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I know that all absolutely continuous functions are differentiable almost everywhere. Is the converse true?

Or a better question: If $f$ is an almost everywhere differentiable function, what is the regularity of $f$?

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    $\begingroup$ Differentiability implies continuity. $\endgroup$
    – c1620
    Sep 5, 2020 at 20:42
  • $\begingroup$ Of corse. But, what about almost everywhere differentiability? $\endgroup$ Sep 5, 2020 at 20:44
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    $\begingroup$ The Heaviside function is differentiable almost everywhere, and definitely not continuous. $\endgroup$ Sep 5, 2020 at 20:55
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    $\begingroup$ en.wikipedia.org/wiki/Absolute_continuity By this the converse is not necessarily true since the set of absolute continuous functions is the subset of the set of almost everywhere differentiable functions, but not vice versa. $\endgroup$
    – c1620
    Sep 5, 2020 at 20:59
  • $\begingroup$ Thanks!...Therefore I conclude that an almost everywhere differentiable function f is continuous almost everywhere. Now, could $f$ be continuous at every irrational numbers and discontinuous at every rational numbers?; I think this not hapen because we need that $\lim_{h \rightarrow 0} \frac{f(x-h)-f(x)}{h}$ has sense when $h$ is small for almost every $x$ in the domine of $f$. Am I right? $\endgroup$ Sep 5, 2020 at 23:02

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I think it is possible for the function to be discontinuous on a dense set, and differentiable almost everywhere. Here is my example.

Let $f_n(x) = 2^{-n} [2^n x]$, that is, the least multiple of $2^{-n}$ less than or equal to $x$. Note that for any $x,y$ we have $|f_n(x) - f_n(y)| \le |x-y| + 2^{-n}$.

Let $$ g = \sum_{n=0}^\infty 2^{-2n} f_n .$$ Clearly $g$ is discontinuous at numbers of the form $\frac{k}{2^n}$.

Let $$ E_n := \bigcup_{k=0}^{2^n-1} [\tfrac k{2^n} + \tfrac1{2^{2n}}, \tfrac {k+1}{2^n} - \tfrac1{2^{2n}}] .$$ Note that the measure of $[0,1] \setminus E_n$ is less than $\frac1{2^{n-1}}$, and hence by the Borel Cantelli Lemma, $E = \liminf E_n$ has full measure in $[0,1]$.

We will show that $f'(x) = 0$ for $x \in E$. Suppose that $x \in E$, that is, for some $n_0 \ge 0$, we have $x \in \bigcap_{m \ge n_0} E_m$. Suppose that $|x - y| < 2^{-2n}$ for $n \ge n_0$. Then there exists an $m \ge n$ such that $2^{-2m-2} \le |x-y| < 2^{-2m}$. Since $x \in E_m$, we have that $f_l(x) = f_l(y)$ for $l \le m$. Then \begin{align} |g(x) - g(y) | &\le \sum_{l \ge m} 2^{-2l} |f_l(x) - f_l(y)| \\& \le \sum_{l \ge m} 2^{-2l}(|x-y| + 2^{-l}) \\& \le 2^{-2m} | x-y | + 2^{-3m} \\ &\le 5 \times2^{-m} |x-y| \\ &\le 5 \times2^{-n} |x-y| .\end{align} To make it discontinuous at the rationals, use $n!$ instead of $2^n$. However, I don't see how to make it differentiable at all the irrationals, only on a set of full measure.

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  • $\begingroup$ Great answer!! I am working with your example. Thanks a lot!! $\endgroup$ Sep 6, 2020 at 1:24

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