Sets with Hausdorff-Measure 0

Solution 1:

(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 $$ H^s(A) \ge H^s(A_{n_0}) = \infty. $$

Solution 2:

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 $H^\beta(K)>0$.