Hausdorff $2$-dimensional measure of $\mathbf{R}$

Question

I know that $\mathcal{H}^2(\mathbf{R}) = 0$, but what is any easy way to see it? I tried coming up with reasonable covers, but they all seem to give upper bounds that are too large.

Answer 1

You can cover by rectangles of small radius. Fix $\delta>0$ and consider covering of $\mathbb{R}$ by intervals of length $\frac{\delta}{n}$, where $n\in \mathbb{N}$. This is possible in view of the divergence of the harmonic series. Namely, ignoring $\delta>0$ for a moment, start with $[0,1]$, then consider $[1, 1 + 1/2]$, $[1+1/2, 1+1/2+1/3]$, ... . Repeat the same construction for the negative axis. We get a countable cover of $\mathbb{R}$ and let $\{I_n\}$ be some enumeration of these intervals (the exact order does not matter). For $n\in \mathbb{N}$, consider a rectangle of height $\delta>0$ whose side length is equal to the length of $I_n$ and which contains the interval $I_n$. Then the union $\bigcup\limits_{n}^\infty I_n$ covers $\mathbb{R}$ with sets of diameter bounded above by $2\delta$. However $$ H^2_{2\delta}(\mathbb{R}) \leq \sum\limits_{n=1}^\infty \mathrm{diam}(I_n)^2 \leq 2\delta^2 \sum\limits_{n=1}^\infty \frac{1}{n^2} \leq 10 \delta^2, $$ hence $H^2(\mathbb{R}) = \lim\limits_{\delta \to 0} H^2_\delta(\mathbb{R}) = 0$.