# 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!

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.
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.