Hausdorff is a metric outer measure I am new to measure& hausdorff measure, when looking at the proof of this property, I have a question :
Given $E_1,E_2 \subset X,X$ is a metric space, we want to prove that
 if $d(E_1,E_2)>0,then  H^s(E_1\cup E_2)=H^s(E_1)+H^s(E_2)$
It is sufices to show that $H^s(E_1\cup E_2) \geq H^s(E_1)+H^s(E_2)$ since the reverse is guaranteed by sub-additivity. So fix $0<\epsilon<  d(E_1,E_2)$. Then given any cover $F_j$of $E_1 \cup E_2$ with the diameter of $F_i=\delta <  \epsilon $, then set 
$$F_j'=E_1 \cap F_j, F_j''=E_2 \cap F_j$$
Then we have $\{F_j'\},\{F_j''\}$ are covers of $E_1,E_2$respectively, and they are disjoint. So we have $H^s_{\delta}(E_1 \cup E_2) \geq H^s_{\delta}(E_1)+H^s_{\delta}(E_2)$..... don't quite understand how that last inequality, does that come from the property of metric space or ?  
 A: Suppose that $\mathcal F$ is a cover of $E_1 \cup E_2$ by sets with diameter less than $\delta$. No set $F \in \mathcal F$ intersects both $E_1$ and $E_2$, since otherwise we would have $\mathrm{dist}(E_1,E_2) \le \delta$. Split $\mathcal F$ into two subfamilies:
$$\mathcal F_i = \{ F \in \mathcal F : E_i \cap F \not= \emptyset\}, \quad i=1,2.$$ Then (this is easy to check) each $\mathcal F_i$ is a cover of $E_i$ by sets of diameter less than $\delta$ so that $$H_\delta^s(E_i) \le \sum_{F \in \mathcal F_i} (\mathrm{diam} F)^s.$$
Since $\mathcal F_1$ and $\mathcal F_2$ are disjoint you have
$$
\sum_{F \in \mathcal F_1} (\mathrm{diam} F)^s + \sum_{F \in \mathcal F_2} (\mathrm{diam} F)^s \le \sum_{F \in \mathcal F} (\mathrm{diam} F)^s,
$$
and inequality is possible since $\mathcal F$ may have contained sets that intersect neither $E_1$ nor $E_2$. Thus
$$
H^s_\delta(E_1) + H^s_\delta(E_2) \le \sum_{F \in \mathcal F} (\mathrm{diam} F)^s.
$$
The cover $\mathcal F$ of $E_1 \cup E_2$ was arbitrary (other than the restriction on diameters). You may take the infimum over all such covers to get
$$
H^s_\delta(E_1) + H^s_\delta(E_2) \le H^s_\delta(E_1 \cup E_2).
$$
