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Is there a measurable set $A$ such that $m(A \cap B) = \frac12 m(B)$ for every open set $B$?
Is there a measurable set $E \subset [0,1]$ such that for any $0 < a < b<1$, the Lebesgue measure $$m(E \cap [a,b])= \frac{b-a}{2}?$$
I am stumped and have no idea.
No, there is no such a set.
Since $E$ is measurable (with positive measure) for all $\epsilon > 0$ there exist a sequence of intervals $(I_n)$ in $[0,1]$ such that $E \subset \bigcup I_n$ and $$ \sum m(I_n) < m(E) + \epsilon. $$ Hence we obtain that $$ \frac{3}{4} \sum m(I_n) < m(E). $$ Thus $$ \frac{3}{4} \sum m(I_n) < m(E) \leq m(E \cap \bigcup I_n) = m(\bigcup E \cap I_n) \leq \sum m(E \cap I_n), $$ which implies that there exists positive integer $k$ such that $$ \frac{3}{4} m(I_k) \leq m(E \cap I_k). $$ The last inequality contradicts your assumption that for any interval $I$ we have $m(E \cap I) = m(I)/2$.
Whats more, in the exactly same way you can prove general fact:
If $E \subset \mathbb{R}$ is a measurable set with positive measure and $0 \leq c < 1$ than there exists an interval $I$ such that $$ m(E \cap I) \geq c \cdot m(I). $$
No there can be no such $E$.
Let $\chi_E$ be the characteristic function of $E$. At every Lebesgue point $x$ of $\chi_E$, we have
$$\chi_E(x) = \lim_{r\to 0}\ \frac{1}{m(B_r)} \int_{B_r(x)} \chi_E(y) \ dy$$
Since almost every point of an integrable function is a Lebesgue point, the set of $x$ for which this limit is not equal to $0$ or $1$ must have measure $0$.
In particular, it cannot be true that
$$m(E\cap[a,b]) = \frac{b-a}{2}$$
for all $a,b$, since that would imply that also
$$\frac{1}{2} = \frac{m(E\cap[x-r,x+r])}{m([x-r,x+r])} = \frac{1}{m(B_r)} \int_{B_r(x)} \chi_E(y) \ dy$$
for all $x$ and all small enough $r>0$. (Which would be in contradiction to the fact about Lebesgue points above.)
