Hi Kevin, welcome to MO! (I know you've been here a while, but I only just saw one of your questions.) I just want to expand a little on the thing about models, because I've asked myself similar questions to yours. Say you've defined a function without using Choice. Explicitly, this means you've defined a set f of ordered pairs, the first components of which come from a set A, and the second from a set B.

Each model of ZF has its own version of the function. This is like defining a function x |--> x^2, which only needs the concept of a binary operation, but in each concrete example the function will look quite different. Actually it's worse, because in the case of ZF you also need to specify the domain of your function. So a more accurate example would be defining the function above but only on the "set of cubes" (relative to the unspecified binary operation). Now the function looks even more different each time: it's just "doubling" in Z/5Z, but in Z/6Z it's the zero map on {0,3}.

So while your function might be measurable in a model where "every set is measurable", when you pass to another model, the ZF-definition may define a different function, which turns out not to be measurable. For example, different models of ZF may have "extra real numbers". (Just as different magmas have distinct sets of cubes.) If your ZF-definition has domain(f) = reals, this will carry over to all the different models, and in each one f will have a different domain. (Hence, more often than not, a different range.)

So you see the situation is quite chaotic a priori. And this is to say nothing of the fact that in each model of ZF "measurability" means something different.

Here's one last, somewhat strained, analogy. Suppose you were looking at structures satisfying the ring axioms (ZF in the analogy), but still interested in our map g above, x |--> x^2, defined on the set of cubes (a ZF-definable map). The ring axioms are silent about whether there are multiplicative inverses for nonzero elements (whether Choice holds). Suppose now someone found an example of a ring, not a field (where Choice failed), s.t. the image of g is exactly the set of fourth powers (is measurable). (E.g. Z/35Z.) Would you be able to conclude that, because your definition of g didn't use inverses (Choice), its range would always be the fourth powers, in any ring?