Rearranging a double series; what's the rigorous argument behind this? Recall that a measure zero subset $A \subset \mathbb{R}$ is one such that for any $\epsilon > 0$ there exists a sequence of intervals $J_n$ whose union cover $A$ and such that $\sum_n l(J_n) < \epsilon$ (here $l$ is the length). In the standard proof that a countable union $A = \cup_i A_i$ of measure zero subsets has measure zero, we pick intervals $J_n^i$ such that $A_i \subset \cup_n J_n^i$ and $\sum_n J_n^i < \epsilon/2^i.$ Then, we conclude that the countable family $J_n^i$ is the required "sequence" of intervals, since it covers $A$ and
$$ \sum_{n,i} l(J_n^i) = \sum_i \sum_n l(J_n^i) < \sum_i \epsilon/2^i = \epsilon. \quad (*)$$
However, it seems to me that to be perfectly rigorous, one would need to carefully justify the first equality in $(*)$. We're looking for a sequence of intervals, after all. My first instinct was to say that one can pick a bijection $\sigma = (\sigma_1,\sigma_2):\mathbb{N} \to \mathbb{N}^2$ and that, since the rearrangement of a convergent series with positive terms does not affect convergence, we have that
$$ \sum_i \sum_n l(J_n^i) = \sum_k l(J_{\sigma_1(k)}^{\sigma_2(k)}) $$
is a converging sequence of lengths of intervals. However that doesn't really make sense, since this isn't even a rearrangement in the usual sense.
In other words: how can I rigorously interpret the sum
$$ l(J_1^1) + l(J_2^1) + \cdots + l(J_1^2) + l(J_2^2) + \cdots $$
as a single series, i.e., a sum of a countable number of lengths?
 A: $\sum_i \sum_n l(J_n^i)$ is interpreted as $\lim_{j\rightarrow\infty}\sum_{i=1}^j\sum_n l(J_n^i)$. Now, you should prove that there's a $\sigma$ such that $\sum_k l(J_{\sigma_1(k)}^{\sigma_2(k)})$ converges. To do that, you take any $\sigma$ and bound the finite sums with $\sum_i \sum_n l(J_n^i)$. As the terms are positive, $\sum_k l(J_{\sigma_1(k)}^{\sigma_2(k)})$ converges.
You must also prove that $\sum_k l(J_{\sigma_1(k)}^{\sigma_2(k)})=\sum_i \sum_n l(J_n^i)$, which you can do by definition.
EDIT: Given $\epsilon>0$, there exists $j_0$ such that $\forall j\ge j_0$, $|\sum_i \sum_n l(J_n^i)-\sum_{i=1}^j\sum_n l(J_n^i)|<\epsilon/2$, and $\forall 1\le i\le j_0$ there exist $m_i$ such that $\forall m\ge m_i$, $|\sum_n l(J_n^i)-\sum_{n=1}^{m} l(J_n^i)|<\frac{\epsilon}{2j_0}$. So if you take a sufficiently large $t_0$ (one such that $\{(n,i)|i\le j_0$ and $n\le m_i\}\subseteq\{\sigma(1),...,\sigma(t_0)\}$), $\forall t\ge t_0$ 
$$\sum_i\sum_n l(J_n^i)-\sum_{k=1}^t l(J_{\sigma_1(k)}^{\sigma_2(k)})\le\sum_i\sum_n l(J_n^i)-\sum_{k=1}^{t_0} l(J_{\sigma_1(k)}^{\sigma_2(k)})\le\sum_i\sum_n l(J_n^i)-\sum_{i=1}^{j_0}\sum_{n=1}^{m_i} l(J_n^i)=\sum_i\sum_n l(J_n^i)-\sum_{i=1}^{j_0}\sum_n l(J_n^i)+\sum_{i=1}^{j_0}\sum_n l(J_n^i)-\sum_{i=1}^{j_0}\sum_{n=1}^{m_i} l(J_n^i)<\epsilon$$
