TITLE: Most functions are measurable
QUESTION [19 upvotes]: My professor once said that if you did not use the axiom of choice to build a function $f : \mathbb{R}^n \to \mathbb{R}^m$, then it is Lebesgue measurable. To what extent this is true?

REPLY [13 votes]: As Henning said, formally speaking, this is not quite true. We can define sets, without using the axiom of choice, which we cannot prove that they are measurable.
For example, every universe of set theory has a subuniverse satisfying the axiom of choice, in a very canonical way, called $L$. We can look at a set of reals which is a Vitali set in $L$, or any other non-measurable set that lives inside $L$. The axiom of choice is not needed for defining this set, however under some assumptions, this set will be an actual Vitali set, and thus non-measurable; and under other assumption it might be a countable set and therefore measurable.
What your professor really meant to say, is that it is consistent that the axiom of choice fails, and every set is Lebesgue measurable. This was proved by Solovay in 1970. So in most cases if you just write a definition of a "reasonable" set, it is most likely measurable. But nonetheless, this is not formally correct. As far as analysis go, though, it is usually the case that "explicitly defined sets" are measurable.