Disclaimer
It is meant to record. See: Answer own Question
It is written as question. Have fun! :)
Reference
It is taken from the original paper: S. Bochner, Integration
It is related to: Bochner Integral: Integrability
Definition
Given a measure space $\Omega$ and a Banach space $E$.
Consider Bochner measurable functions $S_n\to F$.
Suppose Bochner integrability as $\int\|F\|\mathrm{d}\mu<\infty$.
Denote the abstract Bochner integral by $\int F\mathrm{d}\mu$.
Define the Bochner integral axiomatically by: $$\text{(L)}\quad\int(F+G)\mathrm{d}\mu=\int F\mathrm{d}\mu+\int F\mathrm{d}\mu,\,\int\lambda F\mathrm{d}\mu=\lambda\int F\mathrm{d}\mu$$ $$\text{(N)}:\quad\left\|\int F\mathrm{d}\mu\right\|\leq\int\|F\|\mathrm{d}\mu$$ $$\text{(DC)}:\quad\|F_n\|\leq h:\quad F_n\to F\implies\int F\mathrm{d}\mu\to\int F\mathrm{d}\mu\quad\left(\int h\mathrm{d}\mu<\infty\right)$$
And especially for finite measures: $$\text{(C)}:\quad\int C\mathrm{d}\mu=C\mu(\Omega)$$
Construction
This forces the integral for simple functions to become: $$S=\sum_{k=1}^KS_k\chi(A_k):\quad\int S\mathrm{d}\mu=\sum_{k=1}^KS_k\mu(A_k)$$
Now, for Bochner measurable functions Bochner integrability should determine the integral, too: $$S_n\to F\implies\int S_n\mathrm{d}\mu\to\int F\mathrm{d}\mu$$
But how to apply dominated convergence therefore: $$\|F-S_n\|\to0\implies\|S_n\|\leq\|F\|$$ (I got it now; in any way, you are warmly encouraged to do give an answer, too!!)
Discardure
Introduce the restricted integral through $\int_A F\mathrm{d}\mu:=\int\chi_AF\mathrm{d}\mu$.
There is another axiom which seems not independent of the others: $$\text{(S)}:\quad\int_A F\mathrm{d}\mu=\sum_{k\in\mathbb{N}}\int_{A_k}F\mathrm{d}\mu\quad\left(A=\bigsqcup_{k\in\mathbb{N}}A_k\right)$$ (Is it really independent?)
