Prove that if f and g are integrable on [a, b], then so is fg

[a1010711]

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

 

[yyat]

Which definition of integrability are you using?

 

[HallsofIvy]

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

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.

 

[D H]

The Attempt at a Solution

It suffices to show that f^2 is integrable

What if you can find a function f that is integrable on [a,b] but not square integrable?

You can't prove it, it's not true.

That much is true.

For example, if f(x)= 1 if x is rational, -1 if x is irrational ...

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.