## Are there non-measurable sets with smooth boundary?

I learned analysis a while ago, so let me define what I want. Suppose we have a set whose boundary is a closed Jordan curve that has a unique tangent at each point. Is it true that this set is (Lebesgue) measurable? What if we also suppose that the tangents change continuously? I am interested in any modification of the question which is known.

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As was pointed out, the question did not intend the set to be the interior of the curve. The question now boils down to whether a differentiable Jordan curve has measure zero. This follows from the Lebesgue density theorem, see e.g. http://en.wikipedia.org/wiki/Lebesgue's_density_theorem.

If the curve is differentiable, then at each point of the curve the Lebesgue density is zero. By the Lebesgue density theorem, the measure of the curve must be zero.

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This answer should not have been accepted, since the given set is not necessarily the interior, but includes a subset of the Jordan curve itself. Not every Jordan curve has measure zero; there are space-filling Jordan curves. So this proof is not correct. – Michael Beeson Mar 15 2011 at 17:00
dont ask if is true that "A is a measurable (or open) set $\Rightarrow$ A has smooth boundary" but ask if is true that "A is a non measurable set $\Rightarrow$ A has a non-smooth boundary". – Buschi Sergio Mar 15 2011 at 17:15
Thanks Michael Beeson, you are right, I deaccepted the answer. – domotorp Mar 15 2011 at 17:32
Thank you, now I see the proof! – domotorp Mar 16 2011 at 13:55

I would say: The interior is an open set, so is measurable. The boundary (a smooth Jordan curve) has measure zero, so every subset of it is measurable. Our set is the union of the interior and a subset of the boundary, so it is also measurable.

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 I am sure that it is true that smooth Jordan curves have measure zero, but could you please give a proof or a pointer to one? – domotorp Mar 16 2011 at 6:34 As Michael Renardy said in his revised answer: If the curve has a tangent at a certain point, then it has density zero at that point. (Use epsilon-delta definition for tangent.) Therefore by the Lebesgue density theorem, the curve has measure zero. – Gerald Edgar 0 secs ago – Gerald Edgar Mar 16 2011 at 13:30 Thank you! I think it is more fair if I accept his answer as it came first, I hope you don't mind. – domotorp Mar 16 2011 at 14:03

Let $C$ be a Jordan curve, and suppose that (the range of) $C$ has measure zero. The interior of $C$ is an open set and the given set (whose boundary is $C$) then differs from an open set by a set of measure zero (a subset of the range of $C$), so it is measurable.

The question then boils down to, what conditions on $C$ suffice for its range to be of measure zero. It is not hard to see that a bound $M$ on the derivative of $C$ suffices, for then we can cover the range of $C$ with $2\pi/\epsilon$ disks of radius $\epsilon/(2\pi M)$, whose total area is less than $\epsilon$. If the Jordan curve is $C^1$ we have such a bound. However, this approach gets harder if we don't assume $C$ is $C^1$.

So here's another attack: let $C$ be rectifiable, i.e. the limit of polygons of bounded length $L$. Then for each $\epsilon > 0$, the range of $C$ is contained in a union of rectangles, one about each side of an approximating polygon, where each rectangle has height $\epsilon/L$ and the total width of the rectangles is at most $L$. Since $\epsilon$ is arbitrary, the measure of a rectifiable Jordan curve is zero. Rectifiable is more general than $C^1$ so this improves the previous paragraph.

I don't know whether there exists a differentiable but not rectifiable Jordan curve.

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 Thanks! Non rectifiable, differentiable curves exist, something like x^2 sin(1/x^2) should work. – domotorp Mar 15 2011 at 17:30