10.9 Global poles on compact complex manifoldsAC5D271E1A27

Proof. (1) We freeze the coefficients of $\omega $ on small balls. Fix $\varepsilon >0$, a coordinate chart and a compact subset $L$ of it. By uniform continuity of the coefficients of $\omega $ on a compact neighborhood of $L$ in the chart, there is $r>0$ such that, for every $a\in L$, the ball $B(a,r)$ lies in the chart and the form $\gamma _ a=\omega (a)+\tfrac \varepsilon 2\beta $ satisfies $\omega \leq \gamma _ a\leq \omega +\varepsilon \beta $ on $B(a,r)$. Let $g_ a$ be the real quadratic polynomial with $\mathrm{dd}^{\mathrm{c}}g_ a=\gamma _ a$ described at the beginning of this section. If $u\in \mathrm{PSH}(X,\omega )$, then $u+g_ a$ is quasi-plurisubharmonic with $\mathrm{dd}^{\mathrm{c}}(u+g_ a)=\omega _ u+(\gamma _ a-\omega )\geq 0$, so it is plurisubharmonic on $B(a,r)$; in the same way $v+g_ a$ is plurisubharmonic on $D\cap B(a,r)$.

Let $w$ be the function in question: a finite maximum $\max _ iu_ i$, a regularized supremum $(\sup _ iu_ i)^*$, or the pasted function. We show that $w+g_ a$ is plurisubharmonic on $B(a,r)$. In the first case $w+g_ a$ is the maximum of the plurisubharmonic functions $u_ i+g_ a$, which is plurisubharmonic by proposition 5.9.5. In the second case $w+g_ a=(\sup _ i(u_ i+g_ a))^*$, because $g_ a$ is continuous, and it is plurisubharmonic by theorem 5.10.2, since the family $(u_ i+g_ a)$ is uniformly bounded above on $B(a,r)$. In the third case $w+g_ a$ equals $\max \{ u+g_ a,v+g_ a\} $ on $D\cap B(a,r)$ and $u+g_ a$ elsewhere; every boundary point $x$ of $D\cap B(a,r)$ in $B(a,r)$ lies in $\partial D$, and $\varlimsup _{y\to x}(v+g_ a)(y)\leq u(x)+g_ a(x)$ because $g_ a$ is continuous, so proposition 5.9.5 applies on $B(a,r)$. Hence $w$ is quasi-plurisubharmonic on $B(a,r)$ and $\omega +\mathrm{dd}^{\mathrm{c}}w\geq \omega -\gamma _ a\geq -\varepsilon \beta $ there. The balls $B(a,r)$, $a\in L$, cover a neighborhood of $L$, and positivity of currents is a local property, so $\omega +\mathrm{dd}^{\mathrm{c}}w\geq -\varepsilon \beta $ on a neighborhood of $L$. Letting $\varepsilon \downarrow 0$ gives $\omega +\mathrm{dd}^{\mathrm{c}}w\geq 0$ on the interior of $L$, and since the chart and $L$ were arbitrary, $w\in \mathrm{PSH}(X,\omega )$.

(2) We first prove positivity on a coordinate ball $B$ whose closure lies in a chart, choosing $q=C|z|^2$ in (10.9.2). Let $B'\Subset B$ be a smaller concentric ball, let $\rho _ k$ be radial mollifiers whose radii decrease to zero and are smaller than the distance from $B'$ to $\partial B$, and put $u_{l,k}=(u_ l+q)*\rho _ k-q$ on $B'$. Since $\mathrm{dd}^{\mathrm{c}}q$ has constant coefficients and $\int \rho _ k=1$, we have $\mathrm{dd}^{\mathrm{c}}u_{l,k}=(\mathrm{dd}^{\mathrm{c}}u_ l)*\rho _ k$, where the convolution acts on the coefficients, and $(\omega +\mathrm{dd}^{\mathrm{c}}u_ l)*\rho _ k\geq 0$ because mollification preserves positivity (section 5.3). By uniform continuity of the coefficients of $\omega $, there are $\tau _ k\to 0$ with $\omega *\rho _ k\leq \omega +\tau _ k\beta $ on $B'$. Hence the smooth $(1,1)$-forms $\omega _{u_{l,k}}+\tau _ k\beta =(\omega +\mathrm{dd}^{\mathrm{c}}u_ l)*\rho _ k+(\omega +\tau _ k\beta -\omega *\rho _ k)$ are positive, and so is their exterior product, by proposition 5.3.2. Expanding this product of the forms $\mathrm{dd}^{\mathrm{c}}((u_ l+q)*\rho _ k)+(\sigma +\tau _ k\beta )$, we obtain

As $k\to \infty $, the smooth plurisubharmonic functions $(u_ l+q)*\rho _ k$ decrease to $u_ l+q$ by proposition 5.9.2(2), so the Bedford–Taylor products converge weakly by theorem 7.3.7; their masses are bounded on compact subsets of $B'$ uniformly in $k$ by theorem 7.2.5, and the coefficients of $(\sigma +\tau _ k\beta )^{p-|J|}$ converge uniformly to those of $\sigma ^{p-|J|}$. Hence each term converges weakly to the corresponding term of (10.9.2), the product is a weak limit of positive forms on $B'$, and it is positive. Since such balls $B'$ cover $X$, the product is positive on $X$.

Cover $X$ by finitely many such balls $B'$, with $q$ defined on the larger balls $B$. On compact subsets of $B$, theorem 7.2.5 bounds the trace masses of the Bedford–Taylor products in (10.9.2) by a constant times a product of the numbers $\sup _ B|u_ l+q|$, and theorem 5.3.7 bounds the total variations of their coefficient measures in the same way. Since $\sigma $ and $\omega $ are smooth, the measure $(\omega _{u_1}\wedge \cdots \wedge \omega _{u_ p})\wedge \omega ^{n-p}$ is a sum of the coefficient measures of these Bedford–Taylor products, and of a smooth density for $J=\varnothing $, multiplied by bounded smooth functions. Summing over the finite cover gives the asserted bound for the total mass of the trace measure. By corollary 5.47.11(3), every pluripolar set has outer measure zero for the total variations of the coefficient measures of Bedford–Taylor products of locally bounded plurisubharmonic functions; it is also Lebesgue null (remarks following definition 5.47.2). The trace measure is a positive measure bounded on each $B'$ by a finite sum of such total variations and of a multiple of Lebesgue measure, so every pluripolar set has outer measure zero for it.

Finally, let $\mathrm{dd}^{\mathrm{c}}g\geq \omega $ on $D$. We may take $q=g$ in (10.9.2) on $D$. Put $A=\mathrm{dd}^{\mathrm{c}}(u+g)$ and $\gamma =\mathrm{dd}^{\mathrm{c}}g-\omega \geq 0$, so that $\sigma =-\gamma $ and (10.9.2) reads $\omega _ u^ k=\sum _{i=0}^ k\binom ki A^ i\wedge (-\gamma )^{k-i}$, where $A^ i$ is a Bedford–Taylor power. Using $\binom nk\binom ki=\binom ni\binom {n-i}{k-i}$ and $\sum _{m=0}^{n-i}\binom {n-i}m(-1)^ m=0$ for $i<n$, we obtain

The currents $\omega _ u^ k$ are positive and the forms $\gamma ^{n-k}$ are strongly positive, so every term on the left is a positive measure, and $(\mathrm{dd}^{\mathrm{c}}(u+g))^ n=A^ n\geq \omega _ u^ n$.

(3) On a coordinate ball, add a function $q$ as in (10.9.2). In the decreasing case, $u_ l$ is bounded, the functions $u_{l,j}+q$ are plurisubharmonic and decrease to $u_ l+q$, which is bounded below, so $u_ l+q$ is plurisubharmonic and $u_{l,j}+q\to u_ l+q$ in $L^1_{\mathrm{loc}}$ by theorem 5.10.1. Hence $u_ l$ is quasi-plurisubharmonic, and $\omega +\mathrm{dd}^{\mathrm{c}}u_ l$ is the weak limit of the positive currents $\omega +\mathrm{dd}^{\mathrm{c}}u_{l,j}$, so $u_ l\in \mathrm{PSH}(X,\omega )$. The Bedford–Taylor products of the functions $u_{l,j}+q$ converge weakly by theorem 7.3.7. In the increasing case, $u_ l\in \mathrm{PSH}(X,\omega )$ by (1), and $u_ l$ is bounded. The plurisubharmonic functions $u_{l,j}+q$ increase, and $u_ l+q$ is the upper semicontinuous regularization of their limit, because $q$ is continuous; so $u_{l,j}+q\to u_ l+q$ almost everywhere by theorem 5.10.2, and theorem 7.6.1, applied on the ball, gives the weak convergence of the Bedford–Taylor products. In both cases we multiply by the fixed smooth forms $\sigma ^{p-|J|}$ and sum; a partition of unity subordinate to finitely many balls gives weak convergence on $X$.

(4) Choose finitely many coordinate balls $B_ k$, with closures in charts, that cover $X$, concentric balls $B'_ k\Subset B_ k$ that still cover $X$, and smooth functions $q_ k$ on neighborhoods of $\overline{B_ k}$ with $\mathrm{dd}^{\mathrm{c}}q_ k\geq \omega $. For every $u\in \mathrm{PSH}(X,\omega )$ the functions $u+q_ k$ are plurisubharmonic on $B_ k$. Two quasi-plurisubharmonic functions that agree almost everywhere on an open set agree everywhere on it, by lemma 6.14.4(1). We may therefore regard $\mathrm{PSH}(X,\omega )$ as a subset of $L^1(X)$; on compact subsets of a chart, $\omega ^ n$ is bounded above and below by positive multiples of Lebesgue measure.

Continuity of the supremum. Let $u_ j,u\in \mathrm{PSH}(X,\omega )$ with $u_ j\to u$ in $L^1(X)$; then $u_ j+q_ k\to u+q_ k$ in $L^1_{\mathrm{loc}}(B_ k)$ for every $k$. Choose points $x_ j$ with $u_ j(x_ j)=\sup _ Xu_ j$, which exist because $u_ j$ is upper semicontinuous on the compact manifold $X$, and a subsequence along which $\sup _ Xu_ j\to \varlimsup _ j\sup _ Xu_ j$ and $x_ j\to x$. Choose $k$ with $x\in B_ k$ and a real number $c>u(x)$, and a closed ball $K\subset B_ k$ centered at $x$ on which $u<c$. Since $c+q_ k$ is continuous and $u+q_ k<c+q_ k$ on $K$, the second part of theorem 5.10.4 gives $u_ j+q_ k<c+q_ k$ on $K$ for large $j$; hence $u_ j(x_ j)<c$ for large $j$. Letting $c\downarrow u(x)$ we obtain $\varlimsup _ j\sup _ Xu_ j\leq u(x)\leq \sup _ Xu$. Conversely, choose a subsequence along which $\sup _ Xu_ j$ tends to $s=\varliminf _ j\sup _ Xu_ j$ and $u_ j\to u$ almost everywhere. Then $u\leq s$ almost everywhere, so $s>-\infty $ because $u$ is integrable, and $u\leq s$ everywhere by lemma 6.14.4(1), applied to $u$ and the constant $s$. Thus $\sup _ Xu\leq \varliminf _ j\sup _ Xu_ j$, and the supremum is continuous.

Compactness. Let $(u_ j)$ be a sequence in $\mathrm{PSH}(X,\omega )$ with $\sup _ Xu_ j=0$. On each $B_ k$ the plurisubharmonic functions $u_ j+q_ k$ are bounded above by $\sup _{B_ k}q_ k$. Choose $x_ j$ with $u_ j(x_ j)=0$, pass to a subsequence with $x_ j\to x$, and choose $k_0$ with $x\in B'_{k_0}$. On $B_{k_0}$ the functions $u_ j+q_{k_0}$ do not converge locally uniformly to $-\infty $, since for large $j$ their values at the points $x_ j\in \overline{B'_{k_0}}$ are at least $\min _{\overline{B'_{k_0}}}q_{k_0}$. By theorem 5.10.4, a subsequence converges in $L^1_{\mathrm{loc}}(B_{k_0})$.

If $B_ l$ meets a ball $B_ k$ on which $(u_ j+q_ k)$ converges in $L^1_{\mathrm{loc}}(B_ k)$, then on the nonempty open set $B_ k\cap B_ l$ the functions $u_ j+q_ l=(u_ j+q_ k)+(q_ l-q_ k)$ converge in $L^1_{\mathrm{loc}}$ to a locally integrable function, so they do not tend to $-\infty $ locally uniformly on $B_ l$, and a further subsequence converges in $L^1_{\mathrm{loc}}(B_ l)$. Since $X$ is connected, every ball is joined to $B_{k_0}$ by a finite chain of balls, each meeting the next, so after finitely many extractions $(u_ j+q_ k)$ converges in $L^1_{\mathrm{loc}}(B_ k)$ to a plurisubharmonic function $v_ k$ for every $k$ (theorem 5.10.4). Then $(u_ j)$ converges in $L^1(B'_ k)$ for every $k$, hence in $L^1(X)$.

It remains to identify the limit. The quasi-plurisubharmonic functions $v_ k-q_ k$ and $v_ l-q_ l$ agree almost everywhere on $B_ k\cap B_ l$, hence everywhere, so they define a quasi-plurisubharmonic function $u$ on $X$, and $\omega +\mathrm{dd}^{\mathrm{c}}u\geq 0$ as a distributional limit of the positive currents $\omega +\mathrm{dd}^{\mathrm{c}}u_ j$. Thus $u\in \mathrm{PSH}(X,\omega )$, $u_ j\to u$ in $L^1(X)$, and $\sup _ Xu=\lim _ j\sup _ Xu_ j=0$ by the continuity just proved. The set of normalized functions is therefore sequentially compact, hence compact, in the metric space $L^1(X)$, and the continuous function $u\mapsto \| u\| _{L^1(X)}$ is bounded on it. $\square$

Proof. After computing $\mathrm{dd}^{\mathrm{c}}T$ (Step 1), we prove a version of (10.9.5) in which $\mathbf1_ E$ is smoothed by a convex cutoff, integrating by parts against regularizations of $u$ and $v$ (Step 2). The products $T$, $T_{\hat\imath }$ and $T_{\hat\imath \hat\imath '}$ are positive currents: for products with at least one factor this is lemma 10.9.3(2), and the empty product is the constant $1$. (If $n=1$, then $T=1$, $\mathrm{dd}^{\mathrm{c}}T=0$, and both sums in (10.9.6) are empty; the sum over pairs is empty also if $n=2$.)

Step 1. The measure $\mathrm{dd}^{\mathrm{c}}T$. Work on an open subset of a chart with $q$ and $\sigma $ as in (10.9.2). Each factor of $T$ has the form $\Theta _ i+\sigma $, where $\Theta _ i$ is $\mathrm{dd}^{\mathrm{c}}q$, $\mathrm{dd}^{\mathrm{c}}(u+q)$ or $\mathrm{dd}^{\mathrm{c}}(v+q)$; thus $T=\sum _ J\Theta _ J\wedge \sigma ^{n-1-|J|}$, where $\Theta _ J$ is the Bedford–Taylor product of the $\Theta _ i$ with $i\in J$. The currents $\Theta _ J$ are closed of pure bidegree, so $\partial \Theta _ J=\bar\partial \Theta _ J=0$, and hence $\mathrm d\Theta _ J=\mathrm d^{\mathrm c}\Theta _ J=0$, where $\mathrm d^{\mathrm c}$ is the operator (5.2.10). By the Leibniz rule for currents (section 5.3), $\mathrm{dd}^{\mathrm{c}}(\Theta _ J\wedge \sigma ^ m)=\Theta _ J\wedge \mathrm{dd}^{\mathrm{c}}(\sigma ^ m)$. Since the closed $(1,1)$-form $\mathrm{dd}^{\mathrm{c}}q$ satisfies $\partial \mathrm{dd}^{\mathrm{c}}q=\bar\partial \mathrm{dd}^{\mathrm{c}}q=0$, we have $\mathrm d\sigma =\mathrm d\omega $, $\mathrm d^{\mathrm c}\sigma =\mathrm d^{\mathrm c}\omega $ and $\mathrm{dd}^{\mathrm{c}}\sigma =\mathrm{dd}^{\mathrm{c}}\omega $, and the Leibniz rule gives

Expand $T_{\hat\imath }$ and $T_{\hat\imath \hat\imath '}$ in the same way. For a fixed $J$, with $m=n-1-|J|$, there are $m$ indices $i\notin J$ and $m(m-1)/2$ pairs $i<i'$ outside $J$; comparing coefficients, we obtain

In particular $\mathrm dT=\sum _ J\Theta _ J\wedge \mathrm d(\sigma ^{n-1-|J|})$, $\mathrm d^{\mathrm c}T$ and $\mathrm{dd}^{\mathrm{c}}T$ are currents of order zero, being sums of Bedford–Taylor products multiplied by smooth forms. The form $\mathrm{dd}^{\mathrm{c}}\omega $ is real of bidegree $(2,2)$. Since $\partial \omega $ and $\bar\partial \omega $ have odd degree, $\partial \omega \wedge \partial \omega =\bar\partial \omega \wedge \bar\partial \omega =0$ and $\bar\partial \omega \wedge \partial \omega =-\partial \omega \wedge \bar\partial \omega $, so $\mathrm d\omega \wedge \mathrm d^{\mathrm c}\omega =\frac{\mathrm i}{2\pi }\partial \omega \wedge \bar\partial \omega $ is real of bidegree $(3,3)$; these forms vanish if $n<2$ and $n<3$, respectively.

We claim that there is $C$ such that $C\omega ^2\pm \mathrm{dd}^{\mathrm{c}}\omega $ and $C\omega ^3\pm \mathrm d\omega \wedge \mathrm d^{\mathrm c}\omega $ are strongly positive at every point of $X$. Consider $\omega ^2$ and $\mathrm{dd}^{\mathrm{c}}\omega $, with $n\geq 2$; the other case, with $n\geq 3$, is the same. Let $D$ be a coordinate ball whose closure lies in a chart, and let $\mathcal C$ be the cone of strongly positive $(2,2)$-forms on $\mathbb C^ n$. For $x\in \overline D$, a complex linear change of coordinates takes $\omega (x)$ to $\beta $; since such changes preserve strong positivity in both directions, they map the interior of $\mathcal C$ onto itself, and $\beta ^2$ is an interior point of $\mathcal C$ (proposition 5.3.2). Hence $\omega (x)^2$ lies in the interior of $\mathcal C$. As $x\mapsto \omega (x)^2$ is continuous, these forms make up a compact subset of the interior of $\mathcal C$, which has a positive distance $d_0$ from the closed complement of the interior, for a fixed norm on the real $(2,2)$-forms. The coefficients of $\mathrm{dd}^{\mathrm{c}}\omega $ are bounded on $\overline D$, so for $C$ large the forms $\omega (x)^2\pm C^{-1}\mathrm{dd}^{\mathrm{c}}\omega (x)$ lie at distance less than $d_0$ from $\omega (x)^2$, hence in $\mathcal C$, for all $x\in \overline D$. Covering the compact manifold $X$ by finitely many such balls proves the claim.

The currents $T_{\hat\imath }$ and $T_{\hat\imath \hat\imath '}$ are positive, so $-CT_{\hat\imath }\wedge \omega ^2\leq \mathrm{dd}^{\mathrm{c}}\omega \wedge T_{\hat\imath }\leq CT_{\hat\imath }\wedge \omega ^2$, and similarly for the terms with $\mathrm d\omega \wedge \mathrm d^{\mathrm c}\omega $. A real measure $\nu $ with $-\mu \leq \nu \leq \mu $ for a positive measure $\mu $ satisfies $|\nu |\leq \mu $, so (10.9.6), with the constant $2C$, follows from (10.9.7).

Step 2. A convex cutoff. Put $w=v-u$, a bounded Borel function on $X$. Fix a smooth increasing function $\lambda \colon \mathbb R\to [0,1]$ with $\lambda =0$ on $(-\infty ,0]$ and $\lambda =1$ on $[1,\infty )$, and for $\delta >0$ put $F_\delta (t)=\int _0^ t\lambda (s/\delta )\, \mathrm ds$. Then $F_\delta $ is smooth and convex, $F_\delta =0$ on $(-\infty ,0]$, $0\leq F_\delta '\leq 1$, and as $\delta \downarrow 0$ we have $F_\delta (t)\to t^+$ and $F_\delta '(t)\to \mathbf1_{(0,\infty )}(t)$ for every real $t$; indeed $|F_\delta (t)-t|\leq \delta $ for $t>0$. We shall prove that, for every $\delta >0$,

Granting this, let $\delta \downarrow 0$. The integrands converge pointwise and boundedly to $\mathbf1_ E$ and to $(v-u)\mathbf1_ E$; note that $F_\delta '(0)=0$, so the first limit is the indicator of $\{ w>0\} =E$. Both measures have finite total variation, so dominated convergence turns (10.9.8) into (10.9.5).

To prove (10.9.8), choose finitely many coordinate balls $B_ k$ covering $X$, smooth functions $q_ k$ with $\mathrm{dd}^{\mathrm{c}}q_ k\geq \omega $ on a neighborhood of $\overline{B_ k}$, and a smooth partition of unity $(\chi _ k)$ on $X$ with $\operatorname{Supp}\chi _ k\subset B_ k$ (proposition 5.2.6). Fix $k$, write $\sigma _ k=\omega -\mathrm{dd}^{\mathrm{c}}q_ k$, and choose an open set $\Omega _ k$ with $\operatorname{Supp}\chi _ k\subset \Omega _ k\Subset B_ k$. Convolution of $u+q_ k$ and $v+q_ k$ with radial mollifiers whose radii decrease to zero and are smaller than the distance from $\Omega _ k$ to $\partial B_ k$ gives smooth plurisubharmonic functions on $\Omega _ k$, which we write as $u_ l+q_ k$ and $v_ l+q_ k$. They decrease to $u+q_ k$ and $v+q_ k$ as $l\to \infty $, by proposition 5.9.2(2), and they take values in a fixed compact interval $[-M',M']$, because $u+q_ k\leq u_ l+q_ k\leq u_1+q_ k$ and similarly for $v$. Then $w_ l=v_ l-u_ l\to w$ pointwise on $\Omega _ k$, with $|w_ l|\leq 2M'$.

The function $f=F_\delta (w_ l)$ is smooth on $\Omega _ k$, which contains the support of $\chi _ kT$, so the definition of derivatives of currents and the Leibniz rule give

where $\Lambda _ k=\mathrm{dd}^{\mathrm{c}}\chi _ k\wedge T+\mathrm d\chi _ k\wedge \mathrm d^{\mathrm c}T-\mathrm d^{\mathrm c}\chi _ k\wedge \mathrm dT$, and the chain rule gives

The last term is a smooth strongly positive $(1,1)$-form, since $F_\delta ''\geq 0$, and $T$ is positive, so the product of this term with $\chi _ kT$ is a positive measure. Consequently

Now let $l\to \infty $. By Step 1, $\mathrm{dd}^{\mathrm{c}}T$ and $\Lambda _ k$ are measures of finite total variation, so the right side tends to $\int F_\delta (w)(\chi _ k\mathrm{dd}^{\mathrm{c}}T+\Lambda _ k)$ by dominated convergence.

On the left, write $T=\sum _ J\Theta _ J\wedge \sigma _ k^{m_ J}$ as in Step 1, with $q=q_ k$ and $m_ J=n-1-|J|$. The exterior product of the smooth form $\mathrm{dd}^{\mathrm{c}}(v_ l+q_ k)$ with $\Theta _ J$ is the Bedford–Taylor product $\mathrm{dd}^{\mathrm{c}}(v_ l+q_ k)\wedge \Theta _ J$, by (7.2.3), so the left side of (10.9.9) is the sum over $J$ of

Fix $J$ and consider the first integral; the second is treated in the same way. Let $\vartheta >0$. By the Weierstrass approximation theorem there is a real polynomial $P$ of one variable with $|F_\delta '-P|\leq \vartheta $ on $[-2M',2M']$. The weights $P(w_ l)=P\bigl ((v_ l+q_ k)-(u_ l+q_ k)\bigr )$ are polynomials in two decreasing sequences of bounded plurisubharmonic functions on $\Omega _ k$, so corollary 7.3.8, applied on $\Omega _ k$ with the closed positive base current $\Theta _ J$ and the decreasing factors $v_ l+q_ k$, gives the weak convergence of $P(w_ l)\, \mathrm{dd}^{\mathrm{c}}(v_ l+q_ k)\wedge \Theta _ J$ to $P(w)\, \mathrm{dd}^{\mathrm{c}}(v+q_ k)\wedge \Theta _ J$, which we test with the smooth form $\chi _ k\sigma _ k^{m_ J}$ with compact support in $\Omega _ k$. On the other hand, by theorem 7.2.5 and theorem 5.3.7, the total variations on $\operatorname{Supp}\chi _ k$ of the coefficient measures of the positive currents $\mathrm{dd}^{\mathrm{c}}(v_ l+q_ k)\wedge \Theta _ J$ and $\mathrm{dd}^{\mathrm{c}}(v+q_ k)\wedge \Theta _ J$ are bounded uniformly in $l$, so replacing $F_\delta '(w_ l)$ by $P(w_ l)$, or $F_\delta '(w)$ by $P(w)$, changes the integrals by at most a constant times $\vartheta $, with a constant independent of $l$. As $\vartheta $ was arbitrary,

By (10.9.2), $\omega _ v\wedge T=\sum _ J\mathrm{dd}^{\mathrm{c}}(v+q_ k)\wedge \Theta _ J\wedge \sigma _ k^{m_ J}+\sigma _ k\wedge T$ on $B_ k$, and similarly for $u$. Summing over $J$, we conclude that the left side of (10.9.9) tends to $\int \chi _ kF_\delta '(w)\, (\omega _ v-\omega _ u)\wedge T$. Finally we sum the limiting inequalities over $k$. Since $\sum _ k\chi _ k=1$, we have $\sum _ k\Lambda _ k=\mathrm{dd}^{\mathrm{c}}T-\sum _ k\chi _ k\mathrm{dd}^{\mathrm{c}}T=0$, and (10.9.8) follows. $\square$

Proof. We replace the factors $\omega _ v$ of $\omega _ v^ n$ by $\omega _ u$ one at a time, using lemma 10.9.4; the errors are traces of products with fewer factors different from $\omega $, and Steps 1 and 2 bound them by multiples of $\int _ E\omega _ u^ n$. For a product $T$ of $n-1$ factors chosen among $\omega ,\omega _ u,\omega _ v$, let $\mathcal P(T)$ be the set of products obtained from $T$ by deleting one factor and inserting $\omega ^2$, or by deleting two factors and inserting $\omega ^3$. The right side of (10.9.6) has at most $n^2$ terms, each of which is an element of $\mathcal P(T)$, and $0<v-u\leq s$ on $E$. Hence lemma 10.9.4 gives

with a constant $B\geq 1$ depending only on $(X,\omega )$.

Step 1. Pure traces. For $0\leq k\leq n$ put $a_ k=\int _ E\omega _ u^ k\wedge \omega ^{n-k}\geq 0$, and apply (10.9.11) to $T=\omega _ u^ k\wedge \omega ^{n-k-1}$, $0\leq k<n$. The function $2v$ belongs to $\mathrm{PSH}(X,\omega )$, because $\omega +\mathrm{dd}^{\mathrm{c}}(2v)=2\omega _ v-\omega \geq 0$, and it is bounded. Since the products are affine in each potential and $v=\frac12\cdot 0+\frac12\cdot 2v$, we have $\omega _ v\wedge T=\frac12\omega \wedge T+\frac12\omega _{2v}\wedge T$, and $\omega _{2v}\wedge T$ is positive by lemma 10.9.3(2); hence $\int _ E\omega _ v\wedge T\geq \tfrac 12a_ k$. Moreover $\int _ E\omega _ u\wedge T=a_{k+1}$, and the elements of $\mathcal P(T)$ are among the products $\omega _ u^ h\wedge \omega ^{n-h}$ with $h\in \{ k-2,k-1,k\} $. Thus (10.9.11) gives

Choose $s_0>0$ with $s_0B(n+1)4^ n\leq 1/4$; in particular $sB\leq 1$. We prove $a_ k\leq 4a_{k+1}$ for $0\leq k<n$ by induction on $k$. If it holds for all smaller indices, then $a_ h\leq 4^{k-h}a_ k$ for $h\leq k$, so $\sum _{h=0}^ ka_ h\leq (k+1)4^ ka_ k$, and the error term in (10.9.12) is at most $s_0B(n+1)4^ na_ k\leq \tfrac 14a_ k$. Hence $\tfrac 12a_ k\leq a_{k+1}+\tfrac 14a_ k$, that is, $a_ k\leq 4a_{k+1}$. Consequently

Step 2. Mixed traces. For $j,k\geq 0$ with $j+k\leq n$ put $a_{j,k}=\int _ E\omega _ u^ j\wedge \omega _ v^ k\wedge \omega ^{n-j-k}\geq 0$, so that $a_{j,0}=a_ j$. If $k\geq 1$, apply (10.9.11) to $T=\omega _ u^ j\wedge \omega _ v^{k-1}\wedge \omega ^{n-j-k}$. Then $\int _ E\omega _ v\wedge T=a_{j,k}$ and $\int _ E\omega _ u\wedge T=a_{j+1,k-1}$, while every element of $\mathcal P(T)$ has fewer than $j+k$ factors different from $\omega $. Thus

We show by induction on $m=j+k$ that $a_{j,k}\leq C_ m\sum _{h=0}^ ma_ h$ with $C_ m$ depending only on $n$. For $k=0$ this holds with $C_ m=1$. For $k\geq 1$, since $sB\leq 1$, applying the last inequality $k$ times, which replaces the factors $\omega _ v$ one at a time by $\omega _ u$, gives

and by the induction hypothesis the last sum, which has at most $(m+1)^2$ terms, is at most a constant depending only on $n$ times $\sum _{h<m}a_ h$. Together with Step 1 this shows $a_{j,k}\leq C'_ na_ n$ for all $j,k$, where $C'_ n$ depends only on $n$.

Step 3. Conclusion. By symmetry of the products,

since the sum telescopes. Apply (10.9.11) to each of the $n$ products $T=\omega _ v^ i\wedge \omega _ u^{n-1-i}$. The resulting error terms are $sB$ times sums of at most $n^2$ traces $a_{p,q}$, hence at most a constant times $sa_ n$ by Step 2. Since $a_ n=\int _ E\omega _ u^ n$, this proves $\int _ E\omega _ v^ n\leq (1+Cs)\int _ E\omega _ u^ n$ with $C$ depending only on $(X,\omega )$.

Step 4. Positivity. Let $u\in \mathrm{PSH}(X,\omega )$ be bounded, put $c=\inf _ Xu$, fix $s\in (0,s_0)$, and apply the inequality to the constant function $v=c+s$. Then $\omega _ v=\omega \geq \tfrac 12\omega $, and $E=\{ u<c+s\} $ is nonempty by the definition of the infimum and open because $u$ is upper semicontinuous; moreover $0<v-u\leq s$ on $E$ since $u\geq c$. Hence $0<\int _ E\omega ^ n\leq (1+Cs)\int _ E\omega _ u^ n$, and therefore $\int _ X\omega _ u^ n\geq \int _ E\omega _ u^ n>0$. $\square$

Proof. Let $\mathcal F_ A$ be the family of competitors for $T_ A$. It contains the constant $0$, so $T_ A\geq 0$.

We first show that if $\psi _ j\in \mathrm{PSH}(X,\omega )$ and $\sup _ X\psi _ j=0$ for $j\geq 1$, then $\Psi =\sum _{j\geq 1}2^{-j}\psi _ j$ belongs to $\mathrm{PSH}(X,\omega )$. By lemma 10.9.3(4), there is a constant $M$ with $\| \psi _ j\| _{L^1(X)}\leq M$ for all $j$. The partial sums $\Psi _ N=\sum _{j=1}^ N2^{-j}\psi _ j$ decrease with $N$, since $\psi _ j\leq 0$, and $\mathrm{dd}^{\mathrm{c}}\Psi _ N\geq -\sum _ j2^{-j}\omega \geq -\omega $. On a coordinate ball with a function $q$ as in (10.9.2), the functions $\Psi _ N+q$ are plurisubharmonic and decrease, so by theorem 5.10.1 their limit $\Psi +q$ is plurisubharmonic or identically $-\infty $ on the ball. By monotone convergence $\Psi _ N\to \Psi $ in $L^1(X)$ and $\int _ X|\Psi |\, \omega ^ n\leq \sum _ j2^{-j}M\leq M$, so $\Psi $ is finite almost everywhere and $\Psi +q$ is not identically $-\infty $ on any ball. Hence $\Psi $ is quasi-plurisubharmonic, and $\Psi \in \mathrm{PSH}(X,\omega )$ because the inequality $\omega +\mathrm{dd}^{\mathrm{c}}\Psi _ N\geq 0$ passes to the limit.

Suppose now that $\sup \{ \sup _ Xu:u\in \mathcal F_ A\} =\infty $. Choose $u_ j\in \mathcal F_ A$ with $b_ j=\sup _ Xu_ j\geq 2^ j$ for $j\geq 1$, and put $\psi _ j=u_ j-b_ j$. Then $\psi _ j\in \mathrm{PSH}(X,\omega )$, $\sup _ X\psi _ j=0$ and $\psi _ j\leq -2^ j$ on $A$, so $\Psi =\sum _ j2^{-j}\psi _ j$ belongs to $\mathrm{PSH}(X,\omega )$. Moreover $\Psi =-\infty $ on $A$, because every term $2^{-j}\psi _ j$ is at most $-1$ there and all terms are nonpositive. Thus $A$ is globally pluripolar.

Otherwise $T_ A\leq C$ for a constant $C$, and by lemma 10.9.3(1) the regularization $T_ A^*$ belongs to $\mathrm{PSH}(X,\omega )$ and satisfies $0\leq T_ A^*\leq C$. Thus at least one of the alternatives holds. They exclude each other: if $A\subset \{ \Phi =-\infty \} $ for a quasi-plurisubharmonic $\Phi $, choose $t>0$ with $t\Phi \in \mathrm{PSH}(X,\omega )$; then $t\Phi +c\in \mathcal F_ A$, hence $T_ A\geq t\Phi +c$, for every real $c$, so $T_ A^*$ is not bounded.

A quasi-plurisubharmonic function is locally integrable (lemma 6.14.4(1)), so its pole set has Lebesgue measure zero in every chart and contains no nonempty open set. Therefore a set with nonempty interior is not globally pluripolar.

Finally let $A_ j\subset \{ \varphi _ j=-\infty \} $, $j\geq 1$, with $\varphi _ j$ quasi-plurisubharmonic. After multiplying by small positive constants and subtracting constants, we may assume that $\varphi _ j\in \mathrm{PSH}(X,\omega )$ and $\sup _ X\varphi _ j=0$. By the claim proved at the start, $\sum _ j2^{-j}\varphi _ j$ belongs to $\mathrm{PSH}(X,\omega )$; it equals $-\infty $ on every $A_ j$, because one term is $-\infty $ there and all terms are nonpositive. (A finite union is the case in which $A_ j=\varnothing $ for all large $j$.) $\square$

Proof. Put $\rho _ i=r/2+ir/(2n)$ and $B_ i=B(a,\rho _ i)$ for $0\leq i\leq n$, so that $B_0=B(a,r/2)$ and $B_ n=B(a,r)$. For $0\leq i<n$ fix a smooth function $\eta _ i\colon \mathbb R\to [0,1]$ with $\eta _ i=1$ on $(-\infty ,(\rho _ i/r)^2]$ and $\eta _ i=0$ on $[((\rho _ i+\rho _{i+1})/(2r))^2,\infty )$; the numbers $\rho _ i/r$ depend only on $n$ and $i$, so $\eta _ i$ can be chosen depending only on $n$ and $i$. By (10.9.14), the functions $\chi _ i(z)=\eta _ i(|z-a|^2/r^2)$ satisfy $\chi _ i\in C^\infty _ c(B_{i+1})$, $0\leq \chi _ i\leq 1$, $\chi _ i=1$ on $B_ i$ and $-C'r^{-2}\beta \leq \mathrm{dd}^{\mathrm{c}}\chi _ i\leq C'r^{-2}\beta $, where $C'=\max _ iC_{\eta _ i}$ depends only on $n$. For integers $j,m\geq 0$ with $j+m\leq n$ and $0\leq i\leq n$, put

Let $m\geq 1$, $i<n$, and let $R=A^ j\wedge D^{m-1}\wedge \beta ^{n-j-m}$, a closed positive current by proposition 7.2.2. By the definition and the symmetry of Bedford–Taylor products, $D\wedge R=A^ j\wedge D^ m\wedge \beta ^{n-j-m}$ and $D\wedge R-A\wedge R=\mathrm{dd}^{\mathrm{c}}((V-U)R)$, so

Since $R$ is positive, the forms $C'r^{-2}\beta \pm \mathrm{dd}^{\mathrm{c}}\chi _ i$ give positive measures $R\wedge (C'r^{-2}\beta \pm \mathrm{dd}^{\mathrm{c}}\chi _ i)$; as $|V-U|\leq \kappa r^2$ and $\operatorname{Supp}\chi _ i\subset B_{i+1}$, the last integral is at most $C'\kappa \int _{B_{i+1}}R\wedge \beta $ in absolute value. Using $\mathbf1_{B_ i}\leq \chi _ i\leq \mathbf1_{B_{i+1}}$ and the positivity of the currents involved, we obtain

Both terms on the right have one factor $D$ fewer. Applying this inequality $m$ times, starting from $i=0$, we find

Take $m=k-j$ and sum over $0\leq j\leq k$. With $l=j+t$ we have $m-t=k-l$, so every term with $A^ l$ carries the factor $\kappa ^{k-l}$, and the coefficients depend only on $n$. This proves the lemma; for $l=k$ the factor is $\kappa ^0=1$. $\square$

Proof. Steps 1–4 show that $\omega _ u^ n$ vanishes on $X\setminus K$: on small balls outside $K$ we compare $u$ with a homogeneous balayage computed with the coefficients of $\omega $ frozen. Step 5 proves the mass bound.

Step 1. Increasing competitors. The competitors for $T_ K$ are bounded above by $\sup _ Xu$, and $u\in \mathrm{PSH}(X,\omega )$ by lemma 10.9.3(1). Cover $X$ by finitely many coordinate charts. Applying Choquet's lemma (lemma 3.4.22) in each chart, we obtain for each chart a countable family of competitors whose supremum has, on that chart, the same upper semicontinuous regularization as $T_ K$. Let $v_1,v_2,\ldots $ be an enumeration of the union of these families. On each chart, $\sup _ kv_ k$ lies between the supremum of the family chosen for that chart and $T_ K$; since the regularization is computed locally, $(\sup _ kv_ k)^*=u$. The functions $u_ j=\max \{ 0,v_1,\ldots ,v_ j\} $ belong to $\mathrm{PSH}(X,\omega )$ by lemma 10.9.3(1), increase with $j$, vanish on $K$, and are competitors; hence $0\leq u_ j\leq T_ K\leq u$, and $(\sup _ ju_ j)^*=u$, because $\sup _ kv_ k\leq \sup _ ju_ j\leq T_ K$.

Step 2. Frozen balayage. Let $G\subset X\setminus K$ be the domain of a coordinate chart, identified with an open subset of $\mathbb C^ n$, and let $L\subset G$ be compact. Since $X\setminus K$ is open in the compact manifold $X$, it is a countable union of such sets $L$, so it suffices to show that $\omega _ u^ n(L)=0$. Choose open sets $G_2\Subset G_1\Subset G_0\Subset G$ with $L\subset G_2$. For $a\in G_2$ let $h_ a$ be the real quadratic polynomial with $\mathrm{dd}^{\mathrm{c}}h_ a=\omega (a)$ described at the beginning of this section. Choose $r_0>0$ with $2r_0<\operatorname {dist}(G_2,\mathbb C^ n\setminus G_1)$. The coefficients of $\omega $ have bounded first derivatives on $G_1$, so by the mean value inequality on balls there is $C_1$ such that, for $a\in G_2$ and $0<r\leq r_0$, the ball $B(a,2r)$ lies in $G_1$ and

indeed a Hermitian matrix whose entries are bounded by $\tau $ lies between $-n\tau $ and $n\tau $ times the identity.

Fix such $a$ and $r$, put $q=\pi |z-a|^2$, so that $\mathrm{dd}^{\mathrm{c}}q=\beta $ by (5.2.9), and let $B=B(a,r)$. The functions

are bounded and plurisubharmonic on $B(a,2r)$, since they are quasi-plurisubharmonic and $\mathrm{dd}^{\mathrm{c}}U_ j=\omega _{u_ j}+(\omega (a)-\omega +\delta \beta )\geq 0$. Let $V_ j$ be the homogeneous balayage of $U_ j$ in $B$ given by corollary 7.7.5, applied on $B(a,2r)$: it equals $U_ j$ outside $B$, satisfies $V_ j\geq U_ j$ and $(\mathrm{dd}^{\mathrm{c}}V_ j)^ n=0$ in $B$, and its restriction to $B$, being a member of the family described there, has upper boundary limits at most $U_ j$ on $\partial B$.

We compare $V_ j$ with $u$. Put $\rho =q-\pi r^2$, which is nonpositive on $B$ and vanishes on $\partial B$, and let $Z_ j=V_ j-h_ a-\delta q+2\delta \rho $ on $B$. Then $Z_ j$ is quasi-plurisubharmonic and

so $Z_ j\in \mathrm{PSH}(B,\omega )$, and for $x\in \partial B$ we have $\varlimsup _{B\ni y\to x}Z_ j(y)\leq U_ j(x)-h_ a(x)-\delta q(x)=u_ j(x)$. By lemma 10.9.3(1), the function equal to $\max \{ u_ j,Z_ j\} $ on $B$ and to $u_ j$ on $X\setminus B$ belongs to $\mathrm{PSH}(X,\omega )$. It vanishes on $K$, because $B\cap K=\varnothing $, so it is a competitor for $T_ K$; hence $\max \{ u_ j,Z_ j\} \leq T_ K\leq u$ on $B$. Therefore

The balayage is increasing in the data, so $V_ j$ increases with $j$. Let $V$ be the upper semicontinuous regularization of $\lim _ jV_ j$ on $B$. Since $U_1\leq V_ j\leq u+h_ a+\delta q+2\pi \delta r^2$, the $V_ j$ are uniformly bounded, $V$ is bounded and plurisubharmonic, and $V_ j\to V$ almost everywhere by theorem 5.10.2; hence $(\mathrm{dd}^{\mathrm{c}}V)^ n=0$ on $B$ by theorem 7.6.1. Put $U=u+h_ a+\delta q$. Since $\sup _ jU_ j=\sup _ ju_ j+h_ a+\delta q$, $h_ a+\delta q$ is continuous and $(\sup _ ju_ j)^*=u$, upper regularization gives $V\geq (\sup _ jU_ j)^*=U$. The upper bound for $V_ j$ is upper semicontinuous, so it also holds for $V$. As $-2\delta \rho \leq 2\pi \delta r^2$ on $B$, we conclude

Step 3. A local mass estimate. Put $A=\mathrm{dd}^{\mathrm{c}}U$ and $D=\mathrm{dd}^{\mathrm{c}}V$ on $B$. The function $g=h_ a+\delta q$ satisfies $\mathrm{dd}^{\mathrm{c}}g=\omega (a)+\delta \beta \geq \omega $ on $B(a,2r)$, so lemma 10.9.3(2) gives $A^ n=(\mathrm{dd}^{\mathrm{c}}(u+g))^ n\geq \omega _ u^ n$ on $B$. By (10.9.14), with a fixed function $\eta $ equal to $1$ on $(-\infty ,1/16]$ and to $0$ on $[1/5,\infty )$, there is $\chi \in C^\infty _ c(B(a,r/2))$ with $0\leq \chi \leq 1$, $\chi =1$ on $B(a,r/4)$ and $-C_2r^{-2}\beta \leq \mathrm{dd}^{\mathrm{c}}\chi \leq C_2r^{-2}\beta $, where $C_2$ does not depend on $a$ and $r$. By the definition and the symmetry of Bedford–Taylor products, $A^ n-D^ n=\sum _{j=0}^{n-1}(A-D)\wedge A^ j\wedge D^{n-1-j}=\mathrm{dd}^{\mathrm{c}}\bigl ((U-V)\sum _{j=0}^{n-1}A^ j\wedge D^{n-1-j}\bigr )$. Using $D^ n=0$, integration by parts against $\chi $, the positivity of the currents $A^ j\wedge D^{n-1-j}$ and (10.9.18), we obtain

Apply lemma 10.9.15 with $k=n-1$ and $\kappa =2\pi \delta =2\pi C_1r$; for $r\leq 1/(2\pi C_1)$ we have $\kappa \leq 1$, and the last integral is at most $C_ n\sum _{l=0}^{n-1}\int _{B(a,r)}A^ l\wedge \beta ^{n-l}$.

To sum these estimates we replace $A$ by a current that does not depend on $a$ and $r$. Choose $C_0$ such that $C_0\beta \geq \omega $ on $G_0$ and $C_0\beta \geq \omega (a)+\delta \beta $ for all $a\in G_2$ and $r\leq r_0$. Then $\Gamma =\mathrm{dd}^{\mathrm{c}}u+C_0\beta =\mathrm{dd}^{\mathrm{c}}(u+C_0\pi |z|^2)$, where $u+C_0\pi |z|^2$ is bounded and plurisubharmonic on $G_0$ because $\mathrm{dd}^{\mathrm{c}}(u+C_0\pi |z|^2)=\omega _ u+(C_0\beta -\omega )\geq 0$. On $B$ the difference $\Gamma -A$ is $\mathrm{dd}^{\mathrm{c}}$ of the smooth function $C_0\pi |z|^2-h_ a-\delta q$, and it equals the form $\gamma =C_0\beta -\omega (a)-\delta \beta \geq 0$ with constant coefficients. By (10.9.1), $\Gamma ^ l=\sum _{i=0}^ l\binom liA^ i\wedge \gamma ^{l-i}$, and all terms are positive currents, so $A^ l\wedge \beta ^{n-l}\leq \Gamma ^ l\wedge \beta ^{n-l}$. Altogether, for $a\in G_2$ and $r\leq \min \{ r_0,1/(2\pi C_1)\} $,

with $C_3=2\pi C_1C_2C_ n$ independent of $a$ and $r$.

Step 4. Zero mass off $K$. Let $r$ be as above and also smaller than $4\operatorname {dist}(L,\mathbb C^ n\setminus G_2)$, and consider the cubical grid $h\mathbb Z^{2n}\subset \mathbb C^ n=\mathbb R^{2n}$ of mesh $h=r/(4\sqrt{2n})$. The grid points $a$ within distance $r/4$ of $L$ are finitely many and belong to $G_2$. Every point of $\mathbb C^ n$ lies within distance $\sqrt{2n}\, h/2=r/8$ of a grid point, so the balls $B(a,r/4)$ centered at these grid points cover $L$. The centers of the balls $B(a,r)$ containing a given point $x$ lie in the cube of side $2r$ centered at $x$, which contains at most $(2r/h+1)^{2n}=(8\sqrt{2n}+1)^{2n}=N_ n$ grid points; so each point lies in at most $N_ n$ of these balls. Summing the local estimate over these balls, which lie in $G_1$, gives

The integrals on the right are finite by theorem 7.2.5, applied in $G_0$, and do not depend on $r$. Letting $r\downarrow 0$ gives $\omega _ u^ n(L)=0$. Hence $\omega _ u^ n$ is carried by $K$.

Step 5. The mass bound. Since $T_ K\geq 0$, we have $0\leq u\leq S$. The function $v=u/S$ satisfies $0\leq v\leq 1$ and belongs to $\mathrm{PSH}(X,\omega )$ because $S\geq 1$: indeed $\omega +\mathrm{dd}^{\mathrm{c}}v=S^{-1}\omega _ u+(1-S^{-1})\omega \geq 0$. Since $v=S^{-1}u+(1-S^{-1})\cdot 0$ and the products are affine in each potential, expanding $\omega _ v^ n$ factor by factor gives $\omega _ v^ n=\sum _{k=0}^ n\binom nkS^{-k}(1-S^{-1})^{n-k}\omega _ u^ k\wedge \omega ^{n-k}$, and all terms are positive measures; hence $\omega _ v^ n\geq S^{-n}\omega _ u^ n$. As $\omega _ u^ n$ is carried by $K$,

Finally $\int _ X\omega _ u^ n>0$ by proposition 10.9.10. $\square$

Proof. The compact manifold $X$ has finitely many connected components, which are open, and it suffices to construct $\Phi $ on each of them, since the intersection of $A$ with a component is pluripolar in that component. On a component of dimension zero, which is a point, pluripolar sets are empty and $\Phi =0$ works. We may therefore assume that $X$ is connected of dimension $n\geq 1$, and we fix a Hermitian form $\omega $ on $X$ (one exists by section 5.6), so that the notation of this section applies.

Every point of $X$ has a coordinate ball $U$ and a plurisubharmonic function on $U$ whose pole set contains $A\cap U$, by definition 5.47.2; choose a smaller ball $U_1\Subset U$ around the point. By compactness finitely many of the balls $U_1$ cover $X$. By the last assertion of lemma 10.9.13 it therefore suffices to prove the following: if $U$ is a coordinate ball, $\varphi $ is plurisubharmonic on $U$ and $U_1\Subset U$ is a smaller ball, then $E=\{ \varphi =-\infty \} \cap U_1$ is globally pluripolar.

If $E$ is empty, there is nothing to prove, so let $E\neq \varnothing $. Choose balls $U_1\Subset U_2\Subset U_3\Subset U$ and subtract a constant from $\varphi $ so that $\varphi \leq 0$ on $U_3$. Suppose, to reach a contradiction, that $E$ is not globally pluripolar. By lemma 10.9.13, there is a constant $C$ with $0\leq T_ E^*\leq C$. From the extremal functions of compact sublevel sets of regularizations of $\varphi $, which have small capacity, we shall obtain a bounded function in $\mathrm{PSH}(X,\omega )$ with zero Monge–Ampère mass.

Step 1. Sublevel sets and their extremal functions. Convolution with radial mollifiers of radii decreasing to zero and smaller than the distance from $U_2$ to $\partial U_3$ gives smooth plurisubharmonic functions $\varphi _ j$ on $U_2$ that decrease to $\varphi $ as $j\to \infty $, by proposition 5.9.2(2). Since $\varphi \leq 0$ on $U_3$, we have $\varphi \leq \varphi _ j\leq 0$ on $U_2$, so $\| \varphi _ j\| _{L^1(U_2)}\leq \| \varphi \| _{L^1(U_2)}<\infty $. For integers $N,j\geq 1$ put

Then $K_{j,N}$ is compact and contains the open set $O_{j,N}$. Fix $x_0\in E$. Since $\varphi _ j(x_0)\downarrow -\infty $, there is $j_0(N)$ such that $x_0\in O_{j,N}$ for $j\geq j_0(N)$; we choose $j_0(N)$ nondecreasing in $N$. For these $j$ the set $K_{j,N}$ has nonempty interior, so $u_{j,N}=T_{K_{j,N}}^*$ is a bounded nonnegative function in $\mathrm{PSH}(X,\omega )$ by lemma 10.9.13. Every competitor for $T_{K_{j,N}}$ is nonpositive on the open set $O_{j,N}$; hence so is $T_{K_{j,N}}$, and its upper semicontinuous regularization satisfies $u_{j,N}=0$ on $O_{j,N}$.

For fixed $N$, the sets $K_{j,N}$ increase with $j$, because $\varphi _ j$ decreases. A competitor for $T_{K_{j+1,N}}$ is a competitor for $T_{K_{j,N}}$, so $u_{j,N}$ decreases with $j\geq j_0(N)$; let $u_ N$ be its limit. By lemma 10.9.3(3), $u_ N\in \mathrm{PSH}(X,\omega )$, and $0\leq u_ N\leq u_{j_0(N),N}$ is bounded. If $x\in U_1$ and $\varphi (x)<-N$, then $x\in O_{j,N}$ for all large $j$, so $u_ N(x)=0$. In particular $u_ N=0$ on $E$, so $u_ N$ is a competitor for $T_ E$, and

For fixed $j$, the sets $K_{j,N}$ decrease as $N$ increases, so $u_{j,N}\leq u_{j,N+1}$ for $j\geq j_0(N+1)$; letting $j\to \infty $ gives $u_ N\leq u_{N+1}$. Let $u_\infty $ be the upper semicontinuous regularization of $\lim _ Nu_ N$. By lemma 10.9.3(1), $u_\infty \in \mathrm{PSH}(X,\omega )$ and $0\leq u_\infty \leq C$.

Step 2. Convergence of the masses. By lemma 10.9.3(3), applied to the uniformly bounded decreasing sequence $(u_{j,N})_{j\geq j_0(N)}$ and to the uniformly bounded increasing sequence $(u_ N)_{N\geq 1}$, and testing the weak convergence with the constant function $1$ on the compact manifold $X$, we obtain

Moreover $\int _ X\omega _{u_\infty }^ n>0$ by proposition 10.9.10, since $u_\infty $ is bounded.

Step 3. The capacity of $K_{j,N}$. Choose a smooth function $h$ on $U_2$ with $\mathrm{dd}^{\mathrm{c}}h\geq \omega $ and $\sup _{U_2}|h|<\infty $, for instance a large multiple of $|z|^2$. If $v\in \mathrm{PSH}(X,\omega )$ and $0\leq v\leq 1$, then $v+h$ is plurisubharmonic on $U_2$ with $|v+h|\leq 1+\sup _{U_2}|h|$, and $\omega _ v^ n\leq (\mathrm{dd}^{\mathrm{c}}(v+h))^ n$ on $U_2$ by lemma 10.9.3(2). Since $K_{j,N}\subset \overline{U_1}$ and $-\varphi _ j\geq N$ on $K_{j,N}$, while $-\varphi _ j\geq 0$ on $U_2$, the weighted Chern–Levine–Nirenberg inequality (theorem 7.4.1), applied in $U_2$ to $\varphi _ j$, the compact set $\overline{U_1}$ and a compact neighborhood $L\subset U_2$ of $\overline{U_1}$, gives

The constant $C'$ depends on $\| \varphi _ j\| _{L^1(L)}\leq \| \varphi \| _{L^1(U_2)}$ and on the bound for $|v+h|$, but not on $v$, $j$ or $N$. Taking the supremum over $v$, we obtain $c_\omega (K_{j,N})\leq C'/N$.

Step 4. Conclusion. By proposition 10.9.16,

For fixed $N$, the uniformly bounded functions $u_{j,N}$ decrease to $u_ N$, so $u_{j,N}\to u_ N$ in $L^1(X)$ by dominated convergence, and $\sup _ Xu_{j,N}\to \sup _ Xu_ N$ by lemma 10.9.3(4). Letting $j\to \infty $ and using $u_ N\leq C$, we obtain $\int _ X\omega _{u_ N}^ n\leq \max \{ 1,C\} ^ nC'/N$. Letting $N\to \infty $ gives $\int _ X\omega _{u_\infty }^ n=0$, which contradicts Step 2. Hence $E$ is globally pluripolar, and the theorem is proved. $\square$