4.4 Perron theory and fine boundary limits5CFAC22DC258

Proof. The finite superharmonic members of an upper class form a nonempty saturated family whenever that class has a member finite somewhere in the component: minimum and ball lowering preserve the boundary inequalities and the lower bound. The saturated-family lemma gives its harmonic or $-\infty $ infimum; if there is no such member the infimum is infinity. For $u$ in the lower and $v$ in the upper class, $v-u$ is superharmonic, bounded below, with nonnegative boundary lower limits, including where $f=\pm \infty $ (use the global bounds to avoid undefined subtraction). The bounded minimum principle gives $u\le v$. Order, constant translation, positive scaling, sign reversal and upper subadditivity follow from the defining classes, whenever sums are defined. For finite resolutive functions these prove linearity. A uniform change of boundary data by at most $\varepsilon $ changes each solution by at most $\varepsilon $, hence uniform limits of resolutive data are resolutive. The checked Foundations perron resolutivity supplies equality of the continuous upper and lower subsolution envelopes for continuous $f$. These classes are contained in our classes, so their common harmonic value lies between our lower and upper solutions; the reverse inequalities from class inclusion force equality here. Equivalently, approximate continuous data by smooth functions and split each smooth function as a difference of continuous subharmonic functions by adding a large multiple of $|x|^2$.

We supply the required monotone limit argument. If $f_ j\uparrow f$ and $\overline H_{f_1}>-\infty $, then $\overline H_{f_ j}\uparrow \overline H_ f$. Work on one component. If the limit is infinity, order proves the claim. Otherwise the limit $h$ is harmonic by monotone harmonic convergence. At a fixed $x_0$ choose $v_ j\in \mathcal U_{f_ j}$ with $v_ j(x_0)<\overline H_{f_ j}(x_0)+\varepsilon 2^{-j}$. The nonnegative superharmonic series

is finite at $x_0$, hence superharmonic, and dominates every $v_ j$. It is bounded below by one $v_ j$ and has boundary lower limits at least $f$, so belongs to $\mathcal U_ f$. Its value at $x_0$ is at most $h(x_0)+\varepsilon $. Let $\varepsilon \downarrow 0$; harmonicity and the minimum principle, or repetition at each $x_0$, gives the claim. The negative-data version gives the corresponding decreasing limit. If all $f_ j$ are resolutive, their increasing limit has equal upper/lower solutions, and is resolutive when the common value is finite.

For continuous data, $f\mapsto H_ f(x)$ is positive linear, sends $1$ to $1$, and has norm one. Compact-space Riesz representation gives the unique probability $\omega _ x$. Decreasing continuous approximations to indicators of closed subsets, followed by the monotone limit just proved, give $H_{1_ F}(x)=\omega _ x(F)$ for closed $F$. The class of boundary sets for which this equality and resolutivity hold contains the closed sets, is closed under proper differences and increasing unions, and therefore contains their generated sigma algebra. For clarity, the elementary pi-lambda argument proves the last assertion: first fix a closed set and close the class of sets whose intersections with it lie in the class; then repeat with an arbitrary member. This gives closure under finite intersections, hence under finite unions and, with increasing unions, under all countable unions. Thus all Borel indicators, bounded simple functions, and by monotone approximation all nonnegative Borel functions have the integral formula.

For a Borel null set $N$ at $x$, positivity and harmonicity show $H_{1_ N}=0$ on its component. Every subset of $N$ has upper and lower solutions zero there, by order. Thus the formula extends to the completion of $\omega _ x$, and componentwise to the intersection of all these completed sigma algebras. Simple approximation proves the nonnegative formula; positive/negative parts prove sufficiency for integrable data. A positive harmonic function is finite everywhere in a component if finite at one point, so integrability at one point per component suffices.

For necessity suppose $f$ is resolutive and work on one component. Choose decreasing $v_ j\in \mathcal U_ f$ whose values at $x_0$ tend to $H_ f(x_0)$, and put $F_ j(z)=\liminf _{D\ni x\to z}v_ j(x)$. These are lower semicontinuous extended Borel functions, bounded below, decreasing and at least $f$. Their harmonic solutions satisfy $H_ f\le H_{F_ j}\le v_ j$; they are finite at $x_0$ and hence finite harmonic everywhere. Similarly lower-class approximants yield increasing Borel integrable $G_ j\le f$, with $H_{G_ j}(x_0)\to H_ f(x_0)$. Let $F=\inf F_ j$, $G=\sup G_ j$. The inequalities $G_1\le G\le f\le F\le F_1$ imply that $F,G$ are integrable for every $\omega _ x$. Monotone integration gives $\int (F-G)d\omega _{x_0}=0$. The nonnegative harmonic integral of $F-G$ is then zero throughout the component. Thus $F=f=G$ almost everywhere for every $\omega _ x$, proving completed measurability and integrability.

Finally Harnack applied to $H_{1_ E}$ gives $C^{-1}\omega _ y(E)\le \omega _ x(E)\le C\omega _ y(E)$ for fixed interior $x,y$ in one component. Integration yields $C^{-1/p}\| f\| _{p,y}\le \| f\| _{p,x}\le C^{1/p}\| f\| _{p,y}$. Equivalent norms give the same continuous dual and weak topology; strong topologies also agree. These statements apply to the $p$th powers of norms, which are harmonic integrals; the norms themselves need not be harmonic. Weak continuity of measures follows since $\int f d\omega _ x=H_ f(x)$ is continuous for every continuous $f$. $\square$

Proof. For any interior $a$, choose an entire superharmonic polar function infinite on the exceptional set and finite at $a$, adding a constant to make it nonnegative on $\overline D$. Add $\varepsilon $ times it to every upper competitor. This is a competitor for either set of boundary data; evaluation at $a$ and $\varepsilon \downarrow 0$ gives equality. Negate for lower solutions. For restriction, an upper competitor dominates $H_ f$ in $D$, so its boundary lower limits on interior boundary points of $W$ dominate $H_ f$ by harmonic continuity; on the remaining points they dominate $f$. Thus the restricted upper solution is at most $H_ f$, while the lower solution is at least $H_ f$. Bounded-domain comparison forces equality. $\square$

Proof. The convex function $|x-a|$ is subharmonic by Jensen on harmonic coordinate functions, so lies in the lower class for $m$. Thus $H_ m\ge |x-a|>0$. Let $w$ be a weak local barrier, and choose a small sphere of radius $r$ centered at $a$ within its domain. Put $M=\max _{\partial D}|z-a|$. Choose compact $F\subset D\cap \partial B(a,r)$ whose complement in that open spherical set has normalized area less than $r/M$; if the open set is empty no $F$ is needed and comparison directly gives $H_ m\le r$ inside. Otherwise $k=\inf _ Fw>0$ (if $F$ is empty choose it nonempty by taking a smaller error). Let $b$ be the ball Poisson integral of boundary data $M$ on $(D\cap \partial B)\setminus F$ and zero elsewhere. It is nonnegative, continuous near the center, with $b(a)<r$. For every lower competitor $v$ for $m$, bounded comparison on $D\cap B$ gives

On $F$ the $w$ term dominates the bound $v\le M$; on the remaining open sphere the Poisson limit is $M$; on the original boundary $v$ has upper limit at most $m\le r$. These verify every boundary piece. Taking the supremum and then approaching $a$ gives $\limsup H_ m\le 2r$. Let $r\to 0$. The global barrier and its positive-away-from-$a$ lower bound now follow from $H_ m\ge |x-a|$. For bounded-above data with local limsup $L$, the function $L+\varepsilon +cH_ m$ belongs to the upper class when $c$ is large enough to control the data outside a small neighborhood. Its limit at $a$ is $L+\varepsilon $. The case $L=-\infty $ is handled with any finite upper target instead. Negation gives the lower inequality. Conversely, regularity applied to continuous $m$ makes $H_ m$ a weak barrier. $\square$

Proof. Restrict a weak barrier for inheritance. For an exterior ball tangent only at $a$ use minus its radial fundamental function after subtracting its value on the sphere; it is positive on the exterior and tends to zero at $a$. A smaller internally tangent ball removes other contact points. For a planar exterior segment choose the logarithm branch cut along its ray; on a sufficiently small slit ball $-\operatorname {Re}(1/\log (z-a))=-\log |z-a|/|\log (z-a)|^2$ is a positive harmonic barrier tending to zero.

For a cone, shrink it and enlarge the domain locally to the truncated cone complement $D_\rho =B(a,\rho )\setminus C$. All its boundary points other than $a$ have exterior balls, hence are regular. On $D_{2\rho }$ let $u$ be the Perron solution for $|z-a|$. On the outer sphere it equals $2\rho $; on lateral cone boundary it equals $|z-a|$. It is bounded between zero and $2\rho $ and is strictly less than $2\rho $ inside. On the compact radius-$\rho $ spherical section (including its regular lateral endpoints) its maximum is $b<2\rho $. Choose $q$ with $\max \{ 1/2,b/(2\rho )\} <q<1$. On the boundary of $D_\rho $ except $a$, $u(x)\le q u(a+2(x-a))$: on the outer section use $b<q2\rho $ and on the lateral section use the factor two. The difference is bounded harmonic; the polar boundary minimum principle permits the singleton exception at $a$. The inequality holds inside. Its finite nonnegative limsup $L$ at $a$ satisfies $L\le qL$, hence is zero. This is a weak local barrier and proves cone regularity.

For exhaustion we give details of the finite-ball construction. Cover any compact $K$ in the open set by finitely many balls of a common radius $r$, with centers in $K$ and closures still inside the open set; choose the cover so a small interval of larger common radii still covers $K$ and has closures inside. Avoid finitely many critical radii: at a boundary point of the union, if the active outward normals had zero in their convex hull, that point would be the circumcenter in the affine span of some affinely independent subset of at most $d+1$ centers, by a minimal convex representation. Each such subset has at most one circumradius, giving a finite forbidden list. At an allowed radius the convex hull of the active normals avoids zero. Its nearest point to zero gives a direction with strictly positive scalar product with every active normal. A small solid cone in this direction avoids all active balls; the other finitely many balls have a positive gap locally. Thus every boundary point of the union has an exterior cone, and the union is bounded regular. Apply this to a compact exhaustion, including the closure of the previously constructed set at each step, to obtain the desired increasing exhaustion. $\square$

Proof. For a bounded open set, continuous boundary data $\Phi _ a$ have a finite harmonic Perron solution, and $G(a,\cdot )=\Phi _ a-H_{\Phi _ a}$. Indeed $\Phi _ a$ is an upper competitor, so the right side is nonnegative and admissible, hence at least $G$. Writing $G=\Phi _ a+h_ a$, the function $-h_ a$ is a lower competitor: it is bounded above by $\Phi _ a$ and bounded on a pole neighborhood by harmonicity, so bounded above throughout the bounded set. Its boundary upper limits are at most $\Phi _ a$. Thus $-h_ a\le H_{\Phi _ a}$, giving the opposite inequality. At regular points the formula gives zero boundary limit; conversely such a Green function is positive harmonic near a boundary point and supplies a weak local barrier.

For arbitrary Greenian $\Omega $, exhaust by bounded regular open sets $D_ j\Subset \Omega $ containing a fixed pole $a$. Extend $G_{D_ j}(a,\cdot )$ by zero outside $D_ j$. On the complement of a small closed pole ball these are nonnegative continuous subharmonic functions: they are harmonic inside, zero outside, and at a boundary point their value zero is at most every sufficiently small sphere average. They increase to the zero extension $g$ of $G_\Omega (a,\cdot )$, bounded above there by the fixed-pole kernel bound. The checked subharmonic envelope/polar discrepancy theorem says $g=g^*$ outside a polar set. On $\partial \Omega $, $g=0$ and $g^*\ge \limsup _{\Omega \ni x\to z}G_\Omega (a,x)\ge 0$, so that limsup is zero off a polar set.

Choose one pole in each component; there are at most countably many components. The union of their exceptions is polar. Harnack chains in the pole variable make the zero limit uniform on any compact subset of poles in a component. A compact subset meets only finitely many components, so the limit is uniform on every compact subset of $\Omega $. Integration against a finite compactly supported measure proves admissibility with one common exceptional set. The bounded-density statement is then precisely the compact-tail estimate already proved. On a bounded connected set, regularity and the fixed-pole condition are equivalent by the first part, so the irregular set is contained in that pole's polar exception. These arguments remove the temporary admissibility hypothesis in the capacity theorem without using that theorem to prove the boundary discrepancy. $\square$

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. $\square$

Proof. Invert about the sphere and apply Kelvin covariance. For $d\ge 3$, the Kelvin transform of bounded $h$ is bounded in absolute value by a multiple of the fundamental pole near $a$. Adding a sufficiently large multiple of that pole makes it nonnegative, so the proved Bôcher lemma applies. Its singular coefficient is $\alpha R^{d-2}$, by taking the limit back at infinity. Subtracting this singularity leaves a harmonic function on $B$; its continuous boundary data identify it by the ball Poisson formula, giving the displayed representation. In dimension two the Kelvin transform is bounded near the center and extends harmonically there, with center value $\alpha =m$.

For integrable $f$, define the right side with $\alpha =m$. The inverted ball Poisson integral is harmonic away from $a$ and its derivatives are controlled on compact subsets by the integrable boundary density. The inversion distance identity gives precisely the displayed kernel. Its first term is positive; for $d\ge 3$ so is the correction $R^{2-d}-r^{2-d}$, and for $d=2$ that correction is zero. Evaluating at $f=1$ shows the total kernel mass is one. Its limit at infinity is $m$, and the ball Poisson continuity-point estimate, together with the vanishing correction at $r=R$, proves the boundary assertion.

The Kelvin identity for concentric spherical means gives, for a general finite-limit $h$,

and the constant $m$ if $d=2$. For the chosen $\alpha =m$ these means are constant, so their radial flux is zero. The divergence theorem on the region between any large sphere containing $\overline B$ and a larger concentric sphere transfers this zero flux to the arbitrary sphere. The annular mean derivative formula then makes its subsequent spherical means constant; their limit is the limit at infinity, namely $m$. Thus every exterior-ball mean equals $h(\infty )$, as required. Conversely Helms's mean condition for the concentric spheres forces $\alpha =m$. $\square$

Proof. At finite points the minimum-set argument is the checked Euclidean one. If the minimum is attained at infinity, each large sphere mean is at least the minimum and at most the value there. Lower semicontinuity and positivity of the surface measure of open patches force equality at every point of each such sphere. Thus the minimum set contains an exterior neighborhood and is open and closed. For a proper open set each component has a nonempty boundary in the connected compact ambient space. Extend by boundary liminf; compactness forces a negative minimum to be attained inside if any negative value exists. The strong minimum principle would make the component a negative constant, contradicting its nonnegative boundary limits. For the entire compact space a minimum is attained and the function is constant; one cannot infer a nonnegative constant from an empty boundary.

For algebraic operations use the ordinary local result and the linear sphere inequalities at infinity; nontriviality follows from finite almost everywhere values on each Euclidean component. For exterior lowering, increasing continuous minorants of the lsc sphere data and the positive exterior Poisson kernel give a harmonic minorant bounded above by $u$ via compactified comparison. Finite interior values exclude an infinite limit. The same boundary-liminf/pasting proof as for interior lowering makes the replacement superharmonic at finite points. Near infinity it is the exterior harmonic solution with constant large-sphere means equal to its value there. For every other allowed sphere center, the finite-point superharmonic mean is concave as a function of $r^{2-d}$ (or $-\log r$ in the plane), and is constant at that value for all sufficiently large radii, by the zero-flux argument above. Concavity bounds its earlier means by the same value, giving the required inequality at infinity. Directed harmonic limits reduce to the Euclidean result away from infinity and the positive exterior kernel on a fixed sphere near infinity; directed monotone integration identifies the value and harmonicity there. The saturated-family proof then applies in both kinds of neighborhoods, exactly as for finite balls, with nonemptiness explicitly required.

For $d=2$, $\log |x-a|$ on an exterior neighborhood, with value infinity at infinity, is a superharmonic pole function there. For $d\ge 3$, concentric sphere means of a superharmonic function on an exterior are finite and concave functions of $t=r^{2-d}$. A supporting affine upper bound at an interior $t$ bounds them above as $t\downarrow 0$. If the value at infinity were infinite, lsc would instead force all sufficiently distant sphere means to tend to infinity. This contradiction proves nonpolarity. $\square$

Proof. On compact subdomains avoiding the center, transform harmonic comparison functions, use positivity of the Kelvin multiplier and its involutivity, and transform the comparison back. This proves the superharmonic comparison property for the transformed lsc function. In the normalized case the transform is harmonic near infinity with zero large-sphere means: the original function is harmonic near $a$ with zero center value, so the concentric mean identity and the flux transfer in the exterior theorem apply. In particular its value at infinity is zero and it is continuous there. For any other exterior ball in the domain, its sphere means are concave in $r^{2-d}$ (or $-\log r$ in the plane), and are zero for all sufficiently large radii. Concavity makes them nonpositive at the smaller allowed radii. This is exactly the supermean condition at infinity. $\square$

Proof. Choose a closed ball disjoint from $\overline\Omega $ and invert about its sphere, sending $\Omega $ into a bounded ball. Continuous data, after subtracting a constant and a uniform approximation if necessary, can be assumed zero near infinity. The transformed data, divided by the positive Kelvin multiplier, then extend continuously with value zero near the inversion center. If infinity is interior, the center is separated from the transformed boundary and the same prescription is possible. Approximate uniformly by smooth functions $\varphi $ zero near that center. Such smooth approximation follows by extending a continuous function from the compact boundary, multiplying by a cutoff, and convolving; the uniform error on the boundary is controlled by uniform continuity.

These approximants are differences of superharmonic functions harmonic and zero near the center. To verify this, take $q(x)=-A(|x-a|-r_0)_+^4$, with the support of $\Delta \varphi $ separated from $\overline{B(a,r_0)}$. Its Laplacian is nonpositive and strictly negative on that support; sufficiently large $A$ makes both $q$ and $\varphi +q$ superharmonic. Their normalized Kelvin transforms are superharmonic even at interior infinity by the normalized Kelvin lemma. On the exterior ball the Kelvin multiplier is bounded by one (it is one in the plane), so uniform approximation survives. Each transformed approximant is a difference of bounded superharmonic traces on a compact neighborhood of $\overline\Omega $; bounded subharmonic trace resolutivity, proved by comparing its lower solution with the upper class, applies with the compactified minimum principle. Uniform stability of the Perron classes proves continuous-data resolutivity. Positive-functional representation gives a probability harmonic measure on the compact boundary, including when the pole is infinity.

For a polar boundary set, transform its finite points into the bounded ball and split it into countably many closed shells avoiding the center. The prescribed-point polar construction gives positive superharmonic pole functions finite at the center and at the chosen transformed interior point. Lower each harmonically on a sufficiently small center ball, which avoids that shell, and subtract its center value. Its normalized Kelvin transform is superharmonic at infinity and has the same required boundary blow-up. It is defined on a neighborhood of the compact exterior, and hence has a finite lower bound there by lower semicontinuity. Add a constant to make it positive. Choose summable positive coefficients giving a finite sum at the prescribed point (and at infinity if interior). Their sum is the required function. In dimension two, if the polar set contains infinity, add a positive multiple of $\log (|x-a|/R)$ on the exterior. In higher dimensions infinity cannot belong to a polar set. The same construction on a finite boundary singleton and epsilon comparison shows that its harmonic measure is zero.

A local finite-point weak barrier can be upgraded on the bounded set $D=\Omega \cap B(a,r)$ by the proved bounded-domain weak-barrier theorem. Its strong barrier $g$ has a positive lower bound away from $a$. On a cutting sphere inside $B(a,r)$ choose $c>0$ below that lower bound; paste $\min (g,c)$ inside to the constant $c$ outside. They agree on a collar, so the pasted function is superharmonic, vanishes at $a$, and is bounded away from zero off every neighborhood of $a$. If infinity is interior, its value $c$ dominates every exterior sphere mean, since it is globally at most $c$. Thus the pasting remains valid there. The usual upper/lower barrier comparison proves regularity. Conversely use a bounded continuous boundary function $m\ge 0$ vanishing only at $a$. Its harmonic solution is positive on every component: otherwise its probability harmonic measure would be concentrated on the zero-mass singleton. At a regular point it tends to zero, giving a weak barrier. This avoids unbounded distance data on a boundary containing infinity.

For a boundary infinity in $d\ge 3$, the fundamental function with any fixed finite pole is an explicit local weak barrier, positive and bounded away from zero on every bounded portion away from infinity. In dimension two inversion has multiplier one and transfers the finite-point local barrier criterion. At all finite points inversion preserves weak barriers locally. Thus the bounded-domain Kellogg theorem transfers to the inverted domain; polar sets transfer by Kelvin transformation. In $d\ge 3$ the possible extra center corresponds to an already regular infinity. This proves polar irregularity in all cases under the present nondensity hypothesis. $\square$

Proof. Only the dense case remains. Work in a connected component, choose $a\in \Omega $, and concentric closed balls with $0<r_0<r_1<r_2<r_3$, all contained in $\Omega $. Removing the smallest ball gives a nondense domain. Its pole function $v$ can be harmonically lowered on the annulus $r_0<|x-a|<r_3$, so it is finite harmonic near the two middle spheres. This preserves its true-boundary blow-up and positivity. Write $O=\Omega \setminus \overline{B(a,r_1)}$. This set is connected: any path in $\Omega $ entering the removed ball can be rerouted through the contained annulus. The harmonic measure $Q$ of its inner sphere is strictly below one on $O$. Indeed the original nonpolar boundary contains a regular point, since the nondense Kellogg exception is polar; $Q$ tends to zero there. The strong maximum principle then gives $q=\max _{|x-a|=r_2}Q(x)<1$.

Set $w_0=0$. Given $w_{j-1}$, let $u_ j$ be the ball solution in $B(a,r_2)$ with data $v+w_{j-1}$ on its sphere. Let $w_ j$ be the bounded solution on $O$ with data $u_ j-v$ on the inner sphere and zero on the true boundary. These are continuous boundary data on the nondense domain. For consecutive differences the ball maximum principle and the bound by $Q$ give

Thus the corrections converge in geometric series, uniformly on the respective boundaries and hence throughout their domains. Their limits $u,w$ are harmonic; $w$ is bounded. On both annulus boundary spheres, $u=v+w$, so annular uniqueness gives this equality throughout the overlap. Paste $u$ inside to $v+w$ outside. The result is superharmonic, harmonic near $a$, bounded below, and has the same polar boundary blow-up as $v$. Adding a constant makes it positive. This proves the assertion including finiteness at the prescribed center. Componentwise construction proves the disconnected version.

For bounded harmonic $h$ with zero boundary limits outside a polar $Z$, compare $\pm h+\varepsilon p$ with zero, where $p$ is the positive blow-up function, finite at a prescribed interior point. Compactified boundary comparison applies even on $Z$, and yields $|h(a)|\le \varepsilon p(a)$. Let $\varepsilon \downarrow 0$ and choose $a$ arbitrarily. The same epsilon argument permits polar exceptions in bounded Perron competitors. $\square$

Proof. For a dense domain remove an interior closed ball. Solve the continuous datum $f$ on the true boundary and the constant $M+1$ on the artificial sphere, where $|f|\le M$. Paste $M+1$ inside by taking the minimum with that constant on a collar. The resulting bounded superharmonic function is a global upper competitor. The lower construction uses $-M-1$. At every regular true boundary point of the punctured nondense domain, these give the corresponding boundary limits for the global upper/lower solutions. Their difference is bounded harmonic, with zero boundary limits outside the polar Kellogg set, and is zero by the preceding lemma. Thus $f$ is resolutive. Those same upper/lower bounds show each such punctured-domain regular point is regular for $\Omega $, proving Kellogg for the general domain. Local barriers and the capped-constant construction give the equivalent local criterion as before.

The full measurable-data proof now uses precisely the steps established for bounded domains: closed boundary indicators by continuous decreasing approximation, the pi-lambda argument, nonnegative simple approximation, and monotone continuity of the upper envelope via summable positive errors at a fixed pole. The only local harmonic operations needed are lowering and monotone convergence, available for both interior and exterior balls. For clarity, the necessity step also survives unchanged: countably many upper/lower competitors approximating a finite solution at a pole give Borel boundary liminf/limsup sandwiches. Their integrals have the same value at that pole, so their difference is zero almost everywhere there; positive harmonicity and Harnack transfer this common null set throughout the component. The sandwiches identify the datum in the completed measure class and give its absolute integrability. Conversely positive and negative integrable parts follow from monotone simple functions, and null changes are removed by the sandwich comparison. Apply this separately in the countably many components. Epsilon comparison with the constructed pole functions proves polar boundary nullity. This establishes the same full equivalence as the bounded Perron theorem, now with the compactified boundary condition at infinity included explicitly. $\square$

Proof. The integral solution is bounded above by $u$, which is an upper competitor, and so is finite almost everywhere in each component. Positive harmonic alternatives make it finite everywhere there. Paste $H_{u_0}$ off $F$ to $u$ on $F$ and regularize. The ball comparison argument of the relatively compact sweeping theorem applies: restriction consistency compares on the intersection of each interior ball with $\Omega \setminus F$, and regular boundary limits follow by increasing continuous minorants of $u_0$. The exceptional set is precisely among the irregular interior boundary points, independently of $u$. The regularized paste is therefore superharmonic and majorizes $u$ on $F$ quasi-everywhere. Every defining majorant for $R_ Fu$ is an upper Perron competitor off $F$; the two inequalities identify the reduction. Linearity of nonnegative boundary integrals gives additivity off $F$; on $F$ both sides equal the original sum quasi-everywhere. Canonical ball averages remove this polar exception from the identity everywhere. $\square$

Proof. Shrink a ball so the strict-gap criterion is represented by $G_ B\nu +h$ there, with $\nu $ finite on a larger compact neighborhood. The harmonic part has no effect on the gap. Finiteness at $a$ implies no atom there. For restrictions $\nu _ j=\nu |_{B(a,\varepsilon _ j)}$, their potentials $p_ j$ have $p_ j(a)\downarrow 0$. The difference $G_ B\nu -p_ j$ is harmonic near $a$, so every $p_ j$ retains the same strict limiting gap $\delta >0$ along $E$. Choose a subsequence with $\sum p_ j(a)<\infty $. Its sum is superharmonic, finite at $a$, and its first $m$ terms have limiting lower bound at least $m\delta $. This proves amplification. On every sufficiently small sphere, $E$ is contained in the Borel set $\{ w\ge M\} $. The supermean inequality bounds its relative outer area by $w(a)/M$; let $M\to \infty $.

Represent an amplified barrier locally as a fundamental potential of a finite compactly supported measure plus a harmonic function. The latter can be discarded. Push the measure forward by the Lipschitz map $T$. The decreasing fundamental kernel gives $U(T_\# \nu )(Ty)\ge U\nu (y)$, whereas equality holds at $Ta$ by the radial-distance hypothesis. The transformed potential is finite there. Any sequence in $T(E)$ tending to $Ta$ has preimages tending to $a$, again by radial distance. Thus its values tend to infinity, proving thinness. In dimension two the compact supports make all negative parts bounded; add a local constant as needed. $\square$

Proof. Remove $a$ from $E$, which does not alter regularized reduction. Outside $E$ raw and regularized reductions agree. A deficit at $a$ therefore gives a defining superharmonic majorant $v$ with $v(a)<u(a)$. Lower semicontinuity of $u$ and $v\ge u$ on $E$ give the strict-gap criterion. Conversely a thinness barrier can be chosen bounded near $a$ by truncation above its value there; local exact extension followed by compact reduction gives a bounded nonnegative superharmonic $v$ on $\Omega $, with $v(a)<\alpha <\liminf _ Ev$. On a neighborhood $V$ of $a$ we have $v>\alpha $ along $E$. The peaking assumption gives $\gamma =u(a)-\sup _{\Omega \setminus V}u>0$. For sufficiently small $\lambda >0$, the function $u(a)+\lambda (v-\alpha )$ is nonnegative, majorizes $u$ on all of $E$, and is strictly below $u(a)$ at $a$. This proves the deficit.

Locally work on a ball of radius less than $1/2$. The fundamental potential $w$ of its Lebesgue measure is finite and continuous. For each $a$ in a smaller ball split off the measure of a small ball centered at $a$. Its potential $w_1$ peaks finitely at $a$: inside that ball it is a constant minus $c_ d|y-a|^2/(2d)$, and outside it is strictly radially decreasing by the spherical mean formula. The remaining potential $w_2$ is nonnegative on the original small ball. At a self-thin point of $E$,

The polar discrepancy theorem puts all these points in a polar set. Countably many smaller balls cover the domain. Polarity implies thinness by the prescribed-finite-point pole construction (remove the point itself first); the converse follows from this self-thin point assertion. $\square$

Proof. On $B(a,r)$ the function $u(y)=r-|y-a|$ peaks finitely at $a$. Nonthinness and the preceding theorem give $\widehat R_ Fu(a)=u(a)$. On the complement $w=u-\widehat R_ Fu$ is nonnegative superharmonic, tends to zero at $a$ by lower semicontinuity of the reduction, and is strictly positive: otherwise the strong minimum principle would make it zero on a component, contradicting $\Delta u=-(d-1)/|y-a|<0$ there.

Conversely, if $F$ were thin, its strict gap would give a subharmonic $v$ on a smaller ball with $v(a)=1$ and $v\le -1$ on $F\setminus \{ a\} $. On a still smaller concentric sphere, $v<0$ near $F$, and away from that neighborhood the positive lsc barrier $w$ has a positive lower bound. Choose $A$ so $v-Aw\le 0$ on that sphere. Subtracting a small positive fundamental pole at $a$ makes the boundary upper limit nonpositive at $a$ as well. The bounded-domain maximum principle gives $v\le Aw$ after the pole coefficient tends to zero. The upper limit of $v$ along the complement is at most zero and along $F$ at most $-1$, contradicting the canonical subharmonic value $v(a)=1$. The regularity equivalence now follows from the weak-barrier theorem on $\Omega \cap B(a,r)$ and the locality of finite boundary regularity. $\square$

Proof. Apply $\arctan $ to handle extended limits, and choose intervals $J_ j$ shrinking to the finite transformed limit. There are fine neighborhoods $V_ j$ on which $g\in J_ j$ along $E\setminus \{ a\} $. If infinitely many $V_ j$ are Euclidean neighborhoods the ordinary limit already exists. Otherwise discard finitely many and use amplified positive barriers $v_ j$ for their thin complements on a common ball. Choose $\alpha _ j>0$ with $\sum \alpha _ jv_ j(a)<\infty $. Put $F_ j=(B\setminus V_ j)\cap \{ \alpha _ jv_ j>j\} $ and $F'=\bigcup F_ j$. The sum $v=\sum \alpha _ jv_ j$ tends to infinity along $F'$ at $a$: terms with $j\ge p$ are already above $p$, while each of the finitely many remaining terms is above $p$ sufficiently close to $a$ on its complement. Thus $F'$ is thin. Let $V$ be the fine interior of its complement. Choose an actual Euclidean neighborhood $O_ j$ of $a$ with $O_ j\setminus V_ j\subset \{ \alpha _ jv_ j>j\} $. Then $O_ j\cap V\subset V_ j$, proving the ordinary limit on $E\cap V$.

For the second assertion choose an amplified $u\ge 0$, finite at $a$, infinite along $F$ approaching $a$. If $\liminf _{y\to a,\, y\notin F}(v+u)(y)=\infty $, fine continuity of $u$ implies the fine limit of $v$ is infinity. Otherwise call this liminf $\lambda $ and take $k>\lambda $. Near $a$, $u>k$ on $F\setminus \{ a\} $. Paste $\min (v+u,k)$ off $F$ to the value $k$ on $F\setminus \{ a\} $. The boundary lower limits justify superharmonic pasting. It is bounded below and extends across the polar singleton, with canonical value $\lambda $ at $a$. Its fine neighborhood $\{ w<k\} $ contains $a$; on its complement-of-$F$ portion $w=v+u$. Fine continuity of $w,u$ therefore gives the finite fine limit $\lambda -u(a)$ for $v$. $\square$

Proof. An interior point of a segment is regular for a disk slit along the segment by the explicit exterior-segment barrier. The boundary nonthinness theorem therefore makes the segment nonthin at that point. If it were thin at an endpoint, reflect an amplified barrier across the endpoint and add the two barriers. The doubled segment would then be thin at its interior point, a contradiction. The radial projection $y\mapsto a+|y-a|e_1$ is a contraction preserving distances from $a$. If every sufficiently small circle met a thin complement, its projection would contain a nondegenerate segment ending at $a$, contradicting contraction invariance of thinness. Hence the required circles exist. An exterior continuum containing $a$ and another point $b$ meets every circle of radius less than $|a-b|$, by connectedness of its distance image. This contradicts the circle conclusion if the complement were thin. Nonthinness gives regularity. For the final assertion use the standard planar topological characterization of a simply connected domain: its complement in the sphere is connected. For a bounded domain this complement is a nondegenerate compact continuum containing infinity; its radial distance image at each finite boundary point meets every sufficiently small circle. The same argument applies directly on the sphere, without assuming a finite exterior arc. This planar topological characterization is an explicit topological prerequisite, not a new potential-theoretic premise. $\square$

Proof. We give the kernel and measure details used in both directions. If $U\Subset B$ is open, compact exhaustion and the equilibrium measures give a measure $\mu _ U$ on $\partial U$ with mass $C(U)$ and $G_ B\mu _ U=1$ on $U$. Indeed their masses are bounded by $C(\overline U)$; weak compactness gives a limit supported on $\partial U$, and for each interior $x$ the eventual supports avoid a neighborhood of $x$, so the continuous kernel can be tested there. A cutoff equal to one around $\overline U$ gives the limiting mass. Also, if $\nu $ is supported on a closed $F$ and $G_ B\nu \ge b$ quasi-everywhere on $F$, then $\nu (F)\ge b C_*(F)$. To see this, integrate against each compact equilibrium measure $\mu _ K$, $K\subset F$, use reciprocity and $G_ B\mu _ K\le 1$, and then take the compact supremum.

Here is the uniform separated-shell estimate. For $\Phi (y-a)\in [\lambda ^ j,\lambda ^{j+1}]$ and $\Phi (z-a)\notin (\lambda ^{j-1},\lambda ^{j+2})$,

In $d\ge 3$, the radii of the two shells differ by at least the fixed factor $\lambda ^{1/(d-2)}$; the reverse triangle inequality bounds $|y-z|$ below by a fixed positive multiple of $|z-a|$. In the plane, write the radii as $e^{-t}$. In the outer case their ratio is at most $q=e^{-(\lambda -1)}<1$, hence $\log (1/|y-z|)\le \Phi (z-a)+\log (1/(1-q))$. In the inner case the same estimate with the variables reversed gives $\Phi (y-z)\le \lambda ^{-1}\Phi (z-a)+\log (1/(1-q))$. Since $z\in B$ gives $\Phi (z-a)\ge \log 2$, these are uniform ratio bounds. The same argument applies to shells of width three whose indices are separated by six. If $G_ B\nu (a)<\infty $ and $\nu $ is finite, then $\int \Phi (z-a)d\nu =G_ B\nu (a)+\Phi (1/2)\nu (B)<\infty $. Thus the potential contributed by all sufficiently separated shells is bounded by a constant independent of their index and target point.

Suppose first the series converges. Choose open $U_ j\Supset E_ j$ with compact closure in $B$, avoiding $\{ \Phi >\lambda ^{j+2}\} $, and $C(U_ j)\le C^*(E_ j)+\varepsilon _ j$, where $\sum \lambda ^ j\varepsilon _ j<\infty $. Finite initial indices outside $B$ may be discarded. Their open equilibrium potentials $p_ j$ satisfy $p_ j=1$ on $U_ j$ and $p_ j(a)\le \lambda ^{j+2}C(U_ j)$. Hence $v_ k=\sum _{j\ge k}p_ j$ is superharmonic, finite at $a$, with $v_ k(a)\to 0$, while $v_ k\ge 1$ on all sufficiently near points of $E\setminus \{ a\} $. For large $k$ this is the strict-gap criterion.

Conversely use an amplified thinness potential $v=G_ B\nu $, with $\nu $ finite and $v(a)<\infty $. To obtain it on this fixed ball, first extend the local amplified barrier exactly to an entire superharmonic function, then restrict its Riesz measure to $B$. That measure is finite because the function is defined on a neighborhood of $\overline B$. The discarded harmonic function is bounded near $a$ and has no effect on the infinite approach. Remove $a$ from $E$ and discard empty shells. Set $b_ j=\min (j,\inf _{E_ j}v)$, so $b_ j$ is finite and tends to infinity. The Borel sets

contain $E_ j$ and are capacitable. Choose compact $K_ j\subset U_ j$ with $C(K_ j)>C(U_ j)-\varepsilon _ j$, with summable weighted errors. Fix a residue class modulo six and put $K=\{ a\} \cup \bigcup K_{6j+p}$. This is compact. Its reduction $w=\widehat R_ Kv=G_ B\eta $ has $w(a)\le v(a)<\infty $, is supported on $K$, and $\eta (\{ a\} )=0$. The separated-shell estimate bounds the contribution of all the other $K_{6k+p}$ at points of $K_{6j+p}$ by a fixed $\beta $. Since $w=v>b_{6j+p}-1$ quasi-everywhere on that compact set, for all sufficiently large $j$ its own contribution is at least one quasi-everywhere. The preceding mass-capacity inequality gives $\eta (K_{6j+p})\ge C(K_{6j+p})$. Finally $\int \Phi (z-a)d\eta <\infty $, so

Apply this to all six classes, restore the summable errors, and use $C^*(E_ j)\le C(U_ j)$. This proves necessity. The last assertion is exactly the already proved equivalence between irregularity and thinness of the complement. $\square$