Measure vs outer measure

Here is standard way for constructing measures on a measurable space $(S,\mathcal{S})$ that can for example be used to construct Lebesgue measure; various closely related approaches exist. For simplicity, we look at finite measures only.

You start with an algebra $\mathcal{A}$ on $S$ such that $\mathcal{S}$ is the smallest $\sigma$-algebra that contains all elements of $\mathcal{A}$. You then define a finite, nonnegative and finitely additive set-function $\mu_0$ on $\mathcal{A}$ that is countably additive in the sense that if $(A_n)$ is a sequence of disjoint sets in $\mathcal{A}$ and $\bigcup_n A_n\in\mathcal{A}$, then $\sum_n \mu_0(A_n)=\mu_0\big(\bigcup_n A_n\big)$. This assumption is easier to check than for $\sigma$-algebras since it might well be the case that $\bigcup_n A_n\notin\mathcal{A}$.

Then you define an outer measure $\mu^*$ on all of the powerset $2^S$ by letting $$\mu^*(B)=\inf\bigg\{\sum_{A\in\mathcal{C}}\mu_o(A):\mathcal{C}\textrm{ is a countable subset of }\mathcal{A}\textrm{ such that }B\subseteq\bigcup_{A\in\mathcal{C}}A\bigg\}.$$ Let $$\mathcal{S}_\mu=\Big\{B\in 2^S:\mu^*(A)=\mu^*(A\cap B)+\mu^*(A\cap S\backslash A)\textrm{ for all }A\in 2^S\Big\}.$$ It turns out that $\mathcal{S}_\mu$ is a $\sigma$-algebra and the restriction of $\mu^*$ to it, $\mu^*|\mathcal{S}_\mu$, is a measure. Moreover, $\mathcal{S}_\mu\supseteq \mathcal{S}\supseteq\mathcal{A}$, so $\mu=\mu^*|\mathcal{S}$ is a measure on $\mathcal{S}$ and moreover, $\mu^*|\mathcal{A}=\mu_0$.