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Consider a positively measured space $(S,\Sigma,\mu)$ and a real or complex Banach space $X$. Is it possible to build the Bochner integral and the Bochner spaces $L^p(S,\Sigma,\mu;X)$ without completing the measure $\mu$? Will such a construction provide an integration theory with all the desirable tools of analysis?

Here is an attempt.

Definition 1: A simple map $s : S \rightarrow X$ is a map that is of the form $$ s = \sum_{i=0}^k s_i1_{A_i} $$ where $k \in \mathbb N$, $s_i \in X$, $A_i \in \Sigma$ has finite measure ($\mu(A_i) < \infty$) and $1_{A_i}$ denotes the characteristic function of $A_i$.

Now I would like to define the class of functions I know how to integrate, I will call them by a stupid name to emphasize it is not a standard definition.

Definition 2: A map $f : S \rightarrow X$ is EasyBochnerIntegrable if it is $\Sigma - \mathcal B(X)$ measurable (where $\mathcal B(X)$ stands for the Borel sigma-algebra) and there exists a sequence of simple maps $(s_n)$ such that $s_n \rightarrow f$ everywhere on $S$ together with $$ \int_S \| s_n-f\|_Xd\mu \rightarrow 0. $$

As in the standard Bochner integration theory we can then check that

  • For a simple map we can define its integral using its simple representation made with disjoint sets: $$ \int_S \sum_{i=0}^k s_i1_{A_i} d\mu = \sum_{i=0}^k s_i \mu(A_i) $$

  • With the notations of Definition 2, the sequence $(\int_S s_n d\mu)_{n \geq 0}$ is Cauchy in $X$ whence it converges

  • The previous limit only depends on $f$, we call if $\int_S f d\mu$

  • The set of EasyBochnerIntegrable functions, call it $EBI(S,\Sigma,\mu;X)$, is a vector subspace of the set of $\Sigma - \mathcal B(X)$ measurable maps. The integration is linear $EBI(S,\Sigma,\mu;X) \rightarrow X$.

I don't know how far we could go in this direction, any reference or ideas are welcomed. Also note that I am mostly interested in the case where $S$ is an open subset of $\mathbb R^n$ endowed of the Borel sigma-algebra and the trace of the Lebesgue measure. I would rather not specify what is $X$ because it is sometimes useful to consider $X = L^\infty$ which is not separable, bust most of the cases it turns out to be a separable Hilbert space.

blamethelag
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    yes, you can develop a lot of the basic/standard theory without completeness of the measure, and indeed not having to complete measures is nice especially for things like Fubini’s theorem. – peek-a-boo Jun 03 '23 at 12:40
  • @Balajisb I am not satisfied with the explanations in the link you gave. For my personal use of measure theory I do never use completed measures and yet I am able to identify functions which agree almost everywhere and manipulate properties that hold almost everywhere. Therefore I do not feel the need of using the completion as it brings no extra property. – blamethelag Jun 03 '23 at 13:22
  • See https://math.stackexchange.com/questions/147381/why-do-we-essentially-need-complete-measure-space –  Jun 03 '23 at 13:34

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See also https://mathoverflow.net/questions/11554/whats-the-use-of-a-complete-measure

Singletons need not be measurable in incomplete space, as a result you cannot integrate with countable discontinuous spike points sometimes as mentioned in the above link. We cannot define any delta function w.r.t incomplete measure as singletons need not be measurable.

  • Thank you for the discussion you have linked to, I found it very interesting. I should have clarified this from the beginning but I wish to apply Bochner integration (among others) to analysis and PDEs. Therefore I do not have the need of working with a complete filtration nor do I know what is ergodic theory (unfortunately!). – blamethelag Jun 03 '23 at 14:09
  • Moreover your example of a measured space where singletons may fail to be not measurable is also out of my scope. In practice the Bochner integral is used on intervals endowed of the standard structures hence this problem does not occur. More conceptually, you seem to prefer to enlarge the class of admissible functions to be integrated. I rather prefer to accept the fact that in some spaces, there exist "usual sets" that are not measurable hence I shall not to try measuring them, since I do not need it. – blamethelag Jun 03 '23 at 14:19
  • Said in different words: if you want singletons to me measurable, then start with a measurable space where they are. – blamethelag Jun 03 '23 at 14:20