Nonmeasurable Set with Finite Outer Measure

[Bashyboy]

Homework Statement

Let $E$ be a nonmeasurable set of finite outer measure. Show that there is a $G_\delta$ set $G$ that contains $E$ for which $m^*(E)=m^*(G)$, while $m^*(G-E) > 0$.

Homework Equations

$E$ is a measurable set if and only if there is a $G_\delta$ set $G$ containing $E$ for which $m^*(G-E)=0$.

Let $E$ have finite outer measure. Show that there is an $F_\sigma$ set $F$ and a $G_\delta$ set $G$ such that $F \subseteq E \subseteq G$ and $m^*(F)=m^*(E)=m^*(G)$.

The Attempt at a Solution

I pretty certain it wouldn't follow immediately from the two theorems I quoted in section 2, right? In fact, we could prove an analogous theorem for $F_\sigma$ sets, right?

 

[andrewkirk]

Since E is nonmeasurable, if we can find a measurable set G that contains it and has the same outer measure, the result will follow from the first theorem you quote in section 2.

Are you allowed to use the axiom of choice? If so, I suggest the following strategy.

1. prove that, for any positive integer $n$, there exists a measurable set $G_n\supset E$ such that $m^*(G_n)-m^*(E)<1/n$.
2. find a way to construct a measurable set $G\supset E$ such that $m^*(G)-m^*(E)=0$.

Part 2 is easier than part 1. But I have a feeling that 1 is provable, given some key properties of outer measures and measurable sets.

If you're not allowed to use axiom of choice, I don't know whether it's still true. If it is, a more subtle proof may be needed, as the axiom is needed for the approach I had in mind for step 2. IIRC, the axiom is usually assumed in measure theory, as the construction of the most famous nonmeasurable sets, the Vitali sets, requires it.

 

[Bashyboy]

I don't understand why the axiom of choice is needed. Here is proof I came up with:

Suppose that $E$ is a nonmeasurable set of finite outer measure. Then for every $G_\delta$ set $G$ containing $E$ is such that $m^*(G-E) > 0$ (negation of first theorem. Moreover, since $E$ has finite out measure, there exists a $G_\delta$ set $G$ such that $m^*(G)=m^*(E)$ (second theorem). Hence, we have found a $G_\delta$ set $G$ containing $E$ for which $m^*(G)=m^*(E)$ and $m^*(G-E) > 0$.