How to construct this sequence that converges a.e. in product measure and that has a very particular form?

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,
• $(E, | \cdot |)$ 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 $E$,
• $\mathcal C :=\mathcal A \otimes \mathcal B$ the product $\sigma$-algebra of $\mathcal A$ and $\mathcal B$,
• $\lambda := \mu \otimes \nu$ the product measure of $\mu$ and $\nu$.

I'm trying to prove a claim in this thread.

Theorem Let $f \in L^0 (X \times Y)$. There is a sequence $(f_n) \subset S(X \times Y)$ such that $f_n \to f$ $\lambda$-a.e. and that each $f_n$ is of the form $$ f_n (x, y) = \sum_{i,j=1}^{\varphi_n} e_{n, i,j} 1_{A_{n, i}} (x) 1_{B_{n, j}} (y), \tag{$*$} $$ where $e_{n, k} \in E$ and $(A_{n, i})_{i=1}^{\varphi_n}, (B_{n, j})_{j=1}^{\varphi_n}$ are finite measurable partitions of $X,Y$ respectively.

Such $f_n$ has a very nice form that I could not construct. Could you elaborate on how to get it?

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My failed attempt First, we assume $\mu (X) + \nu (Y) +\sup_{(x, y) \in X \times Y} |f(x, y)| < \infty$. Then $\lambda(X \times Y) < \infty$ and $f \in L^1(X \times Y)$. There is a sequence $(f_n) \subset S(X \times Y)$ such that $f_n \to f$ $\lambda$-a.e. and in $L^1(X \times Y)$. Let $$ f_n (x, y) = \sum_{k=1}^{\varphi_n} e_{n, k} 1_{C_{n, k}} (x, y), $$ where $e_{n, k} \in E$ and $C_{n, k} \in \cal C$ such that $\lambda(C_{n, k}) < \infty$.

Answer 1

$\newcommand\C{\mathcal C}\newcommand\A{\mathcal A}\newcommand\B{\mathcal B}$Using the $\sigma$-finiteness condition, truncation of $f$, and your attempt, we see that without loss of generality $f=1_C$ for some $C\in\C$.

But in this case the result follows by (say) Theorem 1.4 of this paper or its arXiv version. Indeed, by that theorem, every $C\in\C$ can be approximated by sets in the algebra generated by product sets $A\times B$ such that $(A,B)\in\A\times\B$ with respect to the pseudometric $d$ given by the formula $d(E,F):=(\mu\otimes\nu)(E+F)$ for $E$ and $F$ in $\C$, where $+$ denotes the symmetric difference.

Answer 2

1. First, we prove that for all $f \in S(X\times Y)$ and $\varepsilon >0$. There is $f_\varepsilon \in S(X\times Y)$ such that $g$ satisfies $(*)$ and that $\lambda (A) \le \varepsilon$ where $A := \{f \neq f_\varepsilon \} \in \mathcal C$.

a. First, we consider the case $f = 1_C$ for some $C \in \mathcal C$ with $\lambda(C) < \infty$. Let $$ \mathcal E := \bigg \{ \bigcup_{i \in I} A_i \times B_i : I \text{ finite}, A_i \in \mathcal A, B_i \in \mathcal B \bigg \}. $$

Lemma 1 Let $\lambda$ be finite and $\mathcal D \subset \mathcal C$ an algebra generating $\cal C$. Then for each $C \in\cal C$ and $\varepsilon>0$, there is $D \in \cal D$ such that $$ \mu(C\Delta D) := \mu(C\setminus D) + \mu(D \setminus C) < \varepsilon. $$

Then $\cal E$ is an algebra generating $\cal C$. By above Lemma 1, there are $A_{k} \in \cal A$ and $B_{k} \in \cal B$ with finite measures such that $$ \lambda \bigg ( C \Delta \bigg ( \bigcup_{k=1}^{{n}} A_{k} \times B_{k} \bigg) \bigg ) < \varepsilon $$

Lemma 2 Let $(A_k \times B_k)_{k=1}^n$ where $A_k \in \mathcal A$ and $B \in \mathcal B$ with finite measures. Then there is a pairwise disjoint sequence $(A'_k \times B'_k)_{k=1}^{n'}$ such that

• $A'_k \in \mathcal A$ and $B'_k \in \mathcal B$ with finite measures,
• either $A'_{i} \cap A'_j = \emptyset$ or $A'_{i} = A'_j$,
• either $A'_{i} \cap A_j = \emptyset$ or $A'_{i} \subset A_j$,
• either $B'_{i} \cap B'_j = \emptyset$ or $B'_{i} = B'_j$,
• either $B'_{i} \cap B_j = \emptyset$ or $B'_{i} \subset B'_j$, and $$ \bigcup_{k=1}^n A_k \times B_k = \bigcup_{k=1}^{n'} A'_k \times B'_k. $$

WLOG, we assume $(A_{k} \times B_{k})$ satisfies the conclusion of Lemma 2. Then $$ 1_{\bigcup_{k=1}^{{n}} A_{k} \times B_{k}} = f_\varepsilon (x, y):= \sum_{k=1}^{{n}} 1_{A_{k}} (x) 1_{B_{k}} (y). $$

Clearly, $f_\varepsilon \in S(X \times Y)$ has our desired form with $$ A = C \Delta \bigg ( \bigcup_{k=1}^{{n}} A_{k} \times B_{k} \bigg). $$

b. Next we consider the case $f \in S(X\times Y)$. We assume $f$ has a form $f = \sum_{n=1}^m e_n 1_{C_n}$. Let $f_{\varepsilon, n}$ be the approximating function for $1_{C_n}$ as in part (a) such that $\lambda(A_n) < \frac{\varepsilon}{n}$ where $A_n := \{f \neq f_\varepsilon \} \in \mathcal C$. Let $$ f_{\varepsilon, n} (x, y) = \sum_{k=1}^{\varphi_{n}} 1_{A_{n, k}} (x) 1_{B_{n, k}} (y), $$ and $$ f_\varepsilon (x, y) := \sum_{n=1}^m e_n f_{\varepsilon, n} (x, y) = \sum_{n=1}^m \sum_{k=1}^{\varphi_{n}} e_n 1_{A_{n, k}} (x) 1_{B_{n, k}} (y). $$

Clearly, $(A_{n, k} \times B_{n, k})_k$ does not necessarily satisfy our requirement, i.e., it's possible that $$ \emptyset \neq A_{n', k'} \cap A_{n, k} \neq A_{n, k}. $$

However, we can apply Lemma 2 to decompose them into our desired form. Clearly, $f_\varepsilon \in S(X \times Y)$ and $\lambda(A) \le \sum_{n=1}^m \lambda (A_n) \le \varepsilon$.

2. Finally, we come back to our original problem. Let $f \in L^0(X \times Y)$. There is a sequence $(f_n) \subset S(X \times Y)$ such that $f_n \to f$ $\lambda$-a.e. Let $g_n$ be the approximating function of $f_n$ as in part (1) such that $\lambda (A_n) \le 2^{-n}$ where $A_n := \{f_n \neq g_n\}$.

Let $B_n := \bigcup_{k=1}^\infty A_{n+k} \in \mathcal C$. Then $\lambda(B_n) \le 2^{-n+1}$. Let $B := \bigcap_n B_n$. Then $B$ is a $\lambda$-null set. Let $N$ be a $\lambda$-null set such that $f_n \to f$ on $N^c$. We will prove that $g_n \to f$ on $(B \cup N)^c = B^c \cap N^c$. Let $(x, y) \in B^c \cap N^c$. There is $m \in \mathbb N$ such that $(x, y) \in B_m^c = \bigcap_{k=1}^\infty A_{m+k}^c$. Then $(x, y) \in A_{n}^c$ and thus $f_n (x, y) = g_n (x, y)$ for all $n>m$. On the other hand, $(x, y) \in N^c$ and thus $f_n (x, y) \to f(x, y)$. It follows that $g_n (x, y) \to f(x, y)$. This completes the proof.

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Update In part (1), it's obvious that given $p \in [1, \infty)$ we can pick $f_\varepsilon$ that in addition satisfies $\|f -f_\varepsilon\|_{L^p} \le \varepsilon$.