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Convergence of sequence of measurable sets

  1. Apr 24, 2012 #1
    Given a totally finite measure μ defined on a [itex]\sigma[/itex]-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 [itex][A_\alpha][/itex] where [itex]d(A_{\alpha_1},A_{\alpha_2})=0[/itex] 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 [itex]A=\cap_{n=1}^\infty A_n[/itex], but how do I formalize the proof? d(An,A)=μ(An-A)=[itex]\mu(A_n-\cap_{n=1}^\infty A_n[/itex])=...,
    How do I make use of the Cauchy sequence {An}?
    Last edited: Apr 24, 2012
  2. jcsd
  3. Apr 24, 2012 #2
    It turns out the limit is not [itex]A=\cap_nA_n[/itex] (e.g., [itex]A_1=\emptyset,A_n=A\neq\emptyset,n>1[/itex]), unless [itex]A_{n+1}\subset A_n[/itex], in which case
    [itex]\mu(A)=\mu(\cap_nA_n)=\lim_{n\rightarrow\infty}\mu(A_n)[/itex], so that [itex]\lim_{n\rightarrow\infty} d(A_n,A)=\lim_{n\rightarrow\infty}[\mu(A_n)-\mu(A)]=0[/itex].

    In the general case, what would be a suitable candidate limit set?
    Last edited: Apr 24, 2012
  4. Apr 25, 2012 #3
    I have come up with two candidates, define {Bn} and {Cn} where [itex]B_n=\cup_{k=n}^\infty A_k[/itex] and [itex]C_n=\cap_{k=n}^\infty A_k[/itex], it can be shown that {Bn} converges to [itex]B=\cap_{k=n}^\infty B_k[/itex] and {Cn} converges to [itex]C=\cup_{k=n}^\infty C_k[/itex], it can also be shown that [itex]C_n\subset A_n\subset B_n[/itex], 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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