7.11.16 Capacity convergence and truncated measures7449912C4F37

Proof. The two measures agree on the open set $\Omega \setminus K$, because the products are local (proposition 7.2.2). For $0\leq k\leq n-1$ put $S_ k=(\mathrm{dd}^{\mathrm{c}}\varphi )^ k\wedge (\mathrm{dd}^{\mathrm{c}}\psi )^{n-1-k}$. By the symmetry of the products (theorem 7.3.7) and definition 7.2.1,

so the sum over $k$ telescopes and

The coefficient measures of the current $R$ of order zero vanish on $\Omega \setminus K$, so $\operatorname{Supp}R\subset K$. By lemma 2.5.1 there is $\chi \in C^\infty _ c(\Omega )$ with $0\leq \chi \leq 1$ which equals one on a neighborhood of $K$. Then $\mathrm{dd}^{\mathrm{c}}\chi $ vanishes on a neighborhood of $\operatorname{Supp}R$, and therefore

where both integrals are finite. Since $\chi =1$ on $K$ and the two measures agree on $\Omega \setminus K$, it follows that $(\mathrm{dd}^{\mathrm{c}}\varphi )^ n(K)=(\mathrm{dd}^{\mathrm{c}}\psi )^ n(K)<\infty $, and adding the equal masses of $\Omega \setminus K$ gives the last assertion. $\square$

Proof. Since $\varphi $ tends to zero uniformly at $\partial B$, there is an open neighborhood $V$ of $\partial B$ such that $\varphi >-T$ on $V\cap B$; there $\varphi _ T=\varphi $. Thus $\varphi _ T$ is bounded, nonpositive and plurisubharmonic, it has zero boundary values, and it coincides with $\varphi $ outside the compact set $\overline B\setminus V\subset B$. By lemma 7.11.18, the measures $(\mathrm{dd}^{\mathrm{c}}\varphi _ T)^ n$ and $(\mathrm{dd}^{\mathrm{c}}\varphi )^ n$ have the same total mass, which is finite; in particular $\varphi _ T\in \mathcal E_0(B)$.

Put $F_ T=\{ \varphi \leq -T\} $. By maximum locality (proposition 7.3.10), the two measures agree on the complement $\{ \varphi >-T\} $. Since their total masses are equal and finite, they also assign the same mass $m_ T$ to $F_ T$, and Chebyshev's inequality gives

As $\varphi _ T=-T$ on $F_ T$, we get

The two measures differ only on $F_ T$, where each has mass $m_ T$; this proves (7.11.20). Finally, $(\varphi _ T+T)/T$ is plurisubharmonic with values in $[0,1]$, so it is a competitor for $\operatorname {Cap}_ B$ (definition 7.4.2), and therefore $\int _ E(\mathrm{dd}^{\mathrm{c}}\varphi _ T)^ n\leq T^ n\operatorname {Cap}_ B(E)$. Since the two measures agree on $E\setminus F_ T$, we conclude

Proof. (1) Let $h$ be a competitor for $\operatorname {Cap}_ B$, that is, $h\in \mathrm{PSH}(B)$ and $0\leq h\leq 1$. The set $\{ \varphi <-T\} $ is open, because $\varphi $ is upper semicontinuous, and it is relatively compact in $B$, because $\varphi $ has zero boundary values. Hence $\nu =\mathbf1_{\{ \varphi <-T\} }(\mathrm{dd}^{\mathrm{c}}h)^ n$ is a finite measure, and it is dominated by $(\mathrm{dd}^{\mathrm{c}}h)^ n$. The tail estimate (7.11.3) of theorem 7.11.2, applied to $\nu $ and $w=h$, involves the constant $\kappa _{n,p}=2^{n+p}(n+1)!M^ n$, where $M\leq 1$ is the oscillation of $h$ on $B$, but not the mass of $\nu $; it gives

Taking the supremum over $h$ proves (1).

(2) Choose an open set $U$ and compact sets $L,L'\subset B$ with $K\subset U\Subset \operatorname{Int}L$ and $L\subset \operatorname{Int}L'$. By the Chern–Levine–Nirenberg inequality (theorem 7.2.5, with the base current $1$ and the pair $L\subset \operatorname{Int}L'$), there is a constant $C_ L$ with $\int _ L(\mathrm{dd}^{\mathrm{c}}h)^ n\leq C_ L$ for every competitor $h$, since $\| h\| _{L^\infty (L')}\leq 1$. The finite measure $\mathbf1_ L(\mathrm{dd}^{\mathrm{c}}h)^ n$ is dominated by $(\mathrm{dd}^{\mathrm{c}}h)^ n$ and has mass at most $C_ L$, so the constant $A_ p$ of theorem 7.11.2 for this measure and $w=h$ is at most

Hence the moment estimate (7.11.4), applied to $\varphi _ j$, gives

for all $j$ and all competitors $h$.

We pass this bound to the limit $u$. Since the $\varphi _ j$ are nonpositive, $u\leq 0$ almost everywhere, and hence everywhere by the submean inequality. Fix $S>0$. The truncations $\max \{ \varphi _ j,-S\} $ take values in $[-S,0]$ and converge to the plurisubharmonic function $\max \{ u,-S\} $ in $L^1_{\mathrm{loc}}(B)$, since $t\mapsto \max \{ t,-S\} $ is $1$-Lipschitz. Since $\mathbf1_ L(\mathrm{dd}^{\mathrm{c}}h)^ n\leq (\mathrm{dd}^{\mathrm{c}}h)^ n$, lemma 7.11.9 shows that they also converge in $L^1(\mathbf1_ L(\mathrm{dd}^{\mathrm{c}}h)^ n)$. As $t\mapsto t^ p$ is Lipschitz on $[0,S]$ and $(-\max \{ \varphi _ j,-S\} )^ p\leq (-\varphi _ j)^ p$, it follows that

Letting $S\to \infty $, monotone convergence gives the same bound for $\int _ L(-u)^ p(\mathrm{dd}^{\mathrm{c}}h)^ n$. The set $U\cap \{ u<-T\} $ is open, because $u$ is upper semicontinuous, and it contains $K\cap \{ u<-T\} $. By Chebyshev's inequality its $(\mathrm{dd}^{\mathrm{c}}h)^ n$-measure is at most $C_{K,p}E_ p^{p/(n+p)}T^{-p}$. Taking the supremum over $h$ bounds $\operatorname {Cap}_ B(U\cap \{ u<-T\} )$, and hence the outer capacity in (7.11.23). Finally, $K\cap \{ u=-\infty \} $ is contained in $K\cap \{ u<-T\} $ for every $T$, so letting $T\to \infty $ shows that it has zero outer capacity. $\square$

Proof. We estimate the capacities of the sets where the truncation $\max \{ U_ j,-T\} $ lies far below or far above $u_ T$ (Steps 1 and 2) and deduce convergence in capacity (Step 3). The estimate for $(\mathrm{dd}^{\mathrm{c}}u_ T)^ n$ then follows from the convergence of the truncated measures (Step 4). Fix $T>0$ and write $a_ j=\max \{ U_ j,-T\} $ and $b=u_ T$. These plurisubharmonic functions take values in $[-T,0]$, and $a_ j\to b$ in $L^1_{\mathrm{loc}}(B)$ because truncation is $1$-Lipschitz. By theorem 7.11.11, $\varepsilon _ j=\int _ B|U_ j-u|\, \mathrm d\mu _ j\to 0$.

Step 1. The sets where $a_ j<b-\delta $. Let $\delta >0$. The set $\{ a_ j<b-\delta /2\} $ is relatively compact in $B$, because $b\leq 0$ while $a_ j$ tends to zero at $\partial B$. Applying lemma 7.10.1 with $s=t=\delta /2$ and then Chebyshev's inequality, we obtain

The weight $(b-a_ j)_+$ takes values in $[0,T]$, and it is at most $|U_ j-u|$, since truncation does not increase distances. By (7.11.20) with $p=2$, $\| (\mathrm{dd}^{\mathrm{c}}a_ j)^ n-\mu _ j\| _{\mathrm{TV}}\leq 2E_2T^{-2}$, where $\mu _ j=(\mathrm{dd}^{\mathrm{c}}U_ j)^ n$. Hence

and therefore $\operatorname {Cap}_ B(\{ a_ j<b-\delta \} )\leq 2^{n+1}\delta ^{-(n+1)}(\varepsilon _ j+2E_2/T)$.

Step 2. The sets where $a_ j>b+\delta $. Let $L\subset B$ be compact. By theorem 7.5.2, for every $\sigma >0$ there is an open set $G\subset B$ with $\operatorname {Cap}_ B(G)<\sigma $ such that $b$ is continuous on $B\setminus G$. We claim that $L\cap \{ a_ j>b+\delta \} \subset G$ for all sufficiently large $j$. Let $z\in L\setminus G$. Since $b$ is upper semicontinuous, $b<b(z)+\delta /3$ on a closed ball $\overline{B(z,r)}\subset B$. Hartogs' lemma (theorem 5.10.4), applied to the sequence $a_ j$ with the constant majorant $b(z)+\delta /3$, gives $a_ j<b(z)+\delta /3$ on $\overline{B(z,r)}$ for all large $j$. Since $b$ is continuous on $B\setminus G$, after shrinking $r$ we also have $b>b(z)-\delta /3$ on $(L\setminus G)\cap B(z,r)$. Hence $a_ j-b<2\delta /3$ on $(L\setminus G)\cap B(z,r)$ for all large $j$. As $L\setminus G$ is compact, finitely many such balls cover it, which proves the claim. Consequently $\varlimsup _ j\operatorname {Cap}_ B(L\cap \{ a_ j>b+\delta \} )\leq \sigma $ for every $\sigma >0$, that is, $\operatorname {Cap}_ B(L\cap \{ a_ j>b+\delta \} )\to 0$.

Step 3. Convergence in capacity. Let $K\subset B$ be compact, and apply Step 2 to a compact set $L\subset B$ with $K\subset \operatorname{Int}L$. The bounded functions $a_ j-b$ are quasi-continuous by theorem 7.5.2 and the remark following definition 7.5.1. Therefore lemma 7.5.4, together with Steps 1 and 2 applied with $\delta /2$ in place of $\delta $, gives

It remains to remove the truncation. Outside $\{ U_ j<-T\} \cup \{ u<-T\} $ we have $a_ j=U_ j$ and $b=u$. Hence both sets $K\cap \{ U_ j>u+\delta \} $ and $K\cap \{ U_ j<u-\delta \} $ in definition 7.5.5 are contained in

By proposition 7.11.22 with $p=2$, applied in part (1) to $U_ j$, whose energy is at most $E_2$, and in part (2) to the sequence $(U_ j)$,

and the capacity of the open set $\{ U_ j<-T\} $ equals its outer capacity. By subadditivity of the outer capacity, the upper limits as $j\to \infty $ of the outer capacities of the two sets in definition 7.5.5 are bounded by the sum of three terms, each of which tends to zero as $T\to \infty $. Since these upper limits do not depend on $T$, they vanish. Thus $U_ j\to u$ locally in capacity.

Step 4. The truncated limit measure. Fix $T>0$ again. By Step 3 and (7.11.17), the uniformly bounded functions $a_ j$ converge to $b$ locally in capacity, so $(\mathrm{dd}^{\mathrm{c}}a_ j)^ n\to (\mathrm{dd}^{\mathrm{c}}u_ T)^ n$ weakly on $B$ by theorem 7.5.8. By lemma 7.11.19, each $(\mathrm{dd}^{\mathrm{c}}a_ j)^ n$ has total mass $\mu _ j(B)=m$. Testing with functions $\chi \in C_ c(B)$, $0\leq \chi \leq 1$, and letting $\chi $ increase to one, we see that $(\mathrm{dd}^{\mathrm{c}}u_ T)^ n$ has total mass at most $m$. Moreover, (7.11.20) gives

for every real $\chi \in C_ c(B)$ with $|\chi |\leq 1$. Since $\mu _ j\to \mu $ weakly by proposition 7.11.5, letting $j\to \infty $ yields the same bound with $(\mathrm{dd}^{\mathrm{c}}u_ T)^ n$ and $\mu $ in place of $(\mathrm{dd}^{\mathrm{c}}a_ j)^ n$ and $\mu _ j$. Both limit measures are finite, so by the remark following the definition of the total variation distance, the supremum over such $\chi $ equals $\| (\mathrm{dd}^{\mathrm{c}}u_ T)^ n-\mu \| _{\mathrm{TV}}$. This proves (7.11.25). $\square$