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.



A measurable set such that its intersection and difference with every interval have the same measure
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)


