7
$\begingroup$

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 $J$?

$\endgroup$
  • 2
    $\begingroup$ This jstor.org/stable/44153840 seems to contain the answer, if you have access to it. I don't. $\endgroup$ – daw Oct 31 '18 at 14:56
  • 1
    $\begingroup$ @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. $\endgroup$ – Dave L. Renfro Oct 31 '18 at 16:53
  • $\begingroup$ @daw I got access to it. See the link in my answer. $\endgroup$ – Keshav Srinivasan Nov 11 '18 at 3:40
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Why the downvote? $\endgroup$ – Keshav Srinivasan Nov 24 '18 at 0:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.