Let $m^{\ast}$ be the Lebesgue outer measure and $m$ the Lebesgue measure. Let $\phi$ be the Cantor Lebesgue function and let $\psi(x) := x + \phi(x)$. Let $C$ be the standard Cantor set, why does $m^{\ast}(\psi(C^{c})) = 1$?
I know that: If $I = (a, b)$ is an open interval removed in the construction of the Cantor set, $\psi(I) = I + \phi(a)$ and hence $m^{\ast}(\psi(I)) = m^{\ast}(I) = m(I)$. Since $m(C^{c}) = 1$, $$1 = m\left(\bigcup_{I \in C^{c}}I\right) = \sum_{I \in C^{c}}m(I) = \sum_{I \in C^{c}}m^{\ast}(\psi(I))$$ but the expression on the right hand side is $\geq m^{\ast}(\psi(C^{c}))$ by countable subadditivity of $m^{\ast}$.