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Construction of a Borel set with positive but not full measure in each interval was discussed in this post Construction of a Borel set with positive but not full measure in each interval.

Here I am interested in a finer version. If $\mu$ denotes Lebesgue measure, given $0<a<1$, how would one construct a Borel set $A\subset R$ such that $\mu(A\cap I)/\mu(I)=a$ for every open interval $I$ in $R$?

For simplicity, I am interested in the case where the $R$ above is replaced with the unit interval.

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Such a construction contradicts the Lebesgue density theorem, which implies that the limit

$$\lim_{\mu(I) \to 0, x \in I} \frac{\mu(A \cap I)}{\mu(I)} = \chi_A(x)$$

almost everywhere, so $a$ would have to be zero or one.

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