Does there exist a function $F(x)$, so that $F'(x) $ is not Riemann integrable?

Question

Does there exist a function $F(x)$ such that it satisfies the following property?

Let $I=[a,b]$ and $F:I\rightarrow\mathbb{R}$ be a function. $F$ is strictly monotonic and differentiable on $I$, but the derivative $F'(x)$ is not Riemann integrable.

Answer 1

Here's another construction. Let $g : [0,1] \to \mathbb R$ satisfies the following properties: $g$ is continuous on $(0,1]$,

$$g(0) = 0,\ g(2^{-n}) = 2^n,\ \ g \left( \frac{1}{2} \left( \frac{1}{2^n} + \frac{1}{2^{n+1}}\right)\right) = 0 \ \ \forall n\in \mathbb N, \ \ g(x) >0 \text{ elsewhere,}$$ and

$$\int_{2^{-n-1}}^{2^{-n}} \ \ \ g(s) ds\le 2^{-2n-1}$$

for all $n\in \mathbb N$. (This might be a bit hard to write it down, but such a function obviously exist).

Then define

$$f(x) = \int_0^x g(s) ds.$$

Then $f$ is stricly increasing on $[0,1]$ and is differentiable and $f'(x)= g(x)$ on $(0,1]$. To check that $f$ is also differentiable at $0$, consider

$$R(x) = \frac{f(x) - f(0)}{x} = \frac 1x \int_0^x g(s) ds.$$

Now if $ 2^{-m-1} < x\le 2^{-m}$, we have

$$|R(x)| \le 2^{m+1} \int_0^{2^{-m}} g(s) ds \le 2^{m+1}\sum_{k=m}^\infty \frac{1}{2^{2k+1}} = \sum_{k=m}^\infty \frac{1}{2^{2k-m}}\le \frac{1}{2^m}\sum_{k=m}^\infty\frac{1}{2^{k-m}} = \frac{1}{2^{m-1}}$$

Thus $|R(x)| \to 0$ as $x\to 0$. This proves that $f$ is also differentiable at $0$.

Now this function is an example, because $g(x)$ is not bounded (thus not Riemann integrable).

Remark Let me clarify a little bit. First of all, $g$ is not continuous at $0$. However it is Lebesgue integrable (Hence it makes sense to define $f$ as an integral of $g$).