Lebesgue measure and measurable problem Prove that there does not exist a measurable set $A \subseteq \mathbb{R}$ such that $m(A \cap (a,b))=(b-a)/2$ for all $a<b$
Do I prove such $A$ not in $\mathbb{R}$ or nto measurable, or if $A$ is measurable it will not hold the equation?
 A: Sketch of a proof:
Assune $A$ is measurable. The characterisic function of $A$ (i.e. the function that equals $1$ on $A$ and zero otherwise) belongs to $L^1$ hence by the lebesgue differentiation theorem it follows that for almost all $y\in A$: $$\text{lim}_{x\rightarrow y} \frac{m(A \cap[y,x])}{m([y,x])}=1$$  Pick one of those $y$, then for sufficiently small $x$ this is a contradiction to your claim. 
Remark: There is an elementary proof of this fact too. This proof essentially entails approximating $A$ with simpler sets. 
A: It is possible to show that if $A \subseteq \mathbb{R}$ is a set of positive Lebesgue measure, then for all $0 < \alpha < 1$, there exists a bounded interval $I$ such that
$$m(A \cap I) \geq \alpha m(I).$$ To prove this, you can analyse the two cases whether $m(A)$ is finite or not. (First prove the case $m(A) < \infty$ and then reduce the case of $m(A) = \infty$ to the finite one). After having proven the above, it becomes easy to prove your statement. I let you figure out why.
