13.7.7 Ordered error sets, monotone nets and weights64A0C33C8690

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. $\square$

Proof. Lower inequality. Let $B$ be a coordinate ball such that $\overline B$ lies in a larger coordinate ball, add fixed smooth local potentials of the $\theta _ j$, defined near $\overline B$, and denote the resulting plurisubharmonic functions on $B$ by $v_{j,k}$ and $v_ j$. By lemma 13.7.3, $v_{j,k}\to v_ j$ in extended capacity on $B$, and the $v_{j,k}$ are uniformly bounded above on $B$, because the suprema of the $u_{j,k}$ are bounded above; the local products are finite by theorem 13.5.4. For $C,\varepsilon >0$ let $v_{j,k}^ C$, $v_ j^ C$ and $a_ k$, $a$ be the truncations and cutoff weights of lemma 13.7.4; thus $0\leq a_ k,a\leq 1$, $a$ is quasi-continuous and $a_ k\to a$ locally in capacity. We may assume $A>0$. The bounded Borel weights $\chi _ ka_ k$ converge locally in capacity to the bounded quasi-continuous function $\chi a$, because $|\chi _ ka_ k-\chi a|\leq A|a_ k-a|+|\chi _ k-\chi |$, so that the set where $|\chi _ ka_ k-\chi a|>\delta $ is contained in the union of the sets where $|a_ k-a|>\delta /(2A)$ and where $|\chi _ k-\chi |>\delta /2$. The product $\chi a$ is quasi-continuous, since $\chi $ and $a$ are continuous outside open sets of arbitrarily small capacity and capacity is subadditive.

By lemma 13.7.4(2) and $0\leq a_ k\leq 1$,

The functions $v_{j,k}^ C\geq -C$ are uniformly bounded on $B$ and converge to $v_ j^ C$ locally in capacity (lemma 13.7.3). By the weighted capacity continuity of bounded products (theorem 7.5.8), the right side converges weakly to $\chi a\bigwedge _ j\mathrm{dd}^{\mathrm{c}}v_ j^ C=\chi aR$, the last identity by lemma 13.7.4(2). Testing against $g\eta $ for a nonnegative test function $g$ on $B$, we obtain $\int g\chi aR\wedge \eta \leq \varliminf _ k\int g\chi _ kR_ k\wedge \eta $, and we let $\varepsilon \downarrow 0$ and then $C\to \infty $. By lemma 13.7.4(3), the weight $a$ increases to one outside the union of the pole sets, which $R$ does not charge, so monotone convergence gives the local lower inequality, and a finite partition of unity (proposition 5.2.6) gives the global one. The local argument uses only the local upper bounds, the extended capacity convergence and the finiteness of the local products.

Convergence. Under the additional mass hypothesis, theorem 13.7.5 gives $R_ k\to R$ weakly, and in particular $\int _ XR_ k\wedge \eta \to \int _ XR\wedge \eta $. Apply the lower inequality also to the weights $A-\chi _ k$, which converge in capacity to $A-\chi $. Since

this gives the matching upper limit, so $\int _ X\chi _ kR_ k\wedge \eta \to \int _ X\chi R\wedge \eta $.

Every smooth real $(n-p,n-p)$-form $\eta $ on $X$ is the difference of two smooth strongly positive forms. Indeed, if $(\rho _ i)$ is a finite partition of unity subordinate to coordinate charts (proposition 5.2.6), then $\rho _ i\eta =\sum _ lf_ le_ l$ with smooth real functions $f_ l$ supported in $\operatorname{Supp}\rho _ i$ and constant strongly positive forms $e_ l$, which span the real forms by proposition 5.3.2. If $\kappa _ i\geq 0$ is a test function of the chart equal to one on $\operatorname{Supp}\rho _ i$ and $c_ l\geq \sup |f_ l|$, then $f_ le_ l=(f_ l+c_ l\kappa _ i)e_ l-c_ l\kappa _ ie_ l$ is a difference of smooth strongly positive forms. Hence the convergence holds for all smooth real forms, and by linearity for all smooth forms, so $\chi _ kR_ k\to \chi R$ weakly as currents. Since $\chi _ k\leq A$ and the trace masses of $R_ k$ are bounded, the coefficient measures of $\chi _ kR_ k$ have bounded total variation; when $p=n$, approximating continuous functions on the compact manifold $X$ uniformly by smooth ones, using a partition of unity and lemma 2.5.1, upgrades the convergence to weak convergence of Radon measures. Finally, for monotone sequences the extended capacity hypothesis follows from proposition 13.7.8, applied to the local potentials. $\square$