Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.)

share|cite|improve this question
up vote 5 down vote accepted

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!

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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