2
$\begingroup$

If I have understood correctly, a vector valued measure $\mu$ is simply a vector of measures, that is $\mu=(\mu_1,\dots,\mu_n)$, where $\mu_i$ is a possibly signed measure on the measure space $(X,\Sigma)$.

Now, for each $\mu_i$, one can define the total variation of $\mu_i$, thanks to the Hahn decomposition theorem. Indeed, from this theorem, for each $i$, one has two positive measure $\mu_i^+$ and $\mu^{-}_i$ such that $\mu_i=\mu_i^+-\mu_i^-$. At this point, one defines the total variation of $\mu_i$ as $|\mu_i|:=\mu_i^++\mu_i^-$. Is it correct? So we have $$\mu=\mu^+-\mu^-:=(\mu_1^+,\dots,\mu_n^+)-(\mu_1^-,\dots,\mu_n^-).$$

Now I know that one can define also the total variation of $\mu$ and it must be a scalar. So, how can one define the total variation of the vector valued measure $\mu$?

$\endgroup$

1 Answer 1

3
$\begingroup$

The total variation $|\mu|$ of $\mu$ is defined via $$|\mu|(B) = \sup_{\pi} \sum_{A \in \pi} \|\mu(A)\|_{\mathbb{R}^n}$$ where the supremum ranges over all finite partitions of the Borel set $B$ into disjoint Borel sets.

In the scalar case, this coincides with the definition via the Hahn decomposition. In the vector-valued case, however, there is no relation between the Hahn decompositions of $\mu_i$ and $|\mu|$ (or, at least, I am not aware of any connection). Similarly, there is no relation between the total variation $|\mu|$ of the vector measure $\mu$ and the vector of total variations $(|\mu_1|, \ldots, |\mu_n|)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .