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How can I prove that $f(x)=x \sin(1/x), f(0)=0$ has the luzin N property (The image of a null set is a null set): in this link:

Example of a function that has the Luzin $n$-property and is not absolutely continuous.

"Luzin Property follows from the fact that f is locally lispchitz in $\mathbb{R} \setminus \{0\}$"

Suppose that $N$ is a null set such that $0\notin N$: we need to prove that $f(N)$ is also a null set: by definition for every $\epsilon>0$ there exists a countable family of closed intervals such that $N\subset \bigcup_{i=1}^\infty R_i$ and $\sum_{i=1}^\infty m(R_i)<\epsilon$ (where $m$ denotes the lebesgue measure)

Then we have that $m(f(n))<\sum_{i=1}^\infty m(f(R_i))<\sum_{i=1}^\infty f(\xi_i)-f(\eta_i)$ where $\xi_i$ and $\eta_i$ are the points where $f$ attains a maximum and minimum value in the closed interval $R_i$

I'm wondering how can I use the fact that $f$ is locally Lispchitz to conclude that $f(N)$ is a null set

user128422
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1 Answers1

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You should know that Lipschitz maps have (N)-property.

Given a null set $N$ (which we may assume to not contain $0$, since one point doesn't matter), write it as $N=\bigcup_{k=1}^\infty N_k$ where $N_k=N\cap \{x: 1/k\le |x|\le k\}$. Since $f'$ is differentiable, with continuous derivative on $\{x: 1/k\le |x|\le k\}$, it is Lipschitz there. Hence $f(N_k)$ is a null set, which implies $$ f(N) = \bigcup_k f(N_k) $$ is a null set.