1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook