# Is cartesian product of the Cantor set and real numbers ($C\times\Bbb R$) has measure zero in $\Bbb R^2$

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 $$0$$.
• I have difficulty showing that $\textit{S$_n$}$ has measure $0$. Would you be so kind and provide anothet tip? Nov 4, 2020 at 18:46
• @Nitaaa You can cover the cantor set with intervals whose total length is arbitrarily small. use these intervals to make rectangles that cover $S_n$ and have arbitrarily small total area. Nov 4, 2020 at 18:48