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.
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.