Let $s \geq 0$ be a real number and $E \subset \mathbb{R^n}$.
Suppose that for all $x \in E$ there is an open subset $U \subset \mathbb{R^n}$ which contains $x$, and $\mathcal{H^s}(E \cap U)=0 \Rightarrow \mathcal{H^s}(E)=0$.
How to show this implication?
I tried to use:
Since $E \subset \mathbb{R^n}$, it exists a number $s_0=$inf$\lbrace s \in [0, \infty):\mathcal{H^s}(E)=0 \rbrace$.
So $s_0(E)\leq s$ and for all $x \in E$ there is an open subset $U \Rightarrow s_0(E) \geq s$.
Here I don't see how to conclude. It should be possible to use that every open cover of $E \subset \mathbb{R^n}$ has a countable subcover, but how to apply it here?
Is this way correct or can it be proved differently?
$H^{s}$ is countably subadditive and every open cover has a countable subcover. If $E\subset \cup U_i$ with $H^{s} (E\cap U_i)=0$ for all $i$ then $H^{s}(E) \leq \sum H^{s} (E\cap U_i)=0$.