Let $f : [0,1]^k \longrightarrow [0,1]^k$ be an one to one and onto $C^1$-map. We know that the Lebesgue measure of $\partial \left ( [0,1]^k \right )$ is zero. Can we conclude that the Lebesgue measure of $f \left ( \partial \left ([0,1]^k \right ) \right )$ is also zero?
Any help in this regard will be highly appreciated. Thanks in advance.
Yes, any Lispchitz function maps sets of measure $0$ to sets of measure $0$. And any $C^{1}$ map on $[0,1]^{k}$ is Lipschitz.