Below we use Bochner measurability and Bochner integral. Let
- $(X, \mathcal A, \mu)$ be a complete 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 $$ \hat \rho (f, g) := \inf_{\delta >0} \{ \mu (|f - g| > \delta) +\delta \} \quad \forall f, g \in L^0 (X). $$
Then
Lemma $\hat \rho$ is a pseudometric on $L^0 (X)$.
Proof Clearly, $\hat \rho (f, f)=0$ and $\hat \rho (f, g) = \hat \rho (g, f)$. It remains to verify the triangle inequality. Let $f,g,h \in L^0 (X)$. Fix $\varepsilon >0$. There are $\delta_1, \delta_2 >0$ such that $$ \begin{align*} \varepsilon + \hat \rho (f, g) &> \mu (|f - g| > \delta_1) +\delta_1, \\ \varepsilon + \hat \rho (g, h) &> \mu (|g - h| > \delta_2) +\delta_2. \end{align*} $$ Notice that $|f-h| > \delta_1 + \delta_2$ implies $|f-g| > \delta_1$ or $|g-h| > \delta_2$. Then $$ \begin{align*} 2\varepsilon + \hat \rho (f, g) + \hat \rho (g, h) &> \mu (|f - g| > \delta_1) + \mu (|g-h| > \delta_2) +\delta_1 +\delta_2 \\ &\ge \mu (|f - h|> \delta_1 +\delta_2) +\delta_1 +\delta_2 \\ &\ge \hat \rho (f, h). \end{align*} $$ The claim follows by taking the limit $\varepsilon \to 0^+$.
I would like to verify a claim in this thread, i.e.,
Theorem $\hat \rho (f_n, f) \to 0$ IFF $f_n \to f$ in measure, i.e., $$ \mu ( |f_n - f| > \varepsilon) \xrightarrow{n \to \infty} 0 \quad \forall \varepsilon >0. $$
Could you have a check on my below attempt? Thank you so much for your help!
- $\implies$
Assume $\hat \rho (f_n, f) \to 0$. Fix an arbitrary $\varepsilon>0$. We will prove $$ \mu ( |f_n - f| > \varepsilon) \xrightarrow{n \to \infty} 0. $$
Fix $0 < \varepsilon' < \varepsilon$. There is $N \in \mathbb N$ such that $$ \inf_{\delta >0} \{ \mu (|f_n - f| > \delta) +\delta \} < \varepsilon' \quad \forall n >N. $$
Then there is $\delta_n >0$ such that $$ \mu (|f_n - f| > \delta_n) +\delta_n < \varepsilon' \quad \forall n >N. $$
Then for all $n > N$, we have $\delta_n < \varepsilon' <\varepsilon$ and $\mu (|f_n - f| > \delta_n) < \varepsilon'$. Then for all $n > N$, we have $\mu (|f_n - f| > \varepsilon) < \varepsilon'$. Because $\varepsilon'$ can be arbitrarily small, the claim then follows.
- $\impliedby$
Assume $f_n \to f$ in measure. Fix an arbitrary $\varepsilon >0$. It suffices to prove that $$ \limsup_n \hat \rho (f_n, f) = \limsup_n \inf_{\delta >0} \{ \mu (|f_n - f| > \delta) +\delta \} \le \varepsilon. $$
Fix some $\varepsilon' < \varepsilon/2$. Then $\mu ( |f_n - f| > \varepsilon') \xrightarrow{n \to \infty} 0$. Then there is $N \in \mathbb N$ such that $$ \mu ( |f_n - f| > \varepsilon') < \varepsilon' \quad \forall n > N. $$
Then $$ \mu ( |f_n - f| > \varepsilon') + \varepsilon' < \varepsilon \quad \forall n > N. $$
Then $$ \inf_{\delta >0} \{ \mu (|f_n - f| > \delta) +\delta \} \le \varepsilon \quad \forall n > N. $$
Then $$ \limsup_n \inf_{\delta >0} \{ \mu (|f_n - f| > \delta) +\delta \} \le \varepsilon. $$
The claim then follows.
