TITLE: A question about regularity of Borel measures
QUESTION [18 upvotes]: 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.

REPLY [7 votes]: 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
$$\mu \Big(\bigcup_{i<n} F_i\Big) > \mu\Big(\bigcup_{i\in \mathbb N} A_i\Big) - \varepsilon.$$