True or false that all real function is an antiderivative

[kallazans]

is it true or false that all real function is an antiderivative of some real function but neither real function have an antiderivative?

I still have the doubt!

Definition(Louis Leithold,The Calculus with Analytic Geometry)
Antiderivative: F is antiderivative of f in I if F'(x)=f(x) for all x in I.

The question is all f have some F in some I?
The question is all f is a G of some g in some I?
(Real Analysis)

 

[Hurkyl]

By "primitive" do you mean that if f(x) = \int g(x) \, dx then f(x) is a primitive of g(x)? (The usual English word for this is that f(x) is an antiderivative or an integral of g(x))


I'm not entirely sure what you're trying to ask... though it is false that any function is an antiderivative of another function.

 

[HallsofIvy]

It's not clear what you mean by "either" function. If you mean the original function in the question and its anti-derivative, then obviously IF "every function had a anti-derivative", then it wouldn't make sense to say that THAT function did NOT have an anti-derivative.
However, as Hurkyl pointed out, it is not true that every function has a primitive (anti-derivative). For example, the function, f(x)= 1 if x is rational, 0 if x is irrational, does not have an antiderivative.
It IS true that every bounded function whose points of discontinuity form a set of measure 0 is integrable (has an anti-derivative). In particular every continuous function has an anti-derivative as well as every bounded function with only a finite number of points of discontinuity.