Does there exist a Lebesgue measurable set $E$ such that for all $n>0$ $m(E \bigcap [0,n])=\frac{n}{2}$?
I think we can't have such this set, because if so, then we can write $E$ as a countable union of the intersections $E \bigcap [0,n]$ and then pass the measure with monotonicity we get a contradiction.
I am not sure whether my argument is correct, any idea or hint is appreciated
I'm going to assume $n = x$ is real, because otherwise the question is trivial.
First, if you know the Lebesgue differentiation theorem, then the answer is obviously no, because the condition on $E$ implies that $E$ (or equivalently, $\chi_E$) has no Lebesgue points.
Otherwise, you can prove it directly this way. Let's focus attention on $E_1 = (0,1] \cap E$. We have $m(E_1) = 1/2$. Thus $E_1$ is contained in some open set $U \subseteq (0, +\infty)$ such that $m(U) \leq 3/4$ (using the definition of Lebesgue measure via outer measure).
Now write $U$ as a countable disjoint union of intervals $I_k$. We have $m(E \cap I_k) = \frac{1}{2}m(I_k)$ by the hypothesis on $E$. So $$1/2 = m(E_1) = \sum_k m(E_1 \cap I_k) \leq \sum_k m(E \cap I_k) = \sum_k \frac{1}{2}m(I_k) = \frac{1}{2}m(U) \leq 3/8.$$