What does "measurable" mean intuitively? So if we have an outer measure $\mu$ on a set $\Omega$, we defined:
A subset A $\subseteq$ $\Omega$ is called $\mu$-measurable, if for all B $\subseteq$ $\Omega$:
$\mu$(B) = $\mu$(B $\cap$ A) + $\mu$(B \ A).
And i understand the definition, but i always thought we can only measure measurable sets, i.e. $\mu$ is only defined for measurable sets, but it's defined for all subsets of $\Omega$. So why do we define it this way or what's the intuition behind it?
If we measure a non-measurable set, does that mean the value will be "wrong" in a way? Or do I take the word "measurable" too literally?
 A: If $\mu^*$ denotes an outer measure on a set $\Omega$ then $A\subseteq\Omega$ is by definition $\mu^*$-measurable if:$$\mu^*(B)=\mu^*(B\cap A)+\mu^*(B\cap A^{\complement})\text{ for all }B\subseteq\Omega$$
Be aware that an outer measure is defined on the powerset of $\Omega$ so that $\mu^*(B)$ is defined for every $B\subseteq\Omega$. In general an outer measure is not the same thing as a measure.
The sets that are indeed $\mu^*$-measurable constitute a $\sigma$-algebra and the restriction of $\mu^*$ on this $\sigma$-algebra appears to be a (complete) measure.
A: What you're taking about is not a measure, but an outer measure. They are used to construct actual measures, and are defined on the entire power set of the underlying set.
The first step in constructing a measure on $\Omega$ is asking what the measure should actually tell us about the sets it measures, and then constructing such a measure for every subset of $\Omega$. For instance, we could technically define the volume of an arbitrary subset of $\mathbb R^n$ as $\operatorname{vol}(A)=\inf\{\sum_{B\in\mathcal B} \operatorname{vol}B~|~\mathcal B\textrm{ is a collection of boxes covering }A\}$, where the volume $\operatorname{vol}B$ of a box is the product of its side lengths. This works for any subset of $\mathbb R^n$, since the infimum is well-defined. But now we have to make sure that this outer measure (in this case it's the outer Lebesgue measure) is well-behaved, not just well-defined. In particular, we would like the volume of a disjoint union of sets to be the sum of their volumes. So we throw out any set which would break this feature. That is, those sets $A$ which can cut another set $B$ into two pieces $B\cap A$ and $B\backslash A$, such that the volume of their union is not the sum of their volumes. We call those "not measurable", and the rest "measurable". Now the measurable ones form a $\sigma$-algebra, which in retrospect justifies calling them measurable. If we now restrict the outer measure to the measurable sets, we get an actual measure which captures the idea we used to define the outer measure, and behaves nicely as well.
