Let $(S,\Sigma,\mu)$ be a measure space and $(X,\Vert .\Vert_X)$ a Banach space. I got a question concerning the properties of the Bochner Integral (as it is defined here: https://en.wikipedia.org/wiki/Bochner_integral) in the case of an non-complete measure space (for definition and example look here: https://en.wikipedia.org/wiki/Complete_measure).
In the wikipedia article (or in 'Analysis in Banach Spaces - Volume 1' by Hytönen, van Neerven, Vera, Weis as a recent alternative reference) it is mentioned, that for a Bochner-measurable function $f:S\mapsto X$ bochner-integrability is equivalent to the lebesgue-integrability of $\Vert f\Vert_X$.
On page 14. in the mentioned book, the authors claim: 'If f is Bochner integrable and f=g almost everywhere, then g is Bochner integrable and the Bochner integrals of f and g agree.' ($f, g:S\mapsto X$)
Now if $(S,\Sigma,\mu)$ is non-complete, then there exists a set $M\in\Sigma$ with $\mu(M)=0$ and a subset $N\subset M$ such that $N\notin\Sigma$. For this set N, the characteristic function $\chi_N$ equals almost everywhere (on the complement of the nullset M) the Bochner integrable zero function, and hence, by the mentioned equivalence, $\chi_N$ is Lebesgue integrable without even being measurable.
I hope someone can help me in my confusion. Thanks so far PS: One can find alternative examples without using the claimed fact, that Bochner integrability is preserved by an almost everywhere equality.
