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Let $F:\mathbb R \to \mathbb R$ and $f:\mathbb R \to \mathbb R$ be absolutely continuous (a.c.) where $f$ is (not necessarily strictly) monotone. WLOG, we assume $f$ is increasing. Then $c \le d$. In solving this question, I have come across this result, i.e.,

Theorem: $F \circ f : \mathbb R \to \mathbb R$ is a.c.

Could you have a check on my attempt?

Proof: Fix $\varepsilon>0$. Because $F$ is a.c., there is $\delta>0$ such that for every countable collection $\{[a_n, b_n]\}$ of non-overlapping closed intervals, with $\sum_n (b_n-a_n) < \delta$, we have $$ \sum_n |F(b_n) - F(a_n)| < \varepsilon. $$

Because $f$ is a.c., there is $\gamma>0$ such that , for every collection $\{[c_n, d_n]\}$ of non-overlapping closed intervals with $\sum_n (d_n-c_n) < \gamma$, we have $$ \sum_n |f(d_n)-f(c_n)| < \delta. $$

Because $f$ is increasing, $f(c_n) \le f(d_n)$ and $\{[f(c_n), f(d_n)]\}$ is a collection of non-overlapping closed intervals. Hence $$ \sum_n (f(d_n)-f(c_n)) < \delta. $$

It follows that $$ \sum_n |F \circ f (d_n) - F \circ f (c_n)| =\sum_n |F(f(d_n)) - F((c_n))| < \varepsilon. $$

This completes the proof.

Akira
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  • I may be missing something here, but how does it follow that $f$ if locally Lipschitz if $f$ is differentiable? If $f$ is continuously differentiable this is immediate, but if you only know that $f$ is differentiable I'm hesitating to accept this without justification – Thomas Sep 27 '22 at 20:30
  • @Thomas you're right. It seems we need $f$ to be continuously differentiable. https://math.stackexchange.com/questions/1609160/is-there-a-function-on-a-compact-interval-that-is-differentiable-but-not-lipschi – Akira Sep 27 '22 at 20:57
  • You just need the derivative to be bounded. – copper.hat Sep 27 '22 at 21:18
  • @Thomas It seems I have found better assumptions. Could you have a check on my new edit? – Akira Sep 28 '22 at 07:07
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    now it seems correct to me: for any fixed $\epsilon >0$ you prove that exists $\gamma >0$ such that if $\sum_{n}(b_n-a_n)<\gamma $ then $\sum_{n}|F\circ f(b_n)-F\circ f(a_n)|<\epsilon $, so $F\circ f$ is a.c. –  Sep 28 '22 at 08:40

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