I am studying Riemann integration and I need to find a counterexample.I want a function that is bounded,and has an antiderivative on $[a,b]$ but is not Riemann integrable on $[a,b]$.I think it will not be very easy due to the boundedness condition.Otherwise we could have used the function:
$f(x)=2x\sin(1/x^2)-2/x \cos(1/x^2)$ when $x\neq 0$ and $f(0)=0$ on $[0,1]$ would work as $f$ has antiderivative $\phi(x)=x^2\sin(1/x^2)$ when $x\neq 0$ and $\phi(0)=0$ on $[0,1]$.But $f$ is not bounded,so no question of Riemann integrability arises because Riemann integration is defined for bounded functions.
But I want the function to be bounded in my example.I think it will be hard because the set of all points of discontinuity of $f$,viz $D_f$ is a measure non-zero set but at the same time since $f$ has antiderivative,so $f$ is the derivative of some function and hence $D_f$ must also be an $F_\sigma$ meagre set.So,I do not think it is easy to construct it.