TITLE: Example of a Borel measure, which is not Borel-regular
QUESTION [5 upvotes]: I have asked a question to find four types of outer measures here, and I could find three of the four examples. 
We call an outer measure $\mu: \mathcal P(\mathbb R^n) \to [0, \infty]$ Borel, if every Borel set $B \subset \mathbb R^n$ is $\mu$-measurable. We say, that an outer measure $\mu: \mathcal P(\mathbb R^n) \to [0, \infty]$ is Borel-regular, if $\mu$ is Borel and for any subset $A \subset \mathbb R^n$ there is a Borel set $B \supset A$, such that $\mu(A) = \mu(B)$. 
I would like to give an example of a Borel measure, which is not Borel-regular. Can you help me?

REPLY [5 votes]: Take a set $C\subset\mathbb{R}^n$ that is not Borel. For every $X\subset \mathbb{R}^n$, let
$$
\mu(X) = \begin{cases} 
0 & \text{if } X\subset C; \\
\infty & \text{if } X\not\subset C. \\
\end{cases}
$$
This is a measure on the entire $P(\mathbb{R}^n)$. In particular, all Borel sets are measuable.
For every Borel set $B\supset C$ we have
$\mu(B)=\infty> 0=\mu(C)$, so $\mu$ is not Borel-regular.