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 ...
Integrability of a function means that the function is well-behaved and can be approximated by a finite sum of smaller functions. In other words, it is possible to find the area under the curve of the function using integration.
Proving that f and g are integrable on [a, b] means that the product of the two functions, fg, is also integrable on [a, b]. This is important because it allows us to use integration to find the area under the curve of the product function, which may have practical applications in various fields of science and engineering.
A function that is continuous on a closed interval [a, b] is also integrable on that interval. However, the converse is not always true - a function can be integrable without being continuous. This is because integrability relies on the behavior of a function over a larger interval, while continuity only considers the behavior of a function at a single point.
This can be proven using the Riemann sum definition of integrability and the properties of definite integrals. By breaking up the function fg into smaller subintervals and using the properties of definite integrals, we can show that the limit of the Riemann sums for fg exists and is equal to the definite integral of fg on [a, b]. This proves that fg is integrable on [a, b].
Yes, this theorem can be extended to any finite number of integrable functions. If f1, f2, ..., fn are all integrable on [a, b], then their product f1f2...fn is also integrable on [a, b]. This can be proven using the same approach as for the two function case, by breaking up the product function into smaller subintervals and using the properties of definite integrals.