Green function at infinity from the equilibrium potential

By [F3] the equilibrium measure $\mu :={\mu }_K$ is a Borel probability measure on $K$ with $I(\mu )=V_K<+\mathrm{\infty }$ and $\mu \mathrm{\ne }0$; by [F12] it is carried by $K$, and its support $S:=\mathrm{supp}\mu$ is a nonempty compact subset of $K$ with $\mu (\mathbb{C}\setminus S)=0$. By [F2], $V_K=\log \frac{1}{\mathrm{cap}(K)}<+\mathrm{\infty }$. By [F6] $\mathrm{\Omega }$ is the unique unbounded connected component of $\mathbb{C}\setminus K$, every other component is bounded, and $\{\mathrm{\mid }z\mathrm{\mid }>R_0\}\subseteq \mathrm{\Omega }$ whenever $K\subseteq \{z:\mathrm{\mid }z\mathrm{\mid }\le R_0\}$; by [F11] $\mathrm{\Omega }$ is a nonempty complex domain with $\mathrm{\partial }\mathrm{\Omega }\subseteq K$ compact and $\mathrm{\Omega }\cap K=\mathrm{\varnothing }$, so $\mathrm{dist}(z,K)>0$ for every $z\in \mathrm{\Omega }$.

Let $\nu$ be a finite positive Borel measure of compact support $S_{\nu }\subseteq K$. By [F5], whose Countable Choice hypothesis is available by [F3], the function $p_{\nu }$ is harmonic on $\mathbb{C}\setminus S_{\nu }$; by [F9] the function $U^{\nu }=-p_{\nu }$ is harmonic there too. Since $\mathrm{\Omega }\subseteq \mathbb{C}\setminus K\subseteq \mathbb{C}\setminus S_{\nu }$, every such $U^{\nu }$ is harmonic on $\mathrm{\Omega }$; in particular $U^{\mu }$ is harmonic on $\mathrm{\Omega }$.

Far-field expansion. Let $\nu$ be a finite positive Borel measure of compact support $S_{\nu }$; the case $\nu =0$ is trivial, so assume $S_{\nu }\mathrm{\ne }\mathrm{\varnothing }$ and put $\rho :=\max \{\mathrm{\mid }u\mathrm{\mid }:u\in S_{\nu }\}$. For $\mathrm{\mid }z\mathrm{\mid }\ge 2\rho$ and $u\in S_{\nu }$ one has $\mathrm{\mid }u\mathrm{/}z\mathrm{\mid }\le 1\mathrm{/}2$ and therefore $\mid \log \frac{1}{\mathrm{\mid }1-u\mathrm{/}z\mathrm{\mid }}\mid =\mathrm{\mid }\log \mathrm{\mid }1-u\mathrm{/}z\mathrm{\mid }\mathrm{\mid }\le 2\mathrm{\mid }u\mathrm{/}z\mathrm{\mid }\le \frac{2\rho }{\mathrm{\mid }z\mathrm{\mid }}$; integrating against $\nu$ gives $\mid U^{\nu }(z)+\nu (\mathbb{C})\log \mathrm{\mid }z\mathrm{\mid }\mid =\mid \int \log \frac{1}{\mathrm{\mid }1-u\mathrm{/}z\mathrm{\mid }}d\nu (u)\mid \le \frac{2\rho \nu (\mathbb{C})}{\mathrm{\mid }z\mathrm{\mid }},$ so in particular $U^{\nu }(z)=-\nu (\mathbb{C})\log \mathrm{\mid }z\mathrm{\mid }+o(1)$ as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$.

Finite-energy measures annihilate Borel polar sets. Let $E\subseteq \mathbb{C}$ be a Borel capacity-polar set and let $\rho$ be a finite positive Borel measure of compact support with $I(\rho )<+\mathrm{\infty }$; then $\rho (E)=0$. Indeed, suppose $\rho (E)>0$, put $m:=\rho (\mathbb{C})>0$ and apply [F15] to the Borel probability measure $\rho \mathrm{/}m$ on the Polish space $\mathbb{C}$, the Borel set $E$ and $\epsilon :=\rho (E)\mathrm{/}(2m)>0$: there is a compact $K\subseteq E$ with $(\rho \mathrm{/}m)(E\setminus K)<\rho (E)\mathrm{/}(2m)$, so $t:=\rho (K)>\rho (E)\mathrm{/}2>0$. Put $\nu :=\rho {\mathrm{\mid }}_K\mathrm{/}t\in P(K)$ and choose a real $R>\max \{1,\mathrm{diam}(\mathrm{supp}\rho \cup K)\}$; then $k_R\ge 0$ on $(\mathrm{supp}\rho \cup K)\times (\mathrm{supp}\rho \cup K)$ and $\mathrm{supp}\nu \subseteq K$, so by [F1] and $k_R\ge 0$, $\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int k_{R}d\nu d\nu =t^{-2}\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int k_{R}d(\rho {\mathrm{\mid }}_K)d(\rho {\mathrm{\mid }}_K)\le t^{-2}\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int k_{R}d\rho d\rho =t^{-2}(I(\rho )+m^2\log R)<+\mathrm{\infty }$, whence $I(\nu )=\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int k_{R}d\nu d\nu -\log R<+\mathrm{\infty }$. By [F2] $V_K={\inf }_{\sigma \in P(K)}I(\sigma )\le I(\nu )<+\mathrm{\infty }$, so $\mathrm{cap}(K)=\exp (-V_K)>0$ by [F2], contradicting $\mathrm{cap}(K)=0$, which holds because $K$ is a compact subset of the capacity-polar set $E$ by [F7].

Applying step 1.3 to $\nu :=\mu$ and $\nu (\mathbb{C})=1$ from step 1.1, together with $g=V_K-U^{\mu }$, gives $g(z)-\log \mathrm{\mid }z\mathrm{\mid }=V_K-(U^{\mu }(z)+\log \mathrm{\mid }z\mathrm{\mid })⟶V_K(\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty })\mathrm{.}$

The function $g=V_K-U^{\mu }$ is harmonic on $\mathrm{\Omega }$, because the constant $V_K$ and $U^{\mu }$ are harmonic there by step 1.2 and [F9].

By [F4], $U^{\mu }\le V_K$ on all of $\mathbb{C}$; hence $g\ge 0$ on $\mathrm{\Omega }$, and also $U^{\mu }(\xi )\le V_K$ for every $\xi \in \mathrm{\partial }\mathrm{\Omega }$.

Countable unions of Borel polar sets are polar. Let $(E_j{)}_{j\ge 1}$ be a sequence of Borel capacity-polar sets and let $F$ be compact with $F\subseteq {\bigcup }_{j\ge 1}{E}_j$; if $\mathrm{cap}(F)>0$ then by [F2] there is a Borel probability measure $\sigma$ on $F$ with $I(\sigma )<+\mathrm{\infty }$ (choose $\sigma$ with $I(\sigma )<V_F+1$), while step 1.4 gives $\sigma (E_j)=0$ for every $j$, so $\sigma \left({\bigcup }_{j\ge 1}{E}_j\right)=0$ by countable additivity, contradicting $1=\sigma (F)\le \sigma \left({\bigcup }_{j\ge 1}{E}_j\right)$; hence $\mathrm{cap}(F)=0$, and ${\bigcup }_{j\ge 1}{E}_j$ is capacity-polar in the sense of [F7].

$g>0$ on $\mathrm{\Omega }$. Indeed, if $g(z_0)=0$ for some $z_0\in \mathrm{\Omega }$, then $z_0$ is an interior minimum point of the harmonic function $g$ on the domain $\mathrm{\Omega }$, so $g$ is constant on $\mathrm{\Omega }$ by [F8]; but step 2.1 gives $g(z)=\log \mathrm{\mid }z\mathrm{\mid }+V_K+o(1)\to +\mathrm{\infty }$ as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$ along $\mathrm{\Omega }$, a contradiction.

Local boundedness. Let $\xi \in \mathrm{\partial }\mathrm{\Omega }\subseteq K$. By [F1] and steps 1.1 and 2.3, $c:=U^{\mu }(\xi )$ satisfies $-\mathrm{\infty }<c\le V_K<+\mathrm{\infty }$; by [F5] the function $U^{\mu }=-p_{\mu }$ is lower semicontinuous, so $\{U^{\mu }>c-1\}$ is an open set containing $\xi$ and there is a real $r>0$ with $U^{\mu }(z)>c-1$ for all $z\in B(\xi ,r)$. For $z\in \mathrm{\Omega }\cap B(\xi ,r)$ we then have $g(z)<V_K-c+1<+\mathrm{\infty }$ and $g(z)\ge 0$ by step 2.3, so the supremum in property 3 of [F11] is finite.

Regular points and the quasi-everywhere boundary limit. Let $\xi \in \mathrm{\partial }\mathrm{\Omega }$ with $U^{\mu }(\xi )=V_K$. By [F5] $U^{\mu }$ is lower semicontinuous, so ${\mathrm{lim\hspace{0.17em}inf}}_{\mathrm{\Omega }\ni z\to \xi }{U}^{\mu }(z)\ge V_K$, while step 2.3 gives ${\mathrm{lim\hspace{0.17em}sup}}_{\mathrm{\Omega }\ni z\to \xi }{U}^{\mu }(z)\le V_K$; hence $U^{\mu }(z)\to V_K$ and $g(z)\to 0$ as $\mathrm{\Omega }\ni z\to \xi$. By [F4] there is a Borel capacity-polar set $E\subseteq K$ which is a countable union of compact sets of capacity zero with $U^{\mu }=V_K$ on $K\setminus E$; since $\mathrm{\partial }\mathrm{\Omega }\subseteq K$ by [F11], every point of $\mathrm{\partial }\mathrm{\Omega }\setminus E$ is regular, so $g$ has boundary limit $0$ outside the set $\mathrm{\partial }\mathrm{\Omega }\cap E$, which is Borel and, being a subset of the capacity-polar set $E$, capacity-polar by [F7].

The difference of two candidates. Let $\tilde{g}:\mathrm{\Omega }\to \mathbb{R}$ satisfy properties 1-4 of [F11] and put $w:=\tilde{g}-g$. Then: (i) $w$ is harmonic on $\mathrm{\Omega }$, by property 1 for $\tilde{g}$, step 2.2 and [F9]; (ii) $w(z)\to 0$ as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$ with $z\in \mathrm{\Omega }$, because property 2 for $\tilde{g}$ and step 2.1 give $\tilde{g}(z)-\log \mathrm{\mid }z\mathrm{\mid }\to V_K$ and $g(z)-\log \mathrm{\mid }z\mathrm{\mid }\to V_K$; (iii) for every $\xi \in \mathrm{\partial }\mathrm{\Omega }$ there are $r>0$ and a real $M$ with $\mathrm{\mid }w(z)\mathrm{\mid }\le M$ for all $z\in \mathrm{\Omega }\cap B(\xi ,r)$: by property 3 for $\tilde{g}$ and step 3.2 choose $r$ with $\tilde{g}\le M_1$ and $g\le M_2$ on $\mathrm{\Omega }\cap B(\xi ,r)$; then $w=\tilde{g}-g\le \tilde{g}\le M_1$ using $g\ge 0$ from step 2.3, and $-w=g-\tilde{g}\le g\le M_2$ using $\tilde{g}>0$ from property 1, so $\mathrm{\mid }w\mathrm{\mid }\le \max \{M_1,M_2\}$; (iv) writing $E={\bigcup }_{j\ge 1}{E}_j$ as in step 3.3 and $E^{\mathrm{\prime }}$ for the exceptional set of property 4 for $\tilde{g}$, the set $E_0:=(\mathrm{\partial }\mathrm{\Omega }\cap E)\cup E^{\mathrm{\prime }}$ satisfies $E_0={\bigcup }_{j\ge 1}(\mathrm{\partial }\mathrm{\Omega }\cap E_j)\cup E^{\mathrm{\prime }}$, a countable union of Borel capacity-polar sets, because each $\mathrm{\partial }\mathrm{\Omega }\cap E_j$ is a compact subset of the capacity-polar set $E_j$ (a compact set with $\mathrm{cap}(E_j)=0$) and $E^{\mathrm{\prime }}$ is Borel capacity-polar; hence $E_0$ is capacity-polar by step 2.4, and $w(z)\to 0$ as $\mathrm{\Omega }\ni z\to \xi$ for every $\xi \in \mathrm{\partial }\mathrm{\Omega }\setminus E_0$, because then $\xi \mathrm{\notin }{E}^{\mathrm{\prime }}$ gives $\tilde{g}(z)\to 0$ by property 4 and $\xi \mathrm{\notin }\mathrm{\partial }\mathrm{\Omega }\cap E$ gives $g(z)\to 0$ by step 3.3.

The bad set. For $\xi \in \mathrm{\partial }\mathrm{\Omega }$ put $L(\xi ):={\mathrm{lim\hspace{0.17em}sup}}_{\mathrm{\Omega }\ni z\to \xi }\mathrm{\mid }w(z)\mathrm{\mid }\in [0,+\mathrm{\infty })$ and put $P:=\{\xi \in \mathrm{\partial }\mathrm{\Omega }:L(\xi )>0\}$, so that $P={\bigcup }_{m\ge 1}{E}_m$ with $E_m:=\{\xi \in \mathrm{\partial }\mathrm{\Omega }:L(\xi )\ge 1\mathrm{/}m\}$. The function $L$ is upper semicontinuous on $\mathrm{\partial }\mathrm{\Omega }$: if $L({\xi }_0)<c$ pick $c^{\mathrm{\prime }}$ with $L({\xi }_0)<c^{\mathrm{\prime }}<c$ and $r>0$ with $\mathrm{\mid }w(z)\mathrm{\mid }<c^{\mathrm{\prime }}$ for all $z\in \mathrm{\Omega }$ with $0<\mathrm{\mid }z-{\xi }_0\mathrm{\mid }<r$; then for $\xi \in \mathrm{\partial }\mathrm{\Omega }$ with $\mathrm{\mid }\xi -{\xi }_0\mathrm{\mid }<r\mathrm{/}2$ and $z\in \mathrm{\Omega }$ with $0<\mathrm{\mid }z-\xi \mathrm{\mid }<r\mathrm{/}2$ one has $\mathrm{\mid }z-{\xi }_0\mathrm{\mid }<r$, so $L(\xi )\le c^{\mathrm{\prime }}<c$ and $\{L<c\}$ is relatively open in $\mathrm{\partial }\mathrm{\Omega }$. Hence each $E_m$, the intersection of the closed set $\{L\ge 1\mathrm{/}m\}$ with the compact set $\mathrm{\partial }\mathrm{\Omega }$, is compact by [F14]. Finally $E_m\subseteq E_0$ for the capacity-polar set $E_0$ of step 4.1: if $\xi \in \mathrm{\partial }\mathrm{\Omega }\setminus E_0$ then $w\to 0$ at $\xi$ by step 4.1(iv), so $L(\xi )=0$ and $\xi \mathrm{\notin }{E}_m$; hence $\mathrm{cap}(E_m)=0$ for every $m$ by [F7].

Since $(E_m{)}_{m\ge 1}$ is a specified sequence of compact sets with $\mathrm{cap}(E_m)=0$ and union $P$, [F10] applies with its Dependent Choice hypothesis supplied by [F3]: if $P\mathrm{\ne }\mathrm{\varnothing }$ there is a finite positive Borel measure $\sigma$ carried by $P$ with $U^{\sigma }(\xi )=+\mathrm{\infty }$ for every $\xi \in P$; if $P=\mathrm{\varnothing }$ take $\sigma :=0$. Put $m_{\sigma }:=\sigma (\mathbb{C})\ge 0$, so that $m_{\sigma }>0$ exactly when $P\mathrm{\ne }\mathrm{\varnothing }$.

The barrier. Put ${\rho }_K:=\max \{\mathrm{\mid }\zeta \mathrm{\mid }:\zeta \in K\}$ and choose $R_0>1+{\rho }_K$; then $K\subseteq \{z:\mathrm{\mid }z\mathrm{\mid }\le R_0\}$ and $S_{\sigma }:=\mathrm{supp}\sigma \subseteq \bar{P}\subseteq \mathrm{\partial }\mathrm{\Omega }\subseteq K$ is compact with $\mathrm{\mid }\zeta \mathrm{\mid }\le {\rho }_K$ for $\zeta \in S_{\sigma }$. The function $V:=U^{\sigma }+m_{\sigma }g$ is bounded below on $\mathrm{\Omega }$ and finite at every point of $\mathrm{\Omega }$ (where $z\mathrm{\notin }{S}_{\sigma }$): for $z\in \mathrm{\Omega }$ with $\mathrm{\mid }z\mathrm{\mid }\le R_0$ and $\zeta \in S_{\sigma }$ one has $\mathrm{\mid }z-\zeta \mathrm{\mid }\le R_0+{\rho }_K$, so by [F1] $U^{\sigma }(z)\ge -m_{\sigma }\log (R_0+{\rho }_K)$, while $m_{\sigma }g\ge 0$ by step 2.3; and for $\mathrm{\mid }z\mathrm{\mid }\ge R_0$ step 1.3 applied to $\sigma$ (trivially when $\sigma =0$), together with $g(z)-\log \mathrm{\mid }z\mathrm{\mid }\to V_K$ from step 2.1 multiplied by $m_{\sigma }$, gives $V(z)=m_{\sigma }{V}_K+o(1)\ge m_{\sigma }{V}_K-1$ for all sufficiently large $\mathrm{\mid }z\mathrm{\mid }$. Put $c:=1-{\inf }_{\mathrm{\Omega }}V\in \mathbb{R}$ and $Q:=U^{\sigma }+m_{\sigma }g+c=V+c\ge 1\text{on }\mathrm{\Omega }\mathrm{.}$ Then $Q$ is harmonic on $\mathrm{\Omega }$ by steps 1.2 and 2.2 and [F9]; $Q(z)\to m_{\sigma }{V}_K+c\ge 1$ as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$ by steps 1.3 and 2.1; and for every $\xi \in P$ one has ${\mathrm{lim\hspace{0.17em}inf}}_{\mathrm{\Omega }\ni z\to \xi }Q(z)=+\mathrm{\infty }$, because $U^{\sigma }$ is lower semicontinuous by [F5] with $U^{\sigma }(\xi )=+\mathrm{\infty }$ and $m_{\sigma }g+c\ge c$ is bounded below.

Nonpositive boundary behaviour. Let $A>0$ and put $w_A:=w-AQ$, harmonic on $\mathrm{\Omega }$ by steps 4.1 and 7.1 and [F9]. Then: (a) for $\xi \in P$, ${\mathrm{lim\hspace{0.17em}sup}}_{\mathrm{\Omega }\ni z\to \xi }{w}_A(z)\le \mathrm{lim\hspace{0.17em}sup}w+\mathrm{lim\hspace{0.17em}sup}(-AQ)=-\mathrm{\infty }$, because $\mathrm{lim\hspace{0.17em}sup}(-AQ)=-A\mathrm{lim\hspace{0.17em}inf}Q=-\mathrm{\infty }$ by step 7.1 and $\mathrm{lim\hspace{0.17em}sup}w$ is finite by step 4.1(iii); (b) for $\xi \in \mathrm{\partial }\mathrm{\Omega }\setminus P$, ${\mathrm{lim\hspace{0.17em}sup}}_{\mathrm{\Omega }\ni z\to \xi }{w}_A(z)\le \mathrm{lim\hspace{0.17em}sup}w+\mathrm{lim\hspace{0.17em}sup}(-AQ)\le 0-A\cdot 1<0$, because $L(\xi )=0$ forces $w\to 0$ at $\xi$ and $Q\ge 1$ by step 7.1; (c) as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$ with $z\in \mathrm{\Omega }$, $\mathrm{lim\hspace{0.17em}sup}{w}_A\le 0-A(m_{\sigma }{V}_K+c)<0$, because $w\to 0$ by step 4.1(ii) and $Q\to m_{\sigma }{V}_K+c\ge 1$ by step 7.1.

Maximum principle. Fix $A>0$ and choose $R>R_0$ large enough that $w_A(z)<0$ whenever $z\in \mathrm{\Omega }$ and $\mathrm{\mid }z\mathrm{\mid }\ge R$, possible by step 8.1(c) and the fact that $\mathrm{\Omega }$ is unbounded so $\mathrm{\Omega }\cap \{z:\mathrm{\mid }z\mathrm{\mid }>R\}\mathrm{\ne }\mathrm{\varnothing }$. By step 8.1 and compactness of $\mathrm{\partial }\mathrm{\Omega }$, finitely many boundary neighborhoods cover $\mathrm{\partial }\mathrm{\Omega }$ on which $w_A\le 1$; the set $(\bar{\mathrm{\Omega }}\cap \bar{B}(0,R))$ outside their union is compact and lies in $\mathrm{\Omega }$, so continuity bounds $w_A$ there. Together with the negative tail this proves $s:={\sup }_{\mathrm{\Omega }}{w}_A<+\mathrm{\infty }$. Since $\mathrm{\Omega }$ is nonempty and $w_A$ is real-valued, $s\in \mathbb{R}$. Assume for contradiction that $s>0$. For each $\zeta \in \mathrm{\partial }\mathrm{\Omega }$ step 8.1 gives ${\mathrm{lim\hspace{0.17em}sup}}_{\mathrm{\Omega }\ni z\to \zeta }{w}_A(z)\le 0<s\mathrm{/}2$, so there is $r_{\zeta }>0$ with $w_A\le s\mathrm{/}2$ on $\mathrm{\Omega }\cap B(\zeta ,r_{\zeta })$; since $\mathrm{\partial }\mathrm{\Omega }$ is compact by [F11], finitely many of these balls cover $\mathrm{\partial }\mathrm{\Omega }$, and their union $W$ is an open neighbourhood of $\mathrm{\partial }\mathrm{\Omega }$ with $w_A\le s\mathrm{/}2$ on $\mathrm{\Omega }\cap W$. The set $S:=(\mathrm{\Omega }\cap \{z:\mathrm{\mid }z\mathrm{\mid }\le R\})\setminus W$ equals $(\bar{\mathrm{\Omega }}\cap \{z:\mathrm{\mid }z\mathrm{\mid }\le R\})\setminus W$, is closed and bounded, hence compact by [F14], and satisfies $S\subseteq \mathrm{\Omega }$ because $W\supseteq \mathrm{\partial }\mathrm{\Omega }$ and every point of $\bar{\mathrm{\Omega }}\setminus \mathrm{\partial }\mathrm{\Omega }$ lies in $\mathrm{\Omega }$ by [F13]. Every point of $\mathrm{\Omega }\setminus S$ lies in $\mathrm{\Omega }\cap W$ or satisfies $\mathrm{\mid }z\mathrm{\mid }\ge R$, and at such points $w_A\le s\mathrm{/}2<s$; hence ${\sup }_S{w}_A=s$, and the continuous function $w_A$ attains the value $s$ at some $x^{\ast }\in S\subseteq \mathrm{\Omega }$ by [F14]. Since $w_A$ is harmonic, hence subharmonic, on the domain $\mathrm{\Omega }$ by [F9], [F8] forces $w_A$ to be constant on $\mathrm{\Omega }$, contradicting $w_A(z)<0$ for $\mathrm{\mid }z\mathrm{\mid }\ge R$; therefore ${\sup }_{\mathrm{\Omega }}{w}_A\le 0$, that is $w_A\le 0$ on $\mathrm{\Omega }$.

Uniqueness. Step 9.1 gives $w\le AQ$ on $\mathrm{\Omega }$ for every $A>0$, hence $w\le 0$ on $\mathrm{\Omega }$; the same argument with the roles of $g$ and $\tilde{g}$ interchanged gives $-w\le 0$ on $\mathrm{\Omega }$, because $g$ satisfies properties 1-4 of [F11] by steps 2.1, 2.2, 3.1, 3.2 and 3.3 (with the Borel capacity-polar exceptional set $\mathrm{\partial }\mathrm{\Omega }\cap E$ of step 3.3), while $\tilde{g}$ satisfies them by hypothesis, so steps 4.1, 5.1, 6.1, 7.1, 8.1 and 9.1 apply verbatim to $w^{\mathrm{\prime }}=-w$ with the same set $E_0$ and the same barrier $Q$. Hence $w=0$ and $\tilde{g}=g$ on $\mathrm{\Omega }$: a Green function with pole at infinity is unique, and it equals $V_K-U^{{\mu }_K}$.

Assembly. Step 3.1 gives positivity and step 2.2 harmonicity, so $g$ has property 1 of [F11]; step 2.1 gives property 2; step 3.2 gives property 3; step 3.3 gives property 4, with exceptional set $\mathrm{\partial }\mathrm{\Omega }\cap E$ that is Borel (intersection of the compact set $\mathrm{\partial }\mathrm{\Omega }$ with the Borel set $E$ of [F4]) and capacity-polar as a subset of $E$ by [F7]. Step 10.1 shows that every $\tilde{g}$ with properties 1-4 equals $g$, so the notation $g_{\mathrm{\Omega }}(\cdot ,\mathrm{\infty })$ of [F11] is licensed with $g_{\mathrm{\Omega }}(z,\mathrm{\infty })=V_K-U^{{\mu }_K}(z)$ on $\mathrm{\Omega }$. Assertions 1-5 of the Statement are exactly these conclusions.