4.2 Fine topology and balayage8618E9B4AA3D

Proof. Suppose first that $A$ is thin at $x$. If $x$ does not belong to the Euclidean closure of $A\setminus \{ x\} $, some Euclidean neighborhood of $x$ does not meet $A\setminus \{ x\} $. Otherwise let $u$ be a subharmonic function on a Euclidean open neighborhood $V$ of $x$ as in definition 4.2.3, and choose $a\in \mathbb R$ strictly between the upper limit in that definition and $u(x)$. By the definition of the upper limit, after shrinking $V$ we have $u<a$ on $(A\setminus \{ x\} )\cap V$. Hence $\{ y\in V:u(y)>a\} $ is a fine neighborhood of $x$ that does not meet $A\setminus \{ x\} $.

Conversely, let $N$ be a fine neighborhood of $x$. By the description of the basis above, $N$ contains a set

where $V$ is a Euclidean open neighborhood of $x$; adding the function $0$ with the constant $-1$ if necessary, we may assume that $m\geq 1$. Each $u_ i(x)$ is finite, since $u_ i(x)>a_ i$. Put $b_ i=u_ i(x)-a_ i>0$ and

Then $v(x)=0$. We claim that

There is nothing to prove if $x$ is not an accumulation point of $V\setminus N_0$. Otherwise let $y_ k\to x$ be a sequence in $V\setminus N_0$ along which $v(y_ k)$ converges to the upper limit. For each $k$ some index $i$ satisfies $u_ i(y_ k)\leq a_ i$, that is, $(u_ i(y_ k)-u_ i(x))/b_ i\leq -1$; passing to a subsequence, we may assume that this index $i$ is the same for all $k$. For every other index $j$, upper semicontinuity of $u_ j$ at $x$ gives $\varlimsup _ k(u_ j(y_ k)-u_ j(x))/b_ j\leq 0$. Hence $\lim _ kv(y_ k)\leq -1$, which proves the claim.

If $N$ does not meet $A\setminus \{ x\} $, then $(A\setminus \{ x\} )\cap V\subset V\setminus N_0$, and the claim shows that either $x$ is not in the Euclidean closure of $A\setminus \{ x\} $ or the upper limit of $v$ along $A\setminus \{ x\} $ is at most $-1<v(x)$. In both cases $A$ is thin at $x$. For the last assertion, the claim provides a Euclidean open neighborhood $V'\subset V$ of $x$ such that $v<-1/2$ on $V'\setminus N_0$. Thus $\{ y\in V':v(y)>-1/2\} \subset N_0\subset N$, and $v(x)=0>-1/2$. $\square$

Proof. By proposition 4.2.4, $N$ contains a set $\{ y\in V:u(y)>a\} $ with $u\in \mathrm{SH}(V)$ and $u(x)>a$. It suffices to consider balls $B(x,r)$ with $\overline{B(x,r)}\subset V$. The set $E=\{ y\in B(x,r):u(y)>a\} $ is a Borel subset of $N$, because $u$ is Borel measurable. Let $M$ be a finite upper bound for $u$ on $\overline{B(x,r)}$; then $M\geq u(x)>a$. Since $u\leq M$ on $E$ and $u\leq a$ on $B(x,r)\setminus E$, the ball submean inequality of theorem 3.4.5 gives

Hence $|E|\geq |B(x,r)|\, (u(x)-a)/(M-a)>0$. As a singleton has Lebesgue measure zero, it is not a fine neighborhood of its point. $\square$

Proof. We construct a subharmonic function on $\mathbb R^ d$ that is finite at $x$ and equals $-\infty $ on $A\setminus \{ x\} $ near $x$. Fix $x\in \Omega $ and $\rho >0$ with $\overline{B(x,\rho )}\subset \Omega $. For $k\geq 0$ put $r_ k=2^{-k}\rho $, let $S_ k=\{ y:r_{k+1}\leq |y-x|\leq r_ k\} $ be the closed annulus, and let $A_ k=A\cap S_ k$; these sets cover $A\cap \overline{B(x,\rho )}\setminus \{ x\} $. Each $A_ k$ is polar and has compact closure, so proposition 4.1.11 provides a finite positive measure $\nu _ k'$ of compact support with $E_ d*\nu _ k'=-\infty $ on $A_ k$. We restrict it to the open set $U_ k=\{ y:r_{k+2}<|y-x|<2\rho \} $, which contains $S_ k$, and put $\nu _ k=\nu _ k'|_{U_ k}$. For $y\in A_ k$ and $z\in \operatorname{Supp}\nu _ k'\setminus U_ k$ we have $|y-z|\geq \operatorname {dist}(S_ k,\mathbb R^ d\setminus U_ k)>0$, so $E_ d(y-z)$ is bounded on this set of pairs and

Moreover, $\nu _ k$ is supported in $\overline{U_ k}$, which does not meet $B(x,r_{k+2})$, so $p_ k=E_ d*\nu _ k$ is finite at $x$.

Choose $c_ k>0$ with

and put $\nu =\sum _ kc_ k\nu _ k$, a finite measure supported in $\overline{B(x,2\rho )}$, and $p=E_ d*\nu $. Let $C$ be an upper bound for $E_ d(y-z)$ when $|y-z|\leq 3\rho $. For $y\in \overline{B(x,\rho )}$, monotone convergence applied to the nonnegative functions $C-E_ d(y-\cdot )$ gives

where every term on the right is nonpositive. Hence $p(x)$ is finite, and $p=-\infty $ on $A\cap \overline{B(x,\rho )}\setminus \{ x\} $, because there one term equals $-\infty $. Since $p$ is subharmonic on $\mathbb R^ d$ (proposition 3.4.11), the set

is a fine neighborhood of $x$, and $N\cap A\subset \{ x\} $. If $x\notin A$, this shows that $\Omega \setminus A$ is a fine neighborhood of each of its points, so $A$ is finely closed. If $x\in A$, it shows that $\{ x\} $ is relatively open in $A$ for the fine topology. $\square$

Proof. The function $G$ is upper semicontinuous, because $\{ G<a\} =\{ g<a\} \times \mathbb R^{d-k}$ for every real $a$. Since subharmonicity is a local property, it suffices to show that $G$ is subharmonic on $D\times \mathbb R^{d-k}$ for every open ball $D=B(z_0,s)\subset \mathbb R^ k$ with $\overline{B(z_0,2s)}\subset V$. If $g$ is identically $-\infty $ on $B(z_0,2s)$, then so is $G$ on $B(z_0,2s)\times \mathbb R^{d-k}$, and there is nothing to prove. Otherwise $g$ is locally integrable on the ball $B(z_0,2s)$ by proposition 3.4.8.

For $0<\varepsilon <s$, the $k$-dimensional mollification $g_\varepsilon =g*\rho _\varepsilon $ is smooth on $D$ and satisfies $\Delta _ zg_\varepsilon =(\Delta g)*\rho _\varepsilon \geq 0$ there, by part (1) of theorem 3.4.12 and parts (1) and (3) of proposition 2.5.3; moreover, $g_\varepsilon $ decreases to $g$ on $D$ as $\varepsilon \downarrow 0$, by (3.4.13). The function $G_\varepsilon (z,y)=g_\varepsilon (z)$ is smooth on $D\times \mathbb R^{d-k}$, with $\Delta _ xG_\varepsilon (z,y)=\Delta _ zg_\varepsilon (z)\geq 0$. By (3.1.6), its sphere means in $\mathbb R^ d$ increase with the radius, and they tend to its value at the center as the radius tends to zero, because $G_\varepsilon $ is continuous. Hence $G_\varepsilon $ satisfies the submean inequality and is subharmonic on $D\times \mathbb R^{d-k}$. Along the sequence $\varepsilon _ j=s/(j+1)$, the functions $G_{\varepsilon _ j}$ decrease to $G$, which is therefore subharmonic on $D\times \mathbb R^{d-k}$ by theorem 3.4.9. $\square$

Proof. A similarity $x\mapsto a+tx$ with $t>0$ preserves upper semicontinuity and sphere means, and hence maps polar sets to polar sets. It therefore suffices to find one uncountable compact polar subset of $\mathbb R^ d$. For $d=2$ we take the Cantor set $K$ of proposition 4.1.12. It is compact and polar, and it is uncountable because it carries an atom-free probability measure.

For $d>2$, write $x=(z,y)\in \mathbb R^2\times \mathbb R^{d-2}$. By lemma 4.2.7, applied with $k=2$ to the subharmonic function $p$ of proposition 4.1.12, the function $P(x)=p(z)$ is subharmonic on $\mathbb R^ d$. It is finite when $z\notin K$ and equals $-\infty $ on $K\times \mathbb R^{d-2}$. Hence $K\times \{ 0\} $ is an uncountable compact polar subset of $\mathbb R^ d$.

Now let $\Omega \neq \varnothing $, and let $K'\subset \Omega $ be an uncountable compact polar set. By proposition 4.2.6, $K'$ is finely closed, and its induced fine topology is discrete. A closed subspace $F$ of a Lindelöf space $X$ is Lindelöf: extend the members of an open cover of $F$ to open subsets of $X$, add $X\setminus F$, and pass to a countable subcover. On the other hand, the cover of an uncountable discrete space by its singletons has no countable subcover. Hence the fine topology on $\Omega $ is not Lindelöf. Nor is it second countable, because a second countable space is Lindelöf: given an open cover and a countable basis, choose, for each basis element contained in some member of the cover, one such member; these countably many members cover the space. $\square$

Proof. We first prove the gradient identity for the partial sums of a decomposition of $\mu $ and $\nu $ into pieces with bounded Green potentials, to which proposition 4.1.29 applies, and then pass to the limit. Translating, we may assume that $B=B(0,R)$ for some $R>0$, and we choose $r<R$ such that the supports of $\mu $ and $\nu $ lie in $B(0,r)$. By lemma 4.1.14, the function $\kappa (x,y)=G_ B(x,y)-\Gamma _ d(x-y)$ extends to a continuous function on the compact set $\overline B\times \overline{B(0,r)}$; let $C$ be a bound for $|\kappa |$ there. Since $\Gamma _ d=-E_ d$, every finite positive measure $\lambda $ supported in $B(0,r)$ satisfies

and the last integral is a continuous function of $x\in \overline B$ bounded by $C\lambda (B)$.

Step 1: decomposition. Since $\int G_ B\mu \, \mathrm d\mu =I_ B(\mu )<\infty $, the Green potential $G_ B\mu $ is finite $\mu $-almost everywhere, and by the display so is $E_ d*\mu $. By part (2) of theorem 4.1.10, $\mu $ is therefore the sum of its restrictions to countably many disjoint compact sets, each of which has a finite and continuous fundamental potential. Let $\mu _ n$ be the sum of the first $n$ of these restrictions. Since $E_ d*\mu _ n$ is finite and continuous, the display shows that $G_ B\mu _ n$ is finite, continuous and bounded on $B$; this proves the last assertion of the proposition. By part (2) of proposition 4.1.16, $-G_ B\mu _ n$ is a subharmonic function on $B$ with Riesz measure $\mu _ n$, which extends harmonically across $\partial B$ and vanishes on $\partial B$; in particular, it tends to zero at $\partial B$. Thus proposition 4.1.29 applies to the functions $-G_ B\mu _ n$, $n\geq 1$. Hence $\nabla G_ B\mu _ n\in L^2(B)$ and

Step 2: the gradient of $G_ B\mu $. For $m>n$, expanding the square with the last display gives

where the inequality holds because $0\leq \mu _ m-\mu _ n\leq \mu -\mu _ n$ and $G_ B\geq 0$. The measure $\mu -\mu _ n$ is the restriction of $\mu $ to a decreasing sequence of sets whose intersection is $\mu $-null. Since $G_ B$ is nonnegative and integrable for $\mu \otimes \mu $, dominated convergence gives $I_ B(\mu -\mu _ n)\to 0$. Thus $(\nabla G_ B\mu _ n)$ is a Cauchy sequence in $L^2(B)$; let $g$ be its limit.

On the other hand, $G_ B\mu _ n\uparrow G_ B\mu $ pointwise by monotone convergence, and $G_ B\mu \in L^1(B)$. Indeed, for $y\in \operatorname{Supp}\mu $, the bound $|\kappa |\leq C$ and the local integrability of $\Gamma _ d$ (proposition 3.1.13) give

and integrating against $\mu $, by Tonelli's theorem, shows that $\int _ BG_ B\mu \, \mathrm d x$ is at most $\mu (B)$ times the right side. Hence $G_ B\mu _ n\to G_ B\mu $ in $L^1(B)$ by monotone convergence. For a smooth compactly supported vector field $\varphi $ on $B$ we obtain

so $g=\nabla G_ B\mu $. Since $\| \nabla G_ B\mu _ n\| _2^2=I_ B(\mu _ n)\uparrow I_ B(\mu )$ by monotone convergence, we get $\| \nabla G_ B\mu \| _2^2=I_ B(\mu )$.

Step 3: the mutual energy. Construct the partial sums $\nu _ n$ of $\nu $ in the same way. By proposition 4.1.29 and the Cauchy–Schwarz inequality in $L^2(B)$,

As $n\to \infty $, the left side increases to $I_ B(\mu ,\nu )$ by monotone convergence, so the mutual energy is finite, while the integral converges to $\int _ B\nabla G_ B\mu \mathbin {\cdot }\nabla G_ B\nu \, \mathrm d x$ by convergence in $L^2(B)$. This proves the identity for positive measures. For signed measures, we expand both sides bilinearly in the positive and negative parts.

Step 4: positive definiteness. Let $\sigma $ be a signed measure as in the statement, and put $G_ B\sigma =G_ B\sigma ^+-G_ B\sigma ^-$, a function in $L^1(B)$ by Step 2. Then $I_ B(\sigma )=\| \nabla G_ B\sigma \| _2^2\geq 0$. Suppose that $I_ B(\sigma )=0$, so that $\nabla G_ B\sigma =0$ as a distribution on $B$. Since $B$ is connected, corollary 2.5.5 shows that the distribution defined by $G_ B\sigma $ is a constant. On the other hand, $-\Delta G_ B\sigma =\sigma $ in the sense of distributions, by part (2) of proposition 4.1.16. Hence $\sigma ^+$ and $\sigma ^-$, which are Radon measures by proposition 2.2.2, define the same distribution, so $\sigma ^+=\sigma ^-$ by part (4) of proposition 2.5.3, that is, $\sigma =0$. Finally, the Cauchy–Schwarz inequality for $I_ B$ follows from the gradient identity and the Cauchy–Schwarz inequality in $L^2(B)$. $\square$

Proof. By part (3) of proposition 4.1.16, the function $w_ K=-u_ K$ is the Green potential $G_ B\lambda _ K$, and $0\leq w_ K\leq 1$ because $-1\leq u_ K\leq 0$. Since $u_ K$ is the upper semicontinuous regularization of the envelope $f_ K$, which equals $-1$ on $K$, part (2) of theorem 4.1.26 shows that $w_ K=1$ quasi-everywhere on $K$. In particular $I_ B(\lambda _ K)=\int w_ K\, \mathrm d\lambda _ K\leq \lambda _ K(K)$, so $\lambda _ K$ has finite Green energy. As measures of finite Green energy do not charge polar sets, we obtain, with $m=C_ B(K)$,

for every probability measure $\tau $ on $K$ with finite Green energy; in the second identity we used the symmetry of $G_ B$.

Suppose that $m>0$, and let $\tau $ be a probability measure on $K$. If $I_ B(\tau )<\infty $, the Cauchy–Schwarz inequality of proposition 4.2.11 gives $1=I_ B(\lambda _ K,\tau )^2\leq mI_ B(\tau )$; thus $I_ B(\tau )\geq 1/m$ in all cases. Equality holds for $\tau =\lambda _ K/m$, whose energy is $m/m^2=1/m$. Hence the infimum equals $1/m$, which proves the formula. If $\tau _1\neq \tau _2$ were two minimizers, their midpoint would again be a probability measure on $K$, and by positive definiteness

a contradiction. Suppose next that $m=0$. Then $K$ is polar by theorem 4.1.17, so it carries no probability measure of finite energy, and the infimum is $+\infty $ (if $K=\varnothing $, the family is empty).

For the last assertion, recall that balls are regular domains (theorem 3.1.29). The monotonicity of Green functions in the domain (theorem 3.1.34) gives $G_ D\leq G_ B$ on $D\times D$. Hence $I_ B(\tau )\geq I_ D(\tau )$ for every measure $\tau $ supported on a compact set $K\subset D$, and the variational formula, applied in $B$ and in $D$, gives $C_ B(K)\leq C_ D(K)$. Taking suprema over compact subsets, we obtain $C_ B(O)\leq C_ D(O)$ for every open $O\Subset D$. Every such $O$ containing $A$ is also allowed in the infimum defining $C_ B^*(A)$, so $C_ B^*(A)\leq C_ D^*(A)$. $\square$

Proof. Enlarging $A$ or $s$ shrinks the class of admissible functions, which proves the two monotonicity assertions. For $c>0$, multiplication by $c$ is a bijection between the admissible functions for $s$ and those for $cs$; for $c=0$, both sides vanish (with $0\cdot \infty =0$), since the zero function is admissible. If $h_1$ and $h_2$ are admissible for $s$ and $t$ respectively, then $h_1+h_2$ is admissible for $s+t$, so $R_{s+t}^ A\leq R_ s^ A+R_ t^ A$. Monotonicity and homogeneity are preserved by lower semicontinuous regularization. For subadditivity, recall that each reduction agrees with its regularization outside a polar set, hence almost everywhere. Therefore $\widehat R_{s+t}^ A\leq \widehat R_ s^ A+\widehat R_ t^ A$ almost everywhere. Both sides are superharmonic, so the inequality holds everywhere by the ball-average formula above.

Now let $U$ be bounded and $A$ polar. Then $A$ has compact closure, so proposition 4.1.11 provides a finite positive measure $\nu $ of compact support with $E_ d*\nu =-\infty $ on $A$. The kernel $E_ d$ is bounded above on the bounded set of differences $x-y$ with $x\in U$ and $y\in \operatorname{Supp}\nu $, so adding a constant to $-E_ d*\nu $ gives a nonnegative superharmonic function $p$ on $U$ with $p=+\infty $ on $A$; by proposition 3.4.11, $p$ is locally integrable. For every $\varepsilon >0$, the function $\varepsilon p$ is admissible, even at points of $A$ where $s=+\infty $. Hence $R_ s^ A\leq \varepsilon p$ for every $\varepsilon >0$, and so $R_ s^ A=0$ on the set $\{ p<\infty \} $. This set has full Lebesgue measure, because $p$ is locally integrable, and is therefore dense in $U$. Consequently $\widehat R_ s^ A=0$. $\square$

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 $. $\square$

Proof. The function $G_ B\mu $ is nonnegative, superharmonic by part (2) of proposition 4.1.16, and bounded on $A$. Applying theorem 4.2.14 to $s=G_ B\mu $, we see that $\widehat R_{G_ B\mu }^ A$ is the Green potential of a positive measure $\mu ^ A$ on $A$ with finite Green energy, and that $G_ B\mu ^ A=G_ B\mu $ quasi-everywhere on $A$. Since $G_ B\mu $ is itself admissible, $G_ B\mu ^ A=\widehat R_{G_ B\mu }^ A\leq G_ B\mu $. The measure $\mu ^ A$ is unique because $-\Delta G_ B\nu =\nu $ for every measure $\nu $, by part (2) of proposition 4.1.16, so that a measure is determined by its Green potential. Since $-G_ B\mu \leq -G_ B\mu ^ A$ on $B$, the comparison of boundary fluxes in part (2) of proposition 4.1.16, applied with $\lambda =\mu $ and $\lambda '=\mu ^ A$, gives the mass bound $\mu ^ A(A)\leq \mu (B)$.

Let $\tau $ be a positive measure on $A$ with finite Green energy. It does not charge polar sets, so we may integrate the equality $G_ B\mu =G_ B\mu ^ A$, valid quasi-everywhere on $A$, against $\tau $. Both integrals are finite, since $0\leq G_ B\mu ^ A\leq G_ B\mu $ and $G_ B\mu $ is bounded on $A$. By Tonelli's theorem and the symmetry of $G_ B$, the two integrals are $I_ B(\mu ,\tau )$ and $I_ B(\mu ^ A,\tau )$, which proves (4.2.17).

Suppose that $\nu _1$ and $\nu _2$ are positive measures on $A$ with finite Green energy which both satisfy (4.2.17) in place of $\mu ^ A$. Then $I_ B(\nu _1,\tau )=I_ B(\mu ,\tau )=I_ B(\nu _2,\tau )$ for $\tau =\nu _1$ and $\tau =\nu _2$, so $\nu _1=\nu _2$ by the uniqueness criterion at the beginning of this section. If $\mu _1$ and $\mu _2$ satisfy the boundedness hypothesis, so does $\mu _1+\mu _2$, and $\mu _1^ A+\mu _2^ A$ is a positive measure on $A$ with finite Green energy satisfying (4.2.17) for the source $\mu _1+\mu _2$. By the characterization, $(\mu _1+\mu _2)^ A=\mu _1^ A+\mu _2^ A$. The same argument gives $(c\mu _1)^ A=c\mu _1^ A$. $\square$

Proof. We sweep the partial sums of a decomposition of $\mu $ into pieces with bounded Green potentials, show that the swept measures increase to a measure $\nu $ satisfying (4.2.17), and then identify $G_ B\nu $ with the regularized reduction. By the last assertion of proposition 4.2.11, $\mu $ is the sum of its restrictions to countably many pairwise disjoint compact sets, and the partial sums $\mu _ n$ of this series, which satisfy $\mu _ n\uparrow \mu $, have bounded Green potentials. Hence theorem 4.2.16 applies to the measures $\mu _ n$ and to the differences $\mu _ m-\mu _ n$ with $m>n$, whose Green potentials are bounded by $G_ B\mu _ m$.

By additivity, $\mu _ m^ A=\mu _ n^ A+(\mu _ m-\mu _ n)^ A\geq \mu _ n^ A$. Hence the measures $\mu _ n^ A$ increase setwise, and $\nu (E)=\lim _ n\mu _ n^ A(E)$ defines a measure $\nu $ on $A$: if a Borel set $E$ is the disjoint union of Borel sets $E_ k$, the monotone convergence theorem for series gives $\nu (E)=\lim _ n\sum _ k\mu _ n^ A(E_ k)=\sum _ k\nu (E_ k)$. Moreover $\nu (A)=\lim _ n\mu _ n^ A(A)\leq \mu (B)$. For every nonnegative Borel function $f$ we have

This is clear for simple functions, and in general it follows by approximating $f$ from below by an increasing sequence of simple functions $f_ k$ and interchanging the suprema over $k$ and $n$ of $\int f_ k\, \mathrm d\mu _ n^ A$.

Testing (4.2.17) for $\mu _ n$ against $\tau =\mu _ n^ A$ and using the Cauchy–Schwarz inequality, we get

so that $I_ B(\mu _ n^ A)\leq I_ B(\mu _ n)\leq I_ B(\mu )$. We claim that $I_ B(\mu _ n^ A)\uparrow I_ B(\nu )$. The sequence increases and is bounded by $I_ B(\nu )$, because $\mu _ n^ A\leq \mu _{n+1}^ A\leq \nu $ and $G_ B\geq 0$. Conversely, applying (4.2.19) to $f=G_ B(x,\cdot )$, we get $G_ B\mu _ m^ A\uparrow G_ B\nu $ pointwise as $m\to \infty $. For $m\leq n$ we have $I_ B(\mu _ n^ A)\geq \int G_ B\mu _ m^ A\, \mathrm d\mu _ n^ A$, and by (4.2.19) the right side tends to $\int G_ B\mu _ m^ A\, \mathrm d\nu $ as $n\to \infty $; letting $m\to \infty $ and using monotone convergence for the fixed measure $\nu $, we obtain $\lim _ nI_ B(\mu _ n^ A)\geq \int G_ B\nu \, \mathrm d\nu =I_ B(\nu )$. This proves the claim, and hence $I_ B(\nu )\leq I_ B(\mu )<\infty $.

Now let $\tau $ be a positive measure on $A$ with finite Green energy. We have $I_ B(\mu _ n,\tau )=I_ B(\mu _ n^ A,\tau )$ for every $n$. As $n\to \infty $, the left side $\int G_ B\tau \, \mathrm d\mu _ n$ increases to $\int G_ B\tau \, \mathrm d\mu =I_ B(\mu ,\tau )$ by monotone convergence, since $\mu _ n$ is the restriction of $\mu $ to an increasing sequence of sets whose union carries $\mu $, and the right side $\int G_ B\tau \, \mathrm d\mu _ n^ A$ increases to $I_ B(\nu ,\tau )$ by (4.2.19). Hence $I_ B(\mu ,\tau )=I_ B(\nu ,\tau )$, which is (4.2.17) for $\nu $.

By monotone convergence, $G_ B\mu _ n\uparrow G_ B\mu $ pointwise on $B$, and we have seen that $G_ B\mu _ n^ A\uparrow G_ B\nu $. Since $G_ B\mu _ n^ A\leq G_ B\mu _ n$, it follows that $G_ B\nu \leq G_ B\mu $ everywhere. Moreover, $G_ B\mu _ n^ A=G_ B\mu _ n$ on $A$ outside a polar set $E_ n$, so $G_ B\nu =G_ B\mu $ on $A\setminus E$, where $E=\bigcup _ nE_ n$ is polar by proposition 4.1.11.

We now identify $\widehat R_{G_ B\mu }^ A$ with $G_ B\nu $. Since $B$ is bounded and $E$ is polar, proposition 4.2.13, applied with $U=B$ and with $E$ in place of $A$, provides a nonnegative superharmonic function $p$ on $B$, locally integrable and hence finite almost everywhere, with $p=+\infty $ on $E$. For every $\varepsilon >0$, the function $G_ B\nu +\varepsilon p$ is nonnegative and superharmonic, and it is at least $G_ B\mu $ at every point of $A$: on $A\setminus E$ because $G_ B\nu =G_ B\mu $ there, and on $E$ because $p=+\infty $. Hence $R_{G_ B\mu }^ A\leq G_ B\nu +\varepsilon p$. Letting $\varepsilon \downarrow 0$, we get $R_{G_ B\mu }^ A\leq G_ B\nu $ wherever $p<\infty $, that is, almost everywhere. Thus $\widehat R_{G_ B\mu }^ A\leq G_ B\nu $ almost everywhere, and hence everywhere by the ball-average formula preceding proposition 4.2.13, since both functions are superharmonic. Conversely, $G_ B\mu _ n\leq G_ B\mu $, so proposition 4.2.13 gives $\widehat R_{G_ B\mu }^ A\geq \widehat R_{G_ B\mu _ n}^ A=G_ B\mu _ n^ A$ for every $n$, and therefore $\widehat R_{G_ B\mu }^ A\geq G_ B\nu $. Consequently $\widehat R_{G_ B\mu }^ A=G_ B\nu $, and we put $\mu ^ A=\nu $; this measure is unique because a measure is determined by its Green potential.

It remains to collect the other conclusions. The inequality $G_ B\mu ^ A\leq G_ B\mu $, the equality quasi-everywhere on $A$, the mass bound and the orthogonality relation have been proved above. If $\nu _1$ and $\nu _2$ are positive measures on $A$ with finite Green energy satisfying (4.2.17) in place of $\mu ^ A$, then $I_ B(\nu _1,\tau )=I_ B(\mu ,\tau )=I_ B(\nu _2,\tau )$ for $\tau =\nu _1$ and $\tau =\nu _2$, so $\nu _1=\nu _2$ by the uniqueness criterion at the beginning of this section; thus (4.2.17) characterizes $\mu ^ A$. If $\mu _1$ and $\mu _2$ have finite Green energy, so has $\mu _1+\mu _2$, because the mutual energy $I_ B(\mu _1,\mu _2)$ is finite; and $\mu _1^ A+\mu _2^ A$ is a positive measure on $A$ with finite Green energy which satisfies (4.2.17) for the source $\mu _1+\mu _2$. By the characterization, $(\mu _1+\mu _2)^ A=\mu _1^ A+\mu _2^ A$, and in the same way $(c\mu _1)^ A=c\mu _1^ A$ for $c\geq 0$. Taking $\tau =\mu ^ A$ in (4.2.17) gives $I_ B(\mu ,\mu ^ A)=I_ B(\mu ^ A)$, and then

which is nonnegative by proposition 4.2.11. Finally, if $\mu $ is supported on $A$, then $\mu $ itself satisfies (4.2.17), so $\mu ^ A=\mu $ by the characterization. $\square$

Proof. We first check that $\partial W$ is the disjoint union of $\partial A$ and $\partial B$. Since $W=B\cap (\mathbb R^ d\setminus A)$ is the intersection of two open sets, $\partial W\subset \partial B\cup \partial A$, and $\partial A\subset A\subset B$ does not meet $\partial B$. Conversely, no point of $\partial B\cup \partial A$ lies in $W$. A point of $\partial B$ is a limit of points of $B\setminus A=W$, because the compact set $A\subset B$ has positive distance from $\partial B$. Every small ball around a point of $\partial A$ lies in $B$ and contains points outside $A$, that is, points of $W$. Consequently $\partial A$ and $\partial B$ are disjoint compact sets whose union is $\partial W$, and $f_ x$ is continuous on $\partial W$, because $G_ B(x,\cdot )$ is continuous on $B\setminus \{ x\} \supset A$.

Let $H=H_ Wf_ x$ be the solution of the Dirichlet problem on $W$ with boundary values $f_ x$; it is continuous on $\overline W$ since $W$ is regular, and $H\geq 0$ because $f_ x\geq 0$ and $H_ W$ preserves order. The function $G_ B(x,\cdot )=G_ B\delta _ x$ is superharmonic on $B$ by part (2) of proposition 4.1.16, it is continuous on $B\setminus \{ x\} $, and it tends to zero at $\partial B$ (theorem 3.1.34); hence its boundary limits on $\partial W$ are the values of $f_ x$. Therefore the maximum principle (proposition 3.4.7), applied to the subharmonic function $H-G_ B(x,\cdot )$ on the bounded open set $W$, gives $H\leq G_ B(x,\cdot )$ on $W$. Define $r=H$ on $W$ and $r=G_ B(x,\cdot )$ on $A$. Then $r$ is continuous on $B$, because $H$ and $G_ B(x,\cdot )$ have the same values on $\partial A$, and $0\leq r\leq G_ B(x,\cdot )$.

To see that $r$ is superharmonic, we argue as in Step 2 of the proof of theorem 4.2.14. Let $E$ be a ball with $\overline E\subset B$, and let $q$ be harmonic on $E$ and continuous on $\overline E$ with $q\leq r$ on $\partial E$. Then $q\leq G_ B(x,\cdot )$ on $\partial E$, hence on $E$ by superharmonicity of $G_ B(x,\cdot )$; in particular $r\geq q$ on $E\cap A$. On $E\cap W$ the harmonic function $H-q$ has nonnegative boundary limits: at points of $\partial E$ because $H$ extends continuously to $r\geq q$ there, and at points of $E\cap \partial A$ because $H$ tends to $G_ B(x,\cdot )\geq q$ there. The minimum principle gives $H\geq q$ on $E\cap W$. Hence $r\geq q$ on $E$, and $r$ is superharmonic by theorem 3.4.5.

The function $r$ is admissible for $R_{G_ B(x,\cdot )}^ A$: it is nonnegative and superharmonic on $B$ and equals $G_ B(x,\cdot )$ on $A$. Let $h$ be any admissible function. Then $h\geq r$ on $A$. On $W$, the superharmonic function $h$ satisfies $\varliminf _{y\to \zeta }h(y)\geq h(\zeta )\geq f_ x(\zeta )$ at points $\zeta \in \partial A$, by lower semicontinuity, and $\varliminf _{y\to \zeta }h(y)\geq 0=f_ x(\zeta )$ at points $\zeta \in \partial B$, since $h\geq 0$. Applying proposition 3.4.7 to the subharmonic function $H-h$ on $W$, we get $h\geq H$ there. Hence $h\geq r$ on $B$, and $R_{G_ B(x,\cdot )}^ A=r$ pointwise. Since $r$ is continuous, $\widehat R_{G_ B(x,\cdot )}^ A=r$, that is, $r_ x=r$.

On $W$, both $G_ B(x,\cdot )$ and $G_ W(x,\cdot )$ are locally integrable and satisfy $-\Delta =\delta _ x$ there (theorem 3.1.34). Their difference is therefore distributionally harmonic on $W$; by Weyl's lemma (theorem 3.1.11) it has a harmonic representative, which coincides with the difference off $x$. This harmonic function tends to $f_ x$ at $\partial W$, because $G_ W(x,\cdot )$ tends to zero there. By uniqueness of the solution of the Dirichlet problem (proposition 3.1.10), it equals $H$. The harmonic-measure formula is proposition 3.1.33 applied to $H=H_ Wf_ x$.

Finally, the measure $\delta _ x^ A=-\Delta r_ x$ vanishes on $W$, where $r_ x=H$ is harmonic, and on $\operatorname{Int}A$, where $r_ x=G_ B(x,\cdot )$ is harmonic because $x\notin A$. It is therefore supported on $\partial A$, and its mass is at most $\delta _ x(B)=1$ by theorem 4.2.16.

For the last assertion, write $B=B(c,R)$ and let $A=\overline{B(c,\rho )}$ with $0<\rho <R$, so that $W=\{ y:\rho <|y-c|<R\} $ is a bounded open annulus. It is connected: every point of $W$ is joined by a radial segment in $W$ to the sphere $\partial B(c,(\rho +R)/2)$, which is path-connected because $d\geq 2$. By theorem 3.1.29, it suffices to find, for every $\zeta \in \partial W$, a closed ball that meets $\overline W$ only at $\zeta $. If $|\zeta -c|=R$, the closed ball $\overline{B(2\zeta -c,R)}$ meets $\overline{B(c,R)}\supset \overline W$ only at $\zeta $, as noted at the beginning of subsection 3.1.32. If $|\zeta -c|=\rho $, let $m=(c+\zeta )/2$. Every $y\in \overline{B(m,\rho /2)}$ satisfies $|y-c|\leq |y-m|+|m-c|\leq \rho $, with equality only if $y-m=(\zeta -c)/2$, that is, $y=\zeta $. Since $|y-c|\geq \rho $ on $\overline W$, the closed ball $\overline{B(m,\rho /2)}$ meets $\overline W$ only at $\zeta $. Hence every boundary point of $W$ is regular, and $W$ is a regular domain. $\square$

Proof. We first reduce to sets of a special form attached to balls with rational centers and radii; for each such ball, Choquet's lemma then selects countably many of these sets. Let $U=\bigcup _{i\in I}U_ i$. Call a set basic if it has the form $Q\cap \{ u>a\} $, where $Q$ is a ball with rational center and rational radius such that $\overline Q\subset \Omega $, $u$ is subharmonic on an open neighborhood of $\overline Q$ in $\Omega $, and $a\in \mathbb R$. Basic sets are finely open. Every point $x\in U$ lies in a basic set contained in some $U_ i$. Indeed, if $x\in U_ i$, then proposition 4.2.4 provides a Euclidean open set $V$, a function $u\in \mathrm{SH}(V)$ and $a<u(x)$ with $\{ y\in V:u(y)>a\} \subset U_ i$, and we choose a rational ball $Q$ with $x\in Q$ and $\overline Q\subset V$. Let $\mathcal B$ be the family of basic sets contained in some $U_ i$; thus $U=\bigcup \mathcal B$.

Fix a rational ball $Q$, and let $\mathcal B_ Q$ consist of the members of $\mathcal B$ of the form $Q\cap \{ u>a\} $; we may assume that $\mathcal B_ Q$ is nonempty. For every pair $(u,a)$ as in the definition of basic sets with $Q\cap \{ u>a\} \in \mathcal B_ Q$, put

The supremum is less than $+\infty $, because $u$ is upper semicontinuous on a neighborhood of the compact set $\overline Q$ and hence bounded above on $\overline Q$. Thus $v$ is subharmonic on $Q$, $v\leq 1$, and $\{ v>0\} =Q\cap \{ u>a\} $.

Let $F$ be the supremum of all the functions $v$ so obtained. By Choquet's lemma (lemma 3.4.22), there is a sequence $(v_ k)$ among them such that $F'=\sup _ kv_ k$ satisfies $(F')^*=F^*$. By part (2) of theorem 4.1.26, the sets $\{ F\neq F^*\} $ and $\{ F'\neq (F')^*\} $ are polar; let $P_ Q$ be their union. If $y\in Q\setminus P_ Q$ and $v(y)>0$ for some $v$, then $F'(y)=(F')^*(y)=F^*(y)=F(y)>0$, so $v_ k(y)>0$ for some $k$. Hence the union of $\mathcal B_ Q$ is contained in the union of $P_ Q$ and the countably many basic sets $\{ v_ k>0\} $.

Doing this for each of the countably many rational balls $Q$, we obtain countably many members of $\mathcal B$ whose union covers $U$ outside the set $P=\bigcup _ QP_ Q$, which is polar by proposition 4.1.11. For each selected member choose one index $i$ such that the member is contained in $U_ i$. The set $J$ of chosen indices is at most countable, and $U\setminus \bigcup _{j\in J}U_ j\subset P$. This proves the first assertion.

For the second, let $U$ be finely open, and apply the first assertion to the family of all basic sets contained in $U$, whose union is $U$ by proposition 4.2.4. We obtain countably many basic sets whose union $V\subset U$ is such that $U\setminus V$ is polar. Each basic set is Borel, since subharmonic functions are Borel measurable, so $V$ is Borel.

For the measurability assertions, it suffices to show that the polar set $U\setminus V$ is contained in a Borel set $N$ that is null for Lebesgue measure and for every finite positive measure of compact support with finite energy. Since $\mathbb R^ d$ has a countable base, $U\setminus V$ is covered by countably many connected open sets $V_ k$ carrying nontrivial subharmonic functions $\psi _ k$ with $(U\setminus V)\cap V_ k\subset \{ \psi _ k=-\infty \} $. The Borel set $N=\bigcup _ k\{ \psi _ k=-\infty \} $ is polar by proposition 4.1.11; it is Lebesgue null by proposition 3.4.8, and null for measures of finite energy by theorem 4.1.17. $\square$

Proof. We construct $O$ as the union of an open set covering the polar part of $U$ near $L$ and of open sets outside which the functions defining the rest of $U$ are continuous on $L$. By the second assertion of theorem 4.2.22, there are at most countably many sets $V_ k=Q_ k\cap \{ u_ k>a_ k\} \subset U$, with $Q_ k$, $u_ k$ and $a_ k$ as there, whose union $V$ is such that $U\setminus V$ is contained in a polar set $P$. Discarding the empty ones, we may assume that every $V_ k$ is nonempty and that $k$ ranges over a set $\mathcal K\subset \{ 1,2,\ldots \} $. Since $V_ k\neq \varnothing $, we may assume that $u_ k$ is defined on an open ball $Q_ k'\supset \overline{Q_ k}$ and is nontrivial there; by proposition 3.4.8, it is then locally integrable, hence not identically $-\infty $ on any ball in $Q_ k'$. Fix an open set $W_0$ with $L\subset W_0\Subset B$.

First we cover the polar part. The set $P\cap W_0$ is polar and its closure is a compact subset of $B$, so $C_ B^*(P\cap W_0)=0$ by corollary 4.1.24. Hence there is an open set $O_0$ with $P\cap W_0\subset O_0\subset W_0$ and $C_ B(O_0)<\eta /2$.

Next we deal with the discontinuities of the functions $u_ k$. For each $k\in \mathcal K$, the compact set $\overline{Q_ k}\cap L$ lies in the open set $Q_ k'\cap B$. Each of its points therefore lies in a ball $D'$ such that $D'\Subset D\Subset Q_ k'\cap B$ for another ball $D$, and finitely many such balls $D'$ cover $\overline{Q_ k}\cap L$. Enumerate all the pairs of balls obtained in this way, for all $k\in \mathcal K$, as $(D'_ j,D_ j)$ with $j$ in a set $\mathcal J$ of positive integers, and let $k(j)$ be the index $k$ for which the pair $(D'_ j,D_ j)$ was chosen. For each $j\in \mathcal J$, the quasi-continuity theorem (theorem 4.1.30), applied to $u_{k(j)}$ on $Q_{k(j)}'$ with the ball $D_ j\Subset Q_{k(j)}'$ and the compact set $\overline{D'_ j}\cap L\subset D_ j$, provides an open set $O_ j\Subset D_ j$ with $C_{D_ j}(O_ j)<\eta 2^{-j-1}$ such that $u_{k(j)}$ is finite and continuous on $(\overline{D'_ j}\cap L)\setminus O_ j$. By proposition 4.2.12, $C_ B(O_ j)=C_ B^*(O_ j)\leq C_{D_ j}^*(O_ j)=C_{D_ j}(O_ j)<\eta 2^{-j-1}$.

Put $O=O_0\cup \bigcup _{j\in \mathcal J}(O_ j\cap W_0)$. This is an open subset of $W_0$, so $O\Subset B$, and countable subadditivity and monotonicity of the outer capacity (theorem 4.1.19) give

It remains to show that $U\cap F$ is relatively open in $F=L\setminus O$. Since $L\subset W_0$, the set $(U\setminus V)\cap F$ is contained in $(P\cap W_0)\setminus O_0=\varnothing $, so $U\cap F=V\cap F$. Let $y\in V\cap F$. Then $y\in V_ k$ for some $k\in \mathcal K$; since $y\in \overline{Q_ k}\cap L$, we have $y\in D'_ j$ for some $j$ with $k(j)=k$. Since $L\subset W_0$, we have $O_ j\cap L\subset O$, so $D'_ j\cap F\subset (\overline{D'_ j}\cap L)\setminus O_ j$, and the function $u_ k$ is finite and continuous on $D'_ j\cap F$. Hence

is relatively open in $F$, contains $y$, and is contained in $V_ k\cap F$. Therefore $U\cap F=V\cap F$ is relatively open in $F$.

Write $U\cap F=T\cap F$ with $T$ open in $\mathbb R^ d$. Then $(U\cup O)\cap L=(U\cap F)\cup (O\cap L)=(T\cup O)\cap L$ is relatively open in $L$. Conversely, if $(U\cup O)\cap L$ is relatively open in $L$, then its intersection with $F$, which is $U\cap F$, is relatively open in $F$. This proves the equivalence of the two formulations. $\square$

Proof. By example 3.4.15, the function $w\mapsto \log |w|$ is subharmonic on $\mathbb C$. Hence $z\mapsto \log |z_1|$, a function of the first two real coordinates, is subharmonic on $\mathbb R^{2n}$ by lemma 4.2.7; it is finite off $H$ and equals $-\infty $ on $H$. Thus $H$ is polar, and proposition 4.2.6 shows that the classical fine topology induces the discrete topology on $H$.

Now let $x\in H$, and let $N'$ be a neighborhood of $x$ for $\tau $. As in subsection 4.2.1, upper semicontinuity turns every condition of the form $u<b$ into a Euclidean open condition, and restrictions to smaller open sets preserve the complex-line condition; hence $N'$ contains a set

where $V$ is a Euclidean open neighborhood of $x$, the $u_ i$ are upper semicontinuous functions on $V$ satisfying the complex-line condition in the statement, and $u_ i(x)>a_ i$. Choose an affine complex line $\ell \subset H$ through $x$, which is possible since $n\geq 2$, and identify $\ell $ with $\mathbb C$. On the connected component of $\ell \cap V$ containing $x$, each restriction $u_ i|_\ell $ is subharmonic and is not identically $-\infty $, because $u_ i(x)$ is finite. Hence $\{ u_ i|_\ell >a_ i\} $ is open in the planar fine topology, and the intersection of these sets with that component is a planar fine neighborhood of $x$ contained in $N\cap \ell $. By proposition 4.2.5, applied in the plane, it contains points other than $x$. These points lie in $H\cap N'$, so $x$ is not isolated in $H$. $\square$