Is it consistent (with ZFC) that there is a cardinal $\kappa$ and $m_1$ and $m_2$ two measures on $\kappa$, both $\kappa$-additive, such that

  1. $m_1$ is atomless, and
  2. $m_2$ has an atom?

In that case, the existence of $m_1$ implies that the continuum is at least as large as $\kappa$ and the existence of $m_2$ implies that $\kappa$ is inaccessible. That would mean a radical failure of CH. Of course, we know from Easton's Theorem that the size of the continuum can be "almost" anything. Even then, I am curious to know if $m_1$ and $m_2$ can coexist.

  • $\begingroup$ Real valued measures or $\{0,1\}$-valued ? In any case, I think it's rather $m_1$ that implies that $k$ is inaccessible. And we can't prove that it's consistent with ZFC (unless ZFC isn't) because if we could, then we'd know that an inaccessible is consistent, which would mean ZFC is inconsistent $\endgroup$ – Max Apr 26 '18 at 10:39
  • $\begingroup$ m1 is atomless, so must be real-valued. m2 can also be real-valued, but if A is an atom, we can use that to construct a two-valued k additive measure. Agreed, we can't prove its consistency (unless ZFC is inconsistent). My modified question is, "is the coexistence of m1 and m2 provably inconsistent with ZFC". I understand some set theorists hoped that measurable cardinal hypothesis would turn out to be inconsistent with ZFC. $\endgroup$ – Mangesh Patwardhan Apr 26 '18 at 11:03
  • $\begingroup$ How does "atomless" imply real-valued ? If $k$ is measurable it has a two valued atomless measure... Also I don't get how an atom allows a nontrivial two valued measure (any principal ultrafilter yields an atomless measure). Such an m2 exists on any set, that's what I don't understand $\endgroup$ – Max Apr 26 '18 at 11:48
  • $\begingroup$ 1. Let m be a two-valued measure on k and let A be s/t m(A)>0. Then, m(A)=1. For any subset B of A, either m(B)=0, or m(B)=1=m(A). So, A is an atom. So m is not atomless. Thus, "atomless" implies real-valued. 2. Let m be a k-additive non-trivial measure on k with atom A. Define m' on P(k) by m'(x) = m(x intersection A) / m(A). Then m' is a two-valued, k-additive two valued measure on k, as can easily be checked. $\endgroup$ – Mangesh Patwardhan Apr 27 '18 at 5:41
  • $\begingroup$ Ok sorry there was a misunderstanding on my part of what you called atom ! $\endgroup$ – Max Apr 27 '18 at 8:07

I assume you want the measures to be $0$ on singletons. If this is the intention, it is impossible for both measures to coexist, for the reason that you identify: $m_1$ would force $\kappa$ to be (atomlessly) real-valued measurable, which implies that $\kappa\le\mathfrak c=|\mathbb R|$, while $m_2$ would force $\kappa$ to be a measurable cardinal, and therefore strictly larger than $\mathfrak c$.

(Of course, if you do not add the requirement that the measures vanish on singletons, then we have a silly example by starting with a real-valued measurable cardinal and a witnessing measure $m_1$, and letting $m_2$ be the $\{0,1\}$-measure corresponding to a principal ultrafilter on $\kappa$.)

Kanamori's book on large cardinals and Fremlin's survey on real-valued measurability are both good places to read on this topic. Both discuss in an accessible manner the details required to fully understand the first paragraph above.

  • $\begingroup$ Yes, I'm studying Kanamori's book, At times, his proofs are quite short and cryptic, where I refer to Jech's book. But must a measurable cardinal be strictly bigger than c? I thought by Easton's Theorem, any cardinal with an uncountable cofinality can serve as the value of c. $\endgroup$ – Mangesh Patwardhan Apr 30 '18 at 7:12
  • $\begingroup$ Yes, but Easton's theorem does not preserve in general large cardinals. In particular, measurable cardinals are strongly inaccessible, so they are strictly larger than $\mathfrak c $. What Easton's theorem ensures is that if $\kappa $ is measurable, then in a forcing extension $\mathfrak c\ge\kappa $, so in the extension $\mathfrak c $ is large for sure, but $\kappa $, although still regular limit, it is no longer strongly inaccessible. $\endgroup$ – Andrés E. Caicedo Apr 30 '18 at 12:46
  • $\begingroup$ A simpler version of the same issue is that if $\omega_1$ is collapsed, then the old $\omega_1$ is now countable. This does not mean that the $\omega_1$ of the extension is countable. $\endgroup$ – Andrés E. Caicedo Apr 30 '18 at 12:47
  • $\begingroup$ (And, by the way, the result you are recalling is due to Solovay and precedes Easton's theorem.) $\endgroup$ – Andrés E. Caicedo Apr 30 '18 at 12:48
  • $\begingroup$ Ok. So the way I understand it is if we are in the "real" V - the ground model - any "actually" measurable k will always be bigger than the "actual" c. However, we can build a forcing extension V' where the set that V' thinks is c is bigger than the actual k, but at the cost of k losing its measurability w.r.t. V'. $\endgroup$ – Mangesh Patwardhan Apr 30 '18 at 14:59

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