13.7 Capacity convergence and loss of mass52BF527413C6

Proof. For every $x\in K$ choose $r_ x>0$ such that the closed ball $\overline{B(z(x),2r_ x)}$ lies in $z(W\cap W')$. Finitely many of the sets $z^{-1}(B(z(x_ i),r_ i))$, $r_ i=r_{x_ i}$, cover $K$; put $B_ i=B(z(x_ i),2r_ i)$ and $K_ i=K\cap z^{-1}(\overline{B(z(x_ i),r_ i)})$. The map $F_ i=z'\circ z^{-1}$ is a biholomorphism from $B_ i$ onto the bounded open set $B_ i'=F_ i(B_ i)\subset D'$. Let $E\subset K$ and $E_ i=E\cap K_ i$. Every competitor for $\operatorname {Cap}_{D'}$ restricts to a competitor for $\operatorname {Cap}_{B_ i'}$, so $\operatorname {Cap}_{D'}(G)\leq \operatorname {Cap}_{B_ i'}(G)$ for Borel sets $G\subset B_ i'$, and taking infima over open sets, $\operatorname {Cap}^*_{D'}\leq \operatorname {Cap}^*_{B_ i'}$ on subsets of $B_ i'$. By proposition 7.4.3(5) and (4), applied to the ball $B_ i\Subset D$ and the compact set $\overline{B(z(x_ i),r_ i)}\subset B_ i$, there is a constant $C_ i$ with

By subadditivity of $\operatorname {Cap}^*_{D'}$ (proposition 7.4.3), the assertion holds with $C=\sum _ iC_ i$. If $E$ is a Borel set, then so are $E_ i$, $z(E_ i)$ and $z'(E_ i)$, because $z$ and $z'$ are homeomorphisms onto their images. The restriction argument above and proposition 7.4.3(4),(5) also apply to the capacities of Borel sets, so the same chain of inequalities holds with $\operatorname {Cap}$ in place of $\operatorname {Cap}^*$, and the subadditivity of $\operatorname {Cap}_{D'}$ gives the second assertion with the same constant.

For the consequences, let $(W',z')$ be a chart with bounded image and $K'\subset W'$ compact. Cover $K'$ by finitely many compact sets $K_ i'\subset W'\cap W_ i$, where the $(W_ i,z_ i)$ are charts of the given atlas. Suppose that $(f_ a)$ converges to $f$ locally in capacity in the charts $W_ i$. By subadditivity, the outer capacity in $D'=z'(W')$ of the image of $K'\cap \{ f_ a>f+\delta \} $ is at most the sum over $i$ of the outer capacities of the images of $K_ i'\cap \{ f_ a>f+\delta \} $. By the first part, each of these is at most a constant times the corresponding outer capacity in $z_ i(W_ i)$, which tends to zero; similarly for $\{ f_ a<f-\delta \} $.

If $f$ is quasi-continuous in the charts $W_ i$ and $\varepsilon >0$, choose open sets $G_ i\subset z_ i(W_ i)$ of small capacity such that $f$ is continuous on $K_ i'\setminus z_ i^{-1}(G_ i)$. By the first part, $z'(K_ i'\cap z_ i^{-1}(G_ i))$ has outer capacity at most $C_ i\operatorname {Cap}_{z_ i(W_ i)}(G_ i)$ in $D'$, so it is contained in an open set $G_ i'\subset D'$ with $\operatorname {Cap}_{D'}(G_ i')<\varepsilon /N$, where $N$ is the number of sets $K_ i'$, provided the $G_ i$ are chosen small enough. Then $G'=\bigcup _ iG_ i'$ satisfies $\operatorname {Cap}_{D'}(G')<\varepsilon $, and the set $F=K'\setminus z'^{-1}(G')$ is the union of the finitely many relatively closed sets $F\cap K_ i'$, on each of which $f$ is continuous; hence $f$ is continuous on $F$. $\square$

Proof. Fix a compact set $K$ and let $M$ be an upper bound for all $u_ k$ and $u$ on $K$. On $[-\infty ,M]$, the function $\max \{ s,-C\} $ is a Lipschitz function of $\arctan s$: it is constant for $\arctan s\leq \arctan (-C)$, and the derivative of the tangent is bounded on the compact interval $[\arctan (-C),\arctan M]$. Hence extended capacity convergence implies capacity convergence of the truncations. Conversely, $|\arctan s-\arctan \max \{ s,-C\} |\leq \pi /2-\arctan C$ for all $s$, so that

where we also used $|\arctan s-\arctan t|\leq |s-t|$ for finite $s,t$. Choosing $C$ with $2(\pi /2-\arctan C)<\delta /2$, the set $\{ d_*(u_ k,u)>\delta \} $ is contained in $\{ |\max \{ u_ k,-C\} -\max \{ u,-C\} |>\delta /2\} $, which proves the equivalence.

Adding the same bounded real function $h$ to $s$ and $t$ changes $d_*$ by at most a bounded factor: since $1+\sigma ^2\leq 2(1+h^2)(1+(\sigma +h)^2)$, the formula $\arctan t-\arctan s=\int _ s^ t\mathrm d\sigma /(1+\sigma ^2)$ gives $d_*(s+h,t+h)\leq 2(1+h^2)d_*(s,t)$. Now let $(W_ i,h_ i)$ and $(W'_ l,h'_ l)$ be two choices of a finite cover of $X$ by charts with bounded images and of smooth local potentials $h_ i$ of $\theta $ on $W_ i$ and $h'_ l$ on $W'_ l$, and suppose that $u_ k+h_ i\to u+h_ i$ in extended capacity on every $W_ i$. Fix $l$, a compact set $K\subset W'_ l$ and $\delta >0$, and cover $K$ by finitely many compact sets $K_ i\subset W_ i\cap W'_ l$. The smooth function $h'_ l-h_ i$ is bounded on $K_ i$, say by $H_ i$, so the inequality just proved, applied with $h=h'_ l-h_ i$, gives

By lemma 13.7.1, the outer capacity of the left side, computed in the chart $W'_ l$, is at most a constant times the outer capacity of the right side, computed in $W_ i$, which tends to zero. By subadditivity of outer capacity, the outer capacity of $K\cap \{ d_*(u_ k+h'_ l,u+h'_ l)>\delta \} $ in $W'_ l$ tends to zero. Thus extended capacity convergence does not depend on the choice of the cover or of the local potentials.

If $u_ k$ decreases to $u$, then for every $C$ the truncations decrease to the bounded plurisubharmonic function $\max \{ u,-C\} $, so they converge in capacity by proposition 7.5.6; the equivalence just proved gives extended capacity convergence.

For the global assertion, put $m_ k=\sup _ Xu_ k$ and $v_ k=u_ k-m_ k$. By proposition 10.3.1, $\int _ X|v_ k|\, \omega ^ n\leq M_0$ for a constant $M_0$. Suppose that $m_ k\to +\infty $ along a subsequence, to which we pass. By Markov's inequality, the $\omega ^ n$-volume of $\{ u_ k<m_ k/2\} =\{ v_ k<-m_ k/2\} $ is at most $2M_0/m_ k$, which tends to zero. On the other hand, let $(W,z)$ be one of the charts of definition 13.7.2, with smooth local potential $h$ of $\theta $, and let $K\subset W$ be a compact set of positive volume, with $|h|\leq H$ and $u+h\leq M$ on $K$. On $K\cap \{ u_ k\geq m_ k/2\} $ we have $u_ k+h\geq m_ k/2-H$, so for large $k$ this set is contained in $K\cap \{ d_*(u_ k+h,u+h)>\delta _0\} $, where $\delta _0=(\pi /2-\arctan M)/2$. The outer capacity in $z(W)$ of the image of the latter set tends to zero, and hence so does its Lebesgue measure, by proposition 7.4.3(3). Since Lebesgue measure in the chart and the $\omega ^ n$-volume are comparable on $K$, the volume of $K\cap \{ u_ k\geq m_ k/2\} $ tends to zero, whereas by the first observation it tends to the volume of $K$, which is positive. This contradiction shows that the suprema are bounded above. $\square$

Proof. (1) The weights are Borel functions of the Borel functions $v_{j,k}$ and $v_ j$, and they take values in $[0,1]$ because $v_{j,k}^ C+C\geq 0$ and $v_ j^ C+C\geq 0$. Let $K\subset D$ be compact and $\delta >0$. By theorem 7.5.2 there are open sets $G_ j\subset D$ with $\operatorname {Cap}_ D(G_ j)<\delta /p$ such that $v_ j$ is continuous on $K\setminus G_ j$. Then $a$, a continuous function of $v_1^ C,\ldots ,v_ p^ C$, is continuous on $K\setminus \bigcup _ jG_ j$, and $\operatorname {Cap}_ D(\bigcup _ jG_ j)<\delta $ by subadditivity; thus $a$ is quasi-continuous. By lemma 13.7.3, $v_{j,k}^ C\to v_ j^ C$ locally in capacity. Since $s\mapsto (s+C)/(s+C+\varepsilon )$ is Lipschitz with constant $1/\varepsilon $ on $[-C,\infty )$, and since $|\prod _ jb_ j-\prod _ jb_ j'|\leq \sum _ j|b_ j-b_ j'|$ for numbers in $[0,1]$, the set where $|a_ k-a|>\delta $ is contained in the union over $j$ of the sets where $|v_{j,k}^ C-v_ j^ C|>\varepsilon \delta /p$. By subadditivity of outer capacity, $a_ k\to a$ locally in capacity.

(2) Fix $k$, and let $O=\bigcap _ j\{ v_{j,k}>-C\} $, a plurifine open set. For every integer $m\geq C$, the set $O$ is contained in the level-$m$ set $O_ m=\bigcap _ j\{ v_{j,k}>-m\} $ of the tuple $(v_{j,k})_ j$, and on $O$ the truncations $\max \{ v_{j,k},-m\} $ coincide with $v_{j,k}$, hence with $v_{j,k}^ C$. By lemma 13.2.2, $\mathbf1_{O_ m}R_ k$ is the Bedford–Taylor product of the truncations at level $m$ restricted to $O_ m$, and by theorem 5.47.22 this product coincides on $O$ with $\bigwedge _ j\mathrm{dd}^{\mathrm{c}}v_{j,k}^ C$. Hence $\mathbf1_ OR_ k=\mathbf1_ O\bigwedge _ j\mathrm{dd}^{\mathrm{c}}v_{j,k}^ C$. The weight $a_ k$ vanishes outside $O$, because $a_{j,k}=0$ wherever $v_{j,k}\leq -C$, and multiplying by $a_ k$ gives the first identity. The same argument, applied to the tuple $(v_ j)_ j$, gives the second.

(3) Where $v_ j>-C$ we have $v_ j^ C+C>0$, so $a_ j$ increases to $1$ as $\varepsilon \downarrow 0$; where $v_ j\leq -C$ we have $a_ j=0$. Hence $a$ increases to the indicator function of $\bigcap _ j\{ v_ j>-C\} $, and these sets increase to $\bigcap _ j\{ v_ j>-\infty \} $ as $C\to \infty $. $\square$

Proof. Lower inequality. Work on a coordinate ball $B$ 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 by $v_{j,k}$ and $v_ j$. By lemma 13.7.3 (including its independence assertion, which allows us to use the chart $B$), $v_{j,k}\to v_ j$ 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 non-pluripolar products of the $v_{j,k}$ and of the $v_ j$ are the restrictions of $R_ k$ and $R$ to $B$, and they are finite by theorem 13.5.4. For $C,\varepsilon >0$ let $v_{j,k}^ C$, $v_ j^ C$, $a_ k$ and $a$ be as in lemma 13.7.4. By that lemma, the weights take values in $[0,1]$, $a$ is quasi-continuous, $a_ k\to a$ locally in capacity, and

on $B$. Moreover, the plurisubharmonic functions $v_{j,k}^ C\geq -C$ are uniformly bounded on $B$, and they converge to $v_ j^ C$ locally in capacity, by lemma 13.7.3.

Let $\chi \geq 0$ be a smooth function with compact support in $B$. By the weighted capacity continuity of theorem 7.5.8,

where the inequality holds because $0\leq a_ k\leq 1$ and $R_ k\wedge \chi \eta $ is a positive measure. Now let $\varepsilon \downarrow 0$ and then $C\to \infty $. By lemma 13.7.4(3), the weights $a$ increase to the indicator function of the complement of the union of the pole sets of the $v_ j$, which $R$ does not charge; so monotone convergence gives $\int \chi R\wedge \eta \leq \varliminf _ k\int \chi R_ k\wedge \eta $. This is the local lower inequality; it used only the finiteness of the local products, the extended capacity convergence and the local upper bounds. A finite partition of unity by such functions $\chi $ (proposition 5.2.6) gives the global inequality.

Convergence. By (13.5.5), the masses $\int _ XR_ k\wedge \omega ^{n-p}$ are bounded, since the shifted backgrounds depend only on the fixed forms $\theta _ j$. Near a compact subset $L$ of a chart with Euclidean form $\beta $, we have $\beta \leq c\, \omega $ for some $c>0$ (lemma 6.14.1), and hence $R_ k\wedge \beta ^{n-p}\leq c^{n-p}R_ k\wedge \omega ^{n-p}$ by (13.2.8). Therefore the trace measures of the $R_ k$ are locally uniformly bounded, and by theorem 5.3.7 every subsequence of $(R_ k)$ has a weakly convergent subsequence. Its limit $R'$ is positive, and $R'-R$ is positive, by the lower inequality applied to all smooth strongly positive forms $\eta $ on $X$. The additional hypothesis gives $\int _ X(R'-R)\wedge \omega ^{n-p}\leq 0$, because $\omega ^{n-p}$ is a test form on the compact manifold $X$. Hence the positive measure $(R'-R)\wedge \omega ^{n-p}$ vanishes, so the trace measures of $R'-R$, which are bounded on compact subsets of charts by multiples of it, vanish as well, and $R'-R=0$ by theorem 5.3.7. Thus every subsequence of $(R_ k)$ has a further subsequence converging weakly to $R$, which proves $R_ k\to R$. $\square$

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. $\square$