Derivative of the indefinite integral and Lebesgue point

Question

Give an example where the derivative of the indefinite integral exists at point that are not Lebesgue points.

Answer 1

For $f$ to have an indefinite integral, i.e. an antiderivative $F$, on an interval $(a,b)$, it just means that $F' = f$ at every point of $(a,b)$. The derivative of the antiderivative exists everywhere on $(a,b)$. A standard example of a function with a non-continuous derivative is $$F(x) = \cases{x^2 \sin(1/x) & if $x \ne 0$\cr 0 & for $x=0$\cr}$$ having derivative $$f(x) = \cases{2 x \sin(1/x) - \cos(1/x) & if $x \ne 0$\cr 0 & for $x=0$\cr} $$ It's not hard to show that $$\liminf_{\epsilon \to 0} \frac{1}{2\epsilon} \int_{-\epsilon}^\epsilon |f(x) - f(0)|\ dx > 0$$