Proposition 13.5.106A36B9354A5

Proof. Fix $j$, let $R$ be the closed positive current $\bigwedge _{i>j}(\theta _ i+\mathrm{dd}^{\mathrm{c}}\varphi _ i)$, and let $\eta '$ be a smooth closed form of pure complementary bidegree. Locally $\theta _ j=\mathrm{dd}^{\mathrm{c}}h_ j$ with $h_ j$ smooth, and since $R$ is closed, the recursion gives

Here $\mathrm{dd}^{\mathrm{c}}(h_ jR)=\mathrm{dd}^{\mathrm{c}}h_ j\wedge R$ because $h_ j$ is smooth and $R$ is closed of pure bidegree, so that $\partial R=\bar\partial R=0$. Both sides are globally defined, because $\varphi _ j$ is a global bounded function. Moreover $\int _ X\mathrm{dd}^{\mathrm{c}}(\varphi _ jR)\wedge \eta '=\int _ X\varphi _ jR\wedge \mathrm{dd}^{\mathrm{c}}\eta '=0$, since $\eta '$ is closed of pure bidegree, so that $\partial \eta '=\bar\partial \eta '=0$ and $\mathrm{dd}^{\mathrm{c}}\eta '=0$. Hence $\int _ X(\theta _ j+\mathrm{dd}^{\mathrm{c}}\varphi _ j)\wedge R\wedge \eta '=\int _ X\theta _ j\wedge R\wedge \eta '$. Apply this successively for $j=1,2,\ldots ,p$, with $\eta '=\theta _1\wedge \cdots \wedge \theta _{j-1}\wedge \eta $ at the $j$th step, which is smooth and closed. After $p$ steps the integral becomes $\int _ X\theta _1\wedge \cdots \wedge \theta _ p\wedge \eta $. $\square$