Is there a modification of $f$ on a null set such that $F: [0, T] \to L^p ({\mathbb R}^d), t \mapsto f(t,\cdot)$ is Bochner measurable?

Inspired by two answers in this thread, I'm able to drop the assumption on the integrability of $f$, i.e.,

Theorem Let $p \in [1, \infty)$ and $f \in L^0 (Z)$ such that $$ f(x, \cdot) \in L^p(Y) \quad \text{for}\quad \mu \text{-a.e. } x\in X. \tag{$*$} $$ Then $f \in L^0(X, L^p(Y))$.

My below proof adds nothing new to the other answers. I just fill in the details to fix the ideas.

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 $E$,
• $Z := X \times Y$,
• $\mathcal C :=\mathcal A \otimes \mathcal B$ 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 $(Z, \mathcal C, \lambda)$.

First, we have

Lemma 1 If $f \in L^0 (Z)$ then $f(x, \cdot) \in L^0(Y)$ for $\mu$-a.e. $x\in X$.

Lemma 2 Let $f \in S (Z)$ and $\varepsilon>0$. There is $f_\varepsilon \in S(Z)$ of the form $$ f_\varepsilon (x, y) = \sum_{i,j=1}^{n} e_{i,j} 1_{A_{i}} (x) 1_{B_{j}} (y) $$ such that

• $(A_{i})_{i=1}^{n}, (B_{j})_{j=1}^{n}$ are measurable partitions of $X,Y$ respectively,
• $e_{ij} \in E$ and $\mu(A_i)+\nu(B_j) < \infty$ for all $i,j \in \{1, \ldots, n\}$,
• $\|f-f_\varepsilon\|_{L^1(Z)} < \varepsilon$, and
• $\lambda (\{f \neq f_\varepsilon\}) < \varepsilon$.

By modifying $f$ on a null set, we can assume $(*)$ for all $x\in X$.

There is a sequence $(g_n) \subset S(X\times Y)$ such that $\|g_n-f\|_{L^1(Z)} < \frac{1}{n}$ and $g_n \to f$ $\lambda$-a.e. For each $n$, let $f_n$ be the approximating function of $g_n$ given by Lemma 2 such that $\|f_n-g_n\|_{L^1(Z)} < 2^{-n}$ and $\lambda (\{f_n \neq g_n\}) < 2^{-n}$. It's easy to see that $\|f_n-f\|_{L^1(Z)} < \frac{2}{n}$ and $f_n \to f$ $\lambda$-a.e. Let $$ f_n (x, y) = \sum_{i,j=1}^{\varphi_n} e_{n, i,j} 1_{A_{n, i}} (x) 1_{B_{n, j}} (y). $$

1. First, we consider the case $\nu (Y) < \infty$ and $\alpha := \sup_{(x, y) \in X \times Y} |f (x, y)| +1< \infty$. Let $$ e'_{n,i,j} := \begin{cases} e_{n, i, j} &\text{if} \quad |e_{n, i, j}| \le \alpha, \\ 0 &\text{otherwise}, \end{cases} $$ and $$ f'_n (x, y) := \sum_{i,j=1}^{\varphi_n} e'_{n, i,j} 1_{A_{n, i}} (x) 1_{B_{n, j}} (y). $$

Clearly, $f'_n \in S(Z)$ and $f'_n \to f$ $\lambda$-a.e. I already proved that for $\mu$-a.e. $x \in X$: $$ \begin{align*} f'_n(x, \cdot) \xrightarrow{n \to \infty} f(x, \cdot) \quad \nu \text{-a.e.} \end{align*} $$

By dominated convergence theorem, for $\mu$-a.e. $x \in X$: $\|f'_n(x, \cdot) - f(x, \cdot) \|_{L^p(Y)} \to 0$. Let $$ F_n: X \to L^p(Y), x \mapsto f'_n(x, \cdot), \\ F: X \to L^p(Y), x \mapsto f(x, \cdot). $$

Clearly, $F_n \in S(X, L^p(Y))$ for all $n$. The $\mu$-a.e. limit of a sequence of $\mu$-measurable functions is again $\mu$-measurable. It suffices to prove that for $\mu$-a.e. $x \in X$: $\|F_n (x)-F(x)\|_{L^p(Y)} \to 0$, which we have already done.

2. Second, we consider the case $\alpha < \infty$. Because $(Y, \mathcal B, \nu)$ is $\sigma$-finite, there is a countable measurable partition $(Y_n)$ of $Y$ with $\sup_n \nu(Y_n) < \infty$. We apply part (1) for each function $f_n := f1_{X \times Y_n}$ and get $f_n \in L^0 (Z)$. Clearly, $\sum_{k=1}^n f_k \to f$ everywhere. The $\lambda$-a.e. limit of a sequence of $\lambda$-measurable functions is again $\lambda$-measurable. Then $f \in L^0 (Z)$.

3. Finally, we also drop the assumption $\alpha < \infty$. Let $Z_n := \{n \le |f| < n+1 \}$ for $n \in \mathbb N$. It is not necessarily true that $Z_n \in \cal C$, but we always have $Z_n \in \overline{\cal C}$. Clearly, $(Z_n)_n$ is a partition of $Z$. We apply part (2) for each function $f_n := f1_{Z_n}$ and get $f_n \in L^0 (Z)$. Clearly, $\sum_{k=1}^n f_k \to f$ everywhere. The $\lambda$-a.e. limit of a sequence of $\lambda$-measurable functions is again $\lambda$-measurable. Then $f \in L^0 (Z)$.

This completes the proof!