Sets with Hausdorff-Measure 0 The $\alpha$-Dimensional Hausdorff-Measure of a Set A is defined as $H^\alpha (A)=\inf_{A\text{ is countable covering}}\sum_{A'\in A} diam(A')^\alpha$. It is easy to show, that for every set $E\subseteq\mathbb{R}^d$ there exists a unique $\beta\in \mathbb{R}$, so that $H^\alpha(E)=0$ for $\alpha>\beta$ and $H^\alpha(E)=\infty$ for $\alpha < \beta$. This $\beta$ is called the Hausdorff-Dimension $dim(E)$ of E. It is easy to show, that $dim(\mathbb{R}^d)=d$ and $H^d(\mathbb{R}^d)=\infty$. I  wonder if there is any Set A with $dim(A)=\alpha$ and $H^\alpha(A)=0$.
Does anybody have an idea how to approach this?
 A: (See Ullrich's link for corrected definition of Hausdorff measure.)  
Another (softer) approach.  For $n=1,2,3,\dots$ let $A_n \subseteq \mathbb R$ be a set with Hausdorff dimension $1-\frac{1}{n}$.  Then I claim that
$$
A := \bigcup_{n=1}^\infty A_n
$$
satisfies $\dim A = 1$ but $H^1(A) = 0$.  Indeed, for $s\ge 1$ we have $s>1-\frac{1}{n}$ for all $n$ and therefore
$$
H^s(A) \le \sum_n H^s(A_n) = \sum_n 0 = 0 .
$$
But for $s<1$ there exists $n_0$ so that $1-\frac{1}{n_0} > s$ and therefore
$$
H^s(A) \ge H^s(A_{n_0}) = \infty.
$$
A: Your definition of Hausdorff measure is wrong; the actual definition is a little more complicated. See https://en.wikipedia.org/wiki/Hausdorff_measure . 
But what you've defined, something that's sometimes called the Hausdorff "content", is sufficient to give a correct definition of the Hausdorff dimension; the Hausdorff content vanishes if and only if the Hausdorff measure vanishes.
Yes, for $0<\alpha\le d$ there exists $A\subset\Bbb R^d$ with $\dim(A)=\alpha$ and $H^\alpha(A)=0$. Take $d=1$ for (relative) simplicity. The construction is like the construction of the middle-thirds Cantor set, except that at each stage we remove the middle something, not necessarily the middle third.
In particular: Suppose that $\delta_n>0$ is such that $\delta_0=1$ and $$\delta_{n+1}<\frac12\delta_n.$$There exist compact sets $K_n\subset\Bbb R$ with $K_0=[0,1]$, $K_{n+1}\subset K_n$, and such that $K_n$ is the union of $2^n$ disjoint closed intervals $I_1^n,\dots,I_{2^n}^n$, each of length $\delta_n$. We're going to set $K=\bigcap_{n=1}^\infty K_n$. 
Given $\alpha\in(0,1]$ it's possible to choose $(\delta_n)$ in such a way that $H^\alpha(K)=0$ while $H^\beta(K)>0$ for all $\beta\in(0,\alpha)$. 
(It follows then that $K$ has infinite $\beta$-dimensional Hausdorff measure for $0<\beta<\alpha$; note however that the Hausdorff content of any compact set is finite.)
It's enough to obtain $$2^n\delta_n^\alpha\to0$$while$$2^n\delta_n^\beta\to\infty\quad(0<\beta<\alpha).$$ So if we say $\delta_n=2^{-\gamma_n}$ we need $$\gamma_{n+1}-\gamma_n>1,$$
$$n-\alpha\gamma_n\to-\infty,$$and $$n-\beta\gamma_n\to\infty.$$ You can check that all three conditions hold if $$\gamma_n=\frac n\alpha+\sqrt n.$$
Now the fact that $2^n\delta_n^\alpha\to0$ makes it clear that $H^\alpha(K)=0$, by definition. Similarly $2^n\delta_n^\beta\to\infty$ makes it plausible that $H^\beta(K)>0$, although here there's something to be proved.
The proof is by the trivial converse of "Frostman's Lemma". There exists a probability measure $\mu$ supported on $K$ such that $$\mu(I_j^n)=2^{-n}$$for $1\le j\le 2^n$. That is, if $|I|$ denotes the length of the interval $I$, we have $$\mu(I_j^n)=|I_j^n|^{n/\gamma_n}.$$It follows that if $0<\beta<\alpha$ there exists $c$ such that for every interval $I\subset[0,1]$ we have $$\mu(I)\ge c|I|^\beta.$$Hence if $K$ is covered by a collection of intervals $I$ we have $$1=\mu(K)\le\sum_I\mu(I)\le c\sum_I|I|^\beta,$$so $H^\beta(K)>0$.
