Lebesgue measure and area of a set

Question

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.

Well, in my textbook it's defined as the integral of the set's characteristic function over a rectangle superset. — Luka Horvat, Mar 27 '14 at 22:22
What integral? Riemann integral? Lebesgue integral? This assumes the characteristic function is integrable, right? What does that mean, precisely? — Andrés E. Caicedo, Mar 27 '14 at 22:24
If a set $A$ has zero Lebesgue measure then $\int_A d\mu = 0$, where $\mu$ is the Lebesgue measure. So any null set has zero area by your definition.... — Frank, Mar 27 '14 at 23:11
If the characteristic function is intergrable, then the set has an area and it's equal to that integral. It's a Riemann integral. — Luka Horvat, Mar 28 '14 at 07:06

score 0 · Accepted Answer · answered Mar 31 '14 at 08:50

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.

I think the word for "a set that can be closed in a finite sphere" is "bounded". And the word for it's "edge" is "boundary". — Erel Segal-Halevi, Jan 09 '15 at 11:45

Lebesgue measure and area of a set

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