# 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 $M\subseteq\mathbb{R}\cup\left\{ \pm\infty\right\}$.

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For 1), for $f(\cup_i A_i)$ to even make sense we need $\mathcal{A}$ to be closed under countable union :p – uncookedfalcon Apr 22 '12 at 9:19
@uncookedfalcon Granted, but that is still weaker than $\sigma$-additivity (I edited it in my question). – MyCatsHat Apr 22 '12 at 9:39

1. 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.

2. 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.

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As you have written the answer, in a response to uncookedfalcons comment I have also edited my question: Could you please expand your answer of 1) to include also my different approach (i.e. is it meaningful to drop even the property that $\bigcup_i A_i \subseteq \mathcal{A}$ and define $\sigma$-additiviy only for those families of subsets, that do have this property? Also, I didn't quite understood, what you meant with $f(\bigcup_i A_i)$ "being in this class"), before I accept your answer ? – MyCatsHat Apr 22 '12 at 9:54
Concerning 2): Why are then (discrete) probability spaces defined with values only in $[0,1]$ ? (Please bare with me, that I have absolutely know knowledge of measure theory/probability and this question was only motivated by some stuff we did in an introductory probability course that I'm taking now.) – MyCatsHat Apr 22 '12 at 9:56
@user26698: The "$f(\bigcup_i A_i)$" was a typo, I actually gave your extended definition. Probability measures are those (nonnegative) measures $\mu$ with the property that $\mu(X)=1$. If we just require $\mu(X)<\infty$, we get the larger class of finite measures. If we require $X=\bigcup_{i=1}^n A_i$ for some family of measurable sets satisfying $\mu(A_i)<\infty$, we get a $\sigma$-finite measure. There are even weaker forms, but these are the main ones. – Michael Greinecker Apr 22 '12 at 10:00

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.

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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.

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For the second question, $M$ can be an arbitrary subset of $\mathbb{R}$.

You can get some knowledge about the abstract measure spaces.

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