A well-known result of Solovay states that $ZFC$ + "the continuum is real-valued measurable" is equiconsistent with $ZFC$ + "there is a measurable cardinal", over $ZFC$. That the continuum is mentioned here is no coincidence. A theorem of Ulam back in 1930 gave evidence that the continuum is some sort of a dividing line: any real-valued measurable cardinal above the continuum must be measurable. On the other hand, cardinals at or below the continuum cannot be measurable (or even strongly inaccessible), and in general are a lot more tame.

A natural question to follow-up would be, where can the least real-valued measurable cardinal lie, using the continuum as a benchmark? In particular, I would like to ask if each of the following is known to be consistent with $ZFC$ in conjunction with any large cardinal axiom:

  1. the least real-valued measurable cardinal $< \mathfrak{c}$.
  2. the least real-valued measurable cardinal $= \mathfrak{c}$.
  3. the least real-valued measurable cardinal $> \mathfrak{c}$.

Solovay's result means that either 1. or 2. is consistent, but I cannot find any hint as to which of the two holds in his model.

  • 2
    $\begingroup$ I think the following works: Let $\kappa$ be the least measurable cardinal in $V$. For 1), take $\lambda>\kappa$ such that $\lambda^\omega=\lambda$, then adding $\lambda$ many random reals will produce $2^\omega=\lambda$ and $\kappa<\lambda$ is real-valued meas. For 2), you can start with $V=L[U]$ and choose $\lambda=\kappa$ as in the first case. If there is a real-valued meas below $\mathfrak{c}=\kappa$ I think you produce an inner model of measurable cardinal $<\kappa$, this will contradicts uniqueness. For 3), any model of CH works. $\endgroup$
    – Jing Zhang
    May 2, 2018 at 16:55

1 Answer 1


All of them are equiconsistent with the existence of a measurable cardinal. I find that a nice reference for this stuff, in addition to Solovay's article, is Jech's Set Theory: Third Millennium Edition.

I will discuss 3 first. Any measurable cardinal is real-valued measurable (it just happens not to use any of the values between 0 and 1). Solovay proved that if $\kappa$ is real-valued measurable, $\mu$ is a normal measure on it with $I$ the ideal of null sets, then in $L[I]$ the cardinal $\kappa$ is measurable. Section 3.6 of his paper has a reference to a paper by Kunen proving that the generalized continuum hypothesis holds in $L[I]$, so $\mathfrak{c} = \aleph_1$ is not real-valued measurable in $L[I]$ (nor is any smaller cardinal).

Case 2 is what occurs if we do Solovay's construction of adding $\kappa$ random reals, where $\kappa$ is a measurable cardinal (which becomes an atomlessly real-valued measurable in the forcing extension).

Case 1 happens if we take a $\lambda > \kappa$ where $\lambda$ has uncountable cofinality (this is automatic if $\kappa = \lambda$ itself because it is regular, but the extra assumption is needed here), and force $\lambda$ random reals. In the forcing extension, $\mathfrak{c} = \lambda$ and $\kappa$ is atomlessly real-valued measurable, and $\kappa < \lambda$ is unchanged. This, and case 2, are explained in section 4 of Solovay's paper, and also in Chapter 2 of Jech's book, around page 415.

  • $\begingroup$ Regarding 1: If by "real-valued measurable" we mean atomlessly measurable (rather than measurable in the usual sense of large cardinals), then any real-valued measurable cardinal is of size at most the continuum. $\endgroup$ May 2, 2018 at 23:31
  • $\begingroup$ Can you briefly illustrate how adding $\kappa$ random reals in Case 2 would not make any cardinal below $\kappa$ real-valued measurable? $\endgroup$
    – Zoorado
    May 3, 2018 at 2:42
  • 5
    $\begingroup$ @Zoorado If you start with $L[\mu]$, the smallest model with a measurable cardinal $\kappa$ (with $\kappa$ as small as possible) and add $\kappa$ random reals, Solovay's argument shows that in the extension $\kappa$ is real-valued measurable and the continuum. If $\lambda<\kappa$ is also real-valued measurable, Solovay's argument using the null ideal for a witnessing measure on $\lambda$ would give you an inner model with $\lambda$ measurable. This contradicts the minimality of $\kappa$. (The one technical further detail is that minimality of $L[\mu]$ is absolute under forcing extensions.) $\endgroup$ May 3, 2018 at 3:24

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