Let $(X, \mathcal A, \mu)$ be a finite measure space. We define a pseudometric $d_\mu$ by $d_{\mu}(A, B) := \mu(A \triangle B)$. Then $d_\mu$ becomes a metric when $\mathcal A$ is considered modulo the equivalence relation $\sim$ defined by $$ A \sim B \iff \mu(A \triangle B) = 0 \quad \forall A,B \in \mathcal A. $$
I'm trying to fill in the details in this answer.
Theorem: If the $\sigma$-algebra $\mathcal A$ is generated by a countable collection $\mathcal B \subset \mathcal A$, then $d_\mu$ is separable.
Could you have a check on my proof?
My attempt: Let $\mathcal C$ be the smallest algebra (over $X$) containing $\mathcal B$. Because $\mathcal B$ is countable, so is $\mathcal C$. Next we prove that $\overline{\mathcal C} = \mathcal A$, i.e., $\mathcal C$ is dense in $\mathcal A$. Because $\mathcal B \subset \mathcal C$, it is enough to prove that $\overline{\mathcal C}$ is a $\sigma$-algebra over $X$.
Assume $A \in \overline{\mathcal C}$, i.e., there is a sequence $(A_n) \subset \mathcal C$ such that $d_\mu (A_n, A) < 1/n$. Then $(A_n^c) \subset \mathcal C$ and $$ d_\mu (A^c_n, A^c) = \int |1_{A^c_n} - 1_{A^c}| = \int |(1-1_{A_n}) - (1-1_{A})| = \int |1_{A_n} - 1_{A}| = d_\mu (A_n, A) < 1/n. $$
Hence $A^c \in \overline{\mathcal C}$, i.e., $\overline{\mathcal C}$ is closed under taking complement.
Let $(A_n) \subset \overline{\mathcal C}$. We will prove that $A :=\bigcup_n A_n \in \overline{\mathcal C}$. Fix $\varepsilon>0$. There is a sequence $(B_n) \subset \mathcal C$ such that $d_\mu (B_n, A_n) < \varepsilon 2^{-n}$. From this Wikipedia page, we have
$$ \left(\bigcup_{\alpha \in \mathcal{I}} A_{\alpha}\right) \triangle\left(\bigcup_{\alpha \in \mathcal{I}} B_{\alpha}\right) \subseteq \bigcup_{\alpha \in \mathcal{I}}\left(A_{\alpha} \triangle B_{\alpha}\right), $$ where $\mathcal{I}$ is an arbitrary non-empty index set.
It follows that $$ d_\mu \left ( \bigcup_{i=1}^n B_i, \bigcup_{i=1}^n A_i \right ) \le \sum_{i=1}^n d_\mu (B_i, A_i) \le \varepsilon \quad \forall n \in \mathbb N. $$
By continuity of measure from below, we have $$ \lim_n \mu \left ( \bigcup_{i=1}^n A_i \right ) = \mu (A) < \infty. $$
It follows that there is $N$ such that $$ d_\mu \left ( \bigcup_{i=1}^N A_i, A \right ) = \mu \left (A \setminus \left ( \bigcup_{i=1}^N A_i \right ) \right ) < \varepsilon. $$
Hence $$ d_\mu \left ( \bigcup_{i=1}^N B_i, A \right ) \le d_\mu \left ( \bigcup_{i=1}^N B_i, \bigcup_{i=1}^N A_i \right ) + d_\mu \left ( \bigcup_{i=1}^N A_i, A \right ) < 2 \varepsilon. $$
This completes the proof.
