Applying measure zero definition to Cantor sets I just learned about the concept of measure zero in real analysis, i.e. the definition that a set in $\mathbb{R}^n$ has measure zero if for any $\epsilon$ it can be covered by countably many rectangles whose volume sum to $<\epsilon$.
I'm wondering if knowing only this, I can prove that the Cantor-third set has measure zero, while the fat Cantor set doesn't have measure zero.
If the answer is that I should wait until I learn more about measures in general, I will happily do that. I'm just curious whether this definition alone is enough.

Yes, it is enough. If you think about the elimination process to create the Cantor set, after the first stage you see that the Cantor set can be covered by two intervals of length $1/3$ and a bit. After the second stage of middle-third removal,  you see that the Cantor set can be covered by four intervals of length $1/9$ plus a bit, etc. So after the $n^{\rm th}$ stage you see that the Cantor set can be covered in $2^n$ intervals of length $3^{-n}$ (and bit, but is is easier not to write it!). In other words, the outer measure of the Cantor set is less than $(2/3)^n$ for all $n$, and so it has to be zero. 
