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. 



Is the existence of a well-ordering on R independent of ZF?
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.


