Proposition 13.4.73553558C2E27

Proof. Upper bound. The assertion is local, so we may assume $u\leq 0$. The truncations $u_ l=\max \{ u,-l\} $ decrease to $u$ and converge to it in $L^1_{\mathrm{loc}}$ (theorem 5.10.1), so $\mathrm{dd}^{\mathrm{c}}u_ l\to \mathrm{dd}^{\mathrm{c}}u$ weakly. For fixed $k$ and every $l\geq k$, plurifine locality (theorem 5.47.22) and lemma 13.2.1 give $\mathbf1_{\{ u>-k\} }\mathrm{dd}^{\mathrm{c}}u_ k=\mathbf1_{\{ u>-k\} }\mathrm{dd}^{\mathrm{c}}u_ l\leq \mathrm{dd}^{\mathrm{c}}u_ l$, and letting $l\to \infty $ yields $\mathbf1_{\{ u>-k\} }\mathrm{dd}^{\mathrm{c}}u_ k\leq \mathrm{dd}^{\mathrm{c}}u$. Hence the truncated products have locally bounded mass, the non-pluripolar product $S=\langle \mathrm{dd}^{\mathrm{c}}u\rangle $ is locally finite, and $0\leq S\leq \mathrm{dd}^{\mathrm{c}}u$. Since $S$ gives no mass to the pole set of $u$, it follows that $S\leq \mathbf1_{\{ u>-\infty \} }\mathrm{dd}^{\mathrm{c}}u$. It remains to prove the reverse inequality.

One variable. Let $n=1$ and put $\mu =\mathrm{dd}^{\mathrm{c}}u$, a positive measure. Let $V$ be a relatively compact open subset of the domain, and choose a smooth compactly supported function $\kappa \geq 0$ equal to one on $V$. With $\nu =\kappa \mu $ and the fundamental solution $E_2$ of (3.1.14), theorem 3.4.14 gives

where $h$ is harmonic; the factor $4\pi $ appears because $\mathrm{dd}^{\mathrm{c}}u=(\Delta u)/(4\pi )$ as planar measures, while the Riesz measure is $\Delta u$. Put $p=E_2*\nu $ and $\nu _ f=\mathbf1_{\{ p>-\infty \} }\nu $; on $V$ we have $\{ p>-\infty \} =\{ u>-\infty \} $. The kernel $E_2$ is bounded above on the compact set of differences of points of $\operatorname{Supp}\nu $, so removing positive mass from $\nu $ lowers its potential on $\operatorname{Supp}\nu $ by at most a constant: $E_2*\nu _ f\geq p-C$ there. Hence $E_2*\nu _ f$ is finite $\nu _ f$-almost everywhere, and theorem 4.1.10 writes $\nu _ f=\sum _ i\nu _ i$ with compactly supported measures $\nu _ i$ whose potentials are finite and continuous.

For each $N$, the measure $\nu ^{(N)}=\sum _{i\leq N}\nu _ i$ has the continuous subharmonic potential $v_ N=4\pi E_2*\nu ^{(N)}$, with $\mathrm{dd}^{\mathrm{c}}v_ N=\nu ^{(N)}$. On $V$ we have $\mu =\nu \geq \nu _ f\geq \nu ^{(N)}$, so the distribution $u-v_ N$ has positive Laplacian on $V$. By theorem 3.4.12(3), this distribution has a subharmonic representative $s$ on $V$, and by part (2) of that theorem the value of $s$ at each point is the limit of its averages over small discs. These averages are also those of the function $u-v_ N$, and they converge to $u-v_ N$ at every point, by part (2) applied to $u$ and the continuity of $v_ N$. Hence $s=u-v_ N$ everywhere on $V$, and $u-v_ N$ is subharmonic on $V$. Since $v_ N$ is locally bounded, multilinearity (proposition 13.2.6) gives

Letting $N\to \infty $ yields $S\geq \nu _ f=\mathbf1_{\{ u>-\infty \} }\mu $ on $V$, which proves the proposition for $n=1$.

Several variables. Let $n\geq 2$. The assertion is local, so we may work on a polydisc $P$ such that $u\leq 0$ is plurisubharmonic on a neighborhood of $\overline P$. Fix one of the coordinates, denote it by $x$, and write $z\in \mathbb C^{n-1}$ for the others, so that $P=\Delta \times Z$ with a disc $\Delta \subset \mathbb C$ and a polydisc $Z\subset \mathbb C^{n-1}$. Let $\mathrm dV_ z$ be the Euclidean volume form of the $z$-coordinates, viewed as an $(n-1,n-1)$-form, and $\mathrm dA$ the area measure on $\Delta $. We disintegrate $\mathrm{dd}^{\mathrm{c}}u\wedge \mathrm dV_ z$ along the fibers $\Delta \times \{ z\} $ and apply the one-variable case on each fiber. Since $u$ is integrable on $\Delta \times Z$, Fubini's theorem (proposition 2.2.7) shows that $G=\{ z\in Z:\int _\Delta |u(x,z)|\, \mathrm dA(x)<\infty \} $ is a Borel set of full measure in $Z$. For $z\in G$ the function $u_ z=u(\, \cdot \, ,z)$ is integrable, hence not identically $-\infty $, so it is subharmonic on $\Delta $ by proposition 5.9.3; let $\mu _ z=\mathrm{dd}^{\mathrm{c}}_ xu_ z$ be its Hessian measure.

For a test function $a$ on $\Delta $, we have $\int a\, \mathrm d\mu _ z=\int _\Delta u_ z\mathrm{dd}^{\mathrm{c}}_ xa$, and Fubini's theorem, applied to the positive and negative parts of the integrand, shows that this is a Borel function of $z\in G$. Uniform approximation by test functions supported in a fixed compact set extends this to continuous functions $a$ with compact support, since each $\mu _ z$ is finite on compact sets. Hence lemma 13.4.4, with $f=1$, gives a measure $\lambda $ on $\Delta \times Z$ with $\int h\, \mathrm d\lambda =\int _ G(\int h(\, \cdot \, ,z)\, \mathrm d\mu _ z)\, \mathrm dV(z)$ for nonnegative Borel $h$. If $\chi \in C_ c^\infty (\Delta )$ satisfies $0\leq \chi \leq 1$ and equals one on a disc $\Delta '\Subset \Delta $, then $\mu _ z(\Delta ')\leq \int u_ z\mathrm{dd}^{\mathrm{c}}_ x\chi \leq C_\chi \int _\Delta |u_ z|\, \mathrm dA$, which is integrable over $G$; so $\lambda $ is finite on compact sets. For a test function $\varphi $ on $\Delta \times Z$, only the derivative $\partial _ x\partial _{\bar x}\varphi $ survives in $\mathrm{dd}^{\mathrm{c}}\varphi \wedge \mathrm dV_ z$, and Fubini's theorem gives

By the uniqueness in theorem 2.5.6, the positive measure $\mathrm{dd}^{\mathrm{c}}u\wedge \mathrm dV_ z$ equals $\lambda $, that is,

for every nonnegative Borel function $g$ on $\Delta \times Z$.

Now let $g\geq 0$ be a bounded Borel function with compact support in $\Delta \times Z$, and apply lemma 13.4.5, with $r=1$, $Q=\Delta $, $D=Z$ and $\eta =\mathrm dV_ z$, to the bounded function $u_ k=\max \{ u,-k\} $ and the weight $g\mathbf1_{\{ u>-k\} }$:

As $k\to \infty $, the left side increases to $\int gS\wedge \mathrm dV_ z$ by lemma 13.2.2, since $\mathbf1_{\{ u>-k\} }\mathrm{dd}^{\mathrm{c}}u_ k$ is the truncated product $S_ k$ of $u$. For $z\in G$, the inner integral on the right is the integral of $g(\, \cdot \, ,z)$ against the truncated product of $u_ z$ at level $k$; by lemma 13.2.2 and the one-variable case, it increases to $\int g\, \mathrm d\langle \mathrm{dd}^{\mathrm{c}}_ xu_ z\rangle =\int g\mathbf1_{\{ u_ z>-\infty \} }\, \mathrm d\mu _ z$. Since $Z\setminus G$ is a null set, monotone convergence in $z$ and (13.4.9), applied to $g\mathbf1_{\{ u>-\infty \} }$, give $\int gS\wedge \mathrm dV_ z=\int g\mathbf1_{\{ u>-\infty \} }\mathrm{dd}^{\mathrm{c}}u\wedge \mathrm dV_ z$, so that $S\wedge \mathrm dV_ z=\mathbf1_{\{ u>-\infty \} }\mathrm{dd}^{\mathrm{c}}u\wedge \mathrm dV_ z$.

Since $\beta ^{n-1}=(n-1)!\sum _{j=1}^ n\mathrm dV_{(j)}$, where $\mathrm dV_{(j)}$ is the volume form of the coordinates other than the $j$th, summing over the $n$ coordinate directions shows that the trace measures of $S$ and of $\mathbf1_{\{ u>-\infty \} }\mathrm{dd}^{\mathrm{c}}u$ coincide. Thus $\mathbf1_{\{ u>-\infty \} }\mathrm{dd}^{\mathrm{c}}u-S$ is a positive current with zero trace measure, and it vanishes by theorem 5.3.7. This proves (13.4.8); closedness follows from theorem 13.3.1. $\square$