Hausdorff Measure is Non-increasing My teacher was explaining fractals to me and he said "clearly $H_{\delta}^{s}(F)$ is non-increasing with $s$". I do not understand how it is that he has come to this. Could somebody tell me please why it is that it is non-increasing? 
I know this might be a silly question because he said clearly, but I do not see it.
 A: First off, the phrases "clearly" and "it's obvious that..." should be banned from mathematics. These are basically mathematician for one of the following:


*

*I'm too lazy to do it;

*if you can't do it, you are stupid, and unworthy of learning; or

*I don't know how to do it, but it must be true.


Never believe a mathematician who says that something is clear.  It almost never is, and it is insulting to the student when it happens. 
Recall that
$$ \mathcal{H}^{s}_{\delta}(E) := \inf\left\{ \sum_{j=1}^{\infty} |E_j|^s : E\subseteq \bigcup_{j=1}^{\infty} E_j \text{ and } |E_j| < \delta \right\},$$
where $|E_j|$ is the diameter of $E_j$.  Observe that for $s\ge 0$ and $\alpha < 1$, the function $s \mapsto \alpha^s$ is a decreasing function of $s$.  From this, it follows that if $s < t$, then
$$ \sum_{j=1}^{\infty} |E_j|^s \ge \sum_{j=1}^{\infty} |E_j|^t, $$
as long as $|E_j| < \delta < 1$.  But then we have
\begin{align}
\mathcal{H}^{s}_{\delta}(E)
&= \inf\left\{ \sum_{j=1}^{\infty} |E_j|^s : E\subseteq \bigcup_{j=1}^{\infty} E_j \text{ and } |E_j| < \delta \right\} \\
&\ge \inf\left\{ \sum_{j=1}^{\infty} |E_j|^t : E\subseteq \bigcup_{j=1}^{\infty} E_j \text{ and } |E_j| < \delta \right\} \\
&= \mathcal{H}^{t}_{\delta}(E)
\end{align}
whenever $s < t$ and $\delta < 1$.  Hence $\mathcal{H}^{s}_{\delta}(E)$ is non-increasing in $s$ so long as $\delta < 1$.  We can actually deal with $\delta \ge 1$ in most cases, but the arguments are slightly delicate, and rely on properties of the underlying space (I think that separability may be good enough, but don't quote me on that---if we assume $E\subseteq \mathbb{R}^n$, we should be okay).
On the positive side, we are often concerned with the asymptotic behaviour, i.e. with the limit
$$\lim_{\delta\to 0} \mathcal{H}^s_{\delta}(E),$$
hence we can assume $\delta < 1$ with no loss of generality.
