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Consider an infinite cardinal $\kappa$. Is it the case that the existence of a $\kappa$-additive measure on some infinite set implies the existence of such a measure on every infinite set of size greater than $\kappa$ (and never for sets of size less than or equal to $\kappa$, of course)?

If so, it would seem to follow that the real-valued measurable cardinals form an initial segment. Is this a known fact?

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Initial segment of what? – Asaf Karagila Feb 13 '14 at 1:03
It is well known and it can be proven using Ulam matrices that real-valued measurable cardinals are weakly inaccessible, so they cannot gather together into a segment. – Joseph Van Name Feb 13 '14 at 1:37

It is clear that if there is a $\kappa$-additive measure $\mu$ on a set $X$, measuring every subset of $X$, then we can use $\mu$ to construct a $\kappa$-additive measure on any superset $Y\supset X$, in a trivial way, simply by defining $\mu^*(A)=\mu(A\cap X)$. The new measure $\mu^*$ will be $\kappa$-additive and measure every subset of $Y$. In this sense, yes, we get $\kappa$-additive measures on all larger cardinals, and your question has this positive answer.

But these measures do not concentrate on sets of larger size, and so one might regard them as unsatisfactory. What one would really want is a $\kappa$-additive uniform measure on the larger cardinals $\theta$, meaning that the measure concentrates on subsets of $\theta$ of size $\theta$. In this case, the answer is much more interesting, in light of the following:

Theorem. (Ketonen) A cardinal $\kappa$ is strongly compact if and only if there is a $\kappa$-additive uniform 2-valued measure on every regular cardinal $\theta\geq\kappa$.

So one can get such measures on all larger cardinals just in case $\kappa$ is strongly compact, which is a strictly stronger large cardinal property than measurability. In this sense, your question can have a negative answer, for there can be measurable cardinals that are not strongly compact.

But the plot thickens further in light of Magidor's result that it is relatively consistent with ZFC that there are measurable cardinals, yet every measurable cardinal is strongly compact. In such a model, we recover a positive answer to the question again, simply because in this model all measurable cardinals $\kappa$ do have such uniform $\kappa$-complete measures on all larger regular cardinals $\theta$.

(And as pointed out in the comments, your remark on real-valued measurable cardinals is not correct.)

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