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.

  • $\begingroup$ 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

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 $\lambda^*(A)=\lambda(G)$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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