So the question is something like this: If a subset of $\mathbb{R}^2$ has a Lebesgue measure 0 and HAS an area. Does that area have to be 0? Prove or disprove.
I know of counterexamples that don't even have an area defined but I figured that since the area exists there could be some way to connect it to the Heine-Borel theorem and conclude that, if there are countably infinite rectangles with a total area less than epsilon that cover the set, there has to be a finite subset with the same properties. I don't know how to prove that.
A friend knew the answer and it turns out I was right. If the set has an area, it means it's restricted (or what ever the English word is for a set that can be closed in a finite sphere). Then we have two cases. If it's closed, it's compact so Heine-Borel gives us a finite set of rectangles that cover the set.
If it isn't closed, it's "edge" (again, the terminology) has the Lebesgue measure of 0 so the union of the set and it's edge has the measure of 0 and is closed. Again, we can get a finite set of rectangles that cover that union, and then in turn cover the original set.