Let $T>0$ and $p \in [1, \infty)$. Let $f \in L^p ([0, T] \times {\mathbb R}^d)$. By a theorem in [this](https://math.stackexchange.com/questions/4301414/if-f-n-is-cauchy-and-converges-a-e-to-f-there-is-a-cauchy-subsequence) thread, there is a Lebesgue null subset $N$ of $[0, T]$ such that $f(t, \cdot)$ is Lebesgue measurable and that $f(t, \cdot) \in L^p ({\mathbb R}^d)$ for all $t \in [0, T] \setminus N$. For $t \in N$, we don't know the measurability and thus the integrability of $f(t, \cdot)$. However, we can re-define $f$ such that $f(t, \cdot) \equiv 0$ for all $t \in N$. Then writing $f(t, \cdot) \in L^p ({\mathbb R}^d)$ is rigorously valid for *every* $t \in [0, T]$.



> 
> Is there a modification of $f$ on a Lebesgue null subset of $[0, T]$ such that the map
> $$
> F: [0, T] \to L^p ({\mathbb R}^d), t \mapsto f(t, \cdot)
> $$
> is [Bochner measurable](https://www.wikiwand.com/en/Bochner_measurable_function)?
> 
> 
> 


Intuitively, the answer is positive. But I don't know how to prove it...



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?
Let


* $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be complete $\sigma$-finite measure spaces, and $(E, | \cdot |)$ a Banach space.
* $S (X)$ the space of $\mu$-simple 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 $\nu$.


Here we use Bochner integral. Then we have



> 
> Let $p \in [1, \infty)$.
> 
> 
> * [Lemma 1](https://math.stackexchange.com/questions/4737770/tensor-product-of-lp-spaces-is-dense-in-the-product-lp-space) Then
> $$
> \overline{S(X) \otimes S(Y)} = L^p(X \times Y),
> $$
> where the closure is with respect to the norm in $L^p(X \times Y)$. Here $S (X) \otimes S(Y)$ is identified with a subspace of $L^p(X \times Y)$ through the natural embedding
> $$
> f \otimes g \mapsto \big( (x,y) \mapsto f(x) g(y) \big).
> $$
> * [Lemma 2](https://math.stackexchange.com/questions/4301414/if-f-n-is-cauchy-and-converges-a-e-to-f-there-is-a-cauchy-subsequence) Let $(f\_n) \subset S (X \times Y)$ be a Cauchy sequence in $L^p(X \times Y)$ that converges $\lambda$-a.e. to $f$. Then there is a subsequence $\varphi$ of $\mathbb N$ such that for $\mu$-a.e. $x \in X$, we have $(f\_{\varphi (n)} (x, \cdot))\_n \subset S (Y)$ is a Cauchy sequence in $L^p (Y)$ and converges to $f(x, \cdot)$ both $\nu$-a.e. and in $L^p (Y)$.
> 
> 
> 


Let $X := [0, T]$ and $Y := \mathbb R^d$. Let $\mathcal A, \mathcal B$ be the corresponding Lebesgue $\sigma$-algebras of $X$ and $Y$. Let $\mu, \nu$ be the corresponding Lebesgue measures on $X, Y$.


By **Lemma 1**, there is a sequence of $(f\_n) \subset L^p(X \times Y)$ with $f\_n = g\_n h\_n$ for some $g\_n \in S (X)$ and $h\_n \in S (Y)$ such that $\|f\_n-f\|\_{L^p(X \times Y)} \to 0$. Clearly, $(f\_n)$ is a Cauchy sequence in $S(X \times Y)$. Also, each $f\_n$ is of the form
$$
f\_n (t, x) = \sum\_{k=1}^{\varphi\_n} e\_{n, k} 1\_{A\_{n, k}} (t) 1\_{B\_{n, k}} (x)
\quad \forall t\in X, x \in Y,
$$
for some $e\_{n, k} \in \mathbb R$ and $A\_{n, k} \in \mathcal A, B\_{n, k} \in \mathcal B$ such that $\mu(A\_{n, k}) + \nu (B\_{n, k}) < \infty$. Then $f\_n (t, \cdot) \in L^p(Y)$ for all $t \in X$. Let
$$
F\_n : X \to L^p (Y), t \mapsto f\_n (t, \cdot)
\quad \forall n \in \mathbb N.
$$


Clearly, $F\_n \in S(X, L^p(Y)) \subset L^p(X, L^p(Y))$ for all $n \in \mathbb N$. By Fubini's theorem,
$$
\begin{align\*}
\| F\_n - F\_m\|\_{L^p(X, L^p(Y))}^p &= \int\_0^T \mathrm d t \, \|F\_n (t) - F\_m (t)\|\_{L^p (Y)}^p \\
&= \|f\_n-f\_m\|\_{L^p(X \times Y)}^p.
\end{align\*}
$$


Then $(F\_n)$ is a Cauchy sequence in the Banach space $L^p(X, L^p(Y))$. Then there is $F \in L^p(X, L^p(Y))$ such that $\|F\_n-F\|\_{L^p(X, L^p(Y))} \to 0$.


Convergence in $L^p$ implies a.e. convergence of a subsequence, so WLOG we assume $f\_n \to f$ $\lambda$ a.e. and $F\_n \to F$ $\mu$-a.e. By **Lemma 2**, there is a subsequence $\varphi$ of $\mathbb N$ su for $\mu$-a.e. $t \in X$ we have
$$
\|f\_{\varphi (n)} (t, \cdot)-f(t, \cdot) \|\_{L^p (Y)} \xrightarrow{n \to \infty} 0.
$$


On the other hand, for $\mu$-a.e. $t \in X$ we have
$$
\|f\_n (t, \cdot) - F (t)\|\_{L^p(Y)} =\|F\_n (t) - F (t)\|\_{L^p(Y)} \xrightarrow{n \to \infty} 0.
$$


It follows that for $\mu$-a.e. $t \in X$ we have
$$
\|f (t, \cdot) - F(t) \|\_{L^p (Y)} = 0.
$$


The claim then follows.


