Question on a Proof of Rearrangements for Absolutely Converging Double Series In Appendix B of Jameson's The Prime Number Theorem, the author gives a proof of the assertion that given real numbers $\left\{a_{j,k}\right\}_{j,k\ge 1},$ if 
$$\sum_{j=1}^\infty \sum_{k=1}^\infty \left|a_{j,k}\right|$$ then for any bijection $\sigma : \mathbb{N} \to \mathbb{N}^2$ we have that the sums 
$$\sum_{j=1}^\infty \sum_{k=1}^\infty a_{j,k}
= \sum_{k=1}^\infty \sum_{j=1}^\infty a_{j,k} 
= \sum_{n=1}^\infty a_{\sigma(n)}$$ all converge and are equal to one another. 
The proof begins (and this is the part I do not understand) by saying that it suffices to show that the result holds for the case where $a_{j,k} \ge 0$ for all $j,k\ge 1$ because if this is true then the assertion follows for general real valued sequences by writing 
$$a_{j,k} =  a_{j,k}^+ - a_{j,k}^-$$ where for any real $x$ we denote
$$x^+ = \begin{cases} 
x & \text{if} & x\ge 0 \\
0 & \text{otherwise}
\end{cases} $$
and 
$$x^- = \begin{cases} 
-x & \text{if} & x < 0 \\
0 & \text{otherwise.}
\end{cases} $$
How does this substitution actually imply that the result holds for general real-valued sequences once we know that it holds for nonnegative sequences? Any explanation on how this argument works would be greatly appreciated.
 A: These three sums take these numbers in some order. Now consider this:
\begin{align}
\sum_{j,k\,\in\,\mathbb N^2} a_{i,j} = {} & \sup\left\{ \sum_{j,k\,\in\,J} a_{j,k}^+ : J \text{ is a finite subset of } \mathbb N^2 \right\} \\
& {} - \sup\left\{ \sum_{j,k\,\in\,J} a_{j,k}^- : J \text{ is a finite subset of } \mathbb N^2 \right\}
\end{align}
The way I would try to prove the equalities in $(1)$ is by showing that all of the sums in $(1)$ are equal to this "orderless" sum.
With the sum
$$
\sum_{j=1}^\infty \sum_{k=1}^\infty a_{j,k},
$$
write
$$
\sum_{j=1}^\infty \sum_{k=1}^\infty (a_{j,k}^+ - a_{j,k}^-)
$$
and see if you can show that that is
$$
\left(\sum_{j=1}^\infty \sum_{k=1}^\infty a_{j,k}^+\right) - \left( \sum_{j=1}^\infty \sum_{k=1}^\infty a_{j,k}^- \right).
$$
A: Recall that if $\sum_n b_n$ and $\sum_n c_n$ are two absolutely convergent series, then $\sum_n (b_n + c_n) = \sum_n b_n + \sum_n c_n$. Now for each $j$, $\sum_k a_{j,k}^+$ and $\sum_k a_{j,k}^-$ are absolutely convergent. Hence, we have:
$$\sum_j \sum_k a_{j,k} = \sum_j \sum_k (a_{j,k}^+ - a_{j,k}^-) = \sum_j\left(\sum_k a_{j,k}^+ - \sum_k a_{j,k}^-\right) $$
Now $\sum_j \left(\sum_k a_{j,k}^+\right)$ is absolutely convergent because: $$\sum_j \left|\sum_k a_{j,k}^+\right| \le \sum_j \sum_k |a_{j,k}^+| \le \sum_j\sum_k |a_{j,k}| < \infty$$
Similarly $\sum_j \sum_k a_{j,k}^-$ is absolutely convergent. Therefore,
$$\sum_j \sum_k a_{j,k} = \sum_j \sum_k a_{j,k}^+ - \sum_j \sum_k a_{j,k}^- \\ = \sum_k \sum_j a_{j,k}^+ - \sum_k \sum_j a_{j,k}^- \text{ (using the result from the case $a_{j,k}\ge 0$)}$$
Similarly to what we have done above, we prove that:
$$\sum_k \sum_j a_{j,k}^+ - \sum_k \sum_j a_{j,k}^- = \sum_k \sum_j a_{j,k}$$
Try to do the one with $\sigma$ on your own. 
