Proposition 3.4.76D3B30A16AAE

Proof. Let $m=\max _\Omega u$. If $m=-\infty $, then $u$ is identically $-\infty $, so assume that $m$ is finite, and let $E=\{ u=m\} =\{ u\geq m\} $. This set is nonempty, and it is closed in $\Omega $ because $u$ is upper semicontinuous. Let $x\in E$. For every sufficiently small $r>0$, (3.4.3) gives $m=u(x)\leq M_ u(x,r)\leq m$, so

Suppose that $u(x+r\omega _0)<a<m$ for some $\omega _0\in S^{d-1}$ and some real number $a$. By upper semicontinuity, the set $A=\{ \omega \in S^{d-1}:u(x+r\omega )<a\} $ is relatively open in $S^{d-1}$, and it contains $\omega _0$. The cone $\{ t\omega :0<t<1,\ \omega \in A\} $ is the inverse image of $A$ under the continuous map $y\mapsto y/|y|$ on $B(0,1)\setminus \{ 0\} $, so it is a nonempty open set, and therefore $S(A)>0$ by the definition of surface measure in subsection 3.1.1. Since $m-u(x+r\omega )>m-a>0$ for $\omega \in A$, the integral above would be at least $(m-a)S(A)>0$, a contradiction. Hence $u=m$ on every sufficiently small sphere centered at $x$, and $E$ contains a ball around $x$. Thus $E$ is open and closed in $\Omega $, and $E=\Omega $ because $\Omega $ is connected.

The second assertion is the comparison principle on the bounded open set $D=\Omega $ with $h=0$, in the form noted after theorem 3.4.5. $\square$