Below we use Bochner measurability and Bochner integral. Let
- $(X, \mathcal A, \mu)$ be a complete finite measure space,
- $(E, | \cdot |)$ a Banach space,
- $S (X)$ the space of $\mu$-simple functions from $X$ to $E$, and
- $L^0 (X)$ the space of $\mu$-measurable functions from $X$ to $E$.
- $L^1 (X)$ the space of $\mu$-integrable functions from $X$ to $E$.
We define $$ \rho (f, g) := \int_X \min\{|f-g|, 1\} \, \mathrm d \mu \quad \forall f, g \in L^0 (X). $$
Then $\rho$ is a pseudometric on $L^0 (X)$ such that $\rho (f, g) = 0$ IFF $f=g$ a.e. First, we verify that $S(X)$ is dense in $L^0 (X)$ w.r.t. $\rho$. Fix $f \in L^0 (X)$. There is a sequence $(f_n) \subset S(X)$ such that $f_n \to f$ a.e. Then $\min\{|f_n-f|, 1\} \to 0$ a.e. By dominated convergence theorem, $\min\{|f_n-f|, 1\} \to 0$ in $L^1(X)$. Second, we verify that $\rho$ is complete.
Could you have a check on my below attempt? Thank you so much for your help!
Fix a Cauchy sequence $(f_n)$ in $L^0 (X)$. There is $g_n \in S(X)$ such that $\rho(g_n, f_n) < 1/n$. Then $(g_n)$ is a Cauchy sequence in $S(X)$.
Theorem Let $(f_n)$ be a Cauchy sequence in $S (X)$. Then there is a subsequence $(f_{n_k})$ and $f \in L^0(X)$ such that
- $f_{n_k} \xrightarrow{k \to \infty} f$ a.e.
- for each $\varepsilon>0$ there is $A_\varepsilon \in \mathcal A$ such that $\mu(A_\varepsilon) < \varepsilon$ and that $(f_{n_k})$ converges uniformly to $f$ on $A_\varepsilon^c$.
By above Theorem, then there is a subsequence $(g_{n_k})$ and $g \in L^0(X)$ such that $g_{n_k} \xrightarrow{k \to \infty} g$ a.e. and thus $|g_{n_k} -g| \xrightarrow{k \to \infty} 0$ a.e. By dominated convergence theorem, $\min\{|g_{n_k} - g|, 1\} \xrightarrow{k \to \infty} 0$ in $L^1(X)$ and thus $\rho (g_{n_k}, g) \xrightarrow{k \to \infty} 0$. By triangle inequality, $$ \rho(f_{n_k}, g) \le \rho(f_{n_k}, g_{n_k}) + \rho (g_{n_k}, g) \le \frac{1}{n_k} + \rho (g_{n_k}, g) \xrightarrow{k \to \infty} 0. $$
The claim then follows.
