What is the asymptotic density of $A$ and $B$ which partition the reals into subsets of positive measure?

Suppose we partition the reals into two sets $A$ and $B$ that are dense and has 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.

Question: 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 $1/2$?

Attempt: In this construction, we can take pairwise disjoint copies of $A$ and $B$ shifted infinite times (i.e., $x\in A \iff x-\lfloor x\rfloor\in A$ and $x\in B:=[0,1]\setminus A \iff x-\lfloor x\rfloor\in B:=\mathbb{R}\setminus A$).

Edit: According to a user to Reddit, using the construction in the attempt, we get:
$$\lim\limits_{t\to\infty}\lambda(A\cap [-t,t])/(2t)$$ and $$\lim\limits_{t\to\infty}\lambda(B\cap [-t,t])/(2t)$$ both equal $1/2$. We need to adjust the construction so the limits are non-equal.

• @Mathe Actually the answer is incorrect. It doesn’t have positive measure for every non-empty sub-interval (a,b) of R. The user who answered admitted this on Reddit but forgot to delete their answer.

• Oh I see !

• Mathe
+1

Hi, just to be sure, we are looking for: 1. A partition of the real numbers, into two sets A and B, such that 2. both A and B have positives measure, 3. A and B are dense in every subinterval (a, b) and 4. The limits above are both not equal to 1/2

• Yes, that’s what I’m look for.

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Mathe
3.4K
• @Mathe I forgot to say that A and B must have positive measure in every non-empty sub-interval (a,b) of R.

• Oh well, this changes things. This is why I confirmed before what you were looking for.

• I should phrased the question more carefully but I got a better answer here: https://math.stackexchange.com/a/4754256/125918

• Mathe
+1

I think that's a good answer. It built on all the previous approaches and I think it meets all the requirements.

• @Mathe I actually did mention this in the second comment below my post.

• But I should have mentioned this when you asked for clarification.

• Mathe
+1

Yes, the original wording had me confused, so I asked to clarify my understanding.

• @Mathe Is it possible to delete the answer, since someone already answered it on MSE. I could close the question and withdraw the money since no one could answer here.

• I can't delete my answer. I think you could contact support and ask them how to proceed, since I did answer your question based on your clarification.

• They told me I have to dispute if I don’t agree. You can change the answer so I can accept.

• I updated the file.