Problem
To show that a measure $\mu$ defined on a metric space $(S,d)$ is regular.
- outer regularity: approximation by inner closed sets
- inner regularity: approximation by outer open sets
Discussion
Since this problem involves all borel sets $A \in \mathcal{B}(S)$, the direct way $\forall A \in \mathcal{B}(S), \dots$ won't work. We have to use the indirect way: denote \(\mathcal{C} = \lbrace A \in \mathcal{B}(S) \mid \mathinner{\text{desired properties}} \dots \rbrace.\) Show that
- $\mathcal{C}$ is a $\sigma$-algebra;
- $\mathcal{C}$ contains all open sets.
Since $\mathcal{B}(S)$ is generated by open sets, then we get the desired conclusion.
I thought the property in the definition of $\mathcal{C}$ would be either one of the regularity conditions (1) and (2). WLOG, I had chosen (1). However, I was unable to "turn $A$ inside out". (i.e. I couldn't show stability under set complement.)
The trick of preimage by metric $d$ wouldn't help because $A$ isn't necessarily $\mu$-continuous. (i.e. $\mu(\partial A) > 0$)
\[ F_\epsilon = \{x \in S \mid d(x,A^\complement) \ge \epsilon\} \]
In this case, all outer open approximation $F_{1/n}^\complement$ would contain $\partial A$, which we may not want.
\[ \bigcap_{\epsilon>0} F_\epsilon^\complement = \{x \in S \mid d(x,A^\complement) \le 0\} = \overline{A^\complement} \]
By construction of $\mathcal{C}$, each member $A \in \mathcal{C}$ only has outer approximation. When we take complement $A^\complement$, we're turning things inside out, so we need inner approximations, but there's no way to construct this with our metric $d$.
Solution
To overcome this $\mu$-discontinuity problem, we have to include also regularity condition (2) while defining $\mathcal{C}$. In this way, we can sandwich $A^\complement$ by $F$ and $G$. Since $\mu(G \setminus F)$ can be made arbitrarily small, that will approximate $\mu(A^\complement \setminus G^\complement)$ well. However, the price for this is that checking $\mathcal{C}$ is a $\sigma$-algebra requires more work.
Polish our space
If $(S, d)$ is a Polish space, then the measure $\mu$ can be tightened. Since a Polish space is, by definition, separation, proofs often start with dense sequence $(x_k)_k$ in $S$.
One of my classmates asked why we couldn't simply take the compact set to be \(K = \bigcup_{k=1}^{n_p} \bar{B}(x_k,\frac1p)\). My instructor reminded us that a closed unit ball is compact iff the space is finite dimensional.
For example, in $\R^\N$, i.e. the space of real-valued sequences, the orthonormal basis $(e_n)_n$ is a sequence in the closed unit ball centred at the origin having no convergent subsequence. (The $\ell^p$ distance between any two distinct elements is $2^{1/p}$ for any $p>0$.)
To make $K$ compact, we need to make it totally bounded, so that total boundedness and completeness are equivalent to compactness in metric spaces.
\[ K = \bigcap_{p>0}\bigcup_{k=1}^{n_p}\overline{B}\left(x_k,\frac{1}{p}\right) \]
Since that's my personal reminder, I'll leave out the technical details.