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May I ask what is the (historical) reason we adopted the $\sigma$-algebra rhetoric instead of $\sigma$-rings (like used in Halmos)? To my knowledge almost all modern measure theory or real analysis textbooks use $\sigma$-algebra nowadays, but it seems the old theory is perfectly legitimate to teach students basics of Lebesgue theory. In particular when I was learning the material, Halmos' book was very readable. Hence is my question.

Motivation: May need to explain measure theory to undergraduates next semester as I work as a probability TA, though the class itself does not involve measure theory. My advisor told me he believe the old theory still has much of its vitality, and everything works well. So he saw no need to introduce the new theory to (first-time) learners. In fact, he is going to publish a new book in GTM series using the old theory. While this is certainly true, I do not understand what pushed the historical conceptual change. What is the real benefit(further unification of conceptual framework, cleaned up better proofs, generalized easily for non-archemedian fields, etc)?

(For an example of my past unsuccessful effort to explain measure to undergraduates, see here)

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    $\begingroup$ I guess you looking for a integration for addittive measure theory instead $\sigma$-addittive. THere is the book: Theory of Charges by K. P. S. Bhaskara Rao, M. Bhaskara Rao . But Lebesgue theory need $\sigma$-addittive measure and $\sigma$-algebra (measurable set's). In the $\sigma$-addittive theory you have more useful theorem's.... $\endgroup$
    – Buschi Sergio
    Commented Dec 30, 2014 at 11:16
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    $\begingroup$ Halmos uses "sigma-ring" quite a bit. He proposed that set-up for abstract measure theory, and wrote his book using it, but others did not adopt it. $\endgroup$
    – Gerald Edgar
    Commented Dec 30, 2014 at 13:40
  • $\begingroup$ @GeraldEdgar: May I ask what is the reason for the reluctance for acceptance? His theory works fine to me. $\endgroup$
    – Bombyx mori
    Commented Dec 30, 2014 at 13:42
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    $\begingroup$ Oh, my! So it is "sigma-ring", not "semiring"? I was totally perplexed with that typo in the post. Then there is not much difference, really. It is just convenient to assume that the whole space is always measurable and to be able to pass to the complement freely, but otherwise it is more a question about terminology than about substance unless I misunderstand something. $\endgroup$
    – fedja
    Commented Dec 30, 2014 at 22:39
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    $\begingroup$ If you are a probability TA, don't worry about this. In probability theory you always use $\sigma$-algebra. Halmos proposed $\sigma$-rings to be able to say certain very big sets (such as non sigma-finite sets) are "non-measurable". $\endgroup$
    – Gerald Edgar
    Commented Dec 31, 2014 at 16:00

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