Regularity of an outer measure and regularity of a measure with respect to open and compact sets are different conditions, both satisfied here

Assuming the Axiom of Countable Choice, the word regular is carrying two different conditions in this development, and both of them hold for Lebesgue measure. They are not variants of one statement: one is about arbitrary subsets and measurable supersets, the other about measurable sets and topologically distinguished sub- and supersets.

Regularity of an outer measure. Measurable hulls and regular outer measures calls an outer measure ${\mu }^{\ast }$ regular when every subset $E$ of the ambient set has a measurable hull: a Carathéodory measurable $H\supseteq E$ with ${\mu }^{\ast }(H)={\mu }^{\ast }(E)$. This mentions no topology at all, and it is a condition that fails for some outer measures. For ${\lambda }_n^{\ast }$ it holds, with the hull available in the special form $G_{\delta }$: Every subset of ${\mathbb{R}}^n$ has a $G_{\delta }$ measurable hull of the same outer measure.

Regularity of a measure with respect to open and compact sets. Here the statements are that ${\lambda }_n^{\ast }(E)$ is the infimum of ${\lambda }_n(U)$ over open $U\supseteq E$ (Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of ${\mathbb{R}}^n$ is the infimum of the measures of the open sets containing it), and that ${\lambda }_n(E)$ is the supremum of ${\lambda }_n(K)$ over compact $K\subseteq E$ for measurable $E$ (Assuming countable choice, the Lebesgue measure of a measurable set is the supremum of the measures of its compact subsets). Both mention the topology essentially, and the second is restricted to measurable sets, which the first is not.

Why the distinction has to be made rather than left to context. The two conditions have different hypotheses on $E$, different quantifiers, and different witnesses: a measurable hull is a superset with equal outer measure, while outer regularity produces supersets whose measures merely approach the outer measure and are open. The one implies the other only through an argument — here, intersecting a sequence of open supersets, which is exactly the proof of the $G_{\delta }$ hull. Nothing below uses the word regular without saying which of the two is meant.