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

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.

• If I recall correctly, Dunford and Schwartz (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. – TaQ Oct 19 '14 at 13:35
• @TaQ: I’ve looked at Dunford’s and Schwartz’s book, but I don’t see any theorem that talks about vector-valued $L^{2}$-spaces. Thanks for your suggestion, anyway! I intend to offer a bounty on this problem tomorrow. – Transcendental Oct 19 '14 at 17:24

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.
• Dear TaQ, thank you very much for your response. Let me carefully study your solution. I’ll get back to you soon. – Transcendental Oct 22 '14 at 16:29

(edit. I turned the preceding incorrect answer into a partial answer: yes with an ugly technical assumption).

Let $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ be measure spaces, and let $(\mathbb{B},\|\;\|)$ be a Banach space.

Lemma. 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.

The proof follows from an elementary density lemma for sets in measure theory, also quoted in this old answer.

Now let's assume $F:X\times Y\to \mathbb{B}$ is strongly measurable and $F(x,\cdot)\in L^2(X,\mathbb{B})$ for $\mu$-a.e. $x\in X$.

Let $F_n$ a sequence of simple functions as in the above lemma, say converging to $F$ in $(X\times Y)\setminus Z$, for a null set $Z$.

Assume that $F_n$ also satisfy, for a.e. $x\in X$: $\|F_n(x,\cdot)\|$ is dominated by some function $g_x\in L^2(Y,\nu)$. Then the conclusion follows.

Recall that a null subset $Z$ of $X\times Y$ has $\mu$-almost all sections $Z_x:=\{y\in Y: (x,y)\in Z\}$ of $\nu$-null measure. Therefore $F_n(x,\cdot)=\sum_i\big(\sum_j v_{ij}^n\chi_{B_j^n}\big)\chi_{A_i^n}(x)$ is a sequence of simple function on $X$, valued in $L^2(Y,\mathbb{B})$; for $\mu$-a.e. $x\in X$ it converges $\nu$-a.e. to $F(x,\cdot)$, hence in $L^2(Y,\mathbb{B})$ by the Lebesgue dominated convergence because $\|F_(x,\cdot)\|\le g_x$, that is $x\mapsto F(x,\cdot)$ is strongly measurable as a map $X\to L^2(Y,\mathbb{B})$.

• Hi Pietro. Could you explain why $\| {F_{n}}(x,y) \|_{B} \leq \| F(x,y) \|_{B}$ for almost every $(x,y) \in X \times Y$? I have doubts about this point because $F_{n}$ is a $B$-linear combination of indicator functions of measurable rectangles with sides of finite measure. If $F_{n}$ is merely a $B$-linear combination of indicator functions of measurable subsets of $X \times Y$, then there’s no problem. However, if those measurable subsets are required to be measurable rectangles, then I don’t see it. Thank you for devoting your time and energy to answering my question! – Transcendental Oct 21 '14 at 0:36
• yes, that's what I was now thinking too... maybe I made an over-simplification. I'll think a bit; the idea of the proof should be the same (dominated convergence), however. – Pietro Majer Oct 21 '14 at 0:47
• I modified the preceding incorrect proof into a kind of obvious starting remark… I'll go back to this nice problem as soon as I'm more free :) – Pietro Majer Oct 21 '14 at 6:46