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.



author of a paradoxical decomposition of the interval
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)](http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-fmv6i1p27bwm).


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)](https://babel.hathitrust.org/cgi/pt?id=umn.31951d00697922m;view=1up;seq=154): "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)](http://gallica.bnf.fr/ark:/12148/bpt6k3176m/f1543). 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.


