$f: [0,1] \to \mathbb{R}$ absolutely continuous, $f' \in \{0,1\}$ (a.e.), $f(0)=0$. Prove that for some measurable subset $A$, $f(x)=m(A \cap (0,x))$ Problem: Suppose that $f: [0,1] \to \mathbb{R}$ is absolutely continuous, $f' \in \{0,1\}$ (a.e.) and $f(0)=0$. Prove that for some measurable subset $A \subset [0,1]$ and every $x \in [0,1]$ we have $f(x)=m(A \cap (0,x))$
I have proved it on my own. Please help me verify and tell me how much of the problem you think I have solved. (This was an exam problem and I'm trying to verify my given answer). I didn't have time to study derivatives for the real analysis exam and my argument might lack a bit of rigor, I think.
Edit: I'm going to write down my proof as an answer to mark this question solved.
 A: 
Suppose that $f: [0,1] \to \mathbb{R}$ is absolutely continuous, $f' \in \{0,1\}$ (a.e.) and $f(0)=0$. Prove that for some measurable subset $A \subset [0,1]$ and every $x \in [0,1]$ we have $f(x)=m(A \cap (0,x))$

Since $f$ is absolutely continuous, we know that it comes from an indefinite integral like below:
$$f(x) = f(0) + \int_0^xf'$$
Set $A = \{x\in [0,1] : f'(x) = 1\}$, $B = \{x\in [0,1] : f'(x) = 0\}$ and $C = (A\cup B)^c$. By assumption, $m(C)=0$. Moreover, since $f'$  is a measurable function (being the limit of the measurable function $\mathrm{Diff}_{1/n}(f)$ when $n$ goes to infinity), $A$ and $B$ are measurable sets and $C$ is measurable because its measure is $0$ and we're working with Lebesgue measure.
Using the assumptions, we have:
$$f(x) = 0 + \int_0^x f' \times \mathbf{1}_A + \int_0^x f' \times \mathbf{1}_B + \int_0^x f' \times \mathbf{1}_C$$
But the two integrals on the right will vanish and we'll get
$$f(x) = \int_0^x1\times \mathbf{1}_A = m(A \cap(0,x))$$
Remark: $\mathbf{1}_X$ denotes the indicator function on the set $X$.
