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We know that a $\sigma$-algebra is a collection of sets closed under countable set operations. My question is:

how was it determined that this is the right collection?

i.e., how was it determined that these are the sets that reasonably should be measurable.

As an analogue, we know that the definition of a topology comes from the behaviour of the collection of open sets in $\mathbb R^n$, and the fact that continuity may be rephrased as a statement about open sets. That is, the generalization of taking the essential property of behaviour of open sets and continuity is intuitive.

I've never heard nor read an explanation in the same spirit of this one for a $\sigma$-algebra. I'd assume in the same sense, this comes from what subsets of $\mathbb R^n$ we should expect to be Lebesgue Measurable, and then generalizing to arbitrary measures.

Or Is there some way to show this is the maximal collection of sets for any measure?

That is,

does the definition of a $\sigma$-algebra come from some maximality property, or does it come from some generalization of Lebesgue Measurable Sets?

I've done a bit of research on the topic and come up short of an answer, so any discussion of the subject would be appreciated.

Anthony Peter
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    It's just a very natural thing when you start thinking about integrals. At first we integrate over intervals. But it also makes sense to integrate over unions of such things, and also intersections of such things. One then discovers that you can't integrate over all subsets of $\Bbb R$. This leads one to quickly consider the Borel sets as a natural collection of sets to consider integrating over. – Gregory Grant Mar 09 '16 at 01:42
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    Adding to the comment by @Gregory Grant, countable unions and countable intersections of "sets of interest" can arise when you are considering sequences (or infinite sums, as in Fourier series, etc.) of functions. For example, if $f_n$ is a monotone increasing sequence of functions that converges pointwise to $f$ (for example, $f=\Sigma f_n,$ where each $f_n \geq 0),$ then the set of points where the values of $f$ are greater than some given number $c$ is the union of the countably many corresponding sets of points where the values of $f_n$ are greater than $c.$ – Dave L. Renfro Mar 09 '16 at 15:45
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    And adding to the comment of @DaveL.Renfro we could cite the original ideas of Lebesgue on this from his lectures on the subject 1902-03. He said "it is our purpose to associate with every bounded function defined on a finite interval $(a,b)$ a certain finite number $\int_a^b f(x),dx$ which we will call the integral of $f$ on $(a,b)$ and which satisfies the following six properties ..." His final assumed property was that if $f_n\to f$ increasingly then $\int f_n\to \int f$. As soon as you commit to this program of Lebesgue's you find that countable set operations are inevitable. – B. S. Thomson Mar 09 '16 at 16:27

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The issue here is why require countable unions (certainly countable unions of closed sets are not closed) and the answer is related to the countable additivity of the Lebesgue measure, which would be harder to express if the family of sets considered were not a $\sigma$-algebra. Kolmogorov started the trend of assuming that the Lebesgue measure is countably additive, but notice that this is strictly speaking not required for many applications, and the obligatory nature of such an assumption has been questioned in the recent literature. Notice that in some models of the Zermelo-Fraenkel set theory, the Lebesgue measure is not $\sigma$-additive (look for "Feferman-Levy model" here and at MO).

Mikhail Katz
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