0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.