TITLE: Why isn't conditional convergence an issue when defining $\sigma$-additivity of signed measures
QUESTION [5 upvotes]: A signed measure $\mu$ has to satisfy
$$\mu(\bigcup_{n=1}^\infty A_n)=\sum_{n=1}^\infty\mu(A_n)$$
for $A_n$ disjoint and measurable. Clearly the LHS does not depend on rearrangements of the sequence $(A_n)$, but that's not so clear (to me) for the RHS. What happens if you choose $A_n$ so that the RHS converges conditionally? Doesn't Riemann's rearrangement theorem lead to a problem?
Either I'm missing something silly or we just "define away" this possibility. I looked at the Wikipedia page and my probability book (Shiryaev) and neither addresses the issue.

REPLY [3 votes]: By definition, for any $\mu$ that satisfies the definition of a signed measure, such a sum cannot be conditionally convergent. It has to be absolutely convergent, as otherwise — as you observe — the RHS would not be order-invariant by Riemann's rearrangement theorem, while the LHS is.
So if $\mu$ is a signed measure, then for any sequence of disjoint measurable sets $(A_n)_{n\in\mathbb{N}}$, we have that $\left\lvert \sum_{n=1}^\infty \mu(A_n)\right\rvert < \infty$ implies absolute convergence of the series $\sum_{n=1}^\infty \mu(A_n)$.
As mentioned in a comment, this is discussed briefly (but explicitly) for instance in Measure Theory: A First Course of Carlos S. Kubrusly, 2006 (after Definition 2.3, p.24).
Edit: As pointed out by @shall.i.am in another comment (now deleted), this is also shown in Proposition 10.7 of Real Analysis: Theory of Measure and Integration of J. Yeh (2nd edition, 2006).