A derivative which is not Lebesgue integrable on any interval?

If $f=x^2\sin(x^{-2})$, then $f'$ exists everywhere (including $x=0$) but $f'$ is not Lebesgue integrable on $[0,1]$ (precisely because of the singularity at $x=0).$ I'm trying to find a function $f$ such that $f'$ exists everywhere but $f'$ is not Lebesgue integrable on any interval. Perhaps someone can recall a standard example?

Solution 1:

There are no such functions. Since $f$ is continuous, the derivative $$ f'(x) = \lim_{n\to\infty} n(f(x+1/n)-f(x)) $$ is a pointwise limit of continuous functions, i.e., a function of Baire class $1$. Functions of Baire class $1$ have many points of continuity. Every point of continuity has a neighborhood in which the function is bounded; since it's also a Borel function, it is Lebesgue integrable in such neighborhood.