In the measure theory lecture, we defined a measure-theoretic content as follows:
$ \mu: \mathscr{C} \rightarrow [0,\infty]$
with the property being additive on disjoint sets and that the empty set goes to 0 with $\mathscr{C}$ being a set-theoretic ring.
We had a theorem when a content is a premeasure (sigma-additive content) and saw following implications:
$ (1) \Rightarrow ( (2) \Leftrightarrow (3))$
with
$(1)$ For $A_n \searrow A$ with $A, A_n \in \mathscr{C}$ for all $n$: $\lim\limits_{n \to \infty}\mu(A_n)=\mu(A)$ "Continuity from above"
$(2)$ For $A_n \nearrow A$ with $A, A_n \in \mathscr{C}$ for all $n$: $\lim\limits_{n \to \infty}\mu(A_n)=\mu(A)$ "Continuity from below"
$(3)$ $\mu$ is premeasure.
I do understand the argumentation of this proposition. My observation here was that it is not such a big contraint to approximate a set from below. Approximating it from the top is not that easily performed.
Now at the end of the lecture, we had a look inner and outer regular measures. I know that we are now dealing with measures, those are defined on a $\sigma$-Algebra $\mathscr{A}$ instead of a ring. The definitions were as follows:
Inner regular: For all $A \in \mathscr{A}$
$\mu (A) = \sup \{ \mu (F) | F \subseteq A, F \mbox{ compact and measurable} \}$
Outer regular: For all $A \in \mathscr{A}$
$\mu (A) = \inf \{ \mu (G) | G \supseteq A, G \mbox{ open and measurable} \}$
I understand that those two things are something different. Especially that for regular measures, we work with a Hausdorff-space and with a Borel-$\sigma$-Algebra having special topological separation properties. There are examples of inner- but not outer-regular measures and vice versa.
My intuitive problem is now the following: I do associate inner regularity with approximation from below, outer regularity with approximation from the top.
Can someone tell my an idea, why this asymmetry observed with approximation (for premeasures) breaks down when it comes to measures and make those concepts in a way independent (that there are no implications between inner reg. and outer reg.)
Half a month later, I realised the following:
The theorem mentioned above can be extended:
$(1')$ For $A_n \searrow A$ with $A, A_n \in \mathscr{C}$ for all $n$ and $\mu(A_n)<\infty$ for a particular $n_0$ (due to monotonicity: for all $n≥n_0$ holds: $\lim\limits_{n \to \infty}\mu(A_n)=\mu(A)$ "Continuity from above in the finite case"
The implications proven in our script are: $(1) \Rightarrow ((2) \Leftrightarrow (3)) \Rightarrow (1')$ and $(1) \Leftrightarrow (1')$ for finite measures, so all statements are equivalent under that assuption.
When we deal with inner regular measures or outer regular measures, the set $X$ where the measure is defined on is equiped with a topology. Therefore we do have "open" and "compact".
My answer to the question above (why there is no connection when it comes to inner and outer regular measures) is the following:
Open sets and compact sets are in general different concepts. They hardly have anything to do with each other, so only conditions combining those two concepts like locally compact will set up a connection.
When we talk about measures being defined on the Borel-$\sigma$-Algebra, it is common to call it "Borel-measure" having the property (+) that $\mu(K)<\infty$ for all compact sets $K \in \mathfrak{B}$, $\mathfrak{B}$ being the Borel-$\sigma$-Algebra. This is not necessary, but mostly meaningful, because compactness should capture the idea of "close to finiteness". Assuming now that the room is locally compact, we see that the assumption (+) is in a good context.
An outer regular measure being inner-regular on open sets implies that $\mu(B) < \infty$, $B \in \mathfrak{B}$. For a $\sigma$-finite room, we get that the outer regular measure is even regular (which seems to be quite okay, when the room is small enough as the theorem about premeasures gives an intuition about).