# regularity of a measure

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?

## 1 Answer

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.