10

Unlike Lebesgue measurable sets, Jordan measurable sets do not form a Sigma algebra. So my question is, what is the Sigma algebra $J$ generated by Jordan measurable sets?

All intervals are Jordan measurable, so $J$ contains all the Borel sets. But this answer shows that not all Jordan measurable sets are Borel sets, so the Borel Sigma algebra is a proper subset of $J$. And all Jordan measurable sets are Lebesgue measurable, so $J$ is a subset of the Lebesgue Sigma algebra. But are there Lebesgue measurable sets not contained in $J$?

Keshav Srinivasan
  • 10,204
  • 3
    This https://www.jstor.org/stable/44153840 seems to contain the answer, if you have access to it. I don't. – daw Oct 31 '18 at 14:56
  • 1
    @daw: If anyone without access is interested, I have a copy of the original journal volume (from personal subscription) from which I can make photocopy, followed by a .pdf of that photocopy. Send me an email request. My email can be deduced from information in my mathematical stack exchange profile. – Dave L. Renfro Oct 31 '18 at 16:53
  • @daw I got access to it. See the link in my answer. – Keshav Srinivasan Nov 11 '18 at 03:40

1 Answers1

6

I found the answer in this journal paper. The Sigma algebra generated by the Jordan measurable sets is the collection of all sets which can be written as a union of a Borel set and a subset of a measure $0$ $F_\sigma$ set. As a point of comparison, the collection of Lebesgue measurable sets is the collection of all sets which can be written as a union of a Borel set and a subset of a measure $0$ Borel set.

And the paper gives an example of a Lebesgue measurable set which is not in the Sigma algebra generated by Jordan measurable sets. Let $\beta$ be a Bernstein set, i.e. a subset of $\mathbb{R}$ such that both it and its complement intersects every uncountable closed subset of $\mathbb{R}$. (This post describes how to construct such a set using the axiom of choice.) And let $\gamma$ be a dense measure-$0$ $G_\delta$ subset of the fat Cantor set. (This answer describes how to construct such a set.) Then $\beta\cap\gamma$ is a Lebesgue measurable set which is not in the Sigma algebra generated by Jordan measurable sets.

But there may still be unsolved problems about this Sigma algebra, so I just posted a question on MathOverflow about it.

Keshav Srinivasan
  • 10,204