$\sum_{(s,t)\in S\times T} f(s,t)=\sum_{s\in S}\sum_{t\in T} f(s,t)$ via Fubini-Tonelli?

Question

Let $f: S\times T \to [0,\infty[$ be a function. I wish to show that

$$\sum_{(s,t)\in S\times T} f(s,t)=\sum_{s\in S}\sum_{t\in T} f(s,t)$$

I know that in general we have $$\int_X g d\mu =\sum_{x\in X} g(x)$$ for a function $g$ defined on $X$ where $\mu$ denotes the counting measure and the theorem then follows by Fubini-Tonelli if we ask that $S,T$ are countable so that our measures are $\sigma$-finite.

But is there a way to get this result if S or T is uncountable?

Answer 1

If an uncountable sum $\sum a_i$ of non-negative numbers is finite then all but countably many terms are $0$. Using this you can see that if either side of the equation is finite then $f(s,t)=0$ except for countably many pairs $(s,t)$ so the uncountable case can also be proved using Tonelli's Theorem.

Answer 2

If you just want to get the result, then you can prove it directly, without bringing measure theory into it. First, we need a standard non-measure-theoretic definition of the sum: $$ \sum _ { s \in S } f ( s ) : = \sup _ { F \in \mathcal F S } \sum _ { s \in F } f ( s ) \text , $$ where $ \mathcal F S $ is the collection of finite subsets of $ S $, if $ f $ is a positive function. (Then $ \sum f : = \sum f ^ + - \sum f ^ - $ for general functions, as usual. But the proof will focus on positive functions.)

Now let $ F $ be a finite subset of $ S \times T $. Then we have finite subsets $ F _ 1 $ of $ S $ and $ F _ 2 $ of $ T $ with $ F \subseteq F _ 1 \times F _ 2 $, so $$ \sum _ { ( s , t ) \in F } f ( s , t ) \leq \sum _ { ( s , t ) \in F _ 1 \times F _ 2 } f ( s , t ) = \sum _ { s \in F _ 1 } \sum _ { s \in F _ 2 } f ( s , t ) \leq \sum _ { s \in F _ 1 } \sum _ { t \in T } f ( s , t ) \leq \sum _ { s \in S } \sum _ { t \in T } f ( s , t ) \text . $$ Since this holds for all finite $ F $, we've proved one direction of the theorem: $$ \sum _ { ( s , t ) \in S \times T } f ( s , t ) \leq \sum _ { s \in S } \sum _ { t \in T } f ( s , t ) \text . $$ Now let $ F $ be a finite subset of $ S $ and for each $ s \in F $, let $ F _ s $ be a finite subset of $ T $. Then $$ \biguplus _ { s \in S } F _ s : = \bigcup _ { s \in S } ( \{ s \} \times F _ s ) $$ is a finite subset of $ S \times T $, and $$ \sum _ { s \in F } \sum _ { t \in F _ s } f ( s , t ) = \sum _ { ( s , t ) \in \biguplus _ s F _ s } f ( s , t ) \leq \sum _ { ( s , t ) \in S \times T } f ( s , t ) \text . $$ Since this holds for all choices of $ F _ s $, $$ \sum _ { s \in F } \sum _ { t \in T } f ( s , t ) \leq \sum _ { ( s , t ) \in S \times T } f ( s , t ) \text . $$ And since this holds for all $ F $, $$ \sum _ { s \in S } \sum _ { t \in T } f ( s , t ) \leq \sum _ { ( s , t ) \in S \times T } f ( s , t ) \text , $$ the other direction of the theorem. (This all assumes that $ f $ is a positive function, but the theorem follows for any function, since it holds separately for $ f ^ + = \max ( f , 0 ) $ and for $ f ^ - = \max ( - f , 0 ) $.)

Although finite subsets are important here, and although it's true that $ f $ can take positive values only at countably many places (if the sum converges), countability plays no role in the proof.