Let $X$ be a bounded subset in $\mathbb{R}^n$ of Lebesgue measure $0$. Then $X$ may be represented as a countable union of pairwise disjoint sets $X_n$, such that measure of the closure of each $X_n$ is still $0$. Is this right?
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2That doesn't seem right. – Aug 19 '21 at 10:49
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Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos Aug 19 '21 at 10:54
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Can you give a counterexample? – Dekay Aug 19 '21 at 11:01
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@Dekay At the moment, no. I'm thinking about it. – Aug 19 '21 at 11:02
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That is not correct. A set with the quoted property must be of first category. But it is known that $\mathbb{R}^n$ is the union of a zero set and a set of first category (cf this answer.), while $\mathbb{R}^n$ itself is not of first category.
orangeskid
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1Note that such a set is constructed in chapter 1 of Oxtoby’s book “measure and category “ – Henno Brandsma Aug 19 '21 at 12:07
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1@Dekay: Omitting the requirement that the sets $X_n$ are pairwise disjoint (which I think is superfluous, but I haven't given this any thought), one can consider 4 types of "small" sets, each of which forms a distinct $\sigma$-ideal of sets: measure zero, first category, simultaneously measure zero and first category, subset of a countable union of closed measure zero sets. (continued) – Dave L. Renfro Aug 19 '21 at 13:08
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1The first two are orthogonal $\sigma$-ideals (hence in a strong way the third is a proper subcollection of each of the first two), and the fourth (what you're essentially dealing with) is a proper subcollection of the third. This last result means that there exists a set that is both measure zero and first category that cannot be covered by (i.e. a subset of) a countable union of sets of the type you're considering, which incidentally are exactly the discontinuity sets for a Riemann integrable function. – Dave L. Renfro Aug 19 '21 at 13:11
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