Let $A\in\mathcal{B}(\mathbb{R})$, be a set satisfying that $\forall I\in\ \mathcal{I}_\mathbb{R}\colon \lambda(A\cap I)=\lambda(I)/2$. Can such a set $A$ really exist? The answer is no, see this question. I want to find a proof without using the Lebesgue density theorem.
Here is my attempt: Let $A\subset \mathbb{R}^d$ be open. I can create an open subset $B\subset A$ with measure $\lambda(B)\in (0,\lambda(A))$ such that $B$ is dense in $A$.
This can be done by enumerating $\mathbb{Q}\cap A$ by $\{q_n\}_{n\in\mathbb{N}}$ and setting
$B_\epsilon = \bigcup_{n\in\mathbb{N}}(q_n-\epsilon 2^{-(n+1)},q_n+\epsilon 2^{-(n+1)})$
This set is dense in $A$ and since $\lambda(B)\le \epsilon$ by the continuity in $\epsilon$ I can give it any measure $\lambda(B)=\epsilon$ I want.
Now here is the part, where I somehow implicitly have to use the density theorem:
Since $B$ is dense in $A$ we have that $\lambda(A\cap B_\epsilon) = \lambda(B_\epsilon)=\epsilon$ except on the points on the boundary of $B_\epsilon$, we would then have $\lambda(A\cap B_\epsilon)=0$
which is in some way the density theorem. Then I would obtain a contradiction by assuming that such a set $I$ exists (The density would then be $1/2$ for all $B_\epsilon$).
Is this correct so far?
Assume it exists and assume that has finite positive measure.
(If has zero measure then it is obvious that is not true)
Then $\forall \epsilon>0 $,exists a covering $\{I_n:n \in \Bbb{N}\}$ of $A$ such that $\sum_nm(I_n) \leq m(A)+\epsilon$
Thus $A=\bigcup_n(A \cap I_n)$ and $$m(A) \leq \sum_nm(A \cap I_n)=\sum_n\frac{m(I_n)}{2} \leq \frac{m(A)+\epsilon}{2}$$
Choose $\epsilon =\frac{m(A)}{4}$ and you have a contradiction.
if $A$ has infinite measure and we assume that has the property then set $A_m=A \cap[-m,m]$ will have the same property so we can restrict ourselves to every subinterval of $[-m,m]$ so by the previous case you can arrive to a contradiction again