Definition of a signed measure in Folland and its meaning when the signed measure is not finite 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?

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.
