I am reading a brief introduction to Bochner integral, and before introducing it, it is proved the Pettis measurability theorem, which says that a function $f:A \to B$ (with $A$ a measure space with $\sigma$-finite measure $\mu$ and $B$ a Banach space) is strongly $\mu$-measurable (that is that there is a sequence of simple functions which converges pointwise to $f$ almost everywhere) iff its image is almost fully contained in a separable closed subspace of $B$ and $\Lambda \circ f$ is measurable $\forall \Lambda \in \mathcal{L}(A,\mathbb{R})$, where $\mathbb{R}$ is considered to have borelians as measurable subsets. There is a third equivalence, in fact, but I don't think it is important for my question.
For standard Lebesgue integrals, there is a canonical way to construct a sequence of simple functions that tends to $f$ almost everywhere, but this construction uses the ordering of $\mathbb{R}$. With Banach spaces, I suppose that Pettis theorem helps by giving an equivalent condition, given that this construction is not possible (or at least I wouldn't know how to do that). Once you have a strongly $\mu$-measurable function, the other properties should follow like the Lebesgue case. For instance, a function $f$ is Bochner integrable iff its norm is integrable and its integral is finite (and this gives us a useful criterion for Bochner integrability, because I know how to integrate $||f||:A \to \mathbb{R}$ ).
This criterion is very useful and the proof works if you know that $||f||$ is strongly $\mu$-measurable, and this is indeed a corollary of Pettis measurability theorem.
I am wondering though: is this the only way in which Pettis theorem helps in the construction of the integral? I see that it is useful in other ways, but it looks like it is used only to prove the strong $\mu$-measurability of $||f||$.