I reworded the question a bit to fit in the title, but the full question is as follows:
Let (Ω, F, µ) be a measure space with µ(Ω) = 1. For any two $A, B \in \mathcal{F}$ define $d(A, B):=\mu(A△B)$. Note that $d(A, B) = ∫ |1_A(ω) − 1_B(ω)| \mu(dω) = |1_A − 1_B|_{L_1}$ , which shows that d defines a (pseudo–)metric on F. Prove that in the (pseudo–)metric space (F, d) for every Cauchy sequence $(A_n)_{n∈\mathbb{N}}$ there exists $A \in\mathcal{F}$ such that $\lim_{n→∞} d(A_n, A) = 0$.
Now, this is what I have so far:
Let $\{C_n\}$ be a Cauchy sequence in $\mathcal{F}$. That is, $\forall\epsilon>0, \exists N$ such that $\forall n,m>N$, we have $d(C_n,C_m)<\epsilon$. Well, as noted above, $d(C_n,C_m)=\int|1_{C_n}-1_{C_m}|d\mu=|1_{C_n}-1_{C_m}|_{L_1}$, thus $\{C_n\}$ is Cauchy in $L_1$, which itself is complete, thus there exists an $f\in L_1$ such that $\lim_{n\rightarrow\infty}|1_{C_n}-f|_{L_1}=0$. Let $A=\{x|f(x)=1\}$. Then $f=1_A$ $\mu$-a.e., thus $\lim_{n\rightarrow\infty} d(C_n,A)=0$.
My concerns, how do we know $A$ is measurable? I believe this is a direct consequence of knowing that $f$ is measurable, we have $f$ is measurable if and only if $\{x|f(x)\geq(\text{and})\leq r\}$ is measurable for any $r\in\mathbb{R}$, so let $r=1$ and take the intersection of both cases, which must then be measurable by the axioms of $\sigma$-algebras. Also, when I said $f=1_A$ $\mu$-a.e., I certainly believe this, but is there a (necessary) argument for why? Because convergence in $L_1$ implies convergence in measure?