Does there exist a monotone function $f: [0, 1] \to \mathbb R$ that is differentiable everywhere, but its derivative is discontinuous a.e.?
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1$\begingroup$ Volterra's function is differentiable everywhere and discontinuous on a set of positive measure. Since it also has bounded derivatives, you can add a linear function to it to make it monotone. If you work a bit harder probably you can also make it full measure. Certainly without the monotone assumption the result holds (see this MSE post). $\endgroup$– Willie WongCommented Feb 4, 2021 at 16:48
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$\begingroup$ @willie: I guess you want to say, ``and its derivative is discontinuous on a set of positive measure''. $\endgroup$– Dirk WernerCommented Feb 4, 2021 at 17:28
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I think you want the Pompeiu function (Feel free to add questions on the details of the construction).
answered Feb 4, 2021 at 17:05
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$\begingroup$ Thank you! What a nice construction. I assume one has to do some extra work to make the set of discontinuity measure 1? $\endgroup$ Commented Feb 5, 2021 at 5:17
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1$\begingroup$ @Nate River: To get the discontinuity set to have measure $1$ (i.e. the continuity set to have measure $0),$ or even more extreme to get the continuity set to have Hausdorff $h$-measure zero for any specified (in advance) Hausdorff measure function $h$ (in particular, the continuity set can have Hausdorff dimension zero, which is MUCH smaller than having Lebesgue measure zero), you only need to choose $A = \mathbb Q$ and the closed sets $F_1 \subset F_2 \subset \cdots$ (continued) $\endgroup$ Commented Feb 5, 2021 at 11:49
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1$\begingroup$ in Theorem 1 (top p. 56) of this paper so that the $F_k$ are nowhere dense such that their union contains no rational numbers and the complement relative to $[0,1]$ of their union has Lebesgue measure zero (or has Hausdorff $h$-measure zero for the stronger version). Now use the construction in Theorem 2 (bottom p. 56), letting $A$ and $B$ be a partition of $\mathbb Q$ into two disjoint dense sets. Sufficiently "thick" Cantor sets (containing only irrationals) can be used for the sets $F_k.$ $\endgroup$ Commented Feb 5, 2021 at 11:59