Positivity and Interchange of Summation

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?


It's worth knowing that rearrangements can change the value of sums or integrals only if the positive and negative parts both diverge to infinity.

Fubini's theorem says rearrangements are fine if both parts are finite.

Tonelli's theorem says rearrangements are fine if what's being summed or integrated is everywhere non-negative. (It follows that it also works if it's everywhere non-positive, since the minus sign pulls out.)

Putting the two together gives you what I said in the first paragraph.


Yes. Because either way it is equal to $\displaystyle{\sup\limits_{M,N}\;\sum_{n=1}^N\sum_{m=1}^M a_{m,n}}$.


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).