# The measure of the set of discontinuity point

If $f(x)$ is derivable,and $f'(x)$ is Riemann integrable, the continuity point of $f'(x)$ is dense, i.e. the measure of the set of discontinuity point of the second kind is zero. I'm thinking out that is it true that for all the derivable functions, $f'(x)$ is Riemann integrable? If not,can anyone give me a counterexample?

• A function can have more than one discontinuity point (and needs not have any). Do you mean "the measure of the set of discontinuity points"? – hmakholm left over Monica Apr 23 '12 at 10:03
• @HenningMakholm Sorry,I make a mistake. – 89085731 Apr 23 '12 at 10:18

## 1 Answer

Assume $f$ is a function on a compact interval. A function on a compact interval is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure).

Therefore what you are looking for is a function $f$ that is differentiable almost everywhere and its derivative is Lebesgue integrable but not Riemann integrable.

Take a Lebesgue set whose characteristic function is not riemann integrable. Take f to be the antiderivative of this function. Then the Lebesgue Differentiation Theorem does the rest.

• You need to be a bit careful about the choice of the "a Lebesgue set": it shouldn't coincide with a Jordan measurable set up to a null set (e.g. the characteristic function of $\mathbb{Q} \cup [0,1]$ is not Riemann integrable but the derivative of its antiderivative will be). Tage e.g. a fat Cantor set to avoid such silliness. – t.b. Apr 23 '12 at 11:00
• Of course, thank you for correcting me. – fk2012 Apr 23 '12 at 11:11