Question about definition of signed measures [Stein and Shakarchi] In Stein and Shakarchi's Real Analysis, p. 285-6, they define a signed measure $\nu$ on a $\sigma$-algebra $\mathcal M$ of subsets of a set $X$ as a function that


*

*Is extended, in the sense that $\nu$ is a function $\mathcal M\to(-\infty,+\infty]$.

*If $\{E_j\}_{j=1}^\infty$ are disjoint subsets of $\mathcal M$, then
$$\nu\bigg(\bigcup_{j=1}^\infty E_j\bigg) = \sum_{j=1}^\infty \nu(E_j).$$


Then they say, "Note that for this to hold the sum $\sum \nu(E_j)$ must be independent of the rearrangement of terms, so that if $\nu(\bigcup_{j=1}^\infty E_j)$ is finite, it implies that the sum converges absolutely."
This definition seems off to me. I understand that for $\nu(\bigcup_{j=1}^\infty E_j)$ to be well-defined, the right-hand side must be independent of rearrangement, but it doesn't make sense to me to say "if $\nu(\bigcup_{j=1}^\infty E_j)$ is finite, it implies that the sum converges absolutely" because we could take $E_j$ to be disjoint sets of measure $(-1)^{j+1}/j$ for $j=1,2,3,\dots,$ and then we would have
$$
\nu(E_1\cup E_2\cup E_3\dotsb) = 1-\frac{1}{2}+\frac{1}{3}-\dotsb = \ln(2),
$$
so the sum is finite, but the series does not converge absolutely.
So what is the correct way to state this definition? Or where am I misinterpreting the definition?
 A: This answer is a little late (as the OP is by know an expert in analysis) but since it was left unanswered for so long, here is a solution the this question
The reason that $\sum_n\nu(E_n)$ must converge absolutely has to do with the fact that the sum is independent of rearrangements. Here are the missing ingredients:

Theorem: Suppose $g:\mathbb{N}\rightarrow\mathbb{N}$ is a bijective function. For any series $\sum_na_n$ define $b_n=a_{f(n)}$ and consider the series $\sum_nb_n$. If the series $\sum_na_n$ converges absolutely and $\sum_na_n=s$, then $\sum_nb_n$ also converges absolutely and $\sum_nb_n=s$.

and  following remarkable result by Riemann. Recall that a convergent  series $\sum_na_n$ is conditionally convergent if $\sum_n|a_n|$ diverges.

Theorem: Let $\sum_na_n$ be a conditionally convergent series with real-valued terms. Let $x$ and $y$ be given numbers in the extended real line $[-\infty,\infty]$, with $x<y$. Then there exists a rearangement of $\sum_n b_n$ of $\sum_na_n$, that is, $b_n=a_{f(n)}$ for some bijective function $f:\mathbb{N}\rightarrow\mathbb{N}$, such that
$$\liminf_{n\rightarrow\infty}\sum^n_{k=1}b_k=x,\qquad \limsup_{n\rightarrow\infty}\sum^n_{k=1}b_k=y$$

In your example, the series $\log 2=\sum_n\frac{(-1)^n}{n}$ is conditionally convergent (i.e. $\sum_n\frac{1}{n}=\infty$). According to Riemann's theorem, for any real-extended number $x\neq\log(2)$, there is a bijective function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $\sum_n\frac{(-1)^{f(n)}}{f(n)}$ converges to $x$.
Combined, the theorems above imply that

Corollary: A series $\sum^\infty_{n=1}a_n$ converges absolutely with sum $s$ iff for any bijective function $f:\mathbb{N}\rightarrow\mathbb{N}$,  $\sum^\infty_{n=1}a_{f(n)}$ converges with sum $s$.

An nice presentation of the theorems I mention above can be found in  Apostol, T. Mathematical Analysis, 2nd edition,  Addison Wesley,  1974 pp. 187, 196-199.
A: Observe that $\nu(\cup_{k\in\mathbb{N}}E_{2k})=-\infty$ while $\nu(\cup_{k\in\mathbb{N}}E_{2k-1})=\infty$. Measure $\nu$ in the example is not a signed measure in the sense of the definition because it maps into $[-\infty,\infty]$ rather than $(-\infty,\infty]$. Only one of the infinities is allowed but not both, precisely to make countable sums over disjoint events absolutely summable.
