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.

  • 1
    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? – Clement C. Aug 28 '16 at 0:46
  • So I guess the converse of Riemann's theorem implies that the sums have to converge absolutely. – Funktorality Aug 28 '16 at 0:51
  • Yes. See also this (Discussion after Definition 2.3) of Measure Theory: A First Course by Carlos S. Kubrusly. – Clement C. Aug 28 '16 at 0:51
  • 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? – Funktorality Aug 28 '16 at 0:52
  • 1
    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. – Dilemian Aug 28 '16 at 0:54
up vote 2 down vote accepted

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).

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.