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Math Help - Real Valued Lebesgue Intergal

  1. #1
    Member
    Joined
    Oct 2009
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    128

    Real Valued Lebesgue Intergal

    Hello!

    I'm trying to show that if f is a measurable function on some space X, and if

    \int{fd\mu} < \infty

    then \mu\{x \in X : f(x) = \infty\} = 0

    In other words, trying to show f is real-valued (not infinity) mu-almost everywhere.

    I thought maybe the contrapostive would work, ie,
    Presume \mu\{x \in X : f(x) = \infty\} > 0 .
    Call the set \{x \in X : f(x) = \infty\} = B (for "big" valued, hehe)
    Then \int_B{fd\mu} = \int{f\chi_B d \mu}
    If I can just show \int{f\chi_B d \mu} = \infty , then since  B \subset X,
    \int_X{f d \mu} = \int{f d \mu} = \infty as well.

    So I am just wondering if the above looks alright, and what's left is how to possibly show that \int{f\chi_B d \mu} = \infty for \{x \in X : f(x) = \infty\} = B and \mu(B) > 0 ?!
    Thanks very much!
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  2. #2
    Newbie
    Joined
    Apr 2011
    Posts
    14

    Re: Real Valued Lebesgue Intergal

    There's an issue with the way you've set up the argument. You're not assuming that f is a nonnegative function, so it does not follow that the integral over the whole space is necessarily larger than the integral over a measurable subset. This isn't a big issue, but it does require some comment. Otherwise, it looks like you're on the right track. Just remember that in this context, infinity is an extended real number which has a defined set of algebraic rules and is not a limit.
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