There was a similar question on this matter, but "continuous differentiable" is too strong an assumption, so I was wondering if this can be relaxed to "differentiable".
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I got an answer referencing Rudin'n Real and Complex Analysis, Thm 7.21, which the author deleted later because of a mistake. I am grateful that he led me to this theorem.
Assume $f\in BV([0,1])$, then $f=f_1-f_2$, where $f_i$ are bounded increasing, $f'_i$ exists a.e., and $f'=f'_1-f'_2$ a.e.. Since $\int_0^1 f'_i\leq f_i(1)-f_i(0)$, we know $f'_i\in L^1([0,1])$ and thus $f\in L^1([0,1])$. Then we can apply Rudin 7.21 (i.e. If $f$ is differentiable, $f'\in L^1([0,1])$, then $f$ is absolutely continuous)
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1Yes, I got this idea after deleting my answer but I had logged out of MSE. This argument is correct. – Kavi Rama Murthy Jul 09 '20 at 07:32
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Thanks man, I am glad this was solved so quickly. @KaviRamaMurthy – Ryan Jul 09 '20 at 08:29
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By Papa Rudin you mean his Real and Complex Analysis, right? This one is not as famous as Baby Rudin, so wanted to make explicit. – Paramanand Singh Jul 10 '20 at 12:31
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Yes @ParamanandSingh – Ryan Jul 11 '20 at 23:09