The Hahn-Banach theorem implies that Lebesgue measure can be extended give a "measure" on all subsets of [0,1], but this measure is only guaranteed to be finitely additive. It might magically turn out that this measure is countably additive, but this can only happen if the continuum is a real-valued measurable cardinal, a strong set-theoretic assumption. My question is: if it turns out that measure is countably additive on the measure zero sets, does this imply that the measure is countably additive everywhere?
Measure Theory – Extensions of Lebesgue Measure
measure-theoryset-theory
Best Answer
The answer is no. A proof can be found here.