$\sigma$-Algebra: Why do we want it to contain complements as well?

Question

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.

Answer 1

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.

Answer 2

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).