Definition of Hausdorff Measure: example question I am studying the Hausdorff measure and dimension, but I am struggling to understand the reason that the $n$-dimensional Hausdorff measure is zero for a set with Hausdorff dimension $<n$. 
The s-dimensional Hausdorff measure is defined by $H^s (E) = \lim_{\delta\rightarrow 0} \inf\{\sum\limits_{i=0}^{\infty} (diam(U_i))^s : \bigcup_{i=0}^{\infty} U_i \supset E, diam(U_i) < \delta\}$
Consider a a 1D arc (i.e. just a curve), say in $\mathbb{R}^2$. Its 1-dimensional Hausdorff measure is its arclength.  This makes sense to me, as the above definition then looks almost identical to the definition of the Lebesgue measure on R.
However, the 2-dimensional Hasudorff measure for a 1D arc will be zero, as it has no area.   I do not understand how with the above definition, changing s to 2 will give us zero.  All we are doing is squaring the diameter of the covering "balls."  I suppose this is the equivalent of covering the arc with infinitesimally small circles, but it is not clear to me why the sum of the area of all these circles should be zero.
 A: Notice that at scale $\delta$, we have that
$$\sum_{i = 0}^{\infty} \operatorname{diam}(U_i)^2 \le \delta \sum_{i = 0}^{\infty} \operatorname{diam}(U_i)$$
so that the sum of squares is much smaller than the sum without squares. It's never actually zero, but letting $\delta \to 0$ shows that any set with finite $\mathcal{H}^1$ measure zero has zero $\mathcal{H}^2$ measure. A minor refinement of the argument gives that finite $\mathcal{H}^s$ measure implies zero $\mathcal{H}^t$ measure whenever $s < t$.
A: Since the question is about "not understand[ing] how" and presumably why of these facts, I will try this:
If you have a segment of a line in, say R^2, then you can cover it by order N many sets of diameter 1/N. So, when you estimate the 1-Hausdorff content you end up with a finite value. If you use the same covering but you write the estimate for 2-Hausdorff measure, then you raise each diameter to power 2. Thus, you end up with a sum of order N*(1/N)^2 = 1/N. So, as N goes to infinity (diameters of coverings must go to zero), we get 0 for 2-Hausdorff measure. The result would have been the same for any s-Hausdorff for $s>1$.
