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Math Help - Proof critique

  1. #1
    Junior Member
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    Dec 2008
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    Proof critique

    Hello, I am trying to prove that given a proper integral,  \displaystyle \int_a^b \sum_{n=0}^\infty f_n(x) dx = \sum_{n=0}^\infty \int_a^b f_n(x) dx , assuming both converge.

    Here's my attempted proof:

     \displaystyle \sum_{n=0}^\infty f_n(x) = \sum_{n=0}^N f_n(x) + R_N(x) , where  \forall x, \; R_N(x) \to 0 as  N \to \infty .

    Thus  \displaystyle \int_a^b \sum_{n=0}^\infty f_n(x) dx = \int_a^b \sum_{n=0}^N f_n(x) dx + \int_a^b R_N(x) dx = \sum_{n=0}^N \int_a^b f_n(x) dx + \int_a^b R_N(x) dx .

    But  \displaystyle 0 \leq \left| \int_a^b R_N(x) dx \right| \leq \int_a^b \left|R_N(x)\right| dx \leq (a-b)\cdot\max_{a\leq x\leq b}\bigg\{|R_n(x)|\bigg\} \to 0 as  N \to \infty .

    Hence  \displaystyle \int_a^b \sum_{n=0}^\infty f_n(x) dx = \sum_{n=0}^\infty \int_a^b f_n(x) dx .

    Does this look valid?
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  2. #2
    Super Member
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    Quote Originally Posted by mathman88 View Post
    (a-b)\cdot\max_{a\leq x\leq b}\bigg\{|R_n(x)|\bigg\} \to 0 as  N \to \infty .
    There's a mistake in this step, since we don't know that we can bound R_N(x) uniformly on the interval.

    That said what you're trying to prove is false as stated (indeed, if it were true, theorems like dominated and monotone convergence would be unecessary), there is a sequence of functions such that all terms have the same integral and the sequence converges pointwise to zero.
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