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Let $B$ be a Banach space, let $(X,\mathcal{A})$ be a measurable space, and let $\mu:\mathcal{A}\to B$ be a vector-valued measure of bounded variation.

In general, if $B$ doesn't have the Radon-Nikodym property, it is not true that we can express $\mu$ as $$ \mu(A) = \int_A f(x)\,m(dx) $$ for some (real-valued) measure $m$ on $X$ and some Bochner-measurable function $f:X\to B$.

Instead, could we find a (real-valued) measure $m$ on a product $X\times Y$ and a Bochner-measurable function $f:Y\to B$, such that $\mu$ can be written as the following Bochner integral, $$ \mu(A) = \int_{A\times Y} f(y) \,m(dx\,dy) \qquad ? $$

(This seems easy when Radon-Nikodym holds, but what about in the general case?)

geodude
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  • When you write $\mu(A)=\int_Af(x),m(dx)$, should it be understood as a Bochner integral or do you allow it to be a Pettis integral? – P. P. Tuong Apr 22 '24 at 14:18
  • Bochner. Thank you. Let me add that. – geodude Apr 22 '24 at 17:42
  • For given $\mu$, if we want $\mu(A) = \int_A f(x),m(dx)$ perhaps we could take $m = |\mu|$, the variation of $\mu$. – GEdgar Apr 22 '24 at 17:53
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    I think $\mu$ needs to be assumed to have bounded variation to avoid problems with summable but not absolutely summable families in $B$. – P. P. Tuong Apr 22 '24 at 19:06
  • Yes, thank you. Added that as well. – geodude Apr 22 '24 at 21:07
  • If $\mu(A)=\int_{A\times Y}f(y),m(dx,dy)$ for every $A\in\mathcal A$ and if $f(Y)$ is relatively compact in $B$, I wonder if the linear map $\varphi\mapsto\int_{X\times Y}\varphi(x)f(y),m(dx,dy)$ from $L^1(|\mu|)$ to $B$ is weakly compact... Because if it is, the Dunford-Pettis-Phillips theorem gives the existence of a measurable density of $\mu$ with respect to $|\mu|$. And then maybe this could also exist in the case of general $f$ by writing $f$ as an infinite sum of measurable functions with relatively compact images. – P. P. Tuong Apr 23 '24 at 17:23

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