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.



Drawing conclusions by NOT using AC.
With some difficulty, you can define a set of reals which is measurable under
one extra-ZF assumption (the axiom of determinacy) and nonmeasurable under
another (V=L). Under V=L you have a definable well-ordering of the reals, and
this enables you to *define* any of the nonmeasurable functions you normally get using the axiom of choice. On the other hand, under the axiom of determinacy, *every* real function is measurable.


That's the strict answer. However, I think that any subset or function you are *likely*
to define will be Borel, and hence provably measurable in ZF.


