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Math Help - Integration II

  1. #1
    Senior Member slevvio's Avatar
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    Integration II

    Let  \lambda \in \mathbb{R}, f:[a +\lambda , b + \lambda ] \rightarrow \mathbb{R} continuous.

    Is this correct?

     \int_{a+ \lambda}^{b+\lambda} f(s)ds := sup \{ \int_{-\infty}^{\infty} g(s)d(s): g supported on  [a +\lambda , b + \lambda ], g(s) \le f(s) \forall s \in [a +\lambda , b + \lambda ] \} = supA.

    Define  h(s) = g(s+\lambda) . Then h is supported on [a,b],  h(s) \le f(s + \lambda) \forall s \in [a, b] .

    Hence A = \{ \int_{-\infty}^{\infty} h(s)d(s): h supported on  [a , b ], h(s) \le f(s+\lambda) \forall s \in [a , b ] \} = B. [since \int_{-\infty}^{\infty} h(s)d(s) = \int_{-\infty}^{\infty} g(s)d(s) ??? ]

    so sup A = sup B  \iff \int_{a+ \lambda}^{b+\lambda} f(s)ds = \int_{a}^{b} f(t+\lambda)dt . ?

    Thanks for any advice or input
    Last edited by slevvio; December 13th 2009 at 03:56 PM.
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