Let $(X, \tau)$ be a Hausdorff topological space and let $\mu$ be a measure on the $\sigma$-algebra of Borel subsets of $X$. The measure $\mu$ is (usually) defined to be outer regular if measure of any set $B \in \mathcal{B}$ can be approximated by the measures of the open supersets of $B$, i.e.
$$ \mu(B) = \inf \{\mu(U): B \subseteq U \in \tau \}.$$
There is a corresponding (dual) definition for inner regularity, where openness is replaced by closedness (and sometimes compactness) and infimum by supremum.
Question: if I require measurable sets to be approximable by measures of its open subsets, i.e. if I want the measures of measurable (e.g. Borel) sets to be a supremum of measures of its open subsets, do I get some kind of trivial notion?
Under the definitions above, let's call a measure $\nu$ (inner) shregular if for any nonempty measurable set $G \subseteq X$ we have:
$$\nu(G) = \sup\{\nu(V): V \subseteq G \: \& \: V \in \tau\}.$$
Two questions: is the Lebesgue measure $\lambda$ on $\mathbb{R}^n$ shregular?
Are there examples of non-trivial spaces and measures that are not shregular? Or perhaps there is some silly reason I do not notice for which none 'reasonable' measure is shregular?
If this helps, I would ask a similar question for (outer) shregularity w.r.t. closed supersets of a given nonempty $G \subseteq X$.
Best Answer
Taking the Lebesgue measure $\lambda$ on $\Bbb R$ and $A=([0,1]\setminus \Bbb Q)$ we see that $\operatorname{int}(A)=\emptyset$ and $\lambda(A)=1$, so inner regularity wrt open sets is quite hopeless. Similarly $B=[0,1]\cap \Bbb Q$ has $\lambda(B)=0$ while all closed sets that contain $B$ contain $[0,1]$ and so approaching from outside with closed sets we'll never go below $1$.