Theorem 3.4.278B711E470600

Proof. Put $\widetilde u=u$ on $\Omega \setminus A$, and define $\widetilde u$ on $A$ by the displayed formula. Since $A$ has Lebesgue measure zero, it has no interior points, so every point of $A$ is a limit of points of $\Omega \setminus A$; together with the local upper bound for $u$, this shows that $\widetilde u(x)\in [-\infty ,\infty )$ is well defined for $x\in A$. We first show that $\widetilde u$ is upper semicontinuous and locally bounded above. On the open set $\Omega \setminus A$ both properties hold because $\widetilde u=u$ there. Let $x\in A$ and $\overline{B(x,r)}\subset \Omega $. For $y\in A\cap B(x,r)$, the definition gives $\widetilde u(y)\leq \sup _{B(x,r)\setminus A}u$, hence

The right-hand side is bounded above by hypothesis and tends to $\widetilde u(x)$ as $r\downarrow 0$, which proves both claims at $x$.

Next we show that $\widetilde u$ is subharmonic near each point $x\in A$; near points of $\Omega \setminus A$ this holds because $\widetilde u=u$. The idea is to approximate $u$ from below by functions $u+\varepsilon \psi $ that tend to $-\infty $ at $A$. Let $V$ and $\psi $ be as in definition 3.4.26. By proposition 3.4.8, $\psi $ is locally integrable on the connected set $V$, so it is not identically $-\infty $ on any ball in $V$, and its pole set $\{ \psi =-\infty \} $ is a null set. Replacing $V$ by a ball centered at $x$ whose closure lies in $V$, and $\psi $ by $\psi $ minus an upper bound on that ball, we may therefore assume that $V$ is a ball, that $\psi \leq 0$ on $V$, and that $\psi $ is nontrivial on $V$.

For $\varepsilon >0$ define $v_\varepsilon =u+\varepsilon \psi $ on $V\setminus A$ and $v_\varepsilon =-\infty $ on $A\cap V$. On the open set $V\setminus A$, $v_\varepsilon $ is subharmonic. At a point $y\in A\cap V$, $u$ is bounded above near $y$, and $\psi (z)\to -\infty $ as $z\to y$ because $\psi $ is upper semicontinuous with $\psi (y)=-\infty $; hence $v_\varepsilon (z)\to -\infty $ as $z\to y$. So $v_\varepsilon $ is upper semicontinuous at $y$, and the submean inequality at $y$ holds because $v_\varepsilon (y)=-\infty $. Thus $v_\varepsilon \in \mathrm{SH}(V)$.

As $\varepsilon \downarrow 0$, the functions $v_\varepsilon $ increase because $\psi \leq 0$, and they are bounded above by $u$ on $V\setminus A$, hence uniformly bounded above on compact subsets of $V$. By theorem 3.4.23, the regularized supremum $w=(\sup _{\varepsilon >0}v_\varepsilon )^*$ is subharmonic on $V$ and equals $\sup _{\varepsilon >0}v_\varepsilon $ almost everywhere. This supremum equals $u$ on $V\setminus (A\cup \{ \psi =-\infty \} )$, so $w=u$ almost everywhere on $V\setminus A$. Since $w$ and $u$ are subharmonic on the open set $V\setminus A$, they agree everywhere there by theorem 3.4.12(2).

It remains to show that $w=\widetilde u$ on $A\cap V$. Let $y\in A\cap V$. Since $w$ is upper semicontinuous and $w=u$ on $V\setminus A$, we have $w(y)\geq \varlimsup _{V\setminus A\ni z\to y}u(z)=\widetilde u(y)$. Conversely, by (3.4.13), $w(y)$ is the limit of the averages of $w$ over $B(y,r)$. Since $A$ is null and $w=u$ off $A$, these averages are at most $\sup _{B(y,r)\setminus A}u$, which tends to $\widetilde u(y)$ as $r\downarrow 0$. Hence $\widetilde u=w$ on $V$, and $\widetilde u$ is subharmonic on $V$. This proves that $\widetilde u\in \mathrm{SH}(\Omega )$.

For uniqueness, let $u'\in \mathrm{SH}(\Omega )$ agree with $u$ on $\Omega \setminus A$. Since $A$ is null, $u'=\widetilde u$ almost everywhere on $\Omega $, and hence everywhere by theorem 3.4.12(2).

Finally, let $\Omega $ be a domain, and suppose that $\Omega \setminus A$ is the union of two disjoint nonempty open sets $W_0$ and $W_1$. The function $g$ equal to $0$ on $W_0$ and to $1$ on $W_1$ is harmonic, hence subharmonic, and bounded on $\Omega \setminus A$. Its subharmonic extension $\widetilde g$ to $\Omega $ takes values in $[0,1]$ and equals $1$ on $W_1$, so it attains the finite maximum $1$. By proposition 3.4.7, $\widetilde g$ is identically $1$ on $\Omega $, contradicting $\widetilde g=0$ on $W_0$. $\square$