# Any borel measure on a locally compact, sigma-compact, Hausdorff space is regular?

on the book "measure and integration theory" by Heinz Bauer , theorem 29.12, the same proof works for a sigma-compact space instead of a locally compact space with countable basis(which is in particular sigma-compact.).

(or one can even just do the same until (29.12). Then because {open sets} is a under intersection stable generator of the borel sigma algebra, and both measures are sigma finite (since E is sigma compact and both measures are borel measure.), thus these two measures equal on the borel sigma algebra.

but the remark 4. before this theorem gives a example of a borel measure which is not regular, on a compact space.

since I believe the example is correct, so the proof must be somewhere wrong. but i really can't find the error, the proof doesn't need the space to have countable basis, it only needs it to be sigma compact (having countable basis just guarantees this).

Please help me. Thank you very much！

(It's a very simple proof, but i can't find the mistake .Please help me instead of voting a -1. )

(hello, I wanted to add a comment but it didn't work on my browser. it's urgent in the sense that I really feel the urge to know the answer. because I think it's a principle question, so i can't really feel right when going on learning the stuff without knowing the answer(it will sort of disturb my thinking..))

• Including "urgent" in your title tends to be poorly received here. For what reason could this be urgent other than you are seeking help for a timed exam or homework set? – Austin Mohr Jun 9 '16 at 2:46
• Please take the time to make this question self contained. If you're referring to theorems, please write them down. If there is an example, please write that down too. – Alex R. Jun 9 '16 at 3:32
• The example probably is right (there certainly are such measures) so either the proof of 29.12 has a gap, or you have overlooked where it uses the countable basis. Unfortunately, as it stands, the only people who can read the proof to tell which are those who own a copy of the book. But if you can copy the proof into this post (or even an image of the page), you will increase the number of people who could answer. That will help you get an answer faster, which apparently is important to you. – Nate Eldredge Jun 9 '16 at 3:51

## 1 Answer

The existence of a countable basis (second countability) is used right near the beginning of the proof, where we work with an arbitrary open set $U$ and exploit the fact that it is $\sigma$-compact (Bauer calls this "countable at infinity").

In general, it is possible that a locally compact Hausdorff $\sigma$-compact topological space has an open subset which is not $\sigma$-compact. The uncountable ordinal space $[1,\Omega]$ of the previous paragraph is an example; it is compact itself, but the open subset $[1, \Omega)$ is not $\sigma$-compact. So that's where the proof would fail for that space.

So the problem is that $\sigma$-compactness is not generally inherited by open subsets. However, second countability is inherited by open subsets, and it implies $\sigma$-compactness in this setting.

• Do you know if this can occur in left invariant measures in locally compact $\sigma$-compact groups? That is, whether a (non-second-countable) locally compact $\sigma$-compact group $G$ can admit a non-zero locally finite left invariant measure which is not regular. – Cronus Feb 10 '18 at 13:25
• @Cronus: I don't know offhand, maybe you should ask it as a new question. At a guess, I would say the answer is no, and for a possible counterexample I would look at an uncountable product, like $Z_2^\kappa$ or $(S^1)^\kappa$ where $\kappa$ is uncountable. – Nate Eldredge Feb 10 '18 at 16:24
• Thanks! I'll think about it. – Cronus Feb 10 '18 at 21:01