1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $1\ge 2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.