Does there exist a measurable subset $E$ such that $\varepsilon < \frac{|E\cap{I}|}{I} < 1-\varepsilon$ for every finite interval I =[a,b]? Given $0 < \varepsilon < \frac{1}{10}$, does there exist a measurable subset $E \subset \mathbb{R^1}$ such that $\varepsilon < \frac{|E\cap{I}|}{I} < 1-\varepsilon$ for every finite interval I =[a,b]? Prove or disprove it . Here ||denotes the Lebesgue measure.
I think it is exist . But I have no ideal of the example.

No. Note:
$$\varepsilon<\frac{|E\cap I|}{|I|}=\frac{1}{|I|}\int_I \chi_E(s)~ds<1-\varepsilon$$
Then take $$\lim_{I\to x}\frac{1}{|I|}\int_I \chi_E(s)~ds$$
Lebesgue differentiation theorem tells us that a.e., $\lim_{I\to x}\frac{1}{|I|}\int_I \chi_E(s)~ds=\chi_E(x)$
The bounds show that $\varepsilon\le\chi_E(x)\le 1-\varepsilon$ for almost all $x$. However $\chi_E$ can only be $1$ or $0$ a.e. Contradiction.
