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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?

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    $\begingroup$ Anti-foundation won't help you here, since every subset of $\mathbb R$ is in WF anyway, as is the entire Lebesgue measure. $\endgroup$
    – hmakholm left over Monica
    Sep 10, 2019 at 11:28
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    $\begingroup$ @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. $\endgroup$
    – hmakholm left over Monica
    Sep 10, 2019 at 12:34
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    $\begingroup$ @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. $\endgroup$
    – hmakholm left over Monica
    Sep 10, 2019 at 15:08
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    $\begingroup$ @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. $\endgroup$
    – hmakholm left over Monica
    Sep 23, 2019 at 18:51
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    $\begingroup$ 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. $\endgroup$
    – hmakholm left over Monica
    Sep 23, 2019 at 18:52

1 Answer 1

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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 $N$.

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