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Homework Help: Integrable Functions

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data

    I need to find 2 functions on [0,1]: one that is integrable, one that isn't, such that their product is integrable.

    I'd like to use functions that are non-constant, if possible

    2. Relevant equations

    3. The attempt at a solution

    I was thinking of using
    g(x) := x-1
    h(x) := 1/(x-1)

    ... But I'm having trouble proving h(x) is not integrable on [0,1]...

    I used the fact that g was monotone to prove it is integrable and obviously gh is integrable because it's constant. Right now we're looking at Riemann integrals, but I don't know how to state the proof...

    Please help!
  2. jcsd
  3. Feb 27, 2010 #2
    EDIT: Irrelevant since the integral is improper and thus not integrable in the Riemann sense.

    My favorite example of a non-Riemann integrable function is the characteristic function of the rationals over [0,1] which is simply the function whose value is 1 at rational numbers in [0,1] and 0 at irrational numbers in [0,1]. Call this function g. Can you think of an integrable function f for which fg is integrable? (Hint: f is a special case of what is typically the first or easiest example of a Riemann integrable function).
    Last edited: Feb 27, 2010
  4. Feb 27, 2010 #3
    I thought that 1/(x-1) was unbounded, and therefore not integrable? Am I mistaken? I'm still trying to understand this whole chapter...

    Would it work if I had

    f(x): = 0

    g(x): = 1, x is rational
    g(x): = 0, x is irrational

    Because then f is constant ->integrable
    and fg=0 is constant -> integrable

    Am I on the right track? I really wanted to use something non-constant to try to understnad the concept better, but would this at least work?

    Thanks for your help!
  5. Feb 27, 2010 #4
    Sorry, I made a silly oversight. You are right since the integral doesn't converge due to the singularity at x = 1, so we have something improper. So your case worked fine, I had just forgotten to consider improper integrals. But you're also correct in your follow-up of my example. I just chose an example that is in some sense simpler, even if the function is considered pathological.
  6. Feb 27, 2010 #5
    Thank you so much---it worked out better for me because I had to prove to myself that the case worked, which helped me understand so much better.

    Thank you!!!:smile:
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