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 $f$?

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