Why does Hausdorff measure go to zero as diameter power increases? 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?
 A: 
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:
  
  
*
  
*If $\mathscr{H}^s(E) < \infty$ and $t > s$, then $\mathscr{H}^t(E) = 0$, and
  
*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 $t$.

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


*

*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 $\mathscr{H}^s(E) < \infty$).

*Therefore we could equivalently define
$$ \dim_{\mathrm{H}}(E) = \inf\{ s : \mathscr{H}^s(E) = 0 \}, $$
which justifies the dichotomy stated above.

*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 $\varepsilon \in [0,2)$.

A: 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 $2\pi (1/n)^r\to 0.$
