I have recently started to study measure theory and in particular, right now, the Hausdorff measure. This question concerns a pdf-file I am reading:
https://www.math.cuhk.edu.hk/course_builder/1415/math5011/MATH5011_Chapter_3.2014.pdf,
page 17, proposition 3.9 (c).
Proposition 3.9
Let $A\subset\mathbb{R}^n$.
(b) If $\mathcal{H}^s(A)<\infty$ for some $s\in [0,\infty)$, then $\mathcal{H}^t(A)=0$ for all $t>s$.
(c) If $0<\mathcal{H}^s(A)<\infty$ for some $s\in [0,\infty)$, then $\mathcal{H}^t(A)=\infty$ for all $t<s$.
In the paper, they write that (c) follows from (b). I might be stupid but I cannot really see how it follows from (b). Do they mean we just have to apply the statement of (b) to conclude (c) (which I do not see how it does)? Or do they mean that the argument of (b) just have to be tweaked a little to arrive at (c)?
Solution
(b) By assumption $\mathcal{H}^s(A)$ is finite for some $s\in [0,\infty)$. So, for every $\delta>0$ there exist $\{C_j\}_{j\in I}$ with $$ \begin{cases} A\subset \bigcup\limits_j C_j \\ d(C_j)<\delta \\ \sum\limits_j d(C_j)^s\leq \mathcal{H}^s(A)+1. \end{cases} $$ Now $$(1)\quad d(C_j)^s= d(C_j)^{s-t}d(C_j)^t\geq \delta^{s-t}d(C_j)^t.$$ By multiplying both sides of the inequality by $\delta^{t-s}$ we have $$d(C_j)^s\delta^{t-s}\geq d(C_j)^t.$$ Lastly we have $$(2)\quad \mathcal{H}_\delta^t(A)\leq\sum_{_j}d(C_j)^t\leq \delta^{t-s}\sum_j d(C_j)^s\leq\delta^{t-s}(\mathcal{H}^s(A)+1).$$ The above expression approaches $0$ as $\delta\to 0$, since $t>s$ and hence, $t-s>0$.
(c) To prove $\mathcal{H}^t(A)=\infty$, I was thinking I want something like $$H_\delta^t(A)\geq \delta^{t-s}\cdot (\mathcal{H}^s(A)+1)\to\infty,$$ as $\delta\to 0$ since $t-s<0$.
The inequality "$\geq$" in (1) can just be interchanged with "$\leq$" since $t<s$. This implies that every inequality "$\leq$" in (2) can be interchanged with "$\geq$". I think I can switch every inequality except that $$\mathcal{H}_\delta^t(A)\leq \sum_{j} d(C_j)^t$$ can't be interchanged with $$\mathcal{H}_\delta^t(A)\geq \sum_{j} d(C_j)^t,$$ this is because $\mathcal{H}_\delta^t(A)$ is the infimum. And I do not see how I can find a lower bound for $H_\delta^t(A)$ in any way just because it is an infimum.
Do you have any idea how to prove (c) or how it follows from (b)? I would be really happy if you could help me out.
Some Notations:
Let $0\leq s<\infty$ and $A\subset\mathbb{R}^n$. For $0<\delta\leq\infty$, define
$$\mathcal{H}_\delta^s(A)=\inf\Big \{ \sum_{j} d(C_j)^s:A\subset\bigcup_j C_j,\text{ } d(C_j)<\delta,\text{ } C_j\subset\mathbb{R}^n\Big \},$$ where $d(C)$ is the diameter of $C$.
We also define $$\mathcal{H}^s(A)=\lim_{\delta\to 0}\mathcal{H}^s_\delta(A)=\sup_{\delta > 0}\mathcal{H}_\delta^s(A).$$