I was talking with a professor, and he mentioned the following theorem:
Let $(X,\mathcal{F},\mu):=([0,1],\mathcal{B},\lambda)$ be the interval with the usual Lebesgue measure on the Borel sets. For $A,B\in\mathcal{F}$, we declare $A\sim B$ whenever $\mu(A\Delta B)=0$, and so we get an equivalence relation on $\mathcal{F}$. If we define on $\check{\mathcal{F}}:=\mathcal{F}/\sim$ a metric $d$ by setting $d([A],[B]):=\mu(A\Delta B)$, this is a well defined metric space. Until here, there is no problem, however, here is my doubt: According to my professor, $(\check{\mathcal{F}},d)$ is a complete metric space.
I have been struggling a lot to prove this. I imagine one can take an arbitrary Cauchy sequence and construct a limit, but I haven't been succesful there. Could anyone help me with this or give a good place to find a proof of it?
Also, for which other measure spaces $(X,\mathcal{F},\mu)$ can the construction of $(\check{\mathcal{F}},d)$ also be a complete metric space? And where can it be polish? Is there a theorem giving good sufficient conditions for this?
Thanks in advance for any answer.
Let $\{A_n\}$ be a cauchy sequence, then, Let $$B_{m} = A_{m} \setminus \cup_{\ell < m} (A_{m} \bigtriangleup A_{\ell})$$ Let $$A = \cap_{k=1}^{\infty}\cup_{m \geq k}^{\infty} B_{m}$$
$$d(A,A_{r}) = \mu(A \bigtriangleup A_{r}) = \mu\left(\cap_{k=1}^{\infty}\cup_{m \geq k}^{\infty} B_{m} \cap A^c_{r} \right)+ \mu \left(\cup_{k=1}^{\infty}\cap_{m \geq k}^{\infty} B^c_{m} \cap A_{r} \right) = \lim_{k \rightarrow \infty} \mu\left( \cup_{m \geq k}^{\infty} B_{m} \cap A^c_{r} \right)+ \lim_{k \rightarrow \infty} \mu \left(\cap_{m \geq k}^{\infty} B^c_{m} \cap A_{r} \right) \leq \lim_{k \rightarrow \infty} \mu \left(\cap_{m \geq k}^{\infty} B^c_{m} \cap A_{r} \right) \leq \lim_{k \rightarrow \infty } \mu(\cap_{m \geq k} \cup_{\ell < m} ((A_m^c \cap A_{r}) \cup ((A_m \bigtriangleup A_{\ell}) \cap A_{r})) = \lim_{k \rightarrow \infty } \mu(\cap_{m \geq k} (A_m \bigtriangleup A_{r}))) \leq \lim_{k \rightarrow \infty } \mu(A_k \bigtriangleup A_r) \leq \lim_{k \rightarrow \infty } d(A_k,A_r) \leq \epsilon$$
This is because $$ \cup_{\ell < m} ((A_m^c \cap A_{r}) \cup ((A_m \bigtriangleup A_{\ell}) \cap A_{r})) \subseteq A_m \bigtriangleup A_r$$
So it seems $A$ is the limit !