The Dominated Convergence Theorem on Wikipedia does not assume each $f_n$ to be integrable (although it turns out they have to). However, the same theorem appears in Dudley's book seems to require $f_n$ to be integrable in the first place.
Let $f_n$ and $g$ be in $\mathcal{L}^1 \left(X,S,\mu\right)$, $\left|f_n(x)\right| \le g(x)$ and $f_n(x) \to f(x)$ for all $x$. Then $f\in\mathcal{L}^1$ and $\int f_n\,d\mu \to \int f \,d\mu$.
Is Dudley's book making a stronger-than-usual assumption?
As for why I am interested. It is true that if $\int |f|\,d\mu <\infty$, then $\int f \,d\mu <\infty$ and is well-defined. If we do not require $f$ to be integrable in the assumption, then simply consider $f\to f$ and it is dominated by $|f|$, which is integrable. So $\int f \,d\mu$ is integrable. But if we add the assumption that each $f_n$ needs to be integrable, then we cannot invoke this theorem in this problem. Another situation is where I want to explain if $|f| < M$, then it is integrable. I guess in both case I can go back to writing $f$ as $f^+ - f^-$, but just curious on how to use DCT.
The requirement of $f_n$ to be integrable is irrelevant, because the hypothesis $|f_n|\leq g$ with $g$ integrable implies that $f_n$ is integrable. This is because for a positive function its integral is either finite or infinite, but always exists.
For a non-positive function $f$, for its Lebesgue integral to be defined we need that at least one $f_+$ and $f_-$ (its positive and negative parts) is integrable. Then, by definition, if $|f|$ is integrable then so is $f$.
For your final example, bounded implies integrable only on a finite measure space. But, in that case, again you don't need DCT but only the definition of "integrable".