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Math Help - Characteristic Function

  1. #1
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    Characteristic Function

    Let (A_n) be a sequence of measurable sets in X.Define \liminf_j A_j = \bigcup^\infty _{n=1} \bigcap^\infty_{j=n} A_j and \limsup_j A_j = \bigcap^\infty _{n=1} \bigcup^\infty_{j=n} A_j.

    Prove or disprove
    \chi_{\liminf_j A_j} = \liminf_j  \chi_{A_j} and \chi_{\limsup_j A_j} = \limsup_j  \chi_{A_j} .

    Can anyone help me to initialize the proof?
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  2. #2
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    Quote Originally Posted by problem View Post
    Let (A_n) be a sequence of measurable sets in X.Define \liminf_j A_j = \bigcup^\infty _{n=1} \bigcap^\infty_{j=n} A_j and \limsup_j A_j = \bigcap^\infty _{n=1} \bigcup^\infty_{j=n} A_j.

    Prove or disprove
    \chi_{\liminf_j A_j} = \liminf_j  \chi_{A_j} and \chi_{\limsup_j A_j} = \limsup_j  \chi_{A_j} .

    Can anyone help me to initialize the proof?
    A characteristic function can only take the values 0 or 1. Therefore for each x in X, \textstyle\limsup_j  \chi_{A_j}(x) must be 0 or 1. Then it's just a matter of definition-chasing:

     \begin{aligned}x\in \limsup_j A_j &\Longrightarrow\ \forall n\ \exists j\geqslant n\text{ such that }x\in A_j\\ &\Longrightarrow\ \forall n\ \exists j\geqslant n\text{ such that }\chi_{A_j}(x)=1\\  &\Longrightarrow\ \textstyle\forall n\ \sup_{j\geqslant n}\chi_{A_j}(x)=1\\   &\Longrightarrow\ \textstyle\limsup_j \chi_{A_j}(x)=1.\end{aligned}<br />

     \begin{aligned}x\notin \limsup_j A_j &\Longrightarrow\ \exists n\ \forall j\geqslant n\ x\notin {\textstyle\bigcup_{j\geqslant n}}A_j\\ &\Longrightarrow\ \exists n\ \forall j\geqslant n\ \chi_{A_j}(x)=0\\  &\Longrightarrow\ \ldots\ \Longrightarrow\ \textstyle\limsup_j \chi_{A_j}(x)=0.\end{aligned}

    Therefore \chi_{\limsup_j A_j} = \textstyle\limsup_j  \chi_{A_j}

    Similar argument for the liminf.
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