So I'm currently reading about signed measures in "Real analysis" by Folland. In it, he defines a signed measure as follows.

Let $(X,\mathcal{M})$ be a measurable space. Then a signed measure $\nu$ is a function from $\mathcal{M}$ to $[-\infty,\infty]$ satisfying
(i) $\nu(\emptyset) = 0$
(ii) At most one of the values $\pm\infty$ are assumed
(iii) If $A_1, A_2, \dots$ are (pairwise) disjoint elements of $\mathcal{M}$, then $\nu(\cup_iA_i) = \sum_i\nu(A_i)$ where the sum on the RHS converges absolutely if the LHS is finite.

Now suppose we have a sequence $(A_i)$ of sets in $\mathcal{M}$. If $\nu(\cup_iA_i) = \infty$, it is not obvious to me that, under these hypotheses, for any permutation $\sigma$, one automatically has $\sum_i\nu(A_{\sigma(i)}) = \infty$.

Is this in fact automatically true or is it an additional assumption that we bake into the definition?

  • $\begingroup$ That is correct. What happens when you try to prove it? en.wikipedia.org/wiki/Riemann_series_theorem $\endgroup$
    – GEdgar
    Nov 19, 2022 at 18:45
  • $\begingroup$ @GEdgar Could you elaborate? I've tried proving it myself and have frequented that link before posting, but I've not been able to work it out myself. That's why I posted here. I don't see how my two questions are answered by that link and need help. $\endgroup$
    – Simon SMN
    Nov 19, 2022 at 19:22

2 Answers 2


Answer for the original question.

Let $a_n \in [-\infty,\infty]$, for $n=1,2,3,\dots$. Write $\mathbb N = \{1,2,3,\dots\}$. Let $\mathfrak S$ be the set of all bijections $\sigma : \mathbb N \to \mathbb N$. Let $\mathfrak S_0$ be the set of all $\sigma \in \mathfrak S$ such that $\lim_{n \in \mathbb N}\sum_{j=1}^n a_{\sigma(j)}$ exists in $[-\infty,\infty]$. For $\sigma \in \mathfrak S_0$, write $$ A(\sigma) = \lim_{n \in \mathbb N}\sum_{j=1}^n a_{\sigma(j)} . $$ Write $A_+ = \sum_{j : a_j>0} a_j$ and $A_- = \sum_{j : a_j<0} a_j$.

Assume $$ \text{there exists $\sigma_1, \sigma_2 \in \mathfrak S_0$ with $A(\sigma_1) \ne A(\sigma_2)$}. \tag1$$ We claim $A_+ = +\infty$ and $A_- = -\infty$.

First, we eliminate some trivial cases.

$\bullet$ If there exist $j_1, j_2 \in \mathbb N$ with $a_{j_1} = +\infty$ and $a_{j_2} = -\infty$, then $\mathfrak S_0 = \varnothing$. This contradicts $(1)$.

$\bullet$ If there exists $j_1 \in \mathbb N$ with $a_{j_1} = +\infty$, but there is no $j_2$ with $a_{j_2}=-\infty$, then $A(\sigma) = +\infty$ for all $\sigma$. This contradicts $(1)$.

$\bullet$ If there exists $j_2 \in \mathbb N$ with $a_{j_2} = -\infty$, but there is no $j_1$ with $a_{j_1}=+\infty$, then $A(\sigma) = -\infty$ for all $\sigma$. This contradicts $(1)$.

So, we assume from now on that $a_j \in \mathbb R$ for all $j$.

$\bullet$ If $A_+ = +\infty$ and $A_- \ne -\infty$, then $A(\sigma)=+\infty$ for all $\sigma$. This contradicts $(1)$.

$\bullet$ If $A_+ \ne +\infty$ and $A_- = -\infty$, then $A(\sigma)=-\infty$ for all $\sigma$. This contradicts $(1)$.

$\bullet$ If $A_+ \ne +\infty$ and $A_- \ne -\infty$, then the series converges absolutely. THEOREM For an absolutely convergetn series $\sum a_j$, we have $A(\sigma_1) = A(\sigma_2)$ for all $\sigma_1,\sigma_2 \in \mathfrak S$. This contradicts $(1)$.

The only case remaining is: $A_+=+\infty$ and $A_-=-\infty$.

So we see that the only nontrivial case is the THEOREM, which is in every calculus text. We can consider this result an easy restatement of the THEOREM.

For the the revised version of the question, it seems to me there is nothing to do. $$\bigcup_{j=1}^\infty A_j = \bigcup_{j=1}^\infty A_{\sigma(j)}$$ for any permutation, so if the sets are pairwise disjiont, then by (iii) twice we have $$ \sum_{j=1}^\infty \nu(A_j) =\nu\left(\bigcup_{j=1}^\infty A_j\right) = \nu\left(\bigcup_{j=1}^\infty A_{\sigma(j)}\right) =\sum_{j=1}^\infty \nu(A_\sigma(j)) $$ even if the series do not converge absolutely.

  • $\begingroup$ Okay. I modified my question just before your answer to make my question more concise. I'm still left wondering, if $(A_i)$ is a sequence of sets in $\mathcal{M}$ and $\sum_i\nu(A_i) = \infty$, then why is it that $\sum_i\nu(A_{\sigma(i)}) = \infty$ for any permutation $\sigma$? $\endgroup$
    – Simon SMN
    Nov 19, 2022 at 21:01
  • $\begingroup$ I now see your edit where you adress my revised version of the question. From your answer, I gather that $\sum_j\nu(A_{\sigma(j)}) = \infty$ hold for the permuted sequence $(A_{\sigma(j)})$ is part of what we require of $\nu$ in order for it to be called a signed measure. I am somewhat concerned by this because it seems like quite a strong requirement since this is not some inherent property of sequences of real numbers. $\endgroup$
    – Simon SMN
    Nov 20, 2022 at 8:47
  • $\begingroup$ In particular, if I come up with a function $\nu'$ from $\mathcal{M}$ to $[-\infty,\infty]$ that satisfies both (i) and (ii) but for which there exists a sequence of sets $(A_j)$ and a permutation $\sigma$ such that $\sum_j\nu'(A_j) = \infty$ BUT $\sum_j\nu'(A_{\sigma(j)}) \not = \infty$ (perhaps it converges finitely or perhaps it oscillates and converges neither finitely nor infinitely), then $\nu'$ is not a signed measure. Is this correct? $\endgroup$
    – Simon SMN
    Nov 20, 2022 at 8:51
  • $\begingroup$ Correct. Your example does not satisfy (iii), since there is some sequence of sets [Either $(A_j)$ or $(A_{\sigma(j)})$] where (iii) fails. $\endgroup$
    – GEdgar
    Nov 20, 2022 at 11:17
  • $\begingroup$ And $\nu'(\cup_iA_i) = \sum_i\nu'(A_i) = \infty$ but $\sum_i\nu'(A_{\sigma(i)})$ is oscillating can happen. Correct? $\endgroup$
    – Simon SMN
    Nov 20, 2022 at 12:36

Okay, I've done some digging and I think I've formulated my original hesitancy and resolved it.

Let $(X,\mathcal{M})$ be a measurable space and suppose $\nu':\mathcal{M}\to[-\infty,\infty]$. Suppose further that

(i) $\nu'(\emptyset) = 0$
(ii) $\nu'$ assumes at most one of the values $\pm\infty$.

Then, given any sequence $(A_i)$ of sets in $\mathcal{M}$, that (iii) hold for some permutation of the $A_i$'s is in fact equivalent to it holding for all possible permutations of them. The left implication is obvious. For the right, one supposes the negation and, using the Riemann series theorem, shows that this leads to a violation of (ii). QED.

Hopefully, someone can confirm that I'm right. My hope is this answer will be of help to someone else reading in Folland about signed measures.

  • $\begingroup$ I am not really seeing the argument here. Let us assume $\{E_j\}$ is a disjoint sequence of measurable sets. If $\mu(E)=+\infty$, where $E$ denotes the union over $E_j$. Assume also there exist permutations indexed by $k$ and $l$ such that $\sum \mu(E_k)\in\mathbb{R}$, and $\sum \mu(E_l)=+\infty$. From Riemann's Theorem, all we can conclude is $\{\mu(E_j)\}_{j\geq 1}$ is not an $l^1$ series. The contradiction of (ii) requires (iii) to hold, which is what we are trying to make sense of. $\endgroup$
    – Qqqq123123
    Jul 21 at 18:07

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