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This question asks for a set of real numbers that is measure-dense, whose complement is also measure dense. In terms of $[0,1]$, the question asks for an $S$ such that for every open interval $I$ we have $0 < \mu(I \cap S) < 1$.

This question can be made a little more precise: is there a set of real numbers $S$ such that for every open interval $I$, we have $\mu(I \cap S) = \frac{1}{2} \mu(I)$? The Cantor-set construction referenced above doesn't really lend itself to finding the volume of $I$ contained in $S$, just guaranteeing that it isn't zero. You can modify that construction to get a set that has measure $1/2$, but it's not clear at all that it satisfies the property mentioned here.

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Julien Clancy
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1 Answers1

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I remember seeing this question here, but I can't find it now. Here's the proof that no such $S$ exists:

Suppose such $S$ existed. Define the ($\sigma$-finite) Borel measure $\lambda(A)=\mu(A\cap S)$. Since $\lambda$ and $\mu/2$ are equal on the open intervals, which generate the Borel $\sigma$-algebra, then $\lambda=\mu/2$. Let $\chi_S$ be the characteristic function of $S$. Then for every $f\in L^1(\mathbb{R},\mu)$,

$$\int_\mathbb{R} f\chi_Sd\mu=\int_{\mathbb{R}\cap S} fd\mu=\int_\mathbb{R}fd\lambda=\int_\mathbb{R} fd\frac{\mu}{2}=\int_\mathbb{R}\frac{1}{2}fd\mu$$

so $\int(\chi_S-\frac{1}{2})fd\mu=0$ for every $f\in L^1(\mathbb{R},\mu)$, so this means that $\chi_S-1/2=0$ a.e., that is, $\chi_S=1/2$ a.e. (with respect to $\mu$). But $\chi_S$ is either $0$ or $1$, an absurd.

Therefore, no such $S$ exists.

Luiz Cordeiro
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