Showing that the unit circle has measure zero How can  I show that $\{x \in \mathbb{R}^{2}: |x| =1\}$ has measure zero using the definition of measure zero? I don't want to use the interpretation of measure in $\mathbb{R}^2$ as area (and then show this by using the fact that the area of a circle is $\pi r^2$. I was thinking somehow find rectangles to that contain the circumference of this circle and keeping one of the sides of each rectangle free to make $\varepsilon$-small. I'm not able to get ahead with this approach though. Thoughts and ideas?
Note - this question is NOT a duplicate of Graph of real continuous function has measure zero because it focuses on a specific example. The results from the other question might be modifiable to apply here, but look at something different. 
 A: Look at the regular $4n$-gon with $(1,0)$ being a vertex.  We can cover the circle by $4n$ rectangles
$$
\overline{\operatorname{conv}}\left\{\cos\left(\frac{k\pi}{2n}\right),\cos\left(\frac{(k+1)\pi}{2n}\right)\right\}\times
\overline{\operatorname{conv}}\left\{\sin\left(\frac{k\pi}{2n}\right),\sin\left(\frac{(k+1)\pi}{2n}\right)\right\},\quad k=0,1,\dots,4n-1
$$
The sum of measures of these rectangles is
\begin{align*}
&\sum_k\left\lvert\cos\left(\frac{k\pi}{2n}\right)-\cos\left(\frac{(k+1)\pi}{2n}\right)\right\rvert\cdot\left\lvert\sin\left(\frac{k\pi}{2n}\right)-\sin\left(\frac{(k+1)\pi}{2n}\right)\right\rvert\\
&=4\sum_k\left\lvert\sin\left(\frac{\pi}{4n}\right)\sin\left(\frac{(2k+1)\pi}{4n}\right)\right\rvert\cdot\left\lvert\sin\left(\frac{\pi}{4n}\right)\cos\left(\frac{(2k+1)\pi}{4n}\right)\right\rvert\\
&=2\sin^2\left(\frac{\pi}{4n}\right)\sum_k\left\lvert\sin\left(\frac{(2k+1)\pi}{2n}\right)\right\rvert\\
&\leq 2\sin^2\left(\frac{\pi}{4n}\right)\cdot 4n\to 0\quad\text{as }n\to\infty.
\end{align*}
A: I will show that
the difference between
the areas of the
circumscribed and inscribed
$n$-gons goes to zero.
Consider a regular $n$-gon
inscribed in the unit circle.
There are $2n$ triangles
with central angle
$t = \pi/n$
and hypotenuse
$1$,
so the distance to the side
$s_n$ and length $h_n$
satisfy
$s_n = \cos(t)$
and $h_n = \sin(t)$.
The area of each triangle
is thus
$\frac12 s_nh_n
=\frac12\cos(t)\sin(t)
=\frac14\sin(2t)
$
so the area of the
inscribed $n$-gon is
$2n$ times this or
$\frac12n\sin(2t)
$.
Extend the radii
to get the
circumscribed $n$-gon.
There are $2n$ triangles
with base $1$
and height
$g_n$ such that
$g_n = \tan(t)$,
so the area is
$\frac12 g_n
=\frac12 \tan(t)
$.
The total area is thus
$n\tan(t)
$.
Note that
both of these areas go to
$\pi$ as $n \to \infty$
since
$\sin(x) \approx \tan(x)
\approx x$
as $x \to 0$.
However,
the only inequality needed is
$\sin(x) < x$
for $0 < x < \pi/2$.
The difference in
the two areas is thus
$\begin{array}\\
D_n
&=n\tan(t)-\frac12 n\sin(2t)\\
&=n\left(\tan(t)-\frac12 \sin(2t)\right)\\
&=n\left(\dfrac{\sin(t)}{\cos(t)}-\sin(t)\cos(t)\right)\\
&=n\sin(t)\left(\dfrac{1}{\cos(t)}-\cos(t)\right)\\
&=n\sin(t)\dfrac{1-\cos^2(t)}{\cos(t)}\\
&=n\sin(t)\dfrac{\sin^2(t)}{\cos(t)}\\
&=n\dfrac{\sin^3(t)}{\cos(t)}\\
&=n\dfrac{\sin^3(t)}{1-2\sin^2(t/2)}\\
\end{array}
$
We now use
$\sin(x) < x$
for
$0 < x < \pi/2$
so
$\begin{array}\\
D_n
&\lt n\dfrac{t^3}{1-2(t/2)^2}\\
&= n\dfrac{(\pi/n)^3}{1-2(\pi/(2n))^2}\\
&= \dfrac{1}{n^2}\dfrac{\pi^3}{1-\pi^2/(2n^2)}\\
&< \dfrac{2\pi^3}{n^2}
\qquad\text{for } n \ge 4\\
&\to 0
\qquad\text{as } n \to \infty\\
\end{array}
$
