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?

Question

Let $T>0$ and $p \in [1, \infty)$. Let $f \in L^p ([0, T] \times {\mathbb R}^d)$. By a theorem in this 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?

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

Answer 1

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 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 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.

Answer 2

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.

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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!