There is an elementary theory of subsets of $\Bbb{R}^n$ of measure zero, namely one defines the volume of a cube in the obvious way and one says that a subset $A$ has measure zero if given any $\epsilon>0$ there exists a countable number of cubes that cover $A$ and such that the sum of the volumes of the cubes is $\leq \epsilon$.
One can show, with modest effort, that this notion is invariant under diffeomorphisms and thus leads to the notion of subsets of measure zero on a smooth manifold. This notion shows up in Sard's Theorem which says that the set of critical values has measure zero.


Is there an elementary argument why non-empty open subsets do not have measure zero? Evidently it follows from standard measure theory, but for my topology course I would appreciate it if there was an elementary argument, but I can't think of one and I can't find one.


This is stated as an exercise in Lee's book on smooth manifolds, but it's not obvious to me. Note that even $n=1$ seems tricky.



Elementary proof that an open subset of $\Bbb{R}^n$ does not have measure zero?
The first part of my argument is borrowed from Iosif Pinelis, and the second part is different.


Every open set contains a closed cube, so it suffices to show the closed cube does not have measure $0$. In fact, we will show that for a cube of side length $r$, the total volume of open cubes covering it is at least $r^n$.


To do this, assume that it is covered by open cubes of total volume $<r^n$, and note that we can assume there are finitely many cubes in the covering by compactness. Let $\ell\_1,\dots, \ell\_m$ be the side lengths of the cubes.


Say our cube is $[0,r]^n$. Consider only the points of the form $(x\_1,\dots, x\_n)$ where $x\_i \in \{0 , r/N, 2r/N, 3r/N, \dots, r\}$. There are $(N+1)^n$ such points. The maximum number of points of the form $\{0 , r/N, 2r/N, 3r/N, \dots, r\}$ contained in an interval of length $\ell$ is $\leq N \ell+1$, so the number of such points contained in a cube of side length $\ell$ is $\leq (N\ell/r+1)^n$.


Since every point is in one cube, we obtain
$$ \sum\_{i=1}^m (N \ell\_i/r+1)^n \geq (N+ 1)^n$$


Divide both sides by $N^n$ and take the limit as $N$ goes to $\infty$ to obtain


$$ \sum\_{i=1}^m \ell\_i^n / r^n \geq 1$$ or $$ \sum\_{i=1}^m \ell^n \geq r^n,$$ as desired.


