Show that there exists two closed sets $A, B$ such that $m(A)=m(B)=0$ but $m(A+B)>0$.
My question is very similar to this. The difference is that I am required to use $A = \mathcal{C}$ and $\displaystyle B = \frac{\mathcal{C}}{2}$. I'm trying to use the bof's ideia, that is:
Consider the projection
$$\begin{eqnarray*} \mathcal{C}_{n} \times \frac{\mathcal{C}_{n}}{2} & \longrightarrow & [0,1]\\ (x,y) & \mapsto x + y \end{eqnarray*}$$ and to show that is surjective in each $n$, where $n$ is the $n$-th step of construction of the Cantor Set. The problem is, I can see and show that this is true for $n = 1, 2,\dots$; but I'm not able to correctly show that it's worth for all $n$. Can someone help me?
Hint 1: $\mathcal{C} \subset \mathcal{C}_n$ so if the same map restricted to $\mathcal{C} \times \mathcal{C}/2$ is surjective, then so is the original map.
Hint 2: The Cantor set $\mathcal{C}$ is the set of all real numbers in $[0,1]$ that can be expressed using $0$ and $2$ in base $3$. Similarly, $\mathcal{C}/2$ is the set of all real numbers in $[0,1]$ that can be expressed using $0$ and $1$ in base $3$.
Alternatively, if you don't want to use restrictions, then use the following
Hint 3: The set $\mathcal{C}_n$ (resp. $\mathcal{C}_n/2$) is the set of all real numbers in $[0,1]$ that can be expressed using $0$ and $2$ (resp. $0$ and $1$) in base $3$ in the first $n$ decimal places (and any number from the $(n+1)$-th decimal place onwards).