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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?

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  • $\begingroup$ 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. $\endgroup$ – user131781 May 22 at 15:05
  • $\begingroup$ 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? $\endgroup$ – leo monsaingeon May 22 at 19:22
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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.

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  • $\begingroup$ 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. $\endgroup$ – leo monsaingeon May 22 at 11:26

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