TITLE: How does $\sum_{i=1}^\infty \sum_{j=1}^\infty a_{ij} = \sum_{j=1}^\infty \sum_{i=1}^\infty a_{ij}$ follow from the monotone convergence theorem?
QUESTION [5 upvotes]: In Rudin's Real and Complex analysis, he says that the equality
$$\sum_{i=1}^\infty \sum_{j=1}^\infty a_{ij} = \sum_{j=1}^\infty \sum_{i=1}^\infty a_{ij}$$
for $a_{i,j} \ge 0$ follows from this corollary of the monotone convergence theorem (via counting measure on a countable set):
If $f_n: X \to [0, \infty]$ is measurable and $f = \sum f_n$, then
$$\int_X f =\sum_{n=1}^\infty \int_X fn $$
However, I'm having a hard time seeing this. I'm guessing you use indicator functions for each point in the countable set, but I don't see any obvious manipulations to make it true. Any help would be appreciated.

REPLY [3 votes]: Let $X=\mathbb N$ and $S$ be the power set of $X$. Let $\mu$ be the counting measure on $X$. [$\mu(E)$ is the number of points of $E$ which is taken to be $+\infty$ if $E$ is an infinite set].
For any function $g: X \to [0,\infty]$ we have $\int g d\mu= \sum\limits_{k=1}^{\infty} g(k)$.
Now take $f_n(j)=\sum\limits_{i=1}^{n}  a_{ij}$. Then $f_n$ increases to the function $f$ defined by $f(j)=\sum\limits_{i=1}^{\infty} a_{ij}$. Hence $\int f_n  d\mu \to \int f d\mu$. This gives $\lim_n  \sum\limits_{i=1}^{n} \sum\limits_{j=1}^{\infty}  a_{ij}=\lim_n  \sum\limits_{j=1}^{\infty} \sum\limits_{i=1}^{n}   a_{ij}=\lim_n \int f_n d\mu=\int f d\mu=\sum\limits_{j=1}^{\infty} \sum\limits_{i=1}^{\infty} a_{ij}$.