Let $g(x)=x+C(x)$ for $x\in[0,1]$, where $C(x)$ is the Cantor-Lebesgue function.
Let $Z$ be the Cantor Set, which has Lebesgue measure zero.
How would we show that $g(Z)$ does not have measure zero?
What I know is $C(Z)=[0,1]$, i.e. the Cantor function maps the Cantor set to $[0,1]$. So intuitively, I would suppose $g(Z)=[0,2]$?
How would we show that? (I only need the weaker condition that $g(Z)$ has positive measure.)
Thanks!
Best Answer
Denote $\mathcal{C}$ the cantor set.
Note that $[0,1] \setminus \mathcal{C}$ is the disjoint union of line segments of where we define $x\mapsto C(x)$ to be constant (middle thirds). On each of this, $g(x)$ is a function of the form $x\mapsto x + b$ where $b$ is som constant (related to the construction of $x\mapsto C(x)$). Therefore $g$ is an affine map on each of this, and so it preserves their measure (lebesgue measure is translation invariant + the map $x\mapsto x$ is linear with determinant $1$).
That being said: $m \left( g\left([0,1]\setminus \mathcal{C}\right) \right)= m([0,1]\setminus \mathcal{C}) = 1$
Now, $g(x)$ is continuous and stricly increasing (this is obvious since $x\mapsto C(x)$ is non-decreasing and $x\mapsto x$ is stricly increasing), $g(0) = 0$ and $g(1)=2$ and so $g([0,1]) = [0,2]$.
This together with what we showed at the begining gives us the desired result: $g(\mathcal{C}) =1 >0$.