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Not all Borel sets are Jordan measurable. For instance, the set of rational numbers is Borel but not Jordan measurable. But my question is, are all Jordan measurable sets Borel sets?

I know all Jordan measurable sets are Lebesgue measurable, but of course not all Lebesgue measurable sets are Borel sets.

Keshav Srinivasan
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  • Note that any closed continuum-sized set of measure zero in the Lebesgue sense generates a counterexample in the usual way, since all of its subsets are Jordan measurable (they have null closure). You're never going to get a "---measurable implies Borel" kind of result (at least, for any interesting notion of "measurable"): there are simply too few Borel sets, and too many ways to take a non-measurable set and "shrink" it so it becomes negligible in the sense of the measure. (cont'd) – Noah Schweber Oct 31 '18 at 14:00
  • Besides "subsets of the usual Cantor set," another really useful generator of counterexamples is "a disc minus a subset of the boundary," since there are $2^{2^{\aleph_0}}$-many of these and convex sets are usually as nice as can be from a measure-y perspective. – Noah Schweber Oct 31 '18 at 14:06
  • @NoahSchweber By the way, I just posted a follow-up question: https://math.stackexchange.com/q/2979131/71829 – Keshav Srinivasan Oct 31 '18 at 14:25
  • Yeah, I saw it (and upvoted) - it's a great question, and I don't right now see the answer. It might be useful that a set is in the Jordan $\sigma$-algebra iff it is Borel-mod-closure-null, whereas a set is Lebesgue-measurable iff it is Borel-mod-null (where "closure-null" means "has null closure"). – Noah Schweber Oct 31 '18 at 14:31

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