# completeness of $\mathcal M(\Omega)$ without any topological assumptions?

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?

• Two statements here are in need of correction. You say that the dual of a complete vector space is always complete. If you mean that the dual of a complete normed space (aka Banach space) is complete, then this is true but the completeness condition is superfluous. If you mean a complete topological vector space (even lcs), it is false. Secondly, if $X$ is ,say, a completely regular space (in particular, locally compact or metrisable), then the dual space of $C_b(X,V^\ast)$ is indeed a space of measures on the Baire $\sigma$-algebra, but not the one you are interested in. – user131781 May 22 at 15:05
• Thanks for pointing this out, I meant indeed a Banach space, not a general topotlogical vector space. Regarding the second part of your comment: I was not aware of these subtleties. Out of curiosity: what would be the dual space (space of measures on the Baire $\sigma$-algebra) that you mentioned for completely regular spaces? – leo monsaingeon May 22 at 19:22

## 1 Answer

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.

• Yup, that does the job. (Although I find this book is a bit outdated for my taste and hard to read, nothing beats the classic old-timers). Thank you very much @Dieter. – leo monsaingeon May 22 at 11:26