I have the impression that modern texts deal almost excusively with measures on $\sigma$-algebras, while older texts, such as the one of Halmos, deal mainly with measures defined on $\sigma$-rings. I'm curious what motivated this change and in what context are $\sigma$-rings more natural domains for measures?
Why define measures on $\sigma$-rings?
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0I'm pretty sure, you now know something to answer this question yourself. Also, on page 10 of Dubins and Savage gambling book, they mention several paper by De Finetti that compare $\sigma$-additive and finitely-additive approach. – 2013-07-16
2 Answers
Being unfamiliar with the older text, I can only speculate. One explanation is that one prefers to work with sets of $\sigma$-finite measure: those that can be written as a countable union of sets of finite measure. For example, sets of $\sigma$-finite length ($1$-dimensional Hausdorff measure) in the plane form a $\sigma$-ring, not a $\sigma$-algebra. It is rather fruitless to think about 1-dimensional measure of the complement of a line, so removing such sets from consideration seems reasonable.
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0And we should not forget that outer measures are more general than measures. But most textbooks focus on measures. Probably it is just a matter of convenience: we choose the setting which is most frequent in special cases. – 2012-06-25
From Wikipedia:
σ-rings can be used instead of σ-fields in the development of measure and integration theory, if one does not wish to require that the universal set be measurable. Every σ-field is also a σ-ring, but a σ-ring need not be a σ-field.
I guess that interesting measure spaces are always measurable, and that's why modern books tend to use $\sigma$-algebras rather than $\sigma$-rings. For example, on $\sigma$-rings it may be impossible to integrate over the whole space, and this is usually a useless restriction.