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Prove that if f and g are integrable on [a, b], then so is fg

  1. Feb 24, 2009 #1
    3. The attempt at a solution

    It suffices to show that f^2 is integrable, since

    fg= [(f+g)^2-(f-g)^2]/4

    The function x --> x^2 is uniformly continuous on the range of f,


    im not sure how to turn this into a formal proof, im lost


    Riemann integration
     
    Last edited: Feb 24, 2009
  2. jcsd
  3. Feb 24, 2009 #2
    Which definition of integrability are you using?
     
  4. Feb 24, 2009 #3

    HallsofIvy

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    You can't prove it, it's not true. For example, if f(x)= 1 if x is rational, -1 if x is irrational, the f2(x)= 1 so f2(x) is integrable on, say, [0, 1] but f is not. Also, if g(x)= 2, then g2(x)= 4 so is integrable on [0,1] but fg(x)= 2 if x is rational, -1 if x is irrational is not itegrable.
     
  5. Feb 24, 2009 #4

    D H

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    What if you can find a function f that is integrable on [a,b] but not square integrable?

    That much is true.

    That's fine, but the title of the thread is "Prove that if f and g are integrable on [a, b], then so is fg", so you are picking some f that violates the given conditions.
     
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