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Math Help - Convergence of the indictor function

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    Convergence of the indictor function

    Why does  1_{(n,n+1)} \rightarrow 0 ?
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  2. #2
    Super Member girdav's Avatar
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    Re: Convergence of the indictor function

    It depends on the type of convergence that you consider. It's true for the convergence almost everywhere because there is in particular a pointwise convergence (if you take x\in\mathbb R, then for n\geq \lfloor x\rfloor +1 you will have \mathbf 1_{(n,n+1)}(x)=0).
    We don't have unform convergence on the whole real line since for each n, \sup_{x\in \mathbb R}\mathbf 1_{(n,n+1)}(x)=1, but the convergence is uniform on every compact subset of \mathbb{R}.
    You can show that this sequence of function doesn't converge on measure.
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