Motivation: (If you don't need the motivation or attempts, skip to the question.)
Suppose, we partition $\mathbb{R}$ into sets $A$ and $B$ with a positive measure in each non-empty, open interval. I want a simple example of a piece-wise function $f:\mathbb{R} \to \mathbb{R}$, where $\mathscr{G}_1:A \to \mathbb{R}$ and $\mathscr{G}_2:B \to\mathbb{R}$, such that: $$f(x)=\begin{cases} \mathscr{G}_1(x) & x\in A\\ \mathscr{G}_2(x) & x\notin A \Rightarrow x\in\mathbb{R}\setminus A \Rightarrow x\in B \end{cases}$$ where $f$ is simple to average with this paper:
We take chosen sequences of bounded functions converging to $f$ with the same satisfying and finite expected value w.r.t. a reference point, the rate of expansion of a sequence of each bounded function’s graph, and a “measure” of each bounded function's graph involving covers, samples, pathways, and entropy.
I assume the more "uniformally mixed" $A$ and $B$ are in $\mathbb{R}$ (i.e., the next section), the simpler one can average $f$ with this paper.
Attempts to define $A$ and $B$
- I tried partitioning $\mathbb{R}$ into sets $A$ and $B$, such that the Lebesgues measures of $A$ and $B$ in every non-empty open interval has a non-zero constant ratio non-equivelant to one. However, such a constant doesn’t exist, due to Lebesgue’s Density Theorem.
Observe that sets $A$ and $B$ are completely uniformally mixed, when a non-zero constant exists. (This is impossible!)
I also tried partitioning $\mathbb{R}$ into sets $A$ and $B$, such that the Lebesgue measures of $A$ and $B$ in every non-empty open interval have "almost" non-zero constant ratios. This is impossible, due to the Lebesgue differentiation theorem, where the lower bound of the non-constant ratio is always zero and the upper bound is always positive infinity.
Finally, I tried setting the non-empty open intervals to a constant length, such that the Lebesgue measures of $A$ and $B$ in every non-empty open interval have constant ratios w.r.t. the constant length. This is impossible, since the constant ratio w.r.t the constant length has an upper bound of positive infinity.
The last attempt (i.e., the question), similar to the $2$nd and $3$rd attempt, sets the non-empty open intervals to a constant length and the non-constant ratios (w.r.t. the constant length) as close to constant as possible.
Question:
Suppose $\lambda$ is the Lebesgue measure on the Borel $\sigma$-algebra of $\mathbb{R}$: i.e., $\mathcal{B}(\mathbb{R})$.
Does there exist an example of sets $A,B\subset\mathbb{R}$, where:
$A\cup B=\mathbb{R}$
$A\cap B=\emptyset$
For all $\varepsilon>0$, there exists set function $c:\mathbb{R}^{2}\times\mathcal{B}(\mathbb{R})^2\to\mathbb{R}$, where for all $a\in\mathbb{R}$: $$\lambda(A\cap (a,a+\varepsilon))=c(a,\varepsilon,A,B)\cdot\lambda(B \cap (a,a+\varepsilon))$$ where for some $\normalsize{\varepsilon_1} \, \small{\gtrapprox}\, \normalsize{0}$ (e.g., $\varepsilon_1=.01$),
- the set functions $L:\mathbb{R}\times\mathcal{B}(\mathbb{R})^{2}\to\mathbb{R}$ and $U:\mathbb{R}\times\mathcal{B}(\mathbb{R})^2\to\mathbb{R}$ bound: $$L(\varepsilon,A,B)\le c(a,\varepsilon,A,B)\pm \varepsilon_1\le U(\varepsilon,A,B)$$ where for all $\varepsilon>0$, the set functions $L_1:\mathcal{B}(\mathbb{R})^2\to\mathbb{R}$ and $U_1:\mathcal{B}(\mathbb{R})^2\to\mathbb{R}$ bound: $$L_1(A,B)\le U(\varepsilon,A,B)-L(\varepsilon,A,B)\pm \varepsilon_1\le U_1(A,B)$$ such that for all $A,B\in\mathcal{B}(\mathbb{R})$, the real numbers $L_2$ and $U_2$ bound: $$L_2\le U_1(A,B)-L_1(A,B)\pm\varepsilon_1\le U_2$$ where we want $A,B\in\mathcal{B}(\mathbb{R})$, such that:$$L_2=U_1(A,B)-L_1(A,B)-\varepsilon_1?$$
Follow Up Question: If $A$ and $B$ does not exist, how do we change the question so that:
- $A$ and $B$ exist
- The function $f:\mathbb{R}\to\mathbb{R}$ satisfies the motivation
