Motivation:
In Spivak "Calculus on Manifolds" proof of Change of Variable in Integration - Theorem 3-13 (p. 68). He mentions, in a part of the proof, that: \begin{equation} \sum_S \int_{g^{-1}(\text{int}S)} f \circ g |\det g'| \le \int_{g^{-1}(V)} f \circ g |\det g'| \end{equation}
where $V$ is a rectangle and $S$ are all the subrectangles contained in a partition of $V$. And $\text{int}S$ is the interior of this subrectangle. Where $f$ is integrable and $g$ is a 1-1 continuos differentiable function.
Question
Why do we need the above inequality? Because I am of the impression that the equality would always hold in the above equation.
My take on the problem
In more general terms, I am of the impression that for any integrable funciton $h$, given an open set $A$ and open subsets $G_i$ such that $B = (A -\cup_i G_i)$ has measure 0, than: \begin{equation} \int_A h = \sum_i \int_{G_i} h \end{equation}
(tentative) Proof. Assume that we define an admissible open cover to each $G_i$, and define $\Phi_i$ a partition of the unit subordinate to this open cover (the existance of such partition is given by theorem 3-11 of the same book). By the definition of integrable in the extended sense (p.65 of the same book): \begin{equation} \int_{G_i} h = \sum_{\phi_i \in \Phi_i} \int_{G_i}\phi_i h \end{equation}
Now, the set $B = (A -\cup_i G_i)$ has measure 0 and hence can be covered with closed rectangles ${R_1, R_2, \cdots}$ with total volume smaller than any $\epsilon$. For each of these rectangles, we can define an open set inside $A$ covering $R_j$, and associate a function $\psi_j$, such that this function is zero outside of $R_j$, is $C^{\inf}$ and $\sum_j \psi_j(x) = 1$ and $0 \le \psi_j(x) \le 1$ for every $x \in B$. Similarly to theorem 3-11 of the same book.
It follow that $(\cup_i \Phi_i) \cup (\cup_j \psi_j)$ is a partition of the unit subordinate to a admissible cover of $A$ and hence, by definition: \begin{equation} \int_{A} h = \sum_i \sum_{\phi_i \in \Phi_i} \int_{G_i}\phi_i h + \sum_j \int_{R_j}\psi_j h \end{equation} However, we can made the volume of $R_j$ arbitrarily small. And, hence: \begin{equation} \int_{A} h = \sum_i \sum_{\phi_i \in \Phi_i} \int_{G_i}\phi_i h = \sum_i \int_{G}h \end{equation}
