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 $\nu$,
- $(Z, \overline{\mathcal C}, \overline{\lambda})$ the completion of $(Z, \mathcal C, \lambda)$.
Let $\rho_Z$ be a pseudometric on $L^0(Z) := L^0(Z, E)$ defined by $$ \rho_Z (f, g) := \int_Z \min\{|f-g|, 1\} \, \mathrm d \lambda \quad \forall f, g \in L^0 (Z). $$
Then convergence in $\rho_Z$ is equivalent to that in measure. We define $\rho_X$ on $L^0 (X, \mathbb R)$ similarly. I'm trying to verify a result at page 32 of Martin Väth's monograph Ideal Spaces, i.e.,
Lemma 3.1.3. Let $f_n \in L^0(Z)$ for all $n \in \mathbb N$ such that for $\mu$-a.e. $x \in X$ we have $(f_n(x, \cdot))_n \subset L^1(Y)$. We define $$ g_n: X \to \mathbb R, x \mapsto \| f_n (x, \cdot) \|_{L^1 (Y)}. $$ If $\rho_{X} (g_n, 0_\mathbb R) \to 0$ then $\rho_Z (f_n, 0_E) \to 0$.
It has been verified that the map $x \mapsto f_n(x, \cdot)$ belongs to $L^0 (X, L^1(Y))$. Hence $g_n$ indeed belongs to $L^0 (X, \mathbb R)$. 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 $$ c(m) := \inf \{\min\{ \nu (B), 1\} : B \in \mathcal{B} \text{ s.t. } \nu (B) \ge m\}. $$
Then $c$ is non-decreasing and thus Borel measurable. Clearly, $\nu(B) \ge c (\nu (B))$ 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 $\min\{ \nu (B_n), 1\} \xrightarrow{n \to \infty} 0$. 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 $$ \begin{align*} g_n (x) &\ge \| \varepsilon 1_{M_n} (x, \cdot)\|_{L^1 (Y)} \\ &= \varepsilon \nu (\{y \in Y : (x, y) \in M_n\}) \\ &\ge \varepsilon c (\nu ( \{y \in Y : (x, y) \in M_n\} )) \\ &= \varepsilon c \bigg ( \int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu \bigg ). \end{align*} $$
Then $c ( \int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu ) \xrightarrow{n \to \infty} 0$ in $\rho_X$. 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. Dominated convergence theorem 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.
