I'm reading an excellent book, A concise introduction to the theory of integration(second edition) by Daniel Stroock. In this book the writer proves some properties of rectangles first, and then construct the integration theory. A rectangle in $\mathbb R^N$ is defined as set like $\displaystyle\prod_{i=1}^N [a_i,b_i].$ And two rectangles are said to be non-overlapping iff they have disjoint interiors.

The volume of rectangle $I=\displaystyle\prod_{i=1}^N [a_i,b_i]$ is defined as $\mathrm{vol}(I)=(b_1-a_1)\cdot(b_2-a_2)\cdots\cdot(b_N-a_N).$

A lemma states that

If $\mathcal C$ is a non-overlapping, finite collection of rectangles each of which is contained in the rectangle $J$, then $\mathrm{vol} (J) \geqslant \sum_{I\in \mathcal C}\mathrm{ vol} (I).$ On the other hand, if $\mathcal C$ is any finite collection of rectangles and $J$ is a rectangle which is covered by $\mathcal C,$ then $\mathrm{vol} (J) \leqslant \sum_{I\in \mathcal C} \mathrm{vol} (I).$

In the proof of this lemma, the writer states that "assume that rectangles $I_1,\cdots,I_n$ are mutually disjoint, then write $I_\mu$ as $I_\mu=[a_\mu,b_\mu]\times\hat I_\mu,$ here $\hat I_\mu$ is the rectangle in $\mathbb R^{N-1}$,then $\hat I_1,\cdots,\hat I_n$ are mutually disjoint." (this appears in page 3.)

Obviously the writer's statement is wrong, and I tried to save his proof, but failed. Could anyone help to save his proof?

  • $\begingroup$ What's the goal of the proof? $\endgroup$ – user284331 Jan 10 '18 at 4:55
  • $\begingroup$ Well, the proof is about the properties of the volume of rectangles: if some rectangles $I_1,\cdots,I_n$ are mutually non-overlapping, and $I_\mu\subset J,\ \mu=1,2,\cdots,n,$ then $\mathrm{vol}(J)\geqslant \sum_{\mu=1}^n \mathrm{vol}(I_\mu).$ $\endgroup$ – painday Jan 10 '18 at 4:58
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    $\begingroup$ You can read the version before the Springer one. The proof of the previous version is correct. I believe what you wrote here is from the Springer version (I just took a look of it). $\endgroup$ – user284331 Jan 10 '18 at 5:03
  • $\begingroup$ Thanks! I doubt it is a typo or something like that... $\endgroup$ – painday Jan 10 '18 at 5:04

The second inequality sign in the statement of the lemma should be $\le $ reversed. The statement is not correct as it is stated.

For example let $C_1=[1,2]$ and $C_2 =[2,3].$ Then The collection covers $J=[1/2,3/2].$

According to $$ \mathrm{vol} (J) \geqslant \sum_{I\in \mathcal C} \mathrm{vol} (I)$$.we have to have $1\ge 2$.

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  • $\begingroup$ Sorry, the inequality sign was my typo. I've edited the question. $\endgroup$ – painday Jan 10 '18 at 5:54
  • $\begingroup$ This doesn't answer the question though. $\endgroup$ – jgon Jan 10 '18 at 6:08

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