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

DefinitionLet $\mu$ 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 1Let $X$ 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 1Let $X$ 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.

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

Theorem 2Let $X$ 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 2Let $X$ 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:

LemmaA 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 3Let $X$ 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 3Let $X$ 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.

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

QuestionLet $X$ 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.

Measure Theory, if I have understood the hint correctly.) $\endgroup$ – Nate Eldredge Oct 11 '19 at 15:37Measure Theory: www1.essex.ac.uk/maths/people/fremlin/cont43.htm $\endgroup$ – Robert Furber Oct 11 '19 at 17:14