Proposition 4.1.183CC68CCFAF3A

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. $\square$