Questions about Hausdorff measure and general metric outer measures. The Hasudorff measure of a set $A\subset \mathbb R^n$ is defined as 
$$ \mathcal H^s(A) = \lim_{\delta\downarrow 0} \mathcal H^s_\delta(A)$$ 
where 
$$\mathcal H^s_\delta(A) = \inf\left\{\sum_{j=1}^\infty (\text{diam }S_j)^s : \{S_j\}_{j=1}^\infty \subset 2^{\mathbb R ^n} \text{ covers } A \text{ and } \text{diam } S_j < \delta  \right\} $$
It is easy to see that $\mathcal H^s_\delta(A)$ is non-increasing in $\delta$. However, I don't quite understand why $\mathcal H^s_\delta(A)$ is not actually constant in $\delta$ for each $A$; it would seem that the infimum should be achieved as the diameters of the sets $S_j$ get smaller and smaller, meaning that bounding the diameter of $S_j$ above by $\delta$ should do nothing at all to $\inf\sum_{j=1}^\infty (\text{diam }S_j)^s $. Moreover, suppose we approach Lebesgue measure analogously. Let $\mathscr{L^n}(A)$ be $n$-dimensional Lebesgue measure, and let 
$$\mathscr{L_\delta^n}(A) := \inf\left\{\sum_{j=1}^\infty|B_j| : \{B_j\}_{j=1}^\infty \subset 2^{\mathbb R ^n} \text{ is a box cover of } A \text{ and } \text{diam } B_j < \delta  \right\} $$
we find that $\mathscr{L^n_\delta}(A)$ is constant in $\delta$, so that for any $\delta>0$ 
$$\mathscr{L^n}(A) = \mathscr{L^n_\delta}(A)= \lim_{\epsilon\rightarrow 0}\mathscr{L^n_\epsilon}(A)$$
So what about switching from volumes of boxes to diameters of arbitrary sets makes $\mathcal H^s_\delta(A)$ not necessarily constant in $\delta$?
This question hints at a much more general question about the construction of outer measures via the Carathéodory method. A brief recap on this method (as I have learned it in class): Suppose $\mathcal F$ is a collection of subsets of some metric space $X$, let $A\subset X$, and consider the following set
$$\mathcal C_{\delta, \mathcal F}(A) : = \left\{\{S_j\}^\infty\subset \mathcal F \mid \{S_j\}^\infty \text{ covers } A \text{ and } \text{diam }S_j< \delta\right\}$$
and consider any function $\zeta: \mathcal F \rightarrow [0,\infty]$.
Then, the following functions are outer measures:
$$\mu_{\delta, \mathcal F}(A) := \inf_{\{S_j\}^\infty \in \mathcal C_{\delta, \mathcal F}(A)} \sum^\infty_{j=1} \zeta(S_j)$$
$$\mu_{\mathcal F}(A) := \lim_{\delta \downarrow 0}\mu_{\delta, \mathcal F}(A)$$
Notice that the  Hausdorff measure $\mathcal H ^ s(A)$ is the special case where $\mathcal F = 2^X$ and $\zeta(S) = (\text{diam } S)^s$. Under what conditions is $\mu_{\delta,\mathcal F}(A)$ not constant in $\delta$ for every $A$, as is the case for $\mathcal H^s_{\delta}$?
 A: Isn't it perfectly plausible that if you are allowed to use larger sets, then you get a smaller number? (because you are only summing up a function of the diameter of the set, remember)
Note that one reason for this construction is that lots of sets are not $\mathcal{H}_{\delta}^s$ measurable but are $\mathcal{H}^s$ measurable, but this isn't what you asked specifically.
The following is pretty much the simplest situation where something goes wrong: Take $\mathcal{H}_{\delta}^1$ on $\mathbf{R}^2$ for some $\delta > 0$ and then consider 
Let $A := ([0,1) \times \{0\}) \cup ([0,1) \times \{\epsilon\}) \subset \mathbf{R}^2$ for some $\epsilon << \delta$. This is two parallel, horizontal line segments of length $1$, separated by a tiny gap of width $\epsilon$.
Now, how would you efficiently cover this set if you are allowed to use a covering by sets of diameter $\delta$?
You can essentially cover both lines at once, right?  To cover $A$ efficiently you use very thin rectangles whose long sides are just less than delta and whose short sides are $\epsilon$. Summing their diameters gives you something close to 1. (this can be made rigorous) 
(roughly speaking, since $\delta$ is much bigger than $\epsilon$, the outer measure $\mathcal{H}^1_{\delta}$ is too coarse; the set looks like a single line from the point of view of $\mathcal{H}^1_{\delta}$).
But now if you imagine changing $\delta$ so that things are the other way around, i.e. take $\delta << \epsilon$, then of course eventually the $\mathcal{H}^1_{\delta}$ measure has to get close to 2, which is the true $\mathcal{H}^1$ measure.
