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 $\delta >0$, 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 $$ \hat \rho (f, g) := \inf_{\delta >0} \{ \mu (|f - g| > \delta) +\delta \}. $$
Then $\rho$ is an extended pseudometric on $L^0 (X)$. Let $f_n, f \in L^0(X)$. Then $\hat \rho (f_n, f) \to 0$ IFF $f_n \to f$ in measure, i.e., $$ \mu ( |f_n - f| > \delta) \xrightarrow{n \to \infty} 0 \quad \forall \delta >0. $$
I would like to verify a claim in this Wikipedia page, i.e.,
Theorem Assume $\hat \rho (f_n, f) \to 0$. There is a subsequence $\varphi$ of $\mathbb N$ such that $f_{\varphi (n)} \xrightarrow{n \to \infty} f$ $\mu$-a.e.
Could you have a check on my below attempt? Thank you so much for your help!
For each $k \in \mathbb N$, there is $j_k \in \mathbb N$ such that $$ \hat \rho (f_n, f) < 2^{-k} \quad \forall n \ge j_k. $$
WLOG, we can assume $(j_k)$ is strictly increasing in $k$. Let $g_k := f_{j_k}$. Then $$ \hat \rho (g_n, f) < 2^{-n} \quad \forall n \ge 1. $$
There is $\delta_n >0$ such that $$ \mu (|g_n - f| > \delta_n) +\delta_n < 2^{-n}. $$
Then $\delta_n < 2^{-n}$ and $\mu (|g_n - f| > \delta_n) < 2^{-n}$. Let $A_n := \bigcup_{k=n}^\infty \{|g_k - f| > \delta_k\}$. Then $\mu(A_n) \le 2^{-n+1}$. Let $A := \bigcap_{n=1}^\infty A_n$. Then $A$ is a $\mu$-null set. We claim that $g_n \to g$ on $A^c$.
Fix $x \in A^c = \bigcup_{n=1}^\infty A_n^c$. There is some $m \ge 1$ such that $x \in A_m^c = \bigcap_{k=m}^\infty \{|g_k - f| \le \delta_k\}$. Then $\lim_k |g_k(x) - f(x)| \le \lim_k \delta_k \le \lim_k 2^{-k} =0$. This completes the proof.
