I'm trying to prove below statement that appears in this question.
Let $(\Omega,\mathcal{A},\mu)$ be a finite measure space. We define a pseudometric $d_\mu:\mathcal A \times \mathcal A \to [0, \infty)$ 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 $$ A \sim B \iff \mu(A \triangle B) = 0 \quad \forall A,B \in \mathcal A. $$
Theorem: $(\mathcal A, d_\mu)$ is a complete metric space.
Could you have a check on my attempt?
Is there other approach that do not require an excursion through $L_1$ space?
My attempt: Let $L_1 :=L_1(\Omega, \mu, \mathbb R)$. Consider the map $T:\mathcal A \to L_1, A \mapsto 1_A$. It follows from $1_{A \triangle B} = |1_A-1_B|$ that $T$ is an isometry. Because $L_1$ is complete, it suffices to show that $T(\mathcal A)$ is closed in $L_1$.
Let $(1_{A_n}) \subset T(\mathcal A)$ and $f \in L_1$ such that $\|1_{A_n} -f\|_1 \to 0$. There is a subsequence $\varphi$ of $\mathbb N$ such that $1_{A_{\varphi (n)}} \to f$ $\mu$-a.e. Because $\operatorname{im} (1_{A_{\varphi (n)}}) \subset \{0, 1\}$, we have $f(x) \in \{0, 1\}$ for $\mu$-a.e. $x \in \Omega$. Let $A := f^{-1} (1)$. Then $f = 1_A$ $\mu$-a.e. It follows that $\| 1_{A_{\varphi (n)}} - 1_A\|_1 \to 0$. Because convergent sequence is Cauchy, we get $\| 1_{A_{n}} - 1_A\|_1 \to 0$. This completes the proof.
