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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.

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It seems like you should be able to replace $\mu(X) = 1$ by "finite", but that's not so deep. – Dylan Moreland Nov 14 '11 at 16:15
up vote 12 down vote accepted

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.

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Great answer, and thanks for steering me straight. – Byron Schmuland Nov 14 '11 at 17:52
Thanks!${}{}{}$ – t.b. Nov 14 '11 at 17:58
@t.b.: To show that it's closed under countable union one can use the equivalent definition on Wikipedia. Then for each $i$ there are $F_i \subset A_i \subset G_i$ with $\mu(G_i - F_i)$ small enough. Set $G := \cup_i G_i$ and $F := \cup F_i$ then $F \subset A \subset G$ and $\mu(G - F) \leq \sum_i \mu((G_i - F_i)) < \delta$. – Rudy the Reindeer Nov 14 '11 at 22:44
@t.b.: thanks for the nice answer! – Rudy the Reindeer Nov 14 '11 at 22:46
@MattN. Which equivalent definition do you mean? "For each $\varepsilon$ there is a closed set $F$ and an open set $G$ such that..."? 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 15:05

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 $$\mu \Big(\bigcup_{i<n} F_i\Big) > \mu\Big(\bigcup_{i\in \mathbb N} A_i\Big) - \varepsilon.$$

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