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

Question

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?

Answer 1

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.

Answer 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 $A_-=-\infty$.

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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.

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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.