TITLE: Elementary proof that an open subset of $\Bbb{R}^n$ does not have measure zero?
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: It should also be mentioned that $[0,1]$ being not a null Lebesgue set can also be derived by the Lebesgue-Vitali characterization theorem for Riemann integrable function. If  $[0,1]$ were negligible, any bounded function on it would be  Riemann integrable, but we know that for instance the Dirichlet function $\chi_{[0,1]\cap\mathbb{Q}}$ is not.