# Approximation by a $G_{\delta}$ set in Outer Measure implies Measurability

Question: If $E\subseteq {\mathbb R}$ and if there is some $G_{\delta}$ set $G$ such that $E\subseteq G$ and $m^{\ast}(G - E) = 0$ (where this is the outer measure that is used to define the Lebesgue measure), then $E$ is (Lebesgue) measurable.

Motivation: This is part of a question in Royden's book (page 64) which probably should be obvious to me. The way that I wanted to do this question was to use the equivalence that if, given $\epsilon > 0$, there is some open set $U\supseteq E$ such that $m^{\ast}(U-E) < \epsilon$ then $E$ is measurable. Since a $G_{\delta}$ set is not necessarily open, but made up of a countable intersection of open sets, I wanted to make it the limit of some open sets. So, if $U_{i}$ are open, I wanted to say that $G = \bigcap_{i}U_{i}$, and we can find an open set that we need just by taking some finite intersection (say, up to $N$) of the $U_{i}$. I'm not entirely convinced that I'm given these $U_{i}$, though, if I just know that $G$ is a $G_{\delta}$ set.

(Also, I think this question is different from the other few asking about related issues, but if it is not, then I will delete it.)

-

The following steps lead to a solution:

(1) Prove that if $A$ is a null set in $\mathbb{R}^n$, i.e., if $m^{*}(A)=0$, then $A$ is Lebesgue measurable.

(2) We are given that there exists a $\text{G}_{\delta}$ set $G$ such that $E\subseteq G$ and $m^{*}(G\setminus E)=0$. We wish to prove that $E$ is measurable. Note that $G\setminus E$ is measurable by (1). Prove that $E=G\setminus (G\setminus E)$.

(3) Prove that every $\text{G}_{\delta}$ set is measurable.

(4) Prove that the set-theoretic difference of two measurable sets is measurable.

(5) Conclude that $E$ is Lebesgue measurable.

I hope this helps!

-
This certainly does help, and I feel silly for not being able to see the end of part (2). In terms of being given a $G_{\delta}$ set, then, is it not a good idea to write it in terms of an intersection of open sets? Or are we not sure WHICH open sets constitute the $G_{\delta}$ set? –  james Jun 27 '11 at 6:57
Dear james, I think that the most relevant information when we are given a $G_{\delta}$ set $G$ is that there exists a collection $\{U_i\}_{i\in\mathbb{N}}$ of open sets such that $G=\bigcap_{i=1}^{\infty} U_i$. If we are proving general results about $G_{\delta}$ sets, then the exact description of the $U_i$'s should not be essential; the only essential information should be that the $U_i$'s are open and that there are countably many $U_i$'s. However, when we construct $G_{\delta}$ sets, we usually wish to select our $U_i$'s carefully. Does that answer your question? –  Amitesh Datta Jun 27 '11 at 7:40
Thank you, that seems reasonable. I think I understand now! –  james Jun 27 '11 at 22:55