Let $\mu^{\star}$ be an outer measure on $X$.
The book "Real Analysis" by Folland motivates the definition of set $A$ being $\mu^\star$ measurable as follows. It says first that
If $E$ is a "well-behaved" set such that $A \subset E$, then the equation $$ \mu^{\star}(E) = \mu^{\star}(E \cap A) + \mu^{\star}(E \cap A^c) $$ says that the outer measure of $A$, $\mu^{\star}(A)$ is equal to the "inner measure" of $A$, $\mu^{\star}(E)-\mu^{\star}(E\cap A^c)$.
This I understand, but I do not understand the next part.
The leap from "well-behaved" sets $E$ containing $A$ to arbitrary $E$ is a large one, but it is justified by Caratheodory's Theorem that states $$ \text{1) The collection $M$ of $\mu^{\star}$ measurable sets is a $\sigma$ algebra.} \\ \text{2) The restriction of $\mu^{\star}$ to $M$ is a complete measure.} $$
Could anyone explain to me how Caratheodory's theorem fills in this "leap"?
Don't think of it as being an extension to arbitrary $E$; rather regard the theorem as a statement in its own right and Folland's comment to be purely motivational.
If $\mu^*$ is an outer measure on $X$, you may define a set $A$ to be measurable if $\mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c)$ for every $E \subset X$. The Caratheodory theorem states that the collection of measurable sets is a $\sigma$-algebra and that $\mu^*$ is a measure on that $\sigma$-algebra.