# Can an example of a non-measurable set be exhibited with non-well-founded set theories?

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?

• Anti-foundation won't help you here, since every subset of $\mathbb R$ is in WF anyway, as is the entire Lebesgue measure. Sep 10, 2019 at 11:28
• @samerivertwice: I'm saying every non-measurable subset of $\mathbb R$ is well-founded (in the set-theoretic sense), simply by virtue of being a subset of $\mathbb R$ in the first place. Sep 10, 2019 at 12:34
• @samerivertwice: No, I'm pointing out that there is no connection at all between measurability (or lack of same) and well-foundedness, because even if there are non-well-founded-sets, all real numbers and all sets of real numbers will be among those sets that happen to be well-founded. This is not about measurability at all -- merely about what the real numbers are. Sep 10, 2019 at 15:08
• @samerivertwice: I don't know how else to say it. Well-foundedness is completely irrelevant for the existence of non-measurable sets. If you have a set theory without Foundation, all subsets of $\mathbb R$ in that theory will be in WF, and each of them is measurable in the larger model if and only if they are measurable as seen within WF. Speaking about well-foundedness in the context of which sets are measurable is barking up an entirely wrong tree. Sep 23, 2019 at 18:51
• It's like asking how early you need to arrive at the airport to minimize the risk that your flight will delayed by bad weather. Sep 23, 2019 at 18:52

## 1 Answer

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 $$N$$.