I need help to understand the proof below of the following theorem.

Let $f\colon (a,b)\to \mathbb{R}$ be an arbitrary function. If $E=\{x\in (a,b)\mid f'(x)\text{ exists and } f'(x)=0\}$, then $$\lambda(f(E))=0.$$ $\lambda$ denotes the Lebesgue measure.

You pointed out this wonderful answer to me:

More precisely, let $f:\mathbb R\to\mathbb R$ be an arbitrary function, $\Sigma$ is the set of $x\in\mathbb R$ such that $f'(x)$ exists and equals 0. Then $f(\Sigma)$ has measure 0. By countable subadditivity of measure, we may assume that the domain of $f$ is $[0,1]$ rather than $\mathbb R$. Fix an $\varepsilon>0$. For every $x\in\Sigma$ there exists a subinterval $I_x\ni x$ of $[0,1]$ such that $f(5I_x)$ is contained in an interval $J_x$ with $m(J_x)<\varepsilon m(I_x)$. Here $m$ denotes the Lebesgue measure and $5I_x$ the interval 5 times longer than $I_x$ with the same midpoint. Now by Vitali's Covering Lemma there exists a countable collection $\{x_i\}$ such that the intervals $I_{x_i}$ are disjoint and the intervals $5I_{x_i}$ cover $\Sigma$. Since $I_{x_i}$ are disjoint, we have $\sum m(I_{x_i})\le 1$. Therefore $f(\Sigma)$ is covered by intervals $J_{x_i}$ whose total measure is no greater than $\varepsilon$. Since $\varepsilon$ is arbitrary, it follows that $f(\Sigma)$ has measure $0$.

This answer is certainly correct, but for me, who am just a student, it is too lacking in details that I cannot write. I can't translate what is said into symbols. Could someone be kind enough to explain the details of this answer to me?

Prior state of this question.

I was looking for some text or some suggestions to prove this statement using Vitali's lemma, but at first, under the additional nondecreasingness and continuity assumptions for $f$, hoping the proof should prove to be more simplified. I even decided to open a bounty, in the hope that someone will be able to provide me with a proof of this fact with these hypotheses using Vitali's Lemma. But since I didn't receive any answer, I shifted my attention to the more general result.

  • $\begingroup$ Check this: math.stackexchange.com/a/363208/42969 $\endgroup$
    – Martin R
    Commented Jun 13 at 11:20
  • $\begingroup$ This is true without the nondecreasingness or the continuity assumption, see the answer to mathoverflow.net/questions/113991/… $\endgroup$ Commented Jun 13 at 12:51
  • $\begingroup$ Thanks, but on the basis of these hypotheses the result should be simplified, right? Here I am looking for this simplified version. $\endgroup$
    – NatMath
    Commented Jun 13 at 14:11
  • $\begingroup$ The application of Vitali’s Lemma in the link I posted is really as simple as applications of Vitali’s Lemma go. So are you looking for a proof that does not use this lemma? $\endgroup$ Commented Jun 13 at 14:20
  • 2
    $\begingroup$ @NatMath The answer provided in the MO link given by Jonathan Hole proceeded by finding a collection of subintervals in an obvious way, then applying the Vitali covering lemma. It cannot reasonably get any simpler than that. $\endgroup$
    – Divide1918
    Commented Jun 22 at 7:47

1 Answer 1


The following proof is a more explicit version of the one already mentioned above. If anything remains unclear, I’ll be happy to provide clarification.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be an arbitrary function, and $\Sigma$ the set of $x\in \mathbb{R}$ such that $f’(x)$ exists and equals $0$. Then $f(\Sigma)$ has measure $0$.


Since $$m(f(\Sigma))\leq \sum_{n\in \mathbb{Z}} m(f(\Sigma \cap [n, n+1]))$$ (by $\sigma$-subadditivity), we can assume the domain of $f$ to be the interval $[0,1]$, i.e. if the all of the measures of the series are $0$, then so must the measure of $f(\Sigma)$. Fix $\varepsilon >0$, and let $x\in \Sigma$. Since $$ \lim_{y\rightarrow x} \frac{f(y)-f(x)}{y-x}=0, $$ there exists a $\delta>0$ such that $$ \left| \frac{f(y)-f(x)}{y-x} \right|<\frac\varepsilon5,$$ for all $y\in (x-\delta,x+\delta)$. Now define $I_x=(x-\delta/5,x+\delta/5)$. We can choose $\delta$ small enough so that $I_x\subset [0,1]$. Now let $z\in 5I_x$ (if and only if $|z-x|<\delta$). Then $$|f(z)-f(x)|<|z-x|\frac{\varepsilon}5< \frac{\delta\varepsilon}5 .$$ So $$f(z)\in (f(x)-\delta\varepsilon/5,f(x)+ \delta \varepsilon/5)=:J_x,$$ which implies $f(5I_x)\subset J_x$, and importantly $m(J_x)=2\delta\varepsilon/5 = \varepsilon m(I_x)$.

Now by Vitali’s covering lemma there exists a countable collection $\{x_i\}$ such that the intervals $I_{x_i}$ are disjoint and the intervals $5I_{x_i}$ cover $\Sigma$. Since $I_{x_i}$ are disjoint and $I_{x_i}\subset [0,1]$, we have $\sum_i m(I_{x_i})\leq m([0,1])= 1$. Since $\Sigma$ is covered by $\{5I_{x_i}\}$, and $f(5I_{x_i})\subset J_{x_i}$ for each $i$, we have that $f(\Sigma)$ is covered by $\{J_{x_i}\}$. Finally $$ m(f(\Sigma))\leq m\left(\bigcup_i J_{x_i}\right)\\ \leq \sum_im(J_{x_i})=\varepsilon \sum_im(I_{x_i})\leq \varepsilon, $$ but $\varepsilon$ was arbitrary, so $m(f(\Sigma))=0$.

Edit: clarification of the usage of Vitali’s covering lemma. Although Vitali’s (infinite) covering lemma holds for all separable metric spaces, I’ll formulate it in the case of $\mathbb{R}$.

If $\mathbf{F}$ is a family of intervals where $$ \sup \{\ell(I)\mid I\in \mathbf{F}\}<\infty, $$ i.e. the lengths of the intervals are bounded ($\ell(I)$ denotes the length of $I$). Then there exists a countable subfamily of intervals $\mathbf{G}\subseteq \mathbf{F}$ such that the elements of $\mathbf{G}$ are pairwise disjoint, and $$ \bigcup_{I\in \mathbf{F}}I\subseteq \bigcup_{I\in \mathbf{G}} 5I, $$ Where $5I$ denotes the interval of length $5\ell(I)$ with the same midpoint as $I$.

Why does the countable collection $\{x_i\}$ mentioned in the proof exist?

The construction of $I_x$ and $J_x$ is of course for each $x\in \Sigma$. Thus, the family of intervals (with bounded lengths, $\ell(I_x)\leq 1$) $\mathbf{F}=\{I_{x}\mid x\in \Sigma\}$ has a countable subfamily, $\mathbf{G}$, of intervals as described in the lemma. The midpoints of those intervals are, by definition, the $\{x_i\}$.

Why do the intervals $5I_{x_i}$ cover $\Sigma$?

Since $$\Sigma=\{x\in \Sigma\}\subseteq \bigcup_{x\in \Sigma} I_x\subseteq \bigcup_{i} 5I_{x_i},$$ by the construction of the $\{x_i\}$ in Vitali’s covering lemma.

  • $\begingroup$ @NatMath. What is it you don’t understand about the proof? $\endgroup$
    – Jan
    Commented Jul 9 at 10:05
  • $\begingroup$ @JanThanks, The application of Vitali's lemma is not clear to me. $\{x_i\}$ is a Vitali covering of $\Sigma$? Why the intervals $5I_{x_i}$ cover $\Sigma$? $\endgroup$
    – NatMath
    Commented Jul 13 at 12:59
  • $\begingroup$ I don't know this version of Vitali's Lemma. Shouldn't we find closed intervals such that $E$ minus the union of these intervals is less than epsilon? $\endgroup$
    – NatMath
    Commented Jul 13 at 13:10
  • $\begingroup$ $z\in 5I_x$ means that $z$ is in an interval five times longer than $I_x$? $\endgroup$
    – NatMath
    Commented Jul 13 at 13:12
  • 1
    $\begingroup$ Correct, with the same midpoint as $I_x$. I’m currently editing the post for clarification. $\endgroup$
    – Jan
    Commented Jul 13 at 13:13

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