How to prove the measure of $g(C)$ is 1 Let $C$ be the cantor set on [0,1] and $f$ is the cantor Lebesgue function. define $g(x)=x+f(x)$, we know that $g$ is strictly increasing, how to show $m(g(C))=1$?

It may be like this, let $G=\bigcup_{i}I_i$ be the complement of $C$ in $[0,1]$, $I_i$ are open disjoint intervals, $f$ maps each of the $I_i$ to constant, so $g(I_i)$ is merely tranlation of $I_i$ and since $f$ is increasing, $g(I_i)$ are disjoint, hence, $$m(g(G))=\sum m(g(I_i))=\sum m(I_i)= m(G)=1.$$
$$
m(g(C))=2-m(g(G))=1
$$
I do not know if it works or not. Hope it may help.
