Let $(X, \mathcal A, \mu)$ be a measure space and $(E, | \cdot |)$ a Banach space.
$f \in E^{X}$ is called $\mu$-simple if $f = \sum_{k=1}^n e_k 1_{A_k}$ where $e_k \in E$ and $(A_k)_{k=1}^n$ is a finite sequence of pairwise disjoint sets with finite measure in $\mathcal A$. The integral of such $f$ w.r.t. $\mu$ is defined by $\int_{X} f \mathrm d \mu := \sum_{k=1}^n e_k \mu(A_k)$. Let $\mathcal S (X, \mu, E)$ be the space of such $\mu$-simple functions. We define a semi-norm $\|\cdot\|_1$ on $\mathcal S (X, \mu, E)$ by $\|f\|_1 :=\int_{X} |f| \mathrm d \mu$. The notion of Cauchy sequence is thus applicable for $(\mathcal S (X, \mu, E), \| \cdot \|_1)$.
$f \in E^{X}$ is called $\mu$-measurable if $f$ is a $\mu$-a.e. limit of a sequence $(f_n) \subset \mathcal S (X, \mu, E)$. Let $\mathcal L_0 (X, \mu, E)$ be the space of such $\mu$-measurable functions.
$f \in E^{X}$ is called $\mu$-integrable if $f$ is $\mu$-a.e. limit of a Cauchy sequence $(f_n) \subset \mathcal S (X, \mu, E)$. The integral of such $f$ w.r.t. $\mu$ is defined by $\int_{X} f \mathrm d \mu := \lim_n \int_{X} f_n \mathrm d \mu$. Let $\mathcal L_1 (X, \mu, E)$ be the space of such $\mu$-integrable functions. We also define a semi-norm $\|\cdot\|_1$ on $\mathcal L_1 (X, \mu, E)$ by $\| f\|_1 := \int_{X} |f| \mathrm d \mu$.
Previously, I proved that
Theorem 1: Let $(X, \mathcal A', \mu')$ be the completion of $(X, \mathcal A, \mu)$. A function $f \in E^X$ is $\mu$-measurable if and only if $f$ is $\mu'$-measurable.
Now I try to prove below complementary result, i.e.,
Theorem 2: Let $(X, \mathcal A', \mu')$ be the completion of $(X, \mathcal A, \mu)$. A function $f \in E^X$ is $\mu$-integrable if and only if $f$ is $\mu'$-integrable.
Could you have a check on my attempt?
My attempt: Let $\mathcal Z$ be the collection of all subsets of the $\mu$-null subsets of $X$.
Because $\mathcal A \subset \mathcal A'$ and $\mu' \restriction \mathcal A = \mu$, the direction $\implies$ then follows. Let's prove the reverse direction. Fix a $\mu'$-integrable function $f$. There is a Cauchy sequence $(f_n) \subset \mathcal S (X, \mu', E)$ such that $f_n \to f$ $\mu'$-a.e.
Assume $f_n = \sum_{i=1}^{\varphi (n)} e_{ni} 1_{A_{ni}}$ with $(A_{ni})_{i=1}^{\varphi (n)}$ a finite sequence of pairwise disjoint sets with finite measure in $\mathcal A'$. By construction of $\mathcal A'$, there exist $B_{ni} \in \mathcal A$ and $C_{ni} \in \mathcal Z$ such that $A_{ni} = B_{ni} \cup C_{ni}$ and $\mu(B_{ni}) = \mu'(A_{ni}) < \infty$. Let $g_n := \sum_{i=1}^{\varphi (n)} e_{ni} 1_{B_{ni}}$. Then $(g_n) \subset \mathcal S (X, \mu, E)$.
- $g_n \to f$ $\mu$-a.e.
Let $N'$ be a $\mu'$-null set such that $f_n \to f$ on $(N')^c$. Notice that $\{x \in X \mid f_n (x) \neq g_n (x)\} \subset C_n :=\bigcup_{i=1}^{\varphi (n)} C_{ni}$. Let $N'' := N' \cup \bigcup_n C_n$. Then $N''$ is a $\mu'$-null set. It follows that $g_n \to f$ on $A :=(N'')^c$. There exist $B \in \mathcal A$ and $C \in \mathcal Z$ such that $A = B \cup C$ and $\mu(B) = \mu'(A) = 1$. It follows that $g_n \to f$ on $B$.
- $(g_n)$ is a Cauchy sequence.
Let $\| \cdot \|_1$ and $\| \cdot \|_1'$ be the semi-norms of $\mathcal S (X, \mu, E)$ and $\mathcal S (X, \mu', E)$ respectively. Notice that $\mathcal S (X, \mu, E) \subset \mathcal S (X, \mu', E)$, so $(g_n) \subset \mathcal S (X, \mu', E)$. We have $$ \begin{align} \|f_n-f_m\|_1' &= \|(f_n - g_n) - (f_m - g_m) + (g_n-g_m)\|_1' \\ &\le \|f_n - g_n\|_1' +\| f_m - g_m \|_1'+ \|g_n-g_m\|_1' . \end{align} $$ and $$ \begin{align} \|f_n-f_m\|_1' &= \|(f_n - g_n) - (f_m - g_m) + (g_n-g_m)\|_1' \\ &\ge \|g_n-g_m\|_1' - \|f_n - g_n\|_1' - \| f_m - g_m \|_1'. \end{align} $$
We have $$ \|f_n - g_n\|_1' \le \sum_{i=1}^{\varphi (n)} \|e_{ni} 1_{C_{ni}}\|_1' = \sum_{i=1}^{\varphi (n)} |e_{ni}| \mu'(C_{ni}) = 0. $$
Similarly, $\|f_m - g_m\|_1' \le 0$. Also, $$ \|g_n-g_m\|_1' = \|g_n-g_m\|_1. $$
It follows that $\|g_n-g_m\|_1 = \|f_n-f_m\|_1'$. This completes the proof.
