I found this question quite interesting, but its answers were disappointingly non-geometric. I'd be interested to know whether there exists a geometric example.
To be precise about what I mean by a geometric example, recall that an open set $U\subset \mathbb{R}^n$ is called regular if it is equal to the interior of its closure. Here is my question:
Question: Does there exist a regular open set $U\subset \mathbb{R}^n$ whose boundary has nonzero Lebesgue measure?
I would guess that the answer is no in $\mathbb{R}^1$, but it ought to be yes in $\mathbb{R}^2$. Part of why this is interesting is that it seems to me that the Mandelbrot set might fall into this class (although according to this MathOverflow post, it is an open question whether the boundary of the Mandelbrot set has positive Lebesgue measure).
Osgood constructed an example of a Jordan curve of positive area in Trans. Amer. Math. Soc. 4 (1903), 107-112. Knopp's construction of such a curve is the content of a demonstration on Wolfram|Alpha.
By the Schoenflies theorem we can extend such a Jordan curve $\gamma$ to a homeomorphism $\mathbb{C} \to \mathbb{C}$ such that the interior and exterior of the unit circle are sent to the interior and exterior of $\gamma$, respectively. Since homeomorphisms preserve regular open sets, the interior and exterior of such a Jordan curve will be examples of regular open sets whose boundary has nonzero volume.
Added: Brian M. Scott shows in Cantor set is boundary of regular open set how to get a Cantor set as boundary of a regular open set. The argument appears to apply just as well for a fat Cantor set, so it is even possible in one dimension.