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?](https://mathoverflow.net/questions/184649/if-fx-bullet-in-l-inftyg-b-for-all-x-in-g-then-is-x-maps)


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




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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 $.




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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?
> 
> 
> 




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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.



If $ F(x,\bullet) \in {L^{2}}(G,B) $ for all $ x \in G $, then is $ x \mapsto F(x,\bullet) $ strongly measurable?
Taking for granted the



>  **Lemma.** Let $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ be measure spaces, and let $(\mathbb{B},\|\;\|)$ be a Banach space. Let $F: X\times Y\to B$ be strongly measurable, with $\sigma$-finite support. Then there exists a sequence of simple functions $(F\_n)$ converging a.e. to $F$ (w.r.to $\mu\otimes\nu$), and such that $F\_n=\sum\_{ij} v\_{ij}^n\chi\_{A\_i^n\times B\_j^n}$, where, for each $n$, $(v\_{ij}^n)\_{ij}$ is a finite family of elements of $\mathbb{B}$, and $(A\_i^n)\_i\subset \mathcal{A}$ and $(B\_j^n)\_j\subset \mathcal{B}$ are finite measurable partitions of $X$, resp. of $Y$, by subsets of finite measure. 


in Pietro Majer's "a kind of obvious starting remark", the proof can be arranged as follows.


Let $\{\,Y\_l:l\in\mathbb N\_0\,\}$ be a measurable partition of $G$ into disjoint sets of finite measure, and let $\Omega\_i=\{\,(x,y):y\in Y\_{l\_i}$ and $k\_i\le\|\,F(x,y)\,\|<k\_i+1\,\}$ , where $i\mapsto(k\_i,l\_i)$ is some chosen bijection $\mathbb N\_0\to\mathbb N\_0\times\mathbb N\_0$ . Then these sets $\Omega\_i$ are measurable w.r.t. the completion of the product measure. Take $f\_i=\chi\_{\,\Omega\_i}F$ , and apply the lemma to each $f\_i$ separately. Then one can take the bound $k\_i+2$ for the occurring sequence of simple functions so that the dominated convergence theorem can be applied to prove the desired conclusion with each $f\_i$ in place of $F$ separately, i.e. we have $x\mapsto f\_i(x,\bullet)$ strongly measurable $G\to L^2(G,B)$ for every $i\in\mathbb N\_0$ . Since $x\mapsto F(x,\bullet)$ is the pointwise limit in $L^2(G,B)$ of the sequence $\big\langle\,\sum\_{i=0}^kf\_i(x,\bullet):k\in\mathbb N\_0\,\rangle$ for almost all $x\in G$ , the conclusion follows.


Obviously, the assumption that $B$ be separable is superfluous. The above in fact gives a proof for the following 



> **Theorem.** Let $(T,\mu)$ and $(S,\mu\_1)$ be $\sigma\,$−finite measure spaces, and let $\mu\_2$ be the completion of the product measure. Let $1\le p<+\infty$ and let $E$ be any Banach space. Let $F=L^{\kern.4mm p}(\mu\_1,E)$ and let $f:T\times S\to E$ be a $\mu\_2\,$−almost everywhere pointwise limit of a sequence of simple functions such that for $\mu\,$−almost all $t\in T$ the function $S\owns s\mapsto f(t,s)$ defines a vector $[\,f(t,\kern.4mm\cdot\kern.4mm)\,]$ of $F$ . Then the zero extension of the $\mu\,$−almost everywhere defined function $t\mapsto[\,f(t,\kern.4mm\cdot\kern.4mm)\,]$ is a measurable function $(T,\mu)\to F$ in the sense that it is a $\mu\,$−almost everywhere pointwise limit of a sequence of simple functions. 


