I cannot figure out which is the relationship between Radon measures and measures that have bounded variation. I think it's almost a matter of definition, but I cannot find a proper source to which refer to.
In particular, I was given the following definitions: given $\Omega \subset \mathbb{R}^n$ and $\mathcal{B}(\Omega)$ the $\sigma$-algebra of Borel sets and $\Omega$ locally compact separeble space we say
- that a set function $\mu$ is a real Radon measure if it is a measure on the spaces $(K,\mathcal{B}(K))$ $\forall K \subset \Omega \text{ compact } $
- that the same set function $\mu$ is a finite Radon measure if it is also a measure $\mathcal{B}(\Omega)$
First, I assume (for the confidence I place on my professor) that those definitions are equivalent in this setting to the more common ones based on the regularity of the measure.
Moreover, given $\Omega \subset \mathbb{R}^n$ open nonempty, I was given this form of the Riesz's representation theorem:
$\forall L$ linear and bouned functional on $C_0(\Omega;\mathbb{R}^m)$ $\exists!$ $\mathbb{R}^m$-valued Radon measure $\mu$ such that the operatorial norm of $L$ is equal to the total variation of $\mu$ and $$\forall v \in C_0(\Omega;\mathbb{R}^m) \quad Lv=\int_{\Omega}\sum_{j=1}^{m}v_j \ d\mu_j$$
My problem arise when considering the space $\mathcal{M}(\Omega;\mathbb{R}^m)$ of bounded variation measures (I do not report here the definition fot the sake of brevity, but it is the usual one that can be found also here https://en.wikipedia.org/wiki/Vector_measure). My professor states that $\mathcal{M}(\Omega;\mathbb{R}^m)$ is the topological dual of $C_0(\Omega;\mathbb{R}^m)$ by the Riesz's theorem above stated, but this means that every finite Radon measure is bounded variation. How can be this? I know that every real valued measure (no $\pm \infty $ values allowed) is bounded variation (cf. Rudin, RCA, p.118 thm 6.4), but this is not true for $\mathbb{R}^m$-vector valued measures in general (am I wrong?) and so we are asserting that any $\mathbb{R}^m$-vector valued finite Radon measure in the sense above is also bounded variation, so any finite Radon measure is bounded variation? Is this the sense of those definitions?
Thanks for the attention!