Cfrgtkky

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 $A_{sigmadelta}$ denote the collection of countable intersections of countable unions of elements of $A$, and let $A_{deltasigma}$ denote the collection of countable unions of countable intersections of elements of $A$. My question is, for any $Einsigma(A)$, does there exist an $Gin A_{sigmadelta}$ containing $E$ such that $m(F)=m(E)$? And does there exist a $Gin A_{deltasigma}$ 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.

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
$$mu(E) = infBig{sum_{j in mathbb{N}}mu(A_j) : A_j in mathcal{A}, bigcup_{j in mathbb{N}}A_j supseteq E Big}$$
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
$$mu(E) geq sum_{j in mathbb{N}}mu(A^n_j) - 1/n geq mubig(bigcup_{j in mathbb{N}} A^n_j big) - 1/n. $$
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
$$mubig( bigcap_{n in mathbb{N}}bigcup_{j in mathbb{N}} A_j^nbig) leq mubig(bigcup_{j in mathbb{N}} A^n_j big) leq mu(E) + 1/n.$$
Since, this holds for all $n$, it follows that $mubig( bigcap_{n in mathbb{N}}bigcup_{j in mathbb{N}} A_j^nbig) leq mu(E)$. 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.