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.


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .