Below we use Bochner measurability and Bochner integral. Let
- $(Y, d)$ be a separable metric space,
- $\mathcal B$ Borel $\sigma$-algebra of $Y$,
- $\nu$ a $\sigma$-finite Borel measure on $Y$,
- $(Y, \overline{\mathcal B}, \overline{\nu})$ the completion of $(Y, \mathcal B, \nu)$.
- $(E, |\cdot|)$ a Banach space,
- $L^0(Y) := L^0(Y, \nu, E)$ the space of $\nu$-measurable functions from $Y$ to $E$,
- $p \in [1, \infty)$,
- $L^p (Y) := L^p(Y, \nu, E)$ the space of $p$-integrable functions from $Y$ to $E$.
For $\delta >0$ and $f,g \in L^0 (Y)$, we write $$ \begin{align*} \{|f - g| > \delta\} &:= \{y \in Y : |f (y) - g(y)| > \delta\} \in \overline{\mathcal B}, \\ \overline{\nu} (|f - g| > \delta) &:= \overline{\nu} (\{|f - g| > \delta\}). \end{align*} $$
For $f, g \in L^0 (Y)$, we define $$ \hat \rho (f, g) := \inf_{\delta >0} \{ \overline{\nu} (|f - g| > \delta) +\delta \}. $$
Then $\hat \rho$ is an extended pseudometric on $L^0 (Y)$. By Weierstrass criterion for Banach space, we get that $(L^0 (X), \hat \rho)$ is complete. Of course, we can make $\hat \rho$ into a bona fide metric by trimming and taking quotient by the class of functions equal to $0$ $\nu$-a.e. For $f_n, f \in L^0(Y)$, we have $\hat \rho (f_n, f) \to 0$ IFF $f_n \to f$ in measure.
Let $(y_n)$ be a countable dense subset of $Y$. Let $L^p_\text{loc} (Y) := L^p_\text{loc} (Y, \nu, E)$ be the space of all functions $f \in L^0(Y)$ with the norm $$ \|f\|_{L^p_\text{loc}} := \sup_{n \ge 1} \|1_{B(y_n, 1)} f\|_{L^p} < \infty, $$ where $B(y, r)$ is the open ball centered at $y$ with radius $r>0$. Then $L^p_\text{loc} (Y)$ is complete but not separable.
Is $L^p_\text{loc} (Y)$ dense in $(L^0(Y), \hat \rho)$?
Thank you so much for your elaboration!
