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I know that the definition of absolute continuity is

for a real-valued function $f$, which is defined on $[a,b]$, if $\forall\epsilon>0$, $\exists\delta>0$, such that whenever $\{(a_k, b_k)\}_{k=1}^{n}$ are disjoint open intervals on $[a,b]$ and $\sum\limits_{k=1}^{n}|a_k-b_k|<\delta$, $\sum\limits_{k=1}^{n}|f(a_k)-f(b_k)|<\epsilon$.

So here is the question, if I replace the last condition with $|\sum\limits_{k=1}^{n}(f(a_k)-f(b_k))|<\epsilon$, does the function $f$ remain absolutely continuous?

I have tried to prove it or give a counterexample but all these efforts have failed. And I would like to appreciate it if you could help me solve this problem.

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1 Answer 1

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Just apply this new hypothesis to the subcollection of the intervals where $f(b_k) \geq f(a_k)$ and then to the subcollection of the intervals where $f(b_k) < f(a_k)$. You will see that $f$ is necessarily absolutely continuous.

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  • $\begingroup$ Thanks a lot! I was focusing on putting forward a counterexample, and I haven't noticed such a straight-forward idea. $\endgroup$
    – Ding
    Commented Sep 7, 2020 at 13:28

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