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 finite 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 $\nu$,
- $(Z, \overline{\mathcal C}, \overline{\lambda})$ the completion of $(Z, \mathcal C, \lambda)$,
- $L^0 (X) := L^0 (X, \mathbb R)$ the space of measurable functions from $X$ to $\mathbb R$.
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 a $\nu$-null set. Then by dominated convergence theorem (DCT), $$ \|f\|_{L^p_{\text{loc}}} = \sup_{m \in \mathbb N} \|1_{B(y_m, 1)} f\|_{L^p} \quad \forall f \in L^p_{\text{loc}} (Y). $$
Let $\rho_Y$ be a pseudometric on $L^0(Y)$ defined by $$ \rho_Y (f, g) := \int_Y \min\{|f-g|, 1\} \, \mathrm d \nu \quad \forall f, g \in L^0 (Y). $$
We define $\rho_Z$ on $L^0(Z)$ similarly. I would like to prove an analogue of this result, i.e.,
Theorem 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 $(L^p_{\text{loc}} (Y), \| \cdot\|_{L^p_{\text{loc}} (Y)})$. Then $(f_n)$ is a Cauchy sequence in $(L^0 (Z), \rho_Z)$.
Could you have a check on my below attempt? Thank you so much for your help!
For simplicity, we write $$ \nu (|f - g| > \delta) := \nu (\{y \in Y : |f (y) - g(y)| > \delta\}) \quad \forall \delta >0, \forall f,g \in L^0 (Y). $$
Let $\hat \rho_Y$ be another pseudometric on $L^0(Y)$ defined by $$ \hat \rho_Y (f, g) := \inf_{\delta >0} \{ \nu (|f - g| > \delta) +\delta \} \quad \forall f, g \in L^0 (Y). $$
We define $\hat\rho_Z$ on $L^0(Z)$ similarly. We have some useful properties, i.e.,
Lemma 1 $(L^0(Y), \rho_Y)$ is complete.
Lemma 2 $(L^0(Y), \hat \rho_Y)$ is complete.
Lemma 3 The following statements are equivalent:
- (S1) $\rho_Y (f_n, f) \to 0$.
- (S2) Every subsequence of $(f_n)$ has a further subsequence that converges to $f$ $\nu$-a.e.
- (S3) $f_n \to f$ in measure, i.e., $$ \nu (\{y \in Y : |f_n (y)-f(y)| > \varepsilon\}) \xrightarrow{n \to \infty} 0 \quad \forall \varepsilon>0. $$
- (S4) $\hat \rho_Y (f_n, f) \to 0$.
Proof
Now we are going to prove our theorem. Assume the contrary that $(f_n)$ is not a Cauchy sequence in $(L^0 (Z), \rho_Z)$. By above Lemmas $(f_n)$ is not a Cauchy sequence in $(L^0 (Z), \hat \rho_Z)$. There is $\varepsilon >0$ and two subsequences $\varphi, \psi$ of $\mathbb N$ such that $$ \hat \rho_Z (f_{\varphi (n)}, f_{\psi (n)}) \ge \varepsilon \quad \forall n \in \mathbb N. $$
Then $$ \lambda (|f_{\varphi (n)} - f_{\psi (n)}| > \delta) +\delta \ge \varepsilon \quad \forall n \in \mathbb N, \forall \delta >0. $$
We pick $\delta := \varepsilon/2$ and get $$ \lambda (|f_{\varphi (n)} - f_{\psi (n)}| > \varepsilon/2) \ge \varepsilon/2 \quad \forall n \in \mathbb N. $$
Let $g_n := f_{\varphi (n)} - f_{\psi (n)}$. Then $g_n \in L^0(Z)$. We define $$ h_n: X \to \mathbb R, x \mapsto \| g_n (x, \cdot) \|_{L^p_{\text{loc}}}. $$
It has been verified that $h_n$ indeed belongs to $L^0 (X)$. By our assumption, $h_n \to 0$ $\mu$-a.e.
Lemma 4 Let $g_n \in L^0(Z)$ for all $n \in \mathbb N$ such that for $\mu$-a.e. $x \in X$ we have $(g_n(x, \cdot))_n \subset L^p_{\text{loc}} (Y)$. We define $$ h_n: X \to \mathbb R, x \mapsto \| g_n (x, \cdot) \|_{L^p_{\text{loc}}}. $$ If $\rho_{X} (h_n, 0) \to 0$ then $\rho_Z (g_n, 0) \to 0$.
By Lemma 4, $g_n \to 0$ in measure, which is a contradiction. This completes the proof.
