Hints: First show that $1/3\leq\mu(C)\leq2/3$ for some $C$; if not, show that a largest value $s<1/3$ is attained, on a set $B$, and that the complement of $B$ includes an atom. Then repeat the argument for $\mu$ restricted to $C$ and to its complement to get sets of intermediate measure, and iterate to get a dense set of values of $\mu$.
My efforts:
Assume that either $\mu(B)<1/3$ or $\mu(B)>2/3$ for any $B\in\mathcal{S}$.
We claim that a largest value $s<1/3$ is attained on a set $B$. Otherwise, there exists an increasing sequence of set $B_n$ with $\mu(B_n)<1/3$ and $\lim_{n\rightarrow\infty}\mu(B_n)=1/3$. Then $\mu(\bigcup B_n)=\lim_{n\rightarrow\infty}\mu(B_n)=1/3$, which violates the assumption of either $\mu(B)<1/3$ or $\mu(B)>2/3$ for any $B\in\mathcal{S}$.
Now we have a $B\in\mathcal{S}$ with $\mu(B)=s<1/3$ such that $\mu(D)\leq s$ if $\mu(D)<1/3$. Let $X\setminus B$ be the complement of $B$.
We claim that if $E\subset X\setminus B$ and $\mu(E)>0$ then $\mu(E)>2/3$. Since $E\subset X\setminus B$, $E\cap B=\emptyset$. $\mu(E\cup B)=\mu(E)+\mu(B)=\mu(E)+s>s$. By the assumption, $\mu(E\cup B)=\mu(E)+s>2/3$. Then $\mu(E)>2/3-s>1/3$. Again by the assumption, $\mu(E)>2/3$.
Let $E\subset X\setminus B$ and $F\subset E$ with $0<\mu(F)<\mu(E)$. Then we have $\mu(E)>2/3$, $\mu(F)>2/3$ and $\mu(E\setminus F)>2/3$ by the above claim. But $\mu(E\setminus F)=\mu(E)-\mu(F)\leq1-\mu(F)<1/3$, a contradiction. Thus we must have $\mu(F)=\mu(E)$, i.e., $E$ is an atom.
But $\mu$ is nonatomic. So the assumption does not hold.
We have shown that $1/3\leq\mu(C_1)\leq2/3$ for some $C_1$. Then repeat the argument for $\mu$ restricted to $C_1$ and to its complement, we can show that $\mu(C_1)/3\leq\mu(C_2)\leq2\mu(C_1)/3$ for some $C_2$. We can iterate to get a sequence of sets {$C_n$} such that $\mu(C_n)/3\leq\mu(C_{n+1})\leq2\mu(C_n)/3$.
My questions:
How to get a dense set of values of $\mu$ and how to make it fill the whole [0, 1]?
