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 $$ \sum_{j=1}^\infty \epsilon/2^j = \epsilon. $$

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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.


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