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Suppose we partition the reals into two sets $A$ and $B$ that are dense (with positive Lebesgue measure) in every non-empty sub-interval $(a,b)$ of $\mathbb{R}$, where Lebesgue measure $\lambda$ restricts outer measure $\lambda^{*}$ to sets measurable in the Caratheodory sense.

Does there exist an example (similar to this one) where both:

$$\lim\limits_{t\to\infty}\lambda(A\cap [-t,t])/(2t)$$ and $$\lim\limits_{t\to\infty}\lambda(B\cap [-t,t])/(2t)$$ are greater than zero but neither equal $1/2$?

A bounty is also offered for this question here.

Paul F
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    For the first question, could you just pick your sets $A$ and $B$ from before and intersect them with the complement of a fat cantor set (do this on $[0,1]$ and translate to get such a set on $\mathbb{R}$) (or some other nowhere dense set with positive measure)?

    Also, does the $A$ and $B$ in this question correspond to the post you linked or to the example you want in question (1)?

    – 1mdlrjcmed Aug 09 '23 at 05:49
  • You maybe right about your suggested construction. Could you pleas elaborate further? and Yes, both A and B correspond to those the linked post. – Paul F Aug 09 '23 at 17:35

1 Answers1

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An example satisfying Q1 is as follows: \begin{align*} & A=\left\{\bigcup_{n\in\mathbb{Z}}[3n,3n+2)\setminus D'\right\}\cup D, \\ & B=\left\{\bigcup_{n\in\mathbb{Z}}[3n-1,3n)\setminus D\right\}\cup D', \\ & D=\{m/2^n\mid n=1,2,\dots,\,m=\pm 1,\pm 2,\dots,\,m/2^n\notin\mathbb{Z}\}, \\ & D'=\{m/3^n\mid n=1,2,\dots,\,m=\pm 1,\pm 2,\dots,\,m/3^n\notin\mathbb{Z}\}. \end{align*}

We can confirm:

  • $A\cap B=\emptyset$ and $A\cup B=\mathbb{R}$.
  • $A$ and $B$ are dense in any open interval $(a,b)$ since $D\subset A$, $\bar{D}=\mathbb{R}$ and $D'\subset B$, $\bar{D'}=\mathbb{R}$. Here, for $X\subset \mathbb{R}$, $\bar{X}$ represents the closure of $X$ under the usual topology of $\mathbb{R}$. Note that $D$ means binary approximation and $D'$ means ternary approximation, and $D$ and $D'$ do not intersect.
  • We have \begin{align*} \frac{\lambda(A\cap [-3n,3n])}{2\cdot 3n} &= \frac{2n\cdot 2}{6n}=\frac{2}{3}, \qquad n=1,2,\dots \end{align*} then it follows that \begin{align*} \lim_{t\to\infty}\frac{\lambda(A\cap [-t,t])}{2t} &= \frac{2}{3}\neq \frac{1}{2}. \end{align*} By the same manner as above, we see also \begin{align*} \lim_{t\to\infty}\frac{\lambda(B\cap[-t,t])}{2t}=\frac{1}{3}\neq \frac{1}{2}. \end{align*}

Q2: The way of giving meaning is up to you. I'd consider it as normalization over $2ε=λ(B_\varepsilon(x))$, and I would answer that it's the probability of a number chosen from a uniform distribution on $B_\varepsilon(x)$ being contained in $A$ (or $B$).
I have no idea of another interpretation. Other people may have different answers.

noa_jp
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  • Thank you su much. If you are interested, a bounty is offered here: https://matchmaticians.com/questions/kdnngg/kdnngg-what-is-the-asymptotic-density-and-lebesgue-of-a-b – Paul F Aug 12 '23 at 17:14