Proposition 3.4.85975EB5B5D7D

Proof. Let $x\in \Omega $ with $u(x)>-\infty $, and let $\overline{B(x,r)}\subset \Omega $. If $M$ is a real upper bound for $u$ on $B(x,r)$, the ball submean inequality of theorem 3.4.5 gives

Since $M-u\geq 0$, it follows that $u$ is integrable on $B(x,r)$.

Now let $\Omega '$ be a connected component of $\Omega $ on which $u$ is not identically $-\infty $, and let $Y$ be the set of points of $\Omega '$ that have a neighborhood on which $u$ is integrable. Then $Y$ is open, and it is nonempty by the first paragraph. Let $y\in \Omega '$ lie in the closure of $Y$, and choose $r>0$ with $\overline{B(y,3r)}\subset \Omega $. The set $Y\cap B(y,r)$ contains a point $z$, and $u$ is integrable on an open neighborhood $U$ of $z$; hence $u$ is finite almost everywhere on the nonempty open set $U\cap B(y,r)$, which has positive measure, and there is a point $x\in B(y,r)$ with $u(x)>-\infty $. Since $|x-y|<r$, the ball $B(x,2r)$ contains $y$ and its closure lies in $\overline{B(y,3r)}\subset \Omega $. By the first paragraph, $u$ is integrable on $B(x,2r)$, hence $y\in Y$. Thus $Y$ is also closed in $\Omega '$, and $Y=\Omega '$ because $\Omega '$ is connected. Finally, a locally integrable function is finite almost everywhere. $\square$