TITLE: Intuition behind the definition of a measurable set
QUESTION [18 upvotes]: This week I saw the definition of a measurable set for an outer measure.

Let $\mu^*$ be an outer measure on a set $X$. We call $A \subseteq X$ measurable if $$\mu^*(E) = \mu^*(A\cap E) + \mu^*(A^c\cap E)$$ for every $E \subseteq X$.

This is not the first time I've seen this definition. Unlike most other things in mathematics, over time I have gained absolutely no intuition as to why this is the definition. 
The only explanation I've ever seen is that a set is measurable if it 'breaks up' other sets in the way you'd want. I don't really see why this is the motivation though. One reason I am not comfortable with it is that you require a measurable set to break up sets which, according to this definition, are non-measurable; why would you require that? Of course, you can't say what a non-measurable set is without first defining what it means to be measurable so I suppose no matter what your condition is, it will have to apply to all subsets of $X$.

Is there an intuitive way to think about the definition of measurable sets? Is there a good reason why we should use this definition, aside from "it works"?

REPLY [4 votes]: Old question here, but it is often asked by students of measure theory.
In our text Real Analysis (Bruckner${}^2$*Thomson) Example 2.28 in Section 2.7 there is a motivation for this that Andy used in his classes.
You would have seen that the inner/outer measure idea was successful for Lebesgue measure on an interval.  Certainly if you are hoping for additivity of the measure then that is an obvious idea.  Besides, when Lebesgue used it the upper=lower idea was pretty much well-established.
The example Andy gives is of a very simple finite outer measure $\mu^*$ on $X=\{1,2,3\}$ with $\mu^*(X)=2$, $\mu^*(\emptyset)=0$ and otherwise $\mu^*(A)=1$.  You can certainly define an inner measure but $\mu_*(A)=\mu^*(A)$ works for all subsets and yet that measure is not even finitely additive.  Evidently to spot the sets on which the measure will be additive you need to test more than just whether $\mu^*(A)+\mu^*(X\setminus A)=\mu^*(X)$.  Caratheodory came up with the idea of testing all sets not just $X$ itself (since it clearly doesn't work).