Does a measurable $F :[0, T] \to L^p (\mathbb R^d; \mathbb R_{\ge 0})$ have a "flattened" measurable version?

Question

Let $d \in \mathbb N^*,p \in [1, \infty]$ and $T>0$. Let $$ F :[0, T] \to L^p (\mathbb R^d; \mathbb R_{\ge 0}), t \mapsto F_t $$ be measurable. I would like to ask if there is a measurable function $G:[0, T] \times \mathbb R^d \to \mathbb R_{\ge 0}$ such that

• $G(t, \cdot) \in L^p (\mathbb R^d; \mathbb R_{\ge 0})$ for all $t \in [0, T]$.
• $\|G(t, \cdot) - F_t\|_{L^p} = 0$ for a.e. $t \in [0, T]$.

Thank you so much for your elaboration!

Answer 1

$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\LL}{\mathcal L}\newcommand{\si}{\sigma}$The answer is yes, at least for $p\in[1,\infty)$.

Indeed, $L^p(\R^d)$ is a separable metric space. So, for each real $\ep$ there is a countable measurable partition $(B_{\ep,j})$ of $L^p(\R^d)$ such that for each $j$ we have $B_{\ep,j}\ne\emptyset$ and the diameter of $B_{\ep,j}$ is $\le\ep$. Pick any $y_{\ep,j}$ in $B_{\ep,j}$. For $(t,x)\in[0,T]\times\R^d$, let \begin{equation} G_\ep(t,x):=\sum_j y_{\ep,j}(x)\,1(F(t)\in B_{\ep,j}), \end{equation} where $F(t):=F_t$. Then for any real $c$ \begin{equation} \{(t,x)\in[0,T]\times\R^d\colon G_\ep(t,x)>c\} =\bigcup_j F^{-1}(B_{\ep,j})\times y_{\ep,j}^{-1}((c,\infty)) \in\LL([0,T])\otimes\LL(\R^d), \end{equation} where $\LL(\cdot)$ denotes the Lebesgue $\si$-algebra. So, the function $G_\ep$ is measurable, for each $\ep$.

Also, $\|G_\ep(t,\cdot)-F(t)\|_{L^p(\R^d)}\le\ep$ and hence $\|G_\ep(t,\cdot)\|_{L^p(\R^d)}\le\ep+\|F(t)\|_{L^p(\R^d)}$ for each $t\in[0,T]$.

Since $F\colon[0,T]\to L^p(\R^d)$ is measurable and the norm on $L^p(\R^d)$ is continuous and hence measurable, we see that the function $[0,T]\ni t\mapsto w(t):=\dfrac1{1+\|F(t)\|^p_{L^p(\R^d)}}\in[0,\infty)$ is measurable. So, for each real $\ep>0$ we have $G_\ep\in L^p_w([0,T]\times\R^d)$, where $L^p_w([0,T]\times\R^d)$ is the space of all measurable functions $H\colon[0,T]\times\R^d\to\R$ with norm $$\|H\|_{L^p_w([0,T]\times\R^d)}:=\Big(\int_0^T dt\,w(t)\,\|H(t,\cdot)\|_{L^p(\R^d)}^p\Big)^{1/p}<\infty.$$

For all integers $m,n$ such that $m\ge n\ge1$ \begin{equation} \|G_{1/m}-G_{1/n}\|_{L^p_w([0,T]\times\R^d)}^p =\int_0^T dt\,w(t)\|G_{1/m}(t,\cdot)-G_{1/n}(t,\cdot)\|_{L^p(\R^d)}^p\le(2/n)^pT\to0 \end{equation} as $n\to\infty$. So, by the completeness of $L^p_w([0,T]\times\R^d)$, for some sequence $(n_k)$ of natural numbers going to $\infty$ there is a limit $G$ of $G_{1/n_k}$ in $L^p_w([0,T]\times\R^d)$. Clearly, this limit $G$ satisfies your desired conditions. $\quad\Box$

Answer 2

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 $\nu$,
• $(Z, \overline{\mathcal C}, \overline{\nu})$ the completion of $(Z, \mathcal C, \nu)$.

We are going to prove

Theorem Let $p \in [1, \infty)$ and $f \in L^0 (X, L^p(Y))$. Then $f \in L^0 (Z)$.

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1. First, we consider the case $\mu (X) + \nu(Y) < \infty$. There is a sequence $(f_n) \subset S(X, L^p(Y))$ such that $f_n \to f$ $\mu$-a.e. in $L^p(Y)$. Let $$ f_n (x) = \sum_{k=1}^{\varphi_n} 1_{A_{n, k}} (x) h_{n, k}, $$ where $h_{n, k} \in L^p (Y)$ 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, $F_n \in L^0 (X \times Y)$. For $\mu$-a.e. $x \in X$, $$ \|F_n(x, \cdot) - f(x) \|_{L^p(Y)} \xrightarrow{n \to \infty} 0, $$ which implies $(F_n(x, \cdot))_n$ is a Cauchy sequence in $L^p(Y)$.

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 $(L^1 (Y), \| \cdot\|_{L^1 (Y)})$. 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.

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 $\|f_n (x) - f(x)\|_{L^p (Y)} \xrightarrow{k \to \infty} 0$. Convergence in $L^p$ 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 $\|F(x, \cdot) - f(x)\|_{L^p (Y)}=0$.

2. We consider the case $\mu (X) < \infty$. There is a countable measurable partition $(Y_n)$ of $Y$ such that $\sup_n \nu(Y_n) < \infty$. We define $$ f_n:X \to L^p (Y_n), x \mapsto f(x) 1_{Y_n}. $$

Then $f_n \in L^0(X, L^p(Y_n))$. We apply part (1) for $f_n$ and get $f_n \in L^0 (X \times Y_n)$. We have $L^p (Y_n)$ can be considered a closed subspace of $L^p (Y)$. So $L^0(X, L^p(Y_n))$ can be considered a closed subspace of $L^0(X, L^p(Y))$ 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 $f = \lim_n \sum_{k=1}^n f_k \in L^0 (X \times Y)$.

3. We also drop the assumption $\mu (X) < \infty$. There is a countable measurable partition $(X_n)$ of $X$ such that $\sup_n \mu (X_n) < \infty$. We define $$ f_n:X_n \to L^p (Y), x \mapsto f(x). $$

Then $f_n \in L^0(X_n, L^p(Y))$. We apply part (2) for $f_n$ and get $f_n \in L^0 (X_n \times Y)$. We have $L^0(X_n, L^p(Y))$ can be considered a closed subspace of $L^0 (X, L^p (Y))$ 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 $f = \lim_n \sum_{k=1}^n f_k \in L^0 (X \times Y)$.