Does there exist a subset $S$ in $[0,1]$ such that for all interval, $S$ and its complement in the interval has the same Lebesgue masure?

Question

I would want to see a strengthening or a disproof of a result from an exercise from Rudin's Real and Complex Analysis, that is asked before at here Construct a Borel set on R such that it intersect every open interval with non-zero non-"full" measure and here Construction of a Borel set with positive but not full measure in each interval and possibly elsewhere.

Does there exists a subset $S\subseteq [0,1]$, such that for all intervals $(a,b)\subseteq [0,1]$, the subset and its complement are measurable and has the same positive Lebesgue measure in $(a,b)$, that is, $\mu((a,b)\cap S)=\mu((a,b)\setminus S)=\dfrac{b-a}{2}$? Thanks.

I guess you mean $\mu$ is the Lebesgue measure and $$\mu((a,b)\color{red}{\cap} S)=\frac{b-a}{2}$$ — Sangchul Lee, Feb 22 '24 at 23:52
See also this. LDT works but it's a bit overkill - this is a simpler reason. — Izaak van Dongen, Feb 23 '24 at 14:22

score 2 · Accepted Answer · answered Feb 22 '24 at 23:51

Assuming $\mu$ stands for the Lebesgue measure, the existence of such a set $S$ would imply that

$$\mathbf{1}_{S}(x)=\frac{\mathrm{d}}{\mathrm{d}x}\mu(S\cap[0,x])=\frac{1}{2}\quad\text{a.e.}$$

by Lebesgue's differentiation theorem, which is of course a contradiction.

Does there exist a subset $S$ in $[0,1]$ such that for all interval, $S$ and its complement in the interval has the same Lebesgue masure?

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