Let $f : \mathbb{R} \to \mathbb{R}$ be measurable and let $Z = {\{x : f'(x)=0}\}$. Prove that $λ(f(Z)) = 0$.

Question

The following is an exercise from Bruckner's Real Analysis:

Let $f : \mathbb{R} \to \mathbb{R}$ be measurable and let $Z = {\{x : f'(x)=0}\}$. Prove that $λ(f(Z)) = 0$.

For the case $f$ being nondecreasing / nonincreasing, we can defined $g=f^{-1}$ and then $g'$ and then use the following theorem from the book :

Let $f$ be nondecreasing / nonincreasing / of bounded variation on $[a,b]$. Then $f$ has a finite derivative almost everywhere.

Is it possible to "reduce" evaluation of any measurable $f$ to a nondecreasing one or otherwise how the claim can be proved for any measurable $f$?

Answer 1

There was something bothering me in the answers given thus far so I decided to produce a solution of my own. The result of your problem will follow for the next Theorem:

Theorem A: If $f:(a,b)\rightarrow\mathbb{R}$ is measurable, and $f$ is differentiable on a measurable set $C\subset(a,b)$, then $$\lambda^*(f(C))\leq \int_C |f'(x)|\,dx$$

In your case $C=Z=\{x: f'(x)=0\}$ (which is measurable (why?)) and so, $\int_Z|f'(x)|\,dx=0$.

In the rest of this answer I will sketch a proof of this result, which I learned from some notes I took from a course on mathematical economics by Roko Aliprantis (Classical Sard's theorem).

There are some measurability issues that require some attention.

1. If the function $f$ is differentiable on $(a, b)$, then of course $f'(x)$ is measurable (being the limit of measurable functions $\phi_n:x\mapsto n(f(x+\tfrac1n)-f(x))$. However, in the statement of the theorem only differentiability on $C$ is assume, To fixed that, we consider $f'(x):=\limsup_n\phi_n(x)$ which is measurable as an extended real function and coincides with the derivative of $f$ at any point $x$ where $f$ is differentiable.
2. The outer measure $\lambda^*$ is countably sub additive and by construction it is inner and outer regular. This implies that $\lambda^*$ itself satisfies monotone convergence: that is, for any sequence of subsets $(A_n,A:n\in\mathbb{N})$ in $\mathbb{R}$, if $A_n\nearrow A$ as $n\rightarrow\infty$, then $\lim_n\lambda^*(A_n)=\lambda^*(A)$. This can also be explained by noticing that the outer measure of a set is the Daniell mean (associated to the Riemann integral on step functions) of $\mathbb{1}_A$ , and that the Daniell mean is maximal (Bichteler. K., Integration: a functional approach, Birkhauser-Verlag, Section 3.6).

Sketch of proof of Theorem A.

Without loss of generality suppose $(a, b)$ is a bounded interval. For any $\varepsilon>0$, $$\begin{align} C=C\cap\{|f'|<\infty\}=\bigcap_{n\geq1}\{x\in C: (n-1)\varepsilon\leq |f'(x)|< n\varepsilon\}\tag{0}\label{zero} \end{align}$$

Claim: For any set $A\subset (a,b)$ and $c>0$, if $|f'(x)| < c$ for all $x\in A$, then $\lambda^*(f(A))\leq c\lambda^*(A)$. Consider the sets $$\begin{align}A_n=\{x\in A: \text{for any $y$, if}, |x-y|<\frac1n,\,\text{then}\,|f(x)-f(y)|<c|x-y|\}\tag{1}\label{one} \end{align}$$ Although not necessarily Lebesgue measurable, these sets satisfy $A_n\nearrow A$ as $n\rightarrow\infty$. Thus $\lambda^*(A_n)\xrightarrow{n\rightarrow\infty}\lambda^*(A)$, and $\lambda^*\big(f(A_n)\big)\xrightarrow{n\rightarrow\infty}\lambda^*\big(f(A)\big)$. Here is the key part: for each $n\in\mathbb{N}$, there is a sequence of intervals $\{I_{n,k}=(a_{n,k},b_{n,k}]:k\in\mathbb{N}\}$ such that $A_n\subset\bigcup_kI_{n,k}$, and $$\begin{align} \sum_k\lambda(I_{n,k})<\lambda^*(A_n)+\varepsilon\tag{2}\label{two} \end{align}$$ By splitting the intervals $I_{n,k}$ further into finite subintervals of length at most $\frac{1}{n}$, we may assume that $\lambda(I_{n,k})<\frac1n$ for all $k\in\mathbb{N}$. From \eqref{one}, if $x,x'\in A_{n}\cap I_{n,k}$, $|f(x)-f(x')|<c|x-x'|<c\lambda(I_{n,k})$, that is, $f(A\cap I_{n,k})$ is contained in an interval of diameter $c\lambda(I_{n,k})$. Hence $\lambda^*\big(f(A_n\cap I_{n,k})\big)\leq c\lambda(I_{n,k})$. Subadditivity of $\lambda^*$ and \eqref{two} gives $$ \lambda^*\big(f(A_n)\big)\leq \sum_k\lambda^*\big(f(A_n\cap I_{n,k})\big)\leq c\sum_k\lambda(I_{n,k}) < c\big(\varepsilon + \lambda^*(A_n)\big) $$ Letting $n\rightarrow\infty$ and then $\varepsilon\searrow0$ gives the claim: that is, $\lambda^*\big(f(A)\big)\leq c\lambda^*(A)$.

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The conclusion of the proof of theorem A is now routine: Let $C_n=\{x\in C: (n-1)\varepsilon\leq |f'(x)|< n\varepsilon\}$. Clearly $(C_n:n\in\mathbb{N})$ is a pairwise disjoint sequence of measurable set. Hence $$\begin{align} \lambda^*\big(f(C)\big)&\leq \sum_n\lambda^*\big(f(C_n)\big)\leq \sum_n\varepsilon n\lambda(C_n)= \sum_n \varepsilon(n-1+1)\lambda(C_n)\\ &=\sum_n \varepsilon(n-1)\lambda(C_n)+ \sum_n\varepsilon\lambda(C_n)\leq \sum_n\int_{C_n}|f'(x)|\,dx + \varepsilon\lambda(C)\\ &=\int_C|f'(x)|\,dx +\varepsilon\lambda(C) \end{align}$$ The conclusion follows by letting $\varepsilon\searrow0$.

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Final comments:

• There are other approaches to the OP using Vitali's covering lemma. This seems to be more common in the literature (see for example F. Jones, Lebesgue Integration in Euclidean space, Jones & Barlett, section 15. J, where the claim I stated is proven (for $f:U\subset\mathbb{R}^n\rightarrow\mathbb{R}^n$).

• Under some Lipschitz conditions, a proof of the claim in $\mathbb{R}^n$ appears in Evans, C. and Gariepy, R, Measure theory and fine properties of functions, CRC Press, Revised Edition, Section 2.4.

• Under smoothness conditions (for $f:U\subset R^n\rightarrow\mathbb{R}^m$) we have the well know result from Morse-Sard (see for example Milnor, . D. W. Topology from the differentiable viewpoint, The University of Virginia Press, 1965, Chapter 3).

• Other extensions can be found in the following paper by Figalli

Answer 2

Let $Z_n=\{x\in Z |\, \forall y, |x-y|<\frac{1}{n}\implies |f(x)-f(y)|<\epsilon|x-y| \}$. Then $Z\subset\cap_n Z_n$. Take a cover $U_n$ of $Z_n\cap[a,b]$ by intervals of length less than $\frac{1}{n}$. Have $U_n$ be chosen so that $\sum_n \lambda(U_n)$ is less than $2(b-a).$

Now as $f(Z_n\cap[a,b])\subset\cup f(U_n),\lambda(f(U_n))<\epsilon\lambda(U_n)$ we have that $\sum_n \lambda(f(U_n))<\epsilon\sum_n \lambda(U_n)<2\epsilon(b-a).$ This tells us that $\lambda(f(Z_n\cap[a,b]))<2\epsilon(b-a).$ Hence $\lambda(f(Z\cap[a,b]))<2\epsilon(b-a)$ and taking $\epsilon \to 0$ gives us $\lambda(f(Z\cap[a,b]))=0.$ As this holds true for all $a,b$ we have that $\lambda(f(Z))=0.$

Answer 3

A slightly simpler way to proof:

We have that $x\in Z$ if and only if, for all $m$, there is $n$ , such that $\forall y, |x-y| \leq \frac{1}{n} \implies |f(x)-f(y)| \leq \frac{1}{m} |x-y|$.

Let $Z_{m,n}= \{ x \in \Bbb R: \forall y, |x-y| \leq \frac{1}{n} \implies |f(x)-f(y)| \leq \frac{1}{m} |x-y|\}$.

So we have $Z= \bigcap_m \bigcup_n Z_{m,n}$.

Now, for all $p \in \Bbb Z$ and $k \in \{ 0, ... , n-1\}$, we have

$$ \lambda^* \left(f \left( Z_{m,n} \cap \left [p+\frac{k}{n},p+\frac{k+1}{n} \right ] \right ) \right ) \leq \frac{1}{m}\frac{1}{n} $$ where $\lambda^*$ is the Lebesgue outer measure. Since $$f \left( Z_{m,n} \cap \left [p,p+1\right ] \right ) = \bigcup_{k=0}^{n-1}f \left( Z_{m,n} \cap \left [p+\frac{k}{n},p+\frac{k+1}{n} \right ] \right )$$ we have, by sub-additivity,
$$ \lambda^* \left(f \left( Z_{m,n} \cap \left [p,p+1\right ] \right ) \right ) \leq \frac{1}{m} $$

Now, note that, for all $m$ and $n$, $Z_{m,n} \subseteq Z_{m,n+1}$, so $f(Z_{m,n} \cap [p, p+1]) \subseteq f(Z_{m,n+1} \cap [p, p+1]) $. So (see Remark 1),

$$ \lambda^* \left(f \left( \bigcup_n Z_{m,n} \cap \left [p,p+1\right ] \right ) \right )= \lambda^* \left(\bigcup_n f \left( Z_{m,n} \cap \left [p,p+1\right ] \right ) \right ) = \\ = \lim_{n \to \infty} \lambda^* \left(f \left( Z_{m,n} \cap \left [p,p+1\right ] \right ) \right ) \leq \frac{1}{m} $$

Since $Z= \bigcap_m \bigcup_n Z_{m,n}$, we have, for all $p \in \Bbb Z$, $\lambda^* (f(Z \cap [p, p+1])) \leq \lambda^* (f(\bigcup_n Z_{m,n} \cap [p, p+1])) \leq \frac{1}{m}$, for all $m$.

So, for all $p \in \Bbb Z$, $\lambda^* (f(Z \cap [p, p+1])) =0$. It follows that $\lambda^* (f(Z)) =0$. So $f(Z)$ is measurable and $\lambda(f(Z))=0$.

Remark 1: Here, we use the result:

if $ E_{n} \subseteq E_{n+1} $ then $\lim\limits_{n\mapsto \infty} \mu^*(E_n) = \mu^*(\bigcup_{n=1}^\infty E_n) $ even if each $E_n$ is a non-measurable set, where $\mu^*$ is Lebesgue outer measure (or any regular outer measure).

This can be easily proved using measurable covers and it is an exercise in Halmos (section 12 exercise 4).

Remark 2: If you know Lemma 7.10 in Bruckner's Real Analysis: you can use exercise 7.3.1 in the same book to easily prove that $Z = {\{x : f'(x)=0}\}$ is measurable and then the answer to your question becomes a direct trivial application of Lemma 7.10. (However the proof itself of Lemma 7.10 depends on the proof of Lemma 7.9 and exercise 7.3.1 and so it is more complex than what is needed to directly answer the question as posted).