If $\nu(Y) < \infty$ then $F: X \to L^0(Y), x \mapsto f(x, \cdot)$ is $\mu$-measurable

Question

Below we use Bochner measurability and Bochner integral. Let

• $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be complete $\sigma$-finite measure spaces,
• $(E, | \cdot |)$ a Banach space,
• $S (X)$ the space of $\mu$-simple functions from $X$ to $E$,
• $L^0 (X)$ the space of $\mu$-measurable functions from $X$ to $E$,
• $L^1 (X)$ the space of $\mu$-integrable functions from $X$ to $E$,
• $Z := X \times Y$,
• $\mathcal C$ the product $\sigma$-algebra of $\mathcal A$ and $\mathcal B$,
• $\lambda$ the product measure of $\mu$ and $\nu$,
• $(Z, \overline{\mathcal C}, \overline{\lambda})$ the completion of $(Z, \mathcal C, \lambda)$.

For $\delta >0$ and $f,g \in L^0 (Y)$, we write $$ \begin{align*} \{|f - g| > \delta\} &:= \{y \in Y : |f (y) - g(y)| > \delta\}, \\ \nu (|f - g| > \delta) &:= \nu (\{|f - g| > \delta\}). \end{align*} $$

For $f, g \in L^0 (Y)$, we define $$ \hat \rho (f, g) := \inf_{\delta >0} \{ \nu (|f - g| > \delta) +\delta \}. $$

Then $\hat \rho$ is an extended pseudometric on $L^0 (Y)$. For $f_n, f \in L^0(Y)$, we have $\hat \rho (f_n, f) \to 0$ IFF $f_n \to f$ in measure, i.e., $$ \nu ( |f_n - f| > \delta) \xrightarrow{n \to \infty} 0 \quad \forall \delta >0. $$

I'm trying to prove an analogue of the main result in this thread, i.e.,

Theorem Let $f:Z \to E$ be $\lambda$-measurable. Then

1. For $\mu$-a.e. $x \in X$ we have $f(x, \cdot) \in L^0 (Y)$.
2. If $\nu(Y) < \infty$ then $F: X \to L^0(Y), x \mapsto f(x, \cdot)$ is $\mu$-measurable.
3. The map $G: X \to \mathbb R \cup \{+\infty\}, x \mapsto \hat \rho(f(x, \cdot), 0)$ is $\mu$-measurable.

There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?

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We need the following result, i.e.,

Lemma Let $f \in S (Z)$ and $\varepsilon>0$. There is $f_\varepsilon \in S(Z)$ of the form $$ f_\varepsilon (x, y) = \sum_{i,j=1}^{n} e_{i,j} 1_{A_{i}} (x) 1_{B_{j}} (y) $$ such that

• $(A_{i})_{i=1}^{n}, (B_{j})_{j=1}^{n}$ are measurable partitions of $X,Y$ respectively,
• $e_{ij} \in E$ and $\mu(A_i)+\nu(B_j) < \infty$ for all $i,j \in \{1, \ldots, n\}$,
• $\|f-f_\varepsilon\|_{L^1(Z)} < \varepsilon$ and $\lambda (\{f \neq f_\varepsilon\}) < \varepsilon$.

There is a sequence $(g_n) \subset S(Z)$ such that $g_n \to f$ $\lambda$-a.e. For each $n$, let $f_n$ be the approximating function of $g_n$ given by Lemma such that $\lambda (\{f_n \neq g_n\}) < 2^{-n}$. It's easy to verify that $f_n \to f$ $\lambda$-a.e.

1. I proved that for $\mu$-a.e. $x \in X$ we have $f_n (x, \cdot) \to f(x, \cdot)$ $\nu$-a.e. Let $$ f_n (x, y) = \sum_{i,j=1}^{\varphi_n} e_{n, i,j} 1_{A_{n, i}} (x) 1_{B_{n, j}} (y). $$

Clearly, $f_n (x, \cdot) \in S(Y)$. Then for $\mu$-a.e. $x \in X$ we have$f(x, \cdot) \in L^0(Y)$.

2. We define $$ F_n: X \to L^0(Y), x \mapsto f_n(x, \cdot). $$

Then $F_n \in S(X, L^0(Y))$. For $\mu$-a.e. $x\in X$ we have $F_n (x) \to f(x, \cdot)$ $\nu$-a.e. We have $\nu(Y) < \infty$, so $\nu$-a.e. convergence implies that in $\hat \rho$. Then for $\mu$-a.e. $x\in X$ we have $\hat \rho (F_n (x), F(x)) \xrightarrow{n \to \infty} 0$. The $\mu$-a.e. limit of a sequence of $\mu$-measurable functions is again $\mu$-measurable. The second claim then follows.

3. There is a countable measurable partition $(Y_n)$ of $Y$ with $\nu(Y_n) < \infty$ for all $n$. Let $f_n:X \times Y_n \to E$ be the restriction of $f$ to $X \times Y_n$. Then $f_n$ is measurable. We apply part (1) and get that $$ F_n: X \to L^0(Y_n), x \mapsto f_n(x, \cdot) $$ is measurable. Then $$ G_n: X \to \mathbb R, x \mapsto \hat \rho(f_n(x, \cdot), 0) $$ is measurable. We have $G(x) = \sum_{n=1}^\infty G_n (x)$ for all $x\in X$. Then $G$ is measurable.