If $f$ is lipschitzian, then carries null measure set into null measure set. But if $f$ is just continuous?

Question

Let $f: X \rightarrow \mathbb{R}^n$ is a lipschitzian application in the set $X \subset \mathbb{R}^n $. If $medX = 0$ then $med f(X) = 0$. But if $f$ is just continuous, is the same true? or is there a counter example?

Answer 1

Take the function $g: \to [0,1] \to \Bbb{R}$ to be $g(x)=f(x)+x$ where $f$ is the Cantor-Lebesgue function.

Then $m(g(C))>0$ where $C$ is the Cantor set on $[0,1]$