4.1 Potentials, energy and capacity765E6C8792CB

Proof. The function $h_ t^\sigma $ is bounded by $|\sigma |(\mathbb R^ d)$, and it is continuous by dominated convergence. Completing the square gives $|z-x|^2+|z-y|^2=2|z-\tfrac 12(x+y)|^2+\tfrac 12|x-y|^2$. Since $\int _{\mathbb R^ d}\mathrm e^{-4t|w|^2}\, \mathrm dw=(\pi /4t)^{d/2}$, it follows that

We integrate this identity against $\sigma \otimes \tau $. Fubini's theorem applies, because the integral of the absolute value of the integrand with respect to $\mathrm dz\otimes |\sigma |\otimes |\tau |$ is at most $(\pi /4t)^{d/2}|\sigma |(\mathbb R^ d)|\tau |(\mathbb R^ d)$, and integrating first in $x$ and in $y$ gives (4.1.4). For $\tau =\sigma $ it shows that $Q_ t(\sigma )$ is a positive multiple of $\int (h_ t^\sigma )^2\, \mathrm dz$. Hence $Q_ t(\sigma )\geq 0$, and if $Q_ t(\sigma )=0$, then $h_ t^\sigma =0$ almost everywhere, hence everywhere, because $h_ t^\sigma $ is continuous.

Now suppose that $Q_{t_ j}(\sigma )=0$ for a sequence $t_ j\to \infty $. The functions $g_ t(z)=(2t/\pi )^{d/2}\mathrm e^{-2t|z|^2}$ are nonnegative and have integral one, and $\int _{|z|\geq \delta }g_ t\, \mathrm d z\to 0$ as $t\to \infty $ for every $\delta >0$. Hence $g_ t*f\to f$ uniformly on $\mathbb R^ d$ for every $f\in C_ c(\mathbb R^ d)$: in the integral defining $g_ t*f-f$ we use the uniform continuity of $f$ on the region $|z|<\delta $ and the smallness of the Gaussian tail on the region $|z|\geq \delta $. Since $g_{t_ j}*\sigma =(2t_ j/\pi )^{d/2}h_{t_ j}^\sigma =0$, Fubini's theorem and the symmetry of $g_ t$ give

Letting $j\to \infty $, we get $\int f\, \mathrm d\sigma =0$ for every $f\in C_ c(\mathbb R^ d)$. Thus the positive and negative parts of $\sigma $ define the same positive linear functional on $C_ c(\mathbb R^ d)$, and the uniqueness assertion of theorem 2.3.3 shows that they are equal, that is, $\sigma =0$. $\square$

Proof. By (4.1.4), the expression $Q_ t(\sigma ,\tau )$ of lemma 4.1.3 is $(4t/\pi )^{d/2}$ times the $L^2$ inner product of $h_ t^\sigma $ and $h_ t^\tau $. Hence, by the Cauchy–Schwarz inequality in $z$,

for positive measures $\mu $ and $\nu $.

The case $d\geq 3$. Put $a=(d-2)/2>0$ and $c_ a=\int _0^\infty s^{a-1}\mathrm e^{-s}\, \mathrm d s\in (0,\infty )$. The substitution $s=tr^2$ gives

and both sides are $+\infty $ for $r=0$. By Tonelli's theorem,

Applying (4.1.6) and then the Cauchy–Schwarz inequality for the measure $t^{a-1}\, \mathrm dt$, we obtain $I(\mu ,\nu )\leq I(\mu )^{1/2}I(\nu )^{1/2}$. Now let $\sigma $ be a real linear combination of measures on $K$ with finite energy. Its total variation is dominated by a combination of the same measures with nonnegative coefficients, so the inequality just proved shows that $\iint k\, \mathrm d|\sigma |\, \mathrm d|\sigma |<\infty $. Consequently Fubini's theorem applies to the integrand $t^{a-1}\mathrm e^{-t|x-y|^2}$, and

by lemma 4.1.3. If $I(\sigma )=0$, then $Q_ t(\sigma )=0$ for almost every $t>0$, in particular for arbitrarily large $t$, and lemma 4.1.3 gives $\sigma =0$.

The case $d=2$. For $0<r\leq D$ we have

Indeed, the right side vanishes for $r=D$, and differentiation under the integral sign, which is legitimate for $r$ in a compact subinterval of $(0,\infty )$, shows that its derivative in $r$ is $-r\int _0^\infty \mathrm e^{-tr^2}\, \mathrm dt=-1/r$. The integrand is nonnegative, and since $|\mathrm e^{-t r^2}-\mathrm e^{-tD^2}|\leq tD^2$, its integral over $0<t<1$ is at most $D^2$. Since $|x-y|\leq \operatorname {diam}K<D$ for $x,y\in K$, Tonelli's theorem gives

with a nonnegative integrand. As $\int _1^\infty \mathrm e^{-tD^2}\, \mathrm dt/t<\infty $, it follows that $\int _1^\infty Q_ t(\mu )\, \mathrm dt/t<\infty $, and likewise for $\nu $. By (4.1.6) and the Cauchy–Schwarz inequality for the measure $\mathrm dt/t$ on $(1,\infty )$, also $\int _1^\infty Q_ t(\mu ,\nu )\, \mathrm dt/t<\infty $. Since the part of the integral over $0<t<1$ contributes at most $D^2\mu (K)\nu (K)$ to $4\pi I_ D(\mu ,\nu )$, the mutual energy $I_ D(\mu ,\nu )$ is finite.

Now let $\sigma $ be a signed measure on $K$ of total mass zero whose positive and negative parts $\sigma ^\pm $ have finite energy. Then

so Fubini's theorem applies to the integrand $(\mathrm e^{-t|x-y|^2}-\mathrm e^{-tD^2})/t$ with respect to $\mathrm dt\otimes \sigma \otimes \sigma $. Since $\sigma (K)=0$, the term $\mathrm e^{-tD^2}$ integrates to zero, and

In particular $I_ D(\sigma )$ does not depend on $D$. As in the case $d\geq 3$, $I_ D(\sigma )=0$ forces $\sigma =0$ by lemma 4.1.3.

It remains to prove the Cauchy–Schwarz inequality. Let $\mathcal E$ be the real linear span of the measures on $K$ with finite energy if $d\geq 3$, and the set of signed measures on $K$ of total mass zero whose positive and negative parts have finite energy if $d=2$. In the second case $\mathcal E$ is also a vector space. Indeed, let $\sigma ,\tau \in \mathcal E$ and $c\in \mathbb R$. Since $|\sigma +\tau |\leq |\sigma |+|\tau |$, we have $(\sigma +\tau )^\pm =\frac12\bigl (|\sigma +\tau |\pm (\sigma +\tau )\bigr )\leq \sigma ^\pm +\tau ^\pm $, while $(c\sigma )^\pm =c\sigma ^\pm $ if $c\geq 0$ and $(c\sigma )^\pm =|c|\sigma ^\mp $ if $c<0$. A positive measure on $K$ dominated by the sum of two measures with finite energy has finite energy, because the kernel is nonnegative on $K\times K$ and the mutual energy of two measures with finite energy is finite. Hence $\sigma +\tau $ and $c\sigma $ belong to $\mathcal E$.

In both cases, the total variation of an element of $\mathcal E$ is dominated by a sum of measures on $K$ with finite energy, so for $\sigma ,\tau \in \mathcal E$ the kernel is integrable with respect to $|\sigma |\otimes |\tau |$. Consequently $I(\sigma ,\tau )=\iint k\, \mathrm d\sigma \, \mathrm d\tau $ (with $k_ D$ in place of $k$ when $d=2$) is a symmetric bilinear form on $\mathcal E$, and $I(\sigma )=I(\sigma ,\sigma )\geq 0$ by what we have proved. Hence, for $\sigma ,\tau \in \mathcal E$, the quadratic polynomial $s\mapsto I(\sigma +s\tau )=I(\sigma )+2sI(\sigma ,\tau )+s^2I(\tau )$ is nonnegative on $\mathbb R$. Its discriminant is therefore nonpositive, which is the Cauchy–Schwarz inequality $I(\sigma ,\tau )^2\leq I(\sigma )I(\tau )$. $\square$

Proof. Existence. Let $\mu _ j$ be probability measures on $K$ whose energies converge to the infimum $m$, which is finite by hypothesis. By theorem 2.6.5, after passing to a subsequence, $\mu _ j$ converges vaguely to a Radon measure $\mu $ on $\mathbb R^ d$. By proposition 2.6.2, applied to the open set $\mathbb R^ d\setminus K$, we have $\mu (\mathbb R^ d\setminus K)\leq \varliminf _ j\mu _ j(\mathbb R^ d\setminus K)=0$, so $\mu $ is carried by $K$; testing against a function in $C_ c(\mathbb R^ d)$ equal to one on $K$ then shows that $\mu (K)=1$. To prove that $I(\mu )\leq m$, we first show that

for every continuous function $F$ on $\mathbb R^ d\times \mathbb R^ d$.

Given $\delta >0$, cover $K$ by finitely many balls $B(x_ i,\delta )$ with $x_ i\in K$. By lemma 2.5.2, there are functions $0\leq \varphi _ i\in C_ c^\infty (B(x_ i,\delta ))$ with $\sum _ i\varphi _ i=1$ on $K$. For $x,y\in K$, a term $\varphi _ i(x)\varphi _ l(y)$ is nonzero only if $|x-x_ i|<\delta $ and $|y-x_ l|<\delta $, so

where $\omega $ is the modulus of continuity of $F$ on $K\times K$. The double integrals of the finite sums converge, since $\int \varphi _ i\, \mathrm d\mu _ j\to \int \varphi _ i\, \mathrm d\mu $. As all the measures are probability measures carried by $K$ and $\omega (2\delta )\to 0$ as $\delta \downarrow 0$, this proves the asserted convergence.

For $n>0$, the truncated kernel $\min \{ k,n\} $ is continuous on $\mathbb R^ d\times \mathbb R^ d$, including the diagonal, and it increases to $k$ as $n\to \infty $. Applying the convergence just proved to $F=\min \{ k,n\} $, we get

Since $k\geq 0$ on $K\times K$, monotone convergence in $n$ gives $I(\mu )\leq m$. Hence $\mu $ is a minimizer.

Uniqueness. Let $\mu $ and $\mu '$ be minimizers. The signed measure $\sigma =\mu -\mu '$ has total mass zero. Since $|\sigma |\leq \mu +\mu '$, its positive and negative parts satisfy $\sigma ^+=\frac12(|\sigma |+\sigma )\leq \mu $ and $\sigma ^-\leq \mu '$, so they have finite energy, the kernel being nonnegative on $K\times K$. All mutual energies are finite by proposition 4.1.5, so we may expand and obtain

Since $\frac12(\mu +\mu ')$ is a probability measure on $K$, this energy is at least $m$, so $I(\sigma )\leq 0$. By proposition 4.1.5, $\sigma =0$.

The inequality. Let $\nu $ be a probability measure on $K$ with finite energy. For $0<t\leq 1$ the probability measure $(1-t)\mu _ K+t\nu $ competes in the minimization, so

all terms being finite. Subtracting $V(K)$, dividing by $t$ and letting $t\downarrow 0$ gives $0\leq -2V(K)+2I(\mu _ K,\nu )$.

The capacities. Let $d\geq 3$. A point mass has infinite energy, so $K$ is not a single point and $\operatorname {diam}K>0$. The kernel is at least $((d-2)s_{d-1})^{-1}(\operatorname {diam}K)^{2-d}>0$ on $K\times K$, hence $V(K)>0$. Let $d=2$. Replacing $D$ by $D'>\operatorname {diam}K$ changes the energy of every probability measure on $K$ by the same constant $(2\pi )^{-1}\log (D'/D)$. Hence $\mu _ K$ does not depend on the scale, $V_{D'}(K)=V_ D(K)+(2\pi )^{-1}\log (D'/D)$, and $D'\mathrm e^{-2\pi V_{D'}(K)}=D\mathrm e^{-2\pi V_ D(K)}$. $\square$

Proof. The sets $\{ |f|>N\} $ decrease to the $\mu $-null set $\{ |f|=\infty \} $ as $N\to \infty $, and $\mu $ is finite, so we can choose $N$ with $\mu (\{ |f|>N\} )<\varepsilon /3$. The truncation $f_ N=\max \{ -N,\min \{ f,N\} \} $ is bounded, hence belongs to $L^1(\mu )$. By proposition 2.3.5 there are continuous functions $g_ j$, $j\geq 1$, with $\int |g_ j-f_ N|\, \mathrm d\mu <\varepsilon 2^{-2j}/3$. By Chebyshev's inequality, the set $E_ j=\{ |g_ j-f_ N|>2^{-j}\} $ satisfies $\mu (E_ j)<\varepsilon 2^{-j}/3$, so $\mu (\bigcup _ jE_ j)<\varepsilon /3$. On the Borel set $F=\mathbb R^ d\setminus (\{ |f|>N\} \cup \bigcup _ jE_ j)$ we have $f=f_ N$ and $|g_ j-f|\leq 2^{-j}$ for every $j$, so $g_ j\to f$ uniformly on $F$. Since $\mu (\mathbb R^ d\setminus F)<2\varepsilon /3$, inner regularity gives a compact set $L\subset F$ with $\mu (F\setminus L)<\varepsilon /3$. Then $\mu (\mathbb R^ d\setminus L)<\varepsilon $, and $f|_ L$ is finite and continuous as a uniform limit of continuous functions. $\square$

Proof. (1) Since $p$ is upper semicontinuous (proposition 3.4.11), $\varlimsup _{z\to x}p(z)\leq p(x)$; in particular $p$ is continuous at $x$ if $p(x)=-\infty $. Assume that $p(x)$ is finite; we must show that $\varliminf _{z\to x}p(z)\geq p(x)$. For $z\in \mathbb R^ d$ choose a point $z'\in K$ closest to $z$. Then $|z'-x|\leq |z'-z|+|z-x|\leq 2|z-x|$, so $z'\to x$ as $z\to x$, and $|z'-y|\leq |z'-z|+|z-y|\leq 2|z-y|$ for every $y\in K$. Consequently, for $y\in K$,

Fix $\delta >0$ and split $p=p'_\delta +p''_\delta $, where $p'_\delta (w)=\int _{B(x,\delta )}E_ d(w-y)\, \mathrm d\mu (y)$ is the near part and $p''_\delta $ the far part. The far part is continuous at $x$, since $E_ d(w-y)\to E_ d(x-y)$ uniformly for $y\in K$ with $|y-x|\geq \delta $ as $w\to x$, by the uniform continuity of $E_ d$ on compact subsets of $\mathbb R^ d\setminus \{ 0\} $. The near part, restricted to $K$, is continuous at $x$, being the difference of $p|_ K$ and $p''_\delta |_ K$. Since $\mu $ is carried by $K$, we may integrate the inequalities above with respect to $\mu $ over $B(x,\delta )$; letting $z\to x$, we obtain for $d\geq 3$

and for $d=2$ the bound $\varliminf _{z\to x}p(z)\geq p(x)-(2\pi )^{-1}\mu (B(x,\delta ))\log 2$. Since $p(x)$ is finite, the function $E_ d(x-\cdot )$ is $\mu $-integrable and $\mu (\{ x\} )=0$. By dominated convergence and continuity of $\mu $ from above, $p'_\delta (x)\to 0$ and $\mu (B(x,\delta ))\to 0$ as $\delta \downarrow 0$, which proves $\varliminf _{z\to x}p(z)\geq p(x)$.

(2) Let $M\geq 0$ be an upper bound of $E_ d$ on the compact set $K-K$. If $\nu $ is the restriction of $\mu $ to a Borel set, then $p=E_ d*\nu +E_ d*(\mu -\nu )$ and $E_ d*(\mu -\nu )\leq M\mu (K)$ on $K$, so $E_ d*\nu \geq p-M\mu (K)$ on $K$; hence $E_ d*\nu $ is finite $\nu $-almost everywhere.

We construct inductively pairwise disjoint compact sets $K_1,K_2,\ldots \subset K$ such that the residual measures $\nu _ j=\mu |_{K\setminus (K_1\cup \cdots \cup K_ j)}$ satisfy $\nu _ j(\mathbb R^ d)<2^{-j}$, and such that each $E_ d*(\mu |_{K_ j})$ is finite and continuous on $\mathbb R^ d$. Put $\nu _0=\mu $, and suppose $K_1,\ldots ,K_{j-1}$ have been chosen. Apply lemma 4.1.9 to $\nu =\nu _{j-1}$ and $E_ d*\nu $: there is a compact set $L$ with $\nu (\mathbb R^ d\setminus L)<2^{-j-1}$ on which $E_ d*\nu $ is finite and continuous. Since $\nu $ is carried by $K$ and does not charge $K_1\cup \cdots \cup K_{j-1}$, inner regularity gives a compact set $K_ j\subset (L\cap K)\setminus (K_1\cup \cdots \cup K_{j-1})$ with $\nu (\mathbb R^ d\setminus K_ j)<2^{-j}$. Put $\mu _ j=\nu |_{K_ j}=\mu |_{K_ j}$; then $\nu _ j=\nu -\mu _ j$ has mass less than $2^{-j}$.

It remains to show that $E_ d*\mu _ j$ is finite and continuous. On $K_ j$ the two functions $E_ d*\mu _ j$ and $E_ d*(\nu -\mu _ j)$ are upper semicontinuous and less than $+\infty $, and their sum $E_ d*\nu $ is finite and continuous; hence both are finite on $K_ j$. Moreover, for $x\in K_ j$ and $K_ j\ni w\to x$,

so the restriction of $E_ d*\mu _ j$ to $K_ j$ is continuous. The support of $\mu _ j$ is contained in $K_ j$, so part (1), applied to $\mu _ j$ and the compact set $K_ j$, shows that $E_ d*\mu _ j$ is continuous at every point of $K_ j$. At a point $x\notin K_ j$, the function $E_ d*\mu _ j$ is finite and continuous, because $E_ d(w-y)\to E_ d(x-y)$ uniformly in $y\in K_ j$ as $w\to x$, as for the far part in (1). Thus $E_ d*\mu _ j$ is finite and continuous on $\mathbb R^ d$. Finally, $\mu -\sum _{i\leq j}\mu _ i=\nu _ j$ has mass less than $2^{-j}$, so $\mu =\sum _ j\mu _ j$. $\square$

Proof. Fix a bounded open set $W\supset \overline A$; we construct $\nu $ with support in $\overline W$. Let $a\in A$. By definition 3.4.26, after shrinking the neighborhood as explained above, there are an open ball $V\subset W$ centered at $a$ and a nontrivial function $\psi \in \mathrm{SH}(V)$ with $A\cap V\subset \{ \psi =-\infty \} $; by proposition 3.4.8, $\psi $ is locally integrable. Choose an open ball $B_ a$ centered at $a$ with $\overline{B_ a}\subset V$, and a cutoff $\eta \in C_ c^\infty (V)$ with $0\leq \eta \leq 1$ and $\eta =1$ near $\overline{B_ a}$. By the Riesz decomposition (theorem 3.4.14), $\psi =E_ d*(\eta \mu _\psi )+h$ on $B_ a$, where $\mu _\psi $ is the Riesz measure of $\psi $ and $h$ is harmonic, hence finite; therefore $E_ d*(\eta \mu _\psi )=-\infty $ on $A\cap B_ a$. The measure $\eta \mu _\psi $ is finite, positive and supported in $\overline W$. Since $\mathbb R^ d$ has a countable base, countably many of the balls $B_ a$ cover $A$. Let $\nu _ j$ be the corresponding measures.

Choose $c_ j>0$ with $\sum _ jc_ j\nu _ j(\mathbb R^ d)<\infty $, and put $\nu =\sum _ jc_ j\nu _ j$, a finite positive measure supported in $\overline W$. Let $S\subset \mathbb R^ d$ be compact and let $M$ be an upper bound of $E_ d$ on the compact set $S-\overline W$. For $x\in S$, Tonelli's theorem applied to the nonnegative functions $M-E_ d(x-\cdot )$ gives

Thus $E_ d*\nu (x)=-\infty $ whenever $E_ d*\nu _ j(x)=-\infty $ for some $j$; in particular $E_ d*\nu =-\infty $ on $A$. By proposition 3.4.11, $E_ d*\nu $ is subharmonic and locally integrable, so it is not identically $-\infty $.

Now let $A=\bigcup _ kA_ k$ with polar sets $A_ k$, let $x\in \mathbb R^ d$, and let $B'$ be an open ball centered at $x$. Apply the construction above to the bounded polar sets $A_ k\cap B'$, with one and the same bounded open set $W\supset \overline{B'}$. It gives measures $\nu ^{(k)}$ supported in $\overline W$ with $E_ d*\nu ^{(k)}=-\infty $ on $A_ k\cap B'$. Choosing $c_ k>0$ with $\sum _ kc_ k\nu ^{(k)}(\mathbb R^ d)<\infty $, the Tonelli argument above shows that $\nu =\sum _ kc_ k\nu ^{(k)}$ satisfies $E_ d*\nu =-\infty $ on $A\cap B'$. Since $E_ d*\nu $ is subharmonic on $B'$ and not identically $-\infty $ there, $A$ is polar. $\square$

Proof. The set. Put $a_ j=\mathrm e^{-4^ j}$ for $j\geq 1$, and let $K$ be the set of all sums $\sum _{j\geq 1}\varepsilon _ ja_ j$ with $\varepsilon _ j\in \{ 0,1\} $. Since $a_{j+1}/a_ j=\mathrm e^{-3\cdot 4^ j}\leq \mathrm e^{-12}<1/2$, the tails $R_ k=\sum _{j>k}a_ j$ satisfy

In particular $K\subset [0,R_0]\subset [0,2\mathrm e^{-4}]\subset [0,1/10]$. Let $\mathcal S=\prod _{j\geq 1}\{ 0,1\} $ carry the product topology, and let $c\colon \mathcal S\to \mathbb R$ be the coding map $c(\varepsilon )=\sum _ j\varepsilon _ ja_ j$. If $\varepsilon $ and $\varepsilon '$ agree in their first $k$ digits, then $|c(\varepsilon )-c(\varepsilon ')|\leq R_ k$; hence $c$ is continuous, and $K=c(\mathcal S)$ is compact by lemma 2.7.2. If $\varepsilon $ and $\varepsilon '$ first differ at the $k$-th digit, say $\varepsilon _ k=1$ and $\varepsilon '_ k=0$, then $c(\varepsilon )-c(\varepsilon ')\geq a_ k-R_ k>0$; hence $c$ is injective.

For a word $\sigma \in \{ 0,1\} ^ k$, let $K_\sigma $ be the image of the sequences beginning with $\sigma $. The $2^ k$ sets $K_\sigma $ are compact, pairwise disjoint by injectivity, cover $K$, and have diameter at most $R_ k$. Consequently each $K_\sigma $ is open and closed in $K$, and two distinct points of $K$ lie in different sets $K_\sigma $ once $k$ is large; thus $K$ is totally disconnected. Changing the $k$-th digit of $\varepsilon $ produces a point of $K$ different from $c(\varepsilon )$ at distance $a_ k$, so $K$ has no isolated points. Hence $K$ is a Cantor set: a nonempty compact metric space that is totally disconnected and has no isolated points.

The measure. For $t\in [0,1)$ let $\varepsilon _ j(t)$ be its binary digits, using the terminating expansion at dyadic rationals, and put $\Phi (t)=\sum _ j\varepsilon _ j(t)a_ j$. Each $\varepsilon _ j$ is a Borel step function, so $\Phi $ is Borel as a uniform limit of Borel functions. Let $\mu $ be the image of Lebesgue measure on $[0,1)$ under $\Phi $. It is a probability measure carried by the compact set $K$, and it is a Radon measure by proposition 2.2.2. For $\sigma \in \{ 0,1\} ^ k$, injectivity of $c$ shows that $\Phi (t)\in K_\sigma $ if and only if the first $k$ binary digits of $t$ form the word $\sigma $, and this happens exactly on a dyadic interval of length $2^{-k}$. Hence $\mu (K_\sigma )=2^{-k}$. Every $x\in K$ lies in one of the sets $K_\sigma $ with $\sigma $ of length $k$, so $\mu (\{ x\} )\leq 2^{-k}$ for every $k$; thus $\mu $ has no atoms.

The potential. Since $\log |z-a|^2=4\pi E_2(z-a)$, we have $p=4\pi E_2*\mu $, which is subharmonic and locally integrable on $\mathbb C$ by proposition 3.4.11. Let $z_0\notin K$ and $\delta =\frac12\operatorname {dist}(z_0,K)$. For $|z-z_0|<\delta $ and $a\in K$, the kernel and all its derivatives in $z$ are bounded uniformly, so differentiation under the integral sign shows that $p$ is smooth on $B(z_0,\delta )$ with $\Delta p(z)=\int \Delta _ z\log |z-a|^2\, \mathrm d\mu (a)=0$. Thus $p$ is finite and harmonic on $\mathbb C\setminus K$.

Polarity. Let $x\in K$ and $k\geq 1$, and let $K_\sigma $, with $\sigma $ of length $k$, be the piece containing $x$. Since $\operatorname {diam}K\leq R_0<1$, every term $\log |x-a|^2$ with $a\in K$ is nonpositive, and for $a\in K_\sigma $ we have $|x-a|\leq R_ k\leq 2a_{k+1}$. Consequently

The right side tends to $-\infty $ as $k\to \infty $, so $p(x)=-\infty $. Since $p$ is subharmonic on $\mathbb C$ and not identically $-\infty $, the set $K$ is polar. For $D>\operatorname {diam}K$, the logarithmic potential of the probability measure $\mu $ is $U^\mu =(2\pi )^{-1}\log D-p/(4\pi )$, which equals $+\infty $ on $K$. Since $\mu $ is carried by $K$, Tonelli's theorem gives $I_ D(\mu )=\int U^\mu \, \mathrm d\mu =+\infty $. $\square$

Proof. Fix $y\in B\setminus \{ 0\} $. Since $|y^*|=R^2/|y|>R$, the function $x\mapsto \Gamma _ d(x-y^*)$ is harmonic on a neighborhood of $\overline B$. Expanding the square gives

and therefore $|y|^2|x-y^*|^2-R^2|x-y|^2=(R^2-|x|^2)(R^2-|y|^2)$. Each displayed expression, as a function of $x$, is the sum of $\Gamma _ d(x-y)$ and a function harmonic near $\overline B$. By proposition 3.1.13, it is therefore harmonic on $B\setminus \{ y\} $ and locally integrable on $B$, with distributional Laplacian $-\delta _ y$ on $B$. Moreover, the last identity shows that $|y|\, |x-y^*|>R|x-y|$ on $B$, with equality on $\partial B$. Hence the expression is positive on $B\setminus \{ y\} $ and tends to zero at every point of $\partial B$. The ball is a regular domain by theorem 3.1.29, and the uniqueness assertion of theorem 3.1.34 identifies the displayed expression with $G_ B(y,x)$ for $x\neq y$. The same argument applies to $y=0$, where the stated expressions differ from $\Gamma _ d(x)$ by constants.

Let $Q(x,y)$ denote the right side of (4.1.15); it is symmetric in $x$ and $y$, and $Q(x,0)=R^4$. The formulas just proved can be written, for all $x,y\in B$ including $y=0$, as

Both right sides are symmetric in $x$ and $y$, so $G_ B(y,x)=G_ B(x,y)$; this also proves the formulas in the order stated. Finally, $Q(x,y)\geq (R^2-|x||y|)^2>0$ when $|y|\leq r$ and $|x|<R^2/r$, so the correction terms $G_ B(x,y)-\Gamma _ d(x-y)$ in the last display are smooth functions of $(x,y)$ there. $\square$

Proof. (1) Since $K$ is a compact subset of $B$, there is $c>0$ such that $q(x)=c(|x|^2-R^2)\leq -1$ on $K$. The function $q$ is smooth and subharmonic, since $\Delta q=2dc>0$, and it is nonpositive on $B$; hence it is admissible, and $q\leq f_ K\leq u_ K\leq 0$. In particular $f_ K$ and $u_ K$ tend to zero at $\partial B$.

We show that $f_ K$ is the supremum of the continuous admissible functions. Let $v$ be admissible. We first extend $\max \{ v,q\} $ to a subharmonic function on $\mathbb R^ d$: put $w=\max \{ v,q\} $ on $B$ and $w=q$ on $\mathbb R^ d\setminus B$. At $\zeta \in \partial B$ we have $w(\zeta )=0$ and $\varlimsup _{x\to \zeta }w(x)\leq 0$, because $v\leq 0$ on $B$ and $q$ is continuous with $q(\zeta )=0$; so $w$ is upper semicontinuous. We verify the comparison principle of theorem 3.4.5 for $w$ on an arbitrary ball $D$. Let $h$ be harmonic on $D$ and continuous on $\overline D$, with $\varlimsup _{D\ni x\to \zeta }w(x)\leq h(\zeta )$ for $\zeta \in \partial D$. Since $w\geq q$ and $q$ is continuous, $q\leq h$ on $\partial D$, hence on $D$ by the comparison principle for $q$; in particular $w=q\leq h$ on $D\setminus B$. On the bounded open set $D\cap B$, the function $v-h$ has nonpositive boundary upper limits: at points of $\partial D\cap B$ because $v\leq w$, and at points of $\partial B$ because $\varlimsup v\leq 0=q\leq h$ there. By the comparison principle on bounded open sets (the remark after theorem 3.4.5), $v\leq h$ on $D\cap B$. Thus $w\leq h$ on $D$, and the converse part of theorem 3.4.5 shows that $w$ is subharmonic on $\mathbb R^ d$.

Since $w\geq q$, it is locally integrable. Its mollifications $w*\rho _\varepsilon $ are smooth and satisfy $\Delta (w*\rho _\varepsilon )=(\Delta w)*\rho _\varepsilon \geq 0$, so they are subharmonic, and they decrease to $w$ as $\varepsilon \downarrow 0$ by (3.4.13). Fix $\delta >0$. On $B$ we have $w*\rho _\varepsilon \leq \sup _{B(0,R+\varepsilon )}w\leq c((R+\varepsilon )^2-R^2)$, because $w\leq 0$ on $B$ and $w=q$ outside $B$; hence $w*\rho _\varepsilon -\delta \leq 0$ on $B$ for small $\varepsilon $. On $K$ we have $w\leq -1$. Since the continuous functions $w*\rho _\varepsilon $ decrease to $w$, the open sets $\{ w*\rho _\varepsilon <-1+\delta \} $ increase and cover the compact set $K$, so $w*\rho _\varepsilon <-1+\delta $ on $K$ for small $\varepsilon $. For such $\varepsilon $ the function $w*\rho _\varepsilon -\delta $ is continuous and admissible, and it is at least $w-\delta \geq v-\delta $. Since $v$ and $\delta $ are arbitrary, $f_ K$ is the supremum of the continuous admissible functions; in particular $f_ K$ is lower semicontinuous. Every admissible function is at most $-1$ on $K$, so $f_ K=-1$ on $K$.

Let $D$ be an open ball with $\overline D\subset B\setminus K$; we show that $f_ K=u_ K$ on $D$ and that $u_ K$ is harmonic there. For a continuous admissible function $v$, let $v^ D$ be its harmonic lifting on $D$, which equals the Poisson integral of $v|_{\partial D}$ on $D$ and $v$ elsewhere. As noted in subsection 3.1.23, the continuous subsolutions defined there are exactly the real-valued continuous subharmonic functions. Hence lemma 3.1.24 shows that $v^ D$ is continuous and subharmonic and that $v^ D\geq v$. It is admissible: $v^ D\leq 0$ because the Poisson integral preserves order, and $v^ D=v$ on $K$.

Apply Choquet's lemma (lemma 3.4.22) to the family of continuous admissible functions, whose supremum is $f_ K$: there are continuous admissible functions $v_1,v_2,\ldots $ with $(\sup _ jv_ j)^*=u_ K$. The functions $w_ j=\max \{ -1,v_1,\ldots ,v_ j\} $ are continuous, admissible and increasing, and so are the functions $w_ j^ D$. On $D$, the harmonic functions $w_ j^ D$ take values in $[-1,0]$ and increase, so by corollary 3.1.21 they converge locally uniformly to a harmonic function $H$. On the one hand, each $w_ j^ D$ is admissible, so $H\leq f_ K\leq u_ K$ on $D$. On the other hand, $H\geq \sup _ jw_ j$. The upper semicontinuous regularization of $\sup _ jw_ j=\max \{ -1,\sup _ jv_ j\} $ is $\max \{ -1,u_ K\} =u_ K$, so $\sup _ jw_ j=u_ K$ almost everywhere by theorem 3.4.23. Hence $H=u_ K$ almost everywhere on $D$; since both functions are subharmonic and locally integrable, $H=u_ K$ on $D$ by theorem 3.4.12. Therefore $f_ K=u_ K=H$ on $D$. As $D$ is arbitrary, $f_ K=u_ K$ on $B\setminus K$, and $u_ K$ is harmonic there.

(2) Fix $0<r<R$ with $\operatorname{Supp}\lambda \subset B(0,r)$. By the definition of $G_ B$ in theorem 3.1.34 and the symmetry in lemma 4.1.14,

By Fubini's theorem the last integral is the Poisson integral of the continuous function $\zeta \mapsto E_ d*\lambda (\zeta )$ on $\partial B$, so it is harmonic on $B$. Hence $u$ is subharmonic with Riesz measure $\lambda $, by proposition 3.4.11. By lemma 4.1.14, for $y\in \operatorname{Supp}\lambda $ the formula given there for $G_ B(x,y)$ defines a harmonic function of $x$ on the annulus $r<|x|<R^2/r$, which vanishes for $|x|=R$. Since this function is jointly smooth in $(x,y)$, its derivatives in $x$ are bounded uniformly in $y\in \operatorname{Supp}\lambda $ on compact subsets of the annulus. Differentiating under the integral sign, we see that $-\int G_ B(x,y)\, \mathrm d\lambda (y)$ defines a harmonic function on this annulus that extends $u$ across $\partial B$ and vanishes on $\partial B$; we denote the extension again by $u$.

Choose $r<r'<R$ and $\chi \in C_ c^\infty (B)$ with $\chi =1$ on a neighborhood of $\overline{B(0,r')}$. Then $\lambda (B)=\int \chi \, \mathrm d\lambda =\int _ Bu\Delta \chi \, \mathrm d x$. The integrand vanishes outside the annulus $r'<|x|<R$, near whose closure $u$ is smooth and harmonic. The divergence identity (3.1.4) on this annulus, applied to the field $u\nabla \chi -\chi \nabla u$, whose divergence there is $u\Delta \chi $, gives

because $\chi $ and its derivatives vanish near $|x|=R$, while $\chi =1$ near $|x|=r'$; here $\partial _ r$ is the radial derivative. Applying the divergence identity on the same annulus to $\nabla u$, whose divergence vanishes there, we see that this integral equals $\int _{\partial B}\partial _ nu\, \mathrm dS$. This proves the flux identity.

For the comparison of masses, let $u'=-G_ B\lambda '$ be a second function as above with $u\leq u'$ on $B$. Then $u-u'\leq 0$ on $B$ and $u-u'=0$ on $\partial B$, and since $u-u'$ is smooth near $\partial B$, its derivative at a point of $\partial B$ along the outward normal is the limit of difference quotients $((u-u')(\zeta -sn)-0)/(-s)\geq 0$ as $s\downarrow 0$. Hence $\partial _ nu\geq \partial _ nu'$ on $\partial B$, and integration gives $\lambda (B)\geq \lambda '(B)$.

For the converse, let $v$ and $\lambda $ be as stated, extend $\lambda $ by zero to $\mathbb R^ d$, and put $p=E_ d*\lambda $. By proposition 3.4.11, $\Delta (v-p)=0$ on $B$ in the sense of distributions, so by theorem 3.1.11 the locally integrable function $v-p$ agrees almost everywhere on $B$ with a harmonic function $h$. Then $p+h$ is a locally integrable subharmonic function on $B$ that agrees with $v$ almost everywhere, hence everywhere by theorem 3.4.12. The function $p$ is finite and continuous on the neighborhood $\mathbb R^ d\setminus \operatorname{Supp}\lambda $ of $\partial B$: if $x\notin \operatorname{Supp}\lambda $, then $E_ d(w-y)\to E_ d(x-y)$ uniformly in $y\in \operatorname{Supp}\lambda $ as $w\to x$, because $E_ d$ is uniformly continuous on compact subsets of $\mathbb R^ d\setminus \{ 0\} $. As $v$ tends to zero at $\partial B$, the harmonic function $h=v-p$ extends continuously to $\overline B$ with boundary values $-p|_{\partial B}$, and the uniqueness in theorem 3.1.18 gives $h=-H_ B(p|_{\partial B})$. By the formula for $-G_ B\lambda $ proved at the beginning of (2), $v=p-H_ B(p|_{\partial B})=-G_ B\lambda $.

(3) By (1), $u_ K$ is harmonic on $B\setminus K$, so its Riesz measure $\lambda _ K$, a positive Radon measure on $B$ by theorem 3.4.12, is supported on the compact set $K$; in particular it is finite. Moreover $-1\leq u_ K\leq 0$, so $u_ K$ is locally integrable, and $u_ K$ tends to zero at $\partial B$ by (1). The converse in (2) therefore gives $u_ K=-G_ B\lambda _ K$. The flux formula for $C_ B(K)$ is now (2) with $\lambda =\lambda _ K$. $\square$

Proof. Conditions (1)–(3) do not involve $B$, so for their equivalence we may enlarge $R$ and assume that $K\subset B$. We first show, for every ball $B=B(0,R)$ containing $K$, that $K$ is polar if and only if $C_ B(K)=0$. Since $u_ K=-G_ B\lambda _ K$ and $C_ B(K)=\lambda _ K(K)$ (proposition 4.1.16), we have $C_ B(K)=0$ if and only if $u_ K=0$. Suppose $u_ K=0$. Then $f_ K=0$ almost everywhere by theorem 3.4.23. Choose a point $x\in B$ with $f_ K(x)=0$ and admissible functions $v_ j$ with $v_ j(x)>-2^{-j}$. The partial sums $\sum _{j\leq N}v_ j$ are subharmonic and decrease with $N$, and their values at $x$ stay above $-1$. By theorem 3.4.9, the limit $\sum _ jv_ j$ is subharmonic on $B$ and not identically $-\infty $. It equals $-\infty $ on $K$, since every $v_ j$ is at most $-1$ there. Thus $K$ is polar.

Conversely, if $K$ is polar, proposition 4.1.11 gives a function $E_ d*\nu $, not identically $-\infty $, that equals $-\infty $ on $K$; subtracting a constant, we may assume it is negative on $B$. All its positive multiples are then admissible, so $f_ K=0$ wherever $E_ d*\nu >-\infty $, that is, almost everywhere. Hence $u_ K=0$ almost everywhere, and therefore everywhere by theorem 3.4.12.

Suppose that $K$ is not polar, so that $\lambda _ K\neq 0$. By parts (2) and (3) of proposition 4.1.16, $u_ K=E_ d*\lambda _ K+h$ with the harmonic function $h=-H_ B\bigl ((E_ d*\lambda _ K)|_{\partial B}\bigr )$ on $B$. Since $u_ K\geq -1$, it follows that $E_ d*\lambda _ K\geq -1-h>-\infty $ on $K$. Theorem 4.1.10 writes $\lambda _ K$ as a sum of measures with finite continuous potentials, and one of them, $\mu $, is nonzero; so (3) holds. A measure $\mu $ as in (3) has finite energy, because its potential $U^\mu $ is bounded on the compact support of $\mu $; so (3) implies (2).

Next, let $\mu $ be a finite positive measure with compact support such that $E_ d*\mu $ is finite and continuous, and let $\nu $ be any finite positive measure with compact support. We show that $\mu $ gives measure zero to the Borel set $\{ E_ d*\nu =-\infty \} $; by proposition 4.1.11, every bounded polar set lies in a set of this form. By the reciprocity law (4.1.2), $\int (E_ d*\nu )\, \mathrm d\mu =\int (E_ d*\mu )\, \mathrm d\nu $. The right side is finite because $E_ d*\mu $ is continuous, hence so is the left side, and therefore $E_ d*\nu >-\infty $ $\mu $-almost everywhere.

Now let $\mu $ be a finite positive measure with compact support and finite energy. Then $U^\mu <\infty $ $\mu $-almost everywhere, that is, $E_ d*\mu $ is finite $\mu $-almost everywhere, and theorem 4.1.10 writes $\mu $ as a sum of measures with finite continuous potentials. If $A$ is a bounded polar set and $\nu $ is chosen for $A$ as in proposition 4.1.11, the previous paragraph shows that each of these measures, and hence $\mu $, gives measure zero to the Borel set $\{ E_ d*\nu =-\infty \} $, which contains $A$. An arbitrary polar set is a countable union of bounded polar sets, so it is contained in a countable union of Borel sets of $\mu $-measure zero; this proves the final assertion. In particular, a nonzero measure with finite energy cannot be carried by a polar compact set, so (2) implies (1). $\square$

Proof. If $K\subset L$, every function admissible for $L$ is admissible for $K$, so $u_ L\leq u_ K$. By proposition 4.1.16, $u_ L=-G_ B\lambda _ L$ and $u_ K=-G_ B\lambda _ K$, and part (2) of that proposition gives $C_ B(L)\geq C_ B(K)$.

Let $K_ j$ be compact sets decreasing to $K$. Their extremal functions increase with $j$ and are bounded above by $f_ K$. Let $v$ be admissible for $K$ and $\varepsilon >0$. The function $v+\varepsilon (|x|^2-R^2)$ is less than $-1$ on $K$, so by upper semicontinuity it is less than $-1$ on an open neighborhood $U$ of $K$. The compact sets $K_ j\setminus U$ decrease and have empty intersection, so one of them is empty; thus $K_ j\subset U$ for all large $j$. Hence $v+\varepsilon (|x|^2-R^2)$ is admissible for $K_ j$, and therefore at most $f_{K_ j}$, when $j$ is large. Letting $\varepsilon \downarrow 0$ and taking the supremum over $v$, we obtain $f_{K_ j}\to f_ K$ pointwise. Since $f_{K_ j}=u_{K_ j}$ and $f_ K=u_ K$ almost everywhere (theorem 3.4.23) and all these functions take values in $[-1,0]$, bounded convergence gives $u_{K_ j}\to u_ K$ in $L^1(B)$. Choose $\chi \in C_ c^\infty (B)$ equal to one near $K_1$. Since all the Riesz measures are supported in $K_1$,

For strong subadditivity we first prove that $F=f_{K\cup L}+f_{K\cap L}$ satisfies $F\geq f_ K+f_ L$. At a point of $K$ we have $f_{K\cup L}=-1=f_ K$, and $f_{K\cap L}\geq f_ L$ because $K\cap L\subset L$; the case of a point of $L$ is symmetric. So the inequality holds on $K\cup L$. By proposition 4.1.16, $F$ is harmonic on the open set $U=B\setminus (K\cup L)$, lower semicontinuous on $B$, and tends to zero at $\partial B$. Let $v$ and $w$ be continuous admissible functions for $K$ and $L$. At a boundary point $\zeta $ of $U$ in $B$, which lies in $K\cup L$, we have

and at points of $\partial B$ the upper limits of $v+w$ are nonpositive while $F$ tends to zero. By proposition 3.4.7, applied to the subharmonic function $v+w-F$ on the bounded open set $U$, we get $v+w\leq F$ on $U$. Taking the supremum over $v$ and $w$, which by proposition 4.1.16 gives $f_ K$ and $f_ L$, proves the inequality on $U$, hence on $B$.

Since each $f$ equals the corresponding $u$ almost everywhere, we obtain $u_{K\cup L}+u_{K\cap L}\geq u_ K+u_ L$ almost everywhere; taking ball averages and letting the radius tend to zero, (3.4.13) gives this inequality everywhere. The two sides are the negatives of the Green potentials of $\lambda _{K\cup L}+\lambda _{K\cap L}$ and of $\lambda _ K+\lambda _ L$, so part (2) of proposition 4.1.16 gives the asserted inequality of masses. $\square$

Proof. (1) Let $K\subset B$ be compact; we may assume that $K\neq \varnothing $, since $C_ B^*(\varnothing )=0=C_ B(\varnothing )$. For $0<\delta <\operatorname {dist}(K,\mathbb R^ d\setminus B)$, the compact sets $K_ j=\{ x:\operatorname {dist}(x,K)\leq \delta /j\} $ are neighborhoods of $K$ in $B$ that decrease to $K$; let $O_ j$ be their interiors. Every open set containing $K$ has capacity at least $C_ B(K)$, so by proposition 4.1.18

Strong subadditivity. Parts (2) and (3) rest on two inequalities for $C_ B^*$, the first of which is $C_ B^*(A\cup A')+C_ B^*(A\cap A')\leq C_ B^*(A)+C_ B^*(A')$ for $A,A'\Subset B$. Consider first open sets $O,O'\Subset B$, and let $K\subset O\cup O'$ and $L\subset O\cap O'$ be compact. Cover $K$ by finitely many closed balls, each contained in $O$ or in $O'$, and let $K'$ and $K''$ be the intersections of $K$ with the union of those contained in $O$ and in $O'$, respectively. Then $K'\cup L\subset O$ and $K''\cup L\subset O'$ are compact, their union contains $K$ and their intersection contains $L$. By monotonicity and proposition 4.1.18,

Taking the supremum over $K$ and $L$ proves the inequality for open sets. For arbitrary $A,A'\Subset B$, apply it to open sets $O\supset A$ and $O'\supset A'$ with compact closures in $B$, and take the infimum.

An excess inequality. Let $A_ j\subset E_ j$ be sets with $E_1\cup E_2\Subset B$. All the capacities involved are finite. Strong subadditivity applied to $E_1$ and $A_1\cup E_2$, and to $A_1\cup A_2$ and $E_2$, gives

where we have used monotonicity to bound the capacities of the intersections from below. Adding these inequalities and cancelling $C_ B^*(A_1\cup E_2)$ gives the case $N=2$ of

valid whenever $A_ j\subset E_ j$ and $\bigcup _ jE_ j\Subset B$. Induction on $N$, applied to the pairs $\bigcup _{j<N}A_ j\subset \bigcup _{j<N}E_ j$ and $A_ N\subset E_ N$, proves (4.1.20) in general.

(2) and (3). If $O_ j$ are open sets increasing to an open set $O\Subset B$, every compact subset of $O$ lies in some $O_ j$, so $C_ B(O_ j)\to C_ B(O)$. Now let $A_ j$ be arbitrary sets with $A=\bigcup _ jA_ j\Subset B$, fix an open set $W$ with $\overline A\subset W\Subset B$, and let $\varepsilon >0$. Intersecting almost optimal open sets in the definition of $C_ B^*(A_ j)$ with $W$, we find open sets $E_ j$ with $A_ j\subset E_ j\subset W$ and $C_ B(E_ j)<C_ B^*(A_ j)+\varepsilon 2^{-j}$. By (4.1.20) and the result for open sets,

Let $\varepsilon \downarrow 0$. If the $A_ j$ increase, the right side is $\lim _ NC_ B^*(A_ N)\leq C_ B^*(A)$, which proves (3). In general, strong subadditivity gives finite subadditivity, and the displayed inequality extends it to countable subadditivity, which proves (2).

(4) Suppose $C_ B^*(A)=0$. We construct subharmonic functions $v_ j$ on $B$ that equal $-1$ on $A$ and whose Riesz measures have mass less than $2^{-j}$. Their sum will be a nontrivial subharmonic function equal to $-\infty $ on $A$. Fix an open set $W$ with $\overline A\subset W\Subset B$, and choose open sets $O_ j$ with $A\subset O_ j\subset W$ and $C_ B(O_ j)<2^{-j}$. Exhaust $O_ j$ by compact sets $L_{j,m}$, increasing in $m$, whose interiors cover $O_ j$. Choose $c>0$ with $q(x)=c(|x|^2-R^2)\leq -1$ on $\overline W$; then $q$ is admissible for every $L_{j,m}$, so $u_{L_{j,m}}\geq q$. The functions $u_{L_{j,m}}$ decrease as $m$ increases, and their limit $v_ j$ is subharmonic by theorem 3.4.9, lies between $q$ and $0$, and hence tends to zero at $\partial B$. Moreover, $v_ j=-1$ on $O_ j$, since $f_{L_{j,m}}$ and hence $u_{L_{j,m}}$ equal $-1$ on the interior of $L_{j,m}$.

The functions $u_{L_{j,m}}$ converge to $v_ j$ in $L^1(B)$ by bounded convergence, so their Riesz measures, which are supported in $\overline W$, converge in the sense of distributions on $B$, and hence vaguely by proposition 2.6.1, to the Riesz measure $\lambda _ j$ of $v_ j$. By proposition 2.6.2, applied to the open set $B\setminus \overline W$, the measure $\lambda _ j$ is supported in $\overline W$. Testing against a function in $C_ c^\infty (B)$ equal to one near $\overline W$ gives $\lambda _ j(B)=\lim _ mC_ B(L_{j,m})=C_ B(O_ j)$, since every compact subset of $O_ j$ lies in some $L_{j,m}$. The function $v_ j$ is bounded, hence locally integrable, tends to zero at $\partial B$, and has a Riesz measure with compact support in $B$; by the converse in part (2) of proposition 4.1.16, $v_ j=-G_ B\lambda _ j$.

The measure $\nu =\sum _ j\lambda _ j$ is finite, since $\lambda _ j(B)=C_ B(O_ j)<2^{-j}$, and it is supported in $\overline W$. By Tonelli's theorem for the nonnegative kernel $G_ B$, we have $\sum _ jv_ j=-G_ B\nu $. By part (2) of proposition 4.1.16, $-G_ B\nu $ is subharmonic on $B$ with Riesz measure $\nu $, so it is locally integrable and not identically $-\infty $. Since every $v_ j$ equals $-1$ on $A$, the sum $\sum _ jv_ j$ equals $-\infty $ on $A$. Hence $A$ is polar. $\square$

Proof. Capacitability of $f(\widetilde A)$. Replacing $K_{ik}$ by $K_{i1}\cup \cdots \cup K_{ik}$, we may assume that $K_{ik}$ increases with $k$. Put $A=f(\widetilde A)$ and fix $\varepsilon >0$. We choose integers $h_1,h_2,\ldots $ inductively. Put $\widetilde A_0=\widetilde A$, and suppose $\widetilde A_{p-1}$ has been defined. Since $\widetilde A_{p-1}\subset \bigcup _ kK_{pk}$, the sets $\widetilde A_{p-1}\cap K_{ph}$ increase with $h$ and have union $\widetilde A_{p-1}$, so their images under $f$ increase to $f(\widetilde A_{p-1})\subset L$. By part (3) of theorem 4.1.19, we can choose $h_ p$ such that $\widetilde A_ p=\widetilde A_{p-1}\cap K_{p,h_ p}$ satisfies

Hence $C_ B^*(f(\widetilde A_ p))>C_ B^*(A)-\varepsilon $ for every $p$.

The sets $L_ p=K_{1,h_1}\cap \cdots \cap K_{p,h_ p}$ are compact, because compact subsets of the Hausdorff space $X$ are closed; they decrease and contain $\widetilde A_ p$, so the compact sets $f(L_ p)\subset L$ decrease and, by part (1) of theorem 4.1.19 and the monotonicity of $C_ B^*$, satisfy $C_ B(f(L_ p))=C_ B^*(f(L_ p))\geq C_ B^*(f(\widetilde A_ p))>C_ B^*(A)-\varepsilon $. Moreover $\bigcap _ pf(L_ p)=f(\bigcap _ pL_ p)$: if $y\in f(L_ p)$ for every $p$, the sets $f^{-1}(y)\cap L_ p$ form a decreasing sequence of nonempty compact sets, since $f^{-1}(y)$ is closed, and hence have a common point. Since $\bigcap _ pL_ p=\bigcap _ iK_{i,h_ i}\subset \widetilde A$, the compact set $\bigcap _ pf(L_ p)$ is contained in $A$, and by proposition 4.1.18 its capacity is $\lim _ pC_ B(f(L_ p))\geq C_ B^*(A)-\varepsilon $. As $\varepsilon $ is arbitrary, $A$ is capacitable.

Borel sets. If $L=\varnothing $ there is nothing to prove, so let $L\neq \varnothing $. Call a subset of $L$ analytic if it has the form $f(\widetilde A)$ considered above; this is the notion of descriptive set theory and is unrelated to complex analytic sets. In such a representation we may assume that $X$ is nonempty, since the empty set is the image of the empty subset of a one-point space under a constant map. We may also assume that $X=Y\times L$ for a nonempty compact Hausdorff space $Y$ and that $f$ is the projection onto $L$. Indeed, put $Y=X$. The map $g\colon X\to Y\times L$, $g(x)=(x,f(x))$, is continuous and injective, so $g(\widetilde A)=\bigcap _ i\bigcup _ kg(K_{ik})$ with compact sets $g(K_{ik})$, and the projection of $g(\widetilde A)$ onto $L$ is $f(\widetilde A)$.

Analytic sets are closed under countable intersections and countable unions. To see this, let $A_ j=\pi (\widetilde A_ j)$, $j\geq 1$, be analytic, with $\widetilde A_ j\subset Y_ j\times L$ as above and $\pi $ denoting projections onto $L$. The space $X=(\prod _ jY_ j)\times L$ is compact by lemma 2.7.2 and Hausdorff, as a product of Hausdorff spaces. Let $\widehat A_ j\subset X$ be the inverse image of $\widetilde A_ j$ under the continuous map $\pi _ j\colon (y,z)\mapsto (y_ j,z)$. If $\widetilde A_ j=\bigcap _ i\bigcup _ kK^{(j)}_{ik}$ with compact sets $K^{(j)}_{ik}\subset Y_ j\times L$, then $\widehat A_ j=\bigcap _ i\bigcup _ k\pi _ j^{-1}(K^{(j)}_{ik})$. Each $K^{(j)}_{ik}$ is closed, being a compact subset of a Hausdorff space, so $\pi _ j^{-1}(K^{(j)}_{ik})$ is a closed and hence compact subset of $X$. Thus each $\widehat A_ j$, and therefore also $\bigcap _ j\widehat A_ j$, is a countable intersection of countable unions of compact subsets of $X$.

A point $z\in L$ lies in $\bigcap _ jA_ j$ if and only if for every $j$ there is $y_ j\in Y_ j$ with $(y_ j,z)\in \widetilde A_ j$, that is, if and only if $(y,z)\in \bigcap _ j\widehat A_ j$ for some $y=(y_ j)_ j\in \prod _ jY_ j$. Hence the projection of $\bigcap _ j\widehat A_ j$ onto $L$ is $\bigcap _ jA_ j$, and countable intersections of analytic sets are analytic. Since every $Y_ i$ is nonempty, the same reasoning shows that the projection of each $\widehat A_ j$ onto $L$ is $A_ j$.

For unions, write $\widehat A_ j=\bigcap _ n\bigcup _ kH_{jnk}$ with compact $H_{jnk}\subset X$, and consider, in the product of $X$ with the one-point compactification $\{ 1,2,\ldots \} \cup \{ \infty \} $ of the positive integers (a compact Hausdorff space, homeomorphic to $\{ 1/j:j\geq 1\} \cup \{ 0\} \subset \mathbb R$), the set

A point $(x,j)$ belongs to it if and only if $j\neq \infty $ and $x\in \widehat A_ j$, so its projection onto $L$ is $\bigcup _ jA_ j$. Each $H_{jnk}\times \{ j\} $ is compact, so countable unions of analytic sets are analytic.

Closed subsets of the compact metric space $L$ are compact, hence analytic, and an open set $U\subset L$ is the union of the compact sets $\{ z\in L:\operatorname {dist}(z,L\setminus U)\geq 1/n\} $, hence analytic. The family of sets $E\subset L$ such that both $E$ and $L\setminus E$ are analytic is closed under complements, and under countable unions, because the complement of a countable union is the countable intersection of the complements. It is therefore a $\sigma $-algebra containing all open subsets of $L$, so every Borel subset of $L$ is analytic and hence capacitable.

Increasing compact sets. Let $A\subset L$ be Borel. Since $A$ is capacitable, there are compact sets $K_ j\subset A$ with $C_ B(K_ j)\to C_ B^*(A)$. The compact sets $K_1\cup \cdots \cup K_ j\subset A$ increase, and by monotonicity their capacities increase to $C_ B^*(A)$. $\square$

Proof. By theorem 4.1.19, it remains to show that a polar set has outer capacity zero. By proposition 4.1.11, $A$ is contained in the pole set of a function $E_ d*\nu $ with $\nu $ finite, positive and compactly supported. Let $N$ be a compact neighborhood of $\overline A$ in $B$. The set $P=N\cap \{ E_ d*\nu =-\infty \} $ is Borel, being the intersection of $N$ with the open sets $\{ E_ d*\nu <-n\} $, $n\geq 1$. Every compact subset of $P$ is polar and therefore has capacity zero by theorem 4.1.17. By theorem 4.1.23, $C_ B^*(P)=0$, and monotonicity gives $C_ B^*(A)=0$. For the kernel capacities, a compact set has capacity zero if and only if it carries no probability measure of finite energy (theorem 4.1.7), that is, no nonzero positive measure of finite energy; by theorem 4.1.17 this happens if and only if it is polar. $\square$

Proof. Choose a cutoff $\chi \in C_ c^\infty (\Omega )$ with $0\leq \chi \leq 1$ and $\chi =1$ on an open neighborhood $V$ of $\operatorname{Supp}\sigma $, and put $\nu _ j=\chi \Delta v_ j$ and $\nu =\chi \Delta v$. Since $\Delta v_ j\to \Delta v$ distributionally, proposition 2.6.1 shows that $\nu _ j\to \nu $ vaguely, and the total masses $\nu _ j(\Omega )=\int \chi \, \mathrm d\Delta v_ j$ are bounded. By the Riesz decomposition (theorem 3.4.14), $v_ j=E_ d*\nu _ j+h_ j$ and $v=E_ d*\nu +h$ on $V$, with $h_ j$ and $h$ harmonic.

We show that $h_ j\to h$ uniformly on compact subsets of $V$. For a test function $\varphi $, Fubini's theorem gives $\int (E_ d*\nu _ j)\varphi \, \mathrm d x=\int (E_ d*\varphi )\, \mathrm d\nu _ j$, and $E_ d*\varphi $ is continuous by proposition 3.1.13; since the $\nu _ j$ are supported in the fixed compact set $\operatorname{Supp}\chi $, it follows that $E_ d*\nu _ j\to E_ d*\nu $ distributionally, and hence $h_ j\to h$ distributionally on $V$. The functions $h_ j$ are also bounded in $L^1$ on compact subsets of $V$. This holds for the $v_ j$, which converge in $L^1_{\mathrm{loc}}$, and for the $E_ d*\nu _ j$: if $Q\subset V$ is compact and $s>0$ is such that $Q-\operatorname{Supp}\chi \subset B(0,s)$, then Tonelli's theorem gives

and the masses $\nu _ j(\Omega )$ are bounded.

By the interior estimates (3.1.8), the derivatives of the $h_ j$ are therefore bounded on compact subsets of $V$, so the $h_ j$ are locally equicontinuous. Moreover, $h_ j\to h$ pointwise on $V$. Indeed, if $x\in V$ and $\varepsilon >0$ is less than the distance from $x$ to $\mathbb R^ d\setminus V$, then $h_ j(x)=(h_ j*\rho _\varepsilon )(x)$ and $h(x)=(h*\rho _\varepsilon )(x)$ by the last assertion of theorem 3.1.7, and $(h_ j*\rho _\varepsilon )(x)\to (h*\rho _\varepsilon )(x)$ by distributional convergence. To upgrade this to uniform convergence, let $Q\subset V$ be compact and $\theta >0$. By equicontinuity, every point of $Q$ has a neighborhood on which each $h_ j$, and also the continuous function $h$, varies by less than $\theta $. Finitely many of these neighborhoods, centered at points $x_1,\ldots ,x_ N$, cover $Q$, and for all large $j$ we have $|h_ j(x_ i)-h(x_ i)|<\theta $ for every $i$; hence $|h_ j-h|<3\theta $ on $Q$.

Next, the reciprocity law (4.1.2), applied to $\sigma $ and $\nu _ j$ and to $\sigma $ and $\nu $, gives

where the convergence holds because $E_ d*\sigma $ is continuous and the $\nu _ j$ converge vaguely with supports in $\operatorname{Supp}\chi $. All these integrals are finite, since $E_ d*\sigma $ is bounded on $\operatorname{Supp}\chi $. Adding the integrals of $h_ j$, which converge to $\int h\, \mathrm d\sigma $, proves that $\int v_ j\, \mathrm d\sigma \to \int v\, \mathrm d\sigma >-\infty $.

By Hartogs' lemma (theorem 3.4.20), $\varlimsup _ jv_ j\leq v$ everywhere, and the $v_ j$ are bounded above on $\operatorname{Supp}\sigma $ by a common constant for large $j$. Reverse Fatou's lemma therefore gives

Since $v-\varlimsup _ jv_ j\geq 0$ and $\int v\, \mathrm d\sigma $ is finite, it follows that $\varlimsup _ jv_ j=v$ $\sigma $-almost everywhere. $\square$

Proof. (1) Let $N\subset \Omega $ be compact, and take $R$ so large that $N\subset B$. The set $E=\{ x\in N:\varlimsup _ jv_ j(x)<v(x)\} $ is the union, over rational numbers $a<b$ and integers $J\geq 1$, of the Borel sets

so it is Borel. Suppose some compact subset $S$ of $E$ were not polar. By theorem 4.1.17, $S$ would carry a nonzero positive measure $\sigma $ with $E_ d*\sigma $ finite and continuous, and lemma 4.1.25 would give $\varlimsup _ jv_ j=v$ $\sigma $-almost everywhere on $S$, which is impossible. Hence every compact subset of $E$ is polar and has capacity zero. By theorem 4.1.23 and corollary 4.1.24, $E$ is polar. Since countably many compact sets $N$ cover $\Omega $, proposition 4.1.11 completes the proof.

(2) If the family is empty, or if all its members are identically $-\infty $ on a component of $\Omega $, then $F=F^*=-\infty $ there. Let $\Omega '$ be a component on which some member $u_{t_0}$ is not identically $-\infty $; by proposition 3.4.8, $u_{t_0}$ is locally integrable on $\Omega '$. By lemma 3.4.22, applied on $\Omega '$, there are members $u_{t_1},u_{t_2},\ldots $ of the family with $(\sup _{k\geq 1}u_{t_ k})^*=F^*$ on $\Omega '$. Put $w_ j=\max \{ u_{t_0},\ldots ,u_{t_ j}\} $ and $w=\sup _ jw_ j$. Since $\sup _{k\geq 1}u_{t_ k}\leq w\leq F$, we have $w^*=F^*$ on $\Omega '$, and theorem 3.4.23, applied to the family $(w_ j)$, shows that $F^*$ is subharmonic on $\Omega '$ and that $w=F^*$ almost everywhere on $\Omega '$.

We show next that $w_ j\to F^*$ in $L^1_{\mathrm{loc}}(\Omega ')$. As $u_{t_0}\leq w_ j\leq F^*$, the functions $w_ j$ and $F^*$ are locally integrable on $\Omega '$. If $Q\subset \Omega '$ is compact and $M$ is a common upper bound of the $w_ j$ on $Q$, then $0\leq M-w_ j\leq M-u_{t_0}$ on $Q$ and $M-w_ j$ decreases to $M-w$, so dominated convergence gives $w_ j\to w=F^*$ in $L^1(Q)$. Since $\varlimsup _ jw_ j=w$, part (1) shows that $\{ w<F^*\} \cap \Omega '$ is polar, and $\{ F<F^*\} \cap \Omega '$ is contained in it because $w\leq F$. The components of $\Omega $ are pairwise disjoint nonempty open sets, hence of positive Lebesgue measure, so there are at most countably many of them (lemma 2.2.5), and the union over them is polar by proposition 4.1.11. $\square$

Proof. Take $R$ so large that $K\subset B$. Let $\varepsilon >0$, and consider the set $E_\varepsilon =\{ x\in K:U^{\mu _ K}(x)<V(K)-\varepsilon \} $, which is Borel since $U^{\mu _ K}$ is lower semicontinuous. If $C_ B^*(E_\varepsilon )>0$, then by theorem 4.1.23 $E_\varepsilon $ contains a compact set of positive capacity, which is not polar by theorem 4.1.17; that theorem provides a probability measure $\nu $ on this compact set with finite continuous potential, hence with finite energy. Then $I(\mu _ K,\nu )=\int U^{\mu _ K}\, \mathrm d\nu \leq V(K)-\varepsilon $, contradicting theorem 4.1.7. Thus $C_ B^*(E_\varepsilon )=0$ for every $\varepsilon >0$, and countable subadditivity (theorem 4.1.19) and corollary 4.1.24 show that $\{ x\in K:U^{\mu _ K}(x)<V(K)\} $ is polar. The finite-energy measure $\mu _ K$ gives zero mass to this set (theorem 4.1.17), so $U^{\mu _ K}\geq V(K)$ $\mu _ K$-almost everywhere. Since $\int U^{\mu _ K}\, \mathrm d\mu _ K=V(K)$, it follows that $U^{\mu _ K}=V(K)$ $\mu _ K$-almost everywhere.

Let $x\in \operatorname{Supp}\mu _ K$. Every neighborhood of $x$ has positive $\mu _ K$-measure and therefore contains points where $U^{\mu _ K}=V(K)$; by lower semicontinuity, $U^{\mu _ K}(x)\leq V(K)$. To extend this bound to $\mathbb R^ d$, decompose $\mu _ K=\sum _ j\mu _ j$ by theorem 4.1.10, which applies because $E_ d*\mu _ K$ is finite $\mu _ K$-almost everywhere, and let $\pi _ N=\sum _{j\leq N}\mu _ j$. The potential $U^{\pi _ N}$ is continuous on $\mathbb R^ d$ and, by differentiation under the integral sign, harmonic off the compact set $S_ N=\operatorname{Supp}\pi _ N\subset \operatorname{Supp}\mu _ K$. Since the kernel is nonnegative on $K\times K$, we have $U^{\pi _ N}\leq U^{\mu _ K}\leq V(K)$ on $S_ N$.

We claim that $U^{\pi _ N}\leq V(K)$ on $\mathbb R^ d$. If $\pi _ N=0$, this holds because $V(K)\geq 0$, so let $\pi _ N\neq 0$. As $|x|\to \infty $, the Newtonian potential $U^{\pi _ N}(x)$ tends to $0<V(K)$, and the logarithmic potential tends to $-\infty $. Applying the maximum principle (proposition 3.1.10) to $U^{\pi _ N}$ on the bounded open sets $B(0,\rho )\setminus S_ N$ with $\rho $ large, we obtain the claim. Finally, fix $x\in \mathbb R^ d$. The kernel $k(x,\cdot )$ is bounded below on $K$ by a constant $-c_ x$ (with $c_ x=0$ in the Newtonian case), and monotone convergence applied to $k(x,\cdot )+c_ x$ gives $U^{\pi _ N}(x)\to U^{\mu _ K}(x)$. Hence $U^{\mu _ K}(x)\leq V(K)$. $\square$

Proof. We extend $u$ harmonically across $\partial B$, regularize the extension by convolution, and derive the identities from the divergence theorem for the regularizations. Since $u$ is bounded, hence locally integrable, the converse in part (2) of proposition 4.1.16 shows that $u=-G_ B\mu $; likewise $v=-G_ B\nu $. Choose $0<r<R$ with $\operatorname{Supp}\mu \subset B(0,r)$, and put $R'=R^2/r$. By part (2) of that proposition, $u$ extends to a function $\widetilde u$ on $B(0,R')$ that is harmonic on the annulus $r<|x|<R'$ and vanishes on $\partial B$. This extension is subharmonic on $B(0,R')$, because subharmonicity is a local property and the open sets $B$ and $\{ r<|x|<R'\} $ cover $B(0,R')$. Moreover, the distribution $\Delta \widetilde u$ on $B(0,R')$ equals $\mu $, extended by zero: the two agree on $B$, and both vanish on the annulus, so they agree on $B(0,R')$ by the last assertion of lemma 2.5.2.

Fix $r'$ with $r<r'<R$, put $Q=\overline{B(0,r')}$, and let $u_\varepsilon =\widetilde u*\rho _\varepsilon $ for $0<\varepsilon <\min \{ r'-r,R'-R\} $. Since $\widetilde u$ is harmonic on the annulus $r<|x|<R'$, the last assertion of theorem 3.1.7 gives $u_\varepsilon =\widetilde u$ on $r+\varepsilon <|x|<R'-\varepsilon $. Hence $u_\varepsilon $ is smooth on a neighborhood of $\overline B$, equals $u$ on $B\setminus Q$, vanishes on $\partial B$, and satisfies $\Delta u_\varepsilon =(\Delta \widetilde u)*\rho _\varepsilon =\mu *\rho _\varepsilon \geq 0$ on $B$ by proposition 2.5.3(1). In particular, $u_\varepsilon $ is a bounded subharmonic function on $B$ (a smooth function with nonnegative Laplacian is subharmonic, subsection 3.1.23), it tends to zero at $\partial B$, and its Riesz measure $(\mu *\rho _\varepsilon )\, \mathrm dx$ is supported in $\overline{B(0,r+\varepsilon )}\subset Q$. By (3.4.13), applied to $\widetilde u$, the functions $u_\varepsilon $ decrease to $u$ on $B$ as $\varepsilon \downarrow 0$.

For small $\varepsilon ,\delta >0$, the divergence identity (3.1.3) on $B$, applied to the field $u_\varepsilon \nabla u_\delta $, together with Fubini's theorem and the symmetry of $\rho _\delta $, gives

We let $\varepsilon $ and $\delta $ tend to zero. On $\operatorname{Supp}\mu $, the functions $\varphi _{\varepsilon ,\delta }=\widetilde u*\rho _\varepsilon *\rho _\delta $ are bounded by $\sup |\widetilde u|$ near $\overline B$, which is finite. By (3.4.13), $\widetilde u*\rho _\varepsilon $ decreases to $\widetilde u$ as $\varepsilon \downarrow 0$; convolving with $\rho _\delta \geq 0$ preserves order, so $\varphi _{\varepsilon ,\delta }$ decreases as either radius decreases, and $\varphi _{\varepsilon ,\delta }\geq \widetilde u*\rho _\delta \geq \widetilde u$. If $\varepsilon ,\delta \leq \eta \leq \eta '$, then $\varphi _{\varepsilon ,\delta }\leq \varphi _{\eta ,\eta }\leq \varphi _{\eta ',\eta }$, and $\varphi _{\eta ',\eta }\to \widetilde u*\rho _{\eta '}$ as $\eta \downarrow 0$ because $\widetilde u*\rho _{\eta '}$ is continuous. Letting $\eta \downarrow 0$ and then $\eta '\downarrow 0$, we see that $\varphi _{\varepsilon ,\delta }\to u$ pointwise on $\operatorname{Supp}\mu $ as $\max \{ \varepsilon ,\delta \} \to 0$.

By dominated convergence, $\int \nabla u_\varepsilon \mathbin {\cdot }\nabla u_\delta \, \mathrm d x\to -\int u\, \mathrm d\mu $. Expanding $\| \nabla u_\varepsilon -\nabla u_\delta \| _2^2$, we find that the gradients $\nabla u_\varepsilon $ form a Cauchy family in $L^2(B)$. Their limit is the distributional gradient of $u$, as one sees by integrating against test functions and using $u_\varepsilon \to u$ in $L^1_{\mathrm{loc}}(B)$. Hence $\nabla u\in L^2(B)$ and $\| \nabla u\| _2^2=\lim _\varepsilon \| \nabla u_\varepsilon \| _2^2=-\int u\, \mathrm d\mu $.

The function $u+v$ satisfies the same hypotheses, with Riesz measure $\mu +\nu $. Applying the first identity to $u$, $v$ and $u+v$ gives $2\int \nabla u\mathbin {\cdot }\nabla v\, \mathrm dx=-\int u\, \mathrm d\nu -\int v\, \mathrm d\mu $. Since $u=-G_ B\mu $ and $v=-G_ B\nu $, Tonelli's theorem and the symmetry of $G_ B$ (lemma 4.1.14) give

which proves the second line of the display. All these integrals are finite because $u$ and $v$ are bounded. The Cauchy–Schwarz inequality in $L^2(B)$ now gives the Cauchy–Schwarz inequality for the Green energy. Finally, $u_ K$ satisfies the hypotheses by proposition 4.1.16, and since $u_ K\geq -1$ and $\lambda _ K$ has total mass $C_ B(K)$, we get $\| \nabla u_ K\| _2^2=-\int u_ K\, \mathrm d\lambda _ K\leq C_ B(K)$. $\square$

Proof. The case of a bounded Green potential. Suppose first that $u$ is a bounded subharmonic function on $B$ that tends to zero at $\partial B$ and whose Riesz measure $\mu $ has compact support in $B$. Let $Q\subset B$ and $u_\varepsilon $ be the compact set and the regularizations given by the last assertion of proposition 4.1.29. These regularizations are continuous on $\overline B$, decrease to $u$, equal $u$ on $B\setminus Q$, and $\nabla u_\varepsilon \to \nabla u$ in $L^2(B)$. For $\delta >0$, the set $O_{\delta ,\varepsilon }=\{ u_\varepsilon -u>\delta \} $ is open, because $u_\varepsilon -u$ is lower semicontinuous, and it is contained in $Q$, so it has compact closure in $B$. Let $K\subset O_{\delta ,\varepsilon }$ be compact. The function $u_\varepsilon -u$ is a difference of two functions satisfying the hypotheses of proposition 4.1.29, and $\lambda _ K$ is supported on $K$, where $u_\varepsilon -u>\delta $. By the bilinear identity of that proposition, the Cauchy–Schwarz inequality and the bound $\| \nabla u_ K\| _2^2\leq C_ B(K)$,

Hence $C_ B(K)\leq \delta ^{-2}\| \nabla (u_\varepsilon -u)\| _2^2$, and taking the supremum over $K$ gives $C_ B(O_{\delta ,\varepsilon })\leq \delta ^{-2}\| \nabla (u_\varepsilon -u)\| _2^2$.

Choose $\delta _ j\downarrow 0$ and then radii $\varepsilon _ j\downarrow 0$ such that $C_ B(O_{\delta _ j,\varepsilon _ j})<\eta 2^{-j}$, and let $O=\bigcup _ jO_{\delta _ j,\varepsilon _ j}$, which is open with compact closure in $B$. By countable subadditivity (theorem 4.1.19), $C_ B(O)<\eta $. Outside $O$ we have $0\leq u_{\varepsilon _ j}-u\leq \delta _ j$ for all $j$, so the continuous functions $u_{\varepsilon _ j}$ converge uniformly to $u$ on $B\setminus O$. Hence $u$ is finite and continuous on $B\setminus O$, in particular on $L\setminus O$.

The case of a function bounded near $\overline B$. Suppose next that $u$ is bounded on a neighborhood of $\overline B$; subtracting a constant, we may assume that $u\leq -1$ there. Choose $r<R$ with $L\subset B(0,r)$, and $M>0$ with $-M<\inf _ Bu$. Let $q=Mu_{\overline{B(0,r)}}$, where the extremal function $u_{\overline{B(0,r)}}$ was computed in example 4.1.21. Then $q$ is subharmonic on $B$, equals $-M$ on $\overline{B(0,r)}$, is harmonic on $r<|x|<R$, and tends to zero at $\partial B$. The function $w=\max \{ u,q\} $ is subharmonic and bounded on $B$; it equals $u$ on $B(0,r)$, and it equals $q$ near $\partial B$, where $q>-1\geq u$. Hence $w$ tends to zero at $\partial B$, and its Riesz measure has compact support in $B$. The first case, applied to $w$, gives the required open set, since $w=u$ near $L$.

The general case. The function $u$ is bounded above near $\overline B$, so the functions $u_ N=\max \{ u,-N\} $, $N\geq 1$, are subharmonic and bounded near $\overline B$. Fix an open set $W$ with $L\subset W\Subset B$. By the second case, there are open sets $O_ N\subset W$ with $C_ B(O_ N)<\eta 2^{-N-1}$ such that $u_ N$ is continuous on $L\setminus O_ N$ (intersecting with $W$ does not affect this). On $L\setminus \bigcup _ NO_ N$, every $u_ N$ is continuous, and $\mathrm e^{u_ N}$ converges uniformly to $\mathrm e^ u$, because $0\leq \mathrm e^{u_ N}-\mathrm e^ u\leq \mathrm e^{-N}$. Hence $u$ is continuous there as a function with values in $[-\infty ,\infty )$. Let $\Omega _0$ be the connected component of $\Omega $ that contains the connected set $\overline B$. Since $u$ is not identically $-\infty $ on $\Omega _0$, the set $\{ u=-\infty \} \cap \Omega _0$ is polar by the definition of polarity. By corollary 4.1.24, $L\cap \{ u=-\infty \} $ is contained in an open set $O'\Subset B$ with $C_ B(O')<\eta /2$. The open set $O=O'\cup \bigcup _ NO_ N$ has compact closure in $B$ and capacity less than $\eta $ by countable subadditivity, and the restriction of $u$ to $L\setminus O$ is finite and continuous. $\square$