The discussion centers on the claim that if functions f and g are integrable on the interval [a, b], then their product fg is also integrable. A participant argues that it suffices to show f^2 is integrable, using the identity fg = [(f+g)^2 - (f-g)^2]/4. However, another participant counters this by providing a counterexample with the function f(x) = 1 if x is rational and -1 if x is irrational, demonstrating that f can be integrable while fg is not. This indicates that the original claim is false under certain conditions.
PREREQUISITESMathematicians, students studying real analysis, and anyone interested in the properties of integrable functions and their products.
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 f2(x)= 1 so f2(x) is integrable on, say, [0, 1] but f is not. Also, if g(x)= 2, then g2(x)= 4 so is integrable on [0,1] but fg(x)= 2 if x is rational, -1 if x is irrational is not itegrable.a1010711 said: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, I am lost
Riemann integration
What if you can find a function f that is integrable on [a,b] but not square integrable?a1010711 said:The Attempt at a Solution
It suffices to show that f^2 is integrable
That much is true.HallsofIvy said: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 ...