I'm currently teaching a segment on the Bolyai-Gerwein theorem, and I was asked about the following question:
Suppose I have measurable sets $P,Q\subset\mathbb R^2$ such that there exists sets $P_1,\ldots,P_n,Q_1,\ldots,Q_n$ such that
- $\bigsqcup\limits_{i=1}^nP_i=P$, $\bigsqcup\limits_{i=1}^nQ_i=Q$
- For each $i\in\{1,\ldots,n\}$, $P_i$ can be transformed to $Q_i$ via a finite sequence of translations, rotations, and reflections.
In this case, does $\mu(P)=\mu(Q)$?
I seem to recall a Terrence Tao blog post on how a Banach-Tarski esque result is not possible in 2 dimensions due to the lack of degrees of freedom for rotation, but I'm not sure whether that is enough to make this statement true.
The Banach-Tarski paradox in 2 dimensions is not possible because the group generated by reflections, rotations and translations is solvable (of class 3) and hence amenable. It was proved first by von Neumann in 1929. So yes, $\mu(P)=\mu(Q)$. However, if you allow also all area preserving linear maps (that is with $\det = 1$), you can double a disc because $SL(2,\mathbb R)$ contains a free non-abelian subgroup. It was also proved by von Neumann in the same paper, Neumann, J. 1929, Zur allgemeinen Theorie des Masses, Fund. Math., vol. 13, pp. 73116, Available at: http://eudml.org/doc/211921