Proof of the equivalence of definitions of absolute convergence. In Terence Tao's book Analysis I he defines the sum of a countable set as follows.
Let X be a countable set and $f:X\to \Bbb R$ any function.  If there exists a bijection $g:\Bbb N\to X$ such that $\sum_{n=1}^\infty f(g(n))$ is absolutely convergent then we say $\sum_{x\in X} f(x)$ is absolutely convergent, and we define its value by the equation
$$ \sum_{x\in X} f(x) = \sum_{n=1}^\infty f(g(n)) $$
He then states and proves the following theorem which he calls Fubini's theorem for infinite series (well, he "proof sketches" it but I think I've been able to fill in the details of the proof myself, so that's not a concern).
Theorem: Suppose $X$ is a countable set and $f:X\to \Bbb R$ a function, and suppose $\sum_{x\in X}f(x)$ is absolutely convergent.  Then
$$ \sum_{(m,n)\in \Bbb N\times \Bbb N} f(m,n)=\sum_{m=1}^\infty\sum_{n=1}^\infty f(m,n) = \sum_{n=1}^\infty\sum_{m=1}^\infty f(m,n) $$

At least one, and probably several other authors, define the absolute convergence of an iterated sum by the following.
Let $f:\Bbb N\times \Bbb N\to \Bbb R$ be any function.  We say that $\sum_{m=1}^\infty\sum_{n=1}^\infty f(m,n)$ is absolutely convergent if for each fixed $m\in\Bbb N$ the series $b_m=\sum_{n=1}^\infty |f(m,n)|$ converges and $\sum_m b_m$ also converges.

Presumably the two definitions of absolute convergence should be equivalent.  And with all the work I've done so far, I have a proof that if $\sum_{(m,n)\in\Bbb N\times \Bbb N} f(m,n)$ is absolutely convergent then so is $\sum_m\sum_n f(m,n)$.  However, I've been unable to prove the converse.  I also know that most analysis textbooks contain something that almost looks like the converse, but is not the converse.  For instance Rudin proves that if the series $\sum_{m=1}^\infty \sum_{n=1}^\infty f(m,n)$ is absolutely convergent (using the definition for iterated sums this time) then $\sum_{m=1}^\infty \sum_{n=1}^\infty f(m,n)=\sum_{n=1}^\infty\sum_{m=1}^\infty f(m,n)$.  However, I haven't found a way to exploit that to show $\sum_{(m,n)\in \Bbb N\times \Bbb N} f(m,n)$ is absolutely convergent.
Does anyone know if the converse is in fact true, and if so, a proof of it?  To be explicit, the question is to find a proof of the following claim:
If $\sum_m\sum_n f(m,n)$ is absolutely convergent in the sense for iterated sums, then $\sum_{(m,n)\in\Bbb N\times \Bbb N}f(m,n)$ is absolutely convergent in the sense for a sum over a countable set.
Note: I also know that some people try to justify the equality of all of these series as a consequence of the more general measure-theoretic Fubini-Tonelli theorem.  However, most of the textbooks which I've seen prove Fubini-Tonelli only do so after using (sometimes in a very obscured way) the ability to re-index double-sum in order to prove the additivity and/or subadditivity of measures.  Therefore such an argument would be circular.
 A: Suppose $f:\Bbb{N}\times\Bbb{N}\to \Bbb{R}$ is a given double sequence such that the iterated sum $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}|f(m,n)|$ is finite. Fix a bijection $\alpha:\Bbb{N}\to\Bbb{N}\times\Bbb{N}$; we wish to show that $\sum_{i=1}^{\infty}|f(\alpha(i))|$ is finite as well. Let $k\in\Bbb{N}$ be arbitrary. Then, surely there is an $N\in\Bbb{N}$ such that the tuples $\alpha(1),\dots, \alpha(k)$ all lie in the square $[1,N]\times[1,N]$. So,
\begin{align}
\sum_{i=1}^k|f(\alpha(i))| &\leq \sum_{m=1}^N\sum_{n=1}^N|f(m,n)|\\
&\leq \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}|f(m,n)|\\
&<\infty.
\end{align}
The first inequality is simply because every term on the LHS appears atleast once on the RHS. The second inequality should be clear (since we're dealing with non-negative terms), and the final one is by assumption. Thus, the sequence of partial sums $\left\{\sum_{i=1}^k|f(\alpha(i))|\right\}_{k=1}^{\infty}$ is a weakly-increasing and bounded sequence. Therefore, the limit as $k\to \infty$ exists (and in fact equals the supremum over $k$ of that sequence). Thus,
\begin{align}
\sum_{i=1}^{\infty}|f(\alpha(i))|:=\lim_{k\to \infty}\sum_{i=1}^k|f(\alpha(i))| \leq
\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}|f(m,n)|<\infty.
\end{align}
A: Consider the subset $ S_m = \{ 1, \ldots, m \}^2 $.  You can compare the sum over $ S_m $ to $ \sum_{i=1}^{m}{\sum_{j=1}^{\infty}{|f(i,j)|}} $.  The latter quantity remains bounded as $ m\rightarrow \infty $ by assumption.  Since $ \bigcup S_m = \mathbb{N}^2 $, your result follows.
