Proof of Fubini's theorem for infinite sums. In his book Analysis 1, the author Tao states Fubini's theorem as follows 

Let $f:N \times N \rightarrow \mathbb{R}$ be a function such that $\sum_{(n,m)\in N\times N}f(n,m)  $ is absolutely convergent. Then we have
  $$\begin{align*}
\sum_{n=0}^{\infty}\Bigg(\sum^{\infty}_{m=0}f(n,m)\Bigg) &=\sum_{(n,m)\in N \times N}f(n,m) \\
&=\sum_{(m,n)\in N \times N}f(n,m)\\
&=\sum_{m=0}^{\infty}\Bigg(\sum^{\infty}_{n=0}f(n,m)\Bigg)
\end{align*}$$  

He says that the second inequality follows from the rearrangement of absolutely convergent series. But that theorem is for bijective functions $f:N\rightarrow N$.
 How can we use it to obtain the second equality in the theorem stated above ?
 A: I think, regardless of absolute convergence, the meaning of $\sum_{(n,m) \in \Bbb{N}^2} f(n,m) = S$ can be taken as for any $\epsilon > 0$ there exists $K \in \mathbb{N}$ such that for all $N \geqslant K$ and all $M \geqslant $K
$$\left|\sum_{n=0}^N \sum_{m=0}^M f(n,m) - S \right| < \epsilon.$$
and the order of summation in the double partial sum is interchangeable.
In this definition, there is no distinction between $\sum_{(n,m) \in \Bbb{N}^2} f(n,m)$ and $\sum_{(m,n) \in \Bbb{N}^2} f(n,m).$
If $\sum_{(n,m) \in \Bbb{N}^2} |f(n,m)|$ is convergent then, with non-negative terms, it is easy to show by monotone convergence that sums in any order (rows, columns, diagonals, etc.) are identical, and
$$\sum_{(n,m) \in \Bbb{N}^2} |f(n,m)| = \lim_{N \to \infty}\sum_{n=0}^N \sum_{m=0}^\infty |f(n,m)| = \sum_{n=0}^\infty \sum_{m=0}^\infty |f(n,m)|, \\ \sum_{(n,m) \in \Bbb{N}^2} |f(n,m)| = \lim_{M \to \infty}\sum_{m=0}^M \sum_{n=0}^\infty |f(n,m)| =   \sum_{m=0}^\infty \sum_{n=0}^\infty |f(n,m)| $$
Define the positive and negative parts of $f(n,m)$ with the notation $f(n,m)^+$ and $f(n,m)^-,$ where $f(n,m) \geqslant 0 \implies f(n,m)^+ = f(n,m)$ and $f(n,m) < 0 \implies f(n,m)^+ = 0$ etc.
Note that
$$0 \leqslant f(n,m)^+ \leqslant |f(n,m)|,  \\ 0 \leqslant f(n,m)^- \leqslant |f(n,m)|. $$
We can apply a comparison test for series with non-negative terms. The comparison with $\sum \sum|f(n,m)|$ shows that  $\sum \sum f(n,m)^+$ and   $\sum \sum f(n,m)^-$  converge both by rows and columns
$$\sum_{n=0}^\infty \sum_{m=0}^\infty f(n,m)^+ = \sum_{m=0}^\infty \sum_{n=0}^\infty f(n,m)^+ , \\ \sum_{n=0}^\infty \sum_{m=0}^\infty f(n,m)^- = \sum_{m=0}^\infty \sum_{n=0}^\infty f(n,m)^-. $$
Since $\sum \sum f(n,m) = \sum \sum [f(n,m)^+ -  f(n,m)^-]$, it follows that
$$\sum_{n=0}^\infty \sum_{m=0}^\infty f(n,m) = \sum_{m=0}^\infty \sum_{n=0}^\infty f(n,m).$$
