author of a paradoxical decomposition of the interval

Question

I am looking for the original author and the date of publication of the following result.

Theorem There exist subsets $E_i\subset [0,1)$, $i\in {\bf Z}$, pairwise disjoints and real numbers $a_i$ such that $$ [0,1) = \coprod_{i\in {\bf Z}} E_i, \quad [0,2) = \coprod_{i\in {\bf Z}} a_i + E_i. $$ I am pretty sure that this is either Borel or Lebesgue. I remember having seen the result in the collected works of one of these two mathematicians, who advertised it as a reason to reject the axiom of choice.

Answer 1

Your quoted theorem (actually, a generalization where the second set is $[0,n)$ or even $[0,\infty)$) appears as Lemme 32, p. 267 of the Banach-Tarski paper (1924).

This is on the way to their third main result (see Introduction and Thm 35, p. 270): In $\mathbf R^n$ ($n\geqslant 1$) any two subsets $A$, $B$ with nonempty interior are "equivalent by countable decomposition", i.e. $A=\coprod E_i$ and $B=\coprod g_i(E_i)$ for some disjoint $E_i$ and displacements $g_i$.

The only antecedent they quote is by Sierpiński (1918, p. 142): "Let us remark that, by using Mr. Zermelo's axiom, one could decompose a square into countably many sets with which one could then compose (by a suitable translation of each of these sets) a square larger than the given one."

Note. What you saw may well have been Borel's note Les paradoxes de l'axiome du choix (1947). By an argument which is only superficially different, he shows there that Zermelo's axiom implies "euclidean equality of the whole and the part, in a finite domain" (namely $[0,1]$), and concludes: "It seems preferable to me not to admit the axiom". He doesn't cite Banach, Tarski, or Sierpiński.