4.3 Polar sets, capacity and sweepingAEB997E884B4

Proof. We use the checked Foundations polar globalization: every bounded polar set has a finite compactly supported fundamental measure whose potential is infinite on that set. For the prescribed point put $A_ j=A\cap \{ 1/j\le |x-a|\le j\} $. Apply that result and restrict the resulting measure to $\{ |y-a|>1/(2j)\} $. The removed part has a bounded kernel at each point of $A_ j$, so the restricted fundamental potential $p_ j$ is still infinite on $A_ j$ and finite at $a$. In dimension at least three choose $b_ j>0$ so that $\sum b_ j(\mu _ j(\mathbb R^ d)+p_ j(a))<\infty $. Then $\mu =\sum b_ j\mu _ j$ is finite and its positive Newtonian potential is finite at $a$ and infinite on $A=\bigcup A_ j$, by Tonelli.

In dimension two set $q_ j=p_ j-m_ j$, where $m_ j$ is the finite infimum of $p_ j$ on $\overline{B(a,j)}$. Such an infimum is finite by lsc and local integrability. Choose $b_ j>0$ with $\sum b_ jq_ j(a)<\infty $. For each fixed integer $k$, the tail $\sum _{j\ge k}b_ jq_ j$ is a monotone sum of nonnegative superharmonic functions on $B(a,k)$, finite at $a$; it is superharmonic there. Adding the finite initial sum gives an entire superharmonic function, finite at $a$ and infinite on $A$. Without a prescribed point, first choose $a$ outside the polar (hence null) set.

For the last assertion first treat $K$ contained in an interior ball $B$. Its normalized ball capacity is zero by the checked polar zero capacity. Choose decreasing compact neighborhoods $K_ j\Subset B$ with intersection $K$; their capacitary potentials $v_ j$ lie in $[0,1]$, equal one near $K$, and decrease to zero off $K$. Indeed the limit is bounded harmonic off $K$, extends harmonically across $K$ by polar removability, and is dominated by the zero-minorant potential $v_1$, so is zero. On every compact subset of $B\setminus K$ this convergence is uniform by Dini. Pass to a subsequence such that both $\operatorname {cap}_ B(K_ j)<2^{-j}$ and $v_ j<2^{-j}$ on the first $j$ members of a compact exhaustion of $B\setminus K$ (and put every support in one fixed compact subset of $B$). This is possible by decreasing capacity continuity and the preceding uniform convergence.

Write $v_ j=G_ B\mu _ j$, scaling the normalized Riesz measures by $1/c_ d$ for our kernel. Then $\sum \mu _ j(B)<\infty $. The potential of $\mu =\sum \mu _ j$ is $\sum v_ j$, infinite on $K$ and finite off $K$ in $B$. Its fundamental potential differs by a harmonic function on $B$; thus it has exactly the same finiteness set there. Off the common compact support it is finite automatically. The same distributional/harmonic-correction argument transfers it to $G_\Omega \mu $ without changing its poles. For general compact $K\Subset \Omega $, choose finitely many closed balls inside slightly larger interior balls whose interiors cover $K$; apply the construction to the polar compact intersections with these closed balls and add their measures. The sum is finite, compactly supported in $\Omega $, infinite on $K$ and finite at every point outside $K$. $\square$

Proof. For a bounded $\Omega $, a global polar function infinite on $Z$ can be made nonnegative on $\overline\Omega $ by adding a constant. Call it $v$. Then $u+\varepsilon v$ has boundary lower limits at least $\alpha $ at all points, including $Z$, since $u$ is bounded below. The bounded-domain minimum principle gives $u+\varepsilon v\ge \alpha $. Average over a small interior ball; its average of $v$ is finite. Let $\varepsilon \downarrow 0$ and then the radius decrease to zero to obtain $u\ge \alpha $ everywhere. The case with a condition at infinity follows by truncating with large balls, using $u\ge \alpha -\eta $ outside such a ball, and then letting $\eta \downarrow 0$.

We need one planar fact: the boundary of a nonempty Greenian open set is not polar. If $\Omega $ is not dense and its boundary were polar, take an entire superharmonic polar function $v$ infinite there and replace it by infinity on one component of $\Omega $. The result is lsc, locally supermean (the boundary values are infinity), finite somewhere in the open complement, and hence superharmonic on the plane. It cannot be infinity on a nonempty open set, a contradiction. If $\Omega $ is dense, its complement equals its boundary. A fixed-pole Green function is bounded away from its pole, so polar removability extends it harmonically across the complement outside a small pole ball. Its harmonic correction then extends to the whole plane, and the resulting nonnegative fundamental-plus-harmonic function contradicts the whole-plane nonexistence proved earlier. Thus the planar fact holds.

For unbounded planar $\Omega $, choose $z_0\in \partial \Omega \setminus Z$. For any $\eta >0$ choose $r>0$ with $u\ge \alpha -\eta $ on $\Omega \cap \overline{B(z_0,r)}$. On $D=\Omega \setminus \overline{B(z_0,r)}$ the function $v(x)=\log (|x-z_0|/r)$ is nonnegative harmonic and tends to infinity at infinity. Since $u$ is bounded below, $u+\varepsilon v$ satisfies the required boundary inequality $\alpha -\eta $ on a sufficiently large outer sphere and on the inner sphere, and on the original boundary outside $Z$. The bounded-domain result gives $u+\varepsilon v\ge \alpha -\eta $ on every large truncation of $D$. Let the truncation grow, then $\varepsilon \downarrow 0$, then $\eta \downarrow 0$. Together with the inner ball inequality this proves the assertion on $\Omega $. $\square$

Proof. The family of majorants is nonempty (it contains $u$), closed under minimum and locally bounded below. The checked Foundations subharmonic envelope and polar limit discrepancy, applied after sign reversal, give superharmonic regularization and a polar discrepancy set. On $E$ the unregularized infimum equals $u$, and locality gives equality of the regularization on the interior. Outside $\overline E$, the family is saturated under ball lowering, so its infimum is harmonic by Lemma 3.5.2; it is nonnegative, excluding $-\infty $. Order and scaling follow from the defining families. Sums of majorants majorize the sum; the unregularized inequality passes to regularizations almost everywhere by the polar discrepancy assertion, and then everywhere by ball-average representatives.

For bounded $u$ on $K$, in each component meeting $K$ choose a pole $a$. A multiple of $G(a,\cdot )$ majorizes $u$ on that component's compact intersection with $K$, since its positive lsc kernel has a positive minimum there. On all other components use $u$. This competitor shows $\widehat R_ Ku\le C G(a,\cdot )$ on the chosen component, so it has zero greatest harmonic minorant there. Components disjoint from $K$ have zero reduction. The global Riesz theorem makes the reduction a potential. For general $u=p+h$ with $p$ a potential and $h\ge 0$ harmonic, $h$ is bounded on $K$. The reduction of $p$ is dominated by $p$, hence has zero harmonic minorant; that of $h$ is a potential by the bounded case. Subadditivity bounds the reduction of $u$ by their sum, so it too has zero minorant and is a potential. $\square$

Proof. The reduction theorem proves the representation and, since $V_ K$ equals one on the interior and is harmonic outside $K$, its Riesz measure is supported on $\partial K$. If $p=G\mu \le 1$ is supported on $K$ and $v$ is admissible for $W_ K$, then $v-p$ is superharmonic and bounded below outside $K$. Its boundary lower limits are nonnegative: at $K$ use $v\ge 1$, at the ambient boundary use admissibility, outside a polar exception. At infinity use decay when required. The polar boundary minimum principle gives $v\ge p$, including on $K$. Infimize and regularize to get $V_ K\ge p$.

Choose a compact neighborhood $L$ of $K$, with $K\subset \operatorname {int}L$. Its potential $G\mu _ L=1$ on $K$. For measures $\mu ,\nu $ supported on $K$ with $G\mu \le G\nu $, reciprocity gives $\mu (K)=\int G\mu \, d\mu _ L\le \int G\nu \, d\mu _ L=\nu (K)$. Taking $\nu =\mu _ K$ proves the supremum formula and monotonicity.

If $K_ j\downarrow K$, put $g=\lim W_{K_ j}$. Its regularization is a potential dominated by $V_{K_1}$, harmonic off $K$ by harmonic convergence, and at most one. The greatest-potential characterization bounds it by $V_ K$, while monotonicity bounds it below by $V_ K$. Hence $g=V_ K=W_ K$ off $K$, and all $W_{K_ j}=W_ K=1$ on $K$, proving $W_{K_ j}\downarrow W_ K$. For a compact neighborhood $L$ of $K_1$ with its boundary outside $K_1$, reciprocity gives $C(K_ j)=\int W_{K_ j}\, d\mu _ L$. Bounded convergence then gives decreasing compact continuity.

For strong subadditivity first prove $W_{K\cup L}+W_{K\cap L}\le W_ K+W_ L$. On $K\cup L$ it follows from $W_{K\cup L}=1$ and order of the other reductions. On its complement the left side is harmonic. If $v,w$ are majorants for $K,L$, their sum minus that left side has nonnegative boundary lower limits: at the compact boundary, use that $W_{K\cup L}\le 1$ and that $W_{K\cap L}$ is continuous off $K\cap L$, while at points of the intersection it is bounded by one. At the ambient boundary use admissibility and at infinity use decay. Apply the polar minimum principle and infimize over $v,w$. Integrate the inequality against $\mu _ N$ for a larger compact neighborhood $N$; reciprocity changes each integral to its capacity, giving strong subadditivity.

Here are the extension details, to specify the precise Choquet interface. Compact continuity and compact neighborhoods imply $C^*(K)=C(K)$. For open sets strong subadditivity follows by decomposing each compact subset of their union into finitely many compact pieces in the respective opens, adding a compact subset of the intersection, and applying compact strong subadditivity. Taking almost optimal open majorants gives strong subadditivity of $C^*$. Repeated use gives the finite excess inequality

when the right side is finite; if a term is infinite use the equivalent inequality with sums on opposite sides. Continuity on increasing open sets is immediate from compactness. For increasing arbitrary $A_ i$ with bounded capacities choose open $O_ i\supset A_ i$ with $C(O_ i)<C^*(A_ i)+\varepsilon 2^{-i}$. The excess inequality bounds $C(\bigcup _{i\le m}O_ i)$ by $C^*(A_ m)+\varepsilon $. Continuity for open unions and then $\varepsilon \downarrow 0$ proves $C^*(\bigcup A_ i)=\lim C^*(A_ i)$. If the limit is infinite, monotonicity already proves equality. This also gives countable subadditivity.

The capacitability argument is purely axiomatic. In a compact Hausdorff space, a set $D=\bigcap _ i\bigcup _ kK_{ik}$ with compact $K_{ik}$ is capacitable for any such outer capacity: replace each row by increasing unions and successively choose row bounds so that the capacity of the remaining part stays above any fixed number below $C^*(D)$. The resulting decreasing compact row intersections have capacity at least that number, and their intersection lies in $D$. For a continuous map on $D$ into a compact subset of $\Omega $, take its graph in the product of the ambient compact space and that target compact. The graph is relatively closed in $D$ times the target, hence is again of the same compact-intersection form. Pull back capacity by the projection and apply the preceding argument. Continuous images commute with decreasing compact intersections because the closed fibers form nested nonempty compact sets. This proves capacitability of K-analytic sets in compact target neighborhoods; an increasing compact exhaustion and capacity continuity remove that restriction. Countable unions/intersections of these images are again of that form, by the product/equal-image construction for intersections and a disjoint union with one-point compactification for unions. Closed and open subsets of Euclidean space are countable unions of compacts, so the class of sets whose complements are also K-analytic is a sigma algebra containing the open sets. Thus all Borel sets are capacitable.

If $A\Subset \Omega $ is polar, a finite compactly supported fundamental measure producing its poles can be localized to a compact neighborhood in $\Omega $; its Green potential $p$ is infinite on $A$. For $O_ t=\{ p>t\} $ and compact $L\subset O_ t$, reciprocity gives $C(L)\le t^{-1}\int p\, d\mu _ L\le \mu (\Omega )/t$. Thus $C^*(A)=0$, and exhaustions handle arbitrary polar $A$. Conversely, locally separate a set $A$ of outer capacity zero from one interior point $a$ in a connected relatively compact neighborhood. Choose open $O_ j\supset A$ with $C(O_ j)<2^{-j}$, still separated from $a$. Increasing compact exhaustions of $O_ j$ have potentials increasing to $w_ j\le 1$, equal one on $O_ j$. The Green kernel at $a$ is bounded on the separated region, so $w_ j(a)\le M2^{-j}$. Hence $\sum _ jw_ j$ is superharmonic on that component, finite at $a$ and infinite on $A$. To perform the separation for a general local set, split it into its intersections with finitely many smaller closed neighborhoods, each avoiding a different interior ball; then use a countable local covering. Countable unions of polar sets are polar, proving the converse. $\square$

Proof. Work componentwise, choosing the prescribed point in its component and an arbitrary point in every other component outside the polar set. Cover $Z$ in the component by countably many small closed interior balls avoiding that point. The checked bounded-polar globalization with compact cutoffs gives finite measures supported in these balls, whose fundamental potentials are infinite on the covered polar pieces. Their Green potentials have the same poles by the local harmonic-correction identity, and are finite at the chosen point. Weight them so the sum of their masses and their potentials at that point is finite. The sum is a Green potential finite there and infinite on $Z$. Add the constant one for strict positivity, and glue the componentwise functions; superharmonicity is local.

The discrepancy between $R_ Eu$ and its regularization is polar. For $a\notin E$ choose the pole function $p$ for its discrepancy on $E$, finite at $a$. Then $\widehat R_ Eu+\varepsilon p$ is a majorant of $u$ on $E$; evaluating at $a$ and letting $\varepsilon \downarrow 0$ proves equality there. If a polar subset is removed from $E$, adding its $\varepsilon $-pole function to any majorant on the remainder proves the opposite inequality almost everywhere; ball averages make it hold for the regularizations everywhere. For increasing $F_ j$, their increasing regularized reductions have a superharmonic limit $v\le \widehat R_ Fu$. On each $F_ j$ they equal $u$ outside a polar set; the union of these exceptions is polar. Thus $v$ majorizes $u$ on $F$ outside a polar set, and the polar-removal argument gives $v\ge \widehat R_ Fu$.

Put $h=H^ W_ u$, which is finite harmonic because $0\le h\le u$ on $W$, and set $v=h$ on $W$, $v=u$ elsewhere. We justify the pasting even if $u$ has infinite boundary values. For any ball $B\Subset \Omega $, positivity of Perron solutions gives $H^ B_ v\le H^ B_ u\le u$ on $B\setminus W$. On $B\cap W$, compare the boundary values of $H^ B_ v$ and $v$, using the restriction lemma first from $B$ and then from $W$; both domains involved are bounded. This gives $H^ B_ v\le v$ there as well. If $\widehat v$ is the lsc regularization, $H^ B_{\widehat v}\le H^ B_ v\le v$; its harmonic continuity then gives $H^ B_{\widehat v}\le \widehat v$. The ball comparison characterization proves $\widehat v$ superharmonic. On $\partial W$, approximate the lsc boundary function $u$ increasingly by continuous functions. At regular points of $W$, their Perron solutions have the prescribed limits, so $\liminf _ W h\ge u$ there. The irregular set is polar. Hence $\widehat v\ge u$ on $\Omega \setminus W$ outside a polar set, proving $\widehat v\ge \widehat R_{\Omega \setminus W}u$. Conversely any defining majorant is in the upper Perron class for $u$ on $W$, so dominates $v$; regularization gives the reverse inequality. On $W$, $\widehat v=v=h$, proving the formula.

Linearity of integrable boundary solutions proves additivity inside $W$; outside $W$, both reductions equal the original function off a polar set. Ball averages remove that exception from the additive identity. In an exhaustion the reductions decrease by set order. Their interior harmonic limit is nonnegative and at most $u$. Every global harmonic minorant of $u$ is a lower Perron competitor on each $W_ j$, since it is continuous near $\partial W_ j$; it is therefore at most this limit. This identifies the greatest harmonic minorant. The zero characterization is the global Riesz theorem. $\square$

Proof. First let $\mu $ have compact support $K$ and $p$ be continuous. It is bounded on a compact neighborhood of $K$ and bounded outside it by the fixed-pole/Harnack bound, hence globally bounded. On $\Omega \setminus K$, $w-p$ is superharmonic and bounded below. At $K$ its lower boundary limits are nonnegative outside the stated polar exception by continuity of $p$ and lsc of $w$. At the ambient boundary the same holds because compact potentials tend to zero outside a polar set. If $d\ge 3$, at infinity $p\to 0$. The polar boundary minimum theorem gives $w\ge p$ off $K$; the inequality holds almost everywhere everywhere, and canonical ball averages give it at every point.

For general compactly supported $\mu $, the finite potential permits the Lusin construction of pairwise disjoint compact restrictions $\mu _ j$ with continuous potentials and $\mu =\sum \mu _ j$. To obtain them, apply Lusin to each residual potential, choose a compact subset of the residual carrier with arbitrarily small residual mass, and use the support-continuity principle; repeat with errors $2^{-j}$. The sum of the first $j$ restrictions has continuous potential, is dominated by $p$ on $K$, and therefore by $w$ everywhere by the first case. Nonnegative monotone convergence gives $p\le w$. Finally exhaust the support of a general Radon measure by increasing compact subsets. Each restriction has finite potential dominated by $p$, so the compact case applies. Monotone convergence finishes the proof. No convergence of signed differences at infinite values is used. $\square$