Differentiable and Non-decreasing with $f' = 0$ a.e. But Not Constant

Question

Context / Motivation

In this question, the OP asks for a proof that if $f : [0,1] \to [0,1]$ is increasing and differentiable with $f(0) = 0$ and $f(1) = 1$, then the length of the graph $L(f)$ is less than $2$.

As is shown in the answers to that question, if we only know that $f$ is non-decreasing, $f(0) = 0$ and $f(1) = 1$, we may conclude $L(f) \leq 2$. However, additional assumptions are required for the strict inequality.

If we additionally suppose, for example, that $f' > 0$ on some interval in $[0,1]$ (assuming nothing about the differentiability of $f$ outside of this interval), then we may easily obtain the strict inequality $L(f) < 2$.

By contrast, if we instead suppose that $f$ is differentiable almost everywhere on $[0,1]$, we cannot guarantee strict inequality. The famous Cantor function has $f(0) = 0$, $f(1) = 1$, is non-decreasing, and is differentiable almost everywhere, yet $L(f) = 2$.

The above considerations made me curious how weak the hypotheses on $f$ could be. What seems to have gone wrong with the Cantor function is that the function was not differentiable everywhere, only almost everywhere.

This made me wonder whether it would be enough to assume that $f$ is differentiable everywhere. What could go wrong in this case? The only way to still have $L(f) = 2$ would be if $f' = 0$ a.e. But is this possible? Can we have a non-decreasing differentiable function with $f' = 0$ a.e., $f(0) = 0$, and $f(1) = 1$?

My Question

Does there exist a function $f : [0,1] \to [0,1]$ such that

1. $f$ is non-decreasing
2. $f$ is differentiable everywhere (this rules out the Cantor function)
3. $f' = 0$ almost everywhere
4. $f(0) = 0$ and $f(1) = 1$

Note that through re-scaling and translation this is the same as the question in the title. I have only asked it this way here to be consistent with with the context/motivation above.

Answer 1

Rudin, Real and Complex Analysis, Theorem 7.21:

If $f:[a,b]\to\Bbb R$ is differentiable everywhere and $f'\in L^1([a,b])$ then $$\forall x\in[a,b]\quad f(x)-f(a)=\int_a^xf'(t)dt.$$

(I.e.: [...] then $f$ is absolutely continuous.)

Proof