Let $(X, \mathcal{A})$ be a measurable space and $B$ a Banach space. Let $\mu: \mathcal{A} \to B$ be a countably additive vector measure. For a measurable function $f:X \to \mathbb{R}$. I want to define $\int_{X}fd\mu$. It seems like this is "simpler" when the variation $|\mu|$ is a finite measure, but I don't know why.
Let $\mathcal{S}$ denote the set of all simple functions on $(X, \mathcal{A})$ and $\mathcal{S}_b$ the set of all simple functions in $\mathcal{S}$ whose support has finite $|\mu|$-measure. Then $\mathcal{S}_b = \mathcal{S} \cap \mathcal{L}_1(|\mu|)$ is a dense subpace of $\mathcal{L}_1(|\mu|)$. Define $T:\mathcal{S}_b \to B$ to send $f = \sum_{i=1}^na_i1_{E_i}$ to $$T\left(\sum_{i=1}^na_i1_{E_i}\right) = \sum_{i=1}^na_iu(E_i)$$ It is well defined and linear, and $$\lVert T(f)\rVert = \lVert T\left(\sum_{i=1}^na_i1_{E_i}\right)\rVert \leq \sum_{i=1}^n|a_i|\lVert \mu(E_i)\rVert \leq \sum_{i=1}^n|a_i||\mu|(E_i) = \lVert f\rVert_{\mathcal{L}_1(|\mu|)}$$ since we always have $\lVert \mu(E_i)\rVert < |\mu|(E_i)$. Therefore $T$ is a bounded linear map from $\mathcal{S}_b$ to $B$ and therefore has a unique continuous linear extension to $\mathcal{L}_1(|\mu|)$. At what point do I need the fact that $|\mu|$ is a finite measure?