I have the following problem and I have noticed that someone alread answered the question Let $f : \mathbb{R} \to \mathbb{R}$ be measurable and let $Z = {\{x : f'(x)=0}\}$. Prove that $λ(f(Z)) = 0$. But there is something I didn't understand i'll report the part of the text
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 $k\in\mathbb{N}$. 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 gives $$ \lambda^*\big(f(A_n)\big)\leq \sum_k\lambda^*\big(f(A_n\cap I_{n,k})\big)\leq c\sum_k\lambda(I_{n,k}) < c\big(\varepsilon + \lambda^*(A_n)\big) $$
what I didn't understand in this proof is $1$ wh do I have to conside the intersection $\in A_{n}\cap I_{n,k}$ and $2$ why is important $\lambda(I_{n,k})<\frac1n$ I don't think is used
please let me know .Thanks
