Let $a:\mathbb{N^2}\rightarrow\mathbb{R}^+$. Show that $$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} a(n,m) =\sum_{m=0}^{\infty}\sum_{n=0}^{\infty} a(n,m) $$ using the monotone convergence theorem.
So obvioulsy, since $a(n,m)\geq 0$ we can find $\pi_1, \pi_2$ reorderings of $n$ and $m$, such that $a(\pi_1(n), \pi_2(m))$ is monotone and
$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} a(\pi_1(n),\pi_2(m))=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} a(n,m) $$
because it is absolutely convergent. Then
$$\int_{\mathbb{N}}\sum_{m=0}^{\infty} a(\pi_1(n),\pi_2(m))\, \, d\, \lambda(m)=\sum_{m=0}^{\infty}\int_{\mathbb{N}} a(\pi_1(n),\pi_2(m))\, \, d\, \lambda(m)$$
where $\lambda$ is the counting measure.
I'm pretty sure, this is not right. Can you point me to the right direction, please?
Thank you!
Your solution is almost right. I am just refining.
Note that $f_k(n)=\sum_{m=0}^ka(n,m)$ is positive and monotonically increasing since $a(n,m)> 0$ for any $n,m$. Thus, $$\sum_{n=0}^\infty \sum_{m=0}^{\infty}a(n,m)=\int_{\mathbb{N}} \sum_{m=0}^{\infty}a(n,m)d\lambda(n)=\int_{\mathbb{N}}\lim_{k\rightarrow \infty} f_k(n)d\lambda(n)$$
Then by the monotone convergence theorem and linearity, $$\int_{\mathbb{N}}\lim_{k\rightarrow \infty} f_k(n)d\lambda(n)=\lim_{k\rightarrow \infty}\int_{\mathbb{N}} f_k(n)d\lambda(n)=\lim_{k\rightarrow \infty} \sum_{m=0}^{k}\int_\mathbb{N}a(n,m)d\lambda(n)=\sum_{m=0}^{\infty}\sum_{n=0}^\infty a(n,m).$$
Therefore, $$\sum_{n=0}^\infty \sum_{m=0}^{\infty}a(n,m)= \sum_{m=0}^{\infty}\sum_{n=0}^\infty a(n,m). $$