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If $f:\mathbb{R}\to \mathbb{R}$ is an absolutely continuous function and let $C$ be the set of all $x\in\mathbb{R}$ such that $f'(x)=0$, is it necessarily that $f(C)$ is a Lebesgue $0$ set?

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What does $f'$ mean in this level of generality? –  David Speyer Mar 20 '12 at 16:04
Since $f$ is absolutely continuous, its derivative exists almost everywhere. –  Hezudao Mar 24 '12 at 3:11

2 Answers 2

up vote 4 down vote accepted

The set $\{f(x):f'(x)=0\}$ is Lebesgue null for any function $f:\mathbb R\to\mathbb R$ whatsoever; it does not have to be absolutely continuous, or even continuous, or even measurable. (If the derivative does not exist at $x$, the condition $f'(x)=0$ is not fulfilled.)

Proof. It suffices to prove that $\{f(x): x\in [0,1], \ f'(x)=0\}$ is null. Fix $\epsilon>0$. Cover the set $A=\{x\in [0,1], \ f'(x)=0\}$ by all intervals of the form $I(x,\delta)=(x-\delta,x+\delta)$ where $0<\delta<1$ and $\operatorname{diam} f(I(x,5\delta))\le \epsilon \,\delta$. Using Vitali covering lemma, choose a disjoint set of intervals $I(x_j,\delta_j)$ such that the intervals $I(x_j,5\delta_j)$ still cover $A$. Disjointness implies $\sum 2\delta_j\le 3$. The sets $f(I(x_j,5\delta_j))$ cover $f(A)$ and have $$\sum \operatorname{diam} f(I(x_j,5\delta_j)) \le \sum \epsilon\delta_j\le \frac32 \epsilon $$ Since $\epsilon>0$ was arbitrary, $f(A)$ is Lebesgue null. $\Box$

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Same proof works in higher dimensions, proving the result in the answer by Mher Safaryan. –  ˈjuː.zɚ79365 Jun 13 '13 at 13:36

There is such generalization of Sard's theorem in M.D.Guzman, Differentiation of Integrals in $\mathbb{R}^n:$

let $G$ be an open set of $\mathbb{R}^n$. Let $X$ be any set where there is an exterior measure $\nu$ defined on the subsets of $X$. Let $f : G\to X$ be an arbitrary function. A point $x$ of $G$ will be called critical point of $f$ if there is sequence $\{Q_k(x)\}$ of open cubic intervals centered at $x$ contracting to $x$, i.e. with $diam(Q_k(x))\to 0$, such that $$\frac{\nu(f(Q_k(x)))}{mesQ_k(x)}\to 0,\, \text{as}\, k\to\infty.$$

Let $C$ be the set of critical points of $f$. Then $\nu(f(C)) = 0$.

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