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!



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


