Prove that if $\mathcal H^1(F) = 0$ for $F\subset \Bbb R$, then $\Bbb R\setminus F$ is dense

Question

Prove that if $\mathcal H^1(F) = 0$ for $F\subset \Bbb R$, then $\Bbb R\setminus F$ is dense.

I need to prove the above result to understand Proposition $3.5.$ in Falconer's Fractal Geometry. $\mathcal H^1$ is the $1$-dimensional Hausdorff measure on $\Bbb R$.

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My thoughts. If $\Bbb R\setminus F$ is not dense, there exists a non-empty open set $U \subset F$. To find a contradiction, maybe I can show that $\mathcal H^1(U) > 0$? How should I go about that? I'm trying to strictly use the definitions below, and not involve the fact that the Hausdorff measure on $\mathbb R^n$ is (up to a scalar) the same as the Lebesgue measure on $\Bbb R^n$. The definitions are $$\mathcal H^1(F) := \lim_{\delta\to 0} \mathcal H^1_\delta(F)$$ where $$\mathcal H^1_\delta(F) = \inf\left\{\sum_{i=1}^\infty |U_i|: \{U_i\} \text{ is a }\delta\text{-cover of }F \right\}$$ and $|U_i|$ denotes the diameter of the set $U_i \subset \Bbb R$. $\{U_i\}$ is a $\delta$-cover if the diameter of every set $U_i$ in the cover is no larger than $\delta$. How should I proceed? Since $U$ is open, I know that $\dim_H U = 1$, but I'm not sure that helps.

Answer 1

If you are willing to use that $\mathcal{H}$ is translation invariant, there is a simple argument. In particular, the open set $U$ you describe must have positive diameter, meaning that for some countable set $Z$, $\{ U+x : x \in Z \}$ is an open cover. By the $\sigma$-additivity of the measure, this implies that $\mathcal{H}(\mathbb{R}) \leq \sum_{z \in Z} \mathcal{H}(U) = 0$ since $U \subseteq F$. This is clearly a contradiction, so no such $U$ exists.

To clarify, why such $Z$ must exist can be seen as follows. Since $U$ is open, we can find some open ball $(a - \epsilon, a + \epsilon)$ entirely contained in $U$ for some $a \in U$ and $\epsilon > 0$. Shrinking $\epsilon$ if necessary, we can even go further to say that $I = [a - \epsilon, a + \epsilon] \subseteq U$. Then we can tile $\mathbb{R}$ with $I$, or in other words construct a cover $\{ I + 2\epsilon n : n \in \mathbb{Z}\}$. Then it follows that $\{ U + 2\epsilon n: n \in \mathbb{Z}\}$ is an open cover of the whole space. The countable set $Z$ is just $\{ 2\epsilon n : n \in \mathbb{Z} \}$ (this is of course not the only possibility for $Z$, but I chose the notation $Z$ with this example in mind). This approach actually works with certain spaces other than $\mathbb{R}$, although you would need a different construction for $Z$. In $\mathbb{R}^{n}$, for instance you would want to make $I$ be a closed box, and take $Z$ to be an integer lattice.

Answer 2

It is easy to show that $\mathcal{H}^n = m_n$ when $n = 1$. The key observation is that if $F \subset \mathbb{R}$, then $F \subset I = [\inf(F), \sup(F)]$ and $\text{diam}(F) = \text{diam}(I)$.

Now use the fact that non-empty open intervals have positive Lebesgue measure to prove your claim.