Lemma 15.13.1E6E5D71344DD

Proof. Fix $\varepsilon >0$ and set $v_\varepsilon =\max (u,v-\varepsilon )$, which equals $u$ near the boundary. We claim that

throughout $B$. On $\{ u>v-\varepsilon \} $ this follows from locality, since $v_\varepsilon =u$ there. On $\{ u\leq v-\varepsilon \} $ it follows from ordered contact and from the assumed inequality, which applies because this set is contained in $\{ u<v\} $.

Let $q$ be the plurisubharmonic envelope on $B$ of the obstacle $u-v_\varepsilon $, in the sense of (5.27.8). This obstacle is bounded, quasi-continuous and zero near the boundary. Hence it lies between $Ah$ and zero for a sufficiently large $A$, where $h<0$ is a smooth strictly plurisubharmonic defining function for the ball. Thus $q$ tends to zero at the boundary, and (5.27.8) shows that $(\mathrm{dd}^{\mathrm{c}}q)^ n$ is carried by $D=\{ q=u-v_\varepsilon \} $.

Since $q\leq u-v_\varepsilon $ quasi-everywhere, the canonical representatives satisfy $q+v_\varepsilon \leq u$. Ordered contact, followed by the measure inequality above, yields

The positive expansion of $(\theta _{v_\varepsilon }+\mathrm{dd}^{\mathrm{c}}q)^ n$ contains both $\mathrm{MA}_\theta (v_\varepsilon )$ and $(\mathrm{dd}^{\mathrm{c}}q)^ n$. Thus $\mathbf1_ D(\mathrm{dd}^{\mathrm{c}}q)^ n=0$, and since $(\mathrm{dd}^{\mathrm{c}}q)^ n$ is carried by $D$, we get $(\mathrm{dd}^{\mathrm{c}}q)^ n=0$. The classical Bedford–Taylor comparison principle for bounded plurisubharmonic functions with zero boundary values implies $q=0$. Consequently $u\geq v_\varepsilon $, first quasi-everywhere and then everywhere. Finally let $\varepsilon \downarrow 0$. $\square$