# Set of critical values of one-dimensional continuously differentiable function has measure zero.

I am fighting with the following exercise:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuously differentiable function, $A:=\{x \in \mathbb{R} \ \vert \ f^{\prime}(x)=0 \}$ and $\lambda$ the Lebesgue-measure on the real line. Show that $\lambda(f(A))=0$. Hint: The mean-value theorem may turn out to be useful. What can we say about the measurability of $f(A)$?

Its 2 a.m. now and I've tried for the last few hours to find a way to prove the statement, but nothing has worked so far. I don't think that $f(A)$ is always countable and I'm sure that $A$ can sometimes be uncountable. Thanks for any advice!

• $A=(f')^{-1}\{0\}$ is closed because $f'$ is continuous. $f(A)=\cup_{n\in Z} \{f(A\cap [n,n+1])\}$ is an $F_{\sigma}$ set because each $g_n=f(A\cap [n,n+1])$ is the continuous image of a compact set ,so $g_n$ is closed. So $f(A)$ is measurable. – DanielWainfleet Dec 11 '15 at 4:18
• the following question might be helpful as it discusses related ideas math.stackexchange.com/questions/1045382/… – Mirko Dec 11 '15 at 4:22
• See math.stackexchange.com/questions/648237/…, a few details are missing but the answer there gives the general idea. – Alp Uzman Dec 11 '15 at 4:31

A different approach might help your thinking on this. Locally what does a derivative do? If the value $f'(x_0)=\pm 5$, then closed intervals $J$ close to that point get expanded in length by a factor of $5$, i.e,. $f(J)$ is about five times the length of $J$. At least that is roughly how we would explain "what derivatives do" to an elementary student. (We would have to sell the idea that, for continuous functions, $f$ maps a closed interval $J$ to a closed interval $f(J)$.)

At the level of a course in measure theory we shouldn't forget our baby intuition. It is not difficult with the usual apparatus of the Lebesgue outer measure to prove this lemma, which is just the same kind of "expansion" idea, except that rather than "intervals" expanded in length "sets" get expanded in measure.

Lemma. Let $F:\mathbb R\to \mathbb R$ be an arbitrary function that has a derivative at every point of a set $E$ and suppose that $|f'(x)|\leq M$ for every $x\in E$. Then $$\lambda^*(f(E))\leq M \lambda^*(E).$$

[Aside: Actually you don't need the existence of a derivative. It is enough if $-M \leq D_- f(x)$ and $D^+ f(x)\leq M$ where these are, respectively the lower left Dini derivative and the upper right Dini derivative. But assuming a derivative makes it marginally easier to construct a proof.]

Now your problem is a trivial consequence of the lemma, without any fussing over continuity of the derivative and the mean-value theorem which are quite irrelevant to the nature of the observation. If $f'(x)=0$ for all $x\in E$ then $|f'(x)|<\epsilon$ for any $\epsilon>0$ and so $$\lambda^*(f(E))\leq \epsilon \lambda^*(E).$$ It follows that, if $E$ is bounded, then $\lambda^*(f(E))=0$. For the unbounded case just use a sequence $E\cap [-n,n]$.

Note that measurability of $f(E)$ then follows immediately since it has Lebesgue outer measure zero.

• Very nice answer ! To prove your lemma, do you use Vitali coverings, or is there another proof ? – charmd Jan 9 '17 at 19:01
• @charMD It is a fairly straightforward computation and does not need anything deep like Vitali coverings. Let me refer you to one of the masters, a text that you should read in any case if you have ambitions in real analysis: Saks, Theory of the Integral, Lemma (6.3) on page 226. it is public domain now and there are free PDFs floating around somewhere. – B. S. Thomson Jan 10 '17 at 20:01
• Great, I was able to find the text. I asked this question because of hot_queen's post on this link math.stackexchange.com/questions/1045382/…, which uses Vitali coverings . Thank you very much for your answer – charmd Jan 10 '17 at 20:40