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!