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

Question

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:$$L_1=(U(A,B)-L(A,B))-\varepsilon$$

Answer 1

It seems the following is a weaker version of what you are asking:

Can we find Borel-measurable sets $A, B\subseteq \mathbb{R}$, $A\cap B=\emptyset, A\cup B=\mathbb{R}$, and a positive constants $0<c\leq C$, such that for all non-empty, open interval $I\subseteq \mathbb{R}$ we have $$ c \lambda(B\cap I)\leq \lambda(A) \leq C\lambda(B\cap I).$$

If that is indeed a weaker version of your question, then the answer is no. Let's assume that we have

$$ c \lambda(B\cap I)\leq \lambda(A).$$

We can divide by the measure of the interval and rewrite the measures as the integral over the indicator functions to obtain

$$ \frac{1}{\lambda(I)} \int_I 1_B(x)dx =c \frac{\lambda(B\cap I)}{\lambda(I)}\leq \frac{\lambda(A)}{\lambda(I)} =\frac{1}{\lambda(I)} \int_I 1_A(x)dx.$$

By the Lebesgue differentiation theorem we get for almost all $x\in \mathbb{R}$ $$ c 1_B(x)\leq 1_A(x).$$ Hence, for almost all $x\in B$ we have $$ 1_A(x)\geq c 1_B(x)=c>0.$$ Therefore, up to a null set, $B$ would be contained in $A$. As we have assumed that $A$ and $B$ are disjoint, we get that $B$ is a null set. This already tells you, that getting nonzero density is not really possible, as we have $\lambda(B\cap I)=0$ for all open intervals $I$.

If we also assume that $\lambda(A\cap I)\leq C \lambda(B)$, then $A$ must be a null set too, which is a contradiction as $A\cup B=\mathbb{R}$ implies $$ 1 =\lambda((0,1))= \lambda((0,1)\cap A)+\lambda ((0,1)\cap B) =0.$$

Added: By the Lebesgue differentiation theorem, we get for almost every $a\in \mathbb{R}$ $$\lim_{b\rightarrow a^+} \frac{\lambda(A\cap (a,b))}{\lambda((a,b))}=1_A(a), \quad \lim_{b\rightarrow a^+} \frac{\lambda(B\cap (a,b))}{\lambda((a,b))}=1_B(a).$$ As $A\cap B=\emptyset, A\cup B=\mathbb{R}$, we get that $1_A+1_B=1$. This implies that for almost all $a\in A$ we have $$\lim_{b\rightarrow a^+} c(a,b,A,B)=\infty$$ and for almost all $a\in B$ we have $$\lim_{b\rightarrow a^+} c(a,b,A,B)=0.$$