Let $f(m,n)$ be a real valued function for all $m,n\in\mathbb{N}$. Suppose that $$ \sum_{m=1}^\infty\lvert f(m,n)\rvert\le\frac{1}{n^2} $$ for each positive integer $n$. Use Fubini's Theorem to prove $$ \sum_{m=1}^\infty\sum_{n=1}^\infty f(m,n)=\sum_{n=1}^\infty\sum_{m=1}^\infty f(m,n). $$ Proof: In Fubini's Theorem we have two measure spaces $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ with $\nu$ being complete. Then if $f$ is integrable over $X\times Y$ with respect to the product measure $\mu\times\nu$ then $f(x,\cdot)$ is integrable over $Y$ with respect to $\nu$ and
$$ \int_{X\times Y} f\,d(\mu\times\nu)=\int_X \int_Y f(x,y)\,d\nu\,d\mu. $$
1.) So based on the statement of my problem, I want to let both my measure spaces be the counting measure space, which is a complete measure space.
2.) Then I need to show that $f$ is integrable on $X\times Y$ wrt $\mu\times\nu$ in order to conclude that $f(x,\cdot)$ is integrable on $Y$ wrt $\nu$. Do I simply observe that since $$ \sum_{n=1}^\infty\sum_{m=1}^\infty f(m,n)\le\left\lvert\sum_{n=1}^\infty\sum_{m=1}^\infty f(m,n)\right\rvert\le\sum_{n=1}^\infty\sum_{m=1}^\infty \lvert f(m,n)\rvert\le\frac{\pi^2}{6} $$ we have $f(m,n)$ integrable and then proceed to iterate the integral in both orders to show my desired result? It seems like the given condition already implies that $f(x,\cdot)$ is integrable on $Y$ wrt $\nu$.
Since $\nu$ is complete and $f$ is integrable then by Fubini's Theorem, I can show that $$ \int_{X\times Y}f(m,n)\,d(\mu\times\nu)=\int_X\left(\int_Yf(m,n)\,d(\nu)\right)\,d(\mu)=\sum_{m=1}^\infty\sum_{n=1}^\infty f(m,n). $$
Similarly, since $\mu$ is complete and $f$ is integrable then by Fubini's Theorem, I can show that $$ \int_{X\times Y}f(m,n)\,d(\mu\times\nu)=\int_Y\left(\int_Xf(m,n)\,d(\mu)\right)\,d(\nu)=\sum_{n=1}^\infty\sum_{m=1}^\infty f(m,n). $$ Finally since $f$ is integrable we know $ \int_{X\times Y}f(m,n)\,d(\mu\times\nu)<\infty$ so $$ \sum_{m=1}^\infty\sum_{n=1}^\infty f(m,n)=\int_{X\times Y}f(m,n)\,d(\mu\times\nu)=\sum_{n=1}^\infty\sum_{m=1}^\infty f(m,n). $$