I have some doubts about a proof Ive made so I would appreciate if someone gives its opinion or correction. I proved the following result: Let $F:[a,b]\to \mathbb{R} $ be an absolutely continuous function and monotone increasing. Let $A= F (a)$ and $B = F (b)$, and $H:[A,B] \to \mathbb{R}$ an absolutely continuos function, hence $H\circ F$ is absolutely continuous.
The motivation of this is that if I correctly proved this then I can prove the change of variable formula: $$\int_a^b f (F (t))F'(t)dt = \int_A^B f (u)du$$
Being $f \in \mathcal{L}([A,B])$
My proof is the following: 1) Given $\epsilon > 0$ since $H$ is A.C there exists $\eta >0$ such that for all finite collection of disjoint open intervals $(c_i, d_i)$ such that $\sum_{i=1}^N (c_i-d_i) < \eta $ then $\sum_{i=1}^N |H (c_i) - H(d_i)| < \epsilon$
2) Now since $F$ is AC there exists $\delta $ such that for all finite collection of disjoint intervals $(a_i, b_i)$ such that the sum of their lengths is less than $\delta $ then $\sum_{i=1}^N |F (a_i) - F (b_i)| < \eta $.
One would like to apply 1) to the intervals of the form$ (F (b_i), F (a_i))$ but maybe they are not disjoint so there is a little work to do before.
3) Since F is monotone increasing, $F( (a_i,b_i))$ is an interval formed by $F (a_i) $ and $F (b_i) $. Now I consider the collection of intervals of the form $(F (a_i), F (b_i))$ if $F (a_i)\neq F (b_i) $. Since $F $ is increasing and $F (b_i) \neq F (a_i) $ the open intervals considered are disjoint. Moreover, by 2) the sum of their lengths is less than $\eta $ so I can apply 3) to get $$\sum |H (F (b_i))- H (F (a_i))| < \epsilon$$ for $i $ such that $F (b_i) \neq F (a_i) $. To finish, if $i$ does not meet this condition there is mo problem since $F (a_i) = F (b_i)$ implies $|H (F (a_i))- H (F (b_i))|=0 $ and hence it does not contribute to the last sum, so finally we can conclude that $$\sum_{i=1}^N |H (F (a_i)) - H (F (b_i))|< \epsilon$$ and therefore $H \circ F $ is A.C