PomeLemma 7.11.18.E1A777307E3D Let $\Omega \subset \mathbb C^ n$ be open, and let $\varphi ,\psi $ be locally bounded plurisubharmonic functions on $\Omega $ which coincide outside a compact set $K\subset \Omega $. Then $(\mathrm{dd}^{\mathrm{c}}\varphi )^ n$ and $(\mathrm{dd}^{\mathrm{c}}\psi )^ n$ agree on $\Omega \setminus K$ and assign the same finite mass to $K$. In particular $\int _\Omega (\mathrm{dd}^{\mathrm{c}}\varphi )^ n=\int _\Omega (\mathrm{dd}^{\mathrm{c}}\psi )^ n$, where both sides may be infinite.
Lemma 7.11.18E1A777307E3D
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,
\[ \mathrm{dd}^{\mathrm{c}}(\varphi S_ k)=(\mathrm{dd}^{\mathrm{c}}\varphi )^{k+1}\wedge (\mathrm{dd}^{\mathrm{c}}\psi )^{n-1-k}, \qquad \mathrm{dd}^{\mathrm{c}}(\psi S_ k)=(\mathrm{dd}^{\mathrm{c}}\varphi )^ k\wedge (\mathrm{dd}^{\mathrm{c}}\psi )^{n-k}, \]
so the sum over $k$ telescopes and
\[ (\mathrm{dd}^{\mathrm{c}}\varphi )^ n-(\mathrm{dd}^{\mathrm{c}}\psi )^ n=\mathrm{dd}^{\mathrm{c}}R,\qquad R=(\varphi -\psi )\sum _{k=0}^{n-1}S_ k. \]
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
\[ \int \chi \, (\mathrm{dd}^{\mathrm{c}}\varphi )^ n-\int \chi \, (\mathrm{dd}^{\mathrm{c}}\psi )^ n=R(\mathrm{dd}^{\mathrm{c}}\chi )=0, \]
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. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.