Theorem 3.4.231C986491E366

Proof. If the family is empty, then $F=F^*=-\infty $. Otherwise, we replace the family by an increasing sequence, using Choquet's lemma, and identify the regularized limit of this sequence with a canonical representative. The connected components of $\Omega $ are open, because balls are connected; hence they are pairwise disjoint sets of positive Lebesgue measure, and there are at most countably many of them by lemma 2.2.5. As a countable union of null sets is null, it suffices to prove the assertions on each component $\Omega '$. On $\Omega '$ the regularization $F^*$ coincides with the regularization of $F|_{\Omega '}$ in $\Omega '$, because small balls about a point of $\Omega '$ lie in $\Omega '$. If every $u_ t$ is identically $-\infty $ on $\Omega '$, then $F=F^*=-\infty $ there.

Otherwise fix $t_0$ such that $u_{t_0}$ is nontrivial, hence locally integrable, on $\Omega '$. By lemma 3.4.22, applied on $\Omega '$, there are indices $t_1,t_2,\ldots $ with $(\sup _{k\geq 1}u_{t_ k})^*=F^*$ on $\Omega '$. Put $v_ j=\max \{ u_{t_0},u_{t_1},\ldots ,u_{t_ j}\} $ and $v=\sup _ jv_ j$. Since $\sup _{k\geq 1}u_{t_ k}\leq v\leq F$, we still have $v^*=F^*$ on $\Omega '$. The functions $v_ j$ are subharmonic, increase with $j$, are uniformly bounded above on compact sets and satisfy $v_ j\geq u_{t_0}$; in particular, they are locally integrable on $\Omega '$. If $K\subset \Omega '$ is compact and $M$ bounds all $v_ j$ from above on $K$, then $0\leq M-v_ j\leq M-u_{t_0}$ on $K$ with $M-u_{t_0}$ integrable, and $M-v_ j$ decreases to $M-v$. Dominated convergence shows that $v$ is locally integrable and $v_ j\to v$ in $L^1_{\mathrm{loc}}(\Omega ')$, hence in the sense of distributions. As observed after theorem 3.4.12, $v$ therefore has a canonical representative $w$: a subharmonic function on $\Omega '$ with $w=v$ almost everywhere.

We claim that $w=v^*$. For each $j$ we have $v_ j\leq v$, so $v_ j*\rho _\varepsilon \leq v*\rho _\varepsilon =w*\rho _\varepsilon $, and letting $\varepsilon \downarrow 0$ in (3.4.13) gives $v_ j\leq w$ everywhere. Hence $v\leq w$, and $v^*\leq w$ because $w$ is upper semicontinuous. Conversely, by (3.4.13), $w(x)$ is the limit of the averages of $w$ over $B(x,r)$; these coincide with the averages of $v$, which are at most $\sup _{B(x,r)}v$. Letting $r\downarrow 0$ gives $w(x)\leq v^*(x)$. Thus $F^*=v^*=w$ is subharmonic on $\Omega '$.

Finally, $v\leq F\leq F^*=w$ on $\Omega '$, and $v=w$ almost everywhere. The functions $v$ and $w$ are Borel measurable, so the set where $F\neq F^*$ is contained in the Borel null set $\{ v\neq w\} $. Since Lebesgue measure is complete, $F$ is Lebesgue measurable and $F=F^*$ almost everywhere. $\square$