Integrability of the derivative of an increasing function- MIT primes 2020 A2 Problem A2 from the MIT PRIMES Problem Set 2020 goes as following (discussion is already open):
Let $f: [0,1]\to \mathbb{R}$ be a strictly increasing function which is differentiable in $(0,1)$. Suppose that $f(0)=0$ and for every $x\in(0,1)$ we have $$\frac{f'(x)}{x}\ge f(x)^2+1.$$
How small can $f(1)$ be?
I thought the solution idea was pretty straightforward; it follows pretty quickly from integrating the rearranged inequality $\frac{f'(x)}{f(x)^2+1}\ge x$. However, it is somewhat difficult to prove the left side is integrable. We can show that $\arctan(f(x))$ is increasing, and the derivative of an increasing function satisfies using the Lebesgue integral $\int_a^bf'(x)dx\le f(b)-f(a)$.
I was wondering if it is necessarily true that the left hand side is Riemann integrable, and if not, what is a counterexample? If it were Riemann integrable, we could have $\int_a^bf'(x)dx= f(b)-f(a)$, which is slightly stronger.
 A: If a function is differentiable then its derivative is Henstock–Kurzweil integrable.
Since the function is also increasing, then its derivative is non-negative. Therefore $|f'|=f'$ is Henstock–Kurzweil integrable. If $|g|$ is Henstock–Kurzweil integrable, then $g$ is Lebesgue integrable. It follows that $f'$ is Lebesgue integrable.
Both integrals satisfy the fundamental theorem of calculus in this case.
A: There are strictly increasing, differentiable functions $f$ on $[0,1]$ such that $f'(x)=0$ on a dense set. For such functions, $f'$ is not Riemann integrable. Such derivatives are known as Pompeiu derivatives. See https://en.wikipedia.org/wiki/Pompeiu_derivative
On to your problem: Define $g(x)= f(x)-f(0+),x\in (0,1),$ with $g(0)=0.$ Then $g$ is continuous on $[0,1)$ and differentiable on $(0,1),$ with $g'(x)=f'(x).$ Let $h(x)=\arctan (g(x))-x^2/2.$ Then for $x\in (0,1),$
$$h'(x) =\frac{g'(x)}{1+g(x)^2}-x = \frac{f'(x)}{1+g(x)^2}-x\ge \frac{f'(x)}{1+f(x)^2} - x \ge 0.$$
Because $h$ is continuous on $[0,1)$ and differentiable on $(0,1),$  the MVT shows $h$ is increasing on $[0,1).$ Thus $\lim_{x\to 1^-} h(x)\ge 0.$ This implies $\lim_{x\to 1^-} \arctan (g(x)) \ge 1/2,$ and thus $\lim_{x\to 1^-}g(x) \ge \tan (1/2).$ Since $f\ge g,$ we learn $\lim_{x\to 1^-}f(x) \ge \tan (1/2).$ As $f(1)$ is an upper bound for the last limit, we finally arrive at $f(1)\ge \tan (1/2).$
To see $\tan (1/2)$ is sharp as a lower bound, consider $f(x) = \tan (x^2/2).$
