One answer to your question is to remain within the realm of the continuum, and consider what happens when the Continuum Hypothesis fails. In this case, there are cardinals κ below the continuum c = |R|, and one naturally inquires whether these cardinals behave more like aleph<sub>1</sub>, or more like the continuum, with respect to measure and additivity. For example, we may still want to inquire about our favorite measures, but with uncountable cardinals. The answer is saturated with set-theoretic independence. For example, it is known to be consistent with the axioms of set theory that Lebesgue measure can be *better* than countably additive! Under Martin's Axiom (MA), when the Continuum Hypothesis fails, then the union of κ many measure zero sets remains measure zero. There are similar results concerning the additivity of the ideal of meager sets. There is a rich subject investigating this called *cardinal characteristics of the continuum*. I discussed some of the concepts in [this MO answer](http://mathoverflow.net/questions/8972#9027). Another answer to your question is to go well beyond the continuum. The analogue of the theory of cardinal characteristics of the continuum has been carried out for arbitrary cardinals κ, and again, the situation is saturated with independence results, proved by forcing. One interesting theorem, however, is that for certain types of questions, there is no intermediate possibility between the countable and large cardinals. For example, one can view an ultrafilter on a set as an ω-complete 2 valued measure on subsets of ω. (ω complete = additive for unions of size less than ω = finitely additive). Can one have an Aleph<sub>1</sub> additive ultrafilter on a set? Well, if F is such an ultrafilter, then it must also be Aleph<sub>2</sub> additive, Aleph<sub>3</sub> additive, and so on, for quite a long way, up to the least measurable cardinal. That is, it is just not possible to have a δ-additive ultrafilter, when δ is uncountable, unless it is also additive up to measurable cardinal. Other similar phenomenon surround the question: does every κ additive filter extend to a κ additive ultrafilter on a set? This is true for κ = ω, since this just amounts to being finitely additive, which is what being a filter means. But for uncountable κ, it is equivalent to the assertion that κ is strongly compact, which is very high in the large cardinal hierarchy.