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This is a follow up to this question, where I wish to partition the reals into two sets $A$ and $B$ that are dense (with positive measure) in every non-empty sub-interval $(a,b)$ of $\mathbb{R}$.

(In this case, I want to find $\lim_{t\to\infty} \lambda(A\cap [-t,t])/(2t)$ and $\lim_{t\to\infty} \lambda(B\cap [-t,t])/(2t)$, where $\lambda$ is the Lebesgue measure restricting the Lebesgue outer measure $\lambda^{*}$ on sets measurable in the Caratheodory sense.)

I have a rough understanding of the answer; however, I'm unsure of the measures of $A$ and $B$.

Question: What is the measure of $A$ and $B$? If both measures are $1/2$, how do we change the answer so the measures are positive but unequal?

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    $\begingroup$ How could both measures be $1/2$? Doesn't $\mathbb R$ have measure $\infty$? What measure are you using? $\endgroup$
    – bof
    Commented Aug 4, 2023 at 0:43
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    $\begingroup$ Given a number $\varepsilon\gt0$ we can constract an $F_\sigma$ set $A\subset\mathbb R$ so that (i) for every nonempty open set $U\subseteq\mathbb R$, both $U\cap A$ and $U\setminus A$ have positive Lebesgue measure;, and (ii) the set $A$ has Lebesgue measure less than $\varepsilon$. $\endgroup$
    – bof
    Commented Aug 4, 2023 at 0:48
  • $\begingroup$ @bof I made edits. $\endgroup$
    – Arbuja
    Commented Aug 4, 2023 at 0:53

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We can make those limits be any two positive numbers that add to $1$, or we can make them nonconvergent. The reason is that in any interval, we can construct $A$ and $B$ on that interval by using the iterated fat-Cantor set construction (as described here), and we can make $A$ have arbitrarily small measure on that one interval, which is what bof refers to in his comment.

By repeating those small-measure sets on successive intervals, or by using large-measure sets as desired on successive intervals, we can control the limit value of your asyptotic measure so as to realize any desired value or a nonconvergent value.

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