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.,
$\left(f_n\right)$ converges to $f$ in $(L^0 (X), \rho)$ IFF every subsequence has in turn a further subsequence that converges to $f$ a.e.
Could you have a check on my below attempt? Thank you so much for your help!
- $\implies$
Let $f_n, f \in L^0 (X)$ such that $\rho(f_n, f) \to 0$. In particular, $(f_n)$ is a Cauchy sequence. For each $k \in \mathbb N$ there is $j_k \in \mathbb N$ such that $$ \rho (f_n, f_m) \le 2^{-2k} \quad \forall n,m \ge j_k. $$
WLOG, we can assume $(j_k)$ is strictly increasing in $k$. Let $g_k := f_{j_k}$. Then $$ \rho (g_n, g_m) \le 2^{-2n} \quad \forall m \ge n. $$
Let $$ B_n := \{x \in X : 2^{-n} \le |g_{n+1} (x) - g_n (x)|\}. $$
Because $(X, \mathcal A, \mu)$ is complete, $B_n \in \mathcal A$. We have $$ 2^{-n} \mu(B_n) = \int_{B_n} 2^{-n} \, \mathrm d \mu \le \int_{B_n} \min \{|g_{n+1} - g_n |, 1\} \, \mathrm d \mu \le \rho (g_{n+1}, g_n) \le 2^{-2n}. $$
Then $\mu(B_n) \le 2^{-n}$. Let $A_n := \bigcup_k B_{n+k}$. Then $\mu(A_n) \le2^{-n+1}$. Then $A := \bigcap_n A_n$ is a $\mu$-null set. If $x \in A_n^c = \bigcap_k B_{n+k}^c$, then $$ |g_{m+1} (x) - g_m (x)| < 2^{-m} \quad \forall m \ge n $$
By Weierstrass criterion, the series $$ g_0 + \sum_m (g_{m+1} - g_m) $$ on $A_n^c$ converges uniformly in $E$. Then $$ g(x) := \begin{cases} \lim_m g_m (x) &\text{if} \quad x \in A^c,\\ 0 &\text{otherwise}, \end{cases} $$ is well-defined. The a.e. limit of a sequence of $\mu$-measurable functions is again $\mu$-measurable. So $g \in L^0(X)$. Clearly,
- $f_{j_m} \xrightarrow{m \to \infty} g$ a.e.
- for every $\varepsilon>0$, there is $n \in \mathbb N$ such that $\mu(A_n) \le2^{-n+1} < \varepsilon$ and thus $(f_{j_m})$ converges uniformly to $g$ on $A_n^c$.
By dominated convergence theorem, $\min\{|f_{j_m} - g|, 1\} \xrightarrow{m \to \infty} 0$ in $L^1(X)$ and thus $\rho (f_{j_m}, g) \xrightarrow{m \to \infty} 0$. Then $g =f$ a.e.
- $\impliedby$
Let $f_n, f \in L^0(X)$. Let $(f_{n_k})$ be an arbitrary subsequence of $(f_n)$. Let $g_k := f_{n_k}$. By our hypothesis, $(g_k)$ has a subsequence $(g_{k_m})$ such that $g_{k_m} \xrightarrow{m \to \infty} f$ a.e. By dominated convergence theorem, $\min\{|g_{k_m} - f|, 1\} \xrightarrow{m \to \infty} 0$ in $L^1(X)$ and thus $\rho (f_{k_m}, f) \xrightarrow{m \to \infty} 0$. To sum up, any subsequence of $(f_n)$ has a further subsequence that converges to $f$ in $(L^0 (X), \rho)$. Then $(f_n)$ converges to $f$ in $(L^0 (X), \rho)$.
