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