Let $C_0 = [0,1]$ and if $C_n$ is given as a disjoint union of intervals, construct $C_{n+1}$ by removing from each interval $I$ an open interval of length $(n+2)^{-2}|I|$ in the middle of each interval, and then define $C$ as the intersection of all those intervals. I want to show that this set is not Jordan-measurable.

Note that contrary to the "standard" Cantor set, where we remove $1/3 \cdot |I|$ in the middle of each interval, here we remove $(n+2)^{-2} \cdot |I|$ in each stage.

Using some Theory about Lebesgue-measurable sets, it is easy to show this, because counting what we remove we have $$ 1 - \sum_{n=0}^{\infty} \frac{1}{(n+2)^2} = 1 - \left( \sum_{n=1}^{\infty} \frac{1}{n^2} - 1 - 1/2 \right) = \frac{5}{2} - \frac{\pi^2}{6} > 0 $$ and so it has positive Lebesgue-meassure. Also if it is Jordan-measurable, then its Jordan-measure equals its Lebesgue-measure, and so it would have positive Jordan-measure. But because it equals its boundary, its boundary has positive Jordan-measure, which implies it is not Jordan-measurable, and by this contradiction it cannot be Jordan-measurable.

But I want to proof this fact without using the Lebesgue-measure, so do you know a proof?

Some facts about Jordan-measurable (J-measurable for short) sets I know:

  • a set is J-measurable iff its inner and outer J-measure coincide (definition)
  • a set is $M$ J-measureable iff for each $\varepsilon > 0$ there exists sets $S,T$ which could be written as a union of a finite number of intervals such that $$ S \subseteq M \subseteq T, \quad |T| - |S| < \varepsilon $$
  • a set $M$ is J-measureable iff $|\partial M| = 0$.

and also a fact about approximations if we successively partition $\mathbb R$ in intervals of length $1/2^k, k = 1,2,3,\ldots$.

But I do not see a way to use this facts in any useful way here.


The inner Jordan measure of $C$ is zero, since it does not contain any intervals.

Suppose we covered $C$ by finitely many intervals $I_k$. Let $\delta$ be the minimum of the lengths of $I_k$. Choose $n$ such that during the construction of $C$, all gaps removed at the stages $n$ and later are of length less than $\delta$.

Claim: $C_n \subset \bigcup (2I_k)$, where $2I_k$ means the interval with same center and twice the length.

Proof of claim: Indeed, let $x\in C_n$. If $x\in C$, it is covered by some $I_k$. Otherwise, it is within a gap removed at some stage $\ge n$. Let $y$ be an endpoint of this gap. Note that $y\in C$ and $|x-y|<\delta$. There is $k$ such that $y\in I_k$. It follows that $x\in 2I_k$. $\quad \Box$

Since $C_n$ is the finite union of intervals, we know exactly what its Jordan measure is, and it's bounded from below by a constant independent of $n$. This gives a lower bound on $\sum |I_k|$. In conclusion, the outer Jordan measure of $C$ is positive.

  • $\begingroup$ "it's bounded from below by a constant independent of n" .. I think the constant is 0, only. Please help prove otherwise. $\endgroup$ Oct 7 '20 at 13:22

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