This question has been asked here, but I don't believe that we have had a satisfactory answer, so I would like to reformulate that question. Please forgive me if this question turns out to be rather silly.
Let $(X,\mathcal{T})$ be a topological space, $(X,\Sigma,\mu)$ be a measurable space such that $\Sigma$ contains the borel algebra. Usually, $\mu$ is called outer regular if
$$
\mu(A)=\inf\{\mu(G):G\supset A,G\text{ open}\},\quad\forall A\text{ measurable}.
$$
Personally, I would be very tempted to defind a measure to be inner regular if
$$
\mu(A)=\sup\{\mu(F):F\subset A,F\text{ closed}\},\quad\forall A\text{ measurable}.\tag{1}
$$
But it turns out that the definition in reality is
$$
\mu(A)=\sup\{\mu(F):F\subset A,F\text{ compact}\},\quad\forall A\text{ measurable}.\tag{2}
$$
And I'm wondering why.
Since that a measure satisfying $(2)$ is called inner regular, let's call a measure satisfying $(1)$ one that is approximable by closed sets. The two notions can be very different. For a $KC$ space (where every compact set is closed), every inner regular measure is approximable by closed sets, but not vice versa: consider the lower limit topology and the Lebesgue measure $\mu$ over $\mathbb{R}$. Since this topology contains more closed sets than the usual one, $\mu$ is clearly approximable by closed sets, but $\mu$ is not inner regular since every compact set in the lower limit topology is countable. If we remove the $KC$ assumption, then an inner regular measure is not necessarily approximable by closed sets: consider the trivial topology where every set is compact. If $\mu$ is approximable by closed sets, then every measurable proper subset of $X$ has measure $0$, so either $\Sigma=\{\emptyset,X\}$, either $\mu(X)=0$.
So I would like to ask: why do we choose to break the symmetry here in the definition of outer and inner regularity? In other words, why is using closed sets less desirable, while using open sets is acceptable? Or are we generally working in the setting where the two definitions become equivalent?
Best Answer
This answer works in the usual Hausdorff spaces such as Euclidean measure theory the probability theory in polish space.
If $(2)$ holds, clearly $(1)$ does. So demonstrating $(1)$ is less interesting and less useful than demonstrating $(2)$.
$(2)$ is a stronger result when you need to estimate the measures of maybe very large, unbounded sets. We like $(2)$ because compactness is very useful, in practice with finite covers of course and specific estimates being made on each element of the finite cover. There's not to much too it in my opinion; since $(2)$ is true (for Lebesgue measure and many other measures) and $(2)$ is stronger than $(1)$ but equally easy to discuss, we always discuss $(2)$.