# Atomless and with atoms $\kappa$-additive measures on a cardinal $\kappa$

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.

• 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 – Max Apr 26 at 10:39
• 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. – Mangesh Patwardhan Apr 26 at 11:03
• 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 – Max Apr 26 at 11:48
• 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. – Mangesh Patwardhan Apr 27 at 5:41
• Ok sorry there was a misunderstanding on my part of what you called atom ! – Max Apr 27 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 $\kappa$.)
• 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. – Andrés E. Caicedo Apr 30 at 12:46
• 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. – Andrés E. Caicedo Apr 30 at 12:47