Proving that a modified $\sigma$-algebra is a complete metric space

Let $(X,\mathfrak M,\mu)$ be a measure space with $\sigma$-algebra $\mathfrak M$, $\mu$ a positive measure, and $\mu(X)<\infty$. We define $\rho:\mathfrak M\times\mathfrak M\to\mathbb R$ by $$\rho(A,B)=\int_X|\chi_A-\chi_B|\,d\mu,$$ where $\chi_E$ is the characteristic function of the set $E$. We can define a relation $\sim$ by $A\sim B$ if $\rho(A,B)=0$. It is not too hard to see that this is an equivalence relation, so we may consider the set of equivalence classes $M$, and view $\rho$ as a function $\rho:M\times M\to\mathbb R$. We conclude that $\rho$ is a metric on $M$. The trouble comes in showing that $(M,\rho)$ is a complete metric space. Given an arbitrary cauchy sequence $\{A_n\}\subset M$, to what might it converge? My guess, in which I am rather confident, is that this sequence will converge to $$A=\bigcup_{n=1}^\infty\bigcap_{j=n}^\infty A_j.$$ It is clear that $A\in \mathfrak M$, so it is represented in $M$. But I am having trouble showing that $\{A_n\}\to A$ in the metric. I have boiled it down to: \begin{align} \lim_{n\to\infty}\int_X|\chi_A-\chi_{A_n}|\,d\mu&=\lim_{n\to\infty}\int_X|\lim_{m\to\infty}\chi_{\bigcap_{j=m}^\infty A_j}-\chi_{A_n}|\,d\mu\\ &\text{Applying Lebesgue's Dominated Convergence Theorem,}\\ &=\lim_{n,m\to\infty}\int_X\left|\chi_{\bigcap_{j=m}^\infty A_j}-\chi_{A_n}\right|\,d\mu\\ &=\lim_{n,m\to\infty}\int_X\left|\prod_{j=m}^\infty\chi_{A_j}-\chi_{A_n}\right|\,d\mu. \end{align} I do not know where to go from here, or if I am even on the right track. How can I prove $\{A_n\}\to A$?

• Here is a short way to prove that $(M,\rho)$ is a complete metric space. Given an arbitrary cauchy sequence $\{A_n\}\subset \mathfrak M$, we have that $\{\chi_{A_n}\}$ is a cauchy sequence in $L^1(X,\mathfrak M,\mu)$. Since $L^1(X,\mathfrak M,\mu)$ is complete, there is $f\in L^1(X,\mathfrak M,\mu)$ such that $\{\chi_{A_n}\}$ converges to $f$ in $L^1(X,\mathfrak M,\mu)$. It is a known result that, then there is a subsequence of $\{\chi_{A_n}\}$ that converges to $f$ a.e.. So $f$ is the indicator function of some set $A \in \mathfrak M$, that is $f=\chi_A$. – Ramiro Nov 13 '15 at 2:00

Here is a short way to prove that $(M,\rho)$ is a complete metric space. Given an arbitrary cauchy sequence $\{A_n\}\subset \mathfrak M$, we have that $\{\chi_{A_n}\}$ is a cauchy sequence in $L^1(X,\mathfrak M,\mu)$. Since $L^1(X,\mathfrak M,\mu)$ is complete, there is $f\in L^1(X,\mathfrak M,\mu)$ such that $\{\chi_{A_n}\}$ converges to $f$ in $L^1(X,\mathfrak M,\mu)$. It is a known result that, then there is a subsequence of $\{\chi_{A_n}\}$ that converges to $f$ a.e.. So $f$ is the indicator function of some set $A \in \mathfrak M$, that is $f=\chi_A$.