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A cardinal $\kappa$ is real-valued measurable if there is a $\kappa$-additive probability measure on $2^\kappa$ which vanishes on singletons. The existence of measurable $\kappa$ is independent of ZFC.

Question: if $\kappa$ is assumed to be real-valued measurable, does it necessarily follow that $2^\kappa$ is real-valued measurable?

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    $\begingroup$ No. In fact, this almost never holds - if $2^\kappa$ is greater than continuum and real-valued measurable, then it is measurable, but measurables are strong limits. So $2^\kappa$ can be real-valued measurable only if it's continuum. $\endgroup$
    – Wojowu
    Commented Jan 13, 2019 at 22:30
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    $\begingroup$ @Wojowu You should make that an answer. $\endgroup$
    – Noah Schweber
    Commented Jan 13, 2019 at 22:34
  • $\begingroup$ Independent of ZFC... not exactly: the consistency of ZFC doesn't imply the consistency of of ZFC + there exists a measurable cardinal, if I remember correctly. (But it implies the consistency of ZFC + there's no measurable cardinal.) $\endgroup$
    – YCor
    Commented Jan 13, 2019 at 22:57
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    $\begingroup$ FYI: there is an older definition of "real-valued measurable" that only requires the probability measure to be countably additive. In that case, if one cardinal is rvm then so are all larger cardinals. This version of rvm is more common in earlier literature but one needs to be careful. $\endgroup$
    – François G. Dorais
    Commented Jan 14, 2019 at 23:18
  • $\begingroup$ @FrançoisG.Dorais Could you give a reference? If you turn into a full answer I'll be happy to accept. $\endgroup$
    – Aryeh Kontorovich
    Commented Jan 15, 2019 at 5:27

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No, this is not necessarily true. In fact, this is almost never true - it can hold only if $2^\kappa$ is equal to the continuum.

Indeed, assume $2^\kappa$ is strictly greater than the continuum. By standard results (see e.g. Wikipedia), a cardinal greater than continuum is real-valued measurable iff it's measurable. But measurable cardinals are necessarily strong limits, and $2^\kappa$ is, quite blatantly, not one.

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    $\begingroup$ On the other hand, it is consistent to have $\kappa<2^\kappa=\mathfrak c$, both of them real-valued measurable. $\endgroup$
    – Andrés E. Caicedo
    Commented Jan 13, 2019 at 23:30
  • $\begingroup$ @AndrésE.Caicedo Do you get this by starting with two measurable cardinals $\kappa<\lambda$ and adding $\lambda$ random reals? Or is something trickier involved? $\endgroup$
    – Andreas Blass
    Commented Jan 14, 2019 at 1:47
  • $\begingroup$ @Andreas Yes, that's all that's needed. $\endgroup$
    – Andrés E. Caicedo
    Commented Jan 14, 2019 at 1:58
  • $\begingroup$ @Andrés: And there I was, thinking that all you need is love! I'd be interested in hearing that rendition, "All you need is two measurable cardinals (and a lot of random reals)". $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 14, 2019 at 7:20

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