A question about regularity of Borel measures

Let $X$ be a metrizable space and let $\mu$ be a finite Borel measure on $X$.

Let $\Sigma$ be the family of Borel sets $A$ such that $$ \begin{align*} \mu(A) &= \sup{\{\mu(F)\,:\,F \subset A \text{ is closed}\}} = \inf{\{\mu(G)\,:\,G \supset A \text{ is open}\}}. \end{align*} $$ Since for each closed set $A$ in a metrizable space we have $A = \bigcap_{n=1}^{\infty} G_n$ for some sequence of open sets $G_n$, we see that $\Sigma$ contains the closed sets. Clearly $\Sigma$ is closed under complementation and I leave it to you to verify that it is closed under taking countable unions. Therefore $\Sigma$ is the family of Borel sets.

The same line of reasoning goes through in any space in which closed sets are $G_\delta$-sets, e.g. perfectly normal spaces, which are a bit more general than metric spaces.

Further remarks:

• While the definition of regularity used here is probably the most common one nowadays, it is a bit dangerous to speak of regularity without further specifying what is intended. Many authors mean by inner regularity that $\mu(A) = \sup{\{\mu(K)\,:\,K \subset A \text{ is compact}\}}$ (that's probably due to Bourbaki, where only measures on locally compact spaces were considered). This property is often called tightness in modern texts.

• One can show that in a separable, completely metrizable space, every finite Borel measure is automatically tight, see Kechris, Classical Descriptive Set Theory, Theorem 17.11, p.107 for a proof. Without these assumptions this may fail.

• Some assumptions are necessary, but finiteness of $\mu$ is much too strong. One can show (with quite a bit more work) that in a metrizable space, every semi-finite Borel measure (every set of infinite measure contains a set of finite measure) is inner regular with respect to the closed sets. Again, this extends to perfectly normal spaces.

• As for outer regularity, this notion is not particularly useful beyond $\sigma$-finite Borel measures. One result that is useful is that a locally finite Borel measure (every point has an open neighborhood of finite measure) on a separable metrizable space is automatically $\sigma$-finite and regular.

It's a reply to this comment. I'm writing it as an answer since there is a bit too much $\LaTeX$ in it.

Well, for each $\varepsilon$ you need $n, G_i, F_i$ such that: $\mu\Big(\bigcup_{i\in \mathbb N} G_i \setminus \bigcup_{i<n} F_i\Big) < \varepsilon$.

We can equivalently require: $\mu \Big(\bigcup_{i\in \mathbb N} G_i \setminus \bigcup_{i\in \mathbb N} A_i \Big) < \varepsilon$ and $\mu \Big(\bigcup_{i\in \mathbb N} A_i \setminus \bigcup_{i<n} F_i\Big) < \varepsilon$. So actually the problem is the second assertion.

We have $F_i$ such that $\mu(A_i\setminus F_i)< \frac{\varepsilon}{2^{i+2}}$, so $\mu \Big(\bigcup_{i\in \mathbb N} A_i \setminus \bigcup_{i\in\mathbb N} F_i\Big) \leq \sum_{i \in \mathbb N} \frac{\varepsilon}{2^{i+2}} = \frac{\varepsilon}{2}$. Thus we know: $$\mu \Big(\bigcup_{i\in\mathbb N} F_i\Big) \geq \mu\Big(\bigcup_{i\in \mathbb N} A_i\Big) - \frac{\varepsilon}{2}.$$ Finally, we use continuity of measure to conclude that there exists $n$ such that $$\mu \Big(\bigcup_{i<n} F_i\Big) > \mu\Big(\bigcup_{i\in \mathbb N} A_i\Big) - \varepsilon.$$