Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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\} $.

share|improve this question
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). –  user26698 Apr 22 '12 at 9:39
add comment

4 Answers

up vote 1 down vote accepted
  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.

share|improve this answer
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 ? –  user26698 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.) –  user26698 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
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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

You can get some knowledge about the abstract measure spaces.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.