Prove that, there does not exist any $A ⊆ R$ satisfying $m(A ∩ I) = α l(I)$, for all intervals $I ⊂ R$ Let $0 < α < 1$. Prove that, there does not exist any $A ⊆ R$ satisfying $m(A ∩ I) = α l(I)$, for all intervals $I ⊂ R$. Here, $m$ denotes the Lebesgue measure on $R$ and $l$ denotes the length of the interval.
I have been asked to prove this, but I have found a counterexample instead.
$I=(1,3)$ and $α=\frac{1}{2}$, I have found an $A=(2,4)$ such that $m(A ∩ I) = 1 = α l(I)= \frac{1}{2}*2$
I have also been asked a follow up question:
Does there exist $A ⊆ R$ satisfying
$m(A ∩ I) = l(I)$, for all intervals $I ⊂ R?$ and I think a similar example can be constructed. Am I going wrong somewhere?
 A: If such set existed then it would not have zero measure.
Assume that such set $A$ exists with $m^*(A)>0$
Let $\epsilon>0$
There exists a covering $A \subseteq \bigcup_{n=1}^{\infty}$ such that $\sum_{n=1}^{\infty}l(I_n)<m^*(A)+\epsilon$
Now $$0<m^*(A) \leq \sum_n m^*(A \cap I_n)=a\sum_nl(I_n)<a(m^*(A)+\epsilon)$$
Choosing $\epsilon<\frac{1-a}{a}m^*(A)$, we arrive to contradiction.
A: The question clearly says 'for all intervals' and you are giving an example with one fixed interval.
If such  a set exists with finite measure  then $\frac  1 {b-a} \int_a^{b} \chi_A (x) dx =\alpha$ whenever $a <b$. Letting $b \to a$ we get, using Lebesgue's differentiation theorem, $\chi_A(x)=\alpha $ almost everuywhere. But $\chi_A(x)$ takes only the values $0$ and $1$. Hence, such a set does not exist.
For the case when $A$ does not have finite measure consider $A \cap (-1,1)$ and use intervals $I$ conatined in $(-1,1)$ in above proof.
For the second question take $A=\mathbb R$.
Ref: https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem
