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Math Help - lim inf contained in lim sup

  1. #1
    Junior Member TheProphet's Avatar
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    lim inf contained in lim sup

    Hey,

     (A_{n})_{n \geq 1} a sequence of subsets of  \mathcal{A} , a  \sigma -field on  \Omega .

    With
     \lim_{n \to \infty} \inf A_{n} = \bigcup_{n=1}^{\infty} \bigcap_{m\geq n} A_{m}
    and
     \lim_{n \to \infty} \sup A_{n} = \bigcap_{n=1}^{\infty}\bigcup_{m \geq n} A_{m}

    show that  \lim_{n \to \infty} \inf A_{n} \subset \lim_{n \to \infty} \sup A_{n}

    Hints on how to start?
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  2. #2
    Super Member girdav's Avatar
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    Re: lim inf contained in lim sup

    Let x\in\liminf_{n\to +\infty} A_n. We can find n_0 such that for n\geq n_0, x\in A_n. Now, you have to show that for all k\in\mathbb{N}, x\in\bigcup_{j\geq k}A_j.
    "To be in \liminf_{n\to +\infty} A_n" means "to be in all A_n for n large enough" whereas "To be in \limsup_{n\to +\infty} A_n" means "to be in infinitely many A_n".
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  3. #3
    Junior Member TheProphet's Avatar
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    Re: lim inf contained in lim sup

     x \in \lim_{n \to \infty} \inf A_{n} \Rightarrow \exists n_0 : x \in A_{n} \forall n \geq n_{0}

    If  x \in \bigcup_{j \geq k} A_{j} but  x \notin \bigcup_{j \geq k+1} A_{j} this means  x \notin \lim_{n \to \infty} \inf A_{n} which is a contradiction.

    And clearly  x \in \bigcup_{j \geq 1} A_{j} , whence

     x \in \lim_{n \to \infty} \sup A_{n}

     \lim_{n \to \infty} \inf A_{n} \subset \lim_{n \to \infty} \sup A_n{n}

    Is this ok?
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  4. #4
    Super Member girdav's Avatar
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    Re: lim inf contained in lim sup

    Yes, it works.
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