VIII.4]. When $K$ is compact and $\mathrm{cap}_\Omega(K) > 0$, there exists a unique $\mu_{\Omega,K} \in \mathcal{P}(K)$, called the Green equilibrium measure of $K$ in $\Omega$, to meet the infimum in (A.4). It is characterized by the fact that for some constant $C(= 1/\mathrm{cap}_\Omega(K))$, the Green equilibrium potential $g(\mu_{\Omega,K}, \Omega; z)$ satisfies $g(\mu_{\Omega,K}, \Omega; z) \le C$ for $z \in \Omega$ with $g(\mu_{\Omega,K}, \Omega; z) = C$ for $z \in K \setminus E$, where $E$ has Greenian capacity zero; this can be shown as in the Euclidean case [55, Theorems II.5.11 & II.5.12].

In the case of an arbitrary set $B$, the infimum in (A.5) introduces the outer Greenian capacity of $B$ and will serve as a definition of $\mathrm{cap}_\Omega(B)$. However, it may no longer match the supremum (the latter defines the inner Greenian capacity of $B$).

When $\Omega \subset \mathbb{C}$, another notion of capacity is instrumental in this paper, namely the logarithmic capacity defined for a compact set $K \subset \mathbb{C}$ as

$$ (A.6) \qquad \mathrm{cap}(K) := \exp \left\{ - \inf_{\mu \in \mathcal{P}(K)} \int V^\mu(z) d\mu(z) \right\}, $$

where $V^\mu(z)$ is the logarithmic potential of $\mu$ defined earlier in Section A.1. The logarithmic capacity of a Borel subset of $\mathbb{C}$ and the outer logarithmic capacity of an arbitrary subset are defined via the same process as for Greenian capacity, based on the analog of (A.5), see [53, 55] (note that in [53], $\mathrm{cap}(E)$ denotes the inner logarithmic capacity and potentials carry a sign opposite to the current one). If $K$ is compact and $\mathrm{cap}(K) > 0$, then there is a unique $\mu_K \in \mathcal{P}(K)$, called the logarithmic equilibrium measure of $K$, that realizes the infimum in (A.6). It is characterized by the fact that for some constant $C(= -\log \mathrm{cap}(K))$, the logarithmic equilibrium potential $V^{\mu_K}(z)$ is at most $C$ for $z \in K$ and in fact equal to $C$ on $K$ except possibly for a subset of logarithmic capacity zero.

Both the Greenian and logarithmic capacities are right continuous on compact sets, meaning that

$$ \mathrm{cap}(\cap_{j=1}^{\infty} K_j) = \lim_{n} \mathrm{cap}(\cap_{j=1}^{n} K_j) \quad \text{and} \quad \mathrm{cap}_{\Omega}(\cap_{j=1}^{\infty} K_j) = \lim_{n} \mathrm{cap}_{\Omega}(\cap_{j=1}^{n} K_j) $$

if the $K_j$ are compact; see [53, Theorem 5.1.3(a)] for the logarithmic case, the Greenian one being argued the same way with an obvious adaptation of [53, Lemma 3.3.3]. In addition, the (outer) Greenian and logarithmic capacities are left continuous:

$$ \mathrm{cap}(\cup_{j=1}^{\infty} E_j) = \lim_{n} \mathrm{cap}(\cup_{j=1}^{n} E_j) \quad \text{and} \quad \mathrm{cap}_{\Omega}(\cup_{j=1}^{\infty} E_j) = \lim_{n} \mathrm{cap}_{\Omega}(\cup_{j=1}^{n} E_j); $$

for the logarithmic capacity this follows from [53, Theorem 5.1.3(b)] combined with Choquet's theorem, and the Greenian case can be handled similarly, compare to [8, Section VIII.4].

One form of the domination principle for Green potentials says that if $g(\sigma, \Omega; \cdot) \le v$ on supp $\sigma$ (the support of $\sigma$) for some superharmonic function $v$, then $g(\sigma, \Omega; \cdot) \le v$ everywhere on $\Omega$; in fact, we shall state a stronger version in Section A.6. It implies the continuity theorem, saying that if the restriction of $g(\sigma, \Omega; \cdot)$ to supp $\sigma$ is continuous at $z_0 \in \text{supp } \sigma$ then $g(\sigma, \Omega; \cdot)$ is continuous at $z_0$. When $\sigma$ is a positive Borel measure with compact support such that $g(\sigma, \Omega; z) < +\infty$ for $\sigma$-a.e.
### A.4. Capacities.
Given two Radon measures $\sigma_1$ and $\sigma_2$ on $\Omega$, we put

$$ (\sigma_1, \sigma_2)_\Omega := \int g(\sigma_1, \Omega; z) d\sigma_2(z) = \int g(\sigma_2, \Omega; z) d\sigma_1(z), $$

which is either a non-negative number or $+\infty$. The Green energy of $\sigma$ is defined as $I_\Omega(\sigma) := (\sigma, \sigma)_\Omega$. The Greenian capacity relative to $\Omega$ of a compact set $K \subset \Omega$ is a non-negative number

$$ (A.4) \qquad \mathrm{cap}_{\Omega}(K) := \frac{1}{\inf_{\mu \in \mathcal{P}(K)} I_{\Omega}(\mu)}, $$

where $\mathcal{P}(K)$ is the set of Borel probability measures on $K$. The Greenian capacity of a Borel set $B$ is given by

$$ (A.5) \qquad \mathrm{cap}_{\Omega}(B) := \sup_{K \subset B} \mathrm{cap}_{\Omega}(K) = \inf_{U \supset B} \mathrm{cap}_{\Omega}(U), $$

where the supremum is taken over all compact subsets of $B$, the infimum is taken over all open sets containing $B$, and the equality is due to a theorem by Choquet [8, Section VIII.4]. When $K$ is compact and $\mathrm{cap}_\Omega(K) > 0$, there exists a unique $\mu_{\Omega,K} \in \mathcal{P}(K)$, called the Green equilibrium measure of $K$ in $\Omega$, to meet the infimum in (A.4). It is characterized by the fact that for some constant $C(= 1/\mathrm{cap}_\Omega(K))$, the Green equilibrium potential $g(\mu_{\Omega,K}, \Omega; z)$ satisfies $g(\mu_{\Omega,K}, \Omega; z) \le C$ for $z \in \Omega$ with $g(\mu_{\Omega,K}, \Omega; z) = C$ for $z \in K \setminus E$, where $E$ has Greenian capacity zero; this can be shown as in the Euclidean case [55, Theorems II.5.11 & II.5.12].

In the case of an arbitrary set $B$, the infimum in (A.5) introduces the outer Greenian capacity of $B$ and will serve as a definition of $\mathrm{cap}_\Omega(B)$. However, it may no longer match the supremum (the latter defines the inner Greenian capacity of $B$).

When $\Omega \subset \mathbb{C}$, another notion of capacity is instrumental in this paper, namely the logarithmic capacity defined for a compact set $K \subset \mathbb{C}$ as

$$ (A.6) \qquad \mathrm{cap}(K) := \exp \left\{ - \inf_{\mu \in \mathcal{P}(K)} \int V^\mu(z) d\mu(z) \right\}, $$

where $V^\mu(z)$ is the logarithmic potential of $\mu$ defined earlier in Section A.1. The logarithmic capacity of a Borel subset of $\mathbb{C}$ and the outer logarithmic capacity of an arbitrary subset are defined via the same process as for Greenian capacity, based on the analog of (A.5), see [53, 55] (note that in [53], $\mathrm{cap}(E)$ denotes the inner logarithmic capacity and potentials carry a sign opposite to the current one). If $K$ is compact and $\mathrm{cap}(K) > 0$, then there is a unique $\mu_K \in \mathcal{P}(K)$, called the logarithmic equilibrium measure of $K$, that realizes the infimum in (A.6). It is characterized by the fact that for some constant $C(= -\log \mathrm{cap}(K))$, the logarithmic equilibrium potential $V^{\mu_K}(z)$ is at most $C$ for $z \in K$ and in fact equal to $C$ on $K$ except possibly for a subset of logarithmic capacity zero.

Both the Greenian and logarithmic capacities are right continuous on compact sets, meaning that

$$ \mathrm{cap}(\cap_{j=1}^{\infty} K_j) = \lim_{n} \mathrm{cap}(\cap_{j=1}^{n} K_j) \quad \text{and} \quad \mathrm{cap}_{\Omega}(\cap_{j=1}^{\infty} K_j) = \lim_{n} \mathrm{cap}_{\Omega}(\cap_{j=1}^{n} K_j) $$

if the $K_j$ are compact; see [53, Theorem 5.1.3(a)] for the logarithmic case, the Greenian one being argued the same way with an obvious adaptation of [53, Lemma 3.3.3]. In addition, the (outer) Greenian and logarithmic capacities are left continuous:

$$ \mathrm{cap}(\cup_{j=1}^{\infty} E_j) = \lim_{n} \mathrm{cap}(\cup_{j=1}^{n} E_j) \quad \text{and} \quad \mathrm{cap}_{\Omega}(\cup_{j=1}^{\infty} E_j) = \lim_{n} \mathrm{cap}_{\Omega}(\cup_{j=1}^{n} E_j); $$

for the logarithmic capacity this follows from [53, Theorem 5.1.3(b)] combined with Choquet's theorem, and the Greenian case can be handled similarly, compare to [8, Section VIII.4].

One form of the domination principle for Green potentials says that if $g(\sigma, \Omega; \cdot) \le v$ on supp $\sigma$ (the support of $\sigma$) for some superharmonic function $v$, then $g(\sigma, \Omega; \cdot) \le v$ everywhere on $\Omega$; in fact, we shall state a stronger version in Section A.6. It implies the continuity theorem, saying that if the restriction of $g(\sigma, \Omega; \cdot)$ to supp $\sigma$ is continuous at $z_0 \in \text{supp } \sigma$ then $g(\sigma, \Omega; \cdot)$ is continuous at $z_0$. When $\sigma$ is a positive Borel measure with compact support such that $g(\sigma, \Omega; z) < +\infty$ for $\sigma$-a.e.
z, there is an increasing sequence of measures $\sigma_k$ supported on supp $\sigma$, having continuous Green potentials and converging to $\sigma$ in the strong (total variation) sense, such that $g(\sigma_k, \Omega; \cdot)$ increases pointwise to $g(\sigma, \Omega; \cdot)$ on $\Omega$. The proof is *mutatis mutandis* the same as for logarithmic potentials [55, Lemma I.6.10], using the continuity theorem for Green potentials. In particular, if $K$ is compact with $\text{cap}_{\Omega}(K) > 0$, we find upon letting $\sigma$ be the Green equilibrium distribution that there exist nonzero positive measures supported on $K$ whose Green potentials are continuous.

When $\Omega \subset \mathbb{C}$, a subset of $\Omega$ has (outer) Greenian capacity zero if and only if it has (outer) logarithmic capacity zero. Indeed, it is enough to verify this claim on compact sets since capacity is left continuous and a set of outer (Greenian or logarithmic) capacity zero is contained in a Borel (even $G_\delta$) set of capacity zero, by definition. Moreover, by the increasing character of $g_\Omega(z, w)$ with $\Omega$, we may assume that $\Omega$ is simply connected. Then the result follows by comparing the logarithmic kernel $\log(1/|z-w|)$ with the Green kernel $g_\Omega(z, w) = \log|(1-\varphi(z)\varphi(w))/(\varphi(z)-\varphi(w))|$, where $\varphi$ is a conformal map $\Omega \to \mathbb{D}$. A property holding pointwise except on a set of outer Greenian capacity zero (equivalently: logarithmic capacity zero if $\Omega \subset \mathbb{C}$) is said to hold *quasi everywhere*.

### A.5. Fine Topology.
A basis for the fine topology on $\Omega$ is given by all sets of the form

$$ (A.7) \qquad \cap_{i=1}^{m} \{z \in B : v_i(z) < \alpha_i\}, $$

where $B \subset \Omega$ is open, $v_i$ are superharmonic functions on $B$, and $\alpha_i$ are constants. Consequently, all superharmonic functions $\Omega \to (-\infty, +\infty]$ are finely continuous (equivalently: all subharmonic functions $\Omega \to [-\infty, +\infty)$ are finely continuous), and the fine topology is the coarsest with this property because, by the Riesz representation theorem and the monotonicity of Green functions with respect to the domain, each set of the form (A.7) contains one for which $v_i$ are Green potentials. In particular, we may as well require in (A.7) that $v_i$ be defined and superharmonic on the whole of $\Omega$. Hence, the present definition modeled after [29, Definition 6.5.1] (which deals with the Euclidean case) is equivalent to [8, Definition I.1]. It is known that the fine topology on $\Omega$ is locally connected [20, Corollary to Theorem 9.11], and that the fine connected components of a finely open set are finely open [19, Corollary 1]. Moreover, Lipschitz curves are finely connected [19, Theorem 7], so that Euclidean domains are fine domains as well.

As a general convention, we use the prefix “fine” to signify that a notion is understood with respect to the fine topology. This way we distinguish the latter from the classical, Euclidean topology (more precisely: the one induced on $\Omega$ by the Euclidean topology of charts). The fine boundary of a set $S$ is denoted by $\partial_S$, and its fine closure by $\text{clo}_S$.

A set $E \subset \Omega$ is called *polar* (in $\Omega$) if there is a superharmonic function $u \neq +\infty$ on $\Omega$ such that $u(z) = +\infty$ for $z \in E$. Superharmonic functions not identically $+\infty$ are locally integrable, therefore polar sets have area measure zero, see Section A.1. By definition a polar set is contained in a $G_\delta$ polar set, and every $G_\delta$ polar set arises as the $+\infty$-set of a superharmonic function [8, Section VI.9]. If $U$ is a fine domain and $E$ is polar, then $U \setminus E$ is again a fine domain [19, Theorem 6]. In fact, polar sets are exactly the sets of zero outer Greenian capacity (equivalently: of zero outer logarithmic capacity if $\Omega \subset \mathbb{C}$) defined in Section A.4; this is justified in Section A.9, but we take it presently for granted (we stress that [53] defines polar sets as having inner capacity zero, thereby making for a larger class of non-Borel polar sets). One consequence is: if $\Omega' \supset \Omega$ is hyperbolic and $E$ is polar in $\Omega$, then it is polar in $\Omega'$ as well; indeed, since $g_{\Omega'}(\cdot, w) \ge g_{\Omega}(\cdot, w)$, it is clear that $E$ has zero outer Greenian capacity in $\Omega'$ if it does in $\Omega$. Conversely, it is obvious from the definition that $E$ is polar in $\Omega$ if it is polar in $\Omega' \supset \Omega$, therefore we may speak of a polar set without specifying a hyperbolic subset of $\mathcal{R}_*$ in which $E$ is contained.

A countable union of polar sets is polar, for if $E_k$ is included in the $+\infty$-set of a superharmonic function $u_k \neq +\infty$ while $K \subset \Omega$ is compact and of positive Lebesgue measure, then there are $t_k > 0$ such that $u := \sum_k t_k u_k$ is summable on $K$ (therefore $u \neq +\infty$) and is superharmonic with value $+\infty$ at each point of $\cup_k E_k$. So, if $E \subset \Omega$ is polar and $p : \mathcal{R}_* \to \mathbb{C}$ is the natural projection, then
z, there is an increasing sequence of measures $\sigma_k$ supported on supp $\sigma$, having continuous Green potentials and converging to $\sigma$ in the strong (total variation) sense, such that $g(\sigma_k, \Omega; \cdot)$ increases pointwise to $g(\sigma, \Omega; \cdot)$ on $\Omega$. The proof is *mutatis mutandis* the same as for logarithmic potentials [55, Lemma I.6.10], using the continuity theorem for Green potentials. In particular, if $K$ is compact with $\text{cap}_\Omega(K) > 0$, we find upon letting $\sigma$ be the Green equilibrium distribution that there exist nonzero positive measures supported on $K$ whose Green potentials are continuous.

When $\Omega \subset \mathbb{C}$, a subset of $\Omega$ has (outer) Greenian capacity zero if and only if it has (outer) logarithmic capacity zero. Indeed, it is enough to verify this claim on compact sets since capacity is left continuous and a set of outer (Greenian or logarithmic) capacity zero is contained in a Borel (even $G_\delta$) set of capacity zero, by definition. Moreover, by the increasing character of $g_\Omega(z, w)$ with $\Omega$, we may assume that $\Omega$ is simply connected. Then the result follows by comparing the logarithmic kernel $\log(1/|z-w|)$ with the Green kernel $g_\Omega(z, w) = \log|(1-\varphi(z)\varphi(w))/(\varphi(z)-\varphi(w))|$, where $\varphi$ is a conformal map $\Omega \to \mathbb{D}$. A property holding pointwise except on a set of outer Greenian capacity zero (equivalently: logarithmic capacity zero if $\Omega \subset \mathbb{C}$) is said to hold *quasi everywhere*.

### A.5. Fine Topology.
A basis for the fine topology on $\Omega$ is given by all sets of the form

$$ (A.7) \qquad \cap_{i=1}^{m} \{z \in B : v_i(z) < \alpha_i\}, $$

where $B \subset \Omega$ is open, $v_i$ are superharmonic functions on $B$, and $\alpha_i$ are constants. Consequently, all superharmonic functions $\Omega \to (-\infty, +\infty]$ are finely continuous (equivalently: all subharmonic functions $\Omega \to [-\infty, +\infty)$ are finely continuous), and the fine topology is the coarsest with this property because, by the Riesz representation theorem and the monotonicity of Green functions with respect to the domain, each set of the form (A.7) contains one for which $v_i$ are Green potentials. In particular, we may as well require in (A.7) that $v_i$ be defined and superharmonic on the whole of $\Omega$. Hence, the present definition modeled after [29, Definition 6.5.1] (which deals with the Euclidean case) is equivalent to [8, Definition I.1]. It is known that the fine topology on $\Omega$ is locally connected [20, Corollary to Theorem 9.11], and that the fine connected components of a finely open set are finely open [19, Corollary 1]. Moreover, Lipschitz curves are finely connected [19, Theorem 7], so that Euclidean domains are fine domains as well.

As a general convention, we use the prefix “fine” to signify that a notion is understood with respect to the fine topology. This way we distinguish the latter from the classical, Euclidean topology (more precisely: the one induced on $\Omega$ by the Euclidean topology of charts). The fine boundary of a set $S$ is denoted by $\partial_S$, and its fine closure by $\text{clo}_S$.

A set $E \subset \Omega$ is called *polar* (in $\Omega$) if there is a superharmonic function $u \neq +\infty$ on $\Omega$ such that $u(z) = +\infty$ for $z \in E$. Superharmonic functions not identically $+\infty$ are locally integrable, therefore polar sets have area measure zero, see Section A.1. By definition a polar set is contained in a $G_\delta$ polar set, and every $G_\delta$ polar set arises as the $+\infty$-set of a superharmonic function [8, Section VI.9]. If $U$ is a fine domain and $E$ is polar, then $U \setminus E$ is again a fine domain [19, Theorem 6]. In fact, polar sets are exactly the sets of zero outer Greenian capacity (equivalently: of zero outer logarithmic capacity if $\Omega \subset \mathbb{C}$) defined in Section A.4; this is justified in Section A.9, but we take it presently for granted (we stress that [53] defines polar sets as having inner capacity zero, thereby making for a larger class of non-Borel polar sets). One consequence is: if $\Omega' \supset \Omega$ is hyperbolic and $E$ is polar in $\Omega$, then it is polar in $\Omega'$ as well; indeed, since $g_{\Omega'}(\cdot, w) \ge g_\Omega(\cdot, w)$, it is clear that $E$ has zero outer Greenian capacity in $\Omega'$ if it does in $\Omega$. Conversely, it is obvious from the definition that $E$ is polar in $\Omega$ if it is polar in $\Omega' \supset \Omega$, therefore we may speak of a polar set without specifying a hyperbolic subset of $\mathcal{R}_*$ in which $E$ is contained.

A countable union of polar sets is polar, for if $E_k$ is included in the $+\infty$-set of a superharmonic function $u_k \neq +\infty$ while $K \subset \Omega$ is compact and of positive Lebesgue measure, then there are $t_k > 0$ such that $u := \sum_k t_k u_k$ is summable on $K$ (therefore $u \neq +\infty$) and is superharmonic with value $+\infty$ at each point of $\cup_k E_k$. So, if $E \subset \Omega$ is polar and $p : \mathcal{R}_* \to \mathbb{C}$ is the natural projection, then
$p(E)$ is polar. Indeed, for $V \subset \Omega$ a domain such that $p: V \to p(V)$ is a homeomorphism, $v \circ p^{-1}$ is superharmonic on $p(V)$ when $v$ is superharmonic on $V$, and $\Omega \setminus rp(\mathcal{R}_*)$ can be covered with countably many such domains while $rp(\mathcal{R}_*)$ is finite. Conversely, if $V \subset \mathbb{C}$ is a bounded open set and $E \subset V$ is polar, then $p^{-1}(E)$ is polar because $v \circ p$ is superharmonic as soon as $v$ is superharmonic on $V$.

If $u \neq +\infty$ is superharmonic and finite on a polar set $E$, then $E$ has outer $\Delta u$-measure zero [8, Section VI.9, item $\beta$)]. Consequently a Radon measure $\sigma$ of finite Green energy cannot charge a polar set $E$, for we may assume $\sigma$ has compact support (since $\Omega$ is $\sigma$-compact) and as in Section A.4 there is an increasing sequence of measures $\sigma_k$ converging strongly to $\sigma$ with $g(\sigma_k, \Omega; \cdot)$ continuous, whence $\sigma(E) = \lim_k \sigma_k(E) = 0$.

**Removability Theorem.** If $E \subset \Omega$ is a (relatively) closed polar set while $u$ is superharmonic on $\Omega \setminus E$ and locally bounded below in a neighborhood of $E$, then $u$ extends in a unique manner to a superharmonic function on $\Omega$. Moreover, if $u$ is harmonic in $\Omega \setminus E$ and locally bounded in a neighborhood of $E$, then $u$ extends harmonically to $\Omega$.

*Proof.* The proof of the first statement carries over to hyperbolic Riemann surfaces from its planar version, see [53, Theorem 3.6.1]. When $u$ is harmonic in $\Omega \setminus E$ and locally bounded in a neighborhood of $E$, it extends both to a subharmonic and a superharmonic function on $\Omega$ by the first part. Since $\Delta u$ does not depend on the extension because $E$ has Lebesgue measure zero, we deduce that $\Delta u = 0$. $\square$

The removability theorem implies the following result.

**Generalized Minimum Principle.** If $u$ is superharmonic and bounded below on some open set $U \subset \bar{U} \subset \Omega$, and if moreover $\liminf_{U \ni z \to \xi} u(z) \ge 0$ for quasi every $\xi \in \partial U$, then $u \ge 0$ in $U$.

*Proof.* Given $\varepsilon > 0$, let $E_\varepsilon := \{\zeta \in \partial U : \liminf_{z \to \zeta} u(z) \le -\varepsilon\}$. Then $E_\varepsilon$ is a closed polar set and the function $w : \Omega \to (-\infty, +\infty]$ given by $\min(u, -\varepsilon)$ on $U$ and $-\varepsilon$ on $\Omega \setminus (U \cup E_\varepsilon)$ is superharmonic on $\Omega \setminus E_\varepsilon$, by the glueing theorem, see [53, Theorem 2.4.5] for a planar version of this local result. As $w$ is bounded below, it extends to a superharmonic function on $\Omega$. Because $U \cup E_\varepsilon \subset \bar{U} \subset \Omega$, it holds that $\liminf_{z \to \zeta} w(z) = -\varepsilon$ for any $\zeta \in \partial \Omega$. Thus, $w \ge -\varepsilon$ in $\Omega$ by the classical minimum principle contained in the very definition of superharmonic functions. $\square$

The hypothesis “$u$ is bounded below” can be relaxed somewhat: it is enough to assume that $u \ge -g$ where $g$ is a semi-bounded potential, meaning that it is the increasing pointwise limit of a sequence of locally bounded potentials, see [20, Theorem 9.1]. Note that $g(\sigma, \Omega; \cdot)$ is semi-bounded when it is finite $\sigma$-a.e., for we may assume $\sigma$ has compact support (as $\Omega$ is a countable union of compact sets) and then appeal to properties of the measures $\sigma_k$ in Section A.4. In fact, $g(\sigma, \Omega; \cdot)$ is semi-bounded if and only if it is finite $\sigma$-a.e. in $\Omega$, which is also if and only if $\sigma$ does not charge polar sets, see [20, Section I.2.6, Theorem].

**A.6. Thinness.** Fine topology can also be introduced via the notion of thinness. A set $E \subset \Omega$ is said to be *thin* at $\zeta \in \Omega$ if $\zeta$ is not a fine limit point of $E$. Equivalently, $E$ is thin at $\zeta$ if and only if either $\zeta \notin \bar{E}$ or there exists a function $v$, superharmonic in a neighborhood of $\zeta$, such that

$$ (A.8) \qquad \liminf_{E \ni z \to \zeta, z \to \zeta} v(z) > v(\zeta); $$

see [29, Theorem 6.6.3] for a proof of this equivalence in the Euclidean setting, which applies to hyperbolic Riemann surfaces as well and also shows that $v$ in (A.8) may be taken superharmonic on the whole of $\Omega$. Hence, the above definition of thinness (which is local) matches [8, Definition I.2] (whose local character is not immediate, see [8, Theorem VII.1]). Setting $\liminf$ over the empty set to $+\infty$ by convention, (A.8) may still be regarded as characterizing thinness at $\zeta \notin \bar{E}$, upon letting $v \equiv 0$. Note that when the limit inferior in (A.8) is taken over a full Euclidean neighborhood of $\zeta$, superharmonicity of $v$ implies that the inequality gets replaced by an equality. Clearly, $E$ is thin at $\zeta$
$p(E)$ is polar. Indeed, for $V \subset \Omega$ a domain such that $p: V \to p(V)$ is a homeomorphism, $v \circ p^{-1}$ is superharmonic on $p(V)$ when $v$ is superharmonic on $V$, and $\Omega \setminus rp(\mathcal{R}_*)$ can be covered with countably many such domains while $rp(\mathcal{R}_*)$ is finite. Conversely, if $V \subset \mathbb{C}$ is a bounded open set and $E \subset V$ is polar, then $p^{-1}(E)$ is polar because $v \circ p$ is superharmonic as soon as $v$ is superharmonic on $V$.

If $u \neq +\infty$ is superharmonic and finite on a polar set $E$, then $E$ has outer $\Delta u$-measure zero [8, Section VI.9, item $\beta$)]. Consequently a Radon measure $\sigma$ of finite Green energy cannot charge a polar set $E$, for we may assume $\sigma$ has compact support (since $\Omega$ is $\sigma$-compact) and as in Section A.4 there is an increasing sequence of measures $\sigma_k$ converging strongly to $\sigma$ with $g(\sigma_k, \Omega; \cdot)$ continuous, whence $\sigma(E) = \lim_k \sigma_k(E) = 0$.

**Removability Theorem.** If $E \subset \Omega$ is a (relatively) closed polar set while $u$ is superharmonic on $\Omega \setminus E$ and locally bounded below in a neighborhood of $E$, then $u$ extends in a unique manner to a superharmonic function on $\Omega$. Moreover, if $u$ is harmonic in $\Omega \setminus E$ and locally bounded in a neighborhood of $E$, then $u$ extends harmonically to $\Omega$.

*Proof.* The proof of the first statement carries over to hyperbolic Riemann surfaces from its planar version, see [53, Theorem 3.6.1]. When $u$ is harmonic in $\Omega \setminus E$ and locally bounded in a neighborhood of $E$, it extends both to a subharmonic and a superharmonic function on $\Omega$ by the first part. Since $\Delta u$ does not depend on the extension because $E$ has Lebesgue measure zero, we deduce that $\Delta u = 0$. $\square$

The removability theorem implies the following result.

**Generalized Minimum Principle.** If $u$ is superharmonic and bounded below on some open set $U \subset \bar{U} \subset \Omega$, and if moreover $\liminf_{U \ni z \to \xi} u(z) \ge 0$ for quasi every $\xi \in \partial U$, then $u \ge 0$ in $U$.

*Proof.* Given $\varepsilon > 0$, let $E_\varepsilon := \{\zeta \in \partial U : \liminf_{z \to \zeta} u(z) \le -\varepsilon\}$. Then $E_\varepsilon$ is a closed polar set and the function $w : \Omega \to (-\infty, +\infty]$ given by $\min(u, -\varepsilon)$ on $U$ and $-\varepsilon$ on $\Omega \setminus (U \cup E_\varepsilon)$ is superharmonic on $\Omega \setminus E_\varepsilon$, by the glueing theorem, see [53, Theorem 2.4.5] for a planar version of this local result. As $w$ is bounded below, it extends to a superharmonic function on $\Omega$. Because $U \cup E_\varepsilon \subset \bar{U} \subset \Omega$, it holds that $\liminf_{z \to \zeta} w(z) = -\varepsilon$ for any $\zeta \in \partial \Omega$. Thus, $w \ge -\varepsilon$ in $\Omega$ by the classical minimum principle contained in the very definition of superharmonic functions. $\square$

The hypothesis “$u$ is bounded below” can be relaxed somewhat: it is enough to assume that $u \ge -g$ where $g$ is a semi-bounded potential, meaning that it is the increasing pointwise limit of a sequence of locally bounded potentials, see [20, Theorem 9.1]. Note that $g(\sigma, \Omega; \cdot)$ is semi-bounded when it is finite $\sigma$-a.e., for we may assume $\sigma$ has compact support (as $\Omega$ is a countable union of compact sets) and then appeal to properties of the measures $\sigma_k$ in Section A.4. In fact, $g(\sigma, \Omega; \cdot)$ is semi-bounded if and only if it is finite $\sigma$-a.e. in $\Omega$, which is also if and only if $\sigma$ does not charge polar sets, see [20, Section I.2.6, Theorem].

**A.6. Thinness.** Fine topology can also be introduced via the notion of thinness. A set $E \subset \Omega$ is said to be *thin* at $\zeta \in \Omega$ if $\zeta$ is not a fine limit point of $E$. Equivalently, $E$ is thin at $\zeta$ if and only if either $\zeta \notin \bar{E}$ or there exists a function $v$, superharmonic in a neighborhood of $\zeta$, such that

$$ (A.8) \qquad \liminf_{E \ni z \to \zeta, z \to \zeta} v(z) > v(\zeta); $$

see [29, Theorem 6.6.3] for a proof of this equivalence in the Euclidean setting, which applies to hyperbolic Riemann surfaces as well and also shows that $v$ in (A.8) may be taken superharmonic on the whole of $\Omega$. Hence, the above definition of thinness (which is local) matches [8, Definition I.2] (whose local character is not immediate, see [8, Theorem VII.1]). Setting $\liminf$ over the empty set to $+\infty$ by convention, (A.8) may still be regarded as characterizing thinness at $\zeta \notin \bar{E}$, upon letting $v \equiv 0$. Note that when the limit inferior in (A.8) is taken over a full Euclidean neighborhood of $\zeta$, superharmonicity of $v$ implies that the inequality gets replaced by an equality. Clearly, $E$ is thin at $\zeta$
if and only if for some (hence any) chart $(V, \varphi)$ with $\zeta \in V$, the planar set $\varphi(V \cap E)$ is thin at $\varphi(\zeta)$, and a countable union of thin sets at $\zeta$ is again thin at $\zeta$.

A set $V$ is a fine neighborhood of $\zeta \in V$ if and only if the complement of $V$ is thin at $\zeta$, see [8, Theorem I.3]. In particular, if $V$ is finely open and $Z$ is polar, then $V\setminus Z$ is finely open. The points of $E \subset \Omega$ at which $E$ is thin form a polar set, and $E$ is thin at each of its points if and only if it is polar [8, Theorem VII.7 & Corollary]. One consequence of $E$ being thin at $\zeta$ is that, locally in a chart $(V, \varphi)$ with $\zeta \in V$, there are arbitrary small circles centered at $\varphi(\zeta)$ which do not meet $\varphi(E \cap V)$, see [29, Theorem 6.7.9]; in the same vein, $\varphi(V\setminus E)$ contains a segment of the form $[\varphi(\zeta), \varphi(\zeta) + re^{i\theta}]$ with $r = r(\theta) > 0$ for quasi-every direction $e^{i\theta}$ in $\mathbb{T}$, see [53, Theorem 5.4.3]. In particular, a connected set cannot be thin at an accumulation point and therefore polar sets are totally disconnected.

The base $b(E)$ of a set $E$ is the set of points in $\Omega$ at which $E$ is non-thin, and $E$ is called a base if $b(E) = E$. It is known that $b(E)$ is a finely closed $G_\delta$ set, see [8, Proposition VII.8]. We record the following, elementary fact.

**Lemma A.1.** Let $E, F \subset \Omega$ be disjoint finely open sets such that $\Omega\setminus(E \cup F)$ is a base. Then $\Omega\setminus E$ and $\Omega\setminus F$ are bases as well.

*Proof.* If $x \in (\Omega\setminus E) \cap (\Omega\setminus F) = \Omega\setminus(E \cup F)$, the latter set is non-thin at $x$ by assumption and therefore so is $\Omega\setminus E$. If now $x \in (\Omega\setminus E) \cap F = F$, then the fine openness of $F$ implies that $\Omega\setminus F$ is thin at $x$ and so is $E \subset \Omega\setminus F$. However, if $\Omega\setminus E$ were also thin at $x$ in this case, then $\Omega = E \cup (\Omega\setminus E)$ would be thin at $x$ which is impossible. Hence, $\Omega\setminus E$ is non-thin at any of its points, therefore it is a base. $\square$

The notion of a base generates a strong form of the domination principle, see [8, Theorem VIII.4].

**Strong Domination Principle.** Let $v$ be a non-negative superharmonic function in $\Omega$ such that $v \ge g(\sigma, \Omega; \cdot)$ quasi everywhere on a set $E$ such that $\sigma(\Omega\setminus b(E)) = 0$. Then $v \ge g(\sigma, \Omega; \cdot)$ everywhere in $\Omega$.

The fine closure of $E$ is equal to $\text{cl}_f(E) = b(E) \cup i(E)$, where $i(E)$ is the set of finely isolated points of $E$, see [8, Proposition V.10]. The finely closed sets are precisely those for which $b(E) \subseteq E$. Note that $b(E) \cap i(E) = \emptyset$ and therefore, if $V$ is finely open, we have that

$$ (A.9) \qquad b(\Omega \setminus V) = \Omega \setminus V', \quad V' := V \cup i(\Omega \setminus V), $$

where we observe that $V'$ is in turn finely open. For any set $E$, the fine boundary $\partial_f E$ is finely closed and, as shown in [20, Lemma 12.3], it holds that

$$ (A.10) \qquad i(\partial_f E) = i(E) \cup i(\Omega \setminus E) \quad \text{and} \quad b(\partial_f E) = b(E) \cap b(\Omega \setminus E). $$

The next lemma connects fine topologies in $D$ and $\mathcal{R}$ (defined at the beginning of Section 2.1).

**Lemma A.2.** The map $p : \mathcal{R} \to D$ is finely open and finely continuous, that is, $p(V)$ and $p^{-1}(U)$ are finely open when $V \subset \mathcal{R}$ and $U \subset D$ are finely open. Moreover, $i(\mathcal{R}\setminus p^{-1}(U)) = p^{-1}(i(D\setminus U))$.

*Proof.* Let $V \subset \mathcal{R}$ be finely open and $\zeta \in V$. Denote by $\tilde{O}$ a Euclidean disk centered at $p(\zeta)$ of small enough radius so that $O$, the connected component of $p^{-1}(\tilde{O})$ containing $\zeta$, contains no ramification points of $\mathcal{R}$ except possibly $\zeta$ itself (if $m(\zeta) > 1$) and $O \cap p^{-1}(p(\zeta)) = \{\zeta\}$. Since $E := O\setminus V$ is thin at $\zeta$, there exists a superharmonic function $v$ in $O$ for which (A.8) takes place. We claim that

$$ (A.11) \qquad \tilde{v}(z) := \begin{cases} \sum_{w \in O \cap p^{-1}(z)} v(w), & z \in \tilde{O} \setminus \{p(\zeta)\}, \\ m(\zeta)v(\zeta), & z = p(\zeta), \end{cases} $$

is superharmonic on $\tilde{O}$. Indeed, by shrinking $\tilde{O}$ if needed, we may assume that $v$ is the increasing limit of a sequence of continuous superharmonic functions $v_n$ on $O$, see [53, Corollary 2.7.3] for a proof of this fact in the planar case that carries over to $\mathcal{R}$ using local charts. Define $\tilde{v}_n$ similarly to (A.11), only replacing $v$ with $v_n$. Clearly, $\tilde{v}_n$ is superharmonic on $\{z \in \tilde{O} : z \ne p(\zeta)\}$. Since
if and only if for some (hence any) chart $(V, \varphi)$ with $\zeta \in V$, the planar set $\varphi(V \cap E)$ is thin at $\varphi(\zeta)$, and a countable union of thin sets at $\zeta$ is again thin at $\zeta$.

A set $V$ is a fine neighborhood of $\zeta \in V$ if and only if the complement of $V$ is thin at $\zeta$, see [8, Theorem I.3]. In particular, if $V$ is finely open and $Z$ is polar, then $V\setminus Z$ is finely open. The points of $E \subset \Omega$ at which $E$ is thin form a polar set, and $E$ is thin at each of its points if and only if it is polar [8, Theorem VII.7 & Corollary]. One consequence of $E$ being thin at $\zeta$ is that, locally in a chart $(V, \varphi)$ with $\zeta \in V$, there are arbitrary small circles centered at $\varphi(\zeta)$ which do not meet $\varphi(E \cap V)$, see [29, Theorem 6.7.9]; in the same vein, $\varphi(V\setminus E)$ contains a segment of the form $[\varphi(\zeta), \varphi(\zeta) + re^{i\theta}]$ with $r = r(\theta) > 0$ for quasi-every direction $e^{i\theta}$ in $\mathbb{T}$, see [53, Theorem 5.4.3]. In particular, a connected set cannot be thin at an accumulation point and therefore polar sets are totally disconnected.

The base $b(E)$ of a set $E$ is the set of points in $\Omega$ at which $E$ is non-thin, and $E$ is called a base if $b(E) = E$. It is known that $b(E)$ is a finely closed $G_\delta$ set, see [8, Proposition VII.8]. We record the following, elementary fact.

**Lemma A.1.** Let $E, F \subset \Omega$ be disjoint finely open sets such that $\Omega\setminus(E \cup F)$ is a base. Then $\Omega\setminus E$ and $\Omega\setminus F$ are bases as well.

*Proof.* If $x \in (\Omega\setminus E) \cap (\Omega\setminus F) = \Omega\setminus(E \cup F)$, the latter set is non-thin at $x$ by assumption and therefore so is $\Omega\setminus E$. If now $x \in (\Omega\setminus E) \cap F = F$, then the fine openness of $F$ implies that $\Omega\setminus F$ is thin at $x$ and so is $E \subset \Omega\setminus F$. However, if $\Omega\setminus E$ were also thin at $x$ in this case, then $\Omega = E \cup (\Omega\setminus E)$ would be thin at $x$ which is impossible. Hence, $\Omega\setminus E$ is non-thin at any of its points, therefore it is a base. $\square$

The notion of a base generates a strong form of the domination principle, see [8, Theorem VIII.4].

**Strong Domination Principle.** Let $v$ be a non-negative superharmonic function in $\Omega$ such that $v \ge g(\sigma, \Omega; \cdot)$ quasi everywhere on a set $E$ such that $\sigma(\Omega\setminus b(E)) = 0$. Then $v \ge g(\sigma, \Omega; \cdot)$ everywhere in $\Omega$.

The fine closure of $E$ is equal to $\text{cl}_f(E) = b(E) \cup i(E)$, where $i(E)$ is the set of finely isolated points of $E$, see [8, Proposition V.10]. The finely closed sets are precisely those for which $b(E) \subseteq E$. Note that $b(E) \cap i(E) = \emptyset$ and therefore, if $V$ is finely open, we have that

$$ (A.9) \qquad b(\Omega\setminus V) = \Omega\setminus V', \quad V' := V \cup i(\Omega\setminus V), $$

where we observe that $V'$ is in turn finely open. For any set $E$, the fine boundary $\partial_f E$ is finely closed and, as shown in [20, Lemma 12.3], it holds that

$$ (A.10) \qquad i(\partial_f E) = i(E) \cup i(\Omega\setminus E) \quad \text{and} \quad b(\partial_f E) = b(E) \cap b(\Omega\setminus E). $$

The next lemma connects fine topologies in $D$ and $\mathcal{R}$ (defined at the beginning of Section 2.1).

**Lemma A.2.** The map $p : \mathcal{R} \to D$ is finely open and finely continuous, that is, $p(V)$ and $p^{-1}(U)$ are finely open when $V \subset \mathcal{R}$ and $U \subset D$ are finely open. Moreover, $i(\mathcal{R}\setminus p^{-1}(U)) = p^{-1}(i(D\setminus U))$.

*Proof.* Let $V \subset \mathcal{R}$ be finely open and $\zeta \in V$. Denote by $\tilde{O}$ a Euclidean disk centered at $p(\zeta)$ of small enough radius so that $O$, the connected component of $p^{-1}(\tilde{O})$ containing $\zeta$, contains no ramification points of $\mathcal{R}$ except possibly $\zeta$ itself (if $m(\zeta) > 1$) and $O \cap p^{-1}(p(\zeta)) = \{\zeta\}$. Since $E := O\setminus V$ is thin at $\zeta$, there exists a superharmonic function $v$ in $O$ for which (A.8) takes place. We claim that

$$ (A.11) \qquad \tilde{v}(z) := \begin{cases} \sum_{w \in O \cap p^{-1}(z)} v(w), & z \in \tilde{O} \setminus \{p(\zeta)\}, \\ m(\zeta)v(\zeta), & z = p(\zeta), \end{cases} $$

is superharmonic on $\tilde{O}$. Indeed, by shrinking $\tilde{O}$ if needed, we may assume that $v$ is the increasing limit of a sequence of continuous superharmonic functions $v_n$ on $O$, see [53, Corollary 2.7.3] for a proof of this fact in the planar case that carries over to $\mathcal{R}$ using local charts. Define $\tilde{v}_n$ similarly to (A.11), only replacing $v$ with $v_n$. Clearly, $\tilde{v}_n$ is superharmonic on $\{z \in \tilde{O} : z \neq p(\zeta)\}$. Since
it is bounded around $p(\zeta)$ by the continuity of $v_n$, the restriction $\tilde{v}_{n|\tilde{O}\setminus p(\zeta)}$ uniquely extends to a
superharmonic function on $\tilde{O}$ by the *Removability Theorem*. Of necessity, the value at $p(\zeta)$ of this
extension is given by

$$
\liminf_{z \to p(\zeta), z \neq p(\zeta)} \tilde{v}_n(z) = m(\zeta)v_n(\zeta) = \tilde{v}_n(p(\zeta)),
$$

where the first equality comes from the continuity of $v_n$ at $\zeta$. Hence, $\tilde{v}_n(z)$ is superharmonic on $\tilde{O}$ and so is its increasing limit $\tilde{v}$. *This proves the claim.* In another connection, the lower semicontinuity of $v$ shows that the analog of (A.8) holds for $\tilde{v}$ when the limit inferior is taken along $p(E)$. As $\tilde{O}\setminus p(V) \subseteq p(E)$, we get that $D\setminus p(V)$ is thin at $p(\zeta)$ so that $p(V)$ is finely open, as claimed.

To show the second claim, observe that the lift of a function from $D$ to $\mathcal{R}$ preserves superhar-
monicity and that $\mathcal{R}\backslash p^{-1}(U) = p^{-1}(D\backslash U)$. The identity $i(\mathcal{R}\backslash p^{-1}(U)) = p^{-1}(i(D\backslash U))$ is now
straightforward. $\quad \square$

In Lemma A.3 below, we single out for easy reference a basic geometric fact, used at several places
in the paper. We say that a continuous injective map $\gamma : \mathbb{T} \to \mathcal{R}$ is a *parametrized Jordan curve* in
$\mathcal{R}$ and we simply call the image $\gamma(\mathbb{T})$ a (non-parametrized) Jordan curve. On a hyperbolic Riemann
surface $\mathcal{R}$, any Jordan curve $\Gamma$ homotopic to a point is uniquely the boundary of a (topological) disk
$O \subset \mathcal{R}$ [64, Theorem 2.4]; we say that $O$ is the *interior* of $\Gamma$, and we write $O = \text{int} \Gamma$. That $\Gamma$ is
homotopic to a point in particular holds if it is included in a simply connected open set $U$ which is
the domain of a chart. Recall also that $E \subset \mathcal{R}$ is called *schlicht* over $U \subset \mathbb{C}$ if $U \supset p(E)$ and the
restriction $p|_E : E \to U$ is injective.

**Lemma A.3.** Let $E \subset \mathcal{R}$ be schlicht over $D$ and $\xi \in \mathbf{rp}(\mathcal{R})$. There exists a neighborhood $U_\xi$ of $\xi$ such that no Jordan curve in $U_\xi$ contains $\xi$ in its interior and simultaneously is contained in $E$.

Proof. Let $U$ be a simply connected domain of a chart containing $\xi$, so that every Jordan curve $\mathbb{T} \to U$
is homotopic to a point. Let $O \subset D$ be a disk centered at $p(\xi)$ and $O_\xi$ the connected component of
$p^{-1}(O)$ containing $\xi$, with $O$ small enough that $O_\xi \subset U$ and $O_\xi \cap \mathbf{rp}(\mathcal{R}) = \{\xi\}$; this is possible
since $\mathcal{R}_*$ is compact. Then $O_\xi$ is a neighborhood of $\xi$ which is (isomorphic via a biholomorphic
map fixing $\xi$ to) an $m(\xi)$-sheeted cyclic covering of $O$. So, if we identify the homotopy groups
$\pi_1(O_\xi \setminus \{\xi\})$ and $\pi_1(O \setminus \{p(\xi)\})$ with the infinite cyclic group generated by the symbol $a$, the induced
morphism $p_* : \pi_1(O_\xi \setminus \{\xi\}) \to \pi_1(O \setminus \{p(\xi)\})$ is the map $a \mapsto a^{m(\xi)}$. Now, a parametrized Jordan
curve $\gamma : \mathbb{T} \to O_\xi$ containing $\xi$ in its interior is a generator of the fundamental group of $O_\xi \setminus \{\xi\}$.
Hence, $p \circ \gamma$ is the $m(\xi)$-power of a generator of the fundamental group of $O \setminus \{p(\xi)\}$, in particular it
has winding number $\pm m(\xi)$ with respect to $p(\xi)$. However, if $\gamma$ is valued in $E$, then $p \circ \gamma : \mathbb{T} \to O$
is a parametrized Jordan curve because $p$ is injective on $E$, and therefore it has winding number
$\pm 1$ with respect to $p(\xi)$. This contradicts the assumption that $m(\xi) > 1$, showing that $U_\xi := O_\xi$
satisfies our requirements.
$\quad \square$

The next lemma, used in the proof of Lemma 3.11, depends on Lemma A.3.

**Lemma A.4.** If $E \subset \mathcal{R}$ is schlicht over $D$ and $\xi \in \mathbf{rp}(\mathcal{R})$, then $\mathcal{R}\setminus E$ is non-thin at $\xi$.

*Proof.* Let $(U_\xi, \varphi)$ be a chart around $\xi$, with $U_\xi$ as in Lemma A.3 and $\varphi(\xi) = 0$. If $\mathcal{R}\setminus E$ is thin at $\xi$, then there is a circle $\mathbb{T}_r := \{|z|=r\} \subset \varphi(U_\xi)$ such that $\varphi^{-1}(\mathbb{T}_r) \subset E$. As $\varphi^{-1} : \mathbb{T}_r \to U_\xi$ is a Jordan curve in $E$ that contains $\xi$ in its interior, this contradicts Lemma A.3. $\quad \square$

A.7. **Regularity.** Thinness is intimately connected to the notion of a regular boundary point with respect to the Dirichlet problem. Given a Euclidean open set $U$ with $\bar{U} \subset \Omega$, a point $\zeta \in \partial U$ is called *regular* if for any continuous function $\psi$ on $\partial U$ it holds that

$$
\lim_{U \ni z \to \zeta} H_{\psi}(z) = \psi(\zeta),
$$
it is bounded around $p(\zeta)$ by the continuity of $v_n$, the restriction $\tilde{v}_{n|\tilde{O}\setminus p(\zeta)}$ uniquely extends to a superharmonic function on $\tilde{O}$ by the *Removability Theorem*. Of necessity, the value at $p(\zeta)$ of this extension is given by

$$ \liminf_{z \to p(\zeta), z \neq p(\zeta)} \tilde{v}_n(z) = m(\zeta)v_n(\zeta) = \tilde{v}_n(p(\zeta)), $$

where the first equality comes from the continuity of $v_n$ at $\zeta$. Hence, $\tilde{v}_n(z)$ is superharmonic on $\tilde{O}$ and so is its increasing limit $\tilde{v}$. *This proves the claim.* In another connection, the lower semicontinuity of $v$ shows that the analog of (A.8) holds for $\tilde{v}$ when the limit inferior is taken along $p(E)$. As $\tilde{O}\setminus p(V) \subseteq p(E)$, we get that $D\setminus p(V)$ is thin at $p(\zeta)$ so that $p(V)$ is finely open, as claimed.

To show the second claim, observe that the lift of a function from $D$ to $\mathcal{R}$ preserves superharmonicity and that $\mathcal{R}\setminus p^{-1}(U) = p^{-1}(D\setminus U)$. The identity $i(\mathcal{R}\setminus p^{-1}(U)) = p^{-1}(i(D\setminus U))$ is now straightforward. $\square$

In Lemma A.3 below, we single out for easy reference a basic geometric fact, used at several places in the paper. We say that a continuous injective map $\gamma: \mathbb{T} \to \mathcal{R}$ is a *parametrized Jordan curve* in $\mathcal{R}$ and we simply call the image $\gamma(\mathbb{T})$ a (non-parametrized) Jordan curve. On a hyperbolic Riemann surface $\mathcal{R}$, any Jordan curve $\Gamma$ homotopic to a point is uniquely the boundary of a (topological) disk $O \subset \mathcal{R}$ [64, Theorem 2.4]; we say that $O$ is the *interior* of $\Gamma$, and we write $O = \text{int} \Gamma$. That $\Gamma$ is homotopic to a point in particular holds if it is included in a simply connected open set $U$ which is the domain of a chart. Recall also that $E \subset \mathcal{R}$ is called *schlicht* over $U \subset \mathbb{C}$ if $U \supset p(E)$ and the restriction $p|_E: E \to U$ is injective.

**Lemma A.3.** Let $E \subset \mathcal{R}$ be schlicht over $D$ and $\xi \in \text{rp}(\mathcal{R})$. There exists a neighborhood $U_\xi$ of $\xi$ such that no Jordan curve in $U_\xi$ contains $\xi$ in its interior and simultaneously is contained in $E$.

*Proof.* Let $U$ be a simply connected domain of a chart containing $\xi$, so that every Jordan curve $\mathbb{T} \to U$ is homotopic to a point. Let $O \subset D$ be a disk centered at $p(\xi)$ and $O_\xi$ the connected component of $p^{-1}(O)$ containing $\xi$, with $O$ small enough that $O_\xi \subset U$ and $O_\xi \cap \text{rp}(\mathcal{R}) = \{\xi\}$; this is possible since $\mathcal{R}_*$ is compact. Then $O_\xi$ is a neighborhood of $\xi$ which is (isomorphic via a biholomorphic map fixing $\xi$ to) an $m(\xi)$-sheeted cyclic covering of $O$. So, if we identify the homotopy groups $\pi_1(O_\xi\setminus\{\xi\})$ and $\pi_1(O\setminus\{p(\xi)\})$ with the infinite cyclic group generated by the symbol $a$, the induced morphism $p_*: \pi_1(O_\xi\setminus\{\xi\}) \to \pi_1(O\setminus\{p(\xi)\})$ is the map $a \mapsto a^{m(\xi)}$. Now, a parametrized Jordan curve $\gamma: \mathbb{T} \to O_\xi$ containing $\xi$ in its interior is a generator of the fundamental group of $O_\xi\setminus\{\xi\}$. Hence, $p \circ \gamma$ is the $m(\xi)$-power of a generator of the fundamental group of $O\setminus\{p(\xi)\}$, in particular it has winding number $\pm m(\xi)$ with respect to $p(\xi)$. However, if $\gamma$ is valued in $E$, then $p \circ \gamma: \mathbb{T} \to O$ is a parametrized Jordan curve because $p$ is injective on $E$, and therefore it has winding number $\pm 1$ with respect to $p(\xi)$. This contradicts the assumption that $m(\xi) > 1$, showing that $U_\xi := O_\xi$ satisfies our requirements. $\square$

The next lemma, used in the proof of Lemma 3.11, depends on Lemma A.3.

**Lemma A.4.** If $E \subset \mathcal{R}$ is schlicht over $D$ and $\xi \in \text{rp}(\mathcal{R})$, then $\mathcal{R}\setminus E$ is non-thin at $\xi$.

*Proof.* Let $(U_\xi, \varphi)$ be a chart around $\xi$, with $U_\xi$ as in Lemma A.3 and $\varphi(\xi) = 0$. If $\mathcal{R}\setminus E$ is thin at $\xi$, then there is a circle $\mathbb{T}_r := \{|z| = r\} \subset \varphi(U_\xi)$ such that $\varphi^{-1}(\mathbb{T}_r) \subset E$. As $\varphi^{-1}: \mathbb{T}_r \to U_\xi$ is a Jordan curve in $E$ that contains $\xi$ in its interior, this contradicts Lemma A.3. $\square$

A.7. **Regularity.** Thinness is intimately connected to the notion of a regular boundary point with respect to the Dirichlet problem. Given a Euclidean open set $U$ with $\bar{U} \subset \Omega$, a point $\zeta \in \partial U$ is called *regular* if for any continuous function $\psi$ on $\partial U$ it holds that

$$ \lim_{U \ni z \to \zeta} H_\psi(z) = \psi(\zeta), $$
where $H_\psi(z)$ is the Perron-Wiener-Brelot solution of the Dirichlet problem on $U$ with boundary data $\psi$, see [8, Section VI.6, item $\gamma$]) for a description of the Perron-Wiener-Brelot process; other boundary points are called *irregular*. When all its boundary points are regular, we say that $U$ itself is regular. It is known that $\zeta \in \partial U$ is regular if and only if

$$ (A.12) \qquad \lim_{U \ni z \to \zeta} g_U(z, w) = 0 $$

for some (and then any) $w$ in each connected component of $U$. Moreover, $\zeta$ is irregular if and only if the complement of $U$ is thin at $\zeta$ [8, Theorem VII.13]. This entails that regularity is a local notion, in particular each point of $\partial U$ is regular as soon as the latter is locally connected, as follows from the analogous property in a Euclidean space [53, Theorem 4.2.2].

Let $\sigma$ be a finite measure, compactly supported in $U$. As $g_U(z, w)$ is bounded for $w \in \text{supp } \sigma$ and $z$ outside of a neighborhood of the latter, see Section A.2, we get from (A.12) and the dominated convergence theorem that $g(\sigma, U; \cdot)$ extends continuously by zero to the set of regular points of $\partial U$. When $\sigma$ is not compactly supported, a weaker result is stated in Section A.8 (Lemma A.5).

Regular points of *finely open* sets are defined analogously: when $U$ is finely open, $\zeta \in \partial_t U$ is said to be regular if $\Omega \setminus U$ is non-thin at $\zeta$. By (A.10), the set of regular points is then $b(\partial_t U)$, and if $\partial_t U$ is its own base one says that $U$ is regular, see [20, Section IV.12].

The following results are the natural analogs, for Green potentials on regular open sets of hyperbolic surfaces, of their logarithmic counterparts in the plane, see [55, Theorems I.6.8 & I.6.9].

**Principle of Descent and Lower Envelope Theorem.** Let $U$ be a regular open set with compact closure $\bar{U} \subset \Omega$. If $\sigma_n$ are positive measures on $U$ with uniformly bounded masses that converge weak* to some measure $\sigma$ as $n \to \infty$, then

(1) [Principle of Descent]

$$ (A.13) \qquad \liminf_{n \to \infty} g(\sigma_n, U; z_n) \ge g(\sigma, U; z), \quad z_n \to z \in U. $$

(2) [Lower Envelope Theorem]

$$ (A.14) \qquad \liminf_{n \to \infty} g(\sigma_n, U; z) = g(\sigma, U; z) \quad \text{for quasi every } z \in U. $$

*Proof.* The arguments are a minor variation of those used in [55]. For $M > 0$ and $z \in U$, observe from (A.12) that $\varphi_{M,z}(w) := \min\{M, g_U(z, w)\}$ is continuous on $\bar{U}$ and zero on $\partial U$, locally uniformly with respect to $z$. Thus, $\varphi_{M,z}$ lies in the closure of the space $C_c(U)$ of continuous functions on $U$ with compact support endowed with the sup norm. Moreover, $|\varphi_{M,z_n} - \varphi_{M,z}|$ is arbitrary small on $U$ for $n$ large enough, by the minimum principle and the continuity of Green functions off the diagonal. Hence, as $\sigma_n \xrightarrow{w^*} \sigma$, we get that $\lim_n \int \varphi_{M,z_n} d\sigma_n = \int \varphi_{M,z} d\sigma$ and consequently, by monotone convergence, we deduce (A.13) from the relations

$$ g(\sigma, U; z) = \lim_{M \to \infty} \int \varphi_{M,z} d\sigma = \lim_{n \to \infty} \lim_{n} \int \varphi_{M,z_n} d\sigma_n \le \liminf_n g(\sigma_n, U; z_n). $$

Next, assume to the contrary that $g(\sigma, U; z) < \liminf_n g(\sigma_n, U, z)$ for $z \in K$, where $K \subset U$ is such that $\text{cap}_U(K) > 0$. Clearly, we may suppose that $K$ is compact and so we can find a nonzero measure $\sigma_*$, supported on $K$, such that $g(\sigma_*, U; \cdot)$ is continuous on $U$, see Section A.4. Then by Fatou's lemma it holds that

$$ (A.15) \qquad \int g(\sigma, U; z) d\sigma_*(z) < \int \liminf_n g(\sigma_n, U; z) d\sigma_*(z) \le \liminf_n \int g(\sigma_n, U; z) d\sigma_*(z). $$

Moreover, as $g(\sigma_*, U; \cdot)$ extends continuously by zero on $\partial U$, see discussion after (A.12), it lies in the closure of $C_c(U)$. Therefore, by Fubini's theorem,

$$ \lim_n \int g(\sigma_n, U; z) d\sigma_*(z) = \lim_n \int g(\sigma_*, U; z) d\sigma_n(z) = \int g(\sigma_*, U; z) d\sigma(z) = \int g(\sigma, U; z) d\sigma_*(z), $$
where $H_\psi(z)$ is the Perron-Wiener-Brelot solution of the Dirichlet problem on $U$ with boundary data $\psi$, see [8, Section VI.6, item $\gamma$]) for a description of the Perron-Wiener-Brelot process; other boundary points are called *irregular*. When all its boundary points are regular, we say that $U$ itself is regular. It is known that $\zeta \in \partial U$ is regular if and only if

$$ (A.12) \qquad \lim_{U \ni z \to \zeta} g_U(z, w) = 0 $$

for some (and then any) $w$ in each connected component of $U$. Moreover, $\zeta$ is irregular if and only if the complement of $U$ is thin at $\zeta$ [8, Theorem VII.13]. This entails that regularity is a local notion, in particular each point of $\partial U$ is regular as soon as the latter is locally connected, as follows from the analogous property in a Euclidean space [53, Theorem 4.2.2].

Let $\sigma$ be a finite measure, compactly supported in $U$. As $g_U(z, w)$ is bounded for $w \in \text{supp } \sigma$ and $z$ outside of a neighborhood of the latter, see Section A.2, we get from (A.12) and the dominated convergence theorem that $g(\sigma, U; \cdot)$ extends continuously by zero to the set of regular points of $\partial U$. When $\sigma$ is not compactly supported, a weaker result is stated in Section A.8 (Lemma A.5).

Regular points of *finely open* sets are defined analogously: when $U$ is finely open, $\zeta \in \partial_t U$ is said to be regular if $\Omega \setminus U$ is non-thin at $\zeta$. By (A.10), the set of regular points is then $b(\partial_t U)$, and if $\partial_t U$ is its own base one says that $U$ is regular, see [20, Section IV.12].

The following results are the natural analogs, for Green potentials on regular open sets of hyperbolic surfaces, of their logarithmic counterparts in the plane, see [55, Theorems I.6.8 & I.6.9].

**Principle of Descent and Lower Envelope Theorem.** Let $U$ be a regular open set with compact closure $\bar{U} \subset \Omega$. If $\sigma_n$ are positive measures on $U$ with uniformly bounded masses that converge weak* to some measure $\sigma$ as $n \to \infty$, then

(1) [Principle of Descent]

$$ (A.13) \qquad \liminf_{n \to \infty} g(\sigma_n, U; z_n) \ge g(\sigma, U; z), \quad z_n \to z \in U. $$

(2) [Lower Envelope Theorem]

$$ (A.14) \qquad \liminf_{n \to \infty} g(\sigma_n, U; z) = g(\sigma, U; z) \quad \text{for quasi every } z \in U. $$

*Proof.* The arguments are a minor variation of those used in [55]. For $M > 0$ and $z \in U$, observe from (A.12) that $\varphi_{M,z}(w) := \min\{M, g_U(z, w)\}$ is continuous on $\bar{U}$ and zero on $\partial U$, locally uniformly with respect to $z$. Thus, $\varphi_{M,z}$ lies in the closure of the space $C_c(U)$ of continuous functions on $U$ with compact support endowed with the sup norm. Moreover, $|\varphi_{M,z_n} - \varphi_{M,z}|$ is arbitrary small on $U$ for $n$ large enough, by the minimum principle and the continuity of Green functions off the diagonal. Hence, as $\sigma_n \xrightarrow{w^*} \sigma$, we get that $\lim_n \int \varphi_{M,z_n} d\sigma_n = \int \varphi_{M,z} d\sigma$ and consequently, by monotone convergence, we deduce (A.13) from the relations

$$ g(\sigma, U; z) = \lim_{M \to \infty} \int \varphi_{M,z} d\sigma = \lim_{n \to \infty} \lim_{n} \int \varphi_{M,z_n} d\sigma_n \le \liminf_n g(\sigma_n, U; z_n). $$

Next, assume to the contrary that $g(\sigma, U; z) < \liminf_n g(\sigma_n, U, z)$ for $z \in K$, where $K \subset U$ is such that $\text{cap}_U(K) > 0$. Clearly, we may suppose that $K$ is compact and so we can find a nonzero measure $\sigma_*$, supported on $K$, such that $g(\sigma_*, U; \cdot)$ is continuous on $U$, see Section A.4. Then by Fatou's lemma it holds that

$$ (A.15) \qquad \int g(\sigma, U; z) d\sigma_*(z) < \int \liminf_n g(\sigma_n, U; z) d\sigma_*(z) \le \liminf_n \int g(\sigma_n, U; z) d\sigma_*(z). $$

Moreover, as $g(\sigma_*, U; \cdot)$ extends continuously by zero on $\partial U$, see discussion after (A.12), it lies in the closure of $C_c(U)$. Therefore, by Fubini's theorem,

$$ \lim_n \int g(\sigma_n, U; z) d\sigma_*(z) = \lim_n \int g(\sigma_*, U; z) d\sigma_n(z) = \int g(\sigma_*, U; z) d\sigma(z) = \int g(\sigma, U; z) d\sigma_*(z), $$
thereby contradicting (A.15).

□

A.8. **Superlevel Sets of Green Potentials.** In this section we restrict attention to a planar simply connected domain $D$, which is the interior of a Jordan curve $T$. In this case, any conformal map $\phi: D \to D$ extends to a homeomorphism from $\bar{D}$ onto $\bar{D}$ [50, Theorem 2.6], that we continue to denote with $\phi$. Clearly, such a domain $D$ is regular. As mentioned in the previous subsection, if $\sigma$ is a finite Borel measure compactly supported in $D$, then $g(\sigma, D; \cdot)$ continuously extends by zero to $T$. If $\sigma$ is not compactly supported this may not hold, but when $F$ is a relatively closed subset of $D$ with a limit point $\xi \in T$, it was shown in [39] that

$$ (A.16) \qquad \lim_{\epsilon \to 0} \mathrm{cap}_{\mathbb{D}}(F \cap \{|z - \xi| < \epsilon\}) > 0 \Rightarrow \liminf_{F \ni z \to \xi} g(\sigma, \mathbb{D}; z) = 0, $$

and if the rightmost limit holds for every finite measure $\sigma$, then the implication can be reversed. This result is in fact stated in [39] with $g(\sigma, \mathbb{D}; z)$ replaced by $(1 - |z|)g(v, \mathbb{D}; z)$ where $v$ is any measure whose Green potential is not identically $+\infty$, but the latter condition is equivalent to saying that the measure $d\sigma(z) := (1 - |z|)dv(z)$ is finite and then convergence to zero along $F$ of the limit inferior of $(1 - |z|)g(v, \mathbb{D}; z)$ and of $g(\sigma, \mathbb{D}; z)$ are equivalent, see [39, Section 3, Lemma]. It is also pointed out in [39, Equation (2.5)] that the leftmost limit in (A.16) can be equivalently replaced by $\mathrm{cap}_{\mathbb{D}}(F \cap \{|z - \xi| < \epsilon\}) = \infty$ for every $\epsilon > 0$ (note that instead of the Greenian capacity $\mathrm{cap}_{\mathbb{D}}(E)$ that we use, [39] employs the hyperbolic capacity $\exp\{-1/\mathrm{cap}_{\mathbb{D}}(E)\}$). In Lemma A.5 below, we derive a useful consequence of (A.16). With the notation of this lemma, we stress that a stronger conclusion in fact holds quasi everywhere, namely $U_\epsilon$ is thin at quasi every point of $T$ (this can be deduced from general properties of balayage covered in Section A.9). The interest of Lemma A.5 lies with the fact that its conclusion holds at every point of $T$.

**Lemma A.5.** Let $\sigma$ be a finite measure in $D$, and for $\epsilon > 0$ set $U_\epsilon := \{z \in D : g(\sigma, D; z) > \epsilon\}$. Let $\xi \in \bar{U}_\epsilon \cap T$ and $\phi: D \to D$ be a conformal map such that $\phi(1) = \xi$. Then there exists a closed set $R_\epsilon \subset [0, 1]$ such that $R_\epsilon \cap (1 - \delta, 1)$ is non-polar for any $\delta \in (0, 1)$, and for each $r \in R_\epsilon$ one has $U_\epsilon \cap \phi(\{z \in D : |1 - z| = 1 - r\}) = \emptyset$.

*Proof.* Suppose initially that $D = \mathbb{D}$. Without loss of generality, we can assume that $\xi = 1$. First, we shall show that

$$ (A.17) \qquad \lim_{\delta \to 0} \mathrm{cap}_{\mathbb{D}}(U_\epsilon \cap D_\delta) = 0, $$

where $D_\delta := \{z \in D : |z - 1| < \delta\}$. For $\zeta \in D$, let $S(\zeta) := \{z \in D : g_{\mathbb{D}}(z, \zeta) \ge \log 2\}$. It was shown in [39, Section 3, Lemma] that there exists $\delta_0 = \delta(\epsilon) > 0$ for which

$$ \int_{D \setminus S(\zeta)} g_{\mathbb{D}}(z, \zeta) d\sigma(z) \le \epsilon/2, \quad |\zeta| > 1 - \delta_0. $$

In particular, this inequality holds for $\zeta \in D_{\delta_0}$. Hence, for any compact subset $F \subset U_\epsilon \cap D_{\delta_0}$, it holds when $\zeta \in F$ that

$$ (A.18) \qquad h(\zeta) := \int_{S(\zeta)} g_{\mathbb{D}}(z, \zeta) d\sigma(z) = g(\sigma, \mathbb{D}; \zeta) - \int_{D \setminus S(\zeta)} g_{\mathbb{D}}(z, \zeta) d\sigma(z) > \epsilon/2. $$

Assume to the contrary that the limit in (A.17) is larger than $2\eta > 0$ (the limit must exist as $\mathrm{cap}_{\mathbb{D}}(U_\epsilon \cap D_\delta)$ decreases with $\delta$). Since $U_\epsilon$ is an open set, we get from (A.5) that for any $\delta > 0$ there exists a compact set $F_\delta \subset U_\epsilon \cap D_\delta$ for which $\mathrm{cap}_{\mathbb{D}}(F_\delta) \ge \eta$. This entails that there exist a sequence $\delta_n \to 0$ and disjoint compact sets $F_n \subset U_\epsilon \cap D_{\delta_n}$ with $\mathrm{cap}_{\mathbb{D}}(F_n) \ge \eta$. Let $v_n$ be the Green equilibrium distribution on $F_n$ and $F_n^* := \{z \in S(\zeta) : \text{for some } \zeta \in F_n\}$. In view of (A.18),

$$ \epsilon/2 \le \int_{F_n} h(\zeta) dv_n(\zeta) \le \frac{1}{\eta} \sigma(F_n^*) \to 0, $$
thereby contradicting (A.15).

A.8. **Superlevel Sets of Green Potentials.** In this section we restrict attention to a planar simply connected domain $D$, which is the interior of a Jordan curve $T$. In this case, any conformal map $\phi: D \to D$ extends to a homeomorphism from $\bar{D}$ onto $\bar{D}$ [50, Theorem 2.6], that we continue to denote with $\phi$. Clearly, such a domain $D$ is regular. As mentioned in the previous subsection, if $\sigma$ is a finite Borel measure compactly supported in $D$, then $g(\sigma, D; \cdot)$ continuously extends by zero to $T$. If $\sigma$ is not compactly supported this may not hold, but when $F$ is a relatively closed subset of $D$ with a limit point $\xi \in T$, it was shown in [39] that

$$ (A.16) \qquad \lim_{\epsilon \to 0} \mathrm{cap}_{\mathbb{D}}(F \cap \{|z - \xi| < \epsilon\}) > 0 \Rightarrow \liminf_{F \ni z \to \xi} g(\sigma, \mathbb{D}; z) = 0, $$

and if the rightmost limit holds for every finite measure $\sigma$, then the implication can be reversed. This result is in fact stated in [39] with $g(\sigma, \mathbb{D}; z)$ replaced by $(1 - |z|)g(v, \mathbb{D}; z)$ where $v$ is any measure whose Green potential is not identically $+\infty$, but the latter condition is equivalent to saying that the measure $d\sigma(z) := (1 - |z|)dv(z)$ is finite and then convergence to zero along $F$ of the limit inferior of $(1 - |z|)g(v, \mathbb{D}; z)$ and of $g(\sigma, \mathbb{D}; z)$ are equivalent, see [39, Section 3, Lemma]. It is also pointed out in [39, Equation (2.5)] that the leftmost limit in (A.16) can be equivalently replaced by $\mathrm{cap}_{\mathbb{D}}(F \cap \{|z - \xi| < \epsilon\}) = \infty$ for every $\epsilon > 0$ (note that instead of the Greenian capacity $\mathrm{cap}_{\mathbb{D}}(E)$ that we use, [39] employs the hyperbolic capacity $\exp\{-1/\mathrm{cap}_{\mathbb{D}}(E)\}$). In Lemma A.5 below, we derive a useful consequence of (A.16). With the notation of this lemma, we stress that a stronger conclusion in fact holds quasi everywhere, namely $U_\epsilon$ is thin at quasi every point of $T$ (this can be deduced from general properties of balayage covered in Section A.9). The interest of Lemma A.5 lies with the fact that its conclusion holds at every point of $T$.

**Lemma A.5.** Let $\sigma$ be a finite measure in $D$, and for $\epsilon > 0$ set $U_\epsilon := \{z \in D : g(\sigma, D; z) > \epsilon\}$. Let $\xi \in \bar{U}_\epsilon \cap T$ and $\phi: D \to D$ be a conformal map such that $\phi(1) = \xi$. Then there exists a closed set $R_\epsilon \subset [0, 1]$ such that $R_\epsilon \cap (1 - \delta, 1)$ is non-polar for any $\delta \in (0, 1)$, and for each $r \in R_\epsilon$ one has $U_\epsilon \cap \phi(\{z \in D : |1 - z| = 1 - r\}) = \emptyset$.

*Proof.* Suppose initially that $D = \mathbb{D}$. Without loss of generality, we can assume that $\xi = 1$. First, we shall show that

$$ (A.17) \qquad \lim_{\delta \to 0} \mathrm{cap}_{\mathbb{D}}(U_\epsilon \cap D_\delta) = 0, $$

where $D_\delta := \{z \in \mathbb{D} : |z - 1| < \delta\}$. For $\zeta \in \mathbb{D}$, let $S(\zeta) := \{z \in \mathbb{D} : g_{\mathbb{D}}(z, \zeta) \ge \log 2\}$. It was shown in [39, Section 3, Lemma] that there exists $\delta_0 = \delta(\epsilon) > 0$ for which

$$ \int_{\mathbb{D} \setminus S(\zeta)} g_{\mathbb{D}}(z, \zeta) d\sigma(z) \le \epsilon/2, \quad |\zeta| > 1 - \delta_0. $$

In particular, this inequality holds for $\zeta \in D_{\delta_0}$. Hence, for any compact subset $F \subset U_\epsilon \cap D_{\delta_0}$, it holds when $\zeta \in F$ that

$$ (A.18) \qquad h(\zeta) := \int_{S(\zeta)} g_{\mathbb{D}}(z, \zeta) d\sigma(z) = g(\sigma, \mathbb{D}; \zeta) - \int_{\mathbb{D} \setminus S(\zeta)} g_{\mathbb{D}}(z, \zeta) d\sigma(z) > \epsilon/2. $$

Assume to the contrary that the limit in (A.17) is larger than $2\eta > 0$ (the limit must exist as $\mathrm{cap}_{\mathbb{D}}(U_\epsilon \cap D_\delta)$ decreases with $\delta$). Since $U_\epsilon$ is an open set, we get from (A.5) that for any $\delta > 0$ there exists a compact set $F_\delta \subset U_\epsilon \cap D_\delta$ for which $\mathrm{cap}_{\mathbb{D}}(F_\delta) \ge \eta$. This entails that there exist a sequence $\delta_n \to 0$ and disjoint compact sets $F_n \subset U_\epsilon \cap D_{\delta_n}$ with $\mathrm{cap}_{\mathbb{D}}(F_n) \ge \eta$. Let $v_n$ be the Green equilibrium distribution on $F_n$ and $F_n^* := \{z \in S(\zeta) : \text{for some } \zeta \in F_n\}$. In view of (A.18),

$$ \epsilon/2 \le \int_{F_n} h(\zeta) dv_n(\zeta) \le \frac{1}{\eta} \sigma(F_n^*) \to 0, $$
where the second inequality and the fact that $\lim_n \sigma(F_n^*) \to 0$ can be established as in the proof of [39, Theorem 1] (compare to p. 486 of that reference). This contradiction proves (A.17).

Let $\mathcal{T} : D_1 \to (0, 1)$ be defined by $z \mapsto \mathcal{T}z := 1 - |1 - z|$ and put $V_\epsilon := U_\epsilon \cap D_1$. Denoting by $\mathcal{T}V_\epsilon$ the set $\{\mathcal{T}\zeta : \zeta \in V_\epsilon\}$, we claim that

$$ (A.19) \qquad \lim_{\delta \to 0} \mathrm{cap}_{\mathbb{D}}(\mathcal{T}V_\epsilon \cap D_\delta) = 0. $$

Before proving (A.19), let us show why it implies the lemma. For this, consider $R_\epsilon := [0, 1] \setminus \mathcal{T}V_\epsilon$, which is a closed set. If the conclusion of the lemma were not true, there would exist $\delta_0 > 0$ such that $R_\epsilon \cap (1 - \delta_0, 1)$ is polar. By definition of $\mathcal{T}$ this would imply that $\mathrm{cap}_{\mathbb{D}}(\mathcal{T}V_\epsilon \cap D_\delta) = \mathrm{cap}_{\mathbb{D}}((1 - \delta, 1)) = \infty$ for any $\delta < \delta_0$ (the last equality follows at once from the definition of the Greenian capacity), which contradicts (A.19).

We are now left to demonstrate (A.19). Assume to the contrary that it does not hold, i.e., there exists $\eta > 0$ such that for any $\delta > 0$ there is a compact set $F_\delta \subset \mathcal{T}V_\epsilon \cap D_\delta$ for which $\mathrm{cap}_{\mathbb{D}}(F_\delta) \ge \eta$. The previous inequality means that there exists a probability measure $\mu_\delta$ supported on $F_\delta$ such that

$$ \iint g_{\mathbb{D}}(x, y)d\mu_{\delta}(x)d\mu_{\delta}(y) \le \frac{1}{\eta}. $$

Since $U_\epsilon$ is open, so is $V_\epsilon$ and one easily sees that each $z \in V_\epsilon \cap D_\delta$ has a neighborhood, say $O_z$, whose closure is contained in $V_\epsilon \cap D_\delta$ and whose circular projection $\mathcal{T}O_z$ is an open subinterval of $(0, 1)$. These subintervals form an open cover of $F_\delta$, which necessarily has a finite subcover, say $\mathcal{T}O_{z_1}, \dots, \mathcal{T}O_{z_N}$. The closure $K_\delta$ of $O_{z_1} \cup \dots \cup O_{z_N}$ is a compact subset of $V_\epsilon \cap D_\delta$, and clearly $F_\delta \subset \mathcal{T}K_\delta$. In particular, there exists a probability measure $\nu_\delta$ on $K_\delta$ such that $\nu_\delta \mathcal{T}^{-1} = \mu_\delta$, see for example [53, Theorem A.4.4]. Then, by Fubini's theorem, it holds that

$$ \frac{1}{\eta} \ge \iint g_{\mathbb{D}}(x, y)d\mu_{\delta}(x)\mu_{\delta}(y) = \iint g_{\mathbb{D}}(\mathcal{T}z, \mathcal{T}w)dv_{\delta}(z)dv_{\delta}(w), $$

and if we can show that $g_{\mathbb{D}}(\mathcal{T}z, \mathcal{T}w) \ge g_{\mathbb{D}}(z, w)$ then we will deduce from the above estimate that

$$ \mathrm{cap}_{\mathbb{D}}(U_{\epsilon} \cap D_{\delta}) = \mathrm{cap}_{\mathbb{D}}(V_{\epsilon} \cap D_{\delta}) \ge \mathrm{cap}_{\mathbb{D}}(K_{\delta}) \ge \eta > 0 \quad \text{for any } \delta > 0, $$

which of course contradicts (A.17). Hence, the proof has been reduced to the verification of $g_{\mathbb{D}}(\mathcal{T}z, \mathcal{T}w) \ge g_{\mathbb{D}}(z, w)$ for $z, w \in D_1$, that we now carry out. Since $g_{\mathbb{D}}(z, w) = \log|(1-z\bar{w})/(z-w)|$ and $\mathcal{T}$ is real-valued, we need to show that

$$ (A.20) \qquad E := |1 - \mathcal{T}z\mathcal{T}w|^2|z - w|^2 - |1 - z\bar{w}|^2|\mathcal{T}z - \mathcal{T}w|^2 \ge 0. $$

Set $a\xi := 1 - z$ and $b\eta := 1 - w$, where $a, b \in (0, 1)$ and $|\xi| = |\eta| = 1$ with $\mathrm{Re}\xi, \mathrm{Re}\eta \in (0, 1)$. Then

$$ \begin{aligned} E &= |a + b - ab|^2|a\xi - b\eta|^2 - |a\xi + b\bar{\eta} - ab\xi\bar{\eta}|^2|a - b|^2 \\ &= (S + 2abU)(T - 2abV) - (S + 2abW)(T - 2ab) \end{aligned} $$

where $S := a^2 + b^2 + (ab)^2, T := a^2 + b^2, U := 1 - a - b, V := \mathrm{Re}(\xi\bar{\eta})$, and $W := \mathrm{Re}(\xi\eta - a\eta - b\xi)$. Therefore,

$$ \begin{aligned} E &= 2ab(S(1-V) + T(U-W) + 2ab(W-UV)) \\ &= 2ab((S+2abU)(1-V) + (T-2ab)(U-W)) \\ &= 2ab((a+b-ab)^2(1-V) + (a-b)^2(U-W)). \end{aligned} $$

Because $V \le 1$, the above expression can be estimated from below as

$$ \begin{aligned} E &\ge 2ab(a-b)^2(1-V+U-W) \\ &= 2ab(a-b)^2(2-\mathrm{Re}(\xi\bar{\eta}+\xi\eta)-a(1-\mathrm{Re}\eta)-b(1-\mathrm{Re}\xi)), \end{aligned} $$
where the second inequality and the fact that $\lim_n \sigma(F_n^*) \to 0$ can be established as in the proof of [39, Theorem 1] (compare to p. 486 of that reference). This contradiction proves (A.17).

Let $\mathcal{T} : D_1 \to (0, 1)$ be defined by $z \mapsto \mathcal{T}z := 1 - |1 - z|$ and put $V_\epsilon := U_\epsilon \cap D_1$. Denoting by $\mathcal{T}V_\epsilon$ the set $\{\mathcal{T}\zeta : \zeta \in V_\epsilon\}$, we claim that

$$ (A.19) \qquad \lim_{\delta \to 0} \mathrm{cap}_{\mathbb{D}}(\mathcal{T}V_\epsilon \cap D_\delta) = 0. $$

Before proving (A.19), let us show why it implies the lemma. For this, consider $R_\epsilon := [0, 1] \setminus \mathcal{T}V_\epsilon$, which is a closed set. If the conclusion of the lemma were not true, there would exist $\delta_0 > 0$ such that $R_\epsilon \cap (1 - \delta_0, 1)$ is polar. By definition of $\mathcal{T}$ this would imply that $\mathrm{cap}_{\mathbb{D}}(\mathcal{T}V_\epsilon \cap D_\delta) = \mathrm{cap}_{\mathbb{D}}((1 - \delta, 1)) = \infty$ for any $\delta < \delta_0$ (the last equality follows at once from the definition of the Greenian capacity), which contradicts (A.19).

We are now left to demonstrate (A.19). Assume to the contrary that it does not hold, i.e., there exists $\eta > 0$ such that for any $\delta > 0$ there is a compact set $F_\delta \subset \mathcal{T}V_\epsilon \cap D_\delta$ for which $\mathrm{cap}_{\mathbb{D}}(F_\delta) \ge \eta$. The previous inequality means that there exists a probability measure $\mu_\delta$ supported on $F_\delta$ such that

$$ \iint g_{\mathbb{D}}(x, y)d\mu_{\delta}(x)d\mu_{\delta}(y) \le \frac{1}{\eta}. $$

Since $U_\epsilon$ is open, so is $V_\epsilon$ and one easily sees that each $z \in V_\epsilon \cap D_\delta$ has a neighborhood, say $O_z$, whose closure is contained in $V_\epsilon \cap D_\delta$ and whose circular projection $\mathcal{T}O_z$ is an open subinterval of $(0, 1)$. These subintervals form an open cover of $F_\delta$, which necessarily has a finite subcover, say $\mathcal{T}O_{z_1}, \dots, \mathcal{T}O_{z_N}$. The closure $K_\delta$ of $O_{z_1} \cup \dots \cup O_{z_N}$ is a compact subset of $V_\epsilon \cap D_\delta$, and clearly $F_\delta \subset \mathcal{T}K_\delta$. In particular, there exists a probability measure $\nu_\delta$ on $K_\delta$ such that $\nu_\delta \mathcal{T}^{-1} = \mu_\delta$, see for example [53, Theorem A.4.4]. Then, by Fubini's theorem, it holds that

$$ \frac{1}{\eta} \ge \iint g_{\mathbb{D}}(x, y)d\mu_{\delta}(x)\mu_{\delta}(y) = \iint g_{\mathbb{D}}(\mathcal{T}z, \mathcal{T}w)dv_{\delta}(z)dv_{\delta}(w), $$

and if we can show that $g_{\mathbb{D}}(\mathcal{T}z, \mathcal{T}w) \ge g_{\mathbb{D}}(z, w)$ then we will deduce from the above estimate that

$$ \mathrm{cap}_{\mathbb{D}}(U_\epsilon \cap D_\delta) = \mathrm{cap}_{\mathbb{D}}(V_\epsilon \cap D_\delta) \ge \mathrm{cap}_{\mathbb{D}}(K_\delta) \ge \eta > 0 \quad \text{for any } \delta > 0, $$

which of course contradicts (A.17). Hence, the proof has been reduced to the verification of $g_{\mathbb{D}}(\mathcal{T}z, \mathcal{T}w) \ge g_{\mathbb{D}}(z, w)$ for $z, w \in D_1$, that we now carry out. Since $g_{\mathbb{D}}(z, w) = \log|(1-z\bar{w})/(z-w)|$ and $\mathcal{T}$ is real-valued, we need to show that

$$ (A.20) \qquad E := |1 - \mathcal{T}z\mathcal{T}w|^2|z - w|^2 - |1 - z\bar{w}|^2|\mathcal{T}z - \mathcal{T}w|^2 \ge 0. $$

Set $a\xi := 1 - z$ and $b\eta := 1 - w$, where $a, b \in (0, 1)$ and $|\xi| = |\eta| = 1$ with $\mathrm{Re}\xi, \mathrm{Re}\eta \in (0, 1)$. Then

$$ \begin{aligned} E &= |a + b - ab|^2|a\xi - b\eta|^2 - |a\xi + b\bar{\eta} - ab\xi\bar{\eta}|^2|a - b|^2 \\ &= (S + 2abU)(T - 2abV) - (S + 2abW)(T - 2ab) \end{aligned} $$

where $S := a^2 + b^2 + (ab)^2, T := a^2 + b^2, U := 1 - a - b, V := \mathrm{Re}(\xi\bar{\eta})$, and $W := \mathrm{Re}(\xi\eta - a\eta - b\xi)$. Therefore,

$$ \begin{aligned} E &= 2ab(S(1-V) + T(U-W) + 2ab(W-UV)) \\ &= 2ab((S+2abU)(1-V) + (T-2ab)(U-W)) \\ &= 2ab((a+b-ab)^2(1-V) + (a-b)^2(U-W)). \end{aligned} $$

Because $V \le 1$, the above expression can be estimated from below as

$$ \begin{aligned} E &\ge 2ab(a-b)^2(1-V+U-W) \\ &= 2ab(a-b)^2(2-\mathrm{Re}(\xi\bar{\eta}+\xi\eta)-a(1-\mathrm{Re}\eta)-b(1-\mathrm{Re}\xi)), \end{aligned} $$
and since $1 - \text{Re}\xi, 1 - \text{Re}\eta, 1 - a$ as well as $1 - b$ are all positive, it therefore holds that

$$
\begin{align*}
E &\ge 2ab(a-b)^2(\text{Re}\eta + \text{Re}\xi - \text{Re}(\xi\bar{\eta} + \xi\eta)) \\
&= 2ab(a-b)^2(\text{Re}\eta + \text{Re}\xi - 2\text{Re}\eta\text{Re}\xi) \\
&= 2ab(a-b)^2(\text{Re}\eta(1 - \text{Re}\xi) + \text{Re}\xi(1 - \text{Re}\eta)).
\end{align*}
$$

As $\text{Re}\xi, \text{Re}\eta \in [0, 1]$, this establishes (A.20) and completes the proof of the lemma when $D = \mathbb{D}$.

Finally, it remains to reduce the case of a general domain $D$ to the one of the unit disk. Using [53, Theorem A.4.4] once more, let $v$ be a finite measure in $\bar{\mathbb{D}}$ such that $v\phi^{-1} = \sigma$. Then

$$
\begin{align*}
g(\sigma, D; \phi(z)) &= \int g_D(\phi(z), \zeta) d\sigma(\zeta) = \int g_D(\phi(z), \phi(w)) dv(w) \\
&= \int g_{\mathbb{D}}(z, w) dv(w) = g(v, \mathbb{D}; z), \quad z \in \mathbb{D},
\end{align*}
$$

by conformal equivalence of Green functions. Since $\phi : \bar{\mathbb{D}} \to \bar{\mathbb{D}}$ is a bijection, the superlevel set
$\{z \in \bar{\mathbb{D}} : g(v, \mathbb{D}; z) > \epsilon\}$ is equal to $\phi^{-1}(U_\epsilon)$, from which the desired result follows. $\square$

A.9. **Balayage.** Let $v$ be a non-negative superharmonic function on $\Omega$ and $E$ be a subset of $\Omega$.
The balayage function (or *regularized reduction*) of $v$ relative to $E$, denoted by $B_v^E$, is the lower
semi-continuous regularization of

$$
(A.21) \qquad \inf \{ u \mid u \text{ is superharmonic and positive in } \Omega, u \ge v \text{ on } E \},
$$

see [29, Section 5.3] for an account on $\mathbb{R}^n$ that carries over to $\Omega$ without change; in fact, $B_v^E$ coincides
with the infimum in (A.21) except perhaps on a polar set where lower semi-continuous regularization
may modify the value. The balayage function $B_v^E$ is superharmonic in $\Omega$, harmonic in $\Omega \setminus \bar{E}$, and
equal to $v$ on $b(E)$ [8, Section VIII.1]. Clearly, $B_v^E \le v$ everywhere, since $v$ qualifies as one of the
functions $u$ in (A.21). The balayage function $B_v^E$ does not change if $E$ gets replaced by $b(E)$ or by
clos$_f(E)$; in fact, it remains invariant if $E$ is altered by a polar set.

When $E$ is compact, the **Strong Domination Principle** and properties of the Green equilibrium
potential $g(\mu_{\Omega,E}, \Omega; \cdot)$ imply that $B_1^E = \text{cap}_{\Omega}(E)g(\mu_{\Omega,E}, \Omega; \cdot)$. Thus, it follows from the left
continuity of $B_1^E$ with respect to $E$, see [8, Section VI.10 e)], and the monotone convergence theorem
that the outer Greenian capacity of an arbitrary set $E \subset \Omega$ is the mass of $\Delta B_1^E$ (in fact, this is the way
the outer capacity is defined in [8, Section VIII.4]). From this and [8, Theorem VIII.12], we deduce
in particular that $E$ is polar if and only if $\text{cap}_{\Omega}(E) = 0$, justifying a claim made in Section A.5.

If $v$ is the Green potential of a positive Borel measure $\sigma$, then $B_v^E$ is a Green potential as well [8, Section VI.11] and the measure $\sigma^E$ such that $B_v^E = g(\sigma^E, \Omega; \cdot)$ is called the balayage of $\sigma$ relative to $E$. The measure $\sigma^E$ is characterized as the unique measure satisfying

$$
(A.22) \qquad \sigma^E(\Omega \setminus b(E)) = 0 \quad \text{and} \quad g(\sigma, \Omega; z) = g(\sigma^E, \Omega; z), \quad z \in b(E),
$$

see [8, Theorem VIII.3]. From (A.22), one deduces at once that

$$
(A.23) \qquad (\sigma^E)^F = \sigma^F, \quad F \subset E \subset \Omega.
$$

Moreover, it holds by [8, Section VI.12, Equation (13)] that

$$
(A.24) \qquad \sigma^E(B) = \int \delta_x^E(B) d\sigma(x), \quad B \text{ Borel,}
$$

while it follows from [8, Section VI.12, Equation (9)] and Fubini’s theorem that

$$
(A.25) \qquad g(\sigma^E, \Omega; z) = \int g(\sigma, \Omega; x) d\delta_z^E(x).
$$
and since $1 - \text{Re}\xi, 1 - \text{Re}\eta, 1 - a$ as well as $1 - b$ are all positive, it therefore holds that

$$
\begin{align*}
E &\ge 2ab(a-b)^2(\text{Re}\eta + \text{Re}\xi - \text{Re}(\xi\bar{\eta} + \xi\eta)) \\
&= 2ab(a-b)^2(\text{Re}\eta + \text{Re}\xi - 2\text{Re}\eta\text{Re}\xi) \\
&= 2ab(a-b)^2(\text{Re}\eta(1 - \text{Re}\xi) + \text{Re}\xi(1 - \text{Re}\eta)).
\end{align*}
$$

As $\text{Re}\xi, \text{Re}\eta \in [0, 1]$, this establishes (A.20) and completes the proof of the lemma when $D = \mathbb{D}$.

Finally, it remains to reduce the case of a general domain $D$ to the one of the unit disk. Using [53, Theorem A.4.4] once more, let $v$ be a finite measure in $\bar{\mathbb{D}}$ such that $v\phi^{-1} = \sigma$. Then

$$
\begin{align*}
g(\sigma, D; \phi(z)) &= \int g_D(\phi(z), \zeta) d\sigma(\zeta) = \int g_D(\phi(z), \phi(w)) dv(w) \\
&= \int g_{\mathbb{D}}(z, w) dv(w) = g(v, \mathbb{D}; z), \quad z \in \mathbb{D},
\end{align*}
$$

by conformal equivalence of Green functions. Since $\phi: \bar{\mathbb{D}} \to \bar{\mathbb{D}}$ is a bijection, the superlevel set
$\{z \in \bar{\mathbb{D}} : g(v, \mathbb{D}; z) > \epsilon\}$ is equal to $\phi^{-1}(U_\epsilon)$, from which the desired result follows. $\square$

A.9. **Balayage.** Let $v$ be a non-negative superharmonic function on $\Omega$ and $E$ be a subset of $\Omega$.
The balayage function (or *regularized reduction*) of $v$ relative to $E$, denoted by $B_v^E$, is the lower
semi-continuous regularization of

$$
(A.21) \qquad \inf \{ u \mid u \text{ is superharmonic and positive in } \Omega, u \ge v \text{ on } E \},
$$

see [29, Section 5.3] for an account on $\mathbb{R}^n$ that carries over to $\Omega$ without change; in fact, $B_v^E$ coincides
with the infimum in (A.21) except perhaps on a polar set where lower semi-continuous regularization
may modify the value. The balayage function $B_v^E$ is superharmonic in $\Omega$, harmonic in $\Omega \setminus \bar{E}$, and
equal to $v$ on $b(E)$ [8, Section VIII.1]. Clearly, $B_v^E \le v$ everywhere, since $v$ qualifies as one of the
functions $u$ in (A.21). The balayage function $B_v^E$ does not change if $E$ gets replaced by $b(E)$ or by
clos$_f(E)$; in fact, it remains invariant if $E$ is altered by a polar set.

When $E$ is compact, the **Strong Domination Principle** and properties of the Green equilibrium
potential $g(\mu_{\Omega,E}, \Omega; \cdot)$ imply that $B_1^E = \text{cap}_{\Omega}(E)g(\mu_{\Omega,E}, \Omega; \cdot)$. Thus, it follows from the left
continuity of $B_1^E$ with respect to $E$, see [8, Section VI.10 e)], and the monotone convergence theorem
that the outer Greenian capacity of an arbitrary set $E \subset \Omega$ is the mass of $\Delta B_1^E$ (in fact, this is the way
the outer capacity is defined in [8, Section VIII.4]). From this and [8, Theorem VIII.12], we deduce
in particular that $E$ is polar if and only if $\text{cap}_{\Omega}(E) = 0$, justifying a claim made in Section A.5.

If $v$ is the Green potential of a positive Borel measure $\sigma$, then $B_v^E$ is a Green potential as well [8, Section VI.11] and the measure $\sigma^E$ such that $B_v^E = g(\sigma^E, \Omega; \cdot)$ is called the balayage of $\sigma$ relative to $E$. The measure $\sigma^E$ is characterized as the unique measure satisfying

$$
(A.22) \qquad \sigma^E(\Omega \setminus b(E)) = 0 \quad \text{and} \quad g(\sigma, \Omega; z) = g(\sigma^E, \Omega; z), \quad z \in b(E),
$$

see [8, Theorem VIII.3]. From (A.22), one deduces at once that

$$
(A.23) \qquad (\sigma^E)^F = \sigma^F, \quad F \subset E \subset \Omega.
$$

Moreover, it holds by [8, Section VI.12, Equation (13)] that

$$
(A.24) \qquad \sigma^E(B) = \int \delta_x^E(B) d\sigma(x), \quad B \text{ Borel,}
$$

while it follows from [8, Section VI.12, Equation (9)] and Fubini’s theorem that

$$
(A.25) \qquad g(\sigma^E, \Omega; z) = \int g(\sigma, \Omega; x) d\delta_z^E(x).
$$
Since $g(\sigma^E, \Omega; \cdot) \le g(\sigma, \Omega; \cdot)$ for any Radon measure $\sigma$ by the just discussed properties of balayage, it follows from Fubini's theorem that $\int v d\sigma^E \le \int v d\sigma$ for any Green potential $v$. Since any non-negative superharmonic function $v$ is an increasing limit of potentials, see [10, Lemma 1.1], monotone convergence yields that $\int v d\sigma^E \le \int v d\sigma$ remains valid for such functions as well, see also [20, Section I.3]. In particular, the mass of $\sigma^E$ cannot exceed the mass of $\sigma$.

The fine support of a Radon measure $\sigma$, denoted by $\text{supp}_f \sigma$ when it exists, is the smallest finely closed carrier of $\sigma$. A sufficient condition for its existence is that $\sigma$ does not charge polar sets, in which case $\text{supp}_f \sigma$ is its own base, see [8, Theorem VII.12]. If $\sigma$ is admissible, meaning that $g(\sigma, \Omega; \cdot) \ne +\infty$, see Section A.3, and if $\sigma(F) = 0$ for some polar set $F$, then $\sigma^E(F) = 0$ for any $E \subset \Omega$ [19, Theorem 1] (as usual, $\sigma(F)$ means the outer $\sigma$-measure of $F$ when the latter is not Borel). In particular, if $\sigma$ is admissible and $\sigma(b(E)) = 0$, then $\sigma^E$ does not charge polar sets since it is carried by $b(E)$. Thus, $\text{supp}_f \sigma^E$ exists in this case. An important special case is handled by the following lemma, the first item of which follows from the preceding discussion.

**Lemma A.6.** [19, Corollary 1 to Theorem 4] [20, Corollaries 2 & 3 to Theorem 12.7]

(i) If $W$ is finely open and either $z \in W$ or $z \in i(\Omega \setminus W)$, then $\delta_z^{\Omega \setminus W}$ does not charge polar sets.

(ii) Let $V$ be a regular finely open set, $z \in V$, and $V_z$ the fine component of $V$ containing $z$. Then $V_z$ is regular, and $\Omega \setminus V_z$ is largest among all the bases $B$ such that $\delta_z^B = \delta_z^{\Omega \setminus V}$. Moreover, the fine support of $\delta_z^{\Omega \setminus V}$ exists and

$$ (A.26) \qquad \text{supp}_f \delta_z^{\Omega \setminus V} = \partial_f V_z \subset \partial_f V. $$

(iii) Let $U$ be a fine domain and $z \in U$ or $z \in i(\Omega \setminus U)$. Then the fine support of $\delta_z^{\Omega \setminus U}$ exists and

$$ (A.27) \qquad \text{supp}_f \delta_z^{\Omega \setminus U} = b(\partial_f U) = b(\Omega \setminus U) \cap \partial_f U. $$

Let now $O \subset \Omega$ be (Euclidean) open, $z \in O$, and $O_z$ the connected component of $O$ containing $z$. If we let $V$ be the regular finely open set obtained by adjoining to $O$ the polar set $i(\Omega \setminus O)$ and $V_z$ the fine component containing $z$, then we get from [19, Theorem 6] (see discussion there) that $O_z = V_z \setminus i(\Omega \setminus O)$. Thus, since the balayage function remains the same if the set relative to which it is defined is altered by a polar set, Lemma A.6 (ii) implies that

$$ \delta_z^{\Omega \setminus O} = \delta_z^{\Omega \setminus V} = \delta_z^{\Omega \setminus V_z} = \delta_z^{\Omega \setminus O_z} $$

and the latter is carried by the regular points of $\partial O_z$. Moreover, if $O$ has compact closure in $\Omega$, then $\delta_z^{\Omega \setminus O}$ is a probability measure, and for $h$ a harmonic function in $O$ with continuous extension to $\bar{O}$:

$$ (A.28) \qquad h(z) = \int h d\delta_z^{\Omega \setminus O}, \quad z \in O. $$

Indeed, (A.28) follows from [8, Section VI.12, application 1] since $h$ is the Perron-Wiener-Brelot solution of the Dirichlet problem<fnref n="4" /> in $O$ with boundary data $h_{\lfloor \partial O}$; see [8, Section VI.6, item $\gamma$)]. Equality in (A.28) shows that the measure $\delta_z^{\Omega \setminus O}$ does not depend on $\Omega$, provided that the latter is hyperbolic and compactly contains $\bar{O}$. It is called the *harmonic measure* for $O$ (at $z$). More general versions of (A.28), involving the fine Dirichlet problem and cases where $\bar{O}$ is non-compact, are stated in Theorem A.8 and Lemma A.9 further below.

When $\Omega \subset \mathbb{C}$ and $\bar{O}$ is compact in $\Omega$, it follows from [55, Chapter II, Theorem 5.1] that

$$ (A.29) \qquad g(\sigma, O, z) = V^{\sigma}(z) - V^{\sigma^{\Omega \setminus O}}(z), \quad z \in \Omega, $$

<footnote n="4">⁴In fact, the right-hand side of (A.28) is the Perron-Wiener-Brelot solution of the Dirichlet problem on $O$ with boundary data $h$ as soon as the latter is summable against $\delta_z^{\Omega \setminus O}$ for one (and then any) $z$ in each component of $O$.</footnote>
Since $g(\sigma^E, \Omega; \cdot) \le g(\sigma, \Omega; \cdot)$ for any Radon measure $\sigma$ by the just discussed properties of balayage, it follows from Fubini's theorem that $\int v d\sigma^E \le \int v d\sigma$ for any Green potential $v$. Since any non-negative superharmonic function $v$ is an increasing limit of potentials, see [10, Lemma 1.1], monotone convergence yields that $\int v d\sigma^E \le \int v d\sigma$ remains valid for such functions as well, see also [20, Section I.3]. In particular, the mass of $\sigma^E$ cannot exceed the mass of $\sigma$.

The fine support of a Radon measure $\sigma$, denoted by $\text{supp}_f \sigma$ when it exists, is the smallest finely closed carrier of $\sigma$. A sufficient condition for its existence is that $\sigma$ does not charge polar sets, in which case $\text{supp}_f \sigma$ is its own base, see [8, Theorem VII.12]. If $\sigma$ is admissible, meaning that $g(\sigma, \Omega; \cdot) \ne +\infty$, see Section A.3, and if $\sigma(F) = 0$ for some polar set $F$, then $\sigma^E(F) = 0$ for any $E \subset \Omega$ [19, Theorem 1] (as usual, $\sigma(F)$ means the outer $\sigma$-measure of $F$ when the latter is not Borel). In particular, if $\sigma$ is admissible and $\sigma(b(E)) = 0$, then $\sigma^E$ does not charge polar sets since it is carried by $b(E)$. Thus, $\text{supp}_f \sigma^E$ exists in this case. An important special case is handled by the following lemma, the first item of which follows from the preceding discussion.

**Lemma A.6.** [19, Corollary 1 to Theorem 4] [20, Corollaries 2 & 3 to Theorem 12.7]

(i) If $W$ is finely open and either $z \in W$ or $z \in i(\Omega \setminus W)$, then $\delta_z^{\Omega \setminus W}$ does not charge polar sets.

(ii) Let $V$ be a regular finely open set, $z \in V$, and $V_z$ the fine component of $V$ containing $z$. Then $V_z$ is regular, and $\Omega \setminus V_z$ is largest among all the bases $B$ such that $\delta_z^B = \delta_z^{\Omega \setminus V}$. Moreover, the fine support of $\delta_z^{\Omega \setminus V}$ exists and

$$ (A.26) \qquad \text{supp}_f \delta_z^{\Omega \setminus V} = \partial_f V_z \subset \partial_f V. $$

(iii) Let $U$ be a fine domain and $z \in U$ or $z \in i(\Omega \setminus U)$. Then the fine support of $\delta_z^{\Omega \setminus U}$ exists and

$$ (A.27) \qquad \text{supp}_f \delta_z^{\Omega \setminus U} = b(\partial_f U) = b(\Omega \setminus U) \cap \partial_f U. $$

Let now $O \subset \Omega$ be (Euclidean) open, $z \in O$, and $O_z$ the connected component of $O$ containing $z$. If we let $V$ be the regular finely open set obtained by adjoining to $O$ the polar set $i(\Omega \setminus O)$ and $V_z$ the fine component containing $z$, then we get from [19, Theorem 6] (see discussion there) that $O_z = V_z \setminus i(\Omega \setminus O)$. Thus, since the balayage function remains the same if the set relative to which it is defined is altered by a polar set, Lemma A.6 (ii) implies that

$$ \delta_z^{\Omega \setminus O} = \delta_z^{\Omega \setminus V} = \delta_z^{\Omega \setminus V_z} = \delta_z^{\Omega \setminus O_z} $$

and the latter is carried by the regular points of $\partial O_z$. Moreover, if $O$ has compact closure in $\Omega$, then $\delta_z^{\Omega \setminus O}$ is a probability measure, and for $h$ a harmonic function in $O$ with continuous extension to $\bar{O}$:

$$ (A.28) \qquad h(z) = \int h d\delta_z^{\Omega \setminus O}, \quad z \in O. $$

Indeed, (A.28) follows from [8, Section VI.12, application 1] since $h$ is the Perron-Wiener-Brelot solution of the Dirichlet problem<fnref n="4" /> in $O$ with boundary data $h|_{\partial O}$; see [8, Section VI.6, item $\gamma$)]. Equality in (A.28) shows that the measure $\delta_z^{\Omega \setminus O}$ does not depend on $\Omega$, provided that the latter is hyperbolic and compactly contains $\bar{O}$. It is called the *harmonic measure* for $O$ (at $z$). More general versions of (A.28), involving the fine Dirichlet problem and cases where $\bar{O}$ is non-compact, are stated in Theorem A.8 and Lemma A.9 further below.

When $\Omega \subset \mathbb{C}$ and $\bar{O}$ is compact in $\Omega$, it follows from [55, Chapter II, Theorem 5.1] that

$$ (A.29) \qquad g(\sigma, O, z) = V^{\sigma}(z) - V^{\sigma^{\Omega \setminus O}}(z), \quad z \in \Omega, $$

<footnote n="4">⁴In fact, the right-hand side of (A.28) is the Perron-Wiener-Brelot solution of the Dirichlet problem on $O$ with boundary data $h$ as soon as the latter is summable against $\delta_z^{\Omega \setminus O}$ for one (and then any) $z$ in each component of $O$.</footnote>
where $V^\sigma$ is the logarithmic potential of $\sigma$ and the left-hand side is interpreted as 0 for $z \in b(\Omega \setminus O)$.
More general versions when $\infty \in \Omega$ may be found in [55]. If moreover $O$ is a domain with $K \subset O$ a non-polar compact set such that $\Omega \setminus K$ regular and $O \setminus K$ is non-thin at every point of $K$, then

$$ (A.30) \qquad \supp_t \mu_{O,K} = K. $$

Indeed, $K$ is its own base and $K = \partial_t(O \setminus K)$ by assumption, while $\mu_{O,K}$ is the balayage onto $K$ of the equilibrium measure on the plate $\partial O$ of the condenser $(\partial O, K)$ [55, Chapter VIII, Theorem 2.6]. Thus, (A.30) follows from (A.24) and (A.27).

### A.10. Green Potentials in $D$ and on $\mathcal{R}$.
In this subsection, we connect Green functions and potentials on the domain $D$ and surface $\mathcal{R}$ defined in Section 2.1. First, let us show that

$$ (A.31) \qquad g_D(x, y) = \sum_{z \in p^{-1}(x)} m(z)g_{\mathcal{R}}(z, w), $$

where $m(z)$ is the ramification order of $\mathcal{R}$ at $z$ and $w$ is an arbitrary element of the fiber $p^{-1}(y)$. To check (A.31), note that for fixed $y \notin p(\mathbf{rp}(\mathcal{R}))$ and $w \in \mathcal{R}$ with $p(w) = y$, the right-hand side is well defined and harmonic as a function of $x \in D \setminus (p(\mathbf{rp}(\mathcal{R})) \cup \{y\})$. Thus, it is harmonic for $x \in D \setminus \{y\}$ by the Removability Theorem and the continuity of Green functions off the diagonal. Moreover, the right-hand side clearly has a logarithmic singularity at $y$, and since $\lim_{z \to \partial \mathcal{R}} g_{\mathcal{R}}(z, w) = 0$ by the regularity of $\mathcal{R}$ in $\mathcal{R}_*$, its largest harmonic minorant is zero. This proves (A.31) when $y \notin p(\mathbf{rp}(\mathcal{R}))$, and the general case follows by continuity of Green functions off the diagonal. Consequently, if $v$ is a Radon measure on $\mathcal{R}$ and $p_*(v)$ denotes its pushforward under $p$ (the measure on $D$ such that $p_*(v)(B) = v(p^{-1}(B))$ for a Borel set $B$), integrating (A.31) against $p_*(v)$ with respect to $y$ and changing variables yields

$$ (A.32) \qquad g(p_*(v), D; x) = \sum_{z \in p^{-1}(x)} m(z)g(v, \mathcal{R}; z). $$

In the other direction, for a Radon measure $\sigma$ on $D$, let us define $\hat{\sigma}$ by

$$ (A.33) \qquad \hat{\sigma}(B) := \int_{p(B)} \sum_{z \in p^{-1}(x) \cap B} m(z) d\sigma(x), \quad B \subset \mathcal{R}, \quad B \text{ Borel.} $$

As $p(B)$ is Borel when $B$ is Borel, one easily checks that (A.33) defines a Radon measure on $\mathcal{R}$. In fact, one can verify that $\hat{\sigma} = \sigma^* + \sum_{z \in p(\mathcal{R})} m(z)\sigma(\{p(z)\})\delta_z$, where $\sigma^*$ is the pullback measure resulting from Carathéodory's construction as applied to the map $B \mapsto \sigma(p(B) \setminus p(\mathbf{rp}(\mathcal{R})))$ defined on Borel subsets of $\mathcal{R}$, see [16, Theorem 2.10.10].

Partitioning $D \setminus p(\mathbf{rp}(\mathcal{R}))$ into countably many Borel sets $B_k$ such that $p : p^{-1}(B_k) \to B_k$ induces a homeomorphism on each connected component of $p^{-1}(B_k)$, and invoking the Removability Theorem to proceed by superharmonicity from the case where $\zeta \notin p(\mathbf{rp}(\mathcal{R}))$, one deduces from (A.33) and (A.31) that

$$ (A.34) \qquad g(\hat{\sigma}, \mathcal{R}; \zeta) = g(\sigma, D; p(\zeta)), \quad \zeta \in \mathcal{R}. $$

As a consequence of definition (A.33), we claim that if a sequence $\{\mu_n\}$ of finite positive measures supported on a fixed compact set $K \subset D$ converges weak* to $\mu$ on $D$, then the sequence $\{\hat{\mu}_n\}$ converges weak* to $\hat{\mu}$ on $\mathcal{R}$. Indeed, the total mass of $\mu_n$ is necessarily bounded independently of $n$ by some $C > 0$ (this follows from the Banach-Steinhaus principle) and therefore, in view of (A.33), the total mass of $\hat{\mu}_n$ is bounded by $MC$, where $M$ is the number of sheets of $\mathcal{R}$. Hence, an arbitrary subsequence of $\{\hat{\mu}_n\}$ has a subsequence, say $\{\hat{\mu}_{n_k}\}$, that converges weak* on $\mathcal{R}$ to some finite measure, say $s$. It follows from the Lower Envelope Theorem that $\liminf_k g(\hat{\mu}_{n_k}, \mathcal{R}; z) = g(s, \mathcal{R}; z)$ for quasi every $z \in \mathcal{R}$. Similarly, (A.34) and the Lower Envelope Theorem, applied this time to $\{\mu_{n_k}\}$, yield that $\liminf_k g(\hat{\mu}_{n_k}, \mathcal{R}; z) = g(\mu, D; p(z)) = g(\hat{\mu}, \mathcal{R}; z)$ for quasi every $z \in \mathcal{R}$. Thus,
where $V^\sigma$ is the logarithmic potential of $\sigma$ and the left-hand side is interpreted as 0 for $z \in b(\Omega \setminus O)$.
More general versions when $\infty \in \Omega$ may be found in [55]. If moreover $O$ is a domain with $K \subset O$ a
non-polar compact set such that $\Omega \setminus K$ regular and $O \setminus K$ is non-thin at every point of $K$, then

$$
(A.30) \qquad \supp_t \mu_{O,K} = K.
$$

Indeed, $K$ is its own base and $K = \partial_t(O\setminus K)$ by assumption, while $\mu_{O,K}$ is the balayage onto $K$ of the equilibrium measure on the plate $\partial O$ of the condenser $(\partial O, K)$ [55, Chapter VIII, Theorem 2.6]. Thus, (A.30) follows from (A.24) and (A.27).

A.10. **Green Potentials in D and on R.** In this subsection, we connect Green functions and potentials on the domain D and surface R defined in Section 2.1. First, let us show that

$$
(A.31) \qquad g_D(x, y) = \sum_{z \in p^{-1}(x)} m(z)g_R(z, w),
$$

where $m(z)$ is the ramification order of $\mathcal{R}$ at $z$ and $w$ is an arbitrary element of the fiber $p^{-1}(y)$. To check (A.31), note that for fixed $y \notin p(\mathbf{rp}(\mathcal{R}))$ and $w \in \mathcal{R}$ with $p(w) = y$, the right-hand side is well defined and harmonic as a function of $x \in D \setminus (p(\mathbf{rp}(\mathcal{R})) \cup \{y\})$. Thus, it is harmonic for $x \in D \setminus \{y\}$ by the *Removability Theorem* and the continuity of Green functions off the diagonal. Moreover, the right-hand side clearly has a logarithmic singularity at $y$, and since $\lim_{z \to \partial \mathcal{R}} g_{\mathcal{R}}(z, w) = 0$ by the regularity of $\mathcal{R}$ in $\mathcal{R}_*$, its largest harmonic minorant is zero. This proves (A.31) when $y \notin p(\mathbf{rp}(\mathcal{R}))$, and the general case follows by continuity of Green functions off the diagonal. Consequently, if $v$ is a Radon measure on $\mathcal{R}$ and $p_*(v)$ denotes its pushforward under $p$ (the measure on $D$ such that $p_*(v)(B) = v(p^{-1}(B))$ for a Borel set $B$), integrating (A.31) against $p_*(v)$ with respect to $y$ and changing variables yields

$$
(A.32) \qquad g(p_*(v), D; x) = \sum_{z \in p^{-1}(x)} m(z)g(v, \mathcal{R}; z).
$$

In the other direction, for a Radon measure $\sigma$ on $D$, let us define $\hat{\sigma}$ by

$$
(A.33) \qquad \hat{\sigma}(B) := \int_{p(B)} \sum_{z \in p^{-1}(x) \cap B} m(z) d\sigma(x), \quad B \subset \mathcal{R}, \quad B \text{ Borel.}
$$

As $p(B)$ is Borel when $B$ is Borel, one easily checks that (A.33) defines a Radon measure on $\mathcal{R}$. In fact, one can verify that $\hat{\sigma} = \sigma^* + \sum_{z \in rp(\mathcal{R})} m(z)\sigma(\{p(z)\})\delta_z$, where $\sigma^*$ is the pullback measure resulting from Carathéodory's construction as applied to the map $B \mapsto \sigma(p(B) \setminus p rp(\mathcal{R}))$ defined on Borel subsets of $\mathcal{R}$, see [16, Theorem 2.10.10].

Partitioning $D \setminus p(\mathbf{rp}(\mathcal{R}))$ into countably many Borel sets $B_k$ such that $p : p^{-1}(B_k) \to B_k$
induces a homeomorphism on each connected component of $p^{-1}(B_k)$, and invoking the *Removability Theorem* to proceed by superharmonicity from the case where $\zeta \notin \mathbf{rp}(\mathcal{R})$, one deduces from (A.33)
and (A.31) that

$$
(A.34) \qquad g(\hat{\sigma}, \mathcal{R}; \zeta) = g(\sigma, D; p(\zeta)), \quad \zeta \in \mathcal{R}.
$$

As a consequence of definition (A.33), we claim that if a sequence $\{\mu_n\}$ of finite positive measures supported on a fixed compact set $K \subset D$ converges weak* to $\mu$ on $D$, then the sequence $\{\hat{\mu}_n\}$ converges weak* to $\hat{\mu}$ on $\mathcal{R}$. Indeed, the total mass of $\mu_n$ is necessarily bounded independently of $n$ by some $C > 0$ (this follows from the Banach-Steinhaus principle) and therefore, in view of (A.33), the total mass of $\hat{\mu}_n$ is bounded by $MC$, where $M$ is the number of sheets of $\mathcal{R}$. Hence, an arbitrary subsequence of $\{\hat{\mu}_n\}$ has a subsequence, say $\{\hat{\mu}_{n_k}\}$, that converges weak* on $\mathcal{R}$ to some finite measure, say $s$. It follows from the *Lower Envelope Theorem* that $\liminf_k g(\hat{\mu}_{n_k}, \mathcal{R}; z) = g(s, \mathcal{R}; z)$ for quasi every $z \in \mathcal{R}$. Similarly, (A.34) and the *Lower Envelope Theorem*, applied this time to $\{\mu_{n_k}\}$, yield that $\liminf_k g(\hat{\mu}_{n_k}, \mathcal{R}; z) = g(\mu, D; p(z)) = g(\hat{\mu}, \mathcal{R}; z)$ for quasi every $z \in \mathcal{R}$. Thus,
$g(s, \mathcal{R}; \cdot) = g(\hat{\mu}, \mathcal{R}; \cdot)$ quasi everywhere on $\mathcal{R}$, and the claim follows by taking Laplacians on both sides of this equality.

In the previous construction, $\mathcal{R}$ may of course be replaced by another saturated connected bordered surface $S \subset \mathcal{R}_*$ with bounded projection such that $\bar{\mathcal{R}} \subset S$. Therefore,

$$ (\mu_n \xrightarrow{w^*} \mu \text{ in } \bar{D}) \Rightarrow (\hat{\mu}_n \xrightarrow{w^*} \hat{\mu} \text{ in } \bar{\mathcal{R}}), $$

because $\{\mu_n\}$ also converges weak* in $p(S) \supset \bar{D}$ whence the measures $\hat{\mu}_n$ converge weak* on $S$, while having their supports contained in $\bar{\mathcal{R}}$. The “hat measure” constructed in (A.33) is instrumental both in the proof of Lemma 3.9 and of the following technical result, needed in the paper.

**Lemma A.7.** Let $\sigma$ be a measure in $D$. Given $E \subset D$, set $\hat{E} := p^{-1}(E)$. Then it holds that

$$ g(\sigma^E, D; p(z)) = \int g(\sigma, D; p(x)) d\delta_z^{\hat{E}}(x). $$

*Proof.* If $u$ is superharmonic on $D$, it is obvious from (A.34) that

$$ (u(z) \ge g(\sigma, D; z), \quad z \in E) \Rightarrow (u(p(\xi)) \ge g(\hat{\sigma}^E, \mathcal{R}; \xi), \quad \xi \in \hat{E}). $$

As $u \circ p$ is superharmonic on $\mathcal{R}$, this and the definition of balayage imply that

$$ (A.35) \qquad g(\sigma^E, D; p(\zeta)) \ge g(\hat{\sigma}^E, \mathcal{R}; \zeta), \quad \zeta \in \mathcal{R}. $$

Conversely, averaging over $\zeta \in p^{-1}(\{z\})$, equation (A.34) yields that

$$ (A.36) \qquad g(\sigma, D; z) = \frac{1}{M} \sum_{\zeta \in p^{-1}(\{z\})} m(\zeta)g(\hat{\sigma}, \mathcal{R}; \zeta), \quad z \in D, $$

with $M$ being the total number of sheets of $\mathcal{R}$, entailing when $v$ is superharmonic on $\mathcal{R}$ that

$$ (v(\zeta) \ge g(\hat{\sigma}, \mathcal{R}; \zeta), \quad \zeta \in \hat{E}) \Rightarrow \left( \frac{1}{M} \sum_{\zeta \in p^{-1}(z)} m(\zeta)v(\zeta) \ge g(\sigma, D; z), \quad z \in D \right). $$

Now, the function $z \mapsto \sum_{\zeta \in p^{-1}(z)} m(\zeta)v(\zeta)$ is well-defined and superharmonic on $D \setminus p(\mathbf{rp}(\mathcal{R}))$ and therefore on the whole domain $D$ by the *Removability Theorem*. Thus, by the definition of balayage, we obtain when $v$ is superharmonic on $\mathcal{R}$ that

$$ (v(\zeta) \ge g(\hat{\sigma}, \mathcal{R}; \zeta), \quad \zeta \in \hat{E}) \Rightarrow \left( \frac{1}{M} \sum_{\zeta \in p^{-1}(z)} m(\zeta)v(\zeta) \ge g(\sigma^E, D; z), \quad z \in D \right), $$

and taking the infimum over $v$ before taking the lower semi-continuous regularization gives us, by virtue of the *Strong Domination Principle*, that

$$ \frac{1}{M} \sum_{\zeta \in p^{-1}(z)} m(\zeta) g(\hat{\sigma}^E, \mathcal{R}; \zeta) \ge g(\sigma^E, D; z), \quad z \in D. $$

Combining the above estimate with (A.35), we deduce that

$$ g(\sigma^E, D; p(z)) = g(\hat{\sigma}^E, \mathcal{R}; z), \quad z \in \mathcal{R}, $$

and the conclusion now follows from (A.25) and (A.34). □
$g(s, \mathcal{R}; \cdot) = g(\hat{\mu}, \mathcal{R}; \cdot)$ quasi everywhere on $\mathcal{R}$, and the claim follows by taking Laplacians on both sides of this equality.

In the previous construction, $\mathcal{R}$ may of course be replaced by another saturated connected bordered surface $S \subset \mathcal{R}_*$ with bounded projection such that $\bar{\mathcal{R}} \subset S$. Therefore,

$$ (\mu_n \xrightarrow{w^*} \mu \text{ in } \bar{D}) \Rightarrow (\hat{\mu}_n \xrightarrow{w^*} \hat{\mu} \text{ in } \bar{\mathcal{R}}), $$

because $\{\mu_n\}$ also converges weak* in $p(S) \supset \bar{D}$ whence the measures $\hat{\mu}_n$ converge weak* on $S$, while having their supports contained in $\bar{\mathcal{R}}$. The “hat measure” constructed in (A.33) is instrumental both in the proof of Lemma 3.9 and of the following technical result, needed in the paper.

**Lemma A.7.** Let $\sigma$ be a measure in $D$. Given $E \subset D$, set $\hat{E} := p^{-1}(E)$. Then it holds that

$$ g(\sigma^E, D; p(z)) = \int g(\sigma, D; p(x)) d\delta_z^{\hat{E}}(x). $$

*Proof.* If $u$ is superharmonic on $D$, it is obvious from (A.34) that

$$ (u(z) \ge g(\sigma, D; z), \quad z \in E) \Rightarrow (u(p(\xi)) \ge g(\hat{\sigma}^E, \mathcal{R}; \xi), \quad \xi \in \hat{E}). $$

As $u \circ p$ is superharmonic on $\mathcal{R}$, this and the definition of balayage imply that

$$ (A.35) \qquad g(\sigma^E, D; p(\zeta)) \ge g(\hat{\sigma}^E, \mathcal{R}; \zeta), \quad \zeta \in \mathcal{R}. $$

Conversely, averaging over $\zeta \in p^{-1}(\{z\})$, equation (A.34) yields that

$$ (A.36) \qquad g(\sigma, D; z) = \frac{1}{M} \sum_{\zeta \in p^{-1}(\{z\})} m(\zeta) g(\hat{\sigma}, \mathcal{R}; \zeta), \quad z \in D, $$

with $M$ being the total number of sheets of $\mathcal{R}$, entailing when $v$ is superharmonic on $\mathcal{R}$ that

$$ (v(\zeta) \ge g(\hat{\sigma}, \mathcal{R}; \zeta), \quad \zeta \in \hat{E}) \Rightarrow \left( \frac{1}{M} \sum_{\zeta \in p^{-1}(z)} m(\zeta) v(\zeta) \ge g(\sigma, D; z), \quad z \in D \right). $$

Now, the function $z \mapsto \sum_{\zeta \in p^{-1}(z)} m(\zeta)v(\zeta)$ is well-defined and superharmonic on $D \setminus p(\mathbf{rp}(\mathcal{R}))$ and therefore on the whole domain $D$ by the *Removability Theorem*. Thus, by the definition of balayage, we obtain when $v$ is superharmonic on $\mathcal{R}$ that

$$ (v(\zeta) \ge g(\hat{\sigma}, \mathcal{R}; \zeta), \quad \zeta \in \hat{E}) \Rightarrow \left( \frac{1}{M} \sum_{\zeta \in p^{-1}(z)} m(\zeta) v(\zeta) \ge g(\sigma^E, D; z), \quad z \in D \right), $$

and taking the infimum over $v$ before taking the lower semi-continuous regularization gives us, by virtue of the *Strong Domination Principle*, that

$$ \frac{1}{M} \sum_{\zeta \in p^{-1}(z)} m(\zeta) g(\hat{\sigma}^E, \mathcal{R}; \zeta) \ge g(\sigma^E, D; z), \quad z \in D. $$

Combining the above estimate with (A.35), we deduce that

$$ g(\sigma^E, D; p(z)) = g(\hat{\sigma}^E, \mathcal{R}; z), \quad z \in \mathcal{R}, $$

and the conclusion now follows from (A.25) and (A.34). □
**A.11. Dirichlet problem.** The Dirichlet problem on a domain consists in finding a harmonic function in that domain with given boundary data. In the fine Dirichlet problem, one looks for a finely harmonic function on a fine domain to meet prescribed boundary data. A real-valued function $h$ on a fine domain $V$ is finely harmonic if it is finely continuous, and if the fine topology on $V$ has a basis comprised of finely open sets $E$ with $\text{clos}_f(E) \subset V$ such that

$$h(z) = \int hd\delta_z^{\Omega, E} \quad \text{for every} \quad z \in E$$

(in particular $h$ must be integrable with respect to $\delta_z^{\Omega, E}$ for all $z \in E$ and each $E$); one may even assume that $E$ is regular and has compact closure (with respect to the Euclidean topology) in $V$, see [20, Sections 8 & 14]. Note that a function harmonic in a domain is finely harmonic on any fine subdomain, see [20, Theorem 8.7].

If $V$ is a regular finely open set (recall that it means $\Omega \setminus V$ is its own base), then $b(\partial_f V) = \partial_f V$ by (A.10) whence the result below is a special case of [20, Theorem 14.1] and its proof.

**Theorem A.8.** Let $V \subset \Omega$ be a finely open set such that $\Omega \setminus V$ is non-thin at every point of itself, i.e., such that $V$ is regular. If $\psi$ is a finely continuous function on $\partial_f V$, majorized in absolute value there by a finite semi-bounded potential on $\Omega$, say $g$, then

$$(A.37) \qquad h_{\psi}(z) := \int \psi d\delta_z^{\Omega \setminus V} = \int \psi d\delta_z^{\partial_f V}, \quad z \in V,$$

is the unique finely continuous extension of $\psi$ to $\text{clos}_f(V)$ that is finely harmonic in $V$ and is majorized in absolute value there by a semi-bounded potential. In fact, it holds that $|h_{\psi}| \le g$ on $\text{clos}_f(V)$.

The lemma below addresses the question as to when constant functions solve the fine Dirichlet problem on $V \subset \Omega$ or, equivalently by (A.37), as to when the balayage of $\delta_z$ out of $V$, $z \in V$, has unit mass. We recall that an overline, as in $\overline{V}$, or a $\partial$ sign, as in $\partial\Omega$, refer respectively to the closure and boundary with respect to the Euclidean topology induced by the ambient Riemannian manifold ($\mathcal{R}_*$ or $\mathbb{C}$). In contrast, fine closures and fine boundaries as in $\partial_f V$ and $\text{clos}_f(V)$ refer to the fine topology on $\Omega$; thus, $\partial\Omega$ is “invisible” from the point of view of fine topology in $\Omega$, and if $V \subset \Omega$ then $\partial_f V \cap \partial\Omega$ is disjoint from $\partial_f V$ as the latter is included in $\Omega$.

**Lemma A.9.** Let $V$ be a proper nonempty regular fine domain in $\Omega$, which itself is regular within the ambient Riemann surface ($\mathcal{R}_*$ or $\mathbb{C}$). Then it holds for $z \in V$ that

$$(A.38) \qquad \int d\delta_z^{\Omega \setminus V} \begin{cases} = 1 & \text{if } \overline{V} \cap \partial\Omega = \emptyset, \\ < 1 & \text{if } \overline{\partial_f V} \cap \partial\Omega = \emptyset \text{ and } \overline{V} \cap \partial\Omega \neq \emptyset. \end{cases}$$

Moreover, if either condition on the right-hand side of (A.38) holds, then for any harmonic function $h$ on $\Omega$ one has

$$(A.39) \qquad h(z) = \int hd\delta_z^{\Omega \setminus V}, \quad z \in V,$$

provided that $|h|$ is majorized on $V$ by a semi-bounded potential in $\Omega$ when $\overline{V} \cap \partial\Omega \neq \emptyset$.

*Proof.* Let $K \subset \Omega$ be non-polar and compact. Set, for brevity, $g_K := \text{cap}_\Omega(K)g(\mu_{\Omega,K}, \Omega; \cdot)$, where $\mu_{\Omega,K}$ indicates, as in Section A.4, the Green equilibrium distribution on $K$. Since $\mu_{\Omega,K}$ has finite energy, $g_K$ is semi-bounded. As $\partial\Omega$ is regular in the ambient Riemann surface, $g_K$ extends continuously by zero to $\partial\Omega$, see Section A.7. Since $\text{cap}_\Omega(K) > 0$, it holds that $g_K \le 1$ in $\Omega$, see the paragraph after (A.5), and $g_K = 1$ on $b(K)$ because $g_K = \mathcal{B}_1^K$, see Section A.9. Moreover, $g_K < 1$ in each connected component $U$ of $\Omega \setminus K$ such that $\partial U \cap \partial\Omega \neq \emptyset$ by the maximum principle for harmonic functions.

When $\overline{V}$ is compactly included in $\Omega$, we may put $K := \overline{V}$ in what precedes, and then $\text{cap}_\Omega(K) > 0$ as otherwise $V$ itself would be polar and therefore empty, since it is finely open. The infimum of $g_K$
**A.11. Dirichlet problem.** The Dirichlet problem on a domain consists in finding a harmonic function in that domain with given boundary data. In the fine Dirichlet problem, one looks for a finely harmonic function on a fine domain to meet prescribed boundary data. A real-valued function $h$ on a fine domain $V$ is finely harmonic if it is finely continuous, and if the fine topology on $V$ has a basis comprised of finely open sets $E$ with $\text{clos}_f(E) \subset V$ such that

$$h(z) = \int hd\delta_z^{\Omega,E} \quad \text{for every} \quad z \in E$$

(in particular $h$ must be integrable with respect to $\delta_z^{\Omega,E}$ for all $z \in E$ and each $E$); one may even assume that $E$ is regular and has compact closure (with respect to the Euclidean topology) in $V$, see [20, Sections 8 & 14]. Note that a function harmonic in a domain is finely harmonic on any fine subdomain, see [20, Theorem 8.7].

If $V$ is a regular finely open set (recall that it means $\Omega\setminus V$ is its own base), then $b(\partial_f V) = \partial_f V$ by (A.10) whence the result below is a special case of [20, Theorem 14.1] and its proof.

**Theorem A.8.** Let $V \subset \Omega$ be a finely open set such that $\Omega\setminus V$ is non-thin at every point of itself, i.e., such that $V$ is regular. If $\psi$ is a finely continuous function on $\partial_f V$, majorized in absolute value there by a finite semi-bounded potential on $\Omega$, say $g$, then

$$(A.37) \qquad h_{\psi}(z) := \int \psi d\delta_z^{\Omega\setminus V} = \int \psi d\delta_z^{\partial_f V}, \quad z \in V,$$

is the unique finely continuous extension of $\psi$ to $\text{clos}_f(V)$ that is finely harmonic in $V$ and is majorized in absolute value there by a semi-bounded potential. In fact, it holds that $|h_{\psi}| \le g$ on $\text{clos}_f(V)$.

The lemma below addresses the question as to when constant functions solve the fine Dirichlet problem on $V \subset \Omega$ or, equivalently by (A.37), as to when the balayage of $\delta_z$ out of $V$, $z \in V$, has unit mass. We recall that an overline, as in $\overline{V}$, or a $\partial$ sign, as in $\partial\Omega$, refer respectively to the closure and boundary with respect to the Euclidean topology induced by the ambient Riemannian manifold ($\mathcal{R}_*$ or $\mathbb{C}$). In contrast, fine closures and fine boundaries as in $\partial_f V$ and $\text{clos}_f(V)$ refer to the fine topology on $\Omega$; thus, $\partial\Omega$ is “invisible” from the point of view of fine topology in $\Omega$, and if $V \subset \Omega$ then $\partial_f V \cap \partial\Omega$ is disjoint from $\partial_f V$ as the latter is included in $\Omega$.

**Lemma A.9.** Let $V$ be a proper nonempty regular fine domain in $\Omega$, which itself is regular within the ambient Riemann surface ($\mathcal{R}_*$ or $\mathbb{C}$). Then it holds for $z \in V$ that

$$(A.38) \qquad \int d\delta_z^{\Omega\setminus V} \begin{cases} = 1 & \text{if } \overline{V} \cap \partial\Omega = \emptyset, \\ < 1 & \text{if } \overline{\partial_f V} \cap \partial\Omega = \emptyset \text{ and } \overline{V} \cap \partial\Omega \neq \emptyset. \end{cases}$$

Moreover, if either condition on the right-hand side of (A.38) holds, then for any harmonic function $h$ on $\Omega$ one has

$$(A.39) \qquad h(z) = \int hd\delta_z^{\Omega\setminus V}, \quad z \in V,$$

provided that $|h|$ is majorized on $V$ by a semi-bounded potential in $\Omega$ when $\overline{V} \cap \partial\Omega \neq \emptyset$.

*Proof.* Let $K \subset \Omega$ be non-polar and compact. Set, for brevity, $g_K := \text{cap}_{\Omega}(K)g(\mu_{\Omega,K}, \Omega; \cdot)$, where $\mu_{\Omega,K}$ indicates, as in Section A.4, the Green equilibrium distribution on $K$. Since $\mu_{\Omega,K}$ has finite energy, $g_K$ is semi-bounded. As $\partial\Omega$ is regular in the ambient Riemann surface, $g_K$ extends continuously by zero to $\partial\Omega$, see Section A.7. Since $\text{cap}_{\Omega}(K) > 0$, it holds that $g_K \le 1$ in $\Omega$, see the paragraph after (A.5), and $g_K = 1$ on $b(K)$ because $g_K = \mathcal{B}_1^K$, see Section A.9. Moreover, $g_K < 1$ in each connected component $U$ of $\Omega\setminus K$ such that $\partial U \cap \partial\Omega \neq \emptyset$ by the maximum principle for harmonic functions.

When $\overline{V}$ is compactly included in $\Omega$, we may put $K := \overline{V}$ in what precedes, and then $\text{cap}_{\Omega}(K) > 0$ as otherwise $V$ itself would be polar and therefore empty, since it is finely open. The infimum of $g_K$
on $K$ is attained by lower semi-continuity, and it is strictly positive because nonzero Green potentials are never zero. Therefore, if $h$ is harmonic on $\Omega$, the potential $cg_K$ majorizes $|h|$ on $K$ for sufficiently large $c > 0$. The uniqueness part of Theorem A.8 now implies that $h_{|V}$ is the solution of the fine Dirichlet problem with boundary data $h_{|\partial_t V}$. Hence, (A.39) is just (A.37) while the upper equality in (A.38) follows by taking $h \equiv 1$.

Assume next that $\bar{V} \cap \partial\Omega \neq \emptyset$ and $\overline{\partial_t V} \cap \partial\Omega = \emptyset$. Then $K := \overline{\partial_t V}$ is a compact subset of $\Omega$ which is non-polar, for if $\partial_t V$ were polar, then either $V$ or $\Omega \setminus V$ would be polar [19, Theorem 2] and $V$ would be either empty or irregular, a contradiction. Note that a subdomain of $\Omega \setminus K$ is also a fine domain [19, Theorem 2] and thus, if it contains both a point in $V$ and a point in $\Omega \setminus V$, then it must contain a point in $\partial_t V$ which is impossible by the definition of $K$. Hence, $V \setminus K$ is Euclidean open.

Let $U$ be a connected component of $V \setminus K$ such that $\partial U \cap \partial \Omega \neq \emptyset$; it exists because $\bar{V} \cap \partial \Omega \neq \emptyset$. Since $\partial_t V = b(\partial_t V) \subseteq b(K)$ by assumption, $g_K = 1$ on $\partial_t V$ (see the beginning of the proof). Hence, as $g_K$ is a semi-bounded potential on $\Omega$, it follows from Theorem A.8 that $h_1 \le g_K$ in $V$, where $h_1(z)$ is the solution of the fine Dirichlet problem on $V$ with boundary data identically 1 on $\partial_t V$, see (A.37). In particular, $h_1 \le g_K < 1$ in $U$ by the maximum principle for harmonic functions.

Let $F \subset U$ be a closed disk. Observe that $V \setminus F$ is a fine domain as otherwise $F$ would finely disconnect $U$ whereas $U \setminus F$ is a domain and therefore also a fine domain. Put $\delta_z^* := \delta_z^{(\Omega \setminus V) \cup F}$, $z \in V \setminus F$, and observe that $\delta_{z \perp \partial_t F}^*$ is a non-trivial measure by (A.27) and since $\partial_t ((\Omega \setminus V) \cup F) = \partial_t V \cup \partial_t F$, where the union is disjoint and $b(\partial_t F) = \partial_t F = \partial F$. We now get from (A.23) and (A.27), applied to the fine domain $V \setminus F$, that

$$ (A.40) \qquad \delta_z^{\Omega \setminus V} = (\delta_z^*)^{\Omega \setminus V} = (\delta_{z \perp \partial_t V}^*)^{\Omega \setminus V} + (\delta_{z \perp \partial_t F}^*)^{\Omega \setminus V} = \delta_{z \perp \partial_t V}^* + (\delta_{z \perp \partial_t F}^*)^{\Omega \setminus V}, $$

where we observe that the balayage out of $V$ does not change measures supported on $b(\partial_t V) = \partial_t V$ by (A.22). Since $h_1 < 1$ on $\partial_t F$ as $\partial_t F \subset U$, we get from (A.24) and (A.26) that

$$ 0 < (\delta_{z \perp \partial_t F}^*)^{\Omega \setminus V}(\partial_t V) = \int \delta_x^{\Omega \setminus V}(\partial_t V) d\delta_{z \perp \partial_t F}^*(x) = \int h_1(x) d\delta_{z \perp \partial_t F}^*(x) < \delta_{z \perp \partial_t F}^*(\partial_t F) $$

which implies, in view of (A.40), that $h_1(z) = \delta_z^{\Omega \setminus V}(\partial_t V) < \delta_z^*((\Omega \setminus V) \cup F) \le 1$ as claimed.

Finally, let $h$ be a harmonic function on $\Omega$ which is majorized on $V$ in absolute value by a semi-bounded potential. Because $h$ is also finely harmonic on $V$ and finely continuous on $\partial_{\mathrm{clo}}(V)$, we deduce from Theorem A.8 that it is the solution of the fine Dirichlet problem on $V$ with boundary data $h_{|\partial_t V}$ and that (A.39) holds. $\square$

**Lemma A.10.** Let $V \subset \mathcal{R}$ be a proper regular fine domain such that $\overline{\partial_t V} \cap \partial \mathcal{R} = \emptyset$ and $\bar{V} \cap \partial \mathcal{R} \neq \emptyset$ (the lower assumption on the right-hand side of (A.38) when $\Omega = \mathcal{R}$). Let further $h$ be a harmonic function in $\mathcal{R}$ such that $\lim_{z \to \xi} h(z) = 0$ for every $\xi \in \bar{V} \cap \partial \mathcal{R}$. Then identity (A.39) holds.

*Proof.* In view of Lemma A.9, it is enough to show that $h$ is majorized on $V$ by a semi-bounded potential. Note, as in the proof of Lemma A.9, that $\partial_t V$ is non-polar. Let us show that $\partial \mathcal{R} \cap \bar{V}$ consists of a union of connected components of $\partial \mathcal{R}$. Indeed, any such component $\Gamma$ is a 1-dimensional compact topological submanifold of $\mathcal{R}_*$, and as such it has a tubular neighborhood $N$ that may be chosen so thin that $N \cap \overline{\partial_t V} = \emptyset$. Then, if $\zeta_1, \zeta_2 \in \Gamma$ and $\zeta_1 \in \bar{V}$ while $\zeta_2 \notin \bar{V}$, we can find $z_1 \in N \cap V$ close to $\zeta_1$ and $z_2 \in N \cap \Omega \setminus V$ close to $\zeta_2$. The points $z_1, z_2$ can be joined by a smooth arc contained in $N$. However, such an arc is finely connected [19, Theorem 7], but cannot meet $\overline{\partial_t V}$ by construction, a contradiction that proves our claim.

Assume first that $D = \mathbb{D}$ is the unit disk. Any function $u$ harmonic in an annular region $\{r < |z| < 1\}$ that extends continuously to $\mathbb{T}$ by zero can be harmonically extended to $\{r < |z| < 1/r\}$ by reflection, i.e., by setting $u(z) := -u(1/\bar{z})$ for $z \in \{1 < |z| < 1/r\}$. Due to the smoothness of this extension it necessarily holds that $|u(z)| \le C_\rho(1 - |z|)$ for $r < \rho \le |z| \le 1$. As $h$ is harmonic on $\mathcal{R}$, this principle used around each of the finitely many connected components of $\bar{V} \cap \partial \mathcal{R}$ yields
on $K$ is attained by lower semi-continuity, and it is strictly positive because nonzero Green potentials are never zero. Therefore, if $h$ is harmonic on $\Omega$, the potential $cg_K$ majorizes $|h|$ on $K$ for sufficiently large $c > 0$. The uniqueness part of Theorem A.8 now implies that $h_{|V}$ is the solution of the fine Dirichlet problem with boundary data $h_{|\partial_t V}$. Hence, (A.39) is just (A.37) while the upper equality in (A.38) follows by taking $h \equiv 1$.

Assume next that $\overline{V} \cap \partial\Omega \neq \emptyset$ and $\overline{\partial_t V} \cap \partial\Omega = \emptyset$. Then $K := \overline{\partial_t V}$ is a compact subset of $\Omega$ which is non-polar, for if $\partial_t V$ were polar, then either $V$ or $\Omega \setminus V$ would be polar [19, Theorem 2] and $V$ would be either empty or irregular, a contradiction. Note that a subdomain of $\Omega \setminus K$ is also a fine domain [19, Theorem 2] and thus, if it contains both a point in $V$ and a point in $\Omega \setminus V$, then it must contain a point in $\partial_t V$ which is impossible by the definition of $K$. Hence, $V \setminus K$ is Euclidean open.

Let $U$ be a connected component of $V \setminus K$ such that $\partial U \cap \partial \Omega \neq \emptyset$; it exists because $\overline{V} \cap \partial \Omega \neq \emptyset$. Since $\partial_t V = b(\partial_t V) \subseteq b(K)$ by assumption, $g_K = 1$ on $\partial_t V$ (see the beginning of the proof). Hence, as $g_K$ is a semi-bounded potential on $\Omega$, it follows from Theorem A.8 that $h_1 \le g_K$ in $V$, where $h_1(z)$ is the solution of the fine Dirichlet problem on $V$ with boundary data identically 1 on $\partial_t V$, see (A.37). In particular, $h_1 \le g_K < 1$ in $U$ by the maximum principle for harmonic functions.

Let $F \subset U$ be a closed disk. Observe that $V \setminus F$ is a fine domain as otherwise $F$ would finely disconnect $U$ whereas $U \setminus F$ is a domain and therefore also a fine domain. Put $\delta_z^* := \delta_z^{(\Omega \setminus V) \cup F}$, $z \in V \setminus F$, and observe that $\delta_{z \perp \partial_t F}^*$ is a non-trivial measure by (A.27) and since $\partial_t ((\Omega \setminus V) \cup F) = \partial_t V \cup \partial_t F$, where the union is disjoint and $b(\partial_t F) = \partial_t F = \partial F$. We now get from (A.23) and (A.27), applied to the fine domain $V \setminus F$, that

$$ (A.40) \qquad \delta_z^{\Omega \setminus V} = (\delta_z^*)^{\Omega \setminus V} = (\delta_{z \perp \partial_t V}^*)^{\Omega \setminus V} + (\delta_{z \perp \partial_t F}^*)^{\Omega \setminus V} = \delta_{z \perp \partial_t V}^* + (\delta_{z \perp \partial_t F}^*)^{\Omega \setminus V}, $$

where we observe that the balayage out of $V$ does not change measures supported on $b(\partial_t V) = \partial_t V$ by (A.22). Since $h_1 < 1$ on $\partial_t F$ as $\partial_t F \subset U$, we get from (A.24) and (A.26) that

$$ 0 < (\delta_{z \perp \partial_t F}^*)^{\Omega \setminus V}(\partial_t V) = \int \delta_x^{\Omega \setminus V}(\partial_t V) d\delta_{z \perp \partial_t F}^*(x) = \int h_1(x) d\delta_{z \perp \partial_t F}^*(x) < \delta_{z \perp \partial_t F}^*(\partial_t F) $$

which implies, in view of (A.40), that $h_1(z) = \delta_z^{\Omega \setminus V}(\partial_t V) < \delta_z^*((\Omega \setminus V) \cup F) \le 1$ as claimed.

Finally, let $h$ be a harmonic function on $\Omega$ which is majorized on $V$ in absolute value by a semi-bounded potential. Because $h$ is also finely harmonic on $V$ and finely continuous on $\partial_{\partial_t} V$), we deduce from Theorem A.8 that it is the solution of the fine Dirichlet problem on $V$ with boundary data $h_{|\partial_t V}$ and that (A.39) holds. $\square$

**Lemma A.10.** Let $V \subset \mathcal{R}$ be a proper regular fine domain such that $\overline{\partial_t V} \cap \partial \mathcal{R} = \emptyset$ and $\overline{V} \cap \partial \mathcal{R} \neq \emptyset$ (the lower assumption on the right-hand side of (A.38) when $\Omega = \mathcal{R}$). Let further $h$ be a harmonic function in $\mathcal{R}$ such that $\lim_{z \to \xi} h(z) = 0$ for every $\xi \in \overline{V} \cap \partial \mathcal{R}$. Then identity (A.39) holds.

*Proof.* In view of Lemma A.9, it is enough to show that $h$ is majorized on $V$ by a semi-bounded potential. Note, as in the proof of Lemma A.9, that $\partial_t V$ is non-polar. Let us show that $\partial \mathcal{R} \cap \overline{V}$ consists of a union of connected components of $\partial \mathcal{R}$. Indeed, any such component $\Gamma$ is a 1-dimensional compact topological submanifold of $\mathcal{R}_*$, and as such it has a tubular neighborhood $N$ that may be chosen so thin that $N \cap \overline{\partial_t V} = \emptyset$. Then, if $\zeta_1, \zeta_2 \in \Gamma$ and $\zeta_1 \in \overline{V}$ while $\zeta_2 \notin \overline{V}$, we can find $z_1 \in N \cap V$ close to $\zeta_1$ and $z_2 \in N \cap \Omega \setminus V$ close to $\zeta_2$. The points $z_1, z_2$ can be joined by a smooth arc contained in $N$. However, such an arc is finely connected [19, Theorem 7], but cannot meet $\overline{\partial_t V}$ by construction, a contradiction that proves our claim.

Assume first that $D = \mathbb{D}$ is the unit disk. Any function $u$ harmonic in an annular region $\{r < |z| < 1\}$ that extends continuously to $\mathbb{T}$ by zero can be harmonically extended to $\{r < |z| < 1/r\}$ by reflection, i.e., by setting $u(z) := -u(1/\bar{z})$ for $z \in \{1 < |z| < 1/r\}$. Due to the smoothness of this extension it necessarily holds that $|u(z)| \le C_\rho(1 - |z|)$ for $r < \rho \le |z| \le 1$. As $h$ is harmonic on $\mathcal{R}$, this principle used around each of the finitely many connected components of $\overline{V} \cap \partial \mathcal{R}$ yields
that $|h(z)| \le C(1 - |p(z)|)$ for $z \in V$ and some constant $C > 0$. On the other hand, the function $g_r(z) := -\log \max\{|r, |z|\}$ is a continuous (thus bounded and therefore semi-bounded) potential in $D$ for any $r \in (0, 1)$ (this is the Green equilibrium potential of $\{|z| \le r\}$). It can be readily verified that $g_r(z) \ge (1 - |z|)$ in $D$ when $r \le e^{-1}$. Thus, $|h(z)| \le Cg_r(p(z))$, $z \in V$, for any such $r$. As $g_r(p(z))$ is a (bounded) potential on $\mathcal{R}$ by (A.34), the claim of the corollary follows.

In the general case, let $\phi: D \to D$ be a conformal map. Recall that $\phi$ extends to a homeomorphism from $\bar{D}$ onto $D$. Clearly, $\phi$ is also a homeomorphism for the fine topology since $v$ is superharmonic (resp. harmonic) on $D$ if and only if so is $v \circ \phi$ on $D$. Moreover, $g$ is a bounded potential on $D$ if and only if $g \circ \phi$ is such a potential on $D$. Hence, the result just proven on the disk carries over to $D$ by conformal mapping. $\square$

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that $|h(z)| \le C(1 - |p(z)|)$ for $z \in V$ and some constant $C > 0$. On the other hand, the function $g_r(z) := -\log \max\{|r, |z|\}$ is a continuous (thus bounded and therefore semi-bounded) potential in $D$ for any $r \in (0, 1)$ (this is the Green equilibrium potential of $\{|z| \le r\}$). It can be readily verified that $g_r(z) \ge (1 - |z|)$ in $D$ when $r \le e^{-1}$. Thus, $|h(z)| \le Cg_r(p(z))$, $z \in V$, for any such $r$. As $g_r(p(z))$ is a (bounded) potential on $\mathcal{R}$ by (A.34), the claim of the corollary follows.

In the general case, let $\phi: D \to D$ be a conformal map. Recall that $\phi$ extends to a homeomorphism from $\bar{D}$ onto $D$. Clearly, $\phi$ is also a homeomorphism for the fine topology since $v$ is superharmonic (resp. harmonic) on $D$ if and only if so is $v \circ \phi$ on $D$. Moreover, $g$ is a bounded potential on $D$ if and only if $g \circ \phi$ is such a potential on $D$. Hence, the result just proven on the disk carries over to $D$ by conformal mapping. $\square$

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INRIA, PROJECT FACTAS, 2004 ROUTE DES LUCIOLES — BP 93, 06902 SOPHIA-ANTIPOLIS, FRANCE

Email address: laurent.baratchart@sophia.inria.fr

DEPARTMENT OF MATHEMATICAL SCIENCES, INDIANA UNIVERSITY INDIANAPOLIS, 402 NORTH BLACKFORD STREET, INDIANAPOLIS, IN 46202

Email address: maxyatts@iu.edu
[62] H. Stahl. The convergence of Padé approximants to functions with branch points. *J. Approx. Theory*, 91:139–204, 1997. 3, 4

[63] E.M. Stein. *Singular Integrals and Differentiability Properties of Functions*. Princeton Univ. Press, 1970. 15

[64] K. Strebel. *Quadratic Differentials*, volume 5 of *Ergebnisse der Mathematik und ihrer Grenzgebiete (3) [Results in Mathematics and Related Areas (3)]*. Springer-Verlag, Berlin, 1984. 49

[65] J.J. Telega, S. Tokarzewski, and A. Galka. *Numerical analysis and its applications*, chapter Modelling torsional properties of human bones by multipoint Padé approximants, pages 33–38. Lect. Notes in Comp. Sci. Springer, 2001. 3

[66] J.A. Tjon. Operator Padé approximants and three body scattering. In E. B. Saff and R. S. Varga, editors, *Padé and Rational Approximation*, pages 389–396, 1977. 3

[67] J.L. Walsh. *Interpolation and Approximation by Rational Functions in the Complex Domain*. American Mathematical Society Colloquium Publications, Vol. XX. American Mathematical Society, Providence, R.I., fourth edition, 1965. 3, 5

[68] H. Whitney. Analytic extensions of differentiable functions defined in closed sets. *Transactions of the Amer. Math. Soc.*, pages 63–89, 1934. 15

INRIA, PROJECT FACTAS, 2004 ROUTE DES LUCIOLES — BP 93, 06902 SOPHIA-ANTIPOLIS, FRANCE

Email address: laurent.baratchart@sophia.inria.fr

DEPARTMENT OF MATHEMATICAL SCIENCES, INDIANA UNIVERSITY INDIANAPOLIS, 402 NORTH BLACKFORD STREET, INDIANAPOLIS, IN 46202

Email address: maxyatts@iu.edu