A circle is a set of measure zero. Generalizations? When reading something else I came across the claim that the circle $\{x \in \Re^2 \mid \|x\| = 1\}$ is a set of Lebesgue measure zero.
I can give a proof in this special case by squeezing the circle between two polygons. But is there any general theorem that allows us to identify the circle as a "one-dimensional" set and then conclude that its measure is zero?
 A: It may not be the most general statement of the type you desire, but it certainly has the right flavour.

Let $M$ and $N$ be smooth manifolds of dimensions $m$ and $n$ respectively. If $m < n$, for any smooth map $f : M \to N$, the image of $f$ has measure zero.

This statement follow's immediately from Sard's Theorem which states that the set of critical values of a smooth map has measure zero.
The case in your post corresponds to $M = S^1$, $N = \mathbb{R}^2$ and $f$ is the inclusion $S^1 \hookrightarrow \mathbb{R}^2$.
A: See https://web.archive.org/web/20200928114329/http://www1.uwindsor.ca/math/sites/uwindsor.ca.math/files/05-03.pdf.
The set of zeros of a non-zero polynomial $p: \mathbb{R}^n \to \mathbb{R}$ has measure zero.
In your example, $p(x) = x_1^2+x_2^2-1$.
Another possibility would be to use the implicit function theorem.
In this example, using the above $p$, we have the circle $C$ is given by $p(x) = 0$, and for each $x \in C$, we have $Dp(x) =2x^T\neq 0$. Also, $p$ is $C^1$. At any given point, we can parametrize $C$ in a neighbourhood using the implicit function theorem. A compactness argument shows that we only need a finite number of these neighbourhoods and an argument using the Lipschitz nature of the parametrizing function shows that each has measure zero.
This is a bit of overkill, perhaps, but illustrates the 'one-dimensional' nature of $C$.
A: As others have pointed out Sard's theorem gives a general result. In case you are interested in a low-brow reasoning I proffer the following.


*

*If $C$ is a compact subset of $\Bbb{R}^n$, and $f:C\to\Bbb{R}$ is a continuous function, then the graph of $f$, $G=\{(x,f(x))\in\Bbb{R}^{n+1}\mid x\in C\}$, has measure zero. This is because $f$ is necessarily uniformly continuous, and thus for all $\varepsilon>0$ we can find a finite set of boxes that A) their union contains $G$, B) their bases cover $C$ with as little extra as desired, C) their heights are all $\le\varepsilon$. This implies that $m(G)=0$.

*If a set $S$ consists of at most countably infinitely many pieces like the graph $G$ in item 1, then $m(S)=0$. This follows from countable additivity.


So for example the unit circle is the union of two graphs of a continuous function defined on $C=[-1,1]$, and thus has measure zero. Similarly the sine curve in the plane has measure zero as the union of the graphs of $\sin x$ restricted to $C_n=[2n\pi,2(n+1)\pi],n\in\Bbb{Z}$.
