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?
Best Answer
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. .