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 $Y$. The 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 $\rho_Z$ be a pseudometric on $L^0(Z)$ defined by $$ \rho_Z (f, g) := \int_Z \min\{|f-g|, 1\} \, \mathrm d \lambda \quad \forall f, g \in L^0 (Z). $$
Then convergence in $\rho_Z$ is equivalent to that in measure. We define $\rho_X$ on $L^0 (X)$ similarly. I'm trying to prove an analogue of this result, i.e.,
Theorem Let $f_n \in L^0(Z)$ for all $n \in \mathbb N$ such that for $\mu$-a.e. $x \in X$ we have $(f_n(x, \cdot))_n \subset L^p_{\text{loc}} (Y)$. We define $$ g_n: X \to \mathbb R, x \mapsto \| f_n (x, \cdot) \|_{L^p_{\text{loc}}}. $$ If $\rho_{X} (g_n, 0) \to 0$ then $\rho_Z (f_n, 0) \to 0$.
I have verified that $g_n$ indeed belongs to $L^0 (X)$. Could you have a check on my below attempt? Thank you so much for your help!
We define $c:[0, \infty] \to[0, 1]$ by $$ c(m) := \inf \{\min\{ \| 1_{B} \|_{L^p_{\text{loc}}}, 1\} : B \in \mathcal{B} \text{ s.t. } \nu (B) \ge m\}. $$
Then $c$ is non-decreasing and thus Borel measurable. Clearly, $\| 1_{B} \|_{L^p_{\text{loc}}} \ge c (\nu (B))$ for all $B \in \cal B$. Let's verify $c(m)>0$ for all $m>0$. Assume the contrary that $c(m_0)=0$ for some $m_0>0$. There is a sequence $(B_n) \subset \cal B$ such that $\inf_n \nu(B_n) \ge m_0$ and $\| 1_{B_n} \|_{L^p_{\text{loc}}} \xrightarrow{n \to \infty} 0$. Then there is a subsequence $(n_k)$ such that $1_{B_{n_k}} \xrightarrow{k \to \infty} 0$ $\nu$-a.e. By dominated convergence theorem, $\nu (B_{n_k}) \xrightarrow{k \to \infty} 0$. This is a contradiction.
Let $\varepsilon >0$ and $M_n := \{ (x, y) \in Z: |f_n (x, y)| \ge \varepsilon \} \in \overline{\cal C}$. Then $$ \begin{align*} g_n (x) &\ge \| \varepsilon 1_{M_n} (x, \cdot)\|_{L^p_{\text{loc}} (Y)} \\ &\ge \varepsilon c (\nu ( \{y \in Y : (x, y) \in M_n\} )) \\ &= \varepsilon c \bigg ( \int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu \bigg ). \end{align*} $$
Then $c ( \int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu ) \xrightarrow{n \to \infty} 0$ in $\rho_X$. Then $c ( \int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu ) \xrightarrow{n \to \infty} 0$ in measure. For any $\delta>0$, we have $c(\delta)>0$. So $$ \mu \left ( \bigg \{x \in X : c \bigg ( \int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu \bigg ) \ge c(\delta) \bigg \} \right ) \xrightarrow{n \to \infty} 0 \quad \forall \delta >0, $$ which implies by monotonicity of $c$, $$ \mu \left ( \bigg \{x \in X : \int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu \ge \delta \bigg \} \right ) \xrightarrow{n \to \infty} 0 \quad \forall \delta >0, $$ i.e., $\int_Y 1_{M_n} (x, \cdot) \, \mathrm d \nu \xrightarrow{n \to \infty} 0$ in measure. Dominated convergence theorem also holds if almost everywhere convergence is replaced by convergence in measure. Then $$ \int_X\int_Y 1_{M_n} (x, y) \, \mathrm d \nu (y) \, \mathrm d \mu (x) \xrightarrow{n \to \infty} 0, $$ or equivalently $\lambda (M_n) \to 0$. The claim then follows.
