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Math Help - Measurable function

  1. #1
    Senior Member slevvio's Avatar
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    Measurable function

    Hello everyone I was wondering if you could help me with something. Let f=\displaystyle\sum_{n=1}^{\infty}4^n \chi_{E_n} for x \in \mathbb{R} where E_n = (0,2^{-n}] and chi is the characteristic function. Then f is measurable since it is the limit pointwise (almost) everywhere of step functions, but how do I show that \int f = \infty ?

    If I define f_n = \displaystyle\sum_{r=1}^n 4^n \chi_{E_n}, then of course

    \int f_n = \displaystyle\sum_{r=1}^n 2^r.

    But since the limit of this sequence is infinity, does this tell me that the integral of f diverges? Can I say that the integral of a measurable function \int f = \lim_{n\rightarrow\infty} \int f_n?

    The definition of the integral of a non-negative measurable function is that you take the truncation f(x)=n ( if f(x)\ge n) , -n (if f(x) \le -n) and leave it the same otherwise, then
    \int f := \lim_{n \rightarrow \infty} \chi_{[-n,n]}f_n

    so in the context of my question I guess I am trying to show that

      \int f =\lim_{m\rightarrow \infty} \int \chi_{[-m,m]}\left(\displaystyle\sum_{n=1}^{\infty}4^n\chi_{E_  n}\right)_m=\infty,

    but that seems like a giant mess I can't make sense of. Can anyone give me a hand with this. Thanks a lot
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by slevvio View Post
    Can I say that the integral of a measurable function \int f = \lim_{n\rightarrow\infty} \int f_n?
    In general, no. But the famous convergence theorems of Lebesgue give you some conditions under which the answer is yes. The one that you want here is Lebesgue's monotone convergence theorem.
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  3. #3
    Senior Member slevvio's Avatar
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    We studied that only for functions in L^1(R).

    So does the result hold when the sequence of integrals diverges?
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  4. #4
    Super Member girdav's Avatar
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    The result holds true because \int fd\lambda \geq \int f_nd\lambda for all n.
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