Welcome to MathOverflow

Question

2

1

Let E be the Euclidean plane and let M(X) be two-dimensional Lebesgue measure defined for each Borel subset X of E. Suppose that s is an arc in E and that e is a positive real number. Does there always exist a bounded connected open subset Z of E such that (!) s is a subset of Z (2) M(Z)-M(s) is not greater than e (3) Z is the interior of a Jordan curve? It is not hard to show that such a Z exists if only conditions (1) and (2) are required to be satisfied. But how does one show that Z can be the interior of a Jordan curve, or even be simply connected? Remember that s can have an infinity of wiggles and can have positive two-dimensional Lebesgue measure.

flag	asked Nov 13 at 19:05 Garabed Gulbenkian 3,083●6●13

Lasse Rempe-Gillen · Accepted Answer · 2012-11-13 20:12:25Z UTC

4

The answer should be yes.

Consider $E$ to be a subset of the Riemann sphere $\hat{\mathbb{C}}$. The complement $U$ of $s$ is simply-connected, so we can map it conformally to the unit disk (taking $\infty$ to $0$, say).

Take the preimage of a circle of radius $r$, close to $1$, under this conformal map. This gives you a Jordan curve, and the Jordan domain $Z_r$ enclosed by this curve will have area tending to the area of $s$. So for a suitable value of $r$, this domain will have the desired property.

(Am I am missing something?)

answered Nov 13 at 20:12

Lasse Rempe-Gillen
1,265●4●11

	Thanks. I think you may have found the right approach to proving that the answer is "yes", if in fact it is "yes".But I do not see how the complement of an arc in the Euclidean plane can be simply connected. – Garabed Gulbenkian Nov 17 at 19:37
	I didn't say the complement in the Euclidean plane is simply-connected, I said the complement in the Riemann sphere is simply-connected. Indeed, a connected open subset of the sphere is simply-connected if and only if its complement is connected. (The complement in the Euclidean plane is doubly-connected, and conformally isomorphic to the punctured disk.) – Lasse Rempe-Gillen Nov 18 at 15:50
	I'll also mention that the same argument works not just for an arc, but for any compact and connected set that is full (i.e. its complement is connected). – Lasse Rempe-Gillen Nov 18 at 15:51
	You are right and I apologize. I forgot that the theorem about simply connected planar sets being homeomorphic defines simple connectedness in terms of complements taken with respect to the Riemann sphere-rather than the Euclidean plane. – Garabed Gulbenkian Nov 18 at 19:35

A delicate measure-theoretic question about Jordan curves and arcs in the plane.

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.