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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.

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    I would need to read again to be sure and construct counterexamples, but I am almost positive that your main assertion is *not* true in general, unless I've missed a special part of the hypothesis. Alas, I am sleepy. It is a great question, though. Perhaps someone else can refute/support me? – 2017-02-02
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    @TheCount The claim is true by Fubini's theorem. – 2017-02-02
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    @user375366 I was just about to delete my comment, since I realized I was wrong, but now I will leave it there for context. I was thinking of sums where the indices themselves are ordered. Thanks for hopping in. – 2017-02-02
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    Note: everything is positive. – 2017-02-02
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    Take a look at this. [why can we interchange summations](http://math.stackexchange.com/questions/466757/why-can-we-interchange-summations) – 2017-02-02

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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).