Showing that the unit circle has measure zero

Question

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.

Answer 1

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*}

Answer 2

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} $