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I want to generalize the result for $\mathbb{R}^n$, that any non-trivial open set has positive measure.
Open sets and measure zero

Question: For which types of topological measure spaces is it true that every non-trivial (i.e. non-empty) open set has positive measure?

My thinking was that only inner regularity of the measure, combined with local compactness of the space, would be necessary.

There are some problems with this:

(1) We don't just need that every point has a compact neighborhood for this to work -- otherwise we would be able to prove that any set containing a compact set has positive measure.

We need that every point has a compact neighborhood with positive measure. This does not seem to follow from the given candidate assumptions. And

(2) The possible definitions of local compactness vary depending on whether or not the space is Hausdorff. Obviously the topological measure space needs to be Hausdorff in order for it to be inner regular, but there are weaker notions which might be sufficient and which don't require $T_2$.

Apparently we also need left-invariance, i.e. that our topological measure space is in fact a measurable topological group. measure of open set with measure Haar But from the answer given to that question the following is still unclear to me:

Do we really need Hausdorff? And do we really need group structure and left-invariance to conclude that every point has a compact neighborhood with positive measure?

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I am not sure if I understand your first question, but any topological space $X$ can be equipped with the Borel $\sigma$-algebra $\mathcal{B}_X$, i.e., the $\sigma$-algebra generated by the open sets of $X$. Then $(X, \mathcal{B}_X)$ is a measure space, which can be equipped with a Borel measure, which is a positive measure. However, there are Borel measures which are not "strictly positive".

The following might be an answer to your second question regarding the Haar measure:

The Haar measure is a translation-invariant Radon measure, which is a positive measure. Any non-trivial open set has, by definition, a positive measure with respect to the Haar measure. As you point out, the group structure is needed for the translation-invariance. The Hausdorff property is needed in order to prove the existence of the Haar measure. The book Locally Compact Groups by Markus Stroppel discusses properties of locally compact groups in the most general setting, and in general it is not assumed that the topological group is Hausdorff. However, in the proof of the existence of the Haar measure you will see that the Hausdorff property is in fact needed. In case you want to know where the Hausdorff property is exactly needed in the construction of the Haar measure, I highly recommend to read $\S12$ of Stroppel's book.

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    $\begingroup$ This is a very nice answer; I appreciate the good reference. One question: doesn't a Borel measure just have to be defined on open sets? Why does it need to have positive measure for all open sets? $\endgroup$ – Chill2Macht Aug 5 '16 at 19:07
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    $\begingroup$ I think this was the point where I didn't really understand your question. If you take the empty set $\emptyset$, which is open, then $\mu(\emptyset) = 0$, so it can be 0 but not negative. Do you really mean strictly positive? $\endgroup$ – user342207 Aug 5 '16 at 19:12
  • $\begingroup$ Sorry I meant "non-trivial" -- I wrote non-trivial in the line above the yellow but not in the yellow -- I will fix that $\endgroup$ – Chill2Macht Aug 5 '16 at 19:22
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    $\begingroup$ Okay, I think this is exactly what is meant by a "strictly positive measure". See: en.wikipedia.org/wiki/Strictly_positive_measure. To answer your question then: I think you need the Haar measure, which is strictly positive by definition. An example of a Radon measure which is not strictly positive is the Dirac measure. $\endgroup$ – user342207 Aug 5 '16 at 19:29
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    $\begingroup$ Sorry for forgetting to accept your answer earlier -- I don't have any excuse, I just blanked/spaced out and forgot $\endgroup$ – Chill2Macht Aug 6 '16 at 0:26

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