The Banach-valued simple integral is well defined

Form a finite common refinement. [given, L1] Suppose $s={\sum }_{j=1}^m{x}_j{\mathbf{1}}_{A_j}={\sum }_{k=1}^n{y}_k{\mathbf{1}}_{B_k}$ are two representations from [L1]. Because every displayed coefficient is nonzero, both unions ${\bigcup }_j{A}_j$ and ${\bigcup }_k{B}_k$ are the same set $\{\omega :s(\omega )\mathrm{\ne }0\}$. Hence the cells $C_{jk}=A_j\cap B_k$ with $1\le j\le m$ and $1\le k\le n$ partition every $A_j$ and every $B_k$. On every nonempty $C_{jk}$, pointwise equality gives $x_j=y_k$.

Compare the two integral sums. By finite additivity in [L2], step 1.1 gives

Prove linearity. [L1, step 2.1] For integrable $s,t$ and scalars $a,b$, refine their finite level partitions. On each refined cell $as+bt$ has coefficient $a{x}_j+b{y}_k$. Every cell on which this coefficient is nonzero lies in the union of the finite-measure supports of $s$ and $t$, so $as+bt$ is integrable. Applying step 2.1 and distributing the finite vector sum yields ${\int }_E(as+bt)=a{\int }_{E}s+b{\int }_{E}t$.

Prove null-insensitivity. [L2, step 3.1] If integrable simple functions $s,t$ agree off a null set $N$, refine their level partitions as above. A refined cell on which their coefficients differ is contained in $N$, hence has measure zero by [L2]. Its contribution to ${\int }_E(s-t)$ is zero, and linearity from step 3.1 gives ${\int }_{E}s={\int }_{E}t$.

Prove the norm inequality and conclude. [L1, L3, step 4.1] Write $s={\sum }_j{x}_j{\mathbf{1}}_{A_j}$ in its nonzero disjoint-level form. The triangle inequality in $X$, [L3], and the finite-measure support rule give