Proof of a double series equality Let $b_{n_{i}} \in \mathbb{C}$ for $n,i\in \mathbb{N}$. Suppose that
$$\sum_{n=1}^{\infty}\sum_{i=1}^{\infty}|b_{n_i}|<\infty,$$
then 
$$\sum_{n=1}^{\infty}\sum_{i=1}^{\infty}b_{n_i}=\sum_{i=1}^{\infty}\sum_{n=1}^{\infty}b_{n_i}$$
Can I apply the dominated convergence theorem to solve this fact? 
Any idea on how to apply the dominated convergence theorem.
 A: I am using the following fact:

$$\int_{\mathbb N}ad\mu=\sum_{n=0}^{\infty}a(n)\tag{1}$$ 
This fact can be found in every measure theory textbook.

Now lets prove that $\sum_{i=0}^{\infty}\sum_{n=0}^{\infty}b_{n,i}=\sum_{n=0}^{\infty}\sum_{i=0}^{\infty}b_{n,i}$ for $b_{n,i}\in \mathbb C$.
Define $g_i(n):=b_{n,i}$ for $i\in \mathbb N$. Furthermore define
$f_m(n)=\sum_{i=0}^{m}g_i(n)$ and $f(n)=\sum_{i=0}^{\infty}g_i(n)$
By assumption we have that $\int_{\mathbb N}|f_m(n)|d\mu\leq \int_{\mathbb N}\sum_{i=1}^{m}|g_i(n)|\leq\int_{\mathbb N}\sum_{i=1}^{\infty}|g_i(n)|<\infty$, hence we can apply the dominated convergence theorem which yields:
$$\sum_{i=0}^{\infty}\sum_{n=0}^{\infty}g_i(n)=\lim_{m\rightarrow \infty}\sum_{i=0}^{m}\int_{\mathbb N}g_i(n)d\mu=\lim_{m\rightarrow \infty}\int_{\mathbb N}\sum_{i=0}^{m}g_i(n)d\mu=\lim_{m\rightarrow \infty}\int_{\mathbb N}f_m(n)d\mu=\int_{\mathbb N}\lim_{m\rightarrow \infty}f_m(n)d\mu=\int_{\mathbb N}f(n)d\mu=\int_{\mathbb N}\sum_{i=0}^{\infty}g_i(n)d\mu=\sum_{n=0}^{\infty}\sum_{i=0}^{\infty}g_i(n)$$
