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Let $\mathcal{A}$ be a $\sigma$-algebra containing the Borel algebra (everything is in a topological space). Let $m\colon\mathcal{A}\to[0,\infty]$ be a measure.

The standard definition of regularity goes like this: $m$ is regular if, for any $A\in\mathcal{A}$, the measure of $A$ equals the infimum of measures of open sets containing $A$ and also a supremum of measures of closed sets contained in $A$.

For Lebesgue measure in $\mathbb R^n$, there is a known theorem that it is regular in the following way: for every L-measurable set $A$, for every $\varepsilon>0$ there is a closed subset $K$ and open supset $U$ of $A$ such that $\lambda(U\setminus K)<\varepsilon$.

What about the general case - are these two properties equivalent?

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Regularity of $\mu$: $\forall_{A \in \mathcal{A}}:\mu(A) = \inf\{\mu(U)|U \supset A, U \text{ open}\} = \sup\{\mu(K)|K \subset A, K \text{ closed}\}$

By choosing sequences of open and closed sets that converge to the real measure in the limit it easily follows that $\forall_{\varepsilon > 0}$ the required $K$ and $U$ exist such that $\mu(U \setminus K) < \varepsilon$.

If $\forall_{\varepsilon > 0}$ the required $K$ and $U$ exist such that $\mu(U \setminus K) < \varepsilon$, then it easily follows that $\inf\{\mu(U)|U \supset A, U \text{ open}\} = \sup\{\mu(K)|K \subset A, K \text{ closed}\}$. By non-negativity, $\mu(A)$ cannot be anything other than that same value.

So yes, they are equivalent in the general case.

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