Theorem 4.4.92C2E1569F3E2

Proof. A Greenian set has nonpolar boundary by the planar fact in the polar boundary minimum proof, hence nonpolar complement. Conversely, fix $a\in \Omega $. Countably many open disks tangent at $a$ cover the plane minus $a$. If the closed complement is nonpolar, one of its compact intersections $K$ with the corresponding closed disks is nonpolar. Take a larger tangent disk $B$ with $K\Subset B$ and $a\in \partial B$. The ball capacitary potential $p$ of $K$ is nonzero, lies in $[0,1]$, is harmonic off $K$ and zero at $\partial B$. Extend it by zero outside $B$. This is continuous subharmonic on the complement of $K$, by the same boundary mean argument as above. Then $2-p$ is strictly positive superharmonic on $\Omega $ and nonconstant near $a$: it equals two on the outer side of $\partial B$ and is strictly smaller on the adjacent inner component. This formulation also works if $B\setminus K$ has other components where $p=1$.

It remains to obtain a Green kernel from a nonconstant positive superharmonic function $s$, first when $\Omega $ is connected. For any pole $a$, the strong minimum principle gives $b$ with $s(b)<s(a)$. Scale and truncate $s$ so it is at least one on a small closed ball $\overline{B(a,r)}$ and less than one at $b$. The regularized infimum $v$ of nonnegative superharmonic majorants of one on that ball lies in $[0,1]$, equals one inside, is harmonic outside, and is less than one at $b$; these use only saturated families and the checked polar discrepancy theorem, not a Green kernel on $\Omega $. It is continuous on the inner sphere: on a slightly larger concentric annulus its boundary values are continuous on the outer sphere, one on the inner sphere outside a polar exception; Perron comparison ignoring that exception and annulus regularity give inner limit one. Thus $w=1-v$ is continuous subharmonic, zero on the ball, and strictly positive on its complement. That complement is connected: any path crossing the ball may be detoured through a slightly larger sphere inside $\Omega $.

Choose $t>r$ with $\overline{B(a,t)}\subset \Omega $, and $2\alpha =\min _{|x-a|=t}w(x)>0$. Define a subharmonic $q$ by $\log (|x-a|/r)$ on the inner ball, by $\max \{ \log (|x-a|/r),(w(x)/\alpha )\log (t/r)\} $ on the annulus, and by the second term outside the radius-$t$ ball. At the inner sphere the extension dominates the harmonic logarithm with the same value; its sphere submeans follow. Near the outer sphere the second term dominates strictly, so locality proves subharmonicity. It is bounded above and $q+\Phi _ a=-\log r$ near $a$. If $C$ bounds $q$, the nonnegative superharmonic function $C-q$ is $\Phi _ a$ plus a superharmonic correction extending across $a$. Thus the admissible class $\mathcal B_ a$ is nonempty, and the minimal-kernel construction gives a Green function. A disconnected planar open set is already Greenian by the earlier component construction, so the connected argument suffices. $\square$