Lemma 16.27.15A285A0B9F4B7

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 $F(0)=0$. $\square$