If $a_{ij}\geq0$ if $i,j\in \mathbb{N}$, show using LMCT that,
$$\sum_{i=1}^\infty\sum_{j=1}^\infty a_{ij}=\sum_{j=1}^\infty\sum_{i=1}^\infty a_{ij}.$$
Attempt: Simply I can't realize how to use LMCT to prove it. I think that I'm missing an obvious fact.
Let $f_n(i) := \sum_{j=1}^n a_{ij}$ and let $\mu$ be the counting measure on $\mathbb{N}$. $f_n$ monotonically increases to $f(i) := \sum_{j=1}^\infty a_{ij}$. Then by LMCT $$ \lim_{n \to \infty} \int_{\mathbb{N}} f_n(i) d\mu(i) = \int_{\mathbb{N}} \lim_{n \to \infty} f_i(n) d\mu(i) $$ and this is equivalent to the equation you want to prove.