Let $(\Omega,\Sigma)$ be a measurable space (no reference measure is chosen!), and $V$ a finite-dimensional normed vector space. Note carefully that I am not choosing any topology on $\Omega$, so the $\sigma$-algebra $\Sigma$ is a priori not induced by any Borel structure whatsoever.

The total variation $|\mu|$ of a $V$-valued measure is defined as $$ |\mu|(E)=\sup\limits_{\pi}\sum\limits_{E_i\in E}|\mu(E_i)|, $$ where the supremum is taken among all possible disjoint partitions $\pi=\cup E_i$ of the measurable set $E\in\Sigma$. (the set-function $|\mu|$ is always a positive measure, see [Rudin, real and complex analysis]). For an arbitrary measure we denote $$ \|\mu\|:=|\mu|(\Omega). $$ We denote $\mathcal M(\Omega)$ the set of all measures with finite total variation $\|\mu\|<\infty$, and $\mathcal M(\Omega)$ is therefore a normed vector space with the above total variation norm.

Question:is $(\mathcal M(\Omega), \|\cdot\|)$ automatically a Banach space?

When $\Sigma$ is the Borel algebra then this is true of course, because we can identify $\mathcal M(\Omega)$ with the topological dual $C_b(\Omega;V^\ast)^\ast $ and the dual of a Complete vector space is automatically complete (and in fact $\|\mu\|=\sup\limits_\phi \int \phi(x)\cdot d\mu(x)$ with $\cdot$ denoting the finite-dimensional $V,V^\ast$ pairing). However, I've never seen the statement written in this full generality, so I'm wondering whether this is actually true or not?