Boundary of closure of open set in $\mathbb R^2$ has measure zero 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?
 A: 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.
A: 
Edit: In this case, the boundary of the bounded component $U$ has positive, finite measure, but the question was asking about the closure of the open set. I'll think about this more.

Your thinking is in fact correct for the reason you mention: there is a Jordan Curve in the plane with positive measure. See here. Then, by the Jordan Curve Theorem, there is a bounded interior $U$ and an unbounded exterior $V$, which are open. Since the interior is bounded, it must have finite measure and is open, and  its boundary is the Jordan Curve described.
For added fun, see here for a demonstration on such curves.
