Volume of a Rectangle in Stein and Shakarchi 
My confusion is with the proof; I'm not sure why we need to split the rectangle up as it is in the picture. Is it not sufficient just to find the volume of each almost-disjoint rectangle individually (as this is well defined) and take a finite sum? There seems to be no point in extending the sides of the rectangles except in making the proof more complicated.
 A: Let's say that $R=[a,b]\times [c,d].$ Then, by definition, $|R|=(b-a)(d-c).$ You want to show that summing terms of the form $|R_k|=\ell_k \times \textit{w}_k$ gives exactly $(b-a)(d-c),$ where $\ell_k$ and $\textit w_k$ are the lengths and widths, resp. of $R_k.$ While obvious, it requires proof. I think if you work out your own idea carefully, you will just repeat the proof in the book, which goes as follows:
By construction, $(a,b)$ and $(c,d)$ are partitioned into $a<x_2<\cdots <x_{n-1}<b$ and $c<y_2<\cdots <y_{m-1}<d$ in such a way that the rectangles $\tilde R_k$ are pairwise almost disjoint and are of the form $\tilde R_{ij}=[x_i,x_{i+1}]\times [y_j,y_{j+1}]$ and satisfy
$\tag1 \sum_{i=1}^n\sum_{j=1}^m|\tilde R_{ij}|=\sum_{i=1}^n\sum_{j=1}^m(x_{i+1}-x_i)(y_{j+1}-y_j)=(b-a)(d-c)=|R|.$
But also by construction, each $R_k$ is a union of some number of the $\tilde R_{ij}$. Therefore,
$\tag2 |R_k|=\sum_{i_k=1}^{n_k}\sum_{j_k=1}^{m_k}|\tilde R_{i_kj_k}|=\sum_{i_k=1}^{n_k}\sum_{j_k=1}^{m_k}(x_{i_k+1}-x_{i_k})(y_{j_k+1}-y_{j_k})$
Finally, applying $(2)$ and then $(1)$, we have
$\tag3 \sum_{k=1}^n|R_k|=\sum_{k=1}^n\sum_{i_k=1}^{n_k}\sum_{j_k=1}^{m_k}|\tilde R_{i_kj_k}|=\sum_{i=1}^n\sum_{j=1}^m|\tilde R_{ij}|=|R|.$
