Let $X$ be a compact complex manifold of complex dimension $n$ and let $\omega$ be a smooth Kahler form on it. Let $Y \subset X$ be a complex (possibly singular) hypersurface and let $u: X \setminus Y \to \mathbb{R}$ be a smooth function.
Question: If $ \sup_{X \setminus Y} |u| < \infty$ and $\omega + i \partial\bar{\partial} u>0$ on $X\setminus Y$, is it true that $$ \int_{X \setminus Y} (\omega + i \partial\bar{\partial} u)^n = \int_X \omega^n \hspace{8mm} ? $$
Yes, this is true. By the boundedness assumption, $u$ extends to a (bounded) $\omega$-psh function on $X$.
Bedford and Taylor defined in '82 the Monge-Ampère operator of a bounded psh function, which has later been extended to the global quasi-psh case.
Almost by definition of that operator, you have $\int_X(\omega+dd^c \varphi)^n=\int_X \omega^n$ for any bounded $\omega$-psh function $\varphi$ (cf last paragraph), and BT showed that the measure $(\omega+dd^c \varphi)^n$ puts not mass on pluripolar sets (in particular, it puts no mass on analytic subsets like $Y$). You then get $$\int_{X\setminus Y}(\omega+dd^c \varphi)^n=\int_X(\omega+dd^c \varphi)^n=\int_X \omega ^n.$$
To understand why the mass is cohomological, look at $n=2$ where $(dd^c \varphi)^2:=dd^c(\varphi dd^c \varphi)$ by the very definition of BT. In particular, $\int_X(\omega+dd^c\varphi)^2=\int_X\omega^2+2 \int_X\omega\wedge dd^c \varphi+\int_X(dd^c\varphi)^2$ and the last two terms vanish by Stokes theorem.
