# Is scalarwise measurability determined by the strong dual?

Since this question has not received an answer so far, I try to reformulate the question in a simpler manner as follows: Do there exist $E,F,\ell,f$ such that

1. $E$ and $F$ are separable (real) Banach spaces,

2. $\ell:E^{\kern.3mm\ast}\to F^{\kern.3mm\ast}$ is a linear homeomorphism,

3. $f:\mathbb R\to E^{\kern.3mm\ast}$ is a function with $|f(t)(x)|\le\|x\|$ for all $t\in\mathbb R$ and $x\in E$,

4. $\mathbb R\owns t\mapsto f(t)(x)\in\mathbb R$ is measurable for all $x\in E$,

5. $\mathbb R\owns t\mapsto\ell(f(t))(x)\in\mathbb R$ is not measurable for some $x\in F$.

Above $E^{\kern.3mm\ast}=E^{\\,\prime}_\beta$ denotes the strong topological dual of $E$, and necessarily the spaces $E^{\kern.3mm\ast}$ and $F^{\kern.3mm\ast}$ have to be nonseparable. Of course, the spaces $E$ and $F$ cannot be linearly homeomorphic.

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Perhaps use: $l^\infty$ and $L^\infty$ are linearly homeomorphic, but only from the Axiom of Choice. Constructions essentially using the Axiom of Choice are needed to produce a non-measurable function. –  Gerald Edgar May 27 '13 at 13:08
^ Thanks for the suggestion. I think I am able to write the example with $E=L^{\kern.7mm 1}(\mathbb R)$ and $F=\ell^{\kern.7mm 1}(\mathbb N_0)$ as soon as I have time to arrange the ideas properly. I shall give it as an answer to my own question unless someone else succeeds to give a better answer before that. –  TaQ May 28 '13 at 4:32
^ Unfortunately, the preceding comment of mine was premature since I had a false idea about getting a linear homeomorphism between $L^{\kern.4mm +\infty}(\mathbb R)$ and $\ell^{\kern.4mm +\infty}(\mathbb N_0)$. Any more detailed suggestions are still wellcome. –  TaQ May 28 '13 at 9:05
The MO 2.0 update obviously has spoiled the formatting of "3." in my question which should be the following: "3. $f:\mathbb R\to E^{\kern.3mm\ast}$ is a function with $|\,f(t)(x)\,|\le\|x\|$ for all $t\in\mathbb R$ and $x\in E$," –  TaQ Jun 27 '13 at 14:58
Fixed the formatting. That's what you get for trying to put in your own spacing... –  Gerald Edgar Jun 27 '13 at 15:00