Proposition 5.47.65292FEC65E83

Proof. The boundary behavior of $H_ K$ follows from (5.47.3). Since $H_ K$ is bounded, its Monge–Ampère measure is defined (definition 7.2.1).

We first show that $(\mathrm{dd}^{\mathrm{c}}H_ K)^ n$ vanishes on $B\setminus K$. Choose competitors $v_ j$ for $K$ as in (5.47.4), and let $D$ be a ball with $\overline D\subset B\setminus K$. Let $\widehat v_ j$ be the function obtained from $v_ j$ by corollary 7.7.5: it is plurisubharmonic on $B$, equals $v_ j$ outside $D$, and satisfies $\widehat v_ j\ge v_ j$ and $(\mathrm{dd}^{\mathrm{c}}\widehat v_ j)^ n=0$ on $D$. Each $\widehat v_ j$ is again a competitor for $K$. Indeed, $\widehat v_ j=v_ j\le -1$ on $K$. Moreover $\widehat v_ j\le 0$: outside $D$ because $\widehat v_ j=v_ j$, and on $D$ by the maximum principle (proposition 3.4.7), since by upper semicontinuity the upper limit of $\widehat v_ j$ at each point $x\in \partial D$ is at most $\widehat v_ j(x)=v_ j(x)\le 0$.

Since $\widehat v_ j$ is increasing in $v_ j$ by corollary 7.7.5, the functions $\widehat v_ j$ increase with $j$, and $v_ j\le \widehat v_ j\le h_ K\le H_ K$ shows that $\widehat v_ j\to H_ K$ almost everywhere. They take values in $[-1,0]$, so theorem 7.6.1 gives $(\mathrm{dd}^{\mathrm{c}}\widehat v_ j)^ n\to (\mathrm{dd}^{\mathrm{c}}H_ K)^ n$ weakly, and therefore $(\mathrm{dd}^{\mathrm{c}}H_ K)^ n=0$ on $D$. Covering the open set $B\setminus K$ by countably many such balls $D$, we conclude that $(\mathrm{dd}^{\mathrm{c}}H_ K)^ n$ is carried by $K$.

As $h_ K=-1$ on $K$, the Borel set $K\cap \{ H_ K>-1\} $ is contained in $\{ h_ K<H_ K\} $, which has outer capacity zero by (5.47.5). Because $H_ K+1$ is a competitor for $\operatorname {Cap}_ B$, this set is therefore null for $(\mathrm{dd}^{\mathrm{c}}(H_ K+1))^ n=(\mathrm{dd}^{\mathrm{c}}H_ K)^ n$. Thus $(\mathrm{dd}^{\mathrm{c}}H_ K)^ n$ is carried by $K\cap \{ H_ K=-1\} $.

We now prove the capacity formula. As $H_ K+1$ is a competitor for $\operatorname {Cap}_ B$ and $(\mathrm{dd}^{\mathrm{c}}H_ K)^ n$ is carried by $K$, we have $\int _ B(\mathrm{dd}^{\mathrm{c}}H_ K)^ n=\int _ K(\mathrm{dd}^{\mathrm{c}}(H_ K+1))^ n\le \operatorname {Cap}_ B(K)$. Conversely, let $v$ be a competitor for $\operatorname {Cap}_ B$, let $0<\varepsilon <1/2$, and put $w=(1-2\varepsilon )(v-1)-\varepsilon $. Then $-1+\varepsilon \le w\le -\varepsilon $. In particular $w>-1$, so $K\cap \{ H_ K\ge w\} \subset K\cap \{ H_ K>-1\} $; this set has outer capacity zero and is therefore null for $(\mathrm{dd}^{\mathrm{c}}w)^ n=(1-2\varepsilon )^ n(\mathrm{dd}^{\mathrm{c}}v)^ n$. Moreover $H_ K(z)\to 0$ as $z\to \partial B$ while $w\le -\varepsilon $, so $\varliminf _{z\to x}(H_ K(z)-w(z))\ge \varepsilon >0$ for every $x\in \partial B$, and the comparison principle (theorem 7.3.11) applies to $H_ K$ and $w$ on $B$. We obtain

Letting $\varepsilon \downarrow 0$ and taking the supremum over $v$, we get $\operatorname {Cap}_ B(K)\le \int _ B(\mathrm{dd}^{\mathrm{c}}H_ K)^ n$.

Now let $(K_ j)$ be compact subsets of $B$ decreasing to $K$. Since $K\subset K_{j+1}\subset K_ j$, the functions $H_{K_ j}$ increase and are bounded above by $H_ K$. Put $H=(\sup _ jH_{K_ j})^*$. By theorem 5.10.2, $H$ is plurisubharmonic and equals $\sup _ jH_{K_ j}$ almost everywhere, and $H\le H_ K$ because $H_ K$ is upper semicontinuous. For the reverse inequality, let $v$ be a competitor for $K$ and let $0<\varepsilon <1$. The open set $G=\{ v<-1+\varepsilon \} $ contains $K$. The compact sets $K_ j\setminus G$ decrease and have empty intersection, so one of them is empty; hence $K_ j\subset G$ for all large $j$. For such $j$ the function $v/(1-\varepsilon )$ is a competitor for $K_ j$, and hence $v\le (1-\varepsilon )H_{K_ j}\le (1-\varepsilon )H$. Letting $\varepsilon \downarrow 0$ gives $v\le H$, and taking the supremum over all competitors $v$ for $K$ gives $h_ K\le H$. As $H$ is upper semicontinuous, $H_ K\le H$. Therefore $H=H_ K$, and $H_{K_ j}\to H_ K$ almost everywhere.

By theorem 7.6.1, $(\mathrm{dd}^{\mathrm{c}}H_{K_ j})^ n\to (\mathrm{dd}^{\mathrm{c}}H_ K)^ n$ weakly. All these measures are carried by the fixed compact set $K_1$, so testing with a function in $C_ c(B)$ equal to one on $K_1$ shows that their total masses converge. By the capacity formula, $\operatorname {Cap}_ B(K_ j)\to \operatorname {Cap}_ B(K)$, and this convergence is decreasing because capacity is increasing.

Finally, $\operatorname {Cap}_ B(K)\le \operatorname {Cap}_ B^*(K)$ because capacity is increasing. For the converse we may assume $K\neq \varnothing $, since both capacities of the empty set vanish. Choose $\delta >0$ such that the compact sets $K_ j=\{ z:\operatorname {dist}(z,K)\le \delta /j\} $ lie in $B$. They decrease to $K$, and each contains $K$ in its interior. Hence $\operatorname {Cap}_ B^*(K)\le \operatorname {Cap}_ B(\operatorname{Int}K_ j)\le \operatorname {Cap}_ B(K_ j)$, and the right-hand side tends to $\operatorname {Cap}_ B(K)$ by what we have just proved. $\square$