This question is related to something that I asked yesterday: If $ F(x,\bullet) \in {L^{\infty}}(G,B) $ for all $ x \in G $, then is $ x \mapsto F(x,\bullet) $ strongly measurable?

Pietro Majer provided a non-affirmative answer to the former question by constructing a very elementary counterexample.

Let $ (X,\Sigma,\mu) $ be a $ \sigma $-finite measure space and $ B $ a Banach space. A function $ f: X \to B $ is said to be **strongly $ \mu $-measurable** iff it is the *almost-everywhere* pointwise limit of a sequence $ (s_{n}: X \to B)_{n \in \mathbb{N}} $ of integrable simple functions, where an integrable simple function $ s: X \to B $ has the form
$$
s = \sum_{(E,b) \in I} \chi_{E} \cdot b
$$
for some finite subset $ I $ of $ \{ E \in \Sigma \mid \mu(E) < \infty \} \times B $.

Let $ G $ be a second-countable, locally compact Hausdorff group and $ \mu_{G} $ a fixed Haar measure on the Borel $ \sigma $-algebra $ \mathscr{B}(G) $ of $ G $. The second-countability condition implies that $ G $ is $ \sigma $-compact, which ensures that $ (G,\mathscr{B}(G),\mu_{G}) $ is a $ \sigma $-finite measure space.

Let $ B $ be a *separable* Banach space.

Let $ {L^{2}}(G,B) $ denote the set of all (equivalence classes of) square-integrable strongly $ \mu_{G} $-measurable functions from $ G $ to $ B $.

**Note:** $ {L^{2}}(G,B) $ is a separable Banach space, as the algebraic tensor product $ {L^{2}}(G) \odot B $ can be seen as a dense and separable linear subspace.

Question.If $ F: G \times G \to B $ is a strongly $ \mu_{G \times G} $-measurable function where $ F(x,\bullet) \in {L^{2}}(G,B) $ for all $ x \in G $, then is it true that the mapping \begin{align} G & \to {L^{2}}(G,B); \\ x & \mapsto F(x,\bullet) \end{align} is strongly $ \mu_{G} $-measurable?

One strategy is to use Pettis’ Measurability Theorem, as considered by myself and also suggested to me by Pietro, to prove that \begin{align} G & \to {L^{2}}(G,B); \\ x & \mapsto F(x,\bullet) \end{align} is weakly $ \mu_{G} $-measurable instead. However, the one problem with this is that the dual space of $ {L^{2}}(G,B) $ is hard to visualize. I am actually interested in the case when $ B $ is a separable $ C^{*} $-algebra, which makes $ {L^{2}}(G,B) $ a Hilbert $ B $-module. However, even with this extra bit of structure, it appears difficult to exploit Pettis’ Measurability Theorem due to a lack of understanding of the dual space of $ {L^{2}}(G,B) $.

I sincerely appreciate any help because the answers to these questions would help me better understand measurability issues related to the theory of representations of twisted $ C^{*} $-algebraic crossed products on Hilbert $ C^{*} $-modules.

Linear operators, Part I) have a theorem (or maybe its proof) answering the question. I do not have the book at hand to check or give a precise reference. $\endgroup$ – TaQ Oct 19 '14 at 13:35