MHB How to Prove \(\text{vol}_n (B) = \text{vol}_n (B') + \text{vol}_n (B'')\)?

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The discussion centers on proving Proposition 6.1.2 from Multidimensional Real Analysis II, which states that the volume of a closed n-dimensional rectangle B is equal to the sum of the volumes of two sub-rectangles B' and B''. The proof begins by defining the volumes of B', B'', and B in terms of their respective dimensions. By applying the definition of volume, it is shown that the sum of the volumes of B' and B'' equals the volume of B. The conversation emphasizes the importance of understanding the initial steps of the proof for clarity. This rigorous approach aids in confirming the validity of the proposition.
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I am reading Multidimensional Real Analysis II (Integration) by J.J. Duistermaat and J.A.C. Kolk ... and am focused on Chapter 6: Integration ...

I need some help with the proof of Proposition 6.1.2 ... and for this post I will focus on the first auxiliary result ... see (i) ... at the start of the proof ...Near the start of the proof of Proposition 6.1.2 D&K state that :

" ... ... Because b_j - a_j = (b_j - t_j) + (t_j - a_j), it follows straight away that :

[math] \text{ vol}_n (B) = \text{ vol}_n (B') + \text{ vol}_n (B'') [/math]Readers of this post only need to read the very first part of the proof of Proposition 1 (see scanned text below) ... BUT ... I am providing a full text of the proof together with preliminary definitions so readers can get the context and meaning of the overall proof ... but, as I have said, it is not necessary for readers to read any more than the very first few lines of the proof.
Can someone please help me to rigorously prove that [math] \text{ vol}_n (B) = \text{ vol}_n (B') + \text{ vol}_n (B'') [/math] ...Hope someone can help ...

Help will be much appreciated ...

PeterThe proof of Proposition 6.1.2 together with preliminary notes and definitions reads as follows:
Duistermaat & Kolk_Vol II ... Page 423.png

Duistermaat & Kolk_Vol II ... Page 424.png

Duistermaat & Kolk_Vol II ... Page 425.png
Hope that helps,

Peter
 
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Hi Peter,

By definition of volume, $$\text{vol}_n(B') = (t_j - a_j)\prod_{k \neq j} (b_k - a_k)\quad \text{and}\quad \text{vol}_n(B'') = (b_j - a_j) \prod_{k \neq j} (b_k - a_k)$$
Hence the sum $$\text{vol}_n(B') + \text{vol}_n(B) = [(t_j - a_j) + (b_j - a_j)] \prod_{k \neq j} (b_k - a_k) = (b_j - a_j)\prod_{k \neq j} (b_k - a_k) = \prod_k (b_k - a_k) = \text{vol}_n(B)$$ as desired.
 


Preliminary notes and definitions:

Let B = [a_1, b_1] \times [a_2, b_2] \times ... \times [a_n, b_n] be a closed n-dimensional rectangle in \mathbb{R}^n. We define the volume of B as:

\text{vol}_n (B) = (b_1 - a_1)(b_2 - a_2) ... (b_n - a_n)

Now, let t_j \in [a_j, b_j] for j = 1, 2, ..., n. We define the rectangles B' = [a_1, t_1] \times [a_2, t_2] \times ... \times [a_n, t_n] and B'' = [t_1, b_1] \times [t_2, b_2] \times ... \times [t_n, b_n].

Proposition 6.1.2: Let B, B', B'' be as defined above. Then, \text{ vol}_n (B) = \text{ vol}_n (B') + \text{ vol}_n (B'').
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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