I'm working on a problem involving a constructing a homeomorphism that takes the cantor set to a set of positive measure in the image. I'm having trouble showing the image has Lesbugue measure $1$. The question follows:
Call the cantor set $C$. $x \in C$ will have the form $$x=\sum_{j=0}^\infty\frac{a_j}{3^j} \quad a_j=\{0,2\}$$ Let $f: [0,1] \rightarrow [0,1]$ be the cantor function that sends elements $$C \backepsilon x \mapsto \sum_{j=0}^\infty\frac{a_j}{2^j} \quad a_j=\{0,1\}$$ (This function explicitly takes elements of the cantor set to the real line via their "binary" representation). Let $g(x) = f(x) + x$. Show that $m(g(C))=1$.
The cantor set can be understood as countable intersection of closed sets: $C=\bigcap C_n$ where $C_n$ are obtained in the "cantor thirds iteration". I'm trying to examine the image of the map directly - a previous part of the problem shows that $g$ is a homeomorphism, call $g^{-1}=h$. Then $h^{-1}(C)$ is closed. Then: $$m\left(h^{-1}\left(\bigcap C_n\right)\right)=m\left(\bigcap h^{-1}\left( C_n\right)\right)=\lim_{n \rightarrow\infty} m(h^{-1}\left( C_n\right))$$
I'm not sure what to do with the last equality. Perhaps I need to attack it a different way?