If $f \in L^0 (X, L^0 (Y))$, then $f \in L^0 (Z)$

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,
• $\mu(X) < \infty$,
• $(E, | \cdot |)$ 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 If $f \in L^0 (X, L^0 (Y))$, then $f \in L^0 (Z)$.

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

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There is a sequence $(f_n) \subset S(X, L^0 (Y))$ such that $f_n \to f$ $\mu$-a.e. in $(L^0(Y), \hat \rho_Y)$. Let $$ f_n (x) = \sum_{k=1}^{\varphi_n} 1_{A_{n, k}} (x) h_{n, k}, $$ where $h_{n, k} \in L^0 (Y)$ and $A_{n, k} \in \cal A$ with $\mu(A_{n, k}) < \infty$. Let $$ F_n: X \times Y \to E, (x, y) \mapsto f_n(x)(y). $$

Clearly, $F_n \in L^0 (Z)$. For $\mu$-a.e. $x \in X$, $$ \|F_n(x, \cdot) - f(x) \|_{L^0 (Y)} \xrightarrow{n \to \infty} 0, $$ which implies $(F_n(x, \cdot))_n$ is a Cauchy sequence in $L^0 (Y)$. We need some auxiliary results, i.e.,

• Lemma 1 Let $\mu(X) + \nu(Y) < \infty$. 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 $(L^0 (Z), \hat \rho_Z)$.
• Lemma 2 The metric space $(L^0 (Z), \hat \rho_Z)$ is complete.
• Lemma 3 Convergence in $\hat \rho_Z$ implies $\lambda$-a.e. convergence for a subsequence

1. First, we consider the case $\nu(Y) < \infty$. By Lemma 1, $(F_n)$ is a Cauchy sequence in $(L^0 (Z), \hat \rho_Z)$. By Lemma 2, there is $F\in L^0(Z)$ such that $F_n \to F$ in $\hat \rho_Z$. By Lemma 3, there is subsequence (also denoted by $(F_n)$ for simplicity) such that $F_{n} \xrightarrow{n \to \infty} F$ $\lambda$-a.e. I showed that for $\mu$-a.e $x \in X$ we have: (S1) $F_n(x, \cdot) \in L^0 (Y)$ and (S2) $F_n(x, \cdot) \to F(x, \cdot)$ $\nu$-a.e.

We have $f_n \to f$ $\mu$-a.e. This means for $\mu$-a.e. $x \in X$ we have $\|f_n (x) - f(x)\|_{L^0(Y)} \xrightarrow{n \to \infty} 0$. By Lemma 3 again, for $\mu$-a.e. $x \in X$ there is a subsequence $\varphi_x$ of $\mathbb N$ such that $f_{\varphi_x (n)} (x) \xrightarrow{n \to \infty} f(x)$ $\nu$-a.e. and thus $F_{\varphi_x (n)} (x, \cdot) \xrightarrow{n \to \infty}f(x)$ $\nu$-a.e. It follows that for $\mu$-a.e. $x \in X$ we have $F(x, \cdot) = f(x)$ $\nu$-a.e. and thus $\|F(x, \cdot) - f(x)\|_{L^0 (Y)}=0$.

2. We drop the assumption $\nu (Y) < \infty$. There is a countable measurable partition $(Y_n)$ of $Y$ such that $\sup_n \nu(Y_n) < \infty$. We define $$ f_n:X \to L^0 (Y_n), x \mapsto f(x) 1_{Y_n}. $$

Then $f_n \in L^0(X, L^0 (Y_n))$. We apply part (1) for $f_n$ and get $f_n \in L^0 (X \times Y_n)$. We have $L^0 (Y_n)$ can be considered a closed subspace of $L^0 (Y)$. So $L^0(X, L^0 (Y_n))$ can be considered a closed subspace of $L^0(X, L^0 (Y))$ w.r.t. $\hat \rho_X$. Similarly, $L^0 (X \times Y_n)$ can be considered a closed subspace of $L^0(Z)$ w.r.t $\hat \rho_{Z}$. Then $\sum_{k=1}^n f_k\xrightarrow{n \to \infty} f$ $\lambda$-a.e. The $\mu$-a.e. limit of a sequence of $\mu$-measurable functions is again $\mu$-measurable. Clearly, $\sum_{k=1}^n f_k$ is $\mu$-measurable. This completes the proof.