Proving double sums are interchangable

Question

Let $s(n,m):\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{R}$ be a function s.t. the double series $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\big|\hspace{0.05cm}s(n,m)\hspace{0.05cm}\big|$$ converges to $L<\infty$. As I understand it, $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\big|\hspace{0.05cm}s(n,m)\hspace{0.05cm}\big|=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\big|\hspace{0.05cm}s(n,m)\hspace{0.05cm}\big|=L$$ Heuristically, if one considers the operation to be summing over an infinite array of integers, then this identity is conceptually trivial. I have found, however, little by way of proof besides the casual mention of commutativity or the occasional proof by "obvious" observation. If someone would be so kind as to explain the formal logic behind this, ideally at the "$\forall\hspace{1mm} \epsilon>0$" level, I would be very grateful.

Answer 1

Write $a_{n,m}=\lvert s(n,m)\rvert$. By definition $\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}a_{n,m}=\lim_{n\to\infty}\lim_{m\to\infty}\sum_{i=1}^n\sum_{j=1}^m a_{i,j}$ so consider $\lim_{m\to\infty}\sum_{i=1}^n\sum_{j=1}^m a_{i,j}$ as a sequence in $n$. This sequence converges to $L$ so for $\epsilon>0$ we can find $N$ such that $$\lvert\lim_{m\to\infty}\sum_{i=1}^n\sum_{j=1}^m a_{i,j}-L\rvert<\epsilon$$ when $n\geq N$.

Now fix $n\geq N$. Then since $\lim_{m\to\infty}\lvert\sum_{i=1}^n\sum_{j=1}^m a_{i,j}-L\rvert<\epsilon$ we can find $M$ such that $$\lvert\sum_{i=1}^n\sum_{j=1}^m a_{i,j}-L\rvert<2\epsilon$$ when $m\geq M$. Swap these finite sums to get $$\lvert\sum_{j=1}^m\sum_{i=1}^n a_{i,j}-L\rvert<2\epsilon$$ whenever $n\geq N$ and $m\geq M$. Take limits in $n$ and then in $m$ to get $$\lvert\sum_{m=1}^\infty\sum_{n=1}^\infty a_{n,m}-L\rvert\leq2\epsilon.$$

$\epsilon$ is arbitrary so $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}a_{n,m}=L$.

Note that the assumption of non-negativity for each $a_{n,m}$ came in whenever we had to assume limits of sums exist (they do exist since they are increasing and bounded).