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 $E$.
For simplicity, we write $$ \mu (|f - g| > \delta) := \mu (\{x \in X : |f (x) - g(x)| > \delta\}) \quad \forall \delta >0. $$
Notice that the completeness of $(X, \mathcal A, \mu)$ implies $\{x \in X : |f (x) - g(x)| > \delta\} \in \mathcal A$. We define $$ \hat \rho (f, g) := \inf_{\delta >0} \{ \mu (|f - g| > \delta) +\delta \} \quad \forall f, g \in L^0 (X). $$
I would like to verify that
Theorem $(L^0 (X), \hat \rho)$ is complete.
Could you have a check on my below attempt? Thank you so much for your help!
First, we define another pseudometric $\rho$ on $L^0 (X)$ by $$ \rho (f, g) := \int_X \min\{|f-g|, 1\} \, \mathrm d \mu \quad \forall f, g \in L^0 (X). $$
Then
Lemma
- $(L^0(Y), \rho)$ is complete and $S(X)$ is dense in $(L^0 (X), \rho)$.
- The following statements are equivalent:
- (S1) $\rho (f_n, f) \to 0$,
- (S2) Every subsequence of $(f_n)$ has a further subsequence that converges to $f$ $\mu$-a.e.
- (S3) $f_n \to f$ in measure, i.e., $$ \mu (\{x \in X : |f_n (x)-f(x)| > \varepsilon\}) \xrightarrow{n \to \infty} 0 \quad \forall \varepsilon>0, $$
- (S4) $\hat \rho (f_n, f) \to 0$.
Proof The proof of the completeness of $\rho$ can be found here. The proof of (S1) $\iff$ (S2) can be found here. The proof of (S3) $\iff$ (S4) can be found here. The proof of (S1) $\iff$ (S3) can be found here.
It follows from above Lemma that $(L^0(Y), \rho)$ is complete and that $\rho, \hat \rho$ induce the same topology on $L^0 (X)$. Let $(f_n)$ be a Cauchy sequence in $(L^0 (X), \hat \rho)$. It suffices to prove that $(f_n)$ is a Cauchy sequence in $(L^0 (X), \rho)$.
Fix $\varepsilon >0$. There is $N \in \mathbb N$ such that $$ \hat \rho (f_n, f_m) < \varepsilon \quad \forall n > m \ge N. $$
There is $\delta_{n, m} >0$ such that $$ \mu (|f_n - f_m| > \delta_{n, m}) +\delta_{n, m} < \varepsilon \quad \forall n >m \ge N. $$
For all $n >m \ge N$, we have $\delta_{n, m} < \varepsilon$ and $\mu ( |f_n - f_m| > \varepsilon ) < \varepsilon$. For all $n >m \ge N$, $$ \begin{align*} \rho (f_n, f_m) &= \int_X \min\{|f_n-f_m|, 1\} \, \mathrm d \mu \\ &= \int_{\{|f_n - f_m| > \varepsilon\}} \min\{|f_n-f_m|, 1\} \, \mathrm d \mu + \int_{\{|f_n - f_m| \le \varepsilon\}} \min\{|f_n-f_m|, 1\} \, \mathrm d \mu \\ &\le \mu ( |f_n - f_m| > \varepsilon ) + \varepsilon \mu(X) \\ &\le \varepsilon + \varepsilon \mu(X) \\ &= (1 + \mu(X)) \varepsilon. \end{align*} $$
This completes the proof.
