Suppose $\Omega$ is a $\sigma$-finite measure space (I'm happy to take $\Omega = \mathbb{N}$) and let $X$ be a Banach space. It's pretty well known that the Banach dual of $L^\infty(\Omega)$ can be identified with the space of finitely additive measures of bounded variation on $\Omega$, but is there a corresponding result for the Bochner space $L^\infty(\Omega;X)$?

Edit: for clarity, I define $\ell^\infty(\mathbb{N};X)$ to be the space of all bounded functions $\mathbb{N} \to X$. When $X$ is infinite dimensional, the set of simple functions (i.e. functions with finite range) is not dense in $\ell^\infty(\mathbb{N};X)$, because bounded sets in $X$ are not totally bounded. I only know the answer to this question in the `trivial' case where $X = L^\infty(\Omega)$ for some $\Omega$.

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    $\begingroup$ I don't have a copy handy, but I would look in Vector Measures by Dinculeanu. I suspect the answer is going to be "finitely additive, bounded variation, absolutely continuous, $X^*$-valued measures on $\Omega$". There is a general theory of measures taking values in a Banach space, and it generally works like you think it does. $\endgroup$ – Nate Eldredge Sep 25 '16 at 14:12
  • $\begingroup$ Yep, it's in there (Theorem 6, page 157). I didn't know this reference. I'll have to go through the notation and make sure everything corresponds to what I'm used to. Thanks! $\endgroup$ – Alex Amenta Sep 25 '16 at 15:15
  • $\begingroup$ It looks the situation is more complicated than we thought: the result in Dinculeanu is not for $L^\infty(\Omega;X)$ in the sense of essentially bounded measurable functions, but for the space of 'totally measurable' functions: uniform limits of simple functions (which are bounded). Bounded measurable $X$-valued functions need not be totally measurable (according to Dinculeanu's newer book, 'Vector Integration and Stochastic Integration...', page 10, which I don't have full access to). This is causing me some headaches right now $\endgroup$ – Alex Amenta Oct 13 '16 at 13:54
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    $\begingroup$ Yeah, these are the sort of things that one struggles with in Banach-valued integrals. The Bochner / Pettis divide has similar issues, as I recall. But maybe things simplify if your space $X$ is separable...? $\endgroup$ – Nate Eldredge Oct 13 '16 at 14:25
  • $\begingroup$ Maybe so. For now I'm starting to realise that I don't need the full generality of this result for my applications, but funnily enough arguing as if this result were true for $\ell^\infty(\mathbb{N};X)$ suggests a corollary which I can probably prove directly. (incidentally my $X$ is a variable-exponent Lebesgue space $L^{p(\cdot)}$ with unbounded exponent, which is separable but not reflexive, but which has lots of other exploitable structure) $\endgroup$ – Alex Amenta Oct 13 '16 at 15:06

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