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If $X$ denotes a set, let $C(X)$ denote its cardinal number and let $P(X)$ denote its power set. There is a school of thought which considers any set having the cardinal number $C(P(\mathbb{R}))$—where $\mathbb{R}$ denotes the set of real numbers—to be too large for the intuition to grasp. This school mantains that in almost all branches of mathematics except set theory itself, there is no need to require the existence of any set whose cardinal number is greater than $C(\mathbb{R})$.

Although this viewpoint sounds plausible, I wonder whether it might lead to problems in Measure Theory. Suppose one wants to prove in $\mathsf{ZFC}$ that there exist sets of real numbers which are not Lebesgue measurable. Does there exist any set $M$, definable in $\mathsf{ZFC}$, which satisfies the following conditions:

  1. $M$ is a subset of $P(\mathbb{R})$,
  2. It is provable in $\mathsf{ZFC}$ that at least one element of $M$ is not Lebesgue measurable, and
  3. The cardinal number of $M$ is $C(\mathbb{R})$?

I know a number of examples of sets—definable in ZFC—that satisfy (1) and (2), but none of them also satisfies (3).

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    $\begingroup$ I conjecture there's a typo in the question. As it stands, the set of open intervals in R satisfies the stated conditions. $\endgroup$ – Andreas Blass May 2 '14 at 18:55
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    $\begingroup$ Would you like to share which school of thought (or who) claims that cardinals bigger than $2^{\omega}$ are not required in any field of mathematics except set theory itself? $\endgroup$ – Burak May 2 '14 at 19:58
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    $\begingroup$ Should (2) say "...at least one element of $M$ is not Lebesgue measurable"? $\endgroup$ – Trevor Wilson May 3 '14 at 6:38
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    $\begingroup$ @Burak $M$ is required to be definable. There are models of $\mathsf{ZFC}$ where every definable set of reals is Lebesgue measurable. $\endgroup$ – Trevor Wilson May 4 '14 at 0:03
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    $\begingroup$ What is the "first non-measurable set"? If you mean the "L-least non-measurable set in L", then it is consistent with ZFC that this set is countable, since it is consistent that there are only countable many reals in L, and in such a model of ZFC, all your sets in M will be countable and hence ZFC will not prove that your M contains a non-measurable set. $\endgroup$ – Joel David Hamkins May 4 '14 at 1:59
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Here is a positive answer, unfortunately it is predicated on a hypothesis which is widely believed to be false: that the existence of inaccessible cardinals is inconsistent with ZFC. Stated in a more digestible manner, the argument below shows that some large cardinal hypotheses are necessary to obtain a negative answer to the question.

Let $V$ be a model of $ZFC$, let $\mathfrak{c}^V$ denote the cardinality of the continuum in $V$ and let $\aleph_1^V$ denote the first uncountable ordinal in $V$. In $L$, $\aleph_1^V$ must be an uncountable regular cardinal and, since there aren't any inaccessible cardinals, it must be a successor cardinal. Let $\kappa$ denote the cardinal in $L$ such that $(\kappa^+)^L = \aleph_1^V$; note that $\kappa$ is definable in $V$. In $V$, $\kappa$ is countable, so there are reals $r$ in $V$ such that $\kappa$ is countable in $L[r]$ and for such reals we necessarily have $\aleph_1^{L[r]} = \aleph_1^V$. Let $K$ be the set of all reals such that $\aleph_1^{L[r]} = \aleph_1^V$; note that $K$ is definable in $V$ and $K$ has size $\mathfrak{c}^V$. For each $r \in K$, $X_r = \mathbb{R}^{L[r]}$ is a set of reals with a canonical wellordering of order type $\aleph_1^{L[r]} = \aleph_1^{V}$.

Following Raisonnier [A mathematical proof of S. Shelah's theorem on the measure problem and related results, Israel J. Math. 48 (1984), no. 1, 48–56; MR0768265; DOI:10.1007/BF02760523], we can associate to each $r \in K$ a filter $F_r$ and a collection of sets $\mathcal{H}_r$ of size at most $\mathfrak{c}^V$ such that either $F_r$ is rapid and hence not measurable, or at least one element of $\mathcal{H}_r$ is not measurable. (The set $\mathcal{H}_r$ consists of all sets that Raisonnier denotes $\tilde{H}(X_r)$. These are associated to $G_\delta$ sets $H \subseteq \mathbb{R}\times\mathbb{R}$ with null sections; since there are continuum many such $G_\delta$ sets, it follows that $|\mathcal{H}_r| \leq \mathfrak{c}^V$.)

Putting all this together, under the unlikely assumption that inaccessible cardinals provably do not exist in ZFC, in any model $V$ of ZFC, the definable set $$\{F_r : r \in K\} \cup \bigcup_{r \in K} \mathcal{H}_r$$ has size $\mathfrak{c}^V$ and it must contain a non measurable set.

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  • $\begingroup$ As suggested in comments to the question, it seems likely that Solovay's construction assuming the existence of inaccessible cardinals (or perhaps larger cardinals) could be adapted to give a model where no such definable sets exists. The argument above then simply shows that large cardinal hypotheses are indeed necessary for this. $\endgroup$ – François G. Dorais May 4 '14 at 12:00
  • $\begingroup$ To mitigate the shock of the prefacing sentence, here's an alternative formulation for the additional axiom "There are no inaccessible cardinals in $L$". $\endgroup$ – Asaf Karagila May 4 '14 at 12:22
  • $\begingroup$ @AsafKaragila: But that's equivalent. The way I'm reading the question, a positive answer needs to show that every model $V$ of ZFC has such a definable family. It's true I only use that there are no inaccessibles in $L^V$ but this is true for all $V$ if and only if ZFC proves that there are no inaccessible cardinals. $\endgroup$ – François G. Dorais May 4 '14 at 12:29
  • $\begingroup$ Of course. But it makes the rest of the answer somewhat easier to swallow. $\endgroup$ – Asaf Karagila May 4 '14 at 12:31
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    $\begingroup$ @Garabed: This is not the first time I've seen you post comments to answers when you should have posted them on the question, and vice versa. You should try and pay slightly closer attention as to where you're posting the comment -- question, or answer (and which answer, too). Otherwise the odds of the intended recipients seeing the comments and replying are lower, and it's likely they will see it at a later time if they do see it. $\endgroup$ – Asaf Karagila May 4 '14 at 19:56
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This is independent. See: Harvey Friedman, On definability of nonmeasurable sets, Canad. J. Math. 32 (1980), no. 3, 653--656.

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    $\begingroup$ Great! If I'm reading correctly, the answer to the question is, "no" assuming the consistency of inaccessible cardinals with ZFC and (following my answer) "yes" if inaccessible cardinals are inconsistent with ZFC. $\endgroup$ – François G. Dorais May 4 '14 at 17:14
  • $\begingroup$ I slightly edited your post to include a link (using the \cite plugin, courtesy of Scott Morrison). $\endgroup$ – Asaf Karagila May 4 '14 at 19:54
  • $\begingroup$ The consistency of "there is such a definable collection" is relatively easy because everything is definable in L. $\endgroup$ – Ashutosh May 12 '14 at 22:33

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