Partitioning ℝ into sets $A$ and $B$, such that the measures of $A$ and $B$ in each non-empty open interval have an "almost" non-zero constant ratio

Motivation: I want to partition $\mathbb{R}$ into sets $A$ and $B$, where the measures of $A$ and $B$ in each non-empty open interval have positive ratios with an upper and lower bound, such that the bounds have the smallest absolute difference.

In other words, I want to partition $\mathbb{R}$ into sets $A$ and $B$, such that the measures of $A$ and $B$ in each non-empty open interval have an "almost" non-zero constant ratio.

Thereby, suppose $\lambda$ is the Lebesgue measure on the Borel $\sigma$-algebra: i.e., $\mathcal{B}(\mathbb{R})$

Question:

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 the set function $c:\mathbb{R}^{2}\times\mathcal{B}(\mathbb{R})^{2}\to\mathbb{R}$ satisfies: $$\lambda(A\cap I)=c(a,b,A,B)\cdot\lambda(B \cap I)$$ where for some $\normalsize{\varepsilon} \, \small{\gtrapprox}\, \normalsize{0}$ (e.g., $\varepsilon=.01$),

   • the set functions $q:\mathbb{R}^{2}\times\mathcal{B}(\mathbb{R})^{2}\to\mathbb{R}$ and $r:\mathbb{R}^{2}\times\mathcal{B}(\mathbb{R})^{2}\to\mathbb{R}$ bound:$$q(a,b,A,B)\le c(a,b,A,B)\pm \varepsilon\le r(a,b,A,B)$$where for all $a,b\in\mathbb{R}$, the set functions $L:\mathcal{B}(\mathbb{R})^2\to\mathbb{R}$ and $U:\mathcal{B}(\mathbb{R})^{2}\to\mathbb{R}$ bound: $$L(A,B) \le (r(a,b,A,B)-q(a,b,A,B)) \pm \varepsilon\le U(A,B)$$such that for all $A,B\in\mathcal{B}(\mathbb{R})$, the real numbers $L_1$ and $U_1$ bound: $$L_1\le (U(A,B)-L(A,B))\pm\varepsilon\le U_1$$ where we want $A,B\in\mathcal{B}(\mathbb{R})$, such that:$$L_1=(U(A,B)-L(A,B))-\varepsilon$$