Lemma 16.27.548CC217E0795

Proof. We first construct $g$. If $K=X$, take $g=0$. Otherwise fix finitely many coordinate charts $D_ i$ and compact sets $K_ i\subset D_ i$ whose interiors cover $X$, and measure derivatives on $K_ i$ in the coordinates of $D_ i$. Each point of $X\setminus K$ is the center of a coordinate ball whose closure lies in $X\setminus K$; since $X\setminus K$ has a countable basis, countably many of the concentric balls of half the radius, say $B_1',B_2',\ldots $, cover $X\setminus K$. Let $B_ l$ be the corresponding ball of full radius, and choose a smooth function $g_ l\geq 0$ with compact support in $B_ l$ that is positive on $\overline{B_ l'}$ (a smooth cutoff function, lemma 2.5.1). Multiplying $g_ l$ by a small positive constant, we may assume that all its derivatives of order at most $l$ are bounded by $2^{-l}$ on every $K_ i$. Then the series $g=\sum _ lg_ l$ converges uniformly on every $K_ i$ together with all its derivatives, so $g$ is smooth on the interior of every $K_ i$, hence on $X$; it vanishes on $K$, and it is positive on $X\setminus K$, because every point there lies in some $B_ l'$.

Now let $g$ be any smooth nonnegative function with zero set $K$, and put $q_ j=P_\theta (\phi +jg)$. Since $\phi $ is a competitor and every competitor is at most $\phi +j\sup _ Xg$, we have $\phi \leq q_ j\leq \phi +j\sup _ Xg$, so $q_ j\sim \phi $. The obstacle $\phi +jg$ is upper semicontinuous, so $q_ j\leq \phi +jg$ everywhere by proposition 16.11.4, and therefore $q_ j=\phi $ on $K$. In particular $\sup _ Kq_ j\leq 0$, and corollary 10.3.5 bounds $\sup _ Xq_ j$ uniformly in $j$. The envelopes $q_ j$ increase with $j$, since $g\geq 0$, so $S=(\sup _ jq_ j)^*$ is a potential; put $L=\sup _ XS$. Then $L\geq 0$, because $S\geq \phi $ and $\sup _ X\phi =0$. Since $q_ j\preceq \phi $ and $\sup _ Xq_ j\leq \sup _ XS=L$, lemma 16.22.2 gives $q_ j\leq \phi +L$, hence $\phi \leq S\leq \phi +L$. By corollary 5.47.11, applied after adding smooth local potentials of $\theta $, the regularization changes $\sup _ jq_ j$ only on a pluripolar set, so $S=\phi $ quasi-everywhere on $K$. Since $S\sim \phi $, $\int _ X\langle \theta _ S^ n\rangle =m$.

The obstacle $\phi +jg$ is quasi-continuous, and its family of competitors is nonempty and bounded above by $j\sup _ Xg$, so theorem 16.13.4 shows that $\mu _{q_ j}$ is carried by the contact set $\{ q_ j=\phi +jg\} $. At a point of this set where $\phi $ is finite, the bound $q_ j\leq \phi +L$ gives $jg\leq L$. Thus $\mu _{q_ j}$ is carried by $\{ g\leq L/j\} $. By corollary 13.8.9, $\mu _{q_ j}\to \mu _ S$ weakly. If $\chi \geq 0$ is continuous with compact support in $X\setminus K$, then $g$ has a positive lower bound $c$ on the support of $\chi $, so $\int _ X\chi \, \mu _{q_ j}=0$ for $j>L/c$, and in the limit $\int _ X\chi \, \mu _ S=0$. By lemma 16.12.1(2), applied on $W=X\setminus K$ with the zero measure, $\mu _ S(X\setminus K)=0$.

Set $t=1/\max \{ 1,L\} $ and $h=(1-t)\phi +tS-tL$. Then $\phi -1\leq \phi -tL\leq h\leq \phi $, so $h$ is a competitor for $\operatorname {Cap}_{\theta ,\phi }$, and $\mu _ h\geq t^ n\mu _ S$. Hence $\operatorname {Cap}_{\theta ,\phi }(K)\geq t^ n\mu _ S(K)=t^ nm$, which is (16.27.6).

In the Kähler case apply (16.27.4) to the potential $p=S-L$, which has supremum zero. Quasi-everywhere on $K$ we have $-p=L-\phi \geq L$, so $L\mu _ v(K)\leq C_\kappa $ for every competitor $v$ for $\operatorname {Cap}_{\kappa ,\phi }$. If $L\geq 1$, this gives (16.27.7) with $C=C_\kappa $. If $L<1$, use $\operatorname {Cap}_{\kappa ,\phi }(K)\leq \operatorname {Cap}_{\kappa ,\phi }(X)=m\leq \Omega $; the last inequality is the monotonicity of mass, because $\phi \preceq V_\kappa =0$ and the mass of $0$ is $\int _ X\kappa ^ n=\Omega $. Thus $C=\max \{ C_\kappa ,\Omega \} $ works. $\square$