We have a function $f:[0,1] \rightarrow \mathbb{R}$. We know it is continuous on $[0,1]$. Aside from a set $S$ of measure $0$, we can compute its right derivative, and can show that this right derivative is non-decreasing on $[0,1] \setminus S$.
Is this sufficient to show that $f$ is convex on $[0,1]$?
[If possible, it would be helpful to know also if my reasons for thinking $S$ has measure $0$ are correct: we have another function $g:[0,1] \rightarrow [0,1]$ which is monotonically non-decreasing everywhere in $[0,1]$. We can show that $S$ is equal to those locations where $g$ is discontinuous. I believe this is sufficient to show that $S$ has measure $0$ but it would be great if someone could confirm?]
Many thanks for any comments!
Take the Cantor function $c$. This has zero derivative ae. (hence non-decreasing), but is not convex (since it is constant on some intervals).
A real valued monotonic function can only have a countable number of discontinuities, which has measure zero.