# Let $f\colon (a,b)\to \mathbb{R}$ be nondecreasing and continuous. If $E=\{x\in (a,b)\mid f'(x)\text{ exists and } f'(x)=0\}$, then $\lambda(f(E))=0$

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.

• Check this: math.stackexchange.com/a/363208/42969 Commented Jun 13 at 11:20
• This is true without the nondecreasingness or the continuity assumption, see the answer to mathoverflow.net/questions/113991/… Commented Jun 13 at 12:51
• Thanks, but on the basis of these hypotheses the result should be simplified, right? Here I am looking for this simplified version. Commented Jun 13 at 14:11
• 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? Commented Jun 13 at 14:20
• @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. Commented Jun 22 at 7:47

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 $$0$$.

Proof

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.

• @NatMath. What is it you don’t understand about the proof?
– Jan
Commented Jul 9 at 10:05
• @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 $\Sigma$? Commented Jul 13 at 12:59
• 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? Commented Jul 13 at 13:10
• $z\in 5I_x$ means that $z$ is in an interval five times longer than $I_x$? Commented Jul 13 at 13:12
• Correct, with the same midpoint as $I_x$. I’m currently editing the post for clarification.
– Jan
Commented Jul 13 at 13:13