I am reading "An Introduction to Measure Theory" by Terence Tao.
Exercise 0.0.2 (Tonelli's theorem for series over arbitrary sets). Let $A,B$ be sets (possible infinite or uncountable), and $(x_{n,m})_{n\in A,m\in B}$ be a doubly infinite sequence of extended non-negative reals $x_{n,m}\in [0,+\infty]$ indexed by $A$ and $B$. Show that $$\sum_{(n,m)\in A\times B}x_{n,m}=\sum_{n\in A}\sum_{m\in B}x_{n,m}=\sum_{m\in B}\sum_{n\in A}x_{n,m}.$$
(Hint: although not strictly necessary, you may find it convenient to first establish the fact that if $\sum_{n\in A} x_n$ is finite, then $x_n$ is non-zero for at most countably many $n$.)
Definition:
$\dots$
Motivated by this, given any collection $(x_\alpha)_{\alpha\in A}$ of numbers $x_\alpha\in [0,+\infty]$ indexed by an arbitrary set $A$ (finite or infinite, countable or uncountable), we can define the sum $\sum_{\alpha\in A} x_\alpha$ by the formula $$\sum_{\alpha\in A} x_\alpha=\sup_{F\subset A,F\text{ finite}}\sum_{\alpha\in F} x_\alpha.$$
My solution is here:
Let $F$ be a finite subset of $A\times B$.
Let $F_A:=\{n\in A: (n,m)\in F\text{ for some }m\in B\}$.
Let $F_B:=\{m\in B: (n,m)\in F\text{ for some }n\in A\}$.
Since $F$ is finite, $F_A$ and $F_B$ are also finite and $F\subset F_A\times F_B$ holds.
By definition, $\sum_{m\in F_B}x_{n,m}\leq\sum_{m\in B}x_{n,m}$ holds for any $n\in A$.
By definition, $\sum_{n\in F_A}\sum_{m\in B}x_{n,m}\leq\sum_{n\in A}\sum_{m\in B}x_{n,m}$ holds.
Therefore, $$\sum_{(n,m)\in F} x_{n,m}\leq\sum_{(n,m)\in F_A\times F_B} x_{n,m}=\sum_{n\in F_A}\sum_{m\in F_B}x_{n,m}\leq\sum_{n\in F_A}\sum_{m\in B}x_{n,m}\\\leq\sum_{n\in A}\sum_{m\in B}x_{n,m}$$ holds.
So, by definition, $$\sum_{(n,m)\in A\times B} x_{n,m}\leq\sum_{n\in A}\sum_{m\in B}x_{n,m}$$ holds.
Next we prove that $$\sum_{n\in A}\sum_{m\in B}x_{n,m}\leq\sum_{(n,m)\in A\times B} x_{n,m}$$ holds.
My Lemma:
Suppose $F_A$ is an arbitrary finite subset of $A$.
Then, $$\sum_{n\in F_A}\sup_{F_B\subset B,F_B\text{ finite}}\sum_{m\in F_B} x_{n,m}=\sup_{F_B\subset B,F_B\text{ finite}}\sum_{m\in F_B}\left(\sum_{n\in F_A}x_{n,m}\right)$$ holds.
Proof of this lemma:
Suppose $F_B$ is an arbitrary finite subset of $B$.
$$\sup_{F'_B\subset B,F'_B\text{ finite}}\sum_{m\in F'_B} x_{n,m}\geq\sum_{m\in F_B} x_{n,m}$$
holds for any $n\in A$.
Therefore $$\sum_{n\in F_A}\sup_{F'_B\subset B,F'_B\text{ finite}}\sum_{m\in F'_B} x_{n,m}\geq\sum_{n\in F_A}\sum_{m\in F_B}x_{n,m}=\sum_{m\in F_B}\sum_{n\in F_A}x_{n,m}$$ holds.
So, $$\sum_{n\in F_A}\sup_{F_B\subset B,F_B\text{ finite}}\sum_{m\in F_B} x_{n,m}\geq\sup_{F_B\subset B,F_B\text{ finite}}\sum_{m\in F_B}\left(\sum_{n\in F_A}x_{n,m}\right)$$ holds.
Suppose $\epsilon$ be an arbitrary positive real number.
Suppose $n$ be an arbitrary element of $F_A$.
Then, there exists $(F_B)_n$ such that $(F_B)_n$ is a finite subset of $B$ and $$\sum_{m\in (F_B)n}x_{n,m}>\sup_{F_B\subset B,F_B\text{ finite}}\sum_{m\in F_B}x_{n,m}-\frac{\epsilon}{\#F_A}.$$
Let $F_B:=\bigcup_{n\in F_A} (F_B)_n$.
Then, $F_B$ is a finite subset of $B$.
And $$\sum_{m\in F_B}x_{n,m}>\sup_{F_B\subset B,F_B\text{ finite}}\sum_{m\in F_B}x_{n,m}-\frac{\epsilon}{\#F_A}.$$
Therefore $$\sum_{m\in F_B}\sum_{n\in F_A}x_{n,m}=\sum_{n\in F_A}\sum_{m\in F_B}x_{n,m}>\left(\sum_{n\in F_A}\sup_{F_B\subset B,F_B\text{ finite}}\sum_{m\in F_B}x_{n,m}\right)-\epsilon.$$
Therefore $$\sum_{n\in F_A}\sup_{F_B\subset B,F_B\text{ finite}}\sum_{m\in F_B} x_{n,m}=\sup_{F_B\subset B,F_B\text{ finite}}\sum_{m\in F_B}\left(\sum_{n\in F_A}x_{n,m}\right)$$ holds.
Suppose $F_A$ is an arbitrary finite subset of $A$.
Then, $$\sum_{n\in F_A}\sum_{m\in B}x_{n,m}=\sum_{n\in F_A}\sup_{F_B\subset B,F_B\text{ finite}}\sum_{m\in F_B} x_{n,m}=\sup_{F_B\subset B,F_B\text{ finite}}\sum_{m\in F_B}\left(\sum_{n\in F_A}x_{n,m}\right)\\=\sup_{F_B\subset B,F_B\text{ finite}}\sum_{(n,m)\in F_A\times F_B}x_{n,m}\leq\sum_{(n,m)\in A\times B}x_{n,m}.$$
By definition, $$\sum_{n\in A}\sum_{m\in B}x_{n,m}\leq\sum_{(n,m)\in A\times B}x_{n,m}.$$
My first question:
I didn't use Tao's hint.
So, I am worried if my solution is ok.
Especially, I am not sure my lemma is ok or not.
But I cannot specify exactly where I may be wrong.
My second question:
If my solution is ok, I want to know how to use Tao's hint to solve this exercise because my solution didn't use Tao's hint.
My third question:
Why did Tao write $\sum_{(n,m)\in A\times B}x_{n,m}$ instead of $\sum_{(m,n)\in A\times B}x_{m,n}$?
$A$ is smaller than $B$ in alphabetrical order.
$m$ is smaller than $n$ in alphabetrical order.
I have not read your proof but if you have read Fubini-Tonelli theorem for product of $\sigma$-finite spaces the one can follow easily. Give your spaces $A$ and $B$ the counting measure $\mu$ and $\nu$ respectively on the power set sigma algebras $\mathcal P(A)$ and $\mathcal P(B)$. But here comes the obstacle that these spaces may not be $\sigma$-finite if $A$ or $B$ is uncountable. To deal with this, we turn toward his hint to prove that support of $f:A\times B\to [0,\infty)$ defined as $f(a,b)=x_{a,b}$ is a countable set. This can be easily noticed because the set $W=\{(a,b)~:~f(a,b)\neq 0\}$ is nothing but $\cup_{n=1}^\infty W_n$, where $W_n=\{(a,b)~:~x_{a,b}\geq \frac{1}{n}\}$ can be of atmost finite cardinality (or else the total sum would be infinite). Thus we restrict $A$ and $B$ to the projections $\pi_1(W)$ and $\pi_2(W)$ which are countable sets and hence $\sigma$-finite with respect to counting measure, opening the doors to all the goodness of Fubini-Tonelli theorem. Now notice that $f$ is measurable with respect to the product measure $\mu\times \nu$ on the space $\pi_1(W)\times \pi_2(W)$ and the summation $\sum_{A\times B}x_{a,b}$ is nothing but $\int_{\pi_1(W)\times \pi_2(W)}f.d(\mu\times \nu)$. Now applying the Fubini-Tonelli theorem we get
\begin{eqnarray}\sum_{A\times B}x_{a,b}&=&\int_{\pi_1(W)\times \pi_2(W)}f.d(\mu\times \nu)\\&=&\int_{\pi_1(W)}\int_{\pi_2(W)}f.d\mu d\nu&=&\int_{\pi_2(W)}\int_{\pi_1(W)}f.d\nu d\mu\\&=&\sum_{A}\sum_Bx_{a,b}&=&\sum_{B}\sum_Ax_{a,b}\end{eqnarray}