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Math Help - measure space, sequence

  1. #1
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    measure space, sequence

    Let (X, \mathcal{B}, \mu) be a measure space. Prove that if (A_n)_{n \geq 1} is a sequence in \mathcal{B} with \sum_{n=1}^{\infty} \mu(A_n)< \infty then there exists a set E with \mu(E)=0 such that if x \not \in E then x belongs to at most finitely many of the A_n's.

    Here, I really don't see how to answer this question. I am really struggling with this one. Any hints on where to start would be appreciated.
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  2. #2
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    Quote Originally Posted by poincare4223 View Post
    Let (X, \mathcal{B}, \mu) be a measure space. Prove that if (A_n)_{n \geq 1} is a sequence in \mathcal{B} with \sum_{n=1}^{\infty} \mu(A_n)< \infty then there exists a set E with \mu(E)=0 such that if x \not \in E then x belongs to at most finitely many of the A_n's.

    Here, I really don't see how to answer this question. I am really struggling with this one. Any hints on where to start would be appreciated.
    I don't know if this will ruin it for you but the result is Borel-Cantelli lemma. The set E is given by E=\limsup A_n=\bigcap_{n=1}^\infty \bigcup_{m=n}^\infty A_n. The proof comes from the observation that \bigcup_{m=n}^\infty A_n is increasing on n, so \bigcap_{n=1}^N\bigcup_{m=n}^\infty A_n=\bigcup_{m=N}^\infty A_n.

    Use subaditivity on \mu(\bigcup_{m=N}^\infty A_n), then use monotone convergence to get an inequality for the limit.

    You may want to recall that the tail of a convergent sum converges to zero.
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