Convergence of sequence of measurable sets

[sunjin09]

Given a totally finite measure μ defined on a \sigma-field X, define the (pseudo)metric d(A,B)=μ(A-B)+μ(B-A), (the symmetric difference metric), it can be shown this is a valid pseudo-metric and therefore the metric space (X',d) is well defined if equivalent classes of sets [A_\alpha] where d(A_{\alpha_1},A_{\alpha_2})=0 are considered.

How do I show this metric space (X',d) is complete? In other words, given a Cauchy sequence {An}, the limit seems to be given by A=\cap_{n=1}^\infty A_n, but how do I formalize the proof? d(An,A)=μ(An-A)=\mu(A_n-\cap_{n=1}^\infty A_n)=...,
How do I make use of the Cauchy sequence {An}?

 

[sunjin09]

It turns out the limit is not A=\cap_nA_n (e.g., A_1=\emptyset,A_n=A\neq\emptyset,n>1), unless A_{n+1}\subset A_n, in which case
\mu(A)=\mu(\cap_nA_n)=\lim_{n\rightarrow\infty}\mu(A_n), so that \lim_{n\rightarrow\infty} d(A_n,A)=\lim_{n\rightarrow\infty}[\mu(A_n)-\mu(A)]=0.

In the general case, what would be a suitable candidate limit set?

 

[sunjin09]

I have come up with two candidates, define {Bn} and {Cn} where B_n=\cup_{k=n}^\infty A_k and C_n=\cap_{k=n}^\infty A_k, it can be shown that {Bn} converges to B=\cap_{k=n}^\infty B_k and {Cn} converges to C=\cup_{k=n}^\infty C_k, it can also be shown that C_n\subset A_n\subset B_n, the problem is how to show C=B so that {An} has to converge to B or C. Can anybody help? Thanks a lot.