Why dont we only consider complete measure spaces? This question has been asked a few months ago, but in my opinion, it has not yet received a satisfying answer.
So I ask again: Why should one consider non-complete measure spaces instead of requiring every measure space to be complete per definition?
Most answers on similar questions are like:


*

*If $f$ is a measurable function, one wants its codomain to contain fewer measurable sets, so that one finds more measurable functions beside $f$.

*Lebesgue-Lebesgue-measurable functions are pretty useless because not even continous functions have to be L-L-measurable (but diffeomorphisms are).

*If $\Omega$ is a probability space and $X:\Omega \to \mathbb{R}$ a random variable (where $\mathbb{R}$ is endowed with any $\sigma$-Algebra), its distribution (pushforward measure) isnt necessarily a complete measure.


But I see a mistake in those thoughts. The category of measurable spaces (set + $\sigma$-Algebra) is confused with the category of measure spaces (set + $\sigma$-Algebra + measure). Of course, Lebesgue-Borel-measurable functions are important, because they are the subject of separably-valued integration (on $\mathbb{R}^n$). But for a Bochner-measurable function (a.e. pointwise limit of simple functions) that one wants to integrate, it is irrelevant if there is a measure on its codomain. The $L_1$-functor goes
$$ L_1 : \mathsf{MeasureSpaces} \times \mathsf{BanachSpaces} \to \mathsf{BanachSpaces} \ ,\ (X,A,\mu) , E \mapsto L_1(X,\mu,E)$$
so it may be wrong to think of any sigma-Algebra on the codomain of an integrable function.
Also, if $(X,A,\mu)$ is a measure space and $(X,\bar A,\bar \mu)$ is its canonical completion, a function $f:X \to E$ into a Banach space $E$ is Bochner-measurable with respect to $\mu$, if and only if it is Bochner-measurable with respect to $\bar \mu$. So, for Bochner integration theory, non-complete measure spaces are completely redundant.
And concerning the distribution argument: If one has a random variable $X : \Omega \to \mathbb{R}$, I see no reason to postulate a $\sigma$-Algebra on $\mathbb{R}$ and the measurablitiy of $X$. I think it to be more natural to view $\mathbb{R}$ as simply a set and consider every subset of $\mathbb{R}$ to be measurable, whose preimage is measurable. This yields a $\sigma$-Algebra, bears no needless abandonment of per se measure-accessible subsets and if $\Omega$ is complete, the pushforward measure will be too.
 A: Here are two arguments. I don't think you are addressing them (maybe because I don't speak categories and don't know much about Bochner integration). I use $(\Omega, \mathcal{A}, \mu)$ for a probability space, and $X$ for a r.v.
In short, applications drive the theory. I risk repeating the answers you found unsatisfactory, in which case I am asking you to elaborate your critique, as long as it accepts the usefulness of a theory as a criterion.
First, regarding the "distribution argument". Suppose we observe and want to study a normal distribution on $\mathbb{R}$. The co-domain is fixed, while the abstract outcome space $\Omega$ and r.v. $X$ are arbitrary. We are free to choose any combination of $X$, $(\Omega, \mathcal{A}, \mu)$, and we don't care which, as long as the pushforward is the normal distribution. Our application is only concerned with $\sigma$-summability in $\mathbb{R}$, this is why we postulate a $\sigma$-algebra there. In my understanding, the only real purpose of having abstract outcome and event spaces ($\Omega$ and $\mathcal{A}\subset 2^{\Omega}$) is to formalise interactions between random variables when we have more than one. If we start with $(\Omega, \mathcal{A}, \mu)$, $X_i$, and consider pushforward $\sigma$-algebras, as you suggest in the last paragraph, we risk having several incompatible $\sigma$-algebras in $\mathbb{R}$. That is, questions like "Is $X_1\in (a,b)$ more likely than $X_2\in (a,b)$?" aren't even guaranteed to be well-posed.
Second (which is related to the first), we often want $g \circ X$ to be a random variable on the same domain and co-domain $\sigma$-algebras as $X$ for a wide range of transformations $g$ (when we build a regression model, for example). This motivates the second bullet point and possibly others.
