The Hausdorff dimension (implication)

Question

Let: $s \geq 0, s \in \mathbb{R}$ and $E \subset \mathbb{R^n}$.

Suppose that for every $x \in E$ exists an open subset $U \subset \mathbb{R^n}$ which contains $x$

and dim$_\mathcal{H}(E \cap U) \leq s \Rightarrow$ dim$_\mathcal{H}(E) \leq s$.

How to prove that implication?

I tried to use that:

The Hausdorff dimension is given by $s_0=$inf$\lbrace s \in [0, \infty): \mathcal{H^s}(E)=0 \rbrace$, there is a $s=m \in \mathbb{N}$ such that $s_0(E) \leq m$.

 $U \subset \mathbb{R^n}$ is open $\Rightarrow s_0(E) \geq m$

So I got another result than instead of  dim$_\mathcal{H}(E) \leq s$

How to conclude this?

Answer 1

Pick a $t>s$ to be arbitrary. What you need to show now is that $$\mathcal{H}^{t}(E)=0.$$ Now for every $x\in E$ you find a Ball $B_r(x)$, such that $$\mathcal{H}^t(B_r(x)\cap E)=0.$$ Now you can choose a countable subcollection of these balls (see e.g. Besicovitch covering theorem) such that $$E\subset \bigcup_{J\in N}B_{r_j}(x_j)\cap E$$ and $N$ is countable. Since all these balls satisfy the assumption and $\mathcal{H}^t$ is an outer measure you get $$\mathcal{H}^t(E)\leq \sum_{j\in N}\mathcal{H}^t(B_{r_j}(x_j)\cap E) = 0.$$ Hence $dim_{\mathcal{H}}(E)\leq t$. Since $t>s$ arbitrary the result follows.

Answer 2

The Hausdorff measures satisfies that $\mathcal{H}^t(E)=0$ for every $t>d$ if and only if $dim(E)\leq d$. So we have to prove that $\mathcal{H}^r(E)=0$ for every $r>s$. Let $r$ such a number.

We know that for every $x$ in $E$ there exist an open neighborhood $U_x$ of $x$ such that $\mathcal{H}^r(E\cap U_x)=0$. So we have a cover of $E$ by open sets of $\mathbb{R}^n$ and we can take a countable subcover $\{U_n\}_{n \in \mathbb{N}}$.

By $\sigma$-subadditivity we get that $\mathcal{H}^r(E)\leq\sum_{n\in \mathbb{N}}\mathcal{H}^r(E\cap U_n)=0$. And we are done.