### 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.