Caratheodory: Motivation While reading Rudin's real and complex analysis I came across the following nice reasoning:

Reasoning of Variation Measure
Given a complex measure $\mu$ find its variation measure $|\mu|$ that is the smallest measure among all dominating measures.
This variation measure $|\mu|$ especially would have to satisfy:
$$|\mu|(E)=\sum_{k\in\mathbb{N}}|\mu|(E_k)\geq\sum_{k\in\mathbb{N}}|\mu(E_k)|\qquad E=\bigsqcup_{k\in\mathbb{N}} E_k$$
That suggests to consider:
$$|\mu|(E):=\sup_{E=\bigsqcup_{k\in\mathbb{N}}E_k}\sum_{k\in\mathbb{N}}|\mu(E_k)|$$
And this turns out to be the desired variation measure.
Reasoning of Measure Completion
Given a countably additive set function $\mu$ find the extension to a complete measure $\mu_E$ that is its domain extends to a sigma algebra on which the extension becomes a complete measure.
Now, is there a reasoning giving rise to Caratheodory's construction comparable in structure to the above one? I could think of something like:
A new measurable set $A$ in this sense would certainly have to satisfy:
$$\mu_E(A)=\sum_{k\in\mathbb{N}}\mu(E_k)\qquad A=\bigsqcup_{k\in\mathbb{N}}E_k$$
But then this new measurable sets especially also would have to satisfy:
$$\mu_E(A)+\mu_E(X\setminus A)=\mu(X)$$
But how to continue then?
 A: Ok, if you have any class of sets $\mathcal{A} \subset \mathrm{Pot}(X)$ and any function $\mu :\mathcal{A} \to [0,\infty]$ and want to extend it to a measure $\mu^\ast$ on any $\sigma$-Algebra $\Sigma$, it has to satisfy
$$
\mu^\ast (E) \leq \sum_k \mu^\ast (E_k) = \sum_k \mu (E_k)
$$
for each $E \in \Sigma$ and $E_k \in \mathcal{A}$ with $E \subset E_k$.
In this sense, the outer measure $\mu^\ast$ used in the proof of the Caratheodory extension theorem is the "largest" candidate measure.
Now for $E,F \in \Sigma$ would satisfy
$$
\mu^\ast (F) = \mu^\ast(F \cap E) + \mu^\ast(F \cap E^c), \qquad (\dagger)
$$
because $ \mu^\ast$ is finitely additive on $\Sigma$.
There is quite a leap to now require $(\dagger)$ for all subsets $F$ in order to call $E$ "measurable", but this is justified, because (as the proof shows) it "works".
EDIT: For possible future readers: As this post (Caratheodory: Measurability) shows, one can restrict the condition in $(\dagger)$ to only requiring the identity for $F = X$ as long as $\mu^\ast$ is the outer measure induced by some $\sigma$-additive set function $\mu$ defined on an algebra $\mathcal{A}$ if $\mu^\ast (X)$ is finite.
But the (current) proof given there that this condition actually suffices is using Caratheodory's theorem and its proof in which $(\dagger)$ is still required to hold for all $F$.
