If $A, B$ is a non-trivial partition of $S^1$, is it possible that $R_\theta(A) \cap B$ has measure zero for all rotations $R_\theta$?

Question

This was previously posted to Math StackExchange. I was originally unsure whether it is suitable for posting here, but I've yet to get an answer there, so I'm just trying to see if people here can help.

As in the question title, let $A, B$ be a partition of the unit circle $S^1$, equipped with the Haar measure. Here, we do not require $A, B$ to be measurable. Also, assume neither $A$ nor $B$ is of measure zero, so they are either both non-measurable or both of positive measure. (Equivalently, they both have positive outer measure.) Is it possible, then, for $R_\theta(A) \cap B$ to be measurable and have Haar measure zero for all $\theta$, where $R_\theta$ is the rotation by degree $\theta$?

Answer 1

Assuming the Continuum Hypothesis, the answer is yes. On CH, the $\mathbf Q$-dimension of $\mathbf R$ has the cardinality of the first uncountable ordinal $\omega_1$, so we can find a $\mathbf Q$-linear basis $e_\alpha, \alpha \in \omega_1$ of ${\bf R}$, and we can normalize $e_0=1$. Thus every irrational real number $x$ has a unique representation of the form $$ x = \sum_{\alpha \leq \beta} q_\alpha e_\alpha\tag{1}\label{1}$$ where $0 < \beta < \omega_1$ and $q_\alpha, \alpha \leq \beta$ are rational numbers with all but finitely many $q_\alpha$ non-zero, and $q_\beta$ also non-zero. In particular, $q_\beta$ is the final non-zero coefficient of $x$. We then assign $e^{2\pi i x}$ to lie in $A$ if $x$ is rational, or $x$ is irrational with representation \eqref{1} with $\nu_2(q_\beta)$ even, where $\nu_2(q_\beta) \in \mathbf Z$ is the number of times two divides the final non-zero coefficient $q_\beta$, and $e^{2\pi i x}$ to lie in $B$ if $x$ is irrational with representation \eqref{1} with $\nu_2(q_\beta)$ odd. Since the final non-zero coefficient $q_\beta$ is unaffected by shifts by integers (or even rationals) on the irrationals, this is a well-defined partition of $S^1$. By construction, if $z$ is not a root of unity, then $z$ lies in $A$ if and only if $z^2$ lies in $B$. Because of this, neither $A$ nor $B$ can have measure zero (otherwise $S^1 = A \cup B$ would also have measure zero).

On the other hand, if $\theta = e^{2\pi i s}$ is an angle, then $s$ is a finite $\mathbf Q$-linear combination of the $e_\alpha$. Such rotations do not affect the final non-zero coefficient $q_\beta$ in \eqref{1} unless $\beta$ is less than or equal to one of these $\alpha$ (or $x$ is rational), which only occurs countably often. Thus $R_\theta(A) \cap B$ is countable (hence null) for all $\theta$.

Answer 2

This question is explored in great generality by Laczkovich in

Laczkovich, Miklós, "Two constructions of Sierpiński and some cardinal invariants of ideals", Real Anal. Exch. 24(1998-99), No. 2, 663-676 (1999), MR1704742, Zbl 0964.03056 (the paper is Open access at ProjectEuclid).

Sierpiński proved such a partition exists under CH, by an argument which immediately extends to any model where every set of reals of cardinality less than continuum is null (as Terry and Yair observed in the other answer). Laczkovich shows that there is no such partition in random real models (e.g., $L$ adjoined with $\omega_2$ random reals), so this problem is independent from ZFC.

Answer 3

$\newcommand\th\theta$This is a partial answer, showing that, if $R_\th(A)\cap B$ is measurable and has Haar measure $|\cdot|$ zero for all $\th$, then $A$ and $B$ cannot be measurable with $|A|>0$ and $|B|>0$.

Indeed, suppose the contrary (we do not need to assume here that $B$ is the complement of $A$). By the regularity of the Haar measure, there exist (say open) arcs $I$ and $J$ of the circle such that $|A\cap I|>|I|/2$ and $|B\cap J|>|J|/2$. Since $|A\cap I|,|I|,|B\cap J|,|J|$ depend continuously on the endpoints of the arcs $I$ and $J$, without loss of generality (wlog) $|I|$ and $|J|$ are rational numbers, say $p/q$ and $r/s$, respectively, for some natural $p,q,r,s$. Partitioning each of the arcs $I$ and $J$ into arcs of the same length $1/(qs)$, we see that wlog $|I|=|J|=1/(qs)$.

Renormalizing the Haar measure, assume wlog that $|I|=|J|=1$, so that $|A\cap I|>1/2$ and $|B\cap J|>1/2$. Also, the condition $|I|=|J|$ implies that $I=R_{-\th}(J)$ for some real $\th$, so that $|R_\th(A)\cap J|=|A\cap R_{-\th}(J)|=|A\cap I|$. So, $$ \begin{split} |R_\th(A)\cap B|&\ge|(R_\th(A)\cap J)\cap(B\cap J)| \\ &=|R_\th(A)\cap J|+|B\cap J|-|(R_\th(A)\cap J)\cup(B\cap J)| \\ &=|A\cap I|+|B\cap J|-|(R_\th(A)\cap J)\cup(B\cap J)| \\ &\ge|A\cap I|+|B\cap J|-|J| \\ &>1/2+1/2-1=0. \end{split} $$ So, $|R_\th(A)\cap B|>0$, a contradiction. $\quad\Box$