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

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 $f$?

Solution 1:

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 $\lambda(f(Z))=0.$

Solution 2:

There was something bothering me in the answers given thus far that I decided to go through memory lane and produce a proof of my own. I realized that the result of your problem will follow for the next result

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 if 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.
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: If $c>0$, and $|f'(x)| < c$ for all $x\in A$, then $\lambda^*(f(A))\leq c\lambda^*(A)$. Consider the sets $$\begin{align}A_n=\{x\in A: \text{for any $y$, if}, |x-y|<\frac1n,\,\text{then}\,|f(x)-f(y)|<c|x-y|\}\tag{1}\label{one} \end{align}$$ This set, although to necessarily Lebesgue measurable, satisfies $A_n\nearrow A$ sd $n\rightarrow\infty$. 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, $\lambda^*\big(f(A)\big)\leq c\lambda^*(A)$.

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 $\varepsilon\searrow0$.

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 it the claim I stated is proven (for $fU\subset:\mathbb{R}^n\rightarrow\mathbb{R}^n$).

• 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