Regular Borel Measures equivalent definition

Question

Please help me understand how the below definition is equivalent to the standard definition of regularity which says "that a measure is regular if for which every measurable set can be approximated from above by an open measurable set and from below by a compact measurable set." http://en.wikipedia.org/wiki/Regular_measure

Second def. "$\mu$ is regular whenever $A$ is in the domain of definition of Borel algebra and $\epsilon>0$, there are closed and open sets $C$ and $U$ such that $C \subset A \subset U$ and $|\mu|(U \setminus C)<\epsilon$."

Answer 1

For a general topological space, these definitions are not equivalent.

The following counterexample is essentially Example 7.1.6 in Bogachev's Measure Theory. Let $X$ be a Vitali set of outer measure $1$ in $[0,1]$, such that $X$ is not Lebesgue measurable (and in particular does not have Lebesgue measure zero), but every Lebesgue measurable subset of $X$ does have Lebesgue measure zero. JDH explains how to construct such a Vitali set in this question. Equip $X$ with its subspace topology. (In fact, $X$ is a separable metric space, though it is not completely metrizable.) Then the Borel sets in $X$ are of the form $A \cap X$, where $A$ is Borel in $[0,1]$. For such $A \cap X$, define a measure $\mu$ by $\mu(A \cap X) = m(A)$, where $A$ is Borel in $[0,1]$ and $m$ is Lebesgue measure. It's simple to verify that $\mu$ is a countably additive Borel measure on $X$, with $\mu(X) = 1$. Furthermore, $\mu$ is regular in the sense of your second definition: for a Borel set $A \cap X$ as above, since $A$ is Borel in $[0,1]$ and $m$ is regular, for each $\epsilon$ there are sets $C,U \subset [0,1]$ which are respectively closed and open in $[0,1]$, $C \subset A \subset U$, and $m(U \setminus C) < \epsilon$. But then we have that $C \cap X, C \cap U$ are respectively closed and open in $X$, $C \cap X \subset A \cap X \subset U \cap X$, and $$\mu((U \cap X) \setminus (C \cap X)) = \mu((U \setminus C) \cap X) = m(U \setminus C) < \epsilon.$$

On the other hand, $\mu$ is not regular in the sense of your first definition. If $K \subset X$ is compact, then it is also compact in $[0,1]$ and in particular is Lebesgue measurable, so must have $m(K) = 0$. Then $\mu(K) = 0$ as well. In particular, we cannot approximate $X$ (or any other set of positive $\mu$ measure) from below by compact sets.

If your topological space is Polish, then these two definitions are equivalent, and in fact every finite Borel measure on such a space satisfies both. This is Theorem 7.1.7 in Bogachev.

Answer 2

For an alternative and extreme counterexample, you can take (any) Bernstein set in the interval $[0,1]$. This is a set with the property that it intersects any uncountable closed set, but doesn't contain any.

Notice that any compact subset of a Bernstein set is countable, so no continuous (vanishing at points) measure can be approximated from below by compact sets, as all compact sets will have measure zero. On the other hand, it is a metric space, and metric spaces have the property that any finite Borel measure is regular in the first sense you mentioned.

For a more concrete example, you can take the Lebesgue measure restricted to the Bernstein set like in Nate Eldridge's example. It will be regular in the general sense, but not in the latter you mentioned (this approximation by compact sets is often called Radon property, at least for finite Borel measures).

However, both these examples make substantial use of axiom of choice, I'm quite curious as to whether or not there are any counterexamples that do not need it. Nothing like this or what Nate suggested will work, because if I recall correctly, it is consistent with $\mathrm{ZF}$ that every subset of reals is absolutely measurable.