From Wikipedia:
"Any Borel probability measure on any metric space is a regular measure."
I've been thinking about how to prove this. And now I'm wondering whether being Borel isn't enough. Is $\mu(X) = 1$ (or $\mu(X) < \infty$) necessary and does $X$ have to be a metric space? I assume the answer to this is yes but I don't see why. Maybe someone could give me a hint for a proof or a proof?
Thanks for your help.
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.
Then you can use the infinite sum $\bigcup_i G_i$, but you shouldn't use the infinite sum of $\bigcup_i F_i$ - it may not be a closed set. Fortunately we can choose an appropriate finite sum.
– savick01 May 11 '13 at 15:05It'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 $$\mu \Big(\bigcup_{i<n} F_i\Big) > \mu\Big(\bigcup_{i\in \mathbb N} A_i\Big) - \varepsilon.$$
