Partitioning $\Bbb R$ into sets $A$ and $B$, where the measures of $A$ and $B$ in every non-empty interval of constant length have a constant ratio

Question

Motivation: (If you don't need the motivation, skip it.)

I've tried partitioning $\mathbb{R}$ into sets $A$ and $B$, where the Lebesgues measures in every non-empty open interval have a non-zero constant ratio not equal to one. It turns out such a constant doesn't exist.

I also tried partitioning $\mathbb{R}$ into sets $A$ and $B$, such that the measures of $A$ and $B$ in every non-empty open interval have an "almost" non-zero constant ratio. This is also impossible, since the lower bound of the non-constant ratio is always zero and the upper bound is positive infinity.

Hence, I made non-empty open intervals with a constant length, such that the measures of $A$ and $B$ in every non-empty open interval have a constant ratio w.r.t. the constant length.

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Thereby, 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 $\varepsilon>0$, there exists a non-zero constant $c$, where for all $a\in\mathbb{R}$:

$$\lambda(A\cap (a,a+\varepsilon))=c\cdot\lambda(B \cap (a,a+\varepsilon))?$$

Answer 1

If you arrange the quantifier in this way, i.e. the constant does not depend on $a$, then no such pair of sets can exist.

First thing to note is that for any open, nonempty interval $I$ we must have $\lambda(A\cap I), \lambda(B\cap I)>0$. If one of them was zero, then both are zero due to the requirement that $c>0$. However, they cannot both be zero as otherwise we would get, due to $A\cup B=\mathbb{R}$, that $$ 0<\lambda(I) = \lambda(I \cap A)+\lambda(I\cap B)=0. $$

For all $\delta\in [0,1]$ we have $$ c_{1+\delta} \lambda(B\cap (0,1+\delta)) = \lambda(A\cap (0,1+\delta)) \leq \lambda(A\cap (0,1)) + \lambda(A\cap(\delta, 1+\delta)) = c_1 (\lambda(B\cap (0,1))+\lambda(B\cap (\delta, 1+\delta))) \leq 2c_1 \lambda(B\cap (0,1+\delta)). $$

Thus, we get for all $\delta\in[0,1]$ that $$ c_{1+\delta} \leq 2c_1. $$

Furthermore, we get for all positive natural numbers $n$ that $$ c_1 \lambda(B\cap (0,1)) = \lambda(A\cap(0,1)) = \lambda(A\cap (0,1/n)) + \dots + \lambda(A\cap ((n-1)/n,1)) = c_{1/n} (\lambda(B\cap (0,1/n)) + \dots + \lambda(B\cap ((n-1)/n,1))) = c_{1/n}\lambda(B\cap (0,1)). $$ Thus, we get $c_1=c_{1/n}$. In fact, we get for all $\delta>0$ and all positive natural numbers $n$ that $$ c_{\delta} = c_{\delta/n}. $$ Combining everything, we obtain for all $\varepsilon>0$ we have $$ c_\varepsilon \leq 2c_1.$$ However, as the density is now bounded from above, the arguments in your previous question 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 apply and rule out that such a pair of sets can exist.