10
$\begingroup$

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?

$\endgroup$
2
  • 6
    $\begingroup$ Essentially, the topological boundary is a very diffuse thing and only very loosely related to the boundary you would encounter in (differential) geometry, where the claim about the dimension would be correct. $\endgroup$ Jun 7, 2023 at 7:39
  • 15
    $\begingroup$ In general, when faced with a possible proof and a possible counterexample to the same proposition, you can apply your proof to your counterexample and see where things break. If $n=1$ and $E$ is the set of irrationals, what are $\operatorname{int}(E)$ and $\operatorname{Bd}(E)$ and $\bar{E}$, what are their measures, and where does a mistake appear in the proof? $\endgroup$
    – Stef
    Jun 7, 2023 at 8:07

2 Answers 2

23
$\begingroup$

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.

$\endgroup$
18
$\begingroup$

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)

$\endgroup$
1
  • $\begingroup$ What the sentence in parenthesis means is, for an arbitrary subset, such as $\text{Bd}(E)$ here, it may not be well-defined to talk about what the "dimension" of the subset is. Unless you specify precisely the definition of "dimension" you use. (Of course, if the set is a linear or affine subspace, or an embedded manifold, or something, it is clear what the dimension is.) $\endgroup$ Jun 7, 2023 at 23:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .