For a number $0 < \alpha < 1$, a function $f\colon \mathbb R \to \mathbb R^2$ is said to be $\alpha$-Hölder continuous if there is a constant $C$ less than infinity such that $\lvert f (x)-f (y)\rvert \leq C\lvert x-y\rvert ^{\alpha} $ for all $x,y \in \mathbb R $. Prove that if $f $ is $\alpha$-Hölder continuous with $\alpha > 1/2$, then the image of $f $ has measure zero in $\mathbb R^2$.
I'm not sure how to do this here. The question states that it's known that there is a surjective (1/2)-Hölder continuous map $g\colon \mathbb R \to \mathbb R^2$. Any help would be greatly appreciated.
Let $p \in \mathbb{Z}$. I will prove that $\mu_2(f([p,p+1[))=0$. Let $n \in \mathbb{N}$. Then for $0 \le k \le n-1$, the hypothesis on the Holder continuity yields $f\big(\big[p+\frac{k}{n},p+\frac{k+1}{n}\big[\big) \subset B_f\big(f(k/n), \frac{C}{n^{\alpha}}\big)$, and thus $\mu_2\big(f\big(\big[p+\frac{k}{n},p+\frac{k+1}{n}\big[\big)\big) \le \frac{\pi C^2}{n^{2\alpha}}$, so taking the union for $0\le k<p$, we get $\mu_2\big(f([p,p+1[)\big) \le \frac{\pi C^2}{n^{2\alpha-1}}$. Because $\alpha > \frac{1}{2}$, when $n \to \infty$ we get $\mu_2 \big(f([p,p+1[)\big) =0$. Taking the union on $p \in \mathbb{Z}$, we conclude $\mu_2\big(f(\mathbb{R})\big)=0$.