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 f^{2}(x)= 1 so f^{2}(x) is integrable on, say, [0, 1] but f is not. Also, if g(x)= 2, then g^{2}(x)= 4 so is integrable on [0,1] but fg(x)= 2 if x is rational, -1 if x is irrational is not itegrable.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
What if you can find a function f that is integrable on [a,b] but not square integrable?3. The attempt at a solution
It suffices to show that f^2 is integrable
That much is true.You can't prove it, it's not 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.For example, if f(x)= 1 if x is rational, -1 if x is irrational ...