A set has measure zero iff for every $\epsilon>0$ there is a countable covering of open rectangles such that $\sum_{i=1}^\infty v(Q_i)<\epsilon$

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?

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. .
• 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? Commented May 25, 2020 at 12:29
• @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. Commented May 25, 2020 at 12:29
• 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. Commented May 25, 2020 at 12:40
• @AntonioMariaDiMauro We can choose $\epsilon$ so small that $|v(r')-v(Q') | <v(Q')$. We then get $v(R')=[v(R')-v(Q')] +v(Q') <v(Q')+v(Q')=2v(Q')$. Commented May 25, 2020 at 12:44
• 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!!! Commented May 25, 2020 at 22:24