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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.