Let $\lambda$ be the Lebesgue measure on $[a, b]$ and $f:[a, b] \to \mathbb R$. I would like to prove below property of an absolutely continuous function, i.e.,
Theorem If $f$ is absolutely continuous and $f' = 0$ $\lambda$-a.e., then $f$ is constant.
Could you confirm if my below attempt is fine?
Proof Fix $c \in [a, b]$. We want to prove that $f(c)= f(a)$. Fix $\varepsilon>0$. There is $\delta>0$ such that $\sum_{i=1}^{n} |f(y_{i})-f(x_i)| < \varepsilon$ for every finite collection of pairwise disjoint subintervals $\{[x_i, y_i] : i=1, \ldots, n\}$ of $[a, b]$ with $\sum_{i=1}^n (y_{i}-x_i) <\delta$.
Let $D := \{x\in [a, c] : f'(x)=0\}$. For each $x \in D$, there is an interval $[a_x, c_x] \subset [a, c]$ with arbitrarily small length such that $x \in [a_x, c_x]$ and $|f(c_x)-f(a_x)| \le \varepsilon (c_x-a_x)$. Then $\{[a_x, c_x] : x \in D\}$ is a Vitali cover of $D$. By Vitali's covering theorem, there is a finite subcollection $\{[a_i, c_i] : i=1, \ldots, n\}$ of pairwise disjoint intervals such that $$ \lambda \bigg ( D \setminus \bigcup_{i=1}^n [a_i, c_i] \bigg ) < \delta. $$
Because $\lambda ([a, c] \setminus D)=0$, we get $$ \lambda \bigg ( [a, c] \setminus \bigcup_{i=1}^n [a_i, c_i] \bigg ) < \delta. $$
WLOG, we assume $$ a =: c_0 \le a_1 \le c_1 \le a_2 \le c_2 \le \cdots \le a_n \le c_n \le a_{n+1} :=c. $$
Let $$ (x_1, y_1) := (c_0, a_1), (x_2, y_2) := (c_1, a_2), \ldots, (x_{n+1}, y_{n+1}) := (c_n, a_{n+1}). $$
Then $$ \sum_{i=1}^{n+1} (y_i-x_i) = \lambda \bigg ( [a, c] \setminus \bigcup_{i=1}^n [a_i, c_i] \bigg ) < \delta. $$
Then $$ \begin{align} |f(c)-f(a)| &\le \sum_{i=1}^{n+1} |f(y_{i}) - f(x_i)| + \sum_{i=1}^{n} |f(c_{i}) - f(a_i)| \\ &\le\varepsilon + \sum_{i=1}^{n} \varepsilon (c_{i} - a_i) \\ &\le \varepsilon (1 + c-a). \end{align} $$
We take the limit $\varepsilon \downarrow 0$ and obtain the result.
