16.27 Comparison of relative capacitiesDE4415660D60

Proof. Apply theorem 6.17.6 with $\kappa $ as the Kähler form of reference. It gives potentials $\Phi _ j$ decreasing to $p$ with $\kappa +\mathrm{dd}^{\mathrm{c}}\Phi _ j\geq -\varepsilon _ j\kappa $, where $\varepsilon _ j\downarrow 0$, and with gentle analytic singularities of type $(2^{-j},\mathcal I(2^ jp))$. Since $p$ is bounded, $\mathrm e^{-2^ jp}$ is bounded, so $\mathcal I(2^ jp)$ is the unit ideal. In the local presentations $\Phi _ j=2^{-j}\log \sum _ i|g_ i|^2+b$ of definition 6.14.5, the functions $g_ i$ therefore generate the unit ideal and have no common zero, so $\Phi _ j$ is bounded. Hence the smooth function $\exp (2^ j\Phi _ j)$ has no zeros, and $\Phi _ j=2^{-j}\log \exp (2^ j\Phi _ j)$ is smooth. Put $p_ j=\Phi _ j/(1+\varepsilon _ j)$. Then

The bounds $p\leq \Phi _ j\leq \Phi _1$ give $|p_ j|\leq |\Phi _ j|\leq \max \{ \sup _ X|p|,\sup _ X|\Phi _1|\} $, and $p_ j\to p$ pointwise because $\Phi _ j\to p$ and $\varepsilon _ j\to 0$ (unlike the $\Phi _ j$, the $p_ j$ need not decrease). $\square$

Proof. Write $\kappa _ q=\kappa +\mathrm{dd}^{\mathrm{c}}q$. We first prove (16.27.3) when $u=v$ near the poles of $v$ and then reduce to this case; the second assertion follows from the first.

A special case. Assume first that there are $l_0>0$ and an open set $U\supset \{ v<-l_0\} $ on which $u=v$. For $l>l_0$ put $u_ l=\max \{ u,-l\} $ and $v_ l=\max \{ v,-l\} $; these are bounded functions in $\mathrm{PSH}(X,\kappa )$ with $0\leq u_ l-v_ l\leq B$. On $U$ we have $u_ l=v_ l$ and $u=v$, so $\mu _{u_ l}=\mu _{v_ l}$ and $\mu _ u=\mu _ v$ there. The complement of $U$ lies in the plurifine open set $\{ v>-l\} $, on which $u_ l=u$ and $v_ l=v$, because $u\geq v$; by plurifine locality, $\mu _{u_ l}=\mu _ u$ and $\mu _{v_ l}=\mu _ v$ on this set. Hence $\mu _{u_ l}-\mu _{v_ l}=\mu _ u-\mu _ v$. Expanding $\kappa _{u_ l}^ n-\kappa _{v_ l}^ n$ and integrating by parts with lemma 13.5.2, we obtain

For the last inequality write $\mathrm{dd}^{\mathrm{c}}p=(\kappa +\mathrm{dd}^{\mathrm{c}}p)-\kappa $: the first part contributes a nonnegative amount, since $u_ l-v_ l\geq 0$, and the second contributes at least $-B\int _ X\kappa \wedge \kappa _{u_ l}^ j\wedge \kappa _{v_ l}^{n-1-j}=-B\Omega $, by proposition 13.5.1.

The general case. Subtracting a common constant, we may assume $v\leq 0$. For $0<a<1$ put $v_ a=av$ and $u_ a=\max \{ u,av\} $; these belong to $\mathrm{PSH}(X,\kappa )$ because $\kappa +\mathrm{dd}^{\mathrm{c}}(av)=(1-a)\kappa +a\kappa _ v$. Since $v\leq av$, we have $v_ a\leq u_ a\leq v_ a+B$. On the open set $U_ a=\{ v<-B/(1-a)\} $ we have $av-v>B\geq u-v$, so $u_ a=v_ a$ there. With $l_0=aB/(1-a)+1$, the set $\{ v_ a<-l_0\} =\{ v<-l_0/a\} $ is contained in $U_ a$, so the special case applies to $u_ a$ and $v_ a$.

As $a\uparrow 1$, $v_ a$ and $u_ a$ decrease to $v$ and $u$. The masses of $u_ a$ and $v_ a$ agree, and by multilinearity $\int _ X\langle \kappa _{v_ a}^ n\rangle $ is a polynomial in $a$ that tends to $\int _ X\langle \kappa _ v^ n\rangle =\int _ X\langle \kappa _ u^ n\rangle $. Decreasing sequences converge in extended capacity (lemma 13.7.3), so theorem 13.7.5 shows that $\mu _{u_ a}\to \mu _ u$ and $\mu _{v_ a}\to \mu _ v$ weakly along every sequence $a\uparrow 1$. Passing to the limit in (16.27.3) for $u_ a$ and $v_ a$, we obtain (16.27.3) for every continuous, in particular every smooth, $p\in \mathrm{PSH}(X,\kappa )$. For a bounded $p$, take the approximations $p_ j$ of lemma 16.27.1; they converge pointwise with a common bound, and dominated convergence with respect to the finite measures $\mu _ u$ and $\mu _ v$ proves (16.27.3).

The second assertion. Let $\phi $ be a model potential for $\kappa $. Since $V_\kappa =0$, proposition 16.21.3 gives $\mu _\phi \leq \kappa ^ n$. If $\phi -1\leq v\leq \phi $, the first assertion, applied to $\phi \leq v+1\leq \phi +1$, gives for every bounded $p$ with $\sup _ Xp=0$

By proposition 10.3.1, applied on the compact Kähler manifold $(X,\kappa )$ with background $\kappa $, the last integral is bounded by a constant depending only on $X$ and $\kappa $. For an unbounded $p$ with $\sup _ Xp=0$, apply this bound to the truncations $\max \{ p,-l\} $, which are bounded, belong to $\mathrm{PSH}(X,\kappa )$ and have supremum zero, and let $l\to \infty $ by monotone convergence. $\square$

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. $\square$

Proof. Let $\alpha $ and $C_0$ be the constants of corollary 10.3.3. First let $K$ be compact and non-pluripolar, and let $S,L$ be as in lemma 16.27.5. The potential $p=S-L$ has supremum zero and satisfies $p\leq -L$ quasi-everywhere on $K$, because $S=\phi \leq 0$ there. Since pluripolar sets have zero volume, (10.3.4) gives

Write $c=\operatorname {Cap}_{\theta ,\phi }(K)$, which is positive by (16.27.6). By the same inequality,

so (16.27.9) holds for $K$ with $C=C_0\mathrm e^\alpha $. If $K$ is pluripolar, its volume and its capacity both vanish.

Now let $E$ be a Borel set. If $\operatorname {Cap}_{\theta ,\phi }(E)>0$, the estimate for compact subsets $K\subset E$, together with the monotonicity of capacity and the inner regularity of the smooth volume, gives (16.27.9). If $\operatorname {Cap}_{\theta ,\phi }(E)=0$, every compact subset of $E$ has zero capacity, hence is pluripolar by (16.27.6) and has zero volume; by inner regularity, $E$ has zero volume. $\square$

Proof. If $\operatorname {Cap}_{\theta ,\phi }(E)=0$, then $E$ has zero volume by proposition 16.27.8, and both sides vanish. Otherwise let $q=p/(p-1)$. By Hölder's inequality and (16.27.9), the integral is at most

Put $t=(m/\operatorname {Cap}_{\theta ,\phi }(E))^{1/n}$; then $t\geq 1$, since the capacity is at most $\int _ X\langle \theta _\phi ^ n\rangle =m$. The function $t^{nr}\mathrm e^{-\alpha t/q}$ is bounded on $[1,\infty )$: its logarithmic derivative is $nr/t-\alpha /q$, so its maximum is attained at $1$ or at $nrq/\alpha $. Hence $\mathrm e^{-\alpha t/q}\leq A\, t^{-nr}$ for a constant $A$ depending only on $n,r,q,\alpha $, and this proves (16.27.11). $\square$

Proof. All capacities involved are inner regular (proposition 16.25.6), so it suffices to treat compact sets $K$: the inequalities for a Borel set $E$ follow by applying them to the compact subsets of $E$ and taking suprema. If $K$ is pluripolar, all capacities of $K$ vanish. Otherwise fix a smooth function $g\geq 0$ with zero set $K$, and apply lemma 16.27.5 with this $g$ to $(\theta ,\phi )$ and to $(\kappa ,0)$; here $0=V_\kappa $ is a model potential for $\kappa $ of mass $\Omega $, by proposition 16.21.2(4). Denote the resulting envelopes by $S_\phi $ and $S_0$ and their suprema by $L_\phi $ and $L_0$. Every competitor for $P_\theta (\phi +jg)$ is $\kappa $-plurisubharmonic, because $\theta \leq \kappa $, and it is at most $\phi +jg\leq jg$ quasi-everywhere, so $P_\theta (\phi +jg)\leq P_\kappa (jg)$ for every $j$; hence $S_\phi \leq S_0$ and $L_\phi \leq L_0$. The two estimates of lemma 16.27.5 give

This proves (16.27.14) and, for $\theta =\kappa $, the first inequality in (16.27.13).

For the upper bound let $\theta =\kappa $, and choose $b>1$ with $b^ n(\Omega -m)<\Omega $. By theorem 16.22.4, applied to $\phi \preceq 0$, where the mass of $0$ is $\int _ X\kappa ^ n=\Omega $, the envelope $\eta =E_ b(\phi ,0)=P_\kappa (b\phi )$ is a potential, and it satisfies $b^{-1}\eta \leq \phi $; in particular $\eta =-\infty $ at the poles of $\phi $. Write $c=1-b^{-1}>0$. For fixed $j$ and $\delta >0$ consider

a convex combination of $\kappa $-plurisubharmonic functions minus a constant. The function $P_\kappa (jg)$ is upper semicontinuous and vanishes on $K$, because $0\leq P_\kappa (jg)\leq jg$. Hence $P_\kappa (jg)<\delta /c$ on a neighborhood $U$ of $K$, and there $v_ j<b^{-1}\eta \leq \phi $. Outside the poles of $\phi $, the difference $v_ j-\phi $ is bounded above by $cj\sup _ Xg$, and $g$ has a positive minimum on $X\setminus U$. Therefore $v_ j\leq \phi +lg$ for a sufficiently large integer $l$, so $v_ j\leq P_\kappa (\phi +lg)\leq S_\phi $. Letting $\delta \downarrow 0$ and then $j\to \infty $, and using that the increasing limit of $P_\kappa (jg)$ equals $S_0$ almost everywhere, we obtain

almost everywhere, hence everywhere.

Choose a Borel set $G$ of positive volume on which $b^{-1}\eta \geq -D$ for a constant $D$; such a set exists because $\eta $ is finite almost everywhere. By proposition 10.3.1, there is a constant $B$, independent of $K$, with $\int _ X|S_0-L_0|\, \kappa ^ n\leq B$. Hence $S_0\geq L_0-B/\int _ G\kappa ^ n$ at some point $x\in G$, for otherwise $\int _ G(L_0-S_0)\, \kappa ^ n>B$. At this point the last display gives $L_\phi \geq S_\phi (x)\geq cL_0-D'$ with $D'=cB/\int _ G\kappa ^ n+D$, which does not depend on $K$. If $L_0\geq \max \{ 1,2D'/c\} $, then $L_\phi \geq cL_0/2$, and lemma 16.27.5 gives

the last inequality is the lower bound (16.27.6) for the construction with reference potential $0$, which gives $\Omega /L_0^ n\leq \operatorname {Cap}_\kappa (K)$. In the remaining range $L_0<\max \{ 1,2D'/c\} $, the same lower bound shows that $\operatorname {Cap}_\kappa (K)$ is bounded below by a positive constant independent of $K$, and we use $\operatorname {Cap}_{\kappa ,\phi }(K)\leq m$, enlarging $A$ if necessary. The constant $A$ in the upper bound depends on $\phi $. $\square$

Proof. Since $\operatorname {Cap}_{\theta ,\phi }$ does not change when a constant is added to $\phi $ (proposition 16.25.6), we may assume that $\psi \leq \phi $. Let $u$ be a competitor for $\operatorname {Cap}_{\theta ,\psi }$, so that $\psi -1\leq u\leq \psi $, and let $t\geq 1$. We estimate $\mu _ u(E)$ separately on $\{ u>\phi -2t\} $ and on $\{ u\leq \phi -2t\} $.

Put $u_ t=\max \{ u,\phi -2t\} $, so that $\phi -2t\leq u_ t\leq \phi $. The potential $(1-(2t)^{-1})\phi +(2t)^{-1}u_ t$ is a competitor for $\operatorname {Cap}_{\theta ,\phi }$, and its measure is at least $(2t)^{-n}\mu _{u_ t}$. Since $u_ t=u$ on the plurifine open set $\{ u>\phi -2t\} $, it follows that

The potentials $\psi $, $u$ and $\psi -1$ belong to $\mathcal E(X,\theta ,\phi )$. So does the midpoint $q=(u+\phi )/2$: by proposition 16.22.14, $P_\theta [u]=P_\theta [\phi ]$, so by proposition 16.21.2 each mixed mass $\int _ X\langle \theta _ u^ k\wedge \theta _\phi ^{n-k}\rangle $ equals the mass of $\phi $. Expanding $\mu _ q$ by multilinearity, we see that $q$ also has the mass of $\phi $. On the set $\{ u\leq \phi -2t\} $ we have $q-t\geq u\geq \psi -1$, and on $\{ \psi -1\leq q-t\} $ we have $\psi -1\leq (\psi +\phi )/2-t$, that is, $\psi \leq \phi -2t+2\leq \phi -t+1$. Thus

The non-strict comparison in proposition 16.25.4, applied with $p=n$ to $\psi -1$ and $q-t$, together with $\mu _ q\geq 2^{-n}\mu _ u$, gives

Adding the two estimates and taking the supremum over $u$, we obtain

The last term tends to zero as $t\to \infty $, since $\psi $ and $\phi $ are finite outside a $\mu _\psi $-null set.

It remains to replace the right side by a continuous function of the capacity. Put $m=\int _ X\langle \theta _\psi ^ n\rangle $, and let $H(s)$ be the minimum of $m$ and of the infimum over $t\geq 1$ of the right side of (16.27.16) with $\operatorname {Cap}_{\theta ,\phi }(E)$ replaced by $s$. Since $\operatorname {Cap}_{\theta ,\psi }\leq \int _ X\langle \theta _\psi ^ n\rangle =m$, we have $\operatorname {Cap}_{\theta ,\psi }(E)\leq H(\operatorname {Cap}_{\theta ,\phi }(E))$. The function $H$ is nondecreasing, and $H(s)\to 0$ as $s\downarrow 0$. Choose a strictly decreasing sequence $s_ k\downarrow 0$ with $H(s_ k)\leq m2^{-k}$. Define $F(0)=0$, $F(s_ k)=m2^{1-k}$, $F$ linear on each interval $[s_{k+1},s_ k]$, and $F=m$ on $[s_1,\infty )$. On $[s_{k+1},s_ k]$ we have $H(s)\leq H(s_ k)\leq m2^{-k}=F(s_{k+1})\leq F(s)$, and $H\leq m=F$ on $[s_1,\infty )$. Thus $F$ is a continuous nondecreasing majorant of $H$ with $F(0)=0$. $\square$

Proof. Choose $A\geq 1$ with $\theta \leq \kappa =A\omega $. Every competitor $v$ for $\operatorname {Cap}_{\theta ,\psi }$ is a competitor for $\operatorname {Cap}_{\kappa ,\psi }$, and its measure for $\kappa $ dominates its measure for $\theta $: since $\kappa +\mathrm{dd}^{\mathrm{c}}v=(\theta +\mathrm{dd}^{\mathrm{c}}v)+(\kappa -\theta )$ with a smooth closed semipositive form $\kappa -\theta $, multilinearity and positivity of non-pluripolar products give $\langle (\kappa +\mathrm{dd}^{\mathrm{c}}v)^ n\rangle \geq \langle (\theta +\mathrm{dd}^{\mathrm{c}}v)^ n\rangle $; hence $\operatorname {Cap}_{\theta ,\psi }\leq \operatorname {Cap}_{\kappa ,\psi }$, and the mass of $\psi $ for $\kappa $ is at least its mass for $\theta $, hence positive. Adding a constant to $\psi $ does not change the capacities, so we may assume $\psi \leq 0$. Let $\Phi =P_\kappa [\psi ]$. By theorem 16.22.11, $\Phi $ is a model potential for $\kappa $ with $\psi \leq \Phi $ and with the same mass for $\kappa $ as $\psi $. Thus $\psi \in \mathcal E(X,\kappa ,\Phi )$, and lemma 16.27.15 and proposition 16.27.12 bound $\operatorname {Cap}_{\kappa ,\psi }$ by a continuous nondecreasing function, vanishing at zero, of $\operatorname {Cap}_\kappa $.

Finally, rescaling the competitors gives

every competitor for $\operatorname {Cap}_\omega $ is one for $\operatorname {Cap}_\kappa $, with a larger measure by the same argument, and if $v$ is a competitor for $\operatorname {Cap}_\kappa $, then $v/A$ is a competitor for $\operatorname {Cap}_\omega $ with measure $A^{-n}(\kappa +\mathrm{dd}^{\mathrm{c}}v)^ n$. This proves the first inequality.

For the second inequality use the model potential $\phi =P_\theta [\psi ]$ for $\theta $, which has the same positive mass $m=\int _ X\langle \theta _\psi ^ n\rangle $ as $\psi $. By (16.27.14), $\operatorname {Cap}_\kappa ^ n\leq (C^ n/m)\operatorname {Cap}_{\theta ,\phi }$. By theorem 16.26.1 with $a=2$, applied to $\psi $ and $\phi $, whose singularity envelopes agree, the potential $\eta =P_\theta (2\psi -\phi )$ has mass $m$ and satisfies $P_\theta [\eta ]=\phi $ and $(\eta +\phi )/2\leq \psi $.

It remains to bound $\operatorname {Cap}_{\theta ,\phi }$ by a function of $\operatorname {Cap}_{\theta ,\psi }$. Put $u=(\eta +\phi )/2$. Since $P_\theta [\eta ]=P_\theta [\phi ]$, the preservation of mixed masses in proposition 16.21.2 shows that $u$ also has mass $m$. Since $u\leq \psi $, this means that $u\in \mathcal E(X,\theta ,\psi )$, and lemma 16.27.15 bounds $\operatorname {Cap}_{\theta ,u}$ by a continuous nondecreasing function, vanishing at zero, of $\operatorname {Cap}_{\theta ,\psi }$. If $v$ is a competitor for $\operatorname {Cap}_{\theta ,\phi }$, then $h=(v+\eta )/2$ satisfies $u-1/2\leq h\leq u$, so it is a competitor for $\operatorname {Cap}_{\theta ,u}$, and $\mu _ h\geq 2^{-n}\mu _ v$. Thus

Combining these inequalities with $\operatorname {Cap}_\omega \leq \operatorname {Cap}_\kappa $ gives $G$; compositions, positive multiples and $n$th roots preserve continuity, monotonicity and the value zero at the origin. The remaining assertions follow by comparing both capacities with $\operatorname {Cap}_\omega $. $\square$

Proof. Let $v$ be a competitor for $\operatorname {Cap}_\omega $, so that $v\in \mathrm{PSH}(X,\omega )$ and $-1\leq v\leq 0$. Fix $\alpha $ and write $B'_\alpha =B(0,R_\alpha )$ in its coordinates. Since $\overline{B'_\alpha }$ is a compact subset of a chart, there is $A_\alpha >0$ with $\omega \leq A_\alpha \mathrm{dd}^{\mathrm{c}}|z|^2$ on $B'_\alpha $ (lemma 6.14.1). Then $\mathrm{dd}^{\mathrm{c}}(v+A_\alpha |z|^2)=(\omega +\mathrm{dd}^{\mathrm{c}}v)+(A_\alpha \mathrm{dd}^{\mathrm{c}}|z|^2-\omega )\geq 0$, so $v+A_\alpha |z|^2$ is plurisubharmonic on $B'_\alpha $, by lemma 6.14.4(2), applied on $B'_\alpha $ to the form $A_\alpha \mathrm{dd}^{\mathrm{c}}|z|^2=\mathrm{dd}^{\mathrm{c}}(A_\alpha |z|^2)$, and the multilinearity and positivity of Bedford–Taylor products give $(\mathrm{dd}^{\mathrm{c}}(v+A_\alpha |z|^2))^ n\geq (\omega +\mathrm{dd}^{\mathrm{c}}v)^ n$ on $B'_\alpha $. Put $c_\alpha =1+A_\alpha R_\alpha ^2$. The function $w=(v+A_\alpha |z|^2+1)/c_\alpha $ is plurisubharmonic on $B'_\alpha $ with values in $[0,1]$, and $(\omega +\mathrm{dd}^{\mathrm{c}}v)^ n\leq c_\alpha ^ n(\mathrm{dd}^{\mathrm{c}}w)^ n$ there. Hence $\mu _ v(E)\leq c_\alpha ^ n\operatorname {Cap}_{B'_\alpha }(E)$ for every Borel set $E\subset B'_\alpha $, and taking the supremum over $v$ gives $\operatorname {Cap}_\omega (E)\leq c_\alpha ^ n\operatorname {Cap}_{B'_\alpha }(E)$. Applying this to the sets $E\cap B_\alpha $ and using the subadditivity of $\operatorname {Cap}_\omega $ (proposition 16.25.6), we obtain the upper bound.

For the lower bound fix $\alpha $ and write $B_\alpha =B(0,r)$ and $B'_\alpha =B(0,R)$ in the coordinates, with $r<R$. Choose $L>0$ with $L(R^2-r^2)>1$, and then $b>0$ so small that $\omega \geq bL\mathrm{dd}^{\mathrm{c}}|z|^2$ on $B(0,R)$ (lemma 6.14.1) and that we can choose a number $a$ with

Let $w$ be plurisubharmonic on $B(0,R)$ with $-1\leq w\leq 0$, and put $q=bw-bL|z|^2$. Then $\omega +\mathrm{dd}^{\mathrm{c}}q\geq b\mathrm{dd}^{\mathrm{c}}w\geq 0$ on $B(0,R)$. At the points of $B(0,R)$ where $|z|^2>a/(bL)$, which include a neighborhood of the boundary sphere because $a<bLR^2$, we have $q\leq -bL|z|^2<-a$.

Define $v$ to be $\max \{ q,-a\} $ on $B(0,R)$ and $-a$ elsewhere. Then $v=-a$ on the open set $U$ formed by $X\setminus \overline{B(0,R)}$ and the points of the chart where $|z|^2>a/(bL)$. Since $U$ and $B(0,R)$ cover $X$, and $v$ is $\omega $-plurisubharmonic on each of them, $v\in \mathrm{PSH}(X,\omega )$. It takes values in $[-1,0]$, and on $B(0,r)$ we have $q\geq -b(1+Lr^2)>-a$, so $v=q$ there. Since $\omega +\mathrm{dd}^{\mathrm{c}}q$ is the sum of the semipositive form $\omega -bL\mathrm{dd}^{\mathrm{c}}|z|^2$ and $b\mathrm{dd}^{\mathrm{c}}w$, positivity of the mixed terms gives $(\omega +\mathrm{dd}^{\mathrm{c}}v)^ n\geq b^ n(\mathrm{dd}^{\mathrm{c}}w)^ n$ on $B(0,r)$. The capacity $\operatorname {Cap}_{B(0,R)}$ may equally be computed with competitors taking values in $[-1,0]$, so it follows that

Summing over the finitely many balls gives the lower bound.

It remains to prove (1)–(3). Let $F$ and $G$ be the functions of theorem 16.27.17 for $\operatorname {Cap}_{\theta ,\psi }$; since they are continuous and vanish at zero, a sequence of Borel sets has capacities $\operatorname {Cap}_{\theta ,\psi }$ tending to zero if and only if their capacities $\operatorname {Cap}_\omega $ tend to zero.

(1) Pluripolar Borel sets have capacity zero by proposition 16.25.6. Conversely, let $K$ be a compact set that is not pluripolar. Since $\omega $ is Kähler, $0=V_\omega $ is a model potential for $\omega $ of mass $\int _ X\omega ^ n>0$, by proposition 16.21.2, and (16.27.6) for this model potential gives $\operatorname {Cap}_\omega (K)>0$. Since $\operatorname {Cap}_\omega (K)\leq G(\operatorname {Cap}_{\theta ,\psi }(K))$ and $G(0)=0$, it follows that $\operatorname {Cap}_{\theta ,\psi }(K)>0$. The assertion about Borel sets follows from the inner regularity in proposition 16.25.6.

(2) Let $\delta >0$ and $A_ j=\{ |f_ j-f|>\delta \} $. By the upper bound, $\operatorname {Cap}_\omega (A_ j)\leq c\sum _\alpha \operatorname {Cap}_{B'_\alpha }(A_ j\cap B_\alpha )$. Each term is at most the outer capacity $\operatorname {Cap}^*_{B'_\alpha }(\overline{B_\alpha }\cap A_ j)$, which tends to zero because $\overline{B_\alpha }$ is a compact subset of $B'_\alpha $. Hence $\operatorname {Cap}_\omega (A_ j)\to 0$, and therefore $\operatorname {Cap}_{\theta ,\psi }(A_ j)\to 0$.

(3) Let $f$ be quasi-continuous on $X$ and $\eta >0$. For each $\alpha $ there is an open set $W_\alpha \subset B'_\alpha $ with $\operatorname {Cap}_{B'_\alpha }(W_\alpha )<\eta $ such that $f$ is continuous on the compact set $\overline{B_\alpha }\setminus W_\alpha $. Put $W=\bigcup _\alpha W_\alpha $. Then $X\setminus W$ is the union of the finitely many closed sets $\overline{B_\alpha }\setminus W$, on each of which $f$ is continuous, so $f$ is continuous on $X\setminus W$; and $\operatorname {Cap}_\omega (W)\leq \sum _\alpha c\operatorname {Cap}_{B'_\alpha }(W_\alpha )<Nc\eta $, where $N$ is the number of balls. Now let $u\in \mathrm{PSH}(X,\theta ')$. By lemma 16.11.2(1), $u$ is quasi-continuous. Its pole set $P$ is pluripolar, because near each point $u$ is a plurisubharmonic function plus a smooth one (definition 6.14.3). Hence, by lemma 16.11.1(2) there are open sets $W'_\alpha \subset B'_\alpha $ with $P\cap B'_\alpha \subset W'_\alpha $ and $\operatorname {Cap}_{B'_\alpha }(W'_\alpha )<\eta $. Replacing $W_\alpha $, chosen as above for $f=u$, by $W_\alpha \cup W'_\alpha $, we obtain an open set $W$ with $\operatorname {Cap}_\omega (W)<2Nc\eta $ which contains $P$, such that $u$ is continuous on $X\setminus W$ and finite there. In both cases $\operatorname {Cap}_{\theta ,\psi }(W)\leq F(\operatorname {Cap}_\omega (W))\leq F(2Nc\eta )$, which is arbitrarily small when $\eta $ is small. $\square$