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: enter image description here 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

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    $\begingroup$ why the downvote? $\endgroup$ – vidyarthi Mar 6 at 12:32
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    $\begingroup$ You can indeed do such a computation. Any top form defines a measure on the surface, making is a measure space, and the boundaries of the $V_m$s have measure zero. $\endgroup$ – David Roberts Mar 6 at 12:44
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    $\begingroup$ Isn't this a special case of a partition of unity, namely the characteristic functions of the $V_n$? $\endgroup$ – Achim Krause Mar 6 at 12:45
  • $\begingroup$ Are you trying to integrate numerically over a specific Riemann surface, or are you doing something more theoretical? I have some possibly relevant experience for numerical calculations. $\endgroup$ – Neil Strickland Mar 6 at 12:52
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    $\begingroup$ By the characteristic function of $V_n$ I just mean the function that is constantly 1 on $V_n$ and 0 outside of it. That's a partition of 1 if the $V_n$ cover the whole space and don't overlap. $\endgroup$ – Achim Krause Mar 6 at 16:53

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