# Convergence of sequence of measurable sets

1. Apr 24, 2012

### 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}?

Last edited: Apr 24, 2012
2. Apr 24, 2012

### 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?

Last edited: Apr 24, 2012
3. Apr 25, 2012

### 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.

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