I came across this question:
Question: Let $f$ be a real continuous function defined on [0,1] such that $f(0) = 0$ and $f(1) = 1$. Prove or give a counter-example to the following:
a) If $f'$ exists a.e. in $[0,1]$, then $\int_0^1 f' dx = 1$.
b) If $f'$ is absolutely continuous in $[0,1]$, then $\int_0^1 f' dx = 1$.
c) If $f'$ is non-decreasing in $[0,1]$, then $\int_0^1 f' dx = 1$.
Thoughts: I feel like I can just take $f'(x) = x^{1/2}$ for all three parts and thus I get a counter example for all three conditions? Am I missing something or is this correct?
Create a function that is constant on each middle third in the construction of the Cantor set. To make it continuous and yet pass through $(0,0)$ and $(1,1)$, you need to 'slowly' go up. The way to do this is to iteratively build the function by defining it on the middle thirds first, such that the function on each third has value halfway between the value on the adjacent thirds in earlier steps of the Cantor construction. The remaining function values can be obtained by limits.
It then remains to prove that the function solves (a) and (c). On the other hand, (b) is true but involves a fair bit of machinery. How much do you know?