Three Notions of Small

Question

I’m currently learning about the history of the development of the Lebesgue integral in Thomas Hawkins’s book “Lebesgue’s Theory of Integration; It’s Origins and Development”

Hawkins is stressing how much early confusion about the Riemann integral came from conflating three different notions of “small” subsets of reals: two topological notions and one measure theoretic notion. The three notions are:

1. Nowhere dense: A set $X \subseteq \mathbb{R}$ is nowhere dense if and only if $\text{int}(\text{cl}(X)) = \emptyset$.
2. First species: For $X \subseteq \mathbb{R}$, let $X’$ denote the set of limit points of $X$. Let $X^{(n)} = ( \cdots ((X’)’)’ \cdots )’$, where limit points are taken $n$ times. $X$ is first species if and only if $X^{(n)} = \emptyset$ for some $n$.
3. Zero Jordan Outer Content: Let $X \subseteq \mathbb{R}$, and let $\mathcal{I} = \{I_1, \cdots, I_n\}$ be a finite collection of intervals such that $X \subseteq \bigcup_j I_j$. The Jordan outer content of $X$ is $c_o(X) = \inf_{\mathcal{I}} \sum_{I \in \mathcal{I}} \ell(I)$, where $\ell(I)$ is the length of the interval $I$. A set has zero Jordan outer content if and only if $c_o(X) = 0$.

Hawkins’s book gives the following results:

• Dini proved that every first species set has zero Jordan outer content.
• Volterra prove that there are nowhere dense sets with positive Jordan outer content.
• du Bois Reymond independently proved that there are nowhere dense sets that aren’t of first species (this is implied by Volterra + Dini, of course).

This leaves open two possible containments.
(A) Is every first species set nowhere dense?
(B) Is every set of zero Jordan content nowhere dense?
I think both of these should be true, but I’d like to have my understanding checked. Thanks!

Answer 1

For (A), if $X$ isn't nowhere dense, namely $\mathrm{cl}(X)$ contains a nondegenerate interval, then so does $X'$ and every $X^{(n)}$, so $X$ is not of first species.

For (B), if finitely many closed intervals cover $X$ then they also cover $\mathrm{cl}(X)$, so $X$ has zero Jordan content iff $\mathrm{cl}(X)$ does. In particular, if $\mathrm{cl}(X)$ contains an interval, then $X$ cannot have zero Jordan content.

Answer 2

Yes to both questions:

For the first question, let $X \subset \mathbb{R}$ be of first species. I claim that $\mathrm{int}(\mathrm{cl}(X)) \subset X^{(n)}$ for all $n \geq 1$. This can be proved by induction: When $n = 1$, let $X_{\mathrm{iso}}$ be the set of isolated points of $X$. Then $X^{(1)} = X' = \mathrm{cl}(X) \setminus X_{\mathrm{iso}}$. For any $x \in X_{\mathrm{iso}}$, clearly $x$ is isolated in $\mathrm{cl}(X)$ as well, so $x \notin \mathrm{int}(\mathrm{cl}(X))$. Thus, $\mathrm{int}(\mathrm{cl}(X)) \subset \mathrm{cl}(X) \setminus X_{\mathrm{iso}} = X^{(1)}$.

Assume the result has been proved for some $n$. Consider $X^{(n+1)} = (X^{(n)})'$. By induction assumption, $O = \mathrm{int}(\mathrm{cl}(X)) \subset X^{(n)}$. $O$ is open in $\mathbb{R}$, so all points in it are limit points of $O$, therefore also of $X^{(n)}$. Thus, $O \subset (X^{(n)})' = X^{(n+1)}$.

Of course, this implies that if $X^{(n)} = \varnothing$ for some $n$, then $\mathrm{int}(\mathrm{cl}(X)) = \varnothing$.

For the second question, it is not hard to see that in the definition we may assume $\mathcal{I} = \{I_1, \cdots, I_n\}$ is a finite collection of closed intervals. Thus, $\cup_j I_j$ is closed, so if $X \subset \cup_j I_j$, then so does $\mathrm{cl}(X) \subset \cup_j I_j$. As such, if $X$ is of zero Jordan content, then so does $\mathrm{cl}(X)$ and then also $\mathrm{int}(\mathrm{cl}(X))$. If $\mathrm{int}(\mathrm{cl}(X)) \neq \varnothing$, then it contains a nonempty open interval, whence the said nonempty open interval must be of zero Jordan content as well, which is clearly impossible.