# The Hausdorff dimension (implication)

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?

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.
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.