Proposition 13.7.84B60F7F33590

Proof. Equivalence. Fix a compact set $K\subset D$ and $\varepsilon >0$. By theorem 7.5.2 there is an open set $G$ of outer capacity less than $\varepsilon $, containing the poles of $u$ near $K$, such that $u$ is finite and continuous on the compact set $F=K\setminus G$. Then $u$ is bounded on $F$, say $-C\leq u\leq M$. For $\delta >0$, the two continuous positive functions

have a common positive lower bound $\kappa $ on $[-C,M]$. Hence, at a point of $F$, an ordered error of size $\delta $ in either direction implies $d_*(u_ a,u)>\kappa /2$. By subadditivity, the outer capacities of the two ordered error sets in $K$ are at most $\varepsilon $ plus the outer capacity of $K\cap \{ d_*(u_ a,u)>\kappa /2\} $. Thus extended capacity convergence, followed by $\varepsilon \downarrow 0$, gives convergence in the ordered sense. Conversely, $|\arctan s-\arctan t|\leq |s-t|$ for finite $s,t$; if one value is $-\infty $ and the other finite, there is an ordered error for every finite threshold; and equal infinite values have extended distance zero. Hence the set where $d_*(u_ a,u)>\delta $ is contained in the union of the two ordered error sets of size $\delta $, and subadditivity proves the converse.

Decreasing nets. Let $u_ a$ decrease to $u$, so that $u_ a\geq u$. With $F$ and $G$ as above, the functions $u_ a-u$ are finite and upper semicontinuous on the compact set $F$, and decrease pointwise to zero. Hence the closed sets $\{ u_ a-u\geq \delta \} \cap F$ have empty intersection, so finitely many of them already have empty intersection, and since the net is decreasing, all sets with a sufficiently large index are empty. Consequently, for large $a$ the only ordered error in $K$ lies in $G$, whose capacity is less than $\varepsilon $. This proves convergence. Only compactness is used, so the argument applies to nets and not only to sequences.

Increasing nets. Let the net increase, with $u=(\sup _ au_ a)^*$ plurisubharmonic. By theorem 5.10.2, there are countably many indices $b_1,b_2,\ldots $ such that $(\sup _ iu_{b_ i})^*=u$. Since the index set is directed, there is an increasing sequence of indices $a_ k$ with $a_ k\geq b_1,\ldots ,b_ k$, so that $u_{a_ k}\geq \max \{ u_{b_1},\ldots ,u_{b_ k}\} $. Since every $u_ a\leq \sup _ au_ a\leq u$, the regularized supremum of the sequence $(u_{a_ k})$ is $u$, and by theorem 5.10.2 the increasing limit $\lim _ ku_{a_ k}=\sup _ ku_{a_ k}$ equals $u$ almost everywhere. Hence, for every $C>0$, the truncations $\max \{ u_{a_ k},-C\} $, which lie between $-C$ and $\max \{ u,-C\} $, are locally uniformly bounded and increase almost everywhere to $\max \{ u,-C\} $, so they converge in capacity by theorem 7.6.1. By lemma 13.7.3 and the equivalence just proved, the sequence $(u_{a_ k})$ converges to $u$ locally in capacity in the ordered sense. For $a\geq a_ k$, monotonicity gives

so the estimates for the sequence give convergence of the whole net; the indices $a_ k$ need not be cofinal. Finally, suppose that the increasing net converges to a plurisubharmonic function $u$ outside a Lebesgue null set $N$. For $x\notin N$ we have $\sup _ au_ a(x)=\lim _ au_ a(x)=u(x)$, so $u_ a\leq u$ almost everywhere, and hence everywhere: $\max \{ u_ a,u\} $ and $u$ are plurisubharmonic and agree almost everywhere, so they agree everywhere (theorem 5.9.1, theorem 3.4.12). Thus $v=\sup _ au_ a\leq u$, so that $v^*\leq u$, as $u$ is upper semicontinuous. Moreover $v^*$ is plurisubharmonic and equals $v$, hence $u$, almost everywhere, by theorem 5.10.2; therefore $v^*=u$, and the previous case applies. $\square$