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 $n \in \mathbb N$ such that for $\mu$-a.e. $x \in X$ we have $(f_n(x, \cdot))_n \subset L^0 (Y)$. We define $$ g_n: X \to \mathbb R \cup \{+\infty\}, x \mapsto \| f_n (x, \cdot) \|_{L^0(Y)}. $$ If $\| g_n \|_{L^0(X)} \to 0$ then $\| f_n \|_{L^0 (Z)} \to 0$.
I have verified that $g_n$ indeed belongs to $L^0(X, \mathbb R \cup \{\pm\infty\})$. Could you have a check on my below attempt? Thank you so much for your help!
We define $c:[0, \infty] \to[0, 1]$ by $$ c(m) := \inf \{1, \| 1_B \|_{L^0(Y)} : B \in \mathcal{B} \text{ s.t. } \nu (B) \ge m\}. $$
Then $c$ is non-decreasing and thus Borel measurable. Clearly, $\| 1_B \|_{L^0(Y)} \ge c (\nu (B))$ for all $B \in \cal B$. Let's verify $c(m)>0$ for all $m>0$. Assume the contrary that $c(m_0)=0$ for some $m_0>0$. There is a sequence $(B_n) \subset \cal B$ such that $\inf_n \nu(B_n) \ge m_0$ and $\| 1_{B_n} \|_{L^0(Y)} \xrightarrow{n \to \infty} 0$. Then there is a subsequence $\varphi$ of $\mathbb N$ such that $1_{B_{\varphi (n)}} \xrightarrow{n \to \infty} 0$ $\nu$-a.e. By dominated convergence theorem (DCT), $\nu (B_{\varphi (n)}) \xrightarrow{n \to \infty} 0$. This is a contradiction.
Let $\varepsilon >0$ and $M_n := \{ (x, y) \in Z: |f_n (x, y)| \ge \varepsilon \} \in \overline{\cal C}$. Then $$ \begin{align*} g_n (x) &\ge \| \varepsilon 1_{M_n} (x, \cdot)\|_{L^0 (Y)} \\ &\ge \varepsilon c (\nu ( \{y \in Y : (x, y) \in M_n\} )) \\ &= \varepsilon c \bigg ( \int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu \bigg ). \end{align*} $$
Then $\| c ( \int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu ) \|_{L^0 (X)}\xrightarrow{n \to \infty} 0$. Then $c ( \int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu ) \xrightarrow{n \to \infty} 0$ in measure. For any $\delta>0$, we have $c(\delta)>0$. So $$ \mu \left ( \bigg \{x \in X : c \bigg ( \int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu \bigg ) \ge c(\delta) \bigg \} \right ) \xrightarrow{n \to \infty} 0 \quad \forall \delta >0, $$ which implies by monotonicity of $c$, $$ \mu \left ( \bigg \{x \in X : \int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu \ge \delta \bigg \} \right ) \xrightarrow{n \to \infty} 0 \quad \forall \delta >0, $$ i.e., $\int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu \xrightarrow{n \to \infty} 0$ in measure. DCT also holds if almost everywhere convergence is replaced by convergence in measure. Then $$ \int_X\int_Y 1_{M_n} (x, y) \, \mathrm d \nu (y) \, \mathrm d \mu (x) \xrightarrow{n \to \infty} 0, $$ or equivalently $\lambda (M_n) \to 0$. The claim then follows.
Remark Above we apply DCT twice. Such applications are valid because $\mu(X) + \nu (Y) < \infty$.
