# If $R\subseteq\Bbb{R}^n$ is a rectangle then $m(R)=0\Leftrightarrow v(R)=0\Leftrightarrow\overset{°}R=\varnothing$.

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?

• Use the Heine-Borel theorem. Commented May 31, 2020 at 16:04
• @AnginaSeng Perhaps I understood: now I edited the question. Commented May 31, 2020 at 16:10
• @AnginaSeng I did it: reread the question. What can you say? Commented May 31, 2020 at 16:25

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 $$\sum_{i=1}^\infty v(Q_i) but since for Tychonoff theorem $$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) 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.