Fubini theorem for infinite sum Terence Tao I just have a question below. I don't understand why we need this proof below ? By assumption we know that the we have that $\Sigma_{(n,m) \in \mathbb{N}\times\mathbb{N}} f(n,m)$ is absolutely convergent. This tells us that any rearrangements of the sequence above is also absolutely convergent. Don't we get the proof right away ? 
 A: In the corresponding theorem for a single sum, you are somewhat limited in the "types" of rearrangements you can do.  Specifically, a rearrangement of a single sum is required to reach each term in the original sum within a finite number of steps.  For instance, it would be invalid application of the corresponding theorem for a single sum $\sum_{n=1}^\infty a_n$ to write it as $$\sum_{n=0}^\infty a_{2n+1} + \sum_{n=0}^\infty a_{4n+2} + \sum_{n=0}^\infty a_{8n+4} + \cdots$$
since such a rearrangement has infinitely many terms before it even reaches, e.g. $a_2$, and then doubly-infinitely many terms before it reaches $a_{12}$.  This, however, is exactly the type of rearrangement that Fubini's theorem allows us to use.
In this way, we can realize Fubini's theorem for sums as just a natural expansion of the types of rearrangements allowed for single sums, but we still need to prove that such an expansion of the previous theorem is valid.  It doesn't merely follow from that previous theorem.
