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Duistermaat & Kolk ... Vol II ... First aspects of the proof of Proposition 6.1.2 - Another Question

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,938
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 :

\(\displaystyle \text{ vol}_n (B) = \text{ vol}_n (B') + \text{ vol}_n (B'') \)


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 \(\displaystyle \text{ vol}_n (B) = \text{ vol}_n (B') + \text{ vol}_n (B'') \) ...


Hope someone can help ...

Help will be much appreciated ...

Peter


The 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
 
Last edited:

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,942
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.