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Math Help - proof an inequality; measure theory

  1. #1
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    proof an inequality; measure theory

    Hi there


    Let C be compakt
    \mu a radon measure and \mu(C) < \infty.

    Let v \in L^1(\mathbb{R})  , \ 0 < a < b < \infty

    Do you know how to proof the following inequality:

     \int_{a}^b |v| d\mu \le \mu([a,b]) \cdot \|v\|_\infty

    I don't get it.
    By the way I'm not sure about the correctness of this inequaility if \mu is an arbitrary measure. If it's correct where \mu equals the Lebesgue measure, you can use this instead of the Radon measure.

    Any comments are welcome

    Yours Rapha
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  2. #2
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    Presumably \mu is supposed to be a positive Radon measure.

    From the definition of the L^\infty-norm, |v(t)|\leqslant \|v\|_\infty for almost all t in [a,b]. Then \int_a^b\!\!\!|v(t)|\,d\mu(t) \leqslant \int_a^b\!\!\!\|v\|_\infty\,d\mu(t) = \mu([a,b]) \|v\|_\infty (integral of a constant is the constant times the measure of the interval).
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  3. #3
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    Hello Opalg

    thank you very much!

    Kind regards
    Rapha
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