2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $s < t$.

$\endgroup$
  • 1
    $\begingroup$ This is very helpful, thank you--this makes sense for a finite arc, as I asked in my question. As a further question, say my arc has infinite length though, the $\mathcal{H}^2$ measure will be zero, but I don't think your argument applies in that case, because $\sum_{i = 0}^{\infty} \operatorname{diam}(U_i)$ is infinite. Is there some way for me to understand this case as well? $\endgroup$ – user310374 Apr 24 '16 at 1:30
  • $\begingroup$ Right, the argument certainly does not apply to sets of infinite $\mathcal{H}^1$ measure - the flip side of what I wrote in the answer is that if $\mathcal{H}^t(E)$ is positive then $\mathcal{H}^s(E)$ is infinite for $s < t$. $\endgroup$ – user296602 Apr 24 '16 at 1:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.