Non-integrable function that has an antiderivative The wikipedia article on antiderivatives states:

Non-continuous functions can have antiderivatives. [...]
  
  
*
  
*In some cases, the antiderivatives of such pathological functions may be found by Riemann integration, while in other cases these
  functions are not Riemann integrable.
  

I'm looking for an example of such a function - a bounded function $f : [a, b] \to [m, M]$ that has an antiderivative $F$ such that $F' = f$ on its domain, while $f$ is not Riemann-integrable.
 A: Edit: Now that your function is bounded, by Lebesgue's criterion for Riemann integrability, your function must not be continuous a.e. Now this link gives you the whole gamut of differentiable functions that have derivatives that are discontinuous on sets of different measures/sizes. https://math.stackexchange.com/a/423279/231021
You want the one in the last part "Increasing the measure". That will guarantee that your function is discontinuous on a set of Lebesgue measure greater than $0$. Now you have to ensure that it is bounded and the example in the comments is one of those.
Here's another link, where they talk about construction of such functions with bounded derivatives everywhere on$[0,1]$ but such that the derivatives are discontinuous on a set of positive measure. (see 5.5.5.d):
https://books.google.co.uk/books?id=1WY6u0C_jEsC&pg=PA212&lpg=PA212&dq=derivative+discontinuous+measure+bounded&source=bl&ots=JtO97c8t60&sig=5YGv2dnZpJ-OO5Uzdodt7BRwnuk&hl=en&sa=X&ei=XvY2VYy5A4fpaKClgbAO&ved=0CEoQ6AEwBg#v=onepage&q=derivative%20discontinuous%20measure%20bounded&f=false
Old post before the bounded restriction: Try unbounded functions. They cannot have Riemann integrals, yet some are continuous on an open interval. What about $x\mapsto \frac{1}{x}: (0,1) \rightarrow \mathbb{R}$?
$\int_0^1 \frac{1}{x} dx$ does not exist (Riemann integrable functions are by definition bounded, and this function is not, on $(0,1)$) but it has an antiderivative $\ln(x)$ everywhere on $(0,1)$.
