in $\mathbb{R^2}$ , observe that if $A = I \times {\{0}\}$ and $B = {\{0}\} \times I$ (where $I = [0,1]$) then $A + B = I \times I$. Show that there exists closed sets $A$ and $B$ with $m(A) = m(B) = 0$, but $m(A+B) > 0$. where $A = C$ where  $C$ is the cantor set, and $B = C/2$.
(b) in $\mathbb{R^2}$ , observe that if $A = I \times {\{0}\}$ and $B = {\{0}\} \times I$ (where $I = [0,1]$) then $A + B = I \times I$.
attempt: Suppose  in $\mathbb{R^2}$ , observe that if $A = I \times {\{0}\}$ and $B = {\{0}\} \times I$ (where $I = [0,1]$). Then let $z_1 = (x_1,y_1) \in I$, and  $a \in A $ such that $a = z_1 \times I = (x_1,y_1) \times I$ and $b \in B$ such that $b = {\{0}\} \times I =  {\{0}\} \times (x_2 ,y_2) $.
I don't know how to continue. Can someone please help me? Thank you!

Note that $(x,y)+S$ is simply to shift $S$ by $x$ in $x$-axis and $y$ in $y$-axis. Thus we can define
$$
(x,y)+S=\bigcup_{(u,v)\in S}(x+u, y+v)
$$ 
Hence
\begin{align}
A+B&=\bigcup_{(x,y)\in A}(x,y)+B
\\
&=\bigcup_{(x,y)\in A}((x,y)+B)
\\
&=\bigcup_{(x,y)\in A}\bigcup_{(u,v)\in B}(x+u,y+v)
\\
&=\bigcup_{(x,y)\in I\times\{0\}}\bigcup_{(u,v)\in \{0\}\times I}(x+u,y+v)
\\
&=\bigcup_{x\in I}\bigcup_{v\in I}(x+0,0+v)
\\
&=\bigcup_{x\in I}\bigcup_{v\in I}(x,v)
\\
&=I\times I
\end{align}
