I came across the following challenging problem in my self-study:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be continuous. Then the set of points where $f$ is differentiable is a measurable set.

I am having trouble thinking of where to begin in proving this result, and wanted to see if anyone visiting had some suggestions on how to proceed.


4 Answers 4


Here's an attempt to salvage Matthew Pancia's solution, which unfortunately depended on an uncountable union over all possible derivatives.

Given $f$ we can define, in the obvious way, a continuous function $F:\mathbb R\times(\mathbb R\setminus 0)\to \mathbb R$ such that $f$ is differentiable at $x$ exactly when $\lim_{h\to 0} F(x,h)$ exists. The usual formalization of this is $$\exists y:\forall\varepsilon:\exists \delta:\forall h: |h|<\delta\Rightarrow |F(x,h)-y|<\varepsilon$$ Classically all of the variables here are real, but it is easy to see that we can restrict $\varepsilon$ and $\delta$ to $\mathbb Q$ without changing the meaning. We can also restrict $h$ to $\mathbb Q$ because $F$ is continuous. However, it is essential that $y$ can be an arbitrary real, because otherwise we would be looking for points where $f$ is differentiable with rational derivative, which is something quite different.

However, we can also formalize the existence of a limit like $$\forall\varepsilon:\exists \delta:\exists Y:\forall h: |h|<\delta \Rightarrow |F(x,h)-Y|<\varepsilon$$ This works because $\mathbb R$ is complete; it is essentially the same as changing "has a limit" about a sequence to "is Cauchy". The arguments that $\varepsilon$, $\delta$ and $h$ can be restricted to the rationals work as before, but now $Y$ can also be taken to be a rational in each case.

For each particular choice of $\varepsilon$, $\delta$, $Y$, and $h$, the set of $x$ such that $|h|<\delta\Rightarrow |F(x,h)-Y|<\varepsilon$ is open and therefore Borel.

Now handle each of the quantifiers from the inside out: For each choice of $\varepsilon$, $\delta$, and $Y$, the set of $x$ such that $$\forall h: |h|<\delta \Rightarrow |F(x,h)-Y|<\varepsilon$$ is a countable intersection of Borel sets and therefore Borel. For each choice of $\varepsilon$ and $\delta$ the set of $x$ such that $$\exists Y:\forall h: |h|<\delta \Rightarrow |F(x,h)-Y|<\varepsilon$$ is a countable union of Borel sets and therefore Borel. And so forth. At the top we find that the set of points of differentiability is Borel and thus in particular measurable.

  • $\begingroup$ This is a nice way of thinking about it. I like the analogy to the classic result that the set of points where a function is continuous is $G_\delta$ except here we get (assuming I'm counting my $\delta$s and $\sigma$s correctly $G_{\delta \sigma \delta \sigma})$. Makes me curious if that result is sharp in the Borel heirarchy. $\endgroup$ Feb 5, 2012 at 0:50
  • $\begingroup$ Nice proof, but it's not clear why for each particular choice of $\varepsilon$, $\delta$, $Y$, and $h$, the set of $x$ such that $|h|<\delta\Rightarrow |F(x,h)-Y|<\varepsilon$ is open, I think it's close. Isn't? $\endgroup$
    – Tom Ryddle
    Sep 24, 2019 at 1:40
  • $\begingroup$ @TomRyddle: If $|h|\ge\delta$, then the set is $\mathbb R$, which is open. Otherwise it is $\{x: |F(x,h)-Y| < \varepsilon\}$, which is open because $|F(x,h)-Y|$ is a continuous function of $x$. $\endgroup$ Sep 24, 2019 at 1:49

Theorem 1 Let $f$ be a measurable function on $(a,b)$. Then the function $$ g_n(x)=\sup\left\{f(x+h):h\in(0,\frac{1}{n})\cap(0,b-x)\right\} $$ is measurable.

Proof Fix $c\in\mathbb{R}$. We want to prove that the set $A=\{x\in(a,b):g_n(x)>c\}$ is measurable, then $g_n$ will be measurable. Let $x_0\in A$, then there exist $h_0\in(0,\frac{1}{n})\cap(0,b-x_0)$ such that $f(x_0+h_0)>c$. Define $$ \delta_1=\frac{1}{2}\left(\min\left(\frac{1}{n},b-x_0\right)-h_0\right)\qquad \delta_2=\frac{1}{2}\left(\min\left(\frac{1}{n},b-x_0\right)+h_0\right) $$ then for all $x\in(x_0-\delta_1, x_0+\delta_2)$ we have $x_0+h_0\in\{x+h:h\in(0,\frac{1}{n})\cap(0,b-x)\}$. As the consequence $g_n(x)\geq f(x_0+h_0)>c$, so $x\in A$ for all $x\in(x_0-\delta_1, x_0+\delta_2)$. This means that $A$ is open, hence measurable.

Theorem 2 Let $f$ be a mesurable function on $(a,b)$ then the set $$ A=\{x\in(a,b):\exists f'(x)\in\mathbb{R}\} $$ is measurable.

Proof Extend $f$ by equalities: $f(x)=f(a)$ for $x<a$ and $f(x)=f(b)$ for $x>b$. For each $c\in\mathbb{R}$ consider measurable function $\phi(x)=f(x)-cx$. From theorem 1 it follows that $$ \psi_n(x)=\sup\left\{\phi(x+h):h\in(0,\frac{1}{n})\cap(0,b-x)\right\} $$ is measurable. Then we have measurable function $$ \psi_n(x)-\phi(x)=\sup\left\{\phi(x+h)-\phi(x):h\in(0,\frac{1}{n})\cap(0,b-x)\right\}. $$ Hence the set $$ B=\{x\in(a,b):\psi_n(x)-\phi(x)>0\}= $$ $$ \left\{x\in(a,b):\sup\left\{\frac{f(x+h)-f(x)}{h}:h\in(0,\frac{1}{n})\cap(0,b-x)\right\}>c\right\} $$ is measurable. Since $c\in\mathbb{R}$ is arbitrary then the functions $$ f_n(x)=\sup\left\{\frac{f(x+h)-f(x)}{h}:h\in(0,\frac{1}{n})\cap(0,b-x)\right\} $$ are measurable. So we conclude that the function $$ \overline{f}'_+(x)=\overline{\lim\limits_{h\to 0+}}\frac{f(x+h)-f(x)}{h}=\lim\limits_{n\to\infty}f_n(x) $$ is also measurable. Similarly, we can prove that functions $$ \underline{f}'_+(x)=\underline{\lim\limits_{h\to 0+}}\frac{f(x+h)-f(x)}{h}=\lim\limits_{n\to\infty}f_n(x) $$ $$ \overline{f}'_-(x)=\overline{\lim\limits_{h\to 0-}}\frac{f(x+h)-f(x)}{h}=\lim\limits_{n\to\infty}f_n(x) $$ $$ \underline{f}'_-(x)=\underline{\lim\limits_{h\to 0-}}\frac{f(x+h)-f(x)}{h}=\lim\limits_{n\to\infty}f_n(x) $$ are mesurable. Finally, the set $A=\{x\in(a,b):\exists f'(x)\in\mathbb{R}\}$ is measurable because $$ A=\{x\in(a,b):\overline{f}'_+(x)=\underline{f}'_+(x)=\overline{f}'_-(x)=\underline{f}'_-(x)\in\mathbb{R}\} $$


What follows is taken from a sci.math post of mine from 16 May 2006.

Let $f:{\mathbb R} \rightarrow {\mathbb R}$ be arbitrary. Then the set of points where $f$ does not have a finite derivative is a $G_{\delta \sigma}$ set. This is also true for the set of points where $f$ does not have a finite left derivative or the set of points where $f$ does not have a finite right derivative. A proof of this is given after the next paragraph.

Conversely, given any $G_{\delta}$ set $G_1$ and $G_{\delta \sigma}$ set $G_2$ such that $G_2$ has Lebesgue measure zero, then there exists a continuous function $f:{\mathbb R} \rightarrow {\mathbb R}$ whose non-differentiability set is precisely $G_1 \cup G_2.$ This was proved by Zygmunt Zahorski (1914-1998) in 1941 and re-published in French in 1946 (Zbl 61.11302; MR 9,231a). For the 1946 French paper, see http://tinyurl.com/3zwca22.

The first statement above for continuous $f$ follows from the fact that $f$ has a finite derivative at $x$ if and only if $x$ belongs to

$$\bigcap_{k=1}^{\infty} \;\bigcup_{n=1}^{\infty} \;\; \bigcap_{0 < |\eta| < \frac{1}{n}} \;\; \bigcap_{0 < |\delta| < \frac{1}{n}} \; \left\{x:\;\; \left| \frac{f(x + \eta) - f(x)}{\eta} \;\; - \;\; \frac{f(x + \delta) - f(x)}{\delta} \right| \; \leq \frac{1}{k}\right\}$$

To see that this result holds for an arbitrary function $f$ (not necessarily continuous), we relativize the above decomposition to $C(f)$, the continuity set of $f$. That is, consider the sets $\{x \in C(f): |$ stuff $| \leq \frac{1}{k}\}.$ This will show that the set where $f$ has a finite derivative is an $F_{\sigma \delta}$ subset of the space $C(f),$ where $C(f)$ has the subspace topology from ${\mathbb R}.$ ("Subset", because every point at which a finite derivative exists is a point of continuity.) Hence, this set is the intersection of $C(f)$ with an $F_{\sigma \delta}$ set in ${\mathbb R}$ (subspace topology stuff), and therefore this set is $F_{\sigma \delta}$ in ${\mathbb R}$ (because $C(f)$ has a lower Borel class, being $G_{\delta}$ in ${\mathbb R}).$

  • $\begingroup$ While your argument seems to be working nicely for continuous functions, I am not so sure if its extension to arbitrary (or even Borel measurable) functions goes through. Even if you restrict the domain of interest to $C(f)$, the shift by $\eta$ or $\delta$ may bring us out of $C(f)$, where wild things can happen. That is, even if $x\in C(f)$, there is no guarantee that $x+\eta\in C(f)$. For this reason, the map $x\mapsto f(x+\eta)$ need not be continuous even on $C(f)$ and thus $\{x\in C(f)\,|\,|\text{stuff}|\leq k^{-1}\}$ need not be closed. $\endgroup$
    – triple_sec
    Aug 17, 2016 at 22:54
  • 1
    $\begingroup$ @triple_sec: Two mornings ago I looked this over and sketched out some ideas for follow-up clarifying comments (for example, the equivalence of finite differentiability and being in the set above seems to require two cases, one case in which the upper derivative differs from the lower derivative as extended real numbers, and another case in which there is a signed-infinite derivative), and these included replacing $f$ with $f$ restricted to $C(f)$ along with restricting the unions and intersections to be only over values where $x+ \eta$ and $x+ \delta$ belong to $C(f).$ (continued) $\endgroup$ Aug 22, 2016 at 19:03
  • 2
    $\begingroup$ (continuation) But when I looked at it again this morning, things didn't look so well. The result is true -- see Theorem 3.12 on pp. 315-316 of Garg's 1975 Trans. AMS paper On bilateral derivates and the derivative. Casper Goffman's Ph.D. Dissertation also contains the result in a more general form involving limits of interval functions, and surely it's in Hans Hahn's 1921 and 1932 books (haven't looked yet). I'll see if I can disprove the above approach by an example, or repair the approach, when I get time. $\endgroup$ Aug 22, 2016 at 19:11
  • $\begingroup$ Thank you very much for your detailed reply. BTW, I very much like your approach: you overcome the issue of uncountable set operations (which, in general, do not preserve measurability) by exploiting the fact that the innermost “stuff set” is closed and that the union of arbitrarily many closed set is closed. It works nicely for continuous functions, yet my concern is precisely that the approach unfortunately may fail for generic (or even Borel measurable) functions. (continues below) $\endgroup$
    – triple_sec
    Aug 22, 2016 at 22:45
  • $\begingroup$ (continued from above) I skimmed the literature and, indeed, found that the claim is nevertheless true for arbitrary functions, yet the proofs I found are surprising abstruse (one textbook actually states the result as an exercise). To be honest, I think it is such a shame. This result is both surprising and elegant, so I cannot believe it did not have wider coverage in textbooks that are famous for their simple and accessible proofs (although, for instance, Billingsley does prove a less general result for measurable functions whose discontinuity points are co-countable). $\endgroup$
    – triple_sec
    Aug 22, 2016 at 22:49

You can write the set of points where $f$ is differentiable as the set of points where a particular limit exists (namely, the one that defines the derivative). You can codify a limit (using the $\epsilon,\delta$ definition) in terms of intersections/unions of measurable sets.

What you want to say is that the function $f$ is differentiable at $x$ (with value $y$) if for all $\epsilon$ there exists a $\delta$ such that the difference quotient with parameter $\delta$ is $\epsilon$-close to $y$. First fix an $\epsilon$. Then the set of points where this is true for that particular $\epsilon$ is is the union (over the possible $\delta$) of the sets such that the difference quotient is $\epsilon$-close with that fixed $\delta$. This must be true for all $\epsilon$, so we take the intersection of those sets over all $\epsilon$.

The net result is that we are taking $$\cap_\epsilon \cup_\delta V(\epsilon, \delta)$$ where $V(\epsilon, \delta)$ is the set of points such that the difference quotient is less than $\epsilon$ from $y$. Of course, this was just for one possible value of the derivative, so we need to take the union over all $y$ as well, giving us $$\cup_y \cap_\epsilon \cup_\delta V(\epsilon, \delta).$$

For everything involved to be countable, you'd want to choose the $y$, $\epsilon$, $\delta$ to be rational, which is fine because the rationals are dense in $\mathbb{R}$.

  • $\begingroup$ Taking $\epsilon$ and $\delta$ to be rational is fine, but I don't think restricting $y$ to the rationals will work here. $\endgroup$ Feb 5, 2012 at 0:15
  • $\begingroup$ I agree. $\ \ \ \ \ $ $\endgroup$ Feb 5, 2012 at 0:16
  • $\begingroup$ Ahh, yes. This problem was addressed by Henning above. That's what you get for not thinking about your responses for too long, huh? :) $\endgroup$ Feb 5, 2012 at 2:55

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