Below we use Bochner measurability and Bochner integral. Let
- $T>0$ and $p \in [1, \infty)$,
- $X :=[0, T]$ and $Y:= \mathbb R^d$,
- $\cal A$ the Lebesgue $\sigma$-algebra of $X$,
- $\cal B$ the Lebesgue $\sigma$-algebra of $Y$,
- $\mu, \nu$ complete $\sigma$-finite Radon measures on $X, Y$ respectively,
- $\mu, \nu$ absolutely continuous,
- $Z :=X \times Y$,
- $\cal C$ the product $\sigma$-algebra of $\cal A$ and $\cal B$,
- $\lambda$ the product measure of $\mu$ and $\nu$,
- $L^0 (X) := L^0 (X, \mathbb R)$ the space of measurable functions from $X$ to $\mathbb R$.
Clearly, $\mu(X) < \infty$. Let $L^p_{\text{loc}} (Y)$ be the space of measurable functions $f:Y \to \mathbb R$ such that $$ \|f\|_{L^p_{\text{loc}}} := \sup_{y \in Y} \|1_{B(y, 1)} f\|_{L^p} < \infty, $$ where $B(y, 1)$ is the open unit ball centered at $y$. Let $(y_m)$ be a countable dense subset of $Y$. A sphere is $\nu$-null set. Then by dominated convergence theorem (DCT), $$ \|f\|_{L^p_{\text{loc}}} = \sup_{m \in \mathbb N} \|1_{B(y_m, 1)} f\|_{L^p} \quad \forall f \in L^p_{\text{loc}} (Y). $$
I would like to prove an analogue of the main result in this thread, i.e.,
Theorem If $f \in L^0 (X, L^p_{\text{loc}} (Y))$, then $f \in L^0 (Z)$.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
There is a sequence $(f_n) \subset S(X, L^p_{\text{loc}} (Y))$ such that $f_n \to f$ $\mu$-a.e. in $L^p_{\text{loc}} (Y)$. Let $$ f_n (x) = \sum_{k=1}^{\varphi_n} 1_{A_{n, k}} (x) h_{n, k}, $$ where $h_{n, k} \in L^p_{\text{loc}} (Y)$ 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. $x \in X$, $$ \|F_n(x, \cdot) - f(x) \|_{L^p_{\text{loc}} (Y)} \xrightarrow{n \to \infty} 0, $$ which implies $(F_n(x, \cdot))_n$ is a Cauchy sequence in $L^p_{\text{loc}} (Y)$. We need the following result about convergence in measure.
Let $\rho_Z$ be a pseudometric on $L^0(Z)$ defined by $$ \rho_Z (g_1, g_2) := \int_Z \min\{|g_1 - g_2|, 1\} \, \mathrm d \lambda \quad \forall g_1, g_2 \in L^0 (Z). $$ If $\mu (X) + \nu(Y) < \infty$, then
- Lemma 1 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 $(L^p_{\text{loc}} (Y), \| \cdot\|_{L^p_{\text{loc}}})$. Then $(f_n)$ is a Cauchy sequence in $(L^0 (Z), \rho_Z)$.
- Lemma 2 $\rho_Z$ is a complete metric on $L^0 (Z)$.
- Lemma 3 $(f_n)$ converges to $f$ in $(L^0 (Z), \rho_Z)$ IFF every subsequence of $(f_n)$ has in turn a further subsequence that converges to $f$ $\lambda$-a.e.
- First, we consider the case $\nu(Y) < \infty$.
By Lemma 1, $(F_n)$ is a Cauchy sequence in $(L^0 (Z), \rho_Z)$. By Lemma 2, there is $F\in L^0(Z)$ such that $F_n \to F$ in $\rho_Z$. 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 $\|f_n (x) - f(x)\|_{L^p_{\text{loc}}} \xrightarrow{n \to \infty} 0$. Convergence in $L^p_{\text{loc}}$ implies a.e. convergence of a subsequence. 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 $\|F(x, \cdot) - f(x)\|_{L^p (Y)}=0$.
- 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 $$ f_n:X \to L^p_{\text{loc}} (Y_n), x \mapsto f(x) 1_{Y_n}. $$
Then $f_n \in L^0(X, L^p_{\text{loc}} (Y_n))$. We apply part (1) for $f_n$ and get $f_n \in L^0 (X \times Y_n)$. We have $L^p_{\text{loc}} (Y_n)$ can be considered a closed subspace of $L^p_{\text{loc}} (Y)$. So $L^0(X, L^p_{\text{loc}} (Y_n))$ can be considered a closed subspace of $L^0(X, L^p_{\text{loc}} (Y))$ w.r.t. $\rho_X$. Similarly, $L^0 (X \times Y_n)$ can be considered a closed subspace of $L^0(X \times Y)$ w.r.t $\rho_{X \times Y}$. Then $f = \lim_n \sum_{k=1}^n f_k \in L^0 (X \times Y)$. This completes the proof.
