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I have a bounded set $E$ of real numbers. I'm in the process of showing that there is a set $G$ that is a countable intersection of open sets $G_i$ such that $E\subseteq G$ and $E,G$ have the same outer measure.

What I have so far: I know that outer measure is invariant under unions of disjoint sets, so I am characterizing $E$ as a union of open intervals, closed intervals, half-open intervals, and unions of countably many singletons as well as unions of collections of uncountably many singletons.

Open intervals of $E$ can be replaced by $G_i$ for some $i$. Closed intervals of $E$, say $[a,b]$ can be written as $\bigcap _{i=1}^\infty (a-\frac{1}{i},b+\frac{1}{i})$, half-open intervals $[a, b)$ can be written as $\bigcap _{i=1}^\infty(a-\frac{1}{i},b)$ and a similar statement can be made for half-open intervals of the form (a,b].

I'm stuck on the singletons since I have to make sure I use only countably many sets. Any suggestions? I know that the outer measure of a collection of uncountably many points is equal the measure of the smallest open interval containing the points, but I'm having the hardest time given a point determining whether I want it to belong to a countable or uncountable set.

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  • $\begingroup$ How do you know this characterization of $E$? I don't think it true even for Borel sets. $\endgroup$ – Davide Giraudo Aug 8 '12 at 11:13
  • $\begingroup$ @DavideGiraudo The last ingredient in the proposed characterization, "unions of collections of uncountably many singletons", seems to cover all uncountable sets, whether they're Borel or not. $\endgroup$ – Andreas Blass Jun 3 '13 at 13:54
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Since $A$ is bounded, its outer measure is finite, since $$\inf\left\{\sum_{j=1}^{+\infty}\lambda(I_j),A\subset\bigcup_{j=1}^{+\infty}I_j\right\}$$ is finite.

Therefore, for each integer $n$, we can find an open set $G_n$ containing $A$ such that $\lambda(G_n)-\lambda^*(A)\leqslant \frac 1{2^n}.$ Let $G:=\bigcap_{n\geqslant 1}G_n$. We can assume WLOG that $\{G_n\}$ is decreasing. Then $$\lambda(G)=\lim_{n\to +\infty}\lambda(G_n)\leqslant \lambda^*(A).$$ Since $\lambda^*(A)\leqslant \lambda(G_n)$ for all $n$, we have that $\lambda^*(A)=\lambda(G)$.

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  • $\begingroup$ Thank you, this is much better than what I was doing. $\endgroup$ – The Substitute Aug 8 '12 at 12:10

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