Theorem 4.2.1497A5DA659CA1

Proof. We show that $r$ is harmonic off $A$ by replacing admissible functions with their Poisson integrals on balls in $B\setminus A$ (Steps 1–3), and then identify $r$ with the Green potential of its Riesz measure (Step 4).

Step 1: capped admissible functions. Choose a closed ball $L\subset B$ whose interior contains $A$, and a constant $M\geq 0$ with $s\leq M$ on $A$. The relative extremal function $u_ L$ takes values in $[-1,0]$ and tends to zero at $\partial B$ by part (1) of proposition 4.1.16; it equals $-1$ on $\operatorname{Int}L$, because it is the upper semicontinuous regularization of the envelope $f_ L$, which equals $-1$ on $L$. Hence $h_0=-Mu_ L$ is a nonnegative superharmonic function on $B$, bounded by $M$, which majorizes $s$ on $A$ and tends to zero at $\partial B$. If $h$ is admissible, so is $\min \{ h,h_0\} $. Therefore $R_ s^ A$ is also the infimum of the admissible functions $h$ with $h\leq h_0$; we call these capped. In particular $0\leq r\leq R_ s^ A\leq h_0\leq M$, so $r$ is bounded and tends to zero at $\partial B$.

Step 2: Poisson modification. Let $D$ be an open ball with $\overline D\subset B\setminus A$, and let $h$ be a capped admissible function. Its restriction to $\partial D$ is bounded and lower semicontinuous, so by lemma 3.4.4, applied to $-h$, it is the increasing limit of continuous functions $g_ k$ on $\partial D$. By monotone convergence, the Poisson integrals $H_ Dg_ k$ of theorem 3.1.18 increase to

Since $0\leq h\leq M$ and the Poisson kernel has integral one, $0\leq H\leq M$, and $H$ is harmonic by part (2) of corollary 3.1.21. For $\zeta \in \partial D$, lower semicontinuity gives $\varliminf _{y\to \zeta }h(y)\geq h(\zeta )\geq g_ k(\zeta )$, so the comparison principle for the superharmonic function $h$ (theorem 3.4.5) gives $H_ Dg_ k\leq h$ on $D$; hence $H\leq h$ on $D$. On the other hand, $H\geq H_ Dg_ k$, and $H_ Dg_ k$ extends continuously to $\overline D$ with boundary values $g_ k$; letting $k\to \infty $, we get

Define $h^ D=H$ on $D$ and $h^ D=h$ on $B\setminus D$. The last inequality and the lower semicontinuity of $h$ show that $h^ D$ is lower semicontinuous, and $0\leq h^ D\leq h$.

We check that $h^ D$ is superharmonic, using the converse part of theorem 3.4.5. Let $E$ be a ball with $\overline E\subset B$, and let $q$ be harmonic on $E$ and continuous on $\overline E$ with $\varliminf _{E\ni y\to \zeta }h^ D(y)\geq q(\zeta )$ for $\zeta \in \partial E$. Since $h\geq h^ D$, the function $h$ satisfies the same boundary inequality, so $h\geq q$ on $E$. In particular $h^ D=h\geq q$ on $E\setminus D$. On the open set $E\cap D$ consider the harmonic function $H-q$. At boundary points of $E\cap D$ lying on $\partial E$, its lower limits are nonnegative by the hypothesis on $q$. At boundary points $\zeta \in E\cap \partial D$, they are at least $h(\zeta )-q(\zeta )\geq 0$ by the inequality displayed above. The minimum principle (proposition 3.1.10) gives $H\geq q$ on $E\cap D$. Hence $h^ D\geq q$ on $E$, and $h^ D$ is superharmonic. Since $D\cap A=\varnothing $, the function $h^ D$ is again a capped admissible function. Finally, $h^ D$ depends monotonically on $h$, by the integral formula for $H$.

Step 3: harmonicity off $A$. By Choquet's lemma (lemma 3.4.22), applied to the negatives of the capped admissible functions, there is a sequence of capped admissible functions whose infimum has the same lower semicontinuous regularization $r$ as $R_ s^ A$. By theorem 3.4.23, applied to the negatives of this sequence, the infimum equals $r$ almost everywhere. Replacing the functions of the sequence by their successive minima, we obtain a decreasing sequence $(h_ j)$ of capped admissible functions whose limit equals $r$ almost everywhere.

Now fix a ball $D$ as in Step 2. On $D$, the modified functions $h_ j^ D$ are harmonic, decrease, and lie between $0$ and $M$; by part (3) of corollary 3.1.21, they converge on $D$ to a harmonic function $\Phi $. Each $h_ j^ D$ is admissible, so $h_ j^ D\geq R_ s^ A\geq r$; moreover $h_ j^ D\leq h_ j$. Thus $r\leq \Phi \leq \lim _ jh_ j=r$ almost everywhere on $D$. The bounded superharmonic functions $r$ and $\Phi $ therefore agree almost everywhere, hence everywhere on $D$ by theorem 3.4.12. Since such balls cover $B\setminus A$, the function $r$ is harmonic on $B\setminus A$.

Step 4: Green representation. The function $-r$ is bounded and subharmonic on $B$, so $\nu =-\Delta r$ is a positive measure by part (1) of theorem 3.4.12. As $r$ is harmonic on $B\setminus A$, the measure $\nu $ is supported on the compact set $A$ and is therefore finite. The function $-r$ is bounded, hence locally integrable, its Riesz measure $\nu $ has compact support in $B$, and it tends to zero at $\partial B$; so the converse in part (2) of proposition 4.1.16 gives $-r=-G_ B\nu $, that is, $r=G_ B\nu $ on $B$. The reduction $R_ s^ A$ equals $s$ on $A$, and $r=R_ s^ A$ outside a polar set, so $r=s$ quasi-everywhere on $A$. Finally $I_ B(\nu )=\int r\, \mathrm d\nu \leq M\nu (A)<\infty $. $\square$