Pome

Theorem 4.1.19AEE550586C10

Theorem 4.1.19.AEE550586C10 The outer relative capacity has the following properties.

  1. $C_ B^*(K)=C_ B(K)$ for every compact set $K\subset B$.

  2. It is increasing and countably subadditive: $C_ B^*(\bigcup _ jA_ j)\leq \sum _ jC_ B^*(A_ j)$ whenever $\bigcup _ jA_ j\Subset B$.

  3. If $A_1\subset A_2\subset \cdots $ and $A=\bigcup _ jA_ j\Subset B$, then $C_ B^*(A_ j)\to C_ B^*(A)$.

  4. If $C_ B^*(A)=0$, then $A$ is polar.

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

\[ C_ B(K)\leq C_ B^*(K)\leq C_ B(O_ j)\leq C_ B(K_ j)\longrightarrow C_ B(K). \]

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,

\[ \begin{aligned} C_ B(K)+C_ B(L)& \leq C_ B\bigl ((K'\cup L)\cup (K''\cup L)\bigr )+C_ B\bigl ((K'\cup L)\cap (K''\cup L)\bigr )\\ & \leq C_ B(K'\cup L)+C_ B(K''\cup L)\leq C_ B(O)+C_ B(O'). \end{aligned} \]

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

\[ \begin{aligned} C_ B^*(E_1\cup E_2)+C_ B^*(A_1)& \leq C_ B^*(E_1)+C_ B^*(A_1\cup E_2),\\ C_ B^*(A_1\cup E_2)+C_ B^*(A_2)& \leq C_ B^*(A_1\cup A_2)+C_ B^*(E_2), \end{aligned} \]

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

0A0CA0153B81 4.1.20
\begin{equation} C_ B^*\left(\bigcup _{j=1}^ NE_ j\right)-C_ B^*\left(\bigcup _{j=1}^ NA_ j\right) \leq \sum _{j=1}^ N\bigl (C_ B^*(E_ j)-C_ B^*(A_ j)\bigr ), \end{equation}

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,

\[ C_ B^*\left(\bigcup _ j A_ j\right)\leq C_ B\left(\bigcup _ j E_ j\right)=\lim _{N\to \infty }C_ B\left(\bigcup _{j=1}^ NE_ j\right) \leq \lim _{N\to \infty }C_ B^*\left(\bigcup _{j=1}^ N A_ j\right)+\varepsilon . \]

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. $\square$

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