Take the 2-minute tour ×  I am reasonably certain this is the
case, but can't find a reference that actually states this,
although the Wikipedia article states something close.
share|improve this question  2 Answers 2  up vote 5 down vote
accepted  It is possible to have all the subsets of R be
measurable (Solovay, Robert M. (1970). "A model of set-theory in
which every set of reals is Lebesgue measurable". Annals of
Mathematics. Second Series 92: 1–56.) which implies the
nonexistence of a well ordering of R.  share|improve this answer
Alright, so I think I know what the argument should be: a well-
ordering of R induces a well-ordering of the Borel algebra, so
we can bijectively assign to every r a Borel set B(r) and let S
= {r | r \not \in B(r)}). This subset is non-empty because some
B(r) is the empty set, and it is clearly non-measurable. Is this
correct? –  Qiaochu Yuan Nov 11 '09 at 23:18 Well, you should
also make it not differ from any Borel by a null set, and for
this you need to also simultaneously d