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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 $$ \sum_{i=1}^\infty v(Q_i)<\epsilon $$ so that if this inequality holds, we often say that the total volume of the rectangles $Q_1,Q_2,...$ if less than $\epsilon$

Lemma 1

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 $$ \sum_{i=1}^\infty v(Q_i)<\epsilon $$

Lemma 2

Let $Q$ be a rectangle in $\Bbb{R}^n$; let $\{Q_1,...,Q_k\}$ be a finite collection of rectangles that covers $Q$. Then $$ v(Q)\le\sum_{i=1}^k v(Q_i) $$

Statement

For any rectangle $R:=[a_1,b_1]\times...\times[a_n,b_n]$ the following things are equivalent:

  1. $R$ has measure zero;
  2. the volume of $R$ is zero;
  3. the interior of $R$ is empty.

Unfortunately I can't prove the statement so I ask to do it. So could someone help me, please?

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  • $\begingroup$ Use the Heine-Borel theorem. $\endgroup$ May 31 '20 at 16:04
  • $\begingroup$ @AnginaSeng Perhaps I understood: now I edited the question. $\endgroup$ May 31 '20 at 16:10
  • $\begingroup$ @AnginaSeng I did it: reread the question. What can you say? $\endgroup$ May 31 '20 at 16:25
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Let be $R$ a rectangle of measure zero and we suppose that $v(R)>0$ so that for $\epsilon=v(R)$ there exist $\mathcal{Q}:=\{\overset{°}Q_i:i\in\Bbb{N}\}$ such that $$ R\subseteq\bigcup_{i=1}^\infty\overset{°}Q_i $$ and $$ \sum_{i=1}^\infty v(Q_i)<v(R) $$ but since for Tychonoff theorem $R$ is compact (a rectangle is product of compacts) for some $k\in\Bbb{N}$ there must exist $\mathcal{Q}_k=\{\overset{°}Q_i:i=1,...,k\}\subseteq\mathcal{Q}$ such that $$ R\subseteq\bigcup_{i=1}^k\overset{°}Q_i $$ but this would be imply that $$ \sum_{i=1}^k v(Q_i)\le\sum_{i=1}^\infty v(Q_i)<v(R) $$ that clearly for lemma 2 is impossible so that $v(R)=0$.

Now we suppose that $R$ is a rectangle of $\Bbb{R}^n$ such that $v(R)=0$ so that there exist $j=1,...,n$ such that $a_j=b_j$ and so $$ \overset{°}R=(a_1,b_1)\times...\times(a_j,b_j)\times...\times(a_n,b_n)=(a_1,b_1)\times...\times\varnothing\times...\times(a_n,b_n)=\varnothing $$ that proves the second point.

Finallly we suppose that $\overset{°}R=\varnothing$, that is $$ (a_1,b_1)\times...\times(a_n,b_n)=\varnothing $$ so that for the Axiom of Choice there exist $j=1,...,n$ such that $(a_j,b_j)=\varnothing$, that is $a_j=b_j$. Now for convenience we define $$ \nu:=(b_1-a_1)\cdot...\cdot(b_{j-1}-a_{j-1})\cdot(b_{j+1}-a_{j+1})\cdot...\cdot(b_n-a_n) $$ and for $\epsilon>0$ we choice $\delta>0$ such that $\delta<\frac{\epsilon}{\nu}$. So we define for any $i\in\Bbb{N}$ the rectangle $$ Q_i:=[a_1,b_1]\times...\times\Big[a_j-\frac{\delta}{2^{i+1}},a_j+\frac{\delta}{2^{i+1}}\Big]\times...\times[a_n,b_n] $$ so that $v(Q_i)=\nu\cdot\frac{\delta}{2^i}$. So the collection $\mathcal{Q}=\{Q_i:i\in\Bbb{N}\}$ is a rectangular cover of $R$ such that $$ \sum_{i=1}^\infty v(Q_i)=\sum_{i=1}^\infty \nu\cdot\frac{\delta}{2^i}=\nu\delta\cdot\sum_{i=1}^\infty\frac{1}{2^i}=\nu\delta<\epsilon $$ so that $R$ has measure zero.

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