Unambiguous “weak” vector valued $L^{+\infty}$ spaces?

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 $E_\beta^{\prime}\\,$.

• More generally, if $E'$ 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. – Bill Johnson May 25 '13 at 5:45
• ^ Thanks for the hint. I do not have the book of Diestel and Uhl at hand but can see the result by using Proposition 8.15.3 on page 575 in Edwards' book. – TaQ May 25 '13 at 13:31
• Even if the dual of the separable Banach space is not separable, there is a perfectly respectable complete locally convex structure there for which the dual of $L^1(I,E)$ is canonically identifiable with the space of (equivalence classes of) of bounded, measurable functions--- the bounded, weak-star topology. This not unimportant since probably the most significant example is that where $E$ is the space of bounded, linear operators on Hilbert space which is a non-separable dual of the nuclear operators. There this concept of measurability is ubiquitous in spectral theory. – jbc May 26 '13 at 8:09
• By the way, with suitable tools, one can prove the required representation for the dual in a few lines. One can express $E$ as the closure of the union of an increasing sequence $(E_n)$ of finite dimensional subspaces. It follows easily that the required $L^1$ space is the inductive limit (in the category of Banach spaces with linear contractions as morphisms) of the $L^(I,E_n)$. The natural extension of $L^1$ duality theory to functions with values in finite dimensional spaces plus abstract nonsense on duality for the above category produces the stated identification of the dual. – jbc May 26 '13 at 8:19
• "Even if the dual of the separable Banach space is not separable, there is a perfectly respectable complete locally convex structure there for which the dual of $L^1(I,E)$ is canonically identifiable with the space of (equivalence classes of) of bounded, measurable functions--- the bounded, weak-star topology." Assuming that I understand the preceding as intended, so the dual of $L^1(I,E)$ would be represented by a (suitably defined) space $L^{+\infty}(I,E^{\prime}_\sigma)$. This is what has been in my mind already at the moment of posing the original question. – TaQ May 26 '13 at 10:25