Let $A, B$ be sets and $x_{n,m}$ $n \in A, m \in B$ be a doubly infinite sequence of extended non-negative reals indexed by A and B. Show that
$\sum_{(n,m) \in A \times B} (x_{n,m})$ = $\sum_{n \in A} \sum_{m \in B} (x_{n,m})$ = $\sum_{m \in B} \sum_{n \in A} (x_{n,m})$
I started this proof but have one step I am not sure of. To start, we can observe that if A or B is uncountable then either: a) at least one of the sums is $\infty$, so we are done; or b) the number of non-zero elements is at most countable, which puts us in the case of countable sets.
Here's the step I am unsure of: If A and B are both countable, they can each be mapped to the positive integers, so instead of indexing $x_{n,m}$ by $A \times B$ we can consider them to be indexed by the positive integers. Then this is just Tonelli's Theorem in its original form and we invoke it to complete the proof.
I like this approach which avoids repeating all the work to prove Tonelli's theorem directly, but I'm unsure if it is correct. I am concerned that by mapping A and B into the integers, we are imposing an ordering on the $x_{n,m}$ without knowing that we can reorder them and still get the same sum. Then again, A and B were not ordered to begin with, so we are perhaps assuming that the ordering does not matter? Tonelli's theorem itself seems to say that the ordering does not matter if all the terms are positive.
Can anyone enlighten me?
