Below is problem where, in one direction of the proof, I use the Lebesgue Dominated Convergence Theorem. I have, of course, included my attempt at the solution. Any help concerning the "correctness" of my attempt would be very helpful. Thank you!
$\textbf{Problem:}$ Let $X$ be the set of natural numbers, $\textbf{X}$ be the $\sigma$-algebra consisting of all the subsets of $X$, and let $\mu$ be the counting measure on $\textbf{X}$. Show that $f \in L(X,\textbf{X}, \mu)$ if and only if the series $\sum f(n)$ is absolutely convergent, in which case $$\int f d\mu = \sum f(n).$$
$\textbf{Attempt at Solution:}$
$\Rightarrow$ Assume $f \in L(X,\textbf{X}, \mu)$. We need to show $\sum |f(n)|$ converges. Define a sequence of functions $(\phi_{m})$ by the following equation: $$\phi_{m}=\sum^{m}_{n=1} |f(n)|\chi_{\{n\}}.$$ Note that $(\phi_{m})$ is a monotone increasing sequence of functions which converges to $|f|$, which is integrable by our assumption. Hence, the Monotone Convergence Theorem imples that $$\int |f| d\mu = \lim \int \phi_m d\mu.$$ Using our definition of the integral of a simple function, we see that $$\lim \int \phi_m d\mu = \lim \sum^{m}_{n=1} |f(n)|\mu(\{n\})=\lim \sum^{m}_{n=1} |f(n)|=\sum |f(n)|$$ Indeed, $\sum |f(n)|=\int |f| d\mu$, from which it follows that $\sum f(n)$ is absolutely convergent. (Remember, $|f| \in L(X,\textbf{X}, \mu)$ means its integral is a finite real number.)
$\Leftarrow$ Assume $\sum f(n)$ is absolutely convergent; specifically, let $\sum f(n)=k$. Put $g(x)=k$. Then, $g$ is a constant function and integrable. Define a sequence of functions $(\psi_{m})$ by the following equation: $$\psi_m = \sum^{m}_{n=1} f(n)\chi_{\{n\}}$$ This sequence converges to $f$. For each $m$ (and $x \in X)$, $$|\psi_m|\leq \sum^{m}_{n=1} |f(n)\chi_{\{n\}}| \leq k.$$ Finally, by the Lebesgue Dominated Convergence Theorem, we conclude that $f$ is indeed integrable and $$\int f d\mu = \lim \int \psi_{m} d\mu = \lim \sum^{m}_{n=1} f(n) \mu(\{n\}) = \sum f(n). \blacksquare$$
Best Answer
It seems to me that the converse assumes the constant function $g\equiv k$ is integrable, where it really isn't (its integral is infinity, which is okay by itself, but it does exclude $g$ from $L^1(\mu)$).
What we would do instead is show that $|f|\in L^1(\mu)$, using the $\phi_m$'s from the first part and the monotone convergence theorem, and since $|f|$ dominates $f$ and the $\psi_m$'s, the dominated convergence theorem indeed suffices to conclude the proof.