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Let $\Omega = [0,1]$. I want a Lebesgue measurable set $S$ with the following property.

$$ \ell(S \cap I) = \ell(I \backslash S)$$ for every subinterval $I$ of $[0,1]$, where $\ell(A)$ is the Lebesgue measure of $A$.

A friend recently told me that Lusin's theorem says that such a set does not exist. I don't seem to find a result I can quote (and learn from) that says the same though. Is it true that such a set does not exist?

Thanks.

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    $\begingroup$ If such $S$ existed, we would have $\ell(S \cap I) = \tfrac{1}{2} \ell(I)$ for every interval $I$, which contradicts the Lebesgue differentiation theorem for the indicator function of $S$. $\endgroup$
    – Mateusz Kwaśnicki
    May 16, 2019 at 8:30
  • $\begingroup$ Thanks a lot. This is very helpful. $\endgroup$
    – avk255
    May 16, 2019 at 19:06

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Let $\lambda$ denote Lebesgue measure.

The subsets $[0,t]$ generate the $\sigma$-algebra of Borel sets on $\mathbb R_+$. So there is at most one Borel measure $\mu$ on $\mathbb R_+$ with the property that $\mu([0,t])=t/2$ for every $t\in \mathbb R_+$. That measure in fact exists: it is $1/2\cdot \lambda$.

Now apply this result to $\chi_S \cdot \lambda$ where $S$ is your set to derive a contradiction. (Here, $\chi_S$ is the function which is $1$ on $S$ and zero otherwise)

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    $\begingroup$ For uniqueness of measure, it is not enough that the sets generate the $\sigma$-algebra. However, your family $[0,t]$ is a $\pi$-system, so in fact the measure is unique. See math.stackexchange.com/questions/1193970/… ... the point is that the collection of sets where two measures agree need not be a $\sigma$-algebra, but must be a $\lambda$-system. $\endgroup$
    – Gerald Edgar
    May 16, 2019 at 12:25
  • $\begingroup$ Great. I was about to ask this question. Thanks! $\endgroup$
    – avk255
    May 16, 2019 at 12:32

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