I'm trying to prove this well-known result, i.e.,
Theorem If $f:[a, b] \to \mathbb R$ is absolutely continuous, then it has bounded variation.
Could you have a check on my below proof?
My attempt Fix $\varepsilon>0$. By absolute continuity of $f$, there is $\delta>0$ such that for every finite collection of disjoint subintervals $(a_1, b_1), \ldots, (a_n, b_n)$ of $[a, b]$ with $\sum_{i=1}^{n}|a_i-b_i|<2\delta$ we have $\sum_{i=1}^{n}|f(a_i)-f(b_i)|<\varepsilon$.
Let $P_1 = \{a= x_1 < x_2 < \cdots < x_n = b\}$ be a partition of $[a, b]$. WLOG, we assume $x_{i+1} - x_i < \delta/2$ for all $i=1,\ldots, n-1$. Let $m := \left \lfloor \frac{b-a}{\delta} \right \rfloor$ and $$ P_2 := \{ a <a+\delta< a+ 2\delta <\cdots <a+ m\delta < b\}. $$
Let $I := \{i : x_i \in [a, a+\delta]\}$. Let $j_1 := \min I$ and $j_2 := \max I$. Then $$ \sum_{i=j_1}^{j_2-1}|x_{i+1}-x_i| \le \delta < 2\delta. $$
So $$ \sum_{i=j_1}^{j_2-1}|f(x_{i+1})-f(x_i)| <\varepsilon. $$
We have at most $m+1$ such intervals as $I$. Hence $$ \sum_{i=1}^{n-1}|f(x_{i+1})-f(x_i)| \le (m+1) \varepsilon \le \left (\frac{b-a}{\delta}+1 \right) \varepsilon. $$
This completes the proof.
