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.
Best Answer
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.