I'm looking for as much examples as possible of the following.
Consider a set $X$, a ring of subsets $\mathscr{R}\subseteq X$ and a function $\mu:\mathscr{R}\to\mathbb{R}_{\geq0}$ such that: $\mu(\varnothing)=0$ and whenever $\{A_n\}_{n\in\mathbb{N}}\subseteq \mathscr{R}$ is a disjoint sequence such that $\bigcup_n A_n\in\mathscr{R}$, $\mu(\bigcup_n A_n)=\sum_n\mu(A_n)$ (such a function will be called a finite pre-measure). What I need is that the induced outer measure $\mu^*:P(X)\to\mathbb{R}_{\geq0}\cup\{\infty\}$ given by $$\mu^*(A)=\inf\left\{\sum_n\mu(B_n):\{B_n\}_{n\in\mathbb{N}}\subseteq\mathscr{R}\text{ and }A\subseteq \bigcup_n B_n\right\}$$ is actually finite, that is, $\mu^*(X)<\infty$.
The only such construction that I know of is that of the Lebesgue measure, since it uses a ring of subsets of the real line, however, it is clear that that is no example, because there are a lot of sets of infinite outer measure. Other way would be to look for already existing finite outer measures, but often I have found that they are not induced by premeasures on rings...
So yeah, any help or clue as to where to start searching is greatly appreciated. Thanks in advance!
I'm not quite sure if I get your question right, but I dont think that you can conclude $\mu^*(X) < \infty $ under the given assumptions. Since $ \mu$ is $\sigma$-additive it follows that $\mu(A) = \mu^*(A)$ for $A \in \mathscr{R}.$ However, the set $X$ is generally not in $\mathscr{R}$. So as I see to force $X \in \mathscr{R}$ with $\mu(X) < \infty$ might be one way to construct a finite outer measure under this way.
Otherwise you might assume, that $\mathscr{R}$ consists of a sequence $(A_n)_{n \in \mathbb{N}} \subseteq \mathscr{R}$ so that $X = \bigcup_{n=1}^{\infty} A_n \in \mathscr{R}$ and $\mu(A_n) < \infty.$ for any $n \in \mathbb{N}$. To guarantee $ \bigcup_{n=1}^{\infty} A_n \in \mathscr{R}$ the Ring has to be closed under countable unions, called $\sigma$-Ring. So $\mathscr{R}$ being such a $\sigma$-Ring and consiting of a cover for $X$ with finite pre-measure should help out here.