Why does Hausdorff measure go to zero as diameter power increases?

Question

For example, why does the Hausdorff measure of a flat disc go to zero when the power that the diameter is raised to (in the definition of the Hausdorff measure) reaches 3?

Answer 1

Definition: Let $(X,d)$ be a metric space, let $E\subseteq X$, fix $s,\delta > 0$, and let $$ \mathscr{H}^s_{\delta}(E) = \inf\left\{ \sum_{j} r_j^s : E \subseteq \bigcup_{j} A_j, \operatorname{diam}(A_j) = r_j \le \delta \right\}. $$ Then the $s$-dimensional Hausdorff measure of $E$ is defined to be $$ \mathscr{H}^s(E) = \inf_{\delta > 0} \mathscr{H}^s_{\delta}(E) = \lim_{\delta \to 0^+} \mathscr{H}^{s}_{\delta}(X). $$

Technically, $\mathscr{H}^s$ defines an outer measure on $X$, but the restriction of this outer measure to the measurable sets (in the sense of Carathéodory) is a Borel regular outer measure, so let's just assume that we live in a universe where the Hausdorff measure has been restricted to the Hausdorff measurable sets, which includes all of the Borel sets.

Now, the Hausdorff measure just defined has a really interesting property: the $s$-dimensional measure of a set is infinite for small values of $s$, and zero for large values of $s$. Indeed, if we define $$ \dim_{\mathrm{H}}(E) = \sup \{ s : \mathscr{H}^s(E) = \infty \}, $$ then we have $$ \mathscr{H}^s(E) = \begin{cases} \infty & \text{if $s < \dim_{\mathrm{H}}(E)$, and} \\ 0& \text{if $s > \dim_{\mathrm{H}}(E)$.} \\ \end{cases} $$ That is, the $s$-dimensional Hausdorff measure is either infinite or zero almost all of the time, and can only be finite and nonzero for a very special value of $s$, which we call the Hausdorff dimension. In the case proposed by the question, a disk is two-dimensional, which implies that if $s < 2$ then the $s$-dimensional measure of the of the disk is infinite, and if $s > 2$ then the measure is zero. Note that $s$ need not be an integer.

So, why does the Hausdorff measure have this property? It basically comes down to the following proposition:

Proposition: The following hold:

1. If $\mathscr{H}^s(E) < \infty$ and $t > s$, then $\mathscr{H}^t(E) = 0$, and
2. if $\mathscr{H}^s(E) > 0$ and $t < s$, then $\mathscr{H}^t(E) = \infty$.

Proof: We show only the first (which, to be fair, completely answers the original question), and leave the second as an exercise—the proof is nearly identical.

Observe that if $\operatorname{diam}(A) = r < \delta$, then $$ \operatorname{diam}(A)^t = r^t = r^{t-s}r^s < \delta^{t-s} r^s. $$ From this, it follows that \begin{align} \mathscr{H}^t_{\delta}(E) &= \inf\left\{ \sum_{j} r_j^t : E \subseteq \bigcup_j A_j, \operatorname{diam}(A_j) = r_j < \delta \right\} \\ &\le \inf\left\{ \delta^{t-s} \sum_{j} r_j^s : E \subseteq \bigcup_j A_j, \operatorname{diam}(A_j) = r_j < \delta \right\} \\ &= \delta^{t-s} \inf\left\{ \sum_{j} r_j^s : E \subseteq \bigcup_j A_j, \operatorname{diam}(A_j) = r_j < \delta \right\} \\ &= \delta^{t-s} \mathscr{H}^s_{\delta}(E). \end{align} But we have assumed that $\mathscr{H}^s(E) < \infty$ and we have $t-s > 0$, so, taking limits on both sides, we have $$ \mathscr{H}^t(E) = \lim_{\delta \to 0} \mathscr{H}^t_{\delta}(E) \le \lim_{\delta \to 0} r^{t-s} \mathscr{H}^s_{\delta}(E) = \left(\lim_{\delta\to 0} r^{t-s}\right) \mathscr{H}^s(E) = 0.\tag*{$\blacksquare$}$$

Basically, the point of this proof is that if the $s$-dimensional Hausdorff measure of a set is known to be finite, then the $t$-dimensional Hausdorff measure of that set must be zero for all $t$ that are larger than $s$, because we can factor out a $\delta^{t-s}$ when we look at the sums-of-diameters of a $\delta$-covering. If you are looking for an intuition about what is happening, this is likely about as good as it gets—just pay attention to where that $\delta^{t-s}$ is coming from, and what it means about the Hausdorff measure in dimension $t$.

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With a bit of extra work, we get the following interesting facts:

1. By definition, $\dim_{\mathrm{H}}(E) = \sup\{ s : \mathrm{H}^s(E) = \infty \}$. From this and the above proposition, it follows that $$ \mathscr{H}^{t}(E) = 0 \qquad\forall t > \dim_{\mathrm{H}}(E) $$ (apply the proposition with $s = \frac{1}{2}(\dim_{\mathrm{H}}(E) + t)$; observe that $\mathscr{H}^s(E) < \infty$).
2. Therefore we could equivalently define $$ \dim_{\mathrm{H}}(E) = \inf\{ s : \mathscr{H}^s(E) = 0 \}, $$ which justifies the dichotomy stated above.
3. If it is known that $0 < \mathscr{H}^s(E) < \infty$, then we immediately know that the Hausdorff dimension of $E$ is $s$. Since it is not too difficult to show that the Lebesgue measure on $\mathbb{R}^n$ and the $n$-dimensional Hausdorff measure on $\mathbb{R}^n$ agree (up to a constant), it follows that any set of finite, nonzero Lebesgue measure in $\mathbb{R}^n$ must be of Hausdorff dimension $n$. In particular, a disk in $\mathbb{R}^2$ is $2$-dimensional. This means that the $(2+\varepsilon)$-dimensional Hausdorff measure of such a disk must be zero for all $\varepsilon > 0$, and the $(2-\varepsilon)$-dimensional Hausdorff measure of such a disk must be infinite for all $\varepsilon \in [0,2)$.

Answer 2

A flat disc $D$ of radius $1$ has area $\pi.$ For $n\in \Bbb N$ let $S_n$ be a set of open discs, each of diameter $1/n$ or less, with $\cup S_n \supset D$ and $$\sum_{t\in S_n}A(t)<2\pi,$$ where $A(t)$ is the area of $t.$

For $t\in S_n$ let $d(t)$ be the diameter of $t.$ Then $A(t)=\pi d(t)^2/4.$

Let $r>0.$ Then $$\sum_{t\in S_n} \pi d(t)^{2+r}/4=\sum_{t\in S_n}A(t)d(t)^r\leq \sum_{t\in S_n}A(t)(1/n)^r<2\pi(1/n)^r.$$ As $n\to \infty$ we have $2\pi (1/n)^r\to 0.$