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Thread: Convergence of the indictor function

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    Convergence of the indictor function

    Why does $\displaystyle 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 $\displaystyle x\in\mathbb R$, then for $\displaystyle n\geq \lfloor x\rfloor +1$ you will have $\displaystyle \mathbf 1_{(n,n+1)}(x)=0$).
    We don't have unform convergence on the whole real line since for each $\displaystyle n$, $\displaystyle \sup_{x\in \mathbb R}\mathbf 1_{(n,n+1)}(x)=1$, but the convergence is uniform on every compact subset of $\displaystyle \mathbb{R}$.
    You can show that this sequence of function doesn't converge on measure.
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