Let $A$ be an open set of $\mathbb{R}^m$ with tensor metric $g$. Let $G(x)$ be a Gramm Matrix of $g(x)$ in the canonical basis of $\mathbb{R}^m$.Let $X:A\to\mathbb{R}^{m\times 1}$ be the vector field of $C^1$.Let $r\in(\mathbb{R}_+)^m$ and $a\in A$. Consider each face of dimension $m-1$ of the parallelepiped $\overline{P}(a,r)$ oriented by an unitary normal vector field that pints to the exterior of $\overline{P}(a,r)$. Therefore
$\int_{\overline{P}(a,r)}div(X)dV_A=\Phi(\partial\overline{P}(a,r)$.
Proof: For each N \in $\mathbb{N}$, $P(a,r)$ is decomposed in $N^m$ parallelepipeds $(P_{N,p})_{1\leqslant k\leqslant N^m}$ whose edges are of $\frac{r_j}{N}\:\:(1\leqslant j\leqslant m)$. If $P_{N,p}$ and $P_{N,q}$ are two parallelepipeds whosw decomposition has $r$ face in commmom then $(0\leqslant r\leqslant m-1)$, then $\Phi_X(\partial P_{N,p})+\Phi_X(\partial P_{N,q})=\Phi_X(P_{N,p}\cup P_{N,q})$.
I am stuck at this point in the proof because I do not know how can $\Phi_X(\partial P_{N,p})+\Phi_X(\partial P_{N,q})=\Phi_X(P_{N,p}\cup P_{N,q})$ be true.
Question:
Why $\Phi_X(\partial P_{N,p})+\Phi_X(\partial P_{N,q})=\Phi_X(P_{N,p}\cup P_{N,q})$? My intuition tells me $\Phi_X(\partial P_{N,p})+\Phi_X(\partial P_{N,q})=\Phi_X(\partial P_{N,p}\cup \partial P_{N,q})$
Thanks in advance!