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Is a set of Lebesgue measure zero necessarily a countable union of sets of Jordan content zero? This was a question posed by a student in my undergraduate analysis course. I asked an analyst colleague about this and he did not have an answer off the top of his head.

Here are a few thoughts about this question. Since the closure of a set of Jordan content 0 also has content 0 and a compact set has measure 0 iff it has content 0, this question can be rephrased as follows: is any Lebesgue measurable set contained in an $F_\sigma$ set of the same measure?

Off hand this might seem to be rather implausible. However it is true for open sets. They are countable unions of open balls and are contained in the corresponding union of closed balls.

Martin Sleziak
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Zbigniew Fiedorowicz
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  • Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please refer to this MathJax tutorial. – Pantelis Sopasakis May 10 '19 at 17:23
  • This is equivalent to the following question: is any $G_\delta$ set of Lebesgue measure 0 contained in an $F_\sigma$ set of Lebesgue measure 0? – Zbigniew Fiedorowicz May 10 '19 at 22:29
  • The reason you're getting votes to close is that this sounds like a homework problem, and you haven't shown your own efforts or indicated where you are stuck. The people here are passionate about helping others learn mathematics, and the way to do that is to guide them in forming their own solutions, not do the problem for them while they are not engaged. So please edit the question to show what you've done and where you are encountering problems. You will get better help then. – Paul Sinclair May 11 '19 at 03:59
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    @Paul Sinclair: I did a quick check before posting my answer, and the Zbigniew Fiedorowicz who asked this question is (according to Stack Exchange) the same Zbigniew Fiedorowicz who answered this math overflow question nearly 5 years ago, and thus who I would guess is the Zbigniew Fiedorowicz at Ohio State University (Ph.D. in 1975 from Univ. of Chicago). – Dave L. Renfro May 11 '19 at 16:31
  • Well, I apologize for assuming that you were a student. But i was correct as to why you were getting close votes - I was not the only one assuming this was a homework problem. I'm glad that they've been cleared. – Paul Sinclair May 11 '19 at 16:55

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No, in a rather strong way --- Every Jordan content zero set is also a first (Baire) category set, and there exists a Lebesgue measure zero set $Z \subset \mathbb R$ so large that not only is $Z$ not first category (this much alone means it can't be covered by countably many Jordan content zero sets), but in fact $\mathbb R - Z$ is first category. Moreover, there exist sets simultaneously Lebesgue measure zero and first category that cannot be covered by countably many Jordan content zero sets. See my answer to Jordan measure zero discontinuities a necessary condition for integrability.

Dave L. Renfro
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  • Thanks. Actually it was the Riemann integrability theorem which provoked this question. This is usually stated that a necessary and sufficient condition for Riemann integrability is that the set of discontinuities have measure 0. – Zbigniew Fiedorowicz May 11 '19 at 19:13
  • Also your HISTORICAL ESSAY ON F_SIGMA LEBESGUE NULL SETS deserves a better placement than in the obsolete sci.math.forum, which may well disappear in the near future. – Zbigniew Fiedorowicz May 11 '19 at 19:23