TITLE: Drawing conclusions by NOT using AC.
QUESTION [12 upvotes]: The existence of non-measurable subsets and functions on $\mathbb{R}$ require the use of the axiom of choice. That is, there exist models of ZF in which all subsets of (and hence all functions defined on) $\mathbb{R}$ are measurable. Does this mean that if I can define a subset or function on $\mathbb{R}$ without invoking the axiom of choice, it must be measurable?
Let's say I fully accept AC and am just looking for a quick and dirty way of proving a function is measurable.

REPLY [3 votes]: Solovay showed in the 1960's the consistency of  ZF + axiom of Dependent Choice (for countable sets) + "every set of reals is Lebesgue measurable".  .
This is a formal justification of the idea that any set of reals that can be proved to exist in ZFC without making uncountably many choices ---- i.e., a set that can be constructed in ZF + DC ---- cannot be proved to be non-measurable.   
There are also theorems of Shelah, coming out of the analysis of determinacy (AD, Woodin cardinals, etc) to the effect that "any reasonably defined set is Lebesgue measurable".
So it is impossible to produce a Lebesgue non-measurable set without specifically looking for it in the wilds of uncountable AC.
It would be interesting to know what the state of the art is for metatheorems on how broad a range of constructions can be used in ZF+DC, that are guaranteed to stay within the world of provably Lebesgue measurable sets.