For some time, I have been stuck to the problem to be described as follows. The (perhaps not so commonly known) facts given here are taken from R. E. Edwards' *Functional Analysis* (Holt, Rinehart and Winston 1965) pp. 578−589. Let $I=[0,1]$ with the Lebesgue measure, and consider real separable (not necessarily reflexive) Banach spaces $E$ with strong dual $F=E'_\beta$. A certain kind of "$L^{+\infty}(I,F)$", $\Lambda(I,E')$, representing the dual of $L^1(I,E)$ is constructed as follows. Let $Y$ be the vector space of a.e. bounded (i.e. bounded outside some set of measure zero) functions $g:I\to F$ such that ${\rm ev}_x\circ g:I\to\mathbb R$ given by $t\mapsto(g(t))(x)$ is measurable for all $x\in E$. Letting $N_0$ be the subspace formed by $g\in Y$ vanishing a.e., then $\Lambda(I,E')=Y/N_0$ becomes a Banach space when we equip it with the essential supremum norm of representatives $g$ of the equivalence classes $[g]=\{g+h:h\in N_0\}$. Then a linear homeomorphism $\Lambda(I,E')\to(L^1(I,E))'_\beta$ is given by $[g]\mapsto\ell:[f]\mapsto\int_I(g\ .f)$ where $(g\ .f)(t)=(g(t))(f(t))$.

The problem is now the following. Since generally preduals are not unique, there may be different (separable) spaces $E$ having linearly homeomorphic duals $F$. So, at least a priori, we cannot invariantly define some space "$\Lambda(I,F)$" as a certain kind of "$L^{+\infty}(I,F)$". According to this Philip Brooker's answer, there are nonisomorphic separable spaces $C(S)$ having isomorphic duals. One may then ask, whether (1) the corresponding spaces $\Lambda(I,(C(S))')$ are (isometrically) isomorpic or linearly homeomorphic, under the associated "natural" maps. Further, the dual of $L^1(I\times I)$ is represented by $L^{+\infty}(I\times I)$. Since $L^1(I\times I)$ is isomorphic to $L^1(I,L^1(I))$, we see that $L^{+\infty}(I\times I)$ is isomorphic to $\Lambda(I,(L^1(I))')$. One may then ask, whether (2) there are separable Banach spaces $E$ not linearly homeomorphic to $L^1(I)$, but having dual linearly homeomorphic to $L^{+\infty}(I)$ and $\Lambda(I,E')$ not linearly homeomorphic to $L^{+\infty}(I\times I)$ under the associated natural maps.

So, there are two concrete questions (1) and (2) above.

As an explanation of the phrase "natural map" above, I add the following. If $\ell_0$ is a linear homeomorphism $(C(S_1))'_\beta\to(C(S_2))'_\beta$, then the question is about whether a linear homeomorphism $\Lambda(I,(C(S_1))')\to\Lambda(I,(C(S_2))')$ is given by $[g]\mapsto[\ell_0\circ g]$. For the second question, if $\ell_0$ is a linear homeomorphism $E'_\beta\to L^{+\infty}(I)$, then the question is about whether a linear homeomorphism $\Lambda(I,E')\to L^{+\infty}(I\times I)$ is given by $[g]\mapsto[\hat g]$ where $[\hat g(t,\cdot)]=\ell_0(g(t))$ for suitably chosen $\hat g$.

I have above taken the attitude that the isomophism (or linear homeomorphism) class the space $\Lambda(I,E')$ is not solely determined by that of $E'_\beta$ but depends also on $E$. If someone knows the contrary to be true, I am gratefull for a reference or a proof. Also possible couterexamples where for separable Banach spaces $E,E_1$ with $E'_\beta$ and $(E_1)'_\beta$ linearly homeomorphic but $\Lambda(I,E')$ and $\Lambda(I,E_1')$ not, where the spaces $E,E_1$ are not some $C(S)$ or $L^1(I)$ as I suggested above, are wellcome.

**Edited.** (25.5.2013) The question (1) above is settled since in the case where $E=C(S)$ with $S$ a countable ordinal with the order topology has dual isomorphic to $\ell^{\\\,1}(\mathbb N_0)$ using Pettis' theorem and the dominated convergence theorem one can show that measurability of $I\owns t\mapsto g(t)(x)\in\mathbb R$ for all $x\in E$ implies (strong) measurability into $E_\beta^{\prime}\\,$.

`$E'$`

is separable, then weak$^*$ measurability into`$E'$`

gives strong measurability. This is in books; Diestel-Uhl comes to mind. It follows from the fact that the unit ball is weak$^*$ measurable when $E'$ is separable. $\endgroup$ – Bill Johnson May 25 '13 at 5:45