Preimage of measure 0 set under norm function has measure 0

Question

Let $A \subseteq [0,1]$ be a null set with respect to the Lebesgue measure. Is it true that $\{x\in \mathbb R ^k: |x| \in A\}$ is a null set in $\mathbb R^k$, for all natural $k$ ($|(x_1,\dots,x_k)| = \sqrt{x_1^2+\dots+x_k^2}$)? If so, how can this be shown?

Answer 1

In case $|0| \in A$, write $\{x\in \mathbb R ^k: |x| \in A\} =\{x\in \mathbb R ^k\setminus \{0\}: |x| \in A\} \cup \{0\}$.

The euclidean norm $||\cdot||$ is a smooth function outside $0$, so you can write $||\cdot|| \in C^1$ and the set $\lbrace x \, | \, \nabla ||x|| = 0 \rbrace$ has measure zero (try to compute the gradient of the euclidean norm). Hence the result follows by this.