Proposition 16.25.61587DBB4654F

Proof. Every competitor has the singularity type of $\phi $ and hence mass $\int _ X\langle \theta _\phi ^ n\rangle $; since $\phi $ itself is a competitor, the capacity is finite and $\operatorname {Cap}_{\theta ,\phi }(X)=\int _ X\langle \theta _\phi ^ n\rangle $. Monotonicity, countable subadditivity and continuity along increasing sequences hold for each measure $\mu _ v$ and pass to the supremum; for the last property, interchange the supremum over the sequence with the supremum over competitors. Inner regularity follows in the same way from the inner regularity of each finite measure $\mu _ v$ (lemma 16.12.1): $\operatorname {Cap}_{\theta ,\phi }(A)=\sup _ v\sup _ K\mu _ v(K)=\sup _ K\operatorname {Cap}_{\theta ,\phi }(K)$, the inner suprema being taken over the compact sets $K\subset A$. Each $\mu _ v$ gives no mass to pluripolar sets. Adding a constant $c$ to $\phi $ corresponds to the bijection $v\mapsto v+c$ of competitors, which does not change $\mu _ v$.

For the comparison, note first that $\phi \leq \psi +C$ and $\psi \leq \phi +C$ hold almost everywhere, hence everywhere; in particular $\phi $ and $\psi $ have the same pole set. Put $t=(2C+1)^{-1}$. If $\phi -1\leq v\leq \phi $, the potential $q=(1-t)\psi +tv-tC$ satisfies

so it is a competitor for $\operatorname {Cap}_{\theta ,\psi }$, and $\mu _ q\geq t^ n\mu _ v$. This proves the upper comparison, and exchanging $\phi $ and $\psi $ proves the lower one. The last assertion follows from the bijection $v\mapsto v-g$ between the two families of competitors, which preserves the current $\theta +\mathrm{dd}^{\mathrm{c}}v$. $\square$