On wikipedia it says an example of a non-measurable set cannot be exhibited, although they do exist.
I was curious what premises lead to this conclusion, and whether there are ways to exhibit non-measurable sets, for example with less conventional axioms such as anti foundation.
In non-well-founded set theory, sets can EXIST not found in well-founded set theory. But can it also EXHIBIT some well-founded sets which are found in well-founded set theory but cannot be exhibited in well-founded theories?
The issue seems to stem from the ambiguity of "exhibit," so let me first try to pin that down.
Under reasonable hypotheses there is a model $M$ of ZFC such that:
$M$ thinks there is a non-measurable set of reals (this is for free, since $M$ satisfies ZFC), but
Every parameter-freely definable set of reals in $M$ is measurable - more precisely, for each formula $\varphi(x)$ in the language of set theory such that $M$ thinks there is exactly one set of reals satisfying $\varphi$, we have that $M$ thinks that that set is measurable.
In particular, under these reasonable hypotheses there is no formula $\varphi$ such that ZFC proves "$\varphi$ defines a unique set of reals and that set is non-measurable." Moreover, by upgrading those hypotheses we can make $M$ a model of any large cardinal axiom I'm aware of.
Now what about a theory of the form ZFC$^-$ + $A$, where $A$ is some antifoundation axiom (e.g. Aczel's $AFA$)?
The answer, in every instance I'm aware of, is that nothing changes - and this is a consequence of the definability arguments we use to establish the consistency of such a theory.
Specifically, in every case I'm aware of we prove the consistency of ZFC$^-$ + $A$ relative to that of ZFC (perhaps plus some large cardinaly hypothesis $*$) by showing how any model $N$ of ZFC (+ $*$) "definably interprets" a model $N'$ of ZFC$^-$ + $A$, and we get for free (looking at this interpretation) that $WF(N')$ (= the well-founded part of $N'$) is isomorphic to $N$ definably in $N$. The result is that any parameter-freely-definable-in-$N'$ subset of $WF(N')$ is already parameter-freely definable in $WF(N')$ - and since measurability is calculated in the well-founded part, putting all this together we get that a parameter-freely-definable non-measurable set in $N'$ yields (indeed, "is") a parameter-freely definable non-measurable set in $N$.