The following question is exercise 14.3 from section 14.8 in Real Analysis For Graduate Students (2nd Edition) by Richard Bass.
Suppose $f$ is a real-valued continuous function on $[0,1]$ and that $\varepsilon >0$. Prove that there exists a continuous function $g$ such that $g'(x)$ exists and equals $0$ for almost every x and $$ \sup_{x\in [0,1]}|f(x)-g(x)|<\varepsilon. $$
What I think I know: The question gives me the weakest (point-wise) continuity on $f$ in $[0,1]$. Given only this information, I need to build up some kind of bulwark of mathematical theory proving the existence of the desired function of $g$, of which I do not believe I am capable. Or alternatively there may be some way to use Radon-Nikodym or Lebesgue decomposition to infer the conclusion. In either or neither case, I am lost, and would be grateful for two things (1) a straight-forward proof and (2) a separate bit of insight/intuition that can motivate my understanding of the proof.