If $\mu$ is a vector measure and $\nu$ is a measure with $\left\|\mu\right\|\le\nu$ on a generating semiring, are we able to conclude $|\mu|\le\nu$?

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Let

  • $(\Omega,\mathcal A)$ be a measurable space
  • $E$ be a $\mathbb R$-Banach space
  • $\mu$ be a $E$-valued measure on $(\Omega,\mathcal A)$
  • $\nu$ be a nonnegative real-valued measure on $(\Omega,\mathcal A)$
  • $\mathcal E\subseteq2^\Omega$ be a semiring with $\Omega\in\mathcal E$ and $\sigma(\mathcal E)=\mathcal A$

Let $|\mu|$ denote the total variation of $\mu$, i.e. $$|\mu|(A)=\sup\left\{\sum_{i=1}^k\left\|\mu(A_i)\right\|_E:k\in\mathbb N\text{ and }A_1,\ldots,A_k\in\mathcal A\text{ are pairwise disjoint with }\biguplus_{i=1}^kA_i\subseteq A\right\}$$ for all $A\in\mathcal A$.

If $$\left\|\mu(A)\right\|_E\le\nu(A)\;\;\;\text{for all }A\in\mathcal E\tag1\;,$$ are we able to conclude $|\mu|\le\nu$?

My first problem is that even when $A\in\mathcal E$, the supremum in the definition of $|\mu|(A)$ is taken over partitions from all of $\mathcal A$ ... Maybe it's possible to show that in that case $$|\mu|(A)=\sup\left\{\sum_{i=1}^k\left\|\mu(A_i)\right\|_E:k\in\mathbb N\text{ and }A_1,\ldots,A_k\in\mathcal E\text{ are pairwise disjoint with }\biguplus_{i=1}^kA_i\subseteq A\right\}\;.$$

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Not an answer, a long comment:

First, the answer is not clear to me even if $\mu$ is a complex measure; there must be an answer to that out there somewhere.

A Banach-space valued measure can be nastier than I realized until just now. In particular, $|\mu|$ need not be a finite measure; this may make the vector-valued case harder.

Simple example where in fact $|\mu(A)|=\infty$ for every $A$ with $\nu(A)>0$: Let $\nu$ be Lebesgue measure on $\Bbb R$, and for $A\subset\Bbb R$ define $\mu(A)\in L^2(\Bbb R)$ by $$\mu(A)=\frac1{1+|t|}\chi_A(t).$$

Not that that's a counterexample, but if $|\mu|$ can be that bad it's hard to see how you're going to prove it has any sane properties.

Oh, speaking of counterexamples, no you can't redefine $|\mu|$ in terms of sets in that semiring: Let $\mu=\nu=$ Lebesgue measure on $\Bbb R$, and consider the generating semiring consisting of finite unions of half-open intervals. If $A$ is compact and has positive measure and empty interior you're sunk.