Tiling the plane with finitely many congruent pieces

Question

Suppose $A_1,\dots,A_n$ are measurable subsets of the plane that are all related by rigid motions such that $|(A_1 \cup \dots \cup A_n)^c| = 0$ and $|A_i \cap A_j| = 0$ for all $1 \leq i < j \leq n$, where $|S|$ denotes the Lebesgue measure of $S$.

Must each $A_i$ have the property that $|A_i \cap B(r)|/|B(r)| \rightarrow 1/n$ as $r \rightarrow \infty$, where $B(r)$ is the disk of radius $r$ centered at 0?

The answer is clearly “yes” if the $A_i$’s are all obtained from one another by rotation about a point (e.g., consider the Fatou sets for Newton’s algorithm applied to the polynomial $z^3 - 1$). Maybe this is the only way to tile the plane by $n$ congruent pieces, but I don’t see why it should be true.

Answer 1

Write $A_i=T_i(A)$ for $A=A_1$, where each $T_i$ is a rigid motion. For each $i$, we have $|A_i\cap B(r)|=|T_i(A\cap T_i^{-1}(B(r))|=|A\cap T_i^{-1}(B(r))|$. The symmetric difference between this set and $A\cap B(r)$ is contained in the symmetric difference of $T_i^{-1}(B(r))$ and $B(r)$, whose measure is $O(r)$. Hence the measures of $A_i\cap B(r)$ are all equal up to $O(r)$. Since the sum of their measures is $B(r)$, this implies $|A_i\cap B(r)|=B(r)/n+O(r)$ for each $i$.