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?



completeness of $\mathcal M(\Omega)$ without any topological assumptions?
Indeed $(\cal{M}(\Omega),\|\cdot\|)$ is a Banach space. For $V = \mathbb{R}$ or $V = \mathbb{C}$ you can find this result in Dunford/Schwartz (1957), Linear Operators I, ch. III.7.4, in particular p. 161. For arbitrary Banach space $V$ this also holds true, but with a sligthly different norm (see p. 160). For finite dimensional $V$ this norm is equivalent to your norm. Proof as in Lemma III.1.5 of Dunford-Schwartz.


