If $E= E_1 \cup E_2$, and $d(E_1,E_2) > 0$, then the outer measure of $E$ equals the sum of the outer measure of $E_1$ and $E_2$ There is a proof from stein for this assertion,


My question is why $\sum_{j \in J_1}|Q_j|+\sum_{j \in J_2}|Qj| \leqslant \sum_{j =1}^\infty|Q_j|$ ?
I have a feeling they are not equal since the order of addition has been changed, but it is not changed in the way that we can apply the proposition that absolute convergence implies that the order of summation does not matter.
 A: They are not equal because the cubes may not intersect neither $E_1$ nor $E_2$, so you might be counting 'excessive' cubes in the large sum.
Here's a hint to see that the inequality holds, using the following:

Proposition: Let $\{a_n\}$ be a sequence of non-negative terms. Then 
  $$\sum_n a_n<\infty\qquad (*)$$
  if and only if, for some (and hence all) sequences $1=r_0<r_1<r_2<\dots$
  $$\sum_{i=0}^\infty\sum_{n=r_i}^{r_{i+1}}a_n<\infty\qquad (**)$$
  Hint of proof: The partial sums of $(**)$ (with respect to the leftmost series) are nothing more than a subsequence of the sequence of partial sums of $(*)$, which is monotonically increasing. 

In other words, an insertion of parenthesis
$$(a_1+\dots+a_{r_1})+(a_{r_1+1}+\dots+a_{r_2})+\dots$$
does not alter the convergence. 
Using this we can write, for any set of indices $J$:
$$\sum_{n\in J}a_n=\sum_{n=1}^\infty b_n$$
where $$b_n=\begin{cases}a_n & if \ n\in J\\
0 & otherwise\end{cases}$$
Can you see how to proceed?
A: The proof looks unnecessarily complicated. Let $\mu^o$ denote outer measure. 
First. Show that $\mu^o(A)+\mu^o(B)=\mu^o(A\cup B)$ when $A, B$ are non-empty open sets and $d(A,B)>0.$
Second. We have $\mu^o(E_1\cup E_2)\leq \mu^o(E_1)+\mu^o(E_2)$ for any $E_1,E_2.$ And if $\mu^o(E_1)=\infty$ or $\mu^o(E_2)=\infty$ then $\infty=\mu^o(E_1\cup E_2)=\mu^o(E_1)+\mu^o(E_2).$
So it suffices to show $\mu^o(E_1)+\mu^o(E_2)\leq \mu^o(E_1)+\mu^o(E_2)$ when $\mu^o(E_1)$ and $\mu^o(E_2)$ are finite, non-empty, and $d(E_1,E_2)=r>0.$ 
For $i\in \{1,2\}$ let $D_i=\cup_{p\in E_i}B(p,r/4)$ where $B(p,r/4)$ is the open ball centered at $p$ with radius $r/4.$ Then $E_1\subset D_1$ and $E_2\subset D_2$ and $d(D_1, D_2)\geq r/4.$
Now for any $e>0$ let $U_e$ be an open set with $U_e\supset (E_1\cup E_2)$ and $\mu^o(U_e)<\mu^o(E_1\cup E_2)+e.$ (Note the strict inequality is possible because $\mu^o(E_1\cap E_2)\leq \mu^o(E_1)+\mu^o(E_2)<\infty.$) Then $E_i\subset U_e\cap D_i$  for $i\in \{1,2\},$ while  $U_e\cap D_2, U_e\cap D_2$ are  open, with $d(U_e\cap D_1,U_e\cap D_2)\geq r/4>0.$ We have therefore  $$\mu^o(E_1)+\mu^o(E_2)\leq \mu^o(U_e\cap D_1)+\mu^o(U_e\cap D_2)=$$ $$=\mu^o((U_e\cap (D_1)\cup (U_e\cap D_2))=\mu^o(U_e\cap (D_1\cup D_2))\leq$$ $$\leq \mu^o(U_e)<\mu^o(E_1\cup E_2)+e.$$ Since $e$ can be arbitrarily small we have $\mu^o(E_1)+\mu^o(E_2)\leq \mu^o(E_1\cup E_2).$
