TITLE: Paradoxical Decompositions 
QUESTION [17 upvotes]: Question: Does Con($ZF$) imply Con($ZF$ + $DC$ + "there is no paradoxical Banach-Tarski decomposition of the unit ball")?

Here Con($X$) is the consistency of $X$; $DC$ is dependent choice.
Motivation for the Question: Since the "paradoxical" sets in the Banach-Tarski theorem have to be non-measurable, Solovay's model of "every subset of reals is measurable" shows: 
Theorem. Con($ZF$+ Inacc) implies Con($ZF$ + $DC$ + "there is no paradoxical Banach-Tarski decomposition of the unit ball"). 
In the above Inacc is the statement "there is an inaccessible cardinal". 
Note that Shelah's model [Israel Journal of Math, 1984], constructed only
from Con($ZF$), in which $DC$ holds and all sets of reals have the Baire
property, is of no help in answering the above question since Dougherty and
Foreman [ J. Amer. Math. Soc., 1994] have shown that there are paradoxical
decompositions of the unit ball using pieces which have the property of
Baire. I do not know how much choice is needed in their construction.
This question was posed a while ago on an FOM-posting of mine, but remained unanswered.

REPLY [9 votes]: A positive answer is proved in S. Wagon's book "The Banach-Tarski Paradox", Theorem 13.2. Specifically, the statement proved there is:
Con(ZF) $\leftrightarrow$ Con(ZF + DC + GM),
where GM is the existence of an isometry-invariant measure on all subsets of $\mathbb R^n$ taking the value $1$ on the unit cube.