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So I've come across this question:

given $A_{1},A_{2},...\ $ Jordan measurable sets in $\mathbb{R}^{n}$, and given that $A=\cup_{n=1}^{\infty}{A_{n}}$ is bounded, is $A$ J-measurable?.

Now, I can think of $\cup_{q\in[0,1]\cap\mathbb{Q}}{\{q\}}$ as a counterexample for this, but I haven't been able neither to find a counterexample in which at least a numerable quantity of the $A_{i}$ are open (and not the empty set) nor to prove that if those $A_{i}$ are open then $A$ must me measurable, so if anyone knows how to answer in that case it'd be cool. Thanks!

Mohsen Shahriari
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valkyriebel
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1 Answers1

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Let $\mathcal{C}$ be the fat Cantor set. The important properties of $\mathcal{C}$ are the following: $\mathcal{C}$ has Lebesgue measure $1/2$, is a perfect subset of $[0,1]$, and has empty interior. In particular, $\mathcal{C}$ is not Jordan measurable, as having empty interior implies its inner Jordan measure is $0$, while having Lebesgue measure $1/2$ and being closed & bounded implies its outer Jordan measure is $1/2$.

Let $U = (0,1) \setminus \mathcal{C}$, which is a bounded open set. $U$ cannot be Jordan measurable, being the set difference of the Jordan measurable set $(0,1)$ and the Jordan nonmeasurable set $\mathcal{C}$.

But $U$ is still open, so it is a countable union of bounded open intervals, and those open intervals are clearly Jordan measurable. Thus, we have a union of countably many open sets which is bounded but not Jordan measurable.

Hayden
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