Variation of a strongly bounded measure is strongly bounded too Let $\mathcal{A}$ be a field of subsets of a set $\Omega$, $X$ a Banach space and $\mu:\mathcal{A}\rightarrow X$ a finitely additive vector measure. 
The variation of $\mu$ is the extended nonnegative function $|\mu|$ whose value  on a set $E\in\mathcal{A}$ is given by $$|\mu|(E)=\sup_\pi\sum_{A\in\pi}\|\mu(A)\|,$$ where the supremum is taken over all partitions $\pi$ of $E$ into a finite number of pairwise disjoint members of $\mathcal{A}$.
It can be shown that $|\mu|$ is also a finitely additive measure.
A finitely additive measure $\mu$ is said to be exhaustive (or strongly bounded) if for every $(E_n)$ sequence of pairwise disjoint members of $\mathcal{A}$, then $\lim_n\mu(E_n) = 0$. 
It is easy to show that $|\mu|$ exhaustive implies $\mu$ exhaustive for every $X-$valued finitely additive measure. If $\mu$ is real-valued or complex-valued and bounded, it can be shown that $\mu$ exhaustive implies $|\mu|$ exhaustive.
Does a bounded $\mu$ that is exhaustive imply $|\mu|$ exhaustive for an $X-$valued finitely additive measure?
I am unsure of how to prove this and I was wondering if I could get a hint. 
Thanks!
 A: If I understood well, you are supposing that given a bounded exhaustive measure $\mu$, its total variation $\vert\mu\vert$ is also an exhaustive measure. I believe that this is not true for all infinite dimensional spaces.
Firs of all, please note that every exhaustive measure is bounded (see Corollary 19 at page 9 of Vector measure by Diestel and Uhl), so if your conjecture were true then every exhaustive measure would automatically have bounded variation. This last assertion is not true in general, as shown in the example 16 at page 7 of Vector Measure (by Diestel and Uhl).
In any case, if $X$ is a Banach space, $\Sigma$ a field of subsets of $\Omega$ and $\mu\colon \Sigma\to X$ is an exhaustive measure, it can be proved that $\vert\mu\vert$ is exhaustive if and only if $\vert\mu\vert$ is bounded. This means that your conjecture is true whenever $X$ is such that every exhaustive measure has bounded variation, which are just the finite dimensional spaces by the Dvoretsky-Rogers theorem.
I hope this answer your question, even if such a long time has passed.
Niccolò
