Measure of an open set containing a Borel set of measure $1/3$

Question

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.

Answer 1

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 $m(A).$