Lemma 5.27.109149838224AE

Proof. For $n\geq 2$, we use Cegrell's decomposition theorem. On a bounded hyperconvex domain in $\mathbb C^ n$, $n\geq 2$, let $\mathcal E_0$ consist of the bounded negative plurisubharmonic functions with zero boundary values and finite total Monge–Ampère mass. The theorem states that every positive compactly supported measure $\tau $ has a decomposition

where $\tau _ s$ is a positive measure carried by a pluripolar set. If $\tau $ is non-pluripolar, then $\tau _ s=0$, so $\tau \ll (\mathrm{dd}^{\mathrm{c}}v)^ n$.

In dimension one, the assertion follows directly from the Green potential. Let $a\leq 0$ be the Green potential of $\tau $ on the disk, normalized by $\mathrm{dd}^{\mathrm{c}}a=\tau $. Then $v=\mathrm{e}^ a$ is bounded and subharmonic, and

To justify this inequality, set $a_ M=\max (a,-M)$. The chain rule (5.27.9) in dimension one gives $\mathrm{dd}^{\mathrm{c}}\mathrm{e}^{a_ M}\geq \mathrm{e}^{a_ M}\mathrm{dd}^{\mathrm{c}}a_ M$, and the right-hand side is at least $\mathrm{e}^ a\tau $ on $\{ a>-M\} $. Let $M\to \infty $, using the decreasing convergence of the bounded subharmonic functions $\mathrm{e}^{a_ M}$ on the left and monotone convergence on the right. Since $\{ a=-\infty \} $ is polar and $\tau $ charges no polar set, $\mathrm{e}^ a>0$ almost everywhere for $\tau $. Thus $\tau \ll \mathrm{dd}^{\mathrm{c}}v$. $\square$