Can we partiton ℝ into sets $A$ and $B$, where the Lebesgue measures in every non-empty open interval have a non-zero constant ratio not equal to one?

Question

Suppose $\lambda$ is the Lebesgue measure on the Borel $\sigma$-algebra.

Does there exist an example of sets $A,B\subset\mathbb{R}$, where:

1. $A\cup B=\mathbb{R}$
2. $A\cap B=\emptyset$
3. for all non-empty open intervals $I:=(a,b)\subset\mathbb{R}$, such that $c\neq 1$ is a non-zero constant:

$$\lambda(A\cap I)=c\cdot\lambda(B \cap I)?$$

This answer might offer a hint, despite $c$ being non-constant, where $c$ is positive (and):

$$\lim_{t\to\infty}\frac{\lambda(A\cap[-t,t])}{2t}=\frac{2}{3}$$

$$\lim_{t\to\infty}\frac{\lambda(B\cap[-t,t])}{2t}=\frac{1}{3}$$

Answer 1

$$\def\l{\lambda}$$ Check the Lebesgue Density Theorem. If $A$ is measurable then at almost every point $x\in \mathbb R$ we have that $$\lim_{\epsilon\to0} \frac{\l(A \cap B_{\epsilon}(x))}{\l(B_{\epsilon}(x))} $$ exists and is either equal to $1$ or $0$. This tells us that the only $c$ that can work for your example is $c=0$. Even the value $c=1$ that you want to avoid is impossible (since it would lead to a limiting ration of $1/2$).

Of course $A=\mathbb{Q}$, $B=\mathbb{R}\setminus \mathbb{Q}$ achieves $c=0$.