Two equivalent definitions of Radon measure [duplicate]

Let $$\Omega$$ be a (for my purposes: locally compact) Hausdorff space. Let $$\mu$$ be a Borel measure on $$\Omega$$ and let $$A\subset\Omega$$ be a Borel set.

• $$\mu$$ is called locally finite if $$\mu(K)<\infty$$ for all compact $$K\subset\Omega$$.
• $$\mu$$ is called inner regular on $$A$$ if $$\mu(A)=\sup\{\mu(K):K\subset A \ \text{compact}\}.$$
• $$\mu$$ is called outer regular on $$A$$ if $$\mu(A)=\inf\{\mu(U):U\supset A \ \text{open}\}.$$

According to this article, the following two definitions are equivalent:

Definition 1. A borel measure $$\mu$$ is a Radon measure if $$\mu$$ is locally finite and inner regular on all Borel sets.

Definition 2. A Borel measure $$\mu$$ is a Radon measure if $$\mu$$ is locally finite, inner regular on all open sets and outer regular on all Borel sets.

Why are these definitions equivalent? Here is another article discussing these two definitions.

• Take a look at math.stackexchange.com/questions/103208/… Commented Jul 6, 2021 at 23:17
• The second measure described in this answer: math.stackexchange.com/a/215261/30222 seems to satisfy 2 but not 1. The main issue is that you might have a set of infinite measure whose all subsets have either infinite or uniformly bounded measure, even if there is no such open set. You can get rid of this by e.g. assuming more additivity (not just countable), $\sigma$-finiteness or second countability. Commented Jul 6, 2021 at 23:18
• The first example satisfies 1 but not 2, also: you might have a null set whose all open supersets have infinite measure. Commented Jul 6, 2021 at 23:27
• @tomasz At first glance it does indeed answer my question. Thank you! Commented Jul 7, 2021 at 10:17