Is it true that a subset of R with empty interior can contain a subset which is not Lebesgue measurable? Can you give me a justification
Real Analysis: subset of R with empty interior can contain a subset which is not Lebesgue measurable
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1It's true. Half of one way to get this result is knowing that any set of positive measure contains a nonmeasurable set. The other half is knowing that . . . – Dave L. Renfro Jul 01 '21 at 18:32
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3Hint: $\mathbb{Q}$ is dense in $\mathbb{R}$, or equivalently, $\mathbb{R}-\mathbb{Q}$ has empty interior. – user10354138 Jul 01 '21 at 19:05
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@user10354138: Despite writing this answer, it seems that I keep misreading what OP's actually write! This is not the first time I've read "closure has empty interior" when only "empty interior" was written, regardless of what the OP might have intended. I don't know what was intended here, but certainly fewer nontrivial results need to be employed to answer the question as asked than what I was thinking of. – Dave L. Renfro Jul 01 '21 at 20:04