What are possible variations of the definition of $\sigma$-additivity? From what I've read in Wikipedia, $\sigma$-additivity is defined
in the following way: Let $\mathcal{A}$ be a $\sigma$-algebra of
some underlying set $X$ and let $f$ be a mapping $f:\mathcal{A}\rightarrow M\subseteq\mathbb{R}\cup\left\{ \pm\infty\right\} $.
Then $f$ is called $\sigma$*-additive*, if for any countable
set $I$ and a family of pairwise disjoint sets $\left(A_{i}\right)_{i\in I}$,
we have 
$
f\left(\bigcup_{i\in I}A_{i}\right)=\sum_{i\in I}f\left(A_{i}\right).
$
My questions are: 1) Did we even need the fact that $\mathcal{A}$
is a full-blown $\sigma$-algebra for this definition ? 
Besides using the fact that  $\mathcal{A}$ being a $\sigma$-algebra guarantees me that   $\bigcup_{i\in I}A_{i}$ is also in  $\mathcal{A}$, at no
point in this definition are we using the other properties of $\mathcal{A}$,
so we could just as well define $\sigma$-additivity for $f:\mathcal{A}\rightarrow M\subseteq\mathbb{R}$,
where $\mathcal{A}$ is just some system of sets such that  $\bigcup_{i\in I}A_{i}\subseteq\mathcal{A}$, for pairwise disjoint sets $A_i$.
A different approach: We could drop the above property of  $\mathcal{A}$ altogether , so that  $\mathcal{A}$ is just  system of sets  without any additional prperties,  and define $f$ to be $\sigma$-additive only for those $\bigcup_{i\in I}A_{i}$ that are contained in  $\mathcal{A}$ .
2) Is it custom for $M$ to be some certain subset of $\mathbb{R}$,
or can $M$ be an arbitrary one ? I've seen $M=\mathbb{R}\cup\left\{ \pm\infty\right\} $
(on wikipedia) and $M=\left[0,1\right]$ (when dealing with probabilities),
so I'm wondering if in the definition of $\sigma$-additivity it is
ok to require just $M\subseteq\mathbb{R}\cup\left\{ \pm\infty\right\} $.
 A: *

*No. As a matter, there are useful results based on $\sigma$-additivity on a smaller class of sets. In this case, one just requires that the condition holds when $\bigcup_i A_i$ is in the class. So closure under countable disjoint unions is not necessary. For example, a countably additive nonegative function on an algebra can be extended to $\sigma$-algebra generated by this algebra. This is a useful result for constructing measures.

*Measures are usually defined so that they have range in $\mathbb{R}\cup\{\infty\}$. For signed measures, one must rule out that the function can take both the values $(+)\infty$ and $-\infty$, since expressions of the form $\infty-\infty$ are not well defined.
A: Here's something that I found worth mentioning.
Note that $\sigma$-additive set functions that take both positive and negative values usually require an additional assumption, which (surprisingly) was not mentioned in the Wikipedia article.
For $\sum_{i=1}^{\infty}\mu(A_{i})$ to be well defined for arbitrary disjoint collections $\{A_{i}\}$, one must in addition require that either the positive or negative terms of the sum are bounded from above or below (respectively) with a constant. Otherwise the sum may not be defined.
For example, one may have $\mu(A_{1})=\infty$ and $\sum_{i=2}^{\infty}\mu(A_{i})=-\infty$, eventhough the measure itself takes only either (one of) $\infty$ or $-\infty$.
In fact, the definition of a signed measure ($\sigma$-additive set function over a $\sigma$-algebra that gives zero to empty set) has usually this assumption included in the definition, that for any countable disjoint collection the measure is controlled to behave as noted above.
A: First, the fact that $\mathcal{A}$ is a $\sigma$-algebra can guarantee that if $A_i$ is belong to the domain field of $f$ for any $i \in I$,$\bigcup\limits_{i \in I}A_i$ is belong to the domain field of $f$.
Then, in my view,maybe the fact that $\mathcal{A}$ is a $\sigma$-algebra is not necessary.It is only needed that if $A_i$ is belong to the domain field of $f$ for any $i \in I$,$\bigcup\limits_{i \in I}A_i$ is belong to the domain field of $f$.
But, under this condition that $\mathcal{A}$ is a $\sigma$-algebra, maybe it's more conveninent for us to do some research on it later.
A: For the second question, $M$ can be an arbitrary subset of $\mathbb{R}$. 
You can get some knowledge about the abstract measure spaces.
