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I'm looking for two sets in $\mathbb{R}$ which are both uncountable and dense, and where one is the complement of the other. I know the question has already been asked here, but the solutions still didn't feel quite right to me: the sets didn't feel "even" enough in $\mathbb{R}$. So I realized what the question was that I really wanted answered.

Are there two Lebesgue measurable subsets of $\mathbb{R}$, one of which is the complement of the other, which have the same Lebesgue measure on any open bounded subset of $\mathbb{R}$?

The density of each set in $\mathbb{R}$ is as well as their uncountability is obvious, and the sets feel "even" everywhere.

Sambo
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1 Answers1

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As it turns out, a variant of this question has already been answered here: the answer is no. :(

Still, I noticed that this is only for Lebesgue measure. Are there any other types of measure where this is possible?

Sambo
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    I think you need some restrictions for the measure, obviously taking the trivial measure giving 0 everywhere together with the first answer gives an example. (or take too poor sigma algebra like the trivial and assign measure 1 to everything and masure 0 for the empty set) – Yanko Jul 11 '17 at 22:58
  • Yeah, that's a good point. I think the question I referenced in this answer is probably a better formulation: that the measure is equal to half the length of the interval. – Sambo Jul 13 '17 at 15:08