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

#### Peter

##### Well-known member
MHB Site Helper
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:   Hope that helps,

Peter

Last edited:

#### Euge

##### MHB Global Moderator
Staff member
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.

• HallsofIvy and Peter