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Math Help - pulling out constants with the lebesgue integral

  1. #1
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    pulling out constants with the lebesgue integral

    Let f be a positive, measurable function. Let c be a constant. let E be a measurable set.

    I know that the integral of cf over E equals c times the integral of f over E when c is greater than or equal to 0 and less than infinity. I need to prove that this works even when c= infinity. I am a little confused because isn't infinity times a positive value infinity? I suppose 0 times infinity is 0. Does it suffice to just look at these two cases?
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  2. #2
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    Infinity is not even a number. I can't imagine why you would want to multiply a function by infinity (nor how you would do it!).
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Infinity is not even a number. I can't imagine why you would want to multiply a function by infinity (nor how you would do it!).
    I am also sort of skeptical but it is an exercise in Rudin's Real and Complex Analysis text.

    In the extended reals we define infinity, call it q, like this:

    let a be in [0,q]

    then

    a+q=q+a=q for all real numbers a
    aq=qa= q if a is nonzero and 0 if a is zero.

    This way we preserve commutative/associative and distributive properties.

    We have to be careful with the cancellation laws: we can't subtract or divide by q.
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  4. #4
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    Nice book Rudin's!
    You must distinguish 2 cases.

    1. If f is diferent form 0 in a subset F of E of positive measure, then it follows that \int_E f\neq 0 because \mu\{x\in E:f(x)>1/n\}>0 for some n. Hence \infty f=\infty in F and hence \int_E\infty f\geq \int_F\infty f=\infty\mu (F)=\infty=\infty\int_E f. . Thus the inequality becomes an equality.

    1 If f equals 0 almost everyhere in E, then we can take F with \mu(E\setminus F)=0 and f=0 on F. Therefore \int_E \infty f=\int_F \infty f=\int_F 0=0=\infty 0=\infty\int_F f=\infty\int_E f.
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