If $\mu$ is a finite measure and $ν$ is a vector measure with $|\nu|\le C\mu$, are we able to show $\left|\frac{{\rm d}ν}{{\rm d}\mu}\right|\le C$?

Question

Let

• $(\Omega,\mathcal A,\mu)$ be a finite measure space
• $E$ be a $\mathbb R$-Banach space with the Radon-Nikodým property
• $\nu$ be a $E$-valued vector measure on $(\Omega,\mathcal A)$ with $$\left\|\nu(A)\right\|_E\le C\mu(A)\;\;\;\text{for all }A\in\mathcal A\tag1$$ for some $C\ge0$

Note that $(1)$ implies $\nu\ll\mu$ and hence $$\nu(A)=\int_Af\:{\rm d}\mu\;\;\;\text{for all }A\in\mathcal A\tag2$$ for some $f\in L^1(\mu;E)$.

I want to show that $$\left\|f\right\|_E\le C\;\;\;\mu\text{-almost everywhere}\tag3\;.$$ How can we do that?

I've asked this question before for the case $E=\mathbb R$, but none of the provided solutions is generalizable to the case of a general Banach space.

Maybe we can reduce the problem to the real-valued case: Note that $$(\varphi\circ\nu)(A)=\int_A\varphi\circ f\:{\rm d}\mu\;\;\;\text{for all }A\in\mathcal A\tag4$$ for all $\varphi\in E'$. Now, $$\left\|f(\omega)\right\|_E=\sup_{\left\|\varphi\right\|_{E'}\:\le\:1}|(\varphi\circ f)(\omega)|\;\;\;\text{for all }\omega\in\Omega\;,\tag5$$ but we would need to select a common $\mu$-null set to conclude ...

Answer 1

It's way easier than I thought:

First of all, let $\lambda:\mathcal A\to[0,\infty)$ be $\sigma$-additive with $$\lambda\le C\mu\tag{10}$$ for some $C\ge0$. Then, as orole mentioned in a comment below my other question, it's easy to show that $$0\le\frac{{\rm d}\lambda}{{\rm d}\mu}\le C\tag{11}.$$ Returning to the question, let $|\nu|$ denote the modulus (the restriction of the variation of $\nu$ to $\mathcal A$) of $\nu$. By $(1)$, we easily obtain $$|\nu|\le C\mu.\tag{12}$$ Now, the crucial point is that $$\left\|f\right\|_E=\frac{{\rm d}|\nu|}{{\rm d}\mu}\tag{13}$$ (it's not hard to prove, but see, for example, Theorem 2.29 (a) in the book Vector Integration and Stochastic integration in Banach spaces by Nicolae Dinculeanu) and hence we immediately obtain $(3)$ by $(11)$.

Answer 2

Partial answer: Assume $E=X'$ for a $\mathbb R$-Banach space $X$ and let $g\in L^1(\mu;X)$ with $\left\|g\right\|_{L^1(\mu;\:X)}\le1$. It's easy to see that if $g$ is a step function, then $$\left|\int fg\:{\rm d}\mu\right|\le C.\tag6$$ In general, there is a sequence $(g_n)_{n\in\mathbb N}$ of $\mathcal A$-measurable step functions $g_n:\Omega\to X$ with $$\left\|g_n\right\|_X\le\left\|g\right\|_X\;\;\;\text{for all }n\in\mathbb N\tag7$$ and $$\left\|g-g_n\right\|_X\xrightarrow{n\to\infty}0.\tag8$$ Hence, we're able to conclude $(6)$ by showing that $f\in L^\infty(\nu;X')$ and invocation of Lebesgue's dominated convergence theorem.

Now, $(3)$ follows by $$\left\|f\right\|_{L^\infty(\mu;\:X')}=\sup_{\left\|g\right\|_{L^1(\mu;\:X)\:\le\:1}}\left|\int fg\:{\rm d}\mu\right|.\tag9$$

I don't know whether $(3)$ can be shown if $E$ is not a dual space.