Absolute continuity of compositions

Question

Let $f:[a,b] \to [\alpha,\beta]$ be an absolutely continuous function and $g:[\alpha,\beta] \to \mathbb R$ a Lipschitz-continuous function. How can I show that then $g\circ f$ is absolutely continuous again?

Answer 1

Since $g$ is $LC$ on $[\alpha,\beta]$, we can find $M>0$ such that $$x,y\in [\alpha,\beta]\quad \implies\quad |g(x)-g(y)|<M|x-y|.\qquad (*)$$ Let $\epsilon>0$. Since $f$ is $AC$, we can find $\delta>0$ such that $$\sum_{i=1}^n|f(t_i)-f(s_i)|<\frac{\epsilon}{M}$$ if

$\{[s_i,t_i]:i=1,\dots,n\}$ is a finite collection or mutually disjoint sub-intervals of $[a,b]$ with $\sum_{i=1}^n(t_i-s_i)<\delta$.$\qquad (**)$

Now, let us assume that statement $(**)$ holds. Clearly, $f(t_i),f(s_i)\in[\alpha,\beta]$ for each $i=1,\dots,n$. Using statement $(*)$, it follows that $|g(f(t_i))-g(f(s_i))|< M|f(t_i)-f(s_i)|$ for each $i=1,\dots,n$. Thus, \begin{align} \sum_{i=1}^n|(g\circ f)(t_i)-(g\circ f)(s_i)|&=\sum_{i=1}^n|g(f(t_i))-g(f(s_i))|\\ &\leq \sum_{i=1}^n M|f(t_i)-f(s_i)|\\ &<M\frac{\epsilon}{M}=\epsilon. \end{align} This shows that $g\circ f$ is $AC$ on $[a,b]$.