TITLE: Definitions of inner and outer regular measures
QUESTION [7 upvotes]: The definitions can be found here.
I am trying to figure out why it's defined specifically this way. More precisely- Why compact sets are used to define inner regularity while open sets are used for outer regularity?
I thought that maybe it has something to do with the facts that the union of open sets is open and that (in a Hausdorff space) the intersection of compact sets is compact, but I dont see how these matter.
Another thought I had is that maybe the motive comes from Riesz–Markov–Kakutani representation theorem.

REPLY [5 votes]: Terminology differs among authors. Another def'n is that a measure $m$ is Radon-regular iff each measurable set $S$ satisfies $m(S)=\sup \{m(T):S\supset T \land T$ compact $\},$ and that inner-regular means $m(S)=\sup \{m(U):S\supset U=\bar U\}.$ In many cases, Radon-regular and inner-measure co-incide.
The motivation is that we can say much more about regular measures because regularity is a relation between the measure  and the topology of the measure space.
For example a set $S$ is Lebesgue-measurable iff there exists a $G_{\delta}$ set $ S^i$ and an $F_{\sigma}$ set $S^o$ with $S^i\subset S \subset S^o$ and $m^o (S^o$  \ $S^i)=0.$
One of the motives of Lebesgue measure was attempts to extend calculus to  a much broader class of functions,  which resulted in extending the notion of length or volume. ...Note the outer measure involves  intersections of subsets of a (usually infinite ) family of open sets, and the inner measure involves  unions of closed sets.