Theorem 5.11.1C910F1109424

Proof. Define $\widetilde u$ by $u$ on $\Omega \setminus S$ and by the displayed formula on $S$. Since plurisubharmonicity is a local property, it suffices to show that every point $x_0\in S$ has a neighborhood on which $\widetilde u$ is plurisubharmonic; uniqueness is proved at the end. By hypothesis there are a connected neighborhood $W_0$ of $x_0$ with a holomorphic function $f\not\equiv 0$ on $W_0$ such that $S\cap W_0\subset \{ f=0\} $, and a neighborhood $W_1$ of $x_0$ such that $u$ is bounded above on $W_1\setminus S$. Choose a ball $W$ about $x_0$ with $\overline W\subset W_0\cap W_1$. By the identity theorem, $f$ does not vanish identically on $W$, and $u$ is bounded above on $W\setminus S$. Multiplying $f$ by a nonzero constant, we may assume that $|f|\leq 1$, that is, $\log |f|^2\le 0$, on $W$. For $\varepsilon >0$ define $u_\varepsilon $ on $W$ by

Step 1: $u_\varepsilon \in \mathrm{PSH}(W)$. At a point $x\in S\cap W$ we have $f(x)=0$, so $\log |f(y)|^2\to -\infty $ as $y\to x$, while $u$ is bounded above near $x$; hence $u_\varepsilon (y)\to -\infty $ as $y\to x$, and $u_\varepsilon $ is upper semicontinuous at $x$. On $W\setminus S$, $u_\varepsilon $ is upper semicontinuous as a sum of upper semicontinuous functions.

Now let $\Lambda $ be a connected component of $L\cap W$ for an affine complex line $L$. If $f$ vanishes identically on $\Lambda $, then $u_\varepsilon \equiv -\infty $ on $\Lambda $. Otherwise the zeros of $f$ on the connected set $\Lambda $ are isolated, by the identity theorem in one variable. Since $S$ is relatively closed, $A=S\cap \Lambda $ is therefore a closed discrete subset of $\Lambda $, hence countable, because each of the countably many compact sets exhausting $\Lambda $ contains only finitely many of its points. It is polar, because points are polar in the plane and countable unions of polar sets are polar by proposition 4.1.11.

Each connected component of $\Lambda \setminus A$ is a connected subset of $L\cap (\Omega \setminus S)$ and therefore lies in a connected component of this line section, on which the restriction of $u$ is subharmonic or identically $-\infty $. Hence the restriction of $u$ to $\Lambda \setminus A$ is subharmonic, the constant $-\infty $ being subharmonic. The function $\log |f|^2$ is subharmonic on $\Lambda $ by example 3.4.15. Thus $u_\varepsilon $ is subharmonic on $\Lambda \setminus A$, and it is bounded above there by $\sup _{W\setminus S}u$, because $\log |f|^2\leq 0$. By theorem 3.4.27, it has a subharmonic extension to $\Lambda $ whose value at $x\in A$ is $\varlimsup _{\Lambda \setminus A\ni y\to x}u_\varepsilon (y)=-\infty $, which is the value of $u_\varepsilon $ at $x$. Hence the restriction of $u_\varepsilon $ to $\Lambda $ is subharmonic. Finally, $u_\varepsilon $ is not identically $-\infty $, because $u$ is finite almost everywhere on $W\setminus S$ and $\{ f=0\} $ is Lebesgue null by lemma 5.5.1. Thus $u_\varepsilon \in \mathrm{PSH}(W)$.

Step 2: the extension. As $\varepsilon \downarrow 0$ the functions $u_\varepsilon $ increase, because $\log |f|^2\leq 0$, and they are bounded above on $W$ by $\sup _{W\setminus S}u$. Let $F=\sup _{\varepsilon >0}u_\varepsilon $. By theorem 5.10.2, $F^*\in \mathrm{PSH}(W)$ and $F^*=F$ almost everywhere. Since $F=u$ outside the null set $\{ f=0\} $, we get $F^*=u$ almost everywhere on $W\setminus S$; both functions are locally integrable and subharmonic there, so $F^*=u$ everywhere on $W\setminus S$ by theorem 3.4.12.

It remains to show that $F^*=\widetilde u$ on $S\cap W$, which is Lebesgue null as a subset of $\{ f=0\} $. Let $x\in S\cap W$. Since $F^*=u$ on $W\setminus S$, the upper semicontinuity of $F^*$ gives $\widetilde u(x)=\varlimsup _{W\setminus S\ni y\to x}u(y)\leq F^*(x)$. Conversely, for small $r>0$ the ball submean inequality for $F^*$ gives $F^*(x)\leq |B(x,r)|^{-1}\int _{B(x,r)\setminus S}F^*\, \mathrm dV\leq \sup _{B(x,r)\setminus S}u$, because removing the null set $S$ does not change the integral, and letting $r\to 0$ gives $F^*(x)\leq \widetilde u(x)$. Hence $\widetilde u=F^*$ on $W$, and $\widetilde u$ is plurisubharmonic on $W$.

Uniqueness. The set $S$ is Lebesgue null, since it is covered by countably many neighborhoods $W$ as in the hypothesis and $S\cap W\subset \{ f=0\} $ is null. Two plurisubharmonic extensions of $u$ therefore agree almost everywhere, and hence everywhere by theorem 5.9.1. $\square$