Definition. Let $A$ be a subset of $\mathbb{R}^n$. We say $A$ has measure zero in $\mathbb{R}^n$ if for very $\epsilon > 0$, there is a covering $Q_1, Q_2, \dotsc$ of $A$ by countably many reactangles such that $$ \sum_{i=0}^\infty v(Q_i) < \epsilon\,. $$ (b) Let $A$ be the union of the countable collection of sets $A_1, A_2, \dotsc$ If each $A_i$ has measure zero in $\mathbb{R}^n$, so does $A$.

Proof. To prove (b), cover the set $A_j$ by countably many reactangles $$ Q_{1j}, Q_{2j}, Q_{3j}, \dotsc $$ of total volume less than $\epsilon/2^j$. Do this for each $j$. Then the collection of rectangles $\{Q_{ij}\}$ is countable, it covers $A$, and it has total volume less than $$ \sum_{j=1}^\infty \epsilon/2^j = \epsilon. $$

(Original scanned image here.)

I am reading James R. Munkres "Analysis on Manifolds" now.

I think there is a logical gap in the proof of Theorem 11.1(b)(p.91).

He showed the following inequality: $$ \sum_{j = 1}^{\infty} \sum_{i = 1}^{\infty} v(Q_{ij}) < \sum_{j = 1}^{\infty} \frac{\epsilon}{2^j} = \epsilon $$
But I think it was necessary for him to show the following equality: $$ v(Q_{11}) + v(Q_{21}) + v(Q_{12}) + v(Q_{31}) + v(Q_{22}) + v(Q_{13}) + \dotsb = \sum_{j = 1}^{\infty} \sum_{i = 1}^{\infty} v(Q_{ij}) $$

Am I wrong or not?

If I am right, then please show the above equality.

  • 2
    $\begingroup$ It's hard to say whether you are right or not. It is well known that every infinite series of nonnegative terms can be rearranged at will without affecting the limiting value. I think Munkres simply assumed that his readers would know this fact. $\endgroup$ – user1551 Aug 27 '18 at 11:33
  • 1
    $\begingroup$ Related: Rearranging a series of nonnegative terms. $\endgroup$ – user1551 Aug 27 '18 at 11:44

No logical gap here. Pay attention to his first sentence:

Cover the set $A_j$ by countably many rectangles of total volume less than $\varepsilon /2^j$.

Sorry, the statement above may not be on topic. Try search theory about "Double series". You may also prove that the double series is bounded by $\varepsilon$ by truncating the series and take the limit.


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.