Theorem 4.1.306B0925DDDD3A

Proof. The case of a bounded Green potential. Suppose first that $u$ is a bounded subharmonic function on $B$ that tends to zero at $\partial B$ and whose Riesz measure $\mu $ has compact support in $B$. Let $Q\subset B$ and $u_\varepsilon $ be the compact set and the regularizations given by the last assertion of proposition 4.1.29. These regularizations are continuous on $\overline B$, decrease to $u$, equal $u$ on $B\setminus Q$, and $\nabla u_\varepsilon \to \nabla u$ in $L^2(B)$. For $\delta >0$, the set $O_{\delta ,\varepsilon }=\{ u_\varepsilon -u>\delta \} $ is open, because $u_\varepsilon -u$ is lower semicontinuous, and it is contained in $Q$, so it has compact closure in $B$. Let $K\subset O_{\delta ,\varepsilon }$ be compact. The function $u_\varepsilon -u$ is a difference of two functions satisfying the hypotheses of proposition 4.1.29, and $\lambda _ K$ is supported on $K$, where $u_\varepsilon -u>\delta $. By the bilinear identity of that proposition, the Cauchy–Schwarz inequality and the bound $\| \nabla u_ K\| _2^2\leq C_ B(K)$,

Hence $C_ B(K)\leq \delta ^{-2}\| \nabla (u_\varepsilon -u)\| _2^2$, and taking the supremum over $K$ gives $C_ B(O_{\delta ,\varepsilon })\leq \delta ^{-2}\| \nabla (u_\varepsilon -u)\| _2^2$.

Choose $\delta _ j\downarrow 0$ and then radii $\varepsilon _ j\downarrow 0$ such that $C_ B(O_{\delta _ j,\varepsilon _ j})<\eta 2^{-j}$, and let $O=\bigcup _ jO_{\delta _ j,\varepsilon _ j}$, which is open with compact closure in $B$. By countable subadditivity (theorem 4.1.19), $C_ B(O)<\eta $. Outside $O$ we have $0\leq u_{\varepsilon _ j}-u\leq \delta _ j$ for all $j$, so the continuous functions $u_{\varepsilon _ j}$ converge uniformly to $u$ on $B\setminus O$. Hence $u$ is finite and continuous on $B\setminus O$, in particular on $L\setminus O$.

The case of a function bounded near $\overline B$. Suppose next that $u$ is bounded on a neighborhood of $\overline B$; subtracting a constant, we may assume that $u\leq -1$ there. Choose $r<R$ with $L\subset B(0,r)$, and $M>0$ with $-M<\inf _ Bu$. Let $q=Mu_{\overline{B(0,r)}}$, where the extremal function $u_{\overline{B(0,r)}}$ was computed in example 4.1.21. Then $q$ is subharmonic on $B$, equals $-M$ on $\overline{B(0,r)}$, is harmonic on $r<|x|<R$, and tends to zero at $\partial B$. The function $w=\max \{ u,q\} $ is subharmonic and bounded on $B$; it equals $u$ on $B(0,r)$, and it equals $q$ near $\partial B$, where $q>-1\geq u$. Hence $w$ tends to zero at $\partial B$, and its Riesz measure has compact support in $B$. The first case, applied to $w$, gives the required open set, since $w=u$ near $L$.

The general case. The function $u$ is bounded above near $\overline B$, so the functions $u_ N=\max \{ u,-N\} $, $N\geq 1$, are subharmonic and bounded near $\overline B$. Fix an open set $W$ with $L\subset W\Subset B$. By the second case, there are open sets $O_ N\subset W$ with $C_ B(O_ N)<\eta 2^{-N-1}$ such that $u_ N$ is continuous on $L\setminus O_ N$ (intersecting with $W$ does not affect this). On $L\setminus \bigcup _ NO_ N$, every $u_ N$ is continuous, and $\mathrm e^{u_ N}$ converges uniformly to $\mathrm e^ u$, because $0\leq \mathrm e^{u_ N}-\mathrm e^ u\leq \mathrm e^{-N}$. Hence $u$ is continuous there as a function with values in $[-\infty ,\infty )$. Let $\Omega _0$ be the connected component of $\Omega $ that contains the connected set $\overline B$. Since $u$ is not identically $-\infty $ on $\Omega _0$, the set $\{ u=-\infty \} \cap \Omega _0$ is polar by the definition of polarity. By corollary 4.1.24, $L\cap \{ u=-\infty \} $ is contained in an open set $O'\Subset B$ with $C_ B(O')<\eta /2$. The open set $O=O'\cup \bigcup _ NO_ N$ has compact closure in $B$ and capacity less than $\eta $ by countable subadditivity, and the restriction of $u$ to $L\setminus O$ is finite and continuous. $\square$