$\sigma$-Algebra: Why do we want it to contain complements as well? Everybody Hello, I was always wondering:
(Please answers apart from historical reasons)


*

*Why do we want a $\sigma$-Algebra to possess more than just its crucial disjoint $\sigma$-union property? Say, why do we want it to contain the complements as well?

*Would it be fine to restrict for first attempt on the disjoint $\sigma$-union property and then define a measure as a $\sigma$-additive set function?


...moreover, I would be appeased if one requires additionally that it contains the empty set and the whole space...I mean, some authors like Dunford and Schwartz follow this "purer" route.
 A: While not an answer per se, here is an example to get you thinking about why you might want more than countable disjoint unions.
Consider the space $[0,\infty)\subset \mathbb R$, and take as your "algebra" the sets of the form $[0,x)$.  The only step functions on this space contain only one step, and the sum of two step functions will not in general be measurable.
It's one thing to have a theory where a space has very few measurable functions.  It is quite another to have a theory where you can't even add measurable functions.
A: There is a textbook by Halmos, Measure Theory, in which he does as much as possible in the setting of $\sigma$-rings, where complements are not assumed.  But subsequent mathematicians have not adopted that point of view.  
(1) In the Halmos definitions, we can postulate that every measurable set is $\sigma$-finite; then when (rarely) it is necessary, extend things to non-$\sigma$-finite sets.
(2) In the non-Halmos definitions, we postulate that the collection of measurable sets is a $\sigma$-algebra, then when (rarely) it is necessary, state theorems for $\sigma$-finite sets.
Each system has some advantages, some simplifications that occur.  But mathematians have chosen system (2).
A: As preliminary answer we want to deal especially with probability theory and stochastic processes where complements constitute essential parts of the theory ...so when considering measure theory we'd like to include complements right from the beginning ...however for pure integration theory sigma algebras seem unnecessary.
