Consider a compact subset $K$ of $R^n$ which is the closure of its interior. Does its boundary $\partial K$ have zero Lebesgue measure ?
I guess it's wrong, because the topological assumption is invariant w.r.t homeomorphism, in contrast to being of zero Lebesgue measure. But I don't see any simple counterexample.
Best Answer
Construct a Cantor set of positive measure in much the same way as you make the `standard' Cantor set but make sure the lengths of the deleted intervals add up to 1/2, say. Let $U$ be the union of the intervals that are deleted at the even-numbered steps and let $V$ be the union of the intervals deleted at the odd-numbered steps. The Cantor set is the common boundary of $U$ and $V$; their closures are as required.