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

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 $$\mathbb{R}$$.

Question: Does the construction in the succeeding section make sense? If not, how do we correct the construction so $$\lim_{r\to\infty} \lambda(A\cap [-r,r])/(2r)$$ and $$\lim_{r\to\infty} \lambda(B\cap [-r,r])/(2r)$$ are positive but not equal to $$1/2$$; (i.e. where $$\lambda$$ is the Lebesgue measure which restricts Lebesgue outer measure $$\lambda^{*}$$ to sets measurable in the Caratheodory sense)?

Note the construction of $$A$$ and $$B$$ in the following section was inspired by this answer.

Construction of $$A$$ and $$B$$

Suppose we take half-open interval $$I_t=[-t,t)$$, where for every $$t\in\mathbb{N}$$ and stage $$n$$ (for positive integers $$n$$) we'll partition $$I_t$$ into two sets $$A_{n,t}$$ and $$B_{n,t}$$, each a union of finitely many half-intervals. Start with $$A_{1,t} = [-t,0)$$ and $$B_{1,t} = [0, t)$$.

Given $$A_{n,t}$$ and $$B_{n,t}$$: for each interval $$[a,b)$$ in one of these sets of length $$s = b-a$$, remove an interval of length $$2^{-n} s$$ from the centre of the interval and give it to the other set. Thus from $$[-t, 0)$$ in $$A_{1,t}$$, we remove $$[-3t/4, -t/4)$$ and put it in $$B_{2,t}$$, while $$A_{2,t}$$ keeps $$[-t,-3t/4)$$ and $$[-t/4, 0)$$, and from $$[0, t)$$ in $$B_{1,t}$$, we remove $$[t/4, 3t/4)$$ and put it in $$A_2$$, resulting in $$A_2 = [-t,-3t/4) \cup [-t/4,0) \cup [t/4, 3t/4)$$ and $$B_2 = [-3t/4, -t/4) \cup [0, t/4) \cup [3t/4, t)$$.

Note that in going from stage $$n$$ to stage $$n+1$$, the measure of the points transferred is $$2^{-n}$$. Since $$\sum_n 2^{-n}$$ is finite, almost every point is transferred only finitely many times. Since sets of measure $$0$$ are negligible, we'll define $$A_t$$ to consist of the points that are eventually in $$A_{n,t}$$, and $$B_t$$ as its complement the points that are in $$B_{n,t}$$ for infinitely many $$n$$, i.e. $$A_t = \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_{k,t}, \ B_t = [-t,t) \backslash A_t =\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty B_{k,t}$$

where if:

1. $$\forall(t\in\mathbb{N})\left(A_{2^{t-1}}\subset A_{2^{t}}\right)$$

2. $$\forall(t\in\mathbb{N})\left(B_{2^{t-1}}\subset B_{2^{t}}\right)$$

we want $$\bigcup\limits_{t=1}^{\infty}A_{2^{t-1}}=A$$ and $$\bigcup\limits_{t=1}^{\infty}B_{2^{t-1}}=B$$.

• 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 $\lambda A_t = 0$. Aug 6, 2023 at 21:18
• @aschepler How do we change the construction so $\lambda(A_t)$ is positive but not $t$? Aug 6, 2023 at 21:48
• 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 $\frac 12$. Aug 6, 2023 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.) Aug 6, 2023 at 22:29
• @aschepler You might want to look at this answer Aug 6, 2023 at 23:21

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.