Corollary 7.3.12F2BBEF3BF09B

Proof. Choose $R$ with $D\subset B(0,R)$, and for $\delta >0$ put $u_\delta =u+\delta (|z|^2-R^2)$, a locally bounded plurisubharmonic function. Since $u_\delta \le u$ on $D$, the boundary condition still holds for $v$ and $u_\delta $, and the set $A=\{ v<u_\delta \} $ is contained in $\{ v<u\} $. By additivity, positive homogeneity and symmetry of the products (proposition 7.2.2, theorem 7.3.7), $(\mathrm{dd}^{\mathrm{c}}u_\delta )^ n=\sum _{k=0}^ n\binom nk\delta ^ k(\mathrm{dd}^{\mathrm{c}}u)^{n-k}\wedge (\mathrm{dd}^{\mathrm{c}}|z|^2)^ k$ is a sum of positive measures, and its term with $k=n$ is $\delta ^ nn!\, \pi ^{-n}\, \mathrm dV_{2n}$, by (7.2.4). The comparison inequality of theorem 7.3.11, applied to $v$ and $u_\delta $, therefore gives

where the last equality holds because $A\subset \{ v<u\} $ and $u\le v$ almost everywhere for $(\mathrm{dd}^{\mathrm{c}}v)^ n$. Thus $A$ is a Lebesgue null set, and $\max \{ u_\delta ,v\} =v$ almost everywhere. Both functions are plurisubharmonic, so the uniqueness of plurisubharmonic representatives in theorem 5.9.1, applied on each connected component of $D$, gives $\max \{ u_\delta ,v\} =v$ everywhere, that is, $u_\delta \leq v$. Letting $\delta \downarrow 0$ gives $u\le v$. $\square$