# Definition of (Finite) Signed Measure

My class notes define a signed measure on a measurable space $$(X, \mathcal{R})$$ as a $$\sigma$$-additive function $$\nu : \mathcal{R} \to \mathbb{R}$$. (I take this to mean we're only considering finite measures.) I'm confused on what I'm supposed to interpret $$\sigma$$-additive to mean here. My first guess would be that if $$A = \bigcup_nA_n$$ where $$A_1,A_2,\ldots$$ are disjoint, then $$\nu(A) = \sum_{n=1}^\infty \nu(A_n);$$ i.e., just the usual definition of $$\sigma$$-additivity. But this seems problematic when the series on the right doesn't converge absolutely, because then its value depends on the order of the $$A_i$$, which my gut tells me shouldn't be the case. Does $$\sigma$$-additivity here also involve the claim that the series on the right always converges absolutely? The wikipedia page on signed measures does require this, but no other source I found online, or my class notes, explicitly states it.

• @EricTowers That argument works for regular (nonnegative) measures, but in this case how can we say that sequence is monotonically increasing? If even one of the terms in the series is negative that isn't the case. Oct 24, 2021 at 23:09
• I observe the following from the referenced by uncited Wikipedia page: "The series on the right must converge absolutely when the value of the left-hand side is finite." Oct 24, 2021 at 23:14

$$\nu$$ takes only finite values. So disjointness of $$(A_n)$$ implies that $$\sum \nu(A_n)$$ is a convergent series. So is any rearrangement of terms since disjointness still holds. If a series of real numbers converges whenever the terms are permuted then the series is absolutely convergent. Hence $$\sum |\nu (A_n)| <\infty$$.