I have been reading the book Modern Real Analysis by Ziemer and have come to an exercise that I am having trouble with. The exercise is in the chapter on measure theory in the section describing the Hausdorff measure. The exercise statement goes :
Show that $H^{s} \equiv 0$ on $\mathbb{R}^{n}$ for $s > n$.
Here $H^{s}$ is the Hausdorff measure with dimension $s$.
There is a theorem included in the text.
Theorem 4.39
Suppose $A \subset \mathbb{R}^{n}$ and $0 \leq s < t < \infty$. Then :
- If $H^{s}(A) < \infty$ then $H^{t}(A) = 0$
- If $H^{t}(A) > 0$ then $H^{s}(A) = \infty$
There is also the exercise that immediately precedes it.
Exercise 4.7.2
For $A \subset \mathbb{R}^{n}$, use the isodiametric inequality to show that : \begin{equation} \lambda^{*}(A) \leq H^{n}(A) \leq \alpha(n)\left( \frac{\sqrt{n}}{2} \right)^{n} \lambda^{*}(A) \end{equation} Here : \begin{equation} \alpha(s) = \frac{\pi^{\frac{s}{2}}}{\Gamma\left( \frac{\pi}{2} + 1 \right)} \end{equation} Where $\Gamma$ is the gamma function. $\lambda^{*}$ is the Lebesgue outer measure.
Here is my solution so far :
We know by exercise 4.7.2 : \begin{equation} \lambda^{*}(A) \leq H^{n}(A) \leq \alpha(n) \left( \frac{\sqrt{n}}{2} \right)^{n} \lambda^{*}(A) \end{equation} So : \begin{equation} \lambda^{*}(A) = \infty \Rightarrow H^{n}(A) = \infty \end{equation} and : \begin{equation} \lambda^{*}(A) = 0 \Rightarrow H^{n}(A) = 0 \end{equation} and : \begin{equation} \lambda^{*}(A) \in (0,\infty) \Rightarrow H^{n}(A) \in (0,\infty) \end{equation} So by theorem 4.39 : \begin{equation} \lambda^{*}(A) = 0 \Rightarrow H^{n}(A) = 0 \Rightarrow H^{s}(A) = 0 \; \forall s > n \; \checkmark \end{equation} and : \begin{align} \lambda^{*}(A) \in (0,\infty) & \Rightarrow H^{n}(A) \in (0,\infty)\\ & \Rightarrow H^{n}(A) < \infty \\ & \Rightarrow H^{s}(A) = 0 \; \forall s > n \; \checkmark \end{align} I'm not sure how to prove that it works when $\lambda^{*}(A) = \infty$ and $H^{n}(A) = \infty$. Can someone help with this ?