If $ F(x,\bullet) \in {L^{\infty}}(G,B) $ for all $ x \in G $, then is $ x \mapsto F(x,\bullet) $ strongly measurable?

Question

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 and $ {L^{\infty}}(G,B) $ the set of all (equivalence classes of) $ B $-norm essentially bounded strongly $ \mu_{G} $-measurable functions from $ G $ to $ B $.

Note: $ {L^{\infty}}(G,B) $ is a Banach space, and except in trivial cases, it is always non-separable.

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

Thank you all very much for your help!

Answer 1

I think the answer is no, even in the case of $G:=\mathbb{R}$ with the Lebesgue measure and $B: =\mathbb{R}$ as a Banach space.

Let $F: \mathbb{R}\times \mathbb{R} \to \mathbb{R} $ be the characteristic function of the half-plane above the diagonal: $F(x,y):=\chi_\mathbb{{R}_+}(y-x)$. So $F\in L^\infty(\mathbb{R}\times \mathbb{R} )$; however the corresponding map $\tilde F: \mathbb{R} \to L^\infty(\mathbb{R} )$ such that $x\mapsto F(x,\cdot)=\chi_{[x,\infty)}$ is not strongly measurable, because its range is not separable, as it is the discrete uncountable set $\{\chi_{[x,\infty)} : x\in \mathbb{R}\}$ (not even if we modify $\tilde F$ on some Lebesgue null set).

Rmk 1. A Banach space valued strongly measurable map $f$ on a measure space $X$ is necessarily essentially separable , that is, up to removing a null set $N\subset X$, its image $f(X\setminus N)$ is a separable subset of $B$ (indeed it is included in the closure of the countable union of the images of the simple functions that converge pointwise to $f$; and a simple function is exactly a measurable function with finite range).

Rmk 2. A theorem of Pettis states that a Banach space valued map $f$ on a measure space $X$ is strongly measurable if and only if: (i) it is essentially separable; (ii) it is weakly measurable (that is, measurable w.r.to the $\sigma$ algebra generated by the weak topology; (iii) its support is $\sigma$-finite. (See e.g. the Kōsaku Yosida's Functional Analysis. The proof is easy. ).