Vtyjkyuk

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

Definition.
Let $A$ be a subset of $mathbbR^n$.
We say $A$ has measure zero in $mathbbR^n$ 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 $mathbbR^n$?

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

p.97 Exercise 3.

Show that the set $mathbbR^n-1 times 0$ has measure zero in $mathbbR^n$.

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 $mathbfx$ of $mathbbR^n$ such that $a_i leq x_i leq b_i$ for all $i$.

This is his definition of rectangle in his book.

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
$$ sum_i=0^infty v(Q'_i) < epsilon qquadtextandqquad bigcup_i=0^infty Q_i subseteq bigcup_i=0^infty Q'_i $$

I found the following statement:

p.91 Theorem 11.1 (d) if $Q$ is a rectangle in $mathbbR^n$, then Bd $Q$ has measure zero in $mathbbR^n$ 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 $i in 1, cdots, n$.