I'm trying to show that the cartesian product of real numbers and the Cantor set ($C$) has (Lebesgue) measure zero in $\Bbb R^2$, I was successful at showing (I think) that $\Bbb R$ has measure zero in $\Bbb R^2$ by covering it with many "rectangles" with Volume(area) zero. Now my plan was to show that $C\times\Bbb R$ is a subset of the union of many "rectangles" and some set that contains $C$, but I can't seem to find that second set. Now, I know how to show that $C$ has measure zero in $\Bbb R$, but I can't seem to translate this into $\Bbb R^2$.
Could anybody help me? I think that it would be useful to find how to cantor set looks in R$^2$ because co-student told me that it would be the same as in R but with infinite strips going up from the small intervals, but this doesn't fit right with me.
Hint: First show that $S_n = (C \times \Bbb R) \cap ([-n,n] \times [-n,n])$ has measure $0$ for any $n \in \Bbb N$. Using the fact that $C \times \Bbb R = \bigcup_{n \in \Bbb N} S_n$, conclude that $C \times \Bbb R$ has measure $0$.