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What shown below is a reference from "Analysis on manifolds" by James R. Munkres.

Definition

Let $A$ a subset of $\Bbb{R}^n$. We say $A$ has measure zero in $\Bbb{R}^n$ if for every $\epsilon>0$, there is a covering $Q_1,Q_2,...$ of $A$ by countably many rectangles such that $$ \sum_{i=1}^\infty v(Q_i)<\epsilon $$

Theorem

A set $A$ has measure zero in $\Bbb{R}^n$ if and only if for every $\epsilon>0$ there is a countable covering of $A$ by open rectangles $\overset{°}Q_1,\overset{°}Q_2,...$ such that $$ \sum_{i=1}^\infty v(Q_i)<\epsilon $$

Proof. If the open rectangles $\overset{°}Q_1,\overset{°}Q_2,...$ cover $A$, then so the rectangles $Q_1,Q_2,...$ . Thus the given condition implies that $A$ has measure zero. Conversely, suppose $A$ has measure zero. Cover $A$ by rectangles $Q'_1,Q'_2,...,$ of total volume $\frac{\epsilon}2$. For each $i$, chose a rectangle $Q_i$ such that $$ 1.\quad Q'_i\subset\overset{°}Q_i\text{ and }v(Q_i)\le 2v(Q'_i) $$ (This we can do because $v(Q)$ is a continuous function of the end points of the component intervals of $Q$). Then the open rectangles $\overset{°}Q_1,\overset{°}Q_2,...$ cover $A$ and $\sum v(Q_i)<\epsilon$.

So I don't understand why it is possible to make the rectangles $Q_i$ such that they respect the condition $1$ and so I ask to well explain this: naturally I don't understand Munkres explanation and so you can or to explain better what Munkres said or to show another explanation. So could someone help me, please?

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Consider a rectangle $R=[a_1,b_1]\times [a_2,b_2] \times ... \times [a_n,b_n]$. For $\epsilon >0$ sufficiently small $R'=[a_1-\epsilon ,b_1+\epsilon ]\times [a_2-\epsilon ,b_2+\epsilon ] \times ... \times [a_n-\epsilon ,b_n+\epsilon ]$ contains $R$ in its interior and its volume tends to volume of $R$ as $ \epsilon \to 0$. Hence the volume of $R'$ is at most equal to $2v(R)$ for $\epsilon$ sufficiently small. .

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    $\begingroup$ Okay, so we have $v(R')=\prod_{i=1}^n(b_i-a_i+2\epsilon)\le 2\prod_{i=1}^n(b_i-a_i)=2v(Q')$ and so how can I prove that this inequality has at least one real positive soluction? $\endgroup$ Commented May 25, 2020 at 12:29
  • $\begingroup$ @AntonioMariaDiMauro I am just using the fact that if $\phi (\epsilon) \to c>0$ as $\epsilon \to 0+$ then $\phi (\epsilon) <2c$ for some $\epsilon >0$. You can prove this easily from definition of limit. $\endgroup$ Commented May 25, 2020 at 12:29
  • $\begingroup$ Excuse me, but I don't well understand. So I think that $\phi:=v(R')$, and $c:=v(Q')$. So clearly if $\epsilon\rightarrow 0^+$ then $v(R')\rightarrow v(Q')$ and so why $v(R')<2c$? could you explain better? Sorry, forgive my confusion. $\endgroup$ Commented May 25, 2020 at 12:40
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    $\begingroup$ @AntonioMariaDiMauro We can choose $\epsilon$ so small that $|v(r')-v(Q') | <v(Q')$. We then get $v(R')=[v(R')-v(Q')] +v(Q') <v(Q')+v(Q')=2v(Q')$. $\endgroup$ Commented May 25, 2020 at 12:44
  • $\begingroup$ Okay, it is clear: all is consequence of the continuity of $v(R')$, right? Excuse me if I have not immediately accepted the answer: unfortunately I was very busy today. Anyway thanks too much for your assistance!!! $\endgroup$ Commented May 25, 2020 at 22:24

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