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 $\bigcup\limits_{t=1}^{\infty}B_{2^{t-1}}=B$.

  • 1
    $\begingroup$ 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 $\lambda A_t = 0$. $\endgroup$
    – aschepler
    Aug 6, 2023 at 21:18
  • $\begingroup$ @aschepler How do we change the construction so $\lambda(A_t)$ is positive but not $t$? $\endgroup$
    – Arbuja
    Aug 6, 2023 at 21:48
  • 1
    $\begingroup$ 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 $\frac 12$. $\endgroup$
    – aschepler
    Aug 6, 2023 at 22:01
  • $\begingroup$ 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.) $\endgroup$
    – Arbuja
    Aug 6, 2023 at 22:29
  • $\begingroup$ @aschepler You might want to look at this answer $\endgroup$
    – Arbuja
    Aug 6, 2023 at 23:21

1 Answer 1


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.


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