Let $X=\mathbb{N}$, let $\Omega$ be all subsets of $\mathbb{N}$, and let $\mu$ be the counting measure on $\Omega$. Show that $f$ belongs to $L(X,\Omega, \mu)$ if and only if the series $\sum f(n)$ is absolutely convertent, in which case: $$\displaystyle\int fd\mu=\displaystyle\sum\limits_{n=1}^{\infty}f(n)$$
My attempts: Maybe I can use the Lebesgue dominated convergence theorem that says: Let $(f_{n})$ be a sequence of integrable functions which converges almost everywhere to a real-valued measurable function $f$. If there exists an integrable function $g$ such that $|f_{n}|\leq g$ for all $n$, then $f$ is integrable and: $$\displaystyle\int f d\mu=lim\displaystyle\int f d\mu$$ I think that I can use the theorem for show the convergence of the series, then the sum of each sequence will be the integral. Am not sure if I'm in the right way to prove it. Thanks.