Is every finite Borel measure on a locally compact Hausdorff, $\sigma$-compact and separable space automatically regular?

Question

The conditions stated in the question seem mouthful and a bit arbitrary, so let me provide some backgrounds.

Definition

Let $\mu$ be a Borel measure on a topological space. We say:

• $\mu$ is outer regular on a Borel set $E$ if $\mu(E) = \inf\{\mu(U) : E\subseteq U \text{ is open}\}$,

• $\mu$ is inner regular on a Borel set $E$ if $\mu(E) = \sup\{\mu(K) : E\supseteq K \text{ is compact}\}$ (Some authors call this tight),

• $\mu$ is a Radon measure if $\mu$ is finite on all compact sets, outer regular on all Borel sets and inner regular on all open sets,

• $\mu$ is a regular measure if $\mu$ is finite on all compact sets and both outer regular and inner regular on all Borel sets.

The subtle difference between a Radon measure and a regular measure is annoying. Fortunately, every $\sigma$-finite Radon measure on a locally compact Hausdorff space is automatically regular:

Theorem 1

Let $X$ be a locally compact Hausdorff space. Then every Radon measure on $X$ is inner regular on all of its $\sigma$-finite sets. In particular, every $\sigma$-finite Radon measure on $X$ is regular.

Corollary 1

Let $X$ be a locally compact Hausdorff space. A finite Borel measure on $X$ is regular if and only if it is outer regular on all Borel sets and inner regular on all open sets.

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Next, by using Riesz representation theorem on locally compact Hausdorff spaces, one can prove the following:

Theorem 2

Let $X$ be a locally compact Hausdorff space. If every open set in $X$ is $\sigma$-compact, then every Borel measure on $X$ that is finite on compact sets is regular.

A special case is a separable metric space:

Corollary 2

Let $X$ be a locally compact, separable metric space. Then every finite Borel measure on $X$ is regular.

The proof of this corollary relies on the following general topological result:

Lemma

A locally compact metric space is $\sigma$-compact if and only if it is separable, in which case every open set is $\sigma$-compact.

In case local compactness is not given, one still has the following result:

Theorem 3

Let $X$ be a separable complete metric space. Then every finite Borel measure on $X$ is regular.

Note that a locally compact metric space can be given a compatible complete metric, so Theorem 3 also implies Corollary 2.

Another special case of Theorem 2 is a second-countable space:

Corollary 3

Let $X$ be a second-countable and locally compact Hausdorff space. Then every finite Borel measure on $X$ is regular.

Proof: A locally compact Hausdorff space is topologically regular, so by Urysohn's metrization theorem, $X$ is metrizable. On the other hand, a second-countable space is separable. Thus, $X$ is locally compact, separable and metrizable, so Corollary 2 applies.

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Finally, in view of the above results, my question is:

Question

Let $X$ be a locally compact Hausdorff, $\sigma$-compact and separable space, which may or may not be metrizable. Is every finite Borel measure on $X$ regular? If not, please give a counter-example.

Answer 1

No. Counterexamples include $X = \{0,1\}^\kappa$ or $[0,1]^\kappa$ for any $\aleph_0 < \kappa \le \mathfrak{c}$. These spaces are compact Hausdorff (Tikhonov's theorem) and are separable (Hewitt-Marczewski-Pondiczery).

(This special case of H-M-P can also be proved directly. For example, with $X =[0,1]^\kappa$, identify $\kappa$ with a subset of $[0,1]$, so that $X$ is a space of functions from $[0,1]$ to itself; then it is easy to show that the set of polynomials with rational coefficients is dense in $X$.)

Now the uncountable successor ordinal $\omega_1 + 1$, with its order topology, can be embedded in $X = \{0,1\}^\kappa$ or $X = [0,1]^\kappa$. See Fremlin, Measure Theory, 434K(d) (thanks to Robert Furber for the reference). One can also follow a construction similar to the Stone-Čech compactification and embed $\omega_1 + 1$ into $[0,1]^{C(\omega_1+1, [0,1])}$, noting that $|C(\omega_1 + 1, [0,1])| = \mathfrak{c}$. Either way, call the embedding $\phi$.

Let $\nu$ be the famous Dieudonné measure on $\omega_1 + 1$, where $\nu(B) = 1$ if $B$ contains a closed unbounded subset of $\omega_1$ and $\nu(B) = 0$ otherwise. (See Fremlin 411Q and 4A3J for details.) Then the pushforward $\mu = \nu \circ \phi^{-1}$ is a finite Borel measure on $X$ which is not regular.

Specifically, since $\phi$ is an embedding and $\omega_1$ is open in $\omega_1 + 1$, then the image $A = \phi(\omega_1)$ is a relatively open subset of the compact set $\phi(\omega_1 + 1)$, so that $A$ is the intersection of an open set and a closed set in $X$, and in particular is Borel. Moreover $\mu(A) = \nu(\omega_1) = 1$. But if $K$ is any compact subset of $A$, then $\phi^{-1}(K)$ is a compact subset of $\omega_1$, hence $\mu(K) = \nu(\phi^{-1}(K)) = 0$. Thus inner regularity fails.

This example is also Exercise 7.14.130 of Bogachev's Measure Theory.