TITLE: Positivity and Interchange of Summation
QUESTION [17 upvotes]: Suppose I have $\sum\limits_{n = 1}^{\infty}\sum\limits_{m = 1}^{\infty} a_{m, n}$ where $a_{m, n} \geq 0$ for all $m$ and $n$. Can I interchange the two summations? If so why?

REPLY [3 votes]: Here's an elementary proof fleshing out @JonasMeyer's answer.
Claim: If $a_{m,n}\ge 0$ for all $m,n$ then
$$
\sum_{m=1}^\infty\sum_{n=1}^\infty a_{m,n}= \sup_{M,N}\sum_{m=1}^M\sum_{n=1}^N a_{m,n}.\tag1
$$
Proof. Let $S$ denote the RHS of (1). Write $$S_{M,N}:=\sum_{m=1}^M\sum_{n=1}^Na_{m,n}\quad\text{and}\quad S_M:=\lim_N S_{M,N}=\sum_{m=1}^M\sum_{n=1}^\infty a_{m,n}.$$ The limit $S_M$ exists, possibly with value infinity, since $S_{M,N}$ is monotonically increasing in $N$. By definition of sup, we have $S_{M,N}\le S\ \forall M,N$, hence $S_M\le S\,\forall M$, and $\color{green}{\lim_M S_M\le  S}$. But $S_{M,N}\le S_M$ by non-negativity of $a_{m,n}$, so $S:=\sup S_{M,N}\le\sup S_M$. Since $S_M$ is nondecreasing (non-negativity again), this means $\color{green}{S\le\lim_MS_M}$. To finish, check that $\lim_M S_M$ equals the LHS of (1).
Corollary:  If $a_{m,n}\ge 0$ for all $m,n$ then
$$
\sum_{m=1}^\infty\sum_{n=1}^\infty a_{m,n}= \sum_{n=1}^\infty\sum_{m=1}^\infty a_{m,n}.
$$
Proof. First apply the Claim to $a_{m,n}':=a_{n,m}$, then swap indices to obtain
$$
\sum_{n=1}^\infty\sum_{m=1}^\infty a_{m,n}= \sup_{M,N}\sum_{n=1}^N\sum_{m=1}^M a_{m,n}.\tag2
$$
Finally note that the RHS of (1) equals the RHS of (2).