Integrable Functions Homework: Finding Non-Constant f & g

[MatthewD]

Homework Statement

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

Homework Equations

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!

 

[snipez90]

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).

 

[MatthewD]

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!

 

[snipez90]

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.

 

[MatthewD]

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!