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. I already proved that $S (X)$ is dense in $L^0 (X)$ and that $L^0(X)$ is complete. I would like to verify below statement found in this Wikipedia page, i.e.,
$(f_n)$ converges to $f$ in $(L^0 (X), \rho)$ IFF for every $\varepsilon>0$ $$ \mu (\{x \in X : |f_n (x)-f(x)| > \varepsilon\}) \xrightarrow{n \to \infty} 0. $$
Could you have a check on my below attempt? Thank you so much for your help!
WLOG, we assume $\mu$ is a probability measure. Then $$ \mu (\{x \in X : |f_n (x)-f(x)| > \varepsilon\}) \xrightarrow{n \to \infty} 0 \quad \forall \varepsilon>0, $$ is precisely the definition of $f_n \to f$ in probability. It remains to prove that $f_n \to f$ in probability IFF $(f_n)$ converges to $f$ in $(L^0 (X), \rho)$.
- $\implies$
Assume $f_n \to f$ in probability. Let $(f_{n_k})$ be an arbitrary subsequence of $(f_n)$. Let $g_k := f_{n_k}$. Clearly, $g_{k} \to f$ in probability. It's well-known that there is a subsequence $(g_{k_m})$ such that $g_{k_m} \xrightarrow{m \to \infty} f$ a.s. The claim then follows from below result, i.e.,
Lemma $(f_n)$ converges to $f$ in $(L^0 (X), \rho)$ IFF every subsequence has in turn a further subsequence that converges to $f$ a.e.
- $\impliedby$
Assume $(f_n)$ converges to $f$ in $(L^0 (X), \rho)$. Fix $\varepsilon>0$ and let $$ r_n := \mu (\{x \in X : |f_n (x)-f(x)| > \varepsilon\}) \quad \forall n \in \mathbb N. $$
It suffices to prove $r_n \to 0$. Let $(f_{n_k})$ be an arbitrary subsequence of $(f_n)$. Let $g_k := f_{n_k}$. By above Lemma, there is a subsequence $(g_{k_m})$ such that $g_{k_m} \xrightarrow{m \to \infty} f$ a.s. It's well-known that $g_{k_m} \xrightarrow{m \to \infty} f$ in probability and thus $$ r_{n_{k_m}} \xrightarrow{m \to \infty} 0. $$
To sum up, any subsequence of $(r_n)$ has a further subsequence that converges to $0$. Then $(r_n)$ converges to $0$.
