# Existence of antiderivative

Let $f$ be a real-valued function defined on the open unit interval.

What assumptions you have to make about $f$ to be sure that it posseses an antiderivative?

I'm interested in the weakest (most general) possible assumptions, so some nontrivial equivalent condition would be optimal.

• I believe there are (at least) two different questions. The weakest condition to have an antiderivative is to simply be a derivative, and there is a lot of literature on attempts to classify the property of being a derivative (in ways besides just restating "is a derivative"). See Andrew M. Bruckner's book Differentiation of Real Functions for an entry point into this area. The other question has to do with methods for obtaining an antiderivative, and for this various integration methods have been developed -- the Riemann integral, the Lebesgue integral, the Denjoy integral, etc. – Dave L. Renfro May 15 '14 at 19:53
• My question is adressing precisely the first issue. The reference you mentioned is kind of thing I was looking for, but I'm afaid I don't have acces to it. Anyway, thank you! – user150316 May 15 '14 at 20:43
• @DaveL.Renfro In addition and maybe I'm not understanding fully, is it safe to say that if $f$ is continuous on $(a,b)$, there exists another function $F$ which is differentiable on $(a,b)$ down to $f$? While continuity doesn't imply the function itself is differentiable, continuity does imply the existence of another function which is differentiable down to $f$? – DWade64 Oct 26 '18 at 19:54
• @DWade64: Yes, given any continuous function $f$ defined on $(a,b),$ then there exists a function $F$ defined on $(a,b)$ such that $F'(x) = f(x)$ for each $x \in (a,b).$ And you don't need advanced mathematical ideas for this such as Lebesgue integration and measure theory --- for continuous functions $f,$ the corresponding functions $F$ can be defined by using the Riemann integral. This is one half of the usual 2-part version of the Fundamental Theorem of Calculus. See Is every continuous function a derivative? – Dave L. Renfro Oct 26 '18 at 20:30
• @DaveL.Renfro Thank you for the help as well as the answer you gave! – DWade64 Oct 26 '18 at 21:14