# A measurable set such that its intersection and difference with every interval have the same measure [duplicate]

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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.

## marked as duplicate by Mateusz Kwaśnicki, Gerald Edgar, Community♦May 16 at 14:48

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• 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 $S$. – Mateusz Kwaśnicki May 16 at 8:30
• Thanks a lot. This is very helpful. – avk255 May 16 at 19:06

## 1 Answer

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)

• 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. – Gerald Edgar May 16 at 12:25
• Great. I was about to ask this question. Thanks! – avk255 May 16 at 12:32