Is a Radon measure always positive on non-empty open sets? The same question about the Haar measure. A measure is locally finite if it is finite on all compact sets from the underlying $\sigma$-algebra.
A measure is regular if every measurable set $A\in\Sigma$ can be approximated from above by open measurable sets:
$$
 \mu(A)\,=\,\operatorname{inf}\,\left\{\,\mu(O)\;\Big{|}\;A\subseteq O\;,\;\;O\in\Sigma\,,\;\,O\;\,\mbox{open}   \,\right\}\;\;,\;\;
 $$
and from below by compact measurable sets:
$$
 \mu(A)\,=\,\operatorname{sup}\,\left\{\,\mu(C)\;\Big{|}\;C\subseteq A\,,\;\,C\in\Sigma\,,\;\,C\;\,\mbox{compact}\,\right\}\;\;.
 $$
A Radon measure is a Borel measure that is introduced on the Borel algebra of a Hausdorff topological space, and is regular and locally finite.
Question 1: $\;\;$is a non-zero Radon measure always positive on non-empty open sets?
Question 2: $\;\;$the same for a non-zero Haar measure
 A: Unless I'm overlooking something trivial, the answer is no. Look at the "restriction" of Lebesgue measure $\lambda$ on $\Bbb{R}$ to $[0,1]$, by which I mean $\mu(E):= \lambda(E\cap [0,1])$. Then, $\mu$ is a finite Borel measure on $\Bbb{R}$, hence is regular, but the open set $(10,11)$ has measure $0$ with respect to $\mu$.
A: As mentioned in the comments, you should require $\mu \ne 0$. But even then, there are trivial counterexamples.
For example, let $U$ be a non-empty subset of any locally compact Hausdorff space $X$ and $x \notin U$. Then consider the Dirac measure $\delta_x$, which is Radon, and note that $\delta_x(U) = 0.$
A: I deeply thank the colleagues who answered my question about the Radon measure.
I think I now can answer the question about the Haar measure.
Suppose there is a nonempty open set $U$ such that $\mu(U)=0$. Any measurable set $C\in\Sigma$ can be covered by left translates of $U$, the measure of each such translate being $\mu(U)=0$ by the invariance of the Haar measure. Specifically, for a compact $C$, this cover contains a finite subcover, wherefrom we observe that $\mu(C)=0$ for any measurable compact $C$.
At the same time, by inner regularity of $\mu$, any measurable set $A\in\Sigma$ can be approximated from below by compact measurable sets. If all of them are of zero measure, then $\mu(A) =0$ for any $A\in\Sigma$. So $\mu$ is a zero measure, which contradicts the initial assumption.
