Let $\lambda$ be the Lebesgue measure on $[a, b]$ and $f:[a, b] \to \mathbb{R}$ $\lambda$-integrable. We define $F:[a, b] \to \mathbb R$ by $F(x) := \int_a^x f \mathrm d \lambda$ for all $x \in [a, b]$. In order to prove a characterization of absolutely continuous function, I come across below result, i.e.,
Theorem The map $F$ is absolutely continuous.
Could you confirm if my below attempt is fine?
Proof Fix $\varepsilon>0$. Let $\delta>0$ be given by below lemma, i.e.,
Lemma For each $\varepsilon>0$, there is $\delta >0$ such that if $B$ is $\lambda$-measurable with $\lambda (B) < \delta$ then $\int_{B} |f| \mathrm d \lambda < \varepsilon$.
Let $\{ [x_i, y_i] : i= 1, \ldots, n \}$ be a collection of pairwise disjoint subintervals of $[a, b]$ such that $\sum_{i=1}^n (y_i-x_i) < \delta$. This implies $\lambda (B) < \delta$ for $B := \bigcup_{i=1}^n [x_i, y_i]$. Hence $$ \sum_{i=1}^n |F(y_i)-F(x_i)| = \sum_{i=1}^n \left | \int_{x_i}^{y_i} f \mathrm d \lambda \right| \le \sum_{i=1}^n \int_{x_i}^{y_i} |f| \mathrm d \lambda = \int_B |f| \mathrm d \lambda < \varepsilon. $$
This completes the proof.
Proof of the Lemma There is a simple function $g:[a, b] \to [0, \infty)$ such that $0 \le g \le |f|$ and $$ 0 \le \int_a^b (|f| -g) \mathrm d \lambda < \frac{\varepsilon}{2}. $$
We assume $g$ has a form $g = \sum_{i=1}^m a_i 1_{A_i}$ where $a_i \ge 0$ for $i = 1, \ldots, m$ and $(A_i)_{i=1}^m$ is a sequence of pairwise disjoint $\lambda$-measurable sets. We have $$ \int_B g \mathrm d \lambda = \sum_{i=1}^m a_i \lambda (A_i \cap B) \le \lambda(B) \sum_{i=1}^m a_i. $$
If $\lambda(B) \le \dfrac{\varepsilon}{2 \sum_{i=1}^m a_i}$, then $$ \int_B |f| \mathrm d \lambda = \int_B (|f| - g) \mathrm d \lambda +\int_B g \mathrm d \lambda\le \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. $$
