Let f(x) = 1 if x is rational
= 0 if not
let g(x) = 0 fr all x
then f is not integrable, g is and there product is the constant function 0 which is cleArly integrable
I need to find 2 functions on [0,1]: one that is integrable, one that isn't, such that their product is integrable.
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...
Sorry for not being more descriptive, but the functions are not supposed to be constant--I apologize for neglecting to post that!
I see what you're saying: since 1/(x-1) is unbounded, it is not integrable on [0,1]. Correct? Is it enough to say that, or do I need to bring in 2 partitions converging to different points?
Just curious, for the function f(x):=1 if x rat, f(x):=-1 if x irrat, since it is clearly bounded, wouldn't I have to talk about 2 different partitions? It's frustrating because I know it's not integrable, but since we are only working with Riemann integrable functions and their proofs, I'm not sure what I'm allowed to use: right now we're doing everything with partitions... any ideas?
I really appreciate your help!