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.

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    $\begingroup$ But since this has to be true (by definition) for any sequence $A_1,\dots,A_n\dots$, doesn't the definition preclude this? If there was such a possibility for some signed measure $\mu$, i.e. a way to rearrange the $A_i$'s into $A_{\sigma(i)}$'s to change the sum, then applying this condition to the (valid) sequence $A_{\sigma(1)},\dots,A_{\sigma(n)}\dots$ would contradict the requirement (so that $\mu$ doesn't satisfy the definition of a signed measure after all)... doesn't it? $\endgroup$ – Clement C. Aug 28 '16 at 0:46
  • $\begingroup$ So I guess the converse of Riemann's theorem implies that the sums have to converge absolutely. $\endgroup$ – Funktorality Aug 28 '16 at 0:51
  • $\begingroup$ Yes. See also this (Discussion after Definition 2.3) of Measure Theory: A First Course by Carlos S. Kubrusly. $\endgroup$ – Clement C. Aug 28 '16 at 0:51
  • $\begingroup$ Is it a triviality that the signed measures we usually talk about (eg. $A\mapsto\int_Af$) don't have sequences of disjoint measurable sets that have conditionally convergent sums? $\endgroup$ – Funktorality Aug 28 '16 at 0:52
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    $\begingroup$ Convergence (to the same value) of the right hand side for any rearrangement is part of the requirement. This issue does not occur for 'positive' measures since sum of non-negative numbers always converges when we include $\infty$ to our number system. Now like you said, Riemann's theorem implies that the series actually converges absolutely so that's an useful fact we can infer, not a contradiction. $\endgroup$ – Dilemian Aug 28 '16 at 0:54

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).


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