[Math] Boundary of closure of open set in $\mathbb R^2$ has measure zero

geometric-measure-theorylebesgue-measuremeasure-theory

Consider problem 4 on day 2 of this exam.

Suppose that $\mathcal O\subset \mathbb R^2$ is an open set with
finite Lebesgue measure. Prove that the boundary of the closure of
$\mathcal O$ has Lebesgue measure $0$.

I have been stuck on this problem for a while, and I now believe it might be false. It is known that there are Jordan curves in the plane with positive area. Is it true that these indeed are counterexamples to this statement, and if not, how can this problem be solved?

Best Answer

Assume we have a Jordan curve of positive area measure whose complement has connected components $U,V$ as in the Jordan Curve theorem. Let $U$ be the bounded component. Every point in $\partial U = \partial V $ is the limit of a sequence of points in $\overline U $ and the limit of a sequence of points in $V= \overline U ^c.$ Thus the boundary of $\overline U$ contains the Jordan curve (it's actually equal to it), hence the boundary of $\overline U$ has positive area measure.