7.3 Decreasing continuity and symmetry2CE0DCA46BB5

Proof. Both integrals are finite, since $a$ and $b$ are bounded on $B$ and $\mathrm{dd}^{\mathrm{c}}a\wedge R$ and $\mathrm{dd}^{\mathrm{c}}b\wedge R$ are measures of finite total variation near $\overline B$. The function $h=b-a$ is bounded and Borel, and it vanishes on a collar $V$ of $\partial B$; hence its restriction to $B$ vanishes outside the compact set $K_0=\overline B\setminus V\subset B$. On $B$ we have $bR=aR+hR$, and since $R$ is closed, $\mathrm{dd}^{\mathrm{c}}(aR)=\mathrm{dd}^{\mathrm{c}}a\wedge R$ by (7.2.3); thus

where $\mathrm{dd}^{\mathrm{c}}(hR)$ is a signed measure with support in $K_0$. Choose $\chi \in C^\infty _ c(B)$ equal to $1$ on a neighborhood of $K_0$. By the definition of $\mathrm{dd}^{\mathrm{c}}$ on currents,

Consequently

The last assertion holds because $S\wedge \gamma $ is then closed, by the Leibniz rule, and positive of bidimension $(1,1)$; moreover $\mathrm{dd}^{\mathrm{c}}b\wedge (S\wedge \gamma )=(\mathrm{dd}^{\mathrm{c}}b\wedge S)\wedge \gamma $ by proposition 7.2.2(4). $\square$

Proof. Let $\chi \ge 0$ be a continuous function with compact support, let $N$ be a compact neighborhood of $\operatorname{Supp}\chi $, and put $W=\operatorname{Int}N$. The function $f_1$ is upper semicontinuous, hence bounded above on $N$, and $f$ is bounded below on $N$; since $f\le f_ j\le f_1$, there is $c\ge 0$ with $|f_ j|\le c$ on $N$ for all $j$. The positive Radon measures $(f_{j_ k}+c)\mu _{j_ k}$ on $W$ converge weakly to $\nu +c\mu $. By proposition 2.6.1, $\nu +c\mu $ is a positive Radon measure on $W$, and the convergence is vague. Since also $\mu _ j\to \mu $ vaguely, we obtain $\int \psi f_{j_ k}\, \mathrm d\mu _{j_ k}\to \int \psi \, \mathrm d\nu $ for every continuous function $\psi $ with compact support in $W$.

Fix $l$. By lemma 3.4.4, there are bounded continuous functions $g_ m$ on $N$ decreasing to $f_ l$. For $j_ k\ge l$ we have $f_{j_ k}\le f_ l\le g_ m$ on $N$, so

Letting $k\to \infty $ gives $\int \chi \, \mathrm{d}\nu \le \int \chi g_ m\, \mathrm{d}\mu $. Since $-c\le f_ l\le g_ m\le g_1$ on $N$ and $\mu (N)<\infty $, bounded convergence as $m\to \infty $ gives $\int \chi \, \mathrm{d}\nu \le \int \chi f_ l\, \mathrm{d}\mu $, and bounded convergence as $l\to \infty $ gives $\int \chi \, \mathrm{d}\nu \le \int \chi f\, \mathrm{d}\mu $. Since $\chi \ge 0$ was arbitrary, $\nu \le f\mu $. $\square$

Proof. Let $\rho _\varepsilon $ be the mollifier of section 2.5 on $\mathbb {R}^2$; it is a nonnegative even radial probability density supported in the disc of radius $\varepsilon $. Put $M_\varepsilon =\max *\rho _\varepsilon $, that is, $M_\varepsilon (x)=\int \max (x-y)\rho _\varepsilon (y)\, \mathrm dy$, where $\max (s,t)=\max \{ s,t\} $; it is smooth, since $\rho _\varepsilon $ is (section 2.5). Since $\rho _\varepsilon \ge 0$ has integral $1$, and $x\mapsto \max (x-y)$ is convex and increasing in each variable for each $y$, the function $M_\varepsilon $ is convex and increasing in each variable. Since $\rho _\varepsilon $ is even, $\int y_ i\rho _\varepsilon (y)\, \mathrm dy=0$ for $i=1,2$; integrating $\max (s-y_1,t-y_2)\ge s-y_1$ and $\max (s-y_1,t-y_2)\ge t-y_2$ against $\rho _\varepsilon $ therefore gives $M_\varepsilon (s,t)\ge s$ and $M_\varepsilon (s,t)\ge t$, that is, $M_\varepsilon \ge \max $. Since $|\max (x-y)-\max (x)|\le |y|\le \varepsilon $ on the support of $\rho _\varepsilon $, we have $M_\varepsilon \le \max +\varepsilon $. If $s-t>2\varepsilon $ and $|y|\le \varepsilon $, then $(s-y_1)-(t-y_2)\ge s-t-2\varepsilon >0$, so $M_\varepsilon (s,t)=\int (s-y_1)\rho _\varepsilon (y)\, \mathrm dy=s$; similarly $M_\varepsilon (s,t)=t$ if $t-s>2\varepsilon $.

It remains to show that $M_\varepsilon $ increases with $\varepsilon $. For fixed $x\in \mathbb R^2$, the function $g(\varepsilon )=\int \max (x-\varepsilon y)\rho _1(y)\, \mathrm{d}y$, defined for all real $\varepsilon $, equals $M_\varepsilon (x)$ for $\varepsilon >0$ and $\max (x)$ for $\varepsilon =0$; it is convex, since $\varepsilon \mapsto \max (x-\varepsilon y)$ is convex, and even, since $\rho _1$ is even. For $0\le \varepsilon _1<\varepsilon _2$, writing $\varepsilon _1=\lambda (-\varepsilon _2)+(1-\lambda )\varepsilon _2$ with $\lambda \in (0,1]$, we get $g(\varepsilon _1)\le \lambda g(-\varepsilon _2)+(1-\lambda )g(\varepsilon _2)=g(\varepsilon _2)$. This proves the monotonicity, and together with $\max \le M_\varepsilon \le \max +\varepsilon $ this gives $M_\varepsilon \downarrow \max $ as $\varepsilon \downarrow 0$.

Finally, let $a$ and $b$ be smooth and plurisubharmonic, and write $(a_1,a_2)=(a,b)$. Denoting by $\partial _ kM_\varepsilon $ and $\partial _ k\partial _ lM_\varepsilon $ the partial derivatives evaluated at $(a,b)$, the chain rule gives

The first sum is positive because $\partial _ kM_\varepsilon \ge 0$, $M_\varepsilon $ being increasing in each variable, and the forms $\mathrm i\partial \bar\partial a_ k$ are positive, since the smooth plurisubharmonic functions $a_ k$ have positive semidefinite complex Hessians (proposition 5.9.2). For the second, note that the real Hessian $H=(\partial _ k\partial _ lM_\varepsilon )$ is positive semidefinite: for $v\in \mathbb R^2$, the function $t\mapsto M_\varepsilon (x+tv)$ is convex, so its second derivative $v^ tHv$ is nonnegative. Writing $H=\sum _ m\lambda _ mv_ mv_ m^ t$ with $\lambda _ m\ge 0$ and real orthonormal vectors $v_ m$, and putting $\theta _ m=\sum _ k(v_ m)_ k\partial a_ k$, we get $\mathrm i\sum _{k,l}\partial _ k\partial _ lM_\varepsilon \, \partial a_ k\wedge \bar\partial a_ l=\sum _ m\lambda _ m\mathrm i\theta _ m\wedge \bar\theta _ m$, because $\bar\partial a_ l=\overline{\partial a_ l}$ for real $a_ l$; this is a positive $(1,1)$-form. Hence the complex Hessian of $M_\varepsilon (a,b)$ is positive semidefinite, and $M_\varepsilon (a,b)$ is plurisubharmonic by proposition 5.9.2. $\square$

Proof. Let $\varphi $ be a test form on $\Omega $ of complementary degree. Its support is compact, so it is covered by finitely many open sets $U_1,\ldots ,U_ m$ as in the hypothesis. By lemma 2.5.2, there are smooth functions $\eta _1,\ldots ,\eta _ m$ with compact supports in $U_1,\ldots ,U_ m$, respectively, such that $\sum _ i\eta _ i=1$ on a neighborhood of $\operatorname{Supp}\varphi $. Then $\varphi =\sum _ i\eta _ i\varphi $, where $\eta _ i\varphi $ is a test form supported in $U_ i$, and therefore $S_ j(\varphi )=\sum _ iS_ j(\eta _ i\varphi )\to \sum _ iS(\eta _ i\varphi )=S(\varphi )$. $\square$

Proof. We argue by induction on $q$. Step 1 derives the unweighted assertion and the symmetry from the weighted assertion for fewer factors. The other steps prove the weighted assertion on a ball, after a modification near its boundary: every weak limit of the weighted products is at most the expected limit (Step 4), and repeated integration by parts gives the reverse inequality for the total masses (Step 5).

Step 1: the induction. We prove the second (weighted) assertion by induction on $q$, simultaneously for all open sets $\Omega \subset \mathbb C^ n$, all $r\ge q$, all closed positive currents $T$ of bidimension $(r,r)$ on $\Omega $ and all decreasing sequences as in the statement. For $q=0$ it states that $u_{0,j}T\to u_0T$; this follows from dominated convergence with respect to the total variations of the coefficient measures of $T$, since $u_0\le u_{0,j}\le u_{0,1}$. Let $q\ge 1$ and assume the weighted assertion for fewer than $q$ factors. Applying it to the $q-1$ factors $u_{2,j},\ldots ,u_{q,j}$ with the weight $u_{1,j}$, and then applying the weakly continuous operator $\mathrm{dd}^{\mathrm{c}}$, we obtain the unweighted assertion for $q$ factors.

Next we prove symmetry for $q$ factors. Let $u_1,\ldots ,u_ q$ be locally bounded and plurisubharmonic, let $\varepsilon >0$, and choose $\varepsilon _ k\downarrow 0$ with $\varepsilon _ k<\varepsilon $. The regularizations $u_ l*\rho _{\varepsilon _ k}$ are smooth plurisubharmonic functions on $\Omega _\varepsilon $ decreasing to $u_ l$ (proposition 5.9.2). By the unweighted assertion, the product $\mathrm{dd}^{\mathrm{c}}u_1\wedge \cdots \wedge \mathrm{dd}^{\mathrm{c}}u_ q\wedge T$ on $\Omega _\varepsilon $ is therefore the weak limit of exterior products of smooth forms with $T$ (proposition 7.2.2(1)), and these are symmetric in the factors because $2$-forms commute. Since the sets $\Omega _\varepsilon $ exhaust $\Omega $, products with $q$ factors are symmetric. It remains to prove the weighted assertion for $q$ factors; in doing so we may use the unweighted assertion and the symmetry for $q$ factors.

Step 2: reduction to a common collar. By lemma 7.3.6, it suffices to prove the convergence near every point. Fix a ball $B=B(a,R)$, a larger ball $B_1$ with $\overline B\subset B_1\Subset \Omega $, and a compact set $K\subset B$; we prove the convergence on a neighborhood of $K$. Since $u_ l\le u_{l,j}\le u_{l,1}$, all the functions $u_{l,j}$ and $u_ l$ are bounded on $B_1$ uniformly in $j$. Subtracting a constant from $u_{l,j}$ and $u_ l$ changes neither the Hessians nor, since the unweighted assertion is already known, the validity of the weighted assertion. We may therefore assume that $-N\le u_{l,j},u_ l\le -1$ on $B_1$ for all $l$ and $j$, where $N\ge 1$. Put $\psi =|z-a|^2-R^2$, choose $\delta >0$ with $K\subset \{ \psi <-\delta \} $, and replace every function $v$ among the $u_{l,j}$ and $u_ l$ by the function $\max \{ v,A\psi \} $ of (7.3.2), with $U=B_1$ and $A=N/\delta $. The new functions are plurisubharmonic on $B_1$, and the new sequences still decrease to the new limits. The functions are unchanged on the neighborhood $\{ \psi <-\delta \} $ of $K$, and by locality so are their products. From now on, $u_{l,j}$ and $u_ l$ denote the new functions; they take values in $[-N,0)$ on $B$ and coincide with $A\psi $ on the collar $V=\{ z\in B_1:\psi (z)>-1/A\} $ of $\partial B$.

Step 3: smooth approximation with the same collar. Fix $\eta >0$ such that $\{ \psi \le \eta \} \subset B_1$, and let $V'=\{ z:-1/(2A)<\psi (z)<\eta \} $; it is a collar of $\partial B$ contained in $V$. Fix $l$ and $j$, and let $\tilde u$ be the function $u_{l,j}$ before the modification of Step 2, so that $\tilde u\le -1$ on $B_1$ and $u_{l,j}=\max \{ \tilde u,A\psi \} $. For $0<\varepsilon <1/4$ smaller than the distance from $\{ \psi \le \eta \} $ to $\partial B_1$, put

with $M_\varepsilon $ as in lemma 7.3.5. The functions $\tilde u*\rho _\varepsilon $ are smooth and plurisubharmonic on $\{ \psi <\eta \} $ and decrease to $\tilde u$ as $\varepsilon \downarrow 0$ (proposition 5.9.2), so the $u^\varepsilon _{l,j}$ are smooth and plurisubharmonic on $\{ \psi <\eta \} $ by lemma 7.3.5. Since $\tilde u*\rho _\varepsilon \le -1$, on $V'$ we have $A\psi -\tilde u*\rho _\varepsilon >1/2>2\varepsilon $, and therefore $u^\varepsilon _{l,j}=A\psi $. Since $M_\varepsilon $ decreases to $\max $, is increasing in each variable and satisfies $\max \le M_\varepsilon \le \max +\varepsilon $, the functions $u^\varepsilon _{l,j}$ decrease to $\max \{ \tilde u,A\psi \} =u_{l,j}$. In particular $u^\varepsilon _{l,j}\ge u_{l,j}\ge u_ l$, and all the functions $u_{l,j}$, $u_ l$ and $u^\varepsilon _{l,j}$ coincide with $A\psi $ on $V'$.

Step 4: an upper bound for weak limits. Put

By Step 1, $S_ j\to S$ weakly. On $B$ the weights $u_{0,j}$ take values in $[-N,0)$, so the currents $-u_{0,j}S_ j$ are positive there, and by theorem 7.2.5, since $|u_{l,j}|\le N$ on $B$, their masses on compact subsets of $B$ are bounded uniformly in $j$. By the compactness part of theorem 5.3.7, every subsequence of $(u_{0,j}S_ j)$ has a further subsequence converging weakly on $B$. Let $\Theta $ be the limit of such a subsequence. Let $\varphi $ be a strongly positive test form on $B$ of complementary bidegree. The measures $\mu _ j=S_ j\wedge \varphi $ are positive and converge weakly to $S\wedge \varphi $, and along the chosen subsequence $u_{0,j}\mu _ j\to \Theta \wedge \varphi $. Hence lemma 7.3.4, applied on $B$ with the real-valued upper semicontinuous functions $f_ j=u_{0,j}$, gives $\Theta \wedge \varphi \le u_0S\wedge \varphi $. Evaluating this inequality of measures on a nonnegative test function equal to $1$ on $\operatorname{Supp}\varphi $, we get $(u_0S-\Theta )(\varphi )\ge 0$; thus $u_0S-\Theta $ is a positive current on $B$. By locality (proposition 7.2.2), all the currents $u_{0,j}S_ j$, and also $u_0S$, coincide on $V'$ with the fixed current $A\psi \, (A\mathrm{dd}^{\mathrm{c}}\psi )^ q\wedge T$; hence so does $\Theta $, and $u_0S-\Theta $ has compact support in $B$.

Step 5: comparison of masses. Set $\gamma =\beta ^{r-q}$ and

We claim that $I\le I_ j$ for every $j$. Fix $j$ and parameters $\varepsilon _0,\ldots ,\varepsilon _ q$, and abbreviate $w_ l=u^{\varepsilon _ l}_{l,j}$. For $0\le k\le q$ put

so that $J_0=I$. The measure in $J_ k$ is positive and $u_ k\le w_ k$, so $J_ k$ does not decrease if $u_ k$ is replaced by $w_ k$. If $k<q$, symmetry of products with $q$ factors and proposition 7.2.2(4) let us write that measure as $\mathrm{dd}^{\mathrm{c}}u_{k+1}\wedge R_ k$, where

is a closed positive current of bidimension $(1,1)$ on $\{ \psi <\eta \} $; then lemma 7.3.3, with $a=w_ k$ and $b=u_{k+1}$, which both equal $A\psi $ on the collar $V'$, gives

Hence

Now let $\varepsilon _ q\downarrow 0$, then $\varepsilon _{q-1}\downarrow 0$, and so on down to $\varepsilon _1\downarrow 0$, keeping $\varepsilon _0$ fixed. We claim that the right-hand side of the last inequality then tends to

When $\varepsilon _ q\downarrow 0$, the measure is fixed and the bounded weights $w_ q$ decrease to $u_{q,j}$, so dominated convergence applies. When $\varepsilon _ k\downarrow 0$ for some $1\le k\le q-1$, the integrand is, after moving the smooth factors next to $T$ by symmetry,

where $T_ k=\mathrm{dd}^{\mathrm{c}}w_0\wedge \cdots \wedge \mathrm{dd}^{\mathrm{c}}w_{k-1}\wedge T$ is a fixed closed positive current on $\{ \psi <\eta \} $ of bidimension $(r-k,r-k)$. There are $q-k\le q-1$ Hessian factors and the fixed weight $u_{q,j}$. Along any sequence $\varepsilon _ k\downarrow 0$, the functions $u^{\varepsilon _ k}_{k,j}$ decrease to $u_{k,j}$ (Step 3), so the induction hypothesis, applied on $\{ \psi <\eta \} $ with the base current $T_ k$, constant sequences for the other factors and the constant weight $u_{q,j}$, shows that these currents converge weakly; so do their exterior products with $\gamma $.

In each case the convergence holds for the integrals over $B$, and not only against compactly supported test forms, because all the currents involved coincide on the collar $V'$ with the same fixed current: if $\chi \in C^\infty _ c(B)$ equals $1$ on $B\setminus V'$, the integrals against $\chi $ converge, and the integrals against $1-\chi $ do not depend on the parameters.

By lemma 7.3.3, with $a=w_0$ and $b=u_{q,j}$, and by symmetry of products with $q$ factors, the limit integral equals

Hence $I\le \int _ Bu^{\varepsilon _0}_{0,j}S_ j\wedge \gamma $ for every $\varepsilon _0$, and dominated convergence with respect to the finite measure $S_ j\wedge \gamma $ on $B$ gives $I\le I_ j$ as $\varepsilon _0\downarrow 0$.

Step 6: conclusion. Along the subsequence defining $\Theta $, splitting the integrals with a cutoff as in Step 5 shows that $I_ j\to \int _ B\Theta \wedge \gamma $. Therefore

Since $u_0S-\Theta $ is positive with compact support in $B$, its trace measure $(u_0S-\Theta )\wedge \gamma $ is a positive measure of total mass at most zero, hence zero; by the coefficient bounds of theorem 5.3.7, $u_0S-\Theta =0$ on $B$. Thus every subsequence of $(u_{0,j}S_ j)$ has a further subsequence converging weakly to $u_0S$ on $B$. Hence the whole sequence converges to $u_0S$ on $B$: otherwise there would be a test form $\varphi $ on $B$ and a subsequence along which $|(u_{0,j}S_ j-u_0S)(\varphi )|$ stays bounded away from zero, and no further subsequence of it could converge to $u_0S$. The replacement by maxima in Step 2 changes neither the functions nor, by locality, the weighted products on the neighborhood $\{ \psi <-\delta \} $ of $K$, and the subtraction of constants does not affect the convergence; hence the original weighted products converge on this neighborhood. As $K$ and $B$ were arbitrary, this proves the weighted assertion for $q$ factors on $\Omega $ and completes the induction. $\square$

Proof. Since weak convergence is a local property (lemma 7.3.6), it suffices to prove the convergence on every relatively compact open set $W\Subset \Omega $. On $W$ all the functions $v_{i,j}$ and $v_ i$ are bounded, because $v_ i\le v_{i,j}\le v_{i,1}$. Adding a constant $c_ i$ to each $v_{i,j}$ and to $v_ i$ replaces $P$ by the polynomial $P(x_1-c_1,\ldots ,x_ m-c_ m)$; so we may assume that all these functions are nonnegative on $W$. Since the weighted products depend linearly on the weight, it suffices to treat a monomial. A constant weight is covered by the unweighted assertion of the theorem. A monomial of degree $N\ge 1$ is a product $a_1\cdots a_ N$, where each $a_ l$ is one of the $v_ i$, repetitions being allowed; we write $a_{l,j}$ for the corresponding function $v_{i,j}$. For real numbers $x_1,\ldots ,x_ N$ we have the polarization identity

To prove it, expand $(\sum _{l\in I}x_ l)^ N=\sum _\alpha \frac{N!}{\alpha !}x^\alpha $, the sum running over the multi-indices $\alpha \in \mathbb N^ N$ with $|\alpha |=N$ and $\alpha _ l=0$ for $l\notin I$. Fix such an $\alpha $, and let $J=\{ l:\alpha _ l>0\} $ have $k$ elements. The monomial $x^\alpha $ occurs exactly in the terms with $I\supset J$, so its total coefficient on the right-hand side is

This vanishes unless $k=N$, in which case $\alpha =(1,\ldots ,1)$ and the coefficient is $N!$.

Each sum $s=\sum _{l\in I}a_ l$ is a bounded nonnegative plurisubharmonic function on $W$, and so is $s^ N$. For $N=1$ there is nothing to prove; for $N\ge 2$ and smooth $s$ this follows from

in which both terms are positive $(1,1)$-forms, the first because $s\ge 0$ and $s$ has a positive semidefinite complex Hessian, and the second by (7.3.1); hence $s^ N$ is plurisubharmonic, by both directions of proposition 5.9.2. In general, the regularizations $s*\rho _\varepsilon $ are smooth, plurisubharmonic and nonnegative, and they decrease to $s$ (proposition 5.9.2); since $t\mapsto t^ N$ is increasing on $[0,\infty )$, the plurisubharmonic functions $(s*\rho _\varepsilon )^ N$ decrease to $s^ N$, which is therefore plurisubharmonic by theorem 5.10.1. Likewise the sums $s_ j=\sum _{l\in I}a_{l,j}$ are nonnegative and decrease to $s$, so the $s_ j^ N$ are bounded plurisubharmonic functions on $W$ decreasing to $s^ N$. The weighted assertion of the theorem therefore applies with the weights $s_ j^ N$, and by (7.3.9), with $x_ l=a_{l,j}$, it applies to the monomial weights $a_{1,j}\cdots a_{N,j}$.

For the second assertion, let $a$ and $b$ be bounded plurisubharmonic functions on a neighborhood $U$ of $\overline B$ which are equal on a collar $V$ of $\partial B$. Let $d>0$ be smaller than the distances from $\partial B$ to $\mathbb C^ n\setminus V$ and from $\overline B$ to $\mathbb C^ n\setminus U$, and put $V'=\{ z:\operatorname {dist}(z,\partial B)<d/2\} $. For $\varepsilon =\varepsilon _ k\downarrow 0$ with $\varepsilon _ k<d/2$, the regularizations $a_\varepsilon =a*\rho _\varepsilon $ and $b_\varepsilon =b*\rho _\varepsilon $ are smooth and plurisubharmonic on a neighborhood of $\overline B$ and decrease to $a$ and $b$ (proposition 5.9.2); moreover $a_\varepsilon =b_\varepsilon $ on the collar $V'$, because $a=b$ on the ball $B(z,\varepsilon )\subset V$ for $z\in V'$. By lemma 7.3.3, applied to the smooth function $a_\varepsilon $ and the bounded plurisubharmonic function $b_\varepsilon $,

Choose $\chi \in C^\infty _ c(B)$ equal to $1$ on $B\setminus V'$. On $V'$ the measures $a_\varepsilon \mathrm{dd}^{\mathrm{c}}b_\varepsilon \wedge R$ and $b_\varepsilon \mathrm{dd}^{\mathrm{c}}a_\varepsilon \wedge R$ coincide, by locality, so the identity is equivalent to $\int \chi a_\varepsilon \, \mathrm{dd}^{\mathrm{c}}b_\varepsilon \wedge R=\int \chi b_\varepsilon \, \mathrm{dd}^{\mathrm{c}}a_\varepsilon \wedge R$. By the weighted assertion of the theorem with one factor and the fixed base current $R$, this passes to the limit: $\int \chi a\, \mathrm{dd}^{\mathrm{c}}b\wedge R=\int \chi b\, \mathrm{dd}^{\mathrm{c}}a\wedge R$. Since $a=b$ on $V$, the measures $a\, \mathrm{dd}^{\mathrm{c}}b\wedge R$ and $b\, \mathrm{dd}^{\mathrm{c}}a\wedge R$ also coincide on $V'$, and adding the integrals against $1-\chi $ over $B$ gives the identity for $a$ and $b$. $\square$

Proof. The assertion is local, so it suffices to prove it on each $\Omega _{\varepsilon _0}$, $\varepsilon _0>0$. Let $u_ j=u*\rho _{\varepsilon _ j}$ with $\varepsilon _ j\downarrow 0$, $\varepsilon _ j<\varepsilon _0$; these are smooth plurisubharmonic functions on $\Omega _{\varepsilon _0}$ decreasing to $u$ (proposition 5.9.2). Since $u_ j$ is continuous and $v$ is upper semicontinuous, the set $\{ u_ j>v\} $ is open, and on it $w_ j=\max \{ u_ j,v\} $ equals $u_ j$; by locality, $(\mathrm{dd}^{\mathrm{c}}w_ j)^ p\wedge T=(\mathrm{dd}^{\mathrm{c}}u_ j)^ p\wedge T$ there. The nonnegative Borel function $f_ j=w_ j-v$ vanishes outside $\{ u_ j>v\} $, so

The functions $w_ j$ decrease to $w=\max \{ u,v\} $. Both sides of the last identity are differences of weighted products, for instance $f_ j(\mathrm{dd}^{\mathrm{c}}w_ j)^ p\wedge T=w_ j(\mathrm{dd}^{\mathrm{c}}w_ j)^ p\wedge T-v(\mathrm{dd}^{\mathrm{c}}w_ j)^ p\wedge T$. Applying theorem 7.3.7 to them, with the weights $w_ j$ and the constant weight $v$, and with the factors $w_ j$ on the left and $u_ j$ on the right, we obtain $f(\mathrm{dd}^{\mathrm{c}}w)^ p\wedge T=f(\mathrm{dd}^{\mathrm{c}}u)^ p\wedge T$, where $f=w-v\ge 0$. Multiply the coefficient measures of both sides by $1/(f+\varepsilon )$ and let $\varepsilon \downarrow 0$. Since $f/(f+\varepsilon )$ increases to the indicator of $\{ f>0\} =\{ u>v\} $, dominated convergence with respect to the locally finite total variations of the coefficient measures gives the assertion. $\square$

Proof. Let $\varepsilon >0$ and $w=\max \{ u,v-\varepsilon \} $. We show that $(\mathrm{dd}^{\mathrm{c}}w)^ n$ and $(\mathrm{dd}^{\mathrm{c}}u)^ n$ have the same mass near a compact set outside which $w=u$, and then identify $(\mathrm{dd}^{\mathrm{c}}w)^ n$ by proposition 7.3.10. Every $x\in \partial D$ has an open neighborhood $U_ x$ with $u>v-\varepsilon $ on $U_ x\cap D$, and finitely many of them cover the compact set $\partial D$; let $U$ be their union. Then $F=D\setminus U=\overline D\setminus U$ is a compact subset of $D$, it contains $\{ u\le v-\varepsilon \} $, and $w=u$ on $D\setminus F$. For $0\le l\le n-1$ put $R_ l=(\mathrm{dd}^{\mathrm{c}}w)^ l\wedge (\mathrm{dd}^{\mathrm{c}}u)^{n-1-l}$. By additivity and symmetry of the products (proposition 7.2.2, theorem 7.3.7),

and summing over $l$, the right-hand sides telescope:

The current in parentheses vanishes outside $F$, since $w-u$ does. Hence the signed measure $(\mathrm{dd}^{\mathrm{c}}w)^ n-(\mathrm{dd}^{\mathrm{c}}u)^ n$ is supported in $F$, and if $\chi \in C^\infty _ c(D)$ equals $1$ on a neighborhood of $F$, then its total mass is

because $\mathrm{dd}^{\mathrm{c}}\chi =0$ on a neighborhood of $F$. Choose a relatively compact open set $D'\subset D$ containing $F$. Both measures are finite on $D'$, and $\int _{D'}(\mathrm{dd}^{\mathrm{c}}w)^ n=\int _{D'}(\mathrm{dd}^{\mathrm{c}}u)^ n$. By proposition 7.3.10, $(\mathrm{dd}^{\mathrm{c}}w)^ n$ coincides with $(\mathrm{dd}^{\mathrm{c}}v)^ n$ on $\{ u<v-\varepsilon \} $ and with $(\mathrm{dd}^{\mathrm{c}}u)^ n$ on $\{ u>v-\varepsilon \} $; here we use that adding the constant $-\varepsilon $ does not change $(\mathrm{dd}^{\mathrm{c}}v)^ n$. Since $\{ u<v-\varepsilon \} $ and $D'\cap \{ u>v-\varepsilon \} $ are disjoint Borel subsets of $D'$, and $\{ u\le v-\varepsilon \} \subset F\subset D'$,

As $\varepsilon \downarrow 0$, the sets $\{ u<v-\varepsilon \} $ increase to $\{ u<v\} $, and monotone convergence proves the comparison inequality.

The last assertion is the case of corollary 7.3.12 below in which the roles of $u$ and $v$ are interchanged: if $(\mathrm{dd}^{\mathrm{c}}u)^ n=0$, then $v\le u$ holds almost everywhere for $(\mathrm{dd}^{\mathrm{c}}u)^ n$. The proof of that corollary uses only the comparison inequality. $\square$

Proof. Choose $R$ with $D\subset B(0,R)$, and for $\delta >0$ put $u_\delta =u+\delta (|z|^2-R^2)$, a locally bounded plurisubharmonic function. Since $u_\delta \le u$ on $D$, the boundary condition still holds for $v$ and $u_\delta $, and the set $A=\{ v<u_\delta \} $ is contained in $\{ v<u\} $. By additivity, positive homogeneity and symmetry of the products (proposition 7.2.2, theorem 7.3.7), $(\mathrm{dd}^{\mathrm{c}}u_\delta )^ n=\sum _{k=0}^ n\binom nk\delta ^ k(\mathrm{dd}^{\mathrm{c}}u)^{n-k}\wedge (\mathrm{dd}^{\mathrm{c}}|z|^2)^ k$ is a sum of positive measures, and its term with $k=n$ is $\delta ^ nn!\, \pi ^{-n}\, \mathrm dV_{2n}$, by (7.2.4). The comparison inequality of theorem 7.3.11, applied to $v$ and $u_\delta $, therefore gives

where the last equality holds because $A\subset \{ v<u\} $ and $u\le v$ almost everywhere for $(\mathrm{dd}^{\mathrm{c}}v)^ n$. Thus $A$ is a Lebesgue null set, and $\max \{ u_\delta ,v\} =v$ almost everywhere. Both functions are plurisubharmonic, so the uniqueness of plurisubharmonic representatives in theorem 5.9.1, applied on each connected component of $D$, gives $\max \{ u_\delta ,v\} =v$ everywhere, that is, $u_\delta \leq v$. Letting $\delta \downarrow 0$ gives $u\le v$. $\square$