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The rational numbers $\Bbb Q$ are dense in $\Bbb R$, but they are still a set of measure zero, i.e.

$$\begin{align} \mu(\Bbb Q \cap [a,b]) &= 0 \\ \mu((\Bbb R\!\setminus\! \Bbb Q) \cap [a,b]) &= b-a \\ \tag 1 \end{align}$$ for any finite interval $[a,b]$.

Is it possible to have more equally distributed sets, so that neither of them is a set of measure 0, and this holds on any interval similar to (1)?

More specifically, are there decompositions $A, B\subset \Bbb R$ and a measure $\mu$ such that all of the following conditions hold?

$$\begin{align} A\cap B = \emptyset\quad&\text{ and }\quad A\cup B = \Bbb R \\ \mu ([a,b]) &= b-a \\ \mu (A\cap [a,b]) &= (b-a) / 2 \\ \mu (B\cap [a,b]) &= (b-a) / 2 \\\tag 2 \end{align}$$ for any finite interval $[a,b]\subset\Bbb R$? The first line just states that $A,B$ is a decomposition of $\Bbb R$, the second line is a common normalizing condition for $\mu$.

Or, at your option, that $$\begin{align} \mu(A\cap[a,b]) &= (b-a)\kappa \qquad\text{ for some } 0<\kappa<1 \\ \mu(B\cap[a,b]) &= (b-a)(1-\kappa) \end{align}$$ again for any finite interval $[a,b]$. And it might even be in order if $\kappa=\kappa(a,b)$ depends on $a$ and $b$ provided $0<\kappa(a,b)<1$ for finite intervals.

My intuition says that there is no such decomposition, but maybe I am wrong.

emacs drives me nuts
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  • You can't get them to both take up half of each interval - this is a reasonably elementary reason. You can get them both to take up some nonzero amount of each interval (depending on the interval) - eyeballfrog links a question about this below, but my personal favourite construction is this one, with Markov chains. – Izaak van Dongen May 16 '22 at 21:50
  • @Izaak van Dongen: Isn't that answer with the Markov chains a constructions of a measurable set $A$ that has non-zero, non-full measure on any subinterval of $[0,1]$? Then by repeating $[0,1]$ to cover all of $\Bbb R$ would give a similar measurable set over all of $\Bbb R$? – emacs drives me nuts May 17 '22 at 08:29
  • That's right (see the last paragraph of the Markov chains answer) - and if you cover all of $\Bbb R$ in this way, then $A$ and its complement satisfy the "$\kappa(a, b)$" part of your question (if I have understood correctly). – Izaak van Dongen May 17 '22 at 09:13
  • @Izaak van Dongen: Ok bit doesn't that contradict David C. Ullrich's answer, because on the left side of the limit equation there would be $\kappa(a,b)$ or $1-\kappa(a,b)$, and on the right side there is 1? – emacs drives me nuts May 17 '22 at 10:29
  • David C. Ullrich has only addressed the part where you're asking for exactly half-measure in each interval. In his answer, you can't directly evaluate the limit for the $\kappa(a, b)$ version, because the limit simplifies to $\lim_{h \to 0} \kappa(a - h, a + h)$ (which is only absurd if you can show this limit is less than $1$ on some non-null set). He is exploiting the fact that in the half-measure version, $\kappa$ is constant so this limit is that constant value. – Izaak van Dongen May 17 '22 at 13:25

1 Answers1

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No.

First, it's clear that $\mu([a,b])=b-a$, and hence that $\mu$ is just Lebesgue measure. Now the Lebesgue Differentiation Theorem (applied to $\chi_A$) shows that $$\frac12=\lim_{h\to0}\frac{\mu(A\cap[a-h,a+h])}{2h}=1$$for almost every $a\in A$, contradiction.

Vincent
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David C. Ullrich
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