Proof of Fubini's Theorem for Series

Question

I'm currently trying to self study Tao's Analysis I, and I'm really struggling with this proof - as I feel like I'm missing something. He goes really slow over sections I feel are obvious - for example he spends a while trying to get me to justify that for: $$f(x, y) \geq 0, \sum_{(x, y) \in A} f(x, y) \leq \sum_{(x, y) \in \mathbb{N} \times \mathbb{N}} f(x, y),$$ for $A \subset \mathbb{N} \times \mathbb{N}$ which seems intuitive to me. He suggests using a bijection to map it to $\mathbb{N}$ and I don't see why that's necessary. Therefore, I feel like I'm missing something critical in this section. I just started this book again after not reading it for a couple months - could this be why?

Answer 1

I don't have the book, so I don't know how Tao defines these sums, but I suppose it's something like the least upper bound of the set of all finite sums. Then, as you say, the theorem is intuitively clear, but that's not a proof. When you allow negative summands, you quickly get non-intuitive results. For example, a conditionally convergent sequence can be rearranged to converge to any given (finite or infinite) value. When the hypotheses of Fubini's theorem are not satisfied, you get non-intuitive results. (My intuition, at least is that Fubini's theorem for sums should always be true, but of course, that's wrong.)

Perhaps you see the need for proof, but think this particular proof is too obvious to be worth spending time on. Very well, just skim it and move on. On the other hand, if you think the proof is too long and involved, and can be much simplified, then you may well be misunderstanding something. If that's the case, you should post your simpler argument, and ask for verification. You'll either find your mistake, or have the gratification of knowing that you improved a proof by Terry Tao!

Answer 2

If you still want an answer, here how you can prove it using a bijection $h:\mathbb{N}\to\mathbb{N} \times \mathbb{N}$. The need for $h$ is due to the definition of sums over countable sets, definition 8.2.1, also $f(x,y)$ is non-negative and the series converges, so the series is absolutely convergent, hence well-defined.

Then $$L=\sum_{(x,y)\in\mathbb{N}\times\mathbb{N}}f(x,y)=\sum_{k=0}^{\infty}f(h(k)),$$

but since $A\subseteq\mathbb{N} \times \mathbb{N}$ is finite, then $\#(A)=q\geq 0$, now set $h({i\in\mathbb{N}:i\leq q})=A$, then from Proposition 7.2.14(c) for infinite sums, $$\sum_{k=0}^{\infty}f(h(k))=\sum_{k=0}^{q}f(h(k)) + \sum_{k=q+1}^{\infty}f(h(k)),$$ all of the sums are non-negative and convergent, so from the ordering of the real numbers, we have $$\sum_{k=0}^{q}f(h(k)) \leq \sum_{k=0}^{\infty}f(h(k)),$$

which can be mapped back to their respective sets using definition 8.2.1 for the infinite sum, and Proposition 7.1.11(c),(d) for the finite sum to conclude that $$\sum_{(x, y) \in A} f(x, y) \leq \sum_{(x, y) \in \mathbb{N} \times \mathbb{N}} f(x, y).$$

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Another simpler argument which follows the hint, is using the fact that the series is of non-negative terms, so the partial sums are increasing and the series also converges, which from Proposition 6.3.8 implies that the series is bounded above and converges to its suprema,let $g:\mathbb{N} \times \mathbb{N}\to \mathbb{N}$, then each partial sum is less than or equal to $L$,

$$S_N=\sum_{k=0}^{N}f(g^{-1}(k))\leq L \text{ for all } N\in\mathbb{N}.$$

Since $g(A)$ is finite and hence bounded, the there exist $M\in\mathbb{N}$ such that $g(x,y)\leq M$ for all $(x,y)\in A$, therefore $A\subseteq g^{-1}(i\in\mathbb{N}:i\leq M)$, then from proposition 7.1.11 (e) and the fact that all terms are non-negative ,we have

$$\sum_{(x, y) \in A} f(x, y) \leq \sum_{(x, y) \in g^{-1}(i\in\mathbb{N}:i\leq M)} f(x, y)=\sum_{k=0}^{M}f(g^{-1}(k))\leq L.$$