Real-measurable cardinals that are not measurable ones 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?
 A: 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.
A: 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 $\mathbb{R}$.
