Suppose we are given a compact Riemann surface $M$, an open cover $\mathscr{U}=\{U_1,U_2,\dots\}$ of $M$, charts $\{(U_1,\phi_1),(U_2,\phi_2),\dots\}$, holomorphic coordinates, $\phi_m:p\in U_m\mapsto z_m$, holomorphic transition functions $z_m=f_{mn}(z_n)$ on patch overlaps $U_m\cap U_n$ (with appropriate cocycle relations, $f_{mn}\circ f_{n\ell}\circ f_{\ell m}=1$ on triple overlaps $U_m\cap U_n\cap U_\ell$), and a globally defined top form, $\omega=\omega(z,\bar{z})dz\wedge d\bar{z}$.

I want to integrate $\omega$ over $M$,
$$
I= \int_M\omega,
$$
using the above data (without using a partition of unity), and I'm wondering whether my reasoning is correct. In particular, suppose we construct non-overlapping sets $\{V_1,V_2,\dots\}$ such as those shown in the figure:
**My question** is whether I can compute $I$ via a sum of integrals over the $\{V_1,V_2,\dots\}$ (to guarantee no overcounting) without using a partition of unity; i.e., is it true that:
\begin{equation}
\begin{aligned}
I &= \sum_m\int_{V_m}\omega\\
&=\sum_m\int_{V_m}\omega_m(z_m,\bar{z}_m)dz_m\wedge d\bar{z}_m
\end{aligned}
\qquad\qquad\qquad(*)
\end{equation}
In the examples I have considered (such as $\omega=\frac{i}{2\pi}\mathcal{R}_{z\bar{z}}dz\wedge d\bar{z}$ the curvature tensor or $\omega=ig_{z\bar{z}}dz\wedge d\bar{z}$ the volume form) this gives the right answer (e.g. the Euler characteristic in the first example and area in the second) but I would like to know if it is true in general. Thanks!

**Update I:** It should presumably be possible to derive (*) using a partition of unity subordinate to $\mathscr{U}$. If so, how might one derive the last equality in:
$$
I=\int_M\omega=\int_{M} \omega\big(\sum_m\lambda_m\big)=\sum_m\int_{U_m}\omega\lambda_m = \sum_m\int_{V_m}\omega
$$

**Update II:** This question now has a clear and explicit answer at here