About the definition that a subset $A$ of $\mathbb{R}^n$ has measure zero in $\mathbb{R}^n$ in Munkres "Analysis on Manifolds"(p.91)

Question

I am reading James R. Munkres "Analysis on Manifolds" now.

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

Let $Q := [a_1, b_1] \times \cdots \times [c_i, c_i] \times \cdots \times [a_n, b_n]$.

$v(Q) = 0$

Is $Q$ a rectangle in $\mathbb{R}^n$?

If $Q$ is a rectangle, then the following exercise is trivial

p.97 Exercise 3.
Show that the set $\mathbb{R}^{n-1} \times 0$ has measure zero in $\mathbb{R}^n$.

By the way, Munkres wrote as follows in p.29:

For example, suppose $Q$ is the rectangle $$ Q = [a_1, b_1] \times \cdots \times [a_n, b_n], $$ consisting of all points $\mathbf{x}$ of $\mathbb{R}^n$ such that $a_i \leq x_i \leq b_i$ for all $i$.

This is his definition of rectangle in his book.

Answer 1

It's a bit of a corner case that doesn't matter much in the big picture because allowing volume-zero rectangles or not does not change the outcome of the definitions that depend on it.

It's a good exercise to show this explicitly -- for example: If you have $$ \sum_{i=0}^\infty v(Q_i) < \epsilon $$ and some of the $Q_i$s have volume $0$, then there is another sequence $Q'_i$ of rectangles with all positive volumes such that $$ \sum_{i=0}^\infty v(Q'_i) < \epsilon \qquad\text{and}\qquad \bigcup_{i=0}^\infty Q_i \subseteq \bigcup_{i=0}^\infty Q'_i $$

Answer 2

I found the following statement:

p.91 Theorem 11.1 (d) if $Q$ is a rectangle in $\mathbb{R}^n$, then Bd $Q$ has measure zero in $\mathbb{R}^n$ but $Q$ does not.

For example, consider $Q = [0, 1] \times [0, 0]$. Obviously $Q$ has measure zero, so $Q$ is not a rectangle by Munkres' definition in this book.

Munkres' Definition: $Q = [a_1, b_1] \times \cdots \times [a_n, b_n]$ is a rectangle if and only if $a_i < b_i$ for all $i \in \{1, \cdots, n\}$.