1. ## Hausdorff Measure

The question I am trying to answer is as follows

Let F={(a,b): 0< a, b<7}$\displaystyle \subset$ $\mathbb{R}^{2}$. Show that Hs(F)=0 if s>2, 0<H2(F)< $\displaystyle \infty$ and Hs(F) = $\displaystyle \infty$ if s<2. (You may need to use the fact that if U is a subset of $\mathbb{R}^{2}$, then the area of U is at most equation to |U|2 x $\displaystyle \pi$/4

I have managed to show Hs(F)=0 if s>2 but showing 0<H2(F)< $\displaystyle \infty$ if s<2 I do not know how to do or even get started.

Any help greatfully appreciated

2. ## Re: Hausdorff Measure

The following is a simple consequence of the definition of Hausdorff measure: if $\displaystyle H^s(E) > 0$ and $\displaystyle t < s$, then $\displaystyle H^t(E) = \infty$, and if $\displaystyle H^s(E) < \infty$ and $\displaystyle t > s$ then $\displaystyle H^t(E) = 0$. This reduces your present situation to showing that $\displaystyle 0 < H^2(F) < \infty$, which is a straightforward computation.

Edit:

Fix a $\displaystyle \delta > 0$ and let $\displaystyle \{E_j\}_{j=1}^{\infty}$ be any covering of F so that $\displaystyle diam(E_j) < \delta$ for each j. Now $\displaystyle diam(E_j)^2 = diam(E_j)^{2-s} diam(E_j)^s < \delta^{2-s} diam(E_j)^s$.

So by definition of the Hausdorff measure, $\displaystyle H^2_{\delta}(F) \le \sum_{j=1}^{\infty} diam(E_j)^2 \le \delta^{2-s} \sum_{j=1}^{\infty} diam(E_j)^s$.

That is, $\displaystyle \delta^{s -2} H^2_{\delta}(F) \le \sum_{j=1}^{\infty} diam(E_j)^s$.

Inffing over all such coverings, we're led to

$\displaystyle \delta^{s - 2}H^2_{\delta}(F) \le H_{\delta}^s (F)$.

By definition, $\displaystyle H^s(F) = \lim_{\delta \to 0^{+}} H_{\delta}^s(F)$. Take the limit as delta tends to 0 and conclude that since s < 2, the left hand side tends to infinity if the 2 dimensional Hausdorff measure of F is positive.