Below we use Bochner measurability and Bochner integral. Let
- $(X, \mathcal A, \mu)$ be a complete $\sigma$-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 $\delta >0$, we write $$ \begin{align*} \{|f - g| > \delta\} &:= \{x \in X : |f (x) - g(x)| > \delta\}, \\ \mu (|f - g| > \delta) &:= \mu (\{|f - g| > \delta\}). \end{align*} $$
For $f, g \in L^0 (X)$, we define $$ \begin{align*} \rho (f, g) &:= \int_X \min\{|f-g|, 1\} \, \mathrm d \mu,\\ \hat \rho (f, g) &:= \inf_{\delta >0} \{ \mu (|f - g| > \delta) +\delta \}. \end{align*} $$
Then $\rho$ and $\hat \rho$ are extended pseudometrics on $L^0 (X)$. Let $f_n, f \in L^0(X)$. We call
- (S1) $f_n \to f$ in measure, i.e., $$ \mu ( |f_n - f| > \delta) \xrightarrow{n \to \infty} 0 \quad \forall \delta >0. $$
- (S2) $\hat\rho (f_n, f) \to 0$.
- (S3) $\rho (f_n, f) \to 0$.
I already proved that (S1) $\iff$ (S2). For $\delta \in (0, 1)$, we have $$ \begin{align*} \rho (f_n, f) &= \int_{\{|f_n - f| > \delta\}} \min\{|f_n-f|, 1\} \, \mathrm d \mu + \int_{\{|f_n - f| \le \delta\}} \min\{|f_n-f|, 1\} \, \mathrm d \mu \\ &\ge \delta \mu (|f_n - f| > \delta) + \int_{\{|f_n - f| \le \delta\}} \min\{|f_n-f|, 1\} \, \mathrm d \mu. \end{align*} $$
Then (S3) $\implies$ (S1). If $\mu(X) < \infty$ then (S3) $\iff$ (S1).
Is it true that (S1) $\implies$ (S3)? If not, can we "slightly" modify the definition of $\rho$ such that (S1) $\implies$ (S3)?
Thank you so much for your elaboration!
