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

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.

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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?)

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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 '12 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 '12 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 '12 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 '12 at 19:35