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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 $E$,
  • $Z := X \times Y$,
  • $\mathcal C :=\mathcal A \otimes \mathcal B$ the product $\sigma$-algebra of $\mathcal A$ and $\mathcal B$,
  • $\lambda := \mu \otimes \nu$ the product measure of $\mu$ and $\nu$,
  • $(Z, \overline{\mathcal C}, \overline{\nu})$ the completion of $(Z, \mathcal C, \nu)$.

Let $\rho_Y$ be a pseudometric on $L^0(Y)$ defined by $$ \rho_Y (f, g) := \int_Y \min\{|f-g|, 1\} \, \mathrm d \nu \quad \forall f, g \in L^0 (Y). $$

We define $\rho_Z$ on $L^0(Z)$ similarly. Theorem 4.4.1 at page 98 of Martin Väth's monograph Ideal Spaces suggests that

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 $(L^1 (Y), \| \cdot\|_{L^1 (Y)})$. Then $(f_n)$ is a Cauchy sequence in $(L^0 (Z), \rho_Z)$.

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

For simplicity, we write $$ \nu (|f - g| > \delta) := \nu (\{y \in Y : |f (y) - g(y)| > \delta\}) \quad \forall \delta >0, \forall f,g \in L^0 (Y). $$

Let $\hat \rho_Y$ be another pseudometric on $L^0(Y)$ defined by $$ \hat \rho_Y (f, g) := \inf_{\delta >0} \{ \nu (|f - g| > \delta) +\delta \} \quad \forall f, g \in L^0 (Y). $$

We have some useful properties, i.e.,

Lemma 1 $(L^0(Y), \rho_Y)$ is complete.

Lemma 2 $(L^0(Y), \hat \rho_Y)$ is complete.

Lemma 3 The following statements are equivalent:

  • (S1) $\rho_Y (f_n, f) \to 0$.
  • (S2) Every subsequence of $(f_n)$ has a further subsequence that converges to $f$ $\nu$-a.e.
  • (S3) $f_n \to f$ in measure, i.e., $$ \nu (\{y \in Y : |f_n (y)-f(y)| > \varepsilon\}) \xrightarrow{n \to \infty} 0 \quad \forall \varepsilon>0. $$
  • (S4) $\hat \rho_Y (f_n, f) \to 0$.

Proof

  1. The proof of the completeness of $\rho_Y$ can be found here.
  2. The proof of the completeness of $\hat\rho_Y$ can be found here.
  3. The proof of (S1) $\iff$ (S2) can be found here. The proof of (S3) $\iff$ (S4) can be found here. The proof of (S1) $\iff$ (S3) can be found here.

Now we are going to prove our theorem. Assume the contrary that $(f_n)$ is not a Cauchy sequence in $(L^0 (Z), \rho_Z)$. By above Lemmas $(f_n)$ is not a Cauchy sequence in $(L^0 (Z), \hat \rho_Z)$. 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 $$ h_n: X \to \mathbb R, x \mapsto \| g_n (x, \cdot) \|_{L^1 (Y)}. $$

It has been verified that the map $x \mapsto g_n (x, \cdot)$ belongs to $L^0 (X, L^1(Y))$. Hence $h_n$ indeed belongs to $L^0 (X, \mathbb R)$. By our assumption, $h_n \to 0$ $\mu$-a.e.

Lemma 3.1.3. Let $g_n \in L^0(Z)$ for all $n \in \mathbb N$ such that for $\mu$-a.e. $x \in X$ we have $(g_n(x, \cdot))_n \subset L^1(Y)$. We define $$ h_n: X \to \mathbb R, x \mapsto \| g_n (x, \cdot) \|_{L^1 (Y)}. $$ If $\rho_{X} (h_n, 0_\mathbb R) \to 0$ then $\rho_Z (g_n, 0_E) \to 0$.

By Lemma 3.1.3., $g_n \to 0$ in measure, which is a contradiction. This completes the proof.

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