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Below we use Bochner measurability and Bochner integral. Let

  • $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be complete 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 $$ \hat \rho_Y (f, g) := \inf_{\delta >0} \{ \nu (|f - g| > \delta) +\delta \}. $$

Then $\hat \rho_Y$ is an extended pseudometric on $L^0 (Y)$. For $f_n, f \in L^0(Y)$, we have $\hat \rho_Y (f_n, f) \to 0$ IFF $f_n \to f$ in measure. We define $\hat \rho_X$ on $L^0(X, \mathbb R \cup \{\pm\infty\})$ and $\hat \rho_Z$ on $L^0(Z)$ similarly. For simplicity, we write $$ \begin{align*} \| f \|_{L^0(X)} & := \hat \rho_X (f, 0), \\ \| f \|_{L^0(Y)} &:= \hat \rho_Y (f, 0), \\ \| f \|_{L^0(Z)} &:= \hat \rho_Z (f, 0). \end{align*} $$

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

Theorem Let $f_n \in L^0(Z)$ for all $n \in \mathbb N$. Assume that for $\mu$-a.e. $x \in X$, the sequence $(f_n(x, \cdot))_n$ is a Cauchy sequence in $(L^0 (Y), \hat \rho_Y)$. Then $(f_n)$ is a Cauchy sequence in $(L^0 (Z), \hat \rho_Z)$.

Could you have a check on my below attempt? Thank you so much for your help!

Assume the contrary that $(f_n)$ is not a Cauchy sequence in $L^0 (Z)$. There is $\varepsilon >0$ and two subsequences $\varphi, \psi$ of $\mathbb N$ such that $$ \hat \rho_Z (f_{\varphi (n)}, f_{\psi (n)}) \ge \varepsilon \quad \forall n \in \mathbb N. $$

Then $$ \lambda (|f_{\varphi (n)} - f_{\psi (n)}| > \delta) +\delta \ge \varepsilon \quad \forall n \in \mathbb N, \forall \delta >0. $$

We pick $\delta := \varepsilon/2$ and get $$ \lambda (|f_{\varphi (n)} - f_{\psi (n)}| > \varepsilon/2) \ge \varepsilon/2 \quad \forall n \in \mathbb N. $$

Let $g_n := f_{\varphi (n)} - f_{\psi (n)}$. Then $g_n \in L^0(Z)$. We define $$ h_n: X \to \mathbb R, x \mapsto \| g_n (x, \cdot) \|_{L^0 (Y)}. $$

It has been verified that $h_n$ indeed belongs to $L^0 (X)$. By our assumption, $h_n \to 0$ $\mu$-a.e. Because $\mu(X) < \infty$, we get $h_n \to 0$ in measure.

Lemma Let $\mu(X) + \nu (Y) < \infty$. Let $g_n \in L^0(Z)$ for all $n \in \mathbb N$ such that for $\mu$-a.e. $x \in X$ we have $(g_n(x, \cdot))_n \subset L^0 (Y)$. We define $$ h_n: X \to \mathbb R \cup \{+\infty\}, x \mapsto \| g_n (x, \cdot) \|_{L^0 (Y)}. $$ If $\| h_n \|_{L^0(X)} \to 0$ then $\| g_n \|_{L^0 (Z)} \to 0$.

By Lemma 4, $g_n \to 0$ in measure, which is a contradiction. This completes the proof.

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