Measure of an open set containing a Borel set of measure $1/3$ I met this problem preparing for my qual exam.
Given a Borel subset A of the unit unterval with Lebesgue measure 1/3, there exists an open set O containing A with measure 2/3.
Using the regularity of Borel set, it can easily be shown that there exists such open set with measure less than 2/3, but how can I get a set with measure exactly 2/3?
Also, does the number 2/3 here bear any significance?
Any help is appreciated.
 A: Let $B$ be open with $A\subset B$ and $m(B)<2/3$. Let $f(x)=m(B\cup (0,x))$.Then $f$ is continuous with $f(0)<2/3$ and $f(1)\ge 1.$
The only significance of the number $2/3$ is that it's more than $m(A).$
A: It is a little long for a comment.
Let $A \subseteq [0,1]$  be a Borel set such that $\mu(A)=1/3$, where $\mu$ is the Lebesgue measure.
In what follows, we will be considering the topology of $[0,1]$, so whenever we write "open" means open set in $[0,1]$.
Let $\Gamma = \{ B \subseteq [0,1]: B \text{ is open}, A\subseteq B, \mu(B) \leq 2/3\}$. Then:

*

*Using the regularity of $\mu$ on Borel set, it is easy to see that $\Gamma \neq \emptyset$.

*$\Gamma$ is partially ordered by $\subseteq$.

*Given any totally ordered chain $\{B_\alpha\}$ in $\Gamma$, let $B= \bigcup_{\alpha} B_\alpha$, then $A \subseteq B \subseteq [0,1]$ and $B$ is open.
Since, the topology of $[0,1]$ is second countable, we have that there is a countable famyly $\{D_n\}_{n \in \Bbb N}$ of basic open set, such that $B = \bigcup_n D_n$ and, for all $k \in \Bbb N$, there is $\alpha$ such that $\bigcup_{n=0}^k D_n \subseteq B_\alpha$. So,  for all $k \in \Bbb N$, $\mu \left (\bigcup_{n=0}^k D_n \right ) \leq 2/3$. So, $\mu(B) = \mu \left (\bigcup_n D_n \right ) \leq 2/3$. So,  $B \in \Gamma$.

So, we can apply Zorn lemma to $\Gamma$. Let $E$ be a maximal element of $\Gamma$. Then $E \subseteq [0,1]$, $E$ is open, $A \subseteq E$ and $\mu(E) \leq 2/3$ and since $E$ is maximal, we can conclude  that $\mu(E)=2/3$.
In fact, if $\mu(E) < 2/3$, then  using the regularity of $\mu$ on Borel set and the fact that $\mu$ has no atoms, we would have an open set $F$ such that $E\subseteq  F \subseteq [0,1]$ and $\mu(E) < \mu(F)< 2/3$. Clearly $F \in \Gamma$ and contradicts the maximality of $E$.
So we have proved there is $E \subseteq [0,1]$, $E$ is open, $A \subseteq E$ and $\mu(E) = 2/3$.
Remark: In fact, this proof works for any second countable space $X$ with any outer regular measure $\mu$ defined on the Borel $\sigma$-algebra of $X$. That means:

Let $X$ be a second countable topological space and let $\mu$ be an outer regular atomless measure defined on the  Borel $\sigma$-algebra of $X$. Then, given any Borel subset $A$ of $X$, and any $r \in (0,+\infty]$ such that $\mu(a) < r \leq \mu(X)$, there exists an open set $O$ containing A and having measure $r$.

