[This question arises from a look at the paper


* Shing-Tung Yau, "[On The Ricci Curvature of a Compact Kähler Manifold and the Complex Monge-Ampére Equation, I](https://jasonpayne.webs.com/Math5339/On%20the%20Ricci%20Curvature%20of%20a%20Compact%20Kahler%20Manifold%20and%20the%20Complex%20Monge-Ampere%20Equation%20I,%20S.T.%20Yau.pdf)", Comm. Pure Appl. Math., **31** (1978) 339-411, doi:[10.1002/cpa.3160310304](https://doi.org/10.1002/cpa.3160310304), [MR0480350](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0480350), [Zbl 0369.53059](https://zbmath.org/?q=an%3A0369.53059).]


My problem arises from (4.1)


It said that itegrating $(\Omega+\partial \bar{\partial} \varphi)^{m}=(\exp \{F\}) \Omega^{m}(4.1)$ then we get $\int \exp \{F\}=\operatorname{Vol}(M)$ where $\Omega$ is the kahler form.


Does this mean $(\Omega+\partial \bar{\partial} \varphi)^{m}$ is also a volume form? I'm confused this step of integrating (4.1).



A problem of the volume form of Kähler manifold in the paper of Yau's proof of Calabi conjecture
Just to close this off: note that $d=\partial+\bar\partial$ and that $\partial^2=0$ so $\partial\bar\partial=d\bar\partial$, and therefore $\Omega+\partial\bar\partial\varphi=\Omega+d\bar\partial\varphi$ is in the same cohomology class as $\Omega$. Since wedge product of forms descends to the usual product in cohomology, $(\Omega+d\bar\partial\varphi)^n=\Omega^n$ in cohomology, giving the same volume integral over our compact manifold. On the other hand, since the Monge-Ampere equation is elliptic, scalar, determined, it is locally solvable, so $\Omega+d\bar\partial\varphi$ can achieve any multiple of any given volume form, locally, by suitable local choice of $\varphi$. So we cannot guaranteed that $\Omega+d\bar\partial\varphi$ is not zero somewhere, if we allow arbitrary choice of $\varphi$. So we can't be sure that this $(\Omega+\partial\bar\partial\varphi)^n$ is actually a volume form, i.e. a nowhere-zero top-degree form with positive integral. That requires more information.


