Area of Plane Figures

Question

Do all plane figures have some area (may be zero, finite or infinite)? What I mean to say is that, do all subsets of the Cartesian plane have area? By area I mean area in the euclidean sense.

Answer 1

If I understand you correctly, you are essentially asking whether all subsets of the plane are measurable according to the Lebesgue measure.

There are some well-known examples of non-measurable sets. The most common one with an obvious geometric interpretation is probably one of those involved in the Banach–Tarski paradox. That paradox is a description of how one can split a ball into a finite number of sets of points, and then reassemble these using rigid motions to obtain two copies of the original ball. This doesn't make sense if all the point sets had well-defined volume. As this is a ball in 3d, and not a planar figure, it doesn't directly serve as a counterexample to what you are asking. But it demonstrates the kind of thinking you'd likely have to employ. In particular, according to the Wikipedia article on non-measurable sets, the existence of these can only be demonstrated using the axiom of choice, and many constructions building on that axiom are hard to imagine in a geometric fashion.

Following the tags lebesgue-measure and measure-theory (which you might want to add to your question) I found Nonmeasureable subset of ${\mathbb{R}}^2$ such that no three points are collinear? Comments there by PhoemueX indicate that such sets do exist, and that Sierpinski has constructed one. So if you want an explicit example, looking up the literature on this might serve, although there are probably easier examples without the additional non-collinearity constraint.

To summarize: no, not all subsets of the plane have an area.