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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. First, I will verify that $S(X)$ is dense in $(L^0 (X), \rho)$. Fix $f \in L^0 (X)$. There is a sequence $(f_n) \subset S(X)$ such that $f_n \to f$ a.e. Then $\min\{|f_n-f|, 1\} \to 0$ a.e. By dominated convergence theorem, $\min\{|f_n-f|, 1\} \to 0$ in $L^1(X)$. Second, I want to verify that

Theorem Let $(f_n)$ be a Cauchy sequence in $S (X)$. Then there is a subsequence $(f_{n_k})$ and $f \in L^0(X)$ such that

  • $f_{n_k} \xrightarrow{k \to \infty} f$ a.e.
  • for each $\varepsilon>0$ there is $A_\varepsilon \in \mathcal A$ such that $\mu(A_\varepsilon) < \varepsilon$ and that $(f_{n_k})$ converges uniformly to $f$ on $A_\varepsilon^c$.

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 $$ \rho (f_n, f_m) \le 2^{-2k} \quad \forall n,m \ge j_k. $$

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 $$ f(x) := \begin{cases} \lim_m g_m (x) &\text{if} \quad x \in A^c,\\ 0 &\text{otherwise}, \end{cases} $$ is well-defined and $\mu$-measurable. Clearly,

  • $f_{j_m} \xrightarrow{m \to \infty} f$ 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 $f$ on $A_n^c$.
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