The outer approximation theorem states that if $E$ is a Lebesgue measurable, then there exists a $G_\delta$ set $G$ containing $E$ such that the Lebesgue measure of $G$ equals the Lebesgue measure of $E$. And the inner approximation theorem states that if $E$ is a Lebesgue measurable set, then there exists a $F_\sigma$ set $F$ contained in $E$ such that the Lebesgue measure of $F$ equals the Lebesgue measure of $E$.

I'm wondering if something similar is true for arbitrary measure spaces. Let $X$ be a measure space, let $A$ be a collection of subsets of $X$, and let $m$ be a measure defined on $\sigma(A)$, the sigma algebra generated by $A$. Let $A_{\sigma\delta}$ denote the collection of countable intersections of countable unions of elements of $A$, and let $A_{\delta\sigma}$ denote the collection of countable unions of countable intersections of elements of $A$. My question is, for any $E\in\sigma(A)$, does there exist an $G\in A_{\sigma\delta}$ containing $E$ such that $m(F)=m(E)$? And does there exist a $G\in A_{\delta\sigma}$ contained in $E$ such that $m(G)=m(E)$?

If so, is it also true for the $m$-completion of $\sigma(A)$? Because that's when you'd have an analogy to the Lebesgue case; the Lebesgue sigma algebra is the completion of the Borel sigma algebra with respect to Lebesgue measure, and the Borel sigma algebra is the sigma algebra generated by open intervals.

  • $\begingroup$ See the Wikipedia article on "Regular measure" for at least some of your questions. $\endgroup$ – Dave L. Renfro Nov 14 '18 at 13:01
  • $\begingroup$ @DaveL.Renfro Which of my questions does it answer? $\endgroup$ – Keshav Srinivasan Nov 14 '18 at 13:51
  • $\begingroup$ I'm not an expert in measure theory, but I do know that the meaning of "measure" varies in the literature, so it will depend on exactly what definition of "measure" you're using. The main purpose of my comment was to tell you that "regular" is probably the appropriate search word to use. That said, maybe this will help --- For pretty much everything you'd ever want to know about measures, see the various volumes of David H. Fremlin's treatise. $\endgroup$ – Dave L. Renfro Nov 14 '18 at 14:17
  • 2
    $\begingroup$ @DaveL.Renfro Not sure "regular" is the right search term because this question doesn't seem to be primarily concerned with measures defined on topological spaces, but rather arbitrary measure spaces. $\endgroup$ – aduh Nov 14 '18 at 17:28
  • $\begingroup$ @aduh: You're correct, and I overlooked that. The question sounds like something that can be found in some general measure theory texts, but off-hand I don't know the answer. However, in case this could be of additional help, if I needed to look into this issue, I'd probably start by flipping through the treatments in Halmos' book, Dunford & Schwartz (Volume I, chapter on integration and set functions), and Hahn/Rosenthal's Set Functions (the last book is pretty much my go-to book for very abstract measure matters of this type). $\endgroup$ – Dave L. Renfro Nov 14 '18 at 18:25

In the case where $\mathcal{A}$ is an algebra, or even a ring, I believe the answer to the first question is true. It's a direct consequence of the Caratheodory extension procedure using an outer measure. Recall that this guarantees that $$\mu(E) = \inf\Big\{\sum_{j \in \mathbb{N}}\mu(A_j) : A_j \in \mathcal{A}, \bigcup_{j \in \mathbb{N}}A_j \supseteq E \Big\}$$ for all $E \in \sigma(\mathcal{A})$. Now, for each $n \in \mathbb{N}$, we can find a sequence $(A^n_j)_{j \in \mathbb{N}}$ in $\mathcal{A}$ such that $\bigcup_{j \in \mathbb{N}} A^n_j \supseteq E$ and $$\mu(E) \geq \sum_{j \in \mathbb{N}}\mu(A^n_j) - 1/n \geq \mu\big(\bigcup_{j \in \mathbb{N}} A^n_j \big) - 1/n. $$ Since $\bigcup_{j \in \mathbb{N}} A^n_j \supseteq E$ for all $n$, then $\bigcap_{n \in \mathbb{N}}\bigcup_{j \in \mathbb{N}} A_j^n \supseteq E$ and $$\mu\big( \bigcap_{n \in \mathbb{N}}\bigcup_{j \in \mathbb{N}} A_j^n\big) \leq \mu\big(\bigcup_{j \in \mathbb{N}} A^n_j \big) \leq \mu(E) + 1/n.$$ Since, this holds for all $n$, it follows that $\mu\big( \bigcap_{n \in \mathbb{N}}\bigcup_{j \in \mathbb{N}} A_j^n\big) \leq \mu(E)$. And since the reverse inequality is immediate, we can conclude. I think that the similar inner approximation result that you ask about will hold by considering an extension procedure with an inner measure, but I leave the details to you.


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.