Cantor set not being Luzin-N with a particular function Let $N\subset [0,\infty)$ s.t. $\lambda^*(N)=0$ and lets define $N+N:=\{x+y:x,y\in N\}$. Let $C$ be the classic Cantor set. So I want to prove that:
$$C+C=[0,2]$$
(And, thus, it's clear that $\lambda^*(C+C)=\lambda^*([0,2])=\lambda([0,2])=2$, but $\lambda^*(C)=\lambda(C)=0$)
What I've done:

Since $$C=\bigcap_{n\in\Bbb N}C_n\\
\text{where}\;\ C_n\subset [0,1] \;\forall\ n\in\Bbb N\;\text{and}\\
C_n=\text{union of}\;2^n\;\text{closed subintervals of lenght}\;\frac{1}{3^n}\ \forall n\in\Bbb N$$
So
$$C+C=\{x+y:x,y\in C\}$$
But got stuck here. Any ideas would be appreciated.
 A: You can prove $C+C=[0,2]$ by using tenary representation of Cantor set. Let $x,y\in C$ where 
$$
x=\sum_{n=1}^{\infty}\frac{a_n}{3^n},\:a_n=0,2\quad\text{and}\quad y=\sum_{n=1}^{\infty}\frac{b_n}{3^n},\:b_n=0,2
$$
Then it can be proved that $x+y$ can cover all $[0,2]$. Let $z\in [0,2]$. Then $$
z=\sum_{n=0}^{\infty}\frac{c_n}{3^n},\:c_0=0,1,\:c_n=0,1,2,\:n>0
$$
If $c_0=0$, then $z\in [0,1]$. If $c_0=1$, then $z\in [1,2]$.
Given $c_n$, we construct $a_n,b_n\:(n\geqslant1)$ as follow:


*

*If $c_n=0$ and no carry-over from $a_{n+1}+b_{n+1}$, then set $a_n=0, \:b_n=0$. 
If $c_n=0$ and has carry-over from $a_{n+1}+b_{n+1}$, then set $a_n=2, \:b_n=0$.

*If $c_n=1$ and no carry-over from $a_{n+1}+b_{n+1}$, then set $a_n=2, \:b_n=2$. 
If $c_n=1$ and has carry-over from $a_{n+1}+b_{n+1}$, then set $a_n=0, \:b_n=0$. 

*If $c_n=2$ and no carry-over from $a_{n+1}+b_{n+1}$, then set $a_n=2, \:b_n=0$. 
If $c_n=2$ and has carry-over from $a_{n+1}+b_{n+1}$, then set $a_n=2, \:b_n=2$.

*$c_0=0$ or $c_0=1$, depends on whether there is carry-over from $a_{1}+b_{1}$.


Hence we show that
$$
z=\sum_{n=0}^{\infty}\frac{c_n}{3^n}=\sum_{n=1}^{\infty}\frac{(a_n+b_n)}{3^n}=x+y
$$
Since $z$ cover entire $[0,2]$, we have $C+C=[0,2]$. 
A: An answer is already posted using the same ideas that I am about to use, but I could not verify the details on my own (which is not to say that it is incorrect), so I write a different version. 
So take any $c\in[0,2]$. Then $\displaystyle\frac c2\in[0,1]$ and we can write $\displaystyle\frac c2=\sum_{n\ge1}\frac{c_n}{3^n}$ where $c_n\in\{0,1,2\}$. 
If $c_n=0$ then let $a_n=b_n=0$.
If $c_n=1$ then let $a_n=0$, $b_n=2$.
If $c_n=2$ then let $a_n=b_n=2$. 
Since $a_n,b_n\in\{0,2\}$, we have that $\displaystyle a= 
\sum_{n\ge1}\frac{a_n}{3^n}\in C$ and 
$\displaystyle b= 
\sum_{n\ge1}\frac{b_n}{3^n}\in C$.  
Also, $\displaystyle a+b=\sum_{n\ge1}\frac{a_n+b_n}{3^n}=\sum_{n\ge1}\frac{2c_n}{3^n}=2\cdot\frac c2=c$.  
Since every $c\in[0,2]$ is represented as $a+b$ for suitable $a,b\in C$ we have that $[0,2]\subseteq C+C$. On the other hand the inclusion $C+C\subseteq[0,1]+[0,1]=[0,2]$ is obvious, hence $C+C=[0,2]$.  
Note that in a similar way one can prove that $C-C=[-1,1]$. 
