5.11 Extension across closed exceptional setsE197EAFECEC0

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. $\square$

Proof. (1) Suppose that $K$ is not polar. By theorem 4.1.17, $K$ carries a nonzero positive measure of finite energy; dividing it by its mass, we obtain a probability measure $\mu $ on $K$ of finite energy $I(\mu )$ for the Newtonian kernel $k(x,y)=|x-y|^{2-2n}/((2n-2)s_{2n-1})$ of subsection 4.1.1 in $\mathbb R^{2n}$. Its potential $U^\mu $ is lower semicontinuous, and it is finite $\mu $-almost everywhere because $\int U^\mu \, \mathrm d\mu =I(\mu )<\infty $. The sets $E_ M=K\cap \{ U^\mu \le M\} $, $M\in \mathbb N$, are compact, because $U^\mu $ is lower semicontinuous, and they increase to a set of full $\mu $-measure; hence $\mu (E_ M)>0$ for $M$ large enough. Let $\nu $ be the restriction of $\mu $ to $E_ M$. For $a\in E_ M$ and $r>0$,

so $\nu (B(a,r))\le C_ nMr^{2n-2}$ with $C_ n=(2n-2)s_{2n-1}$. Now cover $E_ M$ by balls $B(c_ l,r_ l)$ with $r_ l<1$ and $\sum _ lr_ l^{2n-3}<\eta $, and discard those not meeting $E_ M$. Each remaining ball lies in $B(a_ l,2r_ l)$ for some $a_ l\in E_ M$, so

where we used $r_ l<1$. Since $\eta >0$ is arbitrary, $\nu (E_ M)=0$, a contradiction. Hence $K$ is polar.

(2) Fix $x$. For $m\geq 1$ the set $K_ m=K\cap \{ |y-x|\geq 1/m\} $ is compact and $\mathcal H^{2n-3}$-null, and $K\setminus \{ x\} =\bigcup _ mK_ m$. The direction map $\Phi (y)=[y-x]$ is Lipschitz on $K_ m$. Indeed, $\Phi (y)=[(y-x)/|y-x|]$. The normalization $y\mapsto (y-x)/|y-x|$ has Lipschitz constant at most $2m$ on $\{ |y-x|\geq 1/m\} $: for nonzero vectors $a,b$,

and we apply this with $a=y-x$, $b=y'-x$ and $|a|\geq 1/m$. The Hopf map is Lipschitz with constant $\pi /2$ for the metric $d$. Indeed, let $e,e'$ be unit vectors and $\delta =\arccos \operatorname{Re}\langle e,e'\rangle \in [0,\pi ]$ their spherical distance. Since $d$ is the minimum of the spherical distance over the unit representatives (subsection 2.9.17), we have $d([e],[e'])\leq \delta $. Moreover $|e-e'|^2=2-2\cos \delta =4\sin ^2(\delta /2)$, and $\sin s\geq 2s/\pi $ for $0\leq s\leq \pi /2$, so $|e-e'|\geq 2\delta /\pi $. Hence

Since Lipschitz images of $\mathcal H^{2n-3}$-null sets are $\mathcal H^{2n-3}$-null, the compact set $\Phi (K_ m)$ is $\mathcal H^{2n-3}$-null in $(\mathbb P^{n-1}(\mathbb C),d)$. Cover it by balls of radii $r_ l<1$ with $\sum _ lr_ l^{2n-3}<\eta $. By lemma 2.9.19, and since $r_ l<1$, we have $\sigma (\Phi (K_ m))\leq \sum _ lr_ l^{2n-2}<\eta $, so $\sigma (\Phi (K_ m))=0$. Hence the set of directions $\Phi (K\setminus \{ x\} )=\bigcup _ m\Phi (K_ m)$ is a Borel set of $\sigma $-measure zero.

(3) An orthogonal projection onto a complex hyperplane $H\cong \mathbb R^{2n-2}$ is $1$-Lipschitz. Hence the projection of $K$ is $\mathcal H^{2n-3}$-null in $H$, and therefore Lebesgue null because $2n-3<2n-2$. $\square$

Proof. Let $u$ be a plurisubharmonic function on $\Omega \setminus S$. For every closed ball $\overline B\subset \Omega $, the compact set $S\cap \overline B$ is covered by finitely many neighborhoods $W$ as in the hypothesis, and is therefore $\mathcal H^{2n-3}$-null.

Step 1: $u$ is locally bounded above near $S$. Fix $p\in S$ and $r>0$ with $\overline B=\overline{B(p,r)}\subset \Omega $, and write $B=B(p,r)$. By lemma 5.11.2(2) applied with $x=p$, almost every complex line through $p$ meets $S\cap \overline B$ only at $p$; fix such a line $L=\{ p+te:t\in \mathbb {C}\} $, with $|e|=1$. The circle $\Gamma =L\cap \partial B$ is a compact subset of $\Omega \setminus S$, so the upper semicontinuous function $u$ is bounded above on $\Gamma $ by a finite constant $C$, and the open set $O=\{ y\in \Omega \setminus S:u(y)<C+1\} $ contains $\Gamma $.

Suppose that $u$ is not bounded above near $p$. Then there are points $x_ j\in B\setminus S$ with $x_ j\to p$ and $u(x_ j)\to +\infty $. For each $j$, lemma 5.11.2(2) applied with $x=x_ j$ shows that the directions of the lines through $x_ j$ meeting $S\cap \overline B$ form a $\sigma $-null set. Every nonempty open ball in $\mathbb P^{n-1}(\mathbb C)$ has positive $\sigma $-measure by lemma 2.9.19, so the complement of this null set is dense. We can therefore choose directions $\xi _ j$ outside this null set with $d(\xi _ j,[e])<1/j$. Since $d$ is the minimum of the spherical distance over the unit representatives, $\xi _ j$ has a unit representative $e_ j$ whose spherical distance to $e$ is less than $1/j$, and then $|e_ j-e|<1/j$, because a chord is shorter than the arc. Thus $e_ j\to e$, and the lines $L_ j=\{ x_ j+te_ j:t\in \mathbb {C}\} $ do not meet $S\cap \overline B$.

We claim that $L_ j\cap \partial B\subset O$ for all large $j$. Let $c_ j$ be the point of $L_ j$ closest to $p$, that is, the orthogonal projection of $p$ to $L_ j$; then $|c_ j-p|\leq |x_ j-p|\to 0$. Since $L_ j$ is a complex line, $c_ j=x_ j+\langle p-x_ j,e_ j\rangle e_ j$ and $\langle c_ j-p,e_ j\rangle =0$ for the Hermitian inner product, so $\operatorname{Re}\langle c_ j-p,se_ j\rangle =0$ and $|c_ j+se_ j-p|^2=|c_ j-p|^2+|s|^2$ for $s\in \mathbb {C}$. Hence $L_ j\cap \partial B$ is the circle $\{ c_ j+\rho _ j\mathrm{e}^{\mathrm{i}\theta }e_ j\} $ with $\rho _ j=(r^2-|c_ j-p|^2)^{1/2}\to r$. Since $|c_ j+\rho _ j\mathrm{e}^{\mathrm{i}\theta }e_ j-p-r\mathrm{e}^{\mathrm{i}\theta }e|\leq |c_ j-p|+|\rho _ j-r|+r|e_ j-e|$, these circles converge uniformly in $\theta $ to $\Gamma =\{ p+r\mathrm{e}^{\mathrm{i}\theta }e\} $. The distance from the compact set $\Gamma $ to the closed complement of the open set $O$ is positive, so the circles lie in $O$ for large $j$.

For such $j$, the closed disc $\Delta _ j=L_ j\cap \overline B$ lies in $\Omega \setminus S$, and the restriction of $u$ to the connected component of $L_ j\cap (\Omega \setminus S)$ containing $\Delta _ j$ is subharmonic or identically $-\infty $. Its values on the boundary circle of $\Delta _ j$ are less than $C+1$, so the maximum principle, that is, theorem 3.4.5 with the constant harmonic function $C+1$, gives $u\leq C+1$ on $\Delta _ j$. Since $x_ j\in B\cap L_ j\subset \Delta _ j$, this gives $u(x_ j)\leq C+1$, contradicting $u(x_ j)\to +\infty $. Hence $u$ is bounded above near every point of $S$.

Step 2: the subharmonic extension. By lemma 5.11.2(1), every compact set $S\cap \overline B$ as above is real polar. Countably many such sets cover $S$, so $S$ is real polar by proposition 4.1.11. By Step 1 and the upper semicontinuity of $u$ on $\Omega \setminus S$, the function $u$ is bounded above on $K\setminus S$ for every compact $K\subset \Omega $. Moreover $u$ is subharmonic on $\Omega \setminus S$ by theorem 5.9.1. By theorem 3.4.27, $u$ has a subharmonic extension $\widetilde u$ to $\Omega $, given on $S$ by the upper-limit formula. Since $S$ is Lebesgue null and $u$ is finite almost everywhere on $\Omega \setminus S$, the function $\widetilde u$ is not identically $-\infty $ on the domain $\Omega $, so it is locally integrable.

Step 3: $\widetilde u$ is plurisubharmonic. We verify the hypothesis of lemma 5.8.2 for the locally integrable function $\widetilde u$. Fix a unit vector $a\in \mathbb {C}^ n$ and a nonnegative $\chi \in C^\infty _ c(\Omega )$, and write $z=v+ta$ with $v\in (\mathbb {C}a)^\perp $ and $t\in \mathbb {C}$, as in that lemma. The compact set $S\cap \operatorname{Supp}\chi $ is $\mathcal H^{2n-3}$-null, since it is covered by finitely many of the sets $S\cap \overline B$ of the first paragraph, so by lemma 5.11.2(3) its orthogonal projection onto $(\mathbb {C}a)^\perp $ is Lebesgue null. For $v$ outside this projection, the line $\{ v+ta:t\in \mathbb {C}\} $ meets $\operatorname{Supp}\chi $ only in $\Omega \setminus S$. Thus the open set $G_ v=\{ t:v+ta\in \Omega \setminus S\} $ contains $K_ v=\{ t:v+ta\in \operatorname{Supp}\chi \} $, and its connected components are the connected components of an affine complex-line section of $\Omega \setminus S$, on each of which the restriction of $\widetilde u=u$ is subharmonic or identically $-\infty $. As $a$ and $\chi $ are arbitrary, lemma 5.8.2 gives $\mathrm{dd}^{\mathrm{c}}\widetilde u\geq 0$. Since $\widetilde u$ is a locally integrable subharmonic function, it is the canonical representative of its class, and it is plurisubharmonic by theorem 5.9.1. Uniqueness follows from the same theorem, since two plurisubharmonic extensions agree outside the Lebesgue null set $S$. $\square$