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 $\theta$?



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 $R_\theta$?
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 $\theta$.


