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A cardinal $\kappa$ is real-valued measurable if there is a probability measure on the $\sigma$-algebra of all subsets of $\kappa$ which is zero on singletons and additive on disjoint families of fewer than $\kappa$ subsets.

What if I weaken this to: $\kappa$ is uncountable and there is a finitely additive probability measure which is zero on singletons and such that the union of any family of fewer than $\kappa$ null sets is null. Has this notion been studied? What is the consistency strength of the existence of a cardinal with this property?

Edit: A paper has been written on the project which provoked this question. It has been posted on the arXiv. Ashutosh's nice result is discussed in Section 7. (I am adding an "operator algebras" tag because the paper is about large cardinal aspects of von Neumann algebras.)

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  • $\begingroup$ Some sort of a $\kappa$-Hahn-Banach... $\endgroup$
    – Asaf Karagila
    Commented Jun 1, 2016 at 16:12
  • $\begingroup$ @AsafKaragila: could you elaborate? $\endgroup$
    – Nik Weaver
    Commented Jun 1, 2016 at 16:57
  • $\begingroup$ If my memory serves me right, HB is equivalent over ZF to the existence of finitely additive probability measures. So... $\endgroup$
    – Asaf Karagila
    Commented Jun 1, 2016 at 17:13

1 Answer 1

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Suppose $m:\mathcal{P}(\kappa) \to [0. 1]$ is a finitely additive measure whose null ideal $I$ is $\kappa$-additive. Then $I$ is an $\omega_1$-saturated $\kappa$-additive ideal over $\kappa$. Solovay showed this implies that $\kappa$ is measurable in $L[I]$.

If we start with a measurable cardinal $\kappa$, and force with a finite support iteration of random forcing of length $\kappa$, then in the resulting model there is a finitely additive measure on $\mathcal{P}(\kappa)$ whose null ideal is $\kappa$-additive and there is no real valued measurable cardinal below the continuum. You can find a proof of this fact in A. Kumar, K. Kunen, Induced ideal in Cohen and random extensions, Topology and its Applications, Vol. 174, Sept. 2014, 81-87.

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  • $\begingroup$ Great, that really helps! $\endgroup$
    – Nik Weaver
    Commented Jun 1, 2016 at 22:22
  • $\begingroup$ Ashutosh, I want to use this in a paper and would cite a personal communication from you if you care to tell me your name ... my email is nweaver at math dot wustl dot edu. $\endgroup$
    – Nik Weaver
    Commented Jun 16, 2016 at 9:03
  • $\begingroup$ I coauthored that paper with Kunen so just a reference to the paper should suffice. $\endgroup$
    – Ashutosh
    Commented Jun 16, 2016 at 22:01

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