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Let $f:(a,b) \to \mathbb R$ be Lipschitz.

The derivative $f'$ exists on some set $D \subset (a,b)$ of full measure and is bounded (by Rademacher).

Is $f'$ continuous (or some representative) on the same set $D$ or on some other set of full measure $C \subset D$?
In other words, is $f': C \to \mathbb R$ continuous?

If not, do you have a counterexample?

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  • $\begingroup$ The space $C([0,1])$ is a closed subspace of $L^\infty([0,1])$. The inclusion is proper since $C([0,1])$ is separable while $L^\infty$ is not separable. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 17 at 14:18
  • $\begingroup$ @Liviu Nicolaescu I think this doesn’t answer OP’s question specifically, because he asks for continuity on $D$ itself only - I assume he means continuous as a function $D \to \mathbb R$. $\endgroup$
    – Nate River
    Commented Oct 17 at 14:33
  • $\begingroup$ @NateRiver yes, thank you. I'll edit the question to make it clearer. $\endgroup$
    – PapierFlieger
    Commented Oct 17 at 14:36

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The answer is no, even if you restrict to a full measure subset of $D$. Let $(a, b) = A \cup A^c$ be a partition of $(a, b)$ into subsets with nonzero measure in every interval, and take $f(x) := \int_a^x \mathbf 1_{A}$.

Then a.e. on $A$, we have $f’ = 1$, while a.e. on $A^c$ we have $f’ = 0$ by the Lebesgue differentiation theorem, which shows that this is indeed a counterexample.

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  • $\begingroup$ I don't understand, why $f'$ is a counterexample? It seems to me, that $f'$ is continuous a.e. If we take any $x$ which is not on the boundary of any interval of $A$ or $A^c$, then, since the intervals have nonzero measure, there exists a neighborhood s.t. $f'(x) = f'(y)$ for a.e. $y$ in that neighborhood. $\endgroup$
    – PapierFlieger
    Commented Oct 17 at 14:50
  • $\begingroup$ Since $A^c$ has nonzero measure in every interval, for every $x \in A$ and $\delta > 0$, there exist a nonzero measure set of points $y \in B_\delta (x) \cap A^c$. These points witness that $f’$ is not continuous at $x$. $\endgroup$
    – Nate River
    Commented Oct 17 at 14:55
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    $\begingroup$ Though counterintuitive at first glance, such sets do exist - see for instance this StackExchange post. $\endgroup$
    – Nate River
    Commented Oct 17 at 15:37
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    $\begingroup$ @PapierFlieger: I struggled whether such sets exists --- As one might expect for seemingly exotic objects (e.g. nowhere continuous differentiable functions, nowhere analytic $C^{\infty}$ functions, bounded derivatives that are not Riemann integrable on any interval, strictly increasing functions with zero derivative almost everywhere, many results in convex geometry, etc.), "MOST measurable sets in the sense of Baire category have this property". $\endgroup$
    – Dave L Renfro
    Commented Oct 17 at 21:36
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    $\begingroup$ e.g. nowhere continuous differentiable functions --- This is a bit TOO exotic! Of course I meant to say "nowhere differentiable continuous functions". $\endgroup$
    – Dave L Renfro
    Commented Oct 17 at 22:10

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