The n-Category Café

There are a couple of easy mathematical points. One is that a set of measure zero certainly has empty interior—because if not, it would contain a nontrivial cuboid $(a_1, b_1) \times \cdots \times (a_n, b_n)$. The other is that the converse fails: a set can have positive measure but empty interior (e.g. the irrational numbers).

What I only just learned is that there are compact sets with positive measure but empty interior. (I’ll give an example in a moment.) That might sound like an instantly forgettable technicality, but actually it’s the gnarly fact behind an important distinction. Exaggerating only slightly:

the difference between measure zero and empty interior is the difference between Lebesgue and Riemann integrability.

Before I get to that, here’s the promised example of a compact subset of $\mathbb{R}$ with positive measure but empty interior. It’s easy, but it’s not really the main point of this post, so skip the next two paragraphs if you’re not in the mood for details.

Consider the classic Cantor set, obtained from $[0, 1]$ by removing the middle open $1/3$, then removing the middle open $1/3$ of each of the $2$ resulting intervals, then removing the middle open $1/3$ of each of the $4$ resulting intervals, and so on ad infinitum. This isn’t the example, since the Cantor set has measure zero: the set produced at the $n$th stage has measure $(1 - 1/3)^n = (2/3)^n$, which converges to $0$ as $n \to \infty$.

But we can adapt the construction: at the $n$th stage, instead of removing the middle open $1/3$ from each of the $2^n$ intervals, remove the middle open $x_n$. Here $(x_n)$ is any sequence of numbers between $0$ and $1$. So, the three occurrences of “$1/3$” in the first sentence of the previous paragraph become “$x_0$”, “$x_1$” and “$x_2$”, respectively. The measure of the resulting Cantor-like set is $$\prod_{n = 0}^\infty (1 - x_n),$$ which is nonzero if we choose $(x_n)$ to converge rapidly to $0$. And our set certainly has empty interior, since the $n$th stage contains no intervals of length $> 2^{-n}$. For a particular choice of $(x_n)$, our set is called the fat Cantor set.

What does this have to do with Lebesgue and Riemann integrability? The answer lies in the notion of ‘Jordan measurability’. A subset of $\mathbb{R}^n$ is called Jordan measurable if its characteristic function (indicator function) is Riemann integrable. You might expect this to be called ‘Riemann measurability’, but apparently not. It’s sometimes called ‘Peano–Jordan measurability’ too.

The crucial fact relating Jordan and Lebesgue measurability, at least for compact sets, is this:

A compact set is Jordan-measurable if and only if its boundary has Lebesgue measure zero.

An equivalent statement is that a compact set $A$ is Jordan-measurable if and only if $Vol(A) = Vol(Int A)$, where $Vol$ means Lebesgue measure and $Int$ means interior.

But the boundary of a compact set is compact and has empty interior. So if every compact set with empty interior had measure zero, Jordan measurability would be the same as Lebesgue measurability. Conversely, if Jordan measurability were the same as Lebesgue measurability then every compact set $A$ with empty interior would satisfy $Vol(A) = Vol(Int A) = 0$.

Conclusion:  “Measure zero” is different from “empty interior”—even for compact sets—precisely because Lebesgue measurability is different from Jordan measurability.