I'm reading about symmetric difference operator on a $\sigma$-finite measure space from Wikipedia. Below is my try to prove a statement mentioned in this article.
Let $(\Omega,\mathcal{A},\mu)$ be a finite measure space and $(E, |\cdot|)$ a separable Banach space. We define a pseudometric $d_{\mu}(A, B) := \mu(A \triangle B)$ on $\mathcal A$. Then $d_\mu$ becomes a metric when $\mathcal A$ is considered modulo the equivalence relation $A \sim B \iff \mu(A \triangle B) = 0$. Let $L_1 := L_{1}(\Omega, \mu, E)$ and $\|\cdot\|_1$ be its canonical norm.
Theorem: $d_\mu$ is separable if and only if $\|\cdot\|_1$ is separable.
Could you please have a check on my attempt?
My attempt: Let $S := S(\Omega, \mu, E)$ be the space of simple functions and $D$ a countable dense subset of $E$.
$\implies$ Assume $d_\mu$ is separable and $\mathcal D$ a countable dense subset of $\mathcal A$. Because $S$ is dense in $L_1$. It suffices to show that $S$ is separable. Let $$ S_1 := \left \{ \sum_{i=1}^m a_i 1_{A_i} \,\middle\vert\, m \in\mathbb N^*; a_1, \ldots, a_m \in D; A_1, \ldots, A_m \in \mathcal D \right \}. $$
Then $S_1 \subset S$ is countable. Let's prove that $S_1$ is dense $S$. Fix $\varepsilon >0$ and $f=\sum_{i=1}^m x_i 1_{B_i} \in S$ such that $(x_i)_{i=1}^m \subset E$ is pairwise different and $(B_i)_{i=1}^m \subset \mathcal A$ is pairwise disjoint. Then $$ \begin{align} \left \| f - \sum_{i=1}^m a_i 1_{A_i} \right \|_1 &\le \sum_{i=1}^m \left \| x_i 1_{B_i} - a_i 1_{A_i} \right \|_1 \\ &\le \sum_{i=1}^m \|(x_i-a_i) 1_{A_i}\|_1 + \|x_i (1_{B_i} - 1_{A_i})\|_1 \\ &= \sum_{i=1}^m |x_i-a_i| \mu(A_i) + |x_i| d_\mu (A_i, B_i) \\ \end{align} $$
- There are $A_i \in \mathcal D$ such that $d_{\mu}(A_i, B_i) \max_j |x_j| \le \frac{\varepsilon}{2m}$.
- There are $a_i \in D$ such that $|x_i-a_i| \mu(A_i)< \frac{\varepsilon}{2m}$.
Then $$ \left \| f - \sum_{i=1}^m a_i 1_{A_i} \right \|_1 \le \varepsilon. $$
$\impliedby$ Assume $\|\cdot\|_1$ is separable. Fix some $a \in E \setminus \{0\}$. A subset of a separable metric space is separable, so $\{ a1_A \mid A \in \mathcal A\}$ is separable. Let $S_2$ be a countable dense subset of $\{ a1_A \mid A \in \mathcal A\}$. Let $\mathcal D := \{A \in \mathcal A \mid a1_A \in S_2\}$. Let's prove that $\mathcal D$ is dense in $\mathcal A$. Fix $\varepsilon >0$ and $A \in \mathcal A$. There is $B \in \mathcal D$ such that $$ \|a1_A -a1_B\|_1 = |a| d_\mu(A, B) = |a| \mu(A \triangle B) < \varepsilon |a|. $$
This means there is $B \in \mathcal D$ such that $\mu(A \triangle B) < \varepsilon$. This completes the proof.