In http://en.wikipedia.org/wiki/Bochner_integral a notion of measurability is discussed that depends on the measure $\mu$. Usually measurability does not depend on having a measure anyway. Is this really the right definition? If so, how is it related to the more natural notion of measurability which would be the Borel $\sigma$ algebra from the norm topology of the Banach space?
Since this notion of measurability must be not equivalent to the standard one if the Banach space is $\mathbb{C}$ or $\mathbb{R}$, what kind of theorems can we expect about the Bochner integral still being in some sense an extension of the Lebesgue?
Typically what I have seen is the notion of measurable replaced by strong measurability. We say $f:(X,\mathscr{A})\to (E,||\cdot||)$ where $E$ is a Banach space is strongly measurable if $f$ is $\mathscr{A}/\mathscr{B}(E)$ measurable and $f(X)$ is separable.
The strongly measurable functions are the ones which can be pointwise approximated everywhere by simple functions "nicely" ($f = \lim f_n$ and $|f_n| \leq |f|$). Then we introduce a measure $\mu$ on $(X,\mathscr{A})$ and can define $\int f d\mu$ for all $f$ for which $|f|$ is integrable (in the non-Bochner sense, since $|f|:X \to \mathbb{R}$).
But once we have made this definition, we realize that if $\mu$ ignores sets where $f$ is bad, we may as well define $\int f d\mu$ anyway. Thus if $f=g$ a.e $[\mu]$ and $g$ is strongly measurable, then the only natural candidate is $\int f d\mu = \int g d\mu$ and so we extend our definition to include such $f$. Thus the notion of measurability is inherently still one of $\sigma$-algebras, but depending on the measure we can allow bad stuff on null sets, just like we can for the ordinary Lebesgue integral.
The theory agrees with when $E=\mathbb{R}$ by integrating with respect to the completion $\overline{\mu}$ on the completed $\sigma$-algebra $\mathscr{A}_\mu$. (Except for the case of positive functions which are not integrable, where $\int fd\mu$ cannot be defined in the Bochner sense since the Bochner integral does not know about the order on $\mathbb{R}$).