Theorem 15.8.1F6FD255985FF

Proof. Subtract a constant from $\rho $ so that $\rho +2\leq h$. Then $\rho +2$ is a competitor, and since the obstacle is continuous, the envelope lies below it everywhere; thus

In particular, $p$ is locally bounded off $Z$. We first prove that $p$ is continuous there; this part does not use the contact assertion.

Apply lemma 5.42.10 to $p-\sup _ Xp$ and add back the constant. The transforms $\Phi _{c,\delta }$ of that lemma are defined on $X$ and, with fixed constants $A,K,C>0$ and $\epsilon (\delta )=C\delta $, satisfy

Off $Z$ the infimum defining $\Phi _{c,\delta }$ is attained at a positive radius, and

For small $\delta $ the first inequality gives $\Phi _{c,\delta }\geq \rho +1$, hence $\rho \leq \Phi _{c,\delta }$ on $X$.

Set

The averaging kernels are positive, have integral one and are supported in balls of radius comparable to $t$, so uniform continuity of $h$ gives $m(\delta )\to 0$. Define

By the choice of $s$, the strict barrier and (15.8.2) show that $v_\delta $ is $\theta $-plurisubharmonic. Since $\rho \leq \Phi _{c,\delta }$,

Thus $v_\delta -m(\delta )$ is a competitor for the envelope, and $v_\delta \leq p+m(\delta )$.

Let $L\Subset X\setminus Z$ be compact. On $L$ the difference $p-\rho $ is bounded above. Since $s\leq Cc$, the identity

therefore gives $\Phi _{c,\delta }-p\leq C_ Lc$ on $L$. At a radius $t$ where the infimum is attained,

Since $c\geq \delta $, this yields $t\geq a_ L\delta $ with a fixed $a_ L>0$. Because $R_ tp+Kt^2$ is increasing in $t$, it follows that

Consequently $R_{a_ L\delta }p\to p$ uniformly on $L$. These averages are smooth for positive radius, even though $p$ is unbounded on $X$. Applying this to compact neighborhoods of the points of $X\setminus Z$ proves that $p$ is continuous on $X\setminus Z$.

We turn to the contact assertion. By the continuity just proved, the set $\{ p<h\} \setminus Z$ is open. Let $x$ be one of its points, and choose a small coordinate ball $B_ r$ centered at $x$ whose closure lies in this set. Let $H$ be the homogeneous replacement of theorem 15.7.2 with boundary values $p|_{\partial B_ r}$. Its domination property gives $H\geq p$ on the ball.

We claim that $H\leq h$ on $B_ r$ if $r$ is small enough. Indeed, in the chosen Euclidean coordinates $z$, with Laplacian $\Delta $, the inequality $\theta +\mathrm{dd}^{\mathrm{c}}H\geq 0$ implies a fixed bound $\Delta H\geq -C$. Comparison with the function $\sup _{\partial B_ r}p+C'(r^2-|z-x|^2)$, with $C'$ fixed large enough, gives

As $r\to 0$, the right-hand side tends to $p(x)$, whereas $\inf _{B_ r}h\to h(x)>p(x)$. This proves the claim for sufficiently small $r$.

Now replace $p$ by $H$ on the ball. The resulting function is $\theta $-plurisubharmonic. To see this, fix $a>0$ and take $\max \{ p,H-a\} $ inside the ball and $p$ outside. Since $H$ and $p$ are continuous with equal boundary values, $H-a<p$ on a collar of $\partial B_ r$, so this is an ordinary local maximum construction. Letting $a\downarrow 0$ gives the function equal to $H$ on $B_ r$ and to $p$ outside, with locally uniform convergence near the boundary of the ball. This function is bounded above by $h$ on all of $X$, so the definition of the envelope forces it to be at most $p$. Thus $p=H$ on $B_ r$, and $\mathrm{MA}_\theta (p)$ vanishes there.

These balls cover $\{ p<h\} \setminus Z$. There the extended measure agrees with the local bounded product, and it does not charge the pluripolar set $Z$. This proves the contact assertion on $X$; no bound on the total mass of $\mathrm{MA}_\theta (p)$ was needed. $\square$