Measure Theory - Why doesn't empty interior imply zero measure?

Question

I'm studying for a qualifying exam in measure theory and ended up "proving" a set with empty interior implies zero measure. I know this isn't true (irrational numbers provide a counterexample in $\mathbb{R}$) but can't find my error:

Let $E \subset \mathbb{R}^n$ be a set with empty interior. Note that the closure, $\overline{E} = \text{int}(E) \cup \text{Bd} (E)$, where $\text{int}(E)$ is the interior of $E$ and $\text{Bd}(E)$ is the set of boundary points of $E$. By monotonicity and subadditivity of outer measure, $|E|_e \le |\overline{E}_e| \le |\text{int}(E)|_e + |\text{Bd}(E)|_e=0$ by assumption and because the boundary of $E$ is $(n-1)-$dimensional.

Can anyone find my mistake?

Answer 1

You have no reason to assume that the measure of the boundary is $0$ too. And you provided an example yourself: the irrationals. Its boundary is $\Bbb R$, whose Lebesgue measure is infinity.

Answer 2

As your example with irrational numbers shows, $\text{Bd}$ is not necessary $(n - 1)$-dimensional, it can very well be the entire space.

(when talking about dimension for objects other than vector space it's usually better to specify what exactly dimension you use, but in this case it doesn't matter)