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This is a follow up to this and this post. 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 $\mathbb{R}$.

Suppose $\lambda$ is the Lebesgue measure which restricts the Lebesgue outer measure $\lambda^{*}$ to sets measurable in the Carathéodory sense.

Question: How do we construct $A$ and $B$ so that:

$$\lim_{r\to\infty} \lambda(A\cap [-r,r])/(2r)$$ and $$\lim_{r\to\infty} \lambda(B\cap [-r,r])/(2r)$$ are positive but not equal to $1/2$?

Notice, the construction of $A$ and $B$ should be inspired by this answer.

Arbuja
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    In going from stage $n$ to stage $n+1$, the measure of the points transferred is always $\frac t2$, since the same total length is transferred from $A_{n,t}$ to $B_{n+1,t}$ and from $B_{n,t}$ to $A_{n+1,t}$, so that every $A_{n,t}$ has total length $t$. And it turns out $\lambda A_t = 0$. – aschepler Aug 06 '23 at 21:18
  • @aschepler How do we change the construction so $\lambda(A_t)$ is positive but not $t$? – Arbuja Aug 06 '23 at 21:48
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    You could probably use something where the transfer in one direction is always a larger proportion of each interval than the other direction, but as $n$ increases, those proportions both approach $\frac 12$. – aschepler Aug 06 '23 at 22:01
  • If someone gives another example satisfying what I want, I'll give the bounty. (If someone proves what I want is impossible, I'll also give the bounty.) – Arbuja Aug 06 '23 at 22:29
  • @aschepler You might want to look at this answer – Arbuja Aug 06 '23 at 23:21

1 Answers1

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For simplicity, just partition the unit interval $[0,1)$ and then repeat that partition so $x\in A$ when $x-\lfloor x \rfloor\in A$. This lets us avoid dealing with $t$.

This construction elaborates on aschepler’s comment. I use $2/3$ for simplicity, but any ratio is easily accomplished.

Let $A_1=[0,2/3)$. Let $B_1=(2/3,1]$.

If $A_n$ is a union of intervals, then for each interval cut out the middle $1/2^{n+1}$ of the interval and send it to $B_n$. Similarly, for each interval in $B_n$ cut out the middle $1/2^n$ and send it to $A_n$. Each set less what’s cut out plus what was transferred determines $A_{n+1},B_{n+1}$.

$A_n$ is initially twice as large as $B_n$ ($2/3$ is twice $1/3$). In round $n$, $A_n$ transfers $2/(3 \cdot 2^{n+1})=1/(3\cdot 2^n)$ while $1/(3\cdot 2^n)$ is transferred from $B_n$. Thus, the transfer is equivalent and so the measure of each part of the partition is preserved. Similar to the original construction, the total amount of transferred points converges to $2/3$, so an all but measure zero set of points eventually settle on a partition allowing us to construct a full partition into sets $A$ and $B$ which both have positive measure in every interval. On a sufficiently large range (or just on $[0,1)$), $A$ has $2/3$ of the total measure while $B$ has $1/3$, although this measure is not uniform within the range.

Eric
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