# author of a paradoxical decomposition of the interval

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.

• I took the liberty of adding some tags for extra visibility - please feel free to revert/remove them – Yemon Choi Mar 26 '17 at 21:12
• Seems like Vitai's proof perhaps. Try looking up "Sul problema della misura dei gruppi di punti di una retta" from 1905 and checking if it's there. That would be my first guess. – Asaf Karagila Mar 26 '17 at 21:43
• Also, math.stackexchange.com/questions/85609/… might be of interest. – Asaf Karagila Mar 26 '17 at 21:48

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 $g_i$.
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.