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I'm reading Jech's Set Theory, and in the chapter about measurable cardinals there is a theorem that if $\kappa$ is real-measurable but not measurable then it is $\le 2^{\aleph_0}$ and so and so. (Corollary 10.10)

How can a cardinal number be real-measurable without being measurable? Can't a measure be "destructed" (as in opposite of constructed) into a trivial $0,1$ measure?

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Asaf: If you are interested in real-valued measurability, do hunt down an article by David Fremlin, "Real-valued-measurable cardinals", in Proc. 1991 Bar-Ilan Conference (ed. H.Judah), Amer. Math. Soc., 1993 (Israel Mathematical Conference Proceedings 6), 151-304. – Andrés Caicedo Jul 6 '11 at 16:20
@Andres: To be fair I was just retagging old questions. I asked these questions when I was just starting my way and studied a lot since then. Either way I will be sure to track this paper and add it into my archive of papers I should read someday. – Asaf Karagila Jul 6 '11 at 16:46
up vote 6 down vote accepted

Two-valued measures behave very differently from real-valued measures. For example, suppose $\mathcal{U}$ is a countably complete ultrafilter on a set $X$ and suppose that $f:X\to2^\omega$ is an injection. There is a $b \in 2^\omega$ such that $$B_n = \{ a \in X : f(a)(n) = b(n) \} \in \mathcal{U}$$ for every $n < \omega$. By countable completeness, $B = \bigcap_{n<\omega} B_n \in \mathcal{U}$. But $B$ contains exactly one element (namely $f^{-1}(b)$) since $f$ is an injection. Therefore, $\mathcal{U}$ is a principal ultrafilter.

This argument shows that the first measurable cardinal is larger than $2^{\aleph_0}$. Indeed, a slightly more general argument can be used to show that a measurable cardinal must be inaccessible. However, this argument cannot be carried out with a real-valued measure. In fact, it is possible (assuming the consistency of a measurable cardinal) for Lebesgue measure to be extended to a measure defined on all subsets of $\mathbb{R}$.

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I knew the argument for showing that a measurable cardinal has to be inaccessible. I didn't think of using that to show that an ultrafilter extending the filter of sets of measure $1$ induces an atomic measure. Thanks! – Asaf Karagila Sep 28 '10 at 9:10

It is a good question. The answer is that real-valued measurable cardinal need not be strongly inaccessible, whilst every measurable cardinal is strongly inaccessible. Indeed, it is consistent that the continuum itself is a real-valued measurable cardinal, but the continuum can never be a measurable cardinal, since every measurable cardinal is strongly inaccessible.

Nevertheless, part of what you claim is true: Solovay proved that every real-valued measurable cardinal $\kappa$ is fully measurable (with a two-valued measure) in an inner model of the universe. That is, if $\kappa$ is a real-valued measurable cardinal, then there is a definable transitive class $W$ satisfying ZFC in which $\kappa$ is an actual measurable cardinal. The class $W$ is defined directly from the real-valued measure on $\kappa$, and this provides a sense in which the measure is deconstructed to form a 2-valued measure.

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